Convex Polyhedra

Springer Monographs in Mathematics A.D. Alexandrov Convex Polyhedra With 165 Figures 123 † A.D. Alexandrov Englis...

Author: A.D. Alexandrov | N.S. Dairbekov | S.S. Kutateladze | A.B. Sossinsky

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Springer Monographs in Mathematics

A.D. Alexandrov

Convex Polyhedra With 165 Figures

123

† A.D. Alexandrov English translation by N.S. Dairbekov, S.S. Kutateladze and A.B. Sossinsky Comments and bibliography by V.A. Zalgaller Appendices by L.A. Shor and Yu. A. Volkov The Russian edition was published by Gosudarstv. Izdat. Tekhn.-Teor. Lit., Moscow-Leningrad, 1950

Library of Congress Control Number: 2004117404

Mathematics Subject Classification (2000): 52A99, 52B99 ISSN 1439-7382 ISBN 3-540-23158-7 Springer Berlin Heidelberg New York This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer. Violations are liable for prosecution under the German Copyright Law. Springer is a part of Springer Science+Business Media springeronline.com © Springer-Verlag Berlin Heidelberg 2005 Printed in Germany The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Typeset by the authors using a Springer LATEX macro package Production: LE-TEX Jelonek, Schmidt & Vöckler GbR, Leipzig Cover design: Erich Kirchner, Heidelberg Printed on acid-free paper

46/3142YL - 5 4 3 2 1 0

To Boris Nikolaevich Delaunay, my teacher

Preface to the English Translation

This book was published in Russian in 1950 [A16] and in German in 1958 [A19]. It continues the lines of the classical synthetic geometry in its subject and results. The methods are also synthetic and the book proves their might once again. Seemingly, this is why the book attracted a wide readership and gave an impetus to research for a few generations of mathematicians. In the intervening years, most problems that are listed in the book as unsolved were settled. Some results have acquired a more abstract and refined form or alternative proofs. In the present English edition, these changes are mainly commented in footnotes. The list of references is duly enlarged. The comments were made by V. A. Zalgaller on the author’s advice. The present edition includes the translations of two articles by Yu. A. Volkov and an article by L. A. Shor which present supplements to Chapters 3, 4, and 5. These are placed at the end of the book. The author is grateful to V. A. Zalgaller, who took pains to elaborate the present edition of the book, providing it with updating commentaries and references.

Table of Contents

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Content and Purpose of the Book . . . . . . . . . . . . . . . . . . . . . . . . . . . Order and Character of the Exposition . . . . . . . . . . . . . . . . . . . . . . . Remarks for the Professional . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2

Basic Concepts and Simplest Properties of Convex Polyhedra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Definition of a Convex Polyhedron . . . . . . . . . . . . . . . . . . . . . . 1.2 Determining a Polyhedron from the Planes of Its Faces . . . . 1.3 Determining a Closed Polyhedron from Its Vertices . . . . . . . 1.4 Determining an Unbounded Polyhedron from Its Vertices and the Limit Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 The Spherical Image . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Development . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7 Topological Properties of Polyhedra and Developments . . . . 1.8 Some Theorems of the Intrinsic Geometry of Developments 1.9 Generalizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Methods and Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 The Cauchy Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 The Mapping Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Determining a Polyhedron from a Development (Survey of Chapters 3, 4, and 5) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Polyhedra with Prescribed Face Directions (Survey of Chapters 6, 7, and 8) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Polyhedra with Vertices on Prescribed Rays (Survey of Chapter 9) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Infinitesimal Rigidity Theorems (Survey of Chapters 10 and 11) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 Passage from Polyhedra to Curved Surfaces . . . . . . . . . . . . . . 2.8 Basic Topological Notions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9 The Domain Invariance Theorem . . . . . . . . . . . . . . . . . . . . . . .

1 1 3 5

7 7 16 21 26 39 49 56 72 82 87 87 93 99 106 117 124 136 141 147

X

Contents

Uniqueness of Polyhedra with Prescribed Development . 3.1 Several Lemmas on Polyhedral Angles . . . . . . . . . . . . . . . . . . . 3.2 Equality of Dihedral Angles in Polyhedra with Equal Planar Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Uniqueness of Polyhedra with Prescribed Development . . . . 3.4 Unbounded Polyhedra of Curvature Less Than 2π . . . . . . . . 3.5 Polyhedra with Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Generalizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

155 155

Existence of Polyhedra with Prescribed Development . . . 4.1 The Manifold of Developments . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 The Manifold of Polyhedra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Existence of Closed Convex Polyhedra with Prescribed Development . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Existence of Unbounded Convex Polyhedra with Prescribed Development . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Existence of Unbounded Polyhedra Given the Development and the Limit Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

193 193 202

5

Gluing and Flexing Polyhedra with Boundary . . . . . . . . . . . 5.1 Gluing Polyhedra with Boundary . . . . . . . . . . . . . . . . . . . . . . . 5.2 Flexes of Convex Polyhedra . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Generalizations of Chapters 4 and 5 . . . . . . . . . . . . . . . . . . . . .

229 229 246 261

6

Conditions for Congruence of Polyhedra with Parallel Faces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Lemmas on Convex Polyhedra . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 On Linear Combination of Polyhedra . . . . . . . . . . . . . . . . . . . . 6.3 Congruence Conditions for Closed Polyhedra . . . . . . . . . . . . . 6.4 Congruence Conditions for Unbounded Polyhedra . . . . . . . . . 6.5 Another Proof and Generalization of the Theorem on Unbounded Polyhedra. Polyhedra with Boundary . . . . . . . . . 6.6 Generalizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3

4

7

Existence Theorems for Polyhedra with Prescribed Face Directions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Existence of Polyhedra with Prescribed Face Areas . . . . . . . . 7.2 Minkowski’s Proof of the Existence of Polyhedra with Prescribed Face Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Existence of Unbounded Polyhedra with Prescribed Face Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 The General Existence Theorem for Unbounded Polyhedra . 7.5 Existence of Convex Polyhedra with Prescribed Support Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

163 169 173 182 186

210 212 218

271 271 281 287 291 295 304

311 311 317 321 327 332

Contents

XI

7.6 Generalizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342 8

9

Relationship Between the Congruence Condition for Polyhedra with Parallel Faces and Other Problems . . . . . 8.1 Parallelohedra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 A Polyhedron of Least Area with Fixed Volume . . . . . . . . . . 8.3 Mixed Volumes and the Brunn Inequality . . . . . . . . . . . . . . . .

349 349 359 366

Polyhedra with Vertices on Prescribed Rays . . . . . . . . . . . . 9.1 Closed Polyhedra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Unbounded Polyhedra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Generalizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

377 377 388 395

10 Infinitesimal Rigidity of Convex Polyhedra with Stationary Development . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Deformation of Polyhedral Angles . . . . . . . . . . . . . . . . . . . . . . 10.2 The Strong Cauchy Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 Stationary Dihedral Angles for Stationary Planar Angles . . 10.4 Infinitesimal Rigidity of Polyhedra and Equilibrium of Hinge Mechanisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5 On the Deformation of Developments . . . . . . . . . . . . . . . . . . . 10.6 Rigidity of Polyhedra with Stationary Development . . . . . . . 10.7 Generalizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Infinitesimal Rigidity Conditions for Polyhedra with Prescribed Face Directions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 On Deformations of Polygons . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Infinitesimal Rigidity Theorems for Polyhedra . . . . . . . . . . . . 11.3 Relationship of Infinitesimal Rigidity Theorems with One Another and with the Theory of Mixed Volumes . . . . . . . . . . 11.4 Generalizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Supplements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.1 Supplement to Chapter 3: Yu. A. Volkov. An Estimate for the Deformation of a Convex Surface in Dependence on the Variation of Its Intrinsic Metric . . . . . . . . . . . . . . . . . . . . . 12.2 Supplement to Chapter 4: Yu. A. Volkov. Existence of Convex Polyhedra with Prescribed Development. I . . . . . . . . 12.3 Supplement to Chapter 5: L. A. Shor. On Flexibility of Convex Polyhedra with Boundary . . . . . . . . . . . . . . . . . . . . . . .

403 404 410 415 421 425 429 435

439 439 445 453 459 463

463 492 506

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525 Subject Index

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 537

Introduction

Content and Purpose of the Book The present book is not intended to embrace the whole theory of convex polyhedra. It mainly addresses the following question: which data could determine a convex polyhedron and to what extent? This question splits into two for all data pertinent to a polyhedron, e.g., the lengths of edges, areas of faces, etc. We firstly ask whether the data determine the polyhedron uniquely up to a motion or another trivial transformation (reflection, translation, or similarity) in the same fashion as the lengths of edges determine a triangle up to a motion, and the angles up to a similarity. The question is settled with the general uniqueness theorems for convex polyhedra given some data, uniqueness understood to within a motion or another trivial transformation. We secondly ask about necessary and sufficient conditions for the data to satisfy in order that there exist convex polyhedra with these data. This is done by analogy with the fact that a necessary and sufficient condition for the existence of a triangle with sides a, b, and c is the fulfillment of the three inequalities: the sum of two sides is greater than the third, i.e., a + b > c, b + c > a, and c + a > b. Since in all cases the necessity of the corresponding conditions is easy, the crux of the matter resides always in proving sufficiency, i.e., proving the existence of polyhedra with given data provided that the conditions are satisfied. We thus talk about the general existence theorems for convex polyhedra given some data. Moreover, we consider the data determining a polyhedron, i.e., those for which the corresponding uniqueness theorem holds in the same way as the specification of sides a, b, and c determines a triangle. The first general uniqueness theorem was proved by Cauchy [C] in 1913.1 This theorem reads: two closed convex polyhedra composed of the same number of equal similarly-situated faces are congruent; or, in the form of a unique1

Cauchy’s proof is an exemplar of witticism; however, it contains some mistakes that were later repaired by other geometricians [St1, p. 3]. An unimpeachable and lucid proof is given, for instance, in Hadamard’s book [H, Appendix K]. Claiming that Cauchy’s theorem is the first general uniqueness theorem, we leave aside the trivial theorems like, say, the uniqueness theorem for a closed convex polyhedron with given vertices.

2

Introduction

ness theorem: a closed convex polyhedron composed of given faces in a definite fashion is unique up to a motion or a motion and a reflection.2 In 1897 Minkowski [Min1] proved the existence and uniqueness (up to translation) of a closed convex polyhedron with prescribed face areas and face directions. (The precise statement of Minkowski’s theorem together with the existence conditions for polyhedra with prescribed face directions and face areas will be formulated later in due course.) Roughly speaking, this exhausts the list of general existence and uniqueness theorems.3 I have managed to supplement it with new theorems and handle all these theorems, Cauchy’s and Minkowski’s theorems inclusively, by a unified method. As regards uniqueness theorems, this is precisely the method that Cauchy used for proving his own theorem. Our method concerning existence theorems is an innovation in the theory of polyhedra, since it is based on topology, primarily on the so-called Domain Invariance Theorem. Pursuing a similar approach, S. P. Olovyanishnikov discovered two new existence and uniqueness theorems [Ol1, Ol2]. As a result of our research, the whole stock of theorems on convex polyhedra acquires ampleness and grace, deserving a separate systematic exposition. To accomplish this exposition is the first aim of the present book. To make the book comprehensible and interesting for a wide readership, I include the basics of the theory of convex polyhedra, since the literature lacks any systematic and relatively complete treatment of the subject. Topology, which we have just mentioned, often arouses a feeling of something extremely difficult and abstract in people not familiar with it. However, the background of topology consists mainly in evincing clear spatial ideas by using the language of mathematics. In this sense, topology is part and parcel of general geometry and the methods based on topology should be viewed as geometric, though abstract. To make the application of topology more perceivable, I include the exposition of those elements of topology which are necessary for our purposes. Moreover, the abstract notions of topology are needed only in a few sections of the book. Alongside the main results, I include some related results, for example the theorem on a polyhedron of greatest volume for a fixed area and “rigidity 2

3

We say that two polyhedra “have the same structure” if the faces, edges, and vertices of one polyhedron can be associated with the faces, edges, and vertices of the other so that the corresponding faces (edges) abut to the corresponding edges (vertices). Moreover, if the corresponding edges are equal as well as the angles at the corresponding vertices on the corresponding faces, then the polyhedra are “composed of equal similarly-situated faces.” I mention theorems of the above sort. Another important theorem standing aside from this circle is the existence theorem for closed convex polyhedra with a prescribed structure which was found by Steinitz [St2] in 1915. It falls beyond the scope of out concern, since the structure clearly fails to determine a polyhedron in a unique fashion. The proof of this theorem is reproduced in L. A. Lyusternik’s book [Ly]; also see [BaG].

Order and Character of the Exposition

3

theorems.” (The general notion of rigidity is expounded in Section 2.6 of Chapter 2.) Finally, I supplement the exposition of the main methods with related topics. As a result, appears an almost exhaustive exposition of the subject if we speak about the state of the art until 1950.4 The theory of polyhedra and related geometrical methods are attractive not only in their own right. They pave the way for the general theory of surfaces. Surely, it is not always that we may infer a theorem for curved surfaces from a theorem about polyhedra by passage to the limit. However, the theorems about polyhedra always drive us to searching similar theorems about curved surfaces. Moreover, the case of polyhedra reveals elementarygeometric grounds for more general results. To demonstrate these connections, I supplement almost every chapter with a section devoted to generalizations of the topics of the chapter. Among them appear not only generalizations to curved surfaces but also to polyhedra in hyperbolic space, etc. These generalizations are explained in abridged form, since they are not used in the main text. The above explains the second aim of the book: I have tried to exhibit the abundance of content and intrinsic ties innate in the theory of polyhedra and its geometrical methods.

Order and Character of the Exposition In the first chapter, as follows from its title, we introduce all basic notions and properties of convex polyhedra to be used in the sequel. This chapter presupposes no preliminary knowledge but elementary geometry. The second chapter, “Methods and Results,” explains the grounds of the method we use (Sections 2.1 and 2.2) and surveys the main results (Sections 2.3–2.6) to be derived in the next chapters. I have included a survey (immaterial to further exposition) so that the reader, acquainted with it, can steer through the next chapters, choosing topics that he or she is most interested in. As far as the method is concerned, in Section 2.1 of Chapter 2 we describe Cauchy’s method for proving the uniqueness theorems of Chapters 3 and 6. This method is rather elementary. In Section 2.2 we describe the background and underlying ideas for the method we use in proving the existence theorems of Chapters 4, 5, 7, and 9. This method is based essentially on some notions and results from topology. For convenience of the reader not familiar with these questions, all necessary topological facts are presented in the auxiliary 4

Among the themes available, I only omit the proof of rigidity of closed convex polyhedra by Dehn’s method [De]. However, in Chapter 10 we derive stronger results in another way.

4

Introduction

Sections 2.8 and 2.9. We may thus say that no preliminary knowledge of topology is needed for understanding the book. Furthermore, all chapters except for 4, 7, and 9 do not lean on these elements of topology. The general interdependence of chapters is shown in the following scheme from which we see in particular that Chapters 3, 6, and 9 are fully independent of one another. I

II

III

IV

V

VI

X

IX

VII

VIII

XI

Despite the elements of topology which, as mentioned, are explained in Sections 2.8 and 2.9 of Chapter 2 and used only in Chapters 4, 7, and 9, we manage with the simplest tools (elementary geometry, sums of vectors, and the derivative in Chapters 10 and 11). The only exceptions are Section 7.2 of Chapter 7 and Sections 8.2 and 8.3 of Chapter 8, where we use elements of analysis (an extremum of a function in several variables and integration). The above does not apply to the auxiliary sections “Generalizations” ending almost every chapter. While the exposition in the main text is thorough as far as possible, in these sections we only survey generalizations; moreover, we apply a diversity of tools: additive set functions, differential equations, etc. These sections are immaterial as regards the understanding of the main content of the book. During the exposition, I have formulated a series of problems. Most of them are exercises but some are unsolved problems that may serve as themes of independent research.

Remarks for the Professional

5

Remarks for the Professional This book contains a whole series of results that are published for the first time, to say nothing about modifications and improvements in the theorems and proofs known from the earlier articles.5 These are all theorems on polyhedra with boundary (Section 3.5 of Chapter 3, Chapter 5, Section 6.5 of Chapter 6, and Section 9.1 of Chapter 9), theorems on unbounded polyhedra (Sections 6.4 and 6.5 of Chapter 6 and Sections 7.3 and 7.4 of Chapter 7), and, finally, rigidity theorems (Chapters 10 and 116 ). In my opinion, many of these results are not significant enough, but some of them surely deserve attention. Likewise, the professional might find something new in the sections titled “Generalizations.” I wish to emphasize that until recently unbounded convex polyhedra have never been studied at all. I hope that this book, together with the first articles by Olovyanishnikov and me, shall justify interest in their study alongside closed polyhedra. This study seems all the more promising since, in place of an unbounded polyhedron, we may speak of a bounded polyhedron admitting endless prolongation for which rather simple necessary and sufficient conditions are available (Subsection 1.1.6). Polyhedra with boundary are less studied in view of their natural diversity. As far as I know, the present book provides the first results in this direction. The question of flexibility of these polyhedra, solved almost completely in Section 5.2, seems rather enthralling to my taste. In Section 6.5 of Chapter 6 and Section 7.4 of Chapter 7, I reproduce some elegant results by A. V. Pogorelov who acquainted me with them prior to publication. In a few places I use other apt observations by A. V. Pogorelov and V. A. Zalgaller. The latter and Yu. F. Borisov read the whole manuscript and corrected the errors I had made. Yu. F. Borisov also wrote Section 2.9 of Chapter 2 and Subsection 1.7.9 of Chapter 1. I express my deep gratitude to all these gentlemen for help.

5 6

Alongside the articles by Cauchy, Minkowski, and Olovyanishnikov, I bear in mind my articles [A2], [A6], [A8], and [A9]. An exception is the general rigidity theorem for a closed convex polyhedron. Another proof of this theorem is published in my book [A15].

1 Basic Concepts and Simplest Properties of Convex Polyhedra

1.1 Definition of a Convex Polyhedron 1.1.1 A polyhedron means a body bounded by finitely many polygons as well as a surface composed of finitely many polygons. Speaking of a polyhedron, we mostly mean a polyhedral surface, i.e., a figure formed by finitely many polygons. However, it is sometimes more convenient to regard a polyhedron as a body, for instance when we speak about a point inside a polyhedron or about one polyhedron inside another. In such cases we usually indicate that the polyhedron under consideration is solid. However, the similarity and distinction of the two notions are so transparent that the possible mingling in terminology involves no essential misunderstanding. Moreover, we will mainly deal with polyhedra forming entire surfaces of solid polyhedra, e.g., the complete surface of a cube. In such cases the meaning, for instance, of the expression “inside a polyhedron” is evident without stipulating that the polyhedron be solid. The polygons forming a polyhedron (or bounding a solid polyhedron) are referred to as the faces of the polyhedron, provided that all coplanar polygons with common sides or segments of sides are treated as a sole polygon, thus making a single face. The sides and vertices of the faces of a polyhedron are referred to as the edges and vertices of the polyhedron. Here and elsewhere by a polygon we mean an arbitrary planar domain that is bounded by finitely many (straight line) segments or segments and half-lines; moreover, the boundary of the domain itself is included in the polygon.1 The definition implies that we bear in mind not only bounded polygons as usual but also unbounded polygons,2 provided, however, that a polygon must have finitely many sides and vertices. In contradistinction to a bounded polygon, an unbounded polygon has at least two unbounded sides which are half-lines or lines, or it has one unbounded side which is a straight line. 1

2

The concept of domain implies connectedness, i.e., the possibility of joining every pair of points in a domain by a polygonal line (a piecewise linear curve) in the domain. It is for this reason that, for instance, two triangles touching at a vertex are considered as two polygons rather than a single polygon. That is, those lying in no circle of finite radius. – V. Zalgaller

8


A polyhedron is called unbounded or bounded depending on whether or not it has unbounded faces. By definition, the number of faces of a polyhedron is finite as well as the number of sides and vertices of each face. Therefore, an unbounded polyhedron has finitely many faces, edges, and vertices, just like a bounded one, with some of its faces and edges necessarily unbounded. The simplest example of an unbounded polyhedron is a polyhedral angle whose faces extend to infinity. Intersecting such a polyhedral angle by planes, we easily obtain other examples of unbounded polyhedra (Fig. 1). (An unbounded polyhedron might be thought of as a surface comprising infinitely many polygons; however, we do not consider such polyhedra at all. They are beyond our consideration since they are determined from infinite data.)

Fig. 1

1.1.2 A convex polyhedron. We now define a convex polyhedron, the principal topic of the present book. A convex polyhedron is a figure composed of finitely many planar polygons so that (1) it is possible to pass from one polygon to another through polygons having common sides or segments of sides3 ; (2) the entire figure lies on one side of the plane of each constituent polygon. It is the second condition that defines convexity; the first means that a polyhedron does not split into parts meeting only at vertices or even disjoint from each other. A convex solid polyhedron is defined as a body bounded by finitely many planar polygons so that it lies on one side of the plane of each of the polygons.4 3

4

The polyhedron in Fig. 3 has two faces, one touching the other along a segment of a side rather than a whole side. This is possible only for a convex polyhedron not bounding a solid polyhedron. In the topological language, a body is the closure of a domain, i.e., a domain joined with its boundary. The (complete) surface of a body is its boundary. A figure in the previous definition means nothing else but a set.

1.1 Definition of a Convex Polyhedron

9

It is clear from the comparison of the two definitions that every part of the surface, as well as the whole surface, of a convex solid polyhedron represents a convex polyhedron. This is certainly true provided that we only consider parts that consist of polygons and are not split into disjoint pieces or pieces touching only at vertices of polygons. We can also prove the converse: every convex polyhedron is the whole surface or part of the surface of a convex solid polyhedron. Indeed, consider a convex polyhedron P . Draw the planes of all its faces. By definition, the polyhedron lies on one side of each of these planes Qi . Take those half-spaces bounded by Qi which include the polyhedron P . The intersection of the half-spaces is a convex solid polyhedron, since by construction it lies on one side of each of the planes Qi and is bounded by the polygons cut out from these planes by their mutual intersections. The initial polyhedron P is part of the surface or the complete surface of the so-constructed solid polyhedron.

Fig. 2(a)

The polyhedra in Figs. 1 and 2(a) are convex, whereas those in Figs. 2(b) and 2(c) are not. To clarify the content of the theorems to be proved in this chapter, it is instructive to look at the polyhedra in Figs. 2(b) and 2(c): none of these theorems holds for either of the polyhedra.

Fig. 2(b)

Fig. 2(c)

10


1.1.3 Polyhedra with boundary; closed and unbounded polyhedra. Since a convex polyhedron lies on one side of the plane of each of its faces, it is easy to show that at most two faces meet at any edge and that any interior point of one face cannot belong to another face. A polyhedron representing only a part of the surface of a convex solid polyhedron may be characterized as having edges that belong to a single face. These are the edges bounding the part of the surface represented by the polyhedron in question. For this reason, we say that such a polyhedron has a boundary. The boundary is determined by the “boundary” edges, i.e., the edges which belong to exactly one face (Fig. 3).

Fig. 3

If an edge intersects the boundary in a segment, then this segment is also considered a boundary edge. A convex polyhedron without boundary is the complete surface of a solid polyhedron; therefore, such a polyhedron is naturally called complete. A complete bounded polyhedron, i.e., the boundary of a bounded solid polyhedron, is called closed. The unbounded complete convex polyhedra split into two classes: those with vertices and those with no vertices at all. Every doubly infinite prism, for instance an unbounded rectangular tube, is a polyhedron without vertices. Conversely, if a polyhedron has no vertices, then it is a prism that extends infinitely on both sides. Indeed, every edge of a polyhedron without vertices has no endpoints, and consequently each edge is a straight line. Since there are no vertices, these straight lines do not meet, and therefore the faces of such a polyhedron turn out to be either strips between parallel straight lines or half-planes (the faces of a dihedral angle). It is then clear that such a polyhedron is an infinite prism (possibly with a nonclosed cross-section as in a dihedral angle). Thus, the absence of vertices characterizes infinite prisms. The prisms are also distinguished among complete convex polyhedra by the condition that each of them contains a straight line; i.e., a complete convex polyhedron containing a straight line is a prism.


11

Indeed, let a complete convex polyhedron P contain a straight line L. Since the polyhedron lies on one side of the plane of each of its faces, none of these planes intersects L, i.e., the planes of all faces of P are parallel to L. It is now clear that the polyhedron is a prism with edges parallel to L. An infinite prism is determined by its cross-section. Hence, the properties of prisms are immediate from the properties of polygons. For this reason, we exclude prisms from our consideration, except for a few cases to be specified in due course. We always consider closed polyhedra first, addressing unbounded complete polyhedra with vertices afterwards. The latter will be simply referred to as unbounded, without mentioning their completeness and vertices. We break this rule to avoid confusion when we must distinguish such a polyhedron from a prism or an unbounded polyhedron with boundary. An unbounded polyhedron shall thus mean a complete convex polyhedron distinct from a prism, unless stated otherwise. 1.1.4 Another definition of convexity. As the definition of convexity for a polyhedron we took the condition that the polyhedron lies on one side of the plane of each of its faces. However, another definition of convexity is commonly used for bodies; it is more general as applicable not only to polyhedra. According to this definition, a body (or a set) is convex if, for each pair of points, it contains the entire segment between them. A cube, a ball, and a circular cylinder are particular instances of convex bodies. A disk, a segment, and the entire plane give examples of planar convex figures. The two definitions of convexity are equivalent for solid polyhedra. The proof is based on the following simple but important observation which we call the Separation Lemma: Each point not belonging to a given solid polyhedron convex in the sense of our initial definition is separated from the polyhedron by the plane of some face of the polyhedron (which means that the point in question and all interior points of the polyhedron lie on different sides of the separating plane). Indeed, assume that a point A does not belong to a polyhedron P . Then the segment joining A with an arbitrary interior point B of the polyhedron intersects the surface of the polyhedron and so has a common point with at least one face Q of P . Moreover, the segment AB cannot lie in the plane of Q; otherwise, this plane would pass through the interior point B and the polyhedron would not lie on one side of the plane of the face Q. Consequently, the segment AB has to intersect the plane of the face Q. However, if the polyhedron is convex, then by the definition of convexity it lies on one side of the plane of Q, exactly on the side containing B. Therefore, the plane of Q separates the polyhedron from the point A, which completes the proof of the Separation Lemma.

12


Now, we show that if a solid polyhedron P is convex in the sense that it lies on one side of the plane of each of its faces, then each pair of points in P can be joined by a segment in P . Assume by way of contradiction that the polyhedron P contains two points A and B such that on the segment AB there is a point C not belonging to P . Then by the Separation Lemma the point C is separated from P by some plane, which is impossible since each plane intersecting the segment AB separates the points A and B. We finally prove the converse assertion. Let a polyhedron P be convex in the sense that each pair of points in P can be joined by a segment in P . Assume, however, that the polyhedron P does not lie on one side of the plane of some face Q and hence contains points A and B that are to different sides from Q (see Fig. 4). A

X

Q

B Fig. 4

Connecting these points with all points X of Q, we thus obtain two pyramids with common base Q. Therefore, Q lies inside the polyhedron formed by the pyramids. However, each of the segments AX and BX must lie in the polyhedron P . Hence, Q must lie inside P , failing to be a face; a contradiction. So each of the convexity conditions implies the other, which yields their equivalence. Solid convex polyhedra turn out to be particular instances of convex bodies: they are convex bodies bounded by finitely many polygons. Similarly, “surface” convex polyhedra turn out to be particular instances of convex surfaces, if a convex surface is defined as the entire surface or any part of the surface of a convex body. The same arguments apply to the polygons after replacing faces and their planes with sides and the straight lines containing the latter. In exactly the same way we can prove the equivalence of the following two definitions of convexity for polygons:


13

(1) A polygon is convex if, given any side, the polygon lies in one half-plane bounded by the straight line through this side. (2) A polygon is convex if, for each pair of points, it contains the entire segment between them. The two definitions can be proved to be equivalent to the third one: (3) A polygon is convex if its angles are convex, i.e., less than 180◦. (We leave the proof as an exercise, since the last result is not used below. Another exercise is to prove the corresponding fact for polyhedra: a solid polyhedron is convex if all its dihedral angles are convex.) The following two simple properties of convex figures (sets) will be constantly applied to polyhedra: (1) The intersection F of any collection of convex figures Fs is a convex figure itself. Indeed, if points A and B belong to F , then they belong to each of the figures Fs . By convexity, every figure Fs contains the segment AB, hence the segment lies in their intersection, thus yielding the convexity of F . In particular, the intersection of a convex polyhedron and a plane is convex and is either a point, or a segment (or a half-line), or a convex polygon. If the plane under consideration is the plane of a face, then it cannot enter into the interior of the polyhedron and hence its intersection with the polyhedron is the convex polygon representing the given face. Therefore, all faces of a complete convex polyhedron are convex. (2) Every segment with endpoint an interior point O of a convex body intersects the surface of the body in, at most, one point.

A

B S

O

Fig. 5

Indeed, every interior point O is characterized by the condition that there is a ball S about O lying entirely in the body. Therefore, if a segment with endpoint O intersects the surface of the body at two successive points B and A, then convexity implies that the body includes all the segments that

14


join A to the points of S (Fig. 5). These segments cover an entire cone and therefore all points of the segment OA, except for A, must be interior, contradicting the assumption that B lies on the surface of the body. 1.1.5 Polyhedral angles and spherical polyhedra. A polyhedral angle is a polyhedron with a single vertex and no boundary edges. Hence, its faces are thought of as infinitely extendable. If we draw a sphere of unit radius about the vertex of a polyhedral angle, then the angle cuts out a spherical polygon from the sphere which is bounded by the arcs of great circles. Conversely, if such a polygon is given on the sphere, then the rays drawn from the center through the points of the polygon constitute a polyhedral angle. The concept of convexity is transferred from planar to spherical polygons in a straightforward way by replacing straight lines with great circles, and straight line segments with arcs of great circles not larger than a half-circle. The same three definitions of convexity can be given for spherical polygons.5 The equivalence of the first two definitions can be proved as above in the case of polyhedra. The equivalence of the third definition (claiming that all angles are less than 180◦ ) will be established in Subsection 3.1.2. Q

q

Fig. 6

Furthermore, it is easy to see that the correspondence between polyhedral angles and spherical polygons given by the above construction relates convex polyhedral angles to convex spherical polygons, and vise versa. For example, as clearly seen from Fig. 6, the location of a polyhedral angle on one side 5

Since two antipodal points of the sphere are joined by many shortest arcs (the meridians), the second definition of convexity will look like the following: a polygon on the sphere is convex if, together with every pair of points, it contains at least one shortest arc between them. For this reason, the intersection of two convex polygons on the sphere may happen to be nonconvex: the intersection of two digons may be the pair of their vertices. – V. Zalgaller


15

of the plane of a face Q is equivalent to the location of the corresponding spherical polygon on one side of the great circle q in which the plane Q intersects the sphere. Particular instances of spherical polygons are hemispheres and digons bounded by two great circles that join two antipodal points. Digons correspond to polyhedral angles degenerating into dihedral angles, and hemispheres correspond to polyhedral angles degenerating into planes. 1.1.6 A remark on unbounded polyhedra. The study of unbounded convex polyhedra might seem to have a less intuitive and practical meaning than the study of bounded polyhedra. However, this objection can be completely dismissed, because, instead of an unbounded convex polyhedron, we can always think of a bounded convex polyhedron with boundary such that the infinite extension of its extreme faces creates no intersections of any faces except for the pairs of adjacent faces with a common edge extended to infinity. One can formulate necessary and sufficient conditions on a polyhedron with boundary to admit the infinite extension of its extreme faces without new intersections, i.e., to serve as a “bounded” representative of an unbounded polyhedron together with all the vertices, edges, and faces of the latter. We formulate one condition that consists of two parts: (1) For each boundary vertex, there are exactly two extreme faces touching each other along the edge containing the vertex, while no internal faces are incident to it (i.e., as in Fig. 7 for the vertex A but not B). (2) The edges of each extreme face, if extended beyond the boundary, either diverge or are parallel, but do not get closer to each other.

A

B

Fig. 7

The necessity of both parts of this condition is obvious. The proof of sufficiency is left to the reader as an exercise; we will not use it later. Another condition is the convexity of the spherical image of the polyhedron. The notion of spherical image is introduced in Section 1.5, where we

16


also establish the necessity of the condition. Its sufficiency will be proved in Section 7.5. 1.1.7 Exercises. We add the following two exercises to those given above: 1. Prove that if a bounded solid polyhedron has no “cavities” and its intersection with every support plane6 is contractible, then the polyhedron is convex. (A figure is contractible if there is a continuous deformation that collapses it to a point without going beyond the boundary of the original figure. Two distinct points, a closed polygonal line, or a polygon with “holes” are not contractible.) 2. Prove that if a bounded “surface” polyhedron has the same property, then it is closed and convex. (The faces are assumed to meet pairwise along edges, but the presence of a boundary is not excluded a priori.)

1.2 Determining a Polyhedron from the Planes of Its Faces7

1.2.1 Support planes. A support plane of a figure is a plane having at least one point in common with the figure and such that the entire figure lies in one of the two half-spaces bounded by the plane (Fig. 8). n

Q M

Fig. 8

We assume that each half-space always includes its boundary plane, unless we explicitly refer to the half-space as “open.” The definition implies for instance that, for a planar figure, the very plane containing the figure is a support plane at each point of the figure. 6 7

For the definition of a support plane, see Subsection 1.2.1. Determining a polyhedron in this way is connected with an interesting trick proposed by Yu. G. Reshetnyak [R2] for integration over a convex polyhedron, in particular for finding its volume. – V. Zalgaller

1.2 Determining a Polyhedron from the Planes of Its Faces

17

Using the notion of support plane, we can say that a convex polyhedron is characterized by the condition that the planes of all its faces are support planes. However, a polyhedron obviously has other support planes touching it at its vertices or along its edges. The following simple theorem holds for convex solid polyhedra: Theorem 1. The intersection of a convex solid polyhedron and a support plane is either a point (a vertex of the polyhedron), a segment or half-line (an edge of the polyhedron), or a face. To every vertex, there is a support plane that passes through the vertex and has no other points in common with the polyhedron. To every edge, there is a support plane that passes through the edge and intersects the polyhedron only along that edge. The plane of any face is a support plane and it intersects the polyhedron only in that face. Proof. The intersection of a plane and a convex polyhedron is convex as is the intersection of two convex figures. Therefore, it can be either a point, or a segment (or a half-line), or a convex polygon. (Entire straight lines are eliminated by the convention in Subsection 1.1.3 that a polyhedron includes no straight line.) If a plane shares exactly one point with the polyhedron, then the point is a vertex, since it is obviously impossible for a plane to touch the polyhedron at a single point inside an edge or a face. For similar reasons, a single segment common to a plane and the polyhedron must be an edge. This proves the first claim of the theorem. The plane of an arbitrary face is a support plane by the definition of convex polyhedra. Its intersection with the polyhedron is a convex polygon which is thereby a face. If faces Q1 and Q2 touch along an edge a, then the polyhedron lies between these faces, i.e., inside the dihedral angle between them. Therefore, every plane passing through the edge in the complementary angle will be a support plane along the edge a (Fig. 9).

a Q1

Fig. 9

Q2

18


Given a vertex A, there are at least three faces touching at A. Therefore, the polyhedron lies inside the trihedral or polyhedral angle bounded by the planes of these faces. Clearly, it is always possible to draw a plane through the vertex of the polyhedral angle so that the plane has no other points in common with the angle. It suffices to take a support plane along an edge and slightly turn it about the straight line passing through the vertex and perpendicular to the edge (Fig. 10). This plane will be a support plane of the polyhedron only at the vertex A.

A

Fig. 10

1.2.2 Outward normals. Support numbers. If a figure (a set) M lies on one side of a plane Q, then the unit vector n perpendicular to Q and pointing to the interior of the (open) half-space not containing the points of M is called the outward normal of Q relative to M (Fig. 8). The outward normal to a support plane is defined similarly. Speaking of the direction of an arbitrary plane, we bear in mind the direction of some vector perpendicular to the plane. The direction of a support plane is always the direction of its outward normal. The direction of a face of a polyhedron is the direction of the outward normal vector. n

h< 0

h>0 O n

Fig. 11

1.2 Determining a Polyhedron from the Planes of Its Faces

19

If we choose a point O in space as the origin of coordinates, then the position of an arbitrary “directed” plane is determined by its normal n and the distance h to the origin O. Moreover, the distance is taken positive if the origin lies on that side of the plane which is opposite to the direction of n. Otherwise, the distance is negative (Fig. 11). This signed distance is called the support number of the plane. The support number of the plane through a face of a polyhedron is called the support number of the face. The support numbers of a polyhedron are the support numbers of its faces. We prove that every complete polyhedron is uniquely determined by the planes of its faces with specified outward normals. In other words, a complete polyhedron is determined by its outward normals and support numbers. Theorem 2. A convex solid polyhedron is the intersection of the half-spaces bounded by the planes of its faces. (Since each plane determines two halfspaces, we consider those half-spaces that contain the polyhedron.) Proof. First, by definition every convex solid polyhedron lies on one side of the plane of each of its faces, i.e., it always lies in one of the half-spaces determined by the plane. Therefore, the polyhedron lies in the intersection of these half-spaces. Second, by the Separation Lemma proved in Subsection 1.1.4, each point outside the polyhedron is separated from the latter by the plane of some face, i.e., such a point does not belong to at least one of the half-spaces in question. Hence, the intersection of these half-spaces includes the polyhedron but contains no other points, i.e., the intersection coincides with the polyhedron, which was to be proved. Theorem 3. Every complete convex polyhedron is determined by the planes of its faces and their outward normals. Indeed, the outward normals determine the half-spaces containing the polyhedron. By Theorem 2, the solid polyhedron bounded by the complete polyhedron P is the intersection of these half-spaces and consequently is fully determined by them. Thereby its surface P is determined too, which completes the proof of the theorem.

Fig. 12

20


This cannot be done without the outward normals, since there exist different convex polyhedra for which the planes of all faces coincide. An example is given by two polyhedra into which a tetrahedron is divided by a plane intersecting all of its faces (Fig. 12). Exercise. Find conditions under which the planes of the faces of a given bounded convex polyhedron bound a different bounded convex polyhedron. Solve the same problem for unbounded polyhedra. To get started, consider an analogous problem for polygons. 1.2.3 We will now prove a theorem which is in some sense the converse of Theorem 2. Theorem 4. The intersection of finitely many half-spaces is a convex solid polyhedron under the necessary condition that the intersection has nonempty interior.8 If the outward normals to the boundary planes of these half-spaces do not point into a single half-space,9 then the polyhedron is bounded; otherwise it is unbounded. Let the intersection of half-spaces with boundary planes Q1 , Q2 , . . . , Qm have interior points. Then it represents a body bounded by the polygons cut out from the planes Qi by the straight lines in which the planes intersect one another. Moreover, the body lies on one side of each of the planes Qi and therefore represents a convex solid polyhedron. Denote this polyhedron by P . If the outward normals of the planes Qi point into a half-space R (Fig. 13), then the direction of the outward normal m of R does not form an acute angle with any of them. Therefore, the half-line drawn in this direction from any point of the polyhedron P is disjoint from all the planes Qi and thereby lies inside P . Thus, the polyhedron P contains a half-line and is consequently unbounded. If the outward normals of Qi do not point into a single half-space, then it is impossible for the polyhedron P to be unbounded. Were P unbounded, it would contain a half-line, for instance an unbounded edge. The planes bounding the polyhedron would not meet this edge; in consequence, the outward normals of the planes would point into the half-space bounded by the plane perpendicular to this half-line. This completes the proof of the theorem. 8

9

The intersection of half-spaces may have no interior points; for example, the intersection of two complementary half-spaces with a common boundary plane is the plane itself. Some half-spaces may have no common points at all; in this case their intersection is empty. Normals n1 , . . . , nm point into one half-space R if, issued from an arbitrary point of the boundary plane of R, they lie in R. It is not excluded that some of the vectors belong to the boundary plane. If such vectors are absent, then we say that the normals go inside R. If none of the normals goes inside R, then they all lie in the plane and the polyhedron is an unbounded prism.

1.3 Determining a Closed Polyhedron from Its Vertices

n1

n1 n4

n5

21

n4 n3

n3

n5

R

n2

n2

m

Fig. 13

1.2.4 We established that the intersection of finitely many half-spaces is a convex solid polyhedron, provided that the intersection has nonempty interior. However, some of the planes of the half-spaces can be “superfluous,” i.e., it is possible that they are not the planes of faces of the polyhedron but merely touch the polyhedron at edges and vertices or even have no common points with the polyhedron. For this reason, the following question arises: What are necessary and sufficient conditions for the planes with given outward normals n1 , n2 , . . . , nm and support numbers h1 , h2 , . . . , hm to serve as the planes of faces of some convex solid polyhedron? The theorem answering this question is formulated in Subsection 2.4.6. According to the general scheme outlined in the Introduction, we may say that the question is about an existence theorem for polyhedra with given normals and support numbers (corresponding to the proved uniqueness theorem, which states that a convex solid polyhedron is determined from such data in a unique fashion).

1.3 Determining a Closed Polyhedron from Its Vertices 1.3.1 The convex hull. The convex hull of a set is the intersection of all convex solid polyhedra containing the set. The convex hull of an arbitrary set M is convex, since the intersection of any number of convex sets is a convex set as proved in Subsection 1.1.4. Since the convex hull of a set M is convex and lies in each convex polyhedron containing M , it is in some sense the smallest convex set containing M . It is completely defined by M , because M fully determines the collection of all convex polyhedra containing M . So the convex hull of a given set M is exactly the smallest closed convex set containing M .

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According to our definition, the convex hull is always a closed set. This follows from the general theorem claiming that the intersection of closed sets is closed. We can in general define the convex hull of a given set M as the intersection of all closed convex sets containing M .10 This general definition more adequately reflects the idea of the convex hull of M as the smallest closed convex set containing M . It is equivalent to the previous definition. Strictly speaking, we should consider half-spaces and the whole space as polyhedra, since one can consider sets lying in no convex polyhedron. We will use our former definition as more elementary and quite sufficient for our purposes, since we deal only with convex polyhedra. 1.3.2 Theorem 1. Every convex solid polyhedron is the convex hull of its vertices. To prove the theorem, choose some vertex A of the polyhedron and, using diagonals, partition the faces disjoint from A into triangles with vertices at the vertices of the polyhedron. Connecting A with the points of every such triangle, we obtain a trihedral pyramid, i.e., a tetrahedron (Fig. 14). All these tetrahedra are contained in the polyhedron due to the convexity of the polyhedron, and the polyhedron is the union of these tetrahedra.

A C

D

X

B Fig. 14

The convex hull of the set of vertices contains all segments between the latter, since it is convex. However, along with a segment BC and a point D, it includes the triangle BCD, since this triangle is the union of the segments from D to the points of BC. Finally, along with the triangle BCD and a point A, the convex hull includes the tetrahedron ABCD, since it is the union of the segments from A to the points of the triangle BCD. It is then clear that the convex hull of the set of vertices of the polyhedron includes all tetrahedra into which the polyhedron is split, i.e., the convex hull includes the polyhedron itself. On the other hand, our polyhedron certainly includes the convex hull of the set of its vertices, since by definition the convex hull is included in every 10

Nowadays, we call this the closed convex hull of M . For simplicity, we omit the word “closed” since we never use the “nonclosed” convex hull that is defined as the intersection of all convex sets containing M . – V. Zalgaller


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convex polyhedron containing the vertices. Thus, the polyhedron and the convex hull of the set of its vertices coincide, which was to be proved. Since the convex hull of a set M is determined uniquely from M , Theorem 1 implies that a bounded solid polyhedron and its surface, i.e., a closed convex polyhedron, are determined uniquely from its vertices. We say that a closed convex polyhedron is “spanned” by its vertices. This is the uniqueness theorem for closed convex polyhedra with given vertices. Allowing nonconvex polyhedra, we obtain no such theorem: Fig. 15 displays two polyhedra, one with the edge AB and faces ABC and ABD is convex and the other with the edge CD is nonconvex. It is clear that these polyhedra can be positioned so that their vertices coincide. A

B

C

A

D

B

C D

Fig. 15

1.3.3 Theorem 2. The convex hull of finitely many points A1 , . . . , Am not lying in a single plane is a convex solid polyhedron; the convex hull of a finite number of points lying in a single plane is either a polygon, or a segment, or a point. The vertices of this convex hull (i.e., those of the polyhedron, polygon, or segment) are some (not necessarily all) of the points Ai . A point Ak is a vertex if and only if it is not contained in the convex hull of the set of the remaining points. Proof. A single point A1 is its own convex hull. If there are two or more points all lying on a single straight line, then their convex hull is the segment whose endpoints are contained in the initial collection of points. Therefore, we can assume that the points are not collinear. Assume that the theorem holds for m− 1 points and prove it for m points. Let m points A1 , A2 , . . . , Am be given. Choose one of them, say Am . Let P denote the convex hull of the set of the remaining points A1 , A2 , . . . , Am−1 . By assumption, the latter is either a polyhedron whose vertices are among the points Ai , or a segment, or a single point. We can assume that P is a polyhedron; otherwise, the arguments are even easier. If the point Am is contained in P , then its presence does not change the convex hull. In particular, it is obvious that Am cannot be a vertex. Now, assume that the point Am does not lie in P (Fig. 16). Drawing all straight line segments from Am to every point of P , we obtain a polyhedron P1 which is composed of pyramids with the common vertex Am and with bases

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the faces of the polyhedron P . The polyhedron P1 is certainly contained in the convex hull of the set of all the points A1 , A2 , . . . , Am , since the convex hull includes the polyhedron P and all the segments joining the points of P with Am . It remains to show that the polyhedron P1 is convex. Let B and C be two points of P1 . By construction they lie on the segments Am B0 and Am C0 joining Am to some points B0 and C0 of the polyhedron P (Fig. 17). However, P is convex and therefore includes the segment B0 C0 . But then P1 must include the whole triangle Am B0 C0 and hence the segment BC. This proves the convexity of P1 . Am Am C

B B0 P

Fig. 16

P

C0

Fig. 17

However, if P1 is convex and contains the points A1 , A2 , . . . , Am , then P1 certainly contains their convex hull. As we have already established, P1 is also contained in it. Therefore, P1 coincides with this convex hull. We still need to show that that the point Am is a vertex of P1 . This is obvious, since by the Separation Lemma (Subsection 1.1.4) the convex polyhedron P can be separated from the point Am by a plane so that the segments joining Am to the points of P intersect the plane. Hence, the point Am can lie neither on an edge, nor on a face of P1 , nor in the interior of P1 . The theorem is thus completely proved. Combined with Theorem 1, Theorem 2 readily yields the following result: Theorem 3. There exist a closed convex polyhedron with given vertices A1 , . . ., Am , if and only if the points Ai do not lie in a single plane and none of them lies in the convex hull of the others. Furthermore, a polyhedron with given vertices is unique. Consequently, the theorem answers the questions raised in the Introduction: To what extent and under which conditions does the specification of vertices determine a closed convex polyhedron? Vertices can be specified by their coordinates; hence, the conditions imposed on them in the theorem can be expressed analytically.


25

1.3.4 Completing a polyhedron to a closed polyhedron. Theorems 1 and 2 imply another theorem which we shall have the opportunity to use. Theorem 4. Every bounded convex polyhedron with boundary can be completed to a closed polyhedron without adding new vertices; moreover, the completion is unique. In other words, for every bounded convex polyhedron P , there is a unique closed convex polyhedron P such that P is a part of P and P has no other vertices but the vertices of P . (The vertices of P need not all be vertices of P . An arbitrary polyhedron with boundary has vertices of two types: internal vertices, i.e., those not lying on the boundary, and boundary vertices. All internal vertices remain vertices of P , but this is not always so for boundary vertices. For example, this happens if P is a cube with a polygon cut out from a face.) Proof. Let P be a bounded convex polyhedron and let R be the convex hull of its vertices. Since R contains all vertices of P , it also contains the segments between them, i.e., the sides and diagonals of faces. So R includes all faces of P , and P is certainly contained in R. On the other hand, according to Subsection 1.1.2 the polyhedron P is part of the surface of some convex solid polyhedron Q, where Q contains all the vertices of P and hence their convex hull R. Thus it turns out that, while P is included in R and R is included in Q, the polyhedron P itself lies on the surface of Q. Consequently, P is part of the surface of the polyhedron R, i.e., part of a closed convex polyhedron. (Indeed, if a point X of P does not lie on the surface of R, then it lies in the interior of R, since R contains P . But then, since Q contains R, the point X lies in the interior of Q. This, however, contradicts the fact that P is part of the surface of the polyhedron Q.) According to Theorem 2, R, the convex hull of the vertices of P , is a solid polyhedron whose vertices are those of P . Moreover, there is no other convex solid polyhedron with this set of vertices. Consequently, the same holds for the closed polyhedron forming the surface of the convex hull R, which completes the proof of the theorem. 1.3.5 Exercises. 1. Find necessary and sufficient conditions under which a given boundary vertex of a convex polyhedron with boundary remains a vertex after the completion mentioned in Theorem 4. 2. Prove that the convex hull of an arbitrary set M is the intersection of all half-spaces containing M (provided that M lies in at least one half-space). Prove Theorems 1 and 2 by using this definition of convex hull. 3. Prove that the intersection of all convex sets containing a given set M is the union of all tetrahedra whose vertices are points of M . (The set M is not assumed to contain only finitely many points. The tetrahedra may overlap. Triangles, segments, and points are limiting cases of tetrahedra.)

26


−−→ 4. Given points A1 , A2 , . . . , An , let ai = OAi denote the vectors from the origin O to them. Assign nonnegative numbers mi (masses) to the points Ai . −→ The center of the masses is defined to be the endpoint A of the vector a = OA given by the equation mi ai m1 a1 + m2 a2 + . . . + mn an = a= m1 + m2 + . . . + mn mi (it is assumed that mi > 0, so that not all mi vanish). Prove that the convex hull of the points A1 , A2 , . . . , An is the set formed by the centroids of all possible masses placed at A1 , A2 , . . . , An . Prove that the intersection of all convex sets containing a given set M is the set formed by the centroids of all possible masses placed at all possible quadruples of points in M . (M is not assumed to consist of finitely many points. Compare this exercise with Exercise 3.) 5. The half-space bounded by a plane with the equation Ax + By + Cz + D = 0 is determined by the inequality Ax+ By + Cz + D ≤ 0 (or the opposite inequality Ax + By + Cz + D ≥ 0). Being the intersection of half-spaces, a polyhedron is therefore determined from simultaneous linear inequalities. Given points A1 , . . . , Am with coordinates x1 , y1 , z1 , etc., find simultaneous linear inequalities determining the convex hull of the points. The coefficients must be expressed in terms of the coordinates of the points.

1.4 Determining an Unbounded Polyhedron from Its Vertices and the Limit Angle 1.4.1 The limit angle. An unbounded convex polyhedron, as opposed to a bounded one, is certainly not determined solely by its vertices. For example, a convex polyhedral angle has a single vertex and is determined not by the latter alone but by additionally specifying the directions of its edges. Our purpose is to prove that every unbounded convex polyhedron is likewise determined by its vertices and the directions of its unbounded edges. However, instead of the directions of unbounded edges, it turns out to be more convenient to deal with a polyhedral angle whose edges are parallel to the unbounded edges of the polyhedron. We define this polyhedral angle by the following construction. Let P be an unbounded polyhedron. Take an arbitrary point O and draw from it all half-lines parallel to the half-lines lying in P (and having the same directions). These half-lines fill in some conical surface with vertex O. We call this surface the limit angle of P .11 11

It is also called the recession cone of P .

1.4 Determining an Unbounded Polyhedron

27

It is clear from the definition that replacing O with another point results in a parallel translation of the limit angle. Henceforth we always assume that the limit angle is determined up to translation. We now prove that the so-defined conical surface is in fact a polyhedral angle whose edges are parallel to the unbounded edges of P . Let Q be an arbitrary face of P and let q1 and q2 be the unbounded sides of Q. Every half-line p in Q must lie inside the angle between q1 and q2 (Fig. 18). It is clear that if we draw all half-lines from O that are parallel to the half-lines in Q, then they will cover the angle VQ between the half-lines parallel to the edges q1 and q2 .12 It is natural to call this angle the limit angle of the face Q.

O O

Q

VQ

Q2

Q1 Q

q2

q1

VQ1 VQ2

p V

P

Fig. 18

Fig. 19

It is now obvious that the limit angle of a polyhedron consists of the limit angles of its unbounded faces. Moreover, if faces Q and Q meet in an unbounded edge g, then their limit angles VQ and VQ share the edge parallel to g. Consequently, the limit angle is indeed a polyhedral angle whose edges and faces are parallel to the unbounded edges and faces of the polyhedron (Fig. 19). If the unbounded sides of a face Q are parallel to one another, then there is no half-line on Q with another direction and hence the limit angle of Q is reduced to a single half-line. The corresponding “face” of the limit angle of the polyhedron degenerates into an edge. If all unbounded edges of a polyhedron are parallel to one another, then the entire limit angle degenerates into a halfline (Fig. 20(a)). 12

Since we talk about a convex polyhedron, the face Q is a convex polygon and what was said above is trivial and easy to prove. However, the definition of limit angle is also applicable to nonconvex polyhedra; in this case the picture may change, but in essence our conclusions still remain valid.

28


O O

V P Fig. 20(a)

Fig. 20(b)

If all unbounded edges of a polyhedron are parallel to a single plane, then the limit angle is reduced to a planar angle (Fig. 20(b)). Whenever we talk about some limit angle, these cases are not excluded in advance. If an unbounded polygon undergoes the infinite similarity contraction to some point O, then in the limit it transforms obviously into its limit angle (Fig. 21). Therefore, under an infinite similarity contraction of an unbounded polyhedron P with respect to a point, its unbounded faces transform into their limit angles and the polyhedron itself transforms into its limit angle. Consequently, the limit angle may also be defined as the result of an infinite similarity contraction of the polyhedron. O

VQ

X

Q

Fig. 21

It follows easily that if the vertex O of the limit angle is contained in the polyhedron, then the limit angle itself lies in the polyhedron (provided that the polyhedron is convex). Indeed, under the contraction of the polyhedron P to the point O, all points X of P move along the segments OX. By the convexity of P , the segments OX lie in P and, for this reason, during the


29

contraction the polyhedron P does not protrude from its original position, and therefore the limit angle is included in P . Furthermore, it is easy to prove that the limit angle of a convex polyhedron is also convex. Indeed, take a point O on an unbounded face Q of P and begin contracting P to O. By convexity, the polyhedron always remains on one side of the plane of the face Q. In the limit, we obtain the limit angle V which also lies on the same side of the plane of Q. However, Q is now the plane of the corresponding face VQ of the limit angle. Thereby we proved that the limit angle V lies on one side of the plane of each of its faces, i.e., V is convex. All the conclusions above are summarized in the following theorem on the limit angle of a polyhedron. Theorem 1. Let P be an unbounded convex polyhedron. If we draw all half-lines from some point O that are parallel to the half-lines lying in P , then we obtain the limit angle VP of P . This is a convex polyhedral angle (or a planar angle or a half-line) whose edges and faces are parallel to the unbounded edges and faces of the polyhedron (moreover, the faces of P with unbounded edges not parallel to one another correspond to the faces of VP not degenerating into edges). If the point O lies in P , then the angle VP lies in P (Fig. 22). The angle VP can also be obtained as the result (i.e., the limit) of a certain infinite similarity contraction of P .

Fig. 22

The last characterization rather clearly exposes the essence of the concept of the limit angle as a polyhedral angle that determines the properties of the polyhedron in its “infinite” part when the finite quantities are neglected. A polyhedron, if regarded from an infinitely large distance, looks like its limit angle.

30


1.4.2 Theorem 2. Every convex polyhedral angle is the boundary of the convex hull of its edges and is therefore uniquely determined by specifying its vertex and the directions of its edges. (We say that such an angle is spanned by its edges.) Indeed, if a solid angle is convex, then, together with two rays p and q starting at the vertex, it must contain the planar angle between them, since the latter is covered by the segments between the points on p and q. For the same reason, together with three rays p, q, and r, it also contains the solid trihedral angle with edges p, q, and r. It follows that the convex hull of the set of edges of a polyhedral angle includes all solid trihedral angles with edges on these rays. At the same time, a solid polyhedral angle is itself composed of such trihedral angles. Therefore, it coincides with the convex hull of the set of its edges, which completes the proof of the theorem. Theorems 1 and 2 imply Theorem 3. Specifying the limit angle of an unbounded convex polyhedron amounts to specifying the directions of the unbounded edges of the polyhedron, i.e., the limit angle determines these directions and is itself defined by them as the boundary of the convex hull of the half-lines drawn from some point in the directions of the unbounded edges of the polyhedron. The theorem is immediate, since, first, the edges of the limit angle are parallel to the unbounded edges of the polyhedron and, second, a convex polyhedral angle is determined from its vertex and the directions of its edges. Incidentally, Theorem 3 also implies that the limit angle of an unbounded convex polyhedron can be defined as the polyhedral angle spanned by the half-lines drawn from an arbitrary point in the directions of the unbounded edges of the polyhedron, i.e., as the boundary of the convex hull of the set of these half-lines. Also, observe that the limit angle of an unbounded polyhedron P can be obtained in the following ways: (1) If we take an arbitrary point on P or inside P and draw all half-lines that lie in the body bounded by P , then all of them will cover a solid angle whose boundary is exactly the limit angle of P . (2) If we draw planes through some point O parallel to the planes of the unbounded faces of P , then they will bound a solid angle whose surface is the limit angle of P . More precisely, we mean the solid angle that appears as the common part of those half-spaces, bounded by the drawn planes, whose outward normals are parallel to the outward normals of the unbounded faces of the polyhedron. (Without this condition the definition is ambiguous, since, for example, three planes simultaneously bound eight trihedral angles.) We will not prove that these constructions actually yield the limit angle, since we do not use them. Moreover, the result seems rather evident, and a rigorous proof may serve as a good exercise.


31

1.4.3 In the sequel, we need the following lemma: Lemma. If a convex solid polyhedron P contains a half-line a, then it also contains every half-line b which is parallel to a, has the same direction, and starts at any point B of P . Indeed, let R denote the half-space that is bounded by the plane of some face of P and includes P . The half-line a lies in this half-space and so does the point B. But then the half-line b from B parallel to a lies in R too. Thus, the half-line b lies in every half-space that is bounded by the plane of a face of P and includes P . Since the polyhedron P is the intersection of such half-spaces, the half-line b lies in P , as required. Theorem 4. An unbounded convex polyhedron is uniquely determined from its vertices and limit angle. It is the boundary of the convex hull of the figure formed by the vertices and the limit angle, provided that the vertex of the latter lies inside the polyhedron. According to Theorem 3, the specification of the limit angle of a polyhedron is equivalent to the specification of the directions of its unbounded edges. In consequence, Theorem 4 may be restated as follows: An unbounded convex polyhedron is determined from its vertices and the directions of its unbounded edges. It is the surface of the convex hull of the figure formed by the vertices and the half-lines drawn from each of its points in the directions of the unbounded edges. Proof. Let P be an unbounded convex polyhedron and let R denote the convex hull of the figure formed by the limit angle V and the vertices of P . We assume the vertex of V to lie in P ; then the angle V itself is contained in P or, more precisely, in the solid polyhedron P bounded by the polyhedral surface P . Since the convex hull R of the vertices and the angle V is the intersection of all convex solid polyhedra containing them, R lies in the polyhedron P . A P Y

X

Z

q

Q Fig. 23

Conversely, let us prove that P is contained in R. Since the vertices of P lie in R, the convexity of the body R implies that R also contains all bounded edges of P . Next, R includes the angle V and so all edges of V , i.e., the halflines parallel to the unbounded edges of the polyhedron P . But then, by the

32


above lemma, R includes the parallel half-lines that start at its points. Taking the vertices of P as these points, we see that R also includes all unbounded edges of R. Thus, R includes all edges of P . However, if a convex figure includes all edges of a convex polygon, then this figure obviously includes the polygon itself. (To prove this, let Q be a convex polygon and let A be one of its vertices (Fig. 23). The polygon Q is contained in the angle between the half-lines p and q starting at A along the adjacent edges. Let X be an arbitrary point of Q. Draw a straight line through X intersecting p and q. This line intersects the perimeter of the polygon at two points Y and Z. Therefore, if the convex figure contains all sides of the polygon Q, then it contains the segment Y Z and the point X. Thus, all points X of Q belong to the figure. This remark relies upon the existence of vertices. Without them, it may be false, since a half-plane can contain a straight line that bounds another half-plane not included in the first half-plane.) Hence, with all edges of the polyhedron P , the polyhedron R contains all faces of P , i.e., R contains the entire boundary of P . So R contains the polyhedron P itself. The last assertion is obvious and its rigorous proof is as follows: Take a point X inside the polyhedron P . Draw a plane through X and two arbitrary points on the surface of P . This plane intersects P in a convex polygon whose sides lie on the surface of P and consequently belong to the convex hull R. However, if all edges of a convex polyhedron lie in some convex figure R, then the polyhedron itself is contained in the figure. Thus, every point X of P belongs to R, i.e., P is contained in R. Thus, we have proved that both P contains R and R contains P ; hence, the two sets coincide. Therefore, their boundaries coincide, implying that the polyhedron P is the boundary of the convex hull R of its vertices and the limit angle, which is the required conclusion. 1.4.4 Let us compare the just proved Theorem 4 with Theorem 1 of Section 1.3 claiming that a closed convex polyhedron is determined from its vertices. We may interpret the directions of unbounded edges as “infinitely distant vertices” of an unbounded polyhedron. This is especially true if we represent the unbounded polyhedron as the limit of bounded polyhedra. In particular, as a vertex moves to infinity, the edges adjacent to it become parallel; therefore, it is clear that parallel unbounded edges give a single “infinitely distant vertex” in accordance with the fact that they have the same direction. This observation could be taken as a basis for the proof of Theorem 4. On this basis, we now sketch a proof of the theorem stating that the boundary of the convex hull of the figure formed by finitely many points and a polyhedral angle V is an unbounded convex polyhedron with limit angle V . Given points A1 , . . . , Am , let V be a convex polyhedral angle whose vertex is assumed to be among A1 , . . . , Am (Fig. 24). Draw a plane Q which intersects all edges of the angle V ; it cuts out a pyramid V . Choose the plane Q at


33

a distance from the vertex of V large enough for all the points Ai to lie on the same side of it as the vertex. The plane Q intersects the edges b1 , b2 , . . . , bn of the angle V at points B1 , B2 , . . . , Bn . By Theorem 2 of Section 1.3, the convex hull of the set of all the points Ai and Bi is a convex polyhedron P containing the pyramid V . (In Fig. 24, the edges of the polyhedron P are indicated by dash-lines.) A1

Ak

A2

V Bj

Q

bj

Fig. 24

If we move the plane Q to infinity, then the polyhedron P increases and in the limit gives some polyhedron P including the given angle V . This polyhedron P is the convex hull of the points A1 , . . . , Am and the angle V . When we move the point Bj to infinity along the edge bj of V , each segment Ak Bj beginning at Ak becomes parallel to the edge bj . Hence we can conclude that the unbounded edges of P are parallel to the edges of V , so that V is the limit angle of P , since by Theorem 3 the limit angle is determined exactly by the fact that its edges are parallel to the unbounded edges of the polyhedron. The indicated argument requires clarifying some steps. We have to prove that the polyhedron P , being the limit of convex polyhedra, is itself convex. We also have to prove that, for each edge of V , P has an unbounded edge parallel to it. Leaving these details to the reader, we now give an exhaustive proof of the theorem without using the limit of bounded polyhedra. 1.4.5 Theorem 5. The boundary of the convex hull of the figure formed by finitely many points A1 , . . . , Am and a convex polyhedral angle V is a convex polyhedron P with the limit angle V . (If the angle V degenerates into a planar angle or a half-line, and the points A1 , . . . , Am lie in the same plane as V , then the polyhedron P degenerates into a polygon. To avoid additional hypotheses, we implicitly assume

34


that this case is not excluded. Also, recall that according to Subsection 1.1.3 we only need to consider unbounded polyhedra that do not contain a straight line.) We may assume the vertex of V to be one of the points Ai , adding the vertex to these points if necessary. For convenience, by polyhedra and angles we shall mean solid polyhedra and angles. We carry out the proof by induction on the number of edges of the angle V . First, suppose that V is a half-line a. Take the convex hull of the points A1 , . . . , Am . According to Theorem 2 of Section 1.3, this is a convex polyhedron P0 . We now translate a so that its starting points cover the whole polyhedron P0 ; the translates of a then cover some body P (Fig. 25).

X0

Q

P0

P0

Y0

a Y

a P

X aX

Fig. 25

aY

Fig. 26

Let a point X lie inside the polyhedron P0 . The half-line aX starting from X parallel to a meets the boundary of P0 . It follows that the same body P arises if we translate the starting point of a only to points on the faces of P0 . However, while moving along a face Q, the half-line a covers the infinite prism with base Q and lateral edges parallel to a. The whole body P is composed of such prisms and hence is a polyhedron. The polyhedron P is convex. Indeed, if X and Y are two points of P , then by construction they lie on the half-lines aX and aY starting at some points X0 and Y0 of the polyhedron P0 parallel to the half-line a (Fig. 26). The segment X0 Y0 lies in the polyhedron P0 , since P0 is convex. Therefore, the half-lines starting at the points of this segment parallel to a are contained in P . However, they form a strip between aX and aY . Thus, it turns out that, together with the points X and Y , the polyhedron P contains the segment XY , which proves the convexity of P . The polyhedron P is convex and contains the points Ai and the half-line a; hence, it contains their convex hull R. On the other hand, R obviously contains P0 , since P0 is the convex hull of the points A1 , . . . , Am . Furthermore, according to the lemma of Subsection 1.4.3, if a convex polyhedron contains P0 and the half-line a, then it


35

contains every half-line that is parallel to a and starts at a point of P0 . The convex hull R, which is the intersection of such polyhedra, has the same property, i.e., it contains all half-lines parallel to a and starting at points in P0 . But this means that R includes the whole polyhedron P . Thus, the convex hull R contains the polyhedron P and is itself contained in P ; hence, P is the required convex hull. It is clear from the construction that all unbounded edges of P are parallel to the half-line a. Hence, the limit angle of P is this half-line. The theorem is thus proved in the case when the angle V is a single half-line a. If we have an angle V with edges a1 , a2 , . . . , an , then we construct the required polyhedron P by iteration. First, we translate the initial point of the edge a1 to all points of P0 to obtain a polyhedron P1 ; next, we translate the beginning of the edge a2 to all points of P1 to obtain a polyhedron P2 ; and so forth. Assume that the theorem is valid in the case when the angle has n − 1 edges and prove it for an angle V with n edges a1 , . . . , an . If we draw a plane through the edges a1 and an−1 , then the plane separates the edge an from the other edges, and we obtain an angle V with n − 1 edges a1 , a2 , . . . , an−1 (Fig. 27(a)). Let P stand for the convex hull of the figure formed by the given points A1 , . . . , Am and the angle V . By assumption, P is a convex polyhedron with limit angle V .

O O

S

Q

a1 an

an-1

Fig. 27(a)

an Fig. 27(b)

Translating the starting point of the half-line an , the nth edge of the angle V , to all points of the polyhedron P , we obtain a body P . Repeating the above arguments with P substituted for P0 and an for a, we will prove that the body P is a convex polyhedron representing the convex hull of the figure formed by the polyhedron P and the half-line an .13 However, P is itself the convex hull of the figure formed by the points A1 , . . . , Am and the 13

The polyhedron P is now unbounded; therefore, it is possible that a half-line going from the interior of P does not meet the surface of P . However, this halfline then meets the surface if we extend it in the opposite direction. Otherwise P

36


angle V , whereas V is (by Theorem 2) the convex hull of the set of its edges a1 , a2 , . . . , an−1 . It follows that the polyhedron P is the convex hull of the set of the points A1 , . . . , Am and all the half-lines a1 , . . . , an , i.e., the convex hull of the figure formed by the points A1 , . . . , Am and the angle V . This conclusion is based on the following obvious observation: if G is the convex hull of a figure F and if G is the convex hull of F ∪ F , then G is also the convex hull of the figure G ∪ F . The proof is immediate from the definition of convex hull. We are left with showing that V is the limit angle of P . Since V is the limit angle of P , the corresponding unbounded edges of V and P are parallel to one another. In translating the half-line an to the points of P , there appear no new unbounded edges, except for those parallel to an , since we add to P only prisms with bases on its faces and lateral edges parallel to an .14 Hence, the unbounded edges of the polyhedron P may only be parallel to the half-lines a1 , . . . , an , i.e., the edges of the angle V . It remains to prove that to each half-line aj really corresponds at least one edge of P . Assume, for instance, that no edge of P corresponds to a half-line ak . The polyhedron P which is the convex hull of the set of the points A1 , A2 , . . . , Am and the half-lines a1 , a2 , . . . , an is completely determined by them. Therefore, the order of the half-lines aj is not important in the construction of P . We may thus assume that our ak is in fact an . Let V be the limit angle of P . The edges of V are parallel to the edges of P . Hence, they represent all the half-lines aj , except for some of them, at least except for an (according to the above assumption). Therefore, V is only a part of the angle V with all the edges a1 , . . . , an . The edge an is separated from V by the plane of a face S with edges a1 and an−1 (or some other edges if, say, a1 and an−1 are not edges of V ). By Theorem 1 on the limit angle of a polyhedron, P has a face Q parallel to S (Fig. 27(b)). If we translate the vertex of V onto the face Q, then we shall see that the edge an is outside the polyhedron P . This, however, contradicts the construction of P as the union of half-lines parallel to an . We have thus proved that to every edge of V corresponds a parallel unbounded edge of P . Hence, V is the limit angle of P , which completes the proof of the theorem. Theorem 5a. Under the hypotheses of Theorem 5, the polyhedron P may have vertices only among the given points A1 , A2 , . . . , Am . Each of them is a vertex if and only if it is not contained in the convex hull of the figure would include an entire straight line, which is excluded by assumption. Therefore, all preceding arguments may be repeated verbatim. 14 The polyhedron P is composed of such prisms with bases on the faces of P , but now these prisms may have unbounded bases, the unbounded faces of the polyhedron P .


37

formed by the remaining points Ai and the angle V (where we assume that the vertex of V is among the remaining points). It is clear from the construction of the polyhedron P that its vertices are only among the points A1 , . . . , Am . Indeed, we began the construction with the convex hull P0 of the points A1 , . . . , Am . According to Theorem 2 of Section 1.3, the vertices of P0 are only among these points, and drawing the half-lines in the construction of P yields no new vertices. Let P be the convex hull of the figure formed by the angle V and all the points Ai except for Am . So P is a convex polyhedron with vertices only among the points A1 , . . . , Am−1 . If Am lies in P , then adding it will change nothing. Hence, P coincides with P , and Am is not a vertex of P . Conversely, if Am is not a vertex of P , then it lies in P . Indeed, according to Theorem 4, P is the convex hull of its limit angle V and its vertices. Since Am is not among the latter, P coincides with P . Therefore, P contains the point Am . The proof is complete. Observe that, by Theorem 4, a polyhedron is the convex hull of its vertices and its limit angle V , and this is independent of the exact position of the vertex of V in the polyhedron. The vertices of the polyhedron P in Theorem 5 are among the points Ai . The preceding observation implies that the vertex of V can be any point in P , in particular, a point of the polyhedron P0 presenting the convex hull of all the points Ai . 1.4.6 Theorem 6. The convex hull of a finite collection of half-lines starting at a common point towards the same half-space is a convex solid polyhedral angle. Its edges are some (not necessarily all) of the initial half-lines. One of the half-lines is an edge if and only if it does not belong to the convex hull of the collection of the other half-lines. (If all half-lines lie in a plane, then their convex hull is a planar angle.) Let half-lines a1 , . . . , am start at a point O towards a half-space R whose boundary plane Q passes through O. Then each plane Q in R which is parallel to Q intersects the half-lines a1 , . . . , am at some points A1 , . . . , Am . According to Theorem 2 of Section 1.3, the convex hull of the set of the points O, A1 , . . . , Am is a convex solid polyhedron. This polyhedron is a pyramid with vertex O and base on the plane Q . It is also clear that this pyramid P is the convex hull of the segments OA1 , . . . , OAm . If we now move the plane Q to infinity, then in the limit the pyramid P transforms into a convex solid polyhedral angle V which is the convex hull of the collection of the half-lines a1 , . . . , am . (This is clear since the convex hull must contain all the segments OAi and thereby all the pyramids P .) By Theorem 2 of Section 1.3, a point Ak is a vertex of the pyramid P if and only if it does not belong to the convex hull of the remaining points O, A1 , . . . , Ak−1 , Ak+1 , . . . , Am . Therefore, a segment OAk is an edge of P if

38


and only if it does not lie in the convex hull of the remaining segments OAi . In the limit we reach the same conclusion for the half-line ak . This becomes rather obvious after observing the following. As the plane Q moves to infinity, the pyramid P only undergoes a similarity dilation from the point O, so that each of the edges of P remains an edge while any segment OAk that is not an edge cannot change into an edge. This completes the proof. (If half-lines a1 , a2 , . . . , am do not lie in a half-space, then their convex hull coincides with the whole space. However, if they lie in a half-space R but not in the interior of any half-space, then their convex hull is either the half-space R or a dihedral angle whose edge lies in the boundary plane of R.) 1.4.7 Exercises. 1. Prove the following generalization of Theorems 5 and 6: The convex hull of a finite collection of points Ai and half-lines aj starting at some of the points inward the interior of the same half-space is a convex solid polyhedron. Its limit angle is the convex hull of the collection of the half-lines aj starting at a single point. Its vertices are among the points Ai . Further, one of the points is a vertex if and only if it does not belong to the convex hull of the collection of the other points and the half-lines aj . The half-lines aj may be translated so that their initial points remain within the convex hull of the points Ai without changing the polyhedron. 2. Prove Theorems 2–6 using the definition of the convex hull of a set as the intersection of the half-spaces containing the set. (Cf. Exercise 2 in Section 1.3.) 3. Prove Theorems 5–5a by induction on the number of the points A1 , A2 , . . . , Am . 4. Prove that the definitions of limit angle which were supplemented in Subsection 1.4.2 are equivalent to the main definition (in the case of convex polyhedra). 5. Prove that every unbounded convex polyhedron with boundary can be uniquely completed to become a complete polyhedron with no extra vertices. Find conditions under which a given boundary vertex of the initial polyhedron is not a vertex of the so-constructed complete polyhedron. (Cf. Theorem 4 of Section 1.3 and Exercise 1 there.) 6. Let V be a convex solid polyhedral angle with vertex A. Let a1 , . . . , an be vectors along its edges. Provethat V is the set formed by the endpoints −−→ −→ n X of the vectors x = OX = a + i=1 αi ai , where a = OA is the vector from the origin O to the vertex A and αi are arbitrary nonnegative numbers. 7. Given points A1 , . . . , An and half-lines b1 , b2 , . . . , bk starting at these −→ points, let ai = OAi be the vectors from the origin to the points Ai and let bi be vectors along the half-lines bi . Prove that the convex hull of the collection of the pointsAi and the half-lines bi is the set of the endpoints k n of the vectors −−→ n x = OX = i=1 αi ai + i=1 βi bi , where αi , βi ≥ 0 and i=1 αi = 1. (Cf. Exercise 4 in Section 1.3.)

1.5 The Spherical Image

39

1.5 The Spherical Image 1.5.1 The spherical image. Let F be a convex surface and E, a unit sphere, i.e., a sphere of radius 1. Take some set M of points on F and draw all support planes to F through each point of the set. If we draw the outward normals of these planes from the center of the sphere E, then their endpoints will cover some set on the sphere. This set is called the spherical image of M . In particular, this definition makes sense for convex polyhedra. Moreover, if a polyhedron has a boundary, then the support planes at the boundary points are excluded. If M is the interior of a face of a polyhedron P , then the spherical image of M consists of a single point. If M is an edge with deleted endpoints,15 then its spherical image is the arc of a great circle whose endpoints are the spherical images of the faces touching along M , as one can see from the fact that each plane “between” the planes touching along an edge is a support plane. β W α

A

Fig. 28

If the set M consists of a single vertex A, then its spherical image is the part of the sphere which is cut out by the solid angle formed by the normals to all support planes at the vertex A. It is easy to see that this solid angle W represents a convex polyhedral angle whose edges are perpendicular to the faces of P touching at A (Fig. 28). If Q1 and Q2 are support planes at A, then the polyhedron is contained in the dihedral angle between them. Hence, each “intermediate” plane Q is also a support plane at A. The normals N to these planes cover the angle between the normals N1 and N2 to the planes Q1 and Q2 (Fig. 29). This means that, together with the rays N1 and N2 , the angle W includes the planar angle between them. Therefore, the angle W is convex. 15

The endpoints must be deleted, since they are vertices of the polyhedron and their spherical images are spherical polygons. For the same reason, in the previous assertion we talked about the interior of a face, i.e., about a face whose sides and vertices are deleted.

40


N

N1

N2

A Q Q2

Q1

Fig. 29

The faces of W are formed by the normals of the planes that touch the polyhedron along the edges adjacent to A. Therefore, they are planar angles perpendicular to these edges. Consequently, each dihedral angle βi of the solid angle W is the complement of the angle between the two corresponding edges of the polyhedron, i.e., the planar angle αi of the face at A: αi = π − βi .

(1)

The spherical image of a vertex is thus a convex spherical polygon with angles βi . The vertices of the polygon corresponding to the edges of the angle W are the spherical images of the interiors of the faces of P , and the sides of the polygon corresponding to the faces of W are the spherical images of the edges of P , with endpoints deleted. The spherical image of a vertex is certainly the same as the spherical image of the polyhedral angle formed by the faces touching at the vertex. All our conclusions are true for every convex polyhedral angle. Now, if M is an arbitrary part of a given convex polyhedron P , then its spherical image is composed of the spherical images of the vertices, segments of edges, and parts of the interiors of faces in M . Since the spherical images of edges and faces are contained in the spherical images of the vertices at which they touch, the spherical image of the polyhedron is composed of the spherical images of its vertices and the edges with both endpoints on the boundary.16 Here the boundary vertices and edges are excluded by assumption. For example, the lateral surface of a prism has no vertices in the interior, and its spherical image is a great circle. 16

If a polyhedron has a boundary, then the boundary vertices are excluded. However, on a polyhedron with boundary, there may be edges that are not part of the boundary but whose endpoints lie on the boundary. Then their spherical images do not enter into the spherical images of vertices, and we must include them in the spherical image of the polyhedron by themselves. In this case, it can turn out that the spherical images of some vertices are joined by arcs of great circles which are the spherical images of such edges.


41

A support plane touching the polyhedron at a vertex does not contain other vertices, unless it touches the polyhedron along an edge or a face. Therefore, the polygons representing the spherical images of vertices have no common interior points and may touch one another only along edges and at the vertices corresponding to those edges and faces of the polyhedron to which the vertices in question belong. It follows that the spherical image of the entire polyhedron is the spherical polygon that consists of the spherical images of the vertices of the polyhedron. If a polyhedron is closed, then its spherical image covers the whole sphere and represents a partition of the sphere into convex spherical polygons corresponding to the vertices of the polyhedron. We summarize our conclusions in the following: Theorem 1. The spherical image of a convex polyhedron is composed of the convex spherical polygons S1 , . . . , Sm that are the spherical images of the vertices of the polyhedron. These polygons have no common interior points. Each common side of two polygons Sk and Sl is the spherical image of the interior of the edge between the corresponding vertices. Each common vertex of polygons Sk , . . . , Sq is the spherical image of the interior of the face that contains the vertices of the polyhedron corresponding to these polygons. Each angle of the polygons Sl is the complement of the planar angle at the corresponding vertex of the polyhedron. 1.5.2 The area of the spherical image. Let V be a convex polyhedral angle with n faces and planar angles α1 , . . . , αn . Its spherical image is a spherical n-gon with angles βi = π − αi . However, the area ω of a spherical n-gon can be expressed in terms of its angles by the formula ω=

n

βi − (n − 2)π,

(2)

i=1

i.e., ω equals the excess of the sum of its angles over the sum of the angles of a planar polygon with the same number n of vertices. Proof. First, we will prove this formula for a spherical triangle B1 B2 B3 with angles β1 , β2 , and β3 (Fig. 30). Extending the sides of the triangle, we obtain three great circles whose intersection forms one more triangle B1 B2 B3 symmetric to B1 B2 B3 about the center of the sphere. Each pair of great circles bounds two digons with vertices B1 B1 , B2 B2 , and B3 B3 . The angles of these digons are β1 , β2 , and β3 . The ratio of the area of each digon to the area of the entire sphere is obviously the ratio of its angle to 2π. Therefore, the sum of the areas of all six digons equals 2(β1 + β2 + β3 ) 4π = 4(β1 + β2 + β3 ). 2π

42


B3

B1

B2 β2 β1 β3

B1

B3

B2

Fig. 30

At the same time, it is immediately obvious that our digons cover the entire sphere, where the triangles B1 B2 B3 and B1 B2 B3 are covered three times. Hence, the sum of their areas equals 4π +4ω, where ω is the area of the triangle B1 B2 B3 . Therefore, 4(β1 +β2 +β3 ) = 4π+4ω, i.e., ω = β1 +β2 +β3 −π, as required by formula (2). Now, given an n-gon, we partition it by diagonals into n − 2 triangles. Summing over the areas and angles of these triangles simultaneously, we readily obtain formula (2) for our n-gon. Since βi = π − αi , we have ω = 2π −

n

αi .

(3)

i=1

Since the area ω is positive, the preceding formula implies in particular the well-known fact that the sum of the planar angles of a convex polyhedral angle is always less than 2π. We call the sum of the angles αi the complete angle at the vertex, denoting it by θ. The difference 2π − θ is called the curvature at the vertex. Hence, formula (3) means that the curvature at a vertex of a convex polyhedral angle is equal to the area of the spherical image of the vertex. The curvature of a part of a polyhedron is defined as the sum of the curvatures at the vertices lying in this part but not on its boundary. At the same time, according to Theorem 1, the spherical image of any part of a polyhedron is composed of the spherical images of the vertices belonging to this part. Therefore, the area of the spherical image of a part of a polyhedron is equal to the curvature of that part. If we deform a polyhedron without changing the lengths of curves on it, then the angles on it will not change either, since they are determined by lengths. For instance, the length of a curve drawn on a polyhedron at a small constant distance r from a vertex A obviously equals θr, where θ is the


43

complete angle at A. The lengths of the curves on a polyhedron determine the so-called “intrinsic metric” of the polyhedron, and therefore the curvature and the area of the spherical image depend only on the intrinsic metric of the polyhedron. (We will discuss the notion of intrinsic metric in more detail in Subsection 1.6.2.) The above results may be summarized as follows: Theorem 2. The area of the spherical image of a part of a convex polyhedron equals the curvature of that part. Since the curvature depends only on the intrinsic metric of the polyhedron, so does the area of the spherical image. Therefore, under every deformation of the polyhedron which preserves its convexity and the lengths of all curves on it (i.e., which does not change its metric) the area of the spherical image remains the same. The theorem is an elementary analog of Gauss’s celebrated theorem claiming that the area of the spherical image of a surface remains unchanged under continuous deformations of the surface preserving the lengths of all curves. 1.5.3 The spherical image of an unbounded polyhedron. A closed convex polyhedron has support planes of all possible directions. Hence, its spherical image covers the entire sphere and has area 4π. An unbounded convex polyhedron always contains a half-line. Therefore, given a support plane Q to the polyhedron, we can find a support plane of this half-line which is parallel to Q, by shifting Q inside the polyhedron until Q begins to touch the half-line. Consequently, the spherical image of our unbounded polyhedron lies inside the spherical image of the half-line. The latter clearly covers the hemisphere bounded by the equator perpendicular to the half-line. Hence, the spherical image of an unbounded polyhedron is always contained in a hemisphere and so its area does not exceed 2π. Let us prove a more precise theorem: Theorem 3. The spherical image of an unbounded convex polyhedron coincides with the spherical image of its limit angle. Let P be an unbounded convex polyhedron and let V be the limit angle of P . We choose the vertex of V inside the polyhedron. Then, as shown in Subsection 1.4.1, the angle V lies in the polyhedron. Given a support plane Q of the polyhedron, we can obviously find a support plane of V parallel to Q by shifting Q inside the polyhedron so that it begins to touch the angle V . Hence, the spherical image of the polyhedron lies in the spherical image of its limit angle. It remains to show that the spherical image of the angle V is contained in the spherical image of the polyhedron P . In accordance with Theorem 1, both are spherical polygons; hence, its suffices to prove that every interior point N of the spherical image of V belongs to the spherical image of P .

44


By Subsection 1.5.1, the boundary of the spherical image of a polyhedral angle consists of the spherical images of its edges. Therefore, an interior point N of the spherical image of V corresponds to a support plane Q touching V only at its vertex. So the whole angle V , except its vertex, lies inside the half-space bounded by this plane. Since, by definition, V consists of the half-lines parallel to the half-lines on P , all half-lines on P point into the same half-space R. It follows that the opposite half-space R1 may include only a finite part of P . By shifting the plane Q into the half-space R1 , we obtain a position where the whole polyhedron P lies on one side of the plane, i.e., we create a support plane parallel to Q. Therefore, the point N on the sphere which corresponds to the plane Q belongs to the spherical image of the polyhedron P . Consequently, the spherical image of P contains the spherical image of V . Based on the reverse inclusion established earlier, we conclude that the two spherical images coincide, as claimed. As demonstrated, the spherical image of an arbitrary polyhedral angle is a convex polygon S whose vertices are the spherical images of the interiors of the faces of the angle V . The faces of V are parallel to the unbounded faces of P with nonparallel unbounded edges. Combined with the above-proved equality of the spherical images of P and V , this fact leads us to the following conclusion: The spherical image of a complete unbounded convex polyhedron (without boundary) is a convex spherical polygon S whose vertices correspond to the unbounded faces with nonparallel unbounded edges. The spherical images of the faces with parallel unbounded edges are points in the interiors of the sides of S (corresponding to the edges of the limit angle which are parallel to these edges of the polyhedron). The spherical images of the bounded faces are points in the interior of the polygon S. If the limit angle is a half-line, i.e., if all unbounded edges of the polyhedron are parallel to one another, then the polygon S is a hemisphere. If the limit angle is a planar angle, then the polygon S is a digon. Observe that Theorem 3 gives rise to a new definition of the limit angle of a polyhedron as the polyhedral angle whose spherical image coincides with the spherical image of the polyhedron. This is true because a polyhedral angle is determined by its spherical image up to translation. Indeed, if V is a given polyhedral angle and if W is the polyhedral angle covered by the normals to the support planes of V , then the edges and faces of W are perpendicular to the corresponding faces and edges of V . Therefore, the angles V and W are in a one-to-one correspondence and these angles determine each other. The edges of one of them are perpendicular to the faces of the other, and vice versa. Also, one of them is covered by the normals to the support planes of the other.


45

1.5.4 Polar polyhedra. The above correspondence between the polyhedral angles V and W is curiously connected with the existence of polar polygons and polyhedra. If P is a convex polyhedron and O is a point inside P , then we can construct a convex polyhedron P whose faces are perpendicular to the rays issuing from O through the vertices of P and whose vertices lie on the rays issuing from O and perpendicular to the faces of P . The correspondence between the faces and vertices of the polyhedra P and P is reversible. The edges of one of them correspond to the edges of the other. Namely, the edge between two vertices C1 and C2 of P corresponds to the edge of P along which the faces of P corresponding to the vertices C1 and C2 meet. Polyhedra maintaining such a relation are said to be polar to one another. One of the simplest examples is a cube and the octahedron with vertices the centers of the faces of the cube (Fig. 31). In general, to each polyhedron P circumscribed around a ball corresponds the polar polyhedron P1 inscribed in the ball and having vertices at the points of tangency of the ball and the faces of the polyhedron P .

Fig. 31

Fig. 32

By analogy, we define polar polygons: the vertices of one of them lie on the rays perpendicular to the sides of the other, and vice versa (Fig. 32). The construction of the polyhedron polar to a given polyhedron is carried out by performing a polar transformation that consists in associating planes with points and vice versa. Namely, with each point C we associate the plane −−→ that is perpendicular to the ray OC and lies at the distance from O reciprocal to the distance OC. If c is the vector from O to C, then the equation of the corresponding plane is cx = 1 (or cx = r2 , where r is the radius of the ball with center O which is used for the polar transformation). According to this rule, with each vertex Ci of a given polyhedron having the point O in its interior, we associate the plane of a face of the other polyhedron. If a point B moves on the plane cx = 1 corresponding to a point C, then we always have cb = 1. This means that all the planes bx = 1 corresponding to the points B pass through the point C. Therefore, to the plane cx = 1 corresponds the point C that is the intersection of the planes corresponding

46


to the points of the plane cx = 1. In other words, the incidence of a point and a plane is preserved, whereas the planes and points interchange their roles.17 In exactly the same way, we define a polar correspondence between points and straight lines in the plane. First we will consider the polar correspondence on the plane and examine its properties. Similarly we will then look at the polar correspondence in space. Let Q be a bounded convex polyhedron in some plane R. Take a point O in it and drop a perpendicular OO of unit length to the plane R from O. Drawing half-lines from O to all points of the polygon Q, we obtain a convex polyhedral angle V (Fig. 33).

Q1 O1

O1

R1

Ai

ei

O V

E

O

V O

R Q

O

Fig. 33

ei

ai

Fig. 34

We now construct the polyhedral angle V that is formed by the normals to the support planes of V . Consider the intersection of V with a plane R1 parallel to R. This intersection is the convex polygon Q1 whose vertices are the intersections of the edges of V with the plane R1 . It is easy to see that the extension of the perpendicular OO meets the plane R1 at some point O1 inside the polygon Q1 (since the plane perpendicular to OO at the point O is a support plane of V touching it only at the vertex O ). Let ai be a side of the polygon Q and let ei be a ray starting at O perpendicular to the straight line through ai (Fig. 34). The plane E through the point O and the ray ei is perpendicular to the plane of the face of V corresponding to the side ai . However, to this face of V corresponds a perpendicular edge of V ; consequently, the edge lies in the plane E. Hence, the 17

This polar correspondence is a particular instance of the polar correspondence with respect to an arbitrary surface of second degree. Such general polar correspondences are studied in projective geometry. In our case we treat the polar correspondence with respect to the unit ball.


47

corresponding vertex Ai of the polygon Q1 lies on the ray ei starting at O1 parallel to ei (since the plane E intersects the planes R and R1 in parallel straight lines). If we take the projection of the polygon Q1 to the plane R of the polygon Q, then by the above we obtain a polygon Q whose vertices lie on rays perpendicular to the sides of the polygon Q. However, as we noted at the end of Subsection 1.5.3, the correspondence between the polyhedral angles V and V is one-to-one. In consequence, the correspondences between the polygons Q and Q1 and between Q and Q are also one-to-one. It follows that the vertices of Q lie on the rays from O perpendicular to the sides of Q and the vertices of Q lie on the rays from O perpendicular to the sides of Q . Thus, the two polygons are polar to each other. If the ray OX1 passes through an interior point X1 of the polygon Q1 , i.e., OX1 goes in the interior of the angle V , then the plane perpendicular to the ray is a support plane of V . Therefore, this plane intersects the plane R in a straight line x such that the polygon Q lies on one side of x. If the point X1 lies on the boundary of Q1 , then the corresponding straight line x is a support line of Q. Thus, to the points inside the polygon Q1 , and so to those inside Q , correspond straight lines disjoint from Q, whereas to the points on the boundary of Q correspond the support lines of Q. Let the point O stand for the origin of coordinates in space and let O be the origin of coordinates in the plane R. Let the coordinate axes x1 and x2 be parallel to the plane R, and let the axis x3 be the line O O. The position of a point X in space is determined by a vector (x1 , x2 , x3 ), and the position of a point on the plane R is determined by a vector (x1 , x2 ). The equation of the plane through O perpendicular to the ray O C is c1 x1 + c2 x2 + c3 x3 = 0,

(1)

where c1 , c2 , and c3 are the coordinates of C (the inner product of the vector (c1 , c2 , c3 ) of the point C and the vector (x1 , x2 , x3 ) of a point on the plane is equal to zero). If the point C lies in the plane R, then c3 = 1. Taking the points of the intersection of the so-constructed plane with the plane R1 on which x3 = −1, we obtain instead of (1) (2) c1 x1 + c2 x2 = 1. This is nothing but the equation of the straight line c associated with the point C, since with each point C we associate exactly the line of intersection of the plane R1 with the plane perpendicular to the ray O C , or the projection of this line to the plane R. Therefore, analytically our construction assigns to each point C(c1 , c2 ) of the plane R the straight line with equation (2). By reciprocity, the converse is also true. Observe that when the point C moves on a straight line c , the corresponding straight lines rotate around the point C corresponding to the

48


line c . Indeed, the movement of C along c corresponds to the rotation of the ray O C in the plane (O , c ). Hence, the planes perpendicular to the ray pass always through the ray O C perpendicular to the plane (O , c ). However, this means that the straight lines corresponding to C pass through C . The described correspondence between straight lines and points in the plane R is called the polar correspondence (with respect to the unit circle centered at O). The polar correspondence in space is defined quite similarly: to a point c with coordinates (c1 , c2 , c3 ) we assign the plane with equation c1 x1 + c2 x2 + c3 x3 = 1, i.e., with the endpoint of a vector c from the origin O we associate the plane cx = 1. The latter is perpendicular to c and its distance from the origin is reciprocal to the length of c. To construct this correspondence, we use a similar trick. Consider our space as a three-dimensional hyperplane R in the four-dimensional space. From an origin O lying in R, draw a perpendicular OO to R of unit length and direct the axis x4 from O to O. On the other side of O , draw a threedimensional hyperplane R1 parallel to R and at unit distance from O . To each point C on the “plane” R corresponds a ray O C, and to this ray corresponds the three-dimensional hyperplane perpendicular to it and passing through O . The intersection of this hyperplane with R1 yields a two-dimensional plane whose projection to R is exactly the plane corresponding to C. If P is a convex polyhedron (in the “plane” R) whose interior contains the point O, then the rays from O through the points of P cover a fourdimensional polyhedral angle V . The normals to the support planes of V cover some angle V whose intersection with the “plane” R1 yields the polyhedron P polar to P . It now suffices to take the projection of P to the “plane” R. The relation between the polyhedra P and P is analogous to the relation between the polar polygons Q and Q . To a vertex C of one of them corresponds the plane of a face of the other which is perpendicular to the ray OC, and vice versa. To the edge C1 C2 between two vertices C1 and C2 corresponds the perpendicular edge along which the faces corresponding to the vertices meet. All this is immediate from the mutual relationships between the elements of the (four-dimensional) polyhedral angles V and V . To the three-dimensional faces, two-dimensional faces, and edges of one of them correspond perpendicular edges, two-dimensional faces, and three-dimensional faces of the other. Therefore, for the intersections with the planes R and R1 we obtain an analogous correspondence between the faces, edges, and vertices of the polyhedron P and the vertices, edges, and faces of the polyhedron P . With the polyhedron P defined by its vertices as their convex hull, we associate the polyhedron P defined as the intersection of the half-spaces bounded by the planes of its faces. Exercise. Let P be a closed convex polyhedron, let O be a point in its interior, and let S be the unit sphere with center O. The polyhedron P gives rise to two partitions of the sphere S into convex spherical polygons. One of

1.6 Development

49

them is the partition into the polygons that are the spherical images of the vertices of the polyhedron. The other is the partition into the polygons that are the projections of the faces of the polyhedron to the sphere S under the projection by means of the rays issuing from the center O. Verify that the polyhedron polar to P (with respect to the point O) determines the same partitions, with their roles reversed. To begin with, it may be checked that a completely analogous assertion is valid for polar polygons, where we consider partitions of the unit circle into arcs.

1.6 Development 1.6.1 Developing. The topic of this section is the ordinary gluing of a polyhedron from polygons cut out of paper. A development is a collection of polygons with some prescribed rule for gluing them together along their sides. The rule for gluing consists of instructions that specify which side of each polygon is glued to a side of another or the same polygon and in which direction. We must specify a direction, since two equal segments AB and CD can be superposed in two ways: either the endpoint A coincides with C and the endpoint B with D or A coincides with D and B with C. Gluing the sides of some polygon to one another is called “self-gluing.” This happens, for instance, in the well-known crosslike development of a cube (Fig. 35).

a

a

A

A

Fig. 35

Every polygon bounded by several polygonal lines can be partitioned into so-called simple polygons each of which is bounded by a single simple polygonal line (a piecewise linear curve without selfintersections). Therefore, we will consider only developments that are composed of simple polygons. A simple polygon can be bounded or unbounded; in the latter case it has finitely many bounded and two unbounded sides.

50


We consider an interior point of a side of a polygon as a vertex, thus dividing the side into two sides (see, for example, the development of a cube, Fig. 35). Hence, we loose no generality in assuming that gluing is always performed along entire sides. To state it precisely, a development is defined to be a finite collection of simple polygons, bounded or unbounded, with some prescribed rule for “gluing” them together along sides, where “gluing” means “identification.” To glue two segments together, we establish a one-to-one correspondence between their points and identify the corresponding points, thus regarding them as the same point of the development. The rule for gluing is simply a specification of these correspondences between the points on the sides. Moreover, we naturally assume the following: if a point A is identified with B and B is identified with C, then A is identified with C.18 We shall assume every time that the rule for gluing satisfies the following conditions: (1) The segments can be glued together if they have the same length. (In other words, the correspondence between the identified points of glued sides is length-preserving). (2) It is possible to pass from each polygon to any other polygon by traversing polygons with glued sides. (This condition means that a development never splits into disjoint parts.) (3) Each side of every polygon is either glued to no other side or to exactly one side. (This condition means that we never arrive at a branching polyhedron while gluing.) The sides that are not glued to others form the “boundary” of the development. If there are no such sides, then we say that the development is without boundary or, in other words, a boundaryless development. The sides and vertices of the polygons in a development are referred to as the edges and vertices of the development, with the identified sides and vertices regarded as the same edge and the same vertex of the development. A polyhedron is nothing more than a particular instance of a development: it is composed of polygons, its faces, while the identifications that are only abstractly implied in a development are already past of it. For this reason, the study of developments includes the study of all polyhedra. However, by treating a polyhedron as a development, we absolutely ignore its shape in space. We will not examine the structure of a polyhedron or a development, but we will focus our attention on what is common to all developments of one polyhedron and, in general, to all developments that can be obtained from one another by cutting and gluing. This common feature is their “intrinsic metric” which will be defined in the following subsection. 18

For example, if a side A1 B1 of a polygon Q1 is glued to a side A2 B2 of a polygon Q2 (with the endpoint A1 identified with A2 ) and a side A2 C2 of Q2 is glued to a side A3 C3 of a polygon Q3 (the endpoint A2 is identified with the endpoint A3 ), then the vertices A1 , A2 , and A3 are regarded as the same point of the development.

1.6 Development

51

1.6.2 The intrinsic metric. Let X and Y be two points of a development R. We can join X and Y by a polygonal line in R by passing from one polygon to another through identified points of their boundaries.19 The greatest lower bound for the lengths of such lines is the distance between the points X and Y in the development R, denoted by ρR (X, Y ). The distance regarded as a function of the pair of points X and Y is the metric of the development R. If we have a correspondence between the points of two developments R and R such that the distances between pairs of corresponding points are equal: ρR (X, Y ) = ρR (X , Y ), then this correspondence is called an isometry, and the developments admitting an isometry between them are isometric. (The points identified in a development are, of course, considered as the same point.) We say that a development R is obtained from a development R by cutting and gluing if we can cut the polygons of R into smaller polygons and glue them together along their sides so that we obtain a development equal to R , i.e., comprising the same polygons with the same rule for gluing (the rule for gluing in the new development is natural: we should glue together those sides that are glued in R or were created by cutting). Developments obtained from one another by cutting and gluing are isometric, because these operations do not change polygonal lines drawn in a development, apart from connecting identified points, or disconnecting points along a cut, thus transforming one point into two that are subject to identification. The converse assertion can be proved as well: Isometric developments originate from one another by cutting and gluing. Since we do not use this assertion in its full generality, we leave it without proof, although the proof causes no difficulties. All the facts depending only on the intrinsic metric of a development (in particular, of a polyhedron) constitute the intrinsic geometry of the development (or the polyhedron). Isometric developments have the same intrinsic geometry, as do a development and the polyhedron obtained from it by gluing. Together with the metric ρR itself, the lengths, angles, and areas belong to the realm of intrinsic geometry. The length of a polygonal line in a development is the sum of the lengths of its segments. Each sufficiently small segment XY in a development is a shortest arc between its endpoints and hence has length equal to the distance ρR (X, Y ). Therefore, the intrinsic metric determines length. 19

The possibility of passing from one polygon to another in this manner and, consequently, the possibility of joining every two points X and Y in R are stipulated in the definition of development by condition (2).

52


Since the angles of a planar triangle are determined by the lengths of its sides, the angles can be determined in terms of the metric and are preserved under isometries. A similar remark relates to area. As we have already observed, a polyhedron is nothing more than a development in which the identifications only implied in a development are actually implemented. Hence, we can use the same notion of intrinsic geometry as determined by the distances measured in a polyhedron itself. Given a correspondence between the points of two polyhedra such that the distances between pairs of corresponding points are equal, we call the correspondence an isometry and the polyhedra admitting such a correspondence, isometric. Isometric polyhedra have the same intrinsic geometry. The lateral surfaces of prisms with equal perimeters of bases and equal lengths of lateral edges serve as examples.

Fig. 36

In intrinsic geometry, straight line segments are replaced by shortest curves joining two given points of a polyhedron or a development and lying on the polyhedron or development, respectively. We call such a curve a shortest arc. A shortest arc joining two points of one face is a straight line segment, and a shortest arc joining points of different faces is a polygonal line with segments on several faces. Examples of shortest arcs on a cube are exhibited in Fig. 36. The properties of shortest arcs in developments will be discussed further in Section 1.8. 1.6.3 Gluing a polyhedron. We say that a polyhedron P is produced from a development R by gluing, or that a development R defines a polyhedron P by gluing, if the requirements for gluing (identification) implicit in the development are already implemented in the polyhedron. More precisely, this means that P admits a decomposition R into pieces, “polygons” Q (possibly “folded”), which are bounded by lines, “edges,” touching at the “vertices” of the decomposition R such that the following two conditions hold: (1) The developments R and R have the same structure, i.e., to each polygon, edge, or vertex of R corresponds a “polygon,” “edge,” or “vertex” of

1.6 Development

53

R and this correspondence preserves the incidence relation (of an edge to a polygon and of a vertex to an edge). (2) If a “polygon” Q of R corresponds to a polygon Q of R, then Q can be developed on the plane so that it becomes coincident with Q, with its edges and vertices coinciding with the corresponding edges and vertices of Q. The visually comprehensible operation of developing on the plane is an isometric mapping, i.e., it associates the points of Q with the points of a planar polygon so that the lengths of all curves remain the same. The same operation can be described somewhat differently. Each “polygon” Q is generally composed of several polygonal pieces of the faces of P . Developing Q on the plane consists in arranging these pieces in the plane such that they become to touch one another in the same way as they do in the polyhedron. The condition requiring the possibility of developing Q on the plane means that such arrangement of its pieces is actually realizable in the plane. Since the corresponding polygons of the decomposition R and the development R are isometric and touch one another at the corresponding edges and vertices, the lengths of arbitrary corresponding polygonal lines in R and R are equal. This means that the polyhedron P is isometric to the development R. Therefore, the polyhedra admitting identical developments or, which is the same, produced from the same development are isometric to each other. We can generally define the construction of a polyhedron from a development by gluing as an isometric mapping of the development onto the polyhedron. B 1 A

D 2 C

3

B

B

D

B

4 A

C

A

C

Fig. 37

This definition makes it clear that the polygons and edges of a development do not necessarily correspond to the faces and edges of the polyhedron. Fig. 37 shows two developments of a regular tetrahedron which have the shape of parallelograms partitioned into triangles. The identified vertices are indicated by the same letters. The triangles of the first development do correspond to the faces of the tetrahedron, whereas those of the second do not. The two developments have the same structure.20 The second development can be obtained from the first by attaching the first triangle to the forth, 20

Developments have the same structure if their polygons and edges can be associated in such a way that the corresponding polygons are glued together along their corresponding edges.

54


cutting the triangles 2 and 3 along the line BA, cutting the triangles 4 and 1 along the line DC, and finally gluing the new triangles of the resulting pieces together. Applying a similar operation to the second development and repeating the procedure, we obtain infinitely many distinct developments of a regular tetrahedron which, however, all have the same structure. A

A

a

a

b b

a

b

a

A

b

b

a A

A c

(a)

c

A

A c

c

c

(b)

(c)

Figs. 38(a)–(c)

Fig. 38(a) shows a development of a tetrahedron; the edges denoted by the same symbols must be glued together. In Fig. 38(b) the dotted lines indicate the actual edges of the tetrahedron. Fig. 38(c) shows the tetrahedron produced by gluing from the development. Here we also indicate those lines along which we must cut the tetrahedron in order to obtain the initial development of Fig. 38(a). Besides the dotted lines, we must cut the tetrahedron along the edges touching at the vertex A. This development is curious, because its triangles undergo a rather intricate “self-gluing.” The problems for us to solve are, first, to clarify the extent to which a convex polyhedron is determined from its development and, second, to find necessary and sufficient conditions on a given development for it to define a convex polyhedron by gluing. Recalling the Introduction, we can say that the first problem reduces to establishing uniqueness theorems for convex polyhedra with a given development. This problem is solved in Chapter 3 and in part in Section 5.2. One of the necessary conditions to be imposed on a development is found in the next subsection. Other conditions will be established in the next section. Chapter 4 and Section 5.1 address the question of proving the sufficiency of these conditions, i.e., proving existence theorems for convex polyhedra with a given development. The main results of Chapters 3, 4, and 5 are stated in Section 2.3. 1.6.4 Curvature. The complete angle of a development at an interior point, i.e., a point not lying on the boundary, is defined as the sum of angles touching at this point. If a point lies in the interior of an edge, then the complete angle at it comprises two straight angles and is therefore equal to 2π. The complete

1.6 Development

55

angle at a point in the interior of a face is naturally defined to be equal to 2π. (The complete angle at a point on the boundary of a development is defined by analogy; it equals π unless the point is a boundary vertex.) If we construct circular sectors of a small radius r on the polygons touching at a point A, then their union is a neighborhood of A. If θ is the complete angle at A, then the length of the circumference of this neighborhood is θr (Fig. 39).

A A A

Fig. 39

If θ equals 2π, then such a neighborhood covers a flat disk; hence, the intrinsic metric does not discriminate between a point with complete angle 2π from a point in the interior of a face. This is certain for points in the interior of an edge that does not belong to the boundary. The points at which the complete angle differs from 2π must be vertices of the development. They may be called genuine vertices. The difference 2π − θ, with θ the complete angle at an interior point A of a development, is called the curvature at A. Since θ may differ from 2π only at vertices, there are only finitely many points at which the curvature differs from zero. Since the angles are determined by the intrinsic metric, the concept of curvature also belongs to intrinsic geometry. The curvature of an arbitrary part of a development is defined as the sum of curvatures at the genuine vertices in this part. If the curvature is greater than or equal to zero at all vertices of a development, then we speak of a development , or polyhedral metric, of positive curvature. The sum of planar angles of a convex polyhedral angle is always less than 2π; so the curvature of such polyhedral angle is always positive. Therefore, a convex polyhedron has positive curvature. We have thus found one necessary condition on a development for it to define a convex polyhedron by gluing: The sum of angles touching at any vertex of the development must not exceed 2π.

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(Since a convex polyhedron with boundary is part of a complete convex polyhedron, the sum of the angles touching at a boundary vertex of the polyhedron and, in consequence, at a boundary vertex of a development must not exceed 2π.) 1.6.5 Exercises. 1. Prove that (a) a shortest arc that crosses an edge forms equal (not adjacent) angles with it; (b) if we successively develop on the plane the faces traversed by a shortest arc, then the arc becomes a straight line segment; (c) a shortest arc of a convex polyhedron cannot pass through a vertex. 2. Describe the shortest arcs on the lateral surfaces of prisms and pyramids. 3. Find examples where two points of a polyhedron are joined by more than one shortest arc. 4. Give examples of shortest arcs of nonconvex polyhedra passing through vertices. 5. What properties characterize an arbitrary development of the lateral surface of a prism (pyramid) which consists of a single polygon? 6. Prove that every closed convex polyhedron admits a development that consists of a starlike polygon like a natural development of a pyramid. (The “points” of the star must touch at the same vertex). 7. Prove that every (bounded) polyhedron admits only finitely many developments whose vertices correspond to the vertices of the polyhedron and whose edge lengths are bounded by a fixed number.

1.7 Topological Properties of Polyhedra and Developments 1.7.1 Topology studies those properties of figures that are preserved under arbitrary one-to-one transformations continuous in both directions. If the points of one of two figures can be put into a one-to-one correspondence continuous in both directions21 with the points of the other figure, then the figures are said to be homeomorphic to each other and the correspondence itself is said to be a homeomorphism of one figure onto the other. Clearly, two figures homeomorphic to a third are homeomorphic to one another: it suffices to associate their points that correspond to the same point of the third figure. 21

A correspondence ϕ between the points X of one figure and the points Y = ϕ(X) of the other figure is called one-to-one if to each point X corresponds exactly one point Y and vice versa. The correspondence ϕ is continuous if the fact that points Xn converge to a point X implies that the corresponding points Yn converge to the point Y corresponding to X. The correspondence is continuous in both directions if the same holds for the inverse correspondence ϕ−1 .

1.7 Topological Properties of Polyhedra and Developments

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A property of a figure is said to be topological if it is preserved under arbitrary homeomorphisms; homeomorphic figures share the same topological properties. In this section we reveal the main topological properties of convex polyhedra and their developments. 1.7.2 Theorem 1. Every closed convex polyhedron is homeomorphic to the sphere. Take a point O inside a closed convex polyhedron P and consider a sphere S centered at O. Drawing rays from O in all directions, associate those points of P and S that lie on the same rays. Each ray meets the sphere, as well as the polyhedron, at a single point. Therefore, the correspondence is one-to-one. The correspondence is also continuous in both directions. Indeed, as a point moves continuously on the polyhedron, the ray passing through it rotates continuously around O, and hence the corresponding point moves continuously on the sphere. Conversely, if a point moves continuously on the sphere, then the ray through it also rotates continuously, and therefore the point X at which the ray intersects the polyhedron P moves on P continuously.22 Thus, the direct correspondence between the points of the polyhedron and the points of the sphere, and its inverse are both continuous, which means that the correspondence is continuous in both directions. Every bounded convex polyhedron with boundary is part of a closed polyhedron; therefore, projecting it onto the sphere in the same way, we verify that it is homeomorphic to a spherical polygon. If the boundary of the polyhedron consists of a single closed polygonal line, then the resulting spherical polygon will be bounded by a single closed curve. Such a polygon is homeomorphic to a disk. (We leave this assertion without proof, since it is rather obvious. A rigorous proof should not cause any conceptual difficulties but appears to be time-consuming.) If the boundary of a polyhedron consists of several closed polygonal lines, then the corresponding spherical polygon has a similar structure. It is homeomorphic to a disk with the corresponding number of circular holes. (We leave this assertion without proof, since it will not be used in the sequel.) 1.7.3 Theorem 2. An unbounded convex polyhedron whose limit angle does not degenerate into a flat angle or a half-line can be projected to a plane so that to each point of the plane corresponds exactly one point of the polyhedron. Thereby such a polyhedron is homeomorphic to the plane. 22

This is quite obvious when the point X moves over a single face. When the point crosses edges and vertices, the continuity is not violated since each ray close to a ray intersecting the polyhedron at a point A of an edge or a vertex A must intersect the polyhedron at a point X close to A.

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Proof. Since the limit angle V is not degenerate, its interior contains a halfline a. Any point A of the polyhedron can be regarded as the vertex of V . The entire angle V now lies inside the polyhedron (see Theorem 1 in Section 1.4). Therefore, the half-line starting at A in the direction of a is contained in the polyhedron. It follows that every straight line b parallel to a can intersect the polyhedron only in a single point (since the half-line beginning at a point of intersection and going in the direction of a lies entirely inside the polyhedron). We will prove that every straight line parallel to a has to intersect the polyhedron. Assume the converse, and let a straight line b be parallel to a but disjoint from the polyhedron. The straight line b cannot lie inside the polyhedron, since the polyhedron includes no straight line. Therefore, b lies outside the polyhedron.

b A b a

Fig. 40

The intersection of the polyhedron with the plane passing through some interior point of the polyhedron and the straight line b is a convex polygon which lies on one side of b. It is immediately clear that this polygon has a support line b parallel to b (Fig. 40). Let A be a point where b touches the polygon. By above, the half-line starting at A in the direction of a must lie inside the polyhedron, and therefore cannot be a part of the support line b . This contradiction shows that all straight lines b parallel to a meet the polyhedron. Thus, every straight line parallel to a intersects the polyhedron; moreover, it does so only once. Therefore, projecting the polyhedron to a plane along such a straight line, we obtain the required projection. This completes the proof.


59

A polyhedron whose limit angle degenerates into a flat angle or a half-line cannot be projected onto a whole plane and none of its projections is oneto-one (i.e., some points of the plane correspond to several or even infinitely many points of the polyhedron). If the direction of the projection is parallel to a half-line in the limit angle, then some of the projecting straight lines will lie on unbounded faces of the polyhedron, while others will not meet the polyhedron at all. It is easy to see that in the case of a flat limit angle V the polyhedron is projected onto the strip between the pair of parallel straight lines that are the projections of unbounded faces parallel to the plane of the angle V . In the case when V degenerates into a half-line a, the projection along a yields a convex polygon whose sides are the projections of unbounded faces. In any case, the projection does not give a one-to-one mapping onto the entire plane. Therefore, we need a different construction.

O

Fig. 41

Fig. 42

Consider a polyhedron P whose limit angle is a half-line. All unbounded edges of P are parallel to one another. Drawing a perpendicular plane to them, we cut a semi-infinite prism P out of the polyhedron (we attach the base to the prism). Take a point O inside the prism. Every ray starting at O and intersecting P will meet P , and vice versa (Fig. 41). Associating the intersection points on each ray, we thus obtain a homeomorphism between P and P (the common points of P and P correspond to themselves). Obviously, there is a homeomorphism of the prism P onto a plane. It suffices to develop the lateral faces of P on the plane and stretch them so that they cover the entire plane (Fig. 42). Now, consider a polyhedron P whose limit angle is flat, but is not a halfline. Such a polyhedron has two unbounded faces parallel to the plane of V , with unbounded edges parallel to the edges of V . The edges of all other unbounded faces of P are pairwise parallel to one another.

60


O

Fig. 43

If we intersect P by a plane to cut off all its bounded faces (Fig. 43), we obtain a polyhedron P with one bounded face inside P and with unbounded faces lying on the faces of P . Take a point O inside P . It is easy to see that each ray starting at O and meeting one of the polyhedra also meets the other. So P is homeomorphic to P . Further, by analogy to the first case, it is easy to see that the polyhedron P can be mapped homeomorphically onto the plane. This prove that the polyhedron P is also homeomorphic to the plane. (This conclusion applies to every unbounded polyhedron.) Our arguments lead to the following theorem: Theorem 3. Every unbounded convex polyhedron is homeomorphic to the plane. 1.7.4 The construction of a polyhedron from a development by gluing represents a mapping from the development onto the polyhedron. As it follows from the definition in Subsection 1.6.3, this mapping is one-to-one and continuous in both directions under the necessary condition that the points of the development that must be identified by the “law of gluing” are regarded as the same point of the development. Therefore, a development defining a polyhedron by gluing is necessarily homeomorphic to this polyhedron. In this context, we have to keep in mind that glued points are viewed as a single point. In virtue of this remark, Theorems 1 and 3 imply Theorem 4. A development of a closed convex polyhedron is homeomorphic to the sphere. A development of an unbounded convex polyhedron is homeomorphic to the plane. (We can also conclude that a development of a bounded polyhedron with boundary is homeomorphic to a spherical polygon, and a development of an


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unbounded polyhedron with boundary is homeomorphic to an unbounded polygon.) However, this condition on a development is inconvenient, since we have no simple way to determine whether or not a development is homeomorphic to the sphere. For this reason, we replace this condition by another condition which is easy to check for any development. The condition is provided by Euler’s Theorem, which we are going to prove. 1.7.5 By a net we shall mean an arbitrary finite collection of simple (i.e., non-selfintersecting) open polygonal lines lying on a polyhedron and having no common points except possibly endpoints. Each polygonal line is called an edge of the net and the endpoints of edges are the vertices of the net. In general, a net divides the polyhedron into several regions.23 If a net does not divide the polyhedron at all, we have exactly one region. The simplest example of a net is the net of all edges of a polyhedron; here the regions are the faces of the polyhedron. In the general case, a net may consist of several disjoint parts. A part that cannot be decomposed any further is called a connected component of the net. Theorem 5. (The Generalized Euler Theorem.) Given a net on a closed convex polyhedron, assume that v is the number of vertices, e the number of edges, c the number of connected components, and f the number of regions into which the net divides the polyhedron. Then v − e + f = c + 1.

(1)

In particular, if the net is connected (i.e., it consists of a single component), then v − e + f = 2. The proof is based on the following quite transparent theorem known as Jordan’s Theorem: Every closed polygonal line on a closed convex polyhedron separates it. The transparency of this theorem notwithstanding, a rigorous proof is far from simple; we defer it until the last subsection of this section. Now, taking Jordan’s Theorem for granted, we prove Euler’s Theorem. For the “empty net,” i.e., when we have no net at all, the theorem holds, because in this case v = e = c = 0 and we have exactly one region, i.e., f = 1. Hence, equality (1) is valid. Therefore, if we show that, by successively removing edges, we can eliminate the whole net without changing the value of v − e + f − c, then the proof will be complete. Consider the following three possibilities of removing an edge from the net: 23

Each region comprises all points that can be joined with one another by polygonal lines not intersecting the net.

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(1) Removing an edge with one free endpoint. In this case, the number e of edges and the number v of vertices decrease by one, while the number f of regions and the number c of components obviously remain the same. Therefore, the value of v − e + f − c remains the same. (2) Removing an edge which is the boundary (or part of the boundary) between two regions. Then the regions merge together and the number f of regions decreases by one. The number e of edges also decreases by one. Any edge separating a region has no free endpoints; hence, the endpoints remain and the number v does not change. The number c of components remains the same, because each region is bounded by a closed polygonal line which cannot split into disjoint parts if only one edge is removed. Thus, the value of v − e + f − c does not change. (3) Removing an edge with two free endpoints. Such an edge itself is a connected component of the net. Therefore, the number c decreases by one. The number e of edges also decreases by one. The number v of vertices decreases by two, while the number f of regions remains obviously the same. Hence, the value of v − e + f − c again remains invariant. Now, assume given an arbitrary net. Using the first and third operations, we remove all edges with free endpoints. Any remaining edges form closed polygonal lines and therefore (by Jordan’s Theorem) separate the polyhedron into regions. In this case, the second operation is applicable, leading sooner or later to the creation of free endpoints. Again we remove all edges with free endpoints, and continuing in this manner we eliminate the whole net. Under all the operations the number v − e + f − c does not change, and in the case when there is no net it equals 1. Therefore, the equality v − e + f − c = 1 also holds for the original net, which was to be proved. Now, let R be a development defining a convex polyhedron P by gluing. By definition, this means that we can split P into regions corresponding to the polygons of R so that the segments of the boundaries of the regions correspond to the glued sides of the corresponding polygons. These segments of the boundaries are polygonal lines that constitute a net on P . The number v of vertices, the number e of edges, and the number f of regions in the net are equal to the number of vertices, edges, and polygons in R (where we assume that the identified edges or vertices are considered as one edge or vertex). We agreed earlier that a development consists of simple polygons, i.e., polygons bounded by polygonal circles (closed polygonal lines without selfintersections). We will show that this condition implies the connectedness of the net. Let A and B be arbitrary vertices of the development (net). If they belong to the same polygon, then they are joined by a polygonal line, since the polygon is simple. If they belong to different polygons Q and R, then we can pass from Q to R by traversing polygons with common vertices or edges. Therefore, by traversing the boundaries of these polygons, we connect A to


63

B by a chain of edges. Once every two vertices are joined by a chain of edges, the net is connected. In consequence, the number of components in our net is c = 1, and by Euler’s Theorem v − e + f = 2. This argument yields the following theorem: Theorem 6. If a given development defines a closed convex polyhedron by gluing, then the number f of polygons, the number e of edges, and the number v of vertices in the development satisfy the Euler relation v − e + f = 2. The Generalized Euler Theorem has a topological character: under a topological transformation, the lines forming a net may no longer be piecewise linear but their number, as well as the number of vertices and the number of edges, remains the same. Therefore, the same relation holds between them for every net on an arbitrary surface homeomorphic to a closed convex polyhedron, and thereby on an arbitrary surface homeomorphic to the sphere. In this context, Theorem 6 can be restated in topological form: The equation v − e + f = 2 holds for every development homeomorphic to the sphere. Moreover, the polygons of the development can be subjected to any topological transformations and thus can be thought of as “curved polygons.” 1.7.624 Theorem 7. (The Converse Euler Theorem.) If a development without boundary consists of bounded polygons and satisfies the Euler relation v − e + f = 2, then it is homeomorphic to the sphere. In the proof we shall admit topological transformations of polygons. Hence, nothing matters but the number of their sides and the rules for gluing. If a development consists of more than one polygon, then by “gluing” one polygon to another, next gluing a third polygon to them, and so on, we obtain a development that consists of a single polygon. Doing so, we always perform the gluing along a pair of sides as prescribed in the development. Therefore, with every gluing, the number of polygons and the number of edges decrease each by one and the number of vertices remains the same; hence, the characteristic v − e + f does not change, remaining equal to 2. In the final development we have f = 1, and therefore v = e + 1. The last polygon has 2e sides, because the sides are glued pairwise and each pair yields one edge. The number of vertices of the polygon is the same as the number of sides, i.e., it also equals 2e. (Do not confuse the vertices of a polygon with the vertices of a development! The vertices of a polygon which are subject to gluing are treated as a single vertex of the development.) We will show that our polygon has a pair of sides a, b that have a common endpoint and must be glued to one another; moreover, in gluing a to b, their 24

All conclusions of Subsections 1.7.6–1.7.8 are only of importance for Chapter 4.

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common endpoint corresponds to itself and yields exactly one vertex of the development. (If we indicate the directions of these sides, then these directions must be opposite; Fig. 44.) Indeed, if such a pair of sides did not exist, neither endpoint of any pair of glued sides would correspond to itself (the sides either are nonadjacent or are adjacent but have the same orientation, as in Fig. 4525 ). a

A b B

B

A

b

a

A

A

Fig. 44

Fig. 45

Therefore, the endpoints of all sides, i.e., all 2e vertices of the polygon are glued at least pairwise, thus yielding at most e vertices of the development. Consequently, the number of vertices obeys the inequality v ≤ e. However, we have v = e + 1. Hence, there must exist a pair of adjacent sides with opposite orientations which are subject to gluing. A A a

b

a b B

B Fig. 46

If we implement the gluing of two such sides a and b (this is possible, because we admit any topological deformation of polygons), we obtain a new polygon with fewer sides. Obviously, for this polygon we again have v = e +1, and the same arguments apply to it. Repeating the operation, we finally come to a “polygon” with only two sides a and b (Fig. 46). We immediately see that, gluing these sides together, we transform the polygon into a surface 25

Such a gluing seems impossible; however, we must remember that all rules for gluing are given in advance and, in consequence, even these gluings cannot be excluded a priori.


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homeomorphic to the sphere (Fig. 46). This can be treated as the gluing of the sides of some cut made along an arc. The proof of the theorem is thus complete. Combining this theorem with the topological version of the Euler Theorem given at the end of the previous subsection, we obtain the following: A development without boundary, composed of finitely many bounded polygons, is homeomorphic to the sphere if and only if it satisfies the Euler relation v − e + f = 2. This condition describes the situation in which a surface homeomorphic to the sphere can be glued from simple pieces. 1.7.7 Developments with boundary. Theorem 7 can be supplemented with the following assertion: For every development without boundary which is composed of finitely many bounded polygons, the Euler characteristic v − e + f is at most 2. If we assume v − e + f > 2, then, repeating the procedure at the beginning of the proof of Theorem 7, we obtain a development that consists of a single polygon. For the latter, we will have f = 1 and e > k + 1. Continuing in the same way, we find a pair of adjacent sides a, b having different orientations and subject to gluing (Fig. 44). Gluing them together and repeating the arguments, we finally arrive at a polygon with two sides, which must be glued to get a single edge, and with two vertices, i.e., we obtain a development with f = 1, e = 1, and v = 2; therefore, v − e + f = 2. Since all operations preserve the Euler characteristic, this characteristic was 2 from the very beginning, contradicting our assumption. The same result is also valid for developments with boundary. In the general case we can formulate the following theorem: Theorem 8. For every development composed of finitely many bounded polygons, we have v − e + f ≤ 2 − h, where v, e, and f have the same meaning as above, while h is the number of polygonal circles forming the boundary of the development.26 Moreover, if v − e + f = 2 − h, then the development is homeomorphic to a spherical polygon bounded by h polygonal circles; conversely, v − e + f = 2 − h for every such development. 26

Saying that the boundary of a bounded development is formed by polygonal circles, we bear it in mind that, while proceeding from one boundary edge to another edge through a common vertex (a vertex can be common as a result of gluing the corresponding sides of the polygons of the development together), etc., we obtain a closed polygonal line. This is obvious, since the number of edges is finite and such a chain of sides must close. If there are several polygonal circles of this type, then they are disjoint, because only two boundary edges can touch at a single boundary vertex; for this reason, any chain of successive adjacent boundary edges is uniquely determined by each of these edges.

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In particular, we have h = 0 for a development without boundary; hence, the theorem includes all previous results. Proof. For developments without boundary (h = 0), this theorem is contained in the assertions proved above. Consider a development with boundary. We can glue a simple polygon to an arbitrary closed polygonal line bounding the development. We obtain a new development for which the number of vertices and the number of edges are the same, the number of polygons increases by one, and the number of boundary polygonal lines decreases by one, i.e., the new development has the same value of v − e + f + h. Doing this for every boundary polygonal circle, we come to a development without boundary for which v − e + f + h ≤ 2 (h = 0). Since this number remains each time the same, this inequality is also valid for the original development. If the original development has v − e + f = 2 − h, then, proceeding in this way, we arrive at a development without boundary for which v − e + f = 2. By Theorem 7, such a development is homeomorphic to the sphere; therefore, the original development is homeomorphic to a spherical polygon. Conversely, each development homeomorphic to a spherical polygon can be patched in the above way so that we obtain a development homeomorphic to the sphere, for which we therefore have v − e + f = 2 − h. The theorem is thus completely proved. 1.7.8 Unbounded developments. A similar relationship between the topological structure and the Euler characteristic can be established for unbounded developments. Since unbounded convex polyhedra, as well as their developments, are homeomorphic to the plane, we consider the Euler formula for such developments. Theorem 9. In order that a development without boundary, containing unbounded polygons, be homeomorphic to the plane, it is necessary and sufficient that the Euler characteristic of the development be 1: v − e + f = 1. Necessity: Map the development to the plane and cut out from it a bounded part circumscribed by a single polygonal circle. Moreover, do it so that this bounded part contain all vertices, all bounded edges and polygons, one segment per each unbounded edge, and one bounded piece per each unbounded polygon (Fig. 47). We thus obtain a bounded development with the same number f of polygons. Also, the number v of vertices and the number e of edges increase by the number of vertices of the boundary polygonal circle and the number of its edges. However, the number of vertices in a polygonal circle equals the number of its edges; hence, our bounded development has the same Euler characteristic v − e + f as the original development.


67

Fig. 47

Our new development is bounded by one polygonal circle and is homeomorphic to a polygon. According to Theorem 8, we therefore have v − e + f = 2 − 1 = 1, as required. Sufficiency: Assume given an unbounded development without boundary which has the Euler characteristic v − e + f = 1. Take an unbounded polygon Q1 of the development, then take an unbounded polygon Q2 glued to Q1 along an unbounded side L1 , and so forth. Since the number of all polygons is finite while the sides are glued pairwise, the resulting chain must close, so that its last polygon Qn must be glued to Q1 along the unbounded side Ln . We thus obtain a cyclic sequence of unbounded polygons Q1 , Q2 , . . . , Qn adjacent to one another along the unbounded sides L1 , L2 , . . . , Ln . In particular, the polygon Q1 can be glued to itself, but then the sequence consists of Q1 alone. If other unbounded polygons still remain in the development, then we choose one of them. Starting from it, we construct another cyclic sequence. Repeating the construction, we divide the collection of all unbounded polygons into such sequences: R1 , R2 , . . . , Rm . Each of them can be regarded as representing one unbounded end of the development. Take points A1 , A2 , . . . , An on the edges L1 , L2 , . . . , Ln along which the polygons of the first sequence R1 touch each other and connect them successively by polygonal lines in the polygons Q1 , Q2 , . . . , Qn . We thus obtain a polygonal circle M 1 which cuts out the unbounded parts of these polygons (Fig. 48; the picture is very relative, since gluing is carried out only abstractly). Performing the same operation over all sequences Ri , we obtain a bounded development with the same number f of polygons. The number v of its vertices increases by the number of vertices of the polygonal circles M i , and the number e of edges increases by the number of their sides. Any polygonal circle has the number of vertices the same as the number of sides.

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L3 L2 A2

L1

Ln

A1

An Qn

Q2

A3

Q1

Fig. 48

Therefore, the Euler characteristic of the new development is the same as that of the original: v − e + f = 1. By Theorem 8, this may happen for a development with boundary if and only if the development is homeomorphic to an ordinary polygon bounded by a single polygonal circle. Hence, there is only one polygonal circle M and, accordingly, there is only one sequence of unbounded polygons of the original development. Now, we map our bounded development onto a convex polygon. From the vertices of this polygon, we draw half-lines that divide its exterior into unbounded polygons. For each of them, there is a mapping carrying onto it that unbounded part of the corresponding polygon of the original development which was cut out while drawing the polygonal circle M . As a result of this, we obtain a mapping from the original development onto the plane. Thereby the development is homeomorphic to the plane, as claimed. From this consideration, we see that the cut-out of the development has the Euler characteristic v − e + f = 1 if and only if it is bounded by a single polygonal circle M . This corresponds to the fact that the original unbounded development has exactly one “unbounded end.” By above, the equality v−e+f = 1 is necessary and sufficient for an unbounded development to be homeomorphic to the plane. Therefore, the condition of having a single unbounded end is also necessary and sufficient for this to hold. So we arrive at the following theorem: Theorem 10. A development without boundary is homeomorphic to the plane if and only if it has one unbounded end (i.e., it contains at least one unbounded polygon, and all its unbounded polygons form a cyclic sequence if glued together along their unbounded edges as prescribed in the development). Observe that the construction of polygonal circles M i reduces the case of an unbounded development with m “unbounded ends” Ri to the case of


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a development with m boundary polygonal circles. This observation readily allows us to obtain an analog of Theorem 8 for unbounded developments: For every development without boundary, having unbounded polygons, we have v − e + f ≤ 2 − m, where m is the number of “unbounded ends.” Moreover, v − e + f = 1 if and only if the development is homeomorphic to the plane (has one unbounded end).27 Further, the following more general theorem can be derived: For an unbounded development with boundary, we have v − e + f ≤ 2 − m − h, where h is the number of closed or unbounded polygonal lines forming the boundary. However, we need neither of the last two results. 1.7.9 Jordan’s Theorem. This theorem reads: Each simple closed curve on a surface homeomorphic to the sphere divides it into two domains. We will not prove this lucid theorem in full generality. It suffices for us to prove it only in the case when the curve in question is a simple closed polygonal line on a closed convex polyhedron. Thus, assume given a closed convex polyhedron P and a closed polygonal line C on it without multiple points. Take an arbitrary point O inside P and draw a ray L from O which is disjoint from C. It is obvious that a “sufficiently acute” triangular pyramid, having the vertex at O and including a segment of the ray L, intersects P and is still disjoint from C. Taking this pyramid to be long enough for its base T to lie outside P , we construct a truncated pyramid P with base T which includes the polyhedron P (Fig. 49(a)). Consider the projection of P onto P from the point O. The polygonal line C goes into a polygonal line C on P . Moreover, by construction, the polygonal line C is disjoint from the triangle T . Obviously, our theorem will be proved if we establish that C divides the polyhedron P with the face T deleted (T is deleted with all its edges) into two domains. Consider the projection of the truncated pyramid P , with base T deleted, onto the plane E of its second base from some point O inside T (the stereographic projection, Fig. 49(b)). 27

Adding one abstract vertex for each unbounded end and regarding the unbounded edges of such an end as touching at this vertex, we obtain an abstract development without boundary which has m abstract vertices corresponding to m ends. This remark also yields a new proof of the above generalization of Theorem 9.

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O T L P O C

P

E Fig. 49(a)

Fig. 49(b)

Since this projection is obviously a one-to-one mapping from the polyhedron P with a hole onto the entire plane, it remains to prove that the simple closed polygonal line C representing the projection of C divides the plane E into two domains. To this end, we must indicate some partition of the points of the plane E (except for the points of C ) into two classes I and II such that every two points of one class can always be joined by a polygonal line disjoint from C , while no two points of different classes can be joined by such polygonal line. Choose a direction on the plane parallel to none of the sides of the polygonal line C and define the classes I and II as follows: a point belongs to the class I if the ray starting at it in the chosen direction intersects C evenly many times (if the ray passes through a vertex of C , then the vertex is counted as an intersection point only if the segments of C touching at it lie on the different sides of the ray); otherwise the point belongs to the class II. Prove that the classes I and II enjoy the required property. First of all, observe that both the classes are certainly nonempty. To check this, it suffices to draw a straight line in the chosen direction through an interior point m of a side of C . The points of this straight line on the different sides of m and sufficiently close to m belong to different classes. Now, let two points p and q be joined by a polygonal line having no common points with C . Moving continuously along this line, we go from p to q never meeting C . However, during this movement the number of intersections of the ray starting at the moving point in the chosen direction and the polygonal line C can change only if the ray encounters new vertices of C or ceases to pass through some former vertices. It is easy to check that the number of intersections changes only at those vertices for which the adjacent segments of C lie to one side of the ray; moreover, when the ray passes by each of these vertices, the number of intersections changes by ±2


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(Fig. 50).28 Thus, the parity of the number of intersections does not change and the points p and q belong to the same class. This proves that the points in different classes cannot be joined by a polygonal line disjoint from C . C

Fig. 50

It remains to prove that every two points in one class can always be joined by a polygonal line disjoint from C . Take points p and q in one class. Join them by the straight line segment pq. If in pq there are vertices of C for which the adjacent segments of C lie on one side of pq, then we replace a sufficiently small part of pq passing through such a vertex by two straight line segments disjoint from C . Similarly, we replace those parts of pq that happen to lie on segments of C . We thus obtain a polygonal line R that joins p to q and has no common points with C but the points of genuine intersection when a segment of R intersects a segment of C transversally (Fig. 51(a)). B

A pp p

C

q q q

a p

p

pq

q

q A

Fig. 51(a)

B Fig. 51(b)

Let p be the first and q be the last of these points of intersection. On the part pp of R, take a point p close to p and draw a polygonal line R that starts at p and goes “along C ” in one of the two possible directions (this polygonal line results from small parallel shifts of segments of the polygonal line C ). By construction, the polygonal line R has no 28

Here we lean upon the fact that exactly two segments issue from each vertex, i.e., we use the fact that the polygonal line in question is closed.

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common points with C and, being close to C , intersects R at a point q close to q . Consider the polygonal line R composed of the part pp of R, the part p q of R , and the part q q of R (see Fig. 51(a)). This polygonal line might have only one intersection point with C , namely, q in case the latter were inside the straight line segment q q.29 But then, while we traverse from q to q, the number of intersections of the ray issuing from these points and the polygonal line C would change its parity, and the points q and q would belong to different classes. However, q and p belong to one class, implying that p and q would also belong to different classes, contradicting our assumption. Hence, R has no common points with C at all, which completes the proof of the theorem.

1.8 Some Theorems of the Intrinsic Geometry of Developments30

1.8.1 Shortest arc. A curve in a development R whose length is least among the lengths of all curves in the development with the same endpoints is called a shortest arc. (A) Each shortest arc has finitely many straight line segments in the polygons of the development. It traverses an interior point of an edge “without refraction,” i.e., the angles that it forms with the edge are equal from the two sides (Fig. 52). A shortest arc cannot pass through a point with the complete angle less than 2π. (A shortest arc can pass through a point with the complete angle greater than 2π, as we can easily see from simplest examples. This case is not of interest for us, since we deal with developments of positive curvature.) Proofs of these assertions are based on the following trivial observation: if A is a point on a shortest arc, then some small subarc AB of the shortest arc is a straight line segment in a polygon of the development. This is so, because a neighborhood about a point in a development is composed of the 29

This case could seem impossible; however, the fact that the straight line segment q q is disjoint from C is no more evident than Jordan’s Theorem itself expressing a certain topological property of the sphere. It is instructive to verify that this may be otherwise on surfaces of another topological type. Fig. 51(b) shows a development of the M¨ obius band (the opposite sides of the rectangle are glued together so as to make the directions of arrows coincide) for which our construction yields a configuration in which the segment q q intersects the polygonal line C at q . (Since the sides of the rectangle are glued to each other so as to make the arrows coincide, the dotted lines actually form a single line.) 30 These theorems are used only in Chapters 4 and 5.

1.8 Some Theorems of the Intrinsic Geometry of Developments

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angles of polygons or, which is even simpler, is a part of the interior of some polygon. If the complete angle at A were less than 2π, then the adjacent segments AX and AY of the shortest arc form an angle less than π from at least one side. Then points X1 and Y1 lying on these segments and close to A might be connected from that side by a straight line segment (Fig. 53) and thereby the shortest arc might be shortened, which is impossible by its definition. Construction of the segment X1 Y1 can naturally be referred to as the straightening of the angle at A.

X

Y Y1

X1 A

Fig. 52

Fig. 53

If, passing through an interior point A on an edge, a shortest arc refracted, then the segments touching at A would form an angle less than π from one side. Straightening this angle, we would diminish the length of the shortest arc, which is impossible. Now, let L be a segment of a shortest arc which lies in a given polygon of a development.31 Each of its points has a neighborhood in which L is represented by a straight line segment; whence it is seen that L in the whole is a straight line segment. To prove that a shortest arc intersects each polygon finitely many times, suppose that it intersects some polygon Q infinitely many times. Then the lengths of its segments lying in Q must tend to zero, since it would have an infinite length otherwise. These segments must then condense at one of the vertices of Q. This vertex, serving as their limit point, must belong to the shortest arc in question. However, in a neighborhood about a vertex, a shortest arc is a straight line segment and cannot intersect the polygon Q infinitely many times. This contradiction shows that the number of segments of a shortest arc in each polygon is finite, completing the proof of the theorem. 31

It is possible that a shortest arc traverses a polygon several times and therefore has several segments in it. An example of a shortest arc traversing a polygon several times is easy to construct for a development consisting only of a single polygon as a crosslike development of the cube (Fig. 35).

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(B) If a development has boundary, then a shortest arc can partly lie on the boundary (as the line CB in Fig. 54). Moreover, its segments may form angles only at those vertices of the boundary at which the angles of the development are greater than π.

α B

A C

Fig. 54

Indeed, let two segments AX and AY of the same shortest arc touch at a boundary point A of a development and form some angle. If this angle α from the interior of the development were less than π, then it would be possible to straighten α (Fig. 53), thus replacing the shortest arc by an arc with smaller length, which is impossible. Consequently, α > π, and hence the complete angle at A is greater than π. (C) In a development of positive curvature, each shortest arc going in the interior of the development has a neighborhood developable on the plane; moreover, the shortest arc then goes into a straight line segment. (Developing on the plane is thought of as an isometry.) If some endpoint of a shortest arc is a point O with the complete angle less than 2π, then none of its neighborhoods is developable on the plane. However, making a cut along a straight line segment issuing from O in its neighborhood, we can develop the neighborhood with this cut on the plane. The possibility of developing the neighborhoods of endpoints of a shortest arc on the plane is understood in exactly this way. Theorem C follows from Theorem A. Indeed, according to Theorem A, a shortest arc cannot pass through points with complete angles less than 2π. Therefore, in a development of positive curvature, each interior point of a shortest arc has a neighborhood developable on the plane. By Theorem A, in each of these neighborhoods the shortest arc is a straight line segment. Choosing a finite number of such neighborhoods so as to cover the whole shortest arc, we obtain a neighborhood of the arc which can be developed on the plane; moreover, the shortest arc as a whole is transformed into a straight line segment.


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(D) In a development of positive curvature, two shortest arcs AB and AC, starting at one point A and terminating at different points B and C, either are disjoint or one of them is simply a part of the other. To prove this, assume that shortest arcs AB and AC have a common point D but do not overlap along the segment AD (Fig. 55). It is possible that D is an endpoint of one of them, say D = B. Then the segments AD of these shortest arcs have the same length. Otherwise, replacing the longer of them with the shorter, we would reduce the length of a shortest arc. Therefore, we obtain two shortest arcs between A and C that have the common segment CD but further diverge along different segments AD. B D

C

A Fig. 55

But then their branches form three angles at the divergence point D, two of which are equal to π as the angle between AD and DC. The complete angle at D turns out to be greater than 2π, which is excluded by the positive curvature condition. Therefore, the shortest arcs must overlap along AD. For perfectly the same reasons, these arcs cannot diverge father. Thereby they overlap completely, so that one of them is part of the other. (E) In every development, each pair of points can be joined by a shortest arc. Let A and B be two given points of a development R. Let some polygonal line of length l join them. Obviously, a shortest arc AB should be sought only among shorter curves. Let G stand for the set of all points X of R at distance ρR (AX) ≤ l from A. Obviously, each point of the development, in particular each point of G, has a neighborhood in which there is a shortest arc between every pair of points, presenting a straight line segment. Each such neighborhood may be thought of as consisting of polygons that are parts of the polygons of the development. By virtue of the well-known Borel Lemma, we can choose finitely many such neighborhoods covering G. They all can be split into triangles; we denote the latter by Ti . Thus, the set G is covered by disjoint triangles Ti in each of which there is a shortest arc between every pair of points, presenting a straight line segment. Obviously, it suffices to seek a shortest arc AB among the polygonal lines whose segments lie in different triangles Ti . (A shortest arc AB cannot

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leave G, for the length of AB would be otherwise greater than l, because all points whose distance from A is at most l belong to G.) The length of a polygonal line whose segments lie in triangles is obviously a function of a finite number of variables ranging over a bounded closed domain.32 Therefore, it attains a minimum, i.e., a shortest arc AB exists. The above results A–E are summarized in the following theorem: Theorem 1. In every development of positive curvature, each pair of points can be joined by a shortest arc. Such an arc consists of straight line segments lying in polygons of the development. In the interior of the development, a shortest arc traverses edges “without refraction” and has a neighborhood developable on the plane; moreover, a shortest arc then transforms into a straight line segment. If a development has boundary, then a shortest arc, approaching the boundary, may refract only at the boundary vertices with complete angles at them greater than π. 1.8.2 Curvature of a geodesic polygon. A geodesic polygon is thought of as an arbitrary part of a development which is bounded by a closed curve composed of shortest arcs and which is homeomorphic to a disk, i.e., admits a homeomorphism onto a disk like a simple planar polygon, certainly with the identifications (gluings) prescribed by the development taken into due account. Since shortest arcs consist of straight line segments lying in the polygons of the development, a geodesic polygon is cut out of the development by a polygonal line composed of such straight line segments and therefore itself represents some development consisting of pieces of polygons of the original development. However, as opposed to an ordinary polygon, the interior of a geodesic polygon may contain genuine vertices of the development. The geodesic triangle ABC in Fig. 56 can serve as an example, where the role of a development is played by an ordinary cube. 32

This function can be rigorously defined, for example, as follows: Determine the position of a point on a side aj of the triangles Tj by the distance xj from one of the endpoints of aj . Consider those polygonal lines joining A to B which intersect the sides of the triangles Ti in a given order aj1 , aj2 , aj3 , . . . (if a polygonal line overlaps with a side ak over some segment, then we do not count it as intersecting ak but regard as intersecting the sides touching at the endpoints of ak ). The length of such a polygonal line is clearly a quite definite function of the variables xj1 , xj2 , . . . and therefore attains a minimum. Consider all possible sequences aj1 , aj2 , . . . of sides that can be intersected by a polygonal line AB. To each of these sequences corresponds its own function and minimum. The smallest of these minima yields the shortest arc AB. The fact that we speak about several functions does not contradict the proposition we made in the beginning that the length of the polygonal line AB is a single continuous function, because any number of continuous functions can be combined into one function but in a greater number of variables. For example,f (x) (a ≤ x ≤ b) and g(x) (c ≤ x ≤ d) may be treated f (x) (a ≤ x ≤ b, y = y0 < c), as the function ϕ(x, y) = g(y) (c ≤ y ≤ d, x = x0 < a).


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Since a geodesic polygon is itself a development, its angles are defined in the same way as the angles at the boundary vertices of a development (i.e., the angle at a vertex is the sum of the adjacent angles of the planar polygons forming the development which represents the geodesic polygon in question). A B

C

Fig. 56

Theorem 2. The excess of the sum of angles of a geodesic n-gon (in comparison to the sum of angles of a planar n-gon) is equal to the curvature of the n-gon, i.e., if α1 , . . . , αn are the internal angles of a geodesic n-gon and ω1 , . . . , ωm are the curvatures at the vertices of the development lying inside the n-gon, then n m αi − (n − 2)π = ωj . (1) i=1

j=1

To prove this, partition the given geodesic polygon P into triangles Ti each of which lies in some polygon of the development. It can turn out that some vertices of these triangles which lie on the boundary of P are not vertices of P . However, we will treat them as vertices of P , which does not influence the quantity on the left-hand side of (1), since the angle at such a vertex equals π. In much the same way, some vertices of the triangles Ti which lie inside P may fail to be vertices of the development. However, “cutting” the development into the triangles Ti , we make such vertices into vertices of the development. This does not influence the sum of curvatures either, since the complete angle at each new vertex equals 2π and the curvature at the vertex equals zero. Thus, all vertices of the triangles Ti which lie on the boundary of P can be considered as vertices of P , while all vertices inside P can be treated as vertices of the development. Since each triangle Ti lies in some polygon of the development and is therefore an ordinary planar triangle, the sum of its angles is π. If the number of all the triangles Ti is f , then the sum of their angles is f π. On the other hand, this equals the sum of the angles at all vertices lying on the boundary and inside P . Hence, n m αi + θj , (2) fπ = i=1

j=1

78


where θj is the complete angle at a corresponding vertex inside P . If e is the number of sides of the triangles Ti and v is the number of vertices, then by Euler’s Theorem f − e + v = 1 (Theorem 8, Section 1.7). At the same time, the number of all vertices equals the sum of the number of vertices lying inside P and on its boundary, i.e., v = m + n. Therefore, f − e + m + n = 1.

(3)

Finally, each triangle has three sides; exactly n sides on the boundary belong to one triangle each; and the internal sides belong to two triangles each. Hence, 3f = 2e − n. (4) Multiplying both sides of (3) by 2 and considering that 2e = 3f + n in view of (4), we obtain f = 2m + n − 2. (5) Inserting this expression for f in (2), we find that 2πm + π(n − 2) =

n

αi +

i=1

or

n

αi − (n − 2)π =

i=1

m

m

(2π − θj ) =

j=1

θj

j=1 m

ωj ,

j=1

for we have ωj = 2π − θj by the definition of curvature. This completes the proof. Theorem 3. The total curvature of a development homeomorphic to the sphere is equal to 4π. To prove this, cut out a polygon P inside one of the polygons of a development R. If α1 , . . . , αn are the angles of P , then n

αi = (n − 2)π.

(6)

i=1

If the development is homeomorphic to the sphere, then the complement P = R \ P (with its boundary adjoined) is a geodesic polygon. By construction, P contains all vertices of the development and its total curvature equals therefore the total curvature of the development. Hence, if Ω is the total curvature and αj , . . . , αn are the angles of the geodesic polygon P , then by Theorem 2 n αi − (n − 2)π = Ω. (7) i=1


79

However, each vertex of P is simultaneously a vertex of P , and the complete angle at it equals 2π, since the vertex lies inside a polygon of the development. Consequently, αi + αi = 2π, whence

n

αi +

i=1

or, in view of (6),

n

n

αi = 2πn

i=1

αi = (n + 2)π.

i=1

Inserting this in (7), we obtain Ω = 4π, as claimed. 1.8.3 A polygon of zero curvature. Theorem 4. If a geodesic triangle does not contain interior points of nonzero curvature and its sides lie inside the development, then the triangle is developable on the plane, i.e., it is isometric to an ordinary planar triangle.33 Let a geodesic triangle ABC contain no vertices whose curvature differs from zero and let its sides lie inside the development. By Theorem 1, the triangle has no other angles than those at its vertices. By Theorem 2, the sum of its angles is π; hence, each angle is less than π. By Theorem 1, a neighborhood of the side AB can be developed on the plane; moreover, AB then goes into a straight line segment. Since the angle B is less than π, it follows that from the triangle ABC we can cut out some narrow triangle ABX with vertex X on the side BC, so that ABX can be developed on the plane. However, the same argument applies to the remaining triangle AXC. Therefore, we can shift the point X towards C, thus developing a larger and larger part of ABC on the plane until all of ABC becomes developable on the plane. A rigorous exposition of this evident reasoning causes no difficulties.34 33

If we divide a development into parts, then it is always developable on the plane. We speak here about the developing on the plane without any cuts, so that all gluings prescribed in the development must be actually implemented rather than treated abstractly. 34 Prove that the limit of triangles developable on the plane is again a triangle developable on the plane. Then there is an extreme point X0 such that the triangle ABX0 is developable on the plane. The point X0 must coincide with C; otherwise, applying the above argument, we would obtain a larger triangle developable on the plane.

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Theorem 5. If the interior of a geodesic polygon Q contains no points of nonzero curvature, then Q can be split into triangles by disjoint diagonals. We treat the polygon Q as a development, and by a diagonal we mean a shortest arc going inside Q from one of its vertices to another. (If Q is part of a development R, then a shortest arc in Q may fail to be a shortest arc in R. For example, take the arc AB in Fig. 57; it is shortest in the piece Q composed of the faces Q1 , Q2 , and Q3 , but is not shortest in the whole polyhedron.)

A

Q1 Q3

B Q2

Fig. 57

To prove the theorem, observe that the curvature of the polygon Q is zero. Therefore, by Theorem 2, the sum of its angles α1 , . . . , αn is the same as that in a planar polygon with the same number n of vertices, i.e., α1 + α2 + . . . + αn = (n − 2)π. Therefore, at least three of its angles are less than π. B Q

F

E D A Fig. 58(a)

Fig. 58(b)

Let A, B, and C be vertices with angles at them less than π. Obviously, some two of them are not adjacent. (If all the three vertices were adjacent to one another, then the polygon would be itself a triangle and there would be no need to split it into triangles.) Assume the vertices A and B are nonadjacent. They divide the perimeter of the polygon into two polygonal lines on which we also have other vertices (Figs. 58).


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Take any two of these vertices, D and E, on different polygonal lines and connect them by a shortest arc DE in Q. By Theorem 1, DE cannot pass through the vertices A and B, since the angles at A and B are less than π. Therefore, DE cannot always be in contact with the perimeter of the polygon, but must go inside the polygon, backing from some vertices. For example, in Fig. 58(a), F serves as such a vertex. Then the segment EF is a diagonal splitting Q into two polygons with fewer vertices. Applying the same arguments to each of these polygons, we split them by diagonals and continue the procedure until we come to a partition of Q into triangles, thus completing the proof of the theorem. According to Theorem 4, every triangle into which we split Q is developable on the plane. Developing all these triangles on the plane and adjoining them to each other along the sides in which they touch in the polygon Q, we thus develop the polygon Q on the plane. However, after this procedure of successive adjoining, the triangles can overlap one another, so that the developed polygon Q can overlap itself (see Fig. 58(b)). Taking stock of the above, we can formulate the following theorem. Theorem 6. A geodesic polygon without interior points of nonzero curvature is developable on the plane but possibly with overlapping.

L Q

P

Fig. 59

In Section 1.3, we proved that every bounded convex polyhedron P with boundary can be uniquely completed to a closed polyhedron P without adding new vertices (Theorem 4 of Section 1.3). If the boundary of P represents a single polygonal line L, then the closed polyhedron P consists obviously of the polyhedron P and the polyhedron Q adjoining P along L (Fig. 59). The polyhedron Q has only boundary vertices and is therefore developable on the plane by Theorem 6. If the polyhedron P is bounded by several polygonal lines, then the same assertion is valid for each of the pieces Qi that complete P to a closed polyhedron. The result is expressed in the following theorem which we formulate in obvious terms.

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Theorem 7. Every bounded convex polyhedron with boundary can uniquely be completed to a closed polyhedron by gluing polyhedra to the boundary polygonal lines. The glued polyhedra have no interior vertices and so are developable on the plane (possibly with overlapping). 1.8.4 Exercises. 1. Prove that the total curvature (i.e., the sum of curvatures at all vertices) of every bounded development is expressed by the formula n (π − αi ), ω = 2πχ − i=1

where χ = v − e + f is the Euler characteristic of the development and αi are the angles at its boundary vertices. For developments without boundary, ω = 2πχ. (The positivity of curvature is not assumed. Test this on examples like Fig. 2(b).) 2. Prove that the total curvature of an unbounded development homeomorphic to the sphere does not exceed 2π. Generalize this result by showing that ω ≤ 2πχ for every unbounded development without boundary. What inequality (involving the angles at boundary vertices) holds in the presence of a boundary? (This is a fairly hard exercise.) 3. Give an example of shortest arcs on a nonconvex polyhedron which issue from a single point, overlap over some segment, and further diverge forming a “fork.” 4. Cut out a geodesic polygon (without internal vertices) from a cube so that it be developable on the plane with overlapping. 5. Find an example of a planar polygon with overlapping which cannot be superposed on any convex polyhedron. (A due example will be presented in Subsection 5.1.8.)

1.9 Generalizations 1.9.1 Convex bodies. The notion of convex body was given in Section 1.1. Some notions and results of this and forthcoming sections can be generalized to arbitrary convex bodies. For instance, a convex body can be proved to have a support plane at each point of its surface, and conversely: a body possessing this property is convex. Each point not belonging to a convex body can be separated from the body by some support plane. Therefore, a convex body is the intersection of the half-spaces bounded by its support planes. Conversely, the common part of any number of half-spaces is a convex body (provided that it has interior points). Convex bodies may be unbounded or bounded. For an unbounded convex body, the limit cone (an analog of the limit angle of a polyhedron) can be defined as the surface of the body formed by the rays issuing from some of

1.9 Generalizations

83

its points and lying entirely in the convex body in question. Under a similarity contraction, the surface of the body transforms into the limit cone. The spherical image of the limit cone coincides with the closure of the spherical image of the surface of the body. One of the main methods for studying convex bodies is the approximation by convex polyhedra. Taking several points on the surface of a body and considering their convex hull, we arrive at a convex polyhedron inscribed in the body. If we increase the number of points to infinity, placing them more and more densely, then the corresponding convex polyhedra will converge to the body. It is easy to prove that, in the limit, the support planes of these polyhedra give the support planes of the body. Whence we can easily deduce that, for every point on the surface of the body, there is a support plane passing through this point. A convex surface may be incomplete or complete, coinciding with the whole surface of a convex body. A complete surface may be closed, bounding some bounded convex body, or open but containing no straight line, or an unbounded cylinder. A closed surface is homeomorphic to the sphere, and an open surface (not a cylinder) is homeomorphic to the plane (the proof of the first assertion is analogous to that for polyhedra; the proof of the second assertion is slightly different). An incomplete convex surface can be included in a complete surface by taking as such the surface of the convex hull of the initial surface. The concept of development is inapplicable to general convex surfaces, since a general surface is not composed of polygons. However, in a perfect analogy to the case of polyhedra, the concept of intrinsic metric can be introduced for arbitrary surfaces; the metric is determined by the distances between the points of a surface measured on the surface. The collection of the properties of a surface and figures on it which depend only on the intrinsic metric comprises the intrinsic geometry of the surface. The intrinsic geometry of general convex surfaces is studied in my book [A15]. There, among other things, we discuss questions similar to those treated here for polyhedra. An abstract specification of the intrinsic metric of a surface then plays the role of development. 1.9.2 Polyhedra in n-dimensional space. The geometry of the ndimensional Euclidean space can be expounded analytically, in full analogy with the conventional analytic geometry, or synthetically, using some axioms naturally generalizing the axioms of three-dimensional Euclidean space.35 Pursuing a synthetic approach, we obtain the foundations of n-dimensional 35

Introduce the notion of an m-dimensional plane P m , 0 ≤ m ≤ n − 1: a zerodimensional plane is a point; one-dimensional, a straight line; two-dimensional, an ordinary plane; three-dimensional, an ordinary three-dimensional space; etc. The axioms of coincidence: (1) For arbitrary m + 1 points “in general position” (i.e., lying in no (m−1)-dimensional plane) there is a unique m-dimensional plane passing through them. (2) If two planes P m and P k meet, then their common part

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geometry by analogy to the foundations of three-dimensional geometry. In this way, there is no difficulty in laying the foundations of the theory of polyhedra in n-dimensional Euclidean space. The definition of a solid polyhedron goes from m to m+1: an (m+1)-dimensional (solid) polyhedron is a body lying in an (m+1)-dimensional plane, having interior points in it, and bounded by finitely many m-dimensional polyhedra. An n-dimensional polyhedron is a body bounded by finitely many (n − 1)-dimensional polyhedra. Each of these (n − 1)-dimensional polyhedra is called a face, provided that we adjoin to it all other (n − 1)-dimensional polyhedra (if they exist) which lie in the same (n − 1)-dimensional plane and have with it, as well as with each other, common (n − 2)-dimensional faces. If we call an (n − 1)-dimensional plane simply a plane, which is routine, then the notion of convex polyhedron is defined word for word as in Section 1.1. Also, in exactly the same way we establish that the two definitions of convexity are equivalent. Further, all theorems of Sections 1.2–1.5 and their proofs are translated to n-dimensional convex polyhedra without changes. We should only mean by a plane an (n − 1)-dimensional plane and keep in mind that there are also planes of lesser dimensions. For example, the limit angle of a polyhedron may degenerate into a flat angle of any dimension from n − 1 to 1. The notions of a half-space, a perpendicular to a plane, and parallel planes are defined in the same way as in three-dimensional space. The spherical image is defined as a subset of the (n − 1)-dimensional sphere, the latter defined as the set of points equidistant from a given point, the center of the sphere. By analogy to great circles, the sphere has “great spheres” of dimensions from n − 2 to 1 which appear as its sections by the planes of dimensions from n − 1 to 2 passing through the center. A one-dimensional sphere is a circle. The notion of a spherical polyhedron is a natural generalization of the notion of a spherical polygon. Here we once again meet an analogy with usual geometry. Basing on these general observations, we can generalize all results of Sections 1.1–1.5 without effort to polyhedra in n-dimensional space. The notions of development and intrinsic geometry can be generalized as well. However, the matter is essentially more complicated here, since now it is not only vertices that may have neighborhoods not isometric to an (n − 1)dimensional ball (an analog of a disk). This property can be shared with points lying on faces of any dimension from n − 2 to 0 (i.e., up to vertices). In a convex polyhedron, the points lying inside (n − 2)-dimensional faces never have such neighborhoods. This results in complicating the notion of is a plane P l of dimension satisfying the inequality l ≥ m + k − n. (For example, in a four-dimensional space, two-dimensional planes may meet at a single point, l = 0, or in a straight line, l = 1.) (3) In the space, there are at least n + 1 points in general position (not lying in any plane). The remaining axioms (of order, congruence, continuity, and parallelism) are the same as in three-dimensional space. See, for instance, [E1, Chapter V, Section 14].

1.9 Generalizations

85

curvature. There appear, so to speak, curvatures of different dimensions from 1 to n − 1, corresponding to faces of dimensions from n − 2 to 0. Explicating the concept of these curvatures is possible in an elementary level, but we shall not dwell upon this. 1.9.3 Polyhedra on the n-dimensional sphere and in hyperbolic space. A three-dimensional spherical space is nothing more than a sphere in four-dimensional space, and so convex polyhedra are naturally defined in it. Replacing straight line segments by arcs of great circles and replacing planes by “great” two-dimensional spheres, we easily generalize all conclusions of Sections 1.1–1.3 and 1.6–1.7 to a such space. Unbounded polyhedra are absent. The notion of spherical image can be translated to spherical space, despite the absence of parallel straight lines. This is achieved by assigning to a great sphere all its centers (on the whole three-dimensional sphere) assigning to a point the great sphere with center this point. To avoid the ambiguity due to the fact that a great sphere, like a great circle, has two antipodal centers, we can identify antipodal points. We then obtain the so-called Riemann elliptic space. When we draw rays from the center of the sphere in four-dimensional space through the points on its surface, the polyhedra on the sphere are put in correspondence with polyhedral angles. Therefore, the theory of polyhedra in spherical space is equivalent to the theory of polyhedral angles in fourdimensional Euclidean space. Much the same arguments are applicable to n-dimensional spherical space which is nothing more than a sphere in (n + 1)-dimensional Euclidean space. Hyperbolic (Lobachevski˘ı) space differs from Euclidean space by the absence of the axiom of uniqueness of a straight line passing through a given point in parallel to a given straight line. Therefore, all conclusions not leaning upon the concept of parallelism (explicitly or implicitly) can be translated to hyperbolic space word for word. In the same manner, we can translate the notion of convex polyhedra and all results about them which do not lean upon parallelism, i.e., all but the notions of the limit angle and spherical image of a polyhedron. To elucidate the idea of convex polyhedra and arbitrary convex bodies in hyperbolic space, we use the interpretation of hyperbolic space as the interior of some ball E in Euclidean space, where the hyperbolic straight lines are displayed by segments of Euclidean straight lines. Then the straight line segments in the sense of hyperbolic geometry and those in Euclidean geometry coincide, and therefore every convex body in hyperbolic space is displayed by a Euclidean convex body with the points beyond the ball E excluded. Conversely, every Euclidean convex body H of this kind represents a convex body H in hyperbolic space. The fact that the body H approaches the surface of the ball E means that the body H is unbounded. Since H can approach the surface of E several

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times, an unbounded convex body in hyperbolic space can have several (and even infinitely many) “unbounded ends.” A hyperbolic plane is displayed by the piece of a plane which is cut out by the ball E. Therefore, convex polyhedra of hyperbolic space are displayed by Euclidean convex polyhedra without the points lying beyond E. Unbounded convex polyhedra of hyperbolic space may have any finite number of “unbounded ends,” whereas unbounded convex polyhedra in Euclidean space have one end or two ends (if they are prisms). This circumstance essentially enriches and complicates the theory of unbounded convex polyhedra in hyperbolic space. Bounded polyhedra of hyperbolic space are displayed by bounded Euclidean polyhedra and their theory brings about almost nothing new. The notion of development is defined in hyperbolic space in the same way as in Euclidean space; we should only speak of polygons on a hyperbolic plane. The results of Sections 1.1–1.3, 1.6, 1.8, and 1.7 in regard to bounded polyhedra are translated to hyperbolic space verbatim. As for unbounded polyhedra and generalizations of the notion of spherical image, they are scantily studied as yet, and their thorough exploration can serve as a theme of purely geometric research promising, as it seems, some interesting and profound results.

2 Methods and Results

2.1 The Cauchy Lemma 2.1.1 In Chapters 3 and 4 a unified method is used for proving all the uniqueness theorems, i.e., theorems claiming the congruence of polyhedra with coinciding data; these theorems are stated in Sections 2.3 and 2.4 below. This is the very method by which Cauchy proved his congruence theorem for closed convex polyhedra consisting of the same number of equal similarly-situated faces. Cauchy’s method is one of the most impressive arguments in geometry; this is undoubtedly the prevailing opinion. The method relies upon a certain topological lemma, which will be referred to as the Cauchy Lemma. We first prove the lemma; next, we demonstrate how Cauchy applied it himself, thus clarifying the general scheme used below for proving uniqueness theorems. The Cauchy Lemma. Suppose that some edges of a closed convex polyhedron are labeled by plus or minus signs. Sign changes may occur in the labeled edges around a vertex. The claim is as follows: It is impossible to have at least four sign changes at every vertex. (See Fig. 60 as an illustration.)

+

_+

+

_ +

+ _ _

+

+

+

Fig. 60

The lemma can be rephrased in purely topological form: Suppose that a “net of edges” is given on a surface homeomorphic to the sphere, i.e., suppose that finitely many “edges” (each of which is homeomorphic to a straight line segment) are given, and these edges are pairwise disjoint except possibly at their endpoints, the “vertices of the net.” (Such is

88


the net of labeled edges of a polyhedron.) Assume further that none of the regions separated by the net on the sphere is bounded by only two edges. (This condition reflects the fact that no two vertices of a polyhedron are joined by two edges.)1 Then it is impossible to assign pluses and minuses to the edges of the net so that at least four sign changes occur around each vertex. In fact, we shall prove a somewhat sharper assertion: Denote by N the total number of sign changes at all vertices and let V stand for the total number of vertices. Then N ≤ 4V − 8. (1) The claim of the lemma follows immediately from this inequality. 2.1.2 We reproduce Cauchy’s proof of inequality (1). It is based on counting the sign changes that occur as one moves around each of the regions separated on the surface by the edges of the net. If two edges are adjacent at a vertex A, then they are also adjacent in moving around the region to whose boundary they belong. The converse is true as well. Therefore, if we count the sign changes occurring as one moves around the regions, then their total number will again equal N . A B

G

E D

C

Fig. 61

This simple observation requires some justification. Orient the surface and start going around the regions of the surface in the direction prescribed by this orientation. Suppose that, while going along the contour of a region G, we passed an edge AB in the order A to B. Now, we pass by the vertex B, starting from the edge AB and going around G to the next edge BC (Fig. 61). This edge follows AB in the order around the region G as well as in the order around the vertex B. Thus, the sign change from AB to BC is counted in both cases. 1

The lemma fails without this condition. Indeed, it suffices to take an even number 2n > 2 of meridians on a sphere and label them cyclically by pluses and minuses. More complicated examples can also be given.

2.1 The Cauchy Lemma

89

We keep going around the region G further along the edge BC from B to C, etc. We continue the procedure until we arrive at some edge that was already counted and moreover, were the motion continued, the edge would be passed in the same direction as it was at the first time. This is because the net can contain some edges that do not separate G from other regions since the edge in question can, for instance, have a free endpoint in G or can join two closed contours that bound G (see Fig. 61). Every such edge will be passed twice: first, in one direction and, next, in the opposite direction. Thus, we can assume that each edge either belong to the boundary between different regions or is counted twice in the boundary of a single region where it is passed in opposite directions. Therefore, if two edges, say, AB and BC separate the regions G and H, then in G they are passed from AB to BC and in H from BC to AB, which is the same as the total trip around the vertex B. If a vertex D is a free endpoint of the edge DE, then while moving around G we pass the vertex D and return to the same edge DE. This is the same as making a trip around D. When we say that a region has n edges, we shall bear in mind that each edge not separating the region in question from another region is counted twice. The number of sign changes in moving around a region cannot be greater than the number n of its edges; furthermore, this number is always even since, when we complete this trip, we return to the initial sign. Therefore, if Fn is the number of regions with n edges, then for the total number N of sign changes we obtain the estimate N ≤ 2F3 + 4F4 + 4F5 + . . . .

(2)

Here we have used the fact that there is no region bounded by two edges, i.e., that F2 = 0. Now we transform estimate (2) by applying the generalized Euler formula: If V , E, and F are the numbers of vertices, edges, and regions in a net, then V − E + F ≥ 2. (3) In accordance with Euler’s Theorem, V − E + F = 2 for a connected net, while V − E + F = 1 + H for a net with H connected components, as proved in Section 1.7 of Chapter 1 (Theorem 5). Since each edge either belongs to two regions or is counted twice for a single region, we have 2E = nFn . (4) n

The total number F of regions is F =

n

Fn .

(5)

90


Formula (3) implies

4V − 8 ≥ 4E − 4F. Substituting E and F from (4) and (5) in this inequality, we obtain 4V − 8 ≥ 2(n − 2)Fn = 2F3 + 4F4 + 6F6 + . . . .

(6)

n

The right-hand side of (6) is not less than the right-hand side of (2); therefore, for n ≥ 3 the number 2(n − 2) is not less than the even number nearest to n from below. Consequently, (2) and (6) imply N ≤ 4V − 8,

(1)

as desired. From inequality (1) we can infer a slight strengthening of the Cauchy Lemma: Under the conditions of the Cauchy Lemma, it is impossible to have at least four sign changes for all vertices except three of them with two sign changes, or except two of them, one with no sign changes and the other with exactly two sign changes, or except one of them with the number of sign changes less than four. Were it otherwise, we would have N ≥ 4V −6, contradicting inequality (1). We shall have the opportunity to use this refined version of the Cauchy Lemma later. 2.1.3 The above proof of the Cauchy Lemma, computational by nature, looks somewhat mysterious and does not reveal the visual-geometric basis of the lemma. It therefore seems that a purely geometric proof of the lemma will not be amiss. The proof is by contradiction. Suppose that some edges in a net on a sphere are labeled by pluses or minuses so that at each vertex there are at least four sign changes. _

_ + _

+ _

_

+ G +

_ + A

B _

_

_

_ + _

_

_

_

O + _

+ C

Fig. 62

D

+ _

2.1 The Cauchy Lemma

91

Take an edge AB of the net labeled by a plus. The vertex B also has other incident edges so that there are at least four sign changes around B. Hence, there is at least one edge BC incident to B, labeled by a plus, and such that, together with AB, it separates edges incident to B and labeled by minuses. The vertex C too has an incident edge CD labeled by a plus which, together with BC, separates edges labeled by minuses. Continuing in this manner, we go from AB along the edges BC, CD, etc. Since the number of edges is finite, eventually we return to a previously visited vertex O and thus obtain a closed contour L (see Fig. 62). In particular, O may coincide with A. The contour L divides the sphere into two regions; consider one of them, G. All edges of the contour L are labeled by pluses, and each pair of adjacent edges separates edges labeled by minuses, with the possible exception of edges incident to O. Therefore, the region G includes edges which are labeled by minuses and are incident to each of the vertices, except possibly O. Let us prove that the region G contains at least one vertex of the net. Were it otherwise, the endpoints of any edge in G would be at the vertices of G. However, all vertices of G, except possibly O, have incident edges going inside G; but then two neighboring vertices must be joined by such an edge. (To verify this, it is sufficient to imagine G as a polygon; the edges inside G are disjoint lines connecting its vertices. However, disjoint diagonals can be drawn from all but two of the vertices. If the edges go from all but one of the vertices, then among them there is an edge joining two neighboring vertices. A rigorous proof is carried out by induction on the number of vertices.) If two neighboring vertices X and Y of the contour L are joined by an edge inside G, then we obtain a digon bounded by two edges XY . Furthermore, if there are no vertices inside G, then we obtain a digon in our net. By assumption, the net has no such digons. Consequently, the assumption that there are vertices of the net inside G is invalid. Let A1 be a vertex inside G; there is an incident edge A1 B1 labeled by a plus. Arguing as above, we conclude that the vertex B1 also has an incident edge B1 C1 labeled by a plus which, together with A1 B1 , separates some edges incident to B1 and labeled by minuses. Thus, we move along the edges A1 B1 , B1 C1 , C1 D1 , etc. until we arrive at a previously visited vertex within G or reach the boundary of the region G at some vertex N . In the first case we obtain a closed contour L1 that cuts out a new simply connected region G1 from G. To this region G1 we can apply the same arguments as to G. In the second case, we move from the vertex N in the opposite direction towards the initial vertex A1 and beyond it. If the resulting path is closed, we obtain the first case again. But if the path terminates at some vertex M of the region G, we obtain a line M N that divides G into two simply connected parts G and G .

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If the vertices M and N differ from O, then there are edges labeled by minuses, issuing from each of these vertices inside G, i.e., inside G or G . Therefore, at least one of the regions G and G is such that edges with minuses issue from all but possibly one of its vertices. (Three possible situations are shown in Figs. 63(a), 63(b), and 63(c); a region with the required properties and its exceptional vertex are indicated below each separate picture.) If, say, N coincides with O, then for G1 we take one of the regions G and G that includes an edge incident to M and labeled by a minus. Thus, in each case we can cut out a region G1 from G which enjoys the same properties as G. But then, repeating our arguments, we can cut out a region G2 from G1 , etc. ad infinitum. This is, however, impossible since the number of edges in the net is finite. Consequently, there is no net whatsoever with edges labeled as assumed, which was to be proved. N

N +_ _ + G O

_

+

N

_

_ _ +

G

G"

_

+ M

Fig. 63(a) G = G , exceptional point: O

+

_

_

+

+ _

_

_

_ G"

O

+

+

+

_

+

+

M

Fig. 63(b) G = G , exceptional point: N

G

+

G" _

+ _

_ + + M

Fig. 63(c) G = G , exceptional point: M

2.1.4 Now we show how the Cauchy Lemma applies to the proof of his theorem of the congruence of closed convex polyhedra composed of the same number of equal similarly-situated faces. Cauchy proves the following lemma: If two convex polyhedral angles consist of the same number of corresponding planar angles of equal measure which follow each other in the same order, then either these polyhedral angles are congruent or the differences of the corresponding dihedral angles change sign at least four times around the vertex. (This lemma is proved in Section 3.1 of Chapter 3.) Now, let P1 and P2 be two polyhedra satisfying the assumption of the Cauchy Lemma. If all their dihedral angles are equal, then the polyhedra are congruent. When not all dihedral angles are equal, label by a plus (minus) each edge of P1 at which the dihedral angle is larger (smaller) than the corresponding dihedral angle of P2 . In accordance with the lemma, at each vertex with a labeled edge there are at least four sign changes. This is, however, impossible by the Cauchy Lemma. Hence, it is impossible for the polyhedra P1

2.2 The Mapping Lemma

93

and P2 to have nonequal corresponding dihedral angles, which completes the proof of the Cauchy Theorem.2 Certainly, the application of the Cauchy Lemma is not as straightforward as this in all cases. Nevertheless, simple additional arguments, if needed, always lead to our goal, provided that lemmas ensuring the necessary properties of the labeling of the net in question have been proved. That is why we begin Chapters 3 and 4, which are devoted to theorems concerning the congruence of polyhedra, precisely by proving such lemmas.

2.2 The Mapping Lemma3 2.2.1 All (except the two or three simplest) existence theorems for convex polyhedra to be formulated in Sections 2.3–2.5 will be proved by a unified method based on a certain topological lemma first established by Brouwer in 1912 and known as the Domain Invariance Theorem: If a domain, i.e., an open set G in n-dimensional Euclidean space is mapped homeomorphically onto some subset G of the same space, then G is also a domain.4 From the theorem we infer a lemma that is referred to as the Mapping Lemma. Careful application of this lemma constitutes the method for proving our existence theorems. By an n-dimensional manifold we mean a topological space in which each point has a neighborhood homeomorphic to the interior of an n-dimensional cube and in which any two points possess disjoint neighborhoods.5 Recall that a set M (in a topological space) is said to be connected if it cannot be split into two disjoint nonempty subsets closed in M (i.e., such subsets M i that every point adhering to M i and lying in M belongs to M i as well). 2

3

4 5

Historical information about the Cauchy Theorem can be found in [Fen2], [G], [Co6]. In the last article there is also a discussion of a “smooth analog” of the Cauchy Lemma. Versions of the proof of the lemma on the deformation of convex polyhedral angles can be found in [St1], [Eg], [SZ], [Mi2], [Mi7]. An assertion replacing the Cauchy Lemma on sign disposition is presented in [Tr]. On these grounds, various proofs of the Cauchy Theorem are given; see [P7], [Se1], [Tr]. In a radically different fashion, it is proved in [AS, Appendix]. – V. Zalgaller In this section we use some results from topology. The reader unfamiliar with this field should first look through Section 2.8 of the current chapter, which contains most of the necessary information in simplest form. Incidentally, the Mapping Lemma is used only in Chapters 4, 7, and 9. This theorem is proved in Section 2.9 of the current chapter. However, its proof, as such, is of no importance for us. Connectedness and the possibility of partitioning into simplices is usually included in the concept of manifold. We omit these requirements.

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A connected component of a set M is a connected subset M of M that is not contained in any connected subset other than M itself. Each set M is the union of its disjoint connected components. 2.2.2 The Mapping Lemma. Let A and B be two manifolds of dimension n. Let ϕ be a mapping from A into B satisfying the following conditions: (1) (2) (3) (4)

each connected component of B contains images of points in A; ϕ is one-to-one; ϕ is continuous; if points Bm (m = 1, 2, . . . ) of the manifold B are images of points Am of the manifold A and the sequence Bm converges to a point B, then there is a point A in A whose image is B and for which there exists a subsequence Am i of Am converging to A.

Under these conditions, ϕ(A) = B, i.e., all points of the manifold B are images of some points of the manifold A. By condition (4), the mapping ϕ is continuous in both directions, i.e., ϕ and its inverse are continuous. Indeed, let Bm = ϕ(Am ), B = ϕ(A), and assume that the sequence Bm converges to B (m = 1, 2, . . .). Suppose that the sequence Am does not converge to A. Then there is a neighborhood of A whose complement contains infinitely many points Am , say Am 1 , Am 2 , . . .. We observe that6 Bm j = ϕ(Am j );

Bm j → B;

B = ϕ(A),

and no subsequence consisting of the points Am j converges to A. However, this contradicts condition (4), since the injectivity of ϕ means that the only point mapped to B is the initial point A. Hence, the points Am must converge to A, i.e., ϕ is continuous in both directions. Since the mapping ϕ is one-to-one and continuous in both directions, it is a homeomorphism. The Domain Invariance Theorem now implies that the image ϕ(A) of the manifold A is an open set in B. Indeed, assume B ∈ ϕ(A), i.e., suppose B is the image of some point A in A. Let UB denote a neighborhood of B ∈ B that is homeomorphic to the interior of a cube, i.e., homeomorphic to n-dimensional Euclidean space. By the continuity of ϕ, the point A has a neighborhood VA whose image is contained in UB . The point A also has a neighborhood homeomorphic to an n-dimensional cube; the intersection of this neighborhood with VA is also a neighborhood of A, being mapped into UB . This neighborhood WA is obviously homeomorphic to an open set in n-dimensional Euclidean space. Therefore, the Domain Invariance Theorem implies that the homeomorphic image ϕ(WA ) is an open 6

The arrow indicates convergence: the points Bmj converge to B, i.e., for every, “arbitrarily small,” neighborhood of B one can find an index such that all points Bmj with larger indices j belong to the neighborhood.


95

set in UB , i.e., a neighborhood of B. Hence, every point B ∈ ϕ(A) has a neighborhood contained in ϕ(A), which means that ϕ(A) is an open set. Further, condition (4) implies that ϕ(A) is also closed in B. Indeed, if some points Bn = ϕ(An ) converge to B, then by condition (4) there is a point A in A mapped to B, i.e., B belongs to ϕ(A) and consequently ϕ(A) is closed. However, since ϕ(A) is open and closed in B and has a nonempty intersection with each connected component of B, it must contain all of B. Indeed, since ϕ(A) is open, its complement B \ ϕ(A) is closed. Therefore, the intersections of the sets ϕ(A) and B \ ϕ(A) with any set M in B are closed relative to M.7 Take as M a connected component B of the manifold B. Then the formula B = (ϕ(A) ∩ B ) ∪ ((B \ ϕ(A)) ∩ B ) gives a decomposition of the component B into two sets closed in B . This is impossible because B is connected; therefore, one of the sets must be empty. By condition (1), B contains points in ϕ(A), i.e., B ∩ ϕ(A) is nonempty; thus it is (B \ ϕ(A)) ∩ B that is empty. Hence, B = ϕ(A) ∩ B . Taking the union over all connected components, we obtain B = ϕ(A), which completes the proof. 2.2.3 Now, we demonstrate how the Mapping Lemma is used to prove existence theorems. For example, we sketch the proof of Minkowski’s Theorem on the existence of a closed convex polyhedron with given directions8 and areas of faces. (For a detailed proof see Section 7.1 of Chapter 7.) First we specify the necessary conditions that the unit vectors of outward face normals n1 , n2 , . . . , nn and face areas F1 , F2 , . . . , Fn must satisfy. These conditions are as follows: (1) The vectors ni are not coplanar. (2) All Fi are positive. n (3) i=1 ni Fi = 0. The necessity of the first two conditions is obvious. The third condition means that the vector area of a closed polyhedron equals zero. This is easy to prove. Let T be any plane and let n be a unit normal to it. Then the inner product nni is the cosine of the angle between n and the normal ni to the ith face, the ith face normal. Therefore, (nni )Fi is nothing but the area of the projection of the face to the plane T taken with the appropriate sign depending on the angle between n and ni . 7

8

By definition, each closed set contains all its points of adherence. Therefore, the common part of, say, the sets ϕ(A) and M contains all its adherence points that lie in M; consequently, it is closed relative to M. The direction of a face is determined by the outward unit normal to the face.

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If we consider the projection of a closed convex polyhedron to the plane T , then we see that it covers some polygon twice: “positively” and “negatively.” As a result, the sum of projections of the faces counted with sign equals zero, i.e., n n nni Fi = n ni Fi = 0. i=1

i=1

Since the choice of the plane T and hence of the vector n is arbitrary, we necessarily have n ni Fi = 0. i=1

Minkowski’s Theorem asserts that the three conditions are not only necessary but also sufficient for the existence of a convex polyhedron with prescribed face normals ni and face areas Fi , i.e., given unit vectors ni and numbers Fi satisfying these conditions, there exists a closed convex polyhedron with outward face normals ni and face areas Fi . Denote by B the set of all collections of numbers F1 , F2 , . . . , Fn (for some fixed vectors n1 , n2 , . . . , nn ) that satisfy the stated conditions. The set B can be represented as a subset of the n-dimensional space Rn with coordinates n F1 , F2 , . . . , Fn . The vector relation i=1 ni Fi = 0 is equivalent to three scalar equations, thus determining an (n − 3)-dimensional hyperplane in Rn , the vector space Rn−3 . In this vector space, the inequalities Fi > 0 determine some open set. As the intersection of Rn−3 and the open half-spaces Fi > 0, it is convex and consequently connected. It is this set that we take for the manifold B. (It is assumed that B is nonempty: the existence theorem asserts that if there are Fi ’s satisfying the hypotheses, i.e., if B is nonempty, then a polyhedron exists as well.) Now, consider all closed convex polyhedra P with given face normals ni . Such polyhedra exist. Indeed, by assumption the vectors ni are not coplanar n and there are numbers Fi > 0 satisfying i=1 ni Fi = 0. Therefore, the vectors ni do not point to a single half-space. In this case the planes that have normals ni and are tangent to some ball form a bounded convex (solid) polyhedron. Its boundary is the required closed convex polyhedron with face normals ni . Each polyhedron P is determined by fixing its support numbers h1 , h2 , . . ., hn (see Section 1.2 of Chapter 1). Hence we can represent the set of all polyhedra P as a subset of the n-dimensional space Rn1 with coordinates h1 , h2 , . . . , hn . It is easy to verify that this subset is open, because faces do not disappear under small shifts of the planes of the faces. Now we combine all congruent and parallel polyhedra P into a single class. Since any parallel translation is determined by three parameters, each class is determined by n − 3 variables. (For instance, we can take as the representative of each class a polyhedron with barycenter at the origin.) The set of all these classes A constitutes an (n − 3)-dimensional manifold A.


97

Thus, the manifolds A and B have the same dimension n − 3. Since the face areas of a polyhedron always meet the conditions of the theorem, we have a natural mapping from A into B. If we show that it satisfies the assumptions of the Mapping Lemma, then we can infer that each point of B is the image of some point of A, i.e., for every collection of numbers F1 , F2 , . . . , Fn that meets the conditions of the theorem, there exists a polyhedron with these face areas. Since B is connected, condition (1) of the Mapping Lemma holds automatically. The injectivity of ϕ is a consequence of the following uniqueness theorem, which was also discovered by Minkowski (see a proof in Section 6.3 of Chapter 6): a closed convex polyhedron with face normals ni and face areas Fi is unique up to translation. In the language of the manifolds A and B, this means that to each given point B ∈ B there may correspond only one point A ∈ A. The continuity of ϕ is evident from the continuous dependence of the face areas on the disposition of faces, i.e., on the support numbers. This leaves the fourth condition, which can be proved without much effort (see Section 7.1 of Chapter 7). Thus, we have checked all the conditions of the Mapping Lemma, thus completing the proof of the theorem. 2.2.4 The example considered above reveals the main conditions for the applicability of our method. It applies to the cases in which the existence theorem has the following form: “given a, there is a b; and, conversely, given b, there is an a.”9 In our example, a denotes a closed convex polyhedron and b denotes the collection of numbers Fi and the corresponding unit vectors ni ; moreover, only those Fi and ni are admissible that satisfy the conditions of the theorem. Minkowski’s Theorem asserts that for every polyhedron a there exists a collection b of face areas and face normals and simultaneously for every collection b of numbers Fi and vectors ni meeting the relevant conditions there exists a corresponding polyhedron a. Such assertions are theorems on necessary and sufficient conditions. In our example we talk about necessary and sufficient conditions for given numbers Fi and unit vectors ni to be face areas and face normals of a closed convex polyhedron. The necessity of the conditions is often obvious and the crux of the matter is in proving their sufficiency. This feature is characteristic of all existence theorems or theorems on necessary and sufficient conditions proved in this book. In the proposed method, the proof of the uniqueness theorem always precedes the existence. The actual mapping ϕ between the objects under consideration is usually defined in a natural way. To ensure its injectivity, we must construct the manifold of objects in a special way, relying upon the “precision” of the corresponding uniqueness theorem. To this end, when we 9

This fact, of course, does not provide a one-to-one correspondence between the objects in question.

98


define the elements constituting the manifold, we must unite the appropriate objects into classes or endow them with additional features. For instance, in the proof of Minkowski’s Theorem, the manifold A consists of the classes of polyhedra that differ by translations. Or another example: in the proof of the existence of a polyhedron with a prescribed development, the correspondence between the development and the polyhedron becomes one-to-one in a natural way when we consider polyhedra endowed with the net of edges of some development. We then restrict ourselves to dealing with developments and nets of fixed structure. Finally, in order to make the correspondence one-to-one, we unite into a single class polyhedra superposable with their prescribed nets by a motion or a motion and a reflection. The manifold A under consideration consists of these classes. Here the manifold A may be empty. However, the proof of the first requirement of the Mapping Lemma obviously contains the proof of the nonemptiness of A. At the same time, formally speaking, the empty set is also an n-dimensional manifold, since the definition of manifold does not require nonemptiness. 2.2.5 The general method above can certainly be used not only to prove existence theorems for polyhedra. Using it, we can prove, for example, the fundamental theorem of algebra, which claims the existence of n roots for any polynomial of nth degree.10 For the manifold B we take the manifold of polynomials of nth degree with complex coefficients and leading coefficient equal to 1, excluding polynomials with multiple roots. For A we take the manifold of unordered n-tuples A of complex numbers. The two manifolds are 2n-dimensional. There are other examples of the applicability of this method to algebraic problems. An application to an existence theorem in the theory of functions of a complex variable was given by G. M. Goluzin [Go]. Some other, possibly earlier, applications of this method are probably known. (For example, in the trivial one-dimensional case, the method corresponds to the familiar method of continuation in a parameter already used by S. N. Bernshte˘ın [Be, p. 79].) It is worth noting that so far there are no other methods, except a similar one described in Subsections 2.6.7 and 2.6.8, which makes it possible to prove all the existence theorems of the present book. Incidentally, these are all the general existence theorems of this type, known so far, for convex polyhedra (assuming that we talk about data determining a polyhedron). However, for Minkowski’s theorem and similar ones (Theorems 9 and 11 of Section 2.4) other methods of proof are known.11

10 11

Such a proof is given in my article [A6]. Alternative proofs of some theorems are pointed out in further footnotes. – V. Zalgaller

2.3 Determining a Polyhedron from a Development

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2.3 Determining a Polyhedron from a Development (Survey of Chapters 3, 4, and 5) 2.3.1 We are concerned with gluing a polyhedron from some development as described in Section 1.6 of Chapter 1. Conditions that must be satisfied by the development of a closed convex polyhedron were found in Sections 1.6 and 1.7: (1) “The positive curvature condition”: for each vertex of a development, the sum of the angles glued together at this vertex must be at most 2π. The sum of the angles contiguous to a single vertex of a convex polyhedron is always less than 2π. The possibility of the sum of angles to equal 2π means only that, after gluing, the vertex of the development may become an interior point of a face or an edge rather than a true vertex of the polyhedron. Therefore, the condition is necessary for any development of every convex polyhedron. (2) “The Euler condition”: if f , e, and v denote the number of faces, edges, and vertices in a development, then the equality f − e + v = 2 must hold. The necessity of this condition for each development of a closed convex polyhedron was proved in Theorem 6, Section 1.7 in Chapter 1. (We recall that glued edges or vertices are regarded respectively as the same edge or vertex of a development and that all polygons of a development are simple by assumption.) One might think that we should add the condition that a development include no unbounded polygons and have empty boundary. However, it was established in Theorems 7 and 8 of Section 1.7 in Chapter 1 that for such developments we always have f − e + v < 2, and hence the Euler condition alone guarantees boundedness of the polygons and emptiness of the boundary. 2.3.2 If we regard as a closed convex polyhedron any doubly-covered convex polygon obtained by gluing together two superposed equal polygons, then we can assert the following: Theorem 1. The stated conditions are not only necessary but also sufficient for a given development to define a closed convex polyhedron by gluing. Moreover, there may be only two such polyhedra: one is the mirror image of the other or, which is the same, one of them is the other “inside-out.” (If a polyhedron has a plane of symmetry, then this yields the same polyhedron. Polyhedra differing only in their position in space are not regarded as distinct.) We have to take into consideration the case of a doubly-covered polygon, since a development may consist, for example, of two squares that must be

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glued together along their sides. It is obvious that the conditions stated above are satisfied in that case. Despite the degenerate character of this case, we do not exclude it, since this would complicate the formulation of our result. Further in this section as well as in Chapters 3 and 4 devoted to the proof of this and related results, we regard any doubly-covered convex polygon as a convex polyhedron, unless the contrary is stated explicitly. Actually, we will prove Theorem 1 in a somewhat different formulation. As was proved in Section 1.7 of Chapter 1 (Theorem 6), a development satisfying the Euler condition f − e + v = 2 is homeomorphic to the sphere. Therefore, Theorem 1 can be rephrased as Theorem 1∗ . Each development homeomorphic to the sphere and having the sum of angles at most 2π at each vertex defines a closed convex polyhedron by gluing. Moreover, the polyhedron is unique up to a motion or up to a motion and a reflection. 2.3.3 The advantage of the Euler condition is that, like other conditions imposed on developments by definition, the condition can be readily verified for any given development; as a result, it is always possible to find out whether or not the given development defines a closed convex polyhedron by gluing. Moreover, it is not assumed in advance that some polyhedron is actually produced from the given development by gluing; this can be checked under the indicated conditions. Since there may be only one such polyhedron (up to a motion or a motion and a reflection “turning it inside out”), in the course of gluing together a development the polyhedron appears without extra effort.12 However, we cannot predict how the polyhedron will look. The thing is that the polygons and edges of a development might in no way correspond to the faces and edges of the polyhedron; a priori it is only known that the vertices of the polyhedron result from vertices of the development where the sum of angles is less than 2π. Some examples of developments whose edges do not correspond to edges of the polyhedron are shown in Section 1.6 of Chapter 1 (Figs. 37 and 38). To determine the structure of a polyhedron from a development, i.e., to indicate its genuine edges in the development, is a problem whose general solution seems hopeless. It admits a simple solution only for the tetrahedron.13 The point is that, with the growth of the number of vertices, the number of 12

Given a development of a convex polyhedron, it is of course possible to glue together as many nonconvex polyhedra from it as we please. It suffices, for instance, to bend the polyhedron inward in a neighborhood of one of its vertices. For that reason, the words “without extra effort” should be understood in the sense that, while gluing, one must only control convexity. 13 The edges of a convex polyhedron must be shortest arcs joining the vertices of the development at which the sum of angles is less than 2π. Moreover, the planar angles at one vertex must satisfy the inequalities asserting that each of the angles is less than or equal to the sum of the others. If one knows how to draw shortest arcs in a development (and an explicit method for this can be indicated), then


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possible different polyhedron structures increases extremely rapidly. If n(v) denotes the number of structures of polyhedra with v vertices, then n(4) = 1, n(5) = 2, n(6) = 7, n(7) = 34, n(8) = 257.14 Moreover, when the development varies continuously, the structure of the corresponding polyhedron can change. A simple example is given in Figs. 64(a)–(c), where the polyhedra (b) and (c) can be obtained from (a) by continuously lowering the vertices A and B. The dashed line AB in the polyhedra (b) and (c) corresponds to an edge in their developments which have the same structure as the natural development of the polyhedron (a). A

E

B

E B

A D

(a)

D

C

E

(b)

A

B

D

C

(c)

C

Figs. 64(a)–(c)

2.3.4 Since the development of a polyhedron is not uniquely determined by the polyhedron, it is natural to ask what intrinsic characteristics of a polyhedron are encoded in its development. As already pointed out in Subsection 1.6.2 of Chapter 1, the development determines the “intrinsic metric” of a polyhedron. Given two points X and Y in a polyhedron P , the distance between them in P is defined to be the greatest lower bound of the lengths of polygonal lines in P joining the points. The distance function ρp (XY ) is called the intrinsic metric of the polyhedron P . Let R be a development of a polyhedron P . If X and Y are two points of R, then we can connect them in R by polygonal lines composed of line segments lying in the faces of the development and jointed at the identified points of the boundaries of the faces. It is natural to regard the greatest lower bound for the lengths of such polygonal lines as the distance ρR (X Y ) between the points X and Y in the development R. If, in the course of constructing the polyhedron P from the development R by gluing, the points finitely many cases occur, from which we must choose the one that corresponds to the structure of the polyhedron. Consequently, it is possible in principle to effectively solve the problem of constructing a polyhedron by gluing. However, in my opinion, effectiveness is of no interest; more important is the fact that because of uniqueness, the polyhedron is obtained naturally in the required form by gluing. 14 n(9) = 2606; see [F] for how the subsequent numbers grow. – V. Zalgaller

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X and Y become the points X and Y in P , then obviously ρR (X , Y ) = ρP (X, Y ). In other words, making a polyhedron P from a development R by gluing produces an isometric mapping of R onto P , i.e., a mapping that preserves distances. All developments of the same polyhedron are obviously isometric (i.e., they can be mapped onto each other by distance-preserving maps). A development of a polyhedron can be treated as a concrete way of specifying the intrinsic metric of the polyhedron. As was already mentioned, by virtue of the Euler condition the development of a closed convex polyhedron is homeomorphic to the sphere. By mapping a development R onto a sphere S, we carry over the metric ρR of the development R to S, i.e., to each pair of points X and Y of S, we assign the distance ρR (XY ) equal to the distance between the corresponding points in R. (This distance function on the sphere obviously has nothing in common with the intrinsic metric of the sphere; here the sphere is regarded as an abstract manifold.) In this metric ρR , every point of S possesses a neighborhood which is similar to one about a point of the development, i.e., a neighborhood isometric to a polyhedral angle, in particular to a part of the plane. The latter case holds certainly for points in the interiors of the faces of the development or of its edges: the neighborhood of a point of the second kind is the union of two half-neighborhoods that lie on faces glued together along the edge in question. If the sum of angles at a vertex A of the development differs from 2π, then the neighborhood of A is isometric to a polyhedral angle. From the standpoint of the intrinsic metric, it is fully characterized by the sum of angles at A or, in other words, by the complete angle θ at A. The difference 2π − θ is referred to as the curvature at A. The condition imposed on the angles of a development of a convex polyhedron asserts that the curvature is nonnegative at all vertices.15 A metric definable by means of a development will be called polyhedral , and if the curvature is greater than or equal to 0 at all vertices, then we call the metric in question a metric of positive curvature (excluding the case in which there are no vertices at all with nonzero curvature; certainly, this case is impossible for the developments homeomorphic to the sphere.) The existence theorem for a polyhedron with a given development stated above can now be rephrased as follows: 15

Conversely, if a sphere S is endowed with a metric ρ such that every point of S has a neighborhood isometric to a polyhedral angle and the distance between two arbitrary points X and Y equals the greatest lower bound the lengths of curves joining X and Y (the length being measured in the metric ρ), then such a metric can be defined by means of a development. It suffices to surround each point O by a neighborhood composed of triangles with vertices at O and choose finitely many such neighborhoods covering the entire sphere S. The resulting triangles give a finite number of polygons constituting the faces of the required development.


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Theorem 1∗∗ . For every polyhedral metric of positive curvature given on a sphere, there exists a closed convex polyhedron realizing this metric; moreover, the polyhedron is unique up to a motion or a motion and a reflection. (The assertion that a polyhedron P realizes a metric ρR given on S means that P admits a mapping onto S which is isometric with respect to the metric ρR .) There is no freedom of choice in this formulation: a polyhedron and its metric are in one-to-one correspondence provided that we do not distinguish congruent polyhedra. Thus, Theorem 1∗∗ answers the question that was raised at the beginning of the subsection. What is encoded in each development of a polyhedron and corresponds to the polyhedron in a one-to-one fashion is the intrinsic metric of the polyhedron. 2.3.5 The uniqueness claim of Theorem 1∗ is proved in Section 3.3 of Chapter 3 in the following somewhat stronger form: Theorem 2. Each isometric mapping of one closed convex polyhedron onto another can be realized as a motion or a motion and a reflection. In particular, the claim applies to mappings of a polyhedron onto itself.16 This yields a strengthening of Cauchy’s Theorem mentioned already in Section 2.1: two closed convex polyhedra composed of the same number of equal similarly-situated faces are congruent. The proof of our theorem is reduced to Cauchy’s Theorem with the only generalization that one deals with arbitrary polygons composing the polyhedron rather than with the genuine faces, notwithstanding the fact that some of the polygons might represent parts of faces rather than entire faces. We obtain this generalization of the Cauchy Theorem as a direct corollary of the following theorem: Theorem 3. If two closed convex polyhedra are composed of the same number of similarly-situated planar polygons with corresponding angles of equal measure, then all corresponding dihedral angles of these polyhedra are equal. The reader familiar with Cauchy’s proof can observe that it is the very theorem that Cauchy himself had actually proved, but not formulated, with the only difference that he only considered the faces of the polyhedron. Theorem 3 is of interest in its own right, since polyhedra with equal planar angles 16

We will prove in Section 3.5 of Chapter 3 that the same is valid not only for closed but also for arbitrary convex polyhedra of total curvature 4π (the same curvature as that of a closed polyhedron). For example, an arbitrarily thin slice cut from the surface of a cube, containing all the vertices of the cube, admits no flexes preserving convexity!

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can be far from isometric; a simple example is provided by all rectangular parallelepipeds.17 We shall prove an existence theorem for a closed convex polyhedron with given development in Sections 4.1–4.3 of Chapter 4, relying on Theorem 2 in particular. 2.3.6 Besides the conditions indicated in Subsection 2.3.1 and common to all developments, the development of an unbounded convex polyhedron must satisfy two additional conditions: (1) the development is homeomorphic to the plane; (2) the sum of the angles at any vertex is at most 2π. The necessity of the first condition is obvious from the fact that every unbounded polyhedron is homeomorphic to the plane. The second condition is necessary for the same reason as it is for the development of a closed convex polyhedron. If we adjoin doubly-covered unbounded convex polygons to unbounded convex polyhedra, then we can assert the following: Theorem 4. Each development homeomorphic to the plane and having the sum of angles at most 2π at each vertex defines an unbounded convex polyhedron by gluing. In the language of metrics, the theorem can be reformulated as follows: every complete polyhedral metric of positive curvature given on the plane is realizable as an unbounded convex polyhedron.18 This theorem will be proved in Section 4.4 of Chapter 4. The condition of being homeomorphic to the plane is equivalent to the following two conditions, easily verified for any given development: (1a) the development contains at least one unbounded polygon; (1b) if f , e, and v denote the number of polygons, edges, and vertices in the development, then f − e + v = 1. The equivalence of these two conditions to the requirement that a development be homeomorphic to the plane was established in Section 1.7 of Chapter 1 (Theorem 9). Generally speaking, an unbounded polyhedron is not uniquely determined by each of its developments. This can be seen just from the example of a polyhedral angle: it can always be deformed without changing its faces if there are 17

Stoker [Sto] conjectures the validity of the converse assertion: in the class of convex polyhedra with the same combinatorial structure, if the corresponding dihedral angles are equal, then so are the corresponding planar angles. This conjecture has been established in several particular cases [Kar], [Mi3], [And]. – V. Zalgaller 18 Which means that it admits an isometric embedding into three-dimensional Euclidean space as a convex polyhedron. – V. Zalgaller


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at least three of them. The only case in which uniqueness holds is described by the following theorem: Theorem 5. If the total curvature of a development, i.e., the sum of curvatures at all vertices, equals 2π, then exactly one (up to a motion or a motion and a reflection) unbounded convex polyhedron can be produced from it by gluing. Or: if the curvature of an unbounded convex polyhedron equals 2π, then every isometric mapping of this polyhedron onto another convex polyhedron is a motion or a motion and a reflection. The total curvature of an unbounded convex polyhedron never exceeds 2π. For polyhedra with total curvature less than 2π, we have the following theorem (see the proof in Section 4.5 of Chapter 4): Theorem 6. Let R be a development satisfying the necessary conditions indicated above and having total curvature ω < 2π. Let R be supplied with an orientation. Let L be a ray in one of its unbounded polygons. Let V be a convex polyhedral angle of curvature ω with a given orientation, i.e., an oriented circuit around its vertex, and let L1 be a generatrix of V . Then R defines by gluing a convex polyhedron P with limit angle V such that under the infinite similarity contraction of P to V the line on P corresponding to L transforms into L1 and the orientation on P induced by the orientation of R transforms into the given orientation of V . Such a polyhedron P is unique up to translation if the angle V is fixed in space. Consequently, for the polyhedron to be unique, it is necessary to specify, together with the development R, the limit angle V , the corresponding rays L and L1 , and the corresponding orientations on R and V . The change of orientation on V cannot be achieved simply by reflection, since under reflection of the polyhedron P its limit angle is reflected too, and consequently becomes different from the given angle V . If we fix R, V , and L and rotate L1 around the angle V , then the polyhedron P will, in a certain sense, turn around the angle V ; moreover, its structure will in general change. In exactly the same manner, under a continuous variation of the angle V , the polyhedron P will also deform. The equality between the curvatures of the development R and the angle V is required because of Theorem 3 of Section 1.5 in Chapter 1, according to which the curvature of an unbounded convex polyhedron equals that of its limit angle. Theorems 5 and 6 are due to S. P. Olovyanishnikov [Ol1]. We prove Theorem 5 and the uniqueness part of Theorem 6 in Chapter 3 (Sections 3.3–3.4). In the same chapter (Section 3.2) we prove the following theorem on unbounded polyhedra, which is analogous to Theorem 3: Theorem 7. If two unbounded convex polyhedra are composed of the same number of similarly-situated polygons with corresponding angles of equal

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measure and the limit angles of the polyhedra are equal, then the dihedral angles of the polyhedra are equal as well. (If the limit angle of one of the polyhedra degenerates into a ray, then the second condition is redundant, for in this case the equality of the corresponding angles of the polygons implies the equality of the limit angles.) Finally, we establish similar results for polyhedra bounded by a single polygonal line. In this case, in order to obtain the assertions of Theorems 2 and 3, it suffices to impose the following additional condition on the boundaries of the polyhedra: the corresponding angles between adjacent boundary edges of the two polyhedra must be equal (see Section 3.5 of Chapter 3). However, no effective necessary and sufficient conditions are known under which a development defines by gluing a convex polyhedron with boundary. The trivial condition that the development possess an extension to a development of a closed (or unbounded) polyhedron is ineffective. Nevertheless, in some cases it allows us to formulate easily verifiable sufficient conditions. The corresponding results are presented in Section 5.1 of Chapter 5. The question when is a polyhedron with boundary determined by its development (metric) and when does it admit flexes (i.e., continuous deformations preserving the metric) which preserve convexity is discussed in Section 3.5 of Chapter 3 and Section 5.2 of Chapter 5. A number of interesting results have been obtained in this direction, but an exhaustive solution to the problem is still unavailable.19

2.4 Polyhedra with Prescribed Face Directions (Survey of Chapters 6, 7, and 8) 2.4.1 The direction of a face of a convex polyhedron is determined by the outward normal to the face. We consider faces and support planes of polyhedra as parallel only if so are their outward normals.20 Actually, problems concerning polyhedra with prescribed face directions are nontrivial because the face directions fail to determine the structure of a polyhedron to any extent, as can be seen from the simple example of the polyhedra in Fig. 65. 19

At present, effectively verifiable necessary and sufficient conditions are known for the rigidity of a convex polyhedron with one or several components of the boundary. They were established by L. A. Shor in the article [Sho1] (see the Supplement to Chapter 5 in the present book) and the note [Sho4]. – V. Zalgaller 20 Here and in the sequel, the phrase “vectors are parallel to one another” means that they have the same direction. In this sense, if a polyhedron degenerates into a doubly-covered polygon, then its two faces are not considered parallel to one another, since they have opposite outward normals.

2.4 Polyhedra with Prescribed Face Directions

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Fig. 65

As applied to convex polygons, all theorems stated below are either simply trivial, as our main Theorem 8, or have very simple proofs; with the exception of Theorems 10 and 11, whose proofs are not simple even in the case of polygons. In Sections 6.1–6.3 of Chapter 6 we prove the following: Theorem 8. If we are given two closed convex polyhedra such that to each face of one of them there corresponds a parallel face of the other and vice versa, and moreover the corresponding faces cannot be placed inside one another by translation (i.e., they either protrude from one another or coincide), then the polyhedra are translates of one another (i.e., can be obtained from one another by translation). We can even assume that if one of the polyhedra lacks a face with the same outward normal as some face of the other, then we treat such a face as if it exists but degenerates into an edge lying in the corresponding support plane. In this case the edge in question cannot be placed in the corresponding face of the other polyhedron. This theorem can also be rephrased as follows: Two closed convex polyhedra are either translates of one another or one of them has a face Q for which the other polyhedron possesses a boundary element (face, edge, or vertex) lying in a support plane parallel to Q, and this boundary element can be placed into Q by translation, but will not coincide with Q. The already-mentioned Minkowski Uniqueness Theorem is a partial corollary to Theorem 8: Theorem 9. If the faces two closed convex polyhedra are pairwise parallel and of equal area, then the polyhedra are translates of one another. In other words, a closed convex polyhedron is uniquely determined by its face directions and face areas. Indeed, polygons of equal area cannot be placed in one another (if they are not congruent), and this theorem is therefore contained in Theorem 8.

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One can formulate many such corollaries to Theorem 8. For instance, call a function f (Q) defined on the set of convex polyhedra Q monotone if f (Q1 ) < f (Q2 ) whenever Q1 is contained in but does not coincide with Q2 . If two closed convex polyhedra have corresponding parallel faces and, for each (1) (2) pair of parallel faces Qi and Qi , there is a monotone function fi such that (1) (2) fi (Qi ) = fi (Qi ), then the polyhedra are translates of one another. Indeed, (1) (2) since the functions fi are monotone, the equality fi (Qi ) = fi (Qi ) implies (1) (2) that the parallel faces Qi and Qi cannot be placed in one another, so that the congruence of the polyhedra follows from Theorem 8. Taking area as the value of all the fi ’s, we obtain Theorem 9; taking the perimeter as the value of all the fi ’s, we obtain the theorem asserting that a polyhedron is uniquely determined by the directions and perimeters of its faces, etc. Theorem 8 is applicable to some less general but important problems that will be considered in Chapter 8. For example, from Theorem 8 we will infer Lindel¨ of’s Theorem [Lin]:21 Theorem 10. Among all closed convex polyhedra with given face directions and total area, the greatest volume is bounded by a polyhedron circumscribing a ball. This theorem is a partial corollary to a certain theorem due to Minkowski, which is of prime interest in its own right, a theorem about the so-called mixing of convex bodies and their mixed volumes (see Section 6.2 of Chapter 6 and Sections 7.2 and 7.3 of Chapter 7). Another application of Theorem 8 consists in finding all possible parallelohedra, i.e., convex (solid) polyhedra whose translates tile the entire space if fitted together along their faces like holes in a honeycomb (see Section 8.1 of Chapter 8). 2.4.2 Minkowski proved not only the uniqueness, but also the existence of a closed convex polyhedron with prescribed face directions and face areas: Theorem 11. If the unit vectors n1 , . . . , nm and the numbers F1 , . . . , Fm satisfy the conditions: (1) the vectors n1 , . . . , nm are all distinct and not coplanar; (2) all the Fi are positive; m (3) i=1 ni Fi = 0, then there exists a closed convex polyhedron with outward face normals ni and face areas Fi . 21

Another proof was given by Minkowski in 1897 in the same article [Min1] where he proved Theorems 9 and 11.


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We prove this theorem in Section 7.1 of Chapter 7 by a method based on the Mapping Lemma; in Section 7.2 of Chapter 7 we also reproduce Minkowski’s original proof. Other existence theorems similar to Theorem 11 are not known. For example, we have a uniqueness theorem for polyhedra with given face perimeters; however, we have no corresponding existence theorem for the simple reason that the appropriate necessary conditions for the face perimeters of the closed convex polyhedron are not known. If one manages to find such conditions, our method will probably allow to find the corresponding existence theorem.22 This remark also relates to all other monotone functions on faces. Clearly, some nontrivial conditions must always hold here. A closed convex polyhedron having m faces with given directions is determined by the m distances from the planes of its faces to some point in its interior. However, the specification of, say, m face perimeters l1 , . . . , lm determines a polyhedron only up to translation, which reduces the degree of freedom from m to m − 3, since each translation is determined by the three components of the translation vector. Therefore, the perimeters must satisfy some conditions that, in the m-dimensional domain of positive values l1 . . . , lm , should distinguish an (m − 3)-dimensional subset of actually feasible values for the perimeters of the faces. The main difficulty consists in finding conditions distinguishing this set of feasible values l1 , . . . , lm for the face perimeters. 2.4.3 The congruence condition for closed polyhedra which was stipulated in Theorem 8 becomes meaningless for unbounded polyhedra when applied to their unbounded faces; it fails if the polyhedra are translates of one another, since one of two congruent parallel unbounded polygons can be placed inside the other, as can be seen from Fig. 66. To require the validity of this condition only for bounded faces is obviously insufficient; this is clear from the simplest example shown in Fig. 67.

Fig. 66

22

Fig. 67

An algorithm for finding conditions for given numbers to serve as the face perimeters of a closed convex polyhedron is proposed in [Z4] – V. Zalgaller

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Therefore, an additional condition on unbounded faces is in order: Unbounded polyhedra must have “unbounded parts as translates”, i.e., a (sufficiently large) bounded part can be cut out from each of them so that the remaining parts will coincide after translation. This condition is obviously equivalent to the requirement that the planes of all unbounded faces of polyhedra coincide after a common translation. Introducing this condition, we arrive at a theorem which will be proved in Section 6.4 of Chapter 6: Theorem 12. If two unbounded convex polyhedra have unbounded parts as translates and all pairs of bounded parallel faces are such that no face of the pair can be placed inside the other by translation, then the polyhedra are translates of one another. In the case of bounded nonclosed polyhedra, we obtain similar theorems in Section 6.5 of Chapter 6 by introducing certain conditions on the extreme faces (i.e., those adhering to the boundary) and on the boundary edges. Theorem 12 on unbounded polyhedra can easily be restated for bounded polyhedra if we recall the remark of Subsection 1.1.6 on the replacement of unbounded polyhedra by bounded ones whose extreme faces can be extended. In this connection, it should be noted that theorems of such a type can be regarded as concerning polyhedra whose boundary is fixed and which are subject to some additional conditions. This is nothing more than a “boundary value problem” now posed in the framework of elementary geometry. In boundary value problems, one generally deals with a function (or, perhaps, a surface) which is fixed on the boundary of some domain and must satisfy some condition inside the domain, usually the requirement of being a solution of a differential equation. (The connection between our theorems on polyhedra and such boundary value problems for differential equations will be shown in the sections “Generalizations” of Chapters 6 and 7.) 2.4.4 From Theorem 12 on the congruence of unbounded polyhedra we can infer the following corollary, which is similar to Minkowski’s Theorem 9: If two unbounded convex polyhedra have unbounded parts as translates and if their bounded faces are pairwise parallel to one another and have equal areas, then the polyhedra are translates of one another. We obtain a similar result if we take arbitrary monotone functions on faces instead of area; the argument will be the same as in Subsection 2.4.1 above. For unbounded polyhedra we can also formulate theorems that are analogous to Theorem 11 on the existence of a closed convex polyhedron with prescribed face directions and face areas. To this end, we first clarify the necessary conditions that must be satisfied by the face normals and face areas of an unbounded polyhedron with parallel unbounded edges.


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The unbounded part of such a polyhedron is a semi-infinite convex prism Π. If n is a vector along its edge, pointing to the bounded part of the polyhedron, then the outward normals to bounded faces of the polyhedron must form acute angles with n (Fig. 68). ni Fi n

F

Q

Π

Fig. 68

The projections of the bounded faces to the plane Q perpendicular to n cover the section of the prism Π by this plane. Therefore, if ni and Fi are the outward normals and the areas of the bounded faces, respectively, and F is the area of the section, then F = (nni )Fi , i

since (nni )Fi is exactly the area of the projection of the ith face to the plane Q. It turns out that these conditions are also sufficient for the existence of a convex polyhedron with given unbounded part and prescribed normals and areas of bounded faces. We now state the corresponding theorem, which will be proved in Section 7.3 of Chapter 7: Theorem 13. Given an unbounded convex prism Π, let n be the unit vector along an edge of Π. Let F denote the area of the section of the prism by a plane perpendicular to n. Let n1 , . . . , nm be unit vectors forming acute angles with n. Finally, let F1 , . . . , Fm be positive numbers such that m (nni )Fi = F. i=1

Then there exists a convex polyhedron with unbounded part Π whose bounded faces have outward normals ni and areas Fi .

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By virtue of the Corollary to Theorem 12 above, such a polyhedron is unique up to translation along the vector n, since such a translation maps the prism Π into itself. For unbounded polyhedra with nonparallel unbounded edges, the following much more general result is valid: Theorem 14. Let Q be an unbounded part of some unbounded convex polyhedron with nonparallel unbounded edges (Fig. 69). Let S denote the spherical polygon that is spanned by the spherical images of the faces of Q and let n1 , . . . , nm be vectors from the origin to interior points of S. Finally, let f1 , . . . , fm be arbitrary monotone continuous functions, defined on the set of polygons, which take the values zero and infinity when the area of a polygon becomes zero or infinity. Then, for arbitrary positive numbers a1 , . . ., am , there exists an unbounded convex polyhedron, with unbounded part Q, whose bounded faces have outward normals n1 , . . ., nm and, moreover, the functions f1 , . . . , fm take the values a1 , . . . , am at these faces, so that fi = ai for the face with normal ni . S n2 nG1 nG2 G1

G2

G3

n1 nG3

G4

Q

Fig. 69

From the corollary to Theorem 12 stated at the beginning of this subsection, it follows that such a polyhedron is unique, since no translation is possible when the unbounded part Q is fixed. The necessity of the conditions of Theorem 14 imposed on the normals has been already observed in Section 2.2. The spherical image of an unbounded convex polyhedron is a convex spherical polygon spanned by the spherical images of the unbounded faces of the polyhedron in the sense that its vertices are the ends of the outward normals from the center of the sphere to the faces with nonparallel unbounded edges, whereas the ends of the normals to the faces with parallel unbounded edges lie on its sides. The ends of the normals to the bounded faces lie in the interior of this spherical polygon. Specifying the unbounded part Q is equivalent to fixing the planes of its faces. However, not all planes can serve as face planes of some unbounded polyhedron. For them to do so, in accordance with what was said above, it is necessary that the normals to the faces drawn from the center of the sphere pass through all the vertices and, perhaps, through some points on the


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sides of a certain convex spherical polygon. Moreover, even this is insufficient and additional conditions on the distances from the planes to the origin are needed. They will be specified further in Subsection 2.4.7. At first glance, Theorem 14 is rather surprising owing to its generality. However, it is in fact a much more superficial observation than Theorems 11 and 13, which involve only area. This will become apparent from the proof of Theorem 14 proposed by A. V. Pogorelov and described in Section 7.4 of Chapter 7. Incidentally, we shall obtain a still more general result by waiving the condition that the functions fi vanish together with the area. As to polyhedra with boundary, no similar theorems are known, except for rather trivial consequences of Theorems 8, 13, and 14. The difficulty is in formulating conditions to be imposed both on the areas of the faces and on the boundary of the polyhedron, since there is obviously a dependence between them.23 2.4.5 Theorems 9–14 stated above are all generalized word for word to spaces of arbitrary dimension, although we have to change the method for proving Theorems 9, 10, and 12.24 Theorem 8 on the general congruence condition for closed polyhedra is an exception. The fact that it fails in four-dimensional space is apparent from the following simple example: A cube with edge 2 and a rectangular parallelepiped with edges 1, 1, 3, 3 (in four-dimensional space) have faces that do not fit into one another, but are not congruent. 2.4.6 A closed or unbounded convex polyhedron is completely determined by fixing the planes of its faces: it is the boundary of the intersection of the halfspaces bounded by these planes. Given the unit vectors ni of outward normals to the faces, the face areas will be determined if we specify the distances from the faces to the origin of coordinates; here the distance is considered positive when the direction from the origin to the plane in question coincides with the direction of the outward face normal; in the opposite case, the distance is negative. With this agreement on signs, we call these distances the support numbers of the polyhedron.25 The support number hi is the right-hand side of the normal equation of the face plane: ni x = hi , 23

This problem remains unsolved. – V. Zalgaller Such a general proof for Theorems 9 and 10 was given by Minkowski. It is presented in Section 8.3 of Chapter 8. The general proof for Theorem 12 is due to Pogorelov and will be described in Section 7.5 of Chapter 7. 25 The reader familiar with the theory of convex bodies can readily observe that the support numbers are nothing more than the values of the support function of the polyhedron at the unit vectors of the outward face normals. Their specification replaces that of the whole support function. This is the reason for their names. Minkowski, in his article [Min1], called them tangential parameters. The notion of support function was introduced by Minkowski. 24

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where ni is the unit outward normal and x is the vector from the origin to an arbitrary point of the plane. The half-space containing the polyhedron is determined by the inequality ni x ≤ hi (and not by the opposite one, since ni is the outward normal, i.e., the one pointing to the half-space not containing the polyhedron). The polyhedron itself is the boundary of the intersection of these half-spaces. Support numbers are not arbitrary. Let the vectors of outward normals be n1 , . . . , nm and let h1 , . . . , hm be the support numbers. Suppose that the vector nk is a linear combination of the other vectors ni with nonnegative coefficients: nk = νki ni (νki ≥ 0). (1) i=k

If x is the position vector of a point X in the interior of the kth face, then nk x = hk .

(2)

At the same time, since the point X belongs to the polyhedron and lies on none of the planes of the other faces, for i = k we have ni x < hi . Multiplying these inequalities by the numbers νki ≥ 0, we obtain νki hi , νki ni x < or by virtue of (1) and (2) hk

(5)

i=k

Consequently: in order that the given numbers h1 , . . . , hm be the support numbers of a convex polyhedron with outward normals n1 , . . . , nm , it is necessary that, for every expansion (1) or (4) of each of the vectors nk whose


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coefficients νki are all nonnegative or all nonpositive, inequalities (3) or (5) hold. In Section 7.5 and Section 7.6 of Chapter 7, we shall prove that this condition is also sufficient: Theorem 15. Let n1 , . . . , nm be m distinct noncoplanar unit vectors. Assume that m respective numbers h1 , . . . , hm are associated with the vectors nk so that if a vector nk is represented as a linear combination of the other vectors, nk = νki ni , i=k

with all coefficients nonnegative or nonpositive, then hk < νki hi if νki ≥ 0, i=k

hk >

νki hi

if

νki ≤ 0.

i=k

Then there exists a convex polyhedron with outward face normals n1 , . . . , nm and support numbers h1 , . . . , hm . If all the vectors ni point to a single halfspace, then the polyhedron is unbounded. In the opposite case it is bounded. (We assume that each half-space includes its boundary plane.)26 It suffices to require the validity for only those inequalities on hk that correspond to the expansions of each vector nk in terms of three other vectors ni . The last remark is essential for the following reason. If a vector nk admits an expansion in terms of four or more given vectors with positive (negative) coefficients, then it admits infinitely many such expansions. At the same time, there can be only one expansion in terms of three vectors. Hence, the last remark allows us to confine our consideration to finitely many inequalities on the numbers hk , thus making the conditions of the theorem effectively verifiable. The proof of Theorem 15 might seem rather simple at first sight: “Intersect the half-spaces ni x ≤ hi and the desired polyhedron results.” However, to 26

In my article [A6] the formulation of the theorem is erroneous: the inequalities for hk corresponding to expansions of the vectors nk with nonpositive νki ’s were forgotten. The inequalities with nonnegative νki ’s alone are insufficient as can be seen from the following simple example. Take the vectors n1 , n2 , n3 , and n4 of the outward face normals of a right tetrahedron. Each of them expands in terms of the others with only negative coefficients. Therefore, the conditions on hk with positive coefficients νki are left out and, dropping the conditions with negative νki , we impose no conditions whatever. At the same time, it is easy to translate the face planes of the right tetrahedron so that they bound no polyhedron with the corresponding outward normals.

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claim this would be rather hasty, since we do not know if this intersection is nonempty.27 This is the first difficulty in proving Theorem 15. 2.4.7 From Theorem 15 it is easy to deduce a necessary and sufficient condition for given planes to be the planes of the unbounded faces of some convex polyhedron. Theorem 16. In order that the planes P1 , . . . , Pm with normals n1 , . . . , nm be the planes of the unbounded faces of some convex polyhedron (with outward face normals n1 , . . ., nm ), it is necessary and sufficient, first, that the normals n1 , . . . , nm point to vertices and, perhaps, to points on the sides of a convex spherical polygon (namely the polygon which is the spherical image of the entire polyhedron with given unbounded faces) and, second, that the support numbers h1 , . . . , hm of the planes P1 , . . . , Pm satisfy the conditions of Theorem 15. We already know that these conditions are necessary. They are also sufficient. Indeed, under the assumptions of Theorem 15, there exists a polyhedron with normals ni and support numbers hi . It will be unbounded, since the normals ni point to a single half-space; and all its faces will be unbounded, since the normal to a bounded face goes to the interior of the spherical polygon spanned by the ends of the normals to unbounded faces, whereas all the normals ni lie on the boundary of the polygon by assumption. If the vectors ni point to the boundary of the convex polygon or, equivalently, lie on the surface of a convex polyhedral angle, then it is obvious that none of them has an expansion in terms of the others with negative coefficients. For this reason, the corresponding condition of Theorem 15 simply disappears. The expansion of a vector n in the three vectors n1 , n2 , and n3 with positive coefficients means that n lies inside the trihedral angle with edges n1 , n2 , and n3 . Therefore, for the case in which all the vectors ni lie on the surface of a convex polyhedral angle, none of them can be expanded with positive coefficients in terms of three (or more) other vectors. Only one possibility remains open: a vector nk lies in the same plane as two other vectors ni and nj and is located within the angle between them. Then it is a vector that expands in the others with positive coefficients. This means that the end of the vector nk lies in the interior of a side of the spherical polygon spanned by the ends of the three vectors under consideration. Such a vector, as we know, is a normal to an unbounded face with parallel unbounded edges. In consequence, the assumptions of Theorem 15 become much simpler and in this case need to be verified only for vectors pointing to the interiors 27

This aspect is essential not only for geometry. It constitutes an important step in solving many problems of linear programming. We shall return to this question in a more detailed setting in Section 7.5 of Chapter 7. – V. Zalgaller

2.5 Polyhedra with Vertices on Prescribed Rays

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of the sides of the spherical polygon. If there are no such vectors at all, then none of the expansions mentioned in Theorem 15 for the vectors ni is possible, and therefore all the restrictions are absent, i.e., any collection of planes with these normals ni will be the planes of the unbounded faces of some polyhedron. An exceptional case occurs when all the vectors ni are coplanar, so that the spherical polygon spanned by their ends turns out to be a hemisphere. This corresponds to the case in which the unbounded part of the polyhedron is a prism. Here all the conditions on the numbers hi must be taken into account. However, we can regard the hi ’s as the support numbers of a polygon which is a section of the prism. Then in Theorem 15 it suffices to take expansions in two vectors. Theorem 15 and 16 can be generalized to spaces of arbitrary dimension n with the only provision that in n-dimensional space one considers expansions in terms of n vectors.

2.5 Polyhedra with Vertices on Prescribed Rays (Survey of Chapter 9) 2.5.1 Take some point O in space and draw rays l1 , . . . , lm from it which do not all go inside a single half-space. We will consider closed convex polyhedra with vertices on these rays. The point O must lie inside such a polyhedron by assumption. Such polyhedra exist: for instance, any polyhedron inscribed in a ball centered at O. Let r1 , . . . , rm stand for the distances from the vertices of such a polyhedron to O. The numbers r1 , . . . , rm are not arbitrary, as established in the following theorem: Theorem 17. The following condition is necessary and sufficient for the positive numbers r1 , . . . , rm to serve as distances from the vertices to O for . . . , lm : a closed convex polyhedron with vertices on the rays l1 , If ei are the unit vectors along the rays li and ek = νki ei is an expansion of some ek in terms of the other vectors with nonnegative coefficients, then νki 1 < . rk ri It suffices that these inequalities be satisfied for similar expansions of each of the vectors ek in terms of three other vectors. The proof of the theorem is elementary. Taking points Ai on the rays, consider their convex hull. This is the required polyhedron P . A point Ai is a vertex of P if and only if it lies beyond the convex hull of the other points.

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As will be shown in Section 9.1, this yields the necessity and sufficiency of the conditions of the theorem. There is a formal analogy between Theorem 17 and Theorem 15 of the previous section. This analogy has a simple geometric foundation: a polar transformation takes the polyhedron with vertices on the rays li to a polyhedron with faces having outward normals ei . Here, if ri are the distances from the vertices of the given polyhedron to the point O, then the support numbers of the latter polyhedron are hi = 1/ri . (Polar transformations were considered in Subsection 1.5.4.) However, a complete analogy between Theorem 17 to Theorem 15 is impossible, since in Theorem 17 all the ri are possible, whereas in Theorem 15 negative values of hi are admissible. 2.5.2 We raise the following questions: If l1 , . . . , lm are rays using from O and not contained inside a single half-space, then what conditions are necessary and sufficient for given numbers ω1 , . . . , ωm to serve as the curvatures (the areas of spherical images) at the vertices of a closed convex polyhedron with vertices on the rays l1 , . . . , lm ? (The polyhedron is assumed to have no other vertices.) To what extent is a polyhedron determined by the conditions that all its vertices lie on given rays and have given curvatures? The answer to the first question is provided by the following theorem. Theorem 18. Suppose the rays l1 , . . . , lm issue from a point O and do not lie inside a single half-space. Let Ωi1 ,...,ik be the area of the spherical image of the solid angle presenting the convex hull of the rays li1 , . . . , llk . Then the following conditions are necessary and sufficient for a set of numbers ω1 , . . . , ωm to serve as the areas of the spherical images of the vertices of a convex polyhedron with vertices on the rays l1 , . . . , lm : (1) all the ωi are positive; m (2) i=1 ωi = 4π; (3) for every collection of rays li1 , . . . , lik we have ωjp > Ωi1 ,...,ik , where the sum is taken over all rays ljp lying beyond the convex hull of the rays li1 , . . . , lik . Of course, we only consider those collections of rays li1 , . . . , lik whose convex hull does not coincide with the whole space, since the last condition no longer makes sense otherwise. The necessity of conditions (1) and (2) is obvious: condition (2) means that the spherical image of the closed convex polyhedron covers all of the sphere. Let us prove the necessity of the third condition. Let P be a closed convex polyhedron with vertices on the rays l1 , . . . , lm . Take rays li1 , . . . , lik and consider their convex hull V , assuming that V does not cover the whole


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space. The convex hull of every finite collection of rays issuing from a single point O is a convex solid polyhedral angle with vertex O. Suppose that the rays lj1 , . . . , ljl and accordingly the vertices Aj1 , . . . , Ajl lie outside V . Each support plane of V intersects the polyhedron P , and an appropriate parallel translation of the plane from the point O takes it to a position where it is the support plane at some vertex of P that lies outside V . (Fig. 70 shows a similar situation in the case of a polygon.) Consequently, each support plane of V has a translate which is a support plane of P at one of the vertices Aj1 , . . . , Ajl . Actually, there are other support planes at these vertices, for instance the planes of faces disjoint from V . Therefore, the spherical image of V is contained in the spherical image of the vertices Aj1 , . . . , Ajl , while differs from the latter. This yields the inequality lp=1 ωjp > Ωi1 ,...,ik .

P

li 3 li 1 li 2

V

Fig. 70

The sufficiency of the conditions of Theorem 18 will be proved in Section 9.1 of Chapter 9; there we prove the existence of a convex polyhedron whose vertices lie on given rays and have given areas of spherical images. Under a positive dilation (a homothety transformation) from O, the vertices remain on the same rays and their spherical images obviously remain the same. Other transformations are impossible, i.e., a polyhedron is determined from the rays li through its vertices and the curvatures ωi at the vertices to within a similarity transformation with center the common origin of the rays. This is a consequence of the following general theorem: Theorem 19. If the vertices of two bounded convex solid polyhedra lie on the same rays issuing from their common interior point O, then either the polyhedra are homothetic to each other with positive coefficient and center O, or they have a pair of vertices lying on the same ray and the spherical image of one of them is a proper subset of the spherical image of the other. If the areas of the spherical images are equal, then the second possibility is excluded and the polyhedra must be similar. In place of area, we can take

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an arbitrary monotone function of spherical polygons (or, equivalently, of polyhedral angles). The proof of Theorem 19 is very simple. Suppose that the polyhedra P1 and P2 have vertices on common rays issuing from a common interior point O. Contract P2 homothetically to O so that P2 is taken inside P1 . The spherical images do not change. Now, begin increasing the polyhedron P2 homothetically from the point O until P2 touches P1 at some point. If this point is not a vertex, then P1 and P2 touch at a face or an edge (the polyhedra do not cross, since P2 is still inside P1 ). However, in that case, recalling that the vertices of P1 and P2 lie on the same rays, we see that P1 and P2 have at least one common vertex belonging to the face or edge at which the polyhedra touch. Thus, at the moment when the polyhedra P1 and P2 touch, some their vertices A1 and A2 coincide. Since P2 is inside P1 , the polyhedral angle V2 at A2 is contained in the polyhedral angle V1 at A1 . Therefore, every support plane of V1 is a support plane of V2 , i.e., the spherical image of the vertex A2 includes the spherical image of the vertex A1 . If the areas of all the spherical images are assumed equal, then the spherical images of the vertices A2 and A1 coincide. Therefore the polyhedral angles V1 and V2 also coincide, i.e., the edges of P1 and P2 touching at the common vertex A1 = A2 overlap. Since the endpoints of the edges lie on common rays, the endpoints of these edges coincide as well. Applying the same arguments to these endpoints and continuing in a similar way, we see that the polyhedra P1 and P2 coincide. Hence, before the transformation the polyhedra were already similar, with O the center of similarity. This concludes the proof. 2.5.3 We now turn to unbounded polyhedra, assuming them to be solid. We could consider unbounded convex polyhedra with vertices on given rays issuing from some point O; however, we restrict ourselves to polyhedra with vertices on given parallel rays; this case corresponds to an infinitely distant point O. All results that will be obtained can easily be converted to the case of a finite point O. For an infinitely distant point O, the requirement that O lie inside the polyhedron amounts to the condition that every straight line parallel to the given rays is either disjoint from the polyhedron or has a common half-line with it. Specifying parallel rays through the vertices of a polyhedron is equivalent to specifying a plane T perpendicular to the rays together with the projections A1 , . . . , Am of the vertices to this plane. The polyhedron P in question must be situated so that every straight line perpendicular to T is either disjoint from P or has a common half-line with P . Then the projection of P to T either covers all of T once or covers some (bounded or unbounded) convex polygon in T whose interior is covered once, while its sides are the projections of the faces of P perpendicular to T .


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Since to determine an unbounded convex polyhedron we need to know not only its vertices, but also its limit angle, the limit angle of the polyhedron must be involved in analysis as well. Here we also treat the limit angle of a polyhedron as a solid polyhedral angle. (As mentioned above, this angle may degenerate into a half-line or a flat angle.) The limit angle of a polyhedron is defined up to a parallel translation. To be definite, we assume that the vertex of the angle coincides with the vertex O1 of the polyhedron that projects to the point A1 . The limit angle V of the polyhedron P is situated relative to the plane T just like P itself: every straight line perpendicular to T either is disjoint from V or has a common half-line with V . Indeed, suppose that a straight line L is perpendicular to T and has a nonempty intersection with V . Since the vertex O1 of V belongs to the polyhedron, we can homothetically contract P to O1 , obtaining polyhedra P contained in P and containing V . In the limit, the polyhedra P converge to V . Since the straight line L meets V , it meets all the polyhedra P and hence intersects each of them in a half-line. (Otherwise, increasing P to P and accordingly translating the straight line L, we would obtain a straight line L1 whose intersection with P is not a half-line, in contradiction to the condition imposed on the disposition of the polyhedron P .) The limit of these half-lines is a half-line common to the angle V and the straight line L. From now on the conditions imposed on the disposition of the polyhedron and its limit angle relative to the plane T are assumed satisfied. We now state a theorem similar to Theorem 18: Theorem 20. Let points A1 , . . . , Am with associated numbers ω1 , . . . , ωm be given on a plane T . The following conditions are necessary and sufficient for ω1 , . . . , ωm to serve as the areas of the spherical images of the vertices of an unbounded polyhedron whose vertices project to the points A1 , . . . , Am : (1) all the ωi are positive and less than π; (2) m i=1 ωi ≤ 2π. The necessity of the first condition is evident. That of the second was proved in Subsection 1.5.3: the spherical image of an unbounded polyhedron is included in a hemisphere and hence its area is at most 2π. Theorem 20 is interesting because the curvatures at the vertices need only obey the above trivial conditions. Under a translation of a polyhedron in a direction perpendicular to the plane T , the projections of the vertices of the polyhedron and their spherical images remain the same. However, in general, these data do not determine the polyhedron uniquely up to translation, as is seen from the following theorem. Theorem 21. An unbounded convex polyhedron for given A1 , . . . , Am and m ω1 , . . ., ωm is unique up to translation if and only if i=1 ωi = 2π, i.e., if the area of the spherical image of the polyhedron equals 2π.

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It was proved in Subsection 1.5.3 that the spherical image of the limit angle of a polyhedron coincides with the spherical image of the polyhedron. ω = 2π, then the limit angle degenerates into a half-line Therefore, if m i i=1 and, consequently, is given in advance. If the area of the spherical image of a polyhedron, and hence of its limit angle, is less than 2π, then the limit angle may still vary. This leads to the following general theorem: Theorem 22. Let points A1 , . . . , Am with associated numbers mω1 , . . . , ωm be given on a plane T so that (1) 0 < ωi < 2π for all i and (2) i=1 ωi ≤ 2π. Further, assume that V is a convex polyhedral angle for which the area of ω the spherical image is equal to m i=1 i and which is positioned so that every straight line perpendicular to T is either disjoint from V or has a common half-line with V .28 Then there is an unbounded convex polyhedron P for which: (1) the points A1 are the projections of its vertices; (2) the numbers ωi are the areas of the spherical images of the corresponding vertices of P ; (3) the angle V is the limit angle of P . Such a polyhedron is unique up to translation in a direction perpendicular to the plane T . Theorem 22 is proved in Section 9.2 of Chapter 9. The uniqueness part of Theorem 22 is a consequence of a certain general theorem similar to Theorem 19, and has an equally simple proof (see Section 9.2). Analogous existence and uniqueness theorems also hold for polyhedra with boundary. 2.5.4 If the limit angle V and the projections A1 , . . . , Am of the vertices of a polyhedron to a plane T are given, then, to determine the polyhedron, it only remains to specify the heights p1 , . . . , pm of the vertices above the plane T , the heights being positive from one side of T and negative from the other. Conditions that are necessary and sufficient for numbers p1 , . . . , pm to serve as the heights of the vertices of a polyhedron given the limit angle V and the projections A1 , . . . , Am of the vertices are rather cumbersome and will not be given here. They can easily be derived from the fact that a point Bk of height pk above Ak is a vertex of a polyhedron if and only if Bk lies beyond the convex hull of the limit angle and the other vertices Bi . 2.5.5 All the theorems formulated in this section can be generalized to convex polyhedra in a space of arbitrary dimension together with their proofs: it suffices to replace 4π and 2π by the area of a sphere and the area of a hemisphere of radius one in that space. In particular, these theorems are 28

V may be a half-line or a flat angle.


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valid for convex polygons upon replacement of 4π and 2π by 2π and π respectively. The proofs of Theorems 20, 21, and 22 for unbounded polygons are rather simple. The limit angle of a polygon determines the directions of the unbounded edges; hence, the polygon can be constructed immediately from the angles αi at its vertices and the projections of vertices to a straight line T . (Of course, here the role of curvatures ωi is played by the quantities π − αi .) Theorem 18 for bounded polygons is far from being so simple, and we know no other proof but the one duplicating what was done in the case of polyhedra. 2.5.6 Finally, we return to the remark concerning polar transformations of polyhedra that was made in the beginning of the section. The polar transformation with respect the unit sphere centered at an interior point O of a convex polyhedron P associates some convex polyhedron P with P . The faces of P are perpendicular to the rays issuing from O through the vertices of P and conversely: the vertices of P lie on the rays issuing from O and perpendicular to the faces of P (see Subsection 1.5.4). The relationship between the faces and vertices of the two polyhedra is reciprocal: to each face of one of the polyhedra corresponds a vertex of the other and vice versa; moreover, to vertices belonging to a given face correspond faces touching at the corresponding vertex. Take the projection of the polyhedron P from the point O to the sphere S. The sphere splits into convex spherical polygons Si , which are the projections of the faces Q of P . The vertices of these polygons are the projections of the vertices of P . (See Fig. 32, in which polar polygons are shown for simplicity; for them the unit circle centered at O is taken instead of the sphere.) The rays issuing from O and passing through the vertices of some face Q of P are the normals to the faces of P touching at the corresponding vertex Ai . Therefore, Si is the convex polygon spanned by the endpoints of the normals to the faces touching at Ai . In other words, Si is exactly the spherical image of the vertex Ai . Thus, the projection of a face of P to the sphere S is the spherical image of the corresponding vertex of P . The roles of P and P can be interchanged because of their reciprocity. After these remarks, each theorem about polyhedra with vertices on given rays can be related to some theorem about polyhedra with faces perpendicular to the rays. For example, Theorem 18 becomes the following theorem. Theorem 23. Suppose that rays l1 , . . . , lm issuing from a point O are not contained in a single half-space. Let Ωi1 ,...,ik denote the area of the solid angle which is the convex hull of the rays li1 , . . . , lik . Then the following conditions are necessary and sufficient for the existence of a closed convex polyhedron with given areas ω1 , . . . , ωm of the central projections of its faces to the unit sphere centered at its interior point O and with faces perpendicular to the rays li :

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(1) all the ωi are positive; m (2) i=1 ωi = 4π; (3) j ωj > Ωi1 ,...,ik for every collection of rays li1 , . . . , lik going inside one half-space, where the sum is taken over all j corresponding to rays lj not contained in the convex hull of the rays li1 , . . . , lik . Theorem 19 becomes the following assertion: Theorem 24. If two closed convex polyhedra with a common interior point have pairwise parallel faces, then either the polyhedral angles projecting the parallel faces of the polyhedra from a common interior point O coincide and the polyhedra are homothetic with center of homothety at the point O, or among these angles there is one which projects a face of one of the polyhedra and contains as a proper subset the angle projecting the parallel face of the other polyhedron. In particular, the data of Theorem 23 determine a polyhedron uniquely up to similarity. Certainly, Theorems 23 and 24 may be proved directly, in perfect analogy with Theorems 18 and 19. The theorems on unbounded polyhedra with vertices on given parallel rays do not admit such polar transformation, since the center O in that case is at infinity. If we take a finite point O rather than one at infinity, we obtain theorems for unbounded polyhedra that admit a polar transformation.29

2.6 Infinitesimal Rigidity Theorems (Survey of Chapters 10 and 11) 2.6.1 Suppose that a polyhedron P0 is subject to some deformation, i.e., assume given a continuous family of polyhedra Pt depending on a parameter t; moreover, let the value t = 0 correspond to the initial polyhedron P0 . It is convenient to think of the parameter t as time, and so we shall speak of the 29

An unbounded polyhedron with vertices on given parallel rays can be defined by an equation z = f (x, y) over a plane (x, y) perpendicular to these rays. Instead of a polar transformation, we can apply the Legendre transformation (i.e., polarity with respect to an elliptic paraboloid) to such a polyhedron. To a plane z − z0 = p(x − x0 ) + q(y − y0 ) and a point u0 = p, v0 = q, w0 = px0 + qy0 − z0 in it, the Legendge transformation assigns the point (x0 , y0 , z0 ) and the plane w − w0 = x0 (u−u0 )+y0 (v −v0 ). As all polarities, the Legendre transformation maps convex polyhedra to convex polyhedra, so that to each theorem on convex polyhedra given by the equation z = f (x, y) there corresponds some “dual” theorem on polar polyhedra. – V. Zalgaller

2.6 Infinitesimal Rigidity Theorems

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motion of elements of P0 leading to its deformation, e.g. the motion of its vertices, the rotation of its faces, etc.30 If x is some quantity connected with a polyhedron, for example, the area of faces, the length of edges, etc., then under deformation this x becomes a function of t. The derivative dx of x. The dt is called the rate of variation quantity x is stationary if the initial rate of its variation is zero: dx dt t=0 = 0. We will study “initial deformations” of polyhedra, i.e., first order infinitesimal deformations at the initial moment, rather than finite deformations. In other words, we are interested in the principal terms of the first order in t of the deformations rather than the entire deformations of some elements of a polyhedron or related quantities: in the differentials dx, or equivalently in , the initial rates dx dt t=0 rather than in the increments ∆x. A polyhedron P0 is infinitesimally rigid under some conditions if any initial deformation of P0 subject to these conditions is reduced, as regards velocity, to motion of the polyhedron as a rigid body (or, in a more general case, to some other trivial transformation). This means that under the given conditions all elements of the polyhedron are stationary, provided that we exclude any motion, for example, by fixing one of the vertices of the polyhedron together with the direction of an edge issuing from it and fixing the direction of a face containing the edge. Rigidity theorems are assertions of the following kind: a polyhedron is infinitesimally rigid if, under its deformations of some type, certain data pertinent to the polyhedron are stationary. For example, in essence Cauchy proved but did not explicitly state the following theorem31 : if all faces of a closed convex polyhedron are infinitesimally rigid, then the polyhedron itself is infinitesimally rigid. That is, assume a closed polyhedron is deformed so that its structure remains the same and the planes of its faces move (undergo translations and rotations) with some velocity vectors. If all the faces are infinitesimally rigid, i.e., the lengths of edges and the angles on faces are stationary, then the obtained deformation of the polyhedron is given by a rigid motion with the same velocity vectors. This theorem is analogous to Cauchy’s Theorem on the congruence of polyhedra with equal similarly-situated faces. The analogy becomes particularly clear if we observe that the definition of rigidity can be rephrased as follows: a polyhedron P0 is infinitesimally rigid if, under the imposed conditions, every deformed polyhedron Pt coincides with P0 to within first order 30

The main notions of the general theory of flexes and infinitesimal flexes are described, for instance, in [E2], [E3], [IS]. The last two surveys treat infinitesimal flexes of various orders in more detail. – V. Zalgaller 31 This theorem was first explicitly stated and proved by M. Dehn [De]. Dehn overlooked the fact that the rigidity theorem follows by a word for word repetition of Cauchy’s arguments yielding the congruence theorem for polyhedra with equal faces. Dehn’s method differs drastically from that of Cauchy. Another proof was given by H. Weyl [We].

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infinitesimals in t.32 The latter definition is obviously equivalent to the former. At the same time, it shows that rigidity theorems are theorems on congruence to within first order infinitesimals. This observation impels us to seek a rigidity theorem for each congruence theorem. Certainly, such passage from some theorems to others is in general far from trivial. First, rigidity theorems are assertions on congruence to within first order infinitesimals rather than claims on exact congruence. Second, the assumptions of rigidity theorems include the stationarity of elements of polyhedra, i.e., the congruence of the elements of P0 and Pt to within first order infinitesimals in t, rather than their exact congruence. Nevertheless, to each congruence theorem (or, which is the same, to each uniqueness theorem) formulated in Sections 2.3– 2.5, corresponds an analogous rigidity theorem with the assumptions and claim appropriately modified. Below we formulate as examples some rigidity theorems which will be proved in Chapters 10 and 11. 2.6.2 The following statement corresponds to Theorem 2 of Section 2.4: Theorem 25. If a closed convex polyhedron is deformed so that the intrinsic distances between its vertices are stationary, then the initial deformation is given by a motion. In other words, a closed convex polyhedron with stationary intrinsic metric is infinitesimally rigid. This result may be rephrased as follows: Theorem 25∗ . Assume that a development R0 of a closed convex polyhedron P0 be given on P0 so that all vertices of R0 are among the vertices of P0 or lie on edges of P0 . If the polyhedron P0 is deformed in such a way that no new vertices appear on it, except possibly vertices of R0 lying on “old edges,” then the development R0 also undergoes a deformation33 and if the lengths of the edges and the angles of the polygons of R0 are stationary, then the polyhedron P0 is infinitesimally rigid, i.e., the initial deformation of P0 is given by a rigid motion. In other words, a closed convex polyhedron with stationary development is infinitesimally rigid. In Theorems 25∗ and 25, we a priori exclude neither foldings of faces or edges nor even the violation of convexity of the given polyhedron. However, 32

That is, there are polyhedra P0τ congruent to P0 and such that the maximum of the distances of their vertices from the corresponding vertices of Pt is an infinitesimal of higher order than t at small t. 33 We shall prove that on a polyhedron Pt close to P0 there always exists a development close to R0 . It is for this reason that we may speak of the deformation of a development. The deformation of a polyhedron P0 obeys certain general constraints that are not formulated here. In short, they are reduced to the requirement that there appear no new vertices other than the vertices of the development.


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new vertices may appear only on “old edges” of the polyhedron. If such new vertices are present, then in Theorem 25 they are taken into account among the vertices. The violation of convexity of the polyhedron may be caused by “folding the polyhedron inward” at a new vertex. The above-mentioned theorem on the infinitesimal rigidity of a polyhedron with infinitesimally rigid faces is a corollary to Theorem 25∗ (or 25). Theorem 25∗ is surely more general, since it allows foldings of faces. For example, a cube with one face deleted is infinitesimally rigid provided that all faces are infinitesimally rigid, and it is nonrigid in case foldings of faces along diagonals are permitted. (The first assertion is straightforward from the obvious rigidity of trihedral angles with stationary planar angles. The second is verified directly.) Besides the general differences of all rigidity theorems from congruence theorems, Theorem 25∗ differs from the theorem on congruence of polyhedra with equal developments in the following aspects: (1) In the congruence theorem, the condition that the vertices of the development are among the vertices of the polyhedron or lie on the edges is absent: under gluing, these vertices may appear inside faces. In Theorem 25∗ , however, this requirement is necessary: the theorem fails without it, as will be shown in Section 10.1 of Chapter 10. (2) The congruence theorem is valid for polyhedra degenerating into polygons, while Theorem 25∗ fails for them. (This is trivial when we allow the violation of convexity: it suffices to fold a polygon along a diagonal. However, all doubly-covered polygons (except triangles) are also nonrigid without violating convexity. For instance, take a doubly-covered square ABCD with side a and lift the vertex D to height h = vt from the plane of the square, where v is the velocity of motion of D. We obtain a tetrahedron with edge CD =

1 v2 2 t . a2 + h 2 ≈ a + 2 a

We see that the variation of the distances between vertices is a second order infinitesimal in t, i.e., these distances are stationary, although the vertex D moves with a positive speed. A similar argument is obviously applicable to an arbitrary polygon with more than three sides.) (3) In the congruence theorem, the polyhedra in question are assumed convex, whereas in Theorem 25 or 25∗ violation of the convexity under deformation is not excluded: it may be caused by foldings of faces when new vertices appear on edges. (However, the theorem asserts that if a development is stationary, then the movement of a point inside the polyhedron has zero initial velocity.) In this aspect, Theorem 25 is more general than Theorem 2. By analogy to Theorem 3 of Section 2.3 on the equality of dihedral angles in polyhedra with equal planar angles, we have the following

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Theorem 26. If a closed convex polyhedron is deformed so that the angles on its faces are stationary, then the dihedral angles of the polyhedron are also stationary. Here by faces we may mean not only genuine faces but the parts into which they may be split by line segments not interesting in the interiors of genuine faces. The deformation of a polyhedron consists in moving the planes of these parts of faces. For unbounded polyhedra, there also exist rigidity theorems analogous to congruence theorems stated in Section 2.3. Roughly speaking, their statements are obtained by simply substituting stationarity for congruence; the theorems will be formulated precisely in Chapter 10. 2.6.3 Theorem 26 admits of the following mechanical interpretation: Theorem 27. Assume given a system of rectilinear rods joined by hinges at their endpoints and constituting the edges of a closed convex polyhedron. Then the rods of the system are not strained provided that there are no external forces, the rods are not bent, and the system is in equilibrium. An example of a strained system without external forces can be obtained from three matches and three rubber threads (Fig. 71). The match-sticks are put on the sides of an equilateral triangle and the rubber threads connect the vertices to the center of the triangle. The rubber threads are fastened at the center and are strained. Therefore, in the system there are strains, although it is in equilibrium without the action of external forces. Theorems 27 asserts that such a strained state of equilibrium is impossible for the system of edges of a closed convex polyhedron.

Fig. 71

The relationship between Theorems 26 and 27 will be completely revealed in Section 10.4 of Chapter 10, in which we also establish similar results for some other systems of hinged rods. 2.6.4 In Section 2.4 we presented some general theorems on the congruence of convex polyhedra with parallel faces. To these theorems certain rigidity theorems will correspond. To state the latter, we introduce the notion of


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essentially monotone function of polygons. We say that a function f (Q) of a convex polygon Q is essentially monotone if, under the translation of any side of Q outside the polygon, the function f varies at a rate greater than zero, i.e., whenever hi is the distance from a given point inside Q to the ∂f straight line through the ith side, we have ∂h > 0. i The area and perimeter of a polygon serve as examples. We now formulate a rigidity theorem similar to Theorem 8 of Section 2.4 on the congruence of closed convex polyhedra: Theorem 28. If a closed convex polyhedron P is deformed by translations of the planes of the faces of P such that some essentially monotone function (in general, its own for each face) is stationary at every face of P , then the initial deformation of P is given by a translation. As opposed to the corollary to Theorem 8 indicated in Section 2.4, here the function in faces is assumed essentially monotone rather than simply monotone. Without this condition Theorem 28 fails. For example, put f (Q) = [F (Q) − 1]3 , where F (Q) is the area of the polygon Q. It is easy to see that f is monotone: if Q1 contains Q2 , then F (Q1 ) > F (Q2 ) and consequently f (Q1 ) > f (Q2 ). However, since F (Q) = 1, the function f is not essentially monotone, because ∂F ∂f = 3[F (Q) − 1]2 ∂fi ∂hi and consequently ∂f = 0 at F = 1. hi Hence, if we take a cube with unit edge and push some of its faces outwards with arbitrary speed, then the function f is stationary at all faces, although the deformation is not reduced to a translation even to within first order infinitesimals. Rigidity theorems similar to Theorem 28 are also valid for unbounded polyhedra. There the additional condition requires that the planes of the unbounded faces be invariant (or, at least, stationary), while the stationarity condition for the appropriate essentially monotone functions need be imposed only on bounded faces. These theorems will be precisely stated and proved in Chapter 11. 2.6.5 The congruence theorems formulated in Sections 2.3 and 2.4 are proved, as we pointed out, by using the Cauchy Lemma. It turns out that this lemma is inapplicable to the rigidity theorems in the general setting in which we prove them, because its main condition (forbidding only two sign changes around a vertex) cannot be guaranteed in advance. For example, when the

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planar angles of a convex polyhedral angle are stationary, the number of sign changes of the rate of variation of its dihedral angles may equal two, provided that there are dihedral angles of measure π. This corresponds exactly to the fact that we allow a genuine face of the polyhedron under study to be partitioned into separate pieces. The possibility of exactly two sign changes in this situation can be easily seen from examples (see Section 10.1). However, we will prove that in this case the sign distribution still obeys an additional condition which suffices to ensure the result of the Cauchy Lemma: in passing labeled edges around a vertex, it is impossible to have at least four sign changes or two sign changes under the above-mentioned additional condition. This “Strong Cauchy Lemma” is proved in Section 10.2 of Chapter 10 and on its basis we prove all the rigidity theorems of Chapters 10 and 11. 2.6.6 A rigidity theorem also corresponds to the general Theorem 19 on the similarity of polyhedra with vertices on given rays and spherical images that cannot be placed in one another. To formulate it, we introduce the notion of essentially monotone function of spherical polygons by analogy to the corresponding notion for plane polygons. Namely, a convex spherical polygon is bounded by great circles through the sides of the polygon and a rotation of such a great circle about two antipodal points lying on it but not belonging to a side brings about a deformation of the polygon. A function f of spherical polygons is said to be essentially monotone if, under such a rotation of an arbitrary side of the polygon with positive velocity directed out of the polygon, the function f increases at a positive rate. In particular, area is easily seen to possess this property. The rigidity theorem corresponding to Theorem 19 is: Theorem 29. If the deformation of a closed convex polyhedron is due to the motion of its vertices along given rays issuing from an interior point of the polyhedron and such that some essentially monotone function is stationary at the spherical image of each vertex, then the initial deformation of the polyhedron is given by a similarity transformation with center O. The proof of this theorem is almost as simple as that of Theorem 19 (see Section 9.1). Similar theorems are valid for unbounded polyhedra as well, but under the additional assumption of stationarity of the limit angle of the polyhedron (see Section 9.2). 2.6.7 We now establish a deep connection between the infinitesimal rigidity theorems on the one hand and the existence and uniqueness theorems on the other hand. It turns out that we can deduce, under rather general conditions, existence and uniqueness theorems from infinitesimal rigidity theorems.34 34

Observe one more connection between rigidity and infinitesimal rigidity. As was shown by N. V. Efimov [E4], first or second order rigidity implies that a sur-


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Let a1 , . . . , an be some parameters determining a polyhedron up to some trivial transformation and let b1 , . . . , bn be other parameters related to the polyhedron, for which we wish to prove that the polyhedron is determined up to a transformation of the same kind (i.e., an existence and uniqueness theorem holds for polyhedra with data b1 , . . . , bn ). For example, if we consider closed polyhedra with n+ 1 vertices lying on given rays issuing from the same point O, then we can take the parameters a1 , . . . , an to be the ratios of the distance of the n vertices from the point O to the distance of the (n + 1)th vertex from O. These ratios determine the polygon up to homothety of center O. For the parameters b1 , . . . , bn , we can take the curvatures at the n vertices, i.e., the areas of the spherical images of the vertices; the curvature at the (n + 1)th vertex is determined from the condition that the total curvature of the closed polyhedron equals 4π. Since the parameters ai determine a polyhedron, the quantities bi are functions of the ai : bi = fi (a1 , . . . , an )

(i = 1, . . . , n).

(1)

We assume that the functions fi are differentiable; hence, dbi =

∂bi ∂bi da1 + . . . + dan ∂a1 ∂an

(i = 1, . . . , n).

(2)

(The differentiability condition is always satisfied for parameters bi = fi (a1 , . . . , an ) of common interest. In particular, in the above example the curvatures at vertices are easily seen to be differentiable functions of the distances of vertices from O.) The theorem on infinitesimal rigidity of a polyhedron P0 at stationary bi obviously consists in the assertion that if all the dbi vanish, then dai = 0 at the values ai = a0i corresponding to P0 . In other words, the homogeneous system (2) has the trivial solution only. This is equivalent to the fact that the determinant of system (2), i.e., the Jacobian of the system of functions (1), differs from zero. In that case, the functions (1) are well known to be invertible in a neighborhood of the values ai = a0i and bi = b0i (i = 1, . . . , n) face does not admit a continuous flex analytic in a parameter even in the class of nonconvex surfaces. The following generalizations of this fact are known: If a surface admits only one linearly independent field of first order infinitesimal flex, then rigidity of order 3 implies the above-mentioned analytic rigidity [E4]; see [S4] for n ≥ 3 and C 1 -smoothness. In the case of a polyhedron, existence of a continuous flex implies the existence of an analytic continuous flex. This algebraically natural result seems to have been first formulated by Gluck [Gl1]. It implies in particular that a first or second order infinitesimally rigid polyhedron does not admit a continuous flex. Using this result, Connelly [Co4] proved that an arbitrarily triangulated convex polyhedron with the new faces rigid does not admit a continuous flex in the class of all (not necessarily convex) surfaces. As was noted by I. Kh. Sabitov [IS, pp. 153–154], this result can be deduced from A. V. Pogorelov’s theorem on the rigidity of general convex surfaces in the exterior of an open planar domain. – V. Zalgaller

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corresponding to P0 . Consequently, given bi close to b0i , we can uniquely find ai close to a0i ; therefore, there is a polyhedron with data bi which is close to P0 (up to some trivial transformation); moreover, such a polyhedron is unique up to a trivial transformation. Thus, any rigidity theorem implies existence and uniqueness theorems in a small neighborhood of the polyhedron P0 under study. 2.6.8 The next two simple lemmas provide conditions under which the above local result can be extended to global existence and uniqueness theorems. Lemma A. Let A and B be two manifolds and let ϕ be a mapping from A to B satisfying the following conditions: (1) if the point B in B is the image of a point A in A, B = ϕ(A), then there is a neighborhood of B admitting the inverse mapping ϕ−1 into A; (2) every connected component of B contains images of some points in A; (3) ϕ(A) is closed, i.e., if the points Bn converge to B, then B is the image of some point in A. Then ϕ is a mapping from A onto B. By condition (1), if B ∈ ϕ(A), then there is a neighborhood of B contained in ϕ(A). Hence, ϕ(A) is open. Combining this result with conditions (2) and (3), we see that ϕ(A) = B in exactly the same way as in the Mapping Lemma.35 If A is the manifold of all polyhedra considered to within a trivial transformation and B is the manifold of data B(b1 , . . . , bn ) to which the rigidity theorem in question applies, then this rigidity theorem yields condition (1) of Lemma A. Therefore, whenever the other two conditions are satisfied, a global existence theorem holds. Such a proof of Theorem 1 on the existence of a closed convex polyhedron from a development appears in Chapter 6 of my book [A15]. Lemma B. Let A, B, and ϕ have the same meaning as in Lemma A, and let ϕ be a mapping from A onto B (which holds under the assumptions of Lemma A, for instance). Suppose that the following conditions are satisfied: (1) if a point B in B is the image of some point A in A, then there is a neighborhood U of A and a neighborhood V of B such that ϕ is a homeomorphism from U onto V ; 35

This method is often described as the method of continuous extension. Suppose that B0 ∈ ϕ(A) and let B1 be a point such that there is a continuous curve Bt (0 ≤ t ≤ 1) joining B0 to B1 . Then condition (1) of Lemma A implies that Bt ∈ ϕ(A) for small t. Let T be the least upper bound of those t at which Bt ∈ ϕ(A). Then BT ∈ ϕ(A) by condition (3). If we had T < 1, the condition (1) would give us a value t > T for which Bt ∈ ϕ(A). Consequently, T = 1 and B1 ∈ ϕ(A). Condition (2) guarantees that every point B in B is joined to some point in ϕ(A).


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(2) the manifold A is connected and the manifold B is simply connected36 , i.e., every closed curve in B is contractible to a single point. Then the mapping ϕ is injective. This lemma is well known in topology. A mapping ϕ satisfying the conditions of Lemma B is the covering mapping of A onto B. When B is simply connected, such a mapping is well known to be a homeomorphism; in particular, it is injective. Also, Lemma B can easily be proved without appealing to the theory of covering manifolds as follows: Suppose that ϕ is not injective. Then there is a point B in B which is the image of two different points A0 and A1 in A. Since the manifold A is connected, there exists a continuous curve L joining A0 to A1 , i.e., a point A(t) can be assigned to each t in the closed interval [0, 1] in such a way that A(t) depends continuously on t and A(0) = A0 and A(1) = A1 . The image of L is some curve M = ϕ(L) with points B(t) = ϕ(A(t)); it is closed since ϕ(A0 ) = ϕ(A1 ) = B. By condition (2), the curve M may be contracted to some point B0 , i.e., there exists a continuous family of curves M (s) (0 ≤ s ≤ 1) such that M (0) = M and M (1) = B0 . Here the points B(t, s) of the curves M (s) satisfy the two following conditions: (1) B(t, s) depends continuously on the two arguments 0 ≤ t ≤ 1, 0 ≤ s ≤ 1; (2) at fixed s B(t, s) is a parametric representation of the curve M (s). By virtue of the first condition of the lemma, the mapping ϕ is a homeomorphism in some neighborhood of each point A ∈ A. By Borel’s Lemma, the curve L can be covered by finitely many such neighborhoods U1 , . . . , Um and split into arcs L1 , . . . , Lm each lying in U1 , . . . , Um respectively: to each arc Li there corresponds a subinterval τi = [ti−1 , ti ] of the interval [0, 1] such that if t ∈ τi then A(t) ∈ Ui . Accordingly, the curve M will be covered by the neighborhoods νi = ϕ(Ui ) and split into arcs Mi = ϕ(Li ). Assume s0 so small that the curve M (s0 t0 ) lies in the neighborhood ν = ∪i νi of the curve M = M (0). Take an arbitrary value of t; it belongs to some interval τi . Then B(t, s0 ) ∈ νi = ϕ(Ui ), and to B(t, s0 ) we can associate a well-defined inverse image under ϕ; denote it by A(t, s0 ). As the result, to the arcs of M (s0 ) corresponding to the intervals τi , we associate some curves Li (s0 ) in A. These curves have consecutive common endpoints belonging to the same neighborhoods over which ϕ is injective. Therefore, the curves Li (s0 ) all together constitute a single curve L(s0 ) whose endpoints are obviously the points A0 and A1 . The image of L(s0 ) is the curve B(t, s0 ). Proceeding similarly with the curve L(s0 ) replacing At and so forth, we eventually come to a situation in which to each curve M (s) we assign some 36

It suffices for instance to “wind” an open rectangular strip A around the lateral surface of a cylinder B in such a way that the edges of the strip overlap, and as the result we obtain a noninjective mapping that satisfies all the assumptions of the lemma except the simple connectedness of B.

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curve L(s) that joins A0 to A1 and is mapped onto M (s).37 However M (1) is the point B0 and L(1) is a curve joining the two points A0 and A1 . Therefore, the mapping ϕ from L(1) onto M (1) = B0 is not injective in any neighborhood of each point, for example the point A0 . This contradicts the assumptions of the lemma. Hence, the supposition that ϕ takes two points A0 and A1 into one point B0 is invalid. Thus, the mapping ϕ is one-to-one, as claimed. 2.6.9 Let A be the manifold of polyhedra considered up to trivial transformations and let B be the manifold of data B(b1 , . . . , bn ) to which the rigidity theorem applies. We have already observed that this theorem implies the first condition of Lemma B; therefore, if A is connected and B is simply connected, then the mapping ϕ from A onto B is one-to-one, which implies that the uniqueness theorem up to an appropriate trivial transformation holds. For example, if A is the manifold of polyhedra with vertices on given rays considered up to a similarity and B is the manifold of feasible values for the areas ωi of spherical images, then A is obviously connected while B is simply connected, being convex (which follows from the defining conditions indicated in Theorem 18). A similar situation occurs also in other cases, for example in Minkowski’s Theorem 11, where A is the manifold of polyhedra with given face directions and B is the manifold of feasible values of areas of faces; here B is also convex and therefore simply connected. (However, I was unable to establish the simple connectedness of the manifold of feasible developments for existence theorems of polyhedra determined by a development.) We thus have the following general method for proving existence and uniqueness theorems: 1. Prove the corresponding rigidity theorem and from it deduce existence and uniqueness “in small neighborhoods.” 2. Establish the assumptions of Lemma A, thus obtaining the global existence theorem. 3. Also establish the assumptions of Lemma B, thus obtaining the global uniqueness theorem. This method has the following advantages: First, as shown above, the rigidity theorem is reduced to a linear problem (namely to the study of equations (2)) and therefore may be easier, especially if the question is treated analytically. This circumstance also facilitates generalizing the method to problems concerning curved surfaces, where the manifolds A and B are infinite-dimensional spaces.38 Second, in proving existence, the global uniqueness theorem becomes redundant, as opposed to what is required by the method based on the Mapping Lemma. 37

A rigorous justification of the assertion is as follows: for small s, the assertion is true. Let S be the least upper bound for such values of s. Then, as s → S − 0, we obtain a curve L(s) with the same arguments applicable to it. Hence, S = 1. 38 This is the method that H. Weyl [W1] used to solve the existence problem of a closed convex surface with given metric.


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However, if the relevant existence theorem admits a direct proof, then the Mapping Lemma provides a shorter way to proving existence. Moreover, from a purely geometric standpoint, rigidity theorems look somewhat artificial since they involve notions and theorems of the differential calculus. 2.6.10 We return to the fact that the rigidity theorem fails for polyhedra with a stationary development. We have shown that this theorem fails for polyhedra degenerating into polygons even if we require that the convexity of the polyhedron be preserved. Now, we verify this by another argument. Let B be the manifold of developments with a given number of vertices and let A be the manifold of closed convex polyhedra considered up to a motion but not up to a reflection. Since the same development may be obtained by gluing two polyhedra symmetric to each other, the mapping from A onto B is two-to-one rather than one-to-one. The violation of injectivity due to the presence of symmetry elements in a nondegenerate polyhedron can be eliminated by labeling the vertices of this polyhedron according to the vertices of the development.39 However, this is impossible for degenerate polyhedra: the reflection in the plane of a degenerate polyhedron P0 takes all the vertices of P0 into themselves. Arbitrarily small shifts of vertices of P0 from different sides of the plane of P0 lead to symmetric polyhedra, i.e., they induce the passage from one sheet of the manifold A to the other. The set of degenerate polyhedra is mapped into B bijectively and is the branching set of √ the mapping from A onto B just like the branching point of the function x + iy. Therefore, the mapping from A onto B cannot be injective in neighborhoods of degenerate polyhedra; hence, the rigidity theorem cannot hold for them. Proving the existence of polyhedra with a given development on the base of the rigidity theorem, we must thus exclude degenerate polyhedra and their developments from our considerations. 2.6.11 Closing the section, let us make one more remark. Suppose that, as above, A and B are the manifold of polyhedra A determined by the parameters a1 , . . . , an and the manifold of data B(b1 , . . . , bn ). Let a mapping ϕ from A onto B be represented by continuously differentiable functions (1). The set of those A at which the rigidity theorem fails is the set of zeros of the Jacobian of this function system. Hence, this set is closed. Whenever it has an interior point A, the functions (1) are dependent in some neighborhood 39

If a polyhedron P0 has a symmetry plane and if X and Y are symmetric vertices of P0 , then symmetric shifts of X and Y produce a pair of mutually symmetric polyhedra close to P0 . However, the reflection in the plane of symmetry of P0 takes the vertex X into Y and vice versa; therefore, if we only allow vertices with the same label to be superposed, then such reflections are excluded. An analogous remark relates to the case of other types of symmetry. We can make the mapping from A onto B one-to-one by endowing developments with an orientation and thereby doubling B.

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of A and consequently ϕ is not one-to-one. Therefore, if ϕ is one-to-one, i.e., the uniqueness theorem holds, then the set on which the rigidity theorem fails is a closed nowhere dense set.40

2.7 Passage from Polyhedra to Curved Surfaces 2.7.1 The general theorems on convex polyhedra discussed in Sections 2.3– 2.6 are of value not only in their own right but also because they admit generalizations to more or less arbitrary convex surfaces. In studying problems of the theory of convex surfaces, it is fruitful to pose and solve them first for polyhedra. In some cases, this enables us to solve the original problem by passing to the limit from polyhedra to general convex surfaces. In other cases, this approach fails, but at least we are left with some understanding of the question. Due to lucidity and clarity of the properties of polyhedra, this approach turns out particularly useful in questions of geometry “in the large” or in applications to irregular surfaces, when the ordinary analytical methods of the theory of surfaces are no longer automatic.41 Since our main subject is the theory of convex polyhedra, we confine ourselves to formulating the corresponding theorems for other convex surfaces in the section Generalizations concluding almost every chapter. We leave the theorems without proofs, but give references where the proofs can be found. Here we begin with some general remarks, first of all, about what data of a general convex surface correspond to the appropriate data of polyhedra considered above. 2.7.2 In Section 2.3 we have already established that specifying a development of a polyhedron is equivalent to specifying the intrinsic metric of the polyhedron. For a general convex surface, we should speak of the intrinsic metric of the surface, i.e., of the function of pairs of points on the surface, given by the distance between the points along the surface. By definition, this distance equals the greatest lower bound for the lengths of the curves lying on the surface and joining the points. Given a continuous function ρ(X, Y ) of pairs of points on the sphere S (or another manifold), we say that ρ(X, Y ) is the metric of the surface F , 40

Implicitly, this assertion already contains the well-known theorem by Gluck [Gl1] claiming that almost all closed polyhedra in R3 homeomorphic to the sphere do not admit continuous flexes. A generalization of this result to closed polyhedra of ` G. Poznyak [Poz] proved arbitrary topological type is given in [Fog]. Earlier E. that almost all polyhedra homeomorphic to the sphere are infinitesimally rigid; it was not known then that this fact implies the absence of continuous flexes. – V. Zalgaller 41 The successive approximation by polyhedra in the applications to the intrinsic geometry of arbitrary convex surfaces is described in my book [A15].

2.7 Passage from Polyhedra to Curved Surfaces

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or that F realizes the metric ρ(X, Y ), if there is a homeomorphism h from S onto F such that ρ(X, Y ) = ρF (h(X)h(Y )) for every pair of points X and Y , where ρF is the intrinsic metric of the surface F . Let Fi and ni be the areas and outward normals of the faces of a polyhedron P . Consider a unit sphere S on which we construct the spherical image of P . If M is some subset of E, then FP (M ) = Fi ni ∈M

is the area of the subset of P which is the inverse image of M under the spherical mapping. (The sum is taken over all faces whose normals point to M .) FP (M ) is a set function on E, called the area function of P . Specifying the face areas and the face normals of P is obviously equivalent to specifying the area function of P . The area function of an arbitrary convex surface Φ is defined in exactly the same manner: FΦ (M ) is the area of the set N on Φ which is the inverse image of M under the spherical mapping, i.e., at least one support plane whose outward normal points to M passes through each point of N . (It can be proved that FΦ (M ) is a countably additive function defined on at least all Borel sets M .)42 Let ei be rays issuing from a point O or, equivalently, points on the unit sphere E about O. Let ωi be the areas of the spherical images of the vertices of a polyhedron P whose vertices lie on the rays ei . If M is a subset of E then ωi ωP,O (M ) = ei ∈M

is the area of the spherical image of the subset of P whose central projection is the set M . The set function ωP,O (M ) depends not only on the polyhedron P but also on the choice of the point O (or of the sphere E). Therefore, we call ωP,O (M ) the spherical area function relative to E. Specifying the rays ei and the areas ωi of the spherical images of vertices is obviously equivalent to specifying the spherical area function. The spherical area function of an arbitrary convex surface Φ is defined similarly: ωP,O (M ) is the area of the spherical image of the subset of Φ whose central projection from O to the sphere E is the set M . (It can be proved that ωP,O (M ) is a countably additive set function defined for all Borel sets at least.)43 For an unbounded convex surface Φ, we similarly define the function ωΦ,T (M ), the area of the spherical image relative to a plane T : given a set M on T , ωΦ,T (M ) is the area of the spherical image of that subset of Φ whose projection is M . Finally, for arbitrary convex surfaces, we must consider their support functions H(u) instead of the support numbers considered for polyhedra. 42 43

See [A3: I]. It is in this article that the notion of area function was first introduced. See [A15, § 2 and § 4 of Chapter 5].

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For a unit vector u, the value of H(u) is the distance from the origin O to the support plane Q with outward normal u. Moreover, the distance is assumed positive if the direction from O to Q coincides with the direction of the vector u and is negative otherwise. If u is not a unit vector, then u H(u) = |u| · H . |u| We also note that, for a general convex surface, instead of the limit polyhedral angle we must consider the “limit cone” defined in exactly the same way. Using the above general notions, it is easy to formulate uniqueness and rigidity theorems for general convex surfaces; they are analogous to the corresponding theorems for polyhedra. Stating existence theorems requires imposing appropriate conditions on data. For the functions H(u), FΦ (M ), ωΦ,E (M ), and ωΦ,T (M ), the conditions can readily be derived from the conditions of the theorems on polyhedra. They are presented in the last sections of Chapters 7 and 9. The metric ρ(X, Y ) is an exception: the necessary conditions to be imposed on it cannot be obtained by simply reformulating the conditions of theorems on polyhedra.44 2.7.3 The uniqueness and rigidity theorems cannot be carried over from polyhedra to general convex surfaces just by passing to the limit, the theorems must be supplemented by certain estimates for the deformation of a polyhedron depending on the variation of the data involved in the theorems, and these estimates must be independent of the number of vertices, or at least must not degenerate as the number of vertices tends to infinity. For example, consider Theorem 3 on the congruence of isometric closed convex polyhedra. To this theorem must correspond the general theorem, not proved as yet, on the congruence of isometric closed convex surfaces.45 An approach to the proof of the latter theorem via estimates for the deformation of a polyhedron depending on the variation of the intrinsic metric of the polyhedron was indicated by S. Cohn-Vossen [C-V] and consists in the following: Let rij and ρij be the distances between the ith and jth vertices of a polyhedron, measured in the space and on the polyhedron respectively. Given two closed convex polyhedra P1 and P2 whose vertices are in a oneto-one correspondence, denote the above-mentioned distances between their (1) (2) (1) (2) vertices by rij , rij , ρij , and ρij . Then (1) (2) ∆r = max rij − rij 44

These conditions can be found in the same book [A15] or, in another form, in my article [A12]. 45 This general theorem is now proved by A. V. Pogorelov see [P5]. Another proof was given by Yu. A. Volkov in [Vo6]. – V. Zalgaller

2.7 Passage from Polyhedra to Curved Surfaces

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obviously characterizes the magnitude of the spatial deformation upon passage from P1 to P2 , while (1) (2) ∆ρ = max ρ − ρ ij

ij

characterizes the magnitude of the intrinsic deformation. We must establish the estimate ∆r ≤ f (∆ρ), with a continuous function f , vanishing at ∆ρ = 0, which is independent of the polyhedra P1 and P2 or depends on rather general characteristics of their shapes, for instance on the radii of balls containing P1 and P2 and of balls contained in them. Imagine this estimate established. Let F1 and F2 be two isometric closed convex surfaces. Take sufficiently dense nets of points A1i and A2i that correspond to each other under the isometric mapping of F1 onto F2 . Inscribe closed convex polyhedra P1 and P2 in F1 and F2 with vertices at these points. It turns out that, the denser the net of points Ai is, the closer the distance between the points measured on the polyhedron approaches the distance measured on the surface. Since the surfaces are isometric, the quantity ∆ρ for the polyhedra P1 and P2 tends to zero as the net of points Ai becomes infinitely dense. Using estimate (1) with the required properties, we see that ∆r → 0 as (1) (2) ∆ρ → 0. However, ∆r = max rij − rij is nothing but the maximum of the differences of the distances in space between the corresponding points Ai and Aj of the surfaces F1 and F2 . Therefore, we must have ∆r = 0, i.e., the surfaces F1 and F2 are congruent.46 A similar approach is a priori possible for other uniqueness theorem. However, to my knowledge, no estimate of the desired type has been proved for any of them. For this reason, such an approach to carrying uniqueness theorems from polyhedra to general convex surfaces still remains purely hypothetical.47 The situation is the same with rigidity theorems. For this reason, proofs of these theorems for convex surfaces are carried out by other methods. Moreover, the theorems on uniqueness and rigidity of a closed convex surface given a (stationary) intrinsic metric are not proved as yet without extra assumptions on the regularity of the surface under study.48 Thus, in the questions of uniqueness and rigidity, theorems on polyhedra may only be regarded as guidelines for the results that we could try to attain in passing to arbitrary convex surfaces. 2.7.4 The matter is simpler by far for existence theorems: all the above-stated existence theorems for convex polyhedra can be carried over to general convex 46

It is the method that was implemented by Yu. A. Volkov in [Vo6]. – V. Zalgaller At present, the articles [Vo5], [K], [P10, Chap. 7], [Di] should be indicated in addition to [Vo6]. – V. Zalgaller 48 Now these theorems have been proved by A. V. Pogorelov [P10, Chap. 4]. 47

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surfaces by passing to the limit. Without going into details, this is done as follows: Suppose that we wish to prove the existence of a convex surface with given data B0 . Assume known the corresponding existence theorem for convex polyhedra. Then we prove the following convergence theorem: if convex polyhedra Pn converge to a surface F , then their data BPn converge, in an appropriate sense, to the data BF of F . Here the convergence of data is only required to be single-valued, i.e., the sequence Bn cannot converge to two different limits. Next, we construct a sequence of data Bn pertinent to polyhedra which converges to B0 . By the existence theorem for polyhedra, there are convex polyhedra Pn with data Bn . We extract a converging subsequence Pni from the sequence Pn ; fortunately this is always possible. The limit of the subsequence is some convex surface. By the convergence theorem, Bni converge to BF . At the same time, Bni converge to B0 by construction. Since the data Bni may have only one limit, it follows that BF = B0 , i.e., the surface F has the data B. If the data B0 represent the support function H(n), then the convergence of data Bm to B0 is simply the convergence of support functions or the convergence of support numbers hi to the values of the support function H(n) at the vectors ni since the set of points ni on the sphere becomes infinitely dense. If B0 is a metric ρ(X, Y ) on the sphere S, then we can approximate B0 by polyhedral metrics of positive curvature. The theorem on convergence of metrics is as follows: if closed or unbounded complete convex surfaces Φn converge to Φ, then ρΦn (Xn , Yn ) converges to ρΦ (X, Y ). If B0 is one of the set functions F (M ) and ω(M ) on the unit sphere E, then the convergence should be treated as weak. If ϕn (M ) and ϕ(M ) are countably additive set functions on E, then the functions ϕn converge weakly to ϕ provided

f (X)ϕn (dM ) = f (X)ϕ(dM ) lim n→∞

E

E

for every continuous function f (X) on E, with the integral understood in the Radon sense.49 The given functions F (M ) and ω(M ) should be approximated, in the sense of weak convergence, by “discrete” functions Fn (M ) and 49

This notion of convergence, corresponding to the notion of weak convergence of functionals, can be characterized by simple geometric properties. For example, if ϕn (M ) ≥ 0 for all M (which is the case for our functions F and ω), then the functions ϕn converge weakly to ϕ if and only if lim ϕn (E) = ϕ(E)

n→∞

and

lim ϕn (M ) ≤ ϕ(M )

n→∞

for every closed M . Proofs of this and other characteristics of weak convergence for set functions appear in my article [A10], see the résumé and Sections 15–17. Proof of the theorem on weak convergence of the functions F and ω for convergent convex surfaces is given in the articles [A5] and [A6], also see the book [A15].

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ωn (M ) with finite weights at specified points and vanishing for all M not containing any of these points. The method of passing to the limit from polyhedra to other surfaces was first used by Minkowski in his proof of the existence closed convex surfaces Φ of given Gaussian curvature K(n) as a function of the outward normal. If M is a subset of the sphere E, then

dω , F (M ) = K(n) M where dω is the area element on E, is exactly the area of that subset of Φ whose spherical image is M . Consequently, the theorem of Minkowski mentioned above is a particular instance of the general existence theorem for a closed convex surface given the area function F (M ).

2.8 Basic Topological Notions 2.8.1 Intuitive background. Topology can roughly be defined as the part of general geometry where, when studying a figure, we only take account of the mutual adherence of parts of the figure. The general concept of “adherence” is based on the intuitive idea which lies behind the following propositions: the circle adheres to the interior of the disk; one hemisphere adheres to the other, so that the sphere may be regarded as composed of two adhering hemispheres; however, if we delete the equator, then the hemispheres no longer adhere to one another: they appear to be cut off from one another by a slit along the equator. The author has preferred to introduce the topological structure via adherence points, guided by the perfect clarity of the main definitions – of a continuous mapping, boundary, connectedness, etc. – in this approach. Let us try to analyze our intuitive idea of adherence. Every geometric figure is a set or, as one says in elementary geometry, a “set of points”; and in this set there is a visually clear notion of adherence of parts to each other. The principal concept is that of adherence of a point to a set. In our intuitive understanding, a point adheres to a set if it is infinitely close to the set, as for instance a vertex of a square to the interior of the square. If a point belongs to a set, then we also consider it as adherent. Thinking further about the notion of adherence, we see that two sets, two figures or two parts of one figure, adhere to one another if and only if at least one of them contains a point adhering to the other. Looking at the above examples, we readily convince ourselves that it is in this sense that the circle adheres to the interior of a disk and that two hemispheres with deleted equator do not adhere to one another (since each point of one hemisphere is at a positive distance from the points of the other hemisphere not less than the distance from the point to the equator.)

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2.8.2 General topological spaces. The abstract generalization of the intuitive ideas described above leads to the notion of general topological space. Namely, assume given some set R; we call the elements of R points. The set R is a general topological space if to each part of R, i.e., to each subset M of R, we assign, according to some rule, points adherent to M ; these points are also referred to as adherent points of M , the points of M being also considered as adherent points of M . Also, we add the following natural condition: when a set increases, the number of adherent points does not decrease, i.e., speaking rigorously, if M1 is contained in M2 , then each adherent point of M1 is an adherent point of M2 . We say that two sets in a general topological space adhere to one another if at least one of them contains adherent points of the other. Clearly, these definitions simply repeat, in slightly different words, what was explained above about the adherence of figures and points. The basic example of a general topological space is the n-dimensional “arithmetical” space or Euclidean space.50 The points of this space are all the n-tuples (x1 , x2 , . . . , xn ) of real numbers. A point a = (a1 , a2 , . . . , an ) is an adherent point of a set A if there are points in A arbitrarily close to a, i.e., for every ε > 0, the set A contains a point x = (x1 , x2 , . . . , xn ) such that |x1 − a1 | < ε, . . . , |x1 − a1 | < ε. In that case, either a belongs to the set A or A contains a sequence of points xi that converge to a, i.e., the differences between the coordinates of the points xi and the point a tend to zero. The real axis is nothing but the one-dimensional arithmetical space. Similarly, the plane is the two-dimensional arithmetical space. If we impose the condition that all the numbers xi lie within given limits, for example 0 < xi < 1 (i = 1, . . . , n), then the resulting space is called an ndimensional (arithmetical) cube; more precisely, the open cube, as opposed to the closed cube which contains all its faces and is determined by the inequalities 0 ≤ xi ≤ 1. Three-dimensional Euclidean space is also a topological space, with the following definition of adherent points: a point a is an adherent point of a set A if A contains points that are arbitrarily close to a. An arbitrary development also provides an example of a general topological space, with the agreement that two points that will be glued are treated as the same point. If M is a subset of a development, then a point a of the development is adherent to M if there are points of M arbitrarily close to a or to some point that will be glued to a. Each subset A of a general topological space R may itself be treated as a general topological space, by using the following definition: if M is a subset of A, then a point a of A is an adherent point of M relative to A if a is adherent to M in R. We say that A is a subspace of R. 50

We call the space Euclidean, abstracting however from all of its properties except the topological ones.


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In this sense, every figure A in Euclidean space is a general topological space. A point a of A is adherent to a set M contained in A if there are points of M at an arbitrarily small distance from a. In the same way, every part of n-dimensional arithmetical space is a general topological space. 2.8.3 Closed sets and open sets. Boundary. Here we discuss sets in an arbitrary topological space. A set is closed if it contains all points adherent to it. A set is open if its complement (i.e., the set of all points not belonging to it) is closed. This definition is equivalent to the following: a set M is open if no point of M is an adherent point of the complement of M (since in that case the complement of M itself contains all its adherent points, i.e., the complement is closed). A point is a boundary point of a set M if it is adherent to M as well as to the complement of M . The boundary of a set M is the set of all boundary points of M . A point of a set is called interior if it does not lie on the boundary of the set, i.e., it is not adherent to the complement of the set. Clearly, the definition implies that an open set is characterized by the condition that it consists only of interior points. Examples. A half-space with its bounding plane is a closed set; without the plane, it is open; the plane is the boundary of the half-space. The cube in n-dimensional space is defined by the inequalities 0 ≤ xi ≤ 1 (i = 1, . . . , n). The interior of the cube consists of the points for which 0 < xi < 1. The boundary of the cube consists of all the points for which 0 ≤ xi ≤ 1 (i = 1, . . . , n) and at least once we have the equal sign. A neighborhood of a point means, as a rule, an arbitrary open set containing the point. Thus a neighborhood surrounds a point in such a way that the point is not adherent to the complementary set. This corresponds to the intuitive meaning of the term “neighborhood.” 2.8.4 Connectedness. A set is connected if it never splits into two parts that do not adhere to one another. By the definition of adherence, this means that the set never splits into two parts neither of which contains adherence points of the other. If the set M itself is considered as a space (a subspace of the space under study), then relatively closed sets and relatively open sets are naturally defined in M ; briefly, they are the closed sets and the open sets in M . The following theorem holds: A set M is connected if and only if M never splits into two nonempty parts which are both closed and open in M .

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Indeed, suppose that M is disconnected, i.e., M splits into two parts M1 and M2 neither of which contains adherent points of the other. Then M1 and M2 themselves contain the points that are adherent to them in M . Hence, M1 and M2 are closed. Since they are complementary to each other, they are also open in M . Now, suppose that M splits into two parts M and M closed in M . This means that M and M themselves contain their adherent points in M , i.e., M and M do not adhere to one another, and M is therefore disconnected. Since M and M are complementary to each other with respect to M , the fact that M is closed implies that M is open, and M is open for a similar reason. Consequently, splitting into two closed sets is also splitting into open sets. The converse is proved in exactly the same manner: splitting into open sets is also splitting into closed sets. This completes the proof of the theorem. A subset M1 of a set M is a connected component of M if M1 is connected and is not contained in any connected subset of M but for M1 itself. Any set is connected or is the union of disjoint connected components. The proof is based on the following theorem: If connected sets Mξ have at least one common point a, then their union is a connected set. First, we prove the latter assertion. Denote the union of the sets Mξ by M . Were M disconnected, it would split into parts M and M which do not adhere to one another. Since the union of M and M is M , each set Mξ either lies entirely in one of these sets M and M or has common points with both the sets. The second possibility is excluded. Indeed, if some Mξ contains two pieces Mξ and Mξ which belong to M and M , the pieces Mξ and Mξ cannot adhere to one another since M and M do not. As a result, Mξ would be disconnected, contradicting the assumption. Since the second possibility is excluded, each Mξ lies entirely in M or M . Because all the sets Mξ have a common point, M and M would also have a common point, i.e., would adhere to one another. This contradiction proves that M never splits into parts M and M which do not adhere to one another. Hence, M is connected. Now, we prove the former assertion on the decomposition into connected components. A one point set is connected, because it never splits at all. Take some point x of a given set M and consider the union of all connected sets containing x and contained in M . There are such sets, for the point x itself is one of them. This union, denote it by M (x), is connected by virtue of the theorem just proved above. By construction, M (x) is not a proper subset of any connected set; hence, it is a connected component. If M (y) is a set constructed similarly for another point y, then M (y) is connected also. If M (y) has common points with M (x), then the union M (x) ∪ M (y) is connected by the same theorem. This union contains x and


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therefore must be contained in M (x), since by construction M (x) includes all connected sets that contain the point x. The same holds for M (y) as well. Therefore, if the sets M (x) and M (y) have common points, then they coincide. Defining the sets M (x) for all points x in M , we see that every two of these sets either coincide or are disjoint. The set M thus splits into disjoint connected components M (x), as claimed. We also indicate the following two propositions: 1. If, for every two points in a set M , there is a connected subset of M containing the points, then the set M itself is connected. 2. Every two points in a connected open set of Euclidean space (or in ndimensional arithmetical space) can be joined by a polygonal line contained in the set. These propositions will not be used in the sequel and we leave their proofs to the reader. 2.8.5 Mappings. Let X and Y be two sets in the same space or in two different spaces. The assignment to each point x in X of some point y in Y is a mapping from X into Y . Denoting the mapping by ϕ, we write y = ϕ(x). To each part X of X there corresponds a part Y of Y , i.e., Y = ϕ(X ). The set Y is the image of X and X is a preimage of Y . A function y = f (x) of a real variable x, defined on an interval X, is nothing but a mapping from the interval into the real axis. Similarly, a function of n variables is a mapping from a set of n-dimensional space into the real axis. A mapping is continuous if it does not violate the adherence relation (every violation is a discontinuity). In other words, a mapping ϕ is continuous if it preserves the adherence of points, i.e., if x is an adherent point of X , then ϕ(x) is an adherent point of ϕ(X ). In an arithmetical space, a point x adheres to X if in X there are points x whose coordinates x1 , . . . , xn are arbitrarily close to the coordinates of the point x. This is equivalent to the fact that in X there is a sequence of (k) points x(1) , x(2) , . . . converging to x in the sense that the differences xi − xi (between the coordinates of the points x(k) and of the point x) tend to zero. Therefore, a mapping ϕ from one arithmetical space into another is continuous if and only if x(k) → x implies that ϕ(x(k) ) → ϕ(x), which is in perfect agreement with the usual definition of the continuity of a function. A mapping ϕ is said to be one-to-one, or injective, if not only to each point x ϕ assigns a unique point y = ϕ(x), but also to different x’s it assigns different y’s. In other words, to each y there may correspond only one x. In this case, the inverse mapping ϕ−1 associating x’s with y’s is single-valued. If a mapping ϕ and its inverse ϕ−1 are continuous, then ϕ is said to be continuous in both directions.

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A one-to-one mapping continuous in both directions is called a homeomorphism. If a set X admits a homeomorphism onto a set Y , then X is homeomorphic to Y . Since any homeomorphism is one-to-one and continuous in both directions, both sets X and Y play the same role here, and so we say that they are homeomorphic to each other. The one-to-one property of a mapping means that points under such a mapping neither split nor merge together. Continuity in both directions means that adherences neither disappear nor appear. Hence, from the standpoint of adherences alone (abstracting from other properties of elements and sets) two homeomorphic sets have the same structure. They are absolutely the same topologically. All topological conclusions, i.e., conclusions relying only upon the adherence relation and valid for one of the sets are also valid for the other. This is quite similar to the fact that two congruent figures (i.e., figures admitting a distance-preserving mapping between their points) are geometrically the same. The geometric properties based on the concept of length are the same for congruent figures; similarly, the topological properties based on the concept of adherence are the same for homeomorphic sets. 2.8.6 Manifolds. By an n-dimensional manifold we mean a topological space in which each point has a neighborhood homeomorphic to the interior of an n-dimensional cube and in which any two points have disjoint neighborhoods.51 In other words, a manifold is a space each of whose points has a neighborhood into which coordinates x1 , . . . , xn that vary in some interval may be introduced, say a − ε < xi < a + ε (i = 1, 2, . . . , n), and the correspondence between the points of the neighborhood and their coordinates is one-to-one, onto, and continuous in both directions. We also assume that any two points possess nonintersecting neighborhoods. Every open set of n-dimensional arithmetical space is an n-dimensional manifold. Indeed, coordinates have already been introduced in such a space. An open set is characterized by the requirement that none of its points is an adherent point of the complement. This means that all points whose coordinates are close to the coordinates of a point a of the set are also contained in the set. That is, along with any point a(a1 , . . . , an ) the set contains all points x with coordinates x1 , x2 , . . . , xn such that |ai − xi | < ε, or a − ε < xi < a + ε (i = 1, . . . , n). By definition, this means that the set is an n-dimensional manifold. An arbitrary development without boundary provides an example of a two-dimensional manifold. Each point of the development has a neighborhood homeomorphic to a square or, which is equivalent, to a disk. This is obvious for points inside the polygons of the development. A neighborhood of a point on an edge is the union of two “half-neighborhoods” on polygons that 51

It is connectedness and the existence of a covering of the space by at most countably many neighborhoods each homeomorphic to the interior of a cube (as well as partitioning into simplices) which is usually included in the notion of manifold. We do not require these conditions.

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are to be glued together along the edge. Finally, a neighborhood of a vertex is the union of angular sectors touching at the vertex in accordance to the “rules for gluing” of the polygons. By gluing, i.e., identifying, we can build three-dimensional manifolds. For example, take a cube and identify the symmetric points of its opposite faces. Some three-dimensional manifold results. This method applies to constructing n-dimensional manifolds: they originate by gluing from “developments” formed by n-dimensional polyhedra. The topological structure of such a manifold is determined by the law for gluing of the development. Compare this with Euler’s Theorem expressing the characteristic property of the law for gluing when the sphere is glued together from a development. In our book, manifolds will not be defined by means of such a construction, but in accordance with the general definition, by introducing coordinates in a neighborhood of each element (point of the manifold). For example, all closed convex polyhedra with v vertices from a 3v-dimensional manifold. Each vertex is determined by three coordinates and all v vertices, by 3v coordinates. A closed convex polyhedron is determined by its vertices, and hence by the same 3v coordinates. Given a polyhedron P0 , sufficiently small shifts of its vertices never take one of the vertices into the convex hull of the others. Therefore, shifted vertices also span a polyhedron P with v vertices. Consequently, the coordinates of vertices may be changed in a neighborhood of the point corresponding to the polyhedron P0 , and thus the “neighborhood” of the polyhedron is mapped onto a 3v-dimensional cube. The adherence relation in this manifold of polyhedra is defined in the natural way. A polyhedron P0 “adheres to” a set of polyhedra P if, among the latter, there are polyhedra whose vertices are arbitrarily close to the vertices of P0 . Manifolds defined in a similar manner will be used in the proofs of existence theorems in Chapters 4, 7, and 9.

2.9 The Domain Invariance Theorem 2.9.1 We prove the following theorem, on which the Mapping Lemma is based: Under a homeomorphic mapping of an n-dimensional manifold R1 onto a subset of an n-dimensional manifold R2 , the image of every open set in R1 is an open set in R2 .52 52

This theorem might seem trivial, since a homeomorphic mapping of one space onto another always takes open sets into open sets by the definition of homeomorphism. However, what was said above implies only that the image of an open set is open relative to the image of the space R1 in R2 . For example, under the identical

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An open set is characterized by the requirement that each of its points is interior, i.e., may be surrounded by a neighborhood homeomorphic to an n-dimensional ball and contained in the set under consideration. For such a neighborhood, we can take a convex n-dimensional polyhedron with the least possible number of vertices. Such a polyhedron is called an ndimensional simplex. It has n + 1 vertices. For n = 0, 1, 2, 3 simplices are respectively the point, the segment, the triangle, and the tetrahedron. Each (n − 1)-dimensional face of an n-dimensional simplex is obviously an (n − 1)dimensional simplex. Obviously, it suffices to prove that: Under a homeomorphic mapping ϕ of an n-dimensional simplex T onto a subset of n-dimensional Euclidean space E, every interior point p of T goes to an interior point of the set Φ = ϕ(T ) in E. The proof consists in characterizing the interior points of a simplex by a property which is shown to be invariant under homeomorphisms. First of all, we introduce the notion of a closed covering of a closed set: a finite collection of closed sets A1 , A2 , . . . , As is a closed covering of a closed set Φ if the union of the sets Ai is the set Φ. A number k is the multiplicity of a covering if there exists a point of the set Φ which simultaneously belongs to k sets Ai and, at the same time, none of the points of Φ simultaneously belongs to a greater number of sets Ai . If the diameters53 of all sets Ai are all less than a number ε > 0, then we are dealing with an ε-covering. By using these notions, we give a topologically invariant characteristic for interior points of a closed set in n-dimensional Euclidean space: For a point p of a closed set Φ in n-dimensional Euclidean space to be an interior point of Φ, it is necessary and sufficient that there exist a closed covering α = { A1 , . . . , As } of Φ of multiplicity n + 1 such that: (1) p is the only point of Φ belonging to n + 1 sets Ai ; (2) every covering α of Φ differing from α only in a sufficiently small neighborhood U (p) of p54 is of multiplicity mapping of a straight line into a plane, the image of the straight line obviously is not an open set relative to the plane, although it is open relative to itself. The assertion that we are going to prove consists in the fact that under the conditions of the theorem the image of every open set is an open set open relative to the whole space R2 . It is easy to verify that this assertion may be rephrased in the following simpler form: the homeomorphic image of a manifold R1 in a manifold R2 has no boundary. 53 The diameter of a set is the least upper bound of the distance between two points of the set. The distance between two points is understood, as usual, to be the square root of the sum of squares of the differences between the coordinates of the points. 54 Let a covering α consist of the sets A1 , . . . , As and a covering α , of the sets A1 , . . . , At , with s = t in the general case. Denote by Ai (Ai ) the part of the set Ai (Ai ) which has no common points with U (p). We say that α differs from α only in U (p), provided that each Ai coincides with some Ai and vice versa.


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no less than n + 1 (here by U (p) we mean a neighborhood of p relative to Φ, i.e., the intersection of a neighborhood of p in n-dimensional Euclidean space with the set Φ. The assertion is equivalent to the theorem that we are proving. To verify this, it suffices to demonstrate that all the above conditions characterizing interior points of an n-dimensional simplex are invariant under homeomorphisms. It is readily seen that the notions of multiplicity of a covering and of a “sufficiently small” neighborhood U (p) are topologically invariant. It remains to show that any homeomorphism takes a closed covering of a simplex T to a closed covering of the image Φ = ϕ(T ) of T and vice versa. Since the elements of a closed covering of a simplex, as well as the simplex itself, are closed and bounded sets of Euclidean space, we need only prove the following assertion: the homeomorphic image of a closed and bounded set in n-dimensional Euclidean space is closed and bounded. The proof is very simple: the boundedness of the image is immediate from the Weierstrass Theorem claiming that a continuous function given on a closed and bounded set is bounded above and below. The fact that the image is closed follows from the Bolzano–Weierstrass Theorem. (Let m be an accumulation point55 of the image. To a sequence of points converging to m in the image corresponds an infinite sequence of points in the preimage, and this sequence has an accumulation point by the Bolzano-Weierstrass Theorem. By the continuity of homeomorphisms, this accumulation point is mapped into the point m, and therefore the latter belongs to the image. Thus, the image contains all its accumulation points and hence is closed.) 2.9.2 Returning to the property of interior points stated above, we first prove that each noninterior point does not to possess this property. To this end, we need the following lemma: Lemma 1. For an arbitrarily small ε > 0, there exists a closed ε-covering of the boundary T of T of multiplicity not exceeding n. We begin the proof by indicating an ε-covering of multiplicity n+1 (with ε arbitrarily small) for n-dimensional space E. For n = 1, the required covering is constructed by partitioning the real axis into equal intervals. For n > 1, the required covering is constructed by induction: n-dimensional space is partitioned into sufficiently thin parallel layers by (n− 1)-dimensional planes. Let us partition one of these planes into equal (n − 1)-dimensional cubes in the manner provided by the induction hypothesis and orthogonally project this partition onto the other (n − 1)-dimensional planes. Then each layer is partitioned into n-dimensional parallelepipeds. Shifting neighboring layers in parallel, we reach a position in which each vertex of every parallelepiped lies inside an (n − 1)-dimensional face of a parallelepiped of the neighboring layer 55

A point m is an accumulation point of a set M if each neighborhood of m contains infinitely many points of M other than m.

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(the case n = 2 is shown in Fig. 72). Then, obviously, the multiplicity of the covering of the space by such parallelepipeds is greater by 1 than the multiplicity of the covering of the (n − 1)-dimensional plane and thus equals n + 1.

Fig. 72

Now, let T be an arbitrary n-dimensional simplex in E. Consider the constructed ε-covering of E. It is seen from its construction that the distances between the points of E simultaneously belonging to n + 1 cubes of the covering are greater than some positive number δ. Suppose that there are such points on the boundary T of T . Choose a direction in E which is not parallel to any of the (n − 1)-dimensional faces of T and perform a small parallel shift of T in this direction so that none of the points of E belonging to n + 1 cubes of the partition and not contained in T is not shifted into T. (Points not lying on T while belonging to n + 1 cubes of the covering cannot be arbitrarily close to T; otherwise, they would have an accumulation point and hence, among them, there would exist points arbitrarily close to each other.) Moreover, since the direction of the shift is chosen so that each point that belonged to T is shifted outside T, after the shift there are no points on T that can possibly belong to n + 1 cubes of the covering of E. But then the intersections of the cubes of the covering with T, being closed sets (because the intersection of any number of closed sets is itself closed) of diameter < ε, form the required covering of the boundary T, completing the proof of the lemma. Suppose that p is not an interior point of Φ. Consider a closed covering α = { A1 , . . . , As } of multiplicity n + 1 of Φ. Assume that p is the only point of Φ belonging to n + 1 sets Ai ; let us show that we can construct another closed covering α of multiplicity n which coincides with α everywhere except in an arbitrarily small neighborhood U (p) of the point p (here U (p) is a neighborhood relative to Φ). Construct an n-dimensional simplex T whose interior contains the point p and whose intersection with Φ is contained in U (p). Since p is the only point of Φ belonging to n + 1 sets Ai , none of the points of the boundary T of T belongs to more than n sets Ai . It follows that, for a sufficiently small ε > 0, each set M of diameter < ε which lies in T has common points with at most n sets Ai . Indeed, if this is not so, then we can choose a sequence of positive numbers εj (j = 1, 2, . . .) converging to zero and, for each εj ,


151

find a set Mj on T, of diameter < εj , having common points with n + 1 sets Ai1 , . . . , Ain+1 . Since finitely many sets Ai may form only finitely many different combinations of n + 1 sets, we can choose n + 1 sets A1 , . . . , An+1 for which there are arbitrarily large indices j such that all the sets A1 , . . . , An+1 have common points with the sets Mj , and consequently for each such index j there exists a point bj that belongs to T and whose distance from each of the sets A1 , . . . , An+1 does not exceed εj .56 Since all the points bj lie in a bounded part of the space (namely, in T), the Bolzano-Weierstrass Theorem implies that the set of points bj has an accumulation point b in T. In an arbitrarily small neighborhood of b, there are points bj having arbitrarily large indices and therefore arbitrarily close to each of the sets A1 , . . . , An+1 . Hence, b adheres to each of the sets A1 , . . . , An+1 . Since these sets are closed, b belongs simultaneously to all of them, which is excluded by assumption. Thus, we can choose an ε > 0 such that the subsets of T of diameter less than ε cannot simultaneously intersect more than n sets Ai . Using Lemma 1, let us construct an 2ε -covering β = { B1 , . . . , Bm } of the boundary T of the simplex T . Since p is not an interior point of Φ, there is a point O inside T which does not belong to Φ. Consider the join of the set Bj (j = 1, . . . , m) with the point O, i.e., the set Bj which contains the corresponding set Bj and all the straight line segments (with their endpoints included) that connect the points of the set Bj to the point O.

Φ p Ai O Bj Bj

T

Fig. 73

Now, we construct a closed covering αβ of the set Φ and the simplex T as follows: first, we replace each set Ai by the part of Ai not containing interior points of the simplex T (if Ai lies entirely inside T , then we simply exclude it); obviously, each Ai is thus replaced by some closed set Ai . We replace the part of the covering inside the simplex T by the covering of T by the “pyramids” Bj (Fig. 73). Next, we “glue” the pyramids Bj to the sets Ai according to the following rule: if Bj has common points with some sets Ai , then we glue Bj to one and only one of these sets; if Bj has common points 56

The distance from a point to a set is the greatest lower bound for the distances from the point to the points of the set.

152


with none of the sets Ai , then we leave Bj without changes. Afterwards, we define the covering αβ as follows: the elements of the covering are the sets obtained from the sets Aj by taking the union of Aj and all the sets Bj glued to Aj (if nothing is glued to Aj , then Aj is included in the covering αβ without changes) and also the sets Bj that are not glued at all (Fig. 73). We assert that each point other than O belongs to at most n sets of the covering αβ . Indeed, an arbitrary point lying beyond T can belong to n+1 sets in αβ only if it belongs to n + 1 corresponding sets in α, which is impossible since the point p lies inside T . Indeed if such a point b existed inside T , its projection from O to T would simultaneously belong to n + 1 sets Bj , which cannot possibly occur. It remains to consider the points of T. Let a be such a point. Since the diameter of each set Bj is less than ε/2, the diameter of the union of all sets Bj containing the point a (by construction, there are at most n such sets) is less than ε. Therefore, this union has common points with at most n sets Ai . Thus, the number of pyramids Bj containing the point a and the number of sets Aj to which the pyramids can be glued in constructing the covering αβ is less than n. But since each pyramid Bj is glued to at most one set Aj , the number of sets of the covering αβ which contain the point a is also at most n. Now, replacing each set of the covering αβ by its intersection with Φ, we obtain a covering α of Φ which differs from the original covering α only inside U (p) and is of multiplicity not exceeding n (since the point O lies outside of Φ and is excluded from our considerations when we pass from αβ to α ). We have thus established that our test for interior points is sufficient, since it obviously fails for points that are not interior. It remains to prove that the test is necessary, i.e., it holds for every interior point of the set Φ. 2.9.3 So, let p be an interior point. Take a simplex T in Φ whose interior contains the point p. Draw the (n − 1)-dimensional planes through p parallel to the (n − 1)-dimensional faces of T .

Ai

p

Fig. 74

T Φ


153

Each such plane divides the space, and the set Φ as well, into two parts. One of these parts contains exactly one vertex of the simplex T , while the other contains all the remaining vertices. Denote by A1 , . . . , An+1 those parts which contain exactly one vertex of the simplex. These sets obviously form a closed covering α of the set Φ and p is the only point of Φ which belongs to n + 1 sets of the covering (Fig. 74). Let us prove that if a neighborhood U (p) of p is so small that it is disjoint from the boundary of the simplex T , then it is impossible to obtain a covering of lesser multiplicity by modifying α in U (p). Suppose that the covering α has been changed in U (p) somehow. In the general case, the change consists in replacing the sets Ai by other sets coinciding with the former outside U (p) and, possibly, in adding new sets that lie entirely in U (p). Adjoin each of these “new” sets to one of the deformed i . We now show that sets Ai and denote the sets obtained in this way by A i . To this end, we there is a point simultaneously belonging to all the sets A make use of the following lemma: The Sperner Lemma. Let K be an arbitrary triangulation57 of an ndimensional simplex T with vertices e1 , e2 , . . . , en+1 . Suppose that to each vertex ek of the triangulation K we assign a vertex S(ek ) of the simplex T which lies on a face of T containing ek and having the least possible dimension (the simplex itself is its own n-dimensional face; therefore, to an interior point ek there may correspond an arbitrary vertex of the simplex). Then there is an n-dimensional simplex Ti of K whose vertices are mapped to different (and hence all) vertices of the simplex T . A simplex Ti whose vertices are mapped into different vertices of the simplex T is said to be normal. The lemma will be proved, if we establish that the number of normal simplices is odd. We prove this by induction on n. For n = 0 the assertion is trivial. Assume it valid for n−1. Take the face |e1 . . . en | of T with vertices e1 , . . . , en . Given a simplex Ti , an (n − 1)-dimensional face of Ti whose vertices are mapped into e1 , . . . , en will be called distinguished. Obviously, if a simplex Ti has distinguished faces but is not normal, then Ti has exactly two such faces. For this reason, were the number of normal simplices even, the total number of distinguished faces of all simplices Ti would be even as well. Since each distinguished face lying inside T belongs to two simplices, the total number of such faces is also even. It follows that the number of distinguished faces lying on (n − 1)-dimensional faces of the simplex T would be even, too. By the definition of the mapping S, such faces may only lie on faces with vertices e1 , . . . , en . However, this is impossible by the induction hypothesis, since the distinguished faces lying on the face 57

A triangulation of a simplex T is a partition of T into simplices Ti such that the intersection of two simplices Ti and Tj is a face (of any dimension) of each of them, or the empty set. The vertices of the simplices Ti are called vertices of the triangulation.

154


|e1 , . . . , en | play the role of normal simplices under the restriction of the mapping S to the face |e1 , . . . , en |. This completes the proof of the lemma. i have a common point. To this end, take an We now prove that the sets A arbitrary triangulation K of our simplex T . Define some mapping from the vertices of the triangulation K to the vertices of the simplex T , satisfying the condition that a vertex of T and the corresponding vertex of the triangulai containing a given i . By construction, the set A tion belong to the same set A vertex has no points in common with the opposite face; therefore, the mapping S satisfies the conditions of the Sperner Lemma and consequently there is a simplex whose vertices, being mapped into different vertices of T , belong n+1 (because each vertex of T belongs to exactly one 1 , . . . , A to all the sets A set Ai ). Since the triangulation K can be arbitrarily fine, from the fact that i are closed we easily infer that they have a common point. Howthe sets A i , we can only diminish the multiplicity of the ever, by passing to the sets A covering. Therefore, the original covering obtained from α by an arbitrary change in the neighborhood U (p) has multiplicity ≥ n + 1. This concludes the proof of the theorem.

3 Uniqueness of Polyhedra with Prescribed Development

3.1 Several Lemmas on Polyhedral Angles 3.1.1 Proofs of uniqueness theorems for convex polyhedra with prescribed development are based on the Cauchy Lemma (Section 2.1 of Chapter 2). To use the latter, we must, however, establish some lemmas on convex polyhedral angles. If we draw a unit sphere around the vertex of a polyhedral angle V , then the angle V cuts out a polygon V from the sphere; the sides and angles of V are equal to the planar and dihedral angles of V respectively. If the polyhedral angle V is convex, then so is the spherical polygon V , i.e., V lies on one side of every great circle containing a side of V . Conversely, given a convex spherical polygon one obtains a convex polyhedral angle by projecting from the center of the sphere. For this reason, we shall consider convex spherical polygons rather than polyhedral angles. We admit angles that are equal to π, not treating the two sides forming such an angle as a single side. This amounts to the presence of dihedral angles equal to π in a polyhedral angle. At the same time, we exclude spherical polygons with exactly two angles other than π (i.e., those representing digons) from our consideration everywhere except for Lemma 1. This amounts to excluding the degenerate polyhedral angles which are in fact dihedral angles. Further in this section, we often call a spherical polygon simply a polygon. (This is all the more permissible, since the results that will be obtained are also valid for planar polygons.) 3.1.2 Lemma 1. If two convex spherical polygons P and P1 touch along an edge AM and the sums of their angles at each of the vertices A and M do not exceed π, then P and P1 jointly form a convex polygon P ∪ P1 . Let AB and M L be the edges of P adjacent to AM (Fig. 75). Each of the three great circles AM , AB, and M L determines some hemisphere that contains the polygon P . The intersection of these hemispheres is a spherical triangle T that contains P . Similarly, the polygon P1 is contained in the triangle T1 bounded by the side AM and the prolongations of the sides AB1 and M L1 . Since the sums of the angles of the triangles T and T1 at each of

156


the vertices A and M do not exceed π, the triangle T1 is in turn contained in the triangle T2 bounded by the side AM and the prolongations of the sides AB and M L. The triangles T and T2 jointly form the digon T ∪ T2 which contains the polygon P ∪ P1 . T B P A

L M

P1 B1

T 1 L1 T2

Fig. 75

We now draw a great circle through some edge of P other than AM . Since the polygon P is convex, P lies to one side of this great circle. This great circle has a common point with at least one side (other than AM ) of the triangle T . The circle either goes along one of the sides or crosses both. Therefore, it cannot enter into T2 . Otherwise, it would meet each side of T2 (other than AM ) and we would obtain a great circle that meets at least one side of the digon T ∪ T2 twice, which is impossible because a great circle may meet a great half-circle (a side of the digon T ∪ T2 ) only at a single point. Hence, the triangle T2 and, all the more, the polygon P1 , lie on one side of such a great circle; namely, on the same side as the polygon P . For the same reasons, the polygon P lies on the side of the great circle containing an arbitrary edge of P1 other than AM . Consequently, the polygon P ∪ P1 lies on one side of every great circle passing through its side, which means that it is convex. The lemma implies the following theorem: Theorem 1. If none of the angles of a spherical polygon exceeds π, then the polygon is convex. The proof proceeds by induction on the number of angles. For a triangle, the assertion is trivial. Assume the assertion valid for polygons whose number of angles is less than n (n > 3) and prove it for an n-gon. Given an n-gon P , partition P by a diagonal into two polygons P1 and P2 .1 If every angle of P is less than or equal to π, then so is every angle 1

It is simple to prove the possibility of partitioning a polygon with angles less than π by a diagonal. For instance, we can argue as in the proof of a similar assertion

3.1 Several Lemmas on Polyhedral Angles

157

of P1 and P2 . At the same time, each of the polygons P1 and P2 has less than n angles. Therefore, P1 and P2 are convex polygons by the induction hypothesis. However, the polygon P is such that the sum of the two angles at every common vertex of P1 and P2 does not exceed π, since this sum represents an angle of P itself. But then the polygon P is convex by Lemma 1, which is the desired conclusion. Ak-1

Ak-1

Ak Ak

P

A2

P

A2

Ak+1

Ak+1 A1

A1

An

An

Fig. 76

3.1.3 Theorem 2. Let P and P be two convex polygons with the same number of vertices A1 , A2 , . . . , An and A1 , A2 , . . . , An (Fig. 76). Suppose that the following conditions are satisfied: (1) all their corresponding sides except An A1 and An A1 are of equal length, i.e., (1) A1 A2 = A1 A2 , . . . , An−1 An = An−1 An ; (2) the angles between these sides in the first polygon do not exceed those in the second, i.e., ∠A2 ≤ ∠A2 , . . . , ∠An−1 ≤ ∠An−1 ,

(2)

and the strict inequality holds at least once. Then the “exceptional” side of the first polygon is less than the “exceptional” side of the second, i.e., An A1 < An A1 . (3) Stated somewhat differently and possibly in a more visual form this theorem reads: If a convex polygon P is obtained from a convex polygon P by a deformation that preserves the lengths of all sides except an “exceptional” one and increases the angles between these sides (or only decreases none of the angles for polygons in a development. We tacitly assume that the polygon P is simple, bounded by a single polygonal line; this suffices. A ring-shaped polygon may fail to be partitioned by a diagonal; however, we can partition it by several diagonals and proceed with the subsequent arguments based on Lemma 1.

158


and increases at least one of them), then the “exceptional” side lengthens so that it becomes longer in P than in P . We conduct the proof by induction on the number of vertices in the polygons. For triangles, the theorem amounts to the assertion that if two sides of one triangle are equal to two sides of another triangle (A1 A2 = A1 A2 and A2 A3 = A2 A3 ) and the angles between these sides differ (∠A2 < ∠A2 ) then the third side is shorter in the triangle in which the angle is smaller (A3 A1 < A3 A1 ). For planar triangles, this familiar theorem is proved in every textbook on elementary geometry. The same proof applies to spherical triangles word for word. Thus, our theorem is true for triangles. Supposing it is valid for (n − 1)gons, we now exhibit a proof for n-gons (n > 3). Let n-gons P and P satisfy the conditions of the theorem. There are two possibilities: (1) all angles A2 , . . . , An−1 are strictly less than the angles A2 , . . . , An−1 ; (2) among those angles, some are equal, say ∠Ak = ∠Ak . We begin by examining the second case. To this end, we draw the diagonals Ak−1 Ak+1 and Ak−1 Ak+1 and cut off the triangles T = Ak−1 Ak Ak+1 and T = Ak−1 Ak Ak+1 from our polygons. These triangles are obviously congruent (since their angles Ak and Ak are equal by assumption, while the sides forming these angles are equal by the conditions of the theorem). It follows that the polygons Q and Q that remain after cutting off the triangles enjoy the same properties as the original polygons P and P . Indeed, the sides Ak−1 Ak+1 and Ak−1 Ak+1 of Q and Q are equal because the triangles T and T are congruent. The angles at the vertices Ak−1 and Ak+1 in the polygon Q are not greater than the angles at the corresponding vertices in the polygon Q , since these angles are obtained from the angles of P and P by subtracting equal angles of the triangles T and T . All the remaining angles and sides in Q and Q are the same as in P and P and hence have the same ratios. Thus, the polygons Q and Q meet the same conditions but have one vertex less. Hence, the theorem is true for them by the induction hypothesis. Consequently, An A1 < An A1 , which was to be proved. Now, let us examine the case in which the angles A2 , . . . , An−1 of the polygon P are strictly less than the corresponding angles of the polygon P . Take some vertex Ak of P not lying on the prolongation of the side A1 An (since we admit angles that equal π, it might occur that the vertex A2 or An−1 lies on the prolongation of the side A1 An ). Let us construct the triangle T = Ak A1 An (Fig. 77).


159

Ak

R Q

T

A1

An

Fig. 77

In general, the polygon P splits into three parts: the triangle T and the polygons Q and R touching T at the sides Ak A1 and Ak An . (One of the polygons Q and R may degenerate into a segment.) Start deforming T continuously so that the angle Ak increases, while the lengths of the sides Ak A1 and Ak An remain the same. Then the theorem on triangles mentioned implies that the side A1 An lengthens. After the deformation, we shall thus have A1 An < A1 An .

(3 )

When we move apart the sides Ak A1 and Ak An , we also move the polygons Q and R containing these sides. As a result, the whole polygon P = Q ∪ T ∪ R is deformed so that only its angle Ak increases. Obviously, we can increase the angle Ak to any value ≤ π. In particular, we can make it equal to the angle Ak of the polygon P . If the polygon P resulting from this deformation of P is convex, then the same arguments apply to P . However, now P has the angle Ak equal to the angle Ak of P , i.e., we come to the case examined above, that of the presence of equal angles. Using what was proved in that case, we can assert that A1 An < A1 An .

(4)

Combined with (3), this yields A1 An < A1 An , which is the desired conclusion. All of the above is valid, however, only if the polygon P remains convex. But it may turn out that the deformation violates convexity before the angle Ak reaches the required value. We now clarify what violation of convexity may occur. 2 By Theorem 1, a polygon whose angles do not exceed π is convex. Therefore, the only possible cause for the violation of convexity is the fact that some angles, while increasing, become greater than π. The deformation only 2

It is curious that Cauchy, to whom the theorem under proof belongs, overlooked this possibility, and so actually did not give a complete proof of the theorem. Later this gap was filled in by others [St1, p. 34].

160


changes the angles Ak , A1 and An . The angle Ak cannot become greater than π, since it increases only to the value of the angle Ak of the convex polygon P . However, the angle A1 or An can increase too and at some moment it can reach the value π, further becoming greater than π.3 (The fact that this is really possible is seen from Fig. 78, where planar polygons are displayed for simplicity. The situation is similar for spherical polygons.)

Ak

Ak A2

A2 A1

A1

An An

An

Fig. 78

If the angle A1 at some moment becomes equal to π, then at this moment we stop deforming the polygon P . (The angle A1 may equal π in the original polygon P . In this case we proceed with the subsequent construction immediately.) Now, we have a polygon P whose angle A1 equals π. Therefore, A1 A2 and A1 An jointly form a single side. Let us construct the triangle Ak A1 An and start increasing its angle Ak in the same way as we did above for the triangle Ak A1 An . Then the side A2 An lengthens. We shall place the constant side A2 A1 on A2 An so that the side A1 An will always become longer. The angle Ak increases during this deformation, whereas the angles A2 and An may increase or decrease. For this reason, there are several possibilities: 1. The angle Ak can be brought to the value of Ak without violating the convexity of the triangle while keeping the angle A2 less than A2 . Then we again come to the case in which equal angles exist. 2. The angle A2 , while increasing, becomes equal to A2 earlier than Ak reaches the value of Ak . Here also we come to the case in which equal angles exist. 3. At some moment convexity fails: the angle An turns into π and during the subsequent deformation becomes greater than π. At the moment when ∠An = π, we construct the triangle Ak A2 An−1 and repeat the same arguments for this triangle. But now, no violation of 3

If, in the triangle Ak A1 An , the angle Ak increases while the lengths of the sides Ak A1 and Ak An remain constant, then the angle A1 (An ) decreases or increases depending on whether the angle An (respectively A1 ) of the triangle is acute or obtuse. This is immediate from Fig. 78.


161

convexity can occur, since we can bring each of the angles Ak , A2 , and An−1 to the value of the corresponding angle in the convex polygon P . In any case, we can thus construct a convex polygon P at least one of whose angles A i equals the corresponding angle Ai . Then, as has already been shown, we can conclude that A1 An < A1 An . Since in passing from P to P the side A1 An always becomes longer, we have A1 An < A 1 An . Consequently, A1 An < A1 An . This concludes the proof of the theorem. (Actually, we have proved that the polygon P can be deformed continuously without violating its convexity, so that the angles A2 , . . . , An−1 do not decrease and at least one of them reaches the required value. By iterating such deformations, we can transform P into P . Moreover, the assumptions of the theorem always remain valid and the side A1 An gradually lengthens.) 3.1.4 Lemma 2. If two convex spherical polygons have the same number of corresponding sides of equal length but not all the corresponding angles are equal, then there are at least four sign changes in the differences between the corresponding angles as we go around the polygons. In other words, if A1 A2 = B1 B2 , . . . , An A1 = Bn B1 , then either all the differences δi = ∠Ai − ∠Bi vanish, or there are at least four sign changes in the finite sequence δ1 , δ2 , . . . , δn , δ1 . If not all the differences δi are equal to zero, then there must exist changes of sign. Indeed, if, for instance, ∠A2 ≤ ∠B2 , . . . , ∠An−1 ≤ ∠Bn−1 and at least once we have a strict inequality, then by Theorem 2 we must have A1 An < B1 Bn , contradicting the assumption. If there are sign changes, then their number is even because, when we complete our round trip, we return to the difference with which we had started. Consequently, it suffices to show that exactly two sign changes are impossible. Assume to a contradiction that we have exactly two sign changes; for instance, the angles A1 , . . . , Am are greater than the angles B1 , . . . , Bm , whereas the angles Am+1 , . . . , An are less than the angles Bm+1 , . . . , Bn (of course, some angles may be equal). Am

L

A2 A1 K

Am+1

Bm

An

Fig. 79

B2

N

B1

Bm+1

M

Bn

162


Bisect the sides A1 An , Am Am+1 , B1 Bn , and Bm Bm+1 by the points K, L, M , and N and draw KL and M N (Fig. 79). Comparing the polygons A1 . . . Am LK and B1 . . . Bm N M , we see that all their sides, except LK and M N , are equal by assumption and the angles A1 , . . . , Am are greater than the angles B1 , . . . , Bm . Therefore, by Theorem 2 LK > M N . However, if we compare the polygons Am+1 , . . . An KL and Bm+1 . . . Bn M N , then for the same reason we must have LK < N M . This contradiction shows that it is impossible to have exactly two sign changes. Hence, there are at least four sign changes. Lemma 2 can easily be restated for polyhedral angles. In this situation, together with ordinary polyhedral angles we consider “degenerate” ones, i.e., doubly-covered flat angles less than π. Lemma 2a. If two convex polyhedral angles, distinct from dihedral angles and possibly degenerate, have corresponding planar angles of equal measure while not all of their dihedral angles are equal, then there are at least four sign changes in the differences between the corresponding dihedral angles as we go around the vertices. (As indicated at the beginning of this section, dihedral angles equal to π are allowed. In the lemma, by planar angles we mean the angles between all adjacent edges, including the edges of the dihedral angles.) Indeed, if two polyhedral angles are nondegenerate, then, by taking spheres of the same radius centered at their vertices, we obtain spherical polygons which satisfy the conditions of Lemma 2. Hence, in this case Lemma 2a is simply a paraphrase of Lemma 2. Now if at least one of the polyhedral angles is degenerate, we denote this angle by V1 and the other by V2 . Let p1 and q1 be the edges of V1 at which the dihedral angles are equal to zero. If in V2 the dihedral angle at the edge p2 corresponding to p1 is zero, then so is the dihedral angle at the edge q2 . In this case the angles V1 and V2 are congruent. Finally, suppose that the dihedral angle at p2 differs from zero. Then so does the dihedral angle at q2 . So the dihedral angles at p2 and q2 are greater than the dihedral angles at p1 and q1 . At the same time, the sums of the planar angles of V2 between p2 and q2 are the same from both sides, since this is so for the angle V1 . Therefore, in the angle V2 , on both sides between p2 and q2 there are edges at which the dihedral angles are less than π. However, the dihedral angles at the corresponding edges of V1 are equal to π since V1 is a degenerate angle. Thus, in V2 on both sides between the edges p2 and q2 there are edges at which the dihedral angles are less than the angles at the corresponding edges of V1 . Because the angles at p2 and q2 are greater than the angles at p1 and q1 , we obtain exactly four sign changes, thus completing the proof of the lemma.

3.2 Equality of Dihedral Angles in Polyhedra with Equal Planar Angles

163

Exercise.4 We suggest that the reader use the Cauchy Lemma to prove the following assertions about convex polyhedra homeomorphic to a disk whose faces are as shown in Figs. (a), (b), and (c). α

O B

(a)

A

(b)

(c)

In case (a), the polyhedron admits no flex with its faces rigid and convexity preserved. In case (b), under flexes of the type indicated, the minimal value of the angle α is attained when the dihedral angle at the edge AB becomes equal to π. In case (c), if all the faces are regular polygons, then the polyhedron is symmetric about an axis passing through the point O. (Many such applications of the Cauchy Lemma may be found in [Z2].)

3.2 Equality of Dihedral Angles in Polyhedra with Equal Planar Angles 3.2.1 Here we also include among convex polyhedra doubly-covered convex polygons, i.e., “polyhedra” consisting of two superposed congruent polygons.5

Y X

Fig. 80 4 5

Added by V. Zalgaller. The points inside such polygons, although they coincide, are regarded as if they were distinct. It is convenient to regard them as lying on different sides of the “doubly-covered” polygon formed by two superposed polygons. The distance between points on “different sides” of a doubly-covered polygon is measured along the line going to the boundary of the polygon and back (Fig. 80).

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Moreover, we shall admit the partition of faces of a polyhedron into finitely many smaller polygons, each of which is counted as a “face.” In this connection, a polyhedron may have two sorts of edges and vertices: “genuine” and “fictitious.” Each edge is a common side of two “faces.” A fictitious edge is an edge at which the dihedral angle equals π, i.e., is actually a plane. A fictitious vertex is a vertex at which the polyhedral angle is in fact dihedral, in particular a plane when the vertex lies inside a genuine face of the polyhedron. Such an extension of the meaning of the notions of face, edge, and vertex turns out to be convenient and is further used without additional mention. We say that two polyhedra have the same structure if, between the elements of the polyhedra (faces, edges, and vertices), a one-to-one correspondence preserving incidence may be established. For example, if an edge a (a vertex A) of a polyhedron P belongs to a face Q (an edge q) then the corresponding edge a1 (vertex A1 ) of the polyhedron P1 belongs to the corresponding face Q1 (edge q1 ). 3.2.2 Theorem 1. If two closed convex polyhedra have the same structure and the corresponding planar angles on corresponding faces are equal, then the dihedral angles at the corresponding edges are also equal. Let P1 and P2 be two polyhedra satisfying the conditions of the theorem. Assign a plus or minus sign to those edges of P1 at which the dihedral angle increases or decreases, respectively, in going from P1 to P2 . No sign is ascribed to those edges at which there is no change. Let us count the number of sign changes in going around some vertex of P1 . We shall distinguish genuine vertices from fictitious ones. Let A1 and A2 be corresponding vertices of P1 and P2 . As we know, a genuine vertex is characterized by the condition that the sum of the planar angles touching at it is less than 2π. By assumption, the corresponding planar angles on P1 and P2 are equal; therefore, the vertices A1 and A2 are either both genuine or both fictitious. If A1 and A2 are genuine vertices, then we can use Lemma 2a of Section 3.1 asserting that the equality of planar angles implies that the differences between the dihedral angles either all vanish or are involved in at least four sign changes. If A1 and A2 are fictitious vertices, then we consider two cases: (1) There are genuine edges incident to both vertices, but to a genuine edge incident to A1 corresponds a fictitious edge incident to A2 . (2) This case contains all the other possibilities: for at least one of the vertices there are no genuine incident edges, or all genuine incident edges correspond to each other. We will show that in the first case the number of sign changes around the vertex A1 equals four, while in the second case the vertices A1 and A2 can be entirely removed.


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(1) Consider the first case. Let a1 and a1 be genuine edges incident to A1 and let b2 and b2 be genuine edges incident to A2 (Fig. 81). Since the polyhedral angle at a fictitious vertex is in fact dihedral, the two genuine edges issuing from this vertex prolong each other. Therefore, if to the genuine edge a1 of P1 corresponds a fictitious edge a2 of P2 , then the edge a2 corresponding to a1 is fictitious, too. In that case, the edges corresponding to the genuine edges b2 and b2 of P2 are fictitious edges b1 and b1 of P1 . The dihedral angle at a genuine edge is less than π and one at a fictitious edge equals π. Therefore, the signs of the differences between the angles at the edges a1 and a2 , b1 and b2 , a1 and a2 , and b1 and b2 are successively −, +, −, +, which yields exactly four sign changes. (There are no sign assignments for the other edges incident to A1 and A2 , since all of them (at the vertex A1 as well as at the vertex A2 ) are fictitious and so the angles at these edges are equal.)

a1 A1

a2 b2

A2 b1

b2

b1 a1

a2

Fig. 81

(2) In the second case, the following three possibilities may occur: (2a) neither A1 nor A2 has incident genuine edges; (2b) only one of the vertices has such edges; (2c) both vertices have such edges, which correspond to each other (Fig. 82). (2a) When all the edges incident to A1 and A2 are fictitious, all the dihedral angles touching at A1 and A2 are equal to π. Therefore, none of the edges incident to A1 is labeled. Hence, the vertices A1 and A2 and all edges touching at these edges can be removed.

a1

a2

A1 b1

l1

a1

A2

l2

Fig. 82

a2

b2

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(2b) and (2c) Suppose, for instance, that the vertex A1 has a genuine incident edge a1 . Then A1 also has a genuine incident edge a1 which is the prolongation of a1 . Let a2 and a2 be the corresponding edges incident to A2 . If there are any genuine edges incident to A2 , then they are precisely a2 and a2 (case (2c)). The other edges b1 , . . . , l1 incident to A1 are fictitious and the edges b2 , . . . , l2 corresponding to them are fictitious. Therefore, the edges b1 , . . . , l1 are not labeled at all and can be removed. The edges a2 and a2 are certainly prolongations of each other, since the planar angles between them are the same as those between a1 and a1 , i.e., are equal to π. The dihedral angles at the edges a2 and a2 are equal; hence, the edges a1 and a1 are labeled by the same sign. Hence, we can remove the vertices A1 and A2 when we assign the edge a2 ∪ a2 to the edge a1 ∪ a1 ; here the edge a1 ∪ a1 will obviously be labeled by the same sign as that assigned to the edges a1 and a1 . Thus, all the fictitious vertices for which the second case occurs can be removed. From what was proved above, we then infer the following: were there sign assignments on P1 , there would exist at least four sign changes around each vertex contained in some labeled edge. However, the latter is impossible by the Cauchy Lemma (Section 2.1 of Chapter 2). Hence, there are no labeled edges at all, which means that the corresponding dihedral angles of P1 and P2 are equal, and the theorem is thus completely proved. 3.2.3 Theorem 2. Let P1 be an unbounded convex polyhedron none of whose vertices has more than one incident unbounded edge even after all or some of the fictitious vertices of the polyhedron are removed (by uniting into one edge the two edges separated by such a fictitious vertex).6 If a polyhedron P2 has the same structure as P1 and the corresponding planar angles on the corresponding faces of P1 and P2 are equal, then the dihedral angles at the corresponding edges are also equal. Before starting the proof, some remarks are in order. If all the unbounded edges of P1 are parallel to each other, then, certainly, at most one unbounded edge can issue from any vertex. Therefore, for such pairs of polyhedra Theorem 2 asserts that the equality of the angles on the faces implies that the dihedral angles are equal. A polyhedron with parallel unbounded edges is characterized by the condition that its total curvature is 2π, because the fact that the unbounded edges are parallel implies that the spherical image of the 6

This stipulation is necessary, since we admit fictitious vertices which partition edges. Take as an example any polyhedral angle P . Obviously, such an angle admits a deformation that does not change its planar angles but changes the dihedral angles. Take a point x on each edge of P , declare them (fictitious) vertices, and connect them by (fictitious) edges. Then each vertex Ai has exactly one incident unbounded edge, while only bounded edges OAi issue from the genuine vertex O of the angle. Only after removing the fictitious vertices Ai will the vertex O have more than one incident unbounded edge.


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polyhedron covers a whole hemisphere. Hence, Theorem 2 has the following consequence. Theorem 3. If two unbounded convex polyhedra of the same structure and total curvature 2π have equal corresponding angles on corresponding faces, then the dihedral angles at the corresponding edges of the polyhedra are also equal. If an unbounded polyhedron has total curvature less than 2π, then there are nonparallel unbounded edges and several unbounded edges can issue from a single vertex. In this case Theorem 2 is inapplicable and we must impose additional conditions; this will be done in Section 3.4. The example of polyhedral angles, which can always be deformed without changing the planar angles, shows that in this case the equality of the planar angles does not imply that of the dihedral angles. Now let us prove Theorem 2. Let P1 and P2 be polyhedra satisfying the conditions of Theorem 2. Let us add an improper vertex to each of the polyhedra and consider the unbounded edges of the polyhedra as incident to these improper vertices. We obtain abstract polyhedra P 1 and P 2 that have the same structure and are homeomorphic to the sphere. Every pair of vertices of the polyhedron P1 is joined by at most one edge. This is obvious for the proper vertices of P1 . If an edge AB joins a proper vertex A to the improper vertex B, then AB is simply an unbounded edge of P1 . By assumption, at most one unbounded edge of P1 may issue from the vertex A. Hence, AB is the only edge joining the vertices A and B. However, given a polyhedron (homeomorphic to the sphere) any two vertices of which are joined by at most one edge, we can apply the Cauchy Lemma to the net of edges of the polyhedron. Label the edges of P1 with the signs of the differences between the corresponding dihedral angles of P1 and P2 . Arguing in the same manner as in the proof of Theorem 1, we see that, first, there are at least four sign changes around each genuine vertex contained in a labeled edge. Second, for a fictitious vertex, we have the following alternatives: either there are exactly four sign changes around such a vertex or the vertex can be removed. Thus, there are at least four sign changes around each remaining vertex with an incident labeled edge. By the assumption of the theorem, the removal of fictitious vertices does not violate the condition that at most one unbounded edge issues from any vertex of P1 . Therefore, after we remove such vertices, the Cauchy Lemma is still applicable to the remaining net on the polyhedron P 1 obtained from P by adjoining an improper vertex. On the polyhedron P 1 , there are at least four sign changes around each of the remaining vertices (with some labeled edge coming to it). An exception may only occur at an improper vertex: the number of sign changes around it might even equal zero. However, as was observed in Section 2.1 of Chapter 2, the Cauchy Lemma admits the following refinement: its claim is also valid

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in the presence of one vertex with less than four sign changes. Applying this refined version of the Cauchy Lemma, we see that the disposition of signs obtained is impossible. Hence, there are no sign assignments at all, and the differences between the dihedral angles of the polyhedra P1 and P2 are equal to zero. 3.2.4 In perfect analogy with Theorem 2, we have Theorem 4. Let P1 be a bounded convex polyhedron whose boundary consists of a single closed polygonal line and each of whose internal vertices, i.e., vertices not lying on the boundary, is joined to the boundary by at most one edge. If a polyhedron P2 has the same structure as P1 and the planar angles at the corresponding internal vertices on the corresponding faces of P1 and P2 are equal, then the dihedral angles at the corresponding edges of P1 and P2 are also equal. Indeed, adding an improper vertex to P1 we can identify the whole boundary of P1 with this vertex. Then all the edges coming to the boundary from the interior of P1 are incident to this improper vertex. As a result, we obtain an abstract “polyhedron” homeomorphic to the sphere. A word for word repetition of all the arguments in the proof of Theorem 2 yields the proof of Theorem 4.7 Observe that Theorem 4 contains Theorem 1 as well as Theorem 2. If the polyhedra P1 and P2 are closed and the corresponding angles on their faces are equal, then, by straightening the angles at one pair of corresponding vertices, we arrive at two polyhedra P1 and P2 satisfying the conditions of Theorem 4. If the polyhedra P1 and P2 are unbounded, then, cutting their unbounded parts off without touching their vertices, we again come to polyhedra P1 and P2 satisfying the conditions of Theorem 4. For polyhedra bounded by several closed polygonal lines, we will have to introduce several improper vertices, but then the refined Cauchy Lemma will give nothing. Here some additional conditions are needed, but exactly which is an open question.

7

It is only the convexity of polyhedral angles that matters in conclusions about the number of sign changes. For this reason, it suffices to require only that all polyhedral angles of P1 and P2 be convex. The polyhedra themselves may be nonconvex. Such polyhedra can be called locally convex. A complete locally convex polyhedron is certainly convex, but such a polyhedron with boundary may fail to be convex as a whole.

3.3 Uniqueness of Polyhedra with Prescribed Development

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3.3 Uniqueness of Polyhedra with Prescribed Development 3.3.1 In this section, as in the previous one, we include doubly-covered convex polygons among convex polyhedra. If the points X and Y lie on different sides of such a degenerate polyhedron P , then the distance between them is measured by the length of the shortest polygonal line joining X to the boundary of P and continuing further to Y (see Fig. 80). Thus, if X and Y coincide but lie on different sides of P , then the distance between them is nonzero. Before proving the congruence of polyhedra with the same development, we establish some preparatory lemmas. In contrast to the preceding section, in these lemmas by a face of a polyhedron we mean its genuine face. A degenerate polyhedron can consist of only two coinciding faces. Lemma 1. If L is a shortest arc on a convex polyhedron P (i.e., L is the shortest among all lines joining the given points), then L is a polygonal line with at most one segment on each face of P .8 Indeed, let the points X and Y of a shortest arc L lie on a face Q. Then the rectilinear line segment XY is contained in Q by the convexity of Q. This segment obviously provides a shortest join between the points X and Y . Were the subarc XY of L different from the segment XY , replacing it by this segment would shorten L, which contradicts the condition that L is a shortest arc. This clearly implies that the entire part of L contained in Q is a single line segment. Lemma 2. If each segment of a ray L lying in an unbounded convex polyhedron P is a shortest line, then from some point onward L is a ray contained in some unbounded face of P . Indeed, by Lemma 1 any part of L, no matter how long, may have at most one straight line segment on each face. Since the number of faces is finite, L must lie within a single face Q from some point onward. Since L is shortest for each subsegment and unbounded, the part of L on Q is a ray. Lemma 3. If the polyhedra P1 and P2 are produced from the same development R by gluing, then there is an isometry, i.e., a length-preserving mapping, between them. Indeed, to each point X of the development R, there correspond points X1 and X2 of the polyhedra P1 and P2 . This yields a natural correspondence 8

Here P is understood to be a complete polyhedron or at least one all of whose faces are convex. Otherwise, the claim of the lemma might be false as we can see from simple examples.

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between the points X1 of P1 and the points X2 of P2 . The correspondence is isometric, because two lines L1 and L2 on P1 and P2 corresponding to each other correspond to the same line L in the development R and hence have the same length. By Lemma 3, the congruence of polyhedra produced from the same development by gluing is equivalent to the congruence of isometric polyhedra. Lemma 4. Let ϕ be an isometry of a convex polyhedron P1 onto a convex polyhedron P2 and let ϕ−1 be the inverse of ϕ. The images of the faces of P1 under ϕ cover the faces of P2 and thus partition them into “new” faces. In the opposite direction, the images of the faces of P2 under ϕ−1 similarly partition P1 into “new” faces. Then as the result the polyhedra P1 and P2 consist of the same number of equal similarly-situated new faces. Indeed, let Q1 be a face of P1 and let ϕ(Q1 ) be the image of Q1 under ϕ. Suppose that ϕ(Q1 ) at least partly covers a face Q2 of the polyhedron P2 , i.e., the intersection S2 = R2 ∩ ϕ(Q1 ) is nonempty. Let us show that S2 is a convex polygon (Fig. 83).

X2

S2 ϕ (Q1) Y2

R2

Fig. 83

Let X2 and Y2 be points in S2 and let X1 and Y1 be their inverse images in Q1 under the mapping ϕ. By the convexity of Q1 , the segment X1 Y1 belongs to Q1 ; therefore, its image ϕ(X1 Y1 ) belongs to ϕ(Q1 ). Since the segment X1 Y1 is a shortest arc in P1 and ϕ is an isometry, ϕ(X1 Y1 ) is a shortest arc between the points X2 and Y2 in P2 . However, the points X2 and Y2 lie on the same face Q2 and hence the shortest arc between them is the line segment X2 Y2 . Thus, the segment X2 Y2 lies both in ϕ(Q1 ) and in R2 and therefore is contained in S2 , which proves the convexity of S2 . The figure S2 = R2 ∩ ϕ(Q1 ) is cut from the face R2 by the images of the edges of P1 bounding Q1 . The edges of a polyhedron are always shortest arcs in it. Hence, the images of the edges of P1 under the isometry ϕ are shortest arcs in P2 . The images of unbounded edges are unbounded lines shortest over their bounded subarcs. By Lemmas 1 and 2, the part of the

3.3 Uniqueness of Polyhedra with Prescribed Development

171

image of an arbitrary edge of P1 which lies on the face R2 is a straight line segment or a half-line (if the edge is unbounded). Consequently, the figure S2 is cut from R2 by finitely many segments, and possibly half-lines, and therefore is a bounded or unbounded polygon. Since P1 and P2 have finitely many faces, there are finitely many intersections of the faces of P2 with the images of the faces of P1 . We have thus proved that each such intersection is a convex polygon; hence the polyhedron P2 is partitioned into finitely many new faces. Considering the inverse mapping ϕ−1 of the polyhedron P2 onto P1 , we conclude in the same way that the polyhedron P1 is partitioned into finitely many new faces. We will show that the mapping ϕ takes the new faces of P1 to the new faces of P2 . Let S1 be a new face on P1 ; it is the intersection of some face Q1 of P1 with the image of some face R2 of P2 under the mapping ϕ−1 , i.e., S1 = Q1 ∩ ϕ−1 (R2 ). But then ϕ(S1 ) = ϕ(Q1 ) ∩ R2 , i.e., S1 is mapped to the new face S2 in P2 , which is the intersection of the face R2 of P2 with ϕ(Q1 ). Thus, the mapping ϕ takes the new faces of P1 to the new faces of P2 . Since ϕ is an isometry, P1 and P2 consist of the same number of congruent similarly-situated faces. In view of Lemma 4, the congruence of isometric polyhedra is equivalent to the congruence of polyhedra consisting of equal similarly-situated “new” faces. 3.3.2 Theorem 1. Every isometry ϕ of a closed convex polyhedron onto another can be realized as a motion or a motion and a reflection, i.e., there is a motion, or a motion followed by a reflection, which takes each point to its image under the mapping ϕ.9 (According to Lemma 3, this theorem contains the theorem on the congruence of closed convex polyhedra with the same development.) Assume that ϕ is an isometric mapping of a closed convex polyhedron P1 onto a closed convex polyhedron P2 . In view of Lemma 4, P1 and P2 consist of equal similarly-situated new faces corresponding to each other under the mapping ϕ. Since these new faces are equal, so are their angles. Then, by Theorem 1 of Section 3.2, the dihedral angles at the corresponding (“new”) edges are also equal. Whence we can easily infer (see Lemma 5 below) that the polyhedra are congruent and that their corresponding elements (new faces, edges, and vertices) are mapped to each other by a motion or a motion and a reflection. However, the corresponding elements are exactly the elements that are put in correspondence by the mapping ϕ, because ϕ just takes the new faces of P1 into the new faces of P2 . Consequently, the mapping ϕ can 9

This theorem is slightly sharper than if it simply claimed the congruence of P1 and P2 . Clearly, if P1 and P2 are congruent, then they are superposable by a motion or a motion and a reflection; however, it is not known a priori whether the superposed points are the points corresponding to each other under the mapping ϕ.

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be realized by a motion or a motion and a reflection. The following lemma provides an exhaustive basis for the last conclusion. Lemma 5. If ϕ is an isometry of a convex polyhedron P1 onto a convex polyhedron P2 and ϕ takes the “new” faces of P1 to the “new” faces of P2 so that the dihedral angles at the corresponding (new) edges are equal, then ϕ can be realized by a motion or a motion and a reflection. Let Q1 be some new face of P1 and let Q2 be the corresponding face of P2 . Since ϕ is an isometry, it follows that the polygons Q1 and Q2 are congruent by a motion that to each point of Q1 assigns its image under ϕ in Q2 .10 As a result, the edges of Q1 will coincide with the corresponding edges of Q2 . If the polyhedra P1 and P2 happen to be on different sides of the plane of their common face Q1 = Q2 , then we take the reflection of P1 in this plane. Since all the dihedral angles at the corresponding edges are equal it follows that all the faces of P1 adjacent to Q1 will be placed on the corresponding faces of P2 and will even coincide with them. Indeed, the common corresponding edges and vertices of these faces and the face Q1 = Q2 are equal and the faces lie to the same side of such edges. But then all other corresponding elements of these faces must also coincide by the isometry of the mapping ϕ. Now, passing from these faces to adjacent faces and so forth, we can ensure that all elements of P1 coincide with the corresponding elements of P2 . Thus, ϕ has been realized by a motion or a motion and a reflection. 3.3.3 Theorem 2. If ϕ is an isometry of an unbounded convex polyhedron P1 of total curvature 2π onto a convex polyhedron P2 , then ϕ can be realized by a motion or a motion and a reflection. Since the total curvature is not changed under an isometric mapping, the total curvature of the polyhedron P2 also equals 2π. Lemma 4 implies that P1 and P2 consist of the same number of equal similarly-situated new faces. Consequently, the corresponding planar angles on the new faces of the polyhedra are equal. Now, Theorem 3 of Section 3.2 implies that the dihedral angles at the corresponding edges are equal as well. Applying Lemma 5, we infer that the given mapping from P1 onto P2 is realized by a motion or a motion and a reflection. Certainly, in Theorems 1 and 2 we can consider, as particular cases, selfmappings of a polyhedron. For instance, it follows that a closed convex polyhedron admitting a nonidentical isometry onto itself is necessarily symmetric.

10

This stipulation is not redundant when the face Q1 admits self-motions.

3.4 Unbounded Polyhedra of Curvature Less Than 2π

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3.4 Unbounded Polyhedra of Curvature Less Than 2π 3.4.1 Since curvature is fully determined by planar angles, the equality of the planar angles of two polyhedra implies that their curvatures are equal. Hence, if one of the polyhedra has curvature less than 2π, then so does the other. In this situation, the equality of planar angles alone is insufficient for the dihedral angles to be equal, as we see from the example of any flexible polyhedral angle. As an additional condition, we might take the equality of dihedral angles at the corresponding unbounded edges. However, this condition is too stringent and we can require slightly less. Before specifying this less stringent condition, some remarks are in order. As in Section 3.2, we shall assume that the genuine faces of a polyhedron are split into finitely many polygons, and it is precisely these polygons that are referred to as “faces.” Accordingly, among vertices and edges there may exist “fictitious” ones. Let P be an unbounded convex polyhedron. Divide the collection of its unbounded edges into classes by including two parallel edges q and r in one class if q and r belong to one face or there is a sequence of parallel edges between them such that every two neighboring edges belong to one face (Fig. 84). If there is no such sequence, then q and r belong to different classes, disregarding the fact that the edges might be parallel to one another. It can be shown that the latter case is possible only for fictitious edges lying on parallel faces; however, this plays no role in what follows. P

Q R q r

Fig. 84

Let α1 , . . . , αn be the dihedral angles at all the edges p1 , . . . , pn of some class. Assign the number α=π−

n

(π − αi )

(1)

i=1

to this class, treating α as the dihedral angle of the class. Obviously, here π − αi is nothing more than the angle between the normals to the faces touching at the edge pi and π − α is the angle between the normals to the

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external faces containing the extreme edges of the class (e.g. the faces Q and R in Fig. 84). The angle α itself is the angle between these faces. The equality π − α = (π − αi ) follows from the fact that all edges pi are parallel to one another and consequently all the normals involved are coplanar. Under the infinite similarity contraction of P to some point, the polyhedron transforms into its limit angle V . Moreover, parallel edges merge and edges of the same class yield a single edge p on V . The faces of V corresponding to the faces of P “externally containing” the extreme edges of the class will touch at this edge p. The dihedral angle between these faces is the same α. Consequently, the “dihedral angle” of a class of edges is nothing but the corresponding dihedral angle of the limit angle V . The limit angle V can degenerate into a flat angle if the polyhedron P has parallel faces. In that case we regard V as doubly-covered and distinguish edges on different sides of V . (We refrain from any rigorous justification of this remark, because in the sequel it suffices to bear in mind only the formal definition of the dihedral angle of a class of unbounded edges.) If two unbounded polyhedra P1 and P2 have the same structure and the corresponding planar angles of P1 and P2 are equal, then to unbounded edges of P1 lying in one class there correspond unbounded edges of P2 also lying in one class and vice versa. Thus, the classes of edges, and therefore the edges of the limit angles V1 and V2 , are in a one-to-one correspondence. The additional condition ensuring that the equality of planar angles implies that the dihedral angles are equal can be stated as follows: The dihedral angles of the corresponding classes of edges, i.e., of the corresponding edges of limit angles, must be equal.11 If there are no parallel unbounded edges on the polyhedra under study, then this condition is simply reduced to the equality of the dihedral angles at all corresponding unbounded edges. 11

The angles on the faces of the limit angle V of a polyhedron P are determined by the planar angles of P . They are equal to the angles between the unbounded edges on the corresponding faces of the polyhedron. Therefore, the equality of planar angles of P1 and P2 implies the equality of the planar angles on the faces of the limit angles V1 and V2 of P1 and P2 . Then the equality of the dihedral angles in V1 and V2 leads to the congruence of V1 and V2 themselves. At the same time, since V1 and V2 have the same planar angles, the dihedral angles of V1 and V2 are not all independent. For the equality of all the dihedral angles, it suffices to require that all but three neighboring ones be equal. (This is easy to verify by using the convexity of V1 and V2 .) Thus, the condition stated above can be slightly relaxed. In particular, it becomes absolutely redundant in the case of trihedral limit angles. Therefore, if the polyhedra P1 and P2 have only three classes of edges, then the equality of all planar angles alone suffices to ensure that all the dihedral angles are equal.


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3.4.2 We now prove the sufficiency of the above condition. Theorem 1. If two unbounded convex polyhedra of the same structure have equal corresponding planar angles and the dihedral angles of the corresponding classes of unbounded edges are also equal, then all the corresponding dihedral angles of the polyhedra are equal. Suppose that the polyhedra P1 and P2 satisfy the assumptions of the theorem. Let us add improper vertices to the polyhedra, considering the edges of the same class to be touching at one improper vertex. Treat these improper vertices as adjacent if they correspond to classes of edges which are neighboring as we move around the unbounded part of a polyhedron. Connect the adjacent improper vertices by improper edges. Assign a plus or minus sign to those edges of P1 where the dihedral angle increases or decreases, respectively, when we pass from P1 to P2 . No sign is given to the edges with equal dihedral angles and to the improper edges. Also, call a vertex labeled if some labeled edge issues from it. In much the same way as in Theorem 1 of Section 3.2, we must distinguish between proper vertices of two types: genuine and fictitious. Repeating the arguments in the proof of that theorem, we can remove all fictitious vertices at which the number of sign changes is less than four, as well as all “unlabeled” vertices. Once this is done, there will be at least four sign changes at each of the remaining proper vertices. The improper vertices (which remained unlabeled) may also be removed. We will show that, when we turn “half-way around” a labeled improper vertex from one improper edge to another, there is at least one change of (1) (1) sign. Indeed, let α1 , . . . , αn be the dihedral angles at the edges touching (2) (2) at such vertex A1 of P1 and let α1 , . . . , αn be the corresponding dihedral angles in the polyhedron P2 . In accordance with formula (1), the dihedral angles of the classes of these edges are (1) (2) π − αi , α(2) = π − π − αi . α(1) = π − By the assumptions of the theorem, these angles are equal, α(1) = α(2) . Hence, (1) (1) (2) (2) αi − αi = 0. αi , or αi = (1)

(2)

Thus, either all the differences αi − αi vanish or there is at least one sign change. We now take a copy P1 of P1 . Label the edges of P1 by signs opposite to the signs at the corresponding edges of P1 . Identifying the improper edges and vertices of the polyhedra P1 and P1 , we obtain an abstract “polyhedron” P1 ∪ P1 homeomorphic to the sphere. On this polyhedron there are at least four sign changes at each labeled proper vertex, since this is so on the polyhedron P1 . We know that there is at least one sign change at each improper

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vertex of P1 and that the edges of P1 are labeled by opposite signs. Therefore, as we move around an improper vertex on the “polyhedron” P1 ∪ P1 , we also have at least four sign changes: two when moving along P1 and P1 , and two more when traveling from P1 to P1 and back. Thus, there are at least four sign changes at each labeled vertex. However, this is impossible by the Cauchy Lemma. Hence, there were no labeled vertices at all. We only need to verify that the polyhedron P1 ∪ P1 meets the assumptions of this lemma. There are two such assumptions: the polyhedron must be homeomorphic to the sphere and no pair A, B of its vertices may be joined by two edges. The first condition is satisfied. The second holds trivially if the vertices A and B are both proper or both improper. If A is proper and B improper, then the edge AB is an unbounded edge of the polyhedron P1 (or P1 ) and only edges of the same class may touch at the vertex B. These edges are parallel to each other and hence it is impossible to have two different edges connecting A to B. The Cauchy Lemma is thus applicable. Applying it, we complete the proof of the theorem. 3.4.3 We now turn to the congruence condition for isometric unbounded polyhedra of curvature less than 2π. This condition is absolutely analogous to the condition of Theorem 1. (As will be shown in Section 5.2 of Chapter 5, every unbounded polyhedron of curvature less than 2π always admits continuous flexes. For this reason, some additional condition is always necessary to guarantee the congruence of such isometric polyhedra.) Let P1 be an unbounded convex polyhedron and let ϕ be an isometry of P1 onto a convex polyhedron P2 . When we map P1 onto P2 and back, the images of (genuine) faces of P1 split the (genuine) faces of P2 and vice versa. According to Lemma 4 of Section 3.3, we obtain polyhedra consisting of the same number of equal similarly-situated new faces. In particular, their unbounded (new) faces, and consequently unbounded (new)12 edges, are in one-to-one correspondence. From the equality between the corresponding faces, it is clear that for faces with parallel edges the corresponding faces will have parallel edges too. Therefore to each class of parallel edges on P1 there corresponds some class of parallel edges on P2 . Recall that the dihedral angle of a class of such edges is defined as the angle between the external faces containing the extreme edges of the class (as the faces Q and R in Fig. 84). The above-mentioned additional condition then reads: the dihedral angles of the corresponding classes of unbounded (new) edges of the polyhedra P1 and P2 must be equal. In particular, if the polyhedra P1 and P2 possess no parallel edges, then the condition amounts to requiring that the isometry under study take un12

Of course, old faces may be among the new faces when the former are not split. In contrast, the unbounded parts of the old edges, if not the old edges themselves, are always among the unbounded edges.


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bounded edges of P1 to unbounded edges of P2 and that the dihedral angles at the corresponding unbounded edges be equal. Later we shall present another more visual form of this condition. Now we prove its sufficiency. Theorem 2. Every isometry of an unbounded convex polyhedron onto another such that the dihedral angles of the corresponding classes of unbounded (new) edges are equal can be realized by a motion or a motion and a reflection. By Lemma 4 of Section 3.3, an isometric mapping of one polyhedron onto another implies that the polyhedra consist of the same number of equal similarly-situated new faces. In particular, the angles on the new faces are equal. Hence, granted the equality between the dihedral angles of the corresponding classes of new edges, we can apply Theorem 1 and infer from it that all dihedral angles at the corresponding edges are equal. Using Lemma 5 of Section 3.3, we conclude that the given isometric mapping is realizable by a motion or a motion and a reflection. 3.4.4 We now present the condition of Theorem 2 in another form. To this end, we consider the limit angles V (P1 ) and V (P2 ) of two isometric polyhedra P1 and P2 . Given a class of edges on the polyhedron Pi , we have the corresponding edge p on the angle V (Pi ); it is the edge into which all the edges of the class are transformed under the infinite similarity contraction of the polyhedron to its limit angle. The faces of the polyhedron “externally containing” the extreme edges of this class are transformed into the faces of the limit angle touching at the edge p, so that the corresponding dihedral angles between the former and latter faces are equal. Therefore, the condition of Theorem 2 means that the dihedral angles at the corresponding edges of the limit angles V (P1 ) and V (P2 ) must be equal. Of course, we have to keep in mind that to a new edge of Pi lying on an unbounded face a generator of V (Pi ) lying on the corresponding face is assigned, rather than a genuine edge of V (Pi ). However, when the dihedral angles are equal, then to each genuine edge of V (P1 ) corresponds a genuine edge of V (P2 ) and vice versa. The dihedral angles at fictitious edges are always equal to π. Hence, it suffices to consider only genuine edges of the limit angles V (P1 ) and V (P2 ). The faces of V (Pi ) that meet at a genuine edge of V (Pi ) are those which correspond to the faces of Pi with nonparallel unbounded edges. Therefore, the condition of equality of the dihedral angles at such edges is simply reduced to the requirement that the faces of the polyhedra Pi with nonparallel unbounded edges correspond to each other and form equal dihedral angles. The correspondence between faces is established by the isometric mapping ϕ: the unbounded face Q2 with nonparallel edges on P2 corresponds to a face Q1 on P1 if the image of Q1 covers the face Q2 , but possibly only in part. The partial covering of the face Q2 by the image of the face Q1 may be caused by

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the fact that, a priori, some part of Q2 can correspond to a face Q1 adjacent to Q1 and having parallel unbounded edges. However, since by Theorem 2 the polyhedra are in fact congruent, such a partial covering never occurs. 3.4.5 We now present the condition of Theorem 2 in one more new form; it is in this form that the condition will be used further in Chapter 4. Having this in mind, we begin with some definitions. Let P be an unbounded convex polyhedron. A ray on P is defined to be a line in P , unbounded in one direction and shortest over each of it bounded subarcs. Such lines were already considered in Section 3.3 and it was proved in Lemma 2 from that section that, from some point onward, such a line is a half-line contained in an unbounded face of P . Let L and L be two rays on P . Fix a direction of going around the unbounded faces of P . Let us develop these faces onto a plane by successively adjoining them to each other in the chosen direction, starting from the face containing the ray L (or at least its unbounded part). Then the rays L and L give two half-lines on the plane (Fig. 85). Call the angle between these half-lines, counted from L to L in the chosen direction, the angle between the rays L and L (more precisely, this is the angle from L to L given the chosen direction; if another direction is chosen, this angle may differ).

L L

L L

Fig. 85

Clearly, the notions of ray and of the angle between two rays belong to the intrinsic geometry of a polyhedron. Therefore, under an isometry of one polyhedron onto another, rays are taken to rays and the angles between them are preserved. We prove the following lemma: Lemma 1. Let P be an unbounded convex polyhedron and let V (P ) be the limit angle of P . Then: (1) Under the infinite similarity contraction of P to V , each ray L on P is transformed into a generator of V , i.e., into a half-line on V (P ) beginning at the vertex. (We call this generator the limit generator of L.)


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(2) To each ray there corresponds a limit generator and, conversely, each generator is the limit generator of some ray (and of all parallel rays). (3) The angle between rays is equal the angle between their limit generators. Proof. From some point onward, the ray L is a half-line lying in some unbounded face of P . Under the infinite similarity contraction to the vertex of the angle V (P ), every bounded part of the polyhedron converges to the vertex of V (P ) and every half-line converges to a parallel half-line beginning at the vertex. Therefore, the ray L also converges to such a half-line, which proves the first assertion of the lemma. For each face of V (P ) there is a parallel unbounded face of P , and to each half-line on the former we can assign a parallel half-line on the latter. For this reason, each generator v of V (P ) corresponds to a parallel half-line L on P . Obviously, v is the limit generator of L. This proves the second assertion of the lemma concerning the inverse correspondence between the generators and rays. Finally, let L and L be two rays on P . If we develop the unbounded faces of P onto a plane and perform the infinite similarity contraction to a point, then each face will be transformed into the angle corresponding to a face of the limit angle. At the same time, the rays L and L are transformed into their limit generators, while the angle between them obviously remains the same. Thus, the angles between the rays are equal to the angles between their limit generators, and the latter angles can be measured on the angle V itself (or, which the same, on a development of V ) provided that the chosen direction on the polyhedron is carried over to the limit angle V . This completes the proof. From the lemma we easily infer the following statement: Lemma 2. Each isometric mapping from one unbounded polyhedron P1 onto another P2 naturally induces an isometric mapping from the limit angle V (P1 ) onto the limit angle V (P2 ). Indeed, an isometric mapping carries rays into rays and preserves angles between rays. By Lemma 1, to rays on P1 and P2 correspond their limit generators on V (P1 ) and V (P2 ), and the angles between the generators are equal to the angles between the corresponding rays. Thus, if we assign the generators of V (P1 ) to the generators of V (P2 ) in such a way that the corresponding rays on the polyhedra are mapped to each other by the given isometry, then this assignment of the generators will preserve the angles between them. Clearly, the indicated assignment yields an isometric mapping from V (P1 ) onto V (P2 ), which concludes the proof of Lemma 2. Using the preceding result, we can restate Theorem 2 in an essentially new way:

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Theorem 3. For an isometric mapping ϕ from one unbounded convex polyhedron P1 onto another P2 to be realizable by a motion or a motion and a reflection, it is necessary and sufficient that the isometric mapping, induced by ϕ, from the limit angle V (P1 ) onto the limit angle V (P2 ) be realizable as a motion or a motion and a reflection. The necessity of the condition is obvious, because the limit angle V (P1 ) may be rigidly connected to the polyhedron P1 and after the motion superposing P1 and P2 the angle V (P1 ) will become the limit angle of P2 . To prove sufficiency, observe that for each unbounded edge of the polyhedron Pi there is a parallel edge of the limit angle V (Pi ). From the definition of the induced isometric mapping between the angles V (P1 ) and V (P2 ), it is clear that their corresponding edges or, more generally, their generators correspond to the appropriate (possibly new) edges of the polyhedra P1 and P2 . Since the angles V (P1 ) and V (P2 ) are superposable by a motion, their dihedral angles at the corresponding edges are equal. As mentioned, these dihedral angles are equal to the dihedral angles of the classes of edges of the polyhedra P1 and P2 . Therefore, the condition of Theorem 2 is satisfied and the mapping ϕ can be realized by a motion or a motion and a reflection.13 Visually, if we place the vertices of the angles V (P1 ) and V (P2 ) at the corresponding vertices of the polyhedra and move P1 so that V (P1 ) coincides with V (P2 ), then the polyhedron P1 itself will coincide with P2 . It is worth observing that Theorem 3 cannot be given the following simple form: two isometric polyhedra P1 and P2 are congruent if their limit angles are congruent. Such a claim is simply false. Indeed, we shall show in Section 5.2 of Chapter 5 that every unbounded convex polyhedron admits a continuous flex such that its limit angle only rotates but undergoes no other transformation. Such flexes produce polyhedra which are isometric to the initial polyhedron, have the same limit angle, but are not congruent to it. Polyhedra given by polyhedral angles are the only exception since they coincide with their limit angles. This observation makes it clear that it is not in vain that we constantly stipulate that elements and limits angles of polyhedra correspond to each other precisely under the given isometric mapping. 3.4.6 Concluding the section, we briefly indicate one more form of the condition of Theorem 2. It uses the spherical images of isometric polyhedra. 13

If the angles V (P1 ) and V (P2 ) can be superposed without reflection, then so are the polyhedra P1 and P2 , since a reflection of a polyhedron yields a reflection of its limit angle. Polyhedra whose limit angles degenerate into flat angles are the only exception. The reflection of such an angle in its plane not only superposes the angle upon itself but does not even permute its edges.


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N1

p

N1 p

N2

N2

q

q

S

N3 N3 Fig. 86

The spherical image of an unbounded convex polyhedron P is a convex spherical polygon S. The boundary of S is formed by the spherical images of the unbounded edges of P . The spherical images of parallel edges are arcs of the same great circle and hence form a single side of the polygon S (Fig. 86). The vertices of S are the spherical images of the faces of P with nonparallel edges. All this is absolutely clear if we recall that the spherical image of a polyhedron coincides with the spherical image of its limit angle. The length of a side of the polygon S measures the angle between the normals to adjacent faces of the limit angle, i.e., the angle between the normals to the corresponding faces of P . However, the angle between the normals is equal to π minus the dihedral angle. Therefore, the sides of the polygon S also determine the dihedral angles between the faces. An isometric mapping ϕ from a polyhedron P1 onto P2 naturally gives rise to a correspondence ψ between the spherical images of P1 and P2 : the spherical image of a point X1 of P1 is assigned to the spherical image of the point ϕ(X1 ) of P2 . The spherical image of a point may fail to be a single point; hence this correspondence may fail to be single-valued in either direction. However, it preserves area, because the area of the spherical image remains the same under an isometric transformation of the polyhedron. Since the lengths of the sides of the polygons S1 and S2 which are the spherical images of the polyhedra P1 and P2 are determined from the dihedral angles between the faces of P1 and P2 with nonparallel unbounded edges, the condition of Theorem 2 is reduced to the following: the polygons S1 and S2 must have corresponding sides of equal length. Moreover, the correspondence between sides is defined from the isometric mapping of P1 onto P2 as indicated above. The spherical image S of a polyhedral angle V determines this angle uniquely (the edges of V are perpendicular to the faces of the angle S projecting S from the center of the sphere and conversely: the edges of the angle S are perpendicular to the faces of V ). Therefore, the limit angles of

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the polyhedra are equal if and only if the spherical images of the polyhedra are congruent, and Theorem 3 can be given the following form: Theorem 4. Let ϕ be an isometric mapping from one unbounded convex polyhedron P1 onto another P2 . It determines a transformation ψ of the boundary of the spherical image S1 of the polyhedron P1 into the boundary of the spherical image S2 of the polyhedron P2 . The mapping ϕ can be realized by a motion if and only if the mapping ψ can be realized by a motion.

3.5 Polyhedra with Boundary14 3.5.1 First of all, we indicate some interesting corollaries which can be derived from the theorems of Sections 3.3–3.4 for convex polyhedra with boundary. By definition, each such polyhedron is part of a closed or unbounded convex polyhedron. In particular, its total curvature is less than or equal to 4π, never exceeding 2π for the case in which the polyhedron is unbounded. However, not only a closed convex polyhedron may have total curvature 4π. Indeed, the total curvature is the sum of curvatures at all vertices; therefore, cutting away from a closed convex polyhedron any part that contains no vertices either in its interior or on its boundary, we obtain a convex polyhedron of the same total curvature 4π. We only need the remaining part of the polyhedron to be connected; otherwise, this part will not be a polyhedron by the very definition of polyhedra. It turns out that the remaining part of the polyhedron is still rigid with convexity preserved, irrespectively of how large the part cut away was! Namely, the following theorem holds: Theorem 1. If the total curvature of a convex polyhedron P1 is 4π, then every isometric mapping ϕ of P1 onto a convex polyhedron P2 can be realized by a motion or a motion and a reflection.

A1

P1 P1-P1

Fig. 87 14

See also Chapter 5. – V. Zalgaller

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Proof. Let us construct the convex hull of P1 and take its surface, obtaining a closed convex polyhedron P 1 . To P1 we thereby add the set P 1 \ P1 which may have several components. Since the total curvature of P1 is 4π, the added part P 1 \ P1 does not contain any vertices and consequently each of its connected components is developable on a plane. For the proof (see Subsection 1.8.3 of Chapter 1) it suffices to develop successively on a plane the parts of the faces that are contained in each connected component of P 1 \ P1 (Fig. 87). Since the polyhedron P2 is isometric to P1 , it also has total curvature 4π and hence the same arguments apply to P2 . We similarly complete P2 to a closed polyhedron P 2 ; here also each connected component of the set P 2 \P2 is developable on a plane. The isometric mapping ϕ induces a certain isometric correspondence between the boundaries of P1 and P2 . The latter in turn gives rise to an isometric correspondence between the boundaries of the sets P 1 \ P1 and P 2 \ P2 , and so their corresponding segments have equal lengths. We shall prove that the corresponding angles on the boundaries of these figures are equal, provided of course that the angles are measured on the polyhedra or, which amounts to the same, on the developed sets P 1 \ P1 and P 2 \ P2 rather than in space. Consider a pair of corresponding points A1 and A2 on the boundaries of the polyhedra P1 and P2 . None of these points may be a vertex of P 1 or P 2 . Otherwise, in passing from P1 to P 1 (or from P2 to P 2 ), we would add the curvature at the vertex A1 to the total curvature of P1 , and the total curvature of P 1 would be greater than 4π. Recall that if a point on a polyhedron is not a vertex, then the compete angle at the point equals 2π. The complete angles at the points A1 and A2 on the polyhedra P 1 and P 2 are composed of the angles on the polyhedra P1 and P2 and the angles on their complements P 1 \ P1 and P 2 \ P2 . The angles on the polyhedra P1 and P2 are equal by isometry. Consequently, the angles on P 1 \ P1 and P 2 \ P2 are also equal. Thus, the sets P 1 \ P1 and P 2 \ P2 are composed of pieces developable on the plane and having corresponding sides of equal length. Therefore, these sets are isometric. Since P1 and P2 are also isometric, it follows that the closed polyhedra P 1 and P 2 are also isometric. By Theorem 1 of Section 3.3, the isometry between them can be realized by a motion or a motion and a reflection. Moreover, the elements of P 1 and P 2 that correspond to each other under the isometry between P1 and P2 or between P 1 \ P1 and P 2 \ P2 coincide. The isometry between P1 and P2 has thus been realized by a motion or a motion and a reflection. In the same way, using Theorem 2 of Section 3.3, we can prove the following

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Theorem 2. If the total curvature of an unbounded convex polyhedron with boundary is 2π, then every isometric mapping of this polyhedron onto another convex polyhedron can be realized by a motion or a motion and a reflection. Finally, the congruence theorem for isometric closed convex polyhedra implies the following result: Let us call a convex polyhedron with flat boundary a polyhedral cap if the orthogonal projection of the polyhedron on the plane of the boundary is one-to-one (i.e., every straight line perpendicular to the plane intersects the polyhedron only at a single point or merely touches it). Then: Theorem 3. Isometric polyhedral caps are congruent. Each isometric mapping of one polyhedral cap onto another can be realized by a motion or a motion and a reflection. To prove this, we add symmetric caps to the boundaries of the given polyhedral caps. We obtain closed convex polyhedra. With the caps isometric, the entire polyhedra are also isometric. Applying Theorem 1 of Section 3.3 about closed polyhedra, we now infer Theorem 3. 3.5.2 Here we establish a general condition for the congruence of isometric polyhedra with boundary, a condition not involving the total curvature of polyhedra. Theorem 4. If two convex polyhedra each bounded by a single closed polygonal line have the same structure, have equal corresponding angles on corresponding faces, and have equal angles between the boundary edges at all the boundary vertices, then the corresponding dihedral angles of the polyhedra are also equal. As in Theorem 2, fictitious vertices and edges are permissible here, so that the “faces” may merely be parts of genuine faces. In the same way as before, from this theorem and Lemma 4 of Section 3.3 we can deduce the following Theorem 5. Suppose an isometric mapping from a convex polyhedron bounded by a single closed polygonal line onto another convex polyhedron is given. If this mapping preserves the angles between boundary edges at all boundary vertices, then it can be realized by a motion or a motion and a reflection. We only need to prove Theorem 4. Let P1 and P2 be two polyhedra satisfying the conditions of Theorem 4. Label the edges of P1 by signs according to the familiar rule that we have already used repeatedly. Of course, the boundary edges remain unlabeled.

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As in the proof of Theorem 1 of Section 3.2, exclude all fictitious vertices at which there is no sign change as well as all unlabeled vertices in general. There are at least four sign changes at each of the remaining internal vertices. Let A1 and A2 be boundary vertices of P1 and P2 corresponding to each other. Draw planar angles over the boundary edges a1 , b1 and a2 , b2 touching at these vertices. By the convexity of the polyhedra, we obtain convex polyhedral angles at the vertices A1 and A2 . Their corresponding planar angles are equal by the assumptions of the theorem. Comparing their dihedral angles, let us write pluses and minuses around the vertex A1 according to our rule. By Lemma 2a, we must have at least four sign changes around A1 provided that the vertex A1 is labeled. However, while writing signs on the polyhedron P1 , we did not label the boundary edges a1 and b1 at all. The elimination of the signs at these edges decreases the number of sign changes at most by two. Therefore, when we go half way around the vertex A1 on P1 from the edge a1 to the edge b1 , we must have at least one change of sign. We now take a copy P1 of the polyhedron P1 . Assign to its edges the signs opposite to the signs at the corresponding edges of P1 . Identifying the corresponding boundary edges and vertices of the polyhedra P1 and P1 , we obtain an abstract “polyhedron” P1 ∪ P1 homeomorphic to the sphere. There are at least four sign changes at each of the labeled vertices of this polyhedron. Thus, the edges of the polyhedron P1 ∪ P1 are labeled by plus and minus signs so that there are at least four sign changes at each labeled vertex. However, this is impossible by the Cauchy Lemma. Consequently, there were no labeled edges on P1 at all, and Theorem 4 is thus proved.15 In Theorems 4 and 5, the polyhedra are assumed to be bounded by a single polygonal line each, and so they are homeomorphic to a disk. If this is not the case, then our arguments are not applicable. In this connection, we can pose the problem of generalizing Theorems 4 and 5 to convex polyhedra of other topological structure. 3.5.3 As was already pointed out in Section 1.1.6. of Chapter 1, the study of unbounded convex polyhedra is equivalent to that of bounded convex polyhe15

This theorem can be slightly generalized in two directions. First, instead of convex polyhedra, we can consider only locally convex ones, i.e., polyhedra with convex polyhedral angles at all internal and boundary vertices (provided that the boundary vertices are completed by the planar angles between the adjacent boundary edges). Second, it suffices to require that the condition of equality for the angles between the boundary edges be satisfied for all but possibly three of the boundary vertices. This strengthening of the result can be obtained easily by applying the refined version of the Cauchy Lemma. This definitely makes sense. Indeed, by using Euler’s Theorem, if V , V , and E are the numbers of all vertices, of boundary vertices, and of all edges, then 3V − 3 = E + V . However, a polyhedron is determined up to a motion by 3V − 6 parameters (3V coordinates of the vertices minus 6 parameters fixing the position in space). A development is determined by E edges. Since 3V − 6 = E + V − 3, it follows that in order to determine a polyhedron, we need V − 3 rather than V extra parameters; thus we need to specify angles at only V − 3 rather than V boundary vertices.

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dra with boundary whose extreme faces admit infinite prolongations without the appearance of new intersections. In that section necessary and sufficient conditions were also indicated. In this connection, the theorems concerning unbounded polyhedra proved in Sections 3.2–3.4 can be carried over automatically to the unbounded polyhedra with boundary. Besides, theorems analogous to Theorems 3–5 of the current section also hold for unbounded convex polyhedra, each of which is bounded by a single polygonal line rather than being complete. However, if the unbounded edges are not parallel, then it is necessary to introduce the additional condition of equality for the dihedral angles between the unbounded faces in the same way as in Theorems 1 and 2 of Section 3.4. The statements and proofs of the theorems on unbounded polyhedra with boundary are left to the reader.

3.6 Generalizations 3.6.1 The theorems proved in this chapter can be generalized in several directions: (1) The study of isometric mappings from convex polyhedra to arbitrary convex surfaces rather than only polyhedra. (2) The study of isometric mappings between convex surfaces. (3) The translation of theorems to non-Euclidean spaces (hyperbolic and spherical). (4) Generalizations to spaces of higher dimension. In all these cases we keep the convexity requirement. Had we abandoned it, we would come to a huge field of problems. Nevertheless, I would like to mention one such problem, which seems to be the most interesting one among all rigidity problems: Are there closed polyhedra admitting continuous flexes without folding any faces or do not any closed polyhedra admit such continuous flexes? (All previously known examples of flexible polyhedra are theoretical: they have self-intersections and, if implemented physically, would not actually flex.)16 Further (in Subsections 3.6.2–3.6.5) we successively consider generalizations of the theorems of the current chapter in the four directions indicated above. 3.6.1a The first examples concerning the above problem are closed nonconvex polyhedra admitting two different shapes upon preservation of rigid faces and 16

See Subsections 3.6.1a and 3.6.1b below, which are added as comments to the translation. – V. Zalgaller

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simultaneous convexity or concavity at the corresponding edges [AV]. The existence of such pairs is due to the existence of nonconvex polyhedra failing to be infinitesimally rigid [AV]. Many of such examples have appeared recently [J, Wu1–Wu7]. At present, continuously flexible closed nonconvex polyhedra are well known to exist. The merit of discovering such polyhedra is that of R. Connelly [Co2]. Their description can be found in several articles, for instance in [Ku, Z2]. Later V. A. Aleksandrov [A-v2] constructed a continuously flexible polyhedron which has the topology of a torus. 3.6.1b Connelly’s discovery raises many new questions. The flexible polyhedron in the first example by Connelly had 18 vertices. The question of diminishing the number of vertices was posed in [Ku]. Examples with fewer vertices were constructed in [Co3], [Me5]. Steffen constructed such a polyhedron with 9 vertices (see [Co3]). I. G. Maksimov [Ma] proved that the number of vertices must be at least 8. There are strong reasons to believe that this number cannot equal 8. A similar question (diminishing the number of vertices) may also be posed for immersed flexible closed polyhedra of various topological types. In all known examples, continuously flexible immersed closed polyhedra preserve volume during flex. By the Connelly–Sullivan conjecture, this property must be enjoyed by all flexible polyhedra. Studies in the Connelly– Sullivan conjecture ended with the following general theorem by I. Kh. Sabitov [S5, S6]: Under an immersion in three-dimensional space, even with selfintersections, of an arbitrary closed oriented simplicial two-dimensional complex given by a 2-polyhedron with flat embedding of every face, the oriented volume V of the polyhedron is a root of some polynomial whose coefficients depend only on the edge lengths and the combinatorial structure of the complex. A slightly different proof of this result was given in [CSW]. Since the roots of a polynomial constitute a discrete set, this theorem implies that the volume V does not change during any continuous flex with edge lengths and combinatorial structure preserved. Although flexible polyhedra are in a sense exceptional (see the footnote on p. 136), for them one would like to know the topological structure of the set of isometric polyhedra, i.e., the “configuration space” of all polyhedra with a fixed development. Clearly, this space is an algebraic variety. Interest in the structure of this variety was probably first expressed in [Gl1]. The simplest example of possible nontrivial structures for the positions of a closed planar quadrilateral with preassigned side lengths is examined in [S5, pp. 249–250]. Configuration spaces for degenerate “suspensions” and Bricard’s octahedra were considered in [S2] and [BS]. Some general facts on the topology of the set of all configuration spaces were established in [Ka]. Finally, the question of distinguishing classes of nonconvex polyhedra admitting no continuous flexes naturally arises. We indicate two relevant articles: [Co1] and [S3].

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3.6.2 The question of isometric mappings from closed or unbounded convex polyhedra onto arbitrary convex surfaces was first solved in its full generality by S. P. Olovyanishnikov [Ol1, Ol2]. From his articles we can extract the following results: (1) Every isometric mapping from a bounded convex polyhedron P of total curvature 4π onto a convex surface F is given by a motion or a motion and a reflection, so that F is necessarily a polyhedron congruent to P . (2) Every isometric mapping from an unbounded convex polyhedron P of total curvature 2π onto a convex surface F is given by a motion or a motion and a reflection, so that F is necessarily a polyhedron congruent to P . (3) Suppose P is an unbounded convex polyhedron of curvature less than 2π, V (P ) is the limit angle of P , F is an unbounded convex surface, and V (F ) is the limit cone of F (i.e., the result of the infinite similarity contraction of F to some point). An arbitrary isometric mapping ϕ from P onto F induces some isometric mapping ψ from V (P ) onto V (F ). The mapping ϕ is given by a motion, or a motion followed by a reflection, if and only if so is ψ. (4) Theorem 3 of Section 3.5 about polyhedral caps admits the following generalization: A convex surface with flat boundary which is isometric to a polyhedral cap is itself a polyhedral cap.17 The proofs of these theorems are in two parts: first, one proves that, under the stipulated conditions, a surface F isometric to a polyhedron P is itself a polyhedron; after that, it suffices to refer to the proved theorems about mappings between polyhedra. The first part of the proof is based on the fact that an isometric mapping takes the faces of a polyhedron onto developable surfaces, so that the support planes to the surfaces necessarily pass through points corresponding to the vertices of the polyhedron.18 With this fact established, it is comparatively easy to prove that under due conditions the surface F is a polyhedron. 3.6.3 The question of the rigidity of closed convex surfaces has a long history. As far back as in 1838, Minding conjectured the rigidity of the sphere. In 1954 Gillett published an erroneous proof of this conjecture, and only in 1899 it was proved independently by Liebmann and Minkowski in the following form: every sufficiently regular surface (twice continuously differentiable in 17

In 1967 Yu. A. Volkov proved the following exceptionally strong result: he gave an estimate for the variation of the external shape of a polyhedral cap depending on the variation of its intrinsic metric. This yields an independent proof of the stated assertion. – V. Zalgaller 18 Otherwise, the area of the spherical image of the surface into which the faces with excluded vertices are transformed would be, as is easily seen, positive. At the same time, every isometric mapping between convex surfaces preserves the area of the spherical image. This general result is established in my book [A15, Chap. 5]. Olovyanishnikov had managed without it, proving the indicated assertion on support planes directly.

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Minkowski’s proof) isometric to a sphere is a congruent sphere. Another proof was given by Hilbert in 1901. Later the question of rigidity of closed convex surfaces was addressed by Liebmann, Weyl, and Blaschke, but only for infinitesimal flexes. Finally, in 1927 S. Cohn-Vossen proved in general that any two isometric analytic closed convex surfaces whose curvature is positive everywhere are congruent. Later he managed to relax the analyticity requirement to thrice continuous differentiability. Cohn-Vossen’s method is in fact an analog of Cauchy’s method. In 1946 Herglotz gave a new very simple proof to the same theorem, simultaneously removing the requirement that the curvature be everywhere positive.19 All the above results have one common shortcoming. It is always assumed that the given surface and the surface isometric to it are sufficiently regular from the standpoint of their spatial shapes rather than only intrinsically. Therefore, none of the theorems even allows us to assert that every convex surface isometric to a sphere is itself a sphere irrespectively of its presupposed regularity. The first general results free of this shortcoming, and, moreover, applicable not only to closed convex surfaces, were established by A. V. Pogorelov in 1948. Let us say that a surface is of bounded specific curvature if the ratio of the excess of the sum of angles of a geodesic triangle over π, i.e. α + β + γ − π, to the area of this triangle is bounded above by the same number for all triangles. It is noteworthy that this condition, while not too restrictive, is of an intrinsic nature. Pogorelov’s results assert that Theorems 1–4 formulated in the preceding subsection for polyhedra are also valid for convex surfaces of bounded specific curvature. In their statements, we only have to replace ”convex polyhedron” by “convex surface of bounded specific curvature,”20 Pogorelov’s proofs are based on a combination of rather subtle arguments which are quite different from the arguments of Cohn-Vossen and others. However, the methods are considerably simplified when we confine ourselves to regular surfaces. The question of isometric mappings between general convex surfaces, distinct from polyhedra and failing to be of bounded specific curvature, still remains open. Nevertheless, there are reasons to believe that all four Theorems 1–4 on polyhedra are valid for arbitrary convex surfaces as well.21 19

See [Hr] or N. V. Efimov’s articles [E2], [E3]. See [P3]. 21 At present, this is known. A. V. Pogorelov was the first to prove that general closed convex surfaces are uniquely determined by their intrinsic metric [P5], also see [P10, Chap. 4]. Yu. A. Volkov obtained this result independently as the consequence of the estimate he derived for the distortion of the shape of a convex surface depending on the distortion of the intrinsic metric of the surface. The problem of deriving such an estimate was already raised by Cohn-Vossen [CV]. A. V. Pogorelov also proved analogs of the results of Section 3.4 for general unbounded convex surfaces [P10, Chap. 3] as well as that general closed convex surfaces are uniquely determined by their metrics in spaces of constant positive 20

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3.6.4 As far as non-Euclidean spaces are concerned, we can observe that the lemmas of Section 3.1 do not use the parallel axiom at all, since they deal with spherical polygons. The Cauchy Lemma, topological by nature, relates in no way to the type of geometry in space. Therefore, Theorem 1 of Section 3.2 and Theorem 1 of Section 3.3 on closed convex polyhedra can be carried over to to non-Euclidean space (hyperbolic or spherical) word for word together with their proofs, except for an additional condition in the case of spherical space. The matter is that the Cauchy Lemma essentially uses the absence of digons among the faces of a polyhedron. In hyperbolic space, as in Euclidean space, there are no digons and so no difficulties arise. In contrast, digons occur in spherical space; these are ordinary digons bounded on a sphere by two great circles. In this connection, polyhedra in spherical space may have digons as faces. However, it can be shown that a convex polyhedron in spherical space has no digonal faces unless it consists of digons only. Then it has exactly two vertices, with all digonal faces touching at these vertices, while adjoining each other along sides. (The limit case is the sphere, i.e., the “plane” of spherical space.) It is easy to verify that such figures with two vertices (except for the limit case) are flexible, implying that Theorem 1 of Section 3.2 and Theorem 1 of Section 3.3 fail for them. The theorems of Section 3.5 on bounded nonclosed polyhedra can also be translated word for word to non-Euclidean spaces (hyperbolic and spherical) provided that we exclude polyhedra whose completions are digons. Quite different is the case of theorems on unbounded polyhedra, since parallelism plays an essential role in them. These theorems, nevertheless, can also be carried over to hyperbolic space provided that parallelism is interpreted in the hyperbolic sense: two unbounded edges are regarded as parallel to one another if they lie in the same plane and asymptotically converge. Using this definition, we can carry over to hyperbolic space Theorem 2 of Section 3.2 and Theorem 2 of Section 3.3 on polyhedra with parallel unbounded edges and Theorems 1 and 2 of Section 3.4 on polyhedra with nonparallel unbounded edges. All arguments in their proofs remain valid. Similarly, Theorem 3 of Section 3.2 is can be reworded to hyperbolic space without changes. These theorems deal with polyhedra homeomorphic to the plane. However, in hyperbolic space an unbounded convex polyhedron with at least one vertex (and thereby not reduced to an infinite prism) can fail to be homeomorphic to the plane. Such a polyhedron can be homeomorphic to an arbitrary finitely connected domain on the plane, i.e., to a domain with any finite number of “holes.”22 Hence the scope for all possible theorems on unbounded curvature [P10, Chap. 5, § 6]. In the case of hyperbolic spaces, this was proved for closed convex surfaces by A. D. Milka [Mi5], while G. Gayubov [Gay] settled the case of unbounded complete surfaces. – V. Zalgaller 22 To prove this, take the well-known model of hyperbolic space given by the interior of a Euclidean ball B. Straight lines in this model are the chords of B and planes are the disks through great circles of B. Since here the hyperbolic and Euclidean straight lines coincide, a solid body contained in the ball B and convex in the

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convex polyhedra in hyperbolic space is considerably wider than that of Euclidean space. Apparently, detailed research into the theory of unbounded convex polyhedra in hyperbolic space promises many interesting results.23 As a problem on polyhedra homeomorphic to the plane, we can propose that of their limit angle. Since there is no similarity contraction in hyperbolic space, it follows that given a polyhedron P , the limit angle V (P, O) of P with vertex O should be defined as the boundary of the solid angle filled by all rays issuing from O and contained in the solid body bounded by the polyhedron P . Does this angle depend on the point O? What are the relations between the faces and edges of this angle and the unbounded faces and edges of the polyhedron? 3.6.5 Generalizing the theorems of the current chapter to higher-dimensional polyhedra makes no sense, since even a single convex polyhedral angle in four-dimensional space is rigid. To prove this, draw a three-dimensional sphere around the vertex of such an angle V . The angle V cuts out a convex polyhedron P from the sphere. The polyhedron P is a closed polyhedron in three-dimensional spherical space. However, as mentioned, the rigidity theorem for closed convex polyhedra is also valid in spherical space. Therefore, the angle V is also rigid. Of course, the same holds in spaces of higher dimension. We conclude that no convex polyhedron in a space of dimension greater than three admits nontrivial isometric mappings. The question of the equality of dihedral angles with equal planar angles on two-dimensional faces turns out to be trivial for the same reasons.

Euclidean sense is convex in the hyperbolic sense and vice versa. The boundary of the ball B consists of the infinitely distant points of hyperbolic space. Take a convex polyhedron in B whose vertices A1 , . . . , An lie on the boundary sphere of B and vertices B1 , . . . , Bn lie in the interior of B. This polyhedron is an unbounded convex polyhedron in hyperbolic geometry. Its unbounded edges touch at infinitely distant points Ai . Obviously, the polyhedron is homeomorphic to the sphere with n deleted points A1 , . . . , An and therefore is homeomorphic to a planar domain with n − 1 “holes.” 23 Some studies in this direction will be outlined in footnotes on pages 263 and 343. Here we indicate the articles [Ver3, Ver4] describing various approaches to specifying unbounded convex polyhedra in hyperbolic space. – V. Zalgaller

4 Existence of Polyhedra with Prescribed Development

The first three sections of the current chapter are devoted to proving the existence of closed convex polyhedra given their development. The proof is based on the Mapping Lemma (Section 2.1 of Chapter 2). In Section 4.1 we define and study the manifold M of developments of fixed structure. In Section 4.2 we consider the manifold P of polyhedra produced from developments of given structure by gluing. The proof is accomplished in Section 4.3, where we apply the Mapping Lemma to a certain mapping from P into M. The concluding two sections (4.4 and 4.5) deal with unbounded polyhedra. Generalizations of the results of this chapter in the spirit of the last section of Chapter 3 occupy Section 5.3 of Chapter 5.

4.1 The Manifold of Developments 4.1.1 We consider developments homeomorphic to the sphere and having the sum of angles at each vertex less than or equal to 2π. Our final goal is to prove that such a development defines a closed convex polyhedron by gluing. This goal is achieved in Section 4.3. In the current section we obtain some auxiliary results about developments. Since each bounded polyhedron can be partitioned into triangles, we may confine our exposition to developments composed of triangles only. In Sections 4.1–4.3 we only consider developments homeomorphic to the sphere and composed of triangles, and simply call them developments for short. A development with the sum of angles less than or equal to 2π at each vertex is referred as a development of positive curvature. If the sum of angles at each vertex is strictly less than 2π, then we say that the development has no redundant vertices or that it is a development of strictly positive curvature. As was indicated in Section 1.6 of Chapter 1, each development has its intrinsic metric. In this metric, we can draw a shortest arc between any two points (Theorem 1 in Section 1.8 of Chapter 1). Such an arc consists of straight line segments on the polygons of the development, where the passage from one segment of a shortest arc to another is made through identified boundary points of the polygons. We further use some simple theorems on

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shortest arcs and curvatures in developments. These theorems were proved in Section 1.8 of Chapter 1. It is convenient to view a development as mapped onto some sphere S so that the metric and shortest arcs of the development are carried over to S. If two developments can be mapped onto one another in such a way that the distances between corresponding pairs of points are equal, then we say that the developments are isometric.1 If a development defines a closed convex polyhedron by gluing, then we say that the development is realizable. Obviously, each development isometric to a realizable development is realized by the same polyhedron. Owing to this circumstance, a development can always be replaced by an isometric development. 4.1.2 Lemma 1. Every development has at least three vertices at which the sum of angles is less than 2π. If a development has exactly three such vertices and the sum of angles at each of the remaining vertices (if the latter exist) is equal to 2π, then the development is realizable as a doubly-covered triangle. Indeed, the total curvature of a development homeomorphic to the sphere is 4π (see Theorem 3 in Section 1.8 of Chapter 1), while the curvature at one vertex is less than 2π. The first assertion of the lemma follows. Now assume that a development R has exactly three vertices A, B, and C at each of which the sum of angles is less than 2π, while the sum of angles at each of the remaining vertices equals 2π. Map the development onto a sphere S and connect the vertices A, B, and C by shortest arcs. The sphere splits into two triangles, each of which contains no interior points with the sum of angles other than 2π. Therefore, the triangles are developable on a plane (Theorem 4 in Section 1.8 of Chapter 1). Superposing them so that the corresponding sides coincide, we obtain a doubly-covered triangle that realizes the development R. Lemma 2. Every development of positive curvature is isometric to a development satisfying the following conditions: (1) the development is composed of triangles; (2) the development has no “redundant” vertices, i.e., vertices where the sums of angles equal 2π; and (3) no two vertices of any triangle of the development are glued together, i.e., correspond to the same vertex of the development. Proof. Given a development R, denote by A1 , . . . , Av all its nonredundant vertices. By Lemma 1, v ≥ 3. Map the development R to a sphere S. Connect the vertex A1 with the other vertices A2 , . . . , Av by shortest arcs and cut the sphere S, or equivalently the development R, along these shortest arcs (see Fig. 88(a) which displays the situation “topologically”). 1

It is easy to verify that if two developments are isometric, then we can pass from one to the other by finitely many operations of cutting and gluing.

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195

Since shortest arcs issuing from the same point never meet again, after making cuts we obtain a (geodesic) polygon Q homeomorphic to a disk. The interior of Q contains no vertices. Hence, in accordance with Theorem 5 of Section 1.8, the diagonals split Q into triangles each of which is developable on a plane. 1

2

A1

A1 Aq T

Ap

A1

A2 A 3

A4

Fig. 88(a)

3

A1 Q

6

A1

5

A1 A4 1

Fig. 88(b)

However, we wish to prove that Q splits by diagonals into triangles so that no two vertices of any triangle in the resultant development are glued together, i.e., correspond to the same vertex of the development. The polygon Q has vertices of two types: v − 1 vertices correspond to the vertices A2 , A3 , . . . , Av of the development, while the other v − 1 vertices cor, respond to the single vertex A1 (Fig. 88(b)). These vertices A11 , A21 , . . . , Av−1 1 called vertices of the second type, alternate on the perimeter of Q. Since only vertices of the second type correspond to the same vertex of the development, we must prove the following: The polygon Q may be split by diagonals into triangles in such a way that at most one vertex of each triangle is a vertex of the second type. To prove this, observe that the sum of angles at the vertices of the “second type” is nothing but the complete angle at A1 and so it is less than 2π. Therefore, the angle at a vertex can be greater than or equal to π for at most one vertex of the second type. Assume that the angle at the vertex A11 is less than π. Let Ap and Aq be the vertices adjacent to A11 . Draw the segment Ap Aq and consider the triangle T = A11 Ap Aq (Fig. 88(b)). Since the angle at A11 in Q is less than π, this angle is also an angle of the triangle T . Hence, the intersection T ∩ Q of T and Q is a polygon; in particular, this polygon may coincide with T . In this case, removing T from Q, we obtain a polygon Q \ T which has one vertex of the second type. Otherwise, we draw the shortest arc that consists of polygonal lines joining Ap to Aq in the polygon T ∩ Q (Fig. 89).

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4 Existence of Polyhedra with Prescribed Development 1

A1 TQ Ap

Aq

Fig. 89

Clearly,2 the polygonal line Ap Aq is shorter than the polygonal line formed by the sides Ap A11 and Aq A11 : Ap Aq < Ap A11 + A11 Aq .

(1)

Were there some vertex Aj1 of the second type on the arc Ap Aq , we would have Ap Aj1 + Aj1 Aq < Ap A11 + A11 Aq . In this case, at least one of the segments Ap Aj1 and Aj1 Aq of the arc would be shorter than the corresponding segment Ap A11 or A11 Aq , say, Ap Aj1 < Ap A11 . However, the side Ap A11 is by construction a shortest arc between the vertices Ap and A11 in the development under consideration. The line Ap Aj1 also joins Ap and A1 . This line cannot be shorter than the shortest arc; hence, (2) and (1) are both impossible. Thus, the assumption that some vertex of the second type lies on Ap Aq is false. We have thus proved that the polygon separated from Q by the polygonal line Ap Aq contains only one vertex A11 of the second type. Cutting off this polygon, we are left with a subset of Q constituted by polygons with fewer vertices of the second type. The above arguments apply to each of these polygons. So from each of them we can cut off polygons with only one vertex of the second type, continuing the procedure until we split the whole development into polygons each of which has at most one vertex of the second type. Thus we have subdivided each of these polygons into triangles and come to the desired result. In view of Lemmas 1 and 2, we can restrict the exposition to developments composed of triangles and satisfying the following conditions: (1) the 2

Here and below AB denotes both a polygonal line with extremities A and B and its length; similarly, AB denotes the line segment with endpoints A and B as well as its length. – Transator’s Note.


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number v of vertices is greater than 3, (2) there are no redundant vertices, and (3) no vertices of any constituent triangle are glued together. We now agree that these conditions hold without additional mention. 4.1.3 Ignoring the metric relations in the development and focusing our attention only on its combinatorial structure, i.e., considering only identifications of sides and vertices of triangles, we transform a development into a complex. Thus, a complex is a collection of finitely many abstract triangles with identifications of the sides and vertices satisfying all the conditions appearing in the definition of development. This definition is a bit more general than the one adopted in combinatorial topology. Namely, we waive the requirement that at most one side (vertex) of one of two arbitrary triangles be identified with a side (respectively vertex) of the other. However, self-gluings of a triangle are prohibited by the agreement made at the end of the preceding subsection. Since our development is homeomorphic to the sphere we see that if the triangles are regarded as arbitrary homeomorphic images of ordinary triangles, our complex turns out to be homeomorphic to the sphere. To put it briefly, a complex is a collection of topological triangles with sides and vertices identified, which as a whole is homeomorphic to the sphere. With the triangles of a complex K given as planar triangles and the identifications of sides preserving lengths, we actually specify a development. In this case we say that K transforms into a development with structure K. Given a complex K, we denote by f , e, and v the numbers of triangles, edges, and vertices of K. Planar triangles are determined uniquely from their side lengths. So, to transform a complex K into a development, it suffices to assign lengths to the edges of K. Consequently, a development with structure K is determined by specifying k parameters, the edge lengths r1 , . . . , rk . The set of all developments with structure K can be represented as some domain M0 in k-dimensional space. The only conditions to be met by the edge lengths are the “triangle inequalities”: the sum of lengths of every two sides of a triangle must be greater than the length of the third side3 : ri + rj > rl . These inequalities are linear, and so each of them specifies some halfspace. The intersection of these half-spaces is the domain M0 . (Incidentally, we see that M0 is the interior of a convex polyhedral angle with vertex the origin.) The domain M0 is certainly nonempty: if we assign the same length to every edge, then all the triangle inequalities will be satisfied. Thus, for every complex K, there is a development with structure K consisting of equilateral triangles. 3

The positivity of lengths is a consequence of these inequalities (ri + rj > rl and ri + rl > rj , implying 2ri + rj + rl > rj + rl , whence ri > 0).

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The collection of developments of strictly positive curvature is represented 0 by some subset M of M . The sum j ϕij of angles touching at the ith vertex is a continuous function of the edge lengths, where j enumerates the angles at the ith vertex. The set M is determined by the inequalities ϕij < 2π (i = 1, 2, . . . , v). j

Therefore M is open. We call M the manifold of developments of strictly positive curvature. The manifold M may be empty. However, in that case there is nothing for us to speak about, since our aim is to prove that, given a development of positive curvature with structure K, a convex polyhedron is produced from it by gluing. For this reason, we assume M nonempty.4 4.1.4 Lemma 3. If a complex K has more than three vertices, then the manifold M is a proper subset of the domain M0 and consequently has a boundary in M0 . C

A

B

D Fig. 90

We first prove that, in a complex with more than three vertices, there are at least three triangles touching at a single vertex. To this end, it suffices to show that if at most two triangle touch at each vertex, then the complex has at most three vertices. Let ABC be a triangle in such a complex. Assume that some triangle ABD is glued to the side AB of ABC (Fig. 90). Now two triangles touch at 4

The emptiness of M means that K can be transformed into a development with the sum of angles less than 2π at each vertex. Examples of such complexes are easy to construct if we allow triangles with vertices glued together at a single point. However, these “self-gluings” are prohibited by the agreement made at the end of Subsection 4.1.2. It seems plausible that a complex without self-gluings can always be transformed into a development of strictly positive curvature, implying that M is in fact nonempty. However, the proof in the general case evades us. We can only settle the case in which every two triangles are glued together along at most one side or at most one vertex.


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the vertices A and B. So, if there are no other triangles touching at A and B, then the sides AC and BC are obviously glued to AD and BD. (It could happen that BC is glued to AC and BD to AD, but then the vertices A and B of ABC would be identified, which is prohibited by the agreement of Subsection 4.1.2.) So only three vertices remain, and nothing can be added to the complex since there are no free sides. This completes the proof of our assertion. Now, assume that K has more than three vertices. The above then shows that there is a vertex A in K such that at least three triangles touch at A. Assign equal lengths to all edges of K, i.e., take a development consisting of equilateral triangles. Take the same fraction of the lengths of all edges issuing from A, while keeping the other edges constant. Continuing the procedure, observe that the angles touching at A increase and can be made arbitrarily close to π. Since there are at least three such angles, their sum becomes greater than or equal to 2π. We thus arrive at developments not contained in M, which completes the proof of the lemma. The boundary of M consists of the points corresponding to developments in which the sum of angles at each vertex is less than or equal to 2π and which have at least one vertex with this sum equal to 2π. Indeed, the manifold M is determined by the inequalities ϕij < 2π (i = 1, 2, . . . , v),

(3)

j

with ϕij denoting the jth angle at the ith vertex. The angle ϕij is a continuous function of the edge lengths. Therefore, on the boundary of M none of these inequalities can fail and at least one of them must be an equality. The above also implies that the boundary of M is composed of pieces of v surfaces F1 , F2 , . . . , Fv determined by the equations ϕij = 2π (i = 1, 2, . . . , v), (4) j

where the angles ϕij should be regarded as functions of the edge lengths, i.e., as coordinates in k-dimensional space.5 We now prove our last lemma, which will play a crucial role in Section 4.3 in the proof of the theorem on the realization of developments of positive curvature. 5

Although we call the sets Fi that are determined by equations (4) surfaces, some of them may be empty. Namely, if only two triangles, and so only two angles, touch at the ith vertex, then the sum of these angles is clearly always less than 2π. (In all other cases equation (4) really defines a (k − 1)-dimensional surface in k-dimensional space. However, we do not need this assertion and thus omit its proof.)

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Lemma 4. The boundary of each connected component of M contains a point in some neighborhood of which there are no points of the other connected components of M.6 Proof. Let M be some connected component of M. Take a point X0 on the boundary of M which belongs to the least number of surfaces F1 , F2 , . . . , Fv , say F1 , F2 , . . . , Fl . Let us show that this point satisfies the assertion of the lemma.7 The point X0 represents a development R0 for which the sum of angles at each of the vertices A1 , . . . , Al equals 2π, whereas this sum is less than 2π at each of the other vertices. Clearly, we can indicate an ε > 0 so small that the sum of angles at each of the vertices Ai (i > l) never become equal to 2π under any change less than ε in the lengths of edges in R0 . We consider only such changes. This means that we consider a neighborhood W about X0 such that W contains no points of any surface Fi other than F1 , . . . , Fl . r

U2 X2

X0 X1

W U1

Fig. 91(a)

Consider the triangles of R0 touching at the vertex A1 . Let r be the edge of one of them subtended by A1 . Since the angle in a triangle is an increasing function of the side that it subtends, the sum ϕ1j of the angles at A1 is also an increasing function in the length of r. (The edge r may also be subtended 6

7

Actually, almost every point of the boundary enjoys this property. This is easily seen from the analyticity of the function ϕij . However, we use only elementary tools and the result obtained is quite sufficient for us. We do not know whether the manifold M is always connected. Had we succeeded in proving connectedness, the whole proof would have become much simpler. In fact, this least number of surfaces equals one; moreover, almost every point of the boundary lies on exactly one surface Fi . We do not prove this assertion since we can manage without it.


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by A1 in another triangle. In that case our conclusion on the sum of angles remains valid. The edge r cannot contain A1 . Otherwise, two vertices of the first triangle would coincide with A1 , contradicting the condition imposed on the developments under consideration.) Shortening the edge r, we obtain a development R1 with the sum of angles at A1 less than 2π. Lengthening r, we obtain a development R2 with the sum of angles at A1 greater than 2π. X1 and X2 . The developments R1 and R2 , correspond to some points Take neighborhoods U1 and U of these points so as to have ϕ1j < 2π 2 at every point of U1 and ϕ1j > 2π at every point of U2 . Clearly, we may choose U1 and U2 as translates of one another so that the translation taking X1 to X2 carries U1 to U2 . (In Fig. 91(a), U1 and U2 are displayed as squares for simplicity.) To the edge r corresponds the r-axis of the k-dimensional space in which the edge lengths serve as coordinates. Any change of only one edge r yields a shift along the r-axis. Since U1 and U2 are translates of one another, the segments joining their corresponding points Y1 and Y2 are parallel to the r-axis. All such segments cover some neighborhood V of X0 which has the shape of a prism. We shall prove that it is precisely in V that there are no points of the connected components of M other than M . r V2 F1 X0 Y Y1

Z Z1

V1

Fig. 91(b)

At the point Y1 (Y2 ), the sum ϕij is less (greater) than 2π. When we Therefore, there is move from Y1 to Y2 , this sum increases monotonically. a unique point Y on the segment Y1 Y2 at which ϕij = 2π. This means that each segment Y1 Y2 with endpoints in U1 and U2 intersects the surface F1 at a single point. Thus, the whole “prism” V is divided by F1 into two pieces V1 and V2 .

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In V2 , we have ϕij > 2π; hence, no points of V2 can belong to M. However, X0 lies on the boundary of the connected component M of M. Consequently, there are points of M near X0 and so in V1 . We shall now prove that M contains the whole domain V1 . Assume the contrary, and let Y be a point in V1 not belonging to M (Fig. 91(b)). Let Z be some point of V1 that belongs to M . By the construction of the “prism” V , the points Y and Z lie on segments issuing from some points Y1 and Z1 of U1 . Therefore, drawing the segment Y1 Z1 , we obtain a continuous polygonal line L = Y Y1 ∪ Y1 Z1 ∪ Z1 Z that joins Y to Z within V1 . Since Z lies in the component M while Y lies outside of it, the polygonal line L must intersect the boundary of M . The intersection point, being a point of the boundary, must lie on some surface Fi . Belonging to V1 , this point cannot lie on F1 . Recall that the region W was chosen so as to contain no points of the surfaces Fi other than F1 , . . . , Fl . Therefore, the intersection point can lie only on the surfaces F1 , . . . , Fl . However, by the definition of X0 , no point of the boundary of M belongs to less than l surfaces Fi . We come to a contradiction, which shows that V1 is entirely contained in the component M . At the same time, V2 contains no point of M at all. Hence, the “prism” V is the required neighborhood of X0 without any points of any components of M other than M . The proof of the lemma is complete.

4.2 The Manifold of Polyhedra 4.2.1 Together with the manifold of developments with given structure K, we consider convex polyhedra produced from developments with the structure K by gluing. Of course, at this point we do not know whether there is at least one such polyhedron, even if there are developments of positive curvature with structure K. Nevertheless, in the current subsection we proceed as though such polyhedra exist. In the next subsection we shall prove that in fact they always do. Thus, we consider convex polyhedra produced by gluing from developments of positive curvature with structure K. By definition (Subsection 1.6.3), a polyhedron P is produced from a development R by gluing if it admits a decomposition into (not necessarily planar) triangles such that (1) the complex of these triangles has the same structure as K and (2) each triangle of the decomposition can be developed on the plane in such a way that the sides of the triangle become equal to the sides of the corresponding triangle of the development R. In Subsection 1.6.3, we also noticed that gluing is nothing but an isometric mapping of a development onto a polyhedron. Thus, constructing a polyhedron P from a development R by gluing, we define a mapping from R onto P and thereby a mapping of K onto P . A polyhedron

4.2 The Manifold of Polyhedra

203

P with a mapping from some development (and so from some complex) onto it will be called a K-triangulated polyhedron. The complex of triangles on P induced by such a mapping is known as a K-triangulation of P . Graphically, a K-triangulated polyhedron is a polyhedron with a plotted K-triangulation. We consider only developments of strictly positive curvature; for them the sum of angles is less than 2π at each vertex and so the vertices of every K-triangulation are necessarily vertices of a polyhedron. Two K-triangulated polyhedra are equal if there is a motion or a motion followed by a reflection which maps one of the polyhedra to the other so that the K-triangulation of the former becomes the K-triangulation of the latter. In this case, the mappings of K onto these polyhedra also coincide.8 Thus, the set of all K-triangulated polyhedra decomposes into classes of equal polyhedra. We intend to make this set of classes into a manifold by suitably introducing neighborhoods. The result will be the manifold P of K-triangulated polyhedra. 4.2.2 In the following lemma we additionally allow triangulations to have vertices other than those of the polyhedron. We need this strengthening only in Section 4.3. In the current subsection, except for Lemma 1, we assume that the vertices of a triangulation are those of the polyhedron. Lemma 1. Let T0 be a triangulation of a convex polyhedron P0 such that each triangle of the triangulation is developable on a plane. Then it is possible to find an ε > 0 such that if the vertices of some polyhedron P are at distances less than ε from the vertices of T0 (with exactly one vertex of P corresponding to each vertex of T0 ), then there is a unique triangulation of P close to T0 that has the same structure. Proof. We consider P as resulting from small displacements of vertices of the triangulation T0 (i.e., as the surface of the convex hull of the displaced vertices). Under such displacements, the faces of the initial polyhedron P0 may fold in general. The number of different ways in which the faces of P0 fold, yielding polyhedra P of different structures, is clearly finite. Consider polyhedra P of some fixed structure. Triangulate all faces of these P in some unified manner by taking the vertices of the polyhedron as vertices of the triangulation and triangulate P0 in similar way (here the vertices of the triangulation are those 8

If a development R admits different mappings onto some polyhedron, then to each of the mappings there corresponds a K-triangulated polyhedron. For example, given a regular tetrahedron, we can produce different K-triangulated tetrahedra by changing the correspondence between the vertices of the tetrahedron and the vertices of the developments. Fig. 37 (p. 53) displays different developments of a regular tetrahedron with the same structure. To these developments correspond different K-triangulated tetrahedra, which, moreover, are not equal.

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in T0 ). As a result, we can regard the polyhedra P and P0 as composed in a unified manner of triangular faces, some of which may lie in a single plane. Develop some edge AB of the triangulation T0 on a plane together with all traversed faces of P0 . Denote the resulting development by Q (Fig. 92). Note that when we pass to the polyhedron P , the displacements of the vertices of T0 cause the faces of Q to vary. However, for continuous displacements of the vertices, the faces will vary continuously. Therefore, there is a δAB > 0 such that under displacements less than δAB the development Q changes so little that the segment AB remains inside Q.

A

B Q Fig. 92

Taking ε less than all such δ’s for all edges of T0 and for all possible structures of P , we come to the required ε. Indeed, under displacements less than ε, the vertices of the triangulation T never traverse its edges (as they never traverse AB under displacements less than δAB ). Hence, the edges bounding some triangle in T continue to bound a triangle. Therefore, the triangulation structure is preserved. Besides, there is only one modified triangulation, for the simple reason that only one segment joins A to B in the modified development Q. (On the polyhedron P there may exist other geodesic segments AB, but they cannot be arbitrarily close to the fixed segment. In fact, they lie at some finite distance from the segment, this distance depending on the “width” of the development Q.) Thus, on each polyhedron P with vertices close to the vertices of T0 there is a unique triangulation T with the same structure and close to T0 as regards both its disposition in space and its edge lengths. 4.2.3 We now define neighborhoods in the manifold of K-triangulated polyhedra. Let P0 be a K-triangulated polyhedron and let A, B, and C be the three vertices of P0 corresponding to the vertices A, B, and C of the complex K. Choose a Cartesian coordinate system x, y, z and consider the motion of P0 that takes A to the origin, places B on the positive x half-axis, and sends C to the half-plane y > 0 of the xy-plane. If P0 does not degenerate into a doubly-covered polygon, then we choose one more vertex D of P0 not lying in the xy-plane and send D into the half-space z > 0 by reflecting in the plane z = 0 if need be. When the position of the polyhedron is specified in this manner, no motions or motions and reflections are possible. Therefore, if we place another K-triangulated polyhedron P1 equal to P0 in the same way, then P1 and P0 will coincide, and so will their K-triangulations.


205

Indeed, if P0 and P1 have equal K-triangulations, then some mapping from P1 onto P0 takes the K-triangulation of P1 to that of P0 . This mapping is an isometric transformation of P1 onto P0 . By Theorem 1 of Section 3.3, it is given by a motion or a motion and a reflection. However, under the conditions on the vertices A, B, C, and D, neither motion nor reflection is possible. Therefore, the coincidence of the polyhedra P1 and P0 implies that their K-triangulations coincide. Hence, each class of equal K-triangulated polyhedra has a unique representative meeting the above conditions on the vertices A, B, C, and D. We can thus speak of the manifold of polyhedra rather than the manifold of their classes. Suppose that the polyhedron P0 does not degenerate into a polygon. Applying sufficiently small (but in other aspects arbitrary) displacements of its vertices preserving the above conditions on the vertices A, B, and C, we obtain new convex polyhedra P . Each polyhedron P is the boundary of the convex hull of the displaced vertices. According to Lemma 1, if the displacements of vertices are sufficiently small, then the polyhedron P admits a unique K-triangulation close to the K-triangulation of P0 . Thus, the polyhedra P sufficiently close to P0 become K-triangulated. They are all different, since under the conditions imposed on their position, to each class of equal polyhedra there corresponds exactly one polyhedron. (Observe that under small displacements of vertices, the vertex D remains in the half-space z > 0.) We define the neighborhood of P0 as the collection of these polyhedra P . Now, assume that P0 degenerates into a doubly-covered polygon. Then P0 lies on the plane z = 0 and we can distinguish between its upper side looking at the half-space z > 0 and the lower side looking at the half-space z < 0. One part of the complex K is mapped onto the upper side and the other onto the lower side. Therefore these sides of a degenerate K-triangulated polyhedron are distinguished by the corresponding parts of K. Under small displacements of vertices preserving the conditions on the position of A, B, and C, the polyhedron P0 is taken to close polyhedra P . If the displacements are small enough, then the faces rotate slightly and none of them becomes perpendicular to the plane z = 0. Therefore, on each polyhedron P , we can distinguish between its upper part looking at z > 0 and its lower part looking at z < 0. By Lemma 1, the polyhedron P admits a Ktriangulation close to the K-triangulation of P0 . We choose this triangulation so that the part of K corresponding to the upper (lower) side of P0 is mapped onto the upper (lower) side of P . The K-triangulated polyhedra P thus obtained are all different. Indeed, by the conditions on the position of A, B, and C, two equal K-triangulated polyhedra may fail to coincide only if one of them is symmetric to the other with respect to the plane z = 0. However, under reflection in this plane the upper and lower sides interchange, but different parts of the complex K correspond to them by construction. Hence, although symmetric polyhedra

206


are equal, the mappings of the complex K to them differ, and therefore the polyhedra belong to different classes of equal K-triangulated polyhedra. We define the neighborhood of a degenerate polyhedron P0 as the collection of all polyhedra P obtained in this way. In short, a neighborhood of a K-triangulated polyhedron P0 is the collection of all K-triangulated polyhedra close to P0 together with their Ktriangulations. 4.2.4 We have demonstrated that all polyhedra P with vertices close to the vertices of P0 admit close K-triangulations and differ from one another. If v is the number of vertices in the complex K and so in our polyhedra, then we have 3v variable coordinates of vertices. However, the vertices A, B, and C are constant in the third, second, and first coordinate respectively. Therefore, we have 3v − 6 variable coordinates. Their variations are arbitrary within a sufficiently small range and yield different K-triangulated polyhedra. This means that each K-triangulated polyhedron P0 has a neighborhood homeomorphic to a (3v − 6)-dimensional cube. Therefore the collection of all K-triangulated polyhedra is actually a manifold of dimension 3v − 6. (Recall that, of course, equal K-triangulated polyhedra are not distinguished.) Lemma 2. If v is the number of vertices of the complex K and e is the number of its edges, then 3v−6 = e. Therefore the dimensions of the manifold P of polyhedra and the manifold M of developments coincide. Indeed, let f be the number of triangles in K. Each triangle has three sides and the sides are identified pairwise, so that 3f = 2e. By Euler’s Theorem, f − e + v = 2, or 3f − 3e + 3v = 6. Since 3f = 2e, it follows that ev − 6 = e. 4.2.5 Lemma 3. Let a sequence of K-triangulated polyhedra Pn have developments Rn converging to some development R.9 Then from the sequence Pn we can extract a subsequence of polyhedra Pni which converge to some K-triangulated polyhedron P together with their K-triangulations (i.e., the K-triangulation of P is the limit of K-triangulations of Pni .) In this lemma, polyhedra are considered up to a motion. Otherwise the lemma is false: polyhedra Pn going to infinity could be chosen. In this connection, we may assume that all polyhedra Pn contain the origin. Since the developments of Pn converge, the distances between their vertices are uniformly 9

The convergence of developments is the convergence of their edge lengths: this follows from the definition of the manifold of developments M. Developments of positive curvature certainly converge to a development of nonnegative curvature. Generally speaking, some fictitious vertices can occur in the development R and the limit K-triangulation of P . In essence, this does not affect the proof of the lemma. However, in what follows we need this lemma only in the case when the limit development has positive curvature.


207

bounded and the polyhedra themselves are enclosed in some ball centered at the origin. Let us enumerate the vertices of the complex K. Then the vertices of the polyhedra Pn (and of their developments Rn ) are also enumerated. Choose a subsequence of polyhedra for which the first vertices converge; then, from this subsequence, choose another for which the second vertices converge; and so forth. As a result, we obtain a sequence in which all vertices converge. For brevity, we denote the polyhedra of this sequence again by Pn and denote their developments by Rn . The convex polyhedron P spanning the limits of the vertices (i.e., the boundary of their convex hull) is the limit of the polyhedra Pn . We shall prove that P satisfies all the claims of the lemma. To this end, we first show that if some edge of Pn does not coincide with any edge of Rn , then the total number of intersections of this edge with all edges of Rn cannot exceed some number N0 , the same for all edges of all polyhedra Pn . (Here the development Rn is regarded as mapped onto Pn and is therefore its K-triangulation Rn .) Let L be the least upper bound of the edge lengths of Pn and let h be the greatest lower bound of all the altitudes of the triangles of the developments Rn . Observe that h > 0, because the developments Rn converge to some development R. Finally, let m be the largest number of angles contiguous to a single vertex of Rn (this m is the same for all Rn , since the latter have the same structure K.) I assert that for the required N0 we may take N0 =

2Lm + m. h

(1)

Assume the contrary. Then on some polyhedron Pn there is an edge a for which the number of intersections with the edges of Rn is N>

2Lm + m. h

The edge a is partitioned by N intersection points into N + 1 segments. Let us show that among these segments there are at least m successive segments each of length less than h/2. Otherwise, in every collection of m successive segments there would exist a segment of length greater than or equal to h/2. The total number of collections of m successive segments among all N + 1

+1 +1 (the integer part of the ratio Nm ). By (2) we segments is equal to Nm would then have N +1 N 2L 2L +1 > . ≥ ≥ m m h h Therefore, if each such collection contained at least one segment of length greater than or equal to h/2 then the

total length of these segments would +1 h be greater than or equal to Nm 2 which is greater than L; but this is impossible, because the length of the edge a is at most L. Consequently, there are m successive segments each of length less than h/2.

208


Let EF , the first of m successive segments of length less than h/2, lie in a triangle ABC of the development Rn (Fig. 93(a)). The endpoints of EF are at a distance from A less than half of the sides AB and AC. Hence, they are closer to A than to B and C. To check this, drop a perpendicular F H from F to AB. Since F H ≤ EF , it follows that EF is less than half of the altitude dropped from C only if F A < 12 AC. A similar argument holds for the point E. A

E

H F

B C Fig. 93(a)

Leaving ABC, the edge a enters a neighboring triangle ACD where it again has a segment less than h/2. Hence, in this triangle as well, it passes closer to the vertex A than to the others. Since we have a succession of m such segments, while at most m angles touch at A, it follows that the edge a, going around A, returns to the side AB and even intersects it again (Fig. 93(b)). a A

b

E

F

B

D C Fig. 93(b)

This is, however, impossible. Indeed, A is a vertex of the development Rn , and hence a vertex of the polyhedron Pn . Therefore, A must be incident to some edge b of Pn joining it to another vertex. However, by what was


209

established above, the edge b must intersect a, which is impossible, for edges meet only at vertices. We have thus proved that the number of intersections of an arbitrary edge of any polyhedron Pn with the edges of the development Rn does not exceed the number N0 of (1). It is also clear that, conversely, the number of intersections of an arbitrary edge of any development Rn with the edges of the polyhedron Pn does not exceed some number, the same for all polyhedra Pn .10 Now, let us enumerate the edges of each development Rn , assigning the same numbers to corresponding edges. Take the first edges of all developments. If infinitely many of them are edges of the corresponding polyhedra Pn , then we may choose a sequence of polyhedra Pni in which these edges constitute a converging sequence. If infinitely many of them merely intersect edges of Pn , then, since the number of intersections is finite, we may choose a sequence Pni in which the intersection points converge to some limit positions. Then the segments between adjacent intersection points also converge, and hence the first edges of the developments converge again. Applying the same arguments to the second edges, then to the third, etc., we eventually obtain a sequence of polyhedra Pnj such that all edges of the developments Rnj converge; moreover, the lengths of these edges converge as well. The limits of these edges form a net of the same structure on the limit polyhedron P and split P into triangles whose interiors do not contain vertices of the polyhedron. Indeed, the triangles of the developments Rnj are composed of pieces of faces of the polyhedra Pnj which touch one another at segments of edges. These pieces of faces converge to some polygons (possibly degenerate) on the faces of the limit polyhedron P , and the pieces of P composed of these polygons are bounded by the limits of the edges of the developments Rnj . Developed on the plane, these pieces become the limits of triangles of the developments Rnj , i.e., they are triangles themselves. So a development turns out to be plotted on P , with the same structure as the developments Rnj and with edge lengths equal to the limits of the edge lengths of these developments. Hence, the development plotted on P is exactly the limit development R. The elements (vertices, edges, and triangles) of the developments Rnj plotted on the polyhedra Pnj converge to the corresponding elements of the development R plotted on P . This means that the K-triangulated polyhedra Pnj converge to the K-triangulated polyhedron P together with their K-triangulations Rnj .

10

Indeed, the number of edges in an arbitrary polyhedron Pn does not exceed some number N1 that depends only on the number of vertices, which is the same for all polyhedra Pn . Therefore, the total number of intersections of all edges is at most N0 N1 and so there are at most N0 N1 intersections for each edge of the development.

210


4.3 Existence of Closed Convex Polyhedra with Prescribed Development Theorem. Each development homeomorphic to the sphere with the sum of angles at most 2π at each vertex defines some closed convex polyhedron by gluing.11 According to Lemma 2 of Section 4.1, each such development is isometric to a development of “positive curvature” in which the sum of angles is less than 2π at each vertex. We may thus restrict the exposition to developments of the latter kind. Moreover, by Lemma 1 of Section 4.1, a development with only three vertices is realizable by a doubly-covered polygon. For this reason, we may carry out a proof by induction on the number of vertices in the development: assuming the theorem proved for all developments with number of vertices less than v (v > 3), we shall prove it for developments with v vertices. Fixing v, we may finally inspect developments of an arbitrary given structure. It remains to prove the following claim: Let K be a complex with v > 3 vertices admitting a transformation into a development of positive curvature.12 If every development of positive curvature with less than v vertices is realizable, then every development R of positive curvature with structure K is also realizable. Consider the manifold M of developments R and the manifold P of Ktriangulated polyhedra P . By definition, to each K-triangulated polyhedron P corresponds a development R plotted on P . Hence, we have a natural mapping ϕ from P into M. We must prove that ϕ is a mapping onto M. This will establish that some polyhedron P is indeed produced from the given development R by gluing. By Lemma 2 of Section 4.2, the manifolds M and P have the same dimension. Therefore, we may use the Mapping Lemma (Section 2.2 of Chapter 2). By this lemma, to prove that ϕ takes P onto M, it suffices to verify the following conditions: (1) Each connected component of M contains images of points in P, i.e., realizable developments R. (2) ϕ is one-to-one. (3) ϕ is continuous. 11

In [15] the author proves this theorem by another method. Other proofs appeared later. One of them is due to L. A. Lyusternik (see [E2], [E3, § 13]). It consists in approximating the metric (on the sphere) of the given development by smooth Riemannian metrics of positive curvature and realizing the latter by following Weyl. The direct proof given by Yu. A. Volkov [Vo2] is more interesting. It is based on the general idea described below in Section 7.2.4 presented in more detail in the Supplement to Chapter 3. – V. Zalgaller 12 As we have seen from the example in Section 4.1, not every complex enjoys this property.

4.3 Existence of Closed Convex Polyhedra

211

(4) If points Rn of M are images of points Pn of P and converge to a point R (R ∈ M), then there is a point P in P whose image is R and there is a subsequence Pnj converging to P . In other words, if the developments Rn converge to R and the polyhedra Pn are produced from Rn by gluing, then there exist a K-triangulated polyhedron P and a subsequence Pnj of the sequence Pn such that the polyhedra Pnj converge to P together with their K-triangulations Rnj . The last condition is the content of Lemma 3 of Section 4.2; hence, we need only verify the first three conditions. In Section 4.2 we have already noticed that the existence of polyhedra admitting developments of structure K is not available a priori, i.e., the manifold P may be empty, which makes the mapping ϕ meaningless. However, the first of the above conditions requires that each connected component of M contain realizable developments. Clearly, this also implies the existence of K-triangulated polyhedra. Consequently, validating the first condition also includes proving that P is nonempty. Nevertheless, we begin by verifying the second and third conditions for ϕ, because they are immediate from the results already obtained. The mapping ϕ is one-to-one. Indeed, the single-valuedness of ϕ is implied by its definition, since ϕ assigns to each K-triangulated polyhedron a development serving as the K-triangulation of the polyhedron. At the same time, if two polyhedra P1 and P2 have the same development R, then there is an isometric correspondence between them taking the development R plotted on P1 to the development R plotted on P2 . By Theorem 1 of Section 3.3 in Chapter 3, such an isometric transformation of P1 onto P2 can be realized by a motion or a motion and a reflection. By the very definition of equality between K-triangulated polyhedra, this means that the K-triangulated polyhedra P1 and P2 are equal. Hence, to different K-triangulated polyhedra correspond different developments, and so the mapping ϕ is one-to-one. The mapping ϕ is continuous. In fact, this property of ϕ is implicit in the definition of neighborhoods in the manifold P. According to this definition, a neighborhood of a K-triangulated polyhedron P0 consists of polyhedra with close K-triangulations. Hence, to close K-triangulated polyhedra there correspond close developments, i.e., ϕ is continuous. It remains to prove that each connected component of M contains realizable developments. To prove this, we consider the manifold M0 of all developments with structure K as in Section 4.1. By Lemma 3 of Section 4.1, the manifold M is a subset with boundary of the manifold M0 (since the number v of vertices in the complex K is greater than 3 by assumption). The boundary of M consists of developments with the sum of angles less than or equal to 2π at each vertex and equal to 2π at some vertices. By Lemma 2 of Section 4.1, such a development is isometric to a development with the number of vertices less than v, which is realizable by the induction

212


hypothesis. Consequently, the boundary of M in M0 consists of realizable developments. Let M be some connected component of M. In accordance with Lemma 4 of Section 4.1, the boundary of M contains a point (a development R0 ) whose small neighborhood U contains no points (developments) of the other connected components of M. Since R0 lies on the boundary of M, this development is realizable as some convex polyhedron P0 . However, not all vertices of R0 are those of P0 , since the sum of the angles at some of them equals 2π. On the polyhedron P0 , to these vertices correspond some points A1 , . . . , Al lying inside the edges or faces of P0 . Move these points outside at a sufficiently small distance and consider the convex hull of the collection of these displaced points and of the vertices of the polyhedron P0 . The boundary of this convex hull is a convex polyhedron P with vertices close to the vertices of the development R0 plotted on P0 . Moreover, now genuine vertices of P correspond to the points A1 , . . . , Al . According to Lemma 1 of Section 4.2, the polyhedron P admits a K-triangulation R close to R0 . However, all vertices of this K-triangulation are at genuine vertices of the polyhedron P . Hence, the sum of angles at each vertex of R is strictly less than 2π. Therefore, R belongs to the manifold M. Since R is close to R0 and near R0 there are no other developments of M except those belonging to the connected component M under consideration, it follows that R belongs to M . However, the development R is realized by the polyhedron P . Thus, each connected component of M contains realizable developments. All conditions of the Mapping Lemma are thus verified, and the application of that lemma completes the proof of the theorem on the existence of a convex polyhedron with given development.

4.4 Existence of Unbounded Convex Polyhedra with Prescribed Development 4.4.1 As we have shown in Sections 1.6 and 1.7 of Chapter 1, a development of an unbounded convex polyhedron must satisfy the following two conditions: (1) “the positive curvature condition”: the sum of angles at each vertex of the development does not exceed 2π and (2) the development is homeomorphic to the plane. According to Theorem 10 of Section 1.7, the second condition is equivalent to the following two: (2a) the development contains at least one unbounded polygon; (2b) the development has one unbounded end, i.e., in going from an unbounded polygon to the one glued to the first along an unbounded side, then going on to the next polygon, etc., we return to the initial polygon, having traversed all unbounded polygons of the development. (In a particular case, the development may have only one unbounded polygon. In that case the

4.4 Existence of Unbounded Convex Polyhedra

213

above cyclic sequence consists of this polygon only, and the latter is glued to itself.) Conditions (1), (2a), and (2b) are easy to verify for any given development. We now prove that these conditions are sufficient for a development to define an unbounded convex polyhedron by gluing. Theorem. If a development satisfies conditions (1), (2a), and (2b), then it defines some unbounded convex polyhedron by gluing. 4.4.2 We proceed with the proof of the theorem. Let R be a development satisfying conditions (1), (2a), and (2b). Consider some unbounded polygon Qi of R. If the unbounded sides of Qi are not parallel to one another, then prolong them until they intersect and denote by αi the angle formed at the intersection containing the unbounded part of Qi . Call αi the angle between the unbounded sides of Qi . When the unbounded sides of Qi are parallel to one another, put αi = 0. If αi ≥ π, then draw a halfline from an appropriate vertex in Qi so as to divide Qi into two polygons Qi and Qi with vertex angles αi and αi equal to α/2 (Fig. 94). Dividing all the polygons Qi with αi ≥ π in the same way, we come to a development in which all unbounded polygons have angles αi less than π. Hence, in the sequel we can assume that the development R possesses this property.

Qi

α1 Qi

Fig. 94

4.4.3 Suppose that the unbounded sides of the polygon Qi are parallel to each other. If they are glued to sides of other polygons rather than to one another, then, gluing Qi to one of these latter polygons Qj , we obtain the polygon Qi ∪ Qj . The angle between the unbounded sides in Qi ∪ Qj is the same as in Qj . (In gluing Qi to Qi , the bounded parts of Qi and Qi may overlap, but we can eliminate overlapping by cutting off these parts and regarding them as new polygons of the development.) Carrying out such gluings for all unbounded polygons with parallel unbounded sides, we come to a development R such that even if there is an unbounded polygon in R with parallel unbounded sides, then these sides are identified to one another. However, since all unbounded polygons form a single cyclic sequence when

214


traversed through their identified unbounded sides, we may assume that R has no unbounded polygon with parallel sides or R has exactly one such polygon and no other unbounded polygons. We study the two cases separately. 4.4.4 As a preliminary, we prove a simple lemma which will also be used in Section 5.1 of Chapter 5. Lemma. Let a development R homeomorphic to the sphere consist of two parts R1 and R2 with common boundary L. Assume that there is an isometry of R onto itself, interchanging R1 and R2 and identical on L. Then the closed convex polyhedron P obtained from R by gluing has a symmetry plane containing the polygonal line L. Indeed, the polyhedron P consists of two parts P1 and P2 that correspond to the developments R1 and R2 . By the assumptions of the lemma, P admits an isometry onto itself which interchanges the corresponding points of the parts P1 and P2 and does not move the points of their common part L. Every isometry of a closed convex polyhedron is realized by a motion or a motion and a reflection. In our situation, a motion is excluded, since all points of the closed polygonal line L separating P1 and P2 remain in place. Consequently, P1 and P2 are interchanged by a reflection in some plane. Since it is only the points of the reflection plane that are fixed, it follows that the polygonal line L lies in this plane, which completes the proof of the lemma. 4.4.5 Suppose that there is exactly one unbounded polygon Q in R with parallel sides a and a and let A and A be the vertices from which a and a issue. Fix equal segments A B and A B on a and a of length large enough for the segment B B to lie inside the polygon Q developed on a plane. Cutting away from Q the unbounded part separated by the segment B B , we obtain a development R1 that consists of bounded polygons. In R1 , the only side not identified with another side is B B . a

B

B

b c c b

R1 a

R2 B

B

Fig. 95

Take a second copy R2 of the same development and identify the free sides B B of R1 and R2 , simultaneously matching the corresponding vertices. We obtain the development R1 ∪ R2 , which is symmetric about the line B B


215

(Fig. 95). The sum of angles contiguous to the vertex B (= B1 = B2 ) is 2π. Moreover, the development R1 ∪ R2 is obviously homeomorphic to the sphere. Thus, R1 ∪ R2 satisfies all the sufficient conditions imposed on the developments of a closed convex polyhedron and such a polyhedron is produced from R1 ∪ R2 by gluing. This polyhedron consists of two parts P1 and P2 that correspond to the two halves R1 and R2 of the development. By the preceding lemma, the polyhedron P1 ∪ P2 is symmetric about some plane T containing the closed polygonal line L = B B (the points B and B are glued to one point B). The complete angle at each point of L is 2π. Therefore, the plane T contains no vertices of the polyhedron P1 ∪P2 and only some edges of P1 ∪P2 can intersect T . By the symmetry of P1 ∪ P2 in T , these edges are perpendicular to T . Therefore, we can erase P2 and prolong the faces of P1 with points on T to infinity. The unbounded prism thus adjoined to P1 is glued precisely from the strip cut away from Q earlier, as can easily be verified. The unbounded polyhedron obtained from P1 by the prolongation of its faces realizes the development R. 4.4.6 Now, suppose that R has no unbounded polygons with parallel sides. Let A1 , . . . , An be the vertices from which the unbounded edges of R issue. On these edges, fix equal segments Ai Bi of length l large enough to ensure that in the unbounded polygons Qi each segment Bi Bi+1 lies entirely within the corresponding polygon Qi . Cut away the unbounded parts of Qi that are separated in Qi by the segments Bi Bi+1 , obtaining a development that consists of bounded polygons. In this development, only the sides Bi Bi+1 are not identified with other sides (Fig. 96). B2 B2 A1 Q2 A2 B3 B3

B1

Q1

Q3

Q4

A3 B4

B1

A4

B4

Fig. 96

Let Qi be some bounded part of Qi separated by the segment Bi Bi+1 . As we increase the length l of the sides Ai Bi and Ai+1 Bi+1 and simultaneously subject Qi to a similarity contraction such that the lengths of the sides Ai Bi

216


and Ai+1 Bi+1 remain the same, the polygon Qi converges to an isosceles triangle. Therefore we see that, as the length l increases, the angles at the vertices Bi and Bi+1 in the polygon Qi converge to the angles of such an isosceles triangle and so, at some moment, they become less than some α0 < π/2. Therefore, if we choose l sufficiently large, then the angles at the vertices Bi in the polygons Qi are less than π/2 and the sum of these angles at each common vertex Bi is less than π. Now, we fix such an l and consider the corresponding development R1 . Let us take a second copy R2 of R1 and identify the free edges Bi Bi+1 of R1 and R2 . We thus obtain the development R1 ∪ R2 , which is symmetric in the line B1 B2 . . . Bn composed of these edges. Since in R1 the angles at the vertices Bi are less than π, in R1 ∪ R2 these angles are less than 2π. Finally, it is clear that R1 ∪ R2 is homeomorphic to the sphere.13 Therefore, R1 ∪ R2 defines a closed convex polyhedron by gluing. The latter consists of two parts P1 and P2 that correspond to the two parts R1 and R2 of R1 ∪ R2 . As before, the symmetry between these parts of the development implies that the polyhedron P1 ∪ P2 has a symmetry plane separating P1 from P2 . Repeating this procedure for larger and larger values of l (recall that l is the length of the segments chosen on the unbounded edges of the development), we obtain polyhedra P1 ∪P2 ; we will only take their parts P1 . Visually it is almost obvious that as l → ∞ these parts converge to an unbounded polyhedron that realizes the development R. We must only show that, from the collection of polyhedra P1 appearing with the increase of l, we can choose a sequence of polyhedra which converges to some limit polyhedron P . (The metric of P will certainly coincide with the metric of R, since the metrics on the polyhedra P1 must converge both to the metric on P and to the metric in R.) Let us examine the possible structure of the polyhedra P1 ∪ P2 in more detail. Clearly, P1 has as many interior vertices as the original development R has essential vertices Ai . Denote this number by v. The boundary of P1 is a polygon in the plane T . This polygon has vertices of two types. The first consists of points Di that are not vertices of the polyhedron P1 ∪ P2 . Such a point lies on some edge of P1 ∪ P2 perpendicular to T due to symmetry (Fig. 97). The edge must end at one of the v vertices Ai , which means that the are at most v such edges. The points of second type are the vertices Cj of P1 ∪ P2 . Such a point may arise only from some point Bi of the development; therefore, the number of vertices of the second type does not exceed the number of unbounded polygons Qi in R. Denote the latter number by m. 13

If f , e, and v are the numbers of polygons, edges, and vertices in the development R and n is the number of unbounded polygons in R, then the numbers of polygons, edges, and vertices in R1 ∪ R2 are f = 2f , e = 2e + n, and v = 2v + n; whence f − e + v = 2(f − e + v) = 2.


P1 D

C1

C2

217

T

P2

Fig. 97

Some edges issuing from vertices of the second type can go inside P1 , but in any case their total number does not exceed vm. Now, let P1n be the polyhedra obtained for larger and larger values of ln as ln → ∞. Move the polyhedra P1n to positions such that one vertex, say An1 , and the outward normals of the planes Tn are the same for all of them. Then all the vertices of P1n ∪ P2n interior to P1n belong to a bounded part of space; hence it is clear that there is a subsequence of polyhedra P1n (for simplicity, we preserve their numbering) such that, first, all vertices Ani converge to some points Ai ; second, the edges Ani Din perpendicular to Tn , if they exist, issue from the converging vertices Ani and therefore converge to parallel rays issuing from the points Ai ; and, third, the edges Ani Cjn exist only for the same pairs (i, j) and their directions converge to some limit rays issuing from the points Ai .

Fig. 97(a)

The reader may check that the polyhedra P1n then converge to the convex hull of the collection of the limit vertices Ai and the limit rays issuing from them. This convex hull is the polyhedron P that realizes the development R. By way of illustration, Fig. 97(a) displays the polyhedron obtained by passage to the limit from the polyhedra of Fig. 97.

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4.5 Existence of Unbounded Polyhedra Given the Development and the Limit Angle 4.5.1 In Section 3.3 of Chapter 3 we proved that an unbounded convex polyhedron of curvature 2π is determined uniquely from its development (up to a motion or a motion and a reflection). Hence, if a development has curvature 2π, then we can add nothing to the theorem of the previous section. Observe that a development has curvature 2π if and only if all unbounded polygons in it have parallel unbounded sides. This corresponds to the first case of the previous section. However, if the curvature of the development is less than 2π, then it turns out that we can obtain infinitely many unequal polyhedra with different limit angles from the development by gluing. Before stating the corresponding theorem, we recall some definitions. A ray in a development R or on a polyhedron P is a half-bounded line which is shortest for each of its bounded subarcs. Under the infinite similarity contraction to some point, the unbounded convex polyhedron P transforms into some polyhedral angle, the limit angle V of P . If the curvature of P is equal 2π, then V degenerates into a half-line; we exclude this case from further consideration. In general, V may even be a doubly-covered flat angle. Under the contraction of P to V , each ray L on P transforms to a generator of V , the limit generator of L. If P is oriented, then it induces an orientation on V . (See Lemma 1 in Subsection 3.4.5, Chapter 3.) Theorem. Suppose that a development R is of curvature less than 2π and satisfies all the necessary conditions imposed on developments of unbounded convex polyhedra. Assume that R is oriented and some ray L is fixed in Q. (Starting from some point onward, L is a half-line on some unbounded polygon of Q.)14 Finally, let V be a convex polyhedral angle with curvature equal to that of R, with a chosen direction of rotation about the vertex and with a chosen generator L. Then R defines by gluing some unbounded convex polyhedron P with V as the limit angle, for which the line on P corresponding to L has L as its limit generator and the orientation on P induced by the orientation of R coincides with the orientation of V . Such a polyhedron P is unique up to translation. (We assume that the curvature of the development R is greater than zero. Otherwise the assertion concerning uniqueness is false.) The uniqueness part of the theorem is contained in Theorem 3 of Section 3.4. Indeed, that theorem claims that an isometric mapping from one polyhedron P onto another inducing a congruence between limit angles is realizable by a motion or a motion and a reflection. Since in our case the development, the limit angle V , the orientation of V , and the limit generator 14

See Lemma 2 in Section 3.3.

4.5 Unbounded Polyhedra Given the Development and the Limit Angle

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of some ray L are given, a mapping from P onto another polyhedron P1 with the same data is necessarily isometric, inducing the identical transformation of the limit angle V . This follows from the fact that the limit angle V , the limit generator, and the orientation of V are preserved by the mapping from P onto P1 . Therefore, the isometry from P onto P1 is realized by a motion or a motion and a reflection. Any reflection is excluded by the requirement that the orientation of P1 must coincide with the given orientation of V . Any rotation is excluded, since it induces a rotation of the limit angle. We are left with a translation, because the limit angle is determined by the polyhedron up to translation. The above theorem thus presents a complete system of data which determine the polyhedron P . Also, observe that the specification of a ray L and its limit generator L obviously amounts to the specification of another ray L1 and another limit generator L1 provided the angles between L and L1 and between L and L1 (measured in accordance with the orientations of the development R and of the angle V ) are equal. 4.5.2 Given the development R, it is clear that the limit angle V and the generator L may be chosen almost arbitrarily. Hence, changing them, we shall obtain new polyhedra with the same development. Thus, the existence of flexes of unbounded polyhedra of curvature less than 2π is already implicit in this theorem. We discuss this question in more detail in Section 5.2 of Chapter 5. Now, we exhibit an elegant example of a flexible unbounded convex polyhedron. Let Q be a regular n-gon and let V a regular n-hedral angle V whose vertex is the center of Q and whose symmetry axis is perpendicular to the plane of Q (Fig. 98). Q

Q P

V

Fig. 98

The surface of the convex hull of the figure Q ∪ V is an unbounded polyhedron P with limit angle V and vertices only at the vertices of Q (Theorems 5 and 5a of Section 1.4 in Chapter 1). The figure Q ∪ V has an axis of symmetry of degree n; hence, so does P , regardless of how we rotate V around its axis. The sum of curvatures at the vertices of P is equal the curvature of its limit angle V (Theorem 3 of

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Section 1.5, Chapter 1). Since the vertices of P are mapped to one another under rotations about the symmetry axis, the curvature at each angle is one nth of the curvature of V , irrespectively of rotations of V . When we rotate V around the axis, the polyhedron P varies as well. However, its face Q remains in place and, as shown above, the curvatures at the vertices of P do not change. Therefore, the complete angles on the unbounded part of P as well as those on its bounded sides are constant. This means that, although the unbounded part changes its shape, its development remains the same. To see this, it suffices to make a cut along a ray and develop this part on a plane. An unbounded polygon with given angles and given bounded sides then appears. Thus, to a revolution of V around its axis corresponds a flex of P under which the face Q is fixed. It is interesting to picture this flex visually. Here the limit angle and the orientation do not change, but the correspondence between the rays on V and on the unbounded part of P changes. We come to the same result if for Q we take a convex cap with symmetric base and assume that the angle V is not necessarily regular, but still has the same symmetry axis as the base of the cap. 4.5.3 The above general theorem is due to S. P. Olovyanishnikov [Ol1]. Olovyanishnikov’s proof, like the proof of the theorem in the previous section, relies on the existence of a closed convex polyhedron with prescribed development. However, the theorem is a straightforward consequence of the Mapping Lemma. Here we present that independent proof, although Olovyanishnikov’s proof is somewhat simpler (because it assumes the theorem about closed polyhedra). In conclusion, we show how, proceeding on similar lines, an independent proof of the existence of an unbounded convex polyhedron with a given development of curvature 2π can be obtained. 4.5.4 We consider developments satisfying the necessary conditions imposed on developments of unbounded convex polyhedra of curvature ω satisfying ω > 0 and ω < 2π. Such developments are homeomorphic to the plane. Given such a development R, map R onto a (topological) plane E or, in other words, imagine R abstractly glued. Let A1 , . . . , Av be the points of E corresponding to those vertices of the development at which the sum of angles is less than 2π. If there is only one such point, then by cutting E (i.e., the development R glued together) along a ray issuing from this point, we obtain a flat angle from which we can construct (by gluing) any polyhedral angle of curvature ω. Hence, our theorem is valid for v = 1 and we can proceed with the proof by induction on the number of vertices. 4.5.5 Assume that v > 1. Draw a ray a from A1 and draw the shortest arcs A1 Ai to the other points A2 , . . . , Av . Cutting E along these lines, we obtain an unbounded (abstract) polygon Q. It has two vertices A1 with incident unbounded sides a. Join these points by a shortest arc in Q (Fig. 99).


A1

A1

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A1

a

a Q

Fig. 99

According to Theorem 1 of Section 1.8, this arc is a geodesic polygonal line with vertices only at vertices of the polygon or, in a particular case, with no vertices at all. Moreover, this arc cuts out from Q its bounded part, which possibly consists of several, rather than one, polygons. We split each of these pieces by diagonals into triangles (using Theorem 5 of Section 1.8). As a result, we obtain a development R1 which is isometric to R and satisfies the following conditions: (1) R1 has no “redundant” vertices at which the sum of angles equals 2π and (2) R1 consists of triangles and one unbounded polygon. So we can restrict our considerations to developments without “redundant” vertices. A A

a1

a

a1

a

Fig. 100

Ignoring the metric relations, we transform our development into a complex homeomorphic to the plane. This complex consists of topological triangles and one topological polygon open from one side. The reverse transformation of such a complex into a development is reduced to specifying (1) the side lengths of the triangles and (2) the angles of the unbounded polygon (then the angles and sides of the polygon will be fixed, so that the polygon is determined completely). However, if we change the angles of the unbounded

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polygon at the vertices from which unbounded sides issue, keeping the sum of these angles the same, then we obtain isometric developments.15 Therefore, it seems more convenient to glue the unbounded sides together. Then the unbounded polygon is transformed into a figure homeomorphic to an infinite cone with a neighborhood of the vertex deleted. We call this figure a funnel. A funnel is completely defined by specifying its sides and angles, because, making a cut in a funnel along some ray, we obtain an unbounded polygon. Thus, we consider a complex K homeomorphic to the plane and consisting of topological triangles and of a topological funnel. The transformation of the complex K into a development requires specifying (1) the k lengths r1 , . . . , rk of the edges of K and of the sides of the triangles, (2) the n angles α1 , . . . , αn of the funnel. We have k + n variables in all. If R is a development of positive curvature, i.e., if the sum of angles at each of its vertices is less than 2π, then this condition is not violated for small variations of ri and αi . In other respects, variations of these parameters can be arbitrary. This is obvious for the edge lengths. To change the angle αj , we draw some ray from the corresponding vertex of the funnel and make a cut in the funnel along this ray. To both edges of the cut we can glue the sides of an arbitrary flat angle. Therefore it is clear that the angles admit arbitrary increments. Developing the cut funnel on a plane, we obtain an unbounded polygon (possibly with self-intersections). The unbounded sides of this polygon are not parallel to one another; otherwise the curvature of the development would equal 2π. Therefore, from this polygon we can cut off an infinite angular sector and thereby decrease the angle αj of the funnel. Thus, the set of all developments R of positive curvature with the given structure K represents, in a natural way, some (k + n)-dimensional manifold M. 4.5.6 However, in addition to the development, we must specify the polyhedral angle V which will serve as the limit angle of the corresponding polyhedron. We consider all possible convex polyhedral angles with a given number h of edges, having in mind only genuine edges, i.e., edges at which the dihedral angles are less than π. When the vertex is fixed, such an angle is determined by directions of its edges, i.e., by 2h variables. Excluding rotations of the polyhedral angle around its vertex as a solid body, we are left with 2h − 3 variables. Finally, to the given development we must assign a polyhedral angle of the same curvature ω. This excludes one more variable, and we are left with 2h − 4 variables. Thus, we obtain the (2h − 4)-dimensional manifold V of convex polyhedral angles of given curvature ω. Further, in the development R we must choose a direction for going around its funnel and, on the angle V , choose a direction for going around its vertex. 15

This is clear from Fig. 100. A change of the angles at the vertices A and A preserving their sum results in cutting off the angle formed by a and a1 and applying it to the side a. Instead of a polygon with (unbounded) sides a and a , we obtain a polygon with sides a1 and a1 .


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Also, to the ray L in the development R, we must assign a generator L of the polyhedral angle V which will be the limit generator for the corresponding ray on the polyhedron P produced from R by gluing. Observe, however, first, that the assignments will be the same for rays L parallel to one another and, second, the choice of a ray L and a generator L is equivalent to specifying another ray L1 and another generator L1 , provided that the angles between L and L1 and between L and L1 are equal. Of course, we assume that the angles are measured in accordance with the chosen orientations of the development R and the angle V . Therefore, we can speak simply about establishing a correspondence between the directions of rays in the development R and those on the angle V . So, when the ray L is fixed, we can choose from a one-parameter family of generators L. It follows that the arbitrariness in establishing the correspondence between directions in the development R and on the angle V yields one more variable parameter. Thus, we consider the manifold of oriented developments R and angles V with given correspondences between directions. We denote this manifold by MV. By the above calculations, the dimension of MV is (k + n) + (2h − 4) + 1 = k + n + 2h − 3.

(1)

4.5.7 Let R0 and V0 be a development and an angle of the same curvature ω. Let L and L be the corresponding rays in R0 and on V0 . We thus have a “point” (R0 , V0 ) of our manifold MV. There is no loss of generality in assuming that the ray L issues from some vertex A of the funnel of R0 . Cut the funnel along this ray and start pasting in some sectors S with continuously increasing angles into the resulting cut until the development R0 transforms into a development R1 in which the sum of angles at the vertex A equals 2π. The curvature of the development decreases by 2π − θ, where θ is the sum of angles at the vertex A in the original development R0 . However, θ cannot equal zero, because by assumption the number of vertices in R0 is greater than one. Changing R in this manner, we may continuously change the angle V so that the curvature of V remains equal to the curvature of the development. To this end, it suffices for instance to continuously increase the planar angles of V by the same coefficient. We thus obtain an angle V1 with the same curvature as the development R1 . The correspondence between the rays L and L1 can be preserved: for L we should always take one edge of the cut made in the original development R0 and take for L1 any generator on the varying angle V provided this generator varies continuously. Starting from an arbitrary point (R0 , V0 ) of MV, we can thus arrive at a point (R1 , V1 ) that “lies on the boundary” of MV; here the only angle R1 will be equal to 2π at the given vertex of the funnel and therefore near (R1 , V1 ) there are no points of the connected components of MV other than the connected component containing (R1 , V1 ). To prove this, represent “points” (R, V ) close to (R1 , V1 ) by points in (k + n + 2h − 3)-dimensional space, one

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coordinate of which is the angle α of the funnel at the vertex A. Observe that we now consider all developments R, rather than those of positive curvature. The boundary of the manifold MV is now given by the equation α + i ϕi = 2π, where ϕi are the other angles touching at A. This surface dividesthe neighborhood of the point into ϕi two parts: in one of them α > 2π − and in the other α < 2π − ϕi . If points (R , V ) and (R , V ) lie in the same part, then in the neighborhood they can be joined by a continuous path lying totally on one side of the surface α = 2π − ϕi . If the neighborhood is small enough, then the sums of angles at all vertices other than A remain less than 2π and the conditions imposed on the variables determining the convex polyhedral angle are not violated. Therefore, the above-mentioned path lies in the manifold MV, which means that all points of MV close to (R1 , V1 ) belong to the same connected component. The development R1 may now be replaced by a development without the redundant vertex A. By the induction hypothesis, the theorem is valid for developments with fewer vertices. Therefore, R1 defines by gluing a convex polyhedron P1 with limit angle V . To the vertex A corresponds a point on P1 lying in the interior of a face or an edge. Move this point out at a sufficiently small distance and construct the polyhedron P with this moved-out vertex, with all the vertices of the polyhedron P1 , and with the same limit angle V . It is easy to see that each polyhedron P , when close enough to P1 , admits a development R which is close to R1 (we can prove this in the same way as in Lemma 1 of Section 4.2). However, developments close to R1 correspond to “points” (R, V ) of the given connected component of the manifold MV. Consequently, each connected component of MV contains “realizable points” (R, V ), i.e., developments R that define polyhedra with limit angle V by gluing. 4.5.8 Now consider the set of all unbounded convex polyhedra that are produced by gluing from developments with structure K and have limit angles with h edges. Given one of these polyhedra, we draw its development and thus obtain a “K-triangulated” polyhedron P . Together with the Ktriangulation, the orientation of the development induces an orientation of the polyhedron P . So we can speak of an oriented K-triangulated polyhedron. Two K-triangulated polyhedra are equal if their K-triangulations, as well as the polyhedra themselves, can be made to coincide by a motion or a motion and a reflection. We do not distinguish equal K-triangulated polyhedra. An unbounded convex polyhedron P is determined by specifying its vertices and the limit angle, the latter being given up to translation. For polyhedra with v vertices, the specification of vertices requires v variable coordinates. The specification of a limit angle with h edges requires 2h variables more to fix the directions of edges. Since the polyhedra are considered to within a motion (and a reflection), we must exclude 6 variables. So we have 3v + 2h − 6 variables determining the polyhedron P .


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We make the set of all K-triangulated polyhedra P with h-hedral limit angles into a manifold P by defining neighborhoods as follows: a neighborhood of a polyhedron P0 consists of all polyhedra P close to P0 together with their K-triangulations. Two K-triangulated polyhedra P are close if their vertices and directions of edges of limit angles are close. Every polyhedron P close to a given polyhedron P0 admits a unique K-triangulation which is close to the Ktriangulation of P0 . This assertion is proved by using the same obvious arguments as those in the proof of an analogous assertion for closed polyhedra in Section 4.2 (Lemma 1). Moreover, if edge lengths in K-triangulations of close polyhedra are close, then obviously the angles of their funnels are also close. Thus, a small variation of all 3v + 2h − 6 variables produces K-triangulated polyhedra which are close to the given polyhedron together with their Ktriangulations. These polyhedra are all different also when the polyhedron P0 degenerates into a doubly-covered polygon. Any two polyhedra of the latter type could only be equal if they are symmetric to one another with respect to the plane of P0 . However, in that case their orientations differ, and so the polyhedra are not equal because of the difference in orientation.16 Thus, P is really a manifold of dimension 3v + 2h − 6. 4.5.9 To each polyhedron in P correspond uniquely: (1) its development (which is its K-triangulation), (2) its limit angle V , (3) the assignment of the directions of its rays to the generators of the angle V established by the infinite similarity contraction of the polyhedron to the angle V . Therefore there is a natural one-to-one mapping ϕ from the manifold P into the manifold MV. To prove our theorem, it therefore suffices to check all the conditions of the Mapping Lemma. First of all, we shall show that the dimensions of P and MV are the same. Let f , e, and v be the numbers of triangles, edges, and vertices in the complex K. By Euler’s formula, f − e + v = 1.

(2)

The complex K has a funnel with n sides, the latter identified with n sides of triangles in K. The other sides of the triangles are identified with one another pairwise. Therefore, 3f = 2e − n. (3) It follows from (2) and (3) that e + n = 3v − 3. Hence, for the dimension of the manifold MV (see (1)), we obtain 16

The orientation in P0 induces an orientation of close polyhedra in an obvious way. In particular, if P0 degenerates, then we must remember the presence of two sides to which two sides of a close polyhedron correspond. As in the case of closed polyhedra, we can manage, however, without invoking orientation.

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e + n + 2h − 3 = 3v + 2h − 6; i.e., it is equal to the dimension of the manifold P. We have already proved that each connected component of the manifold MV contains images of points in P. The other properties of the mapping ϕ are verified exactly as in Section 4.3 in the case of closed polyhedra. The continuity of ϕ is already implicit in the definition of neighborhoods in the manifold P. The fact that ϕ is one-to-one follows from Theorem 3 of Section 3.4 in Chapter 3. By this theorem, two isometric polyhedra are equal if their limit angles are the same, and the correspondence between the directions of rays on the polyhedra and the generators of the limit angles is the same. Finally, the last required property of ϕ follows from a lemma that is quite similar to Lemma 1 of Section 4.2. The changes in statement and supplements in proof are connected with the fact that, together with its edge lengths, the development is also given by the angles of the funnel. All these modifications are so clear that we feel that it is needless to dwell on them. Thus, all the conditions of the Mapping Lemma are valid, and the proof of our theorem is complete. 4.5.10 The same method applies to proving the existence of an unbounded convex polyhedron with a given development of curvature 2π. Here we must consider only developments of curvature 2π. The funnels now transform into “tubes” because the unbounded polygons of such a development have parallel unbounded sides. The previously used arguments enable us to eliminate from the development all the redundant vertices (where the sum of angles equals 2π). A development with a given structure K is now determined by k edge lengths and the n angles of the tube, where the latter angles are connected by the relation n αi = nπ. i=1

As a result, we obtain an (e+n−1)-dimensional manifold M of developments of positive curvature with given structure K. There are no developments with a single vertex, since the curvature at a vertex is less than 2π. Developments with exactly two vertices are obviously realizable as doubly-covered plane strips. Therefore, the existence of a polyhedron with a prescribed development can again be proved by induction on the number of vertices. The manifold P of K-triangulated polyhedra is defined as before. The variables determining a polyhedron consist of 3v coordinates of vertices plus 2 variables specifying the direction of the limit half-line, minus 6 variables connected with the motion of a polyhedron as a solid body. Thus, P is a (3v − 4)-dimensional manifold. An application of Euler’s formula f − e + v = 1 yields the relation 3v − 4 = e + n − 1;


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i.e., M and P are of the same dimension. Next, we define the natural mapping ϕ from P into M. The fact that ϕ is one-to-one follows from the uniqueness of polyhedra with prescribed development of curvature 2π (Theorem 2 of Section 3.3, Chapter 3). Proofs of the other properties of ϕ requested in the Mapping Lemma are analogous to those in the case of closed polyhedra. As a result, all assumptions of the Mapping Lemma are fulfilled, whence we infer the existence of a polyhedron with arbitrary prescribed development of curvature 2π.

5 Gluing and Flexing Polyhedra with Boundary

5.1 Gluing Polyhedra with Boundary 5.1.1 The sides of polygons in each of the developments we have talked about in Chapter 4 are glued to one another pairwise. So such a development has no boundary. However, if some sides of polygons in the development are glued to no other sides, then the development has a boundary. In that case, we refer to the free sides of polygons in a development as the boundary edges of the development. Regarding as identified the sides and vertices that are to be glued, we may thus conclude that the boundary of the development under study consists of finitely many closed polygonal lines, each formed by boundary edges of the development, and, when the development has unbounded polygons, also of unbounded polygonal lines, each constituted by finitely many bounded boundary edges and two unbounded ones. In particular, the boundary of a development may lack bounded boundary edges. A polyhedron produced from such a development by gluing has the corresponding boundary. Since we cannot fully answer the question of finding conditions under which a given development with boundary defines a convex polyhedron by gluing, we choose to restrict our exposition to general ideas and a few isolated and somewhat special results.1 Further we give examples of developments that define no convex polyhedron by gluing, although they satisfy certain obvious necessary conditions. The following theorem provides an approach to solving the problem we have raised. Theorem 1. Let the boundary of a development R consist of closed or unbounded polygonal lines L1 , . . . , Ln . For R to define a convex polyhedron by gluing, it is necessary and sufficient that there exist developments R1 , . . . , Rn , each of which is bounded by a single polygonal line, so that the identification 1

At present, several effectively verifiable necessary and sufficient conditions are known. They provide criteria for the existence of the “patches” mentioned in Theorems 1 and 7. These conditions are presented in L. A. Shor’s article, whose translation is given in the Supplement to Chapter 5 of the present book. – V. Zalgaller

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of their boundaries with the polygonal lines R1 , . . . , Rn yield a development R ∪ R1 ∪ . . . ∪ Rn that defines a convex polyhedron by gluing. The development R ∪ R1 ∪ . . . ∪ Rn is now without boundary and hence the question of finding conditions under which R ∪ R1 ∪ . . . ∪ Rn defines a convex polyhedron by gluing can be answered by using the theorems of Sections 4.1–4.4. Proof. The argument is quite straightforward. If such developments R1 , . . ., Rn exist, then we obtain a convex polyhedron with development R by producing a convex polyhedron from R ∪ R1 ∪ . . . ∪ Rn by gluing and choosing the part that corresponds to R. This proves the sufficiency of the condition. On the other hand, assume that R defines a convex polyhedron P by gluing. By definition, this means that P is a part of a complete convex (closed or unbounded) polyhedron P without boundary. Then the complement P \ P , taken with the boundary, consists of one or several convex polyhedra P1 , . . . , Pn . Together with R, their developments constitute the development R ∪ R1 ∪ . . . ∪ Rn of the complete polyhedron P ; this proves the necessity of the condition and finishes the proof of the theorem. Theorem 1 reduces the possibility of gluing a convex polyhedron from a development with boundary to that of selecting appropriate developments R1 , . . . , Rn . In some cases, this selection may be carried out. To this end, the development R must satisfy the following three necessary conditions: (1) The development R is homeomorphic to a planar (bounded or unbounded) polygon whose boundary may consist of several polygonal lines.2 (2) The sums of angles at all internal and boundary vertices of R are less than or equal to 2π. (3) The curvature of R (i.e., the sum of curvatures at all internal vertices) is less than or equal to 4π if the development is bounded and less than or equal to 2π if the development is unbounded. The necessity of these conditions is obvious, because every convex polyhedron with boundary possesses these properties. An unbounded complete polyhedron P is homeomorphic to the plane and has curvature at most 2π. Therefore, the polyhedron P , being a part of P , is homeomorphic to a polygon and 2

This condition can be restated without appealing to the notion of homeomorphism. Namely, if f , e, and v are the numbers of polygons, edges, and vertices in R and n is the number of boundary polygonal lines, then we must have f − e + v = 2 − n in the case of a bounded development and f − e + v = 1 − n in the case of an unbounded development. To prove the equivalence of the two conditions, it suffices to observe that, attaching a polygon bounded by a single polygonal line to each polygonal line bounding the given development, we obtain a development R without boundary for which f − e + v = 2 or f − e + v = 1, so that R is homeomorphic to the sphere or the plane, respectively.

5.1 Gluing Polyhedra with Boundary

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has curvature at most 2π. A closed polyhedron P is homeomorphic to the sphere and has curvature 4π; however, the sphere with one point deleted can be mapped onto the plane. So the polyhedron P , as a part of P , is homeomorphic to a planar polygon and its curvature does not exceed that of P , being at most 4π. In turn, each development Ri , attached to R, must possess the same properties (2) and (3) and be homeomorphic to a polygon bounded by a single closed or unbounded polygonal line. As a result, the assumptions of Theorem 1 are reduced to requiring that the sums of angles at the identified points of the boundaries of the developments R and Ri be at most 2π. Since this requirement must be independently satisfied for each polygonal line Li on the boundary of R and of the corresponding development Ri , the main question turns out to be that of gluing a polyhedron from a development bounded by a single polygonal line. Such a development is homeomorphic to a disk or to a half-plane depending on whether it is bounded or unbounded. 5.1.2 The results that follow are obtained by the “method of gluing,” which is based on Theorem 1. From now on, all the developments under consideration are assumed to satisfy the above conditions (1)–(3) without explicit mention. By the angle at a boundary vertex of a development, we mean the sum of angles of the polygons of the development at this vertex. It is convenient here to imagine that the development has been glued, at least abstractly. Recall that by a (polyhedral) cap we mean a convex polyhedron bounded by a planar closed polygonal line whose projection to the plane coincides with the polygon bounded by this line on the plane. Theorem 2. A development R homeomorphic to a disk defines a cap by gluing if and only if the angle at each boundary vertex of R is at most π. Proof.3 Let us prove the necessity of the condition. Let P be a cap. Attaching a cap symmetric with respect to the plane of the bounding polygonal line to P , we obtain a closed convex polyhedron P ∪P1 (as in Fig. 97, p. 217). Each boundary vertex of P is either a vertex of this polyhedron or a point on an edge. The sum of angles at such a vertex in P ∪ P1 is at most 2π. This sum comprises angles both on P and P1 . By the symmetry between P and P1 , these angles are equal, and so the sum of angles at each boundary vertex of P is at most π. Conversely, let R be a development satisfying the conditions of the theorem. Take a copy R1 of R. Identifying the corresponding elements of the boundaries of these developments, we obtain the development R ∪ R1 homeomorphic to the sphere. Since the sums of angles at the boundary vertices in R and R1 are at most π, the sum of angles at each vertex of the development 3

Another direct proof of this theorem appears in [Vo4] (also see the Supplement to Chapter 4 at the end of this book). – V. Zalgaller

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R ∪ R1 is at most 2π. Therefore, this development defines by gluing a closed convex polyhedron P consisting of two parts P and P1 corresponding to R and R1 . By the Lemma in Subsection 4.4.4 of Chapter 4, the polyhedron P is symmetric with respect to the plane containing the common boundary of its parts P and P1 . Clearly, P is a cap. Since R defines P by gluing, the proof of the theorem is complete. Recall that in accordance with Theorem 3 of Section 3.5, a cap with a prescribed development is unique up to a motion and a reflection. Hence, a development R can actually define by gluing only one cap or the cap symmetric to it. Defining an unbounded cap to be an unbounded convex polyhedron whose boundary is a single unbounded planar polygonal line and which projects onto the polygon bounded by this polygonal line, we can state the following theorem: Theorem 3. A development R homeomorphic to a half-plane defines an unbounded cap by gluing if and only if the angles at its boundary vertices are at most π. The proof almost repeats that of Theorem 2. Some modification is required in the final part. There we shall need the uniqueness of the polyhedron P ∪ P1 determined by the development R ∪ R1 . If the curvature of the development is less than 2π, then we know that such a polyhedron is not unique. As proved in Section 4.5 of Chapter 4, the limit angle of a polyhedron can be chosen arbitrarily with the only requirement that its curvature be equal to that of the polyhedron. Therefore, we can choose the limit angle of P ∪ P1 in advance so that its parts V and V1 corresponding to the parts P and P1 of the polyhedron be symmetric to each other. Interchanging P and P1 then yields a mapping of P ∪ P1 onto itself which is the identity on the limit angle V ∪ V1 as well as on its generators corresponding to the common unbounded edges of P and P1 . Hence, by Theorem 3 of Section 3.4, the interchange of P and P1 must be realized by a reflection. Therefore, P is an unbounded cap. The uniqueness of unbounded caps with prescribed development is not true in general. The question of finding when uniqueness still holds or finding extra data completely determining an unbounded cap is left to the reader. The answer may easily be found by using the theorems about unbounded polyhedra (Sections 3.3 and 3.4 of Chapter 3 and Section 4.5 of Chapter 4). Theorems 2 and 3 completely describe developments of caps by means of conditions on the angles at the boundary vertices. The next theorem indicates one more characterization of such developments. Theorem 4. For the angles at the boundary vertices of a development to be at most π, it is necessary and sufficient that every two interior points of the


233

development be connected by a shortest arc contained in the interior of the development. (We may say that such a development is “convex in itself.”) It is convenient here to represent a development as having been glued at least abstractly. A shortest arc joining two interior points in a development may approach the boundary only at a vertex whose angle is strictly greater than π. Otherwise, the arc could be replaced by a shorter line that straightens the angle that it forms (Theorem 1 of Section 1.8 in Chapter 1, Fig. 53). In connection with Theorems 2–4, consider the following exercise: A development in which all angles on the boundary are at most π may fail to be homeomorphic to a disk or a half-plane. Restricting the question to developments homeomorphic to part of the plane, prove that only developments of lateral surfaces of closed or open right prisms are not homeomorphic to a disk or a half-plane. In the second case (of an open prism) the development is homeomorphic to a planar strip between a pair of parallel straight lines. 5.1.3 In this subsection and further, in Subsections 5.1.5–5.1.7, we present some sufficient conditions for a given development to define a convex polyhedron by gluing. Sometimes these conditions are also necessary. It is always easy to check whether or not they hold for any concrete development. The entire consideration is based on the following construction. A1 A1 2π-αn α2 A2

An

2π-α2

Fig. 101(a)

Let R be bounded by a single closed polygonal line L with vertices A1 , . . . , An and angles α1 , . . . , αn . On the plane construct a polygonal line L0 whose segments have lengths equal successively to those of the edges A1 A2 , . . . , An−1 An , An A1 and the angles between these segments on one side are equal to α1 , . . . , αn (Fig. 101(a)). On the opposite side, the angles are βj = 2π − αj . Call this side the “exterior side” of L0 , since it corresponds to the exterior side of the polygonal line L with respect to R. (In Fig. 101(a) this “exterior side” of L0 looks at the bounded part of the plane.) The polygonal line L0 is not closed in general, although the endpoints of L0 correspond to the same vertex A1 of L. At the same time, L0 could be closed and even self-intersecting (Fig. 101(b)). The construction of L0 can naturally be called the unrolling of L torn at A1 . Similarly, we can unroll a polygonal line L torn at any point A of L.

234


A2

A1

A4

A1

A1 A3 An A2

A3

Fig. 101(b)

If the boundary of R is an unbounded polygonal line L, then the unrolling of L is defined in exactly the same way. In this case there is no need to tear L, since L is not closed. Theorem 5. For a bounded development R with boundary polygonal lines L1 , . . . , Ln to define a convex polyhedron of curvature 4π by gluing, it is necessary and sufficient that the unrolled polygonal lines L01 , . . . , L0n be closed and bound some polygons Q1 , . . . , Qn on the plane which lie on the “exterior sides” of the polygonal lines. (Here by a polygon we mean a development without internal vertices. Successively developing the triangles constituting it on the plane, we place the whole development on the plane. It may self-overlap, so that the “polygon” in this generalized sense may be self-overlapping.) The proof of Theorem 5 is evident. Indeed, let the assumptions of the theorem be satisfied. Then, attaching the polygons Qi to R, we obtain the development R = R ∪ Q1 . . . ∪ Qn which is homeomorphic to the sphere. The angles of the polygons Qi complete the angles of R to 2π. It follows, first, that the condition on the sum of angles is satisfied and so P defines a closed convex polyhedron P by gluing. Second, this shows that no point of P corresponding to points in the interior or on the boundary of Qi is a vertex of P . Otherwise, the sum of angles at this point would be less than 2π. Consequently, all the vertices lie in the part P of P that corresponds to the original development R. The curvature of P is thus equal to that of P , i.e., it is 4π. This proves the sufficiency of the condition of the theorem. To prove its necessity, suppose that R defines a convex polyhedron P of curvature 4π by gluing. The polyhedron P is a part of a closed convex polyhedron P . Since the curvature of P is 4π, it follows that P has no vertices other than the internal vertices of P . Therefore, the completion of P to P consists of pieces Qi developable on the plane. The sums of angles at the common vertices of the polyhedron P and the polygons Qi equal 2π, since these vertices are not vertices of P . This means exactly that the angles β of


235

Qi complete the angles α on P to 2π, i.e., the boundaries of Qi result from the unrolling of the polygonal lines bounding P .4 Recall that, in accordance with Theorem 1 of Section 3.5, a convex polyhedron of curvature 4π is determined from its development uniquely up to a motion or a motion and a reflection. Hence, under the conditions of Theorem 5, the development R in fact defines a unique polyhedron by gluing (or a polyhedron symmetric to it). Theorem 6. For an unbounded development R with boundary polygonal lines L1 , . . . , Ln to define a convex polyhedron by gluing, it is sufficient, and also necessary when the curvature of the development equals 2π, that the unrolled polygonal lines L01 , . . . , L0n be closed and bound polygons that lie on their “exterior sides.” The proof is essentially the same as that of Theorem 5. If the curvature of the development is 2π, then uniqueness holds by Theorem 2 of Section 3.5. Given R, unrolling its boundary polygonal lines on the plane raises no difficulties. Therefore, the conditions of Theorem 5 and 6 are always easy to check. 5.1.4 The more general Theorem 1, on which grounds the further results are based, admits a refinement which is especially useful in studying flexes of convex polyhedra, the topic of the next section. Theorem 7. A development R, bounded by polygonal lines L1 , . . . , Ln , defines a convex polyhedron by gluing if and only if there are planar, possibly self-intersecting, polygons Q1 , . . . , Qn whose sides are equal to the corresponding segments of L1 , . . . , Ln and whose angles do not exceed the complements of the angles at the corresponding vertices of L1 , . . . , Ln . Here we are allowed to introduce as many additional vertices on L1 , . . . , Ln as we like. The polygonal lines Li and the corresponding polygons Qi may also be unbounded. (By the agreement stated at the beginning of Subsection 5.1.2, R is assumed homeomorphic to a planar polygon, and the sum of angles at each vertex of R is at most 2π.) The sufficiency of the condition of the theorem is obvious because, by identifying the segments of the polygonal lines Li with the sides of the polygons Qi , we obtain a development in which the angles at all vertices are at most 2π (by the condition on the angles of Qi ). This development R has no 4

An exceptional case appears if some polygonal lines Li bounding P have common vertices. They may then be viewed as a single polygonal line, and the corresponding polygon Q will be composed of the polygons Qi . Strictly speaking, this case should have been explicitly mentioned in the conditions of the theorem.

236


boundary and defines a convex polyhedron by gluing,5 which means that the given development R defines a convex polyhedron by gluing as well. To prove necessity, assume that R defines a convex polyhedron P by gluing. Let P be the boundary of the convex hull of P . So P is a closed or unbounded polyhedron without boundary which contains P . Since the convex hull is fully determined by the vertices alone, the complement P \ P contains no vertices; they all lie in the interior or on the boundary of P . The set P \ P splits into parts Q1 , . . . , Qn corresponding to the polygonal lines L1 , . . . , Ln bounding R. Since there are no vertices inside Qi , the parts Qi are all developable on a plane. If A is some vertex on the boundary of P and α and β are angles at A on P and on the corresponding polygon Qi , then α + β ≤ 2π by the main property of the sums of angles at vertices. Therefore, developed on the plane, the polygons Qi become polygons whose sides are equal to the segments of the polygonal lines Li and whose angles β are no greater than 2π − α, as claimed. Theorem 7 reduces the question of defining a polyhedron from a development by gluing to the plane geometry problem of the existence of a polygon with prescribed sides and prescribed upper bounds of angles. However, the solution of this problem still evades us, except for the particular cases covered already by the theorems above. The problem seems rather intriguing but hard.6 In any case, Theorem 7 gives us almost nothing as regards conditions for producing a convex polyhedron from a given development by gluing. Here we only mean the conditions that can be verified effectively by a geometric construction or calculations with real numbers. We indicate one more result that the reader may derive from Theorem 7: Let a development R be bounded by closed polygonal lines Li . Suppose that all angles in R are no greater than π except possibly two angles αi and βi in each polygonal line Li whose vertices Ai and Bi divide the corresponding polygonal line Li into parts of equal length. Then R defines a convex polyhedron by gluing. 5.1.5 Theorem 8. Let R be a bounded or unbounded development with closed boundary polygonal lines L1 , . . . , Ln . Let L01 , . . . , L0n be the polygonal lines that result from unrolling the polygonal lines Lk torn at arbitrary vertices Ak with angles δk . For R to define a convex polyhedron by gluing it is sufficient and also necessary (when all angles on the boundary are greater 5

6

This is known for the case in which R is homeomorphic to the sphere or the plane. If the development R and the polygons Qi are bounded, then the development R = R ∪ Q1 ∪ . . . ∪ Qn is homeomorphic to the sphere. If R is unbounded or some of the polygons Qi are unbounded, then we can show that R is homeomorphic to the plane, except for the case in which R is a development of an unbounded prism. We leave this remark without proof. This problem was solved in [Sho1]. – V. Zalgaller


237

than or equal to π and each Lk has at least one angle greater than π) that every Lk satisfy the following conditions: (A) The polygonal line L0k together with the straight line segment Ak Ak joining its endpoints bounds a (possibly self-overlapping) polygon Q that lies on the “exterior side” of L0k . (B) An isosceles triangle T with base a = Ak Ak can be constructed so that T lies in Q and the angles γ1 and γ2 in the polygon Q \ T at the endpoints of a satisfy the inequality γ1 + γ2 ≤ 2π − δk (Fig. 102(a)). 0

Lk

Q γ1 Ak

γ2

T a

Ak

Fig. 102(a)

If the polygonal line L0k is closed and there is no segment Ak Ak , then conditions (A) and (B) transform to the following: (A) The polygonal line L0k bounds a polygon Q which lies on the “exterior” side of L0k . (B) In this polygon, the angle γ at the vertex corresponding to the coinciding endpoints of L0k (i.e., to the vertex Ak of the polygonal line Lk ) must not exceed 2π − δk (Fig. 102(b)).

Q γ Ak=Ak Fig. 102(b)

238


For simplicity, we focus our attention on the case of a development bounded by a single polygonal line L. Let A1 , . . . , An be the vertices of L; α1 , . . . , αn , the angles of L; and A1 , the vertex where L is torn for the unrolling. Assume that the conditions of the theorem hold. By the construction of L0 , the angles of L0 on the exterior side are equal to β2 = 2π − α2 ,

,...,

, βn = 2π − αn .

(1)

If L0 is not closed, then we have the polygon Q = Q \ T with angles γ1 and γ2 at the vertices A1 and A2 . Moreover, by assumption γ = γ1 + γ2 ≤ 2π − α1 .

(2)

Identifying the sides of Q which are the lateral sides of the triangle T , we obtain a development Q with one internal vertex. The angle at this vertex is clearly less than 2π. In view of (1) and (2), for the angles on the boundary of Q we have γ ≤ 2π − α1 , β2 = 2π − α2 , . . . , βn = 2π − αn . Therefore, identifying the boundary edges of this development (i.e., the segments of L0 ) with the corresponding boundary edges of the development R, we obtain a development R ∪ Q homeomorphic to the sphere for which the sums of angles at all vertices are less than or equal to 2π. Such a development defines a convex polyhedron by gluing. Thereby the given development R defines a convex polyhedron by gluing as well. If the polygonal line L0 is closed, then we take the polygon Q bounded by L0 for the development Q . Identifying the sides of Q with the edges of R, we then obtain the same result. The sufficiency of the conditions of the theorem is thus proved. If the development is bounded by several polygonal lines, then we only have to attach all the corresponding developments Qi . To prove the necessity of the conditions of the theorem, let us assume all angles on the boundary of R greater than or equal to π and at least one of them greater than π. Suppose that R defines a convex polyhedron P by gluing. This polyhedron is a part of some closed or unbounded convex polyhedron P = P ∪ Q, where Q is the completion of P to P taken with the boundary. Clearly, Q is a convex polyhedron. We never distinguish between Q and some of its developments in what follows. If R is bounded by several polygonal lines, then P \ P splits into several polyhedra; in the sequel by Q we mean any one of them. Let us prove that 2π bounds from above the sum of curvatures at those vertices of P that lie inside or on the boundary of Q. Denote this sum by ω(Q) to distinguish it from the curvature ω(Q) of Q, the latter standing for the sum of curvatures at the internal vertices of Q. (Observe that ω(Q) is the curvature of the closure of Q on P .) If β1 , . . . , βn are angles on the boundary of Q, then for ω(Q) we have


ω(Q) = 2π −

n

(π − βi ).

239

(3)

i=1

The complete angle at the ith vertex of the boundary is composed of the angle βi on Q and the angle αi in P or, which is the same, in the development R. Therefore, the curvature at the vertex is ωi = 2π − αi − βi and the sum of the curvatures is n ωi = (2π − αi − βi ). (4) i

i=1

Combining (3) and (4), we express the sum of curvatures at all vertices as follows ω(Q) = 2π − (π − βi ) + (2π − αi − βi ) = 2π − (αi − π). (5) i

i

All the angles αi are greater than or equal to π by assumption and at least one of them is greater than π. Therefore, ω(Q) < 2π.

(6)

Since the complete angle at each vertex of the convex polyhedron P is at most 2π while the angles on the boundary of P are at least π, it follows that all angles of Q are, on the contrary, at most π. Hence, by Theorem 4, every two interior points of Q are joined by a shortest arc in Q traversing the interior of Q. Using the same trick of “straightening” an angle less than π as in the proof of Theorem 4, it is easy to verify that the shortest arc AB in Q joining two points A and B either lies entirely in the interior of Q (except possibly for its endpoints if they are on the boundary) or coincides with a segment of the boundary of Q. As shown in Subsection 1.8.1, a shortest arc in the interior of a development cannot pass through a point with the sum of angles less than 2π. If an interior point X of the shortest arc AB lies on the boundary of Q, then the sum of angles at X cannot be less than 2π either. Indeed, the angle at X on the side of Q would otherwise be less than π, since the angle at X on the side of P is no less than π by assumption. In this case we could straighten the angle at X in Q, thereby shortening AB, which contradicts the definition of shortest arc. Thus, all angles on the boundary of Q are at most π and every pair of points in Q can be joined by a shortest arc in Q not passing through vertices of P . Let A and B be two points of Q and vertices of P and let ωA and ωB be the curvatures at those points. We have shown that the sum of curvatures at all vertices belonging to Q is less than 2π. Therefore, ωA + ωB < π. 2

(7)

240


Draw a shortest arc AB in Q. Consider a plane and take two equal triangles A B C and A B C with bases A B = A B = AB and angles

ωA ωB , ∠B = ∠B = . (8) 2 2 Such triangles exist because 12 (ωA + ωB ) < π. Making a cut in P ∪ Q along the line AB, glue the bases of the triangles A B C and A B C to both edges of the cut so that the vertices A , A and B , B coincide with A and B respectively. Glue the other sides of the triangles to one another: A C to A C and B C to B C . Instead of P ∪ Q, we thus obtain a new development P ∪ Q or R ∪ Q . The complete angles at A and B increase by angles equal to the curvatures ωA and ωB , i.e., exactly by the amounts needed to complete A and B to 2π. This implies that A and B are no longer vertices. However, we now have the new vertex C = C = C with curvature ωC equal to the sum of ωA and ωB . Indeed, by (8) ∠A = ∠A =

∠C = ∠C = π −

ωA + ωB . 2

The sum of these angles is the complete angle at C, i.e., 2π − ωC . Hence, ωC = ωA + ωB .

(9)

Since the sums of angles at all interior points of the shortest arc AB equal 2π, gluing the triangles A B C and A B C to the cut along AB yields no new vertices on the cut. Thus, the sums of angles at all points are less than or equal to 2π. Since the angles on the boundary of P are greater than or equal to π, those in Q are at most π. Our development Q has one vertex less as compared to Q. At the same time, by (9) the curvature of Q equals that of Q and the angles on the boundary are less than or equal to π. If Q has more than one vertex, then we can repeat the construction and come to a development Q . Continuing in the same way, we obtain a development S that has only one point Q presenting an “essential” vertex of the development R ∪ S (or, equivalently, a vertex of the polyhedron produced from the development by gluing; such a polyhedron exists, since the sums of angles at the vertices of the development are at most 2π). The development R undergoes no changes in this construction. Assume that O is in the interior of S. Then in the development R ∪ S, the sums of angles at all points on the common boundary L of R and S are equal to 2π, implying that the angles on S are given by β1 = 2π − α1 , . . . , βn = 2π − αn , where αi are the angles on R.

(10)


241

Draw a shortest arc OA1 that joins O to the vertex A1 of the boundary of S. Since the development S has no essential vertices but O, by cutting S along OA1 we can place S on a plane and obtain a polygon S (Fig. 103(a)). Its angles are β2 , . . . , βn and the angles β1 and β1 at the vertices A1 and A1 , and (11) β1 + β1 = β1 = 2π − α1 . It follows from (10) that the polygonal line A1 A1 is nothing more than the polygonal line L0 obtained by unrolling the polygonal line L bounding R. Therefore, drawing the segment A1 A1 , we obtain a polygon S bounded by this segment and the polygonal line L0 . So we have checked condition (A) of the theorem in this case. If the polygon S is obtained from S by attaching the triangle T = OA1 A1 (Fig. 103(a)), then, first, S = S \T and, second, formula (11) means that condition (B) of the theorem about the sum of the angles γ1 = β1 and γ2 = β1 of the polygon S = S \ T at the vertices A1 and A1 is satisfied.

S O β1

T

A1

S =S O

S β1

A1

A1

O

A1=A1

Fig. 103(b)

Fig. 103(c)

A1

Fig. 103(a)

T

If the polygon S is obtained from S by subtracting the triangle OA1 A1 (Fig. 103(b)), then the angles γ1 and γ2 at its vertices A1 and A1 are less than those in S . In the extreme case when OA1 A1 degenerates into a segment (Fig. 103(c)), S and S coincide. In any case, γ1 ≤ β1 ,

γ2 ≤ β1

and, by (11), γ1 + γ2 ≤ 2π − α1 . So, even without cutting out any triangle T from S , we infer that condition (B) of the theorem is satisfied. (Of course, we can cut out an arbitrarily narrow triangle from S ; the angles γ1 and γ2 only diminish.) We have thus proved that if the essential vertex O lies in the interior of S, then conditions (A) and (B) of the theorem are satisfied. Now, assume that O lies on the boundary of S. Then there are no essential vertices at all in the interior of S, and S itself is therefore a polygon with angles βi = 2π − αi at all boundary vertices but A1 = O. The angle at

242


A1 = O is clearly γ ≤ 2π − α1 . Therefore, we arrive at the case in which the polygonal line L0 is closed. This line L0 bounds the polygon S and the angle at the coinciding endpoints is γ ≤ 2π − α1 . The proof of the theorem is thus complete.7 It is worth noting that conditions (A) and (B) of the theorem are easy to check for each development R. This is obvious for condition (A), as the construction of the polygonal line L0 raises no difficulties when the boundary edges and the angles αi are given. For condition (B), this is equally simple. Indeed, suppose that (A) is satisfied, so that we have a polygon Q bounded by the polygonal line L0 and the segment A1 A1 . In Q, draw the perpendicular to this segment from its midpoint, take the first point B of its intersection with the polygonal line L0 . Construct the triangle BA1 A1 . It is an isosceles triangle T contained in Q. If the angles at the vertices A1 and A1 in Q \ T are such that their sum γ1 + γ2 is less than or equal to 2π − α1 , then condition (B) is satisfied. However, if γ1 + γ2 > 2π − α1 , then this condition cannot hold, for in every smaller triangle T the sum of angles γ1 + γ2 is still greater. 5.1.6 The next theorem provides sufficient conditions for a given development to define by gluing a convex polyhedron which is a part of an unbounded convex polyhedron. These conditions are in some sense opposite to those of Theorem 8. Theorem 9. Let R be a development homeomorphic to a disk; L, the polygonal line bounding R; and L0 , the polygonal line obtained by unrolling L torn at some vertex A1 . Let α1 be the angle at A1 . Suppose that two rays l1 and l2 can be drawn from the endpoints of L0 so that (A) together with L0 they determine an unbounded (possibly self-overlapping) polygon Q lying on the exterior side of L0 and (B) the sum of the angles of Q at the origins of l1 and l2 is no greater than 2π − α1 . Then R defines by gluing a convex polyhedron that is a part of an unbounded convex polyhedron. Let α1 , . . . , αn be the angles of R at the vertices of the polygonal line L. These angles are also angles on the internal side of L0 . The angles of the polygon Q are β2 = 2π − α2 , . . . , βn = 2π − αn , γ1 , and γ2 ; moreover, γ1 + γ2 ≤ 2π − α1 (Fig. 104). Identifying the unbounded sides l1 and l2 of Q 7

The trick (used in the proof) of reducing two vertices to a single one by attaching triangles turns out to be profitable in other questions. For example, using this trick, we can prove the following theorem: Assume given a cap of curvature ω < 2π. Let its boundary edges be a1 , . . . , an and its angles be α1 , . . . , αn . Then there exists a pyramidal cap (i.e., the lateral surface of some pyramid) with data a1 , . . . , an and α1 , . . . , αn ; moreover, this cap has greatest area among all caps with the same data. Whence we can derive the next theorem: The lateral surface of a regular (n − 1)-hedral pyramid and, not counting isometric polyhedra, this surface only has the largest area among all convex polyhedra homeomorphic to a disk, given the curvature ω < 2π, the perimeter of the bounding polygonal line, and the number of vertices not exceeding a fixed n. The proof is left to the reader.


243

with one another and identifying the bounded sides of Q with the corresponding boundary edges of R, we arrive at a development R ∪ Q homeomorphic to the plane. By the conditions imposed on the angles of Q, in this development the sums of angles at all vertices are less than or equal to 2π, and the development defines a convex polyhedron by gluing. The part of this polyhedron corresponding to R is a convex polyhedron produced from R by gluing. This completes the proof of the theorem. A3

A1 γ1

A2

β3

β2= 2π-α2 l1

γ2

A1

l2

Fig. 104

The conditions of Theorem 9 admit a straightforward verification. Unrolling the polygonal line L, draw a ray l1 from one of its endpoints so that the angle γ1 on the exterior side of L0 is the smallest angle for which l1 (the exterior side of L0 ) and at least one ray l issuing from A1 bound some (possibly self-overlapping) domain. (The possibility of drawing such a ray l1 can be tested in finitely many steps, because we need to examine only finitely many combinations of rays l1 and l going to the vertices of L0 .) Next, draw a ray l2 from A1 forming an angle of γ2 = 2π − α1 − γ1 with the exterior side of L0 . If the rays l1 and l2 and the exterior side of the polygonal line L0 determine an unbounded, although self-overlapping, polygon Q (Fig. 105(a)), then the conditions are satisfied. However, the required polygon Q may not exist. This happens for several possible reasons: (1) The rays l1 and l2 and the exterior side of L0 determine a bounded polygon (Fig. 105(b)). (2) The line L0 self-intersects so that L0 can bound no polygon Q lying on the exterior side of L0 (Fig. 105(c)). (3) The ray l1 or l2 intersects L0 so that, together with L0 , it cannot bound a domain lying on the exterior side of L0 (Fig. 105(d)). In all cases (1)–(3), the conditions of Theorem 9 fail for another choice of the ray l1 issuing from A1 either. Indeed, we cannot decrease γ1 . At the same time, increasing γ1 , we must decrease γ2 , which, however, leads to eliminating none of the cases (1)–(3).

244

5 Gluing and Flexing Polyhedra with Boundary 0

L A1

l2

A1

A1

A1

γ2

γ1 l1

A1

l2

l1

l1

A1

l2

A1

A1 l1

l2

Q

(a)

(b)

(c)

(d)

Figs. 105(a)–(d)

Since the resulting polyhedron in Theorem 9 is a part of an unbounded polyhedron, its curvature is at most 2π. Therefore, the conditions of the theorem apply only to developments of curvature less than or equal to 2π. The question of the extent to which the conditions of Theorem 9, combined with the constraint on the curvature, are necessary remains open, except for the following simple case. Suppose that all angles αi of R, possibly except one angle αj , do not exceed π. In this case, the polygonal line L0 is convex towards the exterior side, and γ1 can be taken to be zero. The only obstacle to the construction of the polygon Q required in Theorem 9 may be the situation in case (1) of Fig. 105(b). In the polygon formed by the polygonal line L0 and the rays l1 and l2 , the angles are γ1 = 0, γ2 = 2π − α1 , then 2π − α1 , . . . , 2π − αn , and finally the angle ϕ between n l1 and l2 . The sum of all the angles of the (n + 2)-gon is nπ. Therefore, i=1 (2π − αi ) + ϕ = nπ and, since ϕ > 0, we have n

(π − αi ) < 0.

(12)

i=1

At the same time, if ω denotes the sum of curvatures at the internal vertices of R, then by Theorem 2 of Section 1.8 n (π − αi ) = 2π − ω. i=1

Together with (12), this yields ω > 2π. The construction is thus impossible only if ω > 2π. So, we have the following theorem. Theorem 10. If in a development homeomorphic to a disk all but possibly one boundary angles are not greater than π while the sum of curvatures at the internal vertices is at most 2π, then this development defines by gluing a convex polyhedron which is a part of an unbounded convex polyhedron.


245

5.1.7 Theorems 9 and 10 cannot be generalized to developments R bounded by several polygonal lines Li . Otherwise, attaching an appropriate polygon Qi to each polygonal line Li , we would obtain an unbounded development not homeomorphic to the plane. Certainly, such a development does not define any convex polyhedron by gluing except possibly an unbounded prism. However, in that case the development R is itself a part of such a prism and has curvature equal to zero. For the same reason, Theorems 9 and 10 cannot be generalized to unbounded developments other than those produced from parts of an unbounded prism by gluing. Further, it can be proved that, with the exception of developments representing parts of an unbounded prism, the conditions of Theorems 9 and 10 are never satisfied by a development not homeomorphic to a disk. Nevertheless, in the case of developments bounded by several closed or unbounded polygonal lines Li , for each of these polygonal lines we can use the sufficient conditions of Theorems 4–10 and so obtain a general theorem incorporating the sufficient conditions of all these theorems. 5.1.8 In conclusion, we present some simple examples of developments that do not define any convex polyhedra by gluing, although they satisfy the necessary conditions (1)–(3) of Subsection 5.1.1.

a

Fig. 106

The polygon Q in Fig. 106 is self-overlapping. Cutting it into triangles, we obtain a development which, however, does not define any convex polyhedron by gluing if the side a is short enough. Indeed, had the polygon Q with an arbitrarily short side a defined a convex polyhedron by gluing, the polygon Q obtained in the limit as the side a tends to zero would yield a region on some convex surface. This is certainly impossible since the sum of angles at the point a in this region would be greater than 2π.8 The curvature of Q is zero. Therefore, the example shows that even a development of zero curvature does not always define a convex polyhedron by 8

It is interesting to check directly that a polygon Q with a relatively short side a cannot be superposed on any convex polyhedron. How small should a be for this phenomenon to hold?

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gluing. The question of finding necessary and sufficient conditions for this to hold is wide open. Fig. 107(a) shows a regular tetrahedron with a triangle T cut out from one of its faces. If we attach new faces to the edges of the cut, for instance, as indicated in Fig. 107(b), then we obtain a nonconvex polyhedron P that cannot be bent into a convex polyhedron. Therefore, the development of this polyhedron defines no convex polyhedron by gluing. The proof is based on the fact that a tetrahedron with the triangle T cut out admits no flexes, i.e., there is no convex polyhedron isometric to it but noncongruent. This easily verifiable assertion is a particular instance of the theorem that will be proved in the next section (Theorem 2 of Section 5.2). At the same time, if a convex polyhedron isometric to P existed, its part corresponding to the tetrahedron with cut T would be a convex polyhedron isometric but not congruent to a tetrahedron with a cut.

Fig. 107(a)

Fig. 107(b)

Theorem 3 of Section 5.2 enables us to construct any number of analogous examples of developments that do not define any convex polyhedron by gluing.

5.2 Flexes of Convex Polyhedra 5.2.1 Calling a convex polyhedron P flexible, we mean that there is a continuous family, a flex, of convex polyhedra Pt each of which is isometric but noncongruent to P . Otherwise, the polyhedron is rigid. More precisely, the above family must possess the following properties: (1) to each value of t in some interval, say 0 ≤ t ≤ 1, corresponds a convex polyhedron Pt and an isometric mapping ϕt of P onto Pt ; moreover, P0 = P , the mapping ϕ0 is the identity transformation of P onto itself, and ϕt is not a motion for t = 0; (2) the family of the polyhedra Pt is continuous, i.e., the points ϕt (X) of the polyhedra Pt corresponding to any given point X of P converge to the point ϕt0 (X) of the polyhedron Pt0 as t → t0 . From now on, the term “flex” is always understood in this sense; so flexes violating convexity or transforming a polyhedron into a set which is not a poly-

5.2 Flexes of Convex Polyhedra

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hedron are excluded from consideration. Every convex polyhedron admits a continuous flex that violates convexity: to see this, it suffices to continuously bend the polyhedron inward in a neighborhood of an arbitrary vertex.9 In Sections 3.3 and 3.5 of Chapter 3, we proved that bounded convex polyhedra of curvature 4π and unbounded convex polyhedra of curvature 2π admit no isometric mappings except motions and reflections and are therefore rigid. Thus, we will study only flexes of bounded polyhedra of curvature less than 4π or those of unbounded polyhedra of curvature less than 2π. We will show, however, that such polyhedra may be rigid as well and even admit no isometric transformations except the trivial ones, i.e., motions and reflections (Theorems 3 and 4). In Subsection 4.5.2 we remarked that each complete (i.e., without boundary) unbounded polyhedron with curvature less than 2π admits a continuous flex. We clarify this result in Subsection 5.2.6. Moreover, we prove that each polyhedron which is a part of a flexible polyhedron admits a flex by itself. Therefore, a polyhedron which is a part of a complete unbounded polyhedron with curvature less than 2π is flexible. (Theorem 7 on flexibility of a part of a flexible polyhedron may seem absolutely trivial at first sight. However, to make such conclusion would be hasty, because in general under a given flex some part of P may remain rigid. Appropriate examples are easy: one of them is the flex of an unbounded polyhedron considered in Subsection 4.5.2. The proof of Theorem 7 below turns out to be not simple at all.) The above remarks show that our problem reduces to studying the flexibility of bounded polyhedra of curvature less than 4π and unbounded polyhedra of curvature less than 2π which are parts of complete unbounded polyhedra of curvature 2π. Since the curvature of a polyhedron is concentrated at its internal vertices, we shall consider polyhedra obtained from complete polyhedra by deleting some parts which contain at least one internal or boundary vertex. We do not know effective conditions necessary and sufficient for the flexibility of a convex polyhedron. The above problem will thus be left unsolved, although the results below seem to bring us rather close to a complete solution.10 At first we consider convex polyhedra homeomorphic to a disk, i.e., bounded by a single closed polygonal line without multiple points. This case is not only the most interesting one from the visual intuitive point of view, it is also most typical because, as we shall further demonstrate, every result 9

Let O be a vertex of a polyhedron P0 and let QO be the support plane of P0 touching P0 only at O. By parallel shifts of this plane inward P0 , we obtain certain planes Qt that cut out pyramids Rt from P0 . Reflecting the pyramid Rt in the plane Qt for each t, we obtain a continuous family of polyhedra Pt isometric to P0 , but nonconvex. 10 At present, such conditions are known; see the Supplement to Chapter 5. – V. Zalgaller

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obtained in this case can be carried over to polyhedra of other topological type. 5.2.2 Let P be a convex polyhedron homeomorphic to a disk and let L be the closed polygonal line bounding P . Let Q be a polygon homeomorphic to a disk and such that the boundary polygonal line of Q admits a length-preserving mapping ψ onto L. Clearly, this means that, with some new vertices taken on L and Q if need be, the sides of the polygon become equal to the segments of L. Here and in what follows, a polygon is thought of as a development without internal vertices which, if glued, transforms into a (possibly self-overlapping) planar polygon. Identifying the points of the boundaries of P and Q corresponding to one another under ψ, we arrive at the development P ∪ Q homeomorphic to the sphere; here we do not distinguish P from its developments. The development P ∪ Q defines a convex polyhedron by gluing if and only if the complete angles (i.e., the sums of angles) at all of its vertices are at most 2π. Since this condition certainly holds for the vertices of P , while Q has no internal vertices, it suffices to require that the sums of angles at the identified points of the boundaries of P and Q be at most 2π. If this “angle condition” is satisfied, then we obtain a closed polyhedron P = P ∪ Q , with P isometric to P and Q to Q. We refer to this construction of the polyhedron P as a patch on the boundary of P by means of Q. Since the closed polyhedron P is determined from the development of P uniquely (up to a motion and a reflection), such a patch is uniquely determined (when P is fixed) if we specify the polygon Q and the mapping ψ from the boundary of Q onto the boundary of P . Some patch is always possible. Indeed, it suffices to take as P the boundary of the convex hull of P . Then P has no other vertices but those of P , and the complement P \ P is always developable on a plane. Taking this complement together with its boundary, we obtain a polygon Q. Gluing this polygon to P yields P . We call such a patch trivial. Two patches by polygons Q and Q and mappings ψ and ψ of their boundaries onto the boundary of a polyhedron P are equal if and only if the developments P ∪ Q and P ∪ Q are isometric and, moreover, the isometric mapping ϕ from the development P ∪ Q onto P ∪ Q takes P onto itself and takes Q onto Q . This means that, first, the polygons Q and Q are isometric and the mapping ϕ between them can be regarded simply as placing one of them on the other and, second, this mapping preserves the correspondence between the boundaries of Q and Q and the boundary of P . (The second condition is essential since, for congruent polygons Q and Q , the rules for their gluing to P may differ, i.e., under the mapping ϕ of Q onto Q , the points corresponding (under the mappings ψ and ψ ) to the same point X on the boundary of P may differ.)


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Theorem 1. A convex polyhedron P homeomorphic to a disk admits a nontrivial isometric mapping if and only if P admits a nontrivial patch. (Here we mean an isometric mapping from P onto a convex polyhedron. Such a mapping is regarded as nontrivial if it is not given by a motion or a motion and a reflection.) The condition is necessary. Assume that P admits a nontrivial isometric mapping ϕ onto some polyhedron P . Denote the convex hulls of P and P by P and P , so that P = P ∪ Q and P = P ∪ Q , where Q and Q are developable on a plane. Since P and P are isometric, we may regard the polyhedron P as the result of attaching Q to P . Thereby we identify the points of P and P that correspond to one another under ϕ and so obtain two developments P ∪ Q and P ∪ Q that result from attaching Q and Q to P .11 Arguing by contradiction, assume that the two patches are equal, i.e., there is an isometric mapping from the development P ∪ Q onto P ∪ Q that takes P onto itself. Obviously, this amounts to the existence of an isometric mapping from P = P ∪ Q onto P = P ∪ Q that takes P onto P . Since these polyhedra are closed, such a mapping must be trivial, which implies that so is the mapping from P onto P . However, regarding P and P as developments, we identify their points that correspond to one another under the given isometric mapping ϕ from P onto P . Therefore, it is ϕ that represents the above-mentioned trivial mapping. This contradicts our initial supposition, which concludes the proof of necessity. Assume now that P admits a nontrivial patch, so that we have two closed polyhedra P = P ∪ Q and P = P ∪ Q , where P is isometric to P , Q and Q are developable on a plane, and P is the boundary of the convex hull of P . Since Q is developable on a plane, the polyhedron P has no vertices other than those of P . Hence, P too is the boundary of the convex hull of P . Since the patches under consideration are distinct, the polyhedra P and P are not isometric and hence not congruent. As the congruence of P and P would imply the congruence of their convex hulls, the polyhedra P and P are not congruent either, which is the desired conclusion. 5.2.3 We now define the notion of deformation of a patch. A polyhedron P with the polygon Q as its patch is the development P ∪ Q. By a deformation of this patch we mean a continuous change of the development P ∪ Q such that the polyhedron P is fixed, while the polygon Q changes (remaining, 11

It is especially important here to distinguish the existence of nontrivial mappings of the polyhedra from the condition that they coincide. A given polyhedron P may admit nontrivial isometric mappings onto itself; in this case the polyhedron P could simply coincide with P , although the mapping ϕ would be neither the identical mapping nor another trivial transformation.

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however, a polygon) or Q itself remains the same but the way it is glued to P changes. Rigorously, a deformation of a patch may be defined as follows. Suppose that, for every value of the parameter t in the interval [0, 1], we are given some patch (Qt , ψt ), i.e., a polygon Qt attached to the boundary of the polyhedron P by means of a mapping ψt from the boundary of Qt onto the boundary of P . Assume that for each t0 the polygons Qt converge (X) for every point X on the to Qt0 as t → t0 so that ψt−1 (X) → ψt−1 0 boundary of P , i.e., the points on the boundaries of the polygons Qt tend to the corresponding points on the boundary of Qt0 . If (Q0 , ψ0 ) is the trivial patch and not all patches (Qt , ψt ) are equal, then we say that a deformation of the trivial patch is given. Theorem 2. A convex polyhedron P homeomorphic to a disk is flexible if and only if there exists a deformation of the corresponding trivial patch. Moreover, to each deformation of the trivial patch corresponds a welldefined flex of P and vice versa. The proof is an easy consequence of Theorem 1, by which a deformation of a patch corresponds to a deformation of the polyhedron and vice versa. We only need to prove that the two deformations are continuous simultaneously. We confine our exposition to proving the following: to a continuous deformation of a patch there does correspond a flex of the polyhedron P in the sense of the rigorous definition at the beginning of the current section. The converse assertion is an almost trivial consequence of definitions; it suffices to consider the convex hulls P t = Pt ∪ Qt of the polyhedra Pt given by the flex of P . To exclude any motion of P , choose three vertices A, B, and C and, fixing coordinate axes x, y, and z in space, move P so that A arrives at the origin, B lies on the positive x half-axis, and C belongs to the half-plane y > 0 of the xy plane. If P is a polyhedron isometric to P , then assume P to be placed so that the positions of its corresponding vertices A , B , and C obey the same requirements. To exclude any reflection, orient P by fixing the direction of rotation on some small contour and consider only those polyhedra P isometric to P that result from P under isometric mappings preserving the orientation given by the right-handed screw determined by the fixed direction of rotation and the outward normal. Since a reflection changes the orientation of the screw, this requirement excludes reflections. If the requirement fails for a given polyhedron P , then we replace P by the symmetric polyhedron. Once a motion and a reflection are excluded, we may say that a flex is any genuine deformation of a polyhedron. (It is about these flexes which exclude motion that we assert their one-to-one correspondence to the deformations of a patch.) Suppose that to each t in some interval we assign a patch on P by a polygon Qt so that together these patches constitute a continuous deforma-


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tion of the trivial patch. We then have a continuous family of developments Rt = P ∪ Qt together with the family of closed polyhedra P t = Pt ∪ Qt produced from Rt by gluing; here Pt are isometric to P . Let us prove that the polyhedra Pt converge to Pt0 as t → t0 . Assuming the contrary, we can find a sequence tn → t0 such that the sequence Ptn does not converge to Pt0 . By the continuity of the family of the developments Rt (see Lemma 3 of Section 4.2), from the sequence of the polyhedra Ptn we can extract a subsequence that converges to a polyhedron P with limit development Rt0 . Since closed convex polyhedra with equal developments are congruent, while motions and reflections are excluded, it follows that the polyhedron P coincides with P t0 . This proves that P tn must converge to P t0 , implying that the sequence of Ptn converges to Pt0 also. Moreover, if X is a point of P and ϕt are isometric mappings of P onto Pt , then the sequence ϕtn (X) converges to ϕt0 (X). Indeed, by the condition on the vertices A, B, and C and by the convergence of P tn and P t0 , the points ϕtn (A), ϕtn (B), and ϕtn (C) must converge to ϕt0 (A), ϕt0 (B), and ϕt0 (C). However, fixing three points, we completely determine the position of all the other points of the polyhedron Pt0 . Hence, ϕtn (X) → ϕt0 (X) for every point X. By definition, this means that the family of polyhedra Pt represents a flex of the given polyhedron P . Our conclusion is far from trivial, since it is based on the congruence theorem for closed convex polyhedra with the same development. Without this theorem, we are not in position to assert the convergence of P tn to P t0 and with it the continuity of the family of polyhedra obtained. 5.2.4 Theorem 2 reduces the question of flexibility of a convex polyhedron to the purely plane geometry problem of whether or not it is possible to deform a given polygon so as to preserve the conditions allowing the attachment of a patch, i.e., mainly to preserve the given upper bounds of its angles. We can solve the problem completely for convex polygons, which gives the following Theorem 3. Let a polyhedron P be obtained from a closed convex polyhedron P by cutting out some part that unrolls on the plane, becoming a convex polygon Q when the boundary of the cut is adjoined. (Since Q unrolls on the plane, it has no internal vertices.) (1) If at least two vertices of P lie on the boundary of the cut P \ P , then P is a flexible polyhedron. (In the limit case, we may simply cut P along a geodesic arc between two vertices.) (2) If exactly one vertex A of P lies on the boundary of P \ P and if α and α0 are the angles at A on P \ P and P , then P is a flexible polyhedron if and only if α ≥ α0 or, in other words, if the angle α cut out together with P \ P is not less than half of the complete angle α0 + α at A. Moreover, if α < α0 , then the polyhedron admits no nontrivial isometric mappings at all. (3) If there are no vertices of P on the boundary of P \P , then P is a rigid polyhedron.

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The last item is covered by Theorem 1 of Section 3.5, because in this case the curvature of P equals 4π. Observe that the polyhedron P is the boundary of the convex hull of P since P contains all vertices of P and so P results from attaching the trivial patch Q to P . In this connection, P can be characterized as a polyhedron admitting a trivial patch by a convex polygon. Let us prove the first assertion of the theorem. Let A and B be vertices of Q corresponding to the vertices of P and let α and β be the angles at these vertices on Q (Fig. 108(a)). If α0 and β0 are the angles at these vertices on P , then α + α0 < 2π and β + β0 < 2π, i.e., α < 2π − α0 ,

β < 2π − β0 .

(1)

A α

α0

A

P C

Q

D

Q

β B

B

Fig. 108(a)

Fig. 108(b)

Take two points C and D separating A and B on the boundary of Q (Fig. 108(b)). Draw the convex quadrangle ACBD and start lengthening the diagonal CD while preserving the sides of ACBD.12 In this process, the parts AC, CB, BD, and DA of the boundary of Q move without changing their shapes. Then the angles α and β at A and B increase, while the other angles of Q remain the same. Inequalities (1) are preserved under a sufficiently small deformation and since the other angles never increase, it follows that the patching condition is not violated. Combined with Theorem 2, this proves the first claim of the theorem. The limit case in which Q degenerates into the segment AB is settled similarly. It suffices then to treat AB as a doubly-covered segment and to deform it into a quadrangle. Now let us prove the second claim of the theorem. Let A be the only vertex of Q corresponding to a vertex of P and let α be the angle in Q at A (Figs. 109(a), (b)). If α0 is the angle at this vertex in P , then α < 2π − α0 . 12

(2)

This quadrangle may in fact be a triangle (when A and B are adjacent vertices and D is a point on the side AB.


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Given any other point B on the boundary of Q, for the angle β at B we have β = 2π − β0 . Therefore, under any deformation of Q that does not violate the patching conditions on the angles, none of the angles β increases. At the same time, under a convexity-preserving deformation of a convex polygon, the increments of angles change sign at least four times as we go around the polygon. (Cf. Lemma 2 of Section 3.1; the lemma and its proof can be carried over word for word to planar polygons.) P B

A Q D

A

C B

α Q

B C

D

(a)

A A Y ξ α η

X

C

D

(b)

(c)

Figs. 109(a)–(c)

Any deformation that violates convexity pushes at least one vertex inward and increases the angle at that vertex, for this angle becomes greater than π. That is why inward pushing is not possible for any vertex other than A. Taking points X and Y on the sides of Q incident to A and reflecting the polygonal line XAY in the straight line XY , we obtain a polygon Q with the vertex A pushed in and two new vertices X and Y pushed out (Fig. 109(c)). The angles at them are α = 2π − α;

ξ, η < π.

The angles ξ0 and η0 on P at the points corresponding to X and Y equal π, since these points are not vertices of P . Therefore, ξ+ξ0 < 2π and η+η0 < 2π, i.e., here the patching condition is satisfied. For the point A , we have α + α = 2π − α + α0 . Hence, if α ≥ α0 , then α + α ≤ 2π and the patching condition is satisfied at A . However, if α < α , then α + α > 2π, i.e., the condition fails. Therefore, if α ≥ α0 , then pushing in A into Q continuously (which corresponds to backing the points X and Y away from A continuously), we obtain a deformation of Q that does not violate the patching conditions and, by Theorem 2, this will correspond to a continuous flex of the polyhedron P . Under this flex, two vertices X and Y appear on P which move away from A. In Fig. 109, the flex consists in folding the “peak” ABC; this can easily be seen in Fig. 110.

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B

X Y

C

A D

Fig. 110

If α < α0 , then the patching condition is violated and cannot be restored by any deformation of the polygon Q . Indeed, the angles at all vertices except X and Y must never increase, but then in accordance with Lemma 3 of Section 3.1 the segment XY shortens, implying that the angle α increases. Hence it follows that for α < α0 by Theorem 1 the polyhedron P admits no nontrivial isometric mappings at all. 5.2.5 Theorem 4. Suppose that a polyhedron P is obtained from a closed convex polyhedron P by cutting out some part P \P that contains no internal vertices and therefore unrolls on the plane. Suppose that the resulting polygon Q is nonconvex. Let n denote the number of vertices of P on the boundary of the cut P \ P or, which is the same, on the boundary of P . (1) If n ≥ 2, then P may be flexible or rigid. (2) If n = 1, then P may be flexible or rigid and, in contrast to the case of a convex polygon Q, the polyhedron P may be flexible also for α < α0 , where α and α0 are the angles in Q and P at the vertices corresponding to the only vertex of P on the boundary of P . With a fixed complete angle α + α0 , the polyhedron P may be flexible for arbitrarily small α. (3) If n = 0, then P is rigid. The last claim follows from the rigidity of polyhedra of curvature 4π. C

C D

γ

Q

Q α

δ

β

A

A B

B Fig. 111(a)

Fig. 111(b)

Consider the first claim of the theorem for n ≥ 3. Let A, B, and C be vertices of Q corresponding to the vertices of P on the boundary of P (Fig. 111(a)). In exactly the same way as in Theorem 3, we see that the angles


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α, β, and γ at these vertices can increase without violating the patching conditions. Consider the quadrangle ABCD with vertex D on the perimeter of Q between A and C. Start deforming this quadrangle, and the polygon Q along with it, moving the polygonal lines AB, BC, CD, and DA while preserving their shapes. Then all the angles of the polygon remain constant, except for the angles α, β, γ, and δ at the vertices A, B, C, and D. Within certain limits, the angles α, β, and γ may undergo arbitrary variations. While, for the patching condition to remain valid, the angle δ may only decrease. However, some deformation decreasing δ is possible. For instance, in the case depicted in Fig. 111(b), the angle δ decreases when we shorten the diagonal AC. For another structure of Q, it may happen that shortening AC leads to an increase of δ, but then, conversely, we lengthen AC and the angle δ decreases again. A special case appears when all three points A, B, and C lie on a single straight line. In that case the segments AB and BC remain of constant length and the “diagonal” AC may only shorten. (This is precisely the place where the possibility of rigidity resides; see the Complement to Chapter 5.) In any case, if the points A, B, and C do not lie on a single straight line, then there is a deformation not violating the patching condition. Now, we turn to the case in which there are at most two vertices of P on the boundary of P . It is obvious that the polyhedron P may be flexible if the angles of Q at the corresponding vertices are so large that pushing them inward does not violate the patching conditions as in the case of a convex polygon Q. Therefore, it suffices to indicate an example of a rigid polyhedron P or, on the contrary, a flexible polyhedron but only if the angle at the vertex of Q corresponding the only vertex of P on the boundary of P is arbitrarily small. Consider the cube with a deleted nonconvex polygon Q depicted in Fig. 112(a). This polygon is also shown separately in Fig. 112(b).

E

B F

A

B

G C

α A

D

F

β γ G

Figs. 112(a)–(c)

G

F D

(b)

C

B A

Q ϕ

E

(a)

C

E

D

(c)

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If we vary the angles α, β, γ, and ϕ, without changing the side lengths of Q, so that the angle β decreases, then, as is easy to see, the angles γ and ϕ also decrease, whereas the angle α increases. The patching conditions remain valid for the modified polygon Q (Fig. 112(c)). So the cube with the indicated cut is flexible. Observe that the polygon Q touches only one vertex A of the cube and can have an arbitrarily small angle α at A. Of course, the resulting flex is rather intricate, it involves three new vertices B, F , and G. We suggest two exercises in connection with Theorem 4. (1) Study the flexibility conditions for the polyhedron obtained from a closed polyhedron by cutting out a polygon Q of the shape shown in Fig. 113(a), with the points A and B located at the vertices of the closed polyhedron. (The more general case of a polygon Q of analogous structure is examined similarly.) Here both flexibility and rigidity are possible, all depending on different conditions. (2) Prove that the polyhedron obtained from a closed polyhedron by cutting out a polygon Q of the shape shown in Fig. 113(b) is flexible, provided that A is a vertex of the closed polyhedron P and the prolongation of at least one side of the angle A intersects Q. Moreover, for flexibility, it suffices that such a prolongation of a side still intersects Q after rotating around A through an angle α ≤ 12 {2π − θ, π}, where θ is the complete angle of P at A. A A Q Q

(a)

B

(b)

Figs. 113(a)–(b)

5.2.6 Theorem 5. If a polyhedron P , homeomorphic to a disk, is obtained from a closed convex polyhedron P0 by deleting the part P0 \P whose interior contains at least one vertex, then P is a flexible polyhedron. Suppose that P is obtained from P0 as indicated and let P be the boundary of the convex hull of P . We will show that P has at least one face which is not a prolongation of any face of P . Assume the contrary. Then P is obtained simply by prolonging the extreme faces of P . Since P0 contains P and therefore necessarily contains P , each face of P , being a prolongation of a face of P , must also belong to P0 . However, the prolongation of faces of P uniquely determines the closed convex polyhedron P . Thus all the faces of this closed polyhedron also belong to P0 , and so it is clear that P and P0 coincide. However, the convex hull of P


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has no vertices other than the vertices of P itself. Hence, P and P0 have the same number of vertices, contradicting the assumption that P is obtained from P0 by deleting a domain whose interior contains at least one vertex. Thus, P has at least one face that is not a prolongation of any face of P . Since P and P have the same set of vertices, the vertices of this face of P lie on the boundary of P . Clearly, there are at least three such vertices. Therefore, P results from P by deleting a part P \ P whose boundary contains at least three vertices. So, P is flexible by Theorem 4.13 5.2.7 We now show that it is actually not too restrictive to consider only polyhedra homeomorphic to a disk. Theorem 6. A bounded convex polyhedron with boundary consisting of several polygonal lines is flexible if and only if it is part of a flexible polyhedron whose boundary is only one of these polygonal lines. Assume that the given polyhedron P with boundary polygonal lines L1 , . . . , Ln is rigid. Let P1 = P ∪ Q1 ∪ . . . ∪ Qn be the polyhedron obtained from P by patching all the polygonal lines Li except L1 . For example, P1 could be the boundary of the convex hull of P with the polygon patching the polygonal line L1 removed. However, Q2 , . . . , Qn need not be polygons; it is only important that they be convex polyhedra, each bounded by a single polygonal line Li . Since the polyhedron P is rigid, the only possible flexes of P1 are reduced to flexes of each of the polyhedra Qi that do not move their boundaries. However, such a flex is impossible because if the boundary of Qi is fixed, then the polygon that can be attached to Qi is determined uniquely. (Moreover, Theorem 4 of Section 3.5 asserts that a flex of the polyhedron Qi is impossible if the angles between the boundary edges of Qi only are fixed.) This proves that if P is rigid, then no polyhedron, containing P and having for its boundary exactly one of the polygonal lines L1 , . . . , Ln , may be rigid. Assume now that P is flexible. Let P = P ∪Q1 ∪. . . ∪Qn be the boundary of the convex hull of P . A flex of P yields a deformation of its convex hull and so yields deformations of the polygons Qi .14 Therefore, if all the polygonal lines L1 , . . . , Ln bounding P have no common vertices, then the fact that the patching conditions for an individual polygonal line are independent implies that among the polygons Qi there must exist at least one polygon, say Q1 , 13

The special case of item (1) of Theorem 4 in which three vertices lie on a single straight line is excluded here, because the three vertices in question belong to a single face of P \ P . If unrolled on the plane, they can never lie on a single straight line. 14 Here and in what follows, speaking of the deformation of a polygon, we have in mind a deformation of a patch in the sense of the rigorous definition of Subsection 5.2.3.

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that may be deformed without changing the remaining polygons. Then the polyhedron P1 = P ∪ Q2 ∪ . . . ∪ Qn , bounded only by the polygonal line L1 , is flexible. If some of the polygonal lines L1 , . . . , Ln have common vertices, then our argument fails. In this case we can choose between two points of view: First, we may treat two polygonal lines with a common vertex as a single one. The cut polygon is “squeezed to a point” at this vertex and the vertex should be viewed as two coinciding vertices. Under the deformation of the patch and, accordingly, under the flex of the polyhedron, these vertices may move apart. Second, we may treat two polygonal lines with a common vertex as two different polygonal lines. Under a flex of the polyhedron, such a vertex never splits. It seems very probable that in this case Theorem 6 remains valid. However, an exhaustive proof eludes us. A complete answer to this question is an intriguing problem.15 5.2.8 Theorem 7. If a polyhedron P is a part of a flexible bounded polyhedron P0 , then P is also flexible. (At first glance, the assertion seems rather trivial. However, the matter is not as simple as it seems, because under a given flex of P0 some part P of P0 may remain rigid. Appropriate examples are easy to find.) Let P0 be a flexible polyhedron. We thus have a family of polyhedra Pt isometric to P0 . Arguing by contradiction, assume that P0 has a rigid part P which is consequently a part of each of the polyhedra Pt . Let P and P t be the boundaries of the convex hulls of P and Pt . The convex hull of a polyhedron is determined by its vertices. Therefore, if the boundary P t for some t was different from P , then P t would have vertices other than those of P . But then P could be obtained from the closed polyhedron Pt by deleting a region whose interior contains a vertex and P would be flexible by Theorem 5 (if the boundary of P consists of several polygonal lines, then the conclusion remains valid by virtue of Theorem 6). This contradicts our assumption. Hence, P t and P coincide for all t. Thus, the convex hull of Pt does not change, coinciding with the convex hull of P0 . Therefore, the polyhedra Pt themselves coincide with P0 . Since this conclusion applies to an arbitrary flex, P0 is rigid, contradicting the original assumption. This completes the proof of the theorem. The above result, combined with item (1) of Theorem 3, allows us to conclude that if we delete, from a closed polyhedron P0 , some part Q that unrolls onto a polygon, then the remaining polyhedron is always flexible, provided that Q includes a segment (geodesic arc) between two vertices of P0 . Hence, after deleting a nonconvex polygon Q, the remaining polyhedron may be rigid only if Q includes no segment with this property. 15

See the Supplement to Chapter 5 and the note [Sho4]. – V. Zalgaller


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In view of Theorems 3–7, it remains to answer the question of the rigidity or flexibility of the polyhedron obtained from a convex polyhedron P0 by deleting a part that unrolls onto a nonconvex polygon Q and whose boundary contains at most two vertices of P0 ; moreover, if the boundary contains exactly two vertices, then we consider only cuts that do not include a geodesic arc between these vertices. By virtue of Theorem 2, the problem is reduced to deforming the polygon Q so as to preserve the patching conditions.16 5.2.9 Now, we turn to flexes of unbounded convex polyhedra. As shown in Sections 3.3 and 3.5 of Chapter 3, an unbounded polyhedron of curvature 2π is rigid even if it has a boundary. Therefore, we consider only polyhedra of curvature less that 2π. In this case we must distinguish between two possibilities: (1) the unbounded polyhedron P is a part of a polyhedron of curvature 2π; (2) the unbounded polyhedron P is a complete unbounded polyhedron of curvature less than 2π or a part of such a polyhedron. Since an unbounded polyhedron of curvature 2π, like a closed polyhedron, is rigid, it follows that the methods and results of the preceding subsections apply readily to polyhedra of type (1). The case in which the given unbounded polyhedron is obtained from a complete polyhedron of curvature 2π by deleting an unbounded part will be rather new. As an extreme case, we may consider polyhedra obtained from complete polyhedra by making unbounded slits. If we make a slit along an unbounded ray issuing from a vertex, then the polyhedron becomes flexible. However, if the slit is strongly curved at the origin and straightens only later as it goes at infinity, then the polyhedron may be rigid.17 However, we do not dwell on the study of this case. The notion of deformation of a patch extends naturally to unbounded polygons. At the same time, the method based on such deformations can be carried over to this case as well. Considering unbounded polyhedra obtained from complete polyhedra of curvature 2π by deleting bounded or unbounded parts and using arguments quite similar to those given above in the case of bounded polyhedra, the reader can formulate and prove theorems about the flexibility of such polyhedra similar to Theorems 1–7. In particular, just as asserted in Theorems 3 and 4, such polyhedra may be flexible or rigid. For polyhedra of type (2), there is no alternative: all of them are flexible. All possible flexes of complete polyhedra are described in the following theorem that follows from Olovyanishnikov’s Theorem proved in Section 4.5. 16 17

At present, this problem is completely solved, see [Sho1] – V. Zalgaller Some results on the polyhedra with slits are obtained in [Sho5]. – V. Zalgaller

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Theorem 8. Let P0 be an unbounded complete convex polyhedron of curvature less than 2π. Assume further that V0 is the limit angle of P0 and G0 is a generator of V0 which is the limit generator of some ray L0 on P0 . Let V1 be a polyhedral angle isometric to V0 and let G1 be a generator of V1 . Then (1) there is a flex of V0 to V1 such that G0 goes to G1 ; (2) this flex of V0 uniquely determines a flex of P0 to a polyhedron P1 with limit angle V1 , and under this flex G1 becomes the limit generator of the ray on P1 corresponding to the ray L0 by isometry. There are no other flexes of P0 . When we say that a flex of V0 to V1 uniquely determines a flex of P0 to P1 , we mean that to the family of angles Vt with generators Gt corresponds a family of polyhedra Pt unique up to translations that change neither the limit angle nor the limit generator. The first assertion claiming the flexibility of the polyhedral angle is rather obvious and its rigorous proof causes no difficulties. Let us prove the second assertion. Suppose that a family of polyhedral angles Vt with distinguished generators Gt realizes a flex of the angle V0 with generator G0 to the angle V1 with generator G1 .18 By Olovyanishnikov’s Theorem (the theorem of Section 4.5 and Theorem 3 of Section 3.4), to each Vt with distinguished generator Gt corresponds a unique polyhedron Pt isometric to P0 , with the same orientation as P0 and with the ray L0 corresponding to the generator Gt . (Clearly, the orientation is preserved under a continuous flex.) The fact that the polyhedra Pt form a continuous family is proved in the same way as a similar assertion in Theorem 2, on the basis of the uniqueness theorem (Section 3.4). Conversely, since every flex of a polyhedron gives rise to a flex of the limit angle, P0 admits no other flexes. 5.2.10 Two more questions deserve discussion. It is easy to give simple examples of isometric convex polyhedra with boundary which admit no isometric flexes to one another preserving convexity (find such examples).19 Nevertheless, the following theorem seems rather plausible: given two sufficiently close isometric polyhedra, there is a flex taking one of them to the other. More 18

It may happen that the angles V0 and V1 are congruent but the generators G0 and G1 are positioned differently in V0 and V1 . Then, assuming that V1 coincides with V0 , by a flex of V0 to V1 we mean a flex of V0 to itself such that the generator G0 is mapped to G1 . 19 On the lateral surface of the prism shown in Fig. (a), consider the “belt” F0 “with its buckle fastened.”

5.3 Generalizations of Chapters 4 and 5

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precisely, if convex polyhedra Pi isometric to P converge to P , then there is a flex of P to each of the polyhedra Pi with sufficiently large index.20 The second question relates to nonconvex polyhedra. There is no example of a closed polyhedron admitting a continuous flex without folding faces. (All known examples are polyhedra with self-intersections; clearly, they to not admit flexes that can be implemented in real models.) The most interesting problem of the theory of polyhedra is probably the question of whether all closed polyhedra are rigid if folding their faces is forbidden or it is possible to find examples of flexible polyhedra.21

5.3 Generalizations of Chapters 4 and 5 5.3.1 Carrying over the theorems of Chapters 4 and 5 to non-Euclidean three-dimensional hyperbolic and spherical spaces, we must deal with developments composed of polygons in the sense of the geometry of the corresponding space. For example, polygons in spherical space are ordinary spherical polygons on the sphere whose radius equals that of the given spherical space. Having this condition in mind, the existence theorem for closed convex polyhedra with prescribed development may be formulated in hyperbolic or spherical space without changes. Also, the proof remains almost the same, since in essence it is not connected with the parallel axiom. We leave the appropriate corrections in the proofs of Sections 4.1–4.3 to the reader. Here we only show the places to be changed.

F1

F0

(a)

(b)

As a convex polyhedron with boundary, the belt is isometric to a planar polygon F1 (Fig. (b)). However, there is no continuous flex of F0 to F1 within the class of convex polyhedra with boundary. (Nevertheless, there is a flex within the class of “locally convex ” polyhedra with boundary: it suffices to increase the cross-section of the prism continuously, allowing the diverging parts of the “buckle” of overlap during the flex. – V. Zalgaller 20 This highly probable assertion for polyhedra remains unproved. However, an analogous claim for convex surfaces with boundary is false, see [Sho2]. – V. Zalgaller 21 Such examples exist, see Subsections 3.6.1a and 3.6.1b . – V. Zalgaller

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First of all, we point out that it is much easier to carry over the proof to hyperbolic space than to spherical space. The deep reason for this is that the axioms of spherical geometry differ from those of Euclidean geometry not only by the absence of the parallel axiom, but also by the axioms of incidence and ordering. For example, in two-dimensional spherical space, or simply on the sphere, two antipodal points are joined by infinitely many “straight” lines whose role is played by the arcs of great circles. This leads in particular to the fact that, in spherical space, it is not true that an arbitrary polygon can be split into triangles by diagonals (which is clear from easy examples). Returning to Lemma 1 of Section 4.1, observe the following. The assertion that the development has at least three vertices remains valid in hyperbolic space. Indeed, since hyperbolic space is of negative curvature, the total (or integral) curvature still equals 4π (the total curvature of a development homeomorphic to the sphere is 4π, a fact completely independent of the curvature of the ambient space). Moreover, the total curvature is obtained by summing the positive curvatures at the vertices and the negative curvatures of the polygons of the development. Therefore, the sum of curvatures at vertices is greater than 4π. Since the curvature at each vertex is less than 2π, there must be at least three vertices. In the case of spherical space, the same argument shows that the sum of curvatures at all the vertices is less than 4π. In this case there are developments with no vertices at all (a polyhedron with such development is simply the sphere). Also, there exist developments with exactly two vertices (such a development can be imagined as a digon with sides glued together). It is possible to demonstrate that no development has exactly one vertex. Now, to obtain an assertion replacing Lemma 1 in non-Euclidean space, it suffices to verify directly that all developments with at most three vertices are realizable. In the proof of Lemma 2 we should also distinguish two cases. In hyperbolic space, the proof is repeated word for word. In spherical space, the theorem on the triangulation of polygons is false, as was mentioned. In spite of this, Lemma 2 remains valid; however, its proof requires extra arguments. To look into relevant details is an interesting problem that we leave the reader. The proof of Lemma 3 requires minor changes. The proof of Lemma 4 is based on the fact that the angle of a triangle is a monotone function of the subtended side; this holds true in non-Euclidean spaces. Finally, the uniqueness theorem for polyhedra with prescribed development remains beyond doubt because, as explained in Subsection 3.6.4, the proof of this theorem can be carried over to non-Euclidean spaces word for word, provided that we exclude “digons” from consideration. However, these digons are of no interest to us, since the uniqueness theorem applies only to polyhedra with at least three vertices. Thus, inspecting the proof from beginning to end and performing the appropriate modification, the reader can prove the existence of polyhedra with rescribed development in non-Euclidean spaces as well.


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Along with this theorem, we carry over to these spaces its corollaries from Section 5.1, which concern the gluing of bounded polyhedra with boundary (general Theorems 1 and 7 and Theorem 2 about caps of Section 5.1). The same relates to the theorems of Section 5.2 on flexes of bounded polyhedra (Theorems 1–7 of Section 5.2). Here no changes in proofs are required. In spherical space, there are no unbounded polyhedra. Hence, it is impossible to raise the question of producing unbounded polyhedra by gluing. On the contrary, in hyperbolic space this question is much richer than in Euclidean space. In the latter, every complete unbounded polyhedron, except for prisms, is homeomorphic to the plane. In hyperbolic space, as it has already been indicated in Section 3.6, there are unbounded convex polyhedra homeomorphic to an arbitrary planar domain bounded by finitely many curves. In this connection, the existence theorem for unbounded polyhedra with prescribed development in hyperbolic space can be stated in the following way: Each development without boundary, consisting of bounded and unbounded polygons in the hyperbolic plane and homeomorphic to an arbitrary planar domain,22 defines some convex polyhedron by gluing in the corresponding hyperbolic space.23 The proof of this theorem is based on the corresponding theorem for closed polyhedra. Namely, from the development R we cut out a bounded part R so that R defines a closed polyhedron by gluing. After that, expanding this part R to the whole development R, in the limit we obtain an unbounded polyhedron which is glued from R.24 Olovyanishnikov’s Theorem does not generalize to hyperbolic space. The notion of limit angle of polyhedra in this space is still unstudied. We have already mentioned this problem in Section 3.6 of Chapter 3. Its solution and, all the more, the complete study of flexes of unbounded polyhedra of an arbitrary topological type in hyperbolic space are very interesting open problems.25 5.3.2 There are two ways of generalizing existence theorems for polyhedra with prescribed development to curved surfaces. The first, more specific but less general, consists in replacing the developments composed of planar polygons with the “developments” formed by 22

There is no need to stipulate explicitely that the domain is bounded by finitely many curves: this is immediate from the fact that a development has finitely many polygons by definition. 23 Recall that hyperbolic spaces differ in their curvatures. 24 See [A11] where this approach is generalized to the case of a not necessarily polyhedral metric. 25 Rigidity (flexibility) problems for unbounded polyhedra in hyperbolic space are not completely solved as yet. Partial results can be found in [Da], [Ver], [VoM], [Al1], and [Al2]. – V. Zalgaller

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finitely many pieces of arbitrary convex surfaces. By a “development” we now mean a collection of pieces G1 , . . . , Gn of convex surfaces with the following properties: (1) Each piece Gi is homeomorphic to a bounded or unbounded polygon and has a boundary consisting of rectifiable curves. The boundary of each piece is partitioned into arbitrary segments, the sides of Gi , and the partition points are treated as vertices. (2) The sides of Gi are “glued” together, i.e., identified pairwise with one another so that all identified segments have the same length. Some sides are possibly glued to no other sides, thus forming the boundary of the “development.” In a particular case, these Gi are pieces of a plane. We say that a surface F is produced from the pieces Gi by gluing if F splits into parts Fi so that (1) Each piece Gi can be mapped isometrically onto the corresponding part Fi of F (i.e., so that to each curve on Gi there corresponds a curve of the same length on F ). (2) These mappings take each of the identified segments on the boundaries of the pieces Gi to a common segment of the boundaries of the parts Fi . The definition of the construction of a polyhedron from a development by gluing represents a particular case of the above general definition. Other examples are the construction of cylinders and cones from familiar pieces of planes or that of a spindle-shaped surface from a hemisphere. The problem consists in describing the conditions under which a given “development” defines by gluing a convex surface of some type, for instance a closed surface. A solution of this problem is given by the theorem that I call the “Gluing Theorem.” A precise general statement of this theorem requires a series of definitions. Avoiding this, let us formulate the theorem under the assumption that the curves bounding the pieces Gi are piecewise smooth and have piecewise continuous geodesic curvature. The latter is nothing more than the curvature (regarded as a set function) on the piece Gi . When the boundary of Gi at the given point is outward convex, this geodesic curvature is assumed positive. Otherwise we assume it negative. In order that a “development” R homeomorphic to the sphere produce a closed convex surface by gluing, it is necessary and sufficient that the following conditions be satisfied: (1) for each vertex, the sum of angles of Gi that will meet at the vertex under the identification of boundaries is at most 2π; (2) the sum of geodesic curvatures at the identified points on the boundaries of the pieces Gi is greater than or equal to 0.


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If the sides of the pieces Gi are geodesics, then their geodesic curvature vanishes everywhere. In this case the second condition holds automatically and we are left with the first condition only. Thus, the condition for gluing a whole convex surface from geodesic polygons cut from some convex surfaces does not differ from the condition for gluing a convex polyhedron from planar polygons. Similar theorems could be formulated for “developments” of other topological types. Conditions (1) and (2) remain unchanged. The Gluing Theorem provides a very powerful instrument for studying convex surfaces and especially their flexes.26 Its proof is based, however, on a generalization of theorems about polyhedra in another more abstract direction.27 5.3.3 This second, more abstract but, at the same time, more general point of view consists in the following. As observed in Section 1.6, the specification of a development is equivalent to the specification of the intrinsic metric of the polyhedron. This metric determines distances between points on the polyhedron which are measured on the polyhedron itself. If a development R is, for example, homeomorphic to the sphere, then, mapping R onto a sphere S, we carry over the metric ρ, given initially on R, onto S. Then construction of a polyhedron P from R by gluing amounts to an isometric mapping from the sphere S with metric ρ onto the polyhedron P . In this connection, we can consider the existence problem of surfaces with prescribed intrinsic metric for arbitrary surfaces. Let F be a surface and let X and Y be two points of F . The distance ρF (X, Y ) between X and Y on F is defined as the greatest lower bound of the lengths of curves in F joining X and Y . This distance, viewed as a function of a pair of points X and Y , is the intrinsic metric of F .28 In these terms, the problem of generalizing the existence theorem for polyhedra with prescribed development may be posed as follows: Let ρ(X, Y ) be a continuous function of pairs of points on the sphere S, an abstract metric satisfying the three conditions necessary for any metric.29 The problem is then to describe the additional necessary and sufficient conditions under which the sphere S with metric ρ can be mapped isometrically onto some closed convex surface F . In other words, when does there exist a mapping ϕ from S onto some surface F such that ρ(X, Y ) = ρF (ϕ(X), ϕ(Y )), where ρF is the intrinsic metric of the surface F ? 26

See [A13] and [A15, Chapter VII, § 1, and Chapter IX, § 3]. Using this method, A. V. Pogorelov [P6] studied the life of a local singularity of a convex surface during a flex. – V. Zalgaller 28 The intrinsic metric of a nonsmooth convex surface is not Riemannian. Geometric, and later analytic, methods for studying manifolds with a nonsmooth metric which started with the book [A15] are now highly developed. See [AZ], [R1],[R3], [BN], [BGP]. – V. Zalgaller 29 It is easy to verify that ρF satisfies the standard conditions: (1) ρ(X, Y ) = 0 if and only if X = Y , (2) ρ(X, Y ) = ρ(Y, X), and (3) ρ(X, Y ) + ρ(Y, Z) ≥ ρ(X, Z). 27

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An similar question can be raised for unbounded surfaces F with the metric ρ now given on a plane. These problems are completely solved for general convex surfaces in Euclidean space as well as in hyperbolic or spherical space. However, I will not develop the corresponding results here, referring the reader to my book [A15].30 Restricting the exposition to regular surfaces, let us note that the metric of such a surface is given by line element which has a certain Gaussian curvature K. In this plane, an existence theorem for a closed convex surface with given line element of positive curvature was found by G. Weyl as far back as 1917 and its proof was completed by H. Lewy in 1938.31 Therefore, this special problem is referred to as the Weyl problem.32 We say that a surface with the given metric realizes this metric. We formulate a general theorem on the realization of a metric given by a line element as a convex surface in space of constant curvature K0 (Euclidean space if K0 = 0, hyperbolic space if K0 < 0, and spherical space if K0 > 0). Let a metric be given on a domain G of the sphere by a line element of Gaussian curvature greater than or equal to K0 . Suppose that every two points of G can be joined by a shortest arc in this metric. (In particular, this condition is satisfied with G the whole sphere.) Then in the space of constant curvature K0 there is a convex surface F realizing the metric, i.e., the domain G with this metric admits an isometric mapping onto F . (A complete proof of this theorem remains unpublished.) Proofs of existence theorems for a convex surface with a prescribed metric are primarily based on approximating the given metric by “polyhedral metrics” ρn , i.e., metrics defined by developments composed of planar triangles. 30

A concise exposition is given in the article [A12]. See the book [E2] with a survey of these results. 32 In Weyl’s statement, the problem is reduced to the existence of a solution to the Darboux equation on the sphere, which is actually an elliptic Monge–Ampère equation. According to the method of continuation in a parameter [Be], proving solvability of this equation requires a priori estimates for a solution in the class C 2 . Relevant estimates were derived by Weyl [W1] and, in final form, by Lewy [Lew]; also see [N]. A. V. Pogorelov proved the regularity of convex surfaces with regular metric [P10, Chapter 2]. Combined with Alexandrov’s Theorem, this result also solves the analytic Weyl problem. More details about the degree of smoothness of a convex surface in its dependence on the smoothness of the metric or curvature can be found in the survey [Bu3, Chapter 2, § 3]. (The most complete results appear in [P10], [S1], [Sh1], [NS], and [SS].) Existence theorems for convex surfaces with prescribed metric in spaces of constant curvature and Riemannian manifolds can be found in [P10, Chapters 5 and 6]. The interesting question of the realization of a given two-dimensional Riemannian metric as a convex surface in a Riemannian manifold with curvature −1 (of arbitrary topological type) is discussed in [L]. – V. Zalgaller 31


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Using the existence theorems for polyhedra with developments (metrics) ρn , we can assert the existence of polyhedra Pn with metrics ρn . The limit of a convergent subsequence of these polyhedra yields a surface with the limit metric which is our metric ρ.33 In the case of a metric given by a line element with sufficiently smooth coefficients, the existence problem of surfaces with this metric may be restated as the existence problem of a solution to a certain partial differential equation. It is just in this form that the problem was considered by Weyl [W1] and Lewy [Lew]. 5.3.4 For flexes of convex surfaces, we can prove a series of theorems similar to those of Section 5.2. First of all, observe that Olovyanishnikov’s Theorem of Section 4.5 about unbounded polyhedra can be generalized to arbitrary unbounded convex surfaces as follows: Let F be an unbounded complete convex surface of total curvature ω(F ) < 2π. Let L be a ray on F , i.e., a half-bounded curve shortest between any pair of its points. Let the surface F be oriented by indicating an oriented contour. Fix an arbitrary convex cone K, requiring only that its curvature (the area of the spherical image) be equal to ω(F ). Take an arbitrary generator G of K. Then there exists an isometry of F with a convex surface F such that (1) K is the limit cone of F ; (2) G is the limit generator of the ray L on F corresponding to L under the isometry of F and F ; (3) under the contraction of F to the cone K, the orientation induced from F to F by the isometry between F and F coincides with the given orientation of K. This theorem was proved by Olovyanishnikov [Ol1] by passing to the limit from polyhedra. It contains the assertion that every unbounded convex surface of curvature less than 2π admits nontrivial isometric mappings. However, the question of whether a continuous flex is possible remains unsolved in the general case. A. V. Pogorelov proved that if F has bounded specific curvature (see Subsection 3.6.2), then F admits continuous flexes and all these flexes are determined by flexes of the cone K with distinguished generator G. This is a complete analog of Theorem 8 of Section 5.2. The result is probably true without requiring the boundedness of the specific curvature.34 Among the theorems about flexes of bounded convex surfaces, we only state the following two. If a convex surface homeomorphic to a disk is such that any two of its points can be joined by a shortest arc, then the surface is isometric to a cap 33

In the simplest case of surfaces in Euclidean space and metrics given by line elements, the method is exposed in minute detail in [A8]; the general case is settled in [A15, Chapter VII]. 34 At present, this is proved. See [P10, Chapter 3, § 9]. – V. Zalgaller

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[A15, Chapter IX, § 4, and Chapter VIII, § 2]. (See the definition of caps in Section 3.5 of Chapter 3.) It seems probable that such a surface can always be flexed to a cap. However, we have no proof even in the case of infinitely smooth surfaces.35 If from a closed convex surface we cut out a geodesic polygon whose interior has nonzero integral curvature, then the remaining part of the surface admits nontrivial isometric mappings. It seems probable that such a surface always admits continuous flexes even when we cut out an arbitrary region of nonzero curvature rather than a polygon. However, this is proved only for thrice continuously differentiable surfaces.36 The proofs of the above theorems are based on the Gluing Theorem, i.e., essentially on the same method of patches that was used in Section 5.2. 5.3.5 We now discuss the question of generalizing our theorems to spaces of dimension greater than three. To be definite, we limit our exposition to three-dimensional polyhedra in four-dimensional Euclidean space. Accordingly, the development must consist of three-dimensional polyhedra Qi with identified (“glued”) faces. Several polyhedra Qi touch at edges of such a development, and in order to have a development of a convex polyhedron, it is necessary that the sum of dihedral angles of these polyhedra Qi around each edge be at most 2π. However, this condition is not sufficient for a development (homeomorphic to the sphere) which satisfies this condition to define a closed convex polyhedron by gluing. We now explain the reason for this effect. Let A be a vertex of a development R; it must be the vertex of a polyhedron P produced from R by gluing. The points on P equidistant from A constitute a closed convex polyhedron QA on a three-dimensional sphere centered at A. This polyhedron determines the polyhedral angle of P at A. The development of QA is constituted by the points of R that are equidistant from A. However, the closed convex polyhedron QA is uniquely determined from the development. Therefore, we come to the following result: arbitrarily small neighborhoods of vertices of a development completely determine the polyhedral angles of the polyhedron that the development produces by gluing. If vertices A and B are joined by an edge, then the polyhedral angles at 35

At present, this is proved. See [Mi4]. Also, in [Kl1] and [Kl2] it was shown by analytical methods that it is possible to flex continuously one of any two given isometric strictly convex C 4,α -smooth surfaces homeomorphic to a disk, transforming it into the other, in the class of isometric locally convex surfaces. But this does not imply that a continuous flex is possible in the class of convex surfaces. About flexes, see also the surveys [E3], [Se4], and [S2], the articles [Fo3], [Fo4], [KF], etc. – V. Zalgaller 36 At present, this is proved [Le] in the general case. See [P10, Chapter 3, § 6]. – V. Zalgaller


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A and B have common faces contiguous to this edge. Accordingly, the spherical polyhedra QA and QB must have equal planar and dihedral angles at their vertices corresponding to the edge AB. At the same time, the dihedral angles of a polyhedron are determined by the development as a whole and they will vary when we change a part of the development in a way which, in a sense, has nothing to do with the edge AB at all. This leads us to the conclusion that even a development with boundary, having exactly two vertices, may fail to define a polyhedron by gluing. Easy examples corroborate this conclusion. A development with two vertices appears if we cut out a neighborhood of an edge from an arbitrary development. Hence, even in the neighborhood of an edge, a three-dimensional development may fail to be glued in four-dimensional space. Certainly, the same is true in spaces of higher dimension. On the other hand, it is easy to prove the following theorem: In order that a development homeomorphic to a three-dimensional sphere and composed of tetrahedra (which is obviously the general case) define a convex polyhedron by gluing, it is necessary and sufficient that it may be glued in a neighborhood of each of its edges. Thus, the question is reduced to finding conditions for gluing a development in a neighborhood of an edge. Finding such conditions seems to be a hard, if not hopeless, problem. 5.3.6 For regular (n − 1)-dimensional surfaces in n-dimensional space, the matter is simpler. Conditions for the realization of a small domain of an abstract (n−1)-dimensional Riemannian manifold as a surface in n-dimensional space have been studied for a long time, until exhaustive results were obtained by Nina Arkad evna Rosenson [Ro].37 Also, it is known that, for n > 3, an (n − 1)-dimensional surface in n-dimensional space is in general rigid even in small parts. Whence we derive the following: In order that a Riemannian metric of positive curvature, given on an (n − 1)-dimensional sphere, be realizable in the large as a surface in n-dimensional space, it is necessary and sufficient that this metric be realizable over an arbitrarily small domain. The conditions for realization over small domains have been found. So the question is settled completely. For a manifold other than the sphere, this result could fail, since its derivation makes use of the simple-connectedness of the sphere.

37

The basic facts may be found in the book [Ei].

6 Congruence Conditions for Polyhedra with Parallel Faces

6.1 Lemmas on Convex Polyhedra 6.1.1 We say that a polygon P1 lies in a polygon P2 if every point of P1 belongs to P2 . If P1 lies in P2 and differs from P2 , then we say that P1 lies inside P2 . If some translate of P1 lies in (inside) P2 , then we say that P1 can be placed in (inside) P2 by translation. Throughout the sequel, a “polygon” means a “bounded convex polygon,” except for Lemma 4a, which deals with unbounded convex polygons. We consider pairs of convex polygons and compare their sides with parallel outward normals. For short, we call such sides parallel, and if for some side l of one polygon there is no parallel side of the other, then we still assume that such a side exists and has zero length: it degenerates into a vertex with support line parallel to l. Since we compare only parallel sides of polygons, we often omit any indication of the parallel property. Finally, we say that sides l1 , . . . , ln of one polygon are longer than sides l1 , . . . , ln of the other polygon if l1 ≥ l1 , . . . , ln ≥ ln and li > li at least once. In this case we also say that the sides l1 , . . . , ln are shorter than the sides l1 , . . . , ln . These conventions apply to all sections of the current chapter. 6.1.2 Lemma 1. If all sides of a polygon P1 except possibly one of them, l0 , are shorter than the parallel sides of a polygon P2 , then P1 can be placed inside P2 . Take vertices A1 and A2 of P1 and P2 such that some support lines of P1 and P2 pass through A1 and A2 with outward normals antiparallel to the normal of l0 .1 Consider the parallel translation of P1 that takes A1 to A2 . Let us show that P1 then lies inside P2 . The vertex A = A1 = A2 and the side l0 divide the boundary of P1 into two polygonal lines AB1 and AC1 (Fig. 114). To these lines correspond two polygonal lines AB2 and AC2 on the boundary of P2 . We shall prove that AB1 cannot leave P2 through AB2 . 1

If the support lines contain some sides, then A1 and A2 may be any pair of corresponding vertices.

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A

P1 B1 l0 C1 P2

B2

C2 Fig. 114 (1)

(1)

(2)

(2)

A

1

Let l1 , l2 , . . . and l1 , l2 , . . . be the successive sides of AB1 and AB2 (1) (2) starting from A. Since l1 ≤ l1 and so forth, at least the first side of AB1 lies in P2 . Suppose that the polygonal line AB1 leaves P2 through a point (1) D on the polygonal line AB2 (Fig. 115(a)). Assume that D belongs to lh (2) (1) (2) and lk . Since li and li (i = 1, 2, . . .) are parallel sides and passing from (1) li to li+1 involves a rotation through a certain angle, it follows that lh may (2) intersect lk only if the rotation angle of the former is less than the rotation angle of the latter, i.e., if h < k.

a

1 2 2

2

3

3

3 4

4 B2 B1

5

D

Fig. 115(a)

5

4 D

5 5

Fig. 115(b)

Now, let us project the polygonal lines AB1 and AB2 on the straight line a perpendicular to l0 . Since the two polygonal lines meet at D, the projections of their parts AD are equal. At the same time, we have h < k. Hence, (1) (1) (1) the projection of the part of AB1 composed of the sides l1 , l2 , . . . , lk is larger than the projection of the corresponding part of AB2 . Since the sides

6.1 Lemmas on Convex Polyhedra

273

of the two polygonal lines are parallel to one another, this is possible only if (1) (2) some of the sides li are longer than li , which contradicts the assumptions of the lemma. Therefore, AB1 cannot intersect AB2 . (1) (2) A special case is shown in Fig. 115(b): all sides li and li coincide up to (1) i = k − 1, while lk leaves P2 . In this case we come to the same contradiction: (1) (2) lk turns out to be longer than lk . Thus, the polygonal line AB1 cannot leave P2 intersecting AB2 . For the same reason, AC1 cannot leave P2 and intersect AC2 . These lines cannot intersect the side B2 C2 (parallel to l0 ) either. Otherwise, for example, the projection of the polygonal line AB1 on the straight line a would be larger than the projection of the polygonal line AB2 , contradicting the fact that the sides of AB1 are shorter than those of AB2 . Thus, the polygonal lines AB1 and AB2 lie in P2 . So the whole boundary of P1 lies in P2 . The boundaries of P1 and P2 cannot coincide, since shorter sides must exist in P1 . Hence P1 lies inside P2 , as claimed. O

Q1 Q2

Fig. 116

Lemma 2. Suppose that two convex polygonal lines Q1 and Q2 lie in an angle with vertex O, have endpoints on the sides of the angle, and are convex towards O (Fig. 116). If some ray issuing from O meets Q1 earlier than Q2 , then Q1 has a side shorter than the corresponding side of Q2 . (Here we deal with parallel sides, and the reader should recall the convention about sides of zero length.) To prove this, let us perform a similarity contraction of Q2 to O. The sides of Q2 decrease. At the moment when Q2 lies in the part of the angle bounded by Q1 and still touches Q1 , consider the sides of Q1 and Q2 in which Q1 and Q2 touch one another. In Q2 , each of these sides is not shorter than in Q1 (Fig. 117(a)). So, before contraction, this side of Q2 was longer than the corresponding side of Q1 , as required. It could happen that, at the moment in question, Q1 and Q2 touch only at a vertex (Fig. 117(b)).

274


O

O

Q1

Q1

Fig. 117(a)

Fig. 117(b)

But then, as is obvious from the picture, this vertex is incident to the sides of Q2 having no parallel sides in Q1 or, which is the same, having parallel sides of zero length in Q1 . As a result, these sides of Q2 are longer than the corresponding parallel sides of Q1 . 6.1.3 We need the following lemma to establish congruence conditions for polyhedra in Sections 6.3 and 6.4. Lemma 3. If two convex polygons cannot be placed in one another by parallel translation, then the difference between the lengths of parallel sides of the polygons changes sign at least four times in passing around either of the polygons.

+

_

+

_

_ +

+ P1

P2

_ _

+ +

_ Fig. 118

Let P1 and P2 be our polygons (Fig. 118). Label by a plus (minus) each side of P1 which is longer (shorter) than the corresponding side of P2 (remembering the convention about sides of zero length), while leaving unlabeled the sides for which the corresponding sides of P2 have the same length. Label the sides of P2 by opposite signs. We must prove that there are at least four sign changes in passing around P1 (or P2 ) .


275

Suppose that there are no sign changes at all. In that case, the sides of, say, P1 are shorter than or equal to the sides of P2 .2 By Lemma 1, P1 can be placed in P2 , contradicting the assumption. Hence, the number of sign changes is nonzero. This number is obviously even, and so it remains to prove that exactly two sign changes are impossible. Assume the converse, i.e., that we have exactly two sign changes. Then the boundary of P1 (P2 ) splits into two polygonal lines P1 (P2 ) and P1 (P2 ): the sides of P1 are shorter than those of P2 and the sides of P1 are longer than those of P2 , i.e., the sides of P1 are labeled by minuses and the sides of P1 , by pluses (Fig. 119). In every convex polygon, the sum of angles between outward normals to adjacent sides is 360◦. Splitting P1 into P1 and P1 , we exclude the angles between normals to the touching sides of P1 and P1 . Therefore, at least one of the polygonal lines P1 and P1 has the sum of angles between outward normals less than 180◦ . Suppose that this holds for P1 , and so for P2 , since P1 and P2 have parallel outward normals. In this case, drawing two support lines to P1 , we find P1 convex towards the vertex of the angle between these support lines (Fig. 119). P1

_

_

_

_

_ +

+

+

P1

Fig. 119

If the sum of angles between outward normals to the sides of P1 were greater than 180◦ , then this sum in P1 would be less than 180◦ . Interchanging the indices of the polygons, we would conclude that the sum of outward normals in P1 is less than 180◦ . Hence, we may assume that it is P1 which has this sum less than 180◦ . What matters here is the fact that the sides of P1 are longer than those of P2 . 2

Recall that we compare only sides with parallel outward normals and sides l1 , . . . , ln are shorter than sides l1 , . . . , ln if l1 ≤ l1 , . . . , ln ≤ ln and li < li at least once.

276


The sides of P1 are shorter than those of P2 . Joining the endpoints of P1 by a straight line segment and doing the same for P2 , we obtain two convex polygons. By Lemma 1, a parallel translate of the first lies in the second. Consequently, we may perform a translation of P1 so that the polygonal line P1 remains inside P2 . Then the polygonal line P1 must protrude from P2 , since by assumption no translate of P1 lies in P2 . Now, there is a point A on P1 such that some support line a to P1 at A is disjoint from P2 .3 Moving this point on P1 in both directions from A and turning the support line accordingly, we obtain two straight lines b and c, which are support lines both for P1 and for P2 (Fig. 120). These lines touch P1 at vertices B1 and C1 belonging to P1 , because P1 lies inside P2 . Since the sides and, consequently, the support lines of P1 and P2 are parallel to one another, b and c touch P2 at vertices B2 and C2 belonging to P2 . P2 P1 c C2 C1

b B2 P1

A

B1

a

O

Fig. 120

Thus we have the following situation: the polygonal lines B1 C1 and B2 C2 are parts of P1 and P2 and their endpoints lie on the straight lines b and c, forming an angle with vertex O. Both polygonal lines are convex towards O, because the sums of angles between the outward normals to P1 and P2 are less than 180◦ , so that the angle of rotation of the support line a from b to c is less than 180◦ . The support line a to B1 C1 at A is disjoint from P2 . Hence, the ray issuing from O and passing through A meets B1 C1 earlier than B2 C2 . 3

Since P1 protrudes from P2 , there is a support line d to P2 intersecting P1 . Draw the support line a (to P1 ) which is parallel to d and lies on the side of d not containing P2 .


277

Applying Lemma 2, we conclude that there is a side of B1 C1 shorter than the corresponding side of B2 C2 . This implies that P1 has some side shorter than the corresponding side of P2 , contradicting the assumption that the sides of P1 are longer than the sides of P2 . Hence, the assumption that we have exactly two sign changes is false, which completes the proof of the lemma. 6.1.4 We shall apply the next lemma in Section 6.4 to studying unbounded polygons. Lemma 4. Assume given two convex polygonal lines with common endpoints and outward normals of their sides pointing to the same half-plane4 (Fig. 121). If the polygonal lines differ, then, in passing from one endpoint of the polygonal lines to the other, the difference between the lengths of the parallel sides changes sign at least twice, except only for the case in which the extreme sides are parallel to one another and the other sides of the polygonal lines are pairwise equal (Fig. 121(c)).

_

+

_

L1 L1

+ + +_ _ _ + + _

+

_ A

B

(a)

L2 l1

A

L2 B

A=A

B

l1 B

(b)

A

l1 B

(c)

Fig. 121(a)–(c)

(As before, we assume the following: if to some side of one of the polygonal lines there is no parallel side of the other, we nevertheless treat such a side as existing but of zero length, the sides of the polygonal lines being regarded as parallel to one another if they have parallel outward normals.) Consider two different polygonal lines L1 and L2 that satisfy the assumptions of the lemma. Drawing the straight line segment AB between their endpoints, we obtain two convex polygons P1 and P2 with common side AB. Since L1 and L2 differ, we have only two possibilities for P1 and P2 : (1) Neither one of them lies in the other (Fig. 121(a)). (2) One of them lies in the other (Figs. 121(b), (c)). 4

The half-plane contains the bounding straight line. The polygonal lines may have sides with antiparallel normals directed along this line.

278


In the first case, it is obvious that no translation puts one of the polygons in the other. Therefore, by Lemma 3 the difference between the side lengths of P1 and P2 changes sign at least four times. The side AB is common to the polygons and is unlabeled. Excluding AB, we eliminate at most one possible sign change in traversing this side. We are thus left with at least three sign changes related to the polygonal lines themselves, and in this case the claim of the lemma is proved. Now, suppose that one of the polygons, say P2 , lies in the other, P1 . Going along L2 from its endpoint A, we find a point A of L2 entering into P1 (Fig. 121(c)). Of course, this may happen at A itself, so that A = A (Fig. 121(b)). On the polygonal line L1 , the point A is either an interior point or beginning of some side l1 . In the first case, the corresponding side l2 of L2 is shorter than l1 . In the second case, L2 has no side parallel to l1 , and so we assume that such a side exists but is of zero length. In any case, l1 > l2 . Going along L2 from the other endpoint B, we similarly find a point B and a pair of parallel sides such that the side of L1 is longer than that of L2 , i.e., l1 > l2 . Thus, from both endpoints of L2 , we first come to the sides of L2 labeled by minuses in accordance with the sign of the differences l2 − l1 and l2 − l1 . The sides l1 and l1 differ, and so l2 and l2 differ too. Otherwise, the points A and B would belong to the same side, and the polygonal line l2 simply could not enter into P1 through these points, while still remaining convex. Now, we distinguish between the two cases: (1) The sides l1 and l1 , and so l2 and l2 , are not parallel to one another. (2) These sides are parallel, i.e., their outward normals are antiparallel. O L1 l1 L2

B

l1 A B

A Fig. 122

In the first case we prolong the sides l1 and l1 until they meet (Fig. 122). Then the segments A B of our polygonal lines L1 and L2 will lie in the angle with vertex O, and L1 will be nearer to O than L2 . From Lemma 2 we then infer that, on the segment A B of L2 , there are sides longer than the parallel sides of L1 . According to our rule, these sides on L2 are labeled by pluses.


279

Thus, it turns out that between the extreme sides l2 and l2 labeled by minuses there is at least one side on L2 labeled by a plus, which gives at least two sign changes. It remains to settle the case in which the sides l1 and l1 , as well as l2 and l2 , are parallel to one another. In this case the outward normals of the sides of our polygonal lines point to the same half-plane by assumption. This implies that l1 and l1 (l2 and l2 ) are extreme sides of the polygonal line L1 (L2 ). The sides l1 and l2 have a common origin A, and since l1 > l2 , it follows that l2 is a proper part of l1 . A similar situation holds for the sides l1 and l2 . Let us add to the sides l2 and l2 the least of the differences l1 − l2 and l1 −l2 . Then L2 is replaced by the new polygonal line L2 at least one of whose extreme sides coincides with an extreme side of L1 . Obviously, the number of sign changes of the differences between the side lengths of L1 and L2 does not increase. The following two possibilities arise: (1) The polygonal lines L1 and L2 coincide. This is the exceptional case stipulated in the lemma. (2) The polygonal lines L1 and L2 differ. Now, excluding the coinciding extreme sides, we obtain two polygonal lines with nonparallel extreme sides. But then we may apply the previous arguments and conclude that there are at least two sign changes for the polygonal lines L1 and L2 .5 Since the number of sign changes for L1 and L2 is not less than that for L1 and L2 , this number, all the more, is not less than two, which concludes the proof of the lemma. _

_ +

+ +

+

_ _

Fig. 123

Let us restate this lemma in another form, the form in which the lemma will be used in our consideration of unbounded polygons. Here we deal with unbounded polygons with overlapping unbounded sides. In this case we agree to consider one unbounded side to be shorter than another unbounded side if 5

For polygons bounded by the polygonal lines L1 and L2 and the straight line segment joining their endpoints, consider the same two cases as above: (1) neither of the polygons lies in the other and (2) one of the polygons lies inside the other.

280


the former is a proper part of the latter (Fig. 123). The sign of the difference between unbounded sides is defined in accordance with this agreement. Lemma 4a. Suppose that the unbounded sides of two unbounded convex polygons overlap so that remote parts of the polygons overlap completely. If the polygons differ, then the differences between their parallel sides change sign at least twice, except for the case in which the unbounded sides of the polygons are parallel to one another and the other sides are pairwise equal. In this exceptional case the polygons can be superposed by parallel translation along the unbounded sides. To reduce this lemma to the preceding, it suffices to take some points A and B on the unbounded sides and declare that they are the endpoints of the polygonal lines L1 and L2 . (The converse is obvious: Lemma 4 follows from Lemma 4a: to see this, prolong the extreme sides of L1 and L2 to infinity. Also observe that if the unbounded sides of one of the polygons are not parallel to each other but are parallel to the unbounded sides of the other polygon, then the polygons can always be superposed by parallel translation. However, if the unbounded sides are parallel to each other, then, for the polygons to be superposed by translation, it is necessary that the distance between these sides be the same in both polygons. Accordingly, we may formulate Lemma 4a by requiring only that the sides of the polygons be superposable by translation, rather than requiring that they overlap.) 6.1.5 Remark. Lemma 3, together with its proof, generalizes readily to arbitrary closed convex curves. Let P be a closed convex curve and let l(ϕ) be the length of the arc of P consisting of the points through which we can construct support lines with outward normals pointing to the arc ϕ of the unit circle E, these normals being drawn from the center of E; l(ϕ) is a function of an arc on E. The theorem generalizing Lemma 3 reads: If two closed convex curves P1 and P2 cannot be placed inside one another by translation, then the unit circle may be partitioned into at least four arcs ϕk such that l1 (ϕk ) − l2 (ϕk ) changes sign in going from an arc to the neighboring one. In the case of twice differentiable curves, l(ϕ) is the integral of the curvature radius with respect to dϕ. Therefore, the above theorem readily implies the well-known “four vertex” theorem for ovals: on every oval (i.e., on a closed twice continuously differentiable convex curve) there are at least two maxima and two minima of curvature. To prove this, for P1 take the oval in question and for P2 , a circle of the same length.6 ) 6

See Blaschke’s book [Bl2, § 12]; it also contains references concerning this theorem, which attracted many geometricians in former times.

6.2 On Linear Combination of Polyhedra

281

Similarly to Lemma 3, we can prove the “2n vertex” theorem for ovals: if an oval P intersects a circle at 2n points, then it has 2n “vertices,” i.e., 2n extremum points of curvature. In exactly the same way, using the function l(ϕ), we can establish theorems on convex curves which are quite similar to the other lemmas proved above. For unbounded curves (in the analog of Lemma 4a), we need not require the coincidence of the unbounded arcs. We leave it to the reader the task of finding a weaker necessary condition.

6.2 On Linear Combination of Polyhedra 6.2.1 Choose a coordinate origin O in space and specify the position of an ar−−→ bitrary point X by the vector x = OX drawn from O to X, the position vector of X. Each set M in space is then determined by the set of ends of all position −−→ vectors x = OX of elements X of M . Adding the same vector a to all vectors x, we perform the translation of M by this vector. Multiplying all vectors x by the same positive number λ, we multiply the lengths of all vectors by λ; as the result, M undergoes the similarity transformation with coefficient λ and center of similarity O. We denote the new set by λM . Now, let M1 and M2 be two sets and let λ1 and λ2 be arbitrary positive numbers. Given the position vectors x1 and x2 of some points of M1 and M2 , consider the vector x = λ1 x1 + λ2 x2 . When the ends of the vectors x1 and x2 run independently through M1 and M2 , the end of the vector x covers some set M . The set M is called the linear combination of M1 and M2 with coefficients λ1 and λ2 and is denoted by M = λ1 M1 + λ2 M2 . An especially simple case is that of λ1 + λ2 = 1. Then, as is well known from analytic geometry, the end of the vector x = λ1 x1 + λ2 x2 divides the straight line segment between the ends of x1 and x2 in the ratio λ2 :(1−λ2 ). Therefore, the set M is formed by the points dividing the straight line segments with endpoints in M1 and M2 in the given ratio λ:(1−λ). In particular, if λ = 1/2, then we obtain the “half-sum” of M1 and M2 ; it consists of the midpoints of the straight line segments with endpoints in M1 and M2 (see Fig. 125 below). In the next sections of the current chapter, we need only “half-sums” of “solid” convex polyhedra. We shall prove that such a half-sum is also a convex polyhedron. General linear combinations appear in Section 8.3 of Chapter 8. Linear combinations of arbitrary convex bodies were introduced and studied by Brunn and Minkowski over fifty years ago, and gave rise to a whole theory.

282


The basics of this theory constitute the content of the present section and Section 8.3 of Chapter 8. Just like the linear combination of two sets, we can define the linear combination of an arbitrary number of sets Mi with coefficients λi : M = λ1 M1 + . . . + λn Mn . Here M is the set formed by the ends of the vectors x = λ1 x1 + . . . + λn xn when the ends of the vectors xi run independently through their own sets Mi . By the associativity of the vector sum, the set M can be obtained by successively taking the linear combination, first, λ1 M1 + λ2 M2 = M , next M + λ3 M3 , etc. Therefore, we may confine our exposition to examining only linear combinations of two sets.7 6.2.2 Lemma 1. Changes of origin and parallel translations of the sets M1 and M2 merely result in a parallel translation of M . −−→ When we move the origin from O to O , we add the same vector a = O O to all vectors x1 and x2 . Therefore, to the vector x = λ1 x1 + λ2 x2 we add (λ1 + λ2 )a. But now the vector x + (λ1 + λ2 )a starts from the new origin O , while the vector x + (λ1 + λ2 )a − a issues from the old origin O to the same point. Hence, the whole set M = λ1 M1 + λ2 M2 , constituted by the ends of these vectors, moves by the vector (λ1 + λ2 − 1)a. In particular, if λ1 + λ2 = 1, then M is not changed at all. (This is also obvious from the fact that, as we have seen above, M now consists of the points dividing the segments with endpoints in M1 and M2 in the given ratio.) When we shift M1 by a vector a1 , we add the vector a1 to all vectors x1 . Then to the vectors x = λ1 x1 + λ2 x2 we add the vector λ1 a1 . Therefore, the set M = λ1 M1 + λ2 M2 is shifted by the vector λ1 x1 . When M1 and M2 are shifted by the vectors a1 and a2 , the set M is shifted by the vector a1 + a2 . (Clearly, the same results remain valid for a linear combination of arbitrarily many sets Mi .) When dealing with linear combinations, sets that are translates of one another may be regarded as not being essentially different. Therefore, the “summands” M1 and M2 , as well as their linear combination M = λ1 M1 + λ2 M2 , may be viewed up to arbitrary translations. Lemma 1 exactly shows that 7

It is possible to consider linear combinations with negative λi . The set −M consists of the ends of the vectors −x opposite to the position vectors x of the points of M . Therefore, −M is symmetric to M with respect to the origin. The multiplication by a negative λi consists in the similarity transformation with coefficient |λi | followed by the reflection with respect to O. A linear combination with negative coefficients λi is obtained from the linear combination with coefficients |λi | by reflection with respect to the origin.


283

translations of the origin and of the “summands” do not influence the set M , provided that its translation is ignored. This remark leads to the following visually clear understanding of the linear combination M = λ1 M1 + λ2 M2 . Subjecting M1 and M2 to similarity transformations, we obtain the sets λ1 M1 and λ2 M2 . Taking an arbitrary point A in λ2 M2 , we shift λ2 M2 so that A is taken to various points of λ1 M1 . Then the set M covered by λ2 M2 is exactly λ1 M1 + λ2 M2 (Fig. 124). Indeed, when we shift A to a point X0 of λ1 M1 , the shifted set λ2 M2 is covered by the ends of the vectors λ1 x01 + λ2 x2 . Letting A = X 0 run over all of λ1 M1 , we obtain the whole set M = λ1 M1 + λ2 M2 .

λ

1

x1 0 +λ

2

x2

λ2

A=X

0

x2

λ2

x2

λ2M2

λ1 x 0 1

λ1M1

A=O λ2 M2

Fig. 124

The order of summands in the definition of linear combination is immaterial (by the associativity of the vector sum). Therefore, the sets M1 and M2 in the above construction may be interchanged. (Their linear combination can only undergo translation.) Using this construction, we come to the following simple conclusion. Lemma 2. If M1 and M2 are nonparallel straight line segments, then the set λ1 M1 + λ2 M2 is a parallelogram whose sides are translates of M1 and M2 . If M1 and M2 are parallel straight line segments, then λ1 M1 + λ2 M2 is a straight line segment parallel to them and its length is the linear combination of the lengths of M1 and M2 , i.e., l = λ1 l1 + λ2 l2 . To prove this, it suffices to transform M1 and M2 by similarities and consider the set swept by the straight line segment λ1 M1 as its extremity sweeps through λ2 M2 . 6.2.3 Lemma 3. Any linear combination of convex sets is a convex set. (See Fig. 125.)

284


X1

X

Y1 M1

Y 1/2(M1+M2)

X2

Y2 M2

Fig. 125

Indeed, let X and Y be points of the set M = λ1 M1 + λ2 M2 . These points are linear combinations of some points X1 , X2 and Y1 , Y2 of the sets M1 and −−→ −−→ −−→ M2 (i.e., OX = λ1 OX1 + λ2 OX2 and similarly for Y ). If M1 and M2 are convex sets, then they contain the segments X1 Y1 and X2 Y2 . Then M clearly contains the combination of these segments, which in turn contains the points X and Y . By Lemma 2, each combination of segments is a parallelogram or a segment and so, together with X and Y , it contains the segment between them. Therefore, when M1 and M2 are convex, the set M = λ1 M1 + λ2 M2 contains the segment between any pair of points X and Y , i.e., M is a convex set. (Successively constructing the linear combination of an arbitrary number of convex sets, we similarly deduce convexity of the linear combination λ1 M1 + λ2 M2 + . . . + λn Mn .) 6.2.4 Define a “face” (in inverted commas) of a set M as the intersection of M with an arbitrary support plane of M . For example, a “face” of a convex polyhedron is either a face, an edge, or a vertex of the polyhedron. Lemma 4. Given sets M1 and M2 , let Q1 and Q2 be support planes of M1 and M2 with parallel outward normals and let G1 and G2 be the corresponding “faces” of M1 and M2 . Then Q = λ1 Q1 + λ2 Q2 is a support plane of M = λ1 M1 + λ2 M2 with the same outward normal, and the corresponding “face” G of M is the linear combination of G1 and G2 with the same coefficients: G = λ1 G1 + λ2 G2 . Conversely, each support plane and each “face” of M can be obtained in this way. (Strictly speaking, we should add that the sets considered here are assumed closed.) Under parallel translations of M1 and M2 , the set M only undergoes a translation. We can thus choose parallel translations of M1 and M2 after which the planes Q1 and Q2 coincide. Furthermore, we can shift the origin and put it on the plane Q = Q1 = Q2 . The first claim of the lemma now becomes obvious. The set M = λ1 M1 + λ2 M2 will lie on the same side of Q as the sets M1 and M2 themselves. The intersection of M and Q, i.e., the “face”


285

G, is the linear combination of the “faces” G1 and G2 . For example, in Fig.127 the lower edge of the middle polyhedron is the half-sum of the lower edges of the “summands.” (If we shift M1 and M2 back to their former positions, then the set M , the plane Q, and the “face” G only undergo the corresponding translation.) To prove the converse assertion, take some support plane Q of M and support planes Q1 and Q2 of M1 and M2 with parallel outward normals.8 Perform parallel translations of M1 and M2 so that the planes Q, Q1 , and Q2 coincide. Then just as before, we see that the plane Q and the corresponding “face” G are combinations of the planes Q1 and Q2 and the corresponding “faces” G1 and G2 . The same conclusion holds for planar figures after replacing support planes by support lines and “faces”by “sides,” i.e., by the intersection of a figure with a support line. (For example, in Fig. 126 the side a is the linear combination of the vertex A and the side a.) λ1 T1+λ2 T2

a

λ2 T2

λ 1 T1

a

A Fig. 126

6.2.5 Lemma 5. The linear combination of polygons lying in parallel planes is a polygon in a parallel plane. (In particular cases, one of the polygons may degenerate into a segment or a single point.) We may assume that the polygons in question lie in the same plane, because the parallel translation of one of them could only result in translating their linear combination. Assume given two triangles T1 and T2 . The figures λ1 T1 and λ2 T2 are triangles similar to T1 and T2 . Their linear combination λ1 T1 + λ2 T2 can be constructed by moving λ1 T1 so that one of its vertices runs over the whole triangle λ2 T2 . It is immediate from Fig. 126 that we thus obtain a hexagon. In particular cases, it may degenerate into a pentagon, quadrangle, or triangle (in the last case T1 and T2 are similar; moreover, they are translates of one another). For the proof we can use the last remark at the end of the previous subsection. 8

M1 and M2 have such support planes. If, for instance, M1 did not possess such a support plane, then M1 would contain points that are arbitrarily far from the plane Q in the direction of its outward normal. But then M would obviously contain such arbitrarily remote points, contradicting the fact that M has the support plane Q.

286


The “sides” of λ1 T1 + λ2 T2 are combinations of sides and vertices of T1 and T2 . Since T1 and T2 have finitely many sides and vertices, the set λ1 T1 + λ2 T2 has finitely many “sides.” Therefore, it is a polygon. Since a genuine side may occur only as the combination of a side with a vertex or with a side in a parallel support line, we have at most six different combinations representing the sides of λ1 T1 + λ2 T2 . Now, assume given two polygons M1 and M2 . Split M1 and M2 into triangles T1i and T2i . Clearly, the combination λ1 M1 + λ2 M2 consists of combinations of the various pairs of triangles: λ1 T1i + λ2 T2i . (Of course, these combinations overlap). Hence, λ1 M1 + λ2 M2 is constituted by finitely many polygons and so λ1 M1 + λ2 M2 is a polygon, as claimed. Lemma 6. A linear combination of solid polyhedra is a solid polyhedron. The proof is the same as in Lemma 5. First, we check that a combination of tetrahedra is a polyhedron (Fig. 127). Afterwards, splitting given polyhedra into tetrahedra, we see that the combination of our polyhedra is the union of the combinations of tetrahedra and is therefore a polyhedron. q1

T1

1/2(q1 +q2)

1/2(T1 +T2)

q2

T2

Fig. 127

To visualize the combination of two tetrahedra is rather difficult. We may therefore appeal to Lemma 4 as we did in the proof of Lemma 5. If T1 and T2 are two tetrahedra, then λ1 T1 + λ2 T2 is a convex set by Lemma 3. By Lemma 4, the “faces” of λ1 T1 + λ2 T2 are the combinations of “faces” of T1 and T2 , i.e., of genuine faces, edges, and vertices of T1 and T2 . Since the number of these elements is finite, the number of their combinations is finite as well. Therefore, λ1 T1 + λ2 T2 has finitely many “faces,” and is therefore a polyhedron. (A genuine face of λ1 T1 + λ2 T2 may occur either as the combination of a face of T1 (or T2 ) with a face, edge, or vertex of T2 (or T1 ) or as the combination of two nonparallel edges lying in parallel support planes. In the latter case, Lemma 2 implies that such a face is a parallelogram. When the tetrahedra T1 and T2 are in “general position,” the faces of one tetrahedron cannot be parallel to any faces or edges of the other; edges lying in parallel support planes are not parallel to one another either. Then the faces of λ1 T1 + λ2 T2

6.3 Congruence Conditions for Closed Polyhedra

287

are combinations of faces with vertices and edges with edges (see Fig. 127). To visualize the combination λ1 T1 + λ2 T2 in this general case and to study all of its various degenerations is a good exercise in visual geometry.) 6.2.6 We can now state the result which is the final object of the current section. Theorem. Each linear combination of convex solid polyhedra is a convex solid polyhedron: P = λ1 P1 + λ2 P2 . If Q is a “face” of P (i.e., a face, edge, or vertex) then Q = λ1 Q1 + λ2 Q2 , where Q1 and Q2 are “faces” of P1 and P2 lying in support planes with parallel outward normals. The set Q is a genuine face whenever Q1 and Q2 are either (1) two faces, or (2) a face and an edge, or (3) a face and a vertex, or (4) a pair of nonparallel edges. The set Q is an edge whenever Q1 and Q2 are parallel edges or an edge and a vertex. The proof is immediate from the preceding lemmas. By Lemma 6, P is a polyhedron. By Lemma 3, P is convex. By Lemma 4, each “face” Q of P is the combination of “faces” Q1 and Q2 of P1 and P2 lying in support planes with parallel outward normals, and if Q1 and Q2 are parallel edges or an edge and a vertex, then their combination yields the parallel edge λ1 Q1 + λ2 Q2 . If Q1 and Q2 are vertices, then they produce the vertex λ1 Q1 + λ2 Q2 . In all remaining cases λ1 Q1 + λ2 Q2 is a genuine face (as we see from Lemmas 2 and 5). These cases are exhausted by the four possibilities in the list of the theorem. (In Fig. 127, the triangular faces of the middle polyhedron are combinations of faces and vertices of the tetrahedra T1 and T2 , while the quadrangular faces are combinations of edges.) Successively constructing the linear combination of several polyhedra, we derive an analogous result for any number of polyhedra.

6.3 Congruence Conditions for Closed Polyhedra 6.3.1 We shall consider pairs of closed convex polyhedra or, which is equivalent, pairs of bounded convex solid polyhedra. (In this subsection, the term “convex polyhedron” always means “bounded convex solid polyhedron.”) Support planes of the polyhedra under study are called parallel to one another if and only if the outward normals to these planes are parallel (i.e., have the same direction). Let P1 and P2 be two convex polyhedra. Let Q1 be a face of P1 and Q2 , a “face,” i.e. a face, edge, or vertex of P2 , lying in a support plane parallel to Q1 ; we refer to such a ”face” Q2 as a face of P2 parallel to Q1 . We define faces of P1 parallel to faces of P1 similarly. Thus, two “parallel faces” lie in

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support planes with parallel outward normals and at least one of them is a genuine face, whereas the other may be an edge or a vertex. Theorem 1. If for all pairs of parallel faces of two convex polyhedra neither face can be placed inside the other by parallel translation, then the polyhedra are translates of one another. (In other words, it is impossible that parallel faces cannot be superposed by parallel translation if each member of any pair of parallel faces cannot be placed inside the other.) Since a vertex can always be placed inside a face, it follows that under the assumptions of the theorem both parallel faces in each pair must be either genuine faces or one of them must be a face and the other an edge. Assume that polyhedra P1 and P2 satisfy the assumptions of the theorem. Consider their half-sum 1 P = (P1 + P2 ). 2 By what was proved in the preceding subsection, P is a convex polyhedron whose faces are of the three following types: (1) the half-sum of parallel genuine faces of P1 and P2 ; (2) the half-sum of parallel faces, one of which is a genuine face and the other an edge; (3) the half-sum of nonparallel edges. (As observed above, the half-sum of a face and a vertex disappears by the assumptions of the theorem.) Each edge of P is the half-sum of parallel edges lying on P1 and P2 in parallel support planes or the half-sum of an edge and a vertex. In the latter case, we shall also treat the edge in question as the sum of two edges, one of which is of zero length. (This convention about edges of zero length corresponds to a similar convention about the sides of a polygon made in Section 6.1.) Thus, to each edge of P there correspond an edge on P1 and an edge on P2 . We now label by a plus (or minus) each edge of P depending on whether the corresponding edge on P1 is longer (or shorter) than on P2 . We leave an edge unlabeled if the corresponding edges on P1 and P2 are equal. Let us prove that there must be at least four sign changes as we pass around each face Q of P , provided that there are labeled edges on Q. We successively examine each of the three types of edges listed above. (1) The face Q is the half-sum of two genuine faces Q1 and Q2 . By the assumptions of the theorem, Q1 and Q2 cannot be placed one inside the other by translation. Therefore, Lemma 3 of Section 6.1 implies that there are at least four sign changes in passing around any one of them, and hence around P . The case in which the sides of Q1 and Q2 are equal is excluded, i.e., Q1 and Q2 are parallel translates of one another.

6.3 Congruence Conditions for Closed Polyhedra

289

(2) The face Q is the half-sum of a genuine face Q1 and an edge Q2 (or vice versa). Since by assumption Q2 cannot be placed in Q1 , the sides of Q1 parallel to Q2 are shorter than Q2 . At the same time, on Q1 there must be sides nonparallel to Q2 ; to these sides correspond “sides” of zero length on Q2 . This makes it obvious that in going around Q exactly four sign changes will occur. (3) The face Q is the half-sum of the nonparallel edges Q1 and Q2 . Then Q is a parallelogram whose sides are 12 Q1 and 12 Q2 . To the side 12 Q1 correspond the edge Q1 and an endpoint of Q2 , i.e., a “side” of zero length. A similar situation holds for 12 Q2 . Consequently, the sides of the parallelogram Q are labeled by pluses and minuses alternatively, which yields four sign changes. Q

Q

A

Q

Q

Fig. 128

Thus, in going around each face of P with labeled edges, we have at least four sign changes. Take a point in the interior of each face and join it by lines to all the points chosen in adjacent faces. We then obtain a net P , which is dual to the polyhedron P (Fig. 128). To the face Q on P corresponds the vertex Q on P . To the edge on P between the faces Q and Q corresponds the edge Q Q on P . To the vertex A on P corresponds the face on P bounded by the edges of P corresponding to the edges of P incident to A. To the edges of P belonging to a single face correspond the edges of P issuing from a single vertex. Assigning the same signs to these edges of P , we find at least four sign changes around each vertex of P with at least one labeled edge. Cauchy’s Lemma now implies that there are no labeled edges at all.9 So, there are no labeled edges on the polyhedron P either. This implies the following: If a face Q of P is the half-sum of two genuine faces, then these faces are parallel translates of one another. No face of P is the half-sum of a face and an edge or the half-sum of two edges; otherwise, as shown above, around each such face there would be exactly four sign changes. 9

The conditions of Cauchy’s Lemma (Section 2.1) are satisfied. Indeed, the net P has no digonal regions, since P has no vertices with exactly two incident edges.

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Therefore, P1 and P2 have pairwise congruent faces which are translates of one another. This implies that the polyhedra themselves are translates of one another. 6.3.2 From Theorem 1 we derive the following corollary: Theorem 2. Let f (Q) be a monotone function of a polygon Q, i.e., we have f (Q1 ) > f (Q2 ) whenever Q2 can be placed inside Q1 by translation. If two convex polyhedra satisfy f (Q1 ) = f (Q2 ) for each pair of parallel faces Q1 and Q2 , then the polyhedra are translates of one another. Or in more general form: Let two convex polyhedra P1 and P2 have pairwise parallel faces Q11 , Q12 ; . . . ; Qn1 , Qn2 . If f1 , . . . , fn are monotone functions of a polygon and fi (Qi1 ) = fi (Qi2 ) (i = 1, 2, . . . , n), then P1 and P2 are parallel translates of one another. Indeed, by the monotonicity of the functions fi , the faces Qi1 and Qi2 cannot be placed inside one another. Therefore, the theorem immediately follows from Theorem 1. Examples of monotone functions are area, perimeter, moment of inertia with respect to the barycenter, etc. In particular, if f (Q) is area, then we arrive at Minkowski’s Uniqueness Theorem: Two convex polyhedra with pairwise parallel faces of equal area are parallel translates of one another. In Subsection 2.4.5, we have shown by a simple example that Theorem 1 and, in consequence, Theorem 2, do not generalize to spaces of dimension greater than three.10 The example was a four-dimensional cube with edge 2 and a four-dimensional rectangular parallelepiped with edges 1, 1, 3, 3. Nevertheless, Minkowski proved his theorem in spaces of arbitrary dimension. Minkowski’s proof relies on entirely different arguments and is described in Section 7.2 of Chapter 7. Observe also that, due to the “extra” 8 in the estimate for the number of sign changes in Cauchy’s Lemma, in Theorem 1 it suffices to require that its conditions be fulfilled for all but one pair of parallel faces. Now, Theorem 1 can be restated in the following slightly strengthened form: 10

For a theorem about the congruence of convex n-dimensional polyhedra with parallel two-dimensional faces that cannot be placed inside one another, see [Me4]. – V. Zalgaller

6.4 Congruence Conditions for Unbounded Polyhedra

291

Two closed convex polyhedra are either translates of one another or they have a pair of parallel faces that cannot be placed inside one another. One of the faces must be a genuine face, while the other may be a genuine face, an edge, or a vertex.

6.4 Congruence Conditions for Unbounded Polyhedra 6.4.1 The congruence condition for closed polyhedra which was derived in the preceding subsection becomes meaningless for unbounded polyhedra. When applied to unbounded faces, this condition cannot hold in general for the simple reason that of two unbounded convex polygons one can always be placed inside the other. Nevertheless, applying this condition to bounded faces only, we may impose a certain condition on unbounded faces so that the two conditions together ensure the congruence of unbounded polyhedra. This additional condition is as follows: Unbounded polyhedra must have unbounded parts as parallel translates11 , i.e., a (sufficiently large) bounded part may be cut out from each of them so that the remaining unbounded parts coincide after translation. The condition is obviously equivalent to the requirement that the planes of all unbounded faces of polyhedra may be superposed by a common translation. For bounded faces, we keep the same conditions as in the previous subsection. Two faces are considered parallel to one another if and only if their outward normals have the same direction. If one of the two given polyhedra has no face parallel to some face of the other, then we treat the former face as existent but degenerating into an edge or a vertex lying in a support plane with the same outward normal. It is exactly in this sense that we understand the term “parallel faces” in the sequel. 6.4.2 We now state a theorem on the congruence of unbounded polyhedra. Theorem 1. If two unbounded convex polyhedra have unbounded parts as parallel translates and all pairs of bounded parallel faces are such that no face of the pair can be placed inside the other by translation, then the polyhedra are translates of one another. The term “can be placed inside” is understood in the same sense as in Sections 6.1 and 6.3, i.e., a parallel translate of one of the polygons in question lies in the other but differs from it. 11

As was mentioned in Section 1.1 of Chapter 1, the consideration of unbounded polyhedra amounts to the consideration of bounded polyhedra whose extreme faces can be infinitely extended without new intersections. Therefore, this condition as well as Theorem 1 below can easily be restated for bounded polyhedra.

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Since a point can always be placed inside a face, under the conditions of the theorem, either both parallel faces in a pair are genuine faces or one of the faces is a genuine face and the other is an edge. Let two polyhedra P1 and P2 satisfy the conditions of the theorem. By virtue of the first condition, the unbounded parts of the polyhedra may be assumed to coincide. So the unbounded edges and faces of the polyhedra overlap. Considering P1 and P2 as solid polyhedra, construct their half-sum 1 (P1 + P2 ). 2 Since the unbounded parts of P1 and P2 coincide, the half-sum P has the same unbounded part. No new unbounded faces or edges appear, and the shape of a face can change only in its finite part. As far as the bounded part of P is concerned, the same arguments apply as those at the end of the previous subsection. Therefore, we may readily formulate the result they imply. If we assign to each edge of P the sign of the difference l1 − l2 between the lengths of the corresponding edges l1 and l2 of P1 and P2 , there will be at least four sign changes around each bounded face of P , provided that at least one edge of the face is labeled (unlabeled edges are those for which l1 − l2 = 0). P =

6.4.3 We now look at the unbounded faces. By the assumption on the disposition of the polyhedra P1 and P2 , the unbounded edges of P1 and P2 overlap pairwise. We say that an unbounded edge l1 of P1 is longer than the corresponding edge l2 of P1 if l2 is a proper part of l1 . We define the reverse inequality in a similar way, and edges are considered equal whenever they coincide. In accordance with this agreement, we define the sign of the difference l1 − l2 between unbounded edges and ascribe this sign to the corresponding edge of the “mean” polyhedron P . Only those edges remain unlabeled to which coinciding unbounded edges correspond on P1 and P2 . The above agreement on the sign of the difference between unbounded edges coincides with the agreement on the sign of the difference between unbounded sides of polygons in Lemma 4a of Section 6.1. Further, since the unbounded edges of overlapping unbounded faces of P1 and P2 overlap themselves, the conditions of this lemma are satisfied for each unbounded face. Applying Lemma 4a, we come to the following conclusion about the possible arrangement of signs around an arbitrary face Q of the mean polyhedron Q. (1) No edge of the face Q is labeled. Then the corresponding faces Q1 and Q2 on P1 and P2 coincide. (2) There are at least two sign changes around the boundary of Q. (3) The unbounded edges of Q1 and Q2 are parallel to each other and the bounded edges are pairwise equal. This corresponds to the exceptional case of Lemma 4a, Section 6.1.

6.4 Congruence Conditions for Unbounded Polyhedra

293

6.4.4 Let us examine the exceptional case more closely. Suppose that it holds for the pair of faces Q1 and Q2 of P1 and P2 . The faces Q1 and Q2 are parallel translates of one another since so are their edges. Therefore, for these faces we have the following two possibilities: (3a) The faces Q1 and Q2 coincide, i.e., the first of the above three cases holds. (3b) One of the faces, say Q1 , is shifted inside the other. Let us show that if Q1 and Q2 have no common edges with bounded faces of P1 and P2 , then Q1 and Q2 coincide. Indeed, the common edges of unbounded faces lie in the intersections of the planes of these faces. By the assumption on the disposition of P1 and P2 , the unbounded faces of the polyhedra overlap. Hence, the intersections of the planes of these faces coincide. It follows that if the faces Q1 and Q2 are bounded only by the intersections with the planes of other unbounded faces, then they necessarily coincide. Thus, having excluded the case in which Q1 and Q2 coincide, we arrive at the following situation: one of the faces, say Q1 , is shifted inside the other and at least one of the faces has common edges with bounded faces of its own polyhedron. The corresponding face Q of the mean polyhedron P is the half-sum of Q1 and Q2 (the theorem of Subsection 6.2.6) and since Q1 and Q2 are parallel translates of one another, Q is also a translate of Q1 and Q2 . Since Q1 is shifted inside Q2 , its unbounded edges are shorter by assumption than the edges of Q2 . Therefore, both unbounded edges of Q are labeled by minuses. At the same time, no bounded edge of Q is labeled, since the bounded edges of Q1 and Q2 are equal. By assumption, at least one of the faces Q1 and Q2 , say Q1 , has an edge l1 belonging to a bounded face R1 . Then the corresponding edge l of Q belongs to some bounded face as well; namely, l belongs to the face R which is the “half-sum” of the face R1 and the parallel face R2 of P2 . The edge l is unlabeled and we assign a plus to it. Since the unbounded edges of Q are labeled by minuses, there are two changes of sign around Q. Clearly, the number of sign changes on bounded faces does not decrease.12 Thus, in the exceptional third case (when it differs from the first) we can always introduce extra signs so as to have two sign changes around the unbounded face in question. We make such sign arrangements on all unbounded faces wherever necessary. 6.4.5 Summarizing all our conclusions about sign arrangements on bounded and unbounded faces, we come to the following result: If a bounded face of P has labeled edges, then there are at least four sign changes around this face. If an unbounded face has labeled edges, then there are at least two sign changes around it. 12

The stipulation of boundedness imposed on the face R adjacent to Q along the edge l is essential. Otherwise, it might happen that, applying the same arguments to R, we would have to assign a minus to l.

294


Suppose that some edges on P are labeled. Draw a closed polygonal line on the unbounded faces of P which has vertices inside unbounded edges and separates the unbounded part of the polyhedron. We thus obtain a bounded polyhedron P . Take a second copy P of P with the same arrangement of signs on its edges and identify the corresponding sides and vertices of the polygonal lines bounding P and P .13 After that, let us erase these polygonal lines together with their vertices. We obtain the abstract polyhedron P ∪ P , which is obviously homeomorphic to the sphere and has no vertices incident to only two edges. Some edges on this polyhedron are labeled and if some face has labeled edges, then the number of sign changes around it is at least four. Indeed, for faces corresponding to the bounded faces of P , this is clear from the above. If a face corresponds to an unbounded face of P , then its boundary consists of two equal parts on each of which there are at least two sign changes. In total, this yields at least four sign changes. However, passing to the dual polyhedron in the same way as in Section 6.3, we see that such an arrangement of signs contradicts Cauchy’s Lemma. Therefore, the assumption that some edges of the mean polyhedron P are labeled is false, and so all edges of P must be unlabeled. As in Section 6.3, this implies that the faces of our original polyhedra P1 and P2 are not only parallel to each other but also are congruent. Hence, the polyhedra are parallel translates of one another, which completes the proof of Theorem 1. 6.4.6 The above theorem, such as Theorem 2 of Section 6.3, has the following immediate corollary: Theorem 2. Let f1 (Q), f2 (Q), . . . be functions of a bounded convex polygon such that fi (Q ) > fi (Q ) whenever Q can be placed inside Q . If two unbounded convex polyhedra have unbounded parts as parallel translates and if fi (Qi ) = fi (Qi ) for all pairs Qi , Qi of their parallel bounded faces, then the polyhedra are parallel translates of one another. In particular, taking area for all the functions fi , we obtain the following: If two unbounded convex polyhedra have unbounded parts as parallel translates and each pair of bounded parallel faces consists of faces of the same area, then the polyhedra are parallel translates of one another. But of course we may require the equality of areas of some faces, the equality of perimeters of others, etc.

13

This identification is made only abstractly. However, it is easy to see that P can be cut away from P by some plane, in which case P is obtained from P simply by reflection in this plane.

6.5 Another Proof and Generalization. Polyhedra with Boundary

295

6.5 Another Proof and Generalization of the Theorem on Unbounded Polyhedra. Polyhedra with Boundary 6.5.1 A. V. Pogorelov in [P1] proposed another proof of Theorem 1 of the preceding section. His proof has important advantages: it extends to spaces of arbitrary dimension and allows us to obtain a more general result. Theorem 1. Suppose that the planes of unbounded faces of two unbounded polyhedra coincide and for each pair of their parallel bounded faces the following alternative holds: either the planes of the faces coincide or none of the faces can be placed inside the other by parallel translation. Then the polyhedra coincide. (Speaking of a pair of parallel faces in the case of n-dimensional polyhedra, we mean that at least one of the faces is (n − 1)-dimensional and the other may be of lesser dimension, but must lie in a parallel (n − 1)-dimensional support plane. In the n-dimensional case, the term “can be placed inside” is understood as before.) As compared to Theorem 1 of the previous subsection, the generalization consists not only in considering a space of arbitrary dimension, but also in the alternative for bounded faces. In this connection, the same theorem may be rephrased for polyhedra with boundary. Theorem 1a. Let two bounded convex polyhedra, each with boundary a single polygonal line, have spherical images belonging to one hemisphere. Suppose that the planes of their extreme faces coincide and for the pairs of parallel interior faces, i.e., those not approaching the boundary, the following alternative holds: either their planes coincide or none of the faces can be placed inside the other by translation. Then the planes of all faces coincide and, consequently, the interior faces coincide completely while the extreme faces can be prolonged so that they coincide. Clearly, Theorem 1 on unbounded polyhedra is contained in the above assertion: it suffices to cut out some remote parts from the unbounded polyhedra so as to obtain polyhedra with boundary. Let us check that, conversely, the above assertion follows from Theorem 1. To this end, prolong the extreme faces of the polyhedra so as to obtain unbounded polyhedra. This is possible because the spherical images are contained in a hemisphere. It can happen that unbounded prolongation is possible not for all extreme faces, but this is immaterial. Obviously, the conditions of the preceding theorem are satisfied for the resulting unbounded polyhedra.

296


Observe that, as is easily seen from simple examples, the above result does not extend to bounded polyhedra with boundary whose spherical images do not lie in a hemisphere. So, the essence of the result consists in Theorem 1 about unbounded polyhedra. 6.5.2 Let us prove Theorem 1 for polyhedra whose spherical images belong to a hemisphere. (So, in the three-dimensional case, we exclude not only polyhedra whose unbounded edges are all parallel but also those with parallel unbounded faces.) Under this assumption, the proof is especially simple. The general case will be treated in short separately. Assume the converse, i.e., that there are two different unbounded polyhedra P1 and P2 satisfying the conditions of the theorem and having spherical images belonging to a hemisphere. These conditions imply that the spherical images of the polyhedra coincide and that the normals to the support planes of the polyhedra, regarded as issuing from the same point, lie in the boundary of a convex solid angle V . Denote by h1 (n) and h2 (n) the support functions, i.e., the distances from the support planes of P1 and P2 to the origin, taken with the appropriate signs. These functions are regarded as functions of the unit outward normal vector n. If we had h1 (n) = h2 (n) for all n, then the polyhedra would coincide. Therefore, there is a nonempty set W of vectors n in V for which h1 (n) − h2 (n) > 0 (or < 0). Without loss of generality, we may assume that there are unit vectors in V for which h1 (n) − h2 (n) > 0. Consider the function ϕ(n) defined at a vector n in V by ϕ(n) =

1 . h1 (n) − h2 (n)

(1)

This function may be interpreted geometrically as follows. Given n, take the end of the vector ϕ(n)n drawn from the origin. The resulting set is a polyhedron R with boundary at infinity which lies inside the angle V and outside the ball of radius r=

1 . supn∈W [h1 (n) − h2 (n)]

To prove this, observe that if A is a vertex of P1 (or P2 ) and a is the position vector of A, then, for the normals n to the support planes at A, we have h1 (n) = an. The normals at a vertex determine a convex solid angle. The whole angle V splits into such angles V1i for P1 and V2j for P2 . Consider the partition of V into the angles V k which are the intersections of the angles V1i and V2j . We have h1 (n) = ai1 n in each angle V1i and h2 (n) = aj2 n in V2j . Therefore, in V k = V1i ∩ V2j we have h1 (n) − h2 (n) = ak n, with ak = ai1 − aj2 . If h1 (n) − h2 (n) = 0, then accordingly ϕ(n) = 1/ak n and ak nϕ(n) = 1, i.e., the ends of the vectors ϕ(n)n going into the angle V k belong to the plane


297

ak x = 1.

(∗)

As the result, we see that the set of ends of all these vectors must be a polyhedron with faces on the planes (∗). We are interested in its part R that lies in the angle W , in which h1 (n) − h2 (n) > 0. On the boundary of W we have h1 (n) − h2 (n) = 0 (on the boundary of V this equality follows from the coincidence of the planes of unbounded faces). Therefore, ϕ(n) → ∞ as n approaches the boundary of W , i.e., the boundary of R lies at infinity. The fact that R lies outside some ball is obvious, because the boundedness of the functions h1 (n) and h2 (n) implies that the function ϕ(n) =

1 h1 (n) − h2 (n)

for h1 − h2 > 0 is bounded below by some positive number. Owing to this disposition of the polyhedron R, there exists a plane T not passing through the origin and containing only one vertex of the polyhedron, the whole polyhedron lying on one side of this plane; namely, on the side opposite to the origin (Fig. 129).

R

T

V

O Fig. 129

(Such a plane obviously exists. Indeed, by the assumption made at the beginning of the proof, the spherical images of the polyhedra P1 and P2 lie inside one hemisphere. Therefore, the solid angle V formed by the normals has support planes passing only through the vertex of V . The plane parallel to this plane and supporting R touches R at some bounded “face.” If this “face” is not a vertex, then, slightly rotating the plane, we achieve a position where the plane touches R only at a vertex.) Let m be the normal to the plane T and p, the distance from T to the origin. Then the distance from the origin to T in the direction of the given vector n is p . (2) r(n) = nm

298


Since R lies on the other side of T , we have ϕ(n) ≥ r(n) or, in view of (1) and (2), m h1 (n) − h2 (n) ≤ n · . (3) p At the same time, only one vertex of the polyhedron lies in the plane T ; therefore, the equality in (3) is attained only at the vector n0 that points to this vertex. By parallel translation of P2 , we can transform (3) into h1 (n) − h2 (n) ≤ 0,

(4)

where h2 (n) is the support function of P2 after translation.14 Since for n0 we have h1 (n) = h2 (n), it follows that the support planes of P1 and P2 with normal n0 coincide. Let G1 and G2 be the faces of P1 and P2 lying in this common support plane. A priori they may have any dimension from 0 to n − 1. By (4), h1 (n) < h2 (n) for n = n0 , i.e., the support planes of P1 are shifted inside P2 .15 This obviously implies that, first, the face G1 lies inside G2 and, second, the face G2 must be (n − 1)-dimensional: otherwise, shifting all support planes touching it inside P2 , we would make G2 disappear. All these support planes can be shifted because h1 (n) < h2 (n) for all n = n0 (and close to n0 since all the vectors n for which (4) holds lie in W ). However, the condition of the theorem requires that no (n−1)-dimensional face G2 properly include the face G1 that lies on the parallel support plane. The contradiction obtained shows that there are no vectors n for which h1 (n) = h2 (n). Hence, all support planes of P1 and P2 coincide, i.e., the polyhedra themselves coincide, which completes the proof of the theorem. 6.5.3 We now prove Theorem 1 in three-dimensional space in the general case. As before, assume that there are two different unbounded polyhedra P1 and P2 satisfying the conditions of the theorem. Then the support functions of P1 and P2 differ and we consider the vectors n for which h1 (n) − h2 (n) > 0 (or h1 − h2 < 0 if there are no vectors with h1 − h2 > 0). Again construct the set R of ends of the vectors ϕ(n)n, where ϕ(n) = 14

1 . h1 (n) − h2 (n)

Indeed, it is well known from analytic geometry and easy to verify immediately that the support number of the plane with normal n is h(n) = nx, where x is the position vector of an arbitrary point of the plane. Therefore, if we shift the plane by a vector a, then the new support number is h (n) = nx + na. Shifting P2 by the vector m , we obtain h2 (n) = h2 (n) + n m , and consequently p p h1 (n) − h2 (n) = h1 (n) − h2 (n) − n m ≤ 0 (by virtue of (3). p 15 Here, of course, n is in W . It is important that n0 lies strictly inside W , which implies that our conclusion holds for all support planes close to the plane with normal n0 .


299

Then R is a polyhedron. This is proved in exactly the same way as above. Since h1 − h2 > 0 and the vectors n point to a single (closed) halfspace, the polyhedron R lies on one side of the plane T0 passing through the origin and parallel to the boundary plane of this half-space. Therefore, R has a support plane T parallel to T0 . Let H be the common part of R and T . Proceeding as above (formulas (2)–(4)), we conclude that, for all vectors n close to vectors pointing to H, we have h1 (n) − h2 (n) ≤ 0,

(4 )

where h2 (n) is the support function of P2 after appropriate translation; moreover, the equality is attained only at vectors pointing to H. If H is a single point (or merely a set with an isolated point), then we repeat the previous arguments and come to a contradiction. So we assume that H contains an edge or a face of R. The set H cannot be the whole plane. Otherwise we would have h1 (n) = h2 (n) for all n16 , i.e., the polyhedra P1 and P2 would coincide, contradicting our assumption. Thus, H has boundary. The subsequent argument consists in showing that this leads to a contradiction again. Let A be a point on the boundary of H and let n0 be the vector directed to A. Let L1 and L2 be the support planes of P1 and P2 with the same normal n0 . Then the following cases are possible: (a) The planes L1 and L2 contain only the vertices B1 and B2 of P1 and P2 . (b) The plane L1 contains an edge K1 of P1 and L2 , an edge K2 (or a vertex) of P2 . (c) The plane L1 contains a face G1 of P1 and L2 , a face G2 (or an edge or a vertex) of P2 . All the other cases result from (b) and (c) by interchanging P1 and P2 . (A) In case (a), a neighborhood of A in R is flat. (Indeed, if a vector a1 points to the vertex B1 , then at this vertex h1 (n) = a1 n. Therefore, h1 (n) − h2 (n) = (a1 − a2 )n = an and so anϕ(n) = 1, i.e., the ends of the vectors ϕ(n)n lie on one plane.) Since A belongs to the boundary of H, this excludes case (a). (C) In case (c), observe that (4) implies that the support planes of P2 either coincide with those of P1 or are shifted inside P1 , so that the face G2 is contained in G1 . But then (since we only consider the vectors n for which h1 (n) = h2 (n)) by the conditions of the theorem this is possible only when the faces G1 and G2 coincide. Then for vectors close to n0 we obviously have 16

Indeed, if we had H = T , the whole polyhedron R would coincide with the plane na T and the function nϕ(n) would satisfy the equation of the plane h1 (n)−h = p, 2 (n) where a is some vector and p is a number. In other words, the function h1 (n) − h2 (n) would be linear, so that h1 (n) − h2 (n) would be linear. However, owing to the inequality h1 (n) − h2 (n) ≤ 0, this is possible only if h1 (n) = h2 (n).

300


h1 (n) = h2 (n). Since h2 (n) differs from h2 (n) only by a summand of the form an corresponding to the translation of P2 ; therefore, h1 (n) − h2 (n) = an. As in case (A), this means that a neighborhood of A in R is flat, contradicting the assumption that A belongs to the boundary of H. (B) Case (b) occurs only if the edges K1 and K2 belong to the common support plane. If K1 and K2 are not parallel to one another, then the polyhedra P1 and P2 intersect and the difference h1 (n) − h2 (n) obviously changes sign for vectors n arbitrarily close to n0 , contradicting (4). However, if K1 and K2 are parallel to one another (or one of them is just a point), then they have a common family of support planes. The normals n to these planes lie in a single plane. Accordingly, the ends of the vectors ϕ(n)n pointing to the points of R belong to the same plane S. The intersection of the planes S and T is a straight line, which implies that we have an edge r on the polyhedron R. Thus, the point A on the boundary of H must belong to the interior of the edge r of R. Since A is an arbitrary point on the boundary of H, it follows that the boundary of H contains no vertices and so consists of straight lines. In other words, the edge r of R must be unbounded on both ends. The vectors n terminating at its points cover a half-circle. This means that the normals n to the support planes of P1 and P2 passing through the edges K1 and K2 cover a half-circle. But that case, the polyhedra degenerate into planar figures. We have arrived at a contradiction again, thus completing the proof of the theorem. In the n-dimensional case, using similar arguments, we must examine n possibilities that are similar to the cases (a), (b), and (c). 6.5.4 In Section 6.3, we showed by means of a simple example that the theorem on closed polyhedra proved there does not generalize to higher dimensions. Therefore, this theorem cannot be given a proof analogous to the one above, unless we use some extra arguments characteristic of the threedimensional space. In the above proof, it was essential that the vectors n for which h1 (n) − h2 (n) > 0 (or < 0) point to a single half-space, implying that the polyhedron R necessarily has a support plane. In the case of closed polyhedra P1 and P2 , it is possible that the vectors n with h1 (n) − h2 (n) > 0 do not point to a single half-space, invalidating the arguments from the very beginning. Nevertheless, we are confident that a proof based on considering the difference h1 (n) − h2 (n) of the support functions can be found for the theorem on closed polyhedra as well. This confidence is based, apart from everything else, on the fact that this approach yields a similar theorem for curved surfaces. This theorem is stated in Section 6.6. To carry out such a proof of the theorem on closed polyhedra remains an intriguing problem. The interest here is in improving the method. The general idea is as follows: given two surfaces, construct a third one that somehow reflects their


301

difference and, having established the existence of support planes for this surface, derive a contradiction from this fact. In this general form, the method is also used in the proofs of rigidity theorems for surfaces; it originated from the first article [Li] by Liebmann in 1900 in which the rigidity of a sphere was first proved. The method became especially effective in Pogorelov’s research: it is by this method that he proved his theorems on the rigidity of convex surfaces which were mentioned in Section 3.6 of Chapter 3. In this connection, it seems quite interesting to find a proof of the uniqueness theorem for a closed convex polyhedron with prescribed development which would be analogous to Pogorelov’s proof of the corresponding theorem for curved surfaces of bounded curvature. 6.5.5 We present one more theorem on bounded polyhedra with boundary, requiring that the boundary be a single closed polygonal line without multiple points. Theorem 2. Assume given two bounded convex polyhedra, each with boundary a single closed polygonal line. Suppose that the polyhedra satisfy the following conditions: (1) To each face of one of the polyhedra there corresponds a parallel face of the other and vice versa; moreover, none of these faces can be placed inside the other by translation. (2) The boundary edges of each extreme face Q of any of the given polyhedra form a single polygonal line and the parallel face Q of the other polyhedron either has no boundary edges or has all boundary edges pairwise parallel to the boundary edges of Q, and all boundary edges of Q are not longer or, on the contrary, not shorter than the parallel edges of Q . Then the polyhedra are translates of one another. P1

P2

Q1 Q2

Fig. 130

As always, we call faces parallel to one another if they have parallel outward normals. One of the parallel faces may degenerate into an edge. Degeneration into a vertex is excluded, since each point can always be placed inside any face. In the second condition, by parallel edges we also mean edges with parallel outward normals, but these normals are now taken in the planes of the faces Q and Q . Here if the face Q (or Q) does not have an edge parallel

302


to the given edge of Q (or Q ), then as before we treat the former edge as existing but having zero length. In the second condition, the face Q parallel to Q may degenerate into an edge. In Fig. 130, all the faces of the polyhedron P2 which are parallel to the extreme faces of the polyhedron P1 degenerate into edges. The face Q1 of P1 parallel to the extreme face Q2 has no common points with the boundary of P1 . In this picture, the second condition is satisfied. The first obviously fails. In the simplest case, the extreme faces of polyhedra are pairwise parallel to one another, do not degenerate, and have pairwise parallel boundary edges. It is important that on one face all these edges be not longer (or not shorter) than on the other. In particular, this condition certainly holds if each extreme face has only one boundary edge and such edges on parallel extreme faces are parallel to one another. 6.5.6 We sketch the proof of this theorem, omitting details that can be easily recovered. Let P1 and P2 be polyhedra satisfying the conditions of the theorem. Consider the convex hulls P 1 and P 2 of P1 and P2 . They are convex solid polyhedra and P1 and P2 are parts of their boundaries. Construct the halfsum P of P 1 and P 2 , i.e., P =

1 (P 1 + P 2 ). 2

We are interested in the part P of the surface of P composed of the faces that are half-sums of faces and edges of the original polyhedra P1 and P2 , rather than in the whole polyhedron P . This P is a convex polyhedron bounded by a single closed polygonal line too. All its boundary edges are half-sums of parallel boundary edges (or edges and vertices) of the polyhedra P1 and P2 . The faces of P can only be of the same three types as in Theorem 1 of Section 6.3. Namely, they are (1) the half-sums of parallel genuine faces of P1 and P2 ; (2) the half-sums of a genuine face Q of one of the polyhedra and an edge of the other; moreover, the edge lies in the support plane parallel to the face Q; (3) the half-sum of nonparallel edges lying in parallel support planes (for example, the edges q1 and q2 in Fig. 127, p. 286). These faces are parallelograms. Each edge p of P is the half-sum of parallel edges p1 and p2 of P1 and P2 , but one of these edges may degenerate into a vertex. Label each edge p of P by the sign of the difference p1 − p2 . If p1 − p2 = 0, then leave p unlabeled. Using Lemma 3 of Section 6.1 in the same manner as in the proof of Theorem 1 of Section 6.3, we can verify that if some face on P has labeled edges, then the number of sign changes in going around this


303

face is not less than four. In particular, there always exist four sign changes around each face of the third type. Condition (2) of the theorem states that the boundary edges of an arbitrary extreme face Q of P form a single polygonal line. By the same condition, all boundary edges of an individual extreme face of P1 are not longer or not shorter than the edges of the parallel extreme face of P2 . Therefore, we observe no sign changes in the sequence of boundary edges of Q. All the more, the same holds if Q is a face of the third type; in this case Q is a parallelogram and has only one boundary edge. Now, take a second copy P of P with the same arrangement of signs on its edges. Identify its boundary edges with the corresponding boundary edges of P . After that, erase the identified boundary edges and delete their endpoints, so that each pair of edges of P and P approaching the boundary merge into one edge, if only these endpoints are free, i.e., belong to exactly one interior edge. Obviously, the polyhedron P ∪ P obtained in this way is homeomorphic to the sphere and has no vertex with exactly two edges touching at it. The polyhedron P ∪ P has faces of two kinds: faces corresponding to the faces on P or P without boundary edges and faces each of which consists of two copies of the same extreme face of the polyhedron P , one belonging to P and the other to P ; these copies merge into one face because of the identification and subsequent elimination of boundary edges. Assume that there is a face with labeled edges on P ∪ P . If this face is of the first kind, then we observe at least four sign changes around it. Let this face be of the second kind. We may now denote it by Q ∪ Q , with Q and Q the constituent faces of P and P . In forming Q ∪ Q , we excluded the boundary edges of Q. However, as was observed, no sign changes happen on these edges; therefore, excluding them, we eliminate at most two sign changes: one in passing to these edges and the other in passing from them to the next edges. As the result, at least two sign changes are left in the remaining part of the contour of Q. Since Q is merely a second copy of Q, the same holds for it. Therefore, in going around Q ∪ Q we have at least four sign changes. Thus there are at least four sign changes in going around any face of P ∪P that possess labeled edges. However, considering the dual polyhedron as in the proof of Theorem 1 of Section 6.3, we infer that, by Cauchy’s Lemma, there can be no labeled edges at all, i.e., all edges of the original polyhedra P1 and P2 must be equal. (In particular, there are no faces of the third type formed by the half-sums of nonparallel edges.) It is now immediate that the polyhedra P1 and P2 are translates of one another, which was to be proved. No theorems of this kind for polyhedra with boundary, except for Theorems 1a and 2, are known.

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6.6 Generalizations 6.6.1 In the case of curved surfaces, the theorem corresponding to the theorem of Section 6.3 on closed polyhedra is the following: If two closed analytic surfaces with positive curvature everywhere are such that their Dupin indicatrices at every pair of points with parallel outward normals cannot be placed inside one another by parallel translation, then the surfaces are parallel translates of one another. In other words, two closed analytic surfaces with everywhere positive curvatures either are parallel translates of one another or have points with parallel normals such that the Dupin indicatrices at these points can be placed inside one another by translation. The Dupin indicatrix of a surface F at a point A is, roughly speaking, the figure determined by the intersection of F with a plane parallel and infinitesimally close to the tangent plane at A. The intersection of a polyhedron with a plane parallel and infinitesimally close to the plane of some face is, obviously, this face itself. From this comparison, we clearly see the analogy between the theorem on surfaces with that on polyhedra. The proof of the theorem begins by considering the difference between the support functions of the two surfaces and actually relies on simple geometric arguments which, however, lead to the result only if the analyticity of the surfaces is assumed.17 The problem of freeing the theorem from this very restrictive requirement is important but hard.18 It would be much nicer if 17

See [A7], where the theorem is proved in a slightly different, but equivalent formulation. 18 The major achievement in this direction is the following theorem by A. D. Alexandrov. Let S and S be closed convex surfaces whose support functions H (x, y, z) and H (x, y, z), defined on R3 \ {0}, are C 2 -smooth. For H = H − H , one of the three eigenvalues of the quadratic form d2 H is identially zero, since H and H are homogeneous of degree 1. Denote the other two eigenvalues by R1 (x, y, z) ≥ R2 (x, y, z) and consider them at the points n of the standard sphere x2 + y 2 + z 2 = 1. For almost every n, the support planes with outward normal n meet S and S only at isolated points. If R1 > R2 in at least one of these vectors n or if exactly one of the numbers R1 and R2 vanishes, then, at all points with outward normal n, the Dupin indicatrix of one of the surfaces is taken by an appropriate parallel translation into the Dupin indicatrix of the other surface. Theorem. (See [A20, VII] and the appendix in [A22, VI, p. 18]) If R1 = R2 = 0 or R1 R2 < 0 for almost every n and the function R1 R2 + (∗) R2 R1 is summable on the subset of the sphere in which R1 R2 < 0, then S and S are parallel translates of each other. This theorem relaxes the analyticity requirement to C 2 -smoothness but does not settle the question, since it involves an extra condition on the function (∗). – V. Zalgaller

6.6 Generalizations

305

we could state and prove a theorem valid for all closed convex surfaces without any regularity assumption and containing, as particular cases, the above theorem on analytic surfaces and the theorem of Section 6.3 on polyhedra. The theorem on surfaces can be given a somewhat weaker analytic form. Let f (x, y; n) be a function of a unit vector n and two numerical variables defined in the region x ≥ y and monotone, i.e., for each n, we have f (x1 , y1 ; n) > f (x2 , y2 ; n) whenever x1 > x2 , y1 ≥ y2 or x1 ≥ x2 , y1 > y2 . Let F1 and F2 be two regular closed convex surfaces and let R1 ≥ R1 and R2 ≥ R2 be the radii of principal curvature at the points with parallel outward normals n. If f (R1 , R1 ; n) = f (R2 , R2 ; n) for every n, i.e., for every pair of such points, then the surfaces are parallel translates of one another.19 In other words, a surface is determined from f (R, R ; n) as a function of a normal (i.e., f = g(n) uniquely up to translation). The relationship between this statement and the initial geometric statement is the following. If f (R1 , R1 ; n) = f (R2 , R2 ; n) and, for instance, R1 > R2 , then the monotonicity of f implies that R1 < R2 . However, as is well known, the radii of principal curvature are equal to the squares of the half-axes of the ellipse that represents the Dupin indicatrix. If the differences between the half-axes of two ellipses have opposite signs, then the ellipses cannot be placed in one another by any translation. (It is important that we compare the major half-axis with the major and the minor half-axis with the minor, since by assumption we have R ≥ R .) The converse assertion may fail. Therefore, the geometric statement is much more general than the analytic one. The theorem implies numerous particular theorems specified by special choices of the function f (x, y; n). The first of these consequences was proved by Christoffel in 1865 [Ch]. In it f = x + y, i.e., the theorem deals with the congruence of surfaces with equal sums of principal curvature radii. The next result is Minkowski’s theorem of 1899, where f = xy, i.e., the theorem concerns the congruence of surfaces with equal products of principal curvature radii. This theorem is close to Minkowski’s Uniqueness Theorem for polyhedra with parallel faces of equal areas, since the Gaussian curvature is equal to the ratio between the areas of infinitesimal pieces on the spherical image and on the surface. Consequently, the Gaussian curvature determines the dependence of the areas of parts of the surface on its spherical image; therefore its specification plays the same role as the specification of the area of a face as a function of the normal. There are many later particular results which are, so to speak, swallowed up by our theorem. 19

We may now consider not necessarily convex surfaces such that each of them is the envelope of the family of the planes nx = h(n), with h(n) a function of a unit vector n, defined for all n, single-valued, and analytic. For such a surface, the normals n to its tangent planes cover the sphere once. Therefore, if the surface is not convex, then it has a cuspidal edge. This case is not excluded; the radii of curvature at the corresponding points exist, but at least one of them vanishes.

306


The above analytic statement admits a series of versions depending on the regularity conditions imposed on the function f and surfaces F1 and F2 . If we require the analyticity of F1 and F2 , then the function f (x, y; n) may be absolutely arbitrary, except for the necessary condition of monotonicity in x, y; for example, f may be even nonmeasurable. If we require that f be thrice continuously differentiable and “essentially” ∂f monotone (i.e., ∂f ∂x , ∂y > 0), then the differentiability of the surfaces F1 and F2 four times will suffice.20 I am still convinced that the function f can be absolutely arbitrary while the surfaces should be twice differentiable, if only the principal curvature radii exist. However, all attempts to prove the theorem in this generality have been unsuccessful so far.21 In the analytic statement, the theorem reduces to the question of uniqueness of the solution of a partial differential equation. As the defining function of the surface, we take the support function at the unit vectors, h(n) = h(ξ, η), i.e., the distance from the origin to the tangent plane. We regard it as a function of the normal n or of coordinates (ξ, η) on the unit sphere. The principal curvature radii are expressed in terms of h(ξ, η) and its firstand second-order derivatives.22 Therefore, the specification of f (R, R ; n) as a function of the normal, i.e., f (R, R ; n) = g(n), is equivalent to a certain second-order partial differential equation: Φ(hξξ , . . . , h; ξ, η) = g(ξ, η). The problem reduces to the question of uniqueness of solutions to this equation on the whole sphere up to a summand an corresponding to a parallel translation. This equation is rather general since f must only obey the condi∂f tion of essential monotonicity: ∂f ∂x , ∂y > 0. The latter implies that the equation is of elliptic type.23 6.6.2 For surfaces with boundary and unbounded surfaces, we can formulate some perfect analogs of Theorems 1 and 1a of Section 6.5. For instance, we have the following version of Theorem 1a: Let F1 and F2 be two convex surfaces with the same spherical image lying in a hemisphere. Suppose for these surfaces that (1) the tangent planes at the boundary points coincide and (2) at the interior points with parallel outward normals the Dupin indicatrices cannot be shifted inside one another. Then the surfaces coincide (except possibly for their parts lying in the tangent planes at boundary points). 20

This best of the existing weaker versions of the conditions on surfaces belongs to A. V. Pogorelov; see his note [P2]. Under stronger requirements, the result first appeared in my article [A4]. Another proof can be found in [A14]. 21 See Subsection 6.6.5 below. – V. Zalgaller 22 See, for instance, [Bl2, § 94]. 23 See [P2].

6.6 Generalizations

307

The proof is similar to that of Theorem 1 of Section 6.5. Denote by h1 (n) and h2 (n) the support functions of F1 and F2 and, for each unit vector n, consider the end of the vector ϕ(n)n, where ϕ(n) =

1 . h1 (n) − h2 (n)

The locus R of these ends with ϕ(n) > 0 (or < 0) must have support planes. However, R turns out to be a surface of nonpositive curvature and so cannot have any support planes except for the case in which R degenerates into a plane. But then h1 (n) = h2 (n), and the surfaces coincide. The theorem, as well the sketch of its proof, generalizes word for word to surfaces in n-dimensional space. Unlike the case of a closed surface, the regularity assumptions are now immaterial. We do not know whether this theorem extends to surfaces whose spherical image lies in a closed hemisphere rather than inside it. The theorem is probably false for surfaces whose spherical image does not lie in a hemisphere; however, to show this, an appropriate example must be constructed. 6.6.3 We have not succeeded in generalizing the theorems on closed polyhedra of Subsection 6.6.1 to higher-dimensional spaces. We do not know whether the reasons behind this are the shortcomings of the method or that such a generalization is impossible in principle.24 The impossibility of generalizing the theorem on closed polyhedra of Section 6.3 has already been established by the example of Section 6.3. Nevertheless, as was indicated in the same section, Minkowski’s Uniqueness Theorem for polyhedra with pairwise parallel faces of equal areas remains valid in spaces of arbitrary dimension. No theorems of this type with monotone functions of (n − 1)-dimensional faces other than area are known.25 A rather general theorem can be formulated for closed convex surfaces in n-dimensional space. In the case of regular surfaces, this theorem amounts to the following: 24

An example of an ellipsoid and a hypersphere in R4 whose indicatrices cannot be placed in one another is given in [Me1]. However, the article [Me3] demonstrates that the theorem of congruence up to parallel translation remains valid in Rn under certain more stringent requirements on the indicatrices. – V. Zalgaller 25 A generalization of Minkowski’s Theorem is possible if it is assumed that the polyhedra P and P have the same structure. However, we do not find this result impressive. Here the congruence condition consists in requiring the equality of the mixed areas of the faces of P , P , and given polyhedra P1 , . . . , Pm of the same structure: Fj P, . . . , P P1 . . . Pm = Fj P , . . . , P P1 . . . Pm n−m−1

n−m−1

(see Section 7.2 of Chapter 7). The theorem fails if the coincidence of structures is not assumed.

308


If, for two closed convex surfaces, the given elementary symmetric functions of the radii of principal curvature are equal at the points with parallel outward normals, then the surfaces are parallel translates of one another. (An elementary symmetric function of R1 , R2 , . . . , Rn−1 is a coefficient of the polynomial with roots R1 , R2 , . . . , Rn−1 .) By considering the corresponding set functions on the sphere instead of functions of curvature radii, the theorem may be generalized to convex surfaces under no regularity assumptions. It is in this general form that I first established the theorem.26 In particular, the integral of the product of principal curvature radii of a surface Φ over the set of normals, i.e., over the unit sphere, is the area of the corresponding subset of the surface. Regarded as a set function of the sphere, this function is the “area function” FΦ (M ). The general definition of this function was given in Subsection 2.7.2. The above theorem contains the assertion that arbitrary closed convex surfaces with the same area functions are parallel translates of one another. This is a straightforward generalization of Minkowski’s Theorem about polyhedra to the case of arbitrary convex surfaces.27 Proofs of these theorems rely on the so-called theory of mixed volumes, whose basics are presented in Section 8.3 of Chapter 8. 6.6.4 At first glance, the generalization of theorems from the current chapter to polyhedra in non-Euclidean space seems absolutely meaningless, since no parallel translation exists in such a space. However, if we define the notion of parallelism in projective terms, then the problem is no longer absurd. Pairwise parallelism of faces in the projective language means that the planes of the faces intersect pairwise in straight lines lying in one (infinitely distant) plane. This condition can now be expressed for spherical as well as hyperbolic space. For instance, we may suggest the following question: Suppose that the corresponding faces of two spherical polyhedra obey an appropriate condition on their disposition and have equal areas; is there some simple transformation taking one of the polyhedra to the other in the same way as this is done by translations in Euclidean space? This question seems less meaningless. However, it may turn out that there is no such simple transformation, and then the question would be insignificant. 6.6.528 The method indicated at the end of Subsection 6.6.1 was developed in a series of articles of the author of this book [A20]. The justification of the method is mainly contained in [A20, II]. Let Φ(k1 , . . . , kn , z, x) be a C 1 smooth function in the variables 26

See [A3, II]. An essentially more general result is established in [A3, III]. See [Mi7] for connections between the Minkowski and Cauchy theorems. – V. Zalgaller 28 Subsections 6.6.5–6.6.9 are comments by V. Zalgaller. 27

6.6 Generalizations

309

k1 ≥ k2 ≥ . . . ≥ kn > 0, z, x = (x1 , . . . , xn ), where ∂Φ/∂ki > 0 everywhere for all i. Let S : z = z (x) and S : z = z (x) be two convex C 2 -smooth surfaces in some family {S} (that depends on the problem) of surfaces which is convex in z. Then the difference ∆Φ = Φ(k1 , . . . , kn , z , x) − Φ(k1 , . . . , kn , z , x), where ki and ki are the principal curvatures of S and S at the points (x, z (x)) and (x, z (x)), represents an elliptic linear expression in ∆z = z − z : n n ∆Φ = aij (∆z)ij + bi (∆z)i + c∆z. i,j=1

i=1

If the values of Φ on S and S at the corresponding points are the same, then ∆z is a solution to a homogeneous linear equation of elliptic type. Considering these equations, A. D. Alexandrov established various generalizations of the maximum principle and uniqueness theorem (see [A18], [A20, I], [A22], [A24], [A25], [Ha]). If we succeed, by using a transformation which is trivial for this problem (translation, similarity, etc.), to bring the surfaces S and S to a position where one of them touches the other from the exterior side, while Φ does not change, then after that S and S will coincide by the uniqueness of solutions to the equation ∆Φ = 0. All the uniqueness theorems in [A20] rely on this argument to a certain extent. In particular, in [A20, I, III] the theorem stated at the end of Subsection 6.6.1 is proved. Another proof in the class C 3 can be found in [P10, Chapter VII] and [HW]. If the function Φ is symmetric in n, i.e., Φ(R1 , R2 ; n) = Φ(R1 , R2 ; −n) then the theorem is valid without differentiability assumptions on Φ (see [Me2]). 6.6.6 For convex surfaces with boundary or unbounded surfaces with prescribed behavior at infinity, all theorems of [A20] are multi-dimensional and they hold true not only in Euclidean spaces [A20, III]. 6.6.7 For closed convex hypersurfaces in E n+1 with n > 2 there are no uniqueness theorems analogous to those stated at the end of Subsection 6.6.1 with conditions on the functions Φ that would be as general. However, if Φ is an elementary symmetric function of principal curvature radii R1 , . . . , Rn , then the corresponding uniqueness theorem holds. This theorem was proved by A. D. Alexandrov [A3, II] in 1937. In the particular cases of Φ = R1 +. . .+ Rn (Christoffel’s problem) and Φ = R1 . . . Rn (Minkowski’s problem) it was proved as far back as at the turn of the century by Hurwitz and Minkowski.

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6.6.8 The following classical “soap-bubble” question is well known: Must a C 2 -smooth closed hypersurface be a sphere if its mean curvature is the same at all points? The weakest sufficient conditions for a positive answer were given in A. D. Alexandrov’s articles [A20, V] and [A23].29 At the same time, it is now known that the answer to the question is negative without some extra assumptions. The first example was constructed in 1982 [Hs], [HTY]. 6.6.9 Stability theorems are a further development of uniqueness theorems. They give a proximity estimate for two convex hypersurfaces in Rn , provided that some functions, which uniquely determine the surfaces, are close to each other. Yu. A. Volkov’s early article [Vo5] on Minkowski’s problem pertains to this direction. It corresponds to the pointwise specification of the (n − 1)th curvature function Fn−1 for regular hypersurfaces. V. I. Diskant [Di, p. 110], using other tools, generalized this result to the general case in which Fn−1 , the “area function,” is a set function on the sphere. The proof is also described in [P10, Chapter 7]. Stability in Christoffel’s problem (which corresponds to the function F1 ) was first proved by A. V. Pogorelov [P10, Chapter 7]; also see the articles [VO] and [Se2]. Stability in the case of the function Fn−2 was proved by V. I. Diskant [Di, p. 130]. See [Di, p. 130] about the stability for other functions Fk . An extensive literature can be found in [Di]. The results of the brilliant articles [Vo6] and [K] also pertain to the stability theorems.

29

Many mathematicians have addressed this problem. A simple proof under strong extra conditions can be found, for instance, in [Am]. – V. Zalgaller

7 Existence Theorems for Polyhedra with Prescribed Face Directions

7.1 Existence of Polyhedra with Prescribed Face Areas 7.1.1 Here we prove Minkowski’s Theorem: noncoplanar unit vectors and F1 , . . . , Fm are Theorem. If n1 , . . . , nm are m positive numbers such that i=1 Fi ni = 0, then there exists a closed convex polyhedron whose faces have outward normals ni and areas Fi . (By Theorem 1 of Section 6.3, such a polyhedron is unique up to parallel translation. Here and in what follows, all vectors ni are assumed distinct.) As has already been observed in Subsection 2.2.3, the conditions imposed on the vectors ni and numbers Fi are obviously necessary for these vectors and numbers to serve as the face normals mand face areas of a closed convex polyhedron. In particular, the condition i=1 Fi ni = 0 means that the vector area of the closed polyhedron equals zero. Minkowski’s Theorem asserts the sufficiency of these conditions. Also, notice the following circumstance important for the sequel: the above conditions on the vectors ni and Fi imply that m the vectors ni do not point to a single half-space. Otherwise, the vector i=1 Fi ni could not vanish since all the coefficients Fi are positive. For this reason, in Minkowski’s Theorem we may a priori require that the vectors ni do not point to a single (closed) half-space, which implies that they are not coplanar. 7.1.2 Before beginning the proof of Minkowski’s Theorem, we prove two lemmas. Lemma 1. For arbitrary noncoplanar vectors ni , there exists a convex polyhedron with face normals ni . If the vectors ni do not point to a single closed half-space, then such a polyhedron is closed. Otherwise it is unbounded. Draw rays li from some point in the direction of the vectors ni and consider planes Qi perpendicular to li . These planes bound some solid polyhedron P . However, if the planes Qi are chosen arbitrarily, then some of them

312


may be nontangent to P . If we choose the planes Qi so that they touch some ball centered at O, then each plane Qi will obviously determine a face of P , so that P is circumscribed around a ball and has the face normals ni . Since the vectors ni are not coplanar, the polyhedron is not an unbounded prism. If the vectors ni and thereby the rays li do not point to a single closed half-space, then to each ray l issuing from O there is a ray li forming an acute angle with l. Therefore, the plane Qi intersects l. Thus the polyhedron P is bounded by these planes and contains no ray. Hence, P is closed. If the rays li point to a single half-space, then the ray l, perpendicular to the bounding plane of this half-space and directed towards the opposite half-space, forms acute or right angles with all the rays li . Therefore, no plane Q may intersect l. It follows that the polyhedron P contains the ray l and so P is unbounded. Lemma 2. Each closed convex polyhedron of area not greater than a given number F and with all faces of area not less than a given number f has support numbers bounded by a quantity depending only on F and f , provided that the origin lies inside the polyhedron. This lemma admits various proofs. In particular, it can easily be derived from the isoperimetric property of the ball: any ball has the smallest area among all figures of equal volume.1 However, first, the proof of the indicated inequality is rather involved and, second, the estimate derived from it for the support numbers of a polyhedron as functions of F and f is too rough. For this reason, we obtain Lemma 2 as a corollary to a theorem which, while 1

By this property, we have the following well-known inequality between the area F and volume V of every convex body: 36πV 2 ≤ F 3 .

(∗)

Assume that the origin lies inside the given polyhedron P . Assume further that Fi is the area and hi is the support number of some face of P . Then P contains a pyramid of altitude hi and base Fi , which implies that V ≥ 13 hi Fi . On the other hand, by (∗), V ≤

3/2 F√ . 6 π

Therefore, hi ≤

3/2 F √ , 2 πFi

so that if Fi > f , then

F 3/2 hi < √ 2 πf

(∗∗)

for every i, as desired. From the theorem stated below, we may, however, infer the estimate hi < √F for every i, and so hi < √F or πFi

πf

√ F f √ . hi < πf

(∗ ∗ ∗)

But f < 14 F , since the number of faces is not less than four. Hence, the estimate (∗ ∗ ∗) is more precise than (∗∗). If f is small as compared to F , which is certain when the number of faces is large, then (∗ ∗ ∗) becomes much more precise than (∗∗).

7.1 Existence of Polyhedra with Prescribed Face Areas

313

having elementary proof, gives an exact bound of the support numbers (more precisely, widths) of the polyhedron. Theorem. Define the width of a polyhedron in a given direction to be the distance between the support planes perpendicular to this direction. Let F be the area of a closed convex polyhedron P . Further, denote by Fi the area of some face of P . Finally, let Bi be the width of P in the direction perpendicular to this face. Then the following inequality holds: F − Fi Bi < √ πFi

(1)

(moreover, this inequality is exact in the sense that there are polyhedra for √ √ i Fi which the fraction πB is arbitrarily close to 1). F −Fi

A

A P

P Bi =H Gi

H ai

Pi

Fig. 131

Let Gi be the chosen face of P and let A be a point at which the support plane parallel but opposite to Gi touches P . Construct the pyramid P with base Gi and vertex A (Fig. 131). Clearly, the width Bi of P in the direction perpendicular to Gi is the same as that of the original polyhedron P ; the face Gi is common to P and P , and the area F of P is not larger than that of P . Therefore, if we establish (1) for P , then (1) will hold for P . Thus, it suffices to prove (1) for a pyramid, where by Fi and Bi we mean the area S of its base Gi and altitude H, respectively. Thus, we consider the pyramid P . Let a1 , . . . , an be the sides of its base and let p1 , . . . , pn be the distances from these sides to the projection of the vertex of the the base. Then the height of the ith pyramid to the plane of face equals H 2 + p2i and its area is 12 H 2 + p2i . Therefore, the area of the lateral surface of the pyramid (the total area minus the base S) is 1 ai 2 i=1 n

F −S =

H 2 + p2i ,

(2)

314


whence F −S √ = H S But clearly

n 2 2 H + pi i=1 ai 1 + i=1 ai √ √ = 2H S 2 S

n

n i=1

ai

p2i H2

.

(3)

p2 1 + i2 > ai = L, H i=1 n

with L the perimeter of the base of the pyramid. Now, (3) yields F −S L √ > √ . H S 2 S

(4)

Since for a fixed perimeter the greatest area is bounded by a circle, it follows that L2 = 4πS; we always have2

and (4) implies that

√ L √ > π, 2 S

(5)

√ F −S √ > π, H S

(6)

which is the required inequality (1) for S = Fi and H = Bi . The above argument also shows that in each of the inequalities (4)–(6) we may come as close as we like to equality: it suffices to take a regular pyramid with sufficiently many sides and large enough altitude. Thus we have also proved additionally the exactness of (1). The claim of Lemma 2 now follows easily from inequality (1). Indeed, if the origin lies inside P , then the support number hi of the face Gi is not greater than the width of the polyhedron in the direction perpendicular to Gi . F F Now by (1) we have Bi < √πF , and so hi < √πF . If the areas of all faces i i are at least f , then we conclude that the support numbers of all faces satisfy the inequality F hi < √ , πf claimed by Lemma 2. 2

A rougher estimate is immediate without appealing to the isoperimetric property of a circle. A polyhedron with perimeter L may be placed in a square with side 1 L. Therefore, its maximal width does not exceed a half of the perimeter. For 2 L > 1. This elementary argument yields the area we then have S < 14 L2 or 2√ S F −S √ H S

n−2

> 1. Similarly, in n-dimensional space we have F − S > HS n−1 .

7.1 Existence of Polyhedra with Prescribed Face Areas

315

7.1.3 We now prove Minkowski’s Theorem by using the Mapping Lemma (Section 2.2 of Chapter 2). Let n1 , . . . , nm be the given unit vectors not pointing to a single closed half-space. By Lemma 1 there are closed polyhedra having these vectors as face normals. Each such polyhedron is uniquely determined by its support numbers h1 , . . . , hm and therefore can be represented as a point in mdimensional space Rm with coordinates h1 , . . . , hm . No face of a polyhedron disappears under small parallel displacements of the bounding planes of the polyhedron. Hence, if h01 , . . . , h0m are support numbers of one of the polyhedra, then numbers h1 , . . . , hm sufficiently close to h01 , . . . , h0m are support numbers of some polyhedron as well. Consequently, the collection of all polyhedra under consideration represents an open set P0 in Rm . However, we must not distinguish between polyhedra that are parallel translates of one another, since the face normals and face areas determine a polyhedron up to translation. For this reason, we must consider the manifold P of classes of congruent parallel polyhedra rather than the set P0 itself. We now define this manifold. When we shift a plane with normal n by a vector a, the summand an is added to the distance from the plane to the origin. Therefore, on shifting the polyhedron by the vector a, each support number of the polyhedron increases by the summand ani . Consequently, a class of congruent parallel polyhedra is characterized by the condition that the support numbers of the polyhedra in the class have the form h0i + xn, where h0i are the support numbers of one of these polyhedra and x is an arbitrary vector. The equalities hi = h0i + xni = h0i + xnix + yniy + zniz

(i = 1, . . . , m),

with h0i fixed, determine a three-dimensional subspace in Rm . The whole space Rm stratifies into three-dimensional subspaces which are all parallel to one another and form the (m−3)-dimensional manifold Rm−3 . We may imagine that Rm projects onto Rm−3 along these three-dimensional subspaces.3 The set P0 of polyhedra projects onto the set P of classes of congruent parallel polyhedra. Since P0 is open, P is also open and it is an (m−3)-dimensional manifold in the sense of our definition (adopted in Subsection 2.8.6). It is this manifold P that we shall consider. Of course, instead of working with classes of polyhedra, we may choose one polyhedron in each class and use these chosen polyhedra. Now consider the set of all collections of positive numbers F1 , . . . , Fm such that m Fi ni = 0. (7) i=1 3

Three-dimensional space stratifies into parallel planes in similar way, forming a one-dimensional manifold.

316


This vector equality amounts to three linear equations, distinguishing an (m − 3)-dimensional subspace in m-dimensional space. The condition that all the coefficients Fi are positive distinguishes the positive coordinate angle in m-dimensional space. The intersection of the angle and the subspace is an open convex (m − 3)-dimensional set. This set is the manifold F of admissible values of the numbers F1 , . . . , Fm . Since there are polyhedra with normals ni , while the areas of their faces are positive and satisfy condition (7), it follows that F is nonempty. Moreover, F, being convex, is connected. 7.1.4 The manifolds P and F have the same dimension. Further, there is a natural mapping ϕ from P to F. Namely, ϕ assigns to each polyhedron P (or its class) in P the collection (F1 , . . . , Fm ) of the face areas of P . If we establish that ϕ is a mapping from P onto F, then Minkowski’s Theorem will be proved. To this end, it suffices to verify that ϕ satisfies all conditions of the Mapping Lemma (Section 2.2 of Chapter 2). Let us check these conditions in succession. The first condition requires that each connected component of F contain images of points in P. In our case, F is connected, i.e., it consists of only one component. Hence, the condition is satisfied, because F contains the image of the manifold P. The second condition is injectivity of ϕ. We have proved in Section 6.3 that face directions and face areas determine a closed convex polyhedron uniquely up to translation. This means that the correspondence ϕ between the classes of congruent parallel polyhedra with given normals ni on the one hand and the collections (F1 , . . . , Fm ) of the face areas of these polyhedra on the other is injective. The third condition, the continuity of ϕ, is obvious: under the variation of the polyhedron obtained by continuously shifting the planes of its faces, the face areas change continuously. We are left with the fourth condition: if the points k F k = (F1k , . . . , Fm ) (k = 1, 2, . . .)

of F are images of points P k in P and {F k } converges to F = (F1 , . . . , Fm ), then there is a point P in P whose image under ϕ is F and such that some subsequence P ki of {P k } converges to P . k In other words, if the face areas F1k , . . . , Fm of some polyhedra P k con4 verge to some positive numbers F1 , . . . , Fm , then there is a subsequence {P ki } of {P k } converging to a polyhedron P with face areas F1 , . . . , Fm . This condition is immediate from Lemma 2. Indeed, since the numbers Fik converge to the positive numbers Fi , all of them lie between some positive 4

k k By assumption, the collection (F1 , . . . , Fm ) belongs to F and so all the Fi must be positive. The equality n n F = 0 for the numbers Fi follows from the fact i=1 i i that this equality holds for the numbers Fik converging to Fi .

Minkowski’s Proof of Existence

317

m k bounds. Therefore, the areas of the polyhedra P k , equal to i+1 Fi , are k not larger than some F , while the area Fi of each face is not less than some f > 0. In this case, carrying out a parallel translation of the polyhedra P k so that the origin becomes an interior point of each of them (which is possible since we consider polyhedra up to translation), from Lemma 2 we infer that the support numbers of the polyhedra are uniformly bounded. We may thus choose a convergent subsequence from them. Therefore the planes converge to the faces of the corresponding polyhedra P ki , i.e., the polyhedra P ki themselves converge. Clearly, the limit polyhedron P has the limit values Fi of the face areas. Since all the conditions of the Mapping Lemma hold, we may apply it and see that to each admissible collection of numbers F1 , . . . , Fm corresponds a polyhedron with face areas F1 , . . . , Fm . This completes the proof of Minkowski’s Theorem. 7.1.5 Observe that our argument translates word for word to spaces of arbitrary dimension if we use the uniqueness theorem for polyhedra with prescribed face directions and face areas in n-dimensional space. This ndimensional uniqueness theorem will be proved in Section 8.3.

7.2 Minkowski’s Proof of the Existence of Polyhedra with Prescribed Face Areas 7.2.1. Here we reproduce Minkowski’s own proof of the existence theorem for a closed convex polyhedron with prescribed face directions and face areas. This proof is quite simple and compares favorably with that of the preceding section, since it does not rely on the corresponding uniqueness theorem. The proof remains valid in spaces of arbitrary dimension. However, unlike our method based on the Mapping Lemma, Minkowski’s proof uses the specifics of his theorem, and this makes the application of analogous ideas to other existence theorems a completely open problem. We return to this problem at the end of the current section. 7.2.2 We begin with some simple remarks. At the beginning of Section 7.1, we indicated that in Minkowski’s Theorem we may require that the given vectors ni do not point to a single closed half-space. Therefore, we assume that the given pairwise distinct unit vectors n1 , . . . , nm satisfy this condition. Let us draw rays li from the origin O in the directions of the vectors ni and consider the planes Qi perpendicular to li at positive distances hi from O. As shown in Lemma 1 of Section 7.1, these planes bound some closed solid polyhedron P . Since the rays li are fixed, the polyhedron P is completely determined by the distances hi from the planes Qi to the origin O. Generally

318


speaking, with hi arbitrary, not every plane Qi is tangent to P , so that not all the normals n1 , . . . , nm correspond to faces of P . We consider all such polyhedra P irrespective of whether or not all the normals ni correspond to faces of P . The volume of P is a function of the distances hi from the bounding planes to the origin O. Lemma. The volume V of P is a differentiable function of hi and we have ∂V ∂hi = Fi , where Fi is the area of the corresponding face of P or zero if the plane Qi produces no face on P . Suppose that the plane Qi produces a face of area Fi on P . Shift Qi parallel to itself by a small distance ∆hi , i.e., add an increment ∆hi to hi . If ∆hi > 0, then to the polyhedron we add a solid of height ∆hi which, up to higher-order infinitesimals, is a prism with base Fi . Hence, the principal term of the increment in the volume of the polyhedron is dV = Fi ∆hi . Similarly, if ∆hi < 0, then from the polyhedron we subtract a solid that differs slightly from a prism. Therefore, dV = Fi ∆hi in each case, and so ∂V = Fi . ∂hi If the plane Qi does not touch P at all, then a small displacement of Qi does not change P . Hence, in this case ∂V = 0. ∂hi Now, assume that Qi touches P in an edge L. Then in the shift of Qi away from the polyhedron, i.e., when ∆hi > 0, the polyhedron does not change and ∆V = 0.

Fig. 132

In the shift of Qi towards the polyhedron, i.e., when ∆hi < 0, from the polyhedron we subtract a bar whose length equals that of L and whose crosssection area is of order (∆hi )2 (Fig. 132). Therefore, the volume of the bar, i.e., the decrease in the volume of the polyhedron, is of order (∆hi )2 , implying ∂V that dV = 0. Hence, we obtain ∂h = 0 independently of the sign of ∆hi . i

Minkowski’s Proof of Existence

319

Finally, if the plane Qi touches P at a vertex, then we again have ∆V = 0 for ∆hi > 0, whereas for ∆hi < 0 from the polyhedron we subtract a pyramid of altitude |∆hi | whose volume |∆V | is of order |∆hi |3 . Therefore, in this case ∂V we also have ∂h = 0. i If the plane Qi produces no face on the polyhedron P , then it is natural to assume that such a face exists but has zero area. This agreement makes ∂V = Fi valid without further specification. the formula ∂h i 7.2.3 Now let us return to Minkowski’s Theorem itself. Let n1 , . . . , nm be 0 be given unit vectors not pointing to a single half-space and let F10 , . . . , Fm given positive numbers satisfying the condition m

Fi0 ni = 0.

(1)

i=1

Consider all polyhedra P bounded by the same planes Qi as in Subsection 7.2.2. Assume positive the distances hi from these planes to the origin. Among these polyhedra, distinguish those bounded by the planes Qi with distances hi to the origin satisfying the condition m

hi Fi0 = 1.

(2)

i=1

We now demonstrate that among the latter there exists a polyhedron of greatest volume. m If the numbers hi are sufficiently then i=1 hi Fi0 < 1. If the numm small, bers hi are large enough, then i=1 hi Fi0 > 1 because all the coefficients Fi0 are positive. It is then clear that, as the numbers hi increase continuously from small initial values, we shall arrive at condition (2). Hence, there are polyhedra P satisfying (2). The numbers hi satisfying (2) are bounded. Indeed, together with the Fi0 , they are all positive, so that hi < 1/Fi0 for all i. Therefore, the polyhedra P obeying (2) are uniformly bounded and there exists a least upper bound of their volumes. Since the volume of a polyhedron continuously depends on the polyhedron, i.e., on the numbers hi , it follows that this least upper bound is attained at some polyhedron P 0 . However, it may turn out for the polyhedron P 0 that not all h0i are positive, i.e., some h0i may vanish, implying that P 0 does not belong to the collection of polyhedra under consideration. We now show that the same maximum of the volume is also attained at polyhedra with hi > 0 for all i. To this end, observe that, shifting the bounding planes Q0i of P 0 by some vector a, we shift P 0 by the same vector a while preserving the volume of P 0 . Therefore, choosing the translation vector so that the origin is interior to the shifted polyhedron, we obtain a polyhedron P 1 with the same volume and all h1i positive. Let us prove that P1 belongs to the collection under consideration, i.e., that P1 satisfies (2).

320


Indeed, after the translation of the plane Q0i with normal ni by some vector a, the distance h0i from Qi to the origin changes to h0i +ni ai . Therefore, the sum (2) becomes (h0i + ni ai )F10 = h0i Fi0 + a ni Fi0 . i=1

But 0by 0 (1) we have i hi Fi = 0. Therefore,

i 0 i ni Fi

i

= 0, and since P 0 satisfies (2), we have

(h0i + ni ai )F10 = 1,

i=1

i.e., (2) also holds for the shifted polyhedron P 1 . Thus we have proved that, among the polyhedra P obeying (2) with all hi positive, there exists a polyhedron of greatest volume. By the Lemma of Subsection 7.2.2, the volume of the polyhedron P is a differentiable function of h1 , . . . , hm . So we deal with maximizing the differentiable function V (h1 , . . . , hm ) subject to (2). By the Lagrange multiplier rule, at a maximum point we must have ∂ V (h1 , . . . , hm ) + λ Fi0 hi = 0 ∂hi i

(i = 1, 2, . . . , m),

(3)

where λ is a real number.5 If the numbers Fi1 are the face areas of a polyhedron P 1 of greatest vol∂V = Fi1 (bearing in mind the convention ume, then by the Lemma we have ∂h i about faces of zero area). Then (3) implies the equalities 1 µFi1 = Fi0 (i = 1, . . . , m), where µ = − . λ

(4)

Equalities (4) mean that the face areas of P 1 are proportional to the given numbers Fi0 and so they are all positive. If we perform a homothety of P1 √ with coefficient µ, then the face areas of P1 become equal to µFi1 = Fi0 . √ So the polyhedron P 1 multiplied by µ has the given face areas Fi0 , which completes the proof of the theorem. Obviously, the above arguments remain valid in space of arbitrary dimen√ √ sion: it suffices to replace the coefficient of similarity µ by n−1 µ. 7.2.4 The above idea is used in many proofs of existence. For instance, Riemann applied it to differential equations. Suppose that we are to prove existence for some object A satisfying given conditions B0 . We choose such a function f of A for which the necessary extremum conditions lead in a purely formal way to the fact that if f (A) has an extremum at A = A0 , then A0 5

See, for instance, [Fi, Vol. 1, Section 202].

7.3 Existence of Unbounded Polyhedra with Prescribed Face Areas

321

satisfies the conditions B0 or, as in Minkowski’s Theorem, to some conditions related to B0 in a simple manner (equalities (4)). Then the problem is reduced to checking that f (A) actually attains a maximum (minimum) and that the formal necessary extremum conditions apply to it. For the case in which the object A is determined by finitely many parameters, as a polyhedron for example, the existence of a maximum (minimum) is guaranteed by the continuity of f , and the applicability of the extremum conditions follows from the differentiability of f . However, if the object A is specified by infinitely many parameters (or by a function), as an arbitrary convex body for example, then the proofs of these two assertions are much more involved by far. I generalized Minkowski’s Theorem to arbitrary convex bodies by exploiting exactly this idea [A3, III]. The statement of my result is given further in Subsection 7.6.2. Blaschke and Herglotz [BH] sketched a proof of Weyl’s Theorem (see Subsection 5.3.3) which was based on the same idea. However, they did not succeed in carrying it out: both the above assertions were left unproved.6 The cleverness of the researcher is in selecting an appropriate function f (A), given a domain of objects A and conditions B. An interesting, but apparently hard problem, is to find along these lines some proofs for other existence theorems on convex polyhedra, i.e., in each case find an appropriate function f (A) of the polyhedron. Success in this will make the method of maxima as general as our method of mappings. Here I can conjecture nothing about the solution of this problem. As regards the existence theorem for a polyhedron given a development, the only intimation is the article by Blaschke and Herglotz. However, that article deals with analytic surfaces and uses the techniques of Riemannian geometry. Notions of Riemannian geometry have yet to be ”translated” to the theory polyhedra. The problem of finding such a translation is of prime importance by itself.

7.3 Existence of Unbounded Polyhedra with Prescribed Face Areas 7.3.1 In this subsection and the next ones, we prove existence theorems for unbounded convex polyhedra with prescribed areas of bounded faces and prescribed planes of unbounded faces. The specification of the latter is equivalent to specifying what was called the unbounded part of a polyhedron in Section 6.4, i.e., what is left of the polyhedron after deleting all its bounded faces and arbitrarily large pieces of its unbounded faces. Two polyhedra have the same unbounded part if all their unbounded faces coincide completely in a sufficiently remote domain. Obviously, this holds if and only if the planes of these faces coincide. Conditions under which the given planes are the planes 6

This was done by Yu. A. Volkov, see [Vo2], [Vo4]. – V. Zalgaller

322


of faces of some unbounded polyhedron are clarified in Section 7.5 (Theorem 4 of Section 7.5; they were also formulated in Section 2.4). Here we assume in advance that the given planes are indeed the planes of faces of the unbounded part of some convex polyhedron. We shall thus speak of the existence of a polyhedron with given areas of bounded faces and given unbounded part. As shown in Section 1.5 of Chapter 1, the spherical image of an unbounded convex polyhedron is a convex spherical polygon, with the spherical images of the unbounded faces of the polyhedron lying on the boundary of the spherical polygon. The vertices of this polygon are the spherical images of faces with nonparallel unbounded edges. If all the unbounded edges of the given polyhedron are parallel to one another, then its spherical image is a hemisphere. This observation implies that, if we are given the unbounded part of a polyhedron, we know the boundary of its spherical image and thereby we know the spherical image itself, except for the case in which the boundary of the spherical image is a great circle, because such a circle divides the sphere into two hemispheres, both of which are convex. However, in this case, given the unbounded part of a polyhedron with parallel unbounded edges, we know the direction of these edges to infinity, and so the spherical image of such a polyhedron is the hemisphere to which the vector opposite to the direction of the unbounded edges points. Thus, if we know the unbounded part of a polyhedron then its spherical image is determined, and so we know to what part of the sphere the normals of the bounded faces must point. Here and elsewhere we have in mind the outward normals. Lemma. Let Q be the unbounded part of some convex polyhedron and let S be the spherical image of the polyhedron. Assume further that n1 , . . . , nm are arbitrary pairwise distinct unit vectors pointing to the interior of S. Then there exist convex polyhedra, with unbounded part Q, whose bounded faces have n1 , . . . , nm for their outward normals. Let P 0 be any unbounded polyhedron with unbounded part Q. Since the vectors n1 , . . . , nm point to the interior of the spherical image S of P 0 , it 0 follows that the polyhedron P 0 has support planes R10 , . . . , Rm with normals 0 0 n1 , . . . , nm . Moreover, the planes R1 , . . . , Rm do not touch P 0 in unbounded edges or faces: otherwise, the vectors ni would lie on the boundary of the spherical image S. Therefore, we can shift the plane R10 so far towards the interior of P 0 that it will cut away all the bounded faces of P 0 . As the result, we obtain a polyhedron P1 with the same unbounded part Q1 and only one bounded face, which we denote by G1 , with normal n1 . Since P 1 has the same unbounded part Q, the spherical image of P 1 is the same S. Hence, P 1 has support planes R2 , . . . , Rm with normals n2 , . . . , nm .


323

Shifting R2 towards the interior of P1 without cutting out the face G1 completely, we obtain a polyhedron P 2 that has one more face G2 , with normal n2 . Repeating the procedure, we come to a polyhedron all of whose bounded faces have the normals n1 , . . . , nm and there are no other bounded faces. 7.3.2 In this subsection, we consider polyhedra with parallel unbounded edges. The case of polyhedra with nonparallel unbounded edges is considered in the next subsection. The difference between the results for these two types of polyhedra is essential. Theorem. Assume given the unbounded part of a convex polyhedron with parallel unbounded edges. This part will be a semi-infinite prism Π. Let n be the unit vector opposite to the direction of these edges to infinity and let n1 , . . . , nm be arbitrary distinct unit vectors forming acute angles with n, so that the vectors ni point to the interior of one of the hemispheres bounded by the equator perpendicular to n, namely the one to which the vector n points. Finally, let F1 , . . . , Fm be positive numbers such that m (nni )Fi = F,

(1)

i=1

where F is the area of the cross-section of Π perpendicular to the unbounded edges. Then there exists a convex polyhedron, with unbounded part Π, whose bounded faces have outward normals ni and areas Fi (Fig. 68, p. 111). The necessity of the condition on the vectors ni was already established. The necessity of condition (1) on the numbers Fi is obvious: the inner product nni is the cosine of the angle between the plane of the face with normal ni and the plane of the cross-section of the prism Π. Therefore, (1) expresses the fact that the area F of the cross-section is the sum of areas of the projections of the bounded faces of the corresponding polyhedron. Also observe that, by what was proved in Section 6.4, a polyhedron with the data (Π, ni , Fi ) of the above theorem is unique up to parallel translation along its unbounded edges. We prove our theorem by using the Mapping Lemma. First, we consider all polyhedra P with the given unbounded part Π and given normals ni . Such polyhedra exist by the above lemma. Each of them is determined from the support numbers h1 , . . . , hm of its bounded faces. Since under small parallel displacements of the faces none of them disappears, the numbers h1 , . . . , hm may vary in sufficiently small neighborhoods of their values corresponding to any given polyhedron P . Therefore, the collection of all admissible values of the support numbers h1 , . . . , hm or, equivalently, the collection of all polyhedra P is represented by an open set P0 in m-dimensional Euclidean space with coordinates h1 , . . . , hm . However, we do not distinguish between polyhedra that are translates of one another along the edges of the prism Π, since none of the data Π, ni , or Fi

324


changes under such a translation. For this reason, we consider the manifold P of classes of polyhedra up to such translations, rather than the set P0 . Since each translation in the given direction depends on a single parameter, the translation magnitude, we thus exclude one variable. Hence, the manifold P is (m − 1)-dimensional. The rigorous definition of this manifold repeats that of the corresponding manifold in the proof of Minkowski’s Theorem in Section 7.1.7 We now consider the set of all collections of m positive numbers F1 , . . . , Fm satisfying the condition m (nni )Fi = F (1) i=1

of the theorem. Condition (1) is linear and hence it determines an (m − 1)dimensional plane in m-dimensional space with coordinates F1 , . . . , Fm . The condition that Fi > 0 for all i distinguishes the positive coordinate angle in m-dimensional space. The intersection of this angle with the plane determined by (1) is, obviously, a convex (m − 1)-dimensional set. It is our manifold F of the feasible values of F1 , . . . , Fm . By the Lemma, there exist polyhedra with normals ni and, as shown above, these polyhedra necessarily satisfy condition (1). Therefore the manifold F is nonempty, and since it is convex, it must also be connected. We thus have manifolds P and F of the same dimension m − 1. Assigning to each polyhedron, more precisely to each class of polyhedra, the areas F1 , . . . , Fm of its faces, we obtain a single-valued mapping ϕ from P into F. We must prove that ϕ satisfies the conditions of the Mapping Lemma. (1) Since F is connected, the condition that each connected component of F contains images of elements of P holds automatically. (2) Since, by what was proved in Section 6.4, a polyhedron with given unbounded part and face areas is unique up to translation along unbounded edges, to each collection (F1 , . . . , Fm ) corresponds a unique class of polyhedra in P. The mapping ϕ is thus injective. (3) The continuity of ϕ is obvious, because it amounts to the assertion that the continuous variation of the support numbers results in a continuous variation of the face areas. 7

If a is the translation vector, then, under the translation, the distance from the origin to a plane with normal ni increases by the summand ni a. Therefore, the support numbers h0i change to hi = h0i + ni a. Since we consider only translations in the direction of the given vector n, we have a = an. Hence, hi = h0i + (ni · n)a (i = 1, . . . , m). With hi fixed and a varying, each of these equations determines a straight line in m-dimensional space with coordinates hi . Since the coefficients of a are constant, all these straight lines are parallel to one another. The points of such straight line in P0 represent polyhedra that are translates of one another along n and so must be identified. Clearly, the identification consists in projecting P0 on an (m − 1)-dimensional plane along these straight lines. The projection of an open set is an open set. Hence, the identification yields an open set P in an (m − 1)-dimensional plane, i.e., an (m − 1)-dimensional manifold.


325

i (4) We are left with the fourth condition: if F i = (F1i , . . . , Fm ), the images i 0 0 0 of elements P of P converge to an element F = (F1 , . . . , Fm ), then there is a convergent subsequence of {P i } whose limit is P 0 and ϕ takes P 0 to F0 ; i in other words, if the face areas F1i , . . . , Fm of the polyhedra P i converge to positive limits, then there exists a subsequence of {P i } converging to a polyhedron with the limit face areas. Consider all polyhedra P with unbounded part Π and normals n1 , . . . , nm . Shift all these polyhedra along unbounded edges so that they have a common support plane R perpendicular to their unbounded edges. Then they all are included in the half-bounded prism Π, touching its base G on the plane R. We shall prove that then all their support numbers are uniformly bounded.

G

Rk

Figs. 133

Indeed, since all the polyhedra P lie in the prism Π, the planes of their faces cross Π. At the same time, if one of these planes Rk crosses Π far away, then Rk separates Π from its base G completely (Fig. 133). In that case, a polyhedron with such plane containing its kth face cannot touch G. Therefore, none of the planes Rk crosses Π far away, which means that the distances from these planes to the origin are uniformly bounded, i.e., the support numbers are bounded. But if the support numbers are bounded, then from every sequence of polyhedra P i we can extract a converging subsequence. The face areas of these polyhedra converge and the limit polyhedron has the limit face areas. So the last condition of the Mapping Lemma also holds. Thus, all the conditions of the Mapping Lemma are satisfied, implying its claim: each collection of numbers F1 , . . . , Fm in F corresponds to some class of polyhedra in P. This completes the proof of the theorem. 7.3.3 We now sketch one more proof of the same theorem. This proof relies on the existence of closed polyhedra with prescribed face normals and face areas. As before (see Fig. 68 on p. 111), let Π be an unbounded prism; n, the unit vector antiparallel to the edges of Π; n1 , . . . , nm , unit vectors forming acute

326


angles with n; F , the area of the cross-section of Π; and, finally, F1 , . . . , Fm , positive numbers satisfying the condition m (nni )Fi = F.

(1)

i=1

Also, let m1 , m2 , . . . , mk be normals to the faces of Π and let l1 , . . . , lk be the widths of these faces, i.e., the lengths of the sides of the cross-section of Π. From the fact that the polygon with sides l1 , . . . , lk is closed, it follows that k mj lj = 0. (2) j=1

It is easy to select numbers g1 , . . . , gk so as to satisfy the equality8 m

ni Fi +

i=1

k

mi gi − nF = 0.

(3)

(j = 1, . . . , k),

(4)

j=1

On putting Hj = g j + e j M

with M an arbitrary positive number, from (2) we infer m

ni Fi −

mj Hj − nF = 0.

(5)

i=1

Let us take M so large that all the numbers Hi will be positive. Since the vectors ni , mi , and −n do not point to a single half-space, there exists a closed convex polyhedron PM with face normals n1 , . . . , nm , m1 , . . . , mk , −n and face areas F1 , . . . , Fm , H1 , . . . , Hk , F . This is valid for every sufficiently large M > 0. The polyhedron PM consists of a “cap” with normals n1 , . . . , nm , a base with normal −n, and “lateral” faces with normals m1 , . . . , mk . The lateral faces are parallel to the faces of the prism Π. As M increases to infinity, the lateral faces of the polyhedron become larger. Since the areas of the other faces are fixed, the lateral faces may only become longer. The relative differences between their lengths will tend to zero, so that the ratios between their areas Hi become approximately equal to the ratios between their widths. It follows that the lateral faces of PM converge as M → ∞ to the lateral faces of the prism Π. Hence, if with the increase of M we move the “base” of PM to infinity, then in the limit we obtain an unbounded polyhedron P∞ with 8

Choosing the axes of a rectangular coordinate system to be n,a, and b, we see that by (1) we only need to satisfy two equalities: ani Fi + ami gi = 0 and the same equality with b replacing a. Since the number of faces of Π is k ≥ 3, we thus obtain two equations with at least three unknowns g1 , . . . , gk , so that these equations can be satisfied.

7.4 The General Existence Theorem for Unbounded Polyhedra

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unbounded part Π, with the given normals ni and given areas Fi of bounded faces. The above proof is of interest since it does not rely on the corresponding existence theorem and, together with the generalization of the existence theorem for a closed polyhedron, extends to spaces of arbitrary dimension. We leave it to the reader to carry out this generalization, assuming that the theorems on closed polyhedra are given.

7.4 The General Existence Theorem for Unbounded Polyhedra 7.4.1 In this section we consider unbounded polyhedra with nonparallel unbounded edges. For such polyhedra, we can prove a much more general theorem than the one simply stating the existence of polyhedra with prescribed areas of bounded faces and given unbounded part. Namely, instead of area, we can consider any function of a polygon satisfying the following conditions: (1) The function f (Q) is defined for all bounded convex polygons, including the limit cases of a straight line segment and a single point, and f takes the same values at polygons that are translates of one another. (2) The function f (Q) is monotone, i.e., if Q1 is contained in Q2 , then f (Q1 ) ≤ f (Q2 ).9 (3) The function f (Q) is continuous, i.e., if Qn → Q, then f (Qn ) → f (Q). (4) f (Qn ) → ∞ if the areas of the polygons Q1 , Q2 , . . . , Qn , . . . tend to infinity. Examples of such functions are area, perimeter, diameter (i.e., the greatest distance between the points of a polygon), radius of the circumscribed circle, etc. The last three functions do not vanish when the polygon degenerates into a segment. Perimeter, for instance, is then equal to twice the length of the segment. In this connection, let us agree that if the polyhedron has no faces with given normal n, then we assume that such face exists but degenerates into a straight line segment or a single point. Actually, degeneration into a point may be excluded, since each function f (Q) takes the same value at all single points and we can make this value vanish by subtracting it from the function f . Let Q0 be the unbounded part of some unbounded polyhedron with nonparallel unbounded edges. Taking the prolongations of the faces of Q0 , we 9

Monotonicity in this sense is weaker than that in Chapter 6. In that chapter it was required that f (Q1 ) < f (Q2 ) whenever Q1 is included in Q2 and differs from Q2 . For example, the diameter of a polygon is not a monotone function in this stronger sense (we may cut off a piece of a polygon without diminishing the diameter) but it is a monotone function in the sense of (2).

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obtain an unbounded convex polyhedron Q without bounded faces. Clearly, the specification of Q0 is equivalent to that of Q. Let R be a support plane of Q with normal n, not touching Q along an unbounded edge or along a face. Then R touches Q at a vertex or along a bounded edge. The value of f at the intersection R ∩ Q is referred to as the value of f at Q corresponding to the normal n. If f is area then this value certainly vanishes. If f is perimeter, then this value vanishes whenever R ∩ Q is a vertex, and, for R ∩ Q an edge, f (R ∩ Q) is equal to twice the length of the edge. Theorem. Let Q be an unbounded convex polyhedron without bounded faces. Let n1 , . . . , nm be vectors pointing to the interior of the spherical image of Q. Further, let f1 , . . . , fm be functions of a polygon satisfying the above conditions (1)–(3). Finally, let a1 , . . . , am be the respective values of these functions at Q corresponding to the normals n1 , . . . , nm . Then for arbitrary numbers b1 , . . . , bm such that a1 < b1 , . . . , am < bm there exists a convex polyhedron for which (1) the unbounded part of the polyhedron is the same as that of Q; (2) the bounded faces of the polyhedron have the normals n1 , . . . , nm ; (3) for each i (i = 1, 2, . . . , m), the function fi takes the value bi at the face with normal ni . By the agreement made before the statement of theorem, some faces may degenerate into edges.10 This theorem is due to Pogorelov, and Pogorelov’s proof is rather simple. Let the notations Q, ni , fi , and ai have the same meaning as in the statement of the theorem. Let us draw support planes to Q with normals ni and shift them towards the interior of Q. We then obtain a polyhedron whose bounded faces have normals ni (see the Lemma of Subsection 7.3.1). Under small shifts of the above-mentioned planes, the values of the functions fi at these faces are close to the numbers ai , which is clear from the definition of these numbers and the continuity of the functions fi . It now follows that, for arbitrary numbers bi > ai (i = 1, . . . , m), there exist polyhedra with the same unbounded part Q and bounded faces Qi with normals ni and such that fi (Qi ) ≤ bi

(i = 1, . . . , m).

(1)

Consider all the polyhedra P satisfying these conditions. Each of them is determined by the support numbers of its bounded faces. Therefore, the family of all polyhedra P is represented by a point set P in m-dimensional space 10

Example: suppose that each of the functions fi is perimeter. (In this case, by Theorem 2 of Section 6.4, the polyhedron with given Q, ni , and f (Qi ) = bi is unique.) Take an arbitrary unbounded polyhedron P and adjoin to its normals the normal n of a support plane through one of the edges of P . Considering this edge as a face, we obtain the polyhedron with some Q, ni , and fi (Qi ). By Theorem 2 of Section 6.4, a polyhedron with these data is unique. Consequently, there is no other polyhedron having the same data and a nondegenerate face with normal n.


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with coordinates h1 , . . . , hm . Inequalities (1) and continuity of the functions fi imply that this set is closed. We will show now that it is bounded. Indeed, assume the converse. Then there exist polyhedra P n for which the planes of the faces Qni cross Q in arbitrarily remote parts. However, the polyhedron Q expands at infinity and so such polyhedra P n must have arbitrarily large areas of some faces Qni . But then condition (4) imposed on the functions fi implies that fi (Qni ) → ∞, which contradicts inequalities (1). Hence, P is bounded. Since P is closed and bounded, among the polyhedra P there exists a polyhedron P 0 with least support numbers, i.e., for none of the polyhedra P satisfying conditions (1) we can have the inequalities hi ≤ h0i and the inequality hi < h0i at least once.11 In other words, none of the planes containing faces of P 0 can be shifted towards the interior of P 0 without violating (1). For P 0 the functions fi have the required values bi . Indeed, suppose for example that f1 (Q1 ) < b1 . Then this inequality remains valid under small shifts of Q1 . If we now shift Q1 inside the polyhedron, then the faces adjacent to Q1 become smaller and, by the monotonicity of the functions fi , condition (1) is still met for them. Hence, under small shifts of Q1 towards the interior of P 0 , we obtain a polyhedron P satisfying the same conditions (1). This, however, contradicts the minimality property of P 0 . Thus, a polyhedron P 0 with least support numbers yields the required values fi (Qi ) = bi , which completes the proof of the theorem. We see from the proof that the theorem can be generalized word for word to spaces of arbitrary dimension. 7.4.2 This theorem cannot be proved by means of the Mapping Lemma, for no uniqueness theorem corresponds to it. In the general uniqueness theorem established in Section 6.4, the functions fi must be strictly monotone (i.e., f (Q1 ) < f (Q2 ) whenever Q1 is contained in Q2 and differs from Q2 ), whereas here we only require nonstrict monotonicity. The following example demonstrates the absence of uniqueness under this weaker condition. Define the function f (Q) by setting f (Q) = 1 if the area of the polygon Q is less than 1 and setting f (Q) equal to the area of Q otherwise. Clearly, f (Q) satisfies all the conditions of the theorem, but there is no uniqueness theorem for such a function (i.e., in the case when all the functions fi coincide with f ): for all polyhedra with face areas less than 1, the values of f at faces are the same, namely f (Q) = 1. Let us require that the functions fi in the theorem satisfy two additional conditions: (a) the fi are strictly monotone; (b) if Q is a straight line segment, then fi (Q) = 0. 11

Clearly, P 0 need not be unique.

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By condition (b), all the numbers ai in Theorem 1 vanish. Therefore, the values bi assigned to the functions fi must obey the positivity condition and none of the faces of the polyhedron degenerates. Under such additional conditions, our theorem may be proved by using the Mapping Lemma. Let P be the manifold of polyhedra with given unbounded part Q and normals ni to the bounded faces. To such a P corresponds an open set in m-dimensional space with coordinates h1 , . . . , hm , where hi are the support numbers of the bounded faces. (Polyhedra with faces degenerating into edges are excluded; this is why P is now open.) In m-dimensional space with coordinates b1 , . . . , bm , the collections of positive numbers b1 , . . . , bm determine the positive coordinate angle, i.e., an open convex set. The manifold B of all B = (b1 , . . . , bm ) is m-dimensional and, by convexity, open. We assign to each polyhedron P the collection of numbers (b1 , . . . , bm ) by the following rule: the number bi is the value of the function fi at the face with normal ni . So we obtain a single-valued mapping ϕ from P into B. By the connectedness of B, the first condition of the Mapping Lemma holds automatically. The injectivity of ϕ follows from the uniqueness of polyhedra with given unbounded part Q, normals ni , and values of the functions fi (Theorem 2 of Section 6.4). The continuity of ϕ is obvious, since the continuous variation of support numbers makes the faces vary continuously and the functions fi are continuous by assumption. It remains to verify the last condition of the Mapping Lemma. In this case, it is as follows: let the numbers bj1 , . . . , bjm (j = 1, 2, . . .) be the values of the functions f1 , . . . , fm at the faces of polyhedra P j and suppose they converge to the positive numbers b01 , . . . , b0m . Then from the sequence of polyhedra P j we can extract a converging subsequence P t such that the values of the functions fk at the faces of the limit polyhedron are equal to b01 , . . . , b0m . Since the numbers bj1 , . . . , bjm converge, they are uniformly bounded. In much the same way as in the proof of the theorem above, this fact implies the boundedness of the support numbers of the polyhedra P j . Hence, we can extract a converging subsequence from the sequence of polyhedra P j . By the continuity of the functions fk , their limit values at the faces of the limit polyhedron are fk = b0k . So the last condition of the Mapping Lemma is satisfied, and its application completes the proof of the theorem. 7.4.3 The first proof of our theorem is of course subtler: first, it does not require the knowledge of a uniqueness theorem and, second, it uses nothing but elementary arguments and the basic properties of continuous functions. Therefore, the very theorem turns out almost as elementary as the Intermediate Value Theorem. It is in this aspect that unbounded polyhedra with nonparallel unbounded edges differ essentially from closed polyhedra and polyhedra with parallel


331

unbounded faces. For the latter we obtained much more restrictive theorems in which the functions fi coincide with area, whereas the proofs of these theorems are more involved in principle. Such a striking difference has a very apparent cause. In the case of unbounded polyhedra with nonparallel unbounded edges, the only condition on the preassigned values of the area is that of positivity. In the other two cases, extra conditions are needed. As regards closed polyhedra, such is the condition of being closed, i.e., m

Fi ni = 0.

i=1

By this condition, the number of coefficients Fi to be specified is m − 3 rather than m. This is not surprising, since a polyhedron with m faces is determined by m − 3 parameters up to translation! Therefore, to a closed polyhedron we may arbitrarily assign the values of the function f only for m − 3 faces. For the remaining 3 faces, the values of f must be determined from these m − 3 values. However, if f is taken to be the perimeter, there is no condition which, like (2), would enable us, given the perimeters of m − 3 faces, to specify the perimeters of the three others. The impossibility of finding and formulating such condition explains the absence of an appropriate generalization of the theorem on closed polyhedra. Certainly, such a condition could simply be assumed in the statement of the generalized theorem and then it would have much the same proof as the existence theorem of Section 7.1 for polyhedra with prescribed face areas. Unfortunately, we can exhibit no concrete example of a function of polygons which, while not being a function of area only, possesses the required property, i.e., given the values of such a function at all but three faces, we would be able to calculate the values of the function at these three faces as well. Without such an example, the general theorem is senseless, although we can formulate and prove it. The same applies essentially to a similar generalization of the theorem about unbounded polyhedra with parallel unbounded edges. The above discussion gives rise to the following natural problem: find examples of functions of polygons which are not functions of area alone and for which it is possible to define explicitly an (m − 3)-dimensional manifold of admissible values in the case of a closed polyhedron and an (m − 1)dimensional manifold in the case of an unbounded polyhedron with parallel unbounded edges and m bounded faces. I must confess that I see no way of solving this problem. And it may be that it is only functions of area that admit an explicit analytic definition for the manifolds in question, provided that we impose some simple natural conditions on the definition.12 12

Some progress in this direction is made in A. V. Pogorelov’s article [P9] – V. Zalgaller

332


7.5 Existence of Convex Polyhedra with Prescribed Support Numbers 7.5.1 When the origin of coordinates is fixed in space, the position of an arbitrary plane Q is determined by the unit normal vector n and the support number h. The equation of the plane Q is xn = h, where x is the position vector of a variable point on the plane. The half-space bounded by the plane Q with outward normal n is described by the inequality xn ≤ h, i.e., this half-space consists of the points whose position vectors x satisfy this inequality. When we say that a half-space is bounded by Q, we always have in mind the half-space with the outward normal n. Each convex polyhedron is the intersection of the half-spaces bounded by the planes of its faces (Theorem 2 of Section 1.2). Throughout the current section, by a convex polyhedron we mean a convex solid polyhedron, not excluding a priori the case of a doubly infinite prism (dihedral angles, slabs between pairs of parallel planes, and half-spaces will also be regarded as prisms). The main purpose of this section is to specify conditions under which the intersection of the half-spaces bounded by the given planes Qi is a convex polyhedron with Qi as the planes of its faces. In other words, we are interested in conditions to be imposed on the outward normals ni and support numbers hi so that ni be the outward normals and hi be the support numbers of some convex polyhedron. We need not impose any conditions on the normals ni : by Lemma 1 of Section 7.1, for arbitrary ni there exists a convex polyhedron with outward normals ni . If the vectors ni are coplanar, then the polyhedron is obviously a doubly infinite prism. If all ni point to a single half-space, then the polyhedron is unbounded. Otherwise, it is bounded. Thus, we must specify conditions to be imposed on the support numbers hi corresponding to given outward normals ni . The intersection of the given half-spaces may be empty. If the intersection is nonempty and has nonempty interior, then it is a convex polyhedron (Theorem 4 of Section 1.2). However, in this case it is possible that not all the boundary planes of the given half-spaces are planes of faces of the polyhedron: some of them may be disjoint from the polyhedron or touch it only along an edge or at a vertex. Therefore, our problem splits into two: (1) find conditions guaranteeing that the intersection P of the given half-spaces is nonempty and has nonempty interior and (2) find conditions guaranteeing that all the boundary planes of these half-spaces are planes of faces of the polyhedron P . The answer is given in the following two theorems. They deal with expansions of one of the normal vectors nk in terms of the others:

7.5 Existence of Convex Polyhedra with Prescribed Support Numbers

nk =

νki ni .

333

(1)

i

We consider only expansions in which (1) the vectors ni are linearly independent and (2) the coefficients νki are all positive or all negative. (The expansion of a vector relative to a set of linearly independent vectors is unique. Since the set of the given vectors ni is finite, the number of expansions to be examined is finite as well. At most three vectors may form a linearly independent set in space. Hence, we deal with expansions in terms of at most three vectors. Since the number of expansions (1) to be examined in Theorems 1 and 2 is finite, the number of conditions in the theorems is also finite, so that these conditions can always be effectively verified.) Theorem 1. The intersection of given half-spaces ni x ≤ hi (i = 1, . . . , m) has nonempty interior if and only if, for each normal nk admitting an expansion of the form (1) in terms of the other linearly independent normals ni with negative coefficients νki < 0, the corresponding support numbers satisfy the inequality νki hi . (2) hk > i

(If none of the vectors nk expands in terms of the others with negative coefficients, then the condition disappears, i.e., it holds automatically. In that case the half-spaces have common interior points. This is easy to see when all vectors ni point to the interior of a single half-space. At the same time, if the vectors do not point to the interior of a single half-space, then one of them always expands in terms of the others with negative coefficients. In the limit case, we “expand” in one vector: nk = −nj .) Theorem 2. If the intersection of the half-spaces ni x ≤ hi bounded by the planes Qi has nonempty interior, thus representing a convex polyhedron P , then the planes Qi are the planes of faces of P if and only if, for each normal nk admitting an expansionof the form (1) in terms of the other linearly independent normals ni with positive coefficients νki > 0, the corresponding support numbers satisfy the inequality13 hk < νki hi . (3) i

Theorems 1 and 2 readily solve the main question of the existence of a polyhedron with prescribed support numbers. 13

In Theorem 1 we can assume that among the vectors nj (j = 1, . . . , m) some may be equal (planes with parallel normals). However, this is prohibited in Theorem 2. If, for example, n1 = n2 , then the equality in question may be viewed as the expansion of n2 in terms of n1 , which would imply h2 < h1 , a contradiction.

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Theorem 3. In order that given unit vectors nj (j = 1, . . . , m) and given numbers hj be the outward normals and support numbers of some convex polyhedron, it is necessary and sufficient that the conditions of Theorems 1 and 2 be satisfied. To understand the conditions of Theorems 1 and 2 better, we give a direct proof of their necessity. 7.5.2 Theorem 1a. If the intersection of the half-spaces ni x ≤ hi has nonempty interior then, for every vector nk admitting an expansion with negative coefficients of the form (1) relative to the set of the remaining vectors, the corresponding support numbers satisfy inequality (2). (This assertion claims more than just the necessity of the conditions of Theorem 1, since in it we do not restrict our considerations to expansions relative to a set of linearly independent vectors ni , but admit arbitrary expansions with negative coefficients.) Proof. Let a be the position vector of one of the interior points of the intersection of the given half-spaces. Since the point in question belongs to the interiors of all these half-spaces, for all i we have ni a < hi .

(4)

Suppose that a vector nk expands in the other vectors with negative coefficients: nk = νki ni (νki < 0). (5) i=k

Multiplying (4) by the corresponding coefficients νki and recalling that an inequality reverses under multiplication by a negative number, we obtain the following inequality by summing νki ni a > νki hi i

i

or, in view of (5), nk a >

νki hi .

(6)

i

However, with i = k (4) yields nk a < hk . Therefore, (6) implies hk > νki hi , i

as required. Theorem 2a. If the planes Qi with normals ni and support numbers hi are the planes of faces of a convex polyhedron then, for every expansion of one of


335

the vectors nk relative to the set of the others with negative coefficients νki , the corresponding support numbers satisfy inequality (3), i.e., hk < i νki hi . (This theorem, similar to Theorem 1a, also claims more than just the necessity of the conditions of Theorem 2: here it is no longer required that the expansions use at most three vectors ni .) Proof. Let a be a vector issuing from the origin and ending at an interior point of the kth face of the given polyhedron P . The end of a then touches the plane Qk , so that (7) nk a = hk . At the same time, since the end of a belongs to the interior of the face, it belongs to no other plane Qi , i.e., it lies in the interiors of the half-spaces bounded by these planes. Hence, ni a < hi

for all i = k.

(8)

Suppose that the vector nk expands in some other vectors with positive coefficients: nk = νki ni (νki > 0). (9) i=k

Multiplying the corresponding inequalities (8) by the coefficients νki and summing the results, we obtain νki ni a < νki hi i

i

or, in view of (9), nk a
0, (2) h0 ≥ j νj hj , where hj are the support numbers of the faces with normals nj . (In particular, ifthe plane nx = h0 is a support plane of P , then we have the equality h0 = νj hj .) Let the convex polyhedron P lie in a half-space nx ≤ h0 . By Lemma 1, P then has a support plane Q with outward normal n. Let Q be given by the equation nx = h, so that nx ≤ h determines the half-space bounded by Q. Since n is the outward normal, the half-space contains our polyhedron P . At the same time, P lies in the half-space nx ≤ h0 . Hence, the latter also contains the half-space nx ≤ h. This implies the inequality h ≤ h0 .

(11)

If the plane Q touches P along a face and nj is the normal to this face, then obviously n = nj and h = hj , i.e., by (11) h0 ≥ hj . The relations n = nj and h0 ≥ hj demonstrate that in this case the claim of the lemma holds with the number of normals nj in the expansion of n equal to 1. Consider the general case. Let Q touch P at some point A, irrespectively of whether it is a vertex, or a point on an edge, or a point on a face. Let W be the solid angle spanned by the rays issuing from A along the outward normals to the support planes at A. As was mentioned in Section 1.5, W is a convex polyhedral angle with edges going along the normals of the faces containing to A. If A belongs to the interior of an edge, then W is a flat angle. If A is in the interior of a face, then W degenerates into a ray. The normal n of the support plane Q belongs to the angle W . If W is not a flat angle then, splitting it into trihedral angles by diagonal planes, we see that n belongs to some trihedral angle with edges going along the normals nj1 , nj2 , and nj3 to the faces of the polyhedron. Clearly, each such vector expands in nj1 , nj2 , and nj3 with nonnegative coefficients: n = νj1 nj1 + νj2 nj2 + νj3 nj3

(νj1 , νj2 , νj2 ≥ 0).

(12)

If W degenerates into a flat angle or ray then we arrive at a similar expansion in terms of at most two vectors. If we admit zero values for νji , these cases are included (12).


337

Thus, the required expansion of n has been obtained and it remains to prove the inequality h0 ≥ νj1 nj1 + νj2 nj2 + νj3 nj3 .

(13)

Let a be the position vector of the point A at which the plane Q touches the polyhedron P . The point A belongs to the planes of the faces containing it, as well as to the plane Q; therefore, a satisfies the equations of these planes nj1 a = hj1 ,

nj2 a = hj2 ,

nj3 a = hj3 ,

and na = h.

(14)

Multiplying the first three equations by the corresponding νjp , summing the results, and using (12), we find that hjp νip = a νjp njp = an = h. (15) p

p

Since h ≤ h0 , (15) implies (13), and the lemma is proved. (The last claim of the lemma (the one in parentheses) follows from (15), since if the plane nx = h0 is a support plane, then it coincides with Q, i.e. h = h0 .) 7.5.4 We now prove that Lemma 2 implies the sufficiency of the conditions of Theorem 2, i.e., we prove the following Theorem 2b. If a polyhedron P is the intersection of the half-spaces ni x =hi (i = 1, . . . , m), then, under the conditions of Theorem 2 (i.e., nk = i νki ni with linearly independent vectors ni and all νki positive implies hk < i νki ni ), each plane ni x = hi determines a whole face on P . Assume that P has no faces with normal nk . Since, by the very definition, nk lies in the half-space nk x ≤ hk , it follows from Lemma 2 that there are linearly independent vectors nj which are the face normals of P and satisfy14 nk = νkj nj (νj > 0), (16) j

hk ≥

νkj hj .

(17)

j

However, since (16) holds, the condition of Theorem 2 implies the converse: hk < νkj hj . j

This contradiction shows that all the vectors ni are face normals of P , which completes the proof of the theorem. 14

Here we exclude the exceptional case of Lemma 2 in which the vector n = nk “expands” in only one normal vector, since nk is not a face normal by assumption.

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7.5.5 We now prove the sufficiency of the conditions of Theorem 1. Theorem 1b. If the numbers hi assigned to the vectors ni satisfy the conditions of Theorem 1 (i.e., nk = i νki ni with the vectors ni linearly independent and all the coefficients νki negative implies hk > i νki hi ), then the intersection of the half-spaces ni x ≤ hi has a nonempty interior. If we are given only one vector ni , then the claim of the theorem is trivial. So we shall prove the theorem by induction on the number of the vectors ni , now assuming that the claim is valid for m − 1 vectors. Let n1 , . . . , nm and h1 , . . . , hm be vectors and numbers satisfying the conditions of the theorem. By the induction assumption, the intersection of the m − 1 half-spaces ni x ≤ hi (i = 1, . . . , m − 1) has a nonempty interior. This intersection is a convex solid polyhedron P , which may be a prism or even a half-space. Suppose that the intersection of all m half-spaces satisfying ni x ≤ hi , i ≤ m, i.e., the intersection of P with the mth half-space satisfying nm x ≤ hm has empty interior or even is empty itself. Clearly, this means that P has no points in the interior of the mth half-space given by nm x ≤ hm . Since P has no points in the interior of the half-space given by nm x ≤ hm , it lies in the complementary half-space nm x ≥ hm or, which is the same, −nm x ≤ −hm . This complementary half-space has the outward normal −nm and the distance from its bounding plane to the origin is −hm , and since it contains P , we can apply Lemma 2. By this lemma, there are at most three linearly independent vectors nj such that −nj = νmj nj (νmj > 0), (18) j

−hm ≥

νmj hj .

(19)

j

However, since all coefficients νmj are positive, (18) at the same time yields an expansion of nm in terms of nj with negative coefficients −νmj . But then the conditions of the theorem imply −νmj hj . (20) hm > i

This contradicts (19). Thereby the assumption that the intersection of all m half-spaces ni x ≤ hi has empty interior is false, and this concludes the proof of Theorem 1b.


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7.5.6 Theorems 1 and 2 readily yield conditions that characterize the unbounded part of a convex polyhedron: Theorem 4. The planes ni x = hi are the planes of the unbounded faces of a convex polyhedron with outward normals ni if and only if the following conditions hold: (1) the numbers hi satisfy the inequalities of Theorems 1 and 2, (2) the vectors ni point to a single half-space and, if applied to the same origin, none of them is contained in the interior of the convex polyhedral angle that they span or, equivalently, the ends of all these vectors lie on the boundary of the convex spherical polygon spanned by them; there must be at least three vectors ni , because an unbounded polyhedron (not reduced to a dihedral angle) must have at least three unbounded faces. The necessity of (1) follows from the necessity of the conditions of Theorems 1 and 2, whereas the necessity of (2) is the consequence of the corresponding properties of the spherical image of an unbounded polyhedron (Subsection 1.5.3). The sufficiency of these conditions can be proved along the same lines. Namely, under condition (1), Theorem 3 implies the existence of a polyhedron P for which the planes ni x = hi are the planes of the faces. By condition (2), the vectors ni point to a single half-space. Therefore, the polyhedron P is unbounded. However, none of its faces is bounded, since the spherical image of a bounded face lies in the interior of the spherical image of the polyhedron. The conditions of Theorem 4 may be restated in a form admitting especially simple verification for given vectors ni and numbers hi . (1) For all the vectors ni to point to a single (closed) half-space, it is necessary and sufficient that none of them expand in any three other noncoplanar vectors with negative coefficients. (For all of these vectors to point to the interior of a single half-space, it is necessary and sufficient that none of them expand in any number of the others with negative coefficients.) (2) In order that none of the vectors ni , pointing to a single half-space, belong to the interior of the polyhedral angle spanned by the other vectors, it is necessary and sufficient that none of the vectors ni expand in any three other noncoplanar vectors with positive coefficients. (3) If follows from (1) and (2) that, in Theorem 4, we need to check the conditions of Theorems 1 and 2 only for expansions of ni in terms of at most two of the other vectors. Assertion (1) is proved by induction on the number of vectors ni . Assertion (2) is obvious, because if a vector n belongs to the interior of a polyhedral angle V , then n belongs to one of the trihedral angles into which V splits and so n expands in the vectors along the edges of this trihedral angle with positive coefficients.

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7.5.7 Remarks and exercises. (1) All results of the current section can be extended word for word to spaces of arbitrary dimension. The only difference is that n vectors may be linearly independent in n-dimensional space. Making the corresponding changes in the proofs causes no difficulties. (2) For the intersection P of given half-spaces ni x ≤ hi , the following four cases are possible: (a) P is empty; (b) P is nonempty but its interior is empty; (c) P has nonempty interior and is therefore a convex polyhedron, but not all bounding planes Qi of the half-spaces are the planes of faces of P ; (d) P is a convex polyhedron and all planes Qi are the planes of faces of P . Theorems 1 and 2 yield conditions characterizing the cases (c) and (d). Prove that the cases (a) and (b) are characterized by the following necessary and sufficient conditions and clarify the meaning of these conditions: (a) Among the vectors ni , there is a vector nk that can be expanded in some of the other vectors with negative coefficients: nk = i νki ni (νki < 0), and the corresponding support numbers satisfy the inequality hk < νki hi . (a) (b) None of the inequalities (a) holds but there is a vector nk possessing an expansion of the form nk = νki ni with negative coefficients and the corresponding support number satisfies the equality hk = νki hi . (b) i

These two theorems, taken together with Theorems 1 and 2, allow us to separate the cases (a)–(d). (3) Specify the geometric significance of the cases in which one or several of the inequalities (2) and (3) in Theorems 1 and 2 become equalities. (4) (a) Prove that the given vectors do not point to the interior of a single half-space if and only if at least one of them expands in some of the others with negative coefficients. (b) Prove that the given vectors do not point to a single half-space (not necessarily to its interior) if and only if at least one of them expands in other three linearly independent vectors with negative coefficients. (c) Prove that if a vector n expands in the vectors n1 , . . . , nm (n = λ1 n1 + . . . + λm nm ) with negative (positive) coefficients, then among the vectors n1 , . . . , nm there are linearly independent vectors in which n expands with coefficients of the same sign. (5) Helly’s Theorem. By Theorem 1, a vector nk expands in three linearly independent vectors with negative coefficients νki and hk > νki hi if the corresponding four half-spaces have common interior points. At the same time, by Theorem 1, the validity of these conditions is sufficient for an


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arbitrary number of half-spaces to have common interior points. This readily yields the following theorem: If among finitely many given half-spaces, every four have common interior points, then all the half-spaces have common interior points. Since every convex polyhedron is the intersection of finitely many halfspaces, the following theorem is immediate: If among finitely many convex polyhedra every four have common interior points, then all the polyhedra have common interior points. The same holds in spaces of arbitrary dimension n if we replace the number 4 by n + 1. It is easy to deduce from the above that if, among finitely many convex polyhedra, every four, or n+1, have common points (not necessarily interior), then all polyhedra have at least one common point. Further, if among finitely many convex solids every four, or n + 1, have common points, then all the solids have at least one common point. (It suffices to approximate the given solids by enveloping convex polyhedra closer and closer and the theorem will then follow from the preceding.) The same result can be proved for any infinite set of bounded convex solids: If every four, or n + 1, of these solids have common points, then all the solids have at least one common point. This is Helly’s Theorem. A very simple proof of it may be found in Helly’s article [Hel]. 15 However, the crucial part of the proof resides in the case of finitely many solids. It is easy to see that such a general theorem fails for unbounded convex bodies. (6) Theorem 3 may be proved by using the Mapping Lemma. In this case, the Mapping Lemma is reduced to the following evident assertion: if an open set P of m-dimensional Euclidean space is contained in another open connected set H of the same space and if P is closed relative to H, then P = H. This claim follows from the definition of connectedness (Subsection 4.8.2). For the set H we take the set of points in m-dimensional space whose coordinates h1 , . . . , hm satisfy the conditions of Theorems 1 and 2 for the given vectors n1 , . . . , nm . For the set P we take the set of points (h1 , . . . , hm ) for which the intersection of the half-spaces ni x ≤ hi is a convex polyhedron. We suggest to the reader to carry out such proof of Theorem 3. It involves no difficulties in principle, although its rigorous implementation requires working out some details of interest in their own right.

15

See also [DGK]. – V. Zalgaller

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7.6 Generalizations 7.6.1 All the theorems of the current chapter, as well as their proofs, can be carried over to Euclidean spaces of arbitrary dimension. Further, all of them may be generalized to arbitrary convex bodies in spaces of arbitrary dimension by passing to the limit. Such generalizations presuppose, however, the introduction of general notions for arbitrary convex bodies which replace their elementary counterparts for polyhedra. A convex body is now determined not by finitely many support numbers but by its “support function,” the function introduced and studied by Minkowski [Min2].16 Let ux = H(u) be the equation of the support plane of a convex body H, where u is the outward normal; here we do not require u to be a unit vector. If u = n is a unit vector, then H(u) = h(n) is the distance with appropriate sign from the origin to the plane in question. Clearly, u , H(u) = |u|H |u| where u/|u| is now a unit vector. The function H(u), defined for all vectors u, is the support function of H. (Strictly speaking, H(u) is not defined at u = 0, but it is natural to put H(0) = 0.) Clearly, H(u) is completely determined from its values at the unit vectors n, which are often convenient to view as points of the unit sphere). The following theorem holds: A function H(u) is the support function of a convex body (possibly degenerating into a flat domain, straight line segment, or single point) if and only if the following two conditions are satisfied: (1) H(λu) = λH(u) for all λ ≥ 0 (the “positive homogeneity” condition); (2) H(u + v) ≤ H(u) + H(v) for all u and v (the convexity condition). A convex body with support function H(u) is nondegenerate if and only if (3) for every noncoplanar vectors u1 , u2 and u3 ui . H(ui ) > −H − (Condition (3) is similar to the condition of Theorem 1 of Section 7.5.17 ) Minkowski was the first who proved this theorem. Many authors have reproved it since then by various methods. The theorem can be derived from Theorem 1 of Section 7.5 by passing to the limit. It corresponds to Theorem 3 of Section 7.5, which deals with support numbers, since each support number 16 17

See also [BF]. See the next footnote. It suffices to substitute n for − i ui and replace ui by νi ni with νi < 0.

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is the value of the support function at the normal to the corresponding face. The support function of a convex polyhedron is piecewise linear and so it is completely determined from its values at the normals. Namely, this function is linear in each solid angle spanned by the normals to the support planes containing a fixed vertex. (If a is the position vector of a point A, then the equation of the plane passing through A with normal u is ux = ua, implying that the support function of the point A is H(u) = au, which yields our assertion about the support function of a polyhedron.)18 The conditions of Theorem 3 of Section 7.5, i.e., the conditions of Theorems 1 and 2 considered together (and admitting the equality sign in the inequalities), follow easily from the conditions of the above-stated theorem about support functions.19 However, Theorem 3 of Section 7.5 cannot be treated as a particular instance of the above theorem, at least because, given the numbers hi assigned to the vectors ni , in general we can construct several piecewise functions H(u) with H(ni ) = hi . (Such a function is determined by a decomposition of space into trihedral angles with edges along the vectors ni . Since the function is piecewise linear, in each angle it is determined from its values on the edges of the angle. However, a decomposition of space into trihedral angles with edges along ni is not unique in general. The question now arises of distinguishing a partition for which the function constructed in this way meets the convexity condition.) 7.6.2 Generalizations of the theorems of Sections 7.1–7.4 to arbitrary convex bodies in n-dimensional Euclidean space are formulated in terms of the area function of a convex surface. The definition of this function was given in Subsection 2.7.2. We recall that the area function is a set function on the unit sphere E. It is defined for all (Borel) subsets of E contained in the spherical image of the given convex surface. If the surface is closed, then so are all subsets of E. If the surface is unbounded, then its spherical image S lies in a hemisphere. In that case, the area function must take finite values on sets whose closures belong to the interior of S; its values on other sets may be infinite. 18

See [Ver6] about the specification of a convex surface by its support function in hyperbolic space. – V. Zalgaller 19 Assume that n = i νi ni with all νi < 0. Then by condition (2) H(−n) = H(−νi ni ) ≤ H(−νi ni ), while by condition (1) H(−νi ni ) = νi H(ni ) provided −νi > 0. Hence, H(−n) ≤ −νi H(ni ). (∗) i

By condition (2) H(−n) + H(n) ≥ H(0). However, H(0) = 0 (by condition (1), in which we should with (∗) put λ = 0). Hence, H(−n) ≥ −H(n), which together νi hi , which yields H(n) ≥ i νi H(ni ) or, in the notations of Section 7.5, h ≥ is exactly condition (2a) of Theorem 1 of Section 7.5. To derive the condition of Theorem 2 is even easier.

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The Generalized Minkowski Theorem. A set function F (M ) on the unit sphere (defined for all Borel subsets of E) is the area function of some bounded convex body if and only if F (M ) satisfies the following conditions: (1) F (M ) is finite, nonnegative, and countably additive. (2) E nF (dM ) = 0, with n the unit vector corresponding to a point on E. (3) E (n0 n)F (dM ) > 0 for every unit vector n0 . The second condition asserts that the vector area of any closed surface vanishes. The third means that the area of the projection of a convex body in any direction does not vanish. If a function F (M ) is a combination of finitely many point masses,20 then this general theorem is reduced to the theorem on polyhedra. The face normals will correspond to the points of the sphere at which the masses are located. In this case the third condition means that the normals are noncoplanar. The simplest proof of this theorem is obtained by passing to the limit from polyhedra.21 Generalization of the Theorem of Section 7.3. Let G be a bounded convex domain (with boundary excluded) on a plane T , let n0 be a normal to T , and let F be the area of G. A set function F (M ) on the sphere E, defined for all Borel sets lying in the open hemisphere E(n0 ) with center n0 , is the area function of an unbounded convex body H whose projection onto T covers G if and only if the following conditions are satisfied: (1) F (M ) is nonnegative, countably additive, and finite at every set M whose closure lies in E(n0 ). (2) E(n0 ) (n0 n)F (dM ) = F . The convex body H has a ray in the direction n0 as its limit cone. The projection of H, while covering G, is included in the closure of G. Since F (M ) is defined only in the open hemisphere E(n0 ), the support planes parallel to n0 are excluded from consideration, although H may have such planes, e.g., if it is a semi-infinite cylinder. The value of F (M ) at the whole hemisphere E(n0 ) is nothing but the area of the part of H which has no support planes parallel to n0 . When F (M ) is a combination of finitely many point masses, this general theorem is reduced to the theorem of Section 7.3 about polyhedra. Generalization of the Theorem of Section 7.4. Let M0 be an (open) convex domain on the sphere which differs from a hemisphere. On the boundary of M0 , let the given function h0 (n) be defined as the distance from the support planes of some convex body H0 to the origin (its values equal those 20

That is, there are points n1 , . . . , nm of the sphere such that F (M ) = 0 if M contains none of these points and F (M ) > 0 if M contains at least one of them. 21 See [A5]. Another direct proof of the theorem in its general form, including the case of polyhedra, appears in my article [A3, I].

7.6 Generalizations

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of the support function of H0 at the points on the boundary of M0 ). Finally, let F (M ) be a nonnegative countably additive set function defined for all Borel subsets of M0 and finite for any M whose closure belongs to M0 . Then there exists an unbounded convex body H such that (1) the spherical image of H covers M0 and is included in the closure of M0 ; (2) the values h(ni ) of the support function of H converge to h0 (n) whenever the points ni converge to the point n on the boundary of M0 ; (3) F (M ) is the area function of H (but only outside the region of H where the normals to support planes point to the boundary of M0 ). Here the body H0 plays the role of the unbounded part of H. The body H may consist of a bounded part which is, so to speak, put on its unbounded part common with H0 . The spherical image of this unbounded part then determines the boundary of M0 . In this case the function F (M ) takes a finite value at the whole domain M0 . However, the body H may fail to have such an unbounded part but rather approach it asymptotically. Proofs of the generalizations of the theorems of Sections 7.3 and 7.4 (still unpublished at present) can be carried out by passing to the limit from polyhedra. 7.6.3 The last theorem is less general than the Theorem of Section 7.4 in the sense that here F (M ) is area, whereas in the Theorem of Section 7.4 we consider arbitrary monotone functions. We do not know a full generalization of the Theorem of Section 7.4. For convex bodies with regular surfaces, we put forward the following theorem. Let M0 be a convex domain on the sphere other than a hemisphere. Let h0 (n) be the values of the support function of some convex body H0 at the points n on the boundary of M0 . Let f (x, y; n) be a function depending on a variable point n in M0 and two numerical variables x and y, which is defined in the region x ≥ y and is monotone (i.e., f (x1 , y1 ; n) > f (x2 , y2 ; n) whenever x1 ≥ x2 , y1 ≥ y2 , and the equality fails at least once). Moreover, f is “sufficiently regular” (continuous or even analytic). Then, for every “sufficiently regular” function g(n) given on M0 , there exists an unbounded convex surface H such that (1) the spherical image of H covers M0 and is included in the closure of M0 ; (2) the limit values of the support function of H on the boundary of M0 coincide with the function h0 (n); (3) if n is the normal at a point X of H and R1 ≥ R2 are the radii of principal curvature at X, then f (R1 , R2 ; n) = g(n). This theorem reduces to the solvability of a certain differential equation. Indeed, the principal curvature radii are expressed in terms of the support function h(n) and its first- and second-order derivatives. Therefore, condition (3) is equivalent to the validity of some second-order partial differential equation. The theorem then asserts that such an equation always has a solution in the form of a “convex function” h(n) taking preassigned values h0 (n) on the

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boundary of M0 , provied, of course, that these values satisfy the necessary convexity condition. With the general theorem of Section 7.4 available, the method of approximation by polyhedra inspires the hope that such a theorem does hold. However, the proof eludes us. Going from polyhedra, we encounter an obstacle in “approximating” the function f (x, y; n) by some function of polygons (the faces of the polyhedron). The case of area corresponds to f (x, y; n) = xy, since the product R1 R2 is exactly the limit of the ratio of the area of a domain on the surface to the area of its spherical image. In this case the “approximation” is obvious. It would be extremely interesting to prove the general theorem put forward, at least under some additional constraints on the function f , first of all for the case in which f is independent of n. 7.6.4 No results similar to the theorems of the current chapter are known for polyhedra in hyperbolic or spherical space. Seeking such results is possible along the lines shown at the end of the preceding chapter, namely by replacing the notion of face direction by some notion not relying on parallelism in the Euclidean sense. For example, assume given in hyperbolic space a point O and straight lines l1 , . . . , ln passing through O. Let positive directions be distinguished on these lines so that the lines do not point to a single half-space. Consider the polyhedra P bounded by planes Qi perpendicular to li so that the positive directions on the lines point outside the half-spaces including P . Just as we did for the theorems of Sections 7.5 and 7.1, we can raise two questions: (1) Let h1 , . . . , hm be the distances from the planes Q1 , . . . , Qm to the point O. What are necessary and sufficient conditions on the numbers hi for the planes Qi to be the planes of faces of some polyhedron P ? (Certainly, each distance hi is taken with the appropriate sign.) (2) What are necessary and sufficient conditions on numbers F1 , . . . , Fm so that there exist a polyhedron P with face areas Fi ? Most likely, the first question can be answered without great effort. At the same time, the second seems rather hard. Similar problems can also be stated for unbounded convex polyhedra. Complications arise in considering polyhedra with several unbounded parts. The existence of such convex polyhedra in hyperbolic space was shown in Subsection 3.6.4. 7.6.522 A special class of convex polyhedra in hyperbolic space Λn consists of the fundamental domains of discrete groups of isometries of Λn generated by reflections in hyperplanes [Vi]. In this connection, Andreev [An1, An2, An3] studied convex polyhedra in Λn such that every dihedral angle between adjacent (n − 1)-dimensional faces is at most π/2. He proved [An1] that if 22

Subsections 7.6.5–7.6.7 are comments by V. Zalgaller.

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n ≥ 3, then all vertices of such a polyhedron are of simplicial type and that dihedral angles in every face do not exceed π/2 either. In the same article, Andreev elaborated the necessary system of linear inequalities connecting dihedral angles in a bounded convex polyhedron in Λ3 with dihedral angles ≤ π/2. This system of inequalities depends on the combinatorial structure of the polyhedron. For two polyhedra of this type with the same combinatorial structure, the equality of the corresponding dihedral angles implies the congruence of the polyhedra. Given an admissible abstract combinatorial structure and dihedral angles at edges, the validity of the above-mentioned system of inequalities suffices for the existence of a convex polyhedron in Λ3 having this structure and those dihedral angles. Thus, the combinatorial structure and a system of linear inequalities are necessary and sufficient conditions for the existence and uniqueness (up to a motion) of a convex polyhedron of this class in Λ3 . 7.6.6 In view of Gauss’s theorem for smooth surfaces, the specification of the area function is equivalent to the specification of the Gaussian curvature as a function of the normal n. In its original statement, Minkowski’s problem exactly consists in finding out when a convex surface of given Gaussian curvature K(n) as a function of the normal exists. Curiously, the area of the spherical image was the primary notion for Gauss. Curvature was regarded as a point function rather than a set function. The above general solution of Minkowski’s problem opens the way for further generalizations. However, if K(n) is a given smooth point function, then the question of the existence of a sufficiently smooth solution to Minkowski’s problem remains. There are two approaches to solving this problem: study the degree of smoothness of a weak solution in dependence on the smoothness of K(n) or reduce the problem to the question of existence of a classical solution to the corresponding Monge–Ampère equation. In both cases, the “smooth” Minkowski problem is reduced to deriving a priori estimates for solutions to the above equations; see the books [P10, Chapter 7] and [P11] and the articles [N], [Gl2], [Sh2], and [CY]. In [P10] the smooth Minkowski problem in E 3 was generalized to the case of f (R1 , R2 ) = ϕ(n), with ϕ(n) a given function for f of a rather general form. The results of the articles [Me6], [Me7], and [Mir] pertain to this problem. See [So1], [So2], and [GIS] for the Minkowski problem in pseudo-Euclidean space. 7.6.7 The smooth Minkowski problem in E n+1 admits a generalization in −1 which the function (R1 . . . Rn ) = ϕ(n) is replaced by another symmetric function Sk = iα =iβ Ri1 . . . Rin of the radii of principal curvature. That is, does there exist a convex hypersurface in E n+1 for which we have Sk (n) = ϕ(n) at points with outward normal n, where ϕ(n) > 0 is a given function on the unit hypersphere S n ?

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For each k > 1, this problem is reduced to solving a nonlinear equation on S n in the support function of the required hypersurface. For k = 1, this is the well-known Christoffel problem. It reduces to solving the linear differential equation ∆2 p + np = f (n), where p(n) is the required support function and ∆2 is the Laplace–Beltrami operator on S n . See [P10, Chapter 7], [DM], [O1], [Sap], [Sa], and [Sov] about the solution of the Christoffel problem and its generalizations. All the above problems admit generalizations to the case of elementary symmetric functions of the so-called “relative” (conditional) principal curvature radii; see [BP], [Ko1], and [Ko2].

8 Relationship Between the Congruence Condition for Polyhedra with Parallel Faces and Other Problems

In this chapter, by a polyhedron we mean a bounded convex solid polyhedron. The general congruence condition for polyhedra with parallel faces of Section 6.3 has essential applications to some problems which seem totally irrelevant at first sight. To be more precise, the main applications we have in mind relate to Minkowski’s Uniqueness Theorem for polyhedra with parallel faces of equal areas.1 The first section of the chapter addresses the tiling of space by congruent polyhedra and has no connections with the next sections. Sections 8.2 and 8.3, on the contrary, address the same class of problems which, in the general setting for arbitrary convex bodies, serves as the subject of an elegant theory rich in corollaries, the so-called “theory of mixed volumes.”

8.1 Parallelohedra 8.1.1 A parallelohedron is a convex polyhedron whose translates tile all of the space, fitting together along entire faces. The simplest examples are the cube and the regular hexahedral prism. The problem consists in finding all possible parallelohedra. This first implies finding all possible types of structures of parallelohedra and, second, describing the metric characteristics for such types which, if enjoyed by a polyhedron, ensure that it is a parallelohedron. This problem is interesting not only in itself but also by its connections with crystallography and number theory. It was first solved by the great Russian crystallographer E. S. Fedorov in 1890 [Fe]. However, Fedorov’s solution had an essential gap: Fedorov never proved that each parallelohedron is centrally symmetric, although he relied on this fact. Minkowski filled the gap by applying his uniqueness theorem for polyhedra with parallel faces of equal areas. It is known that Minkowski found his general uniqueness theorem in his 1

This remark is essential because Minkowski’s Theorem is valid in space of arbitrary dimension. Therefore, the corollaries derived here have the same degree of generality.

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8 Relationship Between the Congruence Condition and Other Problems

search for a proof of the fact that each parallelohedron is centrally symmetric, i.e. has a center. We begin by proving the existence of a center for any parallelohedron. We then exhibit the types of parallelohedra by an argument due to B. N. Delaunay. Delaunay’s argument exemplifies a beautiful geometric construction.2 8.1.2 Theorem 1. If all faces of a convex polyhedron are pairwise translates of one another, then the polyhedron, as well as each of its faces, is centrally symmetric.

n

G1

G X

O

X

n G1

G

Fig. 134

Let P be a polyhedron satisfying the assumption of the theorem. Assume further that P is a polyhedron symmetric to P about some point O. Consider two faces G and G1 of P which are translates of one another. Let G and G1 be the corresponding symmetric faces of P . Clearly, G and G1 (as well as G1 and G ) have parallel outward normals (Fig. 134). Furthermore, G and G1 have the same area. Now Minkowski’s Theorem implies that P and P themselves are translates of one another. But the following statement is easy to prove: If a figure F and one of its translates F are symmetric to one another about some point, then F has a center of symmetry.3 Hence, P is centrally 2 3

See [DPA] for the relation of parallelohedra to geometric crystallography. The proof is clear from Fig. 135. Let F and F be translates of one another. Assume further that F and F are symmetric in O. Let a be the translation vector and let O a point such that O O = a/2. Given an arbitrary point X of F , let X be the point of F corresponding to X by symmetry and let X be the point

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symmetric. This implies that the faces G and G not only are translates of one another but also are symmetric with respect to the center of symmetry of P . Now, each face G of P is centrally symmetric by the same argument as above. F O

X

X

O F a X

Fig. 135

From Theorem 1 we easily infer Theorem 2. Every parallelohedron, as well each of its faces, is centrally symmetric. Indeed, consider some tiling of space by parallelohedra. Let G0 be a face of a parallelohedron P0 . The next parallelohedron P1 adjoins P0 along G. The translation taking P0 to P1 carries G0 , which is also a face of P1 , onto some face G1 of the same parallelohedron P1 . Thus, to each face of P1 there corresponds another congruent parallel face. Theorem 1 now implies that P1 , as well as each of its faces, is centrally symmetric. (Since Minkowski’s Theorem is valid in space of arbitrary dimension, so are Theorems 1 and 2.) 8.1.3 To find all types of parallelohedra in three-dimensional space, we must first solve a similar problem on the plane, i.e., find all convex polygons such that the translates of each of them tile the whole plane if fitted together along entire sides. Such polygons are called parallelogons (Fig. 136). Theorem 3. Parallelograms and certrally symmetric hexagons are parallelogons. There are no other types of parallelogons. of F corresponding to X by translation. Then O O = 12 XX . Hence, O O is the midline of the triangle XX X. Therefore, X and X are symmetric in O . We have thus proved that to each point X of F there corresponds a point X of F symmetric to X with respect to O . Hence O is a center of symmetry of F .

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Fig. 136

The fact that every parallelogon has a center of symmetry can be proved along the same lines as Theorem 2. The proof is much simpler, since we can manage without recourse to the general Theorem 1. In fact, it is perfectly obvious that a polygon with pairwise equal parallel sides has a center of symmetry. Consider some tiling of the plane by parallelogons. Let P0 be one of these parallelogons and let P1 adjoin P0 along a side AB (Fig. 137). The parallelogon P1 has a side A1 B1 that is a translate of AB. Since P1 is convex, P1 includes the whole parallelogram ABB1 A1 . For the case in which P1 coincides with ABB1 A1 , we see that P1 is a parallelogon of the theorem.

C a

P0

P2

A

B

E D

P1 A1

B1

Fig. 137

We must now prove that our parallelogon is a hexagon in the converse case. If P1 (and so P0 as well) is not a parallogram, then P1 and P0 have sides AC and AD from A which are not prolongations of one another.4 We will now show that in this case at most one more parallelogon may touch A. Assume the converse. Then some parallelogon P2 adjoins P0 along AC and has some side AE other than AD (see Fig. 137). Draw a prolongation a of AB. The side AE cannot lie on this prolongation: otherwise P2 would not 4

This fact becomes perfectly apparent if we recall that the parallelogons P0 and P1 include parallelograms that adjoin each other along the side AB. This implies that AC and AD may be prolongations of one another only if they are sides of these parallelograms.

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contain a parallelogram which is a translate of ABB1 A1 . Furthermore, AE cannot lie in the same half-plane as AC relative to a: otherwise P2 would not have a side parallel with AB (if it had, it would be nonconvex). Thus, AE lies on the other side of a. However, some parallelogon P3 adjoins P1 along AD. The above arguments also apply to P3 . In consequence, the parallelogons P2 and P3 must overlap, which is possible if and only if P2 and P3 coincide, i.e., if AE coincides with AD. Thus, the parallelogon P2 must adjoin P0 and P1 along AC and AD. Since all parallelogons are translates of each other and have centers of symmetry, we now see that they all are hexagons. (As soon as P2 is a translate of both P0 and P1 , it has angles that correspond to the angles A in P0 and P1 . Since P2 has a center of symmetry, to those angles there correspond symmetric angles. Now, by convexity P2 may have no angles other these six.) The remaining claim, that translates of a parallelogram or a hexagon with a center of symmetry tile the plane, is a matter of straightforward verification. 8.1.4 We now turn to specifying the types of parallelohedra. (A) All faces of a parallelohedron which are parallel to a given edge L form a “closed zone,” i.e., a cyclic sequence of faces adjoining one another along edges parallel to L. The parallel projection of the parallelohedron along L is a polygon whose sides are the projections of the faces of the zone (Fig. 138).

L

Fig. 138

Let L1 be an edge of a parallelohedron P and let G1 be a face with side L1 . Since G1 has a center of symmetry, it also has some side L2 which is a translate of L1 . There is one more face G2 adjoining G1 along L2 . The face G2 also has an edge L3 that is a translate of L2 , and so forth. Since a polyhedron has finitely many faces, the sequence G1 , G2 , . . . must terminate.

354


Under the projection of the parallelohedron to some plane along L, all faces parallel to L are taken onto straight line segments none of which enters the interior of the projection of the parallelohedron5: otherwise the parallelohedron would not lie to one side of the corresponding face, contradicting convexity. Therefore, the projections of the faces of the zone lie on the boundary of the projection of the parallelohedron. Since the zone closes, all sides of the projection are the projections of the successive faces of the zone. (B) For each tiling of the space, the parallelohedra adjacent along the faces of the same zone and its natural continuations form an impenetrable layer, i.e., a layer separating the space. The projection of these layers along the edge of the zone yields a tiling of the projection plane by parallelogons. Consider a tiling of the space with parallelohedra. Take a parallelohedron P0 of the tiling and let L0 be an edge of P0 . By (A), to L0 there corresponds some closed zone of faces. Call the infinite prism bounded by the planes of these faces a zonal prism. The cross-section of a zonal prism by a plane is the parallel projection of the parallelohedron to the plane along L0 . Some parallelohedron P1 adjoins P0 along a face G containing L0 ; another parallelohedron P−1 adjoins P0 along a face parallel with G; next, one more parallelohedra P2 and P−2 adjoin P1 and P−1 along faces parallel with G; and so on. Thereby we obtain a series of parallelohedra and a series R0 of their zonal prisms Πi . The series R0 of zonal prisms separates the space. (See Fig. 139 representing the projections of parallelohedra along L0 .) 0

R P-2

P-1

P0 G

L0

G P1

P2

P0 1

R Fig. 139

For another face G containing L0 , there is a parallelohedron P0 adjoining P0 along G . This parallelohedron also touches P1 along the edge L0 . Since the dihedral angles at the common edge L0 of P0 , P1 , and P0 do not overlap, the zonal prisms do not overlap either. that adjoin P0 along the faces Taking the parallelohedra P1 and P−1 parallel to G and continuing in a similar way, we obtain a new series of parallelohedra together with the corresponding series R1 of their zonal prisms. 5

Here and in what follows, we need not assume that the projection plane is perpendicular to the projection direction.

8.1 Parallelohedra

355

It is easy to check that the series R0 and R1 of zonal prisms touch one another without gaps and overlaps. To this end, consider all parallelohedra touching at L0 . Each of them lies in a dihedral angle whose edge contains L0 . The corresponding zonal prism also lies in this dihedral angle and the planes of the faces of the prism which touch at L0 coincide with the faces of the dihedral angle. Therefore, the zonal prisms are disposed around L0 without gaps and overlaps. Since any parallelohedron possesses a center of symmetry, so does a zonal prism. Therefore, in the cross-section of the prisms by a plane not parallel to L0 we obtain centrally symmetric polygons that are placed around a common vertex without gaps and overlaps. Clearly, the polygons are translates of one another. Now, as we have already seen while describing parallelogons, there are only two possibilities: the polygons are parallelograms or they are hexagons. It is now immediate that the series R0 and R1 of prisms touch one another without gaps and overlaps. Continuing the procedure of applying the series of parallelohedra from both sides of the initial series R0 , we obtain a whole layer of parallelohedra. By what was said above, the zonal prisms cover the whole space without gaps and overlaps. Since to each adjoining of prisms there corresponds adjoining of parallellohedra along faces of zones, the layer of parallelohedra thus constructed is impenetrable. Moreover, we have checked that the projection of the layer to a plane along the edge of the zone yields a tiling of the plane with parallelogons. (C) If two parallelohedra P and P lie in adjacent layers and their zonal projections, i.e., the projections along the edge of the zone, overlap, then the intersection of the projections of P and P is the projection of some common face of P and P . The preceding implies that the whole system of parallelohedra tiling the space stratifies into layers each of which is formed by parallelohedra adjoining one another along the faces of some zone and its natural continuations. Two layers S and S with nonvoid intersection must adjoin each other without gaps: otherwise, such a gap would be filled in by some parallelohedron giving rise to a layer which, due to impenetrability, would separate S and S . Now, suppose that the zonal projections of the two parallelohedra P and P in adjacent layers S and S overlap, and let A be a common interior point of the projections. Since the projections of parallelohedra in one layer do not overlap, the straight line l drawn through A parallel to the edge of the zone intersects each of the parallelohedra but is disjoint from all the other parallelohedra in the same layers. Since the layers adjoin each other without gaps, the straight line l, leaving P in the direction of P , immediately enters P , so that P and P have a common point projecting onto A. Therefore, the intersection of the projections of P and P coincides with the projection of the intersection of P and P . Hence, this intersection is the projection of

356


some common face of P and P , because parallelohedra adjoin one another along entire faces. (D) All types of parallelohedra can be found by “layerwise construction” as follows: Choose some closed zone of a parallelohedron (for convenience, assume the edges of the zone to be vertical). Then all faces of the parallelohedron are of three kinds: the zon and two caps symmetric to one another, the “upper” cap and the “lower” cap. Let S and S be two adjacent layers corresponding to the chosen zone; assume S to lie above S. Let S and S be the zonal projections of S and S to the same plane. By (B), the projections represent two tilings of the plane with the same parallelogons. Take a parallelohedron P in S and let P be the projection of P . Then P is covered by the projections of parallelohedra in S and, by (C), the intersection of P with each of these projections is the projection of some face of P (Fig. 140). A1 A1 P S

Fig. 140

A2

A3 Q1 B

Q3 Q2

A3

A2 Fig. 141

This consideration justifies the following layerwise construction: To find all possible parallelohedra, it suffices to examine possible overlaps of two tilings of the plane by the same parallelogons (which will determine the structure of the cap and so the structure of the zone). 8.1.5 Now, to conclude the specification of all types of parallelohedra, it remains implement the layerwise construction. We separately consider two cases that correspond to two possible types of parallelogons. The first case: the parallelogons are hexagons. On superposing one tiling S over the other, four possibilities may occur depending on the location of a given vertex A of S (Fig. 141), which may be as follows: (I) in the interior of one of the parallelograms Q1 , Q2 , and Q3 ; (II) in the interior of one of the sides BAi of these parallelograms6; (III) at the point B; 6

It is easy to see that if A lies, for example, on the side A1 A1 , then this is equivalent to the case in which A lies on A2 B.

8.1 Parallelohedra

357

(IV) at one of the points A1 , A2 , and A3 . These four cases correspond to the four projections of caps of parallelohedra which are depicted in Fig. 142 and to the four correspoding types of parallelohedra, drawn so that the edge of the zone is vertical. A

A

A

I

A

II

III

IV

IV

II

V

Fig. 142

The second case: the parallelogons are parallelograms. Now obviously there are only three possibilities for superposing one tiling over the other. They are shown in Fig. 142 and denoted by II , IV , and V. The notations II and IV indicate that the corresponding polyhedra have the same structure as the polyhedra II and IV. The only difference is that the projections are taken for zones playing different roles in the combinatorial structure of the polyhedra.

I

II

III

IV

V

Fig. 143

So we have exactly five different types of parallelohedra (Fig. 143): (I) a 14-hedron 8 faces of which are hexagons and 6 are parallelograms; (II) a 12-hedron 4 faces of which are hexagons and 8 are parallelograms; (III) a 12-hedron with all faces parallelograms; (IV) a hexahedral prism; (V) a parallelepiped. (The most symmetric 14-hedral parallelohedron represents a “cubo-octahedron,” i.e., the polyhedron that results by intersecting a cube with an octahedron with axes perpendicular to the faces of the cube; moreover, the hexagons that are cut out from the faces of the octahedron must have centers of symmetry. Now, we may formulate the final result:

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Theorem 4. For a convex polyhedron to be a parallelohedron, it is necessary and sufficient that the structure of the polyhedron be one of the above five types and the polyhedron, as well each of its faces, be centrally symmetric. (The requirement that the polyhedron have a center of symmetry is redundant: this follows from the requirement that every face of the polyhedron be centrally symmetric.) The necessity of the conditions is justified by the above considerations. Sufficiency is evident. Consider, for example, a polyhedron of the first type. Choosing one of its zones, arrange a layer of such polyhedra that adjoin one another along the faces of the zone. We can implement this since the zonal projection yields a hexagon with a center, i.e., a parallelogon. Now, arrange another parallel layer of the same type and place it above the former so that some face of the lower cap of one of its parallelohedra coincides with the corresponding face of the upper cap of some parallelohedron of the first layer. Then the second layer fits perfectly over the first, which is easy to see from the structure of the caps. We continue superposing layers, thus tiling the whole space.7 8.1.6 The above derivation was based on the existence of a closed zone, which follows in turn from the existence of a center for any parallelohedron and each of its faces. However, such a simple conclusion is not possible in a space of dimension greater than three. In this case all faces parallel to some edge need not form a closed zone, i.e., the faces of an (n − 1)-dimensional projection of an n-dimensional parallelohedron along an edge may fail to be the projections of (n − 1)-dimensional faces of the parallelohedron. Nevertheless, B. N. Delaunay succeeded in implementing the layerwise construction in four-dimensional space and found all four-dimensional parallelohedra. It turns out that there are 52 types of such parallelohedra. In spaces of higher dimension, the layerwise construction cannot give all types of parallelohedra. The deepest research concerning n-dimensional parallelohedra was carried out by G. F. Vorono˘ı (1868–1908).8 LetE be a lattice of points (i.e., the set n constituted by the ends of the vectors i=1 xi ai , where ai are given vectors and xi are arbitrary integers). Given a point A of E, arrange the set of nearest points to A, i.e., the set of points X such that the distance AX is not larger than the distance from X to any other point of E. It is easy to prove that the resulting regions are convex polyhedra adjoining one another 7

8

Necessary and sufficient conditions for tiling an n-dimensional space without gaps and overlaps by polyhedra adjoining one another along entire faces are given in the articles [V] and [A17]; they consist in requiring a local tiling without gaps and overlaps in a neighborhood of each (n − 2)-dimensional face. As shown in [A17], these conditions are valid for all simply connected n-dimensional spaces. – V. Zalgaller G. F. Vorono˘ı’s research is described in [D, pp. 268–321], where the reader can also find further references.

8.2 A Polyhedron of Least Area with Fixed Volume

359

along entire faces.9 These regions are called Vorono˘ı regions10 of E. Since the lattice admits parallel translations, all Vorono˘ı regions are parallelohedra. It is easy to prove that a parallelohedron is the Vorono˘ı region of some lattice if and only if the straight lines from its center to the centers of faces are perpendicular to the faces. Vorono˘ı conjectured that every parallelohedron can be made into a Vorono˘ı region by a suitable affine transformation. He proved this only for primitive parallelohedra, i.e., those determining tilings of the space with the least possible number of polyhedra contiguous to a vertex, namely, n+1. For n = 3, only a parallelohedron of type I, a 14-hedron, may be primitive. For n = 4, there are three types of primitive parallelohedra. Vorono˘ı’s Theorem is proved for all three- and four-dimensional parallelohedra by B. N. Delaunay and generalized to a wider class of parallelohedra by O. K. Zhitomirski˘ı [Zh]. However, the theorem still remains unproved in its full generality. Once it is proved, we will have a method for finding all parallelohedra in n-dimensional space, because Vorono˘ı gave a method for finding all Vorono˘ı regions.

8.2 A Polyhedron of Least Area with Fixed Volume 8.2.1 Here we are concerned with the following theorem, first proved by Lindel¨ of in 1870 [Lin]. Theorem 1. Among all convex polyhedra with given face directions and fixed volume, a polyhedron circumscribed about a ball has least area. This theorem is a simple corollary to a more general theorem due to Minkowski. To formulate the latter, we begin with a few preliminary remarks. Generalizing the notion of support number (Section 1.2), we define the support number h(n) of a polyhedron to be the distance from the origin to the support plane of the polyhedron with outward normal n. As usual, the distance is assumed positive if the normal n, when drawn from the origin, points to the support plane; otherwise, we assume the distance negative. By n we always mean a unit vector. Therefore, we may say that h(n) is the right-hand side of the normal equation of the support plane with outward normal n: nx = h(n). (1) 9

They are bounded by the planes perpendicular to the segments joining the given point to the other points of the lattice at the midpoints of these segments. 10 They are also called Dirichlet regions, because Dirichlet considered them earlier in the tree-dimensional case in connection with some problems of number theory. – V. Zalgaller

360


In comparison with the definition of Section 1.2, the support plane with normal n now need not be the plane of a face. It is this generalization of the notion of support number that we bear in mind in this and subsequent sections. We shall always deal with a finite set of pairwise distinct vectors ni which contains the normals to all faces of the polyhedra under consideration. For brevity, we write hi rather than h(ni ). We call the intersection of a polyhedron with its support plane having normal ni the ith face of the polyhedron, although such a face may degenerate into an edge or a vertex. In the nondegenerate case we say that it is a genuine face. We denote the area of the ith face by Fi ; this area vanishes only if the face is degenerate. Lemma 1. If we translate the origin by a vector a, then the support number h(n) changes by the summand −an. By (1), we have h(n) = nx. If h (n) is the support number for the new choice O of the origin, then h (n) = nx , where x is the vector joining O to an arbitrary point X of the support plane. −−→ −−→ −−→ −−→ If O is the old origin, then OO = a and x = O X = O O + OX = −a + x. Therefore, h (n) = nx − na = h(n) − na. Lemma 2. The volume of any polyhedron is given by the formula 1 hi Fi . V = 3 i This formula is obvious: it expresses the volume of the polyhedron as the algebraic sum of the volumes of all pyramids with common vertex the origin, each having a face of the polyhedron as its base. If the origin lies outside the polyhedron, then the volume of any pyramid outside the polyhedron must be subtracted. The formula accounts for the phenomenon, since the support number hi is negative in this case. Lemma 3. For every polyhedron,

i

ni Fi = 0.

This formula is already familiar to us. Incidentally, it follows from Lemmas 1 and 2. Indeed, the volume clearly does not change under translation of the origin by any vector a, whence (hi − ani )Fi = hi Fi − a ni Fi . hi Fi = Since this holds for every vector a, we have ni Fi = 0.


361

Lemma 4. Let P 0 and P 1 be two polyhedra and let n1 , . . . , nr be a set of vectors containing the normals to all genuine faces of P 0 and P 1 . If h0i and Fi1 are the support numbers and the face areas of P 0 and P 1 , then the sum 1 0 1 (2) V011 ≡ V (P 0 P 1 P 1 ) = hi Fi 3 does not change under any translation of the origin and any translations of P 0 and P 1 . Neither does it change when new vectors ni are introduced. Under the translation of the origin by a vector a, by Lemma 1 the sum (2) must change to (h0i − ani )Fi1 = h0i Fi1 − a ni Fi1 . i

i

i

However, the last sum vanishes by Lemma 3, which yields the first claim of the lemma. Under a translation of P 1 , the areas and directions of its faces do not change. Therefore, the sum V011 remains the same. A translation of P 0 only results in the same changes of its support numbers as the translation of the origin by the opposite vector. Therefore, the sum V011 does not change under any translations of P 0 . The fact that V011 does not change when we introduce new vectors ni is obvious because the support planes with these vectors touch P 1 along “faces” for which Fi = 0. Using the close analogy between the expressions for the volume and the sum V011 , we call V011 the mixed volume of P 0 and P 1 . If P 0 is a translate of P 1 , then V011 obviously equals the volume of P 0 . Another meaning of the value V (P 0 P 1 P 1 ) is clarified in Section 8.3, in which we also explain the reasons for the notation. For a temporary explanation, it suffices to observe that under the dilation of P 0 with coefficient λ the value of V (P 0 P 1 P 1 ) changes by a factor of λ, whereas under the same dilation of P 1 it changes by a factor of λ2 . Since in the case when P 0 is a translate of P 1 the value of V (P 0 P 1 P 1 ) coincides with the volume of P 0 , it is natural to denote this volume by V (P 0 P 0 P 0 ). Now we formulate Minkowski’s Theorem. Theorem 2. For arbitrary convex polyhedra P 0 and P 1 , V 3 (P 0 P 1 P 1 ) ≥ V (P 0 P 0 P 0 )V 2 (P 1 P 1 P 1 ). Moreover, the equality holds if and only if P 0 and P 1 are homothetic (i.e., they are mapped onto one another by a translation followed by a dilation). 8.2.2 Theorem 1 is a simple corollary to Theorem 2 due to the following observation: If two polyhedra P 0 and P 1 have the same outward normals to all genuine faces and P 0 circumscribes a ball, then 3V (P 0 P 1 P 1 ) coincides with the area of the surface of P 1 .

362


Indeed, taking the origin to be the center of the ball that P 0 circumscribes, for the support numbers of P 0 we obtain h0i = 1. By the definition (2) of the mixed volume V (P 0 P 1 P 1 ), we then infer that V (P 0 P 1 P 1 ) = Fi1 = F (P 1 ), where F (P 1 ) is the area of the surface of P 1 . Consequently, Theorem 2 implies that (in the abbreviated notation for volumes) F 3 (P 1 ) ≥ 27V (P 0 )V 2 (P 1 ), with the equality holding if and only if the polyhedron P 1 is homothetic to P 0 . This means that the area F (P 1 ) attains a minimum, given the volume V (P 1 ), if and only if P 1 is homothetic to P 0 , i.e., if P 1 circumscribes some ball. Happily, this is the exact claim of Theorem 1. 8.2.3 We now turn to the proof of Theorem 2. Let P 1 be a given polyhedron and let n1 , . . . , nr be a collection of pairwise distinct unit vectors which contains all outward normals of the genuine faces of P 1 . Let P be a variable polyhedron cut out in space by the planes with the normals ni . Our polyhedron P 1 is among the polyhedra P . The variable polyhedron P is the intersection of the half-planes ni x ≤ hi , where hi is the signed distance from the origin to the ith plane Qi . The plane Qi may fail to touch P . Then hi is not a support number of P . Nevertheless, when hi are given, the polyhedron P is completely determined and we may call the numbers hi its “support numbers,” using quotation marks. (If the ith plane fails to touch P , then hi is not determined by the polyhedron.) Lemma 5. The volume V of P is a differentiable function of its “support numbers” hi ; moreover, ∂V = Fi , ∂hi where Fi is the area of the corresponding face. If the ith plane fails to touch P , then the ith face is absent and we assume Fi = 0.) This lemma is quite obvious and was proved in Section 7.2. Clearly, the face areas Fi depend continuously on the “support numbers” hi . Therefore, V has continuous partial derivatives. We now prove an assertion which provides us with a simple formal argument for demonstrating Theorem 2. Theorem 2a. Let P be the variable polyhedron bounded by the planes ni x = hi with given outward normals ni . Let P 1 be one of these polyhedra. Under the condition


Φ(h1 , . . . , hr ) ≡

r

h1i Fi1 = 3V (P 1 ),

363

(3)

i=1

the polyhedron P has greatest volume if and only if it is a translate of P 1 . (Fi1 are the face areas of P 1 and V (P 1 ) is its volume.) The linear form Φ is independent of the choice of the origin and does not change under translations of polyhedra. This is proved in much the same way as Lemma 4. The only difference is that hi may now fail to be genuine support numbers and the terms in the sum Φ are merely nonnegative. Therefore, if the ith face of P does not degenerate, i.e., if Fi > 0, then (3) implies that hi ≤ 3V (P 1 )/Fi1 . This means that all “support numbers” of P , corresponding to the genuine faces of P 1 , are bounded. Now, assume that the ith face of P 1 degenerates, and so Fi1 = 0. Then the corresponding summand in the linear form Φ disappears and no restrictions are imposed on hi . Nevertheless, the genuine ith support number of P is bounded. Indeed, assume the contrary. Then, however far we move the ith plane Qi , this plane will always touch P . Therefore, as Qi goes to infinity, the polyhedron P becomes unbounded. This means that all planes except Qi bound an unbounded polyhedron and the normals of these planes point to the same half-plane. However, in that case these planes cannot bound a bounded polyhedron no matter how they are positioned with respect to the origin. At the same time, if the ith face of P 1 degenerates, then the plane Qi is not involved in bounding this polyhedron: P 1 is already bounded by the other planes. The contradiction thus obtained shows that the genuine ith support number of P must be bounded. Consequently, all genuine support numbers of the polyhedra P with the constraint (3) are bounded, which implies that the polyhedra themselves lie in a bounded part of space. Hence, there is a polyhedron P 0 of greatest volume among them. We are maximizing V under the constraint (3), i.e., dealing with a constrained extremum problem. By Lemma 5, V is a differentiable function of hi ; hence, the well-known method of Lagrange multipliers implies that at a maximum point ∂V ∂Φ =λ (i = 1, . . . , r) ∂hi ∂hi or, since ∂V ∂Φ = Fi , = Fi1 , ∂hi ∂hi we have

Fi = λFi1

(i = 1, . . . , r).

(4)

This means that the face areas of the maximal polyhedron and P 1 are √ proportional. Dilating P 1 with coefficient λ, we obtain the polyhedron

364


√ 1 λP whose face areas are equal to λFi1 . Therefore, √equalities (4) mean that the maximal polyhedron P and the polyhedron λP 1 have pairwise parallel faces of the same area. In that case, Theorem 2 of Section 6.3 implies that the polyhedra are translates of one another. Making them coincide by translation, we infer that their support numbers are equal. Hence, denoting √ the support numbers of the maximal polyhedron by hi , we obtain hi = λh1i . Inserting these values in (3), we derive that λ = 1, i.e., the maximal polyhedron coincides with the given polyhedron P 1 , as claimed. 8.2.4 We now prove Theorem 2: For two arbitrary polyhedra P 0 and P 1 we have V 3 (P 0 P 1 P 1 ) ≥ V (P 0 P 0 P 0 )V 2 (P 1 P 1 P 1 ), with equality holding if and only if P 0 and P 1 are homothetic polyhedra. Let n1 , . . . , nr be all pairwise distinct outward normals to the genuine faces of P 0 and P 1 . Consider the polyhedra P bounded by the planes with normals ni and satisfying r i=1

hi Fi1 =

r

fi1 Fi1 = 3V (P 1 ).

(5)

i=1

(Here and below, summation is taken over all i from 1 to r because the introduction of redundant vectors ni does not change the sums: the corresponding faces degenerate, i.e., Fi1 = 0.) Define µ so as to have µV (P 0 P 1 P 1 ) = V (P 1 ), i.e., in accordance with the formulas (2) and (5) for V (P 0 P 1 P 1 ) and V (P 1 ), 1 1 h F (6) µ = i0 i0 . hi Fi Dilating P 0 with coefficient µ, we obtain the polyhedron µP 0 with support numbers µh0i . The definition of µ, together with (6), yields h1i Fi1 ; µh0i Fi1 = i.e., µP 0 is among the polyhedra P . By Theorem 2a P 1 has greatest volume among the polyhedra P ; therefore, V (µP 0 ) ≤ V (P 1 )

(7)

and V (µP 0 ) = V (P 1 ) if and only if µP 0 is a translate of P 1 . Under the dilation of a polyhedron with coefficient µ, the volume of the polyhedron is multiplied by µ3 . Using the definition of µ, from (7) we infer that V 3 (P 1 ) V (P 1 ) ≥ µ3 = 3 0 1 1 V (P 0 ) V (P P P ) or


365

V 3 (P 0 P 1 P 1 ) ≥ V 2 (P 1 )V (P 0 ). Moreover, in accordance with the conditions for equality in (7), here we have the equality if and only if µP 0 = P 1 , i.e., if P 0 and P 1 are homothetic. This completes the proof of Theorem 2, also finishing the proof of Theorem 1 as explained in Subsection 8.2.2. 8.2.5 The above proof of Theorem 1 might seem too difficult. Would not it be simpler to seek a minimum of area under a fixed volume and given directions of faces? It is comparatively easy to prove that (i) the minimum is attained and, furthermore, it is attained at a polyhedron with all faces of given directions nondegenerate; (ii) the surface area F of a polyhedron with all faces nondegenerate is a differentiable function of the support numbers hi . Applying the method of Lagrange multipliers to a polyhedron of least area, we then obtain ∂V ∂F =λ = λFi ∂hi ∂hi

(i = 1, . . . , r),

with Fi the area of the ith face. However, to prove that these equations are valid only for a polyhedron circumscribing a ball turns out to be rather difficult. Therefore, proving a more general Theorem 2, we actually gain in simplicity, not to mention the fact that, besides Theorem 1, Theorem 2 has some other interesting corollaries, which we will consider in Section 8.3. 8.2.6 All results of this section generalize to spaces of arbitrary dimension. We define the mixed volume V (P 0 P 1 P 1 . . . P 1 ) of polyhedra P 0 and P 1 in n-dimensional space by the formula 1 V (P 0 P . . P 1) = . n−1 times

1 0 1 h F , n i i i

where h0i are the support numbers of P 0 and Fi1 are the face “areas” of P 1 . Theorem 2 is formulated as follows: for two arbitrary convex n-dimensional polyhedra P 0 , P 1 we have V n (P 0 P 1 . . . P 1 ) ≥ V (P 0 . . . P 0 ) · V n−1 (P 1 . . . P 1 ); 0

1

(8)

moreover, the equality holds if and only if P and P are homothetic. In the same way as in Subsection 8.2.2, this Minkowski Theorem readily implies the minimality property of a polyhedron circumscribing a ball. However, generalization of the above proof of Theorem 2 to n-dimensional space would be an empty formality! This is because in n-dimensional space Minkowski’s Uniqueness Theorem (which we use in the proof) relies on the ndimensional Theorem 2. No other proof is available by now.

366


So, in n-dimensional space the relationship between these theorems reverses, of course until we acquire an independent proof of Minkowski’s Uniqueness Theorem for polyhedra. An exhaustive study of connections between the indicated theorems in n-dimensional space and a proof of the ndimensional Theorem 2 (inequality (8)) are conducted in the next section.

8.3 Mixed Volumes and the Brunn Inequality 8.3.1 Here we will consider linear combinations of convex bodies, primarily convex polyhedra, in n-dimensional space (see Section 6.2). The results of Section 6.2 translate almost literally to the case of an arbitrary dimension, and we will not dwell on this. We only recall that if Pi are convex polyhedra, then P = λ1 P1 + . . . + λm Pm is again a convex polyhedron and every face G of P is the same linear combination of the faces Gi of Pi that lie in planes parallel to G. Here and in the sequel, the coefficients λi are always assumed nonnegative. Parallelism of the planes of faces is understood in the sense of parallelism of outward normals. The faces Gi producing a genuine (n−1)-dimensional face G may degenerate into (n − 2)-dimensional faces, etc. For example, in the three-dimensional case nonparallel edges lying in parallel support planes of polyhedra P1 and P2 give a face of P1 + P2 which is a parallelogram. We indicate one more simple fact: If h is the support number of the face G and hi are the support numbers of the faces Gi , then (1) h = λ1 h1 + . . . + λm hm . Indeed, if x1 , . . . , xm are the position vectors of points on G1 , . . . , Gm , then x = λ1 x1 + . . . + λm xm

(2)

is the position vector of a point on G. By the definition of support numbers, we have h = nxi and hi = nxi . Therefore, multiplying (2) by the normal n, we arrive at (1). Theorem 1. Let P 1 , . . . , P m be convex polyhedra in n-dimensional space, and P = λ1 P 1 + . . . λm P m , where λi are variable nonnegative numbers. The volume V (P ) of P is a homogeneous polynomial in λi of degree n. Here we also admit polyhedra P i that degenerate into polyhedra of lesser dimension. Their linear combination can be n-dimensional, as it happens, for example, for the linear combination of noncoplanar straight line segments.

8.3 Mixed Volumes and the Brunn Inequality

367

The proof is carried out by induction on the dimension n. For n = 1 the theorem is obvious, since in P i are segments on a straight line this case i and the length of the segment λi P equals the same linear combination of their lengths. Assume that the theorem is valid in (n − 1)-dimensional space. Let the λi P i is expolyhedra P i lie in n-dimensional space. The volume of P = pressed in terms of support numbers and face areas as follows: V (P ) =

1 hj Fj . n j

(3)

The faces of P are the same linear combinations of parallel faces of the polyhedra Pi . Since parallel translation of “summands” only results in parallel translation of the linear combination, we may assume that each face of P is the linear combination of polyhedra in (n − 1)-dimensional space. Therefore, by the induction hypothesis its (n − 1)-dimensional volume, i.e., the area Fj , is a homogeneous polynomial in λi of degree (n − 1). However, according to (1) the support numbers hj of P are linear functions in λi . It follows that the right-hand side of (3) is a homogeneous polynomial in λi of degree n, which completes the proof of the theorem. It is a common practice to write homogeneous polynomial given by the m the volume of the polyhedron P = i=1 λi P i as V (P ) =

m

λi1 λi2 . . . λin Vi1 i2 ...in ,

(3 )

i1 ,i2 ,...,in =1

where each index ij runs through 1 to m independently of the other indices. Therefore, the same product λi1 . . . λin occurs in (3 ) as often as there are permutations of the indices λi1 . . . λin . The coefficients Vi1 i2 ...in are defined so as to be independent of the order of indices. λi P i , Take some product λi1 λi2 . . . λin . Considering the formula P = insert zero values for all λi except those involved in the chosen product. The corresponding polyhedra P i are now absent in the linear combination. This implies that the coefficient Vi1 ...in of the product λi1 . . . λin depends only on the polyhedra P i1 , . . . , P in . Since λi1 , . . . , λin need not to be all distinct, some of these polyhedra may coincide, i.e., one polyhedron may be repeated several times. In particular, if we take λ1 = 1 and λi = 0 for the other i, then P = P 1 and (3 ) implies that the coefficient V1,...,1 is the volume of P 1 . In general, the coefficient Vi1 ,...,in is called the mixed volume of the polyhedra P 1 , . . . , P m and written V (P i1 . . . P in ). By definition, it is independent of the order of the indices ij , i.e., is a symmetric function of the polyhedra P ij . We confine our exposition to linear combinations of two polyhedra. In this case, (3 ) can be rewritten as

368


V (λ1 P 1 + λ2 P 2 ) =

n k=1

1 2 λn−k λk2 Cnk V (P . . P 1 P . . P 2 ). 1 . . n−k

(4)

k

8.3.2 We now demonstrate that the mixed volume V (P 1 P 2 . . . P 2 ) coincides with the mixed volume introduced in Section 8.2. This is asserted in the next theorem. Theorem 2. If V (P 0 P 1 . . . P 1 ) is the mixed volume defined by (4), Fi1 are the face areas of the polyhedron P 1 , and h0i are the corresponding support numbers of the polyhedron P 0 , then V (P 0 P 1 . . . P 1 ) =

1 0 1 h F . n i i i

(5)

To prove this, note that the volume of the polyhedron P = P 1 + λP 0 is a differentiable function of its support numbers hi and ∂V (P ) = Fi , ∂hi where Fi is the area of the corresponding face (Lemma 5 of Section 8.2). At the same time, according to (1), hi = h1i + λh0i . Therefore, the derivative of V (P ) with respect to λ equals ∂V (P ) ∂V dhi = = h0i Fi . ∂λ ∂hi dλ However, P = P 1 + λP 0 converges to P 1 as λ → 0 and, therefore, Fi → Fi1 . Hence, ∂V = h0i Fi1 . (6) ∂λ λ=0 At the same time, by the definition of mixed volume (see (4)), we have V (P ) = V (P 1 . . . P 1 ) + nλV (P 0 P 1 . . . P 1 ) + . . . , with the dots standing for terms involving the second and higher powers of λ. This formula implies that ∂V = nV (P 0 P 1 . . . P 1 ). ∂λ λ=0 Together with (6), this proves the theorem.


369

8.3.3 Theorem 1, claiming that the volume of a linear combination is a homogeneous polynomial, generalizes easily from polyhedra to arbitrary convex bodies. Let H1 , . . . Hm be given convex bodies and let H = λ1 H1 + . . . + λm Hm be their linear combination with nonnegative coefficients. It is easy to verify that if the polyhedra P1 , . . . , Pm converge to the bodies H1 , . . . , Hm then the linear combination P = λ1 P1 + . . . + λm Pm converges to H. This means that, for any λi , the volume V (P ) tends to the volume V (H). Since V (P ) is a polynomial of a fixed degree, it follows that V (H) is also a polynomial of the same degree. This proves the following Theorem 1a. The volume of the linear combination of convex bodies with nonnegative coefficients is a homogeneous polynomial of degree n in these coefficients. The coefficients of these polynomials are also called mixed volumes and denoted in the same way. In this connection, for many questions concerning mixed volumes, there is no need to distinguish polyhedra among all convex bodies. (Theorem 2 generalizes to arbitrary convex bodies as well, with the sum in (5) replaced by an integral.) 8.3.4 In 1887, the German mathematician Brunn [Br] proved the following theorem which, together with Theorems 1 and 2, marked the beginning of the whole “theory of mixed volumes”: Theorem 3. If H0 and H1 are arbitrary convex bodies, Ht = (1−t)H0 +tH1 , and 0 < t < 1, then the following inequality holds for volumes (the Brunn inequality): n V (Ht ) ≥ (1 − t) n V (H0 ) + t n V (H1 ). (7) Moreover (as was later clarified by Minkowski), the equality holds if and only if the bodies H0 and H1 are homothetic (in that case, Ht is homothetic to them for every t). We recall that Ht is the set of points dividing the line segments whose and H1 in the ratio t : (1 − t). endpoints are arbitrary points of H0 Given H0 and H1 , the quantity n V (Ht ) is a function of t. Drawing its graph (Fig. 144), we see that the inequality (7) means that the graph of the function n V (Ht ) goes above the chord between the endpoints. For arbitrary t1 , t, and t2 satisfying the condition 0 < t1 < t < t2 < 1, we have Ht =

t2 − t t − t1 Ht + Ht , t2 − t1 1 t2 − t1 2

(8)

which is checked by direct computations based on simple algebraic properties of the linear combination of bodies. Therefore, using the same inequality (7)

370


for Ht1 , Ht , and Ht2 , we see that the graph of the function n V (Ht ) is located above the chord between the points corresponding to t1 and t2 , i.e., the graph is located above each of its chords and is therefore a convex curve. ⎯⎯¬ √V(Ht)

n

⎯⎯¬ √V(H1)

n

⎯⎯¬ √V(H0) n

O

1

t

Fig. 144

Thus, Brunn’s Theorem can be rephrased as follows: n V (Ht ), where Ht = (1 − t)H0 + tH1 and 0 ≤ t ≤ 1, is a (downwards) convex function of t which is reduced to a linear function if and only if H0 and H1 are homothetic bodies. We will outline the proof of this remarkable theorem later; incidentally, it is not too difficult. Now, we clarify its connection with Minkowski’s inequality of Section 8.2. 8.3.5 Let us prove that Brunn’s Theorem implies Minkowski’s inequality V n (H1 H0 . . . H0 ) ≥ V (H1 . . . H1 ) · V n−1 (H0 . . . H0 ),

(9)

and the equality holds if and only if H0 and H1 are homothetic. Here V (H0 . . . H0 ) = V (H0 ) and V (H1 . . . H1 ) = V (H1 ) are the volumes of the bodies H0 and H1 . Since the graph of the function n V (Ht ) is located above the chord, we have f (t) = n V (Ht ) − (1 − t) n V (H0 ) − t n V (H1 ) ≥ 0. Since f (0) = 0, it follows that the derivative f (0) is nonnegative and, by the convexity of f , we have f (0) = 0 if and only if f (t) ≡ 0, i.e., if H0 and H1 are homothetic bodies. We now show that the inequality f (0) ≥ 0 is exactly Minkowski’s inequality (9). According to formula (4), which was established for arbitrary convex bodies, the volume of the body Ht = (1 − t)H0 + tH1 equals V (Ht ) =

n

(1 − t)n−k tk Cnk Vk ,

k=1

with Vk = V (H0 . . . H0 H1 . . . H1 ). n−k

k

Using this formula, we may easily calculate f (0). It turns out that


f (0) =

V1 − V0 (n−1)/n

V0

1/n

+ V0

371

− Vn1/n .

Therefore, the inequality f (0) ≥ 0 is equivalent to the inequality V1 ≥ (n−1)/n 1/n V0 Vn , or V1n ≥ V0n−1 Vnn , which is nothing but Minkowski’s inequality in our notations. The equality holds if and only f (0) = 0, i.e., if H0 and H1 are homothetic. We now prove that, conversely, Minkowski’s inequality implies Brunn’s inequality. It follows from the above that Minkowski’s inequality is equivalent to the condition f (0) ≥ 0, where the function f (t) is the difference between the two sides of Brunn’s inequality. This condition means that the graph of the function n V (Ht ) passes above the chord at least at the initial point. However, by (8), for arbitrary t1 < t2 the body Ht with t1 < t < t2 can be represented as the combination of the bodies Ht1 and Ht2 with coefficients whose sum equals 1 and which, therefore, have the form 1 − s, s. Hence, each observation ret1 may serve as the initial point s = 0, i.e., the preceding mains valid for every point of the graph of the function n V (Ht ) (this follows by applying Minkowski’s inequality to the bodies Ht1 and Ht2 ). Therefore, the graph of the function n V (Ht ) passes above the chord everywhere, which is equivalent to the validity of Brunn’s inequality. If Minkowski’s inequality (9) for the bodies H0 and H1 becomes an equality, then f (0) = 0 and, by convexity of the graph of the function n V (Ht ), this graph is a line segment, i.e., we also have the equality in Brunn’s inequality. Minkowski’s inequality was proved for three-dimensional polyhedra in Section 8.2. By virtue of the result established above, we now obtain Brunn’s inequality for such polyhedra and, by an obvious limit argument, for arbitrary three-dimensional convex bodies. However, the case of equality cannot be settled by passage to the limit. 8.3.6 We now describe one more connection between Minkowski’s inequality and Minkowski’s Uniqueness Theorem for polyhedra with faces of equal areas. In Section 8.2, we derived Minkowski’s inequality for three-dimensional polyhedra by utilizing simple arguments about maxima and essentially using the above-mentioned uniqueness theorem for polyhedra. It is easy to see that all conclusions apply to spaces of arbitrary dimension. This allows us to assert that Minkowski’s inequality for n-dimensional polyhedra (together with the condition for equality) follows from the uniqueness theorem for polyhedra with parallel faces of equal areas. Let us prove that, conversely, the latter theorem can be deduced from Minkowski’s inequality supplemented with the condition for equality. Assume that two polyhedra P 0 and P 1 have parallel faces of equal areas, so that Fi0 = Fi1 .

372


Then for the mixed volume V (P 0 P 1 . . . P 1 ) we have (Theorem 2) 1 0 1 1 0 0 V (P 0 P 1 . . . P 1 ) = hi Fi == h F = V (P 0 . . . P 0 ). n i n i i i

(10)

By Minkowski’s inequality, V (P 0 P 1 . . . P 0 ) ≥ V (P 0 . . . P 0 )V n−1 (P 1 . . . P 1 ).

(11)

Using (10), we now see that V (P 0 . . . P 0 ) ≥ V (P 1 . . . P 1 ).

(12)

However, interchanging P 0 and P 1 , in the same way we obtain V (P 1 . . . P 1 ) ≥ V (P 0 . . . P 0 ). Hence, V (P 0 . . . P 0 ) = V (P 1 . . . P 1 ). But then, as can be seen from the derivation of inequality (12), the equality must hold in Minkowski’s inequality (11). Therefore, the polyhedra P 0 and P 1 must be translates of one another. Thus, Brunn’s inequality, via Minkowski’s inequality, leads to the uniqueness theorem for polyhedra with parallel faces of equal areas.11 Via the same Minkowski inequality, it leads to the maximality property of polyhedra circumscribing a ball, as it was shown in Section 8.2. We are left with proving Brunn’s inequality. 8.3.7 The shortest proof of Theorem 3, Brunn’s inequality with Minkowski’s supplement, is carried out by induction on the dimension of space. In onedimensional space, i.e., on a straight line, the theorem is trivial: the inequality is reduced to the equality, which is obvious from the very definition of linear combination of segments. Let us assume the theorem for (n − 1)-dimensional space and prove it for n-dimensional space. To this end, we first establish the following Theorem 3∗ . If two convex bodies H0 and H1 have the same volume, then, for every t, 0 < t < 1, the body Ht = (1 − t)H0 + tH1 has larger volume, except for the case in which H0 and H1 are translates of one another and so Ht is congruent to them. Let us show that the general statement of Theorem 3 follows from this particular one. Let H0 and H1 be arbitrary convex bodies. Then the bodies 11

Along the same lines, we can prove the general uniqueness theorem for convex bodies with the same area functions. If Fi (M ) is the area function of a body Hi , then, by analogy to Theorem 2, V (H0 H1 . . . H1 ) = n1 h0 (n)F1 (dM ). Therefore, the equality F0 (M ) = F1 (M ) implies the equality V0 (H0 H1 . . . H1 ) = V (H0 . . . H0 ). At the same time, by Minkowski’s inequality, V (H0 . . . H0 ) ≥ V (H1 . . . H1 ). Now, we repeat the same arguments as those for polyhedra.


1 K0 = √ H0 n V0

373

1 and K1 = √ H1 n V1

have unit volume. Therefore, by the particular statement of Brunn’s Theorem, we must have (13) V (Ks ) ≥ 1, where Ks = (1 − s)K0 + sK1 , 0 < s < 1. Take 1/n

s= Then Ks =

tV1

1/n

(1 − t)V0

(1 − t)H0 + tH1 (1 −

1/n t)V0

+

1/n tV1

=

1/n

.

+ tV1

(1 −

Ht 1/n t)V0

1/n

.

+ tV1

Therefore, (13) is equivalent to n 1/n 1/n , V (Ht ) ≥ (1 − t)V0 + tV1 which is the general Brunn inequality. Moreover, the equality holds here if and only if it holds in (13), i.e., if K0 and K1 are translates of one another and, in consequence, H0 and H1 are homothetic bodies. 8.3.8 Now, assuming Theorem 3 valid for (n − 1)-dimensional bodies, we prove that Theorem 3∗ is valid for n-dimensional bodies. Let H0 and H1 be convex bodies in n-dimensional space which have the same volume. We assume the volume to be equal to 1. Draw some plane T and translate H0 and H1 so that T becomes a support plane to both of them. (This does not influence the volume of the body Ht = (1 − t)H0 + tH1 .) Introduce the variable v, 0 ≤ v ≤ 1, and associate to each other the crosssections of H0 and H1 by planes parallel to T cutting out equal volumes v from H0 and H1 . The distances from these sections G0 and G1 of the bodies H0 and H1 to the plane T are functions of v, which we denote by x0 (v) and x1 (v). Denote the areas of the cross-sections by F0 (v) and F1 (v). Since v is the volume cut out by G0 and G1 , it follows that F0 dx0 = dv and F1 dx1 = dv; therefore, 1 1 dx1 dx0 = = , . (14) dv F0 dv F1 Clearly, the linear combination Gt = (1 − t)G0 + tG1 of G0 and G1 is contained in Ht . It lies in the plane parallel to the main plane T and is at distance xt (v) from T ; moreover, xt (v) = (1 − t)x0 (v) + tx1 (v).

(15)

374


If Ft (v) is the area of Gt , then for the volume of Ht we have the inequality

1

1 V (Ht ) ≥ Ft (v) dxt (v) = Ft (v)xt (v) dv. (16) 0

0

By the induction hypothesis, Brunn’s inequality is valid for the sections G0 and G1 ; so n−1 1/(n−1) 1/(n−1) (v) + tF1 (v) . (17) Ft (v) ≥ (1 − t)F0 Further, by (14) and (15) we have 1−t dxt t = + . dv F0 F1 Using (17) and (18), from (16) we infer

1 n−1 1 − t t 1/(n−1) 1/(n−1) V (Ht ) ≥ dv. + tF1 + (1 − t)F0 F0 F1 0 Let us prove that the integrand is at least 1, i.e., n−1 1 − t t 1/(n−1) 1/(n−1) (1 − t)F0 + tF1 + ≥1 F0 F1

(18)

(19)

(20)

for all F0 , F1 > 0 and 0 < t < 1, with the equality holding if and only if F0 = F1 . Dividing and multiplying by F0 and setting F1 /F0 = ξ, we infer that the left-hand side of the inequality equals n−1 (1 − t + t/ξ) . f (ξ) = 1 − t + tξ 1/(n−1) If ξ = 0 or ξ = ∞, then f (ξ) = ∞. Hence, f (ξ) has a minimum for ξ > 0. Calculating the derivative, we find n−2 t(1 − t) 1/(n−1) n/(n−1) 1 − t + tξ ξ f (ξ) = − 1 , ξ2 so that for 0 < t < 1 and ξ > 0 the derivative vanishes only at ξ = 1. Consequently, the minimum of f (ξ) corresponds to this value of ξ. Therefore, f (ξ) ≥ 1; moreover, f (ξ) = 1 if and only if ξ = 1. This is equivalent to inequality (20) with the corresponding condition for equality. It follows from the inequality that the integral (19) is at least 1 and therefore V (Ht ) ≥ 1, which was to be proved. We now demonstrate that V (Ht ) = 1 if and only if the bodies H0 and H1 are translates of one another. To this end, we follow the chain of inequalities which led us to the inequality V (Ht ) ≥ 1. If V (Ht ) = 1, then in each inequality of the chain we


375

must have the equality. Therefore, f (ξ) = 1, implying that ξ = 1. In other words, (21) F0 (v) = F1 (v). But then from (14), assuming that x0 (0) = x1 (0) = 0 (i.e., assuming that the bodies H0 and H1 touch the same plane T ), we infer that x0 (v) = x1 (v),

(22)

which means that the planes cutting out equal volumes from H0 and H1 coincide. It follows that the centers of mass of H0 and H1 are situated at the same height above T .12 Therefore, translating H0 so as to put the centers of mass of the two bodies in coincidence, we ensure that T remains a support plane to both bodies. However, the plane T was chosen arbitrarily. Hence, for the case in which the centers of mass coincide, all support planes of H0 and H1 must coincide, which implies that the bodies themselves coincide. Therefore, before translation the bodies were translates of one another. Brunn’s Theorem with Minkowski’s supplement is thus proved. We have thereby also established Minkowski’s inequality, the uniqueness theorem for polyhedra with the same face areas and face directions, and the maximality property of a polyhedron circumscribing a ball; moreover, now all these statements are proved for n-dimensional space. 8.3.9 We derive one more expression for the mixed volume of several polyhedra analogous to the expression (5) for the mixed volume V (P 0 P 1 . . . P 1 ). Theorem 4. Let P 1 , . . . , P n be convex polyhedra and let V (P 1 P 2 . . . P n ) be their mixed volume, i.e., the coefficient of λ1 λ2 . . . λn , divided by n!, in the expression for the volume of the polyhedron P = λ1 P 1 + . . . λn P n . Let Q1i , Q2i , . . . , Qni be the faces of P 1 , . . . , P n whose combination yields an (n − 1)-dimensional face Qi = λ1 Q1i + . . . + λn Qni of P . If F (Q2i , . . . , Qni ) denotes the mixed area (the mixed (n − 1)-dimensional volume) of faces Q2i , . . . , Qni , then V (P 1 . . . P n ) =

1 1 h F (Q2i . . . Qni ). n i i

(23)

Put λ2 P 2 + . . . + λn P n = P ∗ . Then P = λ1 P 1 + P ∗ and V (P ) = V (P ∗ . . . P ∗ ) + nλ1 V (P 1 P ∗ . . . P ∗ ) + . . . .

(24)

By Theorem 2, 12

This is evident from the formula X = of mass.

1 V

xF dx for the coordinate of the center

376


V (P 1 P ∗ . . . P ∗ ) =

1 1 hi F (Q∗i . . . Q∗i ) n

and, since the face Q∗i is the linear combination of the faces Q2i , . . . , Qni ), we have n

V (P 1 P ∗ . . . P ∗ ) =

j1 ,...,jn−1 =2

λj1 . . . λjn−1

1 1 j h F (Qji 1 . . . Qi n−1 ). n i i

(25)

Divided by (n − 1)!, the coefficient of the product λ2 . . . λn equals 1 1 h F (Q2i . . . Qni ). n i i Owing to (24), this expression is also the coefficient of λ1 . . . λn , divided by n!, in the polynomial V (P ) = λj1 λj2 . . . λjn V (P j1 P j2 . . . P jn ), which concludes the proof of (23). In particular, some of the polyhedra P j may coincide; (23) remains valid anyway. Thus, for the simplest mixed volume V (P 0 P 1 . . . P 1 ), we obtain the expression V (P 0 P 1 . . . P 1 ) =

1 0 1 hi F (Q1i . . . Q1i ) = h1i F (Q0i Q1i . . . Q1i ). n i n

(26)

8.3.10 The “theory of mixed volumes” was constructed by Minkowski in his article [Min3]. Alongside the results presented here, this article contains other inequalities between mixed volumes, as well as their applications to extremal problems of the theory of convex bodies and the uniqueness theorems for convex bodies with given values of certain curvature functions. The diversity and originality of the connections revealed between different hard problems make the theory of mixed volumes one of the most attractive fields of geometry. A survey of results obtained until 1934 is given in the book [BF]. The theory was developed further in my article [A3]. (Simultaneously, part of the same results was published by Fenchel [Fen1].)13 Articles in this direction appear occasionally up to date.14 In Section 9.3 of Chapter 9 we apply the theory of mixed volumes to proving rigidity theorems for polyhedra.15 13

At present, various proofs of the Alexandrov–Fenchel inequality for mixed volumes can also be found in [Lei], [BZ], [Grm]. – V. Zalgaller 14 Time has confirmed the variety of connections between the theory of mixed volumes and other fields of mathematics (see [Kh], [Ego], [Grm]). Now, a survey of results in the theory of mixed volumes can be found in the book [Sch]. – V. Zalgaller 15 Studies of stability in certain inequalities [Di], [Gr] sometimes also give new proofs of uniqueness and stability theorems of the theory of surfaces. See [Di] for more details. – V. Zalgaller

9 Polyhedra with Vertices on Prescribed Rays

9.1 Closed Polyhedra 9.1.1 In this section we consider closed convex polyhedra with common interior point O and vertices lying on some rays e1 , . . . , en issuing from O. We assume that each ray contains only one vertex of a polyhedron. Such polyhedra exist for given O and e1 , . . . , en if and only if the rays e1 , . . . , en do not lie inside one (closed) half-space. Indeed, if all rays ei lie inside the same half-space bounded by a plane Q through O, then every polyhedron with vertices on ei must lie on one side of Q, implying that O cannot belong to the interior of the polyhedron. At the same time, if ei do not lie inside one half-space, then, taking points Ai at equal distances from O and taking their convex hull, we obtain a polyhedron P which is inscribed in a ball and has vertices Ai . The point O must be interior to P . Otherwise some plane through O would bound some half-space that includes the polyhedron and all the rays ei would lie inside this halfspace, contradicting the assumption. For this reason, we always assume that the rays ei do not lie inside one half-space. 9.1.2 When the rays ei are fixed, we can completely determine the polyhedron P by using the distances ri from the vertices of P to O. The numbers ri are not arbitrary: a point Ak on a ray ek is a vertex of P if and only if the convex hull of the other points Ai does not contain Ak . This requirement imposes certain constraints on ri . Writing out these constraints explicitly, we agree for convenience that e1 , . . . , en denote not only rays but also the unit direction vectors of the rays. We treat the point O as the origin. Theorem 1. For positive numbers r1 , . . . , rn to serve as the distances from the vertices to O for some convex polyhedron with vertices on prescribed rays e1 , . . . , en , it is necessary and sufficient that the following holds: If (1) ek = αki1 ei1 + αki2 ei2 + αki3 ei3 is an expansion of one of the vectors ei in the three remaining vectors with nonnegative coefficients, then

378


1 αki1 αki2 αki3 < + + . rk ri1 ri2 ri3

(2)

Necessity. Assume that a polyhedron P has all its vertices A1 , . . . , An on the rays e1 , . . . , en . If r1 , . . . , rn are the distances from the vertices to the −−→ origin O, then the vectors OAi are equal to ri ei . Suppose that a vector ek has expansion (1) with nonnegative coefficients α. This means that the ray ek lies inside the trihedral angle spanned by the rays ei1 , ei2 , and ei3 . If the point Ak on the ray ek is a vertex of P , then the plane passing through Ai1 , Ai2 , and Ai3 separates Ak from O. Therefore, if nx = p

(3)

−−→ is the normal equation of the plane Q, then for the vector OAk we have nek > p.

(4)

Inserting the expansion (1) of ek in (4), we obtain αki1 (ei1 n) + αki2 (ei2 n) + αki3 (ei3 n) >

p . rk

(5)

−−−→ Since the point Aij lies on Q, the vector OAij = rij eij satisfies equation (3). Therefore, p rij (neij ) = p or neij = . (6) rij Inserting these expressions for the inner products neij in (5) and cancelling p, we come to (2). Sufficiency. Let the positive numbers r1 , . . . , rn satisfy the conditions of the theorem. Taking the points A1 , . . . , An on the rays e1 , . . . , en and taking their convex hull, we obtain a polyhedron P . Suppose that one of the points, say A1 , fails to be a vertex of P . Then A1 belongs to the convex hull of the other points A2 , . . . , An . Let Ai1 , . . . , Aik (k ≤ n−1) be the list of all vertices of P . By assumption, the rays ei do not lie inside one half-space and O belongs to the interior of P . Hence, P decomposes into tetrahedra with one common vertex O and other vertices the points Ai1 , . . . , Aik . The point A1 lies in one of these tetrahedra, say OA2 A3 A4 . Then A1 and O are situated to one side of the base A2 A3 A4 of −−→ OA2 A3 A4 , which implies that the vector OA1 = r1 e1 satisfies the inequality reverse to (4): (7) ne1 r1 ≤ p. At the same time, the ray e1 lies inside the trihedral angle spanned by the rays e2 , e3 , and e4 and the direction vector e1 therefore expands in the vectors e2 , e3 , and e4 with nonnegative coefficients. Repeating the preceding arguments, from (7) we infer the inequality reverse to (2). This contradicts

9.1 Closed Polyhedra

379

the assumption and so A1 does not belong to the convex hull of the other points Ai and is consequently a vertex of P . Incidentally, Theorem 1 provides a description for the manifold of closed convex polyhedra with vertices on prescribed rays. Indeed, consider the ndimensional space with coordinates the numbers xi = 1/ri reciprocal to the distances from the vertices of a polyhedron to the origin O. In these coordinates, inequalities (2) are linear and, combined with the positivity conditions xi > 0, determine a convex polyhedral angle with vertex the origin. Each interior point of this angle corresponds to some polyhedron. The points corresponding to similar polyhedra with the center of similarity O lie on the same half-line issuing from the vertex of the angle. 9.1.3 We now wish to clarify the extent to which a polyhedron with vertices on prescribed rays is determined by the curvatures at the vertices and find the values of these curvatures for which such a polyhedron exists. The curvature at a vertex is 2π minus the sum of the planar angles contiguous to the vertex. We can also treat the curvature as the area of the spherical image of the vertex (Section 1.5). Therefore, similar polyhedra have the same curvatures at corresponding vertices. The converse is also true: the equality of the corresponding curvatures implies the similarity of polyhedra with vertices on prescribed rays. This claim is a particular instance of the following theorem. Theorem 2. Assume that the vertices of two closed convex polyhedra P and P lie on the same rays issuing from a common interior point O of the polyhedra. Then either the polyhedra are similar (with O the center of similarity) or one of them has a vertex such that the polyhedral angle of the polyhedron at this vertex can be placed by translation inside the polyhedral angle of the other polyhedron at the corresponding vertex. (A polyhedral angle V1 is placed inside another angle V2 if the vertices of V1 and V2 coincide, V2 includes V1 , and V2 differs from V1 . As usual, we consider polyhedral angles as having faces extending indefinitely.) This theorem can be rephrased as follows: if the polyhedral angles at the corresponding vertices of P and P cannot be taken inside one another by translation, then the polyhedra are similar. Owing to the close relationship between a polyhedral angle and its spherical image, the theorem may be restated in terms of the translation of the spherical images of vertices inside one another. (Clearly, the increase of a polyhedral angle is equivalent to the decrease of its spherical image.) The sum of planar angles in the ambient polyhedral angle V2 is greater than the corresponding sum in V1 . On the other hand, the curvature at the vertex is clearly less. So, equality of curvatures at the corresponding vertices of P and P readily implies that the polyhedral angles at the corresponding

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vertices cannot be taken inside one another by translation. Therefore, the similarity of two polyhedra P and P with the same curvatures at corresponding vertices is indeed a particular instance of Theorem 2. The proof of Theorem 2 is extremely simple and has been already described in Section 2.5 of Chapter 2. Let us recall it. Assume that two polyhedra P and P have all vertices on the same rays ei issuing from a common interior point O. This condition is not violated under a similarity transformation of P , and the polyhedral angles of P do not change (but are merely moved by translation). Therefore, we may apply any similarity transformation to P without violating the conditions of the theorem. Contract P similarly to O so that P lie inside P . Now, begin expanding P continuously until one of the vertices of P coincides with the corresponding vertex of P . Since all vertices of P remain in the solid polyhedron P , the polyhedron P itself remains within P because P is the convex hull of its own vertices. If P and P coincide, then this means that P was similar to P before our transformation. Therefore, suppose that P differs from P and merely touches it from within. When P is contained in P and a vertex A of P coincides with a vertex A of P , the polyhedral angle V of P at A will be be inside the polyhedral angle V of P at A. If the angles V and V differ, then we have found an angle of P that can be be taken by translation inside the corresponding angle of P . If V and V coincide, then their edges coincide and so do the vertices of P and P which are the endpoints of the edges, because by assumption these vertices lie on the same rays issuing from O. As regards these vertices, we may repeat the same arguments. Continuing in the same way, we eventually come to some pair of coinciding vertices of P and P at which the angles differ: otherwise all vertices of P and P would coincide, contradicting our assumption. Thus, supposing P and P not similar, we find a pair of corresponding vertices such that the polyhedral angle at the vertex of P can be taken by translation inside the polyhedral angle at the vertex of P . This completes the proof of the theorem. Theorem 2 readily implies the following result. Say that f (V ) is a monotone function of a convex polyhedral angle if f (V ) = f (V ) whenever V and V are translates of each other and f (V ) < f (V ) whenever V can be taken inside V by translation. Theorem 3. If two polyhedra with vertices on given rays are such that, for every pair of corresponding vertices, the values of some monotone function coincide at the polyhedral angles at the vertices, then the polyhedra are similar. Indeed, the monotonicity condition implies that the corresponding polyhedral angles cannot be taken inside one another by translation, and the polyhedra are similar by Theorem 2.


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Here, for a pair of corresponding vertices, we can take a separate monotone function. Besides the sum of planar angles (or curvature), as an example of a monotone function, we can take the area of the spherical polygon cut out by a polyhedral angle from the unit sphere centered at the vertex of the angle. 9.1.4 The following theorem settles the question of allowable values of curvatures at vertices for a polyhedron with vertices on prescribed rays. Theorem 4. Suppose rays e1 , . . . , en issue from a point O and do not lie inside one half-space. Let σi1 ,...,im be the curvature of the polyhedral angle which bounds the convex hull of ei1 , . . . , eim . If there exists a closed convex polyhedron with vertices on the rays ei , then the curvatures ωi at its vertices satisfy the following inequality for every subcollection ei1 , . . . , eim in the initial collection of rays1 : ωi > σi1 ,...,im , (8) where the sum ranges over all rays lying beyond the convex hull of the rays ei1 , . . . , eim . Conversely, if the positive numbers ωi satisfy all inequalities (8) and the sum of the ωi equals 4π, then there exists a closed convex polyhedron whose vertices lie on the rays e1 , . . . , en and whose curvature at the vertex on the ray ei equals ωi for all i. Since it is clear that the positivity condition of ωi and the condition that the sum of ωi equals 4π are necessary, the theorem asserts that these two conditions together with inequalities (8) are necessary and sufficient for the existence of a polyhedron with prescribed curvatures ωi . The necessity of inequalities (8) was proved in Section 2.5. We recall the proof. Assume that a polyhedron P has vertices on given rays ei . Take rays ek1 , . . . , ekm inside one half-space and take their convex hull V . Then V is a solid angle with vertex O. Let the rays ej1 , . . . , ejp and the vertices Aj1 , . . . , Ajp lie outside V . Every support plane to V intersects P , and some translation of the plane from the point O takes the plane into a position in which it is a support plane to P at some vertex that lies outside V . Consequently, every support plane of V has a translate serving as a support plane of P at one of the vertices Aj1 , . . . , Ajp . Moreover, there are other support planes at these vertices, for instance the planes of faces disjoint from V . Therefore, the spherical image of the collection of the vertices Aj1 , . . . , Ajp is larger than that of V , i.e., (8) holds. 1

The rays ei1 , . . . , eim are assumed to lie inside one half-space: otherwise, the inequality (8) is trivial or meaningless (then the convex hull of ei1 , . . . , eim is the whole space).

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9.1.5 We now prove the sufficiency part of Theorem 4, which claims the existence of a polyhedron with prescribed curvatures ωi provided that all conditions of the theorem are satisfied. To this end, we use the Mapping Lemma of Section 2.1. Consider all closed convex polyhedra with vertices on the given rays e1 , . . . , en . Let ri stand for the distance of the vertex on the ith ray from O. As was observed at the end of Subsection 9.1.2, in the n-dimensional space with coordinates xi = 1/ri , the collection of all polyhedra under consideration is a convex solid polyhedral angle with vertex the point (0, . . . , 0). To each class of similar polyhedra, there corresponds a ray in this angle which issues from the vertex. Therefore, the set of all classes of similar polyhedra can be identified with the part of the unit sphere with center (0, . . . , 0) which is cut out by this solid angle. This (n − 1)-dimensional domain represents the manifold P of classes of similar polyhedra. Let K be the manifold of the collections K of n positive numbers (ω1 , . . . , ωn ) whose sum equals 4π and which satisfy the inequalities (8). In the n-dimensional space with coordinates (ω1 , . . . , ωn ), the linear inequalities (8) together with the inequalities ωi > 0 determine some convex set; the intersection of this set with the plane ni=1 ωi = 4π yields the manifold K. Therefore, K is (n − 1)-dimensional and connected. Every polyhedron under consideration has curvatures at vertices which satisfy the conditions for determining the manifold K. Similar polyhedra have the same curvatures. This gives rise to a mapping ϕ from P into K. The mapping ϕ is one-to-one, since the polyhedra with the same curvatures are similar as we established in Subsection 9.1.3. Clearly, ϕ is continuous. Since K is connected and has the same dimension as P, it suffices to check the last condition of the Mapping Lemma. Let a sequence K1 , K2 , . . . converge to a point K; moreover, let Ki = ϕ(Pi ), i.e., let the points Ki of K be the images of the points Pi of P. We must prove the existence of a converging subsequence Pij whose limit P is carried out by ϕ into K: Pij → P , ϕ(P ) = K. Since the manifold P is a domain on the (n − 1)-dimensional sphere, the sequence Pi certainly has a converging subsequence Pij . However, the limit P0 of the sequence Pij could be a point on the boundary of P rather than inside P. We now show that this is impossible. Each point of P represents a polyhedron in which the distances from the vertices to the origin satisfy the condition 1 = 1. x2i = ri2 Therefore, a point on the boundary of P also represents a convex polyhedron that satisfies the same condition but possibly has some “coordinates” xi = 1/ri that vanish. Such a polyhedron P0 is unbounded and the convex hull of the collection of those rays ei to which there correspond the distances ri = ∞ (xi = r1i = 0) is nothing but the limit angle V of P0 . Indeed, suppose


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that r1 , . . . , rm are finite while rm+1 , . . . , rn tend to infinity. A polyhedron P for which all values ri are finite is clearly the convex hull of the collection of vertices on the rays e1 , . . . , em and the polyhedron R which is the convex hull of the set consisting of the point O and the vertices on the rays em+1 , . . . , en . As the latter vertices tend to infinity, the polyhedron R converges to the polyhedral angle V , the convex hull of the collection of rays em+1 , . . . , en . Therefore, P converges to the convex hull of the collection of the vertices on the rays e1 , . . . , em and the angle V . Hence, V is the limit angle of the limit polyhedron P0 . The curvature of P0 equals the curvature of its limit angle V , i.e., the sum of the curvatures at the vertices A1 , . . . , Am of P0 equals the curvature of the convex hull of the collection of rays e1 , . . . , em . In our notation, this is equivalent to the equality m

ωi = σm+1,...,n .

(9)

i=1

However, since the polyhedra Pi converge, the curvatures at their vertices converge to the curvatures at the vertices of the limit polyhedron P0 . By assumption, the curvatures at the vertices of the polyhedra Pij converge to the collection of numbers ω which belongs to the manifold K, i.e., the limit values of the numbers ω satisfy all inequalities (8). This means that inequality (9) is impossible, and so the polyhedron P0 cannot be unbounded. Thus, the sequence of polyhedra Pij converges to a bounded polyhedron P0 with the limit values of curvatures at vertices. The limit collection K0 of the numbers ω corresponds to this P0 . So ϕ(P0 ) = K0 . This proves that ϕ also satisfies the last condition of the Mapping Lemma. Since all conditions of this lemma are valid, so is its claim: the manifold P is mapped onto the whole manifold K, i.e., to each collection of numbers ω satisfying the conditions of the theorem, there corresponds a class of similar polyhedra. The existence of a polyhedron with the required curvatures is thus proved. 9.1.6 Some theorems analogous to Theorems 2–4 can be established for polyhedra with boundary. We state two of the simplest. Assume given a bounded convex polygon Q on a plane T . Choose some points A1 , . . . , Am inside Q. Let L be a closed polygonal line in space which projects onto the boundary of Q. Consider the convex polyhedra possessing the following properties: (1) every polyhedron has boundary L, (2) all inner vertices of every polyhedron project onto the points Ai , and (3) all polyhedra are convex in the same direction (say, towards the plane T or in the opposite direction).

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For these polyhedra we may assert the following Theorem 5. Every two polyhedra satisfying (1)–(3) (if they do not coincide) always have a pair of corresponding vertices at which the polyhedral angles can be taken inside one another by translation. Indeed, let P1 and P2 be two of these polyhedra; moreover, let P1 have points lying “below” the points of P2 with the same projection to the plane T (we imagine T horizontal). Moving P1 “up,” we reach a position in which P1 touches P2 somewhere but lies “above” P2 as a whole. It is now obvious that P1 has a vertex with the angle at it included in the angle at the corresponding vertex of P2 , which concludes the proof of the theorem. Theorem 6. For arbitrary positive numbers ω1 , . . . , ωm whose sum is less than 2π, there exists a unique convex polyhedron of the type indicated above for which the numbers ωi are the curvatures at vertices: namely, ωi is the curvature at the vertex projecting onto Ai . The uniqueness of the polyhedron follows from Theorem 5. Its existence is immediate from the Mapping Lemma. Indeed, consider the m-dimensional manifold K consisting of all collections of positive numbers ω1 , . . . , ωm whose sum is less than 2π. Let P be the manifold of the polyhedra of the type under study; P is m-dimensional too (and P is obviously nonempty). We have a natural mapping from P into K and easily prove that this mapping satisfies all conditions of the Mapping Lemma. As a result, we come to Theorem 6. (The detailed implementation of the above scheme is left to the reader.) In Theorems 5 and 6 we dealt with polyhedra whose vertices lie on parallel rays perpendicular to the plane T and passing through the points A1 , . . . , Am ; these rays “meet at infinity.” In a perfectly similar way, we may consider polyhedra with vertices on prescribed rays issuing from some point O. Likewise, we may formulate analogs of Theorems 5 and 6 for polyhedra with boundary. Various cases are possible here: the rays lie or do not lie inside one half-space, polyhedra are bounded by one or several polygonal lines, etc. (From polygonal lines we must require that they constitute the boundary of at least one convex polyhedron with vertices on given rays.) In the simplest case, all rays lie inside a convex solid angle and the polygonal line L lies on the surface of this angle; moreover, we should separately consider the polyhedra convex towards the vertex of the angle and those convex outwards it. Stating and proving some of the possible theorems could also serve as a good exercise. A topic of further research could consist in generalizing the above to the case in which the rays do not issue from a common vertex but “pass through” a given polygonal line L. 9.1.7 Theorem 2 is worth comparing with theorems on the infinitesimal deformations of polyhedra. Take a point O inside a closed convex polyhedron


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P0 and draw some rays Li from O through the vertices of P0 . The polyhedron P0 deforms when the vertices move along the rays Li . For sufficiently small displacements, none of the vertices enters into the convex hull of the others and so all vertices of P are still vertices of the deformation. We assume that the vertices move along the rays Li with certain speeds as the parameter t varies. Then, obviously, the faces and edges of the polyhedron rotate with some velocities. As the vertices move, the structure of the polyhedron changes in general. This is so for example when we move the vertices A1 and A2 of the cube in Fig. 145 from the center O of the cube outwards, leaving the vertices A3 and A4 in place, or vice versa. In both cases the face A1 A2 A3 A4 folds but in the first case it folds along the diagonal A1 A3 while in the second, along the diagonal A2 A4 . However, under small deformations the actual edges do not disappear but there may appear new edges arising only from the diagonals of the initial faces. Considering polyhedral angles at the vertices of a polyhedron, we treat these diagonals as “new edges” of the angles, distinguishing them from genuine or “old edges”; the “new edges” lie on the “old faces.” Generalizing, we are in a position to admit deformations that destroy convexity of the polyhedron by folding faces along diagonals. For example, moving the vertices A1 and A3 of the cube in Fig. 145 outwards, we may admit the folding of the diagonal A2 A4 , which leads to a nonconvex polyhedron. A3 A4 A2

A1 O

Fig. 145

In this case also, we say that a polyhedron deforms as the result of the motion of its vertices. We shall consider polyhedral angles to within translation and similarity; these transformations change neither the shape nor the size of the polyhedral angles. We say that a polyhedral angle V0 decreases under a deformation keeping the vertex in place if none of the edges moves outwards with a positive velocity and at least one genuine edge rotates, entering into V0 with

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some nonzero velocity. We say that the angle increases if after changing all velocities to the opposite ones the angle becomes decreasing.2 Theorem 7. Assume that a closed convex polyhedron deforms as the result of some motion of its vertices along given rays issuing from an interior point O of the polyhedron. Then either the initial deformation is a similarity with center O, i.e., the initial speeds of the motion of vertices are proportional to their distances from O, or there exists a vertex at which the polyhedral angle decreases and a vertex at which the polyhedral angle increases (recall the agreement that we consider angles to within translation and similarity; so, comparing angles, we may assume that their vertices coincide). The proof is as simple as that of Theorem 2. Define the relative speed of a vertex Ai to be the speed of variation of the distance OAi per unit length, i.e., the quotient vi = r˙i /ri , where ri = OAi . A similarity transformation is characterized by the equality of all vi . Suppose that the deformation does not reduce to a similarity, which means that not all vi are equal. Take a vertex with the greatest value of the relative speed; let it be A1 . Add (−v1 ) to all relative speeds, i.e., add the relative speed of A1 with opposite sign. This results in adding a similarity transformation to the initial deformation, which does not change the deformation of the polyhedral angles of the polyhedron essentially. Since v1 was the greatest relative speed, now all vertices have nonpositive relative speeds while the speed of A1 vanishes, i.e., A1 stays in place and the other vertices either stay in place or move inwards the polyhedron. Furthermore, they cannot stay in place all together, for otherwise their relative speeds vi would be equal before the addition of −v1 , contradicting our assumption. If at least one vertex Ai joined with A1 by an edge moves inside the polyhedron (i.e., if vi = vi − v1 < 0), then this will mean that the polyhedral angle at A1 decreases. If all vertices joined with A1 by edges remain in place, then we take as A1 any of them and proceed similarly. Eventually we arrive at a vertex with a decreasing polyhedral angle, since not all vertices remain in place. In the same way, subtracting the least relative speed from all relative speeds of vertices, we prove the existence of a vertex with an increasing polyhedral angle, and so conclude the proof of the theorem. From Theorem 5 we can deduce an infinitesimal rigidity theorem analogous to Theorem 3 on the similarity of polyhedra. Recall that a quantity x depending on t is called stationary if dx dt t=0 = 0. We call a polyhedron 2

To a rotation of a genuine edge outwards from the angle V , there may correspond no opposite rotation inwards the angle: this is so for rotations in every support plane passing through the edge. For this reason, when considering genuine edges, we cannot impose conditions on the velocity of motion outwards from the angle.


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infinitesimally rigid under given conditions if under these conditions the polyhedron admits no deformations but the trivial ones; observe that we have in mind only the initial velocities of a deformation. In the case under consideration, a trivial deformation is a similarity, i.e., a deformation under which the initial speeds of vertices are proportional to their distances from the point O. Also, we introduce the notion of essentially monotone (increasing) function of a polyhedral angle. By this we mean a quantity f which depends on a polyhedral angle so that f increases with a positive speed as the angle increases, i.e., df dt > 0. Using the notions introduced in connection with Theorem 7, we readily infer the following infinitesimal rigidity theorem: Theorem 8. A closed convex polyhedron is infinitesimally rigid in the above sense if its vertices remain on given rays issuing from some interior point of the polyhedron and some essentially monotone function is stationary at each of its polyhedral angles. Granted the relationship between a polyhedral angle and its spherical image, we may restate Theorems 5 and 6 in terms of spherical images. Towards this end, consider the deformation of the spherical image of a vertex of a polyhedron which results from the deformation of the polyhedron itself when its vertices move along some rays Li issuing from a point O. When vertices move with certain speeds, the edges rotate with definite velocities. Since the sides of the spherical polygon SAi which is the spherical image of a vertex Ai lie in the planes perpendicular to the edges contiguous to Ai , these sides move with definite velocities too. Strictly speaking, we should speak of the rotation of great circles bounding the polygon SAi ; each of these circles is perpendicular to the corresponding edge incident to Ai and rotates together with the rotation of the edge. If the edge moves inwards the polyhedral angle V at a vertex A, then the corresponding side of S moves outwards from S. Therefore, the decrease of the angle V in the sense of Theorem 5 corresponds to the increase of the polygon S in the sense that its sides move outwards with definite velocities owing to the rotations of the great circles on which these sides lie. Defining the “decrease” and “increase” of the spherical image of a vertex of a polyhedron in accordance with the above observation, we may reformulate Theorems 5 and 6 in terms of spherical images. An essentially monotone increasing function of a polyhedral angle corresponds to an essentially decreasing function of the spherical image and vice versa. 9.1.8 In Subsection 2.5.6 we showed the possibility of polar transformations of theorems about polyhedra with vertices on prescribed rays into theorems about polyhedra with faces perpendicular to prescribed rays. There we also stated the theorems polar to Theorems 2 and 4. Theorems 5 and 6 also admit such polar transformation.

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All results of the current section translate word for word to space of arbitrary dimension n ≥ 2 with the only difference that in Theorem 1 we must consider expansions of the vector ek in terms of n other vectors and in Theorem 4 we should replace 4π by the area of the unit sphere in ndimensional space. Furthermore, curvature should be treated as the (n − 1)dimensional area of the spherical image.

9.2 Unbounded Polyhedra 9.2.1 We consider unbounded convex polyhedra each of which lies above some fixed plane T so that (1) every straight line perpendicular to T either is disjoint from the polyhedron or intersects it in an entire half-line and (2) all vertices of the polyhedron project to given points A0 , A1 , . . . , An of T . From now on these conditions are assumed fulfilled without explicit mention. Since a similarity transformation of a polyhedron does not violate the first condition, the limit angle of the polyhedron satisfies this condition as well. We say that the limit angle “projects univalently” on T .3 We do not exclude polyhedra whose limit angle reduces to a half-line; this half-line is then perpendicular to T . In this connection, speaking of a polyhedral angle V which is the limit angle of a polyhedron, we allow V to be a planar angle or a half-line. Also, in the set of unbounded convex polyhedra we adjoin unbounded convex polygons. For example, if the limit angle V reduces to a planar angle and the points A0 , A1 , . . . , An lie on straight line, then the polyhedron P necessarily reduces to a polygon. This possibility is included throughout without specific mention. Assume given arbitrary points A0 , A1 , . . . , An on T and let V be a polyhedral angle projecting univalently on T . We will prove that in this case there exists an unbounded convex polyhedron with limit angle V and vertices projecting to Ai . The proof proceeds by induction on the number of points Ai . For the point A0 , the claim is obvious since the angle V itself with vertex A0 is the required polyhedron. Now, supposing the claim valid for n points, let us prove it for n + 1 points A0 , A1 , . . . , An . First, assume that none of the points A0 , A1 , . . . , An lies in the convex hull of the others. Placing the vertex of V at A0 , take the convex hull of the angle V and the points A0 , . . . , An . By Theorem 5 of Subsection 1.4.5, we obtain a convex polyhedron with limit angle V and vertices the points A0 , . . . , An , i.e., the required polyhedron. 3

The term is in a sense formal, since the angle may have faces perpendicular to T .

9.2 Unbounded Polyhedra

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Now, assume that at least one of the points, say An , belongs to the convex hull of the other n points A0 , . . . , An−1 . By the induction hypothesis, there exists a polyhedron P with limit angle V and vertices projecting to A0 , . . . , An−1 . Since An belongs to the convex hull of A0 , . . . , An−1 , the line L drawn through An perpendicular to T necessarily meets P , entering into it at some point B (Fig. 146). Taking a point B on the line segment An B sufficiently close to B and considering the convex hull of the polyhedron P and the point B , we obtain a new polyhedron P . Its vertex B projects to An . Further, if the displacement from B to B is small enough, then all vertices of P are also vertices of P . As a result, the polyhedron P has limit angle V and vertices projecting to the points A0 , . . . , An , which completes the proof of our assertion.

T A1

A0

An An-1 B B

P

L

Fig. 146

Every unbounded convex polyhedron is the boundary of the convex hull of its vertices and its limit angle, i.e., it is completely determined from the disposition of all of them; moreover, the prescribed limit angle V may be positioned so that its vertex is any prescribed vertex of the polyhedron. Given the projections A0 , . . . , An on T of the vertices, the vertices themselves are completely determined by the distances h0 , . . . , hn from A0 , . . . , An to the plane T . These distances are assumed positive in the direction of the unbounded part of the polyhedron and negative in the opposite direction. Hence, when the projections of the vertices and the limit angle V are fixed, the polyhedron is completely determined by specifying n + 1 numbers h0 , . . . , hn . The latter cannot be arbitrary, since a point Bk may be a vertex of the convex hull of V and all points Bi with projections Ai if and only if Bk does not belong to the convex hull of the other points Bi together with the angle V , provided that the vertex of V is one of these points. Certainly, this imposes certain constraints on the heights hi of the points Bi , which take the form of inequalities. The derivation of these inequalities presents a rather simple exercise. Since the inequalities are not used in what follows, we leave this as exercise to the reader.

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9.2.2 The following theorem is an analog of Theorem 2 of Section 9.1. Theorem 1. If two unbounded convex polyhedra have the same limit angle V and the same projections A0 , . . . , An of vertices on a plane T , then either the polyhedra can be superposed by a translation in the direction perpendicular to T , or one of them has a vertex such that the polyhedral angle of the polyhedron at this vertex can be taken by such translation inside the polyhedral angle of the other polyhedron at the corresponding vertex. In other words, two polyhedra with the same data V and A0 , . . . , An are translates of one another provided that the polyhedral angles at their corresponding vertices cannot be placed inside one another by translation. As in Section 9.1, we consider an angle W placed inside W if W is properly included in the (solid) angle W . The proof of this theorem is as simple as that of Theorem 2 in Section 9.1. Indeed, let two polyhedra P and P have the same data V and A0 , . . . , An . We may assume that the vertices of the limit angles of P and P are at some corresponding vertices. When P translates in the direction perpendicular to the plane T , the limit angle of P undergoes the same translation, while the projections of vertices do not change; hence, we may translate P in this way. Assume P translated so that its vertices occur above the corresponding vertices of P (i.e., hi > hi for all i). Then the limit angle of P turns out to be inside the limit angle of P . Moreover, once an unbounded polyhedron is the convex hull of the collection of its vertices and the limit angle, P itself lies in P . Now, move P towards T until the first moment when one of its vertices, say B , coincides with the corresponding point B of P . The polyhedron P is still contained in P since all its vertices and the limit angle remain in P (more precisely, in the body bounded by P ). This is clear from the fact that a solid polyhedron is the convex hull of the collection of its vertices and the limit angle. Therefore, the polyhedral angle W at the vertex B of P is included in the angle W at the vertex B of P . If the angles W and W differ, then this is exactly the case in which W is placed inside W by translation, i.e., we arrive at the second possibility of the theorem. If the angles W and W coincide, then their edges coincide and, consequently, the vertices of P and P at the ends of these edges coincide, because by assumption these vertices lie on some common perpendiculars to the plane T . We may repeat the same argument for the polyhedral angles at these vertices and proceed in this way until we arrive at a pair of corresponding vertices at which the polyhedral angles differ. Then the angle U at such a vertex of P is placed ins the corresponding angle U . If we do not find a pair of different angles, then this means that all angles coincide, i.e., the polyhedron P coincides with P , implying that the original polyhedron P is a translate of P , i.e., we come back to the first possibility of the theorem. From Theorem 1 we deduce an analog of Theorem 3 of Section 9.1.


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Theorem 2. If two unbounded convex polyhedra with the same limit angle and the same projections of vertices to a plane T are such that, for every pair of corresponding vertices, the values of some monotone function coincide at the polyhedral angles at the vertices, then the polyhedra are translates of one another. In particular, the claim is valid for the case in which the curvatures at the corresponding vertices are equal. 9.2.3 The following theorem is an analog of Theorem 4 of Section 9.1 for unbounded polyhedra. Theorem 3. Let A0 , . . . , An be points on a plane T and let V be a polyhedral angle projecting univalently on T . Then, for the existence of an unbounded convex polyhedron with vertices projecting to the points A0 , . . . , An , having the limit angle V and curvatures ω1 , . . . , ωn at the vertices, it is necessary and sufficient that the numbers ω1 , . . . , ωn satisfy the following conditions: (1) 0 < ωi < 2π (i = 1, . . . , n); (2) the sum of all ωi equals the curvature of V . In accordance with Theorem 2, such a polyhedron is unique up to translation in the direction perpendicular to the plane T . The necessity of the first condition is obvious. That of the second condition becomes apparent if we recall that, as we saw in Subsection 1.5.3, the spherical image of an unbounded convex polyhedron coincides with the spherical image of its limit angle. This also means that, rather than specifying the angle V , we could specify a convex spherical polygon that would serve as the spherical image of the required polyhedron. We are left with proving the sufficiency of the conditions of the theorem, i.e., we must prove the existence of a polyhedron with given curvatures. To this end, we again use the Mapping Lemma. Assume given A0 , . . . , An and V . Consider an unbounded convex polyhedra P with the limit angle V , one vertex the point A0 , and the other vertices projecting to the remaining points A1 , . . . , An . As established in Subsection 9.2.1, the manifold P of these polyhedra is nonempty. Each polyhedron under consideration is determined by the specification of n heights h1 , . . . , hn of its vertices projecting to the points A1 , . . . , An . If h01 , . . . , h0n are the heights of vertices of a polyhedron P 0 , then all sufficiently close numbers h1 , . . . , hn are also the heights of vertices for some polyhedron P . Indeed, the condition on the heights of vertices consists in requiring that no vertex belongs to the convex hull of the collection of the other vertices and the angle V . If this condition is satisfied for some heights h01 , . . . , h0n , then it is also satisfied for sufficiently close values of the heights. Hence, each point P 0 in P has an n-dimensional cubical neighborhood such that P is an open set in the n-dimensional space with coordinates h1 , . . . , hn .

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Now, consider the (n+1)-dimensional space with coordinates (ω0 , . . . , ωn ). Let Ω be the curvature of V . Condition (2) of the theorem yields ω0 + ω1 + . . . + ωn = Ω,

(1)

i.e., it determines some plane in our space. The first condition of the theorem distinguishes the interior of some cube in this space whose intersection with the plane (1) determines the region of values ωi satisfying both conditions of the theorem. This region is the manifold K of admissible collections of numbers ωi . Observe that K is a connected n-dimensional manifold. Since to each polyhedron P in the manifold P corresponds the collection of curvatures at the vertices of P , we have a natural mapping ϕ from the manifold P into the manifold K. Clearly, this mapping is continuous. Since one vertex of P is fixed, by Theorem 3 polyhedra with the same curvatures simply coincide. Hence, the mapping ϕ is one-to-one. Finally, K is a connected manifold of the same dimension as P. We thus see that the first three conditions of the Mapping Lemma are satisfied and so it remains to establish the forth condition. Let a sequence of points Kj of K, representing the collections of curvatures at the vertices of polyhedra Pj , converge to a point K of K: Kj = ϕ(Pj ) → K = (ω0 , ω1 , . . . , ωn ) ∈ K.

(2)

Since the point K = (ω0 , . . . , ωn ) belongs to K, we have ωi > 0 by the first condition of the theorem, i.e., the curvatures at the vertices of Pj converge to positive values. Therefore, there is an ε > 0 such that for all curvatures at the vertices of Pj we have ωij > ε

(i = 0, . . . , n; j = 1, 2, . . .).

(3)

Translate every polyhedron Pj in the direction perpendicular to the plane T so that T becomes a support plane of Pi . Since all vertices of Pj project to the given points Ai , we can choose a sequence of polyhedra Pj such that all polyhedra of the sequence touch T at the same point in the set of points Ai , say An . To simplify notation, we still denote the polyhedra of this sequence by Pj . Let a vertex Bik of a polyhedron Pmk lie at some height hki above the plane T and let ri be the distance between An and the projection Ai of Bik on T . The point An is a vertex of the polyhedron Pmk ; therefore, no support plane of Pmk at Bik intersects the ray from An that passes through Pmk perpendicular to T . By an obvious and elementary argument, we conclude that if θ is the angle between this ray and a support plane at Bik , then tan θ ≤ Fix a θ0 such that

ri . hki

(4)


tan θ0 =

ri . hki

393

(5)

Then (4) implies that the spherical image of the vertex Bik lies in the strip of width θ0 around the equator of the unit sphere if we take the spherical image of the plane T to be a pole. The area of this strip equals 2π sin θ0 . At the same time, by (3) this area is greater than ε. Therefore, 2π sin θ0 > ε

and

tan θ0 >

ε . 2π

(7)

Using (5), from this we infer the inequality hki
0 and It is in this aspect that the above theorem differs from Theorem 4 of Section 9.1 for closed polyhedra: the latter involves the extra condition (8) which we cannot regard as trivial as (1) and (2). This condition relates, however, to the disposition of the rays ei . A natural question arises in this connection: what conditions are necessary and sufficient for given numbers ω1 , . . . , ωn to serve as the curvatures at vertices for some closed convex polyhedron? The 4

Of course, we translate the polyhedra of this sequence backwards, so that they all have one vertex A0 .

394


n conditions (1) 0 < ωi < 2π and (2) i=1 are necessary. Are they sufficient?5 (Conditions (1) and (2) imply that we must have n ≥ 3. If n = 3, then the positive answer is given by the triangle with angles equal to π − ωi /2; this polyhedron, although degenerate, has ωi as the curvatures at vertices.) 9.2.5 Using the notions of Subsection 9.1.6, in particular the decrease and increase of a polyhedral angle, we can formulate a theorem that relates to Theorem 1 in the same manner as Theorem 5 of Section 9.1 relates to Theorem 2 of Section 9.1. Theorem 4. If an unbounded convex polyhedron, with a constant limit angle projecting univalently on a plane T , deforms as the result of some motion of its vertices along lines perpendicular to T , then at least one polyhedral angle of the polyhedron decreases and at least one increases. Further, from Theorem 4 we can derive an infinitesimal rigidity theorem perfectly analogous to Theorem 6 of Section 9.1. Finally, Theorem 4, as well as the infinitesimal rigidity theorem, can be restated in terms of the spherical images of vertices. These facts, like the proof of Theorem 4, are quite similar to those in Subsection 9.1.6. For this reason, we do not dwell on them. 9.2.6 In the preceding subsections, we considered unbounded polyhedra with vertices on prescribed parallel rays. This may be interpreted as the fact that the common origin of all the rays was the point at infinity. We could, however, consider unbounded convex polyhedra with vertices on prescribed rays issuing from a fixed point O lying inside the polyhedron. In this case, it is natural to place the vertex of the limit angle at O. We now may raise questions quite analogous to those settled in Section 9.1 for closed polyhedra: (1) Under what conditions on the disposition of the rays e1 , . . . , en and the angle V are there unbounded convex polyhedra with vertices on the rays e1 , . . . , en and the limit angle V ? (2) What necessary and sufficient conditions determine the region of admissible values for the distances ri from the vertices of a polyhedron to the origin O (provided that the rays ei and the angle V are fixed)? (3) What conditions are necessary and sufficient for numbers ω1 , . . . , ωn to serve as the curvatures at vertices for a polyhedron with given limit angle and vertices lying on given rays? The fact that the rays ei , the curvatures ωi , and the limit angle V determine a polyhedron uniquely to within a similarity transformation with center O is proved in much the same way as the similar result for closed polyhedra. Naturally, in the new case we also have some general theorems that are similar to Theorems 2 and 3 of Section 9.1. 5

A. A. Zilberberg gave a positive answer to this question [Zil2], [Zil3] (see also [Si]). The existence of unimmersed two-dimensional Riemannian manifolds with preassigned curvature is studied in [Bu1], [Bu2], and [KW1–KW3]. – V. Zalgaller

9.3 Generalizations

395

The above problems admit complete solutions by means of the same methods as those of Section 9.1. We therefore, leave it to the reader to formulate and prove the corresponding theorems. 9.2.7 As we have already observed in Section 1.1 of Chapter 1, studying an unbounded convex polyhedron is essentially equivalent to studying an endlessly prolongable bounded polyhedron. Moreover, the condition for endless prolongation without new intersections of extreme faces is the convexity of the spherical image. At the same time, the specification of the limit angle is equivalent to that of the spherical image of the polyhedron. In the light of these observations, we see that our theorems on unbounded polyhedra could be rephrased for polyhedra with boundary.6 9.2.8 In conclusion, we note that the results of this section can be carried over, together with their proofs, to spaces of arbitrary dimension n ≥ 2; only in Theorem 3 we must replace 2π by the area of the hemisphere in the corresponding space.

9.3 Generalizations 9.3.1 We have already noted that it is possible to generalize the results of Sections 9.1 and 9.2 to polyhedra in n-dimensional space using essentially the same words as before, and so there is no need to dwell upon details. With appropriate modifications, the same results are valid for polyhedra in hyperbolic space. However, since similarity and translation are absent in this space, the uniqueness theorems and the conditions of existence theorems are stated differently. The curvature at a vertex is defined as before: it equals 2π minus the sum of the planar angles at the vertex, which in turn equals the value of the solid angle covered by the normals to the supports planes drawn at the vertex. We formulate the result for closed polyhedra: Let l1 , . . . , lm be rays from a point O in the hyperbolic space not lying inside one half-space. Then, for the existence of a closed convex polyhedron with vertices on the rays l1 , . . . , lm and curvatures ω1 , . . . , ωm at the vertices, it is necessary and sufficient that the numbers ωi satisfy the following condi(2) for every collection of rays li1 ,...,lik tions: (1) 0 < ωi < 2π (i = 1, . . . , m); lying inside one half-space, we have ωi > σi1 ,...,ik , where σi1 ,...,ik is the curvature at the vertex of the polyhedral angle which is the convex hull of ωi ranges over all remaining rays; and (3) the rays li1 , . . . , lik and the sum the sum of all ωi is greater than 4π. Moreover, such a polyhedron is unique. 6

About polyhedra with boundary, see [Zil1], [P4, p. 52], [El], [GKS]. – V. Zalgaller

396


Comparing this theorem with Theorems 3 and 4 of Section 9.1, we see that, first, a polyhedron with vertices on prescribed rays and prescribed curvatures at vertices is unique with no appeal to similarity. Second, conditions (1) and (2) of the theorem repeat the corresponding conditions of Theorem 4 (with the only difference that now we should add the necessary condition ωi < 2π which in Theorem 4 of Section 9.1 was a consequence of the other conditions). However, condition (3) differs from the condition that the sum of curvatures at vertices equals 4π (which we have in Euclidean space); in hyperbolic space this sum can be any number greater than 4π. We do not formulate other analogs of Theorems 1 and 2. Their statements and proofs cause no essential difficulties; in particular, all existence proofs appear easily on using the Mapping Lemma. 9.3.2 We now turn to generalizations to arbitrary convex bodies in Euclidean space. Let H be a convex body, let O be a point inside H, and let r(e) be the distance from O to the boundary of H in the direction of a unit vector e. To every vector x let us associate the number D(x) =

|x| , r (x/|x|)

(1)

which is the ratio of the length of x to the distance from O to the boundary of H in the direction of x. (For x = 0, the direction is undefined and we put D(0) = 0.) It was H. Minkowski who introduced this function of x and called this function the distance function of H. The following theorem is comparatively easy: A function D(x) of a vector x is the distance function of some convex body if and only if the following conditions are satisfied: (1) D(λx) = λD(x) for every λ ≥ 0; (2) D(x + y) ≤ D(x) + D(y) for all x and y; (3) D(x) > 0 whenever x = 0. The necessity of conditions (1) and (3) is obvious from the very definition of the distance function. Condition (2) is obvious from the convexity of H. By the first condition, the function D(x) is completely determined from its values at the unit vectors e. Since |e| = 1, definition (1) implies that D(e) =

1 , r(e)

i.e., D(e) is the reciprocal of the distance from O to the boundary of H in the direction of e. From conditions (1) and (2) we easily deduce that if e = α1 e1 + α2 e2 + α3 e3 with all αi > 0, then

9.3 Generalizations

397

D(e) < α1 D(e1 ) + α2 D(e2 ) + α3 D(e3 ) or

1 α1 α2 α3 < + + . r(e) r(e1 ) r(e2 ) r(e3 )

This is nothing but inequality (2) from Theorem 1 in Section 9.1. It is perfectly apparent that Theorem 1 of Section 9.1 is a particular instance of the above stated Minkowski theorem. For polyhedra, there is no need to consider the function D(x): it suffices to take finitely many of its values, the reciprocals of the distances from vertices to the origin O. It is easy to see that the distance function of a polyhedron is piecewise linear. Indeed, the distance function is linear in each solid angle projecting a face of the polyhedron from the point O. Thereby this function is determined from its values at the vectors ei directed to the vertices of the polyhedron. 9.3.3 The formal analogy between the distance and support functions, and accordingly between Theorem 1 of Section 9.1 and Theorem 1 of Section 7.5, has an interesting geometric background. Consider the polar transformation with respect to the unit sphere, i.e., the transformation that takes a point with position vector re, where e is a unit vector and r > 0, to the plane with the normal equation ex = 1/r. It is easy to prove7 that this transformation assigns to each convex body H the convex body H determined by the following rule: every interior point of H corresponds to a plane disjoint from H and every boundary point of H corresponds to a support plane of H . (The body H is the intersection of the half-spaces determined by these support planes.) Since the inverse of the polar transformation is the polar transformation itself, this connection between the bodies H and H is reciprocal. Assume that the origin lies inside the body H; then it also lies inside H . From the definition of polar transformation and the relationship between H and H , we easily derive that the distance function of one of the bodies is the support function of the other and vice versa. If H is a polyhedron, then H is a polyhedron as well; moreover, to the faces of one polyhedron correspond vertices of the other and vice versa. 9.3.4 Let F be a closed convex surface, O a point inside F , and E the unit sphere of center O. Given a set M of points on E, let ω(M ) denote the area of the spherical image of the subset of F whose central projection from O on E is M . We call the set function ω(M ) the translated curvature of F on E or simply the integral curvature of E. This function is defined for every Borel set M on E. Using this notion, we can formulate a theorem that generalizes Theorem 4 of Section 9.1 to arbitrary closed convex surfaces: 7

Cf. Subsection 1.5.4.

398


For a function ω(M ) of a (Borel) set M on a sphere E to be the translated curvature on E for some convex surface F such that the center O of E lies inside F , it is necessary and sufficient that (1) ω(M ) be nonnegative and countably additive; (2) ω(E) = 4π 8 ; and (3) for every spherically convex set M the inequality ω(E − M ) > σ(M ∗ ) hold; here σ(M ∗ ) is the area of the set M ∗ “dual to” M , i.e., the set formed by the ends of the normals to the support planes of the cone with vertex O which cuts out M from E. If the function ω reduces to finitely many “pointwise weights” ω1 , . . . , ωm at the points A1 , . . ., Am (i.e., ω(M ) = 0 if M contains none of the points Ai and ω(Ai ) = ωi ), then this theorem transforms into Theorem 4 of Section 9.1. The proof of the theorem is carried out by passing to the limit from polyhedra (see [A6]). The corresponding uniqueness theorem reads [A9]: If two closed convex surfaces F and F have the same translated curvature on the sphere E, then one of the surfaces can be taken onto the another by a similarity transformation with center of similarity the center O of the sphere E.9 To Theorems 2 and 3 there also correspond general theorems about general convex surfaces that are proved as easily as before. For instance, for Theorem 2 we have the absolutely trivial assertion: if two closed convex surfaces F1 and F2 have a common point inside each of them, then either the surfaces are similar with center of similarity O or, by applying a similarity transformation with center O, we can take F1 into F2 while touching it at some point, or vice versa. The distinction from Theorem 2 of Section 9.1 consists in the fact that in the case of polyhedra we consider vertices and so the theorem becomes slightly less trivial. Passing to spherical images, we may counterpose to Theorem 2 of Section 9.1 the following assertion which is not as evident but has a very easy proof: Let F1 and F2 be closed convex surfaces with a common point O inside each of them. Then the surfaces are either similar with center of similarity O or contain sets M1 and M2 , with the same cone projecting them from O, such that the spherical image of M1 is a proper part of the spherical image of M2 . This assertion implies the theorem that asserts the similarity of surfaces F1 and F2 such that some monotone set function on the sphere takes the same values at the spherical images of sets M1 and M2 with the same projecting 8 9

In n-dimensional space, ω(E) equals the area of the unit sphere. In the article [K], A. E. Kogan obtained estimates for the variation of the shape of a convex polyhedron or a convex surface under the variation of the distribution of the translated curvature on the sphere. Clearly, these estimates also imply the above theorem. – V. Zalgaller

9.3 Generalizations

399

cone. In particular, we obtain the theorem on the similarity of surfaces with the same translated curvature on a sphere E of center O. My note[A9], in which this similarity theorem was established, in fact contains the proof but not statement of this general theorem. 9.3.5 Let F be an unbounded complete convex surface and let e be the direction of some ray inside F . Let T be a plane perpendicular to e and let M be an arbitrary Borel subset of T . Denote by ω(M ) the area of the spherical image of the subset of F that projects (along e) onto M . It can happen that the projection of F does not cover the whole plane. Then not every M may serve as the projection of some part of F . In that event, if M is disjoint from the projection of F , then we put ω(M ) = 0 and if M is partly covered by the projection of F , then we consider only the spherical image of the subset of F whose projection is contained in M . We thus define the set function ω(M ) on the plane T . We call this function the translated curvature of F on T . Using this notion, we can formulate the following generalization of Theorem 4 of Section 9.2 to arbitrary unbounded convex surfaces: For a function ω(M ) of a (Borel) subset of a plane T to be the translated curvature on T for some convex surface with a prescribed limit cone K, it is necessary and sufficient that (1) ω(M ) be nonnegative and countably additive; (2) if M is a single point, then ω(M ) < 2π; and (3) the value of ω on the whole plane is to equal the curvature of the cone K. Here the cone K may degenerate into a doubly-covered planar angle or a half-line. In any case, we assume that every straight line perpendicular to T is either disjoint from K or intersects it in an entire half-line. This theorem is especially interesting because it shows that the curvature of a convex surface, regarded as a set function, must only satisfy trivial conditions (although a rigorous proof is not easy). The theorem results from passage to the limit from polyhedra. The corresponding uniqueness theorem reads: If two unbounded complete convex surfaces whose limit cones are translates of one another have the same translated curvature on a plane T , then the surfaces can be taken to one another by a translation in the direction perpendicular to T (provided that ω(T ) > 0; if ω(T ) = 0, then the surface may be any unbounded cylinder and its limit cone reduces to a dihedral angle). This theorem is a corollary to the general theorem: if two unbounded surfaces with the same limit cones project univalently on a plane T , then they can either be superposed by a translation in the direction perpendicular to T or contain sets M1 and M2 , with the same projection on T , such that the spherical image of M1 is a proper part of the spherical image of M2 . If F is a regular surface given by the equation z = f (x, y), then its translated curvature on the (x, y) plane is

400


ω(M ) =

K dF = M

2 zxx zyy − zxy dxdy; (1 + zx2 + zy2 )3/2

this immediately follows from the familiar expression for the Gaussian curvature K and the area element dF . Therefore, in the regular case, our theorems are equivalent to the existence and uniqueness (to within a summand) theorems for the solution of the equation 2 = (1 + zx2 + zy2 )3/2 g(x, y) zxx zyy − zxy

on the whole plane, with g(x, y) a known function (g = dω/dxdy). Note, however, that our theorems do not imply even that the function z = f (x, y) is twice differentiable. Therefore, the solvability of the equation should be understood in a generalized sense. However, if g(x, y) is differentiable sufficiently many times, then z = f (x, y) is twice differentiable, i.e., it is a solution to the equation in the usual sense. A generalization of Theorem 6 of Section 9.1 reads: Assume given a countably additive nonnegative set function ω(M ) on a convex domain G of a plane T such that ω(G) < 2π. Let L be a closed curve in space which projects onto the boundary of G. Then there exists a convex surface with boundary L for which ω(M ) is the translated curvature on T . There are exactly two such surfaces: they are convex in opposite directions. In the regular case, when translated into the language of differential equations, this assertion amounts to the solvability of the boundary value problem for the above equation in all convex domains and continuous data on the boundary. 9.3.610 Further generalizations are closely connected with the theory of elliptic equations of Monge–Ampère type. These equations are both the object of geometric research and a tool for solving geometric problems. Consider the simplest example of these equations 2 ) = ϕ(x, y), R(zx , zy )(zxx zyy − zxy

(1)

where R and ϕ are given nonnegative functions. Let z(x, y) be a solution to (1); it determines some convex surface Φ : z(x, y). Integrating (1) over 2 is the Jacobian of a set e in the (x, y) plane and observing that zxx zyy − zxy the mapping ψ from the (x, y) plane into the (p, q) plane acting by the rule p = zx and q = zy (this ψ is called the support (sometimes, normal ) mapping of the surface Φ), we obtain

R(p, q) dpdq = ϕ(x, y) dxdy. (2) ψ(e) 10

e

Subsections 9.3.6–9.3.7 are added as comments by V. Zalgaller.

9.3 Generalizations

401

The set function of e on the left-hand side of (2) is referred to as the conditional curvature of Φ. (For R(p, q) = (1 + p2 + q 2 )−3/2 , it coincides with the Gaussian curvature of Φ). Then we can interpret (1) as problem (2) of finding a surface Φ given its conditional curvature translated to Borel sets on a plane or sphere. This problem can be studied by geometric methods developed in the current chapter. This means that we start by solving the problem in the class of polyhedral surfaces and proceed to a passage to the limit. The limit surface serves as a generalized solution to (1). It is along these lines that A. D. Alexandrov [A21] first studied the multi-dimensional Monge– Ampère equation det zij | = ϕ(xi , z, zi ); his student I. Ya. Bakel man [B1] investigated the generalized Dirichlet problem for equation (1) at the same time. The proof of the existence of a regular solution to boundary value problems for the Monge–Ampère equations reduces to establishing the smoothness of the generalized solution. Also, conditions on the boundary data that must be satisfied in the classical sense are found, as well as uniqueness theorems for solutions to boundary value problems; see [P8]. In the rich literature on these questions, let us point out the articles [P10], [B2], [B4], and [AP] which treat the topic systematically. We also indicate the survey [GT]. The articles [P11], [P12], and [B6] address a multi-dimensional analog of the Monge– Ampère equations. Stability questions for generalized and regular solutions are discussed in [Vo3], [Vo5], [K], [B3], and [B5]. Widespread is the reverse application of solvability theorems for Monge– Ampère equations to geometrical problems. Consider the following example. Assume given a sufficiently smooth scalar field K > 0 in E 3 (or in H 3 or in a Riemannian manifold M 3 ). Is there a closed convex surface S whose extrinsic curvature agrees with K at all points? This problem was studied in [Ve], [Fil], and [Ki]. The field K was defined in spherical coordinates (ϕ, θ, ρ) for E 3 and H 3 , and in polar-geodesic coordinates for M 3 ; the surface S was found as the function ρ = ρ(ϕ, θ) from the Monge–Ampère equation ρ11 ρ22 − ρ212 = (1 + ∆1 ρ) det gij K(ϕ, θ, ρ), where ρij are the second order covariant derivatives and ∆1 is the first differential Beltrami parameter. The solution was reduced to the derivation of a priori estimates in C 2 , followed by the application of the Leray–Schauder principle. The above-mentioned articles combine sufficient conditions for solvability and the corresponding uniqueness theorem. The articles [O2] and [Dl] generalize the results of [Ve]. 9.3.7 The problems of reconstructing a convex surface (in particular, a polyhedron) from the extrinsic curvature translated by projection to another surface (a sphere, plane, etc.) and their various generalizations are the topic of many articles; for example, [Ver2], [Ver4], [Ver5], [Zil4], [M1], [BM], [M2], [Kap], [Ar], [Vm], [Mis], [T].

10 Infinitesimal Rigidity of Convex Polyhedra with Stationary Development

Suppose that a polyhedron P is deformed as some parameter t varies. It is convenient to treat t as time. The initial position corresponds to t = 0. During the deformation the edges and faces of the polyhedron may fold with new vertices and edges appearing. In the current chapter we are interested only in the first-order infinitesimal deformations, in other words, the velocities at time t = 0. Further, speaking of deformations, we usually have in mind infinitesimal deformations; the absence of specification will not lead to any misunderstanding. We always relate velocities to the initial time t = 0. If the derivative of some quantity x with respect to t vanishes at t = 0, dx = 0, dt t=0 then we call x stationary. We say that a figure is infinitesimally rigid if (under given conditions) each deformation is necessarily a motion, i.e., all velocities at the initial time are the same as for a rigid body. In Section 2.6 of Chapter 2 we have already explained the general analogy between infinitesimal rigidity theorems and uniqueness theorems. The theorems in the present chapter are analogous to the uniqueness theorems of Chapter 3 and the proofs remain essentially the same. However, there is one fundamental difference: in the infinitesimal rigidity theorems we do not consider a genuine deformation but rather its principal first order term with respect to t. In this connection, circumstances not appearing for finite deformations and, as a result, for uniqueness or congruence theorems, must now be considered. (In particular, in contradistinction to the results of Chapter 3, the theorems we prove are now not valid for polyhedra degenerating into doubly-covered polygons, as we have already noted in Section 2.6. Therefore, we completely exclude these polyhedra from our considerations.)

404

10 Rigidity of Polyhedra with Stationary Development

10.1 Deformation of Polyhedral Angles 10.1.1 Let V be a convex polyhedral angle.1 We allow some faces of V to lie in the same plane, thus forming one “genuine” face. The edges cutting genuine faces are called “fictitious”; the dihedral angles at these edges are all equal to π. Speaking of edges, from now on we have in mind both genuine, or true, and fictitious edges. Also, we allow V to reduce to a dihedral angle but not to a plane. Then all dihedral angles of V are equal to π, except for two of them, which are less than π; the edges of the latter angle form a single straight line. We shall consider deformations of V due to the rotation of its edges around the vertex. With the rotation of edges, the faces of V rotate and also deform. We assume that no new edges arise, but some “fictitious” edges may turn into “genuine” ones. It is not excluded that the angle V ceases to be convex after the deformation. If the deformations are sufficiently small, this is possible only in the case when there are fictitious edges and the genuine faces fold along the edges inwards. Then the dihedral angles at these edges increase and become greater than π. The problem consists in examining the signs of the initial speeds of variation for the dihedral angles of the deformed angle V under the assumption that all planar angles of V are stationary. This problem is analogous to that of Section 3.1 for finite deformations preserving convexity and can be solved by a similar method. However, instead of the increments of angles, we must consider only their principal terms or, which is the same, the derivatives with respect to t at t = 0.2 Nevertheless, we prefer another method, which not only brings us to the aim more rapidly, but also is of interest in its own right.3 10.1.2 Since the motion of V as a rigid body can be excluded, we may assume that some edge of V , say p1 , and one of the planes of faces touching at p1 are fixed. Then the plane of the other face with edge p1 rotates around p1 . If the planar angle on this face is stationary, then (at the initial time) the whole face rotates around p1 as a rigid body. Now, without excluding the rotation of p1 and the face we have fixed, we may say that the deformation of a dihedral angle is nothing but the relative rotation of the planes of its faces around their common edge. This conclusion is certainly true for every pair of faces with a common edge. The rotation of a rigid body is determined by the angular velocity vector directed along the rotation axis so that, together with the direction of 1 2

3

The degeneration of V into a doubly-covered planar angle is forbidden. We shall see later that, even when convexity is preserved, for an infinitesimal deformation we may have a disposition of signs which is impossible for a finite deformation. This shows once again that, drawing the analogy with the results of Chapter 3, we must use derivatives rather than finite increments. The method is not new, but I do not know to whom it is due.

10.1 Deformation of Polyhedral Angles

405

the rotation itself, it forms a right-handed screw. Composing rotations about intersecting axes is well known to result in taking the vector sum of their angular velocities.4 Choose a circuit around the vertex on V (an orientation of V ) and so define the order of succession of edges and faces. Given an edge pi , draw the angular velocity vector wi of the rotation of the succeeding face with respect to the preceding one. We view the vector wi as lying on the edge and so it can run towards the vertex of V or in the opposite direction. The length of wi is nothing but the absolute value of the initial speed ϕ˙ i of variation of the dihedral angle ϕi at the edge pi . The direction of the vector wi determines the sign of ϕ˙ i , because changing the direction of wi changes the direction of rotation.

pi

ϕi

V

Pi

Fig. 147

This connection between the direction of wi and the sign of ϕ˙ i is the same for all edges, i.e., if on one edge the direction of w away from the vertex corresponds to ϕ˙ i > 0 (or vice versa), then the same holds for all other edges. Indeed, intersect V by the unit sphere S centered at the vertex of V . On S we obtain a polygon V with vertices pi and sides Pi corresponding to the edges and faces of V (Fig. 147). The angles ϕi of this polygon are equal to the dihedral angles of V . The orientatin of V induces the orientation of V and so determines the orientation of the sphere S.5 For definiteness, we assume that this orientation together with the direction from the center 4

5

Let O be a fixed point of a rigid body and assume O to be the origin of coordinates. If w is the angular velocity vector and x is the position vector of an arbitrary point X of the body, then the velocity of X is v = w×x. Composing two rotations with angular velocities w1 and w2 with respect to their axes meeting at O, we obtain v = v1 + v2 = w1 × x + w2 × x = (w1 + w2 ) × x, i.e., the angular velocity of the composite rotation is w = w1 + w2 . As usual, the orientation is specified by the circuit of a small contour. Take a small circle C whose center lies on the boundary of V . Draw the contour C that consists of the arc of C inside V and a segment of the boundary of V . The orientation of this contour, and so that of C, is determined by the direction of motion along the boundary of V . (If we move C along the boundary of V , then the continuity argument makes it obvious that the orientation defined in this way is independent of the choice of the point on the boundary of V which serves as the center of C.

406


of the sphere forms a right-handed screw. (Since the edges of V issue from the center of S, this implies that all screws determined by the circuit of V and its edges are right-handed.) For example, let wi point away from the vertex of V . By the very definition of wi , this means that the direction of the edge pi and the direction of rotation of the face Pi relative to Pi−1 form a right handed screw. Therefore, the side Pi of V must rotate with respect to Pi−1 in the direction specified by the orientation of the sphere S, i.e., inwards the polygon V . The speed ϕ˙ i of variation of the angle ϕi is consequently negative. This conclusion applies to all edges. Thereby the relationship between the direction of the vector wi and the sign of the speed ϕ˙ i is the same for all edges. (If we change the orientation of V , then this relationship will simply be reversed: for the vectors wi pointing away from the vertex there will correspond ϕ˙ i > 0.) Thus, the disposition of signs for the speeds ϕ˙ i is determinated by the distribution of directions for the vectors wi . To specify the latter, we prove the following proposition. The sum of all the vectors wi is zero. Indeed, we have mentioned that composing rotations amounts to summing their angular velocity vectors. Therefore, the sum wi +wi+1 determines the rotation of the (i+2)th face relative to the ith face. Summing all velocities wi from the first to last, we see that the result corresponds to the rotation of the first face relative to itself, i.e., to the absence of rotation. Hence, the sum of all vectors wi vanishes. 10.1.36 We now prove a lemma that contains the required result on the disposition of signs of the speeds of variation of the dihedral angles. Lemma 1. Suppose that a convex polyhedral angle V , possibly dihedral but not reducing to a plane, is deformed so that all its planar angles are stationary. Labeling each edge with the sign of the initial speed of variation of the dihedral angle at the edge, leave the edges of the stationary dihedral angles unlabeled. Then there are only four possibilities for the disposition of signs: (1) None of the edges is labeled. (2) The number of sign changes is at most four. (3) There are exactly two sign changes; in this case all labeled edges belong to one genuine face and the two extreme labeled edges have the same sign. (4) There are exactly two labeled edges; they form a straight line and are not only labeled by the same sign but the speeds of variation of the dihedral angles at them are also equal. Clearly, this case is possible only 6

In connection with this subsection, we turn the reader’s attention to [Vo1] and [Mi1]. – V. Zalgaller


407

when the angle V reduces to a dihedral angle, since only under this condition can the two edges form a straight line, namely the edge of the dihedral angle that V reduces to. (The speeds of variation of the dihedral angles at two such edges are equal, whereas the angles at all other edges are stationary because these edges are unlabeled. This fact means that the angle V remains dihedral during the deformation, provided that we ignore quantities of order greater than one.) Using the results of Subsection 10.1.2, we can rephrase this lemma in the language of the vectors wi 7 : Lemma 1a. Let wi be vectors on the edges of a convex polyhedral angle V , possibly reducing to a dihedral angle but not to a plane, such that the sum of all wi vanishes. Then only the following cases for the distribution of directions of these vectors towards or away from the vertex of V are possible: (1) All the vectors are zero. (2) The number of changes in direction is no less than four. (3) There are exactly two changes in direction; in this case all nonzero vectors wi lie on one genuine face and the two extreme nonzero vectors have the same direction (i.e., both point towards the vertex or away from it.). (4) There are exactly two nonzero vectors; then the vanishing of their sum obviously implies that they lie on a single straight line, have the same length, and both point away from the vertex or both towards it. (This case is possible only if V is a dihedral angle.) In view of the results of Subsection 10.1.2, Lemma 1a implies Lemma 1. So it suffices to prove Lemma 1a.8 The zero vectors have no direction; so we may ignore them. Accordingly, we assume that there is at least one nonzero vector, thus excluding the first possibility. The number of changes in direction in a closed circuit is certainly even. Therefore, we must distinguish between three cases: 7

8

All four cases (1)–(4) may occur: the first, when there is no deformation; the fourth, under any deformation of the dihedral angle without folding of its (genuine) faces; the second is easily seen on the example of the deformation of a tetrahedral angle: two edges at which the dihedral angles increase are separated by two edges at which the angles decrease; the third case holds for a fourhedral angle degenerating into trihedral one: the angles at the edges p, q, and r are less than π, the fourth edge s lies on the genuine face (q, r), and the angle at s equals π; when s moves in the plane perpendicular to the ambient face, all planar angles are stationary and so is the angle at the edge p, but the angles at q, r, and s are not stationary, and if the angle at s decreases, then the angles at p and r increase. The two lemmas are equivalent. The fact that Lemma 1 follows from Lemma 1a has been already shown in Subsection 10.1.2. Conversely, assume given vectors wi on the edges of V such that the sum of wi is zero. Taking wi as the angular velocity vectors, we thereby define a deformation of V under which all planar angles are stationary, as it is easy to see.

408


(a) All vectors have the same direction, i.e., all point towards the vertex or in the opposite direction. (b) There are exactly two changes in direction. (c) The number of changes in direction is at least four. To prove the lemma, it therefore remains to show that in case (a) we have the fourth possibility of the lemma and in case (b), the third. 10.1.4 Let us examine case (a). To be definite, we assume that all vectors point away from the vertex (otherwise it suffices to change their signs). Since V is a convex angle, for each face it lies on one side of the plane P containing this face. Therefore, all vectors wi point to a single closed halfspace bounded by P and containing V . If not all these vectors lie in the plane P , then their sum does not vanish. Hence, all vectors wi necessarily lie in P and so they lie on edges belonging to the same genuine face of V . Since any genuine face is convex, it lies in one half-plane P . Let L be the boundary straight line of this half-plane on P . The vectors wi point to the half-plane P and if they do not lie on L, then their sum cannot vanish. Hence, all the vectors wi lie on the straight line L. But then it is obvious that if such vectors exist, then there are only two vectors, both having the same length and pointing in the opposite directions. This proves that for the case in which there no changes in direction the fourth possibility can occur in the lemma. 10.1.5 Now let us examine case (b) in which there are exactly two changes in direction. We must prove that in this case the vectors wi all lie on one genuine face and the two extreme vectors wi have the same direction.

ωn

ωi ωm

ωm+1

q p

Figs. 148

Since there are exactly two changes in direction, we have exactly two sequences of vectors with the same directions. Enumerating the vectors appropriately, we may assume that the vectors wi , . . . , wm point towards the vertex and the vectors w1 , . . . , wn point away from it. Draw the half-lines p and q from the vertex on the faces of V so that p and q separate these


409

sequences of vectors (Fig. 148). The plane (pq) passing through these halflines, either is the plane of one of the faces of V or divides V into two angles V1 and V2 . Since the half-lines p and q separate one sequence of vectors from the other, the vectors of one sequence lie in one closed half-space bounded by the plane (pq) and the vectors of the other lie in the other half-plane. We have such a disposition of vectors when we regard each vector as lying on its own edge. However, if we suppose that all the vectors issue from the vertex of V , then the vectors pointing away from the vertex will point to the same half-space as the vectors moving towards the vertex. Since the sum of all vectors vanishes, it is obvious that none of them points to the interior of the half-space. Hence, all the vectors lie in the plane (pq). But then this plane is the plane of one of the faces of V since it contains the edges of V with vectors wi . Thus we have proved that all the vectors wi lie in the plane of one face. Now, in contradiction to the second part of the assertion that we are proving, suppose that the extreme vectors, which we denote (changing the numeration if necessary) by w1 and wn , are directed differently: the vector w1 points towards the vertex and wn runs in the opposite direction. Then we encounter a change in direction as we pass from wn to w1 . Since we assumed that there are only two such changes, there is exactly one change in direction in the sequence w1 , . . . , wn , say as we pass from wm to wm+1 . ωn ω m+1

O

ω1 ωm L

Figs. 149

Let P be a plane containing all the vectors w1 , . . . , wn . Let L be a straight line in P passing through the vertex O of V and separating the edges supporting the vectors w1 , . . . , wm from the edges supporting the vectors wm+1 , . . . , wn . When the vectors lie on the corresponding edges, they point towards the vertex and in the opposite direction respectively. If assumed issuing from the vertex, they point from the vertex to the same half-plane bounded by L (Fig. 149). However, in this case it is obvious that their sum cannot vanish. Consequently, our assumption that the extreme vectors are directed differently is false: they must have the same direction.

410


Thus, in case (b) we indeed have the third possibility of the lemma; this completes its proof. 10.1.6 If the polyhedral angle V reduces to a plane, then it is quite clear that all the vectors wi may run towards the vertex (or in the opposite direction) although their sum vanishes. Accordingly, all planar angles can be stationary, whereas the dihedral angles are not stationary; and they then all increase or decrease so that the angle V remains convex. This can be checked directly. To this end, fix points Ai on the edges of V at the distance a from the vertex O and move up the vertex above the plane to which the angle V reduces at the initial time. The points Ai are kept fixed. Then it is easy to verify that the lengths of the straight line segments OAi , as well as the planar angles of the triangles OAi Ai+1 , are stationary, whereas the dihedral angles at the edges OAi decrease with nonzero speeds.

10.2 The Strong Cauchy Lemma 10.2.1 Suppose that a closed convex polyhedron P deforms so that some of its faces fold along diagonals. These diagonals are the “new edges” of the polyhedron. Our aim is to prove that if the planar angles of the polyhedron are stationary, then the dihedral angles are stationary as well. This will imply, in particular, that the angles at the “new edges” are stationary, i.e., the initial velocity of the appearance of these edges is zero. This theorem is analogous to the theorem of Section 3.2 which asserts that the equality of the planar angles implies the equality of the dihedral angles. However, now we cannot use the Cauchy Lemma, since the fact that the planar angles are stationary means that only two sign changes around a vertex rather than four can occur as was shown in the preceding section. (In that section one more possibility was indicated: there is some vertex A with no sign changes at all; then exactly two edges are labeled and, moreover, they are labeled by the same sign. This case can be excluded from consideration: it suffices to erase all unlabeled edges and treat the two labeled edges as one; the vertex A simply disappears.) However, by a lemma of Section 10.1, in the case of two sign changes the disposition of signs has extra special properties. It turns out that if these properties are satisfied the assertion of the Cauchy Lemma holds: such a disposition of signs is impossible. This result will be exploited in various cases: in Sections 10.3– 10.5 of the current chapter and in Chapter 11. Therefore, we state and prove it in general form as the “Strong Cauchy Lemma.” Obviously, instead of the disposition of signs on the edges of a polyhedron, we may consider the disposition of signs on segments (“edges”) of some net on a surface homeomorphic to the sphere, subjecting the net to certain conditions. This more abstract point of view is useful in applications.

10.2 The Strong Cauchy Lemma

411

10.2.2 Let a net of lines (“edges”) be drawn on a surface homeomorphic to the sphere. We assume that the edges have no pairwise common points except their endpoints, the “vertices” of the net. We also assume that the net possesses the three following properties: (a) no two vertices are joined by two edges; (b) each of the domains (“faces”) into which the net divides the surface is bounded by one closed chain of edges without multiple points; and (c) the edge joining two vertices of a face belongs to the face. Suppose that on some faces “new edges,” which are disjoint diagonals of the faces, are drawn; further, assume that the conditions (a) and (c) are preserved. These edges split the “old faces” into “new faces.”9 Let some edges, both “old” and “new,” be labeled by plus and minus signs; some edges may remain unlabeled. Call a vertex labeled if at least one of the incident edges is labeled. Let V stand for the number of labeled vertices. Call a vertex special if the disposition of signs on the incident edges satisfies the conditions of case (3) of Lemma 1 of Section 10.1. The number of sign changes around such a vertex equals two and all incident edges are unlabeled except for the edges belonging to one “old face”; moreover, the two extreme ones of these edges are labeled by the same sign. Let V ∗ stand for the number of special vertices. The Strong Cauchy Lemma asserts that the total sum N of sign changes around all vertices satisfies the inequality N ≤ 4V − 2V ∗ − 8.

(1)

This claim implies that it is impossible to have any disposition of signs for which there are least four sign changes around every nonspecial labeled vertex. Indeed, for such a disposition we would have N ≥ 4(V − V ∗ ) + 2V ∗ , since we have two sign changes around each of the V ∗ special vertices. But this would contradict (1). Furthermore, inequality (1) has the number −8 in “reserve.” Therefore, we may admit one vertex with no sign changes or three vertices with two (arbitrary) sign changes: owing to the “reserve,” such a disposition of signs will be impossible anyway. The proof of this Strong Cauchy Lemma is similar to the proof of the ordinary Cauchy Lemma in Section 2.1. 10.2.3 Consider the net formed only by the labeled edges; the unlabeled edges are completely omitted. This net splits the surface into regions. Condition (a) of the lemma implies that no region is bounded by exactly two edges. 9

The conditions certainly hold for the net of the edges of a convex polyhedron and disjoint diagonals of faces of the polyhedron. The word “line” will mean homeomorphic image of a straight line segment.

412


Counting the total number of sign changes as we go around the regions, we obtain the same number N , since passing from one edge to a neighboring edge we proceed both around a vertex and around a region. Moreover, we traverse every edge belonging to a region but not separating it from the other regions twice (see Fig. 61 on p. 84). For this reason, we regard such an edge as two sides of the region. Therefore, if Fn is the number of regions with n sides and E is the total number of edges, then 2E = nFn . (2) n

If V , E, and F are the numbers of vertices, edges, and regions then by the generalized Euler formula V − E + F ≥ 2. Multiplying by 4 and using (2), we see that 4V − 8 ≥ 2(n − 2)Fn .

(3)

(4)

n

Hence, if we prove that the total number of sign changes satisfies the inequality N≤ 2(n − 2)Fn − 2V ∗ , (5) n

then we will have proved (1). We will establish estimate (5) for the total number of sign changes by counting the changes in sign around regions. 10.2.4 Let G be one of the regions of the net of labeled edges. We say that a vertex A of G is a special vertex of G if (1) A is a special vertex in the sense of the definition of the lemma and (2) the sides of G incident to A are exactly the two extreme edges among the labeled edges incident to A which lie on one “old face”; by the definition of a special vertex, these extreme edges are labeled by the same sign. From the properties of special vertices of a net which are listed in the lemma, we easily infer the following properties of special vertices of an arbitrary region G: (1) Every special vertex of the net is a special vertex of at least one region. (2) At a special vertex of G, there is no sign change in going around G. (3) If A is a special vertex of G, then the sides of G incident to A are edges or diagonals of one old face H. In any case they cut out some part H ∗ from H; we call H ∗ a “face” as well. The face H ∗ coincides with H only if both edges are (old) edges of the face H (Fig. 150). The face H ∗ is hatched by horizontal lines; the region G, by vertical lines; and the face H, by oblique lines. The face H ∗ is part of the old face H; the remaining part lies in G.

10.2 The Strong Cauchy Lemma

413

The region G and the face H ∗ lie on different sides of the edges a and b (at least near them10 ); H ∗ is exactly that part of the face H on which the labeled edges incident to A lie; whence it is clear that H is separated from G by the two extreme edges among those labeled edges. (5) If two adjacent vertices A and B of G are special vertices of G, then all three sides AB, AC, and BD of G belong to the same face (Fig. 151). Indeed, as we have just noted, AB and AC belong to one face H separated from G. Similarly, AB and BD belong to one face H separated from G. However, there may be only one face to one side of AB, and so H and H coincide.

a

G

H*

A b

G

A H

B H D

C Fig. 150

Fig. 151

10.2.5 Having made these remarks, we now turn to evaluating the number of sign changes in going around an arbitrary region G. ∗ be the number of special Let vG be the number of vertices of G, let vG vertices of G, and let nG be the number of sign changes in passage around G. Let us show that ∗ − 2). (6) nG ≤ 2(vG − vG Since there are no sign changes at special vertices, we have ∗ nG ≤ vG − vG ;

moreover, nG is certainly even. Therefore, ∗ vG − vG nG ≤ 2 , 2

(7)

where as usual the brackets stand for the integral part of the number. If x ≥ 3, then it is easy to check that x ≤ x − 2. 2 ∗ Hence, if vG − vG ≥ 3, then (7) yields the desired result: 10

It is unknown a priori whether H ∗ is disjoint from G or not. Although H ∗ is cut by “new edges,” some of its pieces can be contained in G. However, this cannot happen near the indicated edges, which is what we assert.

414

10 Rigidity of Polyhedra with Stationary Development ∗ nG ≤ 2(vG − vG − 2).

(8)

∗ Now, suppose that vG −vG < 3, i.e., G has at most two nonspecial vertices. By the above, no region has exactly two vertices. Therefore, vG ≥ 3 and if ∗ ∗ vG − vG < 3, then vG ≥ 1, i.e., G must have special vertices. The number of ∗ ≤ 2. nonspecial vertices is vG − vG Let us prove that the number of nonspecial vertices can never be less than two and if there are exactly two nonspecial vertices, then they are never adjacent. Assume the contrary and let the successive vertices A1 , . . . , Av−1 be special vertices of G, while the two remaining adjacent vertices Av−1 and Av may be nonspecial. Then the last of the properties of special vertices listed above implies that all sides Av A1 , A1 A2 , . . . , Av−2 Av−1 belong to one face H ∗ . In that case, the side Av−1 Av belongs to H ∗ as well, because by condition (c) of the lemma, the edge of the original net joining two vertices of the same face belongs to the face. Therefore, if the side Av−1 Av is an “old edge,” then it is an edge of the face H ∗ , while if Av−1 Av is a “new edge,” then it is a diagonal of H ∗ . In the latter case the side Av−1 Av cuts out some part of H ∗ . In any case it turns out that all sides of the region G bound a domain H ∗∗ which is homeomorphic to a disk. By the main property of special vertices, each of the vertices A1 , . . . , Av−2 must have incident edges lying between the edges Ai Ai+1 Clearly, these edges must be diagonals of H ∗ and hence diagonals of H ∗∗ . By assumption they are disjoint. However, A1 , . . . , Av−2 are all but the two adjacent vertices Av−1 and Av of H ∗∗ and since H ∗∗ is homeomorphic to a disk, it is impossible to draw a diagonal from each vertex so that the diagonals be disjoint.11 (In particular, if v = 3 and we have exactly one special vertex A1 , then there are no diagonals at all.) Hence, if there are two nonspecial vertices, then they are nonadjacent. However, if all vertices but two nonadjacent are special, then all sides of G are incident to special vertices, which implies that they all are labeled by the same sign, i.e., the number of sign changes is zero, nG = 0. At the same time, in this case the difference between the number of all vertices and ∗ = 2. Therefore, the number of special vertices equals two, vG − vG ∗ − 2), nG = 2(vG − vG ∗ i.e., the estimate (6) is also valid when vG − vG < 3.

10.2.6 Summing the numbers nG of sign changes for all regions G, we find the total number N of sign changes. In view of (6), we obtain ∗ N= nG ≤ 2(vG − 2) − 2 vG . (9) 11

A rigorous proof is immediate by induction on the number of vertices.

10.3 Stationary Dihedral Angles for Stationary Planar Angles

415

If Fn is the number of regions G with n vertices, or equivalently n sides, then 2(n − 2)Fn . (10) 2(vG − 2) n

Furthermore, every special vertex of the net is a special vertex of exactly one region. Therefore, the sum of the numbers of special vertices for all regions equals the number of all special vertices of the net, i.e., ∗ = V ∗. (11) vG Hence, (9) may be rewritten as N≤ 2(n − 2)Fn − 2V ∗ . n

This completes the proof of (5) and so concludes the proof of the Strong Cauchy Lemma.

10.3 Stationary Dihedral Angles for Stationary Planar Angles 10.3.1 Let P be convex polyhedron; it may be closed, unbounded, or with boundary. Partition the faces of P into arbitrary pieces by straight line segments (or half-lines in the case of unbounded faces) disjoint in the interiors of the faces. The endpoints of these segments may consequently lie only on edges or at vertices of the polyhedron.12 We treat every piece of a face as a face, the segments or half-lines partitioning the faces as edges, and the endpoints of these segments (or half-lines) as vertices. Thus, “fictitious” edges lying inside genuine faces and “fictitious” vertices lying on genuine edges arise in the polyhedron. In the current and subsequent sections, the faces, edges, and vertices are understood in this generalized sense. We will consider a continuous deformation of P as some parameter t varies; the initial position corresponds to t = 0. We subject the deformation to the following conditions: 12

Without this condition the fact that the planar angles are stationary may no longer imply that the dihedral angles are stationary, as is clear from the remark at the end of Section 10.1. For example, if we partition some face of a cube by drawing two diagonals, then the dihedral angles will not be stationary. [As for partitions with new vertices appearing in the interiors of old faces, see Connelly’s result mentioned in Subsection 3.6.1b .]

416


(1) Every vertex, including the “fictitious” vertices, moves with a definite (possibly variable) velocity, i.e., the coordinates of the vertex are differentiable functions in t. (2) No new vertices and no new edges arise. These conditions readily imply that the genuine faces of the polyhedron fold along the fictitious edges given in advance and that their pieces, treated as faces, rotate around common edges with definite velocities; moreover, the edges themselves rotate and change their length. It is also clear that all elements of the polyhedron, in particular the planar and dihedral angles, change with definite velocities, i.e., they all are differentiable functions in t. We do not assume that the polyhedron remains convex. From now on we have in mind precisely such deformations. We are interested in conditions under which the fact that the planar angles are stationary implies that the same is true for the dihedral angles. We answer this question by combining Lemma 1 of Section 10.1 with the Strong Cauchy Lemma of Section 10.2. In Lemma 1 of Section 10.1, case (4), in which there are no sign changes around a vertex but labeled edges exist, can occur. However, as we saw in the lemma, this case may hold only at the vertex of an angle reducing to a dihedral angle; moreover, during the deformation, the angle remains dihedral (of course, to within the first order). Then it is clear that such vertices can be excluded: all fictitious edges incident to them are unlabeled and may be omitted while the two genuine edges incident to such a vertex form a single edge. Thus, case (4) of Lemma 1 of Section 10.1 is excluded and, except for the case of no labeled edges, we are left with cases (1), (2), and (3) which are settled by the Strong Cauchy Lemma. This leads us to theorems on stationary dihedral angles of convex polyhedra. 10.3.2 Theorem 1. If all angles on the faces are stationary as a closed convex polyhedron is deformed, then all dihedral angles are stationary as well. Label each edge of the polyhedron with the sign of the initial speed of variation of the dihedral angle at the edge, leaving the edges with stationary dihedral angles unlabeled. Since the planar angles are stationary, the disposition of signs around each vertex must satisfy Lemma 1 of Section 10.1. However, by the Strong Cauchy Lemma of Section 10.2, the disposition of signs satisfying such conditions is impossible if there is at least one labeled edge. Therefore, all edges must be unlabeled, which was to be shown.13 13

As opposed to Theorem 1 of Section 3.2 on the equality of dihedral angles in polyhedra with equal planar angles, the theorem just proved fails for polyhedra degenerating into doubly-covered polygons even if a polyhedron remains convex in the deformation (excluding the trivial case of a doubly-covered triangle). Take,


417

10.3.3 Theorem 2. Let P be a closed convex polyhedron, with boundary a single closed polygonal line, which possesses the following property: no internal vertex, i.e., one not lying on the boundary, is joined to the boundary by more than one edge. If all planar angles of P at internal vertices are stationary when P deforms, then all dihedral angles are stationary as well. Identifying imaginatively all points on the boundary of P to one point O, we transform P into a surface P homeomorphic to a sphere. Now, take the net P of internal edges of P , considering all edges approaching the boundary as incident to O. Then P divides P into faces. By the conditions imposed on P , no vertex of P is joined to O by two edges. Therefore, the net on P satisfies all conditions of the Strong Cauchy Lemma. Suppose that P is deformed so that all planar angles at internal vertices are stationary, whereas the dihedral angles are not. Then, labeling its edges in accordance with the variations of the dihedral angles, we obtain dispositions of signs around internal vertices satisfying the conditions stipulated in Lemma 1 of Section 10.1. If we carry over these signs to the edges of P , then it is only around O that the disposition of signs is not subjected to any conditions. Nevertheless, the Strong Cauchy Lemma asserts that even in the presence of one exceptional vertex of this kind the disposition of signs given by the conditions Lemma 1 of Section 10.1 is impossible. Therefore, the dihedral angles must be stationary. Theorem 3. If each vertex of an unbounded convex polyhedron has at most one unbounded incident edge and all planar angles are stationary, then the dihedral angles are stationary as well. To prove this, it suffices to cut some bounded part of the polyhedron so that only unbounded edges approach the boundary and then appeal to Theorem 2. All unbounded edges of an unbounded polyhedron of curvature 2π are parallel and, therefore, no two of them are incident to the same vertex. Hence, Theorem 3 implies the following Theorem 4. If all planar angles are stationary under a deformation of an unbounded convex polyhedron of total curvature 2π, then the dihedral angles of the polyhedron are stationary as well. for example, a doubly-covered square ABCD with side a and lift the vertex D above the plane to height h = vt, where v is the speed of the vertex D, obtaining a tetrahedron with edge CD =

√

a2 + h2 ≡ a +

1 v2 2 t . 2 a

Therefore, we see that the variation of planar angles is of order t2 , i.e., they are all stationary. At the same time, the variation of the dihedral angles is clearly of order t.

418


10.3.4 If an unbounded polyhedron does not meet the assumptions of Theorem 3, then in general the fact that all planar angles are stationary fails to imply that the dihedral angles are stationary. Examples are given by deformations of polyhedral angles. An unbounded convex polyhedron possessing nonparallel unbounded edges has a limit angle not degenerating into a half-line. The edges of the limit angle are parallel to the unbounded edges of the polyhedron and so, when the polyhedron is deformed, they rotate with definite velocities, causing the limit angle to deform accordingly. Theorem 5. If an unbounded convex polyhedron whose limit angle does not degenerate into a half-line is deformed so that all angles on its faces and the limit angle are stationary,14 then the dihedral angles of the polyhedron are stationary as well. Let P be an unbounded convex polyhedron whose limit angle V does not reduce to a half-line. Given an arbitrary unbounded edge of P , consider the collection p1 , . . . , pn of unbounded edges parallel to it, with the given edge included in the collection, numbered so that they belong to successive pairwise adjacent unbounded faces. Regard the edges p1 , . . . , pn as belonging to one class. Then the set of all unbounded edges splits into classes. In particular, a class may contain only one edge if the faces touching at the edge have no other edges parallel to it. We supplement the polyhedron P with improper vertices and edges in accordance with the following rules: (1) All unbounded edges of one class are incident to one improper vertex and edges of different classes are incident to different improper vertices. (2) The improper vertices naturally form a cyclic sequence. Every two successive improper vertices are joined by one improper edge. We denote the resulting polyhedron by P ; it is homeomorphic to a planar polygon: its “boundary” is constituted by improper edges. Every face of P is bounded by a closed polygonal line without multiple points. Namely, the boundaries of unbounded faces with parallel edges are closed at improper vertices and the boundaries of unbounded faces with nonparallel unbounded edges are closed by improper edges. 14

The edges of the limit angle are parallel to the unbounded edges of the polyhedron. Therefore, the fact that the limit angle is stationary amounts to the fact that the directions of the unbounded edges of the polyhedron are stationary. On the other hand, the planar angles of the limit angle are completely determined by the planar angles of the unbounded faces of the polyhedron: they are equal to the angles between the unbounded edges. Therefore, it suffices to require that the dihedral angles of the limit angle be stationary. These angles are understood as dihedral angles at the classes of parallel unbounded edges in the sense of Section 3.4.


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Any pair of vertices A and B of P may be joined by at most one edge and this edge then belongs to a face with vertices A and B. Indeed, if A and B are both proper or both improper, then this claim is obvious. Now, assume that the vertex A is proper and B is improper. Then the edge AB joining them is unbounded. The other incident edges of the improper vertex B must be parallel to AB, but they cannot join B to the same vertex A. Therefore, the edge AB is the only edge between A and B and it belongs to the two faces touching at it. No other faces may contain both A and B since, by construction, the other faces with the improper vertex B have edges parallel with AB and touch at these edges, which obviously implies that none of them may have A as a vertex. We now take a copy P of P and identify all proper and improper edges and vertices of these polyhedra. As a result, we obtain an abstract polyhedron P ∪ P which is homeomorphic to a sphere and whose structure inherits the same features as those just established for P :

(a) Each face of P ∪P is bounded by a single closed polygonal line without multiple points. (b) If two vertices of the same face are joined by an edge, then the edge belongs to that face. Exactly these conditions are imposed on the nets of edges in the Strong Cauchy Lemma. This lemma therefore applies to the polyhedron P ∪ P . If the original polyhedron P is deformed, then we label the edges of the polyhedron P ∪ P by signs according to the following rule: each proper edge of P is labeled by the sign of the initial speed of the dihedral angle at the edge. The corresponding edge on P is labeled by the opposite sign. The edges of stationary dihedral angles, as well as improper edges, remain unlabeled. To prove the theorem, we must show that if all planar angles of P and its limit angle V are stationary, then there are no labeled edges at all. Since the planar angles are stationary, the disposition of signs around each vertex of P ∪ P possesses the properties listed in Lemma 1 of Section 10.1 (only on P are the signs opposite, and this plays no role in the application of the Strong Cauchy Lemma as we clearly see from its statement). We will now prove that all edges are unlabeled around each improper vertex or else there are at least four sign changes. Let A be an improper vertex with incident unbounded edges p1 , . . . , pn of P and p1 , . . . , pn of P . By assumption, all improper edges are unlabeled and so we may exclude them from consideration. All the edges pi are parallel and, under the infinite similarity contraction of P to the vertex of its limit angle V , they yield one edge p on V . The unbounded faces contiguous to the extreme edges p1 and pn are the faces of V that touch at p. Therefore, if α is the dihedral angle at p in V and α1 , . . . , αn are the dihedral angles at p1 , . . . , pn in P , then

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α = α1 + . . . + αn . If the limit angle is stationary then dα dt t=0 = 0, implying that dα1 dαn + ...+ = 0. dt t=0 dt t=0 Hence, all the angles αi are stationary or else there are sign changes in their initial speeds. Since the edges p1 , . . . , pn are labeled by opposite signs, in the first case all edges around A are unlabeled and in the second case there are at least four sign changes (at least one change for each of the sequences p1 , . . . , pn and p1 , . . . , pn and two changes in passing from one of them to the other). Thus, it turns out that, if there were any labeled edges, the disposition of signs on the edges of the polyhedron P ∪P would be forbidden by the Strong Cauchy Lemma. Hence, there are no labeled edges and the theorem is proved. 10.3.5 Theorem 6. If a convex polyhedron whose boundary is a single closed polygonal line is deformed so that all the angles on its faces, as well as the angles between every pair of boundary edges touching at a vertex, are stationary, then all the dihedral angles of the polyhedron are stationary as well. Label each edge of P by the sign of the initial speed of variation of the corresponding dihedral angle, leaving unlabeled the edges with stationary dihedral angles as well as the boundary edges. This disposition of signs around each internal vertex satisfies the conditions of Lemma 1 of Section 10.1. Let A be a boundary vertex and let p and q be boundary edges touching at A. Stretching the planar angle (pq) over them, we obtain a convex polyhedral angle with vertex A in which all planar angles are stationary. Therefore, all edges touching at A are unlabeled or else there are at least two changes in sign around A, including the signs that should now be ascribed to p and q. If the number of sign changes is at least four, then there is at least one sign change on the internal edges of P touching at A. If the number of sign changes equals two, then the disposition of signs must be the one indicated in the third case of Lemma 1 of Section 10.1: all labeled edges lie on one genuine face and the two extreme ones among them are labeled by the same sign. These extreme edges cannot be p and q since there are no edges on the angle (pq) at all. Hence, we again conclude that there is at least one sign change on the internal edges of P touching at A. We now take a copy P of P , label its edges with opposite signs, and identify the corresponding boundary edges of P and P . Arguing as in the proof of Theorem 5, we convince ourselves that the presence of labeled edges yields a disposition of signs on the abstract polyhedron P ∪ P which is impossible by the Strong Cauchy Lemma. Therefore, there are no labeled edges at all, which completes the proof of the theorem.

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Using the “reserve” in the Strong Cauchy Lemma, we may relax the condition in Theorem 6, requiring stationarity from the angles between all but three pairs of adjacent boundary edges. Theorem 6, together with its proof, translates verbatim to “locally convex” polyhedra, i.e., to polyhedra whose polyhedral angles, both at internal and boundary vertices, are convex (in this case, stretching an angle over a pair of boundary edges touching at a vertex, we obtain a closed convex polyhedral angle). 10.3.6 The question remains open of finding the additional conditions guaranteeing that the dihedral angles are stationary for a polyhedron with boundary consisting of several polygonal lines. It would be trivial to require that the planar angles be stationary on the convex hull of the polyhedron: in that case everything would reduce to the theorem for a closed polyhedron. Such a condition is, however, imposed on the whole boundary of the polyhedron, whereas the condition of Theorem 6 is formulated for each boundary vertex separately. The question consists in finding such local conditions rather than conditions “in the large.”

10.4 Infinitesimal Rigidity of Polyhedra and Equilibrium of Hinge Mechanisms 10.4.1 In this section we derive corollaries to the results of Sections 10.1 and 10.3 from the standpoint of mechanics. Theorem 1. A closed convex polyhedron with infinitesimally rigid faces is infinitesimally rigid.15 Infinitesimal rigidity of faces means that the lengths of their edges and their angles are stationary under any deformation of the polyhedron. However, by Theorem 1 of Section 10.3, the fact that all planar angles are stationary implies that so are all dihedral angles. Therefore, the initial deformation of the polyhedron must reduce to a motion. This conclusion is based on the following obvious general lemma: If all edges as well as planar and dihedral angles are stationary under a deformation of an arbitrary polyhedron, then the initial deformation is a motion.16 15

See another proof of this theorem and connections with hinge mechanisms in [De], [Tr], [Wh], [CoW], and primarily in [Co6]. – V. Zalgaller 16 To prove this, take a face Q1 of the given polyhedron P0 and fix some vertex A of Q1 together with the direction of one of the sides AB of Q1 issuing from A and the direction of the plane of the face Q1 itself. Thereby we exclude from the

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In exactly the same manner, the other theorems of Section 10.3 imply the corresponding infinitesimal rigidity theorems for convex polyhedra. 10.4.2 The complete formal analogy between kinematics and the statics of solids is well known. If we associate forces with angular velocity vectors and the momentum of pairs with velocities, then composition of motions corresponds to sum of forces. In particular, conditions for the mutual compensation of motions correspond to equilibrium conditions. (Indeed, a system of forces momenta m is equivalent to the force p = p and the momentum pi and j i m = mj + (pi × ri ), where ri is the vector from the point of application of the force p to the point of application of the force pi . By analogy, the motions with angular velocities the motion wi and linear velocities v j yield with angular velocity w = wi and linear velocity v = vj + (wi × ri ), where ri is the vector from a point on the rotation axis of w to a point on the rotation vector of wi .) This analogy allows us to reformulate kinematic assertions in the language of statics and vice versa. Therefore, our kinematic theorems on stationary dihedral angles or on infinitesimal rigidity of polyhedra with rigid faces may also be interpreted in terms of forces. We present one of these interpretations. In it we consider the hinge mechanism consisting of the edges and vertices of a convex polyhedron rather than the polyhedron with rigid faces itself. In accordance to our convention, we admit fictitious edges which go across genuine faces and join points on their boundaries. The edges are thought of as rods connected at the vertices by hinges. We say that a hinge mechanism is strained without action of external forces if some tensile or compressive forces (loads) act along each bar so that the forces are in equilibrium at the vertices. An example of a strained hinge mechanism was given in Subsection 2.6.3, Fig. 71. Let us show that Theorem 1 is equivalent to the following outset the motion of P0 as a rigid body. Now, since the vertex A and the direction of the side AB are fixed while, as proved above, the length of AB is stationary, it follows that the vertex B is fixed. (By this we mean the fact that its initial velocity is zero). Next, the angle between the side AB and the succeeding side BC is stationary and the whole plane of the face Q1 is fixed, which implies that the direction of the side BC is also fixed. Now, the fact that the length of BC is stationary implies that the vertex C is fixed. Continuing likewise, we verify that all vertices of the face Q1 are fixed. The angle between Q1 and the face Q2 adjoining Q1 along AB is stationary; therefore, the plane of Q2 is fixed; moreover, so are the vertices A and B of Q2 . Applying the preceding arguments to Q2 , we verify that all vertices of Q2 are fixed. Passing to adjacent faces and continuing likewise, we check that all vertices of P0 are fixed. Thus, excluding any motion of P0 as a rigid body, we make all its vertices fixed. Hence, when motion is not excluded, they may move as though the polyhedron is a rigid body, which was to be proved.

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Theorem 2. The hinge mechanism of the edges and vertices of a closed convex polyhedron cannot be strained without the action of external forces. To prove Theorem 2, we recall that, while establishing Theorem 1 of Section 10.3 on stationary dihedral angles, we started by representing the relative rotations of adjacent faces by means of the angular velocity vectors directed along edges. Moreover, the sums of these vectors around each vertex were equal to zero. The proof of Theorem 1 of Section 10.3 was precisely reduced to showing that such system vectors cannot be drawn on the edges of a closed convex polyhedron (unless all vectors vanish). Now interpreting these vectors as loads, we immediately arrive at Theorem 2. The converse is true as well: Theorem 1 of Section 10.3 follows from Theorem 2. 10.4.3 The following general assertion can be formulated: Theorem 3. Let P be an arbitrary simply connected17 bounded polyhedron and let R be the hinge mechanism of the edges of P , except for the boundary edges. Then the infinitesimal rigidity of P under the assumption of the rigidity of its faces is equivalent to the claim that the mechanism R cannot be strained without action of external forces even if the free ends of edges, i.e., the ends of the edges approaching the boundary of P , are clamped so that the loads at them are compensated by the reaction at the support. (We omit the assumption concerning the free ends in the case of a closed polyhedron.) Let us show that the impossibility of straining R implies the infinitesimal rigidity of P . Let P be a polyhedron with rigid faces hinged to one another along edges. Let wi be the angular velocities of rotation of the faces. The condition that the faces are hinged to one another along edges means that the axis of relative rotation of adjacent faces Qj and Qk is the common edge of Qj and Qk , i.e., the difference wj − wk of their angular velocities is directed along the edge. Fix an orientation on P , by indicating the direction of moving around a small contour. If two faces Qj and Qk have the common edge AB and Qj follows Qk as we move around A, then Qj precedes Qk as we move around B. Therefore, agreeing to consider the relative rotation of each succeeding face with respect to the preceding one, on the edge AB we will have the vector wj − wk for going around A and the vector wk − wj for going around B. The former is naturally considered as issuing from A and the latter as issuing from B. 17

That is, a polyhedron on which every contour bounds some region and is the complete boundary of this region. This property is possessed only by polyhedra homeomorphic to the sphere or disk. It is seen from the proof that simple connectedness is essential and examples showing that without this property infinitesimal rigidity may fail to imply the impossibility of strain can easily be exhibited.

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If we now interpret these vectors as forces, then the edge AB will undergo a tensile or compressive load. Passing through a vertex across the faces Q1 , Q2 , . . ., we return to the initial face Q1 . Therefore, the sum of all differences w2 − w1 , w3 − w2 , . . . vanishes. This means that the loads along the edges are in equilibrium at all vertices. This is so for internal vertices. At the same time, by assumption the loads at the boundary vertices are compensated by the reaction at the support. We infer that the hinge mechanism of the edges of the polyhedron is strained without action of external forces. If this is impossible, then all differences wj − wk vanish, i.e., the angular velocities of all faces are equal and the dihedral angles are stationary. So, the impossibility of straining the hinge mechanism of edges implies the infinitesimal rigidity of the polyhedron. Let us prove the converse assertion. Assume that some loads act on the hinge mechanism of edges of P so that they are in equilibrium at the vertices. Then opposite vectors act at the endpoints of each edge. Fix an orientation on P and assume that the face Qj succeeds a face Qk as we go around the vertex A. Denote by ajk the load at A along the common edge AB of Qj and Qk . Then the load at B equals akj = −ajk . Enumerate all faces and ascribe some vector w1 to the face Q1 . If a face Qi adjoins Q1 along an edge, then we assign the vector wi = w1 + a1i to Qi . Similarly, passing from any face Qj to an adjacent face Qk we assign the vector wk = wj − akj to Qk . Let us show that in this way we uniquely assign some vector to each face independently of the path from Q1 to this face. This is equivalent to the fact that in going around a closed contour we always obtain the same vector w for the initial face. However, this fact is obvious for a circuit around a vertex, since the sum of the vectors ajk around the vertex vanishes. At the same time, every closed circuit of a simply connected polyhedron decomposes into the sum of circuits around its vertices. (This is just the place where we use simple connectedness.) Thus, we uniquely assign some vector w to each face. Interpreting this vector as an angular velocity, we conclude that the faces of the polyhedron rotate around common edges. However, this is impossible if the polyhedron is infinitesimally rigid. Thereby the loads are impossible. Theorem 3 allows us to restate all theorems of Section 10.3 in terms of hinge mechanisms, suitably restating all extra conditions such as, for example, the condition of Theorem 5 of Section 10.3 concerning the stationary limit angle. Moreover, unbounded polyhedra are naturally replaced by bounded ones. The precise statements of the corresponding theorems are left to the reader.

10.5 On The Deformation of Developments

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10.5 On The Deformation of Developments 10.5.1 We shall consider convex polyhedra whose faces, edges, and vertices are understood in the generalized sense according to the convention at the beginning of Section 10.3. Every polyhedron has infinitely many distinct developments. We confine our exposition to developments whose vertices coincide with the vertices of the polyhedron: every vertex of a development is a vertex of the polyhedron and vice versa. This condition is always assumed in this section and subsequent ones. If P is an unbounded convex polyhedron, then P has unbounded faces successively adjoining each other along unbounded edges and forming the socalled funnel of P . If unbounded faces also meet at bounded edges, then we make cuts along these edges too. When speaking of a funnel, we always think of these cuts as having been performed in advance, so that the bounded sides of a funnel are not glued to one another. A funnel slit along any unbounded edge is developable on a plane. However, slitting a funnel along an unbounded edge involves arbitrariness. To exclude the latter, we consider the funnel itself, without a slit. From the viewpoint of intrinsic geometry, a funnel is completely determined by specifying the lengths of its boundary sides (edges) and the angles between adjacent sides. Similar considerations apply to any development of an unbounded polyhedron. Therefore, we regard a development of such a polyhedron as consisting of a finite number of bounded polygons and a “funnel.” In particular cases, it may happen that there are no bounded polygons at all, as for instance in the case of the development of a polyhedral angle. A funnel is in general a development consisting of unbounded polygons cyclically glued to one another along unbounded edges. Since all unbounded edges of a funnel are regarded as glued to each other, the developments of polyhedra under consideration have no unbounded edges. By the natural development of a bounded polyhedron we mean a development composed of the faces of the polyhedron. By the natural development of an unbounded polyhedron we mean the development composed of its bounded faces and of the funnel of unbounded faces. We say that the structure of a development is fixed if the development is only given as an abstract complex with the “rule for gluing,” i.e., the indication of the identifications between the sides and vertices of polygons. When the structure of a development is fixed, the development is completely determined by specifying the side lengths and angles of its polygons (including the funnel). If a development comprises only triangles, then the specification of angles is redundant. Granted an arbitrary structure of a development, we can easily choose independent “defining parameters” among the side lengths

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and angles of polygons. However, the choice will play no role for us; therefore, we shall simply speak of the defining parameters, namely the lengths of the sides (edges) and the measures of angles. Preserving the structure of a development, we can continuously change the defining parameters of the development at least within some limits. It is important to keep in mind that, when the structure of a development is fixed, all elements of the development (faces, their sides, vertices) are individualized, for example enumerated. Therefore, we consider the continuous variation of parameters connected with the given elements; for example, the angle at a given vertex in a given polygon. (This remark is essential, since in general a development may admit mappings onto itself which preserve the “law for gluing”; an example is the natural development of a regular polyhedron.) We say that a development is stationary if its defining parameters are stationary, i.e., they are functions of some parameter t and their derivatives with respect to t vanish at the initial moment of time. 10.5.2 Consider an arbitrary development R of a given polyhedron P . Let an edge L of the development join the vertices A and B, traversing some faces of P . Successively develop all these faces on a plane, starting from A to B; they will cover some polygon Q in which L is a diagonal between A and B (see Fig. 92 on p. 204). Certainly, L can cross some faces several times; in that case these faces appear in Q the same number of times. The edge L meets any edge of the polyhedron, and therefore any face, only finitely many times (except the case in which L coincides with an edge of the polyhedron). The proof is straightforward.18 Also, we may refer to the stronger result established in the proof of Lemma 3 in Section 4.2. This implies that every development of a polyhedron results from a natural development by finitely many cuttings and gluings. Therefore, every pair of developments of the same polyhedron can be obtained from each other in a similar way. Abstracting from the polyhedron, we consider two developments R and S that result from one another by finitely many cuttings and gluings along straight line segments. Moreover, we assume that no new vertices appear and the existing vertices do not disappear; for short, the developments have the same vertices. Every edge of S consists of line segments lying on certain faces of R, so that the edge turns into a line segment when these faces are together on the plane. We say that the position of an edge L of a development S is fixed in a development R if the following are specified: (1) the vertices of R that L joins; (2) the polygons of R that L traverses successively; 18

Otherwise, there would exist a point C such that L meets some edge L of the polyhedron in an arbitrarily small neighborhood of C. Since near C the edge L reduces to a straight line segment, in that case L would coincide with L .

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(3) the sides of these polygons that L successively intersects. If the positions of all edges of S are fixed in R, then we say that the position of S is fixed in R. Clearly, fixing the position of S in R completely determines the rule for cutting and gluing that transforms R into S and so determines the structure of S. Lemma 1. Let the developments R0 and S0 be obtained from one another by cutting and gluing and have the same vertices. Keeping the structure of R0 invariant, begin changing the defining parameters of R0 , thus obtaining new developments R. If the variation of the parameters is sufficiently small, then for every development R there exists a unique development S whose position in R is the same as the position of S0 in R0 . The defining parameters of S are differentiable functions in the parameters of R. Take some edge L0 of S0 . If L0 coincides with one of the edges of R0 , then we assign the corresponding edge in R to L0 . Assume that L0 coincides with none of the edges of R0 . Then develop successively on a plane all the polygons of R0 that L0 traverses. The funnel will develop only if we slit it. Therefore, if L0 meets the funnel, then we must slit it. We slit it along the half-line that issues from a given vertex, forming a given angle with a given bounded side of the funnel. Then, as the defining parameters change, the slit funnel changes in a one-to-one fashion. We finally obtain a polygon Q0 in which L0 is a diagonal. Under a small variation of the parameters of R0 , the polygon Q0 changes slightly. Therefore, there exists an εL > 0 such that, for variations of the parameters less than εL , the changed polygon still has a diagonal L joining the same vertices and intersecting the same sides of the polygons constituting R. We assign this diagonal L to L0 ; the position of L in R is the same as that of L0 in R0 . Obviously, there is no other segment with the same position. Taking ε > 0 less than all the εL for all edges L of S0 , we see that if the parameters of R0 change less than ε, then to each edge of S0 we may assign a unique line segment with the same position in the changed development R. Proclaiming that these line segments are the edges of a development S, we infer that the position of S in R is the same as that of S0 in R0 . Since every edge L of S is either an edge of R or a diagonal of Q, the length of L and the angles between L and the sides of Q are differentiable functions in the side lengths and angles of the polygons of R. (The fact that the length of a diagonal and the angles between the diagonal and sides are differentiable functions of the side lengths and the angles of the polygon is quite obvious.) Each angle in every polygon of S is constituted by the angles of the polygons of R and their parts cut out by the edges incident to the corresponding vertex. Therefore, the angles of S are also differentiable functions of the side lengths and angles in R. This finishes the proof of the second claim of the lemma.

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10.5.3 Lemma 2. Let a convex polyhedron P0 be deformed, possibly with violation of convexity, so that neither new vertices nor new edges appear, which implies that the structure of the natural development R0 of P0 is preserved. Let another development S0 be drawn on P0 . If the deformation is sufficiently small, then on the deformed polyhedron P there is a unique development S whose position in the natural development R of P is the same as that of S0 in R0 . The development S is unique among developments sufficiently close to S0 . This lemma is a direct consequence of Lemma 1. The last claim of the lemma is obvious, since the edges of S are the only line segments close to the edges of the initial development S0 (by the very construction in the proof of Lemma 1). Now, assume that a polyhedron P0 is deformed, possibly violating convexity, so that there appear no new vertices but new edges may arise. This means that the bounded faces of P may fold along diagonals, whereas the unbounded faces may fold along some half-lines issuing from vertices. For example, we may shift all vertices of a closed polyhedron P0 so that the boundary of their convex hull be a polyhedron close to P . In that case the vertices lying on one plane could cease to belong to one plane and the corresponding face folds. The number of possible diagonals is finite, hence so is the number of possible structures of the bounded part of the deformed polyhedron P . The folding of unbounded faces along new unbounded edges does not affect the structure of the natural development, since the development includes all the faces that are glued together in the funnel. If the diagonals on P0 along which the faces of P0 fold during the deformation are drawn in advance and proclaimed edges, then the natural development of the deformed polyhedron P has the same the structure as that of the polyhedron P0 with these new edges. According to Lemma 2, for a sufficiently small deformation we can therefore construct a development S on each polyhedron P that corresponds to the prescribed development S0 of P0 . This result may be worded as follows: Lemma 3. If a convex polyhedron P0 is deformed without the appearance of any new vertices and the deformation is sufficiently small, then on each deformed polyhedron P we can construct a unique development S close to the prescribed development S0 of P0 . In particular, if P0 is deformed as some variable t changes, then the defining parameters of S are differentiable functions of t.

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10.6 Rigidity of Polyhedra with Stationary Development 10.6.1 Theorem 1. Any closed convex polyhedron with stationary development is infinitesimally rigid. In more detail: suppose that on a closed convex polyhedron P0 we are given a development S0 whose vertices coincide with the vertices of P0 . Let P0 be deformed, as some variable t changes continuously, without any new vertices appearing. (On P0 , fictitious vertices on the interiors of the edges are allowed.) Then, in accordance with Lemma 3 of the preceding section, on each deformed polyhedron Pt (at least for t close to the initial time t = 0) there is a unique development St close to S0 and having for its defining parameters certain differentiable functions of t. Suppose that these functions are stationary, i.e., their derivatives vanish at t = 0. Then the infinitesimal deformation of P0 reduces to a motion, i.e., the initial velocities of motion of its vertices (and hence the rotation velocities of its edges and faces) are such as they would be were the polyhedron P0 a rigid body. Proof. As P0 is deformed, its faces may fold along diagonals leading to a change in the structure of its natural development. It is possible that, even for t arbitrarily close to zero, the same faces fold along different diagonals. In that case we choose one type of such folding of faces and take into account only the corresponding values of t. If we prove that in this case the polyhedron is infinitesimally rigid, i.e., its infinitesimal deformation reduces to a motion, then we will have proved the theorem, since this conclusion will apply to every type of the folding of faces. Indeed, the interval of variation of t splits into some sets T1 , T2 , . . . , Tm each of which corresponds to its own type of folding of faces. Confining our analysis to the kth type of folding, we restrict the values of t to the set Tk . If x(t) are the coordinates of some vertex of Pt , then we may compute the derivative x(0) ˙ by using only such values of t (provided there are values arbitrarily close to zero among them). If, with three vertices fixed, we prove that the computed derivative is zero for every k, then so will be the ordinary derivative computed without restricting the values of t. Thus, we will have established that the polyhedron is fixed when three vertices are fixed, proving its rigidity. So we assume that the deformed polyhedron Pt has a fixed structure. The new edges arising on Pt correspond to diagonals of P0 . Proclaiming these diagonals to be edges of P0 , we infer that the polyhedra Pt and P0 have natural developments Rt and R0 with the same structure. By assumption, the development S0 of P0 is stationary, i.e., the derivatives of the defining parameters of the corresponding variable development St of Pt vanish at t = 0.

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By Lemma 1 of Section 10.5, the defining parameters of the development Rt are differentiable functions of the defining parameters of the development St corresponding to the given development S0 of P0 . Therefore, the natural development of P0 is stationary, too. However, in that case all angles and side lengths of the faces of the polyhedron are stationary, since its faces are exactly the faces of its natural development. In accordance with Theorem 1 of Section 10.3, the fact that the angles on faces are stationary implies that so are the dihedral angles. Whence the rigidity of the polyhedron follows in view of the lemma at the beginning of Section 10.4. 10.6.2 Theorem 2. Any unbounded convex polyhedron having all unbounded edges parallel is rigid if one of its developments is stationary. The detailed content of this theorem unravels much in the same way as that of Theorem 1. In the proof we preserve the notations of Theorem 1: P0 is the given polyhedron, S0 is the given development of P0 , and Pt and St are the deformed polyhedron and its corresponding development. A0 α0

Vt

A0 β0

V0 Fig. 152

If the faces of P0 fold, then we again choose one type of folding and restrict our exposition to the corresponding values of t. Proclaiming that the diagonals along which the faces of P0 fold are edges of P0 , we infer that the natural developments Rt of the polyhedra Pt have the same structure as the natural development R0 of the polyhedron P0 . Since S is a stationary development and the parameters of the development R are differentiable functions in the parameters of S, it follows that the development R is stationary as well, i.e., the lengths of edges and the measures of angles of bounded faces of the polyhedron P0 are stationary together with the lengths of edges and the angles of the funnel. However, the angles of the funnel at vertices which are endpoints of unbounded edges of the polyhedron are sums of angles of unbounded faces rather than these angles themselves. Therefore, to prove that all angles on unbounded faces are stationary, we need additional arguments.


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Let A0 be one of the vertices of the funnel of P0 and let L0 be an unbounded edge issuing from A0 . Let α0 and β0 be the angles on the unbounded faces at A0 . Also, let At , Lt , αt , and βt stand for the corresponding elements of the deformed polyhedron Pt . Slitting the funnel of Pt along Lt and developing it on a plane, we obtain an unbounded polygon Vt . At t = 0, it is the developed funnel V0 of the polyhedron P0 (Fig. 152). The unbounded sides of V0 are parallel, whereas those of Vt may be nonparallel. We must prove that dα ≡ α(0) ˙ = 0. dt t=0 Arguing by contradiction, suppose that α(0) ˙ = 0; for definiteness, let α(0) ˙ > 0. We consider the deformation of the polygon V0 to the polygon Vt . Excluding any motion of the polygon as a whole, we may assume fixed the vertex A0 as well as the direction of the bounded edge issuing from A0 . We decompose the deformation into two parts: the first order deformation and the higher order deformation. The first order deformation consists in changing the elements of the polygon with constant velocities equal to the initial velocities of the given deformation. The higher order deformation supplements the first order deformation to the true deformation.19 We begin by considering the first order deformation. All side lengths of Vt and all angles of Vt except for α(t) and β(t) are the corresponding elements of the funnel. Therefore, they are all stationary and do not change under the first order deformation. Furthermore, since the vertex A0 and the incident bounded edge are fixed, we conclude that under the first order deformation all vertices of Vt stay in place. Only the unbounded sides of Vt rotate, so that ˙ the corresponding angles change with the initial velocities α(0) ˙ and β(0). ˙ By stationarity, α(0) ˙ = −β(0). Hence, the unbounded edges remain parallel. Thus, the first order deformation transforms the polygon V0 into a polygon Vt with the same vertices and with parallel unbounded sides (Fig. 152). If from Vt we cut off the angle that is added to α0 and attach it on the other side, then we obtain the initial polygon V0 . Hence, Vt is a development of the funnel of P0 which is slit, not along L0 , but along the line Lt corresponding to the unbounded sides of the polygon Vt . The line Lt intersects L0 at some point Bt and since L0 corresponds to an edge of P0 , the segment A0 Bt of L0 is shorter than the segment A0 Bt of Lt . (Both vertices A0 of the polygon V0 represent the same vertex of the funnel; therefore, we do not distinguish between them in this argument.) We now add the remaining higher order deformation to the first order one. Then the polygon Vt transforms into Vt and the unbounded side Lt of Vt transforms into the unbounded side Lt of Vt . If t is small enough, then 19

If x(t) is a quantity associated with the polygon Vt , then we put x(t) = x(0) + x(0)t ˙ + ε(t). Here x(0)t ˙ represents the first order deformation and ε(t) represents the higher order deformation.

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the higher order deformation results in a rotation of Lt so small that Lt still meets L0 at some point Bt ; moreover, the segment A0 Bt of L0 is again shorter than the segment A0 Bt of Lt . However, by assumption the line Lt corresponds to some edge of the polyhedron Pt . This implies that no two points of Lt may be joined by a segment shorter than the segment of Lt itself. We come to a contradiction, which proves that α(0) ˙ can never be nonzero. We have thus proved that every angle on an unbounded face of P0 is stationary. The fact that the angles on bounded faces are stationary was established above. Therefore, using Theorem 4 of Section 10.3, we conclude that all dihedral angles of the polyhedron are stationary. Now, knowing that the side lengths and the angles of faces, as well as all dihedral angles, are stationary, we establish infinitesimal rigidity of the polyhedron by using the same simple argument as in the proof of Theorem 1. 10.6.3 We now turn to unbounded polyhedra with nonparallel unbounded edges. Each of these polyhedra P has the limit angle V that does not degenerate into a half-line. To every edge L of V there corresponds at least one unbounded edge of P which transforms into L under the infinite similarity contraction of P to V . For the rigidity of P , the two stationarity conditions, that of a development of P and that of its limit angle, are now insufficient. The polyhedron P can be deformed so that its development and limit angle do not change at all but the edge corresponding to a given edge L of the limit angle rotates, i.e., the angle between this edge and an adjacent bounded edge changes. The possibility of such deformation was established in Section 4.5 for any polyhedron and illustrated by a very simple example. In this connection, we need a third stationarity condition. We formulate it and prove the sufficiency of the three conditions (taken together) in the next theorem. Theorem 3. Let an unbounded convex polyhedron P with limit angle not degenerating into a half-line be deformed so that the following three stationarity conditions are satisfied: (1) of some development S of P , (2) of the limit angle V of P , and (3) of some edge L of P corresponding to a given edge of V . Under these conditions, the polyhedron P is infinitesimally rigid, i.e., its infinitesimal first order deformation reduces to a motion. By condition (3) the angle that L forms with one of the bounded edges of the funnel of the development S which are incident to the endpoint of L is stationary. 20 The unbounded edge L obviously lies on the funnel and, roughly speaking, we require that it be fixed on this funnel. 20

The role of L may be played by any ray on P corresponding to a given generatrix of V .


433

The proof of the theorem proceeds along the same lines as that of Theorem 2. First of all, as in Theorem 2, the fact that a given development of P is stationary implies that the same is true for the natural development of P . (Here we also confine our considerations to one type of folding of bounded faces along diagonals and proclaim these diagonals edges.) Next, we again observe that the angles of the funnel of the polyhedron are in general formed by several angles of unbounded faces. Therefore, the same problem arises: to prove the stationarity of these angles. By condition (3) of the theorem, the angle between L and some bounded edge of the funnel of S is stationary. Since S and the natural developments are stationary, we easily infer that so are the angles between their edges. Hence, the angles between L and the bounded edges of P incident to the endpoint of L are stationary (of course, we mean the angles measured on the polyhedron P ; they may comprise angles of several faces). Let L1 , . . . , Ln be all the unbounded edges incident to the endpoint A of L; one of them, say Lk , is L. Moreover, incident to A are two bounded edges L0 and Ln+1 of the funnel of the polyhedron. All edges are enumerated in the order of their disposition. Denote by αi the angle between the ith and (i + 1)th edge (Fig. 153). L0 α0

Ln+1 A

β1 α 1

αn

B

α1

A

βm

L1

Q

L1

M Ln L=Lk Fig. 153

Fig. 154

The angles between the edge Lk = L and the edges L0 and Ln+1 are equal to α0 + . . . + αk−1 and αk + . . . + αn . Since they are stationary, we have α˙ 0 + . . . + α˙ k+1 = α˙ k + . . . + α˙ n = 0.

(1)

The unbounded edges Li are parallel to the edges of the limit angle V into which they transform under the infinite similarity contraction. Therefore, the angles between them are equal to the angles between the edges of V . By the assumption of the theorem, the latter are stationary. Hence, α˙ 1 = α˙ 2 = . . . = α˙ n−1 = 0.

(2)

It follows from (1) and (2) that α˙ 0 = α˙ n = 0.

(3)

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This proves that all angles on unbounded faces contiguous to A are stationary. Now, let Q be the unbounded face including the angle α0 (Fig. 154). Let β1 , β2 , . . . , βm be the other angles of Q. Further, let βm be the angle contiguous to an unbounded edge: the unbounded face Q has two unbounded edges: one is L1 with endpoint A and the other is M with endpoint B. All angles βi , except for βm , are at the same time angles of the funnel. Therefore, β˙ 1 = β˙ 2 = . . . = β˙ m−1 = 0. (4) The angle between the edges L1 and M is easy to calculate: it equals α0 + β1 + . . . + βm − mπ. However, this angle is equal to the angle between the corresponding edges of the limit angle V . Since the latter is stationary, we have α˙ 0 + β˙ 1 + . . . + β˙ m = 0.

(5)

Using (3), (4), and (5), we conclude that β˙ m = 0. Thus, all angles of Q are stationary; in particular, so is the angle between the unbounded edge M and the adjacent bounded edge. Therefore, for the angles at B, we may repeat all the above arguments for the angles at A. As a result, it follows that all angles at B are stationary. Continuing in similar way, we conclude that all angles on unbounded faces are stationary. Thus we have proved that the angles on all bounded and unbounded faces are stationary. By the assumption of the theorem, the limit angle is also stationary. Now, Theorem 5 of Section 10.3 implies that all the dihedral angles are stationary. The infinitesimal rigidity of the polyhedron is now proved by a word for word repetition of the arguments appearing at the end of the proof of Theorem 1. 10.6.4 Theorem 4. Let P be a bounded convex polyhedron whose boundary a single closed polygonal line. If some development of P is stationary together with all (spatial) angles between boundary edges, then P is infinitesimally rigid. The proof of this theorem is a word for word repetition of the proof of Theorem 1, the only difference being that we must refer to Theorem 6 of Section 10.3 on stationary dihedral angles of a polyhedron with boundary. (As in Theorem 6 of Section 10.3, here we may speak of a locally convex polyhedron and require stationarity of angles between adjacent boundary edges from all but three pairs of these edges. As in Section 10.3, the question of finding conditions for the infinitesimal rigidity of a polyhedron bounded by several polygonal lines remains open.)

10.7 Generalizations

435

We formulate some more theorems (without proof): Theorem 5. If some development of a cap is stationary and the boundary vertices of the cap may move only in the plane containing the boundary of the cap, then the cap is infinitesimally rigid: it only admits motions along the bounding plane. The proof is based on gluing together two symmetric caps to obtain a single closed polyhedron. Theorem 6. Let P be an unbounded convex polyhedron with parallel unbounded edges and with boundary a single unbounded polygonal line. If some development of P is stationary together with all (spatial) angles between adjacent boundary edges, then P is infinitesimally rigid. Theorem 7. Let P be an unbounded convex polyhedron with nonparallel unbounded edges and with boundary a single unbounded polygonal line. Suppose that the following stationarity conditions are satisfied: (1) of some development of P , (2) of the limit angle of P , and (3) of the spatial angles between adjacent boundary edges. Then the polyhedron P is infinitesimally rigid.

10.7 Generalizations 10.7.1 Until now we have dealt with deformations of polyhedra in which the latter remain polyhedra. However, we may imagine a polyhedron made of an elastic but inextensible material, for example paper. In this case it is natural to consider flexes of the polyhedron in which the faces and edges not only fold but also bend arbitrarily. Nevertheless, owing to the condition of “inextensibility” of the material, the lengths of all curves on faces must remain constant. It is impossible to have such flex of a closed convex polyhedron with convexity preserved; this was shown in Section 3.6. Here we are interested in an infinitesimal rather than a genuine flex, i.e., instead of the invariance of lengths of curves on faces, we merely require their stationarity. An infinitesimal deformation is determined by specifying the velocity v of motion of each point of the polyhedron. The condition of stationarity of length leads to a linear differential equation in the velocity field v. Let dx be the differential of the position vector of a running point on a curve and let dv be the corresponding differential of the velocity. Since the lengths are stationary, we have

436


∂ (dx)2 = 0. ∂t However,

∂ 2 ∂t (dx )

∂ = 2dx ∂t dx = 2dxd ∂x ∂t = 2dxv. Hence,

dxdv = 0,

(1)

i.e., the differential of the radius vector of the surface is perpendicular to the corresponding differential of the velocity. This is the well-known equation of infinitesimal flex. Its intuitive meaning is quite clear: the segment dx has stationary length if and only if the relative velocity of its endpoints is perpendicular to it, i.e., if one endpoint rotates around the other. If we admit bending of faces, then no polyhedron is infinitesimally rigid, since no flat domain, and in consequence no face, is infinitesimally rigid even if we clamp its boundary. Indeed, if at each point of a flat domain G we take the velocity v perpendicular to the plane of G, then equation (1) will certainly hold. At the same time, we can ensure that v vanishes on the boundary of G. We conclude that, in this general setting, the question of infinitesimal rigidity of a polyhedron is always solved in the negative for trivial reasons. Nevertheless, in this case as well we can obtain some nontrivial results. For example, we can prove the following: If a closed convex polyhedron undergoes an infinitesimal flex, i.e., if it is deformed so that its faces bend but the lengths of curves on them are stationary, then the vertices of the polyhedron move as those of a rigid body, i.e., their velocities at the initial time are such as those of some rigid body. (Preservation of convexity is not required.) Let a closed convex polyhedron P undergo an infinitesimal flex. Consider the boundary of the convex hull of the collection of vertices of the polyhedron during the flex. At the initial time, it coincides with P . Therefore, we may treat it as another deformation of the same polyhedron. It is easy to infer from (1) that if the length of a straight line segment is stationary under a deformation, then so is the distance between its endpoints. Therefore, under the conditions of the above-stated theorem, the distances between the points of each face of the polyhedron are stationary. Now, we conclude easily that the deformation of the boundary of the convex hull of the collection of vertices of P represents an infinitesimal flex too. Then by Theorem 1 of Section 10.6, this deformation reduces to a motion, which completes the proof of our theorem. Clearly, similar theorems can be stated for unbounded polyhedra and polyhedra with boundary, of course involving conditions on the limit angle and, respectively, conditions on the boundary. Precise formulations and proofs of these theorems are still an unsolved problem.21 21

At present, this problem is settled in the general case by A. V. Pogorelov [P10, Chapter 4]. – V. Zalgaller


437

10.7.2 Generalization of the results of the present chapter to polyhedra in non-Euclidean spaces, hyperbolic space and spherical space, is carried out word for word whenever one deals with bounded polyhedra (closed or with boundary). Nothing need be changed in the statements and proofs. In the case of unbounded polyhedra, the question remains open, since in hyperbolic space the substitute for the notion of limit angle is still unspecified. It is hardly possible that this causes great problems for polyhedra homeomorphic to the plane. In the general case, the question is fully open.22 10.7.3 Infinitesimal flexes of curved surfaces have been the topic of numerous studies from the standpoint of differential geometry for a long time. An infinitesimal flex of a surface is a deformation of the surface under which the lengths of all curves on the surface are stationary. If a surface admits no infinitesimal flexes other than motions, then it is called infinitesimally rigid. A survey of studies on infinitesimal flexes of curved surfaces can be found in the series of articles [E2], [A4], [C-V] in the “Russian Mathematical Surveys.” 23 We indicate only two results that concern curved surfaces and are analogous to our theorems on polyhedra. (1) A closed convex surface without flat pieces is infinitesimally rigid. This theorem was first proved by Liebmann under slightly more stringent assumptions. At present this theorem is proved for thrice differentiable surfaces. Without smoothness assumptions, it is proved only for surfaces of revolution [A1].24 (2) A convex surface whose boundary is a single closed curve and which projects univalently on some plane T turns out to be (infinitesimally) rigid if we require that the velocities of its boundary points are parallel to T . This theorem is also proved only for thrice differentiable surfaces. It corresponds to our Theorem 5 in Section 6. But there we only consider caps. Thus the problem of establishing an analogous general theorem for polyhedra arises. The proofs of these theorems and a complete survey of other results known by 1948 can be found in the cited book [E2] by N. V. Efimov. 10.7.4 The question of infinitesimal rigidity of convex polyhedra in spaces of dimension greater than three turns out to be trivial for the same reason as the question of uniqueness of a higher-dimensional polyhedron with a prescribed development. Already any single three-dimensional polyhedral angle in four-dimensional space is infinitesimally rigid since its section by a sphere centered at the vertex of the angle is a closed convex polyhedron on 22

See some partial results in [VoM], [Al1], [Al2], [VV]. – V. Zalgaller Also see the surveys [Se4] and [S2] and the articles [Co2], [He], [CoW], and [U] mentioned in the latter survey. – V. Zalgaller 24 The smoothness requirements were first weakened to some extent by Minagawa [Mig]. A. V. Pogorelov [P10, Chapter 4] finally proved that every closed convex surface is infinitesimally rigid in the exterior of an open planar domain. – V. Zalgaller 23

438


a three-dimensional sphere or, at will, in three-dimensional spherical space. As mentioned in Subsection 10.7.2, this polyhedron is infinitesimally rigid; hence, so is the original polyhedral angle.

11 Infinitesimal Rigidity Conditions for Polyhedra with Prescribed Face Directions

In this chapter we consider deformations of a convex polyhedron in which the directions of faces (i.e., their outward normals) are fixed. As in Section 10, we are interested only in first order deformations or, which is the same, the velocities at the initial moment of time. The problem consists in finding conditions under which a polyhedron is infinitesimally rigid. Since the face directions are constant, rotations are excluded and so infinitesimal rigidity means that each infinitesimal deformation reduces to a translation. Infinitesimal rigidity conditions appear in the theorems of Section 11.2. To prove these theorems, we first examine the deformations of faces resulting from parallel displacements of the planes of the faces. Every face of a closed or unbounded convex polyhedron is bounded by the straight lines along which the plane of the face intersects the planes of the other faces. Therefore, we consider deformations of a convex polygon resulting from parallel displacements of the bounding straight lines. If two faces touch only at a vertex, then an arbitrarily small displacement of the plane of one of them may lead to their touching at an edge. In this connection, we must take into account the possibility of new edges appearing.

11.1 On Deformations of Polygons 11.1.1 Let a convex polygon P be cut from a plane by several straight lines touching P at sides or vertices. In the second case we view P as possessing sides of the corresponding directions but of length zero. The distance from the origin to a bounding straight line is called a support number of the polygon. (Of course,the origin is taken on the plane of the polygon and, as usual, we regard the distance from the origin to the straight line negative if the outward normal beginning at the origin points away from the straight line.) We consider polygons arising from P as the result of deformations caused by parallel displacements of the straight lines bounding P . We assume that each of these straight lines does not cease touching the polygon.1 This con1

For this reason, the possible values of the support numbers are not arbitrary; they belong to some closed domain determined by this condition.

440

11 Rigidity of Polyhedra with Prescribed Face Directions

dition makes sense for arbitrarily small deformations of those straight lines that touch the polygon only at vertices at the initial moment of time. Under this condition, not only the deformed polygon is determined by its support numbers, but, conversely, the distances to bounding straight lines are in turn determined by the polygon, since the directions of these straight lines are fixed. In this connection, every function of polygons can be regarded as some function of its support numbers, say, f (h1 , . . . , hn ). Since we admit sides of zero length, whenever we consider a finite number of bounded polygons, we may assume that all their sides are pairwise parallel. Therefore, the function f (h1 , . . . , hn ) is defined for all such polygons, as well as for polygons arising from them as the result of parallel displacements of the bounding straight lines. We call a function f (Q) = f (h1 , . . . , hn ) of polygons essentially monotone if it satisfies the following conditions: (1) f (h1 , . . . , hn ) does not change under parallel displacements of the polygon. ∂f exists and is continuous up to the disap(2) Every partial derivative ∂h i 2 pearance of the ith side. ∂f (3) ∂h > 0 for every i if the ith side has nonzero length. This implies i ∂f that ∂hi ≥ 0 when the ith side disappears. The last condition implies that if a polygon Q1 lies inside another polygon (1) (2) (1) (2) Q2 , then f (Q1 ) < f (Q2 ), since in this case hi ≤ hi for all i and hk < hk at least once. By condition (1), we also have f (Q1 ) < f (Q2 ) when Q1 can be placed inside Q2 by a translation. Area and perimeter provide examples of essentially monotone functions. ∂f If f is area, then ∂h = li , where li is the length of the corresponding side; i therefore, the shift of this side by the distance ∆hi results, to within higher order infinitesimals, in adding the area of the rectangle with height ∆hi and base li to the area of the polygon. Here we encounter the case in which ∂f ∂f ∂hi = 0 if the ith side disappears. If f is perimeter, then ∂hi = const > 0, which is a consequence of the following lemma. Lemma 1. The length of a side of a polygon is a linear function of the support numbers hi . 2

If the ith side disappears at hi = h0i , then by

∂f ∂hi

hi =h0 i

we mean the cor-

responding one-sided derivative. Continuity up to disappearance of the ith side means that if tends to h0i so that the ith side does not degenerate for all hi , hi ∂f ∂f . then ∂hi → ∂hi 0 hi =hi

11.1 On Deformations of Polygons

441

More precisely, if we enumerate all sides3 and denote the angle between the (i − 1)th and ith sides by ϕi , then li =

hi−1 − hi cos ϕi hi+1 − hi cos ϕi+1 + . sin ϕi sin ϕi+1

(1)

To prove this, look at Fig. 155 which shows three successive sides together with the perpendiculars from O to the straight lines on which they rest. The lengths of these perpendiculars are exactly the corresponding support numbers. Dropping a perpendicular from Ai to OAi−1 , we obtain OAi−1 = hi−1 = Ai B sin ϕi + hi cos ϕi . Whence Ai B =

hi−1 − hi cos ϕi . sin ϕi

A similar formula holds for Ai C. Since li = BAi + Ai C, we arrive at the expression (1) for li . (If the angle ϕi or ϕi+1 is obtuse, then the drawing changes but the same arguments are still valid.) B Ai-1

Ai

hi hi-1 ϕ ϕi+1 i

C

hi+1

O Fig. 155

11.1.2 Lemma 2. Let a bounded convex polygon Q be deformed in result of parallel displacements of its bounding straight lines with definite (possibly variable) velocities. Then it follows from Lemma 1 that the lengths of the sides of Q change with definite velocities as well. Label each side of Q by the sign of the initial speed of variation of its length, leaving the edges with stationary length unlabeled. If some strictly monotone function f of Q is stationary under this deformation, then only the following three possibilities for the sign disposition on the sides of Q may occur: (1) No side is labeled. In this case the initial deformation reduces to an infinitesimal translation. (2) There are at least four sign changes. (3) There are exactly two sign changes. In this case, two sides of nonzero length incident to some vertex A are labeled by minuses and some sides of zero length issuing from A are labeled by pluses. 3

The order of enumeration of the sides of zero length corresponds to the order of the normals to the support lines through them.

442


In the last case some bounding straight lines touching Q at A are shifted inside Q so that the corresponding sides of zero length elongate. The polygon Q undergoes no deformation at the initial moment other than possibly a translation of Q as a whole. If hi is the support number corresponding to ∂f ∂f a side of zero length shifted inside Q, then necessary ∂h = 0, i.e., ∂h >0 i i implies that the third case is impossible. This lemma is the analog of Lemma 3 of Section 6.1, and we will prove it by reducing it to the latter.4 If the initial deformation of the polygon is a translation, then the side lengths are stationary and the first possibility occurs. Suppose that the initial deformation is not a translation and let h˙ i be the initial speeds of variation of the support numbers in this deformation. ∂f =0 First, we exclude the elongation of those sides of zero length for which ∂h i (i.e., we move the straight lines, on which the sides rest, together with the vertices through which they pass. Clearly, this does not violate the stationarity ∂f = 0.) of f since ∂h k Take a sufficiently small value t and construct a polygon Q with the support numbers hi + h˙ i t. Then, in accordance to Lemma 1, the variations of the lengths are l˙i t, where l˙i is the initial speed of variation of the length of the ith side. The polygons Q and Q can be equal and have parallel sides; in that case the deformation reduces to a translation. Assume that it is not and that one of the polygons Q and Q can be placed inside the other by translation, say Q can be placed inside Q . Then, composing the corresponding translation with the original deformation, we shall have h˙ i ≥ 0 for all i and h˙ k > 0 for at least one k. However, since now we have ∂f hi > 0 for all i, we obtain f˙ =

∂f h˙ i > 0, ∂hi

which contradicts the stationarity of f . Hence, none of the polygons Q and Q can be placed inside the other by translation. Then, by Lemma 3 of Section 6.1, the differences between the lengths of their parallel sides change sign at least four times. Since these differences are proportional to the initial speeds l˙i , the speeds change sign at least four times as well. 4

However, it is important to note that in Lemma 3 of Section 7.1 we dealt with two different polygons or, equivalently, with a finite deformation of a given polygon, whereas here we speak of a first order infinitesimal deformation, i.e., about initial velocities. This circumstance causes, first, the necessity of considering an essentially monotone rather that a simply monotone function of polygons and, second, the appearance of the third case of disposition of signs, which is clearly impossible under the conditions of Lemma 3 of Section 6.1. If f (Q) is area, then (f − 1)2 is stationary under any deformation of a polygon of unit area. At the same time, (f − 1)2 is a monotone function of polygons.

11.1 On Deformations of Polygons

443

Now, it remains to introduce the elongation of those sides of zero length ∂f for which ∂h = 0, i.e., after the above deformation of the polygon, we shift i the straight lines supporting these sides inside the polygon by the distance h˙ i t. This results in the appearance of new genuine edges. We must distinguish between two cases: (1) before the additional deformation, the polygons Q and Q were translates of one another and (2) the number of sign changes of the speeds l˙i is at least four. When a side li issuing from the vertex A between sides lj and lk of nonzero length is shifted inside the polygon, the sides lj and lk shorten, i.e., the speeds of variation of their lengths diminish. Therefore, if in the first case this happens at least twice at two different vertices, then we obtain at least four changes of sign. If this happens only at one vertex, then we obtain the third possibility for the sign disposition indicated in the lemma. Now, assume that, before shifting the sides of zero length inside the polygon, we had four sign changes. We demonstrate that the number of sign changes does not diminish after this shift. Let one or several new genuine sides appear at the vertex A at which the sides AB and AC touch. Before these new sides appear, the possible dispositions of signs on AB and AC may be following: 1

2

3

4

5

6

7

8

9

AB

0

−

0

−

+

0

+

−

+

AC

0

0

−

−

0

+

−

+

+

Since AB and AC have equal rights, the dispositions 2 and 3, 5 and 6, and 7 and 8 are equivalent. Therefore, it suffices to consider the table I

II

III

IV

V

VI

AB

0

−

−

+

+

+

AC

0

0

−

0

−

+

When the straight lines that support sides of zero length are shifted inside the polygon, the sides AB and AC shorten, i.e., the speeds of variation of their lengths diminish. At the same time, the sides of zero length certainly elongate. Therefore, if both sides AB and AC do not elongate (cases I, II, and III) then they shorten; hence, they are labeled by minuses and the new sides are labeled by pluses. This produces two sign changes. If only one side elongates (cases IV and V), then, after new sides appear, this side can shorten. This might lead to losses of sign changes. However, these losses are compensated by the appearance of new sign changes due to

444


the pluses standing on the new sides between AB and AC. The possible cases are listed on the table: Before the appearance of new sides After that

AB

+

+

AC AB New sides AC

0 0 + −

− 0 + −

+ + −

− + −

+ + −

− + −

From the table we see that no losses of sign changes occur. Finally, if both sides AB and AC elongate, then they are labeled by pluses and in passing from AB to AC there are no sign changes. The new genuine sides that appear are labeled by pluses too; therefore, the changes in sign from plus to minus (in both directions from A) cannot disappear. New sign changes will arise if, for example, the disposition of signs is as in Fig. 156. A B

+

+ C

+

+ _

+

_

+

+ Fig. 156

Thus, in all cases the appearance of new sides does not lead to losses of sign changes, which completes the proof of the lemma. 11.1.3 We now consider deformations of unbounded convex polygons under which the straight lines containing the unbounded sides remain fixed. We say that an unbounded side shortens if its endpoint moves into it with nonzero initial speed and we say that it elongates if the endpoint moves in the opposite direction. Lemma 3. Let an unbounded convex polygon be deformed as the result of parallel displacements of the straight lines passing through the bounded sides of the polygon, including the sides of zero length. Label each side by the sign of the initial speed of variation of its length, bearing in mind the above convention about unbounded sides. Leave the sides of constant length unlabeled. Then there are only three possibilities for the disposition of signs: (1) None of the sides is labeled; the polygon at the initial time is fixed. (2) There are at least two sign changes.

11.2 Infinitesimal Rigidity Theorems for Polyhedra

445

(3) Both unbounded sides are labeled by the same sign and all bounded sides are unlabeled. This case is possible only for a polygon with parallel unbounded sides. It means that at the initial time the polygon moves along its unbounded sides as a rigid body. This lemma is similar to Lemma 4 of Section 6.1 and is proved by reducing it to the latter in the same way as in the proof of Lemma 2. Namely, if Q is the initial polygon, then we construct a polygon Q with support numbers hi + h˙ i t, where hi are the support numbers of Q and h˙ i are the initial speeds of their variation. Then by Lemma 1 the differences between the lengths of the sides of Q and Q are l˙i t, where l˙i are the initial speeds of variation of these lengths. (By the above convention, for an unbounded side l˙i is the speed of the endpoint.) Therefore, applying Lemma 4 of Section 6.1 to Q and Q , we arrive at the desired conclusion.

11.2 Infinitesimal Rigidity Theorems for Polyhedra 11.2.1 A convex polyhedron, whether closed or unbounded, is completely determined by the planes of its faces. When we move these planes, the polyhedron is deformed accordingly. Here we assume that the planes undergo parallel displacements with definite speeds, i.e., their normals are constant or at least stationary and the support numbers are differentiable functions of time t. In this case we say that the polyhedron is deformed by parallel displacements of the planes of its faces. Every face of a polyhedron is bounded by the straight lines along which the plane of the face intersects the planes of the other faces. These straight lines also move in parallel; therefore, the deformations of faces are exactly as in Section 11.1. On these faces new edges may appear. When the directions are fixed, the positions of the straight lines bounding a face on the plane T of the face are determined by the support numbers relative to some origin on T . It is most natural to take this origin O to be the projection to T of the origin O in space. The support numbers of a face (thus defined) are linear functions of the support numbers of the polyhedron. The proof is obvious from Fig. 157, where the planes of adjacent ith and kth faces are drawn in projection along their common edge. Here θik is the angle between the normals to these faces; hi and hk are the support numbers of the faces; Oi and Ok are the origins on faces; and Oi A = hik and Ok A = hki are the support numbers of the common edge, the first on the ith face and the second on the kth face.

446


A hki

hik Oi hi

θik

Ok

hk

O Fig. 157

Projecting Ok on OOi , we obtain OOi = hi = hki sin θik + hk cos θik ; whence hi − hk cos θik hki = . (1) sin θik Thus, hki is linear in hi and hk . This fact implies that, when the support numbers hi change with definite speeds, the numbers hki also have definite speeds of variation. But then, in view of Lemma 1 of Section 11.1, the same holds for the lengths of the edges of the polyhedron. These remarks enable us to apply Lemmas 2 and 3 of Section 11.1. Since the lengths of edges and the support numbers of faces are linear functions in the support numbers of the polyhedron, it follows that as the faces move with constant speeds, the speeds of variation of the lengths of edges and of the support numbers of faces are also constant. However, in rigidity questions we may restrict ourselves to uniform motion of faces, and so to the uniform variations of edges and support numbers; in particular, if they are stationary, then they must simply be constant. For these reasons, from now on we need not fear confusing the notions of stationarity and constancy as regards edges and support numbers. 11.2.2 Theorem 1. If for every face of a closed convex polyhedron its direction and some essentially monotone function are stationary, then the polyhedron is infinitesimally rigid. In other words, suppose that a closed convex polyhedron is deformed by parallel displacements of the planes of its faces. If some essentially monotone function is stationary for every face, in general a different function for each face, then the deformation of the polyhedron reduces to a translation, i.e., the initial speeds of the planes of the faces of the polyhedron are the same as for a rigid body.5 5

The theorem fails if it is not required that the stationary functions be essentially monotone. An example is the function f = (F (Q) − 1)2 , where F (Q) is area. This function is stationary for arbitrary displacements of the faces of a unit cube.


447

The proof is based on combining Lemma 2 of Section 11.1 with the Strong Cauchy Lemma (Section 10.2). Move the planes of the faces of P by small distances proportional to the initial speeds of the planes. We obtain a new polyhedron P on which no edge present in P disappears but new edges possibly arise. The latter means that some faces of P have sides of zero length which elongate in deformation. Label each edge of P by the sign of the initial speed of variation of its length, more precisely, of the length of the edge of P that gives rise to this edge of P , recalling that some edges of P can arise from zero edges of P . Leave unlabeled the edges of stationary lengths. Since some essentially monotone function is stationary for every face, the disposition of signs around each vertex satisfies Lemma 2 of Section 11.1. That is, on every face we have one of the following three possibilities: (1) None of the edges of the face is labeled. (2) There are at least four sign changes around the face. (3) The number of sign changes equals two. In this case the edges arising at one place from edges of zero length are labeled by pluses and the adjacent ordinary edges are labeled by minuses. By the Strong Cauchy Lemma (Section 10.2), similar conditions are imposed on the disposition of signs around vertices. Therefore, choose a point inside each face of P and connect the points in adjacent faces by lines intersecting the corresponding edges. As the result, we obtain a net R on P which is dual to the net of faces of P : to a face on P there corresponds a vertex of R; to an edge, an edge; and to a vertex, a face (see Fig. 128 on p. 289). Proceeding in the same way, let us construct a dual net R on P and label each edge of R with the sign of the corresponding edge of P . Then to the first two possibilities for the disposition of signs around a face of P listed above there correspond the following two possibilities for the disposition of signs around a vertex on the net R : (1) None of the edges incident to the vertex is labeled. (2) There are at least four sign changes around the vertex. We now clarify what happens in the remaining case of two sign changes. In that case, “edges of zero length” on P that transform into genuine edges of P can exist. This means that four or more faces of P touch at some vertex A and cease touching at one vertex after the displacement of their planes (Fig. 158). This looks as though the vertex A splits and new edges between the faces touching at A appear. For the dual nets R and R , this implies the following. The net R has a face GA corresponding to the vertex A and the vertices of GA correspond to the faces of P touching at A. The splitting of A corresponds to the partition of GA by new edges. These new edges correspond to the new edges of the polyhedron P . They join the vertices of GA corresponding to the faces of P that adjoin one another along new edges after deformation.

448


A

A

A

Fig. 158

The above means that GA is partitioned by diagonals, i.e., the passage from R to R consists in drawing diagonals on some faces of R. Thus the net R satisfies the requirements of the Strong Cauchy Lemma. Suppose that there are exactly two sign changes around some face H of P (the third case). Then the face H of P has some vertex A whose “splitting” gives rise to new edges on P . All the labeled edges of H touch at A in P , but only the two extreme edges are genuine. Thus we see that, when we pass to the dual net R , the third possibility for the sign disposition around a face of P becomes the following: (3) The number of sign changes around the vertex (corresponding to the face H ) equals two. In this case the new edges on one “old face” (corresponding to the vertex A) are labeled by pluses and the adjacent “old edges” are labeled by minuses (Fig. 159 exemplifies both dispositions of signs on P and on R .) This is exactly the third possibility appearing in the Strong Cauchy Lemma. A

+

A

A

-

-

A

H

+-

H

P

R Fig. 159

Therefore, if there are labeled edges, then we come to a sign disposition which is impossible by this lemma. Hence, there are no labeled edges at all, i.e., the lengths of all the edges of P are stationary. Now, it is obvious that the deformation of P reduces to a translation, which completes the proof of the theorem.


449

11.2.3 The result may be slightly generalized by admitting deformations of polyhedra involving the appearance of new faces. Namely, we consider a given polyhedron P as bounded by not only the planes of its faces but also by some planes passing through its edges.6 Then the parallel displacements of these planes can yield new faces on P . This can be imagined as though some faces of P degenerate into edges and become genuine faces after the deformation. Such a degenerate face is “bounded” by two coinciding straight lines (containing the edge) together with other straight lines passing through the endpoints of the edge (Fig. 160).

Fig. 160

In this connection, the appearance of a new face in place of an edge is represented as a deformation of a polygon degenerating into a segment caused by the motion of its bounding straight lines. At the initial moment of time, two sides of such a polygon coincide and the other sides are of zero length. Let the polygon Q degenerate into the segment AB. Assume that a several bounding straight lines pass through A. If one of these straight lines moves “outwards” (the support number increases), then it ceases to touch the polygon Q and thereby ceases to be a bounding straight line. Its motion yields no deformation of the polygon and so we may assume this line fixed. For this reason, we exclude any such motion of the bounding straight lines. Lemma. Let a polygon, which degenerates into a line segment at the initial moment of time, be deformed by parallel displacements of its bounding straight lines. If some essentially monotone function of the polygon is stationary, then either the polygon remains a segment or the speeds of variation of the lengths of its sides change sign at least four times as we go around the polygon. 6

The generalization admitting degeneration of faces into vertices is of no interest: its triviality is clear since, as a point transforms into a polygon, all support numbers may be assumed increasing and so either df > 0 or f ceases to be an essentially monotone function for polygons degenerating into a single point in the same way as it does, for instance, for area.

450


Let Q be a polygon degenerating into a straight line segment. Assume that Q is deformed so that some essentially monotone function f (Q) is stationary. Suppose that Q becomes a genuine polygon after the deformation. This means that two bounding straight lines L1 and L2 , coinciding at the initial time, diverge (Fig. 161). Li Q

O

L1 =L2

Li L2 O

Q l1

L1

Fig. 161

Clearly, on both ends of Q there must be some elongating sides of zero initial length. Therefore, for the presence of four sign changes, it is necessary that both sides of nonzero length shorten (clearly, sides of zero length cannot shorten!). Assume, however, that one of these sides, l1 , does not shorten and is either stationary or elongating. Put the origin O at one of the endpoints of l1 . At the initial time, l1 is bounded by the intersection of the straight line L1 with the straight lines supporting the zero sides. Since l1 does not shorten, these lines never come nearer to O. In consequence, the speeds of variation of their support numbers are nonnegative: h˙ i ≥ 0. At the same time, L2 recedes from L1 and so the speed of variation of its support number is positive h˙ 2 > 0. Finally, the line L1 continues passing through the origin and so h˙ 1 = 0. Thus, h˙ i ≥ 0 for all i and h˙ 2 > 0. However, by the essential monotonicity df df df condition, dh ≥ 0 and dh > 0 for sides of nonzero length, i.e., dh > 0. i i 2 Therefore, df df = h˙ i > 0. dt dhi At the same time, f is stationary (by assumption), i.e., df dt = 0. This contradiction shows that the assumption that one of the sides of nonzero length does not shorten is false. Hence, both sides of nonzero length shorten and we thus have exactly four sign changes, completing the proof of the lemma.


451

Now, assume that a polyhedron P is deformed by parallel displacements of the planes of its faces; moreover, at the initial time some faces are degenerate, namely they are edges. Then, labeling the edges of the deformed polyhedron P by the signs of the speeds of variation of their lengths, we come to the same conclusion as in the proof of Theorem 1. We always have four sign changes around a face arising from an edge; therefore, such a face can never appear at all. Therefore, in Theorem 1 we may admit planes of faces which at the initial time pass along edges, i.e., admit faces degenerating into edges. This is similar to the admittance of faces degenerating into edges in the theorems of Chapter 6 on the congruence of polyhedra with parallel faces. As an example, we can state the infinitesimal rigidity theorem for a polyhedron with stationary perimeters of faces. In this case we may admit among the faces those degenerating into edges. The perimeter of such a “face” equals the doubled length of the edge. In the case of stationary area, the generalization is trivial, because a degenerate face has zero area and if the area is stationary, then such face must remain an edge, i.e., here the generalization becomes meaningless. 11.2.4 Theorem 2. Let an unbounded convex polyhedron be deformed by parallel displacements of the planes of its bounded faces; the planes of the unbounded faces are assumed fixed. If for each bounded face some essentially monotone function is stationary, then the polyhedron remains fixed (more precisely, the planes of all its faces are stationary). (Among the faces we admit faces degenerating into edges at the initial time.) This theorem corresponds to Theorem 2 of Section 6.4 on the congruence of unbounded polyhedra; it relates to the latter exactly like Theorem 1 just proved above relates to Theorem 2 of Section 6.3 on closed polyhedra. The proof of Theorem 2 relies on the same arguments. Shifting the planes of faces of a given polyhedron P by small distances proportional to the initial speeds of the planes, we obtain a polyhedron P . Label the edges of P by the signs of the speeds of variation of their lengths. Then for bounded faces we obtain the same three possibilities as in Theorem 1 and for unbounded faces, the possibilities of Lemma 3 of Section 11.1. Take a copy P of P and label its edges with opposite signs. Define the abstract polyhedron P ∪ P by identifying the corresponding unbounded edges of P and P and uniting every pair of corresponding unbounded faces into one face. Next, pass from P ∪ P to the dual net R and for R repeat the same conclusions as in the proof of Theorem 1. Now apply the Strong Cauchy Lemma to the net R , concluding the proof of Theorem 2. A detailed implementation of the plan outlined would simply be a repetition of the proofs of Theorem 1 and Theorem 2 of Section 6.4.

452


11.2.5 Theorem 2 can be refined and extended to n-dimensional space in exactly the same way as its analog from Section 6.5 on the congruence of unbounded polyhedra. We thus obtain Theorem 3. Let an n-dimensional unbounded convex polyhedron be deformed by parallel displacements of the planes of its bounded faces; the planes of unbounded faces are assumed fixed. Suppose that at least one of the following conditions holds for every bounded face: (1) the plane of the face is stationary and (2) some essentially monotone function of the face is stationary.7 Under these conditions, the polyhedron remains fixed, i.e., the planes of all its faces are stationary. This theorem can be rephrased for bounded polyhedra with boundary similary to the analogous congruence theorem of Section 6.5. The proof is perfectly similar to that of the theorem of Section 6.5. Let h(n, t) be the support function of the deformed polyhedron (i.e., the distance from its support plane with normal n to the coordinate origin, regarded as a function of n and the parameter t of the deformation). If its initial speed of time. variation, ∂h ∂t t=0 , vanishes, then the polyhedron is fixed at the initial Therefore, suppose that there is a domain of vectors n in which ∂h ∂t t=0 > 0. Define the function 1 ϕ(n) = ∂h ∂t t=0

in this domain and consider the surface R covered by the ends of the vectors ϕ(n)n. This construction is similar to that in Section 6.5 with the only difference that in place of the difference of support functions we use the derivative ∂h ∂t . Further arguments are analogous to those in Section 6.5 and we omit them. 11.2.6 A polyhedron with boundary is not determined by the planes of its faces alone: the straight lines through its boundary edges are also needed. These lines lie on the planes of the extreme faces. We admit deformations of the polyhedron under which these lines undergo parallel displacements in the planes of the extreme faces (which may undergo parallel displacements themselves). Theorem 4. Let a bounded convex polyhedron whose boundary is a single closed polygonal line be deformed by parallel displacements of the planes of its faces and of the straight lines passing through its boundary edges. Assume that (1) for each face some essentially monotone function is stationary and (2) on each extreme face the initial speeds of variation of the lengths of the 7

An essentially monotone functions of (n − 1)-dimensional polyhedra are defined in exactly in the same way as for a polygon.

11.3 Rigidity Theorems and the Theory of Mixed Volumes

453

boundary edges have the same sign (its own for each face), with the zero value allowed. Under these conditions, the initial deformation of the polyhedron is a translation. This theorem is analogous to Theorem 2 of Section 6.5. The proof is based, like those of Theorems 1 and 2, on the Strong Cauchy Lemma and Lemma 1 of Section 11.1. We again construct a deformed polyhedron P and label its edges by signs. Next, taking a copy P with opposite signs, we define an abstract polyhedron P ∪ P by identifying and then excluding the corresponding boundary edges of the polyhedra P and P . Afterwards, we construct the dual net and apply the same arguments as in the proof of Theorem 1 to it, thus concluding the proof of our theorem.

11.3 Relationship of Infinitesimal Rigidity Theorems with One Another and with the Theory of Mixed Volumes 11.3.1 Theorem 1 of Section 11.2 contains in particular the following Theorem 1. A closed convex polyhedron with stationary face directions and face areas is infinitesimally rigid. We now show that the following proposition can be derived from this theorem as a purely formal corollary. Theorem 2. A closed convex polyhedron with infinitesimally rigid faces is infinitesimally rigid. This theorem was proved in Section 10.4; moreover, the faces were understood there in a certain generalized sense. Here we have in mind ordinary faces. Hence, the result is weaker than in Section 10.4. Nevertheless, the formal connection of Theorem 2 with Theorem 1, on which our derivation is based, is of interest in its own right. By studying it, we will naturally come across a connection between both theorems and the theory of mixed volumes (Sections 7.2 and 7.3). An application of the latter yields a new proof of Theorem 1 and so a new proof of Theorem 2. The above-mentioned connections are of great importance for the derivation of infinitesimal rigidity theorems for curved surfaces. They were first observed for polyhedra by H. Weyl in 1917 in the article [W2].

454


11.3.2 So let us turn to the formal reduction of Theorem 2 to Theorem 1. Let a polyhedron with infinitesimally rigid faces be deformed. Then each face Qm rotates as a rigid body, and this rotation is characterized by the angular velocity vector wi . Since the relative rotation of two adjacent faces Qi and Qj proceeds around their common edge, the difference wi − wj is directed along this edge. This is equivalent to the condition that the difference wi − wj is perpendicular to the (outward) normals ni and nj of the faces, i.e., (wi − wj )ni = 0, (wi − wj )nj = 0, or wi ni = wj ni ,

wi nj = wj nj .

(1)

Put wi ni = pi

(2)

and consider the deformation of the same polyhedron resulting from the parallel displacements of its faces with constant speeds pi in the directions of their normals. Then at time t the support numbers of the polyhedron are equal to (3) hti = hi + tpi . Let us prove that, in view of relations (1), the face areas are stationary under the deformation determined by (3). This reduces Theorem 2 on infinitesimal rigidity of polyhedra with infinitesimally rigid faces to Theorem 1 on the infinitesimal rigidity of polyhedra with stationary face directions and face areas. Indeed, by Theorem 1 the fact that the face directions and face areas are stationary implies that the polyhedron merely undergoes a translation with some velocity a. But then the pi are the normal components of this translation relative to the faces and so pi = ni a. To the deformation of the polyhedron, let us now add its rotation as a whole with angular velocity −a. Then the velocities of rotation of its faces are wi = wi − a, which implies that pi = wi ni = 0 for all i. However, by (1), wi nj = wj nj = wi ni = 0, where nj is the normal to an arbitrary face having a common edge with Qi . Since among these normals there are surely two not lying together with ni in one plane, the above equalities imply that wi = 0. Thus, the new deformation is a motion. Hence, the original deformation was a motion, i.e., we arrive at Theorem 2.


455

11.3.3 Let us prove that relations (1) imply that the face areas are stationary. To this end, consider a given face Qi . For the origin on its plane take the projection of the origin in space to the plane. If hij are the support numbers of the edges of Qi with respect to this origin, then by formula (1) of Section 11.2 we have hj − hi cos θij hj − hi (ni nj ) hij = , (4) = sin θij |ni × nj | where hj and nj are the support number and the normal of the face Qj adjoining Qi along the corresponding edge. By (3), for the deformed polyhedron we obtain htij = hij + tpij , with pij =

pj − pi (ni nj ) . |ni × nj |

(5)

(6)

If we shift a side l of a polygon by a small distance δ, then we add to the polygon the trapezoid with base l and altitude δ. Therefore, the derivative of the area Fi of the face Qi with respect to the support number hij is the length of the corresponding side: ∂Fi = lij . (7) ∂hij Then, in view of (5), ∂Fi dFi = = pij lij . dt t=0 ∂hij t=0 j j

(8)

It remains to show that the sum on the right-hand side vanishes because of (1). To this end, we observe that by virtue of (1) and (2) pi = wi ni ,

pj = wi nj .

(9)

Therefore, (6) implies that pij =

nj − ni (ni nj ) wi = nij wi , |ni × nj |

(10)

nj − ni (ni nj ) |ni × nj |

(11)

where the vector nij =

is easily seen to be nothing but the outward unit normal to the side lij of the face Qi . Hence, the closure relation holds: pij lij = 0. (12) j

Multiplying (12) by wi and using (10), we obtain

456


pij lij = 0;

(13)

j

whence in view of (8) we infer that dFi = 0, dt t=0 which was to be proved. 11.3.4 We now prove Theorem 1 on the infinitesimal rigidity of polyhedra with stationary face directions and face areas by using the theory of mixed volumes. Let the initial speeds of parallel of the faces of a given displacements i polyhedron P be equal to pi , i.e., dh speeds of dt t=0 = pi , and let the initial dhij = variation of the support numbers of the edges of faces be pij , i.e., dt t=0 pij . For the derivative of the face area we have formula (8): dFi = pij lij . dt t=0 j Therefore, the condition that the face areas are stationary yields pij lij = 0.

(14)

j

Consider the polyhedron P with support numbers hi = hi + tpi

(15)

where t is so small that any two faces not adjacent on P are not adjacent on P . The distinction between the structures of P and P can only be caused by the fact that some faces touching at one vertex in P may diverge on P , while other faces touching at a vertex now touch along an edge. In this case we may assume that such edges are contained in P but have zero length. Let Qi and Qi be corresponding faces of P and P . The mixed area of these faces, defined just like the mixed volume (see Lemma 4 in Section 8.2), is equal to 1 F (Qi Qi ) = h lij , (16) 2 j ij or, in view of (5), to F (Qi Qi ) = Whence, by (14),

1 1 hij + t pij lij . 2 j 2 j


F (Qi Qi ) =

1 hij = F (Qi Qi ), 2 j

457

(17)

where F (Qi Qi ) = 12 j hij is nothing else than the area of the face Qi . Let us write out the Minkowski inequality for mixed areas8 F (Qi Qi )2 ≥ F (Qi Qi )F (Qi Qi );

(18)

here the equality sign will occur if and only if Qi and Qi are homothetic. Owing to (17), this implies that F (Qi Qi ) ≥ F (Qi Qi ),

(19)

where the equality sign will occur only if it occurs in (18), i.e., if the faces Qi and Qi are homothetic. However, since in that case (19) reduces to the equality between the areas of Qi and Qi , (19) becomes an equality only if Qi and Qi are parallel translates of each other. Multiplying (17) by hi and summing over all faces, we obtain hi F (Qi Qi ) = hi F (Qi Qi ). (20) i

i

At the same time, multiplying (19) by hi and summing over all faces, we obtain hi F (Qi Qi ) ≥ hi F (Qi Qi ). (21) i

i

However, by Theorem 4 of Section 8.3 (formula (26))9 , we have 8

This inequality is clearly a particular instance of the general inequality derived in Section 8.3 for n-dimensional space. However, for areas its derivation is quite F (QQ )2 simple. We seek a minimum of the ratio Φ = F (QQ)F for fixed Q and varying (Q Q ) Q . It is easy to see that the minimum will be attained. Denoting the support numbers and the side lengths of the polygons Q and Q by hj , hj and lj , lj , we have ) ∂Φ = 0 at minimum points. By (16), ∂F (QQ = 12 lj . Finally, since F (Q Q ) is ∂h h j

the area of a polygon, we have

9

∂F (Q Q ) ∂hj

j

= lj . Calculating the derivatives

∂Φ ∂hj

with

the help of these relations and equating the derivatives to zero, we infer that at (QQ ) a minimum point we have lj = λlj , with λ = FF(Q Q ) . However, if the side lengths are proportional, then the polygons Q and Q are similar, i.e., the minimum of Φ is attained at similar polygons. But then Φ = 1. Hence, Φ ≥ 1 and Φ = 1 if and only if Q and Q are similar. This is exactly inequality (18) together with the condition for equality. Here we could manage without referring to Theorem 4 of Section 8.3 and obtain the same result immediately. Indeed, using (16) (and interchanging Qi and Qi ) we obtain i

hi Fi (Qi Qi ) =

i

hi ·

hi hj − hi cos θij 1 hij lij = lij . 2 j 2 j sin θij i

458


hi Fi (Qi Qi ) =

i

i

hi F (Qi Qi );

hi F (Qi Qi ) =

h1 F (Qi Qi ). (22)

i

Therefore, from (20) we infer that we have the equality in (21). This is possible only if we have the equality in all inequalities (19). But then all the faces Qi and Qi are translates of one another, i.e., so are the polyhedra P and P . By the construction of P , it means that the speeds pi of motion of the faces are such as though the polyhedron P was a rigid body, i.e., if the face areas are stationary, then the polyhedron is infinitesimally rigid. 11.3.5 All of the above can be carried over to continuous flexes of convex surfaces. If a surface undergoes an infinitesimal flex, then each of its infinitely small pieces moves as a rigid body and we can assign to it a definite angular velocity vector w. Thus we define a “rotation field” w(u, v) of the surface x(u, v). Since two neighboring infinitely small pieces of the surface, like infinitely small faces, behave as though they rotate around a common edge, it follows that dw lies in the tangent plane. This means that the surface generated by the end of the mobile vector w(u, v), the so-called “rotation diagram,” has a singularity at every point, or that its tangent plane is parallel to the tangent plane of the given surface. Now, consider the function p(n) = nw, where n is the normal to x and so to w. The value of p(n) is the distance from the origin to the tangent plane of the rotation diagram with normal n. It turns out that the function p(n) satisfies a certain equation which is equivalent to the following condition: if the support function h(n) changes with speed p(n), then the Gaussian curvature is stationary at every point. The fact that, under this condition, the deformation of a closed convex surface reduces to a translation is proved by means of the same formulas from the theory of mixed volumes, only rewritten for regular surfaces rather than polyhedra.10 Here we encounter every edge lij twice: in the face Qi and in the face Qj . Therefore, the same sum can be rewritten as hi hj lij hi hi cos θij − l 2 sin θij 2 sin θij ij hj hi lij hi hi cos θij − l 2 sin θij 2 sin θij ij 1 hj − hi cos θij = hi lij = hi F (Qi Qi ). 2 sin θ ij i j i

=

10

See [A3, Part IV, p. 242], [Bl1, pp. 607–610], and [W2]. Another form of the same arguments is given in [Bl, §93].


459

11.3.6 The above derivation of the infinitesimal rigidity theorem for polyhedra with stationary face directions and face areas translates word for word to spaces of arbitrary dimension n. However, for n > 3, instead of the Minkowski inequality of Section 8.3, we must use another inequality: V (P1 P0 . . . P0 )2 ≥ V (P1 P1 P0 . . . P0 )V (P0 . . . P0 ). This inequality is deduced from Brunn’s inequality. By Brunn’s inequality, f (t) = n V ((1 − t)P0 + tP1 is a (downward) convex function of t. Therefore, f (0) ≤ 0. Calculating f (0), we arrive at the above quadratic Minkowski inequality. Furthermore, we could prove that if polyhedra P1 and P0 have parallel faces of the same structure then the equality sign occurs in it if and only if P1 and P0 are homothetic.

11.4 Generalizations 11.4.1 In contrast with the case of unbounded polyhedra, for closed polyhedra in spaces of arbitrary dimension, it is impossible to obtain a theorem as general as Theorem 1 of Section 11.2. This is seen from the example showing that such a generalization of Theorem 1 of Section 6.3 is impossible (see Subsection 6.3.2). Therefore, a similar infinitesimal rigidity theorem will hold in n-dimensional space with n > 3 only for some special functions of faces whose stationarity is stipulated as a rigidity condition. For example, we can prove that a closed convex polyhedron with stationary face directions and face areas (i.e., (n − 1)-dimensional areas) is infinitesimally rigid in spaces of arbitrary dimension. The proof is based on the theory of mixed volumes and is quite similar to that of the theorem presented in Subsection 11.3.4 for three-dimensional space.11 However, the question of the infinitesimal rigidity of an n-dimensional closed convex polyhedron under the condition of stationarity of the (n − 2)dimensional surface areas of the boundary of its (n − 1)-dimensional faces (for n = 3 these areas are the perimeters) remains open. No analogous infinitesimal rigidity theorems are known for polyhedra in hyperbolic or spherical space, although the question of finding such theorems 11

In a similar way, we could obtain a more general theorem by replacing the face areas by the mixed areas Fi (P, . . . P P1 . . . Pm ) of the faces of the deformed polyhedron P and polyhedra P1 , . . . , Pm similar to P , i.e., having the same structure as P and faces parallel to those of P (as, for example, a cube and rectangular parallelepipeds; the case m = 0 corresponds to stationarity of areas). The proof is based on the general properties of mixed volumes together with the general inequality between them, and can be found in my article cited at the end of Chapter 8. However, in view of the restriction to “similar” polyhedra P, P1 , . . . , Pm , the result does not seem interesting.

460


is not meaningless, unlike the question of uniqueness raised in Subsection 6.6.4.12 11.4.2 For regular surfaces in Euclidean space, we can state infinitesimal rigidity theorems similar to Theorems 1–3 of Section 11.2. Let f (x, y; n) be a function of a unit vector n and numerical parameters x and y, defined and essentially monotone in the domain x ≥ y, i.e., ∂f ∂x and ∂f exist and are continuous and positive functions in that domain. ∂y The condition imposed on the deformations of a surface is the stationarity of the function f (R1 , R2 ; n) at every point of the surface. Here R1 and R2 are the principal curvature radii and n is the normal at the point in question. The surface itself is regarded as the envelope of its tangent planes and we assume its deformation to be caused by parallel displacements of these planes. In other words, the deformed surface is determined as the envelope of the family of the planes nx = h(t, n; t), is the initial where t is the parameter of the deformation, and v(n) = ∂h(n;0) ∂t speed of the motion of the tangent planes. We now state the results. (1) If a closed surface F undergoes a deformation of the above type and some function f (R1 , R2 ; n), satisfying the above conditions, is stationary at every point of F , then the surface F is infinitesimally rigid. This theorem is proved under the assumption that the functions f , h, and v are piecewise analytic. However, these regularity requirements are hardly inherent to the problem and can be considerably relaxed if the curvature of F is everywhere positive. The surface may be nonconvex in general: it suffice to require that the function h(n; t) be defined for all n and piecewise analytic. The proof of the theorem actually appears in my article [A4]. The regularity requirements are relaxed in Pogorelov’s article cited therein. (2) Let a bounded convex surface F with boundary have the spherical image S in the interior of a hemisphere. Suppose that F is deformed by parallel displacements of its tangent planes; moreover, (a) the tangent planes at the boundary points are stationary and (b) some function f (R1 , R2 ; n), satisfying the above conditions, is stationary at every point of F . Then F is motionless at the initial time, i.e., the speed of the tangent planes equals zero, v(n) = 0. This theorem is valid in spaces of arbitrary dimension and can be proved just like Theorem 3 of Section 11.2 by considering the surface generated by 12

There are essential differences between the elliptic (Riemann) and hyperbolic (Lobachevski˘ı) spaces. For the former, the questions are easier. See [P10, Section 5 of Chapter 6], where an important connection is indicated between the infinitesimal flexes in elliptic and Euclidean space. – V. Zalgaller


the ends of the mobile vectors

1 v(n) n,

where v(n) =

∂h(n;t) ∂t

461

is the initial t=0

speed of the support plane. (If the spherical image fails to belong to the interior of the hemisphere but still lies in the hemisphere, then the theorem is probably valid. In other cases it seems to be wrong.) Both theorems (1) and (2) are equivalent to certain theorems about second-order linear partial differential equations. Indeed, the curvature radii R1 and R2 of the deformed surface are expressed in the familiar way through the “support function” h(n; t)13 and its first- and second-order derivatives with respect to the components of the normal n or, which the same, with respect to the coordinates ξ and η on the unit sphere. Therefore, f (R1 , R2 ; n) = Φ(hξξ , hξη , hηη , . . . , h; n). The stationarity condition reads

∂Φ ∂t

= 0.

(1)

t=0

However, ∂Φ ∂hξξ ∂Φ ∂hξη ∂Φ ∂h ∂Φ = + + ... + ∂t ∂hξξ ∂t ∂hξη ∂t ∂h ∂t

and

∂h ∂t

= v. t=0

Therefore, condition (1) is equivalent to a linear partial differential equation in the function v(n) = v(ξ, η). Owing to the conditions ∂f > 0, ∂R1

∂f > 0, ∂R2

the equation turns out to be of elliptic type. Theorem (2) leads in particular to the uniqueness of the solution to this equation satisfying the given data on the boundary of the domain S. 11.4.3 No generalizations of theorem (1) to closed surfaces in n-dimensional space are known. We must restrict the class of admissible functions to special functions of the principal curvature radii. For example, the following theorem holds in n-dimensional space: A closed convex surface is infinitesimally rigid provided that certain elementary symmetric function of its principal curvature radii are stationary, namely the product of all n−1 curvature radii, the sum of products of n−2 of them, etc. The surface is assumed analytic and the curvature radii, nowhere vanishing. These conditions can be relaxed, but it is known that the theorem 13

See, for instance, [Bl2, § 94].

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cannot be generalized to arbitrary convex surfaces (in which case the curvature functions are defined as set functions on the unit sphere). The proof of this theorem was never published, but it can easily be obtained on the grounds of some general results of the theory of mixed volumes which were established in my article cited at the end of Chapter 8. Along the same lines, we can prove the similar theorem on the infinitesimal rigidity of closed convex surfaces under the assumption of stationarity of the “mixed curvature function” of the surface and other given closed convex surfaces. The same article presents all relevant notions and facts.

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12.1 Supplement To Chapter 3: Yu. A. Volkov. An Estimate for the Deformation of a Convex Surface in Dependence on the Variation of Its Intrinsic Metric1 It is well known that if two closed convex surfaces are isometric, then they are congruent. In the general case, i.e., for surfaces subject to no regularity constraints, this result is due to A. V. Pogorelov [P5]. It implies easily that two closed convex surfaces are nearly congruent with any required accuracy whenever they are nearly isometric with a sufficient accuracy.2 The problem of estimating the deviation between two surfaces in dependence on the deviation between their metrics was raised already by S. CohnVossen in 1936 [C-V]. Here we establish some estimate of this type.3 Our arguments do not rely on A. V. Pogorelov’s result; in particular, they give a new proof of the congruence theorem for closed convex surfaces. The article consists of two parts: § 1 and § 2. §1 § 1.1 In § 1 we establish estimates for deformations of convex caps. Such caps are well known to be uniquely determined from their intrinsic metrics. Irrespective of the fact that some estimates for caps can be easily derived from estimates for closed surfaces, it seems fruitful to start from the case of caps, since the direct approach gives more precise estimates. Also, the direct derivation of estimates for caps acquaints us with all the main ideas of our approach in the simpler situation. Closed surfaces are considered in § 2. Our idea is as follows: given two near isometric caps P 0 and P 1 , we define two “mixed” caps P 01 and P 10 , of which the first gathers the intrinsic metric S 0 of P 0 with the heights4 h1 of P 1 and the second, the intrinsic 1 2 3 4

A translation of the article [Vo6]. Rigorous definitions of near congruence and near isometry are given below in § 1.1 and § 2.1. Precise statements of our result are given in § 1.3 and § 2.3. The heights of a cap are the distances from its vertices to the plane containing its boundary.

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metric S 1 of P 1 with the heights h0 of P 0 . We define some characteristic H on the set of “mixed caps”; for an ordinary cap, it equals the integral mean curvature of the closed surface given by the cap with its projection to the plane of its boundary attached. We establish the following properties of H.5 (1) H always decreases as we pass from an ordinary cap P to each of the “mixed caps” P with the same intrinsic metric, and the amount of decrease admits a lower bound through the maximum of increments of the heights of corresponding (by isometry) points of P and P. (2) The difference between the values of H for an ordinary cap P and each of the “mixed” caps P with the same heights admits an upper bound depending only on the deviation between the intrinsic metrics of P and P . The required estimates for differences between the heights of two ordinary caps P 0 and P 1 with close metrics are then obtained as follows: Let H 0 , H 1 , H 01 , and H 10 be the respective values of H for P 0 , P 1 , P 01 , and P 10 . Clearly, (H 0 − H 01 ) + (H 1 − H 10 ) = (H 0 − H 10 ) + (H 1 − H 01 ). Now, by property (1) each of the differences on the left-hand side is estimated from below through the maximum of differences between the heights of P 0 and P 1 . At the same time, by property (2) the right-hand side admits an upper bound through the deviation between the intrinsic metrics of the caps. This yields an estimate for the differences between heights through the deviation between metrics. Obviously, it suffices to derive the required estimates for polyhedral caps, since the estimates for general caps can then be obtained by passing to the limit. Therefore, we usually define the needed notions only for polyhedral caps, although it is no trouble to give the corresponding definitions in the general case. In the next section (§ 1.2) we give precise definitions and formulate Lemmas 1 and 2 underlying the proof of the main result of the first part of the article (Theorem 1). The theorem itself is proved in § 1.3. The subsequent sections are devoted to proving auxiliary propositions (Lemmas 3–9) helping us to prove Lemmas 1 and 2 (§ 1.7). § 1.2 Given a convex polyhedral cap P , denote by S any development of P and by hi (i = 1, . . . , n), the heights of P , i.e., the distances from its vertices Ai (i = 1, . . . , n) to the plane containing its boundary. Speaking of a development of a cap, we usually assume, not specifying this each time, that the development is triangulated and the vertices of the triangles of the triangulation are among the “genuine” vertices of the development, i.e., among the interior points of nonzero curvature and the boundary points with nonzero swerve of the boundary. 5

To avoid overloading the exposition with details inessential for the understanding of the idea of the proof, we take liberties with the very definition of mixed cap, as well as with the formulation of the properties of H. Rigorous statements are given in § 1.2 and § 1.3.

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A development S (a triangulated development of a convex cap) and a collection of numbers hi ≥ 0 (i = 1, . . . , n) assigned to the vertices of S are said to form a generalized cap P = (S, h) if the following conditions are satisfied: (1) the numbers assigned to the boundary vertices of S are all zero: hi = 0; (2) each triangle of the triangulation of S can be placed above the (horizontal) plane so that the distances from its vertices to this plane be the same as the numbers hi assigned to these vertices. The numbers hi are referred to as the heights of the generalized cap P , and S is referred to as the development of this cap. Generalized caps are the main object of our study; therefore, we usually omit the word “generalized” everywhere except when this is absolutely necessary. Placing a triangle T of the development S of the cap P above the horizontal plane with the preassigned heights of the vertices of T also determines the heights h(A) of all points A of T , the projection of T on the horizontal plane, and the polyhedron composed of the straight line segments projecting all the points A. The latter polyhedron is referred to as the projecting prism of T ; we will often omit the word “projecting,” calling it simply “the prism of T .” Observe that we do not exclude the case in which the projecting prism degenerates into a polygon in a vertical plane. The gluing rules for the triangles in S naturally yield the gluing rules for the projecting prisms of these triangles, as well as the gluing rules for their projections. The development S given by these projections with the corresponding gluing rules is called the projection of the development S. Although some triangles of S could degenerate into segments, the proof of Lemma 9 of [Vo4] shows that, for convex caps, no side of these triangles degenerates into a single point, so that the angles of these triangles are defined correctly, implying that the complete angle is well defined at each vertex of S and hence the curvature of S is also well defined. We define the complete dihedral angle of the generalized cap at an edge of the development S as the sum of the dihedral angles at this edge in all the projecting prisms containing the edge. A generalized cap is said to be convex if its dihedral angles are all no greater than π. We define the H(P ) of a generalized cap P = (S, h) to total curvature be the number i ωi hi + k (π − αk )lk . Here the first sum is taken over all internal vertices Ai of S, hi is the height of the vertex Ai , and ωi is the curvature of the projection S at the point below Ai . The second sum ranges over all edges of S, lk is the length of an edge, and αk is the complete dihedral angle at this edge. We now formulate the main lemmas. Lemma 1. Let P 0 = (S 0 , h0 ) and P 1 = (S 1 , h1 ) be convex polyhedral caps; moreover, P 0 is an ordinary cap, P 1 is a generalized cap, and the development S 0 is isometric to S 1 . Then H(P 0 ) − H(P 1 ) ≥ max{ 0, C(∆h)3 }, where

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∆h = maxi (h1i − h0i ), h1i and h0i are the heights of the vertices of S 1 and S 0 corresponding to one another by isometry, and C is a constant depending only on the diameter of the development S 0 . This lemma will be proved in § 1.7. Observe that we establish a quantitative estimate for the difference H(P 0 ) − H(P 1 ) only in the case when some heights in P 1 are greater than those in P 0 . This is sufficient for our purposes. Nevertheless, it can be shown that H(P 0 ) − H(P 1 ) ≥ C{ max |h1i − h0i | }3 . i

0

1

Two developments S and S are said to be ε-close in intrinsic metrics, or ε-near isometric, if there exists a homeomorphism ϕ from S 0 onto S 1 such that for arbitrary two points A and B of S 0 |ρ0 (A, B) − ρ1 (ϕ(A), ϕ(B))| ≤ ε, where ρ0 and ρ1 are the respective distances in S 0 and S 1 . In this case, ϕ is called a near isometry. Lemma 2. Let P 0 = (S 0 , h0 ) be an ordinary convex cap and let S 1 be a development of another convex cap which is ε-close to S 0 in intrinsic metrics. Then there exists a convex generalized cap P 10 such that (1) the development S 10 of P 10 is isometric to S 1 ; (2) the total curvatures H 0 and H 10 of the caps P 0 and P 10 satisfy the inequality H 0 − H 10 ≤ Cεα ; (3) the heights h0 and h10 of arbitrary points of S 0 and S 10 corresponding to one another under the near isometry ϕ satisfy the inequality h10 ≥ h0 − Cεα . Here α is some absolute constant (≥ 1/4) and the constant C depends on the intrinsic diameters of S 0 and S 1 . The proof of Lemma 2 is given in § 1.7. So far we spoke of estimates for differences between the heights of points of two caps corresponding to one another under a near isometry through the deviation between the intrinsic metrics of the caps. With these estimates available, we can easily evaluate the difference in the spatial distances between the pairs of corresponding points of nearly isometric caps. § 1.3 We now prove the main result of the first part of the article. Theorem 1. If the developments S 0 and S 1 of two convex caps P 0 = (S 0 , h0 ) and P 1 = (S 1 , h1 ) are ε-close in intrinsic metrics, then the heights h0 and h1 of the points corresponding to one another under the near isometry differ at most by Cεβ , where β is some absolute constant (≥ 1/12) and the constant C depends on the diameters of S 0 and S 1 .


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Proof. Let H 0 and H 1 be the total curvatures of P 0 and P 1 . Applying Lemma 2 to the cap P 0 and the development S 1 , construct a cap P 10 such that the total curvature H 10 of P 10 satisfies the inequality H 10 ≥ H 0 − Cεα 10 0 α and the heights h10 satisfy the inequality h10 i of P i ≥ hi − Cε . Similarly, 1 0 01 for the cap P and the development S construct a cap P such that H 01 ≥ H 1 − Cεα Then

1 α and h01 i ≥ hi − Cε .

(H 0 − H 10 ) + (H 1 − H 01 ) ≤ 2Cεα .

(∗).

On the other hand, (H 0 − H 10 ) + (H 1 − H 01 ) = (H 0 − H 01 ) + (H 1 − H 10 ). 1 α 10 0 α The inequalities h01 i ≥ hi − Cε and hi ≥ hi − Cε imply that 0 1 0 α ∆01 h = max{ h01 i − hi } ≥ max(hi − hi ) − Cε , i

10

∆ h=

i

max{ h10 i i

−

h1i

} ≥ max(h0i − h1i ) − Cεα . i

01

Therefore, it suffices to estimate ∆ h and ∆10 h from above. If ∆01 h and ∆10 h are nonpositive, then the required estimate is already at hand. If ∆01 h or ∆10 h is positive (for example, ∆01 h > 0), then by Lemma 1 one of the differences H 0 − H 01 and H 1 − H 10 is nonnegative and the other is not less than C(∆01 h)3 , which implies that6 (H 0 − H 01 ) + (H 1 − H 10 ) ≥ C(∆01 h)3 .

(∗∗)

Comparing (∗) and (∗∗), we obtain (∆01 h)3 ≤ Cεα . So, in all cases max |h1i − h0i | ≤ Cεβ , i

α . β= 3

§ 1.4 We now prove several lemmas about developments. Studying deformations of a given polyhedral angle, it is often convenient to consider, instead of this angle, the spherical polygon in which the angle intersects the (unit) sphere centered at its vertex. The lengths of the sides of this spherical polygon are equal to the planar angles of the polyhedral angle, and the angles of the spherical polygon are equal to the dihedral angles of the given polyhedral angle. 6

The letter C in different places denotes generally different constants depending only on the diameters of S 0 and S 1 .

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In perfect analogy, studying the structure of a generalized cap in a neighborhood of its vertex A, it is useful to consider a certain spherical development, called the spherical section of the cap near A, which we now define rigorously. In the prisms of the given cap, take all polyhedral angles contiguous to A and intersect each of them with the unit sphere centered at A. The rules for gluing together the lateral faces of these prisms determine some rules for gluing together the resulting spherical triangles into some spherical development. It is this spherical development that we call the spherical section of the cap near A. Clearly, such a development consists of finitely many triangles in each of which one (basis) side lies on the boundary of the development while the two other (lateral) sides are incident to the only internal vertex of the development. The basis sides correspond to the angles of those triangles of the cap which are contiguous to A and the lateral sides correspond to the angles of the lateral faces of the projecting prisms. In particular, the length of the boundary of the spherical section is equal to the complete angle of the cap at A. The boundary vertices of the spherical section correspond to those edges of the cap which are incident to A, and the angles at these vertices are equal to the dihedral angles of the cap at the corresponding edges. In particular, each angle on the boundary of the spherical section of a convex cap is at most π. Finally, the internal vertex of the spherical section corresponds to the straight line segment projecting A, and the curvature of the section at this vertex is equal to the curvature of the point below A in the projection S of the development S of our cap. If the spherical section has zero swerve at some boundary vertex, then the dihedral angle of the cap at the edge corresponding to this vertex equals π, and the triangles of the cap which adjoin this edge in gluing together their projecting prisms go in one plane. Therefore, studying the arcs of zero swerve on the boundary of the spherical section amounts to studying the “flat angles” of the cap. We prove the following lemma about the possible lengths of such arcs. Lemma 3. If the length of the boundary of the spherical section is less than 2π and the curvature at the vertex of the section is nonpositive, then no boundary segment of length greater than or equal to π may have zero swerve. Proof. Assume, contrary to the assertion, that such a segment exists. From the section, cut away a sector whose vertex is the vertex of the section and whose angle is such that the angle of the remaining sector equals 2π. We can do this so that the boundary will still include a geodesic segment of length at least π (since the angle at which a boundary segment of length π is seen from the vertex is exactly π). Unroll the remaining part of the section on


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the sphere, carrying out all needed gluings. The boundary of the resultant spherical polygon S has an arc of a great circle whose length is at least π; therefore, the total length l of the boundary of S is at least 2π. It remains to observe that the length of the boundary of the section is no less than l, since it can be calculated by replacing the difference (involved in the boundary of S) between the lateral sides of the cut-off sector with the length (obviously greater than this difference) of the remaining part of the boundary of the sector. However, by hypothesis the length of the boundary of the sector is less than 2π. The contradiction obtained proves our claim. Later we need the following result about the curvature at the internal vertex of the spherical section. Lemma 4. For constant lengths of all boundary edges and a positive swerve at every boundary vertex, the curvature ω at the internal vertex O of the spherical section S is a monotone increasing function of the lengths ρk of the internal edges OBk of S. Proof. The claim is proved by direct calculation of the derivatives calculation gives the following result:

∂ω ∂ρk .

This

∂ω sin(α + β) = > 0. ∂ρk sin α sin β sin ρk Here 0 ≤ γk = α + β ≤ π is the complete angle of the section at the boundary vertex Bk and α and β are the parts into which this angle is divided by the arc OBk . Lemma 5. Let S be the spherical section with curvature ω (≤ 0) at the internal vertex O. Let p (≤ 2π) be the length of the boundary of S; γk (≤ π), the angle of the section at the boundary vertex Bk ; and αk and βk (αk + βk = γk ), the parts into which the angle γk is divided by the arc OBk . Then √ sin γk ≥ C −ω, sin αk sin βk sin OBk k

with C an absolute constant. Proof. First of all, we interpret the left-hand side of the inequality geometrically. Make a slit in S along the arc of a great circle joining O to the most distant boundary point and unroll S on the unit sphere. For lucidity, assume that O is the south pole of the sphere. Let K be the cone whose vertex is the center of the sphere and whose direction set is our unrolled section. Consider the i.e., the cone formed by the outward normals to the support “dual” cone K, planes of K. Some clarifications are required for the construction of that part

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of K which corresponds to the edges of the slit. We omit them as evident. Anyway, the rotation around the axis of the sphere, superposing the “edges” of the cone K, must also superpose the “edges” of the cone K. ∗ Consider the two polygonal lines L and L in which K intersects the the sphere and its tangent plane at the north pole. It is natural to call L ∗ boundary of the spherical image of K and L , the boundary of the normal image of K. Direct computations show that the left-hand side of the inequality of the lemma is equal to the length of L∗ . We derive the required estimate for the length of L∗ differently, depending on whether S lies entirely in the south hemisphere or enters also the north hemisphere. In the first case we connect the north pole of the sphere with the endpoints and L∗ on the sphere and on the plane, and glue together the appearing of L on arcs (segments). We then obtain some development S with boundary L ∗ the sphere and some development S with boundary L on the plane. By the isoperimetric inequality, |L∗ |2 ≥ 4π|S ∗ |, where |L∗ | is the length of L∗ and |S ∗ | is the area of S ∗ . Furthermore, it is obvious that |S ∗ | ≥ |S|, is the area of S. Finally, applying the Gauss–Bonnet Theorem to where |S| equals the length p of the boundary S and observing that the swerve of L of S, we obtain = 2π, p + ω + |S|

= 2π − p − ω ≥ −ω. |S|

Altogether this yields |L∗ |2 ≥ 4π(−ω). In the case when S enters into the north hemisphere, we as before connect with the north pole and consider the closed polygthe endpoints of L∗ and L ∗ onal lines L1 and L1 on the plane and on the sphere which are composed of supplemented with the arcs of those sectors with vertex the north L∗ and L pole whose boundaries are the segments (arcs) joining the endpoints of L∗ to the north pole. and L Denote by ρ the radius of the appearing sector on the sphere. Then the radius of the sector on the plane is clearly tan ρ and the angles of both sectors are −ω. From the isoperimetric inequality applied to the domain S1∗ bounded ∗ ∗ by the polygonal line L1 , we obtain |L1 | + tan ρ · (−ω) ≥ 4π|S1∗ |, where |S1∗ | 1 | |S ∗ is the area of S1∗ . Next, |S1∗ | ≥ cos 3 ρ , since the whole domain S1 lies on the sphere outside the circle of radius ρ around the north pole. By the Gauss–Bonnet Theorem applied to S1 (the swerve of the polygonal equals p and the swerve of the arc of the sector, −ω cos ρ) line L p + ω cos ρ + |S1 | = 2π,

|S1 | ≥ (−ω) cos ρ.

Combining all inequalities and observing that the convex polygonal line L∗ surrounds the closing arc of the sector, implying that |L∗ | ≥ (−ω) tan ρ, we obtain (−ω) 2|L∗ | ≥ |L∗ | + (−ω) tan ρ ≥ 4π 2 , cos ρ


or

|L∗ | ≥

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π(−ω).

The proof is over. Later we will need a special continuous deformation of one of the two boundary-convex developments of positive curvature ε-close in intrinsic metrics to the other. Its special character consists in the fact that the deformation can be carried out in a finite number of steps each given by cutting-off (or gluing) of a digon or a “strip,” another standard figure to be described below. We consider one of the possible implementations of such a deformation and estimate the total width of the cut-off or glued digons and “strips” through ε. We assume ε to be small in comparison with the intrinsic diameters d0 and d1 of the deformed developments S 0 and S 1 , and also introduce some λ assumed “small” in comparison with d0 and d1 and “large” in comparison with ε. We partition each of the developments S 0 and S 1 into “strips” of width λ, starting from their boundaries: the nth strip is the set of points whose distance from the boundary is between (n − 1)λ and nλ. Then we replace the boundaries of each strip by inscribed polygonal lines whose vertices correspond to one another under the near isometry and whose segments have lengths of order λ, and partition the arising strips into triangles with sides of order λ. Here it is possible that we cannot manage without the introduction of “a few” highly stretched triangles. Nevertheless, an easy calculation shows that the latter do not violate the order of our estimates. Now, the required deformation proceeds as follows: using the standard trick of cutting-off digons, we “drive” the curvature from all the triangles splitting S 0 and S 1 . The resulting developments S0 and S1 consist of “wide flat strips.” The deformation of the strips of one of these developments to the strips of the other is a nondifficult, though laborious, essentially planar problem. We will not dwell upon its solution. The total width of the cut-off digons is well known to be of order λ(Ω 0 + 1 Ω ), where Ω 0 and Ω 1 are the total curvatures of S 0 and S 1 . In other words, this width has the upper bound Cλ with some absolute constant C. On the other hand, the sides of the triangles of S0 and S1 , having vertices close to points corresponding to one another under the near isometry, differ by a quantity of order ε, whereas these sides themselves are of order λ. Therefore, the angles of these triangles differ by a quantity of order ε/λ. The number of triangles in one strip is of order d/λ (d = max(d0 , d1 )); therefore, if unrolled on the plane and properly situated, the boundaries of the corresponding strips of S0 and S1 will differ by a quantity of order εd2 /λ2 . Observing that the number of strips is in turn of order d/λ, we easily see that, in the deformation of S0 to S1 , the total width of the “narrow” strips glued and cut away near

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the boundaries of the strips of the partitions of our developments is of order εd3 /λ3 . If we choose λ ∼ ε1/4 d2/3 , then the estimate for the total width of digons and strips in our construction has the best order, namely Cε1/4 (C ∼ d3/4 ). 0 Consider the part S of S 0 that survives all cuttings. Making all gluings 0 to this part, we obtain an isometry χ that takes S together with everything 0 glued to it onto the development S 1 . Take an arbitrary point A in S . Then in S 1 we have the points χ(A) and ϕ(A) corresponding to A under the isometry χ and the near isometry ϕ. Let us show that the distance between these points is at most Cε1/4 . Indeed, let T 0 be a triangle containing A in the triangulation of S 0 and let T 1 be the corresponding triangle of the triangulation of S 1 . Preserving the order of our estimates, we can assume that ϕ(A) and χ(A) belong to T 1 (anyway, we can replace them with close points). Cut all digons away from T 1 and view ϕ(A) and χ(A) as points of the arising triangle T1 (if ϕ(A) disappears after cut, then replace it with a close point, i.e., one at a distance no greater than Cε1/4 . The distances from these points to any vertex of T1 are close to each other by smallness of the cut and the fact that ϕ and χ are near isometries. Since T1 is a flat triangle, the fact that two points in it have close disctances to its vertices implies that the points are close themselves. Whence we infer that the points ϕ(A) and χ(A) in the triangle T 1 are close to each other. § 1.5 In this section we study the deformations of a convex generalized cap to the extent we need. We begin with the deformations preserving both the intrinsic metric of the development of the cap and the convexity of the cap. First we prove Lemma 7, establishing the existence of such deformations, and next justify some qualitative and quantitative results (properties 1–9) on the behavior of the curvature of the projection of the cap in these deformations. We make several remarks on the structure of convex generalized caps. Recall that the development S of a convex generalized cap P is assumed triangulated; moreover, the vertices of triangles are among the genuine vertices of the development. Each of the triangles of the development S may be placed above a horizontal plane so that the heights of its vertices above this plane are exactly their heights in the generalized cap P . We say that two triangles T and T ∗ belong to the same face of P if we can find a chain of triangles, starting from T and finishing at T ∗ , such that every two adjacent triangles in the chain are glued together along an edge in the development S and the complete angle of P at this edge equals π. For the triangles belonging to one face, we preserve only the gluings along the edges with complete angle π, so transforming each face into some development. The next lemma is a direct corollary to Lemma 3 in § 1.4.


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Lemma 6. Let F be a face of a convex cap P ; A, a boundary vertex of F ; and A, the projection of A. If the curvature of the lower base at A is nonpositive, then the complete angle of F at A is less than π. Lemma 7. The following hold for every convex polyhedral cap P : 1. If the curvature of the lower base of P at a point below a vertex Ai of the cap is negative, then we can deform P continuously, preserving its intrinsic metric and convexity, so that the height of Ai decreases and the heights of the other (genuine) vertices remain the same. 2. If the lower base of P has points at which the curvature is positive, then we can deform P continuously, preserving its intrinsic metric and convexity, so that the heights increase for those vertices of P that project to points with positive curvature and the heights remain the same for the other vertices. Proof. Validate the first claim of the lemma. If T is a nondegenerate triangular prism, then we can obviously deform it so that the height of one (arbitrary) vertex decreases and the upper base of T rotates as a rigid body around the straight line through the other two vertices. Likewise, in a degenerate prism we may decrease the heights of the vertices that lie on the boundary (but not in the interior) of the polygon representing our degenerate prism. Now, let P be a convex polyhedral cap and let A be a vertex of P such that the curvature of the lower base is negative at the point below A. Assume all faces of P contiguous to A triangulated somehow. Take all projecting prisms of those triangles of the triangulation which touch at A. By the above method we can deform each of these prisms, decreasing the height of A; A, being a genuine vertex of the development, does not belong to a degenerate prism. We now demonstrate that the triangulation of faces can be chosen so that this decrease of the height of A does not violate the convexity of our cap. Let F be one of the faces contiguous to A and let B1 and B2 be the boundary vertices of F adjacent to A. Draw a shortest arc between B1 and B2 in F . By Lemma 6, the angle of F at A is less than π and the shortest arc B1 B2 does not go along the edges B1 A and AB2 . With the boundary of F this shortest arc may have common vertices B3 , . . . , Bk other than B1 and B2 ; at these points the shortest arc is convex towards the interior of the face F or, in other words, towards the vertex A. Therefore, the part of the boundary of F bounded by the edges B1 A and AB2 and the shortest arc B1 B2 is starlike with respect to A and we can split it into triangles by the segments ABi (i = 1, . . . , k). Having triangulated in this manner each of the faces containing A, if we now lower A, then we easily see that the faces fold along the segments of the shortest arc B1 B2 and the edges ABi ; moreover, they fold so that the convexity is preserved. Certainly, the preservation of convexity near the boundary edges of faces which abut on A is ensured by the smallness of the deformation.

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We turn to the second claim of the lemma. A minor generalization of the preceding arguments allows us to prove the possibility of simultaneously decreasing by the same amount the heights of all internal vertices whose projections have nonpositive curvatures and all boundary vertices. First of all, observe that the difficulties related to the lowering of the vertices of degenerate prisms do not arise here, since in this case the vertices are all lowered by the same amount. As far as the preservation of convexity is concerned, we now achieve it as follows: On the boundary of each face, choose polygonal lines of the shape B1 A1 A2 . . . Ai B2 , with all the vertices lowering except for the first one and the last. Connect the vertices B1 and B2 in the face in question by a shortest arc γ : B1 B3 B4 . . . Bk B2 , where B3 , . . . , Bk are the common vertices of the shortest arc and the boundary of the face. All the vertices Bi (i = 1, . . . , k) do not lower: B1 and B2 by their choice while B3 , . . . , Bk due to the fact that the angles of the face at these vertices are at least π and so the curvatures of their projections are positive by Lemma 3. Triangulate the part of the face between the polygonal line B1 A1 . . . Ai B2 and the shortest arc B1 B3 . . . Bk B2 as follows: connect every two successive vertices Bi and Bi+1 of our shortest arc with that vertex Ak for which the area of the triangle Bi Aki Bi+1 is greatest (i.e., with a vertex most distant from the straight line Bi Bi+1 and lying on the other side of this straight line as compared to the points Bi−1 and Bi+2 ); then connect each of the vertices Bi by rectilinear segments with those vertices Ak that lie between the vertices Aki−1 and Aki of the earlier constructed triangles Bi−1 Aki−1 Bi and Bi Aki Bi+1 . It is easy to check that this triangulation provides the preservation of convexity. To prove the second claim of the lemma, it now suffices to lower by the same amount all boundary vertices and the vertices with projections of nonpositive curvature and next raise by the same amount all vertices without exceptions. We now turn to clarifying the dependence of the curvature of the lower base of the convex cap on its heights (under the condition that the intrinsic metric of the cap is constant). First of all, note that the lower base and in consequence its curvatures are uniquely determined from the heights of the cap only if some triangulation of the cap is fixed. However, changing the triangulation inside any face of the cap keeps the lower base invariant. Thus, we assume that some triangulation of the development S of the cap P is fixed. Let Ai be a vertex of S and let Ak (k = 1, . . . , n) be adjacent vertices; hi and hk (k = 1, . . . , n), the heights of these vertices; and ωi and ωk (k = 1, . . . , n), the curvatures of the projection S of S at the points below Ai and Ak (k = 1, . . . , n). ∂ωi i We now formulate the needed properties of the derivatives ∂ω ∂hi and ∂hk : ∂ωk ∂ωi 1. ∂hk = ∂hi .


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2. Let lik be the length of the projection of the edge Ai Ak on S; γik , the dihedral angle of the cap at the edge Ai Ak ; αik and βik , the parts into which γik is divided by the vertical plane containing the edge Ai Ak ; and ρik , the angle between the edge Ai Ak and a vertical line. Then lik

∂ωi sin γik = ≥ 0. ∂hk sin αik sin βik sin ρik

n ∂ωi i 3. ∂ω k=1 ∂hk . ∂hi = − 4. If ωi = 0, i.e., the triangles of S are factually glued together in some plane Q around the projection of the vertex Ai , and if the numbers ξi and ξk (k = 1, . . . , n) are proportional to the signed distances of the projections of Ai and Ak (k = 1, . . . , n) from some straight line in Q, then ∂ωi ∂ωi ξi + ξk = 0. ∂hi ∂hk n

k=1

∂ωi (see the proof of Lemma 4) Prove 1 and 2. A direct computation of ∂ρ ik yields sin γik ∂ωi = . ∂ρik sin αik sin βik sin ρik

Here ρik (in Lemma 4, it was OBk ) is the distance from the internal vertex Oi of the spherical section near Ai to the boundary vertex Bk of the section corresponding to the edge Ai Ak , and αik and βik are the angles between Oi Bk and Bk−1 Bk and between Oi Bk and Bk Bk+1 . In other words, αik and βik are the angles between the faces of the cap touching at the edge Ai Ak and the vertical plane projecting this edge. Next, it is obvious that cos ρik =

hi − hk . |Ai Ak |

Hence, sin ρik

∂ρik 1 = ∂hk |Ai Ak |

and

∂ρik 1 = . ∂hk lik

Thus, ∂ωi ∂ωi ∂ρik sin γik 1 = = · . ∂hk ∂ρik ∂hk sin αik sin βik sin ρik lik Whence it is obvious that ∂ωi ∂ωk = . ∂hk ∂hi Property 3 expresses the fact that the curvatures of the lower base remain the same when all heights of the cap are changed by the same amount. Similarly, property 4 expresses the fact that the curvature ωi is constant when the part of the cap consisting of the triangles contiguous to Ai is rotated as a rigid body around a straight line which lies in the plane of the lower base.

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Grounding on properties 1–4, we proceed with studying the simultaneous ∂ωi linear equations with the matrix ∂hk . 4. The maximum principle. Consider a connected collection of vertices Ak (k = 1, . . . , m+n) of the cap. By a connected collection we mean one in which we can pass from an arbitrary vertex to any other vertex by moving only along genuine edges of the cap with endpoints belonging to the collection. Suppose that the vertices of the collection are divided into two groups: the “internal” vertices Ak (k = 1, . . . , n) and the “boundary” vertices Ak (k = n + 1, . . . , n + m); moreover, each vertex of the cap which is joined to one of the “internal” vertices by a genuine edge belongs to the collection. Finally, assume that some number ξk (k = 1, . . . , n + m) is assigned to each vertex in the collection and for all i = 1, . . . , n n+m k=1

∂ωi ξk = 0. ∂hk

Then either maxk ξk = mink ξk (k = 1, . . . , m + n) or maxk ξk and mink ξk are attained only at “boundary” points. Proof. Assume for example that ξi = maxk ξk for some internal vertex Ai . By property 3, we have ∂ωi ∂ωi ξk = (ξk − ξi ). ∂hk ∂hk k=i

Since ξk − ξi ≤ 0 while, by property 2, Ai by genuine edges, it follows that

∂ωi ∂hk

> 0 for all vertices Ak joined to

∂ωi (ξk − ξi ) = 0 ∂hk k=i

whenever at least one of the differences ξk − ξi differs from zero. This means that maxk ξk is also attained at all vertices Ak adjacent to Ai . Continuing likewise, we prove that all the ξk are the same, i.e., maxk ξk = mink ξk . As direct corollaries to the maximum principle, we derive the following properties: ∂ωi 5. The matrix ∂h , i, k = 1, . . . , n, is nonsingular. k Indeed, nonsingularity of this matrix is immediate from the fact that the homogeneous system n ∂ωi ξk = 0 hk k=1


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has only the zero solution, which follows in turn from the maximum principle applied to the collection of the numbers ξk , where ξk with k = 1, . . . , n constitute a solution of our system andξk =0 for k = n + 1, . . . , n + m. ∂ωi 6. Let Γik be the inverse matrix of ∂h , i, k = 1, . . . , n, and let Γik = 0 k for i or k greater than n. Then (a) Γik = Γki ; (b) mink Γik = Γii < Γik < 0 for k = i, k = 1, . . . , n. Thesymmetry of the matrix Γik follows from the symmetry of the ∂ωi matrix ∂hk . To prove the inequalities Γii < Γik < 0 for a fixed i and k = 1, . . . , i − 1, i + 1, n, we apply the maximum principle to the collection of the numbers ξk = Γik , considering Ai as a boundary vertex. Since on the “boundary” we have ξk = Γik = 0 for k > 0 and ξi = Γii for k = i; therefore, 0 and Γii are exactly maxk Γik and mink Γik . By the definition of Γik , we have 1=

∂ωi k

hk

Γki =

∂ωi (Γki − Γii ). ∂hk k=i

∂ωi Since ∂h > 0 and all the numbers Γki − Γii have the same sign, all these k numbers are positive and Γki > Γii = mink Γik . Therefore, maxk Γik = 0 > Γik , as required. 7. Suppose that the curvature of the projection S of the cap P vanishes everywhere except possibly the point A0 below the vertex A0 . Let Ai be another vertex of the cap; Ai , the projection of Ai ; li0 , the length of the shortest arc A0 Ai in S; and ri0 , the length of the shortest arc A0 Ci which is the prolongation of A0 Ai beyond Ai until the boundary of S. Then

Γi0 − Γ00 −Γ00 ≥ . li0 ri0

(∗)

Before proving this inequality, we give its geometric interpretation. Make a slit in the development S along some shortest arc joining A0 (the only point of possibly nonzero curvature) to the boundary of S and then unroll it on a plane. Considering this plane to be the xy-plane of some Cartesian coordinate system, take the point Bk below Ak with the z-coordinate zk = Γk0 (recall that all Γk0 < 0). Also, construct the cone K with vertex B0 whose direction set is the boundary of S. The left- and right-hand sides of (∗) are the tangents of the slopes that the segment B0 Bi and the generatrix B0 Ci of K lying in the same vertical plane as B0 Bi make with the xy-plane. Therefore, (∗) is equivalent to the fact that Bi lies not lower than the cone K. We now justify this fact. Let Q be an arbitrary support plane of K passing through the generatrix B0 Ci , and let l be the straight line in which Q intersects the xy-plane.

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Assume that the edges of the slit made in S form equal angles with the straight line l (this can be achieved obviously by changing the place of the slit). Denote by ξk the z-coordinates of the intersection points of the plane Q with the vertical lines passing through Ak (k = 1, . . . , n + m). To prove that Bi lies no lower than K, it suffices to check that Bi lies no lower than the plane Q or, in other words, that ξi ≤ Γi0 . We prove this by applying the maximum principle to the numbers ξk −Γk0 . First, we must suitably divide the vertices of the development into “internal” and “boundary” vertices. Choose some sector U in S according to the following rule: If the complete angle at A0 is less than or equal to π, then choose S as a whole. If the complete angle is greater than π, then choose a sector of opening π in S which includes the shortest arc A0 Ci and is bounded by the shortest arcs from A0 to the boundary of S which go, when the sector is unrolled on a plane, onto a straight line l0 parallel to l (recall that l is a support line to the boundary of S at Ci ). Now, call “internal” those internal vertices of the cap whose projections occur inside the sector. Call “boundary” those remaining vertices of the cap each of which is joined to at least one “internal” vertex by a genuine edge. Let us demonstrate that ξk ≤ Γk0 for every “boundary” vertex. An arbitrary “boundary” vertex may be either a vertex on the common boundary of the cap and the sector U or a vertex inside the cap which is joined to at least one “internal” vertex by a genuine edge but has the projection lying beyond U . In the first case ξk ≤ 0, since the whole sector except Ci lies above Q. Once Γkk = 0, we have ξk ≤ Γk0 . In the second case the projection of our boundary vertex lies in that halfplane determined by the straight line l0 which contains no points of U . For such points we have ξk ≥ Γ00 (ξk = Γ00 below A0 and generally below l0 ). At the same time, we always have Γ00 ≤ Γk0 ; whence ξk ≤ Gk0 . Now, observe that for all “internal” vertices ∂ωi ∂ωi Γk0 = ξk = 0 ∂hk ∂hk k

k

(the last equality holds by property 4). Therefore, ∂ωi (Γk0 − ξk ) = 0. ∂hk k

As was shown above, Γk0 −ξk ≥ 0 on the “boundary.” Applying the maximum principle, we conclude that Γk0 ≥ ξk everywhere. In particular, we obtain Γi0 ≥ ξi , which was to be proved. 8. From properties 8 and 2 and Lemma 5 we easily derive the following estimate for Γ00 :


d −Γ00 ≤ C √ −ω0

479

(ω0 < 0).

Indeed, by the definition of Γk0 and property 3 we have 1=

∂ω0 k

Using the estimate

Γk0 −Γ00 lk0

∂hk ≥

Γk0 =

∂ω0 k=0

∂hk

(Γk0 − Γ00 ).

−Γ00 rk0 ,

we obtain 1 ∂ω0 1 = −Γ00 lk0 . rk0 ∂hk k=0

Clearly, rk0 ≤ d, where d is the diameter of the development S. Furthermore, by property 2 ∂ω0 sin γ0k lk0 = . ∂hk sin α0k sin β0k sin ρ0k k=0

k=0

√ By Lemma 5, the last quantity admits the estimate by C −ω0 through the curvature ω0 at A0 , with C an absolute constant. Now, we obtain the following estimate for Γ00 : d −Γ00 ≤ C √ . −ω0 We proceed with considering the deformations of convex polyhedral caps P for which the curvatures of the projection S are everywhere nonpositive. Lemma 6 readily implies that all faces of such a cap are (planar) convex polygons. Further, it is more convenient to deal with the deformations of the inverted cap P ∗ rather than with those of P . The cap P ∗ is defined as follows: place each triangle T of the development S of P on a horizontal plane and above T take a triangle whose vertices are above the vertices of T at the same heights as the heights ascribed to the vertices of T in the cap P . To put it briefly, the cap P ∗ is the cap for which the triangles of the the original cap serve as projections (on the horizontal plane) and which has the same heights as P . Obviously, not only P uniquely determines P ∗ but also, conversely, P is uniquely reconstructed from P ∗ ; moreover, given some P ∗ , the reconstruction of P is possible if and only if the constituent triangles of the development of P ∗ are sloped with respect to the horizontal plane at angles at most π/4. The caps P and P ∗ are convex or nonconvex simultaneously (this follows immediately from Liberman’s theorem claiming the convexity of Cartesian coordinates of a shortest arc on a convex surface as functions of the arclength, but of course can be proved rather elementarily). Finally, to faces of P correspond faces of P ∗ . This implies that if the projection S of the development of P has nonpositive curvatures, then, together with the faces of P , the faces of P ∗ are also planar convex polygons.

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Now, suppose that we must deform some convex cap P with projection S of nonpositive curvature (preserving the convexity and the metric of S) so that the heights of P will receive (small) increments given in advance. The above implies that we can first construct the corresponding deformation of P ∗ (which is obvious) and afterwards return to P . Later we will also need a deformation of a convex cap P with projection S of nonpositive curvature, preserving the convexity and the metric of S, increasing the height of one vertex A0 by a given amount, and not changing the curvatures ωi of S below all other vertices Ai . We now prove that such a deformation exists. By what was proved above, the curvatures ωi of the projection of the deformed cap are well defined for all hi sufficiently close to the heights h0i of the original cap. Now, we choose the heights hi so as to satisfy the equations ∂ωi ωi (hk ) = ωi (h0k ). This is possible, because the Jacobian det ∂h of this k system differs from zero by property 6. We now turn to considering the deformations of the cap P that result from continuous variations of the development S. As before, passage to the inverted cap P ∗ enables us to easily construct a convex cap with the deformed development. Now raising the vertices below which the curvature of the projection happens to be positive, we can make the curvature of the projection to be nonpositive everywhere (a rigorous proof is easy with the help of Lemma 7). We are in a position to prove the following Lemma 8. Let P be a convex polyhedral cap whose projection S has nonpositive curvature below some vertex A0 and has zero curvature below all other vertices. Then (1) If there are nondegenerate prisms containing A0 , then we can continuously deform P so that its convexity and the intrinsic metric are preserved, the height of A0 increases, and the curvatures of S below all vertices but A0 remain zero. (2) In this deformation, −dω0 √ ≥ Cdh0 −ω0 with some constant C depending only on the diameter of the development S of P . The existence of the required deformation has been already established. Now, we have the equality Γ0k dωk = Γ0 dω0 , dh0 = since dωk = 0 for all k = 0. Whence by property 9


481

Cd dh0 ≤ √ (−dω0 ). −ω0 § 1.6 Here we consider the variation of the total curvature of a cap during its deformation. We start with deriving an expression for the total differential of the total curvature of a polyhedral cap. Lemma 9. The following equality holds: ωi dhi + (π − αk )dlk , dH = i

k

where H is the total curvature of the cap P ; hi (i = 1, . . . , n) are the heights of the vertices Ai of P ; ωi are the curvatures of the projection S at the points below Ai ; αk are the complete angles at boundary edges; and lk are the lengths of the boundary edges of P . Proof. Straightforward differentiation of H yields (ωi dhi + hi dωi ) + [(π − αk )dlk − lk dαk ]. dH = i

k

Therefore, to prove the required formula, it suffices to check that hi dωi − lk dαk = 0 i

k

or, using the complete angle θi = 2π − ωi instead of ωi , that hi dθi + lk dαk = 0. i

(∗)

k

Clearly, it suffices to verify (∗) separately for each prism of P since the required result for the whole cap is then obtained by taking the sum. In turn, each prism splits into tetrahedra. Therefore, it suffices to check that (∗) holds for every deformation of an arbitrary tetrahedron. In view of linearity of (∗) in dl and dh, it suffices to verify (∗) for particular types of deformations of a tetrahedron such that any deformation can be obtained as a sum of these deformations. We take these particular deformations to be ones in each of which a single vertex of the tetrahedron moves along some edge abutting on it. Let in a tetrahedron ABCD the vertex A moves along the edge AD. Introduce the following notations for the sides and planar angles of the face ABC: BC = a, AC = b, AB = c, ∠ABC = β, ∠ACB = γ and let θa , θb , and θc stand for the dihedral angles at a, b, and c.

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If ωa is the angular speed of rotation of the plane ABC around a, then dθa = ωa dt. The variations of θb and θc are affected only by the components −ωa cos γ and −ωb cos β of the angular velocity along the edges b and c: dθb = −ωa cos γdt and dθc = −ωa cos βdt. Therefore, adθa + bdθb + cdθc = ωa dt(a − b cos γ − c cos β) = 0, which completes the proof. Let the development S of a convex cap P with projection S of nonpositive curvature be deformed continuously so that in result some digon (with a single internal vertex) of width δ (the width is the distance from this internal vertex to the boundary of the digon) is cut away from S. Assume further that this causes P to deform as described at the end of § 1.5. Let us estimate from above the decrease of the total curvature H of P in this deformation. The deformation of P can be divided into three steps: the first is the cutting-off of a digon, the second is the “leveling” of the edges of the cut and recovering the convexity, the third is the raising of the vertices below which the curvature of the projection is positive. At the third step, the total curvature H can only increase, since ∆H ≈ i ωi ∆hi (Lemma 9) and the summands on the right-hand side are all nonnegative. At the second step, granted that the different edges of the cut have different heights, we consider naturally generalizing H and defined as follows: the characteristic H = H ω i hi + (π − αk )lk + (π − θj )hj . i

k

j

Here the first sum ranges as before over all vertices of the cap which survive the cutting-off of the digon; the second, over all edges, inclusive those included in the edges of the cut (the edges of the cut are assumed to be identified and the angle αk equals the sum of the angles at both edges of the cut); and the third, over all vertical edges projecting the “boundary” vertices that appear transforms on the edges of the cut. After “leveling” the edges of the cut, H into the total curvature H of the cap that results from gluing the edges together. It is easy to show that the “raising of the lower edge” of the cut can be carried out so that hi does not change for all internal vertices below which ωi < 0. At the same time, it turns out that π − θi ≥ 0 exactly at the “lower edge” of the cut. So, raising this edge, we obtain ≈ ∆H (π − θi )∆hi ≥ 0. Finally, the decrease of the total curvature at the first step, more precisely where H is the total curvature of the initial cap and H the difference H − H, is calculated before the “leveling” of the edges, can be represented (simply as H − (π/2)L, where H is by using the explicit expressions for H and H) the ordinary total curvature of the cap for which the cut-off digon plays the role of S and where L is the length of the “cut contour,” i.e., the sum of the length of the contour of the digon, the length of the projection of this


483

contour, and the doubled sum of the heights of the boundary vertices of the digon. The estimation of H − (π/2)L is an elementary problem: split the digon into two “halves,” dividing H − (π/2)L into the corresponding summands, and then deform the “halves,” intending to increase the quantity in question. In the extremal position we obtain an explicit bound like Cδ (δ is the “width” of the digon). In a similar way, we estimate the decrease of the total curvature in all the other needed cases. § 1.7 Proof of Lemma 1. 1. We first prove that every ordinary cap has the greatest total curvature among all convex polyhedral caps with the given intrinsic metric. This will prove in particular that H(P 0 ) ≥ H(P ). By the obvious continuity of the curvature H(P ) and the closure and boundedness of the domain of variation of the heights of a cap with constant intrinsic metric, it suffices to demonstrate that every convex generalized polyhedral cap can be deformed with its convexity and intrinsic metric preserved and its curvature increased. So we let P be a generalized cap with the same metric as for P 0 . By Lemma 7, P can be deformed, with its convexity and intrinsic metric preserved, so that, for every vertex whose height changes, the increment of the height and the curvature of the projection S below the vertex have the same sign. But then (by Lemma 9) ωi ∆hi > 0. ∆H ≈ i

2. Similary arguments (basing on Lemma 7) show that we can carry out a deformation increasing H(P ) by forbidding one of the vertices to lower (until the vertex becomes the only vertex below which the curvature of the lower base differs from zero). This allows us to conclude that, in the set Mi (δ) of convex caps whose developments are isometric to the development S 0 of P 0 and in which the height of one of the vertices Ai satisfies the inequality hi ≥ h0i + δ, where δ is a positive constant, the greatest total curvature H is attained at a cap P for which the curvature of the projection S is zero below all vertices but Ai and for which hi = h0i + δ. 3. The above implies that, under the conditions of the lemma, for the caps P 0 and P with ∆h = maxk (h1k − h0k ) > 0 we have H(P 0 ) − H(P 1 ) ≥ H(P 0 ) − H(P ), where P has the structure described in the preceding paragraph; moreover, we can take i to be one of the indices at which maxk (h1k − h0k ) is attained and take ∆h for δ. 4. We now deform P 0 to P , increasing hi by the law hi (t) = h0i + t∆h, 0 ≤ t ≤ 1, and preserving the zero curvature everywhere in S except for the

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vertex Ai . The existence of such a deformation was established at the end of § 1.5. We are left with estimating the variation of H in this deformation. Using the equality of Lemma 9, we find

1

1

1 0 H(P ) − H(P ) = − dH(t) = − ωi (t) dhi (t) = ∆h [−ωi (t)] dt. 0

0

0

Next, the estimate of Lemma 8 yields 1 −dωi √ ≥ C∆h −ωi dt

− ωi (t) ≥ (C∆h)2 t2 .

and

Hence, 0

0

H(P ) − H(P ) ≥ H(P ) − H(P ) = ∆h

0

1

[−ωi (t)] dt ≥ C∆h3

and so on.

Proof of Lemma 2. We deform the development S 0 to S 1 as described at the end of § 1.4 and deform P 0 by the method described at the end of § 1.5. Let P 10 denote the cap that appears in this deformation at the moment when S 0 transforms into S 1 . The first claim of the lemma holds by the construction of P 10 . Validity of the second claim, i.e. the estimate H 0 − H 10 ≤ Cεα , follows from the fact that the total width of the digons and strips that are cut away and glued in passing from S 0 to S 1 does not exceed Cεα (α ≥ 1/4) (see § 1.4) and, as shown in § 1.6, the decrease of the total curvature in passing from P 0 to P 10 is not smaller than a quantity of the same order. The third claim of the lemma follows from the fact that, in all deformations taking P 0 to P 1 , the heights of all points of S 0 surviving the cuttings can merely increase; moreover (as shown in § 1.4), for arbitrary two points A and B of S 0 and S 1 corresponding to one another under the near isometry ϕ, there are points A∗ and B ∗ that are at distances at most Cεα (α ≥ 1/4) from A and B and that correspond to one another under the isometry arising after the full deformation of S 0 (in other words, exactly the points A∗ and B ∗ such that, after all deformations, the point A∗ of the cap P 0 goes into the point B ∗ of the cap P 10 ). §2 § 2.1 In § 2 we obtain estimates for deformations of closed convex surfaces. For the reader’s convenience, § 2 is organized in the same manner as § 1: the contents of each subsection in § 2 is similar to that of the corresponding subsection in § 1.


485

§ 2.2 Given a closed convex polyhedron P , we let S stand for an arbitrary development of P composed of triangles with vertices Ai (i = 1, . . . , n) and denote by ri (i = 1, . . . , n) the distances from A0 to Ai (i > 0). We say that some generalized polyhedron P is given if (1) specified is some development S of positive curvature which is homeomorphic to the sphere and partitioned into triangles Tk with vertices Ai (i = 1, . . . , n); (2) specified is some collection of numbers ri > 0 (i = 1, . . . , n); (3) the numbers ri and the development S with the fixed triangulation {Tk } are connected as follows: (a) for the vertices Ai joined to A0 by the edges of the triangulation, the assigned numbers ri are exactly the lengths of these edges; (b) for every triangle Tk not having A0 among vertices, it is possible to construct a pyramid with base Tk and the lengths of lateral edges equal to the numbers ri assigned to the vertices of Tk . The numbers ri are referred to as the radii of the polyhedron, and A0 is called the distinguished vertex. We will write P = (S, ri , A0 ) if we want to indicate the fact that S is the development; ri , the radii; and A0 , the distinguished vertex of the generalized polyhedron. The pyramids constructed on the triangles Tk in the above way are referred to as the constituent pyramids of the polyhedron P . The law for gluing together the triangles Tk in the development S naturally gives rise to some law for gluing together the pyramids of our polyhedron P , thereby transforming P into a three-dimensional manifold with boundary which is produced from the pyramids by gluing; moreover, the boundary of this manifold is obviously the initial development S. By the complete dihedral angle αk of P at some boundary edge Ai Aj , we mean the sum of the dihedral angles of the constituent pyramids containing this edge. The curvature ωi of P at an internal edge A0 Ai is defined as the difference 2π − θi , where θi is the sum of the dihedral angles of the pyramids containing the edge A0 Ai . A generalized polyhedron is said to be convex if the dihedral angle at every boundary edge is at most π. The total curvature H(P ) of a generalized polyhedron P is the sum ωi ri + (π − αk )lk , i

k

where the first sum ranges over all internal vertices Ai of the development S of P and the second, over all edges of S; moreover, lk stands for the edge length. We now formulate the main lemmas. Lemma 1. 1. Let P 0 = (S 0 , ri0 , A00 ) and P 1 = (S 1 , ri1 , A10 ) be convex polyhedra; moreover, P 0 is an ordinary polyhedron, P 1 is a generalized polyhedron,

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the developments S 0 and S 1 are isometric, and A00 and A10 correspond to one another by isometry. Then H(P 0 ) − H(P 1 ) ≥ max{ 0, Cλ(∆r)3 }, where ∆r = maxi (ri1 − ri0 ), ri1 and ri0 are the radii of vertices of S 0 and S 1 corresponding to one another by isometry, C is a constant depending only on the intrinsic diameter of S 0 , and λ is the radius of the greatest disk contained in the spherical image of A0 . The proof of Lemma 1 is given in § 2.7. Recall that two developments S 0 and S 1 are said to be ε-close in intrinsic metrics if there exists a homeomorphism ϕ of S 0 onto S 1 such that for arbitrary two points A and B of S 0 we have |ρ0 (A, B) − ρ1 [ϕ(A), ϕ(B)] ≤ ε, where ρ0 and ρ1 are the distances in the developments S 0 and S 1 respectively. The homeomorphism ϕ is called a near isometry. Lemma 2. Let P 0 = (S 0 , ri0 , A00 ) be an ordinary convex polyhedron and let S 1 be a development of positive curvature which is homeomorphic to the sphere and is ε-close to S 0 in intrinsic metrics. Then there exists a convex generalized polyhedron P 10 such that (1) the development S 10 of P 10 is isometric to S 1 ; (2) the total curvatures H 0 and H 10 of the polyhedra P 0 and P 10 satisfy the inequality H 0 − H 10 ≤ Cεα ; (3) the radii r0 and r10 of arbitrary points of S 0 and S 10 corresponding to one another under the near isometry ϕ satisfy the inequality r10 ≥ r0 − Cεα ; Here α is an absolute constant (≥ 1/4) and the constant C depends only on the intrinsic diameters of S 0 and S 1 . The proof of Lemma 2 is given in § 2.7. § 2.3 The main result of § 2 is the following Theorem 1. If the developments S 0 and S 1 of closed convex polyhedra P 0 and P 1 are ε-close in intrinsic metrics, then the spatial distances r0 and r1 between arbitrary points corresponding to one another under the near isometry differ at most by Cεβ , where β is an absolute constant (≥ 1/24) and the constant C depends on the diameters of S 0 and S 1 . Proof. Denote by A0i (i = 0, . . . , n) and A1i (i = 0, . . . , n) the vertices of P 0 and P 1 that correspond to each other under the near isometry and suppose


487

that the enumeration of the polyhedra themselves and their vertices is chosen so that the maximum maxi,j (|A1i A1j | − A0i A0j |) of the differences between the spatial distances is positive and attained at i = 0 and j = 1. Denote this difference by ∆r. Now, deform P 0 , moving the vertex A00 in the direction of one of the outward normals of P0 at A00 until the spherical image of this vertex becomes to include some disk of radius λ. Denote the resulting polyhedron by Pλ0 and denote by Sλ0 any development of Pλ0 . Applying Lemma 2 first to the polyhedron P 1 and the development S 0 , next to P 0 and S 1 , and finally to P 1 and Sλ0 , construct the “mixed” polyhedra P 01 , P 10 , and Pλ01 . The total curvatures H 01 and H 10 of P 01 and P 10 are connected with the total curvatures H 0 and H 1 of P 0 and P 1 via the following inequalities: H 0 − H 1 ≤ Cεα , Therefore,7

H 1 − H 01 ≤ Cεα .

(H 0 − H 10 ) + (H 1 − H 01 ) ≤ Cεα .

On the other hand, the identity holds: (H 0 − Hλ0 ) + (Hλ0 − Hλ01 ) + (Hλ01 − H 01 ) + (H 1 − H 10 ) = (H 0 − H 10 ) + (H 1 − H 01 ).

(∗)

Here Hλ0 and Hλ01 are the total curvatures of the polyhedra Pλ0 and Pλ01 . From the elementary properties of the total curvature H of ordinary polyhedra and the way in which we construct Pλ0 from P 0 , we easily infer the following inequality for the total curvatures Hλ0 and H 0 of Pλ0 and P 0 : Hλ0 ≤ H 0 + Cλ2 ,

or H 0 − Hλ0 ≥ Cλ2 .

Using the explicit expression for the differential of the total curvature (see Lemma 9 of § 2.6), the way in which P 01 and 01 λ are constructed, and the monotone “increase” of the intrinsic metric in passage from S 0 to 0λ , we can easily show that Hλ01 ≥ H 01 , i.e., Hλ01 − H 01 ≥ 0. Finally, the application of Lemma 1 to P 1 and P 10 and to Pλ0 and Pλ01 yields H 1 − H 10 , Hλ0 − Hλ01 ≥ Cλ(∆r − Cλ − Cεα )3 (the extra summands on the right-hand side of the second inequality are due to the change of radii in passing from P 0 to Pλ0 and in passing from P 1 to Pλ01 ). Using these inequalities, we estimate the left-hand side of (∗) from below and, recalling the available estimate for the right-hand side of (∗) from above, obtain −Cλ2 + Cλ(∆r − Cλ − Cεα )3 ≤ Cεα . 7

As usual, we denote by the same letter C positive constants which differ in general from place to place and depend only on the intrinsic diameters of the developments S 0 and S 1 .

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This inequality yields the best estimate for ∆r in terms of ε (i.e., an estimate with greatest exponent) on choosing λ ≈ εα/2 . In this case we obtain ∆r ≤ Cεβ (with β = α/6 ≥ 1/24). § 2.4 We recall that in § 1.4 we studied spherical sections and deformations of ordinary planar developments of positive curvature. All definitions and results of this section translate without changes to the spherical sections near the vertices of generalized polyhedra. Every development of positive curvature homeomorphic to the sphere can easily be divided into finitely many developments homeomorphic to a disk with boundaries of positive swerve. Afterwards, the arguments of § 1 apply. Also recall that at the end of § 1.4 we established the possibility of deforming one of the two given developments of positive curvature ε-close in intrinsic metrics to the other by gluing and cutting-off digons and deforming some “narrow” strips, and we obtained an estimate of the form Cεα (α ≥ 1/4) for the total “width” of digons and strips in such a deformation. In the current section we inspect the deformations of convex generalized polyhedra. We begin with the deformations not changing the intrinsic metrics of the developments of the polyhedra and preserve their convexity. A face of a convex generalized polyhedron is defined in the same way as a face of a generalized cap in § 1. The role of the projection of a cap belongs now to the spherical section SA0 of the polyhedron near the distinguished vertex A0 . The curvatures of SA0 are exactly the curvatures of the polyhedron at the “internal” edges A0 Ai . Lemma 6. Let F be some face of a convex generalized polyhedron P ; A, a boundary vertex of F ; and A, the projection of A to the spherical section SA0 of P near A0 . If the curvature of SA0 at A is nonpositive, then the complete angle of F at A is less than π. Lemma 7. The following hold for every convex generalized polyhedron P : 1. If the curvature of SA0 at the point below a vertex Ai of P is negative, then we can deform P continuously, preserving its convexity and the interior metric of its development, so that the radius ri of Ai decreases, whereas the radii of the other (genuine) vertices remain the same. 2. If there are vertices of positive curvature in SA0 , then we can deform P , preserving its convexity and the intrinsic metric of its development, so that the radii increase for those vertices of P below which the points of SA0 have positive curvature. In the proof of item 2 of Lemma 7 in § 1 we used the deformation of a cap by the simultaneous raising of all vertices by the same amount. This deformation obviously preserves both the convexity and the intrinsic metric.


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For a convex generalized polyhedron, we must use a slightly different deformation instead, namely the deformation that increases the squares of all the radii of the polyhedron by the same (small) amount. The fact that this deformation possesses the required properties will be established later. Also, in § 1.5 we studied how the curvature of the projection of a convex cap depends on on its heights when the intrinsic metric of the development of the cap remains constant. These results apply in the present section immediately, since, by the familiar classical trick, we can reduce the study of the deformations of a convex generalized polyhedron to that of the deformations of some convex generalized cap (see the proof of Lemma 8). We will now give an analog of the “inverted” cap and generalize the corresponding constructions of § 1 where this cap was used. Imagine that the triangles of the development of a polyhedron P lie in the horizontal XY plane and above every point A of each triangle take the point B with the z-coordinate z(B) = r2 (A), where r(A) is the radius of A in P . Direct calculations show that, above each triangle, we thereby obtain a piece of the paraboloid given by an equation of the form z = x2 + y 2 + αx + βy + γ. The rule for gluing together the triangles of the development of P naturally gives rise to some rule for gluing together the paraboloids above them, and the convexity condition of P yields some restriction on the coefficients α and β of adjacent paraboloids. The resultant “piecewise paraboloidal” surface P ∗ plays the same role in our arguments as the “inverted” cap did in the similar arguments of § 1. We now formulate and prove Lemma 8, an analog of Lemma 8 of § 1. Lemma 8. Let P be an ordinary convex polyhedron and assume that the spherical image of the vertex A0 of P contains a disk of radius λ. Let P be a generalized polyhedron with development isometric to that of P and with distinguished vertex A0 . Assume further that the curvature of the spherical section SA0 of P vanishes everywhere except some point A and the radius r(A) of the point A of P above A is greater than the radius r(A) of the corresponding (by isometry) point of P . Then 1. The polyhedron P can be continuously deformed to P so that during the deformation the polyhedron is always convex, the metric of its development is constant, and the curvature of the section SA0 vanishes everywhere except for the projection A of A. 2. The curvature ω(A) of SA0 at A changes so that Cdω dr ≤ − √ , λ −ω where r(A) is the radius of A. The proof can be carried out as follows. Since all “non-Euclideanity” of the deformed polyhedron Pt is concentrated near the edge A0 A, this polyhedron can be unrolled in Euclidean

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space by making a planar cut in it through the edge AA . Place the unrolled polyhedron Pt so that the point A0 occurs at the origin O of the Cartesian coordinates XY Z, the polyhedron itself lies in the half-space z > 0, and all planes of the polyhedral angle at A0 have slopes greater than λ with respect to the XY -plane. The last requirement can be satisfied at the initial moment of the deformation, since the spherical image of A0 contains a disk of radius λ. During the further deformation the development SA0 , together with the “solid” angle at A0 , merely “shrinks” in some sense. Next, at each moment of the deformation we subject Pt to the projective transformation given by the formulas x = x/z, y = y/z, and z = 1/z. This transformation takes Pt to some generalized cap Qt of exactly that type for which in § 1.5 we obtained the estimate Cd ω˙ h˙ ≤ − √ −ω for the speed h˙ of variation of heights through the speed ω˙ of variation of the curvature at the only “non-Euclidean” edge, the curvature ω itself, and the diameter d of the cap. The speeds of the radii of Pt during the deformation are connected in the familiar way with the speeds of the heights of Qt . Connections between the corresponding values of ω and ω˙ can also be found. The diameter of Qt admits an estimate of the form C/λ, with C dependent only on the diameter of Pt . In result, we come to some estimate of the form Cdω . dr ≤ − √ λ −ω § 2.6 The expression for the total differential of the total curvature is derived in the same way as in § 1. Lemma 9. The following equality holds: ωi dri + (π − αk )dlk , dH = i

k

where H is the total curvature of the generalized polyhedron P , ri are the radii (i.e., the lengths of all internal edges) of P , ωi are the curvatures at these edges, lk are the lengths of all boundary edges, and αk are the complete angles at these edges. In § 1.6 we estimated the decrease of the total curvature in the deformations of convex generalized caps resulting from the deformations of their developments. All these arguments translate almost without changes to the case of generalized polyhedra and demonstrate that the decrease of the total curvature admits an estimate of the form Cδ, where δ is the total “width” of the cut-off digons and deformed strips.


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Recall that this conclusion relies only on the following facts: some “additive” decomposition of the total curvature, the “smallness” of terms in this decomposition relating to “narrow” strips and digons, and, finally, the fact that the “convexification,” accompanying the deformation and changing the total curvature everywhere rather than only in the deformation region, does not damage our estimates. § 2.7 The proof of Lemma 1 goes through in the same way as in § 1. Moreover, the qualitative part of the proof does not change at all, whereas the inference of the estimate bases on Lemma 8 of the current section and leads to the appearance in the final inequality of the radius λ of a disk in the spherical image of A0 . The proof of Lemma 2 differs in no way from that of § 1, provided of course that we use the auxiliary results of the corresponding sections of § 2.

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12.2 Supplement to Chapter 4: Yu. A. Volkov. Existence of Convex Polyhedra with Prescribed Development. I1

§1 Here we give a new proof2 to the existence theorem for convex spherical caps3 with given development (subject, of course, to certain constraints); the proof of a similar theorem for closed convex polyhedra will be published in a forthcoming article.4 Our proof bases on the following idea put forward by Blaschke and Herglotz in [BH]: To construct a closed convex surface with a differentialgeometric metric given in advance on the sphere, it suffices to know how to extend this metric from the sphere to the interior of the ball so as to transform the ball into a manifold isometric to a convex body in Euclidean space. A. D. Alexandrov suggested in [A19] that it stands to reason to use polyhedra for implementing this idea. The main theorem of our article leans upon a method analogous to the one proposed by A. V. Pogorelov in [P4] for proving the existence of polyhedra with prescribed curvatures at vertices. We now sketch our arguments without plunging into details. Let S denote the sought convex cap; S, the projection of S on the plane P containing its boundary; and M , the (solid) polyhedron bounded by the surfaces S and S. Imagine the plane P horizontal and S lying above P . The polyhedron M can be regarded as glued from the polyhedra Mi (we call them prisms) bounded from above by the faces Si of S, from below by the projections S i of these faces, and from the sides by the prismatic surfaces projecting the edges of S on the plane P . It is exactly such collection of polyhedra Mi that we will try to construct from the given development. For the prisms of the so-constructed collection to really produce the required polyhedron M by gluing, it is at least necessary that the gluings be realizable around each internal edge of the collection, i.e., around each edge produced from the edges of the lateral prisms by gluing. To this end, we obviously need the following: the sum of the dihedral angles at each of these edges must equal 2π. To meet this condition is the main difficulty in constructing the sought collection of prisms. First, we consider all possible collections of prisms for which the sum of dihedral 1 2 3 4

A translation of the article [Vo4]. Other proofs can be obtained on the grounds of the familiar proofs of the theorem for closed polyhedra (see [A19], [A15], and [E2]). Throughout we adhere the terminology of [A19] unless it is indicated otherwise. The second part was never published. See a remark at the end of the present article. – V. Zalgaller


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angles at each internal edge is at most 2π and all other conditions for gluing are satisfied. Among these collections, we choose one with the maximal sum of the lengths of internal edges and prove that this collection now satisfies all of the conditions. The proof of the main theorem is given in § 4. It mainly relies on Lemmas 1–3 of § 3. The notions we use in § 3 and § 4 are defined in § 2. In § 5 we prove Lemmas 1–3 together with some auxiliary assertions. §2 Let P be a plane (assumed fixed from now on) which we shall regard as horizontal, let T be some triangle above P , and let hi (i = 1, 2, 3) be the distances from the vertices of T to P . Slightly abusing the conventional terminology, the polyhedron M (T, hi ) covered by the straight line segments joining the points of T with their projections to P is referred to as a prism. The triangle T and its projection on P are referred to as the upper and lower bases of the prism, the other faces are called lateral, and the numbers hi are called the heights. If T lies in a vertical plane, then the prism M (T, hi ) turns out to be a planar polygon. We call such a prism degenerate. It is easy to see that every prism is determined from its upper base and heights uniquely up to congruence. Now, assume given some development R composed of triangles and suppose that some number hi is assigned to each vertex Ai of R so that, for every triangle Tk and the numbers hi assigned to its vertices, it is possible to construct a prism Mk with upper base Tk and heights hi . The rule for gluing together the triangles Tk in the development R naturally leads to some rule for gluing together the prisms Mk : the prisms Mk are glued together along the lateral faces so that their upper bases produce R by gluing. The abstract polyhedron M produced from the prisms Mk by gluing is called a cylindrical polyhedron; the development R is called the upper base of M and the numbers hi are called the heights. From now no, by a polyhedron we always mean a cylindrical polyhedron. We call a polyhedron nondegenerate if so are all constituent prisms. The faces of the prisms of a nondegenerate polyhedron which are not glued to others faces (as a whole) and all edges of these faces are called the boundary faces and the boundary edges of the polyhedron. The remaining edges are called internal. The complete angle θ at an edge of a nondegenerate polyhedron is defined as the sum of the dihedral angles at this edge in the prisms of the polyhedron. For an internal edge, the difference 2π − θ is called the curvature at this edge. A nondegenerate polyhedron is said to be convex if the complete angle at each of its boundary edges is at most π. A polyhedron is said to be of nonnegative (zero) curvature if the curvatures are nonnegative (equal to zero) at all internal edges of the polyhedron.

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The convergence of polyhedra of the same structure5 is defined through the convergence of the lengths of the corresponding edges. A convex polyhedron (a polyhedron of nonnegative curvature) is defined to be the limit of a sequence of convex nondegenerate polyhedra (nondegenerate polyhedra of nonnegative curvature). Note that if a prism degenerates then some of its dihedral angles, namely the angles between faces degenerating into segments, can have no limits.6 For this reason, there are some difficulties in defining all dihedral angles of degenerate prisms and, in consequence, all complete angles of polyhedra containing such prisms. That is why we skip the definition. However, whenever the complete angles of a convex polyhedron (a polyhedron of nonnegative curvature) are defined (by continuity), they obviously satisfy the same inequalities that characterize convex nondegenerate polyhedra (nondegenerate polyhedra of nonpositive curvature). The nondegenerate part M 0 of a polyhedron M is defined to be the polyhedron composed of the nondegenerate prisms of M with all their gluings present in M . A polyhedron of zero (nonzero) curvature is defined to be a polyhedron whose nondegenerate part has zero (nonzero) curvature. The main role is further played by the polyhedra M satisfying the following conditions: (1) M is a convex polyhedron of nonnegative curvature. (2) The upper base R of M is a development homeomorphic to a disk, it is composed of triangles, the curvature of R is positive at each internal vertex, and the swerve of the boundary of R is positive at each boundary vertex. (3) The boundary vertices of R have zero heights in M . We call these polyhedra special. §3 Lemma 1. Let R be a development of positive curvature which is homeomorphic to a disk and has a positive swerve of the boundary.7 Assume further that A1 , . . . , An is a collection of points of R containing all genuine vertices.8 5 6

7 8

That is, those with upper bases of the same structure (see [A19]). Observe, however, that the upper bases of the prisms under consideration never degenerate into segments, and even if the lower bases degenerate, they have the angles at them always equal to π/2. Therefore, the only cause of “danger” is the degenerate lateral faces or, which is the same, the vertical segments of upper bases. We prove later that there are no such edges in some special polyhedra we define below. That is, the swerve of the boundary is nonnegative at every point and positive for at least one point. A genuine vertex is an internal vertex at which the curvature of the development differs from zero or a boundary vertex at which the swerve of the boundary is nonzero.


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Then there are only finitely many developments Ri isometric to R with vertices only at the points corresponding by isometry to the points A1 , . . . , An . Lemma 2. If M is a special polyhedron of nonzero curvature with an upper in which (1) the upper base R, then there exists a special polyhedron M is isometric to R and has vertices only at the points corresponding base R by isometry to the vertices of R and (2) the sum of heights is greater than that in M . Lemma 3. Every special polyhedron of zero curvature is isometric to some convex polyhedron in Euclidean space. §4 Theorem. Every development R of positive curvature which is homeomorphic to a disk and has a positive swerve of the boundary defines some convex cap by gluing. Proof. Let A1 , . . . , An be all genuine vertices of R. Consider all developments R1 , . . . , Rm composed of triangles, isometric to R, and having vertices only at the points corresponding by isometry to the points Ai (i = 1, . . . , n). It is well known (see [A19]) that these developments exist and the number of them is finite by Lemma 1. Denote by Ek the set of all special polyhedra in each of which the upper base is the development Rk . All these sets are nonempty. Indeed, for every development Rk we can construct the required polyhedron by choosing the heights h = 0 for all vertices. In each n of the sets Ek , take one of the polyhedra Mk with the greatest sum i=1 hi of heights. The fact that these polyhedra exist follows in a familiar way from the (obvious) closure and boundedness9 of each of the sets Ek . Now, among Mk (k = 1, . . . , m) we choose a polyhedron M with the n greatest sum i=1 hi . Were M not a polyhedron of zero curvature, Lemma 2 would imply that in one of the sets Ek there is a polyhedron with the sum of heights greater than that of M ; this is, however, impossible by the very choice of M . Hence, M is a polyhedron of zero curvature. By Lemma 3, M admits an embedding in Euclidean space. Furthermore, the development R bounding M from above yields the sought cap.

9

Applying the triangle inequality, we can easily see that, for arbitrary two heights hi and hk of a special polyhedron M , we have |hi − hk | ≤ dik , where dik is the distance in R between the vertices Ai and Ak corresponding to the heights hi and hk .

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§5 (A) To prove Lemma 1, it suffices to validate the following Lemma 4. In a development of positive curvature homeomorphic to a disk and having a positive swerve of the boundary, any two (not necessarily distinct) points may be joined only by finitely many geodesics. Clearly, Lemma 4 will be proved if we establish the following three assertions. Lemma 5. Under the conditions of Lemma 4, the lengths of all geodesics in the development R are bounded by the same number L depending only on R. Lemma 6. The number of triangles in R crossed by a geodesic of length at most L does not exceed some constant depending only on L and the development R (each triangle is counted as many times as the geodesic crosses it). Lemma 7. Given points A and B in R, there exist only finitely many geodesics joining them and traversing each of the triangles of the development at most finitely many times N . Lemma 5 is well known (see, for instance, [ASt]). Lemma 6 was proved but not formulated in [A19], p. 193. To prove Lemma 7, consider all triangles containing A and glue them together to make a planar sector with vertex A0 . For every geodesic AB, successively unroll on a plane all the crossed triangles of the development and glue them together along the sides traversed. It is well known that the geodesic AB then unrolls on some straight line segment A0 B0 . Clearly, the number of the possible geodesics AB does not exceed the number of the so-arising rectilinear segments between A0 and B0 or, which is the same, the number of points B0 appearing from B in this way. However, the latter is obviously finite. (B) Before proving Lemma 2, we define the notion of spherical section of a polyhedron and establish one important property of these sections. Let A be one of the vertices of the upper base R of a nondegenerate polyhedron M . Intersect each of the trihedral angles of the prisms of M containing A with the unit sphere centered at A. The spherical triangles arising in the intersections produce some spherical development SA by gluing. We call SA the spherical section of M near A. The vertices of SA correspond to the edges of the polyhedron and the edges of SA correspond to the faces containing A. Moreover, the complete angles of SA at its vertices are equal to the complete angles of M at the edges corresponding to these vertices, and


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the distances between the vertices in SA are equal to the angles between the corresponding edges of M . Lemma 8. Let SA be the spherical section of a convex polyhedron M near a vertex A which is the upper end of the vertical edge AA, let O be the vertex of SA corresponding to AA, let Ci (i = 1, . . . , n) be the other vertices of SA , and let θ be the complete angle of SA at O. Then θ is a nonincreasing function of the lengths pi of the edges OCi of SA . Proof. Enumerate the vertices Ci in the same order as they are encountered as one proceeds around the boundary of SA in a certain direction. Let θi (i = = C1 if 1, . . . , n) be the angle at the vertex O in the triangle OCi Ci+1 (Cn+1 n O lies inside SA ) and let pi be the length of the arc OCi . Since θ = i=1 θi k while ∂θ ∂pi = 0 if k = i, i + 1, we have ∂θ ∂θi ∂θi+1 = + . ∂pi ∂pi ∂pi Straightforward calculations yield −

∂θi cot αi,i−1 = ∂pi sin pi

and

−

∂θi+1 cot αi,i+1 = , ∂pi sin pi

where αi,i−1 and αi,i+1 are the angles at the vertex Ci in the triangles OCi−1 Ci and OCi Ci+1 . Hence, ∂θ sin αi = , ∂pi sin αi,i−1 sin αi,i+1 sin pi where αi = αi,i−1 + αi,i+1 is the complete angle at Ci . Since the polyhedron M is convex, we have αi ≤ π; whence − ∂θ pi ≥ 0. (C) Proof of Lemma 2. can be obtained from M as follows: Retriangulate 1. Let us show that M R duly and repartition the polyhedron M into prisms accordingly. Keeping the degenerate part of M constant, lengthen all internal edges of the nondegenerate part M 0 at which the curvature is positive by the same small ε > 0. If ε is small enough, then this deformation can be implemented for every triangulation of R, since we can make any small deformations of edges in nondegenerate prisms. Since the dihedral angles depend continuously on the edge lengths while ε is small, we can proceed so that at the edges of R neither the complete angles which were less than π before the deformation nor the complete angles at the elongating edges, i.e., the edges where they were less than 2π, will exceed the indicated limits. Therefore, we are left with validating the following for a suitable triangulation of R: (a) The complete angles at the edges of R which were equal to π become less than or equal to π;

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(b) The complete angles at all nonelongating vertical edges of M 0 only decrease. 2. We begin with demonstrating that, for the above-described deformation , the validity of (a) implies that of (b). of M to M Let A be a vertex of M 0 which is an end of a nonelongating vertical edge AA and let SA be the spherical section of M 0 near A. The distances ρi between the vertex O of SA corresponding to the edge AA and the vertices corresponding to the edges of R abutting on A are obviously equal to the angles between the corresponding edges. In our deformation, these angles can only increase, since the edge AA and the edges of R are immovable while the vertical edges abutting on the vertices adjacent to A can only elongate. Now, Lemma 8 implies that the complete angle of SA at the vertex O and, in consequence, the complete angle of M 0 at the edge AA, being equal to the former, can only decrease, which proves our claim. 3. To find a triangulation not violating condition (a), we first list all cases in which this condition breaks down at some edge AB along which the upper bases ABC and ABD of two nondegenerate prisms are glued together. Elementary geometric considerations, simple but bulky and thus omitted, convince ourselves that this happens only if C and D differ and only in one of the following cases (for convenience, we say that some vertex of R raises if the vertical edge abutting on it elongates): (1) at least one of the vertices C and D raises, whereas the vertices A and B do not raise; (2) one of the vertices A and B raises (for definiteness, let it be A) together with one of the vertices C and D (let it be C), B and D lie to one side of the straight line AC (in the plane ABCD), and B is nearer to this line than D; (3) the vertices C and D raise, one of the vertices A and B raises too, and the angle of the quadrangle ABCD at this vertex is less than π. Now let us show that in each of these three cases the diagonal CD shifts inside the quadrangle ABDC, splitting it into the triangles ACD and BCD; then we can make a retriangulation by replacing the triangles ABC and ABD with ACD and BCD. It suffices to demonstrate that each of the angles of the quadrangle ABCD at the vertices A and B is less than π. If A or B raises, then the required inequality holds, because we have the second or third case. If, say, A does not raise, then the prisms containing the vertical edge AA can factually be glued together, since either the curvature at the edge vanishes or the edge lies on the boundary of M 0 . If AA is an internal edge of M 0 , then the polyhedral angle with vertex A in the polyhedron produced from the prisms by gluing is convex owing to the convexity of M 0 . If AA is a boundary edge of M 0 , then the dihedral angles at all edges abutting on A, except for the one at AA, are convex for the same reason, whereas the angle at AA is convex by Lemma 10 (proved below in (D)). The quadrangle ABCD will lie on one of the faces of the convex polyhedral angle with vertex A; whence we conclude that the angle at A in ABCD is less than π.


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Examining all possibilities, we directly verify that in each of the three cases the retriangulation, consisting in the replacement of the triangles ABC and ABD with ACD and BCD, increases the value of Σ defined by Σ = S1 + 2S2 + 3S3 , where S1 , S2 , and S3 are the sums of the areas of those triangles of the development R in which one, two, and three vertices raise, respectively. Thus, if condition (a) is violated, then there is a retriangulation increasing Σ. 4. We now choose such a triangulation of R (or one of such triangulations if there are several) for which Σ has the greatest value. Then no retriangulation will increase Σ and, by the preceding paragraph, none of the above-listed cases is possible. So condition (a) is not violated. (D) Here we make some preparations for proving Lemma 3; the proof itself is given below in (E). 1. First of all, we introduce the needed definitions and notations. The further consideration deals with a special polyhedron which we denote by M . The upper base of M is denoted by R and the boundary of R is denoted by L. By M 0 and M 1 we mean the nondegenerate and degenerate parts of M , i.e., the polyhedra composed of nondegenerate and degenerate prisms of M , respectively, with all the gluings present in M . By a point A of the degenerate part we naturally mean a maximal collection of points Ai (i = 1, . . . , n) in the respective degenerate prisms Mi (i = 1, . . . , n) such that for every i = 1, . . . , n − 1 the prism Mi is glued (along a face) to Mi+1 and the point Ai is glued to Ai+1 . The points Ai are referred to as the representatives of A. We say that a point A of M 1 lies in R if it has a representative Ai in R. As usual, we imagine the lower bases of the prisms as lying on the horizontal plane P , the prisms themselves are above this plane, and the degenerate prisms are in vertical planes. Henceforth we use the terminology that appeals to this vision. We say that points A and B of M 1 lie on the same vertical if they have representatives Ai and Bi belonging to the same prism Mi and lying on the same vertical line in the plane of this prism. Furthermore, if Ai lies above (below) Bi , then we say that A lies above (below) B. By the upper base of the degenerate prism Mi we mean the collection of edges bounding the polygon Mi above. To the upper boundary L2 of the degenerate part we add those edges of R (keeping all present gluings) for which at least one of the prisms glued along the edge together is degenerate and the complete angle α is well defined and satisfies the inequality α < π. Let A be a vertex of the development R. We say that the angles of the upper bases Tk (k = 1, . . . , n) containing A cover the sector σA if every successive base is glued to the preceding one along an edge so that the complete angle of M is either undefined or equals π.

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We say that σA is a sector of the degenerate part if all points of σA belong to M 1 . 2. We now formulate the necessary assertions on the structure of sectors of the degenerate part. Let σA be such a sector and let A be its vertex. Then the following hold: (a) A lies on L1 ; in other words, σA has a boundary and at least one of the boundary edges lies on L1 ; (b) there are no points above A in the triangles of the sector; (c) if A is an internal vertex of R, then the interior of one of the triangles of the sector has points below A; (d) if A lies on the boundary of R, then at least one boundary edge lies on the boundary of R. We prove these assertions later. Now, we apply them to proving the lemmas necessary for the further exposition. Lemma 9. Let A be a point of R belonging to M 1 . Then all points of R lying on the same vertical with A and sufficiently close to A cover some segment; moreover, the segment has points above A if A does not lie on L1 , has points below A if A does not lie on L, and reduces to the point A if A lies both on L1 and on L. Proof. The claim is obvious when A is not a vertex of R, in other words, when A belongs to the interior of the upper base of one of the degenerate prisms or to the interior of an edge along which two such bases are glued together or to the interior of an edge lying on L. Assume that A is a vertex of R. Then A lies on L1 by condition (a). In that case, if A does not lie on L, then the claim under proof follows from (c), while if A lies on L, then there no points below A at all and (d) implies that there no points above A either. The proof of the lemma is over. Now, observe that in the degenerate prisms of a special polyhedron there are no vertical edges of the upper base; otherwise (b) would be violated at the lower end of such an edge. Hence, the lateral faces of the prism do not degenerate into segments and all dihedral angles between them are well defined. Lemma 10. Suppose that, for an internal vertex A of the development R, the degenerate part has a sector σA with vertex A. Then the sum of dihedral angles at the vertical edge AA is at least π in the degenerate prisms whose bases form the sector σA , and this angle is at most π in the nondegenerate prisms. Proof. The first claim follows from the fact that the angle at the edge AA in a prism whose upper base has points below A is obviously equal to π, and by (b) such a prism always exists under the conditions of the lemma. The


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second claim follows from the first and the fact that the complete angle at the edge AA is less than or equal to 2π. Lemma 11. For every vertex of the development R, there are at most two incident edges each of which is either a boundary edge of a sector of the degenerate part or an edge of R not lying on the degenerate part. Proof. If a vertex A belongs to the interior of R, then, to prove the lemma, it clearly suffices to check that there may exist at most two sectors of the degenerate part containing A and if we have two such sectors, then they exhaust all upper bases containing A. This is indeed the case, since by Lemma 10 each of these sectors contributes at least π to the complete angle at AA. If A is a boundary vertex of R, then it has two incident boundary edges, and by (d) the appearance of sectors of the degenerate part does not increase the number of the edges in question. 3. We now prove the properties (a), (b), (c), and (d). Let Ti (i = 1, . . . , n) be the upper bases forming the sector σA with vertex A of the degenerate part of the polyhedron M . We assume them enumerated in the order of their successive gluing. Let A, Bi , and Bi+1 be the vertices of the triangle Ti (it is possible that Bn+1 = B1 ). Glue the triangles Ti together by adjoining them to one another in some plane along the same edges as they are glued in the development R (excepting possibly AB1 and ABn ). Let B1 AB be the resulting planar sector. Over each of the Ti ’s, construct the degenerate prism of M with upper base Ti . Clearly, all these prisms will be situated in one plane and their lower bases will be situated on one “horizontal” line bounding “below” the polygon covered by the prisms. Let lA be the vertical line through A. None of the boundary edges AB1 and ABn of the sector B1 ABn lies on lA . Otherwise the complete angle at such an edge in M would be undefined and so the edge would not be a boundary edge for σA . In that case, it would be glued to the other edge among AB1 and ABn and the angle of the sector B1 ABn would equal 2π. At the same time, we know that there are no such vertices A on the upper base of a special polyhedron. If the boundary edge of the sector B1 ABn bounds this sector below, then it lies on the boundary of R. Otherwise, the complete angle in M at this edge would equal π and the edge would not be boundary for σA . If both edges AB1 and ABn bounded B1 ABn below, then both would lie on the boundary of R and be horizontal, whereas the complete angle of the sector would equal π, which is impossible for a special polyhedron. Thus, we have two possibilities: 1. Both edges AB1 and ABn bound B1 ABn above. In this case (a), (b), and (c) are obvious, while the premise of (d) fails since there are points below A.

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2. One of the edges AB1 and ABn bounds B1 ABn below and the other, above. In this case (a), (b), and (d) are obvious, while the premise of (c) fails. 4. We now clarify the structure of the degenerate part of a special polyhedron. If L is a polygonal line consisting of edges of the upper base R of a polyhedron M , then by the cylinder M (L) projecting L we mean the development composed of those lateral faces of the prisms of M which are bounded above by the edges of L and are glued together along vertical edges abutting on the vertices of L. Lemma 12. The degenerate part M 1 of a special polyhedron is isometric to the cylinder M (L1 ) projecting the upper boundary L1 of this part. Proof. Define a mapping f from the cylinder M (L1 ) into M 1 by sending each point A of M (L1 ) to the point f (A) of M 1 whose representative is A. Let us show that the image of M (L1 ) is the whole degenerate part. Indeed, let B be an arbitrary point of M 1 , let C be the highest point among the points lying on the same vertical with B, and let B0 and C0 be representatives of B and C in the same prism M0 . Clearly, C0 lies on the upper base of M0 , which implies that it belongs to L1 : otherwise, by Lemma 9 there would exist a point above C0 . Therefore, B0 lies on M (L1 ) and at the same time B = f (B0 ). Prove that f is one-to-one. Suppose the contrary and let A1 and A2 be different points of M (L1 ) going to a point A of M 1 . In other words, A1 and A2 are representatives of A. This means that A1 and A2 are at the same height, and since A1 and A2 are different, it follows that so are the points C1 and C2 of L1 below them. Let Mi (i = 1, . . . , k) be the prisms successively glued to each other and containing the glued points Bi (i = 1, . . . , k) such that B0 = A1 and Bk = A2 . On the upper base of every prism Mi , take a segment li consisting of points on the same vertical with Bi . As the prisms Mi glue to each other, the segments li produce by gluing a polygonal line γ that lies on R and joins the points C1 and C2 . Take the lowest point C0 of this polygonal line. Then C0 lies on the boundary L of the development R: otherwise γ would contain a still lower point. Restricting the consideration to a part of γ if need be, we may assume that between C1 and C0 as well as between C0 and C2 there are no points of L. Let γ1 be the part of γ between C0 and C1 and γ2 , the part between C0 and C2 . It is obvious from Lemma 9 that the constituent rectilinear segments of γ1 (γ2 ) do not overlap and therefore the sum of their lengths equals the length of the vertical segment projecting C1 (C2 ) to the lower base. However, this segment is obviously a shortest arc in the polyhedron M , and so γ1 (γ2 ) is a shortest arc between C1 (C2 ) and the boundary L of R. The point C0 is a common point of γ1 and γ2 , whereas C1 and C2 differ. However, this is impossible since if γ1 and γ2 “diverge” at C0 , then the swerve of L at this point would be negative and if they “diverge” at an interior point of R, then the curvature would be negative at this point.


503

The isometry of f is now obvious, since the lengths of curves in M 1 are defined by using representatives of these curves and each representative can be examined in M (L1 ). 5. The topological structure of the upper boundary L1 of the degenerate 0 part and that of the boundaries L0 and L of the upper and lower bases of the nondegenerate part of a special polyhedron is clarified in the following lemma. Lemma 13. If the nondegenerate part M 0 is empty, then L1 is a simple 0 polygonal line. If M 0 is nonempty, then L1 and L are simple closed polygonal lines and L0 contains L1 . the polygonal line that is the union of L1 and L0 . Then L Proof. Denote by L is connected as the image of the boundary of R under the mapping assigning to each boundary point A of R the highest point of the polyhedron above A has no branching points and this mapping is obviously continuous. The line L as it was proved in Lemma 11. Hence, L is a simple polygonal line, closed or coincides with L1 . Moreover, were open. If M 0 is empty, then so is L0 , and L 1 L = L a closed polygonal line, then it would bound some region in R and we could not connect the interior points of this region to the boundary of the development without touching L1 , but since M 0 is empty, all points lie in M 1 and clearly can be connected to L by segments disjoint from L1 . If M 0 is nonempty, then so is the polygonal line L0 . Being the boundary of some development, L0 contains some simple closed polygonal line L∗ . In this case, also contains L∗ , and since L is connected and has no branching points, L ∗ and itself L and L must coincide. At the same time, L0 is contained in L 0 ∗ 0 ∗ contains L ; hence, L and L coincide. As regards the polygonal line L , this line is obviously homeomorphic to L0 . (E) Proof of Lemma 3. Let M be a special polyhedron of zero curvature. If the nondegenerate part M 0 of M is empty, then by Lemma 13 the upper boundary L1 of the degenerate part M 1 is a simple polygonal line. In this case, by Lemma 12 the degenerate part itself is isometric to a planar polygon which is obviously convex. If M 0 is nonempty, then by Lemma 13 L1 is contained in the boundary 0 L of the upper base of M 0 . In this case, M 1 is isometric to some part of the lateral surface of M 0 , and in consequence M is isometric to M 0 . 0 0 In accordance with Lemma 13, the boundary L of the lower base R of 0 the polyhedron M is a simple closed polygonal line. We now prove that the 0 swerve of this polygonal line (from the side of R ) is nonnegative. Indeed, if some vertex A of this line lies below an internal vertex A of the upper base R of M , then this fact follows from the second claim of Lemma 10. If

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A lies on the boundary of R, then the prisms contiguous to A may in fact be glued together. In the resulting polyhedron, the polyhedral angle at the vertex A is convex by the convexity of M and the complete angle of the lower base becomes one of the planar angles of this polyhedral angle and is consequently less than or equal to π. Next, by the conditions of the lemma, 0 0 the development R has zero curvature. Hence, R defines some convex planar polygon by gluing (see [1]). Having performed this gluing, over each triangle 0 of R we construct the prism of M 0 with base this triangle. Clearly, these prisms cover a convex polyhedron of Euclidean space which is isometric to M 0 and so to M . Remark.10 The second part of this article, which should contain a similar proof for the existence of closed convex polyhedra with prescribed development, was never published. As opposed to the first part, in that case the abstract three-dimensional developments are glued together (along whole faces) from pyramids with a common vertex rather than prisms. The bases of pyramids and free (not glued) faces constitute the prescribed development of the sought polyhedron. The choice for the lengths of the “internal” edges of a three-dimensional development is done under the condition that the sum of the dihedral angles at each internal edge of the resulting three-dimensional development be at least 2π. In this compact class of abstract three-dimensional developments (which turns out to be nonempty), there exists a development with the least sum of the lengths of internal edges. For this development, the sum of the dihedral angles around each internal edge is equal to 2π by necessity. Therefore, the development gives the sought polyhedron in E3 . The details of the proof are appreciably more intricate than in the first part. A concise description for all steps of the proof is given in Yu. A. Volkov’s thesis of 1955 [Vo2]. To call up associations, we quote an excerpt from the Introduction to this thesis: “Another approach to solving Weyl’s problem was indicated by Blaschke and Herglotz [BH], relating this problem to some variational problem. My purpose was to use this approach in the case of polyhedra and to clarify on this example the abilities of the variational method for proving the existence theorems in the theory of convex polyhedra. However, I was compelled to deviate from the plan. Blaschke and Herglotz observed that, for the realization of an abstractly given metric (for definiteness, on the sphere), it suffices to extend the metric to the interior of the ball so that the curvature tensor of the resulting Riemannian metric vanish identically. Later they noticed that (as is well known from the articles on relativity theory) the extremality condition for R dv, where R is the scalar curvature and dv is the volume element, consists in the vanishing of the Ricci tensor, Rik = 0, and in consequence the vanishing 10

Added by V. Zalgaller.


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of the curvature tensor in the three-dimensional situation. Therefore, we can seek the required metric among the metrics providing an extremum for this integral. However, this can happen to be unproductive, since (as one can see in simple examples) the value of the integral corresponding to the Euclidean metric may be neither at a maximum nor at a minimum. Let us clarify the relationship between our proof and the above variational idea in the more familiar terms of Riemannian geometry. The corresponding notions for polyhedra will be introduced in due places of the thesis. We confine the exposition to the (“conical”) extensions of a Riemannian metric from the sphere to the interior of the ball which depend only on one function ρ(u, v), where u, v are the coordinates of a point on the sphere and ρ(u, v) is the distance from this point to some fixed point of the ball with respect to the sought metric. We consider only those ρ(u, v) for which the scalar curvature satisfies the inequality R(u, v, ρ, ρu, ρv , ρuu , ρuv , ρvv ) ≤ 0. It turns out that, in order to increase the integral R dv ≤ 0, we have to diminish ρ(u, v) near a point at which we still have R < 0, and the process does not terminate until we reach R = 0 everywhere. In that case, the curvature tensor vanishes as well, since the curvature tensor for the metrics under consideration has exactly one nonzero component, which is reduced to R. As a result, mentioning R dv becomes redundant, and the equation R(u, v, ρ, . . . , ρvv ) = 0 is in fact the familiar Darboux equation. This argument is nothing but another method (yet not implemented) for solving the Darboux equation. It is analogous to the method for solving the Dirichlet problem by means of subharmonic functions. In the thesis, we use such a method to prove the existence of closed convex polyhedra in E3 with prescribed development (and also sketch the proofs for convex caps and polyhedra in hyperbolic space). It is worth noting that the same idea of one-sided approximation to a solution underlies A. V. Pogorelov’s proof [P4, p. 52] of the existence of convex polyhedra with boundary projecting univalently on a given plane with prescribed projections of vertices and curvatures at them.” The same method is used in [VP] to prove the existence of unbounded convex polyhedra with prescribed development and limit angle.

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12.3 Supplement to Chapter 5: L. A. Shor. On Flexibility of Convex Polyhedra with Boundary1

1 Introduction. Statement of the Result 1.1 In [A19], A. D. Alexandrov posed the problem of finding effectively verifiable necessary and sufficient conditions for flexibility of convex polyhedra with boundary in the class of all convex polyhedra. In the same article, A. D. Alexandrov indicated a way to solving the problem and exhibited a series of sufficient conditions for a polyhedron homeomorphic to a disk to be flexible. We recall that a convex polyhedron P is called flexible in the class of all convex polyhedra if the following conditions are satisfied: (1) with each t ∈ [0, 1] we may associate a convex polyhedron Pt and an isometric mapping ft from P onto Pt so that P0 = P , f0 is the identity mapping of P onto itself, and ft is not reduced to a motion for t = 0; (2) if t0 ∈ [0, 1], then the polyhedra Pt converge to Pt0 as t → t0 and limt→t0 ft (X) = ft0 (X) for all X ∈ P . In this article we give effectively verifiable necessary and sufficient conditions for flexibility of a convex polyhedron P homeomorphic to an open disk and having boundary a simple closed polygonal line. We do not exclude flexes in which the points of the boundary Γt are glued to one another. Thus, let P be a convex polyhedron homeomorphic to an open disk and let the boundary of P be a simple closed polygonal line Γ . Denote the boundary of the convex hull2 of P by P . Since the complement P \ P of P with respect to P is homeomorphic to a closed disk and has no internal vertices, it unrolls on a plane, becoming a polygon Q, many-sheeted in general. Denote the boundary of Q by L. The gluing of Q to P yielding P is said to be the trivial patch. It is routing to say that a deformation of the trivial patch (henceforth, simply deformation) is given if for every t ∈ [0, 1] we have some patch (Qt , ψt ) on P by means of a polygon Qt and a mapping ψt of a polygonal line Lt onto Γ so that the following conditions are satisfied: (1) if t0 ∈ [0, 1], then the polygons Qt converge to Qt0 as t → t0 and limt→t0 ψt−1 (X) = ψt−1 (X) for all X ∈ Γ ; 0 (2) not all patches (Qt , ψt ) are identical; (3) the patch (Q0 , ψ0 ) is trivial. A. D. Alexandrov proved that a convex polyhedron homeomorphic to a disk is flexible if and only if there is a deformation of the corresponding 1 2

A translation of the article [Sh1]. The convex hull of a set M is the smallest closed convex set including M .

Supplement to Chapter 5

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trivial patch. His theorem reduces the problem of flexing a convex polyhedron to that of finding a continuous deformation of a planar polygon Q with the perimeter of Q preserved and with given upper bounds of the angles of Q. This article implements this plan of A. D. Alexandrov ([A19], Chapter 5, § 2). 1.2 Let us agree to denote by the same letters the points of L and Γ that correspond to one another in gluing. Denote by Ai (i = 1, 2, . . . , n; n ≥ 0) the vertices of P on Γ and call them the vertices of type A. Let αi1 be the complete angles at these vertices on P and let αi0 be the complete angles at these vertices on Q. Given a set M of points in the (possibly many-sheeted) polygon Q, denote is clearly a polygon. by M the underlying set of points of the plane. Then Q Denote the convex hull of Q by Q. Furnish Q and Q with an orientation, choosing the positive circuit of their boundaries in which the interiors of Q and Q are to the left of the boundaries. Observe that, for this orientation of Q and Q, the direction of traversing some of their common vertices can be different (see Fig. 1). Denote by li1 and li2 the sides of Q incident to a vertex Ai of type A (it is not excluded that the same side of Q is designated as li1 and lj2 (i = j)). We assume that li1 succeeds li2 as one proceeds around Q in the positive direction. ~ A1( A1)

~ A2( A2)

B

θ12

θ11

θ21

a

θ22 α20

α10 l11

C

l12

l21

l22

Q

Fig. 1

Suppose that Q has exactly one vertex A1 of type A; moreover, the point 1 is an essential vertex of Q (see Fig. 2). Denote by l and l the sides of Q A 1 . Assume that l succeeds l in going around Q in the positive incident to A direction. Unless otherwise stated, we agree to mean by the angle formed by one ray with another the angle through which the first ray must be rotated counterclockwise in order to coincide with the second. Denote by θ1 the angle that the prolongation of the side l of Q forms with the side l11 of Q and denote by θ2 the angle that the side l12 of Q forms with the prolongation of the side l of Q (see Fig. 2). Denote by α the angle formed by l with l . Clearly, α = 2π − α10 − θ1 − θ2 .

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~ A1( A1)

θ1

l

α10

l

θ2

l11

α

l12 Q

Fig. 2

Now, suppose that Q has n ≥ 1 vertices Ai (i = 1, 2, . . . , n) of type A; passes through A 1 moreover, if n = 1, then exactly one support line a of Q i differ and lie on one support line of Q and if n > 1, then all the vertices A which we again denote by a (see Fig. 1). Choose a positive direction on a so that the polygon Q lies to the right of a. Denote by θi1 the angle formed by the negative direction of a with the side li1 of Q and denote by θi2 the angle formed by the side li2 with the positive direction of a. 1.3 In the set of convex polyhedra P homeomorphic to a disk and having boundary a simple closed polygonal line Γ , we distinguish three classes K1 , K2 , and K3 (they do not exhaust all convex polyhedra homeomorphic to a closed disk). We let K1 consist of the polyhedra P with no vertices of type A. 1.4 We let K2 consist of the polyhedra having exactly one vertex of type A and satisfying the three extra conditions: is not a multiple point of the polygonal line L 1 of Q 1.4.1 The vertex A unrolled on the plane. 1 is an essential vertex of Q, i.e., α < π.3 1.4.2 The vertex A 1.4.3 The following inequalities hold for the vertex A1 :

3

α10 < α11 ,

(1)

1 min(2π − α10 − α11 , π) < min(θ1 , θ2 ). 2

(2)

Conditions 1.4.1 and 1.4.2 imply that henceforth we can never discriminate be1 . A similar remark applies to the points Ai and A i of the polytween A1 and A hedra in the class K3 (Section 1.5).


509

1.5 We let K3 consist of the polyhedra P having n vertices Ai of type A (i = 1, 2, . . . , n), n ≥ 1, and satisfying the following conditions: i of Q is a multiple point of the polygonal line 1.5.1 None of the vertices A L unrolled on the plane. 1.5.2 If n = 1, then only one support line a of Q passes through A1 . Moreover, for arbitrary points B and C of L that are to the different sides of A1 , the given order of traversing the points A1 , B, and C determines opposite orientations of Q and Q. If n > 1, then all the vertices Ai of Q lie on one support line a of Q. For no three distinct points of L on a two of which are of type A, the given order of traversing these points determines the same orientations of Q and Q (see Fig. 1). 1.5.3 The following inequalities hold for each vertex Ai : 1 (2π − αi0 − αi1 < min(θi1 , θi2 ) (i = 1, 2, . . . , n). 2

(3)

The main result of this article is the following Theorem. A convex polyhedron P homeomorphic to an open disk and having boundary a simple closed polygonal line is rigid (not flexible) in the class of all convex polyhedra if and only if it belongs to one of the classes K1 , K2 , and K3 .4 As was demonstrated by A. D. Alexandrov (see Theorem 1 in Section 5.3), the convex polyhedra of the class K1 do not admit nontrivial isometric mappings and so are rigid. Proving the rigidity of polyhedra in the classes K2 and K3 , we will establish that none of these polyhedra is the limit of convex polyhedra nontrivially isometric to it. In addition to the main result, we thus obtain the following: A convex polyhedron is flexible if and only if it is the limit of convex polyhedra nontrivially isometric to it. A similar assertion fails for convex surfaces with boundary of bounded swerve [Sho2], [Sho3]. We emphasize once again that in this article we never consider the flexes of convex polyhedra homeomorphic to an open disk whose complements to complete convex polyhedra are not homeomorphic to a closed disk.

4

As shown in [Sho6], polyhedra in these classes are also rigid in the class of all convex surfaces. – V. Zalgaller

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2 Auxiliary Propositions To prove the rigidity of polyhedra in the classes K2 and K3 , we need some lemmas. In Lemma 1 we deal with a convex polygonal line on a plane. Lemma 1. Assume that a polygonal line AC and a straight line segment AB meet at A and have equal lengths. Let the swerves of the polygonal line AC at its vertices and the swerve of the first segment of AC at A with respect to AB have the same direction and their sum is ϕ ≤ π (see Fig. 3). Then the following inequality holds: π 1 (π − ϕ) ≤ ∠ABC < . 2 2 Proof. The length of the last segment of AC does not exceed the length of the polygonal line AC which equals the length of the straight line segment AB by hypothesis. Hence, in the triangle ABC the side AC is not longest and π ∠ABC < . 2 We now prove the inequality ∠ABC ≥

1 (π − ϕ). 2

q C

C

B

A

B Fig. 3

Since ϕ ≤ π, the polygonal line AC lies entirely to one side of the straight line AB. Draw the half-line q from A towards the half-plane containing the polygonal line so that the angle between q and the straight line segment AB be equal to ϕ/2. Let AB and AC be the respective projections of the straight line segment AB and the polygonal line AC to q. Since the angles between the segments of the polygonal line AC and q do not exceed ϕ/2, we have AC ≥ AB . Therefore,


∠ABC ≤ ∠ABB =

511

1 (π − ϕ), 2

which completes the proof of the lemma. Henceforth the images of the elements of P during the flex and the images of the elements of Q during the deformation are furnished with the subscript t. Lemma 2. Suppose that the endpoints of an open piece Γ of the boundary Γ of P are not points of type A and let ω(Mt ) denote the curvature (the area of the spherical image) of the subset Mt of P t . Then the following equality holds for the flex of P : (4) lim ω(Γt ) = ω(Γ ). t→0

Proof. Since Γt \ Γt is a closed subset of P t , by the familiar theorem (see [A15], Chapter V, § 2, Theorem 2) we have lim sup ω(Γt \ Γt ) ≤ ω(Γ \ Γ ).

(5)

t→0

Since the complement of Pt with respect to P t has no internal vertices, ω(Pt ) + ω(Γt ) = 4π. Recalling that the curvature of a polyhedron is constant during a flex, we infer that ω(Γt ) = const. By the additivity of curvature, this allows us to rewrite (5) as follows: ω(Γ ) ≤ lim inf ω(Γt ). t→0

Denoting the endpoints of Γ by B and C, we can write ⎫ lim sup ω(Γt ∪ Bt ∪ Ct ) ≤ ω(Γ ∪ B ∪ C), ⎪ ⎪ ⎪ t→0 ⎬ lim ω(Bt ) = ω(B) = 0, t→0 ⎪ ⎪ ⎪ ⎭ lim ω(Ct ) = ω(C) = 0.

(6)

(7)

t→0

Using the additivity of curvature, from (7) we deduce the inequality lim ω(Γt ) ≤ ω(Γ ).

t→0

(8)

Now, (6) and (8) imply (4), completing the proof of the lemma. Lemma 3. Suppose that the endpoints of an open piece L of the boundary L of Q are not points of type A and let ϕ(Lt ) denote the swerve of the polygonal line Lt from the side of Q. Then the following equality holds for the deformation of Q: (9) lim ϕ(Lt ) = ϕ(L ). t→0

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Proof. Denote by Γt the part of Γt corresponding to the polygonal line Lt by gluing. Denote by ϕ(Γt ) the swerve of Γt from the side of the polyhedron Pt . Then ([2], Chapter IX, § 2, Theorem 3) ϕ(Γt ) + ϕ(Lt ) = ω(Γt ) (t ≥ 0).

(10)

Since ϕ(Γt ) = const during the flex of the polyhedron, (4) and (10) yield (9). 3 Rigidity of Polyhedra of the Class K3 3.1 Let P be a polyhedron in the class K3 . For definiteness, we consider the case in which P has at least two vertices of type A. We also assume that the indices of these vertices Ai on Q increase in the positive direction of the straight line a. In this case, by 1.5.2 the indices of Ai increase in the positive direction of the polygonal line L, and the lengths of the rectilinear segments Ai Ai+1 (i = 1, 2, . . . , n; An+1 ≡ A1 ) closing the parts Ai Ai+1 of L satisfy the inequality n−1 Ai Ai+1 = A1 An . (11) i=1

3.2 Suppose for a contradiction that the polyhedron P is flexible. Then there is a deformation of Q. During the deformation of Q the inequality n−1

Ai,t Ai+1,t ≥ A1,t An,t

(12)

i=1

must clearly be preserved. Since the length of a polygonal line equals the length of the closing rectilinear segment if and only if all segments of this polygonal line lie on one straight line and have disjoint interiors, from (11) we infer that at least one of the polygonal lines A1 A2 , . . . , An A1 is deformed during the deformation of Q. Indeed, if none of the polygonal lines A1 A2 , A2 A3 . . . , An−1 An , An A1 is deformed during the deformation of Q, then the angle between the polygonal lines Ai−1 Ai and Ai Ai+1 must change for at least one vertex Ai , i = 1, 2, . . . , n. In that case, the points Ai−1,t , Ai,t , and Ai+1,t cannot lie on one straight line. Hence, the inequality n−1 i=1

Ai Ai+1 =

n−1

Ai,t Ai+1,t > A1,t An,t = A1 An

i=1

must hold, contradicting (11). Let us prove, however, that for a sufficiently small deformation of Q we have (13) Ai,t Ai+1,t ≤ Ai Ai+1 for i = 1, 2, . . . , n − 1,


A1,t ≥ A1 An .

513

(14)

Moreover, if some polygonal line Ai Ai+1 is really deformed, then the corresponding inequality (13) or (14) is strict. Then (11) combined with (13) and (14) will imply n Ai,t Ai+1,t < A1,t An,t , i=1

contradicting (12). 3.3 To prove (13) and (14), we list some properties of Q following from the definition of the class K3 . 3.3.1 Denote by β(X) the angle at which the segment Ai Ai+1 of the straight line a is seen from an arbitrary point X of the polygonal line Ai Ai+1 (i = 1, 2, . . . , n). According to 1.5.2, for i = 1, 2, . . . , n−1 the points of Ai Ai+1 can belong to a only if they are on the segment Ai Ai+1 . At the same time, the points of An A1 can belong to a only if they are outside the segment Ai Ai+1 . It follows that the following inequalities hold for a sufficiently small ε: ε < β(X) ≤ π

for

X ∈ Ai Ai+1

0 ≤ β(X) < π − ε

for

(i = 1, 2, . . . , n − 1),

(15)

X ∈ An A1 ;

moreover, by (3) 1 (2π − zi0 − αi1 ) < min(θi1 , θi2 ) − ε 2

(i = 1, 2, . . . , n).

(16)

3.3.2 By 1.5.1, there is δ1 > 0 such that above the δ1 -neighborhood of the points Ai (i = 1, 2, . . . , n) there are no other points of L but the points of the sides li1 and li2 . 3.4 By 3.3.2, there is δ > 0 such that the following conditions are satisfied: 3.4.1 The 3δ-neighborhood Ωi of the point Ai (i = 1, 2, . . . , n) is seen from all points of L, possibly except the points of the sides li1 and li2 , at an angle less than ε/2. 3.4.2 The neighborhood Ωi is seen from all points of the neighborhood Ωk (k = 1, 2, . . . , n; i = k) at an angle less than ε/2. 3.4.3 Let p denote the perimeter of Q. Then 3.4.1 implies δ/p < ε/24 < ε/2.

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3.5 The swerves of Lt and its parts are counted from the side of Qt (t ≥ 0). ∗ Let "nLi be the part of L above the δ-neighborhood of Ai , and∗ L = L \ i=1 Li . Denote the swerves of the polygonal lines Li,t and Lt by ϕ(Li,t ) and ϕ(L∗t ). Then by Lemma 3 there is τ > 0 such that for t < τ we have |ϕ(Li,t ) − ϕ(Li )| < δ/p (i = 1, 2, . . . , n);

(17)

|ϕ(L∗t ) − ϕ(L∗ )| < δ/p.

(18)

By the patching conditions, during the deformation of P the angles of Q do not increase at any points of L but possibly points of type A. Therefore, (18) implies that (19) 0 ≤ ϕ(L∗t ) − ϕ(L∗ ) < δ/p. For the points Ai,t we have αi0,t ≤ 2π − αi1 . Hence, the swerve of Li at Ai exceeds the swerve of Li,t at Ai,t by at most 2π − αi0 − αi1 . Now, (17), (19), and 3.4.3 imply that the swerves of the open polygonal lines li1,t and li2,t satisfy the inequality 0 ≤ ϕ(li1,t ) + ϕ(li2,t ) < 2π − αi0 − αi1 + ε/2.

(20)

3.6 Suppose that the polygonal line A1 A2 is deformed to the polygonal line A1,t A2,t for some t < τ . At finitely many points Xj,t of the polygonal line A1,t A2,t , its swerves ϕj,t are greater than the swerves ϕj of the polygonal line A1 A2 at the corresponding points Xj (j = 1, 2, . . . , m; m ≥ 1). For definiteness, assume the points Xj to be enumerated in the order of traversing the line from A1 to A2 ; moreover, X1 , X2 , . . . , Xs ∈ L1 ; X1 , X2 , . . . , Xm1 ∈ l11 ; Xm1 +1 , Xm1 +2 , . . . , Xm2 ∈ A1 A2 \ l11 \ l22 ; Xm2 +1 , Xm2 +2 , . . . , Xm ∈ l22 ; 0 < s < m1 < m2 < m. 3.7 Transform the polygonal line A1 A2 to the polygonal line A1,t A2,t by using the following construction (see Fig. 4). Rotate the part X1 A1 of A1 A2 , without changing its shape, clockwise around X1 through the angle ϕ1,t − ϕ1 . Denote the resulting polygonal line by A11 A2 . Then rotate the part X2 X1 A11 of A11 A2 , without changing its shape, clockwise around X2 through the angle ϕ2,t − ϕ2 . Denote the resulting polygonal line by A12 A2 . Perform the same procedure for the points X3 , X4 , . . . , Xm2 . Were this process extended to the points Xm2 +1 , . . . , Xm , the fact that the angles of Q in the δ-neighborhood of A2 can make jumps during the deformation would imply that the point A1m could occur outside the neighborhood Ω1 of A1 , complicating the further estimates.


X3

A1

a

515

A2

A11 A13 A12

X1

X2 Fig. 4

For this reason, we first rotate the part Xm A2 of A1m2 A2 , without changing its shape, counterclockwise around Xm through the angle ϕm,t − ϕm , denoting the so-obtained polygonal line by A1m2 A2m . Next we rotate the part Xm−1 Xm A2m of A1m2 A2m , without changing its shape, counterclockwise around Xm−1 through the angle ϕm−1,t − ϕm−1 . We proceed similarly for the points Xm−2 , . . . , Xm2 +1 . Finishing the construction, we obtain the polygonal line A1,m2 A2,m2 +1 congruent to A1,t A2,t . Pursuing the plan of the proof in 3.2 and grounding on the equality A1,t A2,t = A1,m2 A2,m2 +1 , we shall now prove the inequality A1,m2 A2,m2 +1 < A1 A2 .

(21)

3.8 Draw a straight line q through A1 so that the angle formed by a with q be equal to π2 + 2ε . Recall that, unless otherwise stated, by the angle formed by one straight line with another we mean the angle through which the first line must be rotated counterclockwise in order to coincide with the second. The straight line q splits the plane into two half-planes. Denote the open half-plane containing the positive direction of a by V . Denote the open disk with center A2 and radius A1 A2 by K. Also, denote by U the intersection of the neighborhood Ω1 of A1 with V and K. Let us prove that A1m2 belongs to U . 3.9 We start with proving that the points A1i belong to Ω for i ≤ m2 . For i ≤ s, the length of the segment A1 A1i is less than 2δ, since this segment closes the polygonal line A1 Xi Xi−1 . . . Ai1 whose length is less than 2δ. Estimate the length of the segment A1s A1k (s < k ≤ m2 ) closing the polygonal line A1s A1,s+1 . . . A1k . According to 3.7, we have 1 A1,j−1 A1j = 2Xj A1,j−1 sin (ϕj,t − ϕj ) (j = 1, 2, . . . , m2 ). 2 Taking the inequality Xj A1.j−1 < p and (19) into account, whence we infer A1s A1k ≤

k j=s+1

A1,j−1 A1j < p

k

(ϕj,t − ϕj ) < p(ϕ(L∗t ) − ϕ(L∗ )) < δ.

j=s+1

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These estimates imply that A1i ∈ Ω1 for i = 1, 2, . . . , m2 . 3.10 Demonstrate that A1m1 ∈ U . In view of (20), the swerve of the convex polygonal line Xm1 Xm1 −1 A1m1 is less than 2π−α10 −α11 +ε/2. By Lemma 1, −−−−−→ the angle formed by the side l11 of Q with the vector A1 A1m1 is less than π π 1 ε 2 and greater than 2 − 2 (2π − α10 − α11 ) − 4 . The angle formed by l11 with the positive direction of a equals π − θ11 . From here and (16) it follows −−−−−→ that the angle between the vector A1 A1m1 and the positive direction of a, which is treated as in the theory of projections, is less than π2 − 2ε . Hence, A1m1 ∈ V . Since A1 m1 ∈ Ω1 , we have A1m1 A2 < A1 A2 and so A1m1 ∈ K. These memberships imply A1m1 ∈ U . 3.11 Consider the isosceles triangle Xj A1,j−1 A1j with m1 + 1 ≤ j ≤ m2 . Since both points A1,j−1 and A1j belong to Ω1 , the angle between the vector −−−→ Xj A1 and the bisector of the angle of the triangle at Xj , being defined as in the theory of projections, is less than ε/8. Hence, the angle formed by −−−−−−−→ −−−→ the vector A1 Xj with the vector A1,j−1 A1j is greater than π2 − ε8 and less −−−→ than π2 + 8ε . By (15) the angle formed by the vector A1 Xj with the positive −−−−−−−→ direction of a is in the interval [0, π − ε). It follows that the vectors A1,j−1 A1j (j = m1 + 1, . . . , m2 ) deviate from the positive direction of a clockwise by an angle less than π2 − 2ε and counterclockwise by an angle less than π2 + 2ε . −−−−−−−→ Therefore, all vectors A1,j−1 A1j (j = m1 + 1, . . . , m2 ) look at the open halfplane V and, since A1m1 ∈ V , we have A1m1 +1 , . . . , A1m2 ∈ V . 3.12 We now prove that all points A1m1 +1 , . . . , A1m2 belong to U . By above, A1j ∈ Ω1 ∩ V for j = m1 + 1, . . . , m2 . Therefore, it suffices to prove that A1j A2 < A1 A2 . Two cases are possible: −−−→ −−−−−−→ (1) the angle A2 Xj A1,j−1 formed by Xj A2 with Xj A1,j−1 does not exceed π; (2) the angle A2 Xj A1,j−1 is greater than π (see Fig. 4, the angle A2 X3 A12 ). Examine the first case. From (15) and 3.9, it follow that ∠A2 Xj A1,j−1 > (3/4)ε. After the transformations of 3.7, the polygonal line Xj Xj−1 A1,j−1 rotates around Xj clockwise through an angle less than ε/2. Therefore, ∠A2 Xj A1j < ∠A2 Xj A1,j−1 . Looking at the triangles A2 Xj A1,j−1 and A2 Xj A1j in which A2 Xj is a common side, Xj A1,j−1 and Xj A1j are equal sides, and ∠A2 Xj A1,j−1 > ∠A2 Xj A1j , we see that A2 A1j < A2 A1,j−1 . Examine the second case. Denote the intersection point of the straight lines A2 Xj and q by Y and denote by Z the foot of the perpendicular −−−→ dropped from Xj on q. Since A1,j−1 ∈ Ω1 , the angle formed by Xj A1 −−−−−−→ −−−→ −−→ with Xj A1,j−1 is less than ε/4. The angle formed by Xj A1 with Xj Z is −−−→ −−−−−−→ not less than ε/2. Hence, the angle formed by Xj A2 with Xj A1,j−1 is −−−→ −−→ greater than π but less than the angle formed by Xj A2 with Xj Z. Moreover, A1,j−1 ∈ V ; so this points lies inside the rectangular triangle Xj Y Z.


517

Therefore, A1,j−1 Xj < Y Xj . Since Y A2 ≤ A1 A2 , we finally obtain A1j A2 ≤ A1j Xj + Xj A2 = A1,j−1 Xj + Xj A2 < Y Xj + Xj A2 ≤ A1 A2 . The above-examined cases imply that for j = m1 + 1, m1 + 2, . . . , m2 we always have A1j < A1 A2 or A1j A2 < A1,j−1 A2 . Since A1,m1 A2 < A1 A2 , for j = m1 + 1, . . . , m2 we have A1j A2 < A1 A2 , i.e., A1,m1 +1 , . . . , A1m2 ∈ U . 3.13 By analogy to 3.9 and 3.10, we prove that A2,m2 +1 ∈ Ω2 and that the −−−−−−−→ angle between the vector A2 A2,m2 +1 and the negative direction of a, defined as in the theory of projections, is less than π2 − 2ε . 3.14 In order to prove (21), consider the triangle T = A2 A2,m2 +1 A1m2 . Since A2,m2 +1 ∈ Ω2 , the angle of T at the vertex A1m2 is less than ε/4. By 3.9 and 3.13, the angle of T at A2 is less than 12 π − 34 ε. Hence, the angle of T at A2,m2 +1 is obtuse and therefore A1m2 A2,m2 +1 < A1m2 A2 . By virtue of 3.12, from here we deduce (21). In other words, A1,t A2,t < A1 A2 . 3.15 We have thus proved that, for a sufficiently small deformation of the polygonal line A1 A2 , the corresponding inequality (13) becomes strict. By perfect analogy we prove that, for a sufficiently small deformation of the polygonal lines A2 A3 , . . . , An−1 An and An A1 , the corresponding inequalities (13) and (14) become strict. This concludes the proof of the rigidity of the polyhedra P ∈ K3 . 4 Rigidity of Polyhedra of the Class K2 4.1 Let P be a polyhedron of the class K2 and let Q be the trivial patch of P . Suppose that P is flexible and consider the patch Qt of Pt for an arbitrary but fixed t > 0. At finitely many points Xj,t , the swerves ϕj,t of the polygonal line Lt are greater than the swerves ϕj of L at the corresponding points Xj (j = 1, 2, . . . , m; m ≥ 1). Assume the points Xj to be enumerated in the positive direction of L so that X1 , X2 , . . . , Xm1 ∈ l11 ; Xm1 +1 , . . . , Xm2 ∈ L = L \ l11 \ l12 ; Xm2 +1 , . . . , Xm ∈ l12 (0 ≤ m1 ≤ m2 ≤ m). The angle α10,t satisfies the inequality α10 < α10,t ≤ 2π − α11 . Transform the polygonal line L to Lt . To this end, we first tear L at A1 and then accomplish the construction of 3.7, rotating the polygonal line A1,j−1 Xj−1 Xj (A10 ≡ A1 ) clockwise around all the points Xj (j = 1, 2, . . . , m). The resulting polygonal line is congruent to Lt . Hence, the following inequality must hold: −−−−−→ −−−−−−−→ −−−−−−−→ (22) A1 A1,m1 + A1,m1 A1m2 + A1,m2 A1m = 0. We now prove, however, that if the deformation of L is small enough and preserves the patching conditions, then (22) cannot hold, since the three

518

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−−−−−→ −−−−−−−→ −−−−−−→ vectors A1 A1m1 , A1m1 A1m2 , and A1m2 A1m , at least one of which is nonzero, point to a single open half-plane. For brevity, we henceforth denote the vectors −−−−−−→ −−−−−→ −−−−−−−→ A1 A1m1 , A1m1 A1m2 , and A1m2 A1m by b1 , b, and b2 . 4.2 To prove the falsity of (22), we indicate some properties of the polygon Q. 4.2.1 By 1.4.2 and 1.4.3, there is ε > 0 such that the following inequalities are valid: (23) α < π − ε, α10 < α11 − ε,

(24)

1 min(2π − α10 − α11 , π) < min(θ1 , θ2 ) − ε. 2

(25)

4.2.2 Denote the successive segments of Γ by g1 , g2 , . . . , gk .5 Moreover, assume that the segments g1 and gk issue from A1 and are glued to the respective sides l11 and l12 of Q. Since L is a simple polygonal line, there is δ1 > 0 such that the spatial distances between two arbitrary nonadjacent segments gi and gj of Γ are greater than δ1 . 4.2.3 By 1.4.1, there is δ2 > 0 such that above the δ2 -neighborhood of A1 there are no points of the polygonal line L = L \ l11 \ l12 . Therefore, we can find a positive δ < (1/4) min(δ1 , δ2 ) such that the 3δ-neighborhood of A1 is seen from all points of L at an angle less than ε/2. Denote the part of L above the δ-neighborhood of A1 by L1 and denote L \ L1 by L∗ . 4.3 By continuity of the flex of P and Lemma 3, there is τ > 0 such that the following two conditions hold for t < τ : 4.3.1 The spatial distances between arbitrary polygonal lines gi,t and gj,t appearing from nonadjacent segments gi and gj of Γ during the flex of P is greater than 3δ. 4.3.2 The swerves of the polygonal lines L∗t and L1,t satisfy the inequalities 0 ≤ ϕ(L∗t ) − ϕ(L∗ ) < δ/p, −(δ/p) < ϕ(L1,t ) − ϕ(L1 ) ≤ 0, where p is the perimeter of Q. From now on, we assume that the value of t fixed in 4.1 satisfies the inequality t < τ . 5

There is no loss of generality in assuming k > 3 since, if need be, we can always introduce extra vertices with angles equal to π.


519

4.4 Let us find the direction of the vector b. To this end, we first observe that all the points A1j (j = 1, 2, . . . , m) belong to the 3δ-neighborhood of A1 . The proof of this claim bases on 4.3.2 and proceeds in the same way as in 3.9. Next, like in 3.11, we prove that for j = m1 + 1, . . . , m2 the angle −−−→ −−−−−−−→ formed by the vector A1 Xj with the vector A1,j−1 A1j is greater than π2 − 4ε −−−→ π ε and less than 2 + 4 . Since all the vectors A1 Xj point to the interior of the −−−−−−−→ angle α , the angles formed by the side l of Q with the vectors A1,j−1 A1j (j = m1 + 1, . . . , m2 ) are greater than π2 − 4ε and less than α + π2 + 4ε . −−−−−−−→ Therefore, by (23) all the vectors A1,j−1 A1j with j = m1 + 1, . . . , m2 point to one open half-plane. Hence, the vector b equals zero if and only if all the −−−−−−−−−→ −−−−−−−−−→ vectors A1m1 A1,m1 +1 , . . . , A1m2 −1 A1,m2 are equal to zero, i.e., if m1 = m2 . However, if m1 = m2 , then b = 0 and the angle β formed by l with b meets the inequalities π ε π ε − < β < + α + . (26) 2 4 2 4 4.5 Before finding the directions of the vectors b1 and b2 , we focus the reader’s attention on the following difficulty. If α10,t − α10 > π (which may hold only when α10 + α11 < π), then the swerve of one of the polygonal lines l11,t and l12,t (say, l11,t ) could occur greater than π. Furthermore, a priori it is not excluded, owing to the fact that Qt has many sheets, that the polygonal line l11,t unrolls on the plane with selfintersections. In that case, the point A1m1 of the polygonal line Xm1 Xm1 −1 A1m1 obtained by the construction of 4.1 could occur on the right of l11 if one looks from A1 along l11 . Were this so, the vectors b1 , b2 , and b would not point in general to one open half-plane. To bypass this difficulty, we inspect the polygonal lines l11,t and l12,t . We shall distinguish between two cases: (1) the polygonal line Γt is a simple closed polygonal line and (2) at least two distinct points of Γt are glued to one another. We examine the first case in 4.6 and the second in 4.7. 4.6 Let us prove that in the first case the polygonal lines l11,t and l12,t do not selfintersect when Qt is unrolled on a plane. Assume the converse. Let two distinct points Mt and Mt of l11,t are above the same point of the plane. Then the swerve of the part Mt Mt of l11,t must be at least π. Since the swerve of the segment M M of the side l11 was equal to zero, this fact and 4.3.2 imply that the points M and M belong to the segment l11 of l11 issuing from A1 and having length 2δ. Furnish Qt with the intrinsic metric, setting as usual the distance between two points of Qt to equal the length of the shortest polygonal line joining the points and lying entirely in Qt (since Γt is a simple closed polygonal line, Qt is a polygon homeomorphic to a closed disk). Draw the shortest arc in Qt between Mt and Mt . Since M and M are distinct points, the length of this shortest arc is greater than zero. On the other hand, since the length of is at most 2δ, the length of the shortest arc the polygonal line Mt Mt ⊂ l11,t in Qt between Mt and Mt does not exceed 2δ all the more. The beginning

520

12 Supplements

and end of the shortest arc are above the same point of the plane and its length is greater than zero. Therefore, the swerve of the shortest arc must be at least π. The shortest arc may have swerves only at those points of the boundary of Qt where the swerves of the polygonal line Lt are negative. The polygonal line Lt may have negative swerves only at the points A1,t and the points B1,t , . . . , Bs,t corresponding to those vertices B1 , . . . , Bs of L at which L had negative swerves. Since every essential vertex of L corresponds to some essential vertex of Γ and since the distance between two points of Lt (t ≥ 0) in the metric of Qt is greater than or equal to the spatial distance between the corresponding points of Γt , by 4.3.1 the minimum of the distances in the metric of Qt between the point A1,t and the points Bi,t is greater than 3δ. The distances in the metric of Qt from the points Mt and Mt to the point A1,t do not exceed 2δ; therefore, the distances from the points Mt and Mt to the points Bi,t (i = 1, 2, . . . , s) are greater than δ. Since the length of the shortest arc in Qt between Mt and Mt is at most 2δ, this shortest arc cannot pass through any of the points B1,t , . . . , Bs,t . Hence, the shortest arc has a fracture only at the point A1,t . However, this is impossible, for the swerve of the shortest arc between the points Mt and Mt is at least π, whereas the angle α10,t is less than 2π. Thus, if Γt is a simple closed polygonal line, then l11,t and l12,t are planar polygonal lines without multiple points. Since l11,t and l12,t have positive swerves from the side of Qt at all their vertices, it follows that the point A1m1 lies to the left of l11 and the point A1m2 lies to the right of l12 (when one looks from A1 along l11 and l12 ). 4.7 Examine the second case. Denote by g1 and gk the segments of the sides g1 and gk of Γ which issue from A1 and have length 2δ. We begin with proving that the points of Γt may be glued to one another only if they are on the polygonal lines g1,t and gk,t and moreover only on one of them (g1,t or gk,t ). First of all, observe that the points of nonadjacent polygonal lines gi,t and gj,t cannot be glued to one another, since by 4.3.1 the spatial distances between them are greater than 3δ. Now, prove that two adjacent polygonal lines gi,t and gi+1,t (gk+1,t ≡ g1,t ) have no common points but a common end. Assume for a contradiction that an interior point Mt of gi,t is glued to an interior point Mt of gi+1,t . Since the complete angles of the polyhedron Pt at Mt and Mt are equal to π, for Pt to be convex it is necessary that the segments of gi,t issuing from Mt be glued pairwise to the segments of gi+1,t issuing from Mt . The directions of the glued segments must be opposite; otherwise P t would be unorientable and, in consequence, nonconvex. It follows that the polygonal lines gi,t and gi+1,t are glued to a doubly-covered polygonal line.6 Moreover, either the common vertex of gi−1,t and gi,t occurs on gi+1,t or the common vertex of 6

It is not excluded that some part of one of them is glued to itself.


521

gi+1,t and gi+2,t occurs on gi,t . Anyway we come to a contradiction with 4.3.1 (see a remark in 4.2.2). Now, suppose that two points Mt and Mt of the same polygonal line gi,t are glued to one another. In that case, the swerve of the part Mt Mt of gi,t on the polyhedron P t from the side of Qt must be at least π. Since the swerve of the segment M M of the side gi of Γ on the polyhedron P from the side of Q was equal to zero, from 4.3.2 we infer that such a case is possible only or gk,t . Since only one of the when the two points Mt and Mt belong to g1,t polygonal lines g1,t and gk,t has the swerve on P t from the side of Qt greater or gk,t may be glued to one another. than or equal to π, the points of only g1,t Thus, the claim at the beginning of the current subsection is proved. For example, let points Mt and Mt of g1,t be glued to one another. Then two cases are possible: (1) One of the points Mt and Mt , say Mt , is A1,t . (2) Both points Mt and Mt are interior points of g1,t . Consider the second case. Since the complete angles of Pt at Mt and are equal to π, for Pt to be convex it is necessary that the segments of g1,t issuing from Mt be glued pairwise to the segments issuing from Mt ; moreover, the directions of the glued segments must be opposite. This means is folded at the middle of Mt Mt so that one of that the polygonal line g1,t its parts is glued to the other. The point A1,t is glued to some point Mt of g1,t which is interior for g1 . The above implies that if the polygonal line Γt contains glued points then, on Lt , the point A1,t is glued to some interior point of l11,t or some interior point of l12,t . For example, assume that A1,t is glued to Mt ∈ l11,t . Since in this case the swerves of the polygonal lines l11,t \ A1,t Mt and l12,t are positive and do not exceed π − α10 − α11 , this obviously implies that A1m1 lies either on l11 or to the left of l11 , whereas A1m2 lies to the right of l12 . Similarly, if A1,t is glued to a point Mt ∈ l12,t then A1m2 lies either on l12 or to the right of l12 , whereas A1m1 lies to the left of l11 . Mt

4.8 We recall that by the angle formed by one direction with another we mean the angle through which we must rotate the first direction counterclockwise in order to make it coincide with the second. The directions of the sides of Q and Q incident to the vertex A1 are defined to be the directions of the rays issuing from A1 and containing these sides. Denote the swerves of the open polygonal lines l11,t and l12,t by ϕ(l11,t ) and ϕ(l12,t ). If ϕ(l11,t ) = 0, then m1 = 0, b1 = 0, and, as shown above, the point A1m1 either lies to the left of l11 or is on l11 . Hence, by Lemma 1 the angle β1 formed by the side l11 with the vector b1 satisfies the inequality π 1 π − min(ϕ(l11,t ), π) ≤ β1 < . 2 2 2

(27)

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12 Supplements

Similarly, if ϕ(l12,t ) = 0, then b2 = 0 and the angle β2 formed by the side l12 with the vector b2 satisfies the inequality 1 π π < β2 ≤ + min(ϕ(l12,t ), π). 2 2 2

(28)

The angle formed by the side l of Q with the side l11 of Q equals π − α10 − θ2 . From here and (27) we obtain the following estimate for the angle β1 formed by the side l1 with the vector b1 : 3 1 3 − α10 − θ2 − min(ϕ(l11,t ), π) ≤ β1 < π − α11 − θ2 . 2 2 2

(29)

A similar estimate for the angle β2 formed by the side l with the vector b2 is based on (28) and looks like 3 3 1 π − θ2 < β2 ≤ π − θ2 + min(ϕ(l12,t ), π). 2 2 2

(30)

From (29) and (30) we readily infer that β2 > β1 ,

(31)

and the angle β12 formed by the vector b1 with the vector b2 satisfies the following estimate from above: β12 = β2 − β1 ≤ α10 +

1 min(ϕ(l11,t ), π) + min(ϕ(l12,t ), π) . 2

Using the inequality ϕ(l11,t ) + ϕ(l12,t ) ≤ 2π − α10 − α11 , we reduce the last estimate to the form 1 α11 − α10 . β12 ≤ α10 + (2π − α10 − α11 ) = π − 2 2 Taking (24) into account, we derive that β12 < π − ε/2.

(32)

4.9 Now, let us prove that the vectors b, b1 , and b2 point to one open halfplane. By (31) we have the following three possibilities for the angles β, β1 , and β2 formed by the side l with the vectors b, b1 , and b2 : (1) β1 ≤ β ≤ β2 ; (2) β ≤ β1 < β2 ; (3) β1 < β2 ≤ β. In the first case, the vector b points to the interior of the angle β12 formed by b1 with b2 . In accordance with (32), this angle is less than π. Hence, b, b1 , and b2 point to one open half-plane. In the second case, b1 points to the interior of the angle formed by b with b2 . This angle equals β2 − β. By (26) and (30),


β2 − β < π − θ 2 +

523

1 1 min(ϕ(l12,t ), π) + ε. 2 4

From here and (25) we find that 3 β2 − β < π − ε. 4 Hence, b, b1 , and b2 point to one open half-plane. The third case is settled in the same way as the second. Thus, in all cases the vectors b, b1 , and b2 point to one open half-plane and, since the patch Qt is nontrivial, at least one of them is nonzero. So (22) cannot hold, which concludes the proof of the rigidity of polyhedra of the class K2 . 5 Flexibility of Polyhedra Not Belonging to the Classes K1 , K2 , and K3 If a polyhedron P belongs to none of the classes K1 , K2 , and K3 , then at least one of the nine conditions listed below is obviously valid. In these conditions, Ai , Ak and Al stand for vertices of type A, and in parentheses we indicate the corresponding defining property of the classes K1 , K2 , and K3 which is opposite to the condition in question. (1) At least three vertices Ai , Ak , and Al of Q do not lie on one straight line (1.5.2). (2) At least one of the straight lines Ai Ak passes through interior points of Q (1.5.2). (3) At least one of the sides Ai Ak of Q contains a point B of the boundary of Q such that the given order of traversing the points Ai , Ak , and B determines the same orientations of Q and Q (1.5.2). (4) At least one of the vertices Ai of Q belongs to the interior of Q (1.4.2, 1.5.2). (5) For at least one of the vertices Ai , there is exactly one support line of Q; moreover, there are points B and C of L lying to the different sides of Ai and the given order of traversing the points A, B, and C determines the same orientations of Q and Q (1.5.2). i of Q there is a boundary point B (6) Above at least one of the vertices A of Q other than Ai (1.4.1, 1.5.1). (7) Inequality (1) fails. (8) Inequality (2) fails. (9) At least one of the inequalities (3) fails. Each of the conditions (1)–(9) is sufficient for the flexibility of P in the class of all convex polyhedra.

524

12 Supplements

The sufficiency of (1) and (7) was proved by A. D. Alexandrov ([1], Chapter 5, § 2). We now prove the sufficiency of (5), (6), and (8).7 Conditions (2), (3), and (4) are examined by analogy to (5), and (9) is analogous to (8). Sufficiency of (5). Take a point D on L so that the points Ai and D separate B and C. Start deforming Q, without changing the shapes of the polygonal lines Ai B, BD, DC, and CA, by shortening the diagonal Ai D of the quadrilateral Ai BDC. Then by condition (5) the angles of Q at B, C, and D decrease. Since the angle at Ai may increase within some limits without violating the patching conditions, the sufficiency of condition (5) is established. Sufficiency of (6). The points Ai and B split L into two parts L1 and L2 . Start deforming Q by rotating L1 and L2 , without changing their shapes, around the points Ai and B so that the angle of Q at B decreases. Since a sufficiently small deformation does not lead to violating the patching condition at Ai , the sufficiency of condition (6) is established. Sufficiency of (8). Assume that the angle β1 between the side A1 C of Q and the prolongation of the side A1 B of Q satisfies the inequality θ−1≤

1 min(2π − α10 − α11 , π). 2

Take a point X on the side A1 C and consider the segment A1 X symmetric to A1 X with respect to the straight line A1 B. Next, rotate the part BCX of L around B through the angle XBXt without changing the shape of this part. If the segment A1 Xt is small enough, then the condition θ − 1 ≤ π/2 implies that the resulting closed polygonal line bounds some polygon Qt . The angles of Qt at all points except A1,t do not exceed the angles at the corresponding points of Q. Since α10,t = α10 + 2θ1 ≤ 2π − α11 , the patching condition is satisfied at the point A1,t as well. The sufficiency of condition (8) is established. In conclusion, the author finds pleasure in expressing his gratitude to A. V. Pogorelov, who turned the author’s attention to the above-considered problem and made many apt observations.

7

Condition (8) was formulated by A. D. Alexandrov [A19] as a problem.

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Usmanov Z.D.: Infinitesimal bendings of surfaces of positive curvature with a flat point. Diff. Geom. Banach Center Public. 12 1984, pp. 241–272 (Russian) Venkov B.A.: On one class of Euclidean polyhedra. Vestnik Leningrad. Univ. (1954), No. 2, 11–31 (Russian) Vereschagin B.M.: Reconstruction of a closed convex surface given a function of Gaussian curvature. Problems of Global Geometry. Leningrad. Ped. Inst., Leningrad 1979, pp. 7–12 (Russian) Verner A.L.: Unique determination and rigidity of unbounded complete convex polyhedra in Lobachevski˘ı space. Vestnik Leningrad. Univ. (1960), No. 13, 90–93 (Russian) Verner A.L.: On extrinsic curvature of convex surfaces in spaces of constant curvature. Izv. Vyssh. Ucheb. Zaved. Matematika (1960), No. 1, 58–68 (Russian) Verner A.L.: On unbounded convex surfaces in Lobachevski˘ı space. Izv. Vyssh. Ucheb. Zaved. Matematika (1960), No. 6, 50–61 Verner A.L.: Existence and uniqueness of an unbounded convex surface with prescribed extrinsic curvature in Lobachevski˘ı space. Sibirsk. Mat. Zh. 2 (1961), No. 1, 20–35 (Russian) Verner A.L.: Reconstruction of a complete convex surface in Lobachevski˘ı space given its extrinsic curvature. Uchenye Zapiski Leningrad. Gos. Ped. Inst. 218 (1961), 119–126 (Russian) Verner A.L.: Reconstruction of a convex surface in Lobachevski˘ı space given its support function. Vestnik Leningrad. Univ. (1962), No. 13, 145–148 (Russian) Verner A.L. and Misikov B.R.: Renewal of a convex surface homeomorphic to the “Klein bottle” according to the integral extrinsic curvature in a Riemannian space of constant negative curvature. Global and Riemannian Geometry. Leningrad. Ped. Inst., Leningrad 1983, pp. 3–5 (Russian) Vladimirova S.M. and Volkov Yu.A.: Bending of infinite convex surfaces of Lobachevski˘ı space. J. Sov. Math. 8 (1978), 438-443 ` Vinberg E.B.: Discrete groups generated by reflections in Lobachevski˘ı spaces. Mat. Sb. 72 (1967), No. 3, 471–488 (Russian) Volkov Yu.A.: On deformations of a convex polyhedral angle. Uspekhi Mat. Nauk 11 (1956), No. 5, 209–210 (Russian) Volkov Yu.A.: Existence of a polyhedron with prescribed development. Dissertation. Leningrad. Univ., Leningrad 1955 (Russian) Volkov Yu.A.: An estimate for the variation of a solution to an equation of the form f (z1 , . . . , zn ) det zij = h(x1 , . . . , xn ) in dependence on the variation of its right-hand side. Vestnik Leningrad. Univ. (1960), No. 13, 5–14 (Russian) Volkov Yu.A.: Existence of convex polyhedra with prescribed development. I. Vestnik Leningrad. Univ. (1960), No. 19, 75–86 (Russian). Engl. transl.: this book, pp. 492–505 Volkov Yu.A.: Stability of a solution to the Minkowski problem. Vestnik Leningrad. Univ. (1963), No. 1, 33–43 (Russian) Volkov Yu.A.: An estimate for the deformation of a convex surface in dependence on the variation of its intrinsic metric. Ukrain. Geom. Sb. 5–6 (1967), 44–69 (Russian). Engl. transl.: this book, pp. 463–491 Volkov Yu.A. and Mekhtiev M.G.: Bendings of unbounded convex surfaces in Lobachevski˘ı space. Vestnik Leningrad. Univ. (1968), No. 19, 30–43 (Russian)

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Volkov Yu.A. and Oliker V.I.: On uniqueness of a solution to the Christoffel problem for nonclosed surfaces. Math. Notes 8 (1970), 611–614 Volkov Yu.A. and Podgornova E.G.: Existence of a convex polyhedron with prescribed development. Uchenye Zapiski Tashkent. Gos. Ped. Inst. 85 (1971) (Russian) ¨ Weil H.: Uber die Bestimmung einer geschlossenen konvexen Fl¨ ache durch ihr Linienelement. Vierteljahrsschr. der Naturforschenden Gesellschaft in Z¨ urich (1916) 40–72 ¨ Weil H.: Uber die Starrheit der Eifl¨ achen und konvexen Polyeder. Sitz. Berichte Preuss. Akad. Wiss. 1 (1917), Halbjahr. XVI, 250–266 Whitely W.: Infinitesimally rigid polyhedra. I: Statics of frameworks. Trans. Amer. Math. Soc. 285 (1984), No. 2, 431–465; Infinitesimally rigid polyhedra.II: Modified spherical frameworks. Trans. Amer. Math. Soc. 306 (1988), No. 1, 115–139 Wunderlich W.: Wackeldodekaeder. Ber. Math. Statist. Sekt. Forschungszent. Graz. (1980), No. 140–150 (Ber. 149), 1–8 ¨ Wunderlich W.: Neue Wackelikosaeder. Anz. Osterr. Akad. Wiss. Math.naturwiss. Kl. (1980), No. 1–9, 28–33 ¨ Wunderlich W.: Wackelige Doppelpyramiden. Anz. Osterr. Akad. Wiss. Math.-naturwiss. Kl. 117 (1980), No. 1–9, 82–87 Wunderlich W.: Wackelikosaeder. Geom. Dedic. 11 (1981), No. 2, 137–146 Wunderlich W.: Kipp-Ikosaeder. Elem. Math. 36 (1981), No. 3, 153–158; 37 (1982), No. 3, 84–89 Wunderlich W.: Fast bewegliche Oktaeder mit zwei Symmetrieebenen. Rad. Jugosl. Akad. Znan. Umjet. 428, Mat. Znan. 6 (1987), 129–135 Wunderlich W.: Shaky polyhedra of higher connection. Publ. Math. 37 (1990), No. 3–4, 355–361 Zalgaller V.A.: On deformations of a polygon on a sphere. Uspekhi Mat. Nauk 11 (1956), No. 5, 177–178 (Russian) Zalgaller V.A.: Convex polyhedra with regular faces. Sem. Steklov Math. Inst. Leningrad, Vol. 2. Consult. Bureau, NU 1969 Zalgaller V.A.: A continuously flexible polyhedron. Kvant (1978), No. 9, 13–19 (Russian) Zalgaller V.A.: An algorithm for finding conditions of existence and constructing a polyhedron with prescribed face directions and face perimeters. Sib. Math. J. 37 (1996), No. 6, 1148–1152 Zalgaller V.A.: Some bendings of a long cylinder. J. Math. Sci. 100 (2000), No. 3, 2228–2238 Zhitomirski˘ı O.K.: Versch¨ arfung eines Satzes von Woronoi. Zh. Leningrad. Mat. Obsch. (1929), 131–151 Zil berberg A.A.: On existence of bounded convex polyhedra with prescribed boundary and curvatures at vertices. Izv. Akad. Nauk Kaz. SSR Ser. Mat. Mech. 10 (1962), 30–40 (Russian) Zil berberg A.A.: On existence of closed convex polyhedra with arbitrary curvatures at vertices. Uspekhi Mat. Nauk 17 (1962), No. 4, 119–126 (Russian) Zil berberg A.A.: On existence of closed convex polyhedra with prescribed curvatures at vertices in n-dimensional Euclidean and Lobachevski˘ı spaces. Vestnik Leningrad. Univ. 1 (1963), 48–52 (Russian) Zil berberg A.A.: On existence of a closed convex surface with integral curvature topologically translated to a sphere. Trudy Sector Mat. Mech. Akad. Nauk Kaz. SSR 2 (1963), 73–77 (Russian)

[VP]

[W1]

[W2] [Wh]

[Wu1] [Wu2] [Wu3] [Wu4] [Wu5] [Wu6] [Wu7] [Z1] [Z2] [Z3] [Z4] [Z5] [Zh] [Zil1] [Zil2] [Zil3] [Zil4]

Subject Index

Angle – complete, 42, 54 – – dihedral 465 – limit, 26, 27, 82 – – generator of 178 – polyhedral, 14 – – deformation of 404 – – degenerate 162 – straightening of, 73 Area function 137 – spherical, 137 Blaschke–Herglotz method Boundary 143 Brunn – inequality, 369 – Theorem, 369

– conditional, 401 – integral, 397 – of development, 55 – of geodesic polygon, 76 – of part of polyhedron, 42 – of polyhedron, 42 – total, 82, 105 – translated, 397 Cutting and gluing 51

321

Cap 184, 231 – generalized, 465 – – convex 465 – – height of 465 – unbounded, 232 Cauchy – Lemma, 87, 92 – – Strong 103, 411 – Theorem, 92 Complex 197 Connelly–Sullivan conjecture 187 Continuation in parameter 98 Convex body 11, 82 – approximation of, 83 – limit angle of, 82 Convex hull 21 – closed, 22 Convexity 8, 11, 12 – on sphere, 14 Covering 148 – multiplicity of, 148 Curvature – at vertex, 42, 55, 102

Darboux equation 505 Defining parameters 425 Deformation – of development, 425 – of polygon, 439 – of polyhedral angle, 404 Development 50 – boundary edge of, 229 – boundary of, 50 – boundaryless, 50 – defining parameters of, 425 – deformation of, 425 – edge of, 50 – natural, 425 – of positive curvature, 55, 193 – of strictly positive curvature, 193 – realizable, 194 – stationary, 426 – structure of, 197, 425 – vertex of, 50 Distance 51, 101, 265 Distance function 396 Domain Invariance Theorem 93, 147 Dupin indicatrix 304 Edge 7 – boundary, 10, 229 – bounded, 8 – fictitious, 164 – genuine, 100, 164 – of development, 50

538

Subject Index

– of net, 61 – unbounded, 8 Euler – characteristic, 65 – condition, 99 – Converse Theorem, 63 – formula, 89 – relation, 63 – Theorem, 61 Face 7, 284, 360 – bounded, 8 – can be placed in another face, 107 – direction of, 18 – fictitious, 404 – genuine, 360, 404 – limit angle of, 27 – unbounded, 8 Flex 106, 246 Four vertex theorem 280 Function – essentially monotone, 129, 130, 440 – monotone, 108 Funnel 222, 425 Gluing 52 Gluing Theorem

264

Helly Theorem 340, 341 Homeomorphism 56, 146 Infinitesimal deformation 403 Intrinsic – geometry, 51 – – of development 72 – – of surface 83, 136, 265 – metric, 51, 265 – – of development 51 – – of polyhedra 101 – – of surface 83 Isometry – between developments, 51 – between polyhedra, 52 Jordan Theorem

61, 69

Layerwise construction 356 Legendre transformation 124 Liebmann method 301 Lindel¨ of Theorem 108, 359 Linear combination 281 Manifold

93, 146

– of developments, 197 – of developments of strictly positive curvature, 198 – of polyhedra, 203 Mapping Lemma 93 Maximum principle 309 Metric – intrinsic, 101, 265 – of positive curvature, 102 – of surface, 136 – polyhedral, 102 – – of positive curvature 55 – realization of, 137, 266 Minkowski – Existence Theorem, 96, 108 – inequality, 361 – problem, 347 – Theorem, 311, 361, 365 – Uniqueness Theorem, 97, 107, 290 Mixed volume 361, 365, 367, 369, 376 Monge–Ampère equation 400 Near isometry 466 Net 61 – edge of, 61 – of edges, 87 – region of, 61 – vertex of, 61, 87 Normal 18 – mapping, 400 Parallelogon 351 Parallelohedron 349 Patch 248 – trivial, 248 Polar – correspondence, 46, 48 – polygon, 45 – polyhedron, 45 – transformation, 45 Polygon 7 – bounded, 7 – deformation of, 439 – doubly-covered, 99, 163 – geodesic, 76 – of zero curvature, 79 – placed in another polygon, 271 – spherical, 14 – unbounded, 7 Polyhedron 7 – bounded, 8 – closed, 10 – complete, 10

Subject Index

539

– containing straight line, 10 – convex, 8 – degenerate, 99, 163 – edge of, 7 – face of, 7, 360 – flexible, 187, 246, 506 – generalized, 485 – – convex 485 – in n-dimensional space, 83, 191 – – rigidity of 437 – in hyperbolic space, 85, 190, 261 – – rigidity of 437 – in spherical space, 85, 190, 261 – – rigidity of 437 – infinitesimally rigid, 125, 387 – limit angle of, 26 – locally convex, 168 – polar, 45 – recession cone of, 26 – rigid, 246 – solid, 7 – – convex 8 – support number of, 19, 113, 359 – support plane of, 17 – translate of, 107 – triangulated, 203 – unbounded, 8, 11 – – funnel of 425 – vertex of, 7 – with boundary, 10, 81 – with cut, 259 – without vertices, 10 Position vector 281 Positive curvature condition 99 Prism 10 – projecting, 465 – zonal, 354 Projection of development 465

– of convex polyhedra, 421 – – with given face directions 445 – – with stationary development 429 Rule for gluing 49, 425

Ray 178, 218 Riemann method 320 Rigidity 125, 403

Width

Separation Lemma 11 Shortest arc 52, 72 Simplex 148 Sperner Lemma 153 Spherical image 39 Stationary quantity 125, 386, 403 Support – function, 342 – mapping, 400 – number, 19, 113, 359 – plane, 16 Topological – property, 56 – space, 142 – – subspace of 142 Topology 141 Triangulation 153, 203 Unrolling

233

Vertex 7 – boundary, 25 – complete angle at, 42 – curvature at, 42, 55 – fictitious, 164 – genuine, 55, 164 – internal, 25 – of development, 50 – of net, 61 Vorono˘ı’s Theorem 359 Vorono˘ı–Dirichlet region 359

Zone

313 353

Convex Polyhedra

Convex polyhedra

Convex polyhedra