Differential Geometry and Topology of Curves

DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES Yu Aminov Institute for Low Temperature Physics and Engineering Kharkov, ...

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DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CURVES

Yu Aminov Institute for Low Temperature Physics and Engineering Kharkov, Ukraine

CRC P R E S S Aoca Raton London New York Washington, D.C. © 2000 CRC Press

Copyright (Q 2000 Gordon and Breach Science Publishers imprint.

All rights reserved. No part of this book may be reproduced or utilized in any form or by any means. electronic or mechanical, including photocopying and recording, or by any information storage or retrieval system, without permission in writing from the publisher. Printed in Singapore.

Reprinted 2003 by Taylor & Francis I I New Fetter Lane London

EC4E 4EE

Transferred to Digital Printing 2003 Printed in Great Britain by Biddies Short Run Books. King's Lynn

British Library Cataloguing in Publication Data

A catalogue record for this book i available from the British L1brary. ISBN : 90-5699-091-8

Contents Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Definition of a Curve

...................................

vii

1

Vector-valued Functions Depending on Numerical Arguments . . . . . . .

5

The Regular Curve and its Representations . . . . . . . . . . . . . . . . . . . .

9

Straight Line Tangent to a Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . .

13

Osculating Plane of a Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

17

The Arc Length of a Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

21

The Curvature and Torsion of a Curve . . . . . . . . . . . . . . . . . . . . . . . .

27

Osculating Circle of a Plane Curve . . . . . . . . . . . . . . . . . . . . . . . . . . .

37

Singular Points of Plane Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

41

Peano's Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

49

Envelope of the Family of Curves . . . . . . . . . . . . . . . . . . . . . . . . . . .

51

Frenet Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

53

Determination of a Curve with Given Curvature and Torsion . . . . . . .

59

Analogies of Curvature and Torsion for Polygonal Lines . . . . . . . . . . .

63

© 2000 CRC Press

CONTENTS

vi

15 Curves with a Constant Ratio of Curvature and Torsion . . . . . . . . . .

65

16 Osculating Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

69

17 Special Planar Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

73

18 Curves in Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

83

19 Curve Filling a Surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

89

20 Curves with Locally Convex Projection . . . . . . . . . . . . . . . . . . . . . . .

91

21

Integral Inequalities for Closed Curves . . . . . . . . . . . . . . . . . . . . . . .

97

22

Reconstruction of a Closed Curve with Given Spherical Indicatrix of Tangents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

101

23

Conditions for a Curve to be Closed . . . . . . . . . . . . . . . . . . . . . . . . .

103

24

Isoperimetric Property of a Circle . . . . . . . . . . . . . . . . . . . . . . . . . . .

115

25 One Inequality for a Closed Curve . . . . . . . . . . . . . . . . . . . . . . . . . .

119

26 Necessary and Sufficient Condition of the Boundedness of a Curve with Periodic Curvature and Torsion . . . . . . . . . . . . . . . . . . . . . . . .

121

27 Delaunay's Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

125

28 Jordan's Theorem on Closed Plane Curves . . . . . . . . . . . . . . . . . . . .

133

29 Gauss's Integral for Two Linked Curves . . . . . . . . . . . . . . . . . . . . . .

139

30 Knots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

149

3 1 Alexander's Polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

157

32 Curves in n-dimensional Euclidean Space . . . . . . . . . . . . . . . . . . . . .

169

33 Curves with Constant Curvatures in n-dimensional Euclidean Space . .

177

34 Generalization of the Fenchel Inequality . . . . . . . . . . . . . . . . . . . . . .

183

Knots and Links in Biology and One Mystery . . . . . . . . . . . . . . . . . .

187

35

36 Jones' Polynomial, Its Generalization and Some Applications . . . . . . .

191

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

199

© 2000 CRC Press

Preface Differential geometry is a wide domain of modern mathematics, whose significance is incrcasing at present. One of its origins is in the theory of curves. Everybody who wishes to study geometric problems has to begin by studying the theory of curves, where exact definitions, notions, and invariant characteristics are introduced for the first time. Here the initial geometric intuition is formed and then it is developed in the studying of surfaces theory and the geometry of submanifolds. There exist good and extensive monographs devoted to special curves, but the problems of the general theory are not presented. On the other hand, many interesting and important questions on curves are not discussed, in most cases, in the courses on differential geometry in universities. This book is devoted to the general topics of the geometry of curves as well as to some particular results. Presentation begins with important definitions, including definition of a curve. We also introduce basic notions by using sufficiently accessible language. Next, we discuss properties 'in the large' of the curves in Euclidean space, which were presented earlier in scientific articles only. For a plane curve, the conditions on the curvature of a closed curve as a function of the arc length are well known. Therefore Efimov, Fenchel and other geometers state the following problem: what are the necessary and sufficient conditions on the curvature and torsion of a space curve in order for the curve to be closed? Probably effective conditions do not exist. But this question connects with other interesting questions. In this book we investigate problems for special classes of curves, give the working method to obtain the conditions for closed polygonal curves, and give the proof of the Bakelman-Werner theorem on necessary and sufficient conditions of the boundedness for curves with periodic curvature and torsion. We investigate the question of the connection between curvature and torsion for curves which we know are closed - curves of the trigonometrical type. An important geometrical characteristic of a curve is its indicatrix of tangents which we construct in the following way. Let P be a point on curve r and T(P)a unit vii © 2000 CRC Press

viii

PREFACE

tangent vector of r at point P. We translate T ( P )in such a manner that the origin of T ( P )coincides with the origin 0 of Cartesian coordinates. The set of end points of translated vectors r ( P ) is called the spherical indicatrix of tangents of the curve l?. For a closed regular curve this set is not arbitrary: it cannot lie on a hemisphere. This circumstance was probably first noticed by Poznjak. It was observed that the mentioned necessary condition is also sufficient. In this book we give the proof of the Vygodsky theorem: if y is a closed curve on the unit sphere such that y does not lie on any hemisphere, then y is the spherical indicatrix of some space curve. The short proof given in the book belongs to the well-known mathematician M. Krein. The wonderful French mathematician and astronomer Ch. Delaunay proposed a problem to obtain the curve of constant curvature k = 1 which passes through given two points and has the smallest or the greatest length. We give the solution of this problem by K. Weierstrass. Later Schwarz formulated a theorem that the length of such a curve cannot lie in some interval, but did not publish any proof. Then Schur, motivated by Hilbert, proved this theorem by using the twisting of a plane curve. By the twisting of y Schur meant a transformation of y preserving the curvature and the length of y. So, the twisting here is some process. Schur proved that if a plane curve y with end points P and T forms, together with the span PT, a closed convex curve, then as a result of the twisting the length of the span PT only increases. In 1929, Fenchel proved a theorem that the integral of curvature for a closed curve is not less than 27r. Borsuk proposed that for a knotted curve this integral is not less than 47r. Fery and Milnor proved his assumption almost simultaneously in 19491950. Later this theorem was generalized for n-dimensional submanifolds by Chern, Lachof, Ferus and others. In 1995, V. Gorkavy suddenly obtained a generalization of the Fenchel inequality in a new direction - for the higher curvatures of a curve in n-dimensional Euclidean space. In accordance with the title of the book, we pay much attention to topological questions. First, we discuss Gauss's classical integral for two linked curves. We observe the links between two infinitesimally close curves and prove the formulas of Calagareanu and White, which have application in biology. Here also twisting arose, but as a number characterizing the form of a curve. Later we discuss knots and knot groups, and give proof of the Pontryagin-Frank1 theorem that every knot is the boundary of some oriented surface. In a geometrical way we construct Alexander's polynomial. It is the knot invariant and the proof is founded on the three kinds of changes in the structure of the knot diagram. A list of the simplest knots and their Alexander polynomials are given. In 1983, Jones constructed a new polynomial using the braid theory and some Markov theorems. This polynomial was a great surprise to topologists. In 1985, almost simultaneously six mathematicians constructed a new polynomial depending on three arguments. It was called the HOMFLY polynomial. It is possible to obtain two previous polynomials from the HOMFLY polynomial. We give the method for calculating the HOMFLY polynomial, moving from simple knots and links to more complicated ones. A very interesting direction for the application of differential geometry and knot theory arose in the biology of DNA molecules. Because biologists obtained closed DNA molecules, the following question began to concern them: how could the © 2000 CRC Press

PREFACE

ix

process of replication of DNA molecules take place, if two chains of nucleotides are very closely linked? The chapter devoted to this question and the list of works give readers the possibility of acquainting themselves with important research in this direction. Three chapters are devoted to the theory of curves in n-dimensional Euclidean space. We give the definition and formulas for the calculation of all curvatures of curves, and obtain the canonical form for curves with all constant curvatures. Its behavior is essentially different in spaces of odd and even dimensions. The curves are applied in mathematics and technology, so investigation of them is very relevant at present. I would like to express my thanks to V. Gorkavy for the translation and other assistance. I am very grateful to Dr R. Rennie for his useful indications on the works of Jones and Witten. I am also grateful to Professor Nigel Hitchin who encouraged me to introduce the section on DNA.

© 2000 CRC Press

l Definition of a Curve The notion of a curve is one of the most important notions in differential geometry. In antiquity this notion had no explicit mathematical definition. Euclid, for example, defines a curve as a "length without width". At this time many wonderful and interesting curves were discovered and studied; however, the idea of a general curve remained at a trivial, obvious level. Further technological progress required the development of natural science, especially the evolution of mechanics and mathematics. It was necessary to understand clearly the foundations of mathematics and, in particular, to construct an accurate definition of a curve. The coordinates method proposed by Descartes prepared the way for a general definition of curves; mathematicians contemporary to Descartes defined a plane curve given by an equation @(x,y)= 0 as a set of points such that their Cartesian coordinates satisfy this equation. Another idea arose in mechanics: a curve is imagined as the trace of some moving point, whose coordinates depend on the time t. Jordan proposed the following definition: a space curve is a set of points whose Cartesian coordinates X, y, z are continuous functions

of some parameter t varying inside a real axis segment (a, 6); in other words, a curve is defined as the image of a real axis segment under a continuous map into the space. This definition seemed to be natural, but in 1890 Peano constructed a continuous map of a segment (a, h) into the space such that the image of (a, h) under this map covered the whole square (we will consider Peano's example in one of the following chapters). In 1897 Klein remarked: "What is an arbitrary curve?. . . One may say that at present in mathematics there exists no more dark and more indefinite notion

© 2000 CRC Press

2

DIFFERENTIAL GEOMETRY AND TOPOLOGY O F CURVES

than the mentioned one. The object, which we call a mass curve, is a strip, whose length is sufficiently great with respect to another strip's measures. But for a curve to be a subject of strong mathematical consideration, we must idealize a curve in the same way as a point is idealized. And here some difficulties arise. . . Let us turn to a proposition playing an essential role in Riemann's investigations into foundations of geometry: the space can be viewed as a three-dimensional continuous manifold.. . We start from a construction of some scale on a mass straight line; then we decompose the scaled line into smaller parts and continue this operation until it is realizable. After that we make the most important step from an experience to an axiom: we postulate that the correspondence between points and real numbers is valid not only empirically, hut also absolutely. . . ." We remark that Veronese considered a geometry with the following assumption: on the real axis there exist numbers different from the rational and irrational numbers; but it seems that this supposition does not lead to essential geometrical statements and there is no natural foundation for it at present. Another extreme way of looking at space is proposed by discrete geometry. Riemann noted: "The question of the validity of the assumptions of geometry 'in the small' is closely connected with the question of inner sources of metric relations in the space. Certainly, this question belongs to the theory of space and we must take into account that in the case of a discrete manifold the principle of metric relations is contained in the notion of this manifold, whereas in the case of a continuous manifold we have to seek it in some other place. From this it follows that either the real space is a discrete manifold, or we must explain the appearance of metric relations by something exterior.. . ." In modern differential geometry a slightly modified definition of Jordan is used. First, we will give the definition of'un elementary curve. Let p be a map of a segment (a,h) of the real axis into the Euclidean space; we denote by y the image of (a,b) under p. The map p is called continuous at a point X r (a, b), if for any positive t there exists a positive S such that the following condition -is fulfilled: if a point Y E (a,b) satisfies the inequality IX - YI S, then the distance between the points p(X), p(Y) is less than t. The map p is said to be continuous if it is continuous at each point of the segment (a,h). The map p is called one-to-one if the pre-image of any point P E 7 consists of one point. If y is one-to-one, then one can construct the assigns to a map p-' inverse to p. The domain of definition of p-' is y;the map point P E y its pre-image under the map p; in other words, if P E y and P = p ( X ) , then p P 1 ( P )= X by definition. The inverse map pp' is continuous at P E y if for any positive t there exists a positive S such that the following condition is fulfilled: if Q E y and the distance between P, Q is less than S, then I p p l ( P )- p-'(Q)l 5 t .

/[U,

rill1

= [a, / l ( [ )dt]

8


For example, let us prove the second formula. The first component of the vector

is equal to

where a, are the components of the vector a. We see that on the right-hand side of the last equality we have the first component of the vector

Considering the second and third components similarly, we prove the desired formula.

© 2000 CRC Press

3 The Regular Curve and its Representations A curve y is called ~ " r e ~ u l a riff, there exists a parametrization of y such that each component of the position vector r(t) is a C'-regular function and r: does not vanish. The C'-regular curve is called srnootlz . The condition ri # 0 is essential for the definition of regular curves. For example, consider the planar curve formed by two rays and represented by the position vector

where a, h, C, d are constants. If the vectors { a ,h ) , {c, d ) are linearly independent, the considered line has a singularity at the point (0,O). At the same time, each component of the position vector is C1-smooth everywhere. But ri(0) = 0. If y is one-to-one projected onto a segment [a, h] of the x-axis, then there exists one of the simplest representations:

Indeed, let r(t) = {x(t), y(t), z(t)} be some representation of y. Because y is one-toone projected onto the segment (a,b) of the x-axis, we can assign to each X E (a, 6) a unique value of the parameter t such that the point (X, 0,O) is the image of the point P(t) of y under the projection. Thus t can be viewed as a function of X, i.e. t = t(x). Substituting t(x) into the expressions of the functions y(x), z(x) we obtain:

Thus, the position vector of y parametrized by x has the form (3.1).

© 2000 CRC Press

10


When is the curve y regularly projected onto a segment of the x-axis? Assume that y is Cl-regular and xi # 0. Consider the function x = x(t). Because xi # 0, the function x(t) is monotone on some interval (a, b) of the x-axis, and there exists the inverse function t = t(x). Since t: = l l x i , t ( x ) is Cl-regular. Therefore to each point x E (a,b) one can assign a unique value of t and a unique point P(t) E y.Thus it is easy to see that y is projected onto the segment (a, b ) of the x-axis. We remark that in this case j ( x ) and Z(x) are C'-smooth. Often a planar curve is represented implicitly, i.e. by an equality

that is, this curve consists of the points whose coordinates X , y satisfy equality (3.2). But for some functions @(X, y ) equation (3.2) has no solution or the solution consists of isolated points. So, it is natural to ask: when does equation (3.2) represent a curve (in the sense of chapter l)? Theorem Let @ ( X , y ) be a C'-regular function. Assume that a point P ~ ~ i coth ordinates x ~y~, satisfies the equation

and grad cP = {Q,, Q,) # 0 at P. Then the points satisjying (3.2) and situated in some sufficiently small neighborhood of P ,form a C' -regular curve. Proof We will apply the theorem on an implicit function: Let a function @ ( Xy, ) be defined and C'-smooth in some neighborhood of a point (xo,y o ) Assume that @(Q, yo) = 0 and Q,(xo, yo) # 0. Then there exist S > 0 and a C'-regular function y = y ( x ) defined in the interval -6 xo 5 x 5 S + x~ such that y(xo) = yo and @ ( X , y(x)) = 0. By the assumption, grad @(xo,yo) # 0. We suppose without loss of generality that Qy(xO,yO)# 0. The assumptions of the theorem on an implicit function are fulfilled. Then the points with coordinates ( X ,y(x)) in some neighborhood of ( x o ,yo) form a curve y coinciding with the curve represented implicitly by the equation @ ( X , y ) = 0. The coordinate x is chosen as a parameter on y. Because the function y ( x ) is C'regular and r: = {l,y:) # 0, the curve is smooth. Sometimes it is useful and interesting to consider a family of curves. Here it is convenient to assign to each curve one or more numbers called parameters of the family. Then according to the number of parameters the family is called oneparametric, two-parametric and so on. For instance, in the family of straight lines x = c = const on the ( X ,y)-plane, every straight line is determined by the value of the constant c, i.e. c is the parameter of this family. A second example: the family of circles on the plane is three-parametric. Let H be a set of planar curves such that for any point P of a domain G on the ( U ,v)-plane there exists a unique curve y E H passing through P. We will say that the set H forms a Ck-regular one-parametric,family, @for any point (uo,vo) there exist a neighborhood and a smooth map x(u, v),y(u, v) of'the neighborhood onto the circle

+

© 2000 CRC Press

THE REGULAR CURVE AND ITS REPRESENTATIONS

+

11

x2 y2 < 1 with Jacohian J # 0 such that the curves of H are tmnsj'ormed into the straight lines x = c. The level lines a)(u,v) = c of a smooth function @(U,v ) form a smooth family in a neighborhood of a point (uo,vo), if grad cP(uo,vo) f 0. Indeed, let for instance @,(uo, v") # 0 . The map x = @(U,v), y = v has the non-zero Jacobian J = @,(uo, vo) and transforms the level lines @(U,v ) = c into the straight lines x = c. Problems 1. Find the prqjection of the curve

into the ( X , y)-plane. Find the projection into the 2. Does the curve

(X,z)-plane.

pass through the points ( 1 , l, 1 ) and ( 1 , 0 , 0)? 3. Are the following curves intersecting?

4. Find the point where the curve

intersects the plane z 5. Show that the curve

= 0.

r ( t ) = { a sin2 t , h cos t sin t , c cos t ) is situated on an ellipsoid. 6. Prove that

is the position vector of a circle whose radius is equal to v-.'

© 2000 CRC Press

4 Straight Line Tangent to a Curve Let PO be a point on a curve y and U C y some neighborhood of PO such that PO decomposes U into two half-neighborhoods U l and U2 (see Figure 4.1). Take another point Q E y and consider the ray PoQ, whose origin is PO.Assume that Q , which is situated in U;, tends to PO.Then the limit position of the ray PoQ, if it exists, is called the ray tangent to y at PO with respect to O;. If there exist both rays tangent to y at POand they form a straight line, then this line is called the straight line tangent to y at P. or the tangent to y at PO. Let us explain the notion of the limit position of the ray PoQ. Consider the unit sphere with its center at PO. The ray PoQ intersects this sphere at some point M. Suppose that Q tends to PO.If the corresponding points M converge to some point MO,then we say that the ray PoMo is the limit position of the rays PoQ. Assume that y is a smooth curve with the position vector r = r(t). What is the directing vector of the tangent to y? We denote by a prime the derivative with respect to t . Theorem For any point POo f t h e smooth curve y there. cxists the tangent to y at PO, and the directing vector of the tangent is r'.

FIGURE 4.1

© 2000 CRC Press

14


Proof Let POcorrespond to the value to of the parameter t , and some close point Q correspond to to A t ; we assume that if A t > 0, then Q is situated in U , , and if A t < 0, then Q lies in U2. The directing vector of the ray PoQ is r(to A t ) - r(to). Since y is smooth, we have

+

+

where o ( A t ) is a vector such that

It follows from (4.1) that the vector

r(to

+At) At

-

r(to)

+4 A t )

= rl(to) -A -t-

is collinear to the ray PoQ at A t > 0 and is opposite to this ray if A t < 0. Because y is smooth and property (4.2) holds, the limit of the expression on the left-hand side of (4.3) as A t 4 0 exists and is equal to rl. Thus there exist both rays tangent to y at PO,they form the tangent to y at PO,and r' is the directing vector of the tangent.. For some non-regular curve, both rays tangent at a point Q" exist and coincide. In this case we say that the curve has the halftangent at Q. and Q. is called the cusp. Let us write the equation of the tangent to y at PO.Again we denote by r(t) the position vector of y and by ?(X)the position vector of the tangent. Since the tangent passes through the point PO with position vector r(to) and rl(to) is the directing vector, the position vector of the tangent is

Rewriting equality (4.1) as

r(to

+ A t ) = r(t0) + rl(to)At+ o ( A t ) ,

and setting X = A t , we see that the difference between the position vectors r(to + A t ) and ?(At) is an infinitesimal o(At):

r(to

+ A t ) - ? ( A t )= o ( A t ) .

Therefore if A t is sufficiently small, the curve y is close to the tangent. In other words, the tangent is the first approximation of the curve. When the parameter X varies from -00 to +oo, the corresponding point passes the whole tangent. In terms of Cartesian coordinates the tangent is represented as

© 2000 CRC Press

STRAIGHT LINE TANGENT T O A CURVE

Consider a space curve given by two equations

Assume that the rank of the matrix

is equal to two at a point Po(xo,.yo,zo).If x = x ( t ) , y = y(t), z = z(t) is the position vector of the considered curve, then, substituting these three functions into (4.4), we obtain two equalities 4 ( x ( t ) r y ( t )4, t ) )

=

0,

l i / ( x ( t ) , y ( t )4, t ) ) = 0;

the differentiation of the equalities leads to two additional equalities

Thus, the components { X ' , y', z') of the tangent vector satisfy the system consisting of two equations (4.6), therefore they are proportional to the corresponding minors of matrix (4.5):

Because the rank of matrix (4.5) at point PO is equal to two, some minor of (4.5) is non-zero. If we have a planar curve given by the equations 4 ( x ,y ) = 0 , z = 0 satisfying the condition 4; 4: # 0, then the components of the tangent vector v' = {X', y') is a solution of the linear equation

+

Therefore

and the equation of the tangent is

Let y be a smooth space curve. The plane passing through a point PO t y and orthogonal to the vector tangent to y at PO is called the plane normal to y at PO. © 2000 CRC Press

16


Denote by i the position vector of the normal plane. Because this plane is orthogonal to the vector rf(to)and contains the point with position vector i - r(to),the equation of the normal plane is ( i - r ( t O )r,l ( t O ) )= 0.

The vectors orthogonal to the tangent are called the vectors norrnul to y.

© 2000 CRC Press

5 Osculating Plane of a Curve Let PO be a point of a curve y.Take two points Q , , Q2 t y situated on different sides with respect to PO and construct the plane passing through PO,Q , , Q 2 . If the points Q , , Q2 tend to PO,then the limit position of the plane containing PO,Q , , Q2 is called the osculatingplur~eof the curve y at the point PO.By this definition, if the osculating plane exists, then it is unique. Obviously the osculating plane of y at PO passes through P().Let PO correspond to the value to. Theorem Let y he a C2-regular curve represented ac r = r(t). Assume tlzut at cr point PO the vectors rl(to)and rl'(to)are not collinear. Then there exists the osculating plurze o f y at PO und it i.s syunned by the vectors rl(tO),rl'(lo) (F~gure5.1) Proof Let points Q , correspond to values to + h, of the parameter t. The position vector of PO is r ( f o ) ,and the position vectors of Q , are r(to +h,). Hence the vectors PoQ, are r(fo + h , ) - Y ( ~ o ) .

FIGURE 5.1

17 © 2000 CRC Press

18


Since y is regular, one can write a Taylor expansion of the position vector r(t) at a neighborhood of to:

where o(h:) is a sufficiently small vector such that o(h?)/h; + O as hi --t 0. Using the Taylor expansion, we find that the vectors PoQl and PoQ2 have the following forms:

The vectors PoQl and PoQz span some plane and this plane contains the following linear combinations of P o Q l , f o e z :

Because the points Q,, Q2 are situated on different sides of y with respect to PO,h1 and h2 have different signs. Therefore I h , - h2 I > max( I h1 I , ( h2 I ) and

-

Hence the infinitesimal summed to rl'(to) in the second vector of (5.1) tends to O as Q, PO. Therefore the limit positions of the vectors presented in (5.1) are rl(to), r1I(to).If these vectors are not collinear, they determine uniquely the limit position of the plane passing through Q , , PO, Q 2 , i.e. the osculating plane of y at PO. If the vectors rl(to),r1I(to)are collinear, the limit position of the considering plane is not determined. For instance, take a straight line r(t) = a

+ bt,

where a, b are constant vectors. Then rl(tO= ) b,

r M ( t o= ) 0,

so the osculating plane of the straight line is not determined uniquely. In this case, one can think that any plane containing the straight line is its osculating plane. © 2000 CRC Press

OSCULATING PLANE OF A CURVE

If r r ( t ) , rl'(t) are collinear, then the straighteningpoint of y.We will view any

19

corresponding point of y is called the plane passing through the tangent to y at

this point as the osculating plane of y. We remark that the notion of an osculating plane does not depend on the choice of the parameter on y. Instead, if r = r ( ~ is) another parametrization of y and t = t ( ~ ) , then, differentiating r(t(7)) as a composite function, we obtain

Calculating the vector [ r : , r : ] , we see that it is collinear to the vector [ r : , r i ] :

Because the osculating plane is passing through PO and [ r i ( t o ) , r i ( t o ) ] is its normal vector, the obtained equality means the independence of the osculating plane on the parametrization of y. The osculating plane of a planar curve coincides with the plane containing this curve. Let us consider the Taylor expansion of the position vector r ( t ) at a neighborhood of PO: r(to

The curve

+ A t ) = r(to) + r r ( t o ) A t+ rl'(to)---A2t 2 + o ( A t 2 )

r:

determined by this expansion is situated in the osculating plane of y at PO; the difference between the position vectors of y and 7 is a sufficiently small vector: r(tn

+ A t ) -?(At)= o(at2).

Hence a sufficiently small neighborhood of PO on the space curve y is near to the planar curve 7 situated in the osculating plane of y at PO. Now let us write the equation of the osculating plane of y at PO.Denote by i the position vector of the osculating plane. The vector product [ r l ( t o ) ,r " ( t o ) ] is orthogonal to the osculating plane, and the vector I: - r(to) belongs to the osculating plane, hence the inner product of these vectors is equal to zero: ( I : - to), + ( t o ) , i l ( t O ) )= 0.

© 2000 CRC Press

20

DIFFERENTIAL GEOMETRY A N D TOPOLOGY OF CUKVES

This is just the desired equation. With respect to Cartesian coordinates it has the following form:

With the help of the osculating plane one can distinguish two special straight lines normal to y at PO.The straight line normal to y and situated in the osculating plane of y at PO is called the principal normal straight line to y at PO.It is the intersection of the plane normal to y and the osculating plane at PO. We have proved that the vector [ r l ( t o ) , r " ( t o ) ]is orthogonal to the osculating plane. Because the principal normal straight line is orthogonal to the normal of the osculating plane and to the tangent vector of y,it is collinear up to a sign to the vector

Problems

1. Write the equations of the tangent to the curve x = 3 t - t - 1,

y=3t2,

z=3t+t3

at the point (0,0, 0). 2. Write the equations of the plane normal to the curve considered in Problem 1 at the point (0,0,0). 3. Find the tangent vector and the normal vector to the planar curve

4. Write the equation of the osculating plane of the curve X = P'' COS U ,

y

= P"

sin U , z = e"

5. Prove that the curve having the tangent straight line at each of its points is smooth.

© 2000 CRC Press

6 The Arc Length of a Curve Now we will define the notion of the arc length of a curve. Let y be a space curve given by its position vector u = r(t), t E [U, h]. The length of an arbitrary polygonal line consisting of a finite number of segments is equal to the sum of the lengths of all these segments. Inscribe a polygonal line into the curve y in the following way. Take some points a = to < t l < . - . < t, = h belonging to the segment [a, h]. Some points PO,P I , . . . , P, on y correspond to the chosen values of the parameter t. If we connect

F: FIGURE 6.1


22


POwith P , , P I with P2 and so on by segments, we obtain a polygonal line, which is inscribed into y and whose vertices are P;. Consider the length of the inscribed polygonal line. If we take some new points in [a, h] and the corresponding new points on y, we obtain a new polygonal line inscribed into y, and its length is greater than the length of the preceding inscribed polygonal line. It is clear that if we have chosen a new point Q E y situated between points Pi, Pi+,, the sum of the lengths of the segments PiQ and QPi+l is greater than the length of the segment Pip,+'; hence the length of the new polygonal line P O . .. P;QPi+l . . . P, inscribed into y is greater than the length of the preceding one P O ...P;P;+l . . . P,. The curve y is called rectifiable, if the lengths of all polygonal lines inscribed into y as described is bounded from above. If y is rectifiable, then the supremum of the lengths of inscribed polygonal lines exists by the Weierstrass theorem, and it is called the length of y. We will denote it by s(y). Theorem If y issmooth, then it is rectifiable, and the length of y is equal to the integral

Proof First, we will prove that the smooth curve y is rectifiable. It is necessary to find some estimate of the length of an arbitrary polygonal line y, inscribed into y. Let r(t,) be the position vector of the vertices of 7,. Then the length of y, is equal to the sum of the lengths of vectors A,r. Because y is smooth,

In addition, from the smoothness of y it follows that r: is a continuous vector-valued function defined on the segment [a,h], hence it is bounded. There exists a constant M such that

Therefore we have

thus the length of an arbitrary polygonal line inscribed into y is bounded from above by the constant M(h - a) depending only on the curve y. So, y is rectifiable. Let us prove that the length of y is computed with the help of formula (6.1). Because y is smooth, i.e. the components x(t), y(t), z(t) of the position vector r(t) are C'-smooth, we have © 2000 CRC Press

THE ARC LENGTH OF A CURVE

6

where r: are some points of the segment [ t i P lti]. , In general, the points are different. We will replace r;, r:, r; by one point ~i E [ti-l,ri]and consider the mistake which appears. We have

where the vector a; has the following form:

Since every component of r' is a continuous function defined on the segment [a, h],it is uniformly continuous. Hence for any positive E there exists a positive 6 such that I a; I < E whenever I t , - t,-l 1 5 S. We denote c = rl(r;)A,t.From (6.2) it follows that

Air - c

= a,A,t.

Therefore It is easy to obtain the following estimates:

and

Thus the difference between / Air I and Ir1(ri)lAlt is less than &Ai(. We apply the obtained estimate in order to estimate the difference between the length of y, and the sum C:=,Ir1(ri)/Ait:

4%)

-

The sum

C:=,(rl(ri)lAitis a Riemann sum for the integral

so the difference between this sum and integral is sufficiently small for a suitable choice of the points ti decomposing the interval [a,h]. On the other hand, one can © 2000 CRC Press

24


inscribe a polygonal line y, into the curve y such that the difference between the lengths of y and y, is sufficiently small. Because E is arbitrary, s(y) coincides with integral (6.7). Thus formula (6.1) is proved. From the definition it follows that the length of any curve does not depend on the parametrization. Hence one can use any parameter T in formula (6.1) to compute the length of y. With the help of the notion of the length, one can introduce some special parameter a , which is most naturally connected with y. Fix a point PO t y. Assume that y is parametrized by a parameter t , some value to corresponds to POand t corresponds to P. The parametrisation t gives us some "positive" direction (orientation) on y. To any point P E y we assign a equal to .(POP) in the case when the arc POPhas the positive direction; if the direction of P O Pis negative, we assign a = -s(PoP) to P. Then the parameter a is computed by the formula

1 r i ( t )1 dt.

~ ( t=) 10

Since

the parameter a is a monotone function of t , hence we can consider a as a new parameter on y. This parameter is called the arc length and for sin~plicitywe use the previous notation S. The parametrization of y by the arc length is called natural. Using (6.8) we obtain

Thus the absolute value of the differential of the arc length is equal to the absolute value of the differential of the position vector of y. Therefore

Thus we have the following characterizing property of the natural parametrization: iJ r(.r) is the position vector of'y with respect to the arc length S, then the length of the vector r(,(s) is equal to l :

4f:Jorsome parameter s the length of the tangent vector ri is equal to l identically, then s is the arc length. © 2000 CRC Press

THE ARC LENGTH OF A CURVE

Problems l . Find the arc length of the curve X =

In sin t,

t

y=JZ'

z=-.

t

JZ

2. Find the length of the arc Q 5 t 5 1 of the curve

3. Find the length of the arc 0 5 t X

© 2000 CRC Press

= h(t

-

< 27r of the cycloid

sin t),

y = h(1

-

cost).

7 The Curvature and Torsion of a Curve In this chapter we will define and study the curvature and torsion of curves. Let P be a point of a curve y. For some point Q E y sufficiently near to P we denote by A 0 the angle between the tangents to y at P, Q and by As the length of the arc PQ. The limit

k

=

lim

Q+P

A0

as

-

is called the curvature of the curve y at the point P. We remark that by supposition here both A$ and A s are positive numbers. Denoting by r(s) the position vector of y with respect to the arc length s we will prove a formula for the curvature k. l a r Then the curvature k o f y exists and it can be Theorem Let y be a ~ ~ - r e ~ ucurve. calculated with the help of the jormulu

The vector ry>is called the curvature vector of y. Its length is equal to the curvature. The vector 11 = r:,/k is called the principal normal vector of y;it is defined at points where k # 0 (compare with the definition of the principal normal straight line given in chapter 5).

+

Proof Let points P, Q correspond to values S , s A s of the arc length. We denote the vectors tangent to y at P, Q by r(s), r ( s As). The angle A 0 is the angle between T ( S ) and r ( s + As). Consider the triangle P M N such that the vectors P M , PN are equal to r ( s ) and ~ (+sAs) respectively (see Figure 7.1). This triangle is isosceles, because the vectors r(s), r(s AS)are unit. The altitude P L is the bisectrix

+

+

© 2000 CRC Press


FIGURE 7.1

of the angle / M P N that

=

AQ. It is easy to show, considering the right triangle PLM,

I T(S+ a s ) - T(S) I 2

.

= sin

AQ . 2

-

(7.1)

Now we divide both parts of equality (7.1) by As/2 and set As --t 0. Because y is c2-regular and we have denoted the vector r:.(s) by ~ ( s ) we , obtain

I r;(s + As) - Y:(s) / I riy(s)( = AS+O lim AS

2 sin(AO/2) AS-o As sin(AO/2) A812 AB = lim 2 = lim - = k. AS-o A012 AS AS-o As =

lim

The theorem is proved. Now let us define the torsion of y. Since T = rt is a unit vector, the curvature vector v,: and the principal normal vector U are orthogonal to the tangent vector T. The vector product [T, U ] is called the hinormal of y and is denoted by P. It is obvious that the binormal P is well defined at points where the curvature k is non-zero. Here we suppose that As can be positive and negative.

FIGURE 7.2

© 2000 CRC Press

29

THE CURVATURE AND TORSlON OF A CURVE

Let Ad denote the angle with sign between the binormals P(P), P(Q) at sufficiently near points P, Q E y;this angle is equal to the angle between the osculating planes of y at P, Q (see Figure 7.2). We define the sign of AB as follows. The normal plane N p to y at P is spanned and oriented by the basis vectors u(P),P(P). We translate the vector P(Q) in such a way that its origin is the point P and then we consider the orthogonal projection of the translated vector P(Q) into the normal plane N p . If the length As of the arc PQ is sufficiently small, the vector P(Q) does not vanish. Assume that As > 0; then the sign of AH coincides with the sign of the angle counted from P(P) to with respect to the given orientation of NI.. If As < 0, then the sign of At4 is opposite to the sign of the considered angle between P(P), P(Q). The limit

P(Q)

B(Q)

K =

AH lim -

Q-P

as

is called the torsion of y at P. Theorem Suppose y is C'-regular and the curvature k qf y is not equal to 0 everywhere. Then ,for any point P E y there exists the torsion K of y at P and the fillo wing forrnula holds: ,,l ,.l1 Ill ( .S 2 s s , r.s.7.71 K = k2 '

Proof Consider a triangle PMN C N p such that the vectors P M , PN are equal to the vector P ( P ) and the translated vector P(Q) respectively. This triangle is isosceles. Similarly to the proof of the previous theorem, consider the altitude PL, which is the median and the bisectrix of the triangle PMN. Using relations between elements of the right triangle PLM we get

We divide both parts of this equality by lAs1/2 and set As

thus I /3:(s)

I = I K 1. Consider

the vector

P:.

-+

0. Then we obtain

We have

Since r(s) is smooth of class c3and k # 0 by the assumptions, the curvature k(s) is a C'-smooth function and @(S)is a C' vector-valued function. Then we have

© 2000 CRC Press

30


Both summands situated on the right-hand side of the last equality are vector products; each one of these products contains the vector T = ri as a factor. Hence f l is orthogonal to T. Since P is a unit vector, is orthogonal to P. Therefore P', is collinear to the principal normal vector U . Because the length of 0;.is equal to I K ~ , we get Pi. = f1. 1 ~. To determine the sign, we consider Taylor's expansion of the binormal ,B at a neighborhood of P:

The projection ,&Q) of the vector @(Q)is equal to the vector P(P) + ,Bi(P)As up to an infinitesimal o(As). Take Q such that A s > 0. It is easy to see that the considered angle between @(P)and ,&Q) is positive iff pi(P)As is collinear to -v. Hence

In order to prove the desired formula presenting the torsion we multiply equality (7.2) by v and obtain:

Note that the sign of the torsion does not depend on thc choice of orientation of y; from formula (7.3) it follows that the sign of K does not change if we replace s by -S. It is useful to know the formulas of k and K in the cases when y is parametrized by a parameter t different from the arc length. Viewing the arc length s as a function of t , we differentiate the position vector r of y as a composite vector-valued function and get:

Since r: is unit, from the first equality it follows that Idt/ds( = l / l r i ( . The vectors r: and rtS are orthogonal, hence we have

=

;l3

I [rl r"] l l'

It

in.c11

= ----- .

i r :13

Thus the curvature k of y with respect to an arbitrary parameter t is given by the formula

© 2000 CRC Press

THE CURVATURE AND TORSION OF A CURVE

The formula just proved can be written in the following coordinate form:

Let us now find a formula for the torsion. We will denote the rnixedproduct ([a,h],c) by (a, h, C). It is known that the mixed product is linear with respect to each of its arguments; for example, (a,h c, d ) = (a,6, d ) + (a,c, d ) . Also, the mixed product is equal to zero iff at least two factors are collinear. Using these facts we get

+

Substituting the found expression of k in formula (7.4) and applying the equality Idtldsl = I/lril, we obtain that the torsion of y with respect to an arbitrary parameter t is computed with help of the formula (r'/ > r"I t > r"') l//

/E.=

l E >r312

.

Curvature of planar curves Planar curves are a particular case of space curves, hence formula (7.4) can be used to find the curvature of planar curves; in this case this formula has a simpler form. Let y be a planar curve. If y is situated in the plane z = 0, then z' = z" = 0. Therefore,

k

=

(x'y" - y'x" (xt2+ y

This expression for k is the simplest if x" = 0, hence

X

l

'2 3i2

'

is taken as the parameter on y;then

X' =

1,

For the planar curve y one can define the notion of the curvature with sign. We will denote the curvature with sign by the same notation k. Let r denote the tangent © 2000 CRC Press


FIGURE 7 . 3

vector to y,e be a unit vector orthogonal to the plane n containing y.The vector 7 = [e,T ] is unit and orthogonal to T ; also it lies in the plane T . It is clear that 7 coincides with the principal normal vector v of y or with -U (Figure 7.3). When a point P is moving along y the vector v(P) is varying continuously. The coefficient k given in the formula

is called the curvuture with sign of y.It is different from the curvature defined earlier maybe only in the sign. We remark that using the curvature with sign k one can describe the changing of the convexity direction of y (see Figure 7.3); for k one of the two following formulas holds:

Now let us find a formula for k in terms of the angle between the tangent vector T to y and a fixed vector a. Without loss of generality we can assume that a is collinear to the positive direction of the x-axis. Suppose the arc length s of y is the parameter given in formulas (7.7). Since = T is a unit vector, the components X',, y: of r: have the following form: X:. ( S ) = COS a ( S ) ,

y:. (S) = sin oi ( S ) .

Substituting the derivatives

into formulas (7.5) we obtain

As an example, we find the curvature of a circle of radius R. Denote by p the angle between the x-axis and the position vector of a point of the considered circle. Obviously, ds = R d p . If the circle is oriented counterclockwise, then a = p n/2.

+

© 2000 CRC Press


33

Applying the formula k = dalds, we obtain that the curvature of the circle is equal to k = 1 / R . Assume that y is given by an equation @ ( x , y )= 0. What is the formula for the curvature k? The vector {Q,, Q,,,) is normal to y at the corresponding point. For the unit normal vector T,J we have

where W = \/(B,):

+ (a,):. T =

Hence the unit tangent vector to y is

dx dy

{;i;.5} =

Q,, {-&}'

@

where d/ds is the differentiation with respect to the arc length of y. From the formula T: = krl it follows that /c = - (Tf .\ > 7 ) . We can find the derivative of q with respect to s by differentiating function:

as a composite

Moreover, it is easy to see that

Thus, we can write

+

Since @,dx QYdy = d@ = 0 at the points of y,the last term on the right-hand side of equality (7.9) is equal to zero. Substituting expressions (7.8) for the components of the vector T into equality (7.9) we get

© 2000 CRC Press

34

DlFFERENTIAL GEOMETRY AND TOPOLOGY O F CURVES

It is easy to see that this formula can be rewritten in the following divergent form:

Problems 1. Find the curvature and torsion of the curve

x=acosu,

y=asinu,

z=bu.

2. Find the curvature and torsion of the curve x=t,

y = a t 2,

z=bt2+ct+d.

3. Find the binormal of the curve x=cost,

y=sint,

z=cos2t.

4. Find the curvature of the ellipse

at the vertices. 5. Prove that the curve

is planar. Find the plane containing this curve. 6 . Find the unit tangent vector T and the principal normal vector v to the curve

7. Find the curvature of the conical spiral

8. Prove that any curve with constant torsion

K

# 0 can be presented in the form

where [ = [(t) is a unit vector-valued function. 9. Let some family of regular planar curves satisfy two conditions: (i) the curvature k of any curve of the family is bounded, k > ko > 0, ko = const; (ii) all curves of the family are situated inside a circle of radius R. Prove that R 5 2/ko. © 2000 CRC Press


35

10. A closed convex planar curve y is called an oval. The distance between two straight lines, which are collinear to a fixed direction T and tangent to y,is said to be the width of y with respect to the vector 7 . Prove Barbier's theorem: if the width d of an oval does not depend on the direction 7 (one can say that this oval has the constant width d), then the length of the considered oval is equal to nd.

© 2000 CRC Press

8 Osculating Circle of a Plane Curve Consider a smooth planar curve y with curvature k # 0. For any point PO E y the difference between the behavior of y and the behavior of the tangent to y at P" is negligible at a sufficiently small neighborhood U of PO;we say that the tangent to y at PO is the first approximation of y at U. The second approximation of y at the sufficiently small neighborhood U of PO is a circle called the osculating circle of y at PO.In order to define the osculating circle we fix Cartesian coordinates X, y in such a way that PO is the origin and the x-axis coincides with the tangent to y at PO. Let y = y(x) be the equation for the curve y and y = j ( x ) be the equation of some circle C passing through PO. This circle C is called the osculating circle of y at POif the values of the function y(x), ? ( X ) are equal at x = 0 and the values of their first and second derivatives coincide at x = 0:

From these equalities it follows that Taylor's expansions

are equal up to an infinitesimal 5 ( A x 2 ) . Hence if we consider all circles passing through PO,the osculating circle is the nearest to the curve y. Since the tangent to y at PO is completely determined by the values y(O), yl(0),the tangent to y at PO and the tangent to the osculating circle at PO coincide, so we can say that the osculating circle is tangent to y. Also, the center 0 of the osculating circle is situated on the straight line that contains P,) and is collinear to the principal normal vector v(Po) (see Figure 8.1).

© 2000 CRC Press

DIFFERENTIAL GEOMETRY A N D TOPOLOGY OF CURVES

FIGURE 8.1

The curvature of a planar curve y = , f ( x ) is computed by the formula

Because the first and second derivatives of y , j are equal at PO, we see that the curvature k of y at PO is equal to the curvature of the osculating circle of y at PO. Thus the radius R of the osculating circle is equal to l / k ( P o ) .The center of the osculating circle is called the curvature center of y at PO;the radius of the osculating circle is called the curvature radius of y at PO. Generally, the curvature center depends on the point PO E y. All curvature centers of y form a curve y called the evolute. Let us find the position vector of the evolute. Let r = r(s) denote the position vector of y with respect to the arc length S . If r(so)is the position vector of a point P, then the position vector p(so) of the curvature center 0 of y at P is the sum of the vector r(so) and the vector PO. This last vector is collinear to the principal normal vector v(so)and its length is equal to the curvature radius R(so) = l/k(so).Thus we obtain the position vector of the evolute:

From the formula evolute is

7: =

kv it follows that v', = -kr. Hence the vector tangent to the

Therefore if (ilk): # 0, then the vector tangent to the evolute of y is collinear to the principal normal vector to y (Figure 8.2). Assume that y is the evolute of some planar curve 7; then 7is called the evolvent of y. Let us find the position vector T; = r(s) of the evolvent with respect to the arc length s of y. Since any point of the evolvent is situated on the corresponding straight line tangent to y,we have © 2000 CRC Press

OSCULATING CIRCLE OF A PLANE CURVE

FIGURE 8.2

The curve y is the evolute of 7,hence its tangent vector r is collinear to the principal normal vector to the evolvent. Therefore T is orthogonal to the vector tangent to 7at the corresponding point. We have

This implies that X(S) = c

- S,

C

= const.

Thus the position vector of the evolvent of y is

where c is a constant. It can be demonstrated that for any arbitrary constant c the curve given by the position vector (8.2) is the evolvent of y.Hence there exist many evolvents of the given curve y. Indeed, from formula (8.1) we get

Let S denote the arc length and k be the curvature of the evolvent 7.Since 7; is a unit vector,we have

Suppose that c - s > 0. Then

r-

-1

"

© 2000 CRC Press

l ds "d3

I/ - =

-kT

1 = (c-s)k

1 c-s

7.


FIGURE 8.3

Therefore the curvature k of 7 is equal to I/(c - S ) and v normal vector of 7.Equation (8.2) can be rewritten as

=

-r is the principal

hence y is really the evolute of 7. It is very easy to demonstrate the form of the evolvent. Fix a point Q E y and consider the arc of y with length c; denote by P the end of this arc. Now imagine that a thread covers the arc QP and the origin of the thread is fixed at Q. If we take the end of the thread and reel the thread off the arc QP viewed as a pattern, then the end of the moving thread forms the evolvent of the arc QP of y.Indeed, some part Q A of the thread coincides with Q A C QP; if the length of Q A is equal to S, then the rest of the thread is a segment of the tangent to y and its length is equal to c - ,F. Therefore from (8.2) it follows that the end of the moving thread forms the desired evolvent. The evolutes and evolvents are frequently used in technology.

© 2000 CRC Press

9 Singular Points of Plane Curves Let a plane curve y be given by an equation cp(x, y) = 0 and cp E ck,k > 1. A point M = (xo, yo) of the curve y is called a singularity of y,if the following conditions are fulfilled at this point:

There exist different types of singularities. Assume that M = (xo,yo) is a singularity of y and some second derivatives of the function p do not vanish at M. Let us introduce some notations:

There are three cases with respect to D: (a) if D > 0, then M is called an isolated point of y (Figure 9. la); (h) if D < 0, then M is called a point of self-intersection of y (Figure 9.lb); (c) if D = 0, then M is either an isolatedpoint, or a cusp, which can be of two types, or an osculate point of y (Figure 9. lc).

a

6

C FIGURE 9.1

© 2000 CRC Press

42


The first case in Figure 9. l c shows a curve such that its singularity M is a cusp of the first type (the corresponding branches of the curve are situated on different sides with respect to the straight line containing the tangent ray to this curve at M); in the second case, M is a cusp of the second type (the tangent ray does not decompose the branches); in the third case, M is an isolated point; and in the fourth case, M is a point where two branches are tangent (in particular, these branches can coincide). If we assume that the first, second and third derivatives of the function p(x, y) are continuous, we can write Taylor's formula:

+

where o(Ar2) is an infinitesimal with respect to Ar2 = (x - x ~ )0,~- y0)2. Let us denote Ax = X - xo, Ay = y - y". Since the conditions (9.1) are fulfilled at the singularity (xo,yo), we obtain:

Denoting Ax Ar2, we get

= Ar

cos a, Ay

= Ar

sin a and dividing the equation p(x, y) = 0 by

Consider case (a), D > 0. Suppose that there exists an angle a" such that A cos2a g

+ 2Bcos no sin a0 + csin2a" = 0.

Assume without loss of generality that A > 0 (we can make a rotation in the (X,y)plane in order that A # 0); then dividing the last equation by sin2 ao, we see that tan a,)is a solution of an algebraic equation of the second degree; on the other hand, the discriminant of this equation is equal to -D and by our assumption it is negative, so this equation has no solution. Thus we obtain a contradiction. Therefore, for any value of a the expression A cos2 Q + 2Bcos a sin a

+ csin2a

(9.3)

is greater than some positive number. Suppose now that there exists a sequence of points of y converging to the singularity M; then o(Ar2)/Ar2 + 0. Since expression (9.3) does not converge to 0, we see that equation (9.2) cannot be fulfilled as Ar2 -+ 0. Hence M is an isolated point of y. Consider case (b), shown in Figure 9.1b. Assume that A # 0. We will demonstrate that the value of lAy/Axl cannot be infinitesimally small at points of y near to M. Suppose that there exists a sequence of points M, -+ M such that © 2000 CRC Press

SINGULAR POINTS OF PLANE CURVES

at these points. If we divide equation (9.2) by cos2a , we get

At the points M, we have the inequality Ar2 5 ( l + &Ax2. Hence the sequence of points M, converging to M satisfies the condition

Since the coefficients corresponding to B and C tend to zero as M, + M and A # 0, equation (9.4) cannot be fulfilled for the sequence of points M,. Thus there exists a positive ko such that (AylAxj 2 ko for all points situated in a sufficiently small neighborhood of M. We rewrite this inequality as

Dividing equation (9.2) by sin2cu, we obtain

We solve this equation as an algebraic equation of the second degree with respect to Axl Ay:

Because the points of y satisfy

we have

as A y -+ 0. From this conclusion and from (9.5) it follows that the values of A x l d y at points of y are infinitesimally near to the following two numbers:

© 2000 CRC Press

44


This means that the points of y,which are sufficiently near to M, are sufficiently near to two straight lines

intersecting at M. Hence the singular points M are points where two branches of y intersect. That is why M is called a selj-intersection point. Now let us consider the most intricate case (Figure 9. lc). Using a suitable rotation of Cartesian coordinates in the plane, we can obtain B = 0. Since AC = 0 in this case, AC = 0. We assume that A = 0. Because some second derivatives of p do not vanish, we have C # 0. The behavior of y at the point M depends on the sign of C and on the properties of the infinitesimal o(Ar2).IJ'the signs of C and o(Ar2)are equal, then M is an isolatedpoint o f M . For our next consideration it is necessary to consider Taylor's formula including the third derivatives of p:

where k , are constants. We will prove that

as points of y converge to M. Suppose that there exists a sequence of points M, E y converging to M such that

for some arbitrary fixed positive number k. If we divide equation (9.6) by ( A y ) * ,we obtain

Since

cm.Then from (9.8) it follows that C = 0, at the points M,, o ( A r 3 ) / A y 2-+0 as n contradicting the assumption that C # 0. Therefore property (9.7) holds, i.e. the curve y is tangent to a straight line collinear to the x-axis. Let us rewrite the equation of y:

© 2000 CRC Press

SINGULAR POINTS OF PLANE CURVES

Then from this equation we obtain:

If k l # 0, then the expression under the square root is equivalent to -4CklAx3 as Ar 4 0. In this case the curve is defined either for A x 2 0 or for A x 5 0, depending on the sign of -4Ckl. It is easy to see that the denominator of the expression for A y is equivalent to f The curve y at a small neigborhood of M is similar to the curve

Js.

consisting of two branches situated on different sides with respect to the tangent ray at M. Tlze singularity M is a cusp of the,fir.rt type. If kl -= 0, then the expression under the square root can be equivalent to the infinitesimal o(Ar3).So we must consider the fourth derivatives of the function cp. Here the equation of y can be rewritten as

where li are constants defined by the fourth derivatives of cp at the point M = (xo,yo). We write the left-hand side of this equation as an algebraic equation of the second degree with respect to Ay;

where by cu we denote the infinitesimal

The discriminant of equation (9.9) is

If k: - 4Cll # 0, then the last expression is equivalent to the first summed. In the case k i - 4C11 < 0, the determinant is negative for small A x ; therefore equation (9.9) does not have a solution different from A x = A y = 0. Hence M is an isolutedpoint of y. If k i - 4Cll 2 0, the curve y is defined for small values A x of different signs; here the equation of y is

© 2000 CRC Press


46

The curve y in a small neighborhood of M consists of two mutually tangent branches. In this case M is called an osculate point. If k: - 4Cll = 0, then the discriminant is equivalent to o(Ax4). Assume o ( A x 4 ) > 0 for all sufficiently small values of A x . Then for A y we have two solutions; hence I' consists of two tangent branches and M is an osculate point. If o ( A x 4 ) 0 for sufficiently small values of A x of sorne.fi'xed sign, then M is a cusp of y.In the case k2 # 0, both branches of y are situated in one side with respect to the tangent ray at M; the singularity M is called a cusp ofthe second type. If k2 = 0, then M is either a cusp of the,first type (for example, in the case 12 = 0), or a cusp of the second type (in the case when l2 # 0 and Jo(ax4) = o(Ax")). The considered investigation is valid iff the function p(x, y) is sufficiently smooth at M. If cp is not differentiable at M, then the behavior of y can be very complicated. For example, let p be the function given by the following equality:

>

The curve presented by the equation cp(x,y) = 0 consists of the point 0 = (0,O) and of a spiral going to 0. The point 0 is a singularity of this curve and cp is not differentiable at 0 .

Problems Find the singularities of the curve "y aax2 + X ' . What are the types of these singularities? What are the types of singularities of the curve

called a Descartes leaf? How does the type of the singularity of the curve

vary as the parameters a, h, c vary? Find and investigate the singularities of the curve y2 = ux3 is called the divergent parabola. Find and investigate the singularities of the cissoid

which can be represented in the parametric form

This curve was discovered by Diocles (second century © 2000 CRC Press

BC).

+ bx2 + cx + d, which

SINGlJLAK POINTS OF PLANE CURVES

6. Find the singularity of Maclaurin's curve

presented in the parametric form asin 3t sin t '

= ------

y

a sin 3t

= -----COS

7. Find the singularities of the curve y2 = x4. 8. Draw the tractrix x = a(1og tan p/2 + cos p) + c,

© 2000 CRC Press

t

y = a sin p.

10 Peano's Curve In 1890, Peano constructed the beautiful example of a continuous map of an interval whose image is a square. The construction is the following. Decompose an interval A into four equal intervals A i l ,i l = 1,2,3,4 indexed from left to right. Then decompose a square A into four equal squares A i l ,il = 1,2,3,4, enumerating these small squares in such a manner that consecutive squares have a mutual side. We call the constructed decomposition a first step. Let us assign to each interval Ail the square A,, and denote this correspondence by f i , i.e. A,, = , f l ( A i , ) .Decompose every interval A,, in a similar way into four equal intervals AiIi2.We also decompose every square A,, into four equal squares here we choose the index i2 in such a way that the Ail,, have a mutual side with AiIi,,I and AiI4have a mutual side with A i l + l l We . call this decomposition a second step. Let us assign to each interval A,,,, the square Ai,;* and denote this correspondence by f 2 , i.e. A,,;, =,f2(AiIi2). We continue the decomposition in this way and denote intervals of the n-th step by A,, ..., and squares by A,, ...i,,. If two intervals of the n-th step have a mutual point, then the corresponding

1

1

1

1

1

4, 1

FIGIJRE 10.1


1

1

1

50


squares of the n-th step have a mutual side. The correspondencef, between intervals and squares has the following property: if Ail...;,,c Ail...;,,_l,then fn(Ai,...,,) c j;,-,(Ail. Let us assign to every point of A a point of the square A in the following manner. Each point t E A belongs to an infinite sequence of enclosed intervals A i l ,A;,,,, . . . , A ;,..., , , . . . . The squares corresponding to these intervals form an infinite sequence of enclosed squares:

Because the lengths of the sides of the squares converge to 0, there exists a unique point P E A, belonging to all squares of the sequence. Thus we define the map f : A 4 A, assigning to every point t E A the corresponding point P = f ( t ) . It follows from the definition that if t E Ail...i,, then f ( t ) E ,fn(Ai,...i,,). Let us prove that every point P E A is an image of some point t E A . For any point PO E A there exists at least one infinite sequence of enclosed squares (A;, > A;,,2 > . . . 3 Ai, ...,,,) > . . . such that Po belongs to all squares of the sequence. This sequence corresponds to an infinite sequence of enclosed intervals A,, A;,;,> . . . 3 A ;,...i,,) > . . . such that j;,(Ai,...;,,) = A ,,...i,,, n = l,m, and the intervals of this sequence have a unique mutual point to, hence we obtain by the definition off: .f(to)= .P Therefore the image of the interval A under f is the square A! The mapf is continuous. In order to prove this fact let us take a positive t. When n tends to infinity, the length of the sides of the n-th step's squares converges to zero. Therefore there exists a number n, such that all squares of the n,-th step , which contain the point f ( t o ) , belong to the F-neighborhood of j'(to). Let us take the intervals of the n-th step containing to. To every point of these intervals the map f' assigns a point of the squares of the n-th step belonging to the t-neighborhood of f(to). This means that the map f is continuous. An analog of Peano's curve was constructed in [57] with the help of a complex power series F(z) = C z ocnzr'.A series F(z) was found, with the following properties: inside the circle / z /< 1 it was convergent ; it was continuous inside the closed circle / z I< l; the set of values of the function ~ ( e " )t ,E [O, 2n]contained an open set of the z-plane.

© 2000 CRC Press

11 Envelope of the Family of Curves Above, we considered a regular family of plane curves given by an equation @(x,y) = 0; different curves of the family were mutually disjoint. Now we will consider families of curves such that different curves of a family can be mutually intersecting. Let us take , for example, the family of circles of radius R whose centers are situated on the X-axis.The equation of any such circle is

For different a we have equations of different circles of the family. Two arbitrary sufficiently near circles are intersecting. In general we will assume that an equation

defines a family of plane curves. This means that for every fixed a the equation ( l 1.2) represents some curve y,,of the family.

FIGURE 1 1.1

© 2000 CRC Press

52

DIFFERENTIAL GEOMETRY A N D TOPOLOGY OF CURVES

On examination of a family of plane curves, it can be seen that some plane curves are singular with respect to the family. For example, the lines y = R, y = -R are the singular curves concerning the family given by (1 1.1); at each point these lines are tangent to a circle of the family, i.e. they are the enveloping curves of the family.

Definition A plane curve I' is called an envelope of a family of plane curves, if I' is tangent at each point P to a curve ofthe,family passing through P. We will find the equation of an envelope I' of the family given by (1 1.2). Let P be a point of F with coordinates X, y. Suppose that there exists a unique curve y,,of the family, passing through P. Then we can assign to the point P a unique value of the parameter a ; thus we can consider the coordinates X,y of points of the envelope as functions of a: x = x(a), y = y(a) - this is a parametrization of the envelope by the parameter of the family. Because P lies on the curve y,, we have

Differentiating with respect to

ai

we get

The envelope I' and the curve y,,have a common tangent at the point P. Since y, is given by equation (1 1.2), the normal of y, at P has the coordinates (f,(P), f;(P)). The tangent vector of F at P is (dxlda, dylda), therefore

Hence the points of the envelope must satisfy the two following equations:

Eliminating (if possible) the parameter a from these two equations, we obtain an equation

which represents the envelope.

© 2000 CRC Press

12 Frenet Formulas Let a space curve y be given by its position vector r = r(s) viewed as a vector-valued function of the arc length S. We will denote by T the unit tangent vector r l ( s ) , by v the principal normal, by 0 the binormal of y. Three vectors T , v, P depend on the parameter S, hence we consider these vectors to be vector-valued functions of S . At any point of y the vectors T , v, p are mutually orthogonal and form a basis of Euclidean space. We say that T , v, P.form the natural frame. The derivatives of these vectors are decomposed into linear combinations with respect to the natural frame at the corresponding point:

Let us find the coefficients of the decomposition. It follows from the definition of the principal normal that d2r dr = = kv. ds2 ds Therefore a1 = 0, a2 = k, a3 = 0. When we proved the theorem about torsion (see chapter 7), we proved the formula

hence cl

= 0, c2 =

-K,

c3 = 0. Because v is a unit vector, 62 = (v,VI, ) = '-(V, V): = 0. 2

© 2000 CRC Press

54


Using the decomposition of v,: and the found values of al, c2 we obtain

Thus

These decompositions of the vector ri, v.:, /3: are called Frenetformulas. They have a very important significance in the differential geometry of curves. Let us apply the Frenet formulas in order to investigate the behavior of y at a neighborhood U of some point P E y, where s = so, k(so)# 0 and &(so)# 0. We write a Taylor expansion of the vector-valued function r(s) at U:

According to our notations we have

It is easy to see, with the help of the Frenet frame, that

Using the obtained expressions we can write

We will describe the behavior of y by considering the projections of y into the coordinate planes of the natural frame at P. Assume that Cartesian coordinates x,y, z are fixed in such a way that P is the origin and the x-axis, y-axis, z-axis are collinear to the vectors T , v, p respectively. The projection of y onto the plane spanned by r and v is given by the coordinates x(s), y(s), which are the coefficients at T and v of the Taylor expansion (12.1):

© 2000 CRC Press

FRENET FORMULAS

FIGURE 12.1

So, the projection of y into the the parabola

( 7 ,U)-plane (Figure

12.1) is approximately equal to

y = - kX .2

2

The projection into the (r,O)-plane is a curve given by the coordinates

+ A S )= as+ +S), as3+ o ( A s ~ ) , Z ( S ~+ as)= kK 6 ---

hence it is similar to the cubic parabola

Such a curve with K > 0 is shown in Figure 12.2. The projection into the (v, P)-plane is a curve given by the coordinates

y(so

as2+ o ( A s 2 ) , + As) = k 2

z(so

As3 + As) = k~ --+ o(As". 6

FIGURE 12.2

© 2000 CRC Press


FIGURE 12.3

It is clear that if As is sufficiently small, the coordinate y is positive. Eliminating As, we find that the projection behaves similarly to the curve

Therefore the point P is a cusp of the first type, and both branches of the projection are tangent to the y-axis (Figure 12.3). Using three constructed projections one can find the form of y at a sufficiently small neighborhood U. Assume that the torsion K. is positive. Because the pro-jection

FIGURE 12.4

© 2000 CRC Press

FRENET FORMULAS

57

into the (X,y)-plane is similar to the parabola y = 5x2, z = 0, the curve y is approximately situated on the cylinder y = 5x2 with the generator collinear to the z-axis. It follows from the behavior of the projection into the (x,z)-plane that the branch P A corresponding to As > 0 of y lies above the (X,y)-plane, and the branch PB corresponding to As < 0 is under the (X,y)-plane (Figure 12.4). For some curves y one can define the curvature with a sign. Suppose that there exists a continuous and differentiable field of orthonormal frames e,, e2, e3 satisfying two conditions: (1) el is the tangent vector; (2) a t points where the principal normal is defined, one of the vectors e2, -e2 is equal to the principal normal. Writing the derivatives of e, with respect to the arc length S , one can obtain an analog of the Frenet formula:

where k, are some functions of S . The function k l is called the curvature with sign. It is equal to the curvature k up to the sign; it may be useful when k = 0 at some points and the principal normal passing through such points changes its direction.

© 2000 CRC Press

13 Determination of a Curve with Given Curvature and Torsion Earlier we defined two geometric notions - the curvature and torsion of space curves. The curvature depends on the first and second derivatives of the position vector; the torsion is calculated with the help of the first, second and third derivatives. Using higher derivatives one can construct other geometrical concepts characterizing the behavior of curves. But it is unnecessary, because any curve is completely determined by its curvature and torsion. More accurately, we will prove Theorem Let k(s) and ~ ( sbe) given continuous functions qf'a parameter s E [0,I ] , k(s) is positive everywhere. Assume that a point PO and three orthogonal unit vectors 7 0 , V O , PO = 170, uO]arefixed in Euclidean space. Then there exists a unique C'-regular curve y having the.following properties: ( I ) y passes through PO and s is its arc length countedfrom PO; ( 2 ) 7 0 , U O , PO is the natural frame of y at PO; ( 3 ) k(s) is the curvature and K ( S ) is the torsion of y. ,

Proof Let us consider the system of linear differential equations similar to Frenet formulas:

, are unknown functions. The coefficients k(s), ~ ( sof) this system where [ ( S ) , ~ ( s )tax{ © 2000 CRC Press

CURVES IN MECHANICS

The integration of this equation leads to the following:

where a is a constant vector. From the last equation it follows that p

+ ( a ,r ) = 0.

This means that the trace of a charged particle is situated on a cone with its center at the magnetic pole. Moreover a trace o f a chargedparticle moving in a magnetic field is a geodesic. curve of a cone, i.e. a curve which transforms in a straight line when tlze cone develops onto the plane. Motion of a Charged Particle in a Constant Electromagnetic Field Let E and H be electric and magnetic components of a constant in the space and time electromagnetic field. The equation of motion (18.6) can be rewritten:

where

P

and p are constants. From this equation it follows that

where c is a constant vector. Let us choose Cartesian coordinates as follows: the z-axis is collinear to H and the y-axis is orthogonal to the plane spanned by E and H. Then

and we can write

© 2000 CRC Press

88


From the first and second equations we obtain

where a = % p and p is a constant. The general solution of this equation is X=

-p

+ sintat + a ) A ,

tu = cons!,

A

= corzst.

Substituting the obtained expression of x into the second equation of system (18.7) we can find y=-qt+h-cos(at+or)A,

q=const,

h=consl.

The general solution of the third equation of system (18.8) is

Thus we can represent the motion of a charged particle in the form of a sum of two motions: a uniform motion along a circle X = Asin(at

+ a),

J, =

-Acos(at

+

tu),

and a motion of this circle whose center is moving along a plane curve

Note that the moving circle is parallel to the ( X , y)-plane. In the case of vanishing XI and q the particle moves along a spiral.

© 2000 CRC Press

19 Curve Filling a Surface Some simple curves have a very complicated behavior. In this chapter we will construct a space curve situated on a torus, which fills this torus densely everywhere. First, let us write the equation of a torus. This surface is formed by points of a circle (1, which is moving in such a way that the center of a is moving along a circle P and the planes containing a and are mutually orthogonal. We will write the position vector r of the points of the torus in the form of a vector-function of two variables 4 and H. Let us place the circle P into the (X, y)-plane with the center of P at the origin 0 of Cartesian coordinates. Denote the radius of B by R and the radius of a by p. Let e, be unit coordinate vectors, P an arbitrary point of P and 4 an angle between el and OP. Then O P = R(cos 4 e l

+ sin 4ez).

+

We will denote the unit vector cos 4 el sin 4 e2 by ~ ( 4 )The . circle a is situated in a plane y, which passes through O and is spanned by e3 and OP; the center of cw coincides with P. When the plane y rotates, the points of the circle a form the torus (Figure 19.1). Let Q be a point of a. Then we can represent OQ in the form of the sum OP PQ. The vector PQ is a linear combination of U and e3. Denote an angle between v and PQ by 0; then we have

+

PQ

= p(cos B I /

+ sin H e3).

Thus the position vector of the points of the torus has the following form: r(4,H) = R(cos 4 el + sin 4 e2) p(cos B(cos 4 el sin 4 e2)

+

© 2000 CRC Press

+

+ sin Q e3).

DIFFERENTIAL GEOMETRY AND TOPOLOGY OF ClJRVES

FIGURE 19.1

Now let 0 be a linear function of 4:

where X is an irrational number. When 4 is varying from -m to +m, the points with coordinates 4, 0 = A4 on the torus form some curve r. If X is a rational number p/q then the curve F will be closed after p rotation with respect to 0 and q rotation with respect to 4. Let us fix the circle a0 given on the torus by the formula 4 = 0. If X is irrational then the number of points where F intersects a0 is infinite. We will show that these points (i.e the points with coordinates 4 = 27rn, 0 = A27rn, n E Z ) form a set on cu0 which is dense everywhere. Let us consider the Poincarl map g: a0 --, a0 defined by the formula g(0) = 0 X27r. Denote a composition of k maps g by $. The map g preserves lengths of arcs of tro. Let us take an arbitrary point Q E a0 with angle 00, a positive t , and an arc U c a0 containing Q and whose length is less than 4 2 . Because the arcs U, g ( U ) , . . . ,g k ( u ) ,. . . have an equal length, there exists k > l such that & U ) n g'(U) is not empty. Every gh has an inverse map which we denote by g p k . We can consider the set M = g'-'(U); it is easy to see that g'(A4) = g k ( u ) . Let P E $ ( U ) n g ' ( ~ ) Then . g - ' ( ~ )C M and g - ' ( ~ )C U. Therefore the intersection g k p ' ( u ) n U is not empty and 100 - g"(OO)l< t, where m = k - l. Hence the points 00, g)ll(OO), g2m(00),. . . ,gnm(O()), . . . decompose a0 into equal arcs whose length is less than t. Thus for any fixed arc of a0 we can find a positive t and an integer m such that points gn"'(Oo) are situated inside this arc. This means that the set of points gN(Oo),n E N is dense everywhere on ao. Now it is easy to show that the curve F filling the torus is dense everywhere. Note that the projection of T onto the ( X , y)-plane is represented by the formulas

+

X

= cos4(R+pcosX4),

y

= sin 4 (R

+ p cos X4).

If R > p then this projection is a regular curve filling an annulus densely everywhere between two concentric circles.

© 2000 CRC Press

20 Curves with Locally Convex Projection Let us translate a straight line l parallel to a vector e along a space curve y. A surface formed by the moving line l is called a cylinder. The curve y is called the d k c t r i x of the cylinder and the line 1 is called the generutor of the cylinder. We devote this chapter to curves on the cylinder. Without loss of generality we can suppose that y is a plane curve situated on the ( X , y)-plane and 1 is parallel to the z-axis. Denote by ro(t)the position vector of y,by t the length of arc of y and by e the unit vector spanning l. Let be a regular curve lying on the cylinder. Its position vector has the form r(t) = ro(t) z(t)e, where z(t) is the third coordinate of the points of r. The curve y can be viewed as the projection of r into the ( X y)-plane. , For the torsion r; of l? we have the following expression:

+

K

=

(r', r", r"') zl'lp + z'lpl + z l / p l [ r y 1 2 = I +z12 +z/12p2

where we denote by I/p the curvature of y and by a prime the derivative with respect to t. If p(t) is the angle between the vector tangent to y at point ro(t) and a fixed vector in the ( X , y)-plane, then dtldp = p. Suppose that p > 0, i.e. y is locally convex. By this assumption y is parametrized regularly by p and z' = $(p). Because

we obtain from (20.1):

It is easy to see that ds = p ~ m d is pthe differential of the length of arc s of We will use formula (20.2) to investigate the behavior of l?.

© 2000 CRC Press

r.

DIFFERENTIAL GEOMETRY AND TOPOLOGY OF CUJIVES

02

Lemma Assume that the curvature k and torsion projection y of is locally convex, then

K

of'r are not vanishing. lj the

Proof Rewrite expression (20.2) as follows:

Multiplying both parts of (20.4) by J1+1/,2and integrating with respect to cp we obtain

for any points PO, P I of I? and corresponding points P ~ , of y. Inequality (20.3) B follows from (20.5) immediately.

Corollary Assumr that the curvature k of r is not vanishing, the torsion r; is bounded K > KO = const > 0 and the total length of is infinite. Ifthe curvature I / p of y is bounded l / p 5 const, then the total length o j ' y is infinite. Proof Because 1 S,. K dsJ 2 1 S,. KO dsl = ~ o L ( ris) infinite, we obtain from (20.3) that dt/p is infinite, and it follows from the boundness of l / p that S*, dt is infinite. Now let us consider the following problem: assume that some curve satisfies three conditions: Jy

(a) its total length is infinite; (h) its curvature is positive and its torsion is bounded from below by a positive constant; (c) it has a locally convex projection. When is this curve bounded? This question is not trivial. For example, the curve x = cos p, y = sin p, z = cp - 2 sin cp has infinitely many stationary points with respect to the z-coordinate.

Theorem [5J Suppose that a regular curve r of infinite tolal length has a positive curvature k > 0, a torsion K bounded from below by a positive constant KO > 0 and a closed strongly convex projection y. Then I' is unbounded. Proof First, the curve r is not tangent to the z-axis, because of the convexity of y and k # 0, K # 0. Therefore the function zi is regular. © 2000 CRC Press

CURVES WITH LOCALLY CONVEX PROJECTlON

Rewrite (20.2) in the following way:

+

where Q(p) = & ( l $2 i-(d$/dp)'). It follows from the differential equations theory that li, can be represented by the formula

/

011

$(h)=

sin($

-

i.)Q(p)pdp

+A

COS

Cl + Bsin 80,

0

where A and B are constants. Assume that the origin of Cartesian coordinates is a point of y. The coordinates of the position vector rO(s) are X($) =

S

COS

p(7) dr,

y(s) =

Denote by I the total length of y and set ~ ( s=) Q(p(s)). When a point of the closed curve y goes along y once, a corresponding point of l? goes along l? and a variation of its z-coordinate is

We observed that x d y - y dx = r i do, where cu is the angle between vector ro and the x-axis. Since y is strongly convex, dolds > 0. Thus

where S is the area of the plane domain bounded by y. From the corollary it follows that if a point P goes along the whole curve F, the corresponding point P goes along the whole curve y infinitely many times. Hence the variation of the z-coordinate of P is infinite. Thus, l? is unbounded. It must be pointed out that the condition of strong convexity is essential. One can construst a closed curve r with k > 0 , K KO > 0 and with a locally convex projection. For this purpose let us set

>

+('P) = 1 - 4 cos X'P)

© 2000 CRC Press


p(p)

=

l

+ -21 cos Xp,

-00

< p < +cm,

where X is a rational number such that X2 = 1 + t, 0 < t < 114. We define the curve l? in the following manner: 'P

'P

X =

c

o

s

y =Ipsinpdp, 0

0

z= /+pdp 0

It is easy to obtain that 112 < p 5 312 and

If X2 - 1 < 114, then K is greater than some positive constant Q. On the other hand, because X is rational and X2 # 1, the curve r is closed. But if the constant KO is sufficiently great, we cannot construct such an example. Theorem Assume that a curve r of injinite length is regularly projected into U locully convex curvcj y with curvature Ilp. Let 0 < l/p < const und the torsion K c,f r satisjy the inequu1it.v K sup&. Then l? is unbounded.

>

Proof Let us show that there exists cpl such that the corresponding function $(p) is monotone on the ray p pl. Assume that there exist two points PO= P(po) and PI = P ( p l ) such that W l d p is vanishing at these points. Using (20.5) we have

>

>

Because K sup ,'2~ 4)j,r 1 on the segment [po,p,]. Therefore 4)E 1 on the whole ray [po, cm), or there exists p1 such that is monotone on the ray cp 2 p,. Then y', does not change the sign on some ray p 2 ( ~ 2Let . us take an arbitrary segment A belonging to the ray [p2, m). From the inequality

+

+

+

we have

where KO = inftc, AZ is a variation of z at A. Because one can take A to be an arbitrarily great length, lAz/can be sufficiently great too. Therefore r is unbounded. © 2000 CRC Press

CURVES WITH LOCALLY CONVEX PROJECTION

95

We remark that the value 112 is essential for lcp. If lcp > 112, then z,, does not change sign when s is sufficiently great; but if i n f ~ p< 112 - t, where F is a small positive constant, the function z,, can have the form of an oscillating function.

© 2000 CRC Press

21 Integral Inequalities for Closed Curves Let r' be a closed regular curve in Euclidean space E ~Denote . the length of arc of F by s and the curvature by k(s). We will prove Fenchel's inequality [12]:

and we have the equality in (21.1) ij'and only if F is a convex plane curve. Suppose that the curvature of is positive. We will define a spherical indicatrix of tangents of the curve .'I Let P be a point on r and T ( P )a unit tangent vector of r at point P. We translate r ( P ) in such a manner that the origin of T ( P )coincides with the origin 0 of Cartesian coordinates. Then the end point P* of the translated vector r ( P ) is situated on the unit sphere S'. When P moves along l?, the point P* moves forming a curve 7 on S 2 . This curve is called U spherical indicatrix of tangents of the curve l?. The position vector of 7 is represented by a vector-valued function ~ ( s ) . Using the Frenet formulas we have

Hence (dr/ds,dr/ds) = k2. From the supposition of positiveness of k it follows that y is parametrized regularly by S . Let rr be a length of arc of y.Then the length of arc s of F can be viewed as an increasing function s = s(a) of m. The derivative of T with respect to is a unit vector, therefore from

d r ( s ( a ) )- -&(S) ds da ds drr

© 2000 CRC Press

98

DIFFERENTIAL GEOMETRY AND TOPOLOGY 01: ClJRVES

we obtain drr

= kds.

Thus the integral in formula (21.1) is equal to the length l of y:

So, in order to prove inequality (21.1) for some closed curve we must evaluate the length of its spherical indicatrix of tangents. Assume that r is closed and non-planar. Because is closed, for any plane a there exist points on which are most (or least) distant from a. The tangent vector T of r at each of these points is parallel to cr. Therefore the indicatrix y meets any plane a passing through 0. Thus the spherical indicatrix of tangents of a closed non-planar curve has the following property: this curve intersects any great circle at least in two points and therefore it does not belong inside any hemisphere. Now we will prove that any closed curve, which is situated on S' and whose length is less than 2n, belongs to some open hemisphere. For this purpose we will use the following property of the great circles on S': for two arbitrary points on S2 the shortest curve on connecting these points is an arc of great circle and this arc is defined uniquely if the points are not antipodal; we will call the length of the shortest curve an inner distance between two points on S*. We will expound the proof of the following more general statement (see [ l l]): Theorem Let y be a closed curve on S* and suppose that length L qf y is less than 2n. Then there exists a point Q E S' such that for any point P E y the inner distance between Q and P is not greater than L/4. Proof We will denote by p(*, *) the inner distance on the unit sphere. It is easy to observe that if the inner distance between two points A and B is less than n, then there exists a unique point C on the arc AB such that p(A, C ) = p(C, B) = p(A, B)/2; the point C is called the center of the pair A, B. Lemma Let p(A, B) < n , C be the center of the pair A, B and X a point on the sphere such that p(X, C ) < 71.12. Then 2p(C,X ) = p(A, X ) p(B, X ) .

+

Proof In order to prove the lemma let us consider the great circle connecting C and X, and fix a point X on this circle in such a way that p(C, X ) = p(C, X ) , i.e. X is symmetric to X with respect to C. Because A is symmetric to B with respect to C, we have p ( ~A) , = p(X, B). Since p(C, X ) < n/2, p(X, X ) = p ( ~ C, ) + p(C, X ) = 2p(X, C ) . On the other hand, by the triangle inequality: p(X, X ) 5 p(X, A) + p(A, X ) = p(X, A) p(B, X ) . Thus 2p(C,X ) = p(A, X ) p(B, X ) . Now let us fix two points A, B E y; these points decompose y into two arcs AB and BA of equal lengths L(y) = 2L(AB) = 2L(BA). Because p(A, B) < L(AB) = L(y)/2 < n,there exists a unique center C of the pair A, B. Let X be an arbitrary point of y; we assume without loss of generality that X belongs to the arc AB. Suppose that p(C, X ) < 71.12.Because p(A, X ) is less than the length of the arc AX and p(B, X ) is less than the length of the arc XB, using the proved lemma we obtain 2p(C, X ) < p(A, X)+p(B, X ) < L ( A X ) L(XB) = L(AB) = L(y)/2. Therefore p(C, X ) < L(y)/4.

+

+

+

© 2000 CRC Press

INTEGRAL INEQUALITIES FOR CLOSED CURVES

99

Thus we have proved that if p(C, X ) < L(y)/2, then p(C, X ) 5 L(y)/4. So, for any point X of y two alternatives exist, p(C, X) 5 L(y)/4 and p(C, X ) > 7r/2. Since p(C, X ) , when viewed as a function of X E y, is continuous and p(C, A)= p(A, R)/2 5 L(AB)/2 = L(y)/4, for any point X E y we have p(C, X ) L(y)/4. Because the spherical indicatrix of any closed non-planar curve does not belong to any hemisphere, we can conclude, using the proved theorem, that the length of the indicatrix is greater than 27r; it is easy to complete the proof of Fenchel's inequality. Let us consider knotted curves without self-intersections. By a knotted curve we understand a closed curve which cannot be deformed continuously without selfintersections into the unit circle. Fenchel's inequality is generalized by the FaryMilnor inequality:

0 for any positive function n(u) satisfying a(n) = o(h). On the other hand, ( [ ( U )e) , is a function of 24 with alternating sign, and we can choose the function o*(u)in such a manner that a*(u) is sufficiently small when ( [ ( U )e) , > 0 and sufficiently great when ( < ( U ) , c ) < 0. Then J ' ([, e)o*(u)du< 0 and we obtain a contradiction. Thus the set M contains 0,i.e. there exists a positive function @ ( U ) satisfying @(U) = @(h)and i ( h ) = (@(U) du = 0. This means that a curve with the position vector ?(U)= h f [ @ ( u ) r his closed, and it is obvious that this curve is regularly parametrized by u and that y is its spherical indicatrix of tangents.

h:

© 2000 CRC Press

23 Conditions for a Curve to be Closed Let us consider a planar curve y with position vector r = r(s). Assume that the curvature k(s) of y is a continuous periodic function and its period is equal to T. We have

The vector tangent to y has the following form: r:, = { cos a , sin a ),

where a denotes the angle between rt. and the x-axis. The curvature of the planar curve y,when viewed as the curvature with sign, can be expressed in terms of a:

Therefore u(s) = a0

+

1 S

k(s) ds,

where

a0 = consr.

0

Assume that y is closed and that its length is equal to L. If a point goes around y, then obviously the variation of a is divided by 27r. Thus

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104


where m is an integer. We obtain, after integration of the expression of rr(s):

S

cos ads = 0,

S

sin ads = 0.

0

0

Substituting (23.1) into (23.2) we conclude that satisfies the equalities

if

y is closed, then its curvature k(s)

One can demonstrate that if the curvature of some planar curve satisfies equality (23.3), then this curve is closed. Efimov, Fenchel and other geometers independently studied the following problem: what are the necessary and sufficient conditions on the curvature and torsion o f a space curve in order for the curve to be closed? It is assumed that the curvature k(s) and torsion IF.(S)are functions of the arc length S. Fenchel thought that this question is the most important one in the theory of closed curves. There exists an ineffective way to solve the problem. The reconstruction of a space curve with given curvature and torsion is reduced to one integral Volterra equation. By the Neumann formula a solution of such an equation can be written in the form of a series, any sum of which is determined by the curvature and torsion. Effective conditions may be found in some special cases. We will consider one such case. Let y be a closed space curve with curvature k(s) and torsion IF.(s). Naturally we assume "k tc2 > 0. We replace the arc length s by the parameter

If L denotes the length of y, we set T = t(L) and define a function cp(t) in the following way: cos cp

=

k

sin cp =

,/W'

IF.

JFT-2

It is clear that if the functions cp(t) and t(s) are given, then the curvature k(s) and torsion ~ ( sare ) reconstructed uniquely, and if the curvature k(s) and torsion ~ ( sare ) given, they determine the functions p(t) and t(s) uniquely. (We assume that K is defined with a sign.)

Theorem Let the curvature k(s) and torsion ~ ( sof) a space curve y be such that p = a1 t + az, ai = const. The curve y is closed and its length is equal to L qj'there exist two integers p, q such that a1 = (g - p)/2&Gj and © 2000 CRC Press

CONDITIONS FOR A CURVE 7'0 BE CLOSED

I

sin p ds = 0,

(23.5)

Proof Denote by t l , G, t3the natural frame of the curve y.Frenet equations have the following form:

3=

-

cos p 0 be a fixed positive number, c 2 ( 1 )denote the class of c2-regular closed curves of length l that are situated on the unit sphere S. Assume s denotes the arc length, and k(s) stands for the geodesic curvature of curves from C2(1). Theorem Let F(x, y) be a continuous .function. Then either ( l ) the set of' values of the integral

c :imputed for all cuvves 1, W I C2(l) contazi 7 int~rval, or ( 2 i F(x,y ) p b ) , whei e p(y) u a perio~ fun^. )n M 111 od

--

© 2000 CRC Press

j

~

h

L

CONDITIONS FOR A CURVE TO BE CLOSED

113

Theorem Let F(x, y, z ) and G ( x ,y) be analytical functions defined in the space and in the ( X , y)-plane respectively. Then either ( l ) the set cf values of' the integral

contains an interval, or computed,for all curves.from c2(1) &), where p(y) is a periodic ,function with prriod I, or (2a) F(x,y, z ) (2h) G ( x , y ) cp(y), where PO/) is a periodic filnction with period l, and F ( x , y , z ) = Y'i(x,y). Remark Cases (2) of the stated theorems mean that the considered integral equalities do not depend on the geodesic curvature, hence they are empty. Necessary conditions for space closed curves were investigated in [62] for so-called trigonometric curves. By definition, a space curve y is called trigonometric if each component of its position vector r(s) with respect to the arc length s is a trigonometric polynomial:

where ck are constant complex vectors satisfying the equations c-k = c,:. Of course, any trigonometric curve is closed. So, we can interpret the Efimov-Fenchel problem in this case as follows: what are the properties of the curvature k(s) and the torsion r;(s) of the trigonometic curve y with position vector (23.25)? It is more convenient to (s) of k(s) and r;(s). One can demonstrate consider the functions /c2 and ~ ( s ) k ~ instead easily that if r(s) is a trigonometric polynomial of degree n, then k 2 and r;(s)k2(s)are trigonometric polynomials of degree 2(n - 1) and 3(n - 1 ) respectively:

It is observed that the coefficients of these two expansions are mutually connected.

Theorem I f r(s) is a trigonometric polynomial of degree n with respect to the arc - 1) are connected by three algebraic rrlations length, then the coefficients ao, . . . ,

involving the complex conjugates 5,. © 2000 CRC Press

1 14


Each coefft'cient

pp is connected with

the coefficients a, by an algebraic relation

The last two coeflicients ,L$ are

Here F,, Q iand 0, denote algebraic polynomials with integer coefficients As we have mentioned, the degrees of k2 and &k2 are not arbitrary. Theorem Let k(s)and ~ ( sbe) the curvature and torsion ofa trigonometric curve. Then the degree v of the trigonometric polynomial k2 is even, and the degree p of the trigonometric polynomial &k2is divided by 3. Also, ij'p is equal to 31, then v is equal to 21. It is natural to consider a trigonometric curve whose position vector is a trigonometric polynomial of degree n such that the degree of the corresponding polynomial k 2 is less than 2(n - l) or the degree of the corresponding polynomial &k2is less than 3(n - 1). One can show that if the last 2k coefficients of the polynomial k 2 are equal to 0, then the last 2k vectors cp have a very specific form. First, we say that two complex vectors c, d are collinear if there exists a complex number X such that c = X4 next, any complex vector c = a ib with non-zero real part a and non-zero imaginary part b determine a unique plane in E3 spanned by the vectors a and h; we say that this plane is the plane of the complex vector c. It can be proved that if ~ 2 ( , - 1 ) = . . . = Gt2(,-k) = 0, then the vectors c,, . . . ,c,-k, c] - (c,, e)e, j = n - k - 1 , n - 2k are collinear, where e is a unit real vector orthogonal to the plane of the vector c,.

+

Theorem Let the curvature k of a trigonometric curve y C is an arc qf a circle. Theorem Let the torsion an arc of a circle.

K

he constant. Then y

of a trigonometric curve y C E3 be constant. Then y is

We remark that the condition y C is essential. In Euclidean spaces, whose dimensions are even and greater than three, there exist trigonometric curves with constant curvature and torsion, which are different from an arc of a circle (see chapter 32).

© 2000 CRC Press

24 Isoperimetric Property of a Circle Let r be a closed rectifiable plane curve, L denote the length of T,F stand for the area of a plane domain bounded by F. The length L and the area F are mutually connected by the following isoperimetric inequality:

= 47rF is true ij" is a circle. One can interpret this fact in the The equality following two ways: among closed rectifiable plane curves with the same length, the circle bounds a plane domain of greatest area; among closed rectifiable plane curves bounding plane domains with the same area, the circle has the smallest length. We will sketch the proof of the isoperimetric inequality in the case of smooth curves. Let %(S),j(s) be Cartesian coordinates of the position vector and s the length of arc of r. It is more convenient to use a parameter t = 2 m l L in place of S. Functions x ( t ) = i ( t L / 2 7 ~ y(t) ) , = j(tL/27~)are periodic functions with the period 27~. Since

we have

therefore

© 2000 CRC Press


1 16

The function x(t), y(t) can be represented by Fourier series

+ C ak cos kt + bk sin ktl m

x(t)= an/2

k= l

y ( t ) = co/2 +

C 02

ci, cos kt

+ 4 sin kt

k= l

as well as the derivatives

"

dx dt

x ( - a h sin kt

dy dt

" C(-ck sin kt + dk cos kt)k.

-=

-=

+ bk cos kt)k,

k= l

k= l

It is easy to show, using the orthogonality of functions sin kt, cos kt, sin nzt, cos mt, k # m, that

The area F is computed with the help of a standard formula of analysis:

+ C ( u kCOS kt + bh sin k t ) W

0

(Xi

sin kt

k= l

Therefore

0

© 2000 CRC Press

0

l

+ dk COS kt)k

dt

ISOPERIMETRIC PROPERTY O F A CIRCLE

- 47rF > 0. The Because the right-hand side of this expression is not less than 0, equality appears iff ak = bk = ck = dk-= 0, k 2 2 and a1 = dl, 61 = -cl. In this case the position vector of has the following form:

+

x(t) = a012 ul cost y(t) = col2 - bl cost

+ hl sin t , + a1 sin t.

Obviously it is the representation of a circle. The isoperimetric inequality was proved by Steiner; the proof expounded above was given by Hurwitz.

Problems 1 . Let y be a closed convex planar curve and 0 E y a fixed point. Consider a function h(P) assigning to each point P E y the distance between 0 and the straight line tangent to y at P. This function is called the support function of y. Assume that y is parametrized by an angle a formed by the straight line tangent to y and a fixed straight line. Prove the formula: p(a) = h(o)

+ hrr(cu),

where p(@)denotes the curvature radius of y. 2. Assume that two planar closed curves yl and y2 are mutually tangent at some point P in such a way that they are situated in the same half-plane with respect to the mutual tangent straight line at the point P. Suppose that for any pair of points P I E yl, P2 E 7 2 such that the straight lines 11, 12 tangent to y l , 7 2 at P I , P:, respectively are parallel, the curvatures k l of yl and k2 of y2 satisfy the inequality k l ( P I )2 k2(P2).Demonstrate that yl is situated inside 7 2 . 3. Prove that the area of the domain bounded by a closed planar curve y is equal to

/"

I h($)ds. 2 4. Prove that there does not exist a regular function fik) satisfying the following condition: for any closed planar curve y the integral

where k(s) denotes the curvature of y, is equal to the area of the domain bounded by 7.

© 2000 CRC Press

25 One Inequality for a Closed Curve Let F be a closed curve situated inside a ball of radius R. We will expound the proof of the following inequality demonstrated in [31]:

where s denotes the length of arc, k is the curvature, K is the torsion and L is the length of S. Let r(s) be the position vector and ( S ) , &(S), &(S) the natural frame of r. We define functions xi(s) = (ti(s),r(s)). Using the Frenet formulas, one can see that the functions XI,x2,x3 satisfy the differential equations:

cl

Because S is closed, we have

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120


for arbitrary constant p. Since r is situated inside a ball of radius R, 1x21 5 R. Let us set

Then we can obtain

Substituting (25.3) into (25.2) we obtain the desired inequality (25.1). Since R 5 L/4, a n inequality involving the length, curvature and torsion of l? follows from (25.1):

The integration of the identity I r (r, r')'

-

(v, r") leads to the following inequality:

In some cases inequality (25.1) is more optimal than (25.4), in other cases inequality (25.4) is more optimal.

© 2000 CRC Press

26 Necessary and Sufficient Condition of the Boundedness of a Curve with Periodic Curvature and Torsion Let the curvature k(s) and torsion &(S)of a curve r, when viewed as functions of the length of arc S, be periodic functions with a period T. Denote by r(s) the position vector and suppose that r(0) = 0; by cp we denote a space rotation transforming the natural frame @(O) of at point r(0) into the natural frame @(gat the point with the position vector r(T). We will prove the following: Theorem (Bakel'man and Werner, [16].) A curve r of infinite length is bounded iff the vector r(T) is orthogonal to any eigenvector of cp corresponding to the eigenvalue l. One can reformulate this statement in the following way. Let l be an axis of the rotation cp; this means that the vector a spanning l is a stationary vector of the rotation cp. If @(O)# @(T),then a is unique. The theorem says that r is bounded iff (r(T), a) = 0. If @(O)= @(T), then l? is bounded iff r(T) = 0, i.e. iff is closed. Proof Because k(s) and &(S) are periodic functions, the curve r is formed by congruent parts. The part rocorresponding to 0 5 s 5 T is congruent to the part r, corresponding to T s I 2T. Since @(O)is transformed into @ ( T )by the rotation cp and F is defined uniquely by the curvature and torsion, the composition of cp and the space translation with respect to r(T) transform rointo rl.Consider two vectors r(T) and r(2T) - r(T). We have

d there exists a curve connecting P I and P2, which has a constant curvature k = 1 and length equal to l; this curve is a suitable helix. This means that Delaunay's problem has no solution. If d < 2 there exist two solutions; they are arcs of the circle with radius equal to 1. If we rotate these arcs around the segment P IP2, we obtain two families of solutions. Schwartz proved the following statement in 1884. Theorem 1 Let the distance d between P I and P2 be less than 2. Let cu < n and

p = 27r - 0 be the lengths of' arcs of'the unit circle, which connect P I and P2. Then any other curve, which has curvature k one of two inequalities:

=

1 and connects P I and P*, has length l satisfying

Schwartz did not publish any proof. In 1921 A. Schur, motivated by David Hilbert, proved Theorem 1 with the help of Theorem 2 Let a plane curve y with end points P, T and the span PTjhrm a closed convex curve. I f we twist the curve y , the length oj'the span P T increases. (By the twisting of y Schur meant a transformation of y preserving the curvature and the length of y). We will demonstrate the proof of Theorem 2 in the case of polygonal lines. An analog of the curvature for a polygonal line L is the set of angles between adjacent segments of L. An analog of the torsion is the set of angles between the planes spanned by the pairs of adjacent segments. So, let y be a plane polygonal line PQl . . . QnT which forms together with the span P T a closed convex curve y*.Let p be a straight line containing PT, qi a straight line passing through Qi, Qi+1; by Mi we denote an intersection point of p and qi, if this point exists. Since y* is convex, the points Mi do not belong to the segment PT. A twisting p of y can be represented by a composition of suitable rotations around q,. More accurately: at the first step the segment PQl is rotated around ql and the part Q l Q 2 . .. Q,T is fixed, so we obtain a polygonal line P ' Q I Q2.. . Q,T; at the second step P 1 Q 1 Q 2is rotated around q2 and Q2Q3 . . . Q, T is fixed, so we obtain a polygonal line P'Q; Q2Q3. . . Q, T , and so on; in the same way we consider n - 3 further rotations leading to a polygonal line Q ~ - ' Q ; - ~. . . Q,-1 Q, T , and this line is an image of PQl Q2 . . . Q, T under cp. © 2000 CRC Press


FIGURE 27.1

Let S be the sphere centered at T of radius R equal to the length of the segment PT, S; the sphere centered at M; of radius R;equal to the length of the segment M;P (see Figure 27.1). Assume that the points M 1 ,. . . , Mk are situated on the ray PM1 of p and M ~ + I ., ..,M,-I belong to the ray PM,-I. Then S1 C . . . C Sk and 3 ... 3 S,-I 3 S. Obviously the sphere Si is transformed into itself under the rotation around q;. The point P', which is an image of P under the rotation around g;, belongs to S 1 , therefore lies inside S2. The point p2 is the image of P' under the rotation around 92, hence p2 is situated inside S2. Continuing in the same way, we see that P' is inside Si for any i 5 k, therefore the length of P'T is greater than the length of PT. Because pk is situated outside the image P"+' of Pk under the rotation around qk+l lies © 2000 CRC Press

DELAUNAY'S PROBLEM

FIGURE 27.2

outside Sktl.Continuing, we obtain that the last point P"-' is situated outside S,,-, , hence it is outside S . Thus the length of the segment P"-' T is greater than the length of PT. Let us apply Theorem 2 in order to prove Theorem 1. Assume that there exists a curve y such that its curvature is equal to 1, it connects P and T, and its length l satisfies the inequality a l 5 P. Suppose that the distance between P and T is equal to d. We take a unit circle passing through P, T; the circle is decomposed into two arcs P T and TP, whose lengths are equal to a and 0respectively. Fix a point N such that the length of the arc PTN is equal to l, and a point T* on the circle in such a way that the length of the segment PT* is equal to d. The point N belongs to the arc TT* and it is easy to see that the length d of the segment PN is not less than d (see Figure 27.2). On the other hand, the curve y is the image of the arc PTN under some twisting, hence d > d by Theorem 2. Thus we have obtained a contradiction. So the assumption a 15 p is false.

O, d(yl,R,)=h>O, d ( C L U L H U R,,, 7 2 ) = C > 0 , where R, is the vertical ray opposite to R. with the origin at C, C L is the vertical segment with end points C and L, L H is the part of y , between L and H. Let © 2000 CRC Press

JORDAN'S THEOREM ON CLOSED PLANE CURVES

135

E = min (a,h, c). We inscribe a closed polygonal line g in y such that the lengths of its edges are less than and the points H , L, P, Q are vertices of g. Let us denote by g ) , g2, g* parts of g whose vertices belong to yl , y;?,LH respectively. The distance between any point of g and the curve y is less than (12. One may state:

3 are more complicated and they are not determined completely by homotopical theory. Maybe, to find connections of the metric properties of a curve y with topology, we must consider a threedimensional closed simply-connected manifold M containing y such that this curve can be viewed as a knot in M.

+

+

© 2000 CRC Press

+

34 Generalization of the Fenchel Inequality Let y be a closed C'-regular curve in the n-dimensional Euclidean space E" with length of arc s and position vector p($). We denote the curvatures of y by k ~ ,. . , k , , - ~ . Fenchel's inequality ( considered in chapter 21) holds for curves in the n-dimensional Euclidean space [63]; if y is a closed C2-regular curve with length of arc s and the first curvature k l ( s )in the n-dimensional Euclidean space E", then

and Jy kl ( S ) ds = 27r if and only if y is flat and convex. In this chapter we demonstrate one generalization of Fenchel's inequality obtained by Gorkaviy [64]. Theorem 1 Let y be a C'-regular closed curve in E", n > 4 . Let for some integer 3 5 m 5 n - 1 , r 2 m 3 the curvatures kl(s),. . . ,km(s)of y be positive; s denotes the length of arc. Then

+

Proof Let y be a C'-regular closed curve in E", n 2 4, and suppose the curvatures kl(sj,. . . ,km($ of y are positive functions for some m such that 3 5 m _< n - 1 , r 2 m + 3. From the assumptions it follows that the first m + 1 vectors El(sj,. . . ,Em+,(s)of the Frenet frame for y are uniquely defined as well as the curvature k m ,,(S), which may vanish somewhere on y. Consider the map G, which assigns to a point s the m-dimensional osculating space P ( s ) c E" of y (i.e. spanned and oriented by the basis of vectors El(s),. . . ,Em(s)). G is a map of y into the

© 2000 CRC Press

184


Grassmannian G(m, n) of oriented m-dimensional subspaces of E". The image S of y under G is called the indicatrix of m-dimensional osculating spaces of y; it is a generalization of the indicatrix of tangents (see chapter 21). We use the Plucker embedding p1 of G(m,n) into the Euclidean space EN, N = C,: (see [65]): if the subspace 7f" c E", which corresponds to the point P E G(m,n), is spanned and oriented by the orthononnal basis 'rll,. . . ,vm, then the position vector of $(P) is the . is known that p1 is multivector [rll,. . . ,rim] considered as a unit vector in E ~ (It analytic, isometric, and maps G(m, n) into the unit sphere sNpl C EN.) We consider the image S*c ENof l? under p2; its position vector is of the form

Using the Frenet formula we obtain :

and it follows from the assumptions that S*is a C'-regular closed curve in EN.Let us select the length of arc a(s) on F* as an increasing function. Then from (34.2) we obtain

and further

+

where w = d

Differential Geometry and Topology of Curves

Differential Geometry and Topology of Curves

Differential Geometry and Topology of Curves

Differential Topology and Geometry