SUBPLANE COVERED NETS Norman L. Johnson University ofIowa low Cl& Iowa
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Johnson, NormanL. (Norman Lloyd). Subplane coverednets /Norman L. Johnson. p. cm. (Monographs and textbooks in pure and applied mathematics;222) Includes index. ISBN 0824790081 (acidfree paper) 1.Projectiveplanes. 2. Nets(Mathematics). I. Title. 11. Series. QA471.564 2000 51 4 l . 2 2 3 4 ~ 2 1 99050200
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PREFACE The nature and theory of geometry, finite or infinite, relies on and springs from the ‘interesting’ examples. It is, of course, wellknown that any projective or affine geometry of dimension larger than two corresponds to the lattice of vector subspaces of a vector space over a skewfield. In this context, a geometry of dimension two shall mean simply a projective or affine plane. That such geometries do not or may not correspond to such lattices is precisely what makes the study of projective planes intriguing. But, what are some examples, what makes them interesting and what is to be learned from them? In the early 1960s, when I became interested in the general area of combinatorial geometry and, more particularly, in projective planes, there were very few examples of ‘interesting’ projective planes, particularly in the finite case. In fact, the then current hot research centered on the LenzBarlotti classification of projective planes by the extent of socalled pointline transitivities (see [ 5 ] ) and the possible determination of examples in the various classes or the nonexistence of the same. When I met Ted Ostrom in 1964, he wascreating, to my mind, the most stunning, deep and, at the same time, simple examples of finite projective planes that had ever been seen. In the early 1960s, T.G.Ostrom created the concept of the ‘derivation of an affine plane’. We will get to the precise definition but, basically, this is a procedure by which one finite affine plane is transformed into another by a renaming of what are called ‘lines’. Some of the lines of the new plane iii
iv
PREFACE
are subplanes of the old plane. About this time, two striking results emerged. I might actually use the term ‘shocking results’ as this seemed to be the mood at the time. The famous Hall planes (see [26]) had been constructed many years earlier by Marshall Hall, Jr., who distorted the multiplication of a finite field to create a coordinate structure for an affine plane which was remarkably different from the original plane coordinatized by the associated field. But it was A.A. Albert ([l])who was first able to use the derivation procedure effectively and showed that theHall planes may be realized from Desarguesian planes (field planes) by the Ostrom derivation process. D.R. Hughes [27] had previously constructed a class of finite projective planes containing no pointline transitivities. These were marvelously interesting in their own right, but that they could be seen to be ‘derivable’ was an unbelievably propitious circumstance, since they ‘derived’ a class of finite projective planes admitting exactly one incident pointline transitivitya completely ‘new’ class. Both Ostrom and G. Rosati were working on these planes independently, so these are called the OstromRosati planes (see [63], 1691). I remember talking to Dan Hughes about this during a conference held at University of IllinoisChicago Circle during the summer of 1967, where it seemed that the important open problem on derivation was to determine conditions for a plane to be derivable. But, Hughes mentioned that it was much more interesting and important to ask whether this technique was ‘geometric’ in any sense of the word. He kept insisting that he knew he was missing something; he just couldn’t put his finger on it, so to speak. I guess you could say that the motivation to write this book came from years of trying to put a metaphorical finger on the construction technique of derivation. Perhaps the most productive study of derivable affine planes came from the simple idea of separating the derivable net from the &ne plane which contained it. Now we have a net which is covered by affine Baer subplanes whose parallel classes coincide with the parallel classes of the net. Of course, from this point of view, there is nothing special about having Baer subplanes as opposed to simply having ‘subplanes’ whose parallel classes coincide with the parallel classes of the net. That is, we may consider ‘subplane covered nets’.
PREFACE
V
In this monograph, we attempt to provide a completely selfcontained account of the beautiful geometry that envelops the derivation process and the analysis of subplane covered nets. In fact, the intuition of Hughes was correct; derivation is a geometric process and this book is an attempt to explain how this is so and how to generalize this explanation to understand completely the natureof subplane covered nets. The ideas encountered in this monograph are amalgamations of ideas of T.G. Ostrom, A.A. Albert, D.R. Hughes, G. Rosati, J. Cofman, A. Barlotti, D.A. Foulser, R.C. Bose, R.H. Bruck, M. Hall, J.A. Thas, F. De Clerck, V. Jha, P.J. Cameron, N. Knarr, A. Bruen, J.C. Fisher, G. Lunardon, M. Walker, H. Luneburg, M. Biliotti, and T. Grundhofer, to mention a few who contributed to this area and related disciplines. While I am enormously indebted to each of these mathematicians for their insights, these ideas have also been twisted and interwoven to fit into my scheme of doing things, so I accept full responsibility for any distortions. I would like to thank Brian Treadway for programs that created all of the diagrams found in this book and for all the varied and many things he did in assembling the text. I am most indebted to my wife who, after years of patiently listening to me 'spin' about geometries, still manages to provide continuous and unfailing support. I dedicate this book to my wife, Bonnie L. Hemenover, with gratitude and love. Norman L. Johnson
CONTENTS PREFACE
iii
1 A BRIEF OVERVIEW
1
2 PROJECTIVE GEOMETRIES 2.1 Projective and Affine Geometries. . . . . . 2.2 Projective and Affine Planes . . . . . . . .
........... ...........
7 9 14
3 BEGINNING DERIVATION 3.1 Affine Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.1 Affine Space as a Vector Space. . . . . . . . . . . . . . 3.2 PG(3,q) and the Regulus. . . . . . . . . . . . . . . . . . . . . 3.2.1 The Standard Vector Space Form of a Regulus . . . . . 3.2.2 The Regulus Question . . . . . . . . . . . . . . . . . . . 3.3 Infinite Derivation . . . . . . . . . . . . . . . . . . . . . . . . .
17 19 21 21 22 23 23
4 SPREADS 4.1 Affine Planes with a Transitive Translation Group . . . . . . . 4.2 The Fundamental Theorem of Translation Planes . . . . . . . 4.3 The Basis Decomposition Theorem . . . . . . . . . . . . . . . . 4.3.1 Some Examples . . . . . . . . . . . . . . . . . . . . . . 4.3.2 Partial Spreads. . . . . . . . . . . . . . . . . . . . . . . 4.3.3 Derivable Partial Spread . . . . . . . . . . . . . . . . .
27 27 31 33 37 39 40
vii
.
.
.
.
. . . . . . .
CONTENTS
viii 5 DERIVABLE NETS
41 5.1 Coordinatization . . . . . . . . . . . . . . . . . . . . . . . . . . 42 5.1.1 Coordinates of a Net . . . . . . . . . . . . . . . . . . . 43 5.1.2 The Ternary Function. . . . . . . . . . . . . . . . . . . 43 5.1.3 Addition. . . . . . . . . . . . . . . . . . . . . . . . . . 44 44 5.1.4 Multiplication. . . . . . . . . . . . . . . . . . . . . . . 5.1.5 Extension to an Affine Plane and Linearity. . . . . . . 45 5.1.6 The Question of Linearity. . . . . . . . . . . . . . . . . 45 5.2 Coordinates for Translation Planes and Their Duals. . . . . . 47 5.2.1KernelMappings and Kernel of the Quasifield. . . . . 50 5.2.2 Dual Affine Planes . . . . . . . . . . . . . . . . . . . . . 51 5.2.3 Dual Translation Planes and Derivation. . . . . . . . . 52 5.2.4 Chains and SemiTranslation Planes. . . . . . . . . . . 53
6 THE HUGHES PLANES 6.1 The AffineHughes Planes . . . . . . . . . . . . . . . . . . . . .
57 60
7 DESARGUESIAN PLANES 7.1 Moufang Planes . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.1 Coordinate Structures forMoufang Planes . . . .
65
8 PAPPIAN PLANES
8.1Reguliin
Pappian Spreads.
. ...
...................
67 69
79 80
9 CHARACTERIZATIONS OF GEOMETRIES 85 9.1 The Fundamental Theorem of Projective Geometry. . . . . . 85 9.2 Projective and Affine Geometries. . . . . . . . . . . . . . . . . 88 DERIVABLE 10 NETS AND GEOMETRIES 10.1 The Characterization Theorem. . . . . . . . . . . . . . . 10.1.1 Cofman’s Theorem (for Derivable Nets) . . . . . . 10.1.2 A Quilt of Affine Geometries. . . . . . . . . . . .
.. .. ..
97 . 102 . 104 . 107
11 STRUCTURE THEORY FOR DERIVABLE NETS 113 11.1 The Collineation Group of a Derivable Net . . . . . . . . . . . 114 11.1.1 The Translation Group of a Derivable Net . . . . . . . 116 11.1.2 A Representation for the Translation Group. . . . . . 118
CONTENTS
ix
11.1.3 The Group of QuasiSkew and Skew Perspectivities .
11.2 11.3 11.4
. 122 The Structure Theorem for Derivable Nets . . . . . . . . . . . 126 The Answer to the Regulus Question . . . . . . . . . . . . . . 127 The Solution to the Coordinates Problem . . . . . . . . . . . . 127
DUAL 12
SPREADS AND BAER SUBPLANES 131 12.1 CoordinateFree Derivation and the Extension Question . . . . 137 12.1.1 CoordinateFree Derivation . . . . . . . . . . . . . . . . 141
DERIVATION 13 AS A GEOMETRIC PROCESS 143 13.1 The Annihilator Mapping . . . . . . . . . . . . . . . . . . . . . 13.2 A ‘Geometric’ Interpretation . . . . . . . . . . . . . . . . .
.
143 .145
14 EMBEDDING 149 14.1 Embedding Nets into Affine Space. . . . . . . . . . . . . . . . 151 14.2 Embedding Nets into Projective Space. . . . . . . . . . . . . . 154 14.3 CodimensionTwo and PseudoRegulus Nets . . . . . . . . . . 156 14.3.1 Embedded Nets Are PseudoRegulus Nets . . . . . . . 163 15 CLASSIFICATION OF SUBPLANE COVERED NETS 165 15.1 Sketch of the Embedding . . . . . . . . . . . . . . . . . . . . . 167 15.2 The Share Two Theorem . . . . . . . . . . . . . . . . . . . . . 168 15.3 The Derivable Net Substructures. . . . . . . . . . . . . . . . . 177 15.3.1 The Subnet Theorem . . . . . . . . . . . . . . . . . . .183 15.4 The Parallel Classes Are Affine Spaces. . . . . . . . . . . . . . 186 15.5 The Projective Space. . . . . . . . . . . . . . . . . . . . . . . 191 15.5.1 The Classification Theorem. . . . . . . . . . . . . . . .195 15.5.2 The Theorem on PseudoRegulus Nets . . . . . . . . . 198 16 SUBPLANE COVERED AFFINE PLANES 16.1 Regular Spreads. . . . . . . . . . . . . . . . . . . . . . . . . . 16.2 The Affine Little Reidemeister Condition. . . . . . . . . . 16.3 Embedded Nets . . . . . . . . . . . . . . . . . . . . . . . . . . 16.4 Subplane Covered Translation Planes . . . . . . . . . . . .
201 201
. . 206
..
210 213
........................................................................................ ll.l,l’.,‘
ISMS
X
CONTENTS
17 DIRECT PRODUCTS 215 17.1 The Reconstruction Theorem. . . . . . . . . . . . . . . . . . . 217 17.1.1 Desarguesian Products. . . . . . . . . . . . . . . . . . 225 17.2 Left Spreads Produce Right Partial Spreads. . . . . . . . . . . 228 17.3 Left Spreads Produce Left Partial Spreads. . . . . . . . . . . 229 17.3.1 2fold Products . . . . . . . . . . . . . . . . . . . . . . 231 18
18.1 Desarguesian Parallelisms. . . . . . . . . . . . . . . . . . . . . 233 18.2 When the Vector Space Is 4.dimensional . . . . . . . . . . . . 235 18.3 Desarguesian Left Parallelisms Construct Right Spreads. . . . 237 18.4 An Infinite Class of ‘Derived’ Parallelisms. . . . . . . . . . . . 239 18.5 The Parallelisms in PG(3,2). . . . . . . . . . . . . . . . . . . 243
PARTIAL 19 PARALLELISMS WITH DEFICIENCY 19.1 Deficiency One . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.2 Partial Dual Parallelisms. . . . . . . . . . . . . . . . . . . . .
245 247 249
20 BAER EXTENSIONS 251 20.1 PointBaer Collineations. . . . . . . . . . . . . . . . . . . . . 251 20.2 Notes on the NonCommutative Case. . . . . . . . . . . . . . 256 20.2.1 Left and Right in Direct Product Nets . . . . . . . . . 259 20.2.2 When K Is KO. . . . . . . . . . . . . . . . . . . . . . . 261 20.3 PointBaer Subplanes in Vector Space Nets . . . . . . . . . . . 262 20.3.1 Three PointBaer Subplanes. . . . . . . . . . . . . . .263 20.3.2 When There Are Two PointBar Subplanes. . . . . . 268 20.3.3 The Space Is Decomposable by a PointBaer Subplane 270 21 TRANSLATION PLANES ADMITTING BAER 275 GROUPS 21.1 General Spreads with Baer Groups. . . . . . . . . . . . . . . .275 21.2 Translation Planes with Spreads in PG(3. K). . . . . . . . . . 285 22 SPREADS COVERED BY PSEUDOREGULI 293 22.1 Pseudo.Reguli . . . . . . . . . . . . . . . . . . . . . . . . . . 293 . 22.2 Conical and Ruled Spreads over Skewfields. . . . . . . . . . . 300
CONTENTS
xi
23 CONICAL AND RULED PLANES OVER FIELDS 309 23.1 Conical Flocks. . . . . . . . . . . . . . . . . . . . . . . 23.2 Hyperbolic Flocks. , , . . . . , . . . . . . , . . . . . . 24 SPREADS WHICH ARE DUAL SPREADS 24.1 Chains and SemitranslationPlanes Revisited.
321
. . . . . . . , . 326
PARTIAL 25 FLOCKS OF DEFICIENCY ONE 25.1 PointBaer Subplanes inSpreads of P G ( 3 , K ) . 25.2 DeficiencyOne Partial Conical Flocks. . . 25.3 DeficiencyOne Partial Hyperbolic Flocks. . . .
.. .
SKEWHALL
26
PLANES 26.1 Doublecovers. . . , 26.2 Postscript, , , . . , ,
.
. . . . 309 . . . . 313
329
. . . . . . . . . 329 , ,
. . . . . . . 331 . . . . . . , 336 341
. . . . . . . . . . . . . . . . . . . . . . . 347 . . . . . . . . . . . . . . . . . . . 350 ,
,
BIBLIOGRAPHY
353
INDEX
359
Chapter l
A BRIEF OVERVIEW The reader unfamiliar with the terms used in this small chapter might like to skip this initially andgo on to read the chapters on projective geometries and beginning derivation before returning to read this for a preview of what the book is all about. Let K be a finite affine plane of over q2. An affine subplane of order q is said to be a ‘Baer subplane’. Let D denote a set of q 1 parallel classes such that for any two distinct affine points P and Q, such that the line t p , ~ joining them is an element of one of the parallel classes of D, there exists a Baer subplane K ~ , Qthat contains P and Q such that the set of q 1 parallel classes of K ~ , Qare those of D. In this situation,we call K a ‘derivable’ affine plane. Furthermore, D is called a ‘derivation set’. In the early 60’8, T.G. Ostrom ([62], [Sl]) realized that when such a set D exists, a potentially new affine plane n ( D ) may be constructed from 7r in the following manner: The ‘points’ of n(D)are the points of K and the ‘lines’ of K ( D )are the lines of K which are not in a class of D together with the Baer subplanes
+
+
T‘p,Q
A.A. Albert [l],showed that the derivation process applies to affine planes ‘coordinatized’ by finite fields K of order q2. Considering the affine plane as an analogue to the real affine plane, points are elements (x,y) of K x K and lines are givenby equations y = x m b and x = c where juxtaposition denotes multiplication in K for all m,b , c E K. The set D consists of the slopes (a)such that a is in the unique subfield of K of q
+
1
2
CHAPTER 1. A BRIEF OVERVIE W
elements together with the parallel class (W)containing the lines z = c for c E K. Previously, in 1943, Marshall Hall Jr. had constructed an interesting and important class of affine planes by altering the multiplication in a field K and using this new coordinate structure t o create the plane. The planes derived from the field planes, which we call Desarguesian planes, turn out to be the planes of Hall. There were a number of striking results concerning derivation proved in the early ’60’s but, for us here, the main questions involve the nature of an affine plane which admits aderivation set and the natureof the substructure, called a ‘derivablenet’, which contains the Baer subplanes that become ‘new’ lines of the ‘new’ plane. What makes an afflne plane derivable? How can a derivation set be recognized? Of course, we want to address the fundamental question: Is derivation a geometric process? Considering the last question, one wonders what it means to be geometric. Perhaps a reasonable definition might be that something is ‘geometric’ if it is closelyconnected to some property of an affine or projective geometry. In fact, it turns out that, taking the above definition, derivation can be viewed geometrically. In order for the reader to understand fully how to see this, we provide chapters on projective and affine geometries and a few results on projective and affine planes. Coordinatization is considered so as to better understand how to recognize a derivable affine plane and for use in later chapters on direct product nets and their reconstruction. After focusing initially on the affine plane containing a derivable net, we consider the structure of the net itself. Hence, for such an analysis, we separate the net from the plane and study derivable nets themselves. At this point, we show how to associate a projective geometry with a derivable net. Furthermore, we discuss embedding and extension problems. For example: Is it possible to embed a derivable net into a projective space and what do embeddings imply about the structure of the net? If a derivable net is structurally determined without the ambient d n e plane containing it, must the net necessarily be ex
CHAPTER 1. A BNEF OVERVIEW
3
tendible to an affine plane? firthermore, if a net can be so extended, how many affine planes extend it? Although we initially considered finite affine planes, there is really no good reason to do so apart from the intuition that isobtained. So, we consider general or arbitrary affine planes that could be derivable and arbitrary derivable nets and ask of their structure. For the arbitrary subplane situations, we require possibly infinite dimensional vector spaces over skewfields (structures that do not have necessarily commutative multiplication but otherwise have the properties of a field). Thus, we study projective geometry, beginning derivation and ask what are some classes of derivable affine planes. To provide some examples and forreference to subsequent research discussions, we consider ‘translation planes’ and their duals. It will be seen that many dual translation planes automatically become derivable. The ideas of Cofman [l91 figure prominently into the relationship with derivable nets and affine geometry and these shall be fully developed. The connection between derivable nets and projective geometry is the heart of the book although it is not the most general result that we shall prove. A derivable net is a net which is covered by subplanes of a certain type that we called ‘Baer subplanes’. A natural generalization is the concept of a ‘subplane covered net’. Definition 1.1 A ‘subplane covered net’ is a net with parallel class set D such that given any two distinct points P and Q that are incident with a line of the net, there is an afine subplane TP,Q containing P and Q whose parallel classes are exactly those of D. All of the previous questions now are appropriate for subplane covered nets; can they be embedded into a projective geometry, are there extensions to affine planes? The most important result in the book and the justification for its existence is the complete classification of subplane covered nets which establishes close connections with projective geometries and this shall be given in detail. It is important to point out that herein lie various connections and a p plications to many types of finite geometries, although the results provided do make any assumptions about finiteness.
4
CHAPTER OVERVIEW 1. A BRIEF
When Ostrom was working on finite derivable affine planes, Bruck [14], [l51 had previously been giving the foundational studies for finite nets. In particular, extension problems were extremely important. Bruck was able to show that once a net has more than a certain ‘critical’ number of parallel classes, there is a unique extension to an affine plane provided there is an extension at all. The ‘critical’ number turns out tobe exactly the number of parallel classes in a finite derivable affine plane minus the number of parallel classes in a derivation set; that is, if the order is q2, the critical number is q2  4.
Ostrom [64] was able to show that any finite net of order q2 with q2  q parallel classes has at most two extensions to affine planes and, if two, they are related by the derivation process; one is the derived plane of the other. The work of Bruck was influenced by the work of R.C. Bose [l21 on partial geometries which was very graph theoretic. The ideas of extension came from the formulation of certain ‘cliques’ of the graph associated with a finite net. In certain situations, it is convenient to study ‘dual nets’ as their pointline properties become phrased more closely to the pointline properties of affine or projective geometries. J.A. Thas andF.De Clerck ([21], [75])have provided basic and important contributions to both (finite) partial and semipartial geometries concerning both embedding questions and characterization results. Certain finite dual nets satisfying a pointline property called the axiom of Pasch are, perhaps in somewhat disguised form, the duals of subplane covered nets. Our treatment and characterizationof subplane covered nets in the general case was influenced both by the work of Cofman on an associated affine geometry of three dimensions and the work of Thas and De Clerck on dual nets in the finite case. The geometries associated with derivable nets turn out to be the projective geometries and the derivation process becomes a natural generalization of duality of a projective space. So, ‘derivation’ becomes a geometric process. It is one of the goals of this monograph to show how this is so. The study of derivable nets which may be embedded in a three dimensional projective space leads to the analysis of subplane covered nets which also have a similar embedding into projective space although the dimension in this case can be infinite.
CHAPTER 1. A BRIEF OVERVIEW
5
There aremany affine planes which are covered by derivable nets or more generally covered bysubplane covered nets in various ways. If an affine plane is covered by subplane covered nets, there is an associated projective space and hence an associated vector space and, when this occurs, thereis probable cause to believe that the affine plane is a translation plane. This idea will be explored after the classification theorem for subplane covered nets has been given. When an affine plane which is covered by subplane covered nets is, in fact, a translation plane, there are often other associated geometries. For example, translation planes which are covered by subplane covered nets defined by reguli in PG(3,K ) all of which share a common line correspond to flocks of quadratic cones in a three dimensional projective space. Furthermore, such translation planes are also associated with certain generalized quadrangles. Translation planes which are coveredby derivable nets that share two common lines are connected with flocks of certain Minkowski planes and when the derivable nets are defined by reguli in P G ( 3 , K ) ,the Minkowski plane is classical. There are other connections with partial geometries and affine planes which are subplanecovered. Note that the translationplanes associated with the above mentioned geometries may be derived using one of the derivable nets involved in the covering. In this case, where the derivable nets initially share a common line, there is now a system of derivable nets sharing a Baer subplane of the derived plane so there is not a covering in the usual sense. However, translation planes admitting such a generalized cover correspond to what might be called ‘partial flocks’ of deficiency one in that such planes correspond to partial flocks ‘missing’ one conic. We also consider the ‘direct product’ of two Desarguesian affine planes and show, under an assumption as to how the two sets of parallel classes line up, that such a direct product always produces a derivable net. This leads to theconsideration of sets of such direct productswhich then turns to the study of ‘parallelisms’ in projective spaces. Recalling that a ‘spread’ in the threedimensional projective space PG(3,q) is a set of q2 1 lines that covers the point set, a ‘parallelism’ is a set of 1+q+q2 spreads which covers the line set. In a sense which shall be made clear, a parallelism in PG(3,q) leads to a spreadin PG(7,q) by the notion of direct products of affine planes, As we are emphasizing general theory concerning derivable nets, we,
+
6
CHAPTER l . A BRlEF OVERVIEW
therefore, consider parallelisms in PG(3,K ) where K is a skewfield. There is a wonderful connecting theory where particular families of direct products of &ne Desarguesian planes with spreads in PG(3,K ) are equivalent to certain spreads in PG(7,K). However, since K may not be commutative, special care must be taken and, in fact, it is only true that ‘left parallelisms’ in PG(3,K)produce ‘right spreads’ in PG(7,K). We have noted that there is a rich geometry associated with nets which are subplane covered. When there is at least one subplane of a net,one might consider what, if anything, can be determined as to the structureof the net. Since we are also interested in nets which can be embedded into translation planes and also derivable nets, we consider vector space nets which admit Baer subplanes. Here there is also a nice structure theory and when there are at least three Baer and Desarguesian subplanes, the net turns out to be a derivable net. We have noted above cbpnections to flocks of quadric sets with spreads which are unions of reguli. But, derivable nets, it is shown, do not always correspond to reguli but do, in fact, correspond to close cousins, the ‘pseudoregulus’ nets. We generalize these ideas of flock spreads by the consideration of spreads which are unions of ‘pseudoreguli’. Furthermore, partial flocks, particularly of deficiency one, may be considered which entails an analysis of vector space nets admitting a Baer subplane that is pointwise fixed by particular groups (Baer groups). So, we consider the structure and nature of vector space nets containing one or more Baer subplanes and their Baer groups. All of this leads to a complete theory of the spreads which correspond to partial flocks of quadric sets of deficiency one. Thus, the majortheme running through the text is how the existence of subplanes of a net or affine plane tend to determine the geometry and we formulate a reasonably complete theory in two major situations when the subplane(s) happens to be Baer or when there is a complete covering by subplanes.
Chapter 2
PROJECTIVE GEOMETRIES Prerequisites and Background This book is intendedto be selfcontained in the sense that only a knowledge of beginning group theory and linear algebra is required. However, an acquaintance with infinite dimensional vector spaces over skewfields is assumed. There are, of course, a few theorems which we shall not prove. At one point, we shall require the theorems of ArtinZorn and theBruckKleinfeld/Skornyakov/San Soucie theorem. A good reference for the first theorem is Hall [26]pp. 375382.
Theorem 2.1 (The Theorem of ArtinZorn). A finite alternative division ring is a field. Theorem 2.2 (BruckKleinfeld/Skornyakov/SanSoucie [lS], [71], [70]). A projective plane which is a tmnslation plane with respect to all lines incident with a given point is a tmnslation plane with respect to all lines; the plane is a Moufang plane. We shall be interested in affine planes which are covered by ‘subplane covered nets’ and shallrequire results of Gleason, Luneburg and Kegel which we shall state without proof. We define both affine and projective Fano 7
..
,
. . . , . . .
..,..........
,
8
C H A P T E R 2. PROJECTIVE GEOMETRIES
planes and planes which satisfy the little Reidemeister condition later in the book.
Theorem 2.3 (Gleason [2.4]). A finite Fano plane is Desarguesian. We also require a technical result of Gleason which we shall give in the appropriate chapter.
Theorem 2.4 (Gleason [24], Luneburg [SO], Kegel and Luneburg [55]). A finite projective plane is Desarguesian if and only if it satisfies the little Reidemeister condition. In the few occasions where a complete proof is not given, a sketch is providedfromwhich the reader could easily fillin the details. There are places in the text where a proofby the Klein quadric would provide an alternative proofwhichmaybemoreelegant than the arguments given. However, we have chosen to provide proofs which are more directly related to derivable nets by coordinate and vector space methods and are thus able to avoid the use of the technical results and methods involved in the use of the Klein quadric.
Notation and conventions We have adopted the convention of normally referring to ‘points’ by capital letters P, Q, R etc. and normally referring to ‘lines’ by small letters a, b, c etc. or by the symbol k‘. Furthermore, too shall be used, in particular, to denote the ‘line at infinity’ of an affine plane which is then equivalent to the ‘set of parallel classes’. In addition, we shall use small Greek letters a,P etc. to denote parallel classes of affine planes with the exception that normally we reserve 7r to denote an affine plane and rE the extension to the associated projective plane. In an affine plane, the unique line of the parallel class a incident with the (affine) point P is denoted by aP, We shall normally use capital Greek letters C, ll to denote projective spaces. Furthermore, we shall use the notation PQ to denote a line between two points P and Q of an affine or projective plane or a net when the points are collinear and use a n b to denote the point of intersection of two lines a
2.1. PROJECTIVE AND
AFFINE GEOMETRJES.
9
and b of a projective plane or of an affine plane or a net when there is an intersection. Normally, sets of points, lines etc., will be given calligraphic lettering P, C, etc. We shall employ the symbol 5 ' to denote isomorphic structures. We shall use '0' to denote both a zero vector or zero of a skewfield where context indicates the precise meaning. Occasionally, we shall also use the symbol '0' to denote a zero vector. Finally, we denote the end of a proof by the symbol 0. In this chapter, we provide some of the fundamental background on projective and affine geometries required for the analysis of derivation and subplane covered nets and discuss briefly projective and affine planes.
2.1
Projective andAffineGeometries.
Definition 2.5 ProjectiveGeometry. Let V be a vector space over a skewfield K . W e note that V need not be finite dimensional. The 'projective geometry' PG(V,K ) is defined as the lattice of nonzero vector subspaces of V . W e shall usetheterrninology of 'points'forldimensionalvector subspaces, 'lines'for2dimensionalvector subspaces,and 'planes'for 3dimensional vector subspaces. More generally, an %dimensionalprojective subspace' shall mean a i+ldimensional vector subspace. In the case when V has finite dimension n, we shall use the notation PG(n1, K)interchangeably with PG(V,K)and call PG(V,K )the '(n1)dimensional projective geometry over K '. If K is a finite field with q elements and V has finite dimension n, we shall use the notation PG(n  1,q) to denote PG(V,K)provided the specific field K is not important to the discussion. W e call PG(2,K) the 'projective plane over K' and PG(1,K) the 'projective line over K '. For historical reasons which we shall discuss in the chapter o n Desarguesian Planes, we also refer to PG(2,K ) as the 'Desarguesian projective plane over the skewfield K '. When it becomes important, we shall specih whether the vector space is to be a 'left' or a 'right' vector space. Although this is not important for
10
CHAPTER 2. PROJECTIVE GEOMETRIES
vector spaces over fields, it can be relevant when the vector space is defined over a skewfield. Definition 2.6 CoDimension. Given any vector subspace W of V , it is possible to choose a basis Bw f o r W which extends to a basis Bv for V as is well known from linear algebra. The subspace generated b y Bv  Bw is called a ‘complement’ of W . If a complement of a subspace has dimension IC, we shall say that W has ‘codimension’ k. Equivalently, W has codimension IC if and only if the quotient space VIW has dimension k. A subspace of codimension 1 is called a ‘hyperplane’ both in the projective and vector space senses. Of course, if a vector space has finite dimension n and a subspace has dimension m then the codimension of the subspace is n  m. The reason for the introduction of this terminology is that we will be considering possibly infinite dimensional vector spaces where the dimension of a subspace could be infinite whereas the codimension could be finite. In particular, we shall be considering subspaces of codimension two.
Remark 2.7 If W is a vector subspace of V,we let P ( W ) denote the corresponding projective subspace in PG(V,K).For vector subspaces T and M , T @ M shall denote the external direct sum. We now consider possible intersections.
Proposition 2.8 ( l ) Two distinct projective hyperplanes P ( W ) , P ( S ) intersect in a hyperplane P(W n S) of P ( W ) and/or P ( S ) . (2) An9 two distinct projective subspaces of the same finite dimension n which are contained in a common projective subspace of dimension n 1 intersect in a projective subspace of dimension n  1. (3) Any two distinct projective subspaces of the same finite codimension n which are contained in a common pmjective subspace of codimension n 1 intersect in a projective subspace of codimension n + 1.
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2.1. PROJECTIVE AFFINE GEOMETRIES. AND
11
Proof: ( 1 ) W and S are vector subspaces of codimension 1. This means that the quotient spaces V/W and V / S have dimension 1. Since the subspaces are distinct, there existsa nonzero element W E W  S and a nonzero element S E S  W . Let S1 @ (S) = S and W1 @ (W) = W where S1 and W1 are hyperplanes of S and W respectively. Hence, (W1 @ (W)) @ S = (S1 @ (S)) @ (W) from which it follows that V = ( W n S ) @ (w,s). Hence, V/(Wn S) has dimension 2 and W n S has codimension 2 with respect to V and clearly codimension 1 with respect to either W or S. That is, W n S is a hyperplane of W or S. This completes the proof of (1). Assume the conditions of (2). Let W and S be vector subspaces of dimension n such that (W,S) = T has dimension n 1. By the ranknullity theorem, dim W dim S  dim W n S = n 1 from which it follows that dim W n S = n  1 from which the projective analogue is immediately implied. This proves (2). Assume the conditions of (3). Let W and S be vector subspaces of codimension n such that (W,S) = T has codimension n  1. So, ( W , S ) = W @ ( s ) = S @ ( w ) w h e r e s E S  W a n d w E W  S . By ( l ) ,the codimension of W nS in (W,S) is 2. So, the codimension of W nS is (n  1) 2. This proves (3).0
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Definition 2.9 A B n e Geometry. W be a vector subspace Let PG(V,K) be aprojectivegeometry.Let of V of codimension 1 and let PG(W,K ) denote the associated projective geometry. The‘afine geometry’ AG(V,K) shall be defined asthedeletion of PG(W,K)fram PG(V,K) as follows: The ‘points’ of AG(V,K) are the points of PG(V,K)  PG(W,K), the ‘lines of AG(V,K ) are the lines e of PG(V,K ) which do not lie in W minus the point of intersection e n W . (Note that a line is a 2dimensional vector subspace and a hyperplane is a codimensionl subspace corresponding to a codimensionl vector subspace W . It follows thatenW is a ldimensional vector subspace or rather a point of the projective space.) Further, the ‘planes’ of AG(V,K) are the planes of PG(V,K) which do not lie within PG(W,K ) minus the intersection line and the ‘subspaces’ of
12
C H A P T E R 2. PROJECTIVE GEOMETRIES
AG(V,K ) are the projective subspaces of PG(V,K ) minus the intersecting sets of points, lines, planes, etc. W e shall use the notation AG(n 1,K ) when PG(V,K ) is PG(n 1, K ) and AG(n  1, q) when PG(V,K ) is PG(n  1,q). W e furthermore call A G ( 2 , K ) the ‘afine plane overK and/or the ‘Desarguesian afine plane over the skewfield K W e refer to AG(1,K ) as the ‘afine line over K Definition 2.10 Parallelism in AG(V,K ) . Two hyperplanes of AG(V,K ) correspond to hyperplanes of PG(V,K)’ minus the intersections on a given hyperplane W of PG(V,K ) . W e have noted previously that two distinct hyperplanes intersect in a hyperplane with respect to the subspaces themselves. Hence, if the intersection relative hyperplane of the two projective subspaces lies within W then they are disjoint within AG(V,K ) . W e define two hyperplanes of AG(V,K ) to be ‘parallel’ if and only if they are disjoint. Two aflne subspaces are said to be ‘parallel’ if and only if the projective versions intersect in the same subspace of the hyperplane which is deleted to form the space. I n particular, this means that any two lines are parallel if and only if they lie in a common AG(2,K ) and are disjoint. firthermore, two spaces are parallel if given any line of either space, there is a line of the remaining space which is parallel to it. W e also allow that any afine subspace is parallel to itsel$ Definition 2.11 Semilinear and linear group. Let V be a left vector space over K . By this, we shall mean that scalar multiplication occurs on the left of the vector. Hence, ax shall denote the scalar multiplication of the vector x by the scalar a E K . An additive mapping Q is said to be ‘semilinear’ over K or ‘Ksemilinear’ if and only if ax) = aPu(x)for all a E K and for all x E V where p is an automorphism of K . W e shall use the terminology that Q is ‘Klinear’ or simply ‘linear’ if and only if p = 1. The group of all bijective semilinear additive mappings over K is called the ‘general semilinear group over K’ and is referred to by rL(V,K ) .
2.1. PROJECTIVE AFFINE GEOMETRIES. AND
13
The subgroup of rL(V,K ) consisting of linearmappings is called the ‘general linear group over K and is denoted b y GL(V,K). We note that when V has finite dimension n over the skewfield K then GL(V,K )is isomorphic to theset of nonsingular n x n matrices with entries in K and furthermore GL(V,K ) is denoted by GL(n,K ) . When K has finite cardinality q then GL(V,K ) is denoted by GL(n,q). We note that GL(V,K ) is a normal subgroup of r L ( V ,K ) .
Definition 2.12 When V has finite dimension n, and K is a field, the subgroup of GL(n,K ) consisting of the n x n matrices of determinant 1 is called the ‘special linear group over K ’and is denoted b y S L ( n ,K ) . It is easy to verify that SL(V,K ) is a characteristic subgroupof GL(V,K ) SL(V,K ) is also a normal subgroup of r L ( V ,K ) .
so that
Definition 2.13 Projective semilinear group. The ‘projective semilineargroup’ is the group induced on PG(V,K ) by r L ( V ,K ) and is denoted by PrL(V,K ) . Clearly, the projective general semilinear group is r L ( V ,K ) modulo the subgroup of r L ( V ,K ) which leaves invariant each ldimensional vector subspace and hence leaves invariant each vector subspace. Proposition 2.14 The subgroup of r L ( V ,K ) which .fixes eachldimensiona1 subspace is the set of mappings of the form:
Furthermore, 2 = (96 ; S E K } is a normal subgroup of r L ( V ,K), Proof: Simply note that ( x ) = (ax) for any a! E K  (0). Hence, any mapping g which fixes eachldimensional subspace must map x onto &,x for some S, E K which possibly depends on x. However, x + y must map onto S,+,(x y) = S,S~Z S+,y, which must be S,x Syy since g is additive, so it clearly follows that S, = S for all x E V . Since the indicated groupis the kernel of the group induced on PG(V,K ) , it follows that the subgroup is normal.0
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14
CHAPTER 2. PROJECTIVE GEOMETRIES
Remark 2.15 Note that 2 n GL(V,K ) = (96 ; S E Z ( K ) } where Z ( K ) denotes the centerof K (the set of elements which commute withall elements of K). Hence, we may make the following definitions. Definition 2.16 The ‘projective general semilinear group’ PI’L(V, K ) is the group l?L(V,K ) / Z . The ‘projective general linear group’ PGL(V,K) is GL(V,K ) / ( Z n G L ( V , K )The ) . ‘projective special linear group’PSL(V,K) is SL(V,K ) / ( 2fl S L ( V , K ) ) .
2.2
ProjectiveandAffinePlanes.
In theprevious section, we have defined projective and affine geometries and furthermore defined projective and affine planes over skewfieldsas PG(2,K ) and AG(2,K ) respectively where K is a skewfield.
Remark 2.17 In PG(2,K ) , every two distinct points are contained in a unique line and every two distinct lines intersect in a unique point. In AG(2,K ) , every two distinct points are contained in a unique line and every two distinct lines intersect in at most one point. Furthermore, two lines of AG(2,K ) are ‘parallel’ if and only if they are disjoint or equal. However, pointline incidence structures satisfying the properties listed in the previous remark are not necessarily projective or affine geometries of the type PG(2,K ) or AG(2,K). Rather than attempt to deal with specific examples in this chapter, we close with two definitions and some fundamentals.
Definition 2.18 Projective plane. A projective plane isa set of ‘points’ and a set of ‘lines’ with an incidence relation E such that the following three properties are satisfied: ( l ) Any two distinct points arq, incident with a unique line, (2) any two distinct lines are incident with a unique point, and (8) there exist at least four points no three of which lie on a common line.
2.2. PROJECTIVE PLANES. AFFINE AND
15
Definition 2.19 Afine Plane. Let T be a projective plane and let Em denote a specific line of 7r. An afine plane A(n)i s a set of ‘points’, ‘lines’ with an incidence relation and defined as follows: The boints’ of the afine planeare the points of 7r which are not incident with the lineE , and the ‘lines’ of the afine planeare the lines of IT not equal t o Em with the corresponding intersection points removed. Incidence in the afine plane is inherited fmm the incidence relation in 7r. firthemnore, the points of E , are called the ‘infinite points’ and the line itself, the ‘line at infinity’ of A(n). If there are finitely many points, the projective plane is said betoa ‘finite projective plane’ and the corresponding afine planes, ‘finite afine planes’. Remark 2.20 It is straightforward to show that any finite projective plane with n 1 points on a given line has n + 1 points on any line. We say that the finite projective plane has ‘order n’ in this case. A projective plane of order n has then n2 n + 1 points and the same number of lines. A corresponding afine plane, also said to be of order n, has n2 points, n2 in lines, n 1 parallel classes and n points per line.
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Definition 2.21 Let 7r1 and ~2 denotetwoprojectiveplanes,twoafine planes, two projective geometries or two afine geometries. The two structures are said to be ‘isomorphic’ if and only if there is a bijection CT from the points of 7r1 onto the points of n2 which induces a bijection from the lines of 7r1 ontothelines of thatpreservesincidence. if and only if u(P),c(Q)are That is, P,& are points on the line E of points on the line .(E) of 7r2. If 7r1 = 7rp, the isomorphism is called a ‘collineation’. forms a group The set of all collineations under composition mapping called the ‘collineation group’. The following is basic to our discussions: Theorem 2.22 Let 7ri f o r i = 1,2 be projective planes and A(7ri)f o r i = 1,2 the corresponding afine planes obtained by the deletion of lines &,,i of 7ri f o r i = 1,2 respectively.
16
CHAPTER 2. PROJECTIVE GEOMETRIES
Then A ( q ) N A ( 7 4 if and only if there exists an isomorphism r l o n t o 7r2 which maps &,,l onto &,a.
U
from
Proof: The reader is invited to complete the proof.U
Remark 2.23 Given an incident pointline pair(P,E ) , we shall adopt standard usage and say that the point P is ‘on’ the line E and the line E is ‘thru’ the point P. In the next chapter, we begin the discussion of derivation dealing first with certain derivable planes that fall into particular classes of a classific& tion of projective planes by pointline transitivities dueto Lenz and Barlotti (see Barlotti [5]).
Definition 2.24 If G is a group acting as a permutation group on a set X and x E X , then G, is the subgmup which f i e s x. G is said to be ‘transitive’ o n X if and only if for any pair x,y E X, there exists an element g of G such that g(x) = y. Definition 2.25 By a ‘(P,!)transitivity’ of a projectiveplane,we shall mean a collineation group that f i e s P, all points on E , all lines thru P and acts transitively o n the nonked points on any line thru P.
Chapter 3
BEGINNING DERIVATION The concept of a finite derivable &ne plane was conceived by T.G. Ostrom in the early 1960's (see [61]and [62])and has been arguably the most important construction procedure of affine planes developed in the last thirtyfive years. Certain finite affine planes may be 'derived' to produce other affine planes of the same order. For example, the Hall planes of order q2 originally constructed by Marshall Hall Jr. [26]by coordinate methods were shown by Albert [l] to be constructible from any Desarguesian &ne plane of order q2 by the method of derivation. The Hughes planes [27]of order q2 were shown to be derivable and the projective planes constructed were the first examples of finite projective planes of LenzBarlotti class 111 (there is a single, incident, pointline transitivity). The planes obtained were independently discovered by T.G. Ostrom [63] and L.A. Rosati [69]and are called the 'OstromRosati planes'. The description of a finite derivable affine plane is as follows: Definition 3.1 Let T denote a finite afine plane of order q2 and let rE denote the projective extensionof T b y the adjunctionof the settw of parallel classes as a line. Let l& denote a subset of q 1 points of too.V m is said to be a 'derivation set' if and only it satisfies the following property: If A and B are any two distinct points of T whose join in T~ intersects Ew in Vw then there is an afine subplaneTA,B of order q containing A and B and whose q + 1 infinite points are exactly those of Dm.
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C H A P T E R 3. BEGINNING DERIVATION
Given any derivation set there is a corresponding set B of q2(q+ 1) afine subplanes of order q each of which has V , as its set of infinite points. T is said to be ‘derivable’ if it contains a derivation set. The main result of Ostrom is
Theorem 3.2 (Ostrom [62]). Let T be a finite derivable afine plane of order q2 with derivation set Let B denote the associated set of q2(q+ 1) subplanes of order q each of which has as its set of infinite points. Form the following incidence structure n(’D,): The ‘points’ of n(’D,) are the points of T and the ‘lines’ of .(V,) are the lines of 71 which do not intersect in the projective extension and the subplanes of B. Then T(’D,) is an afine plane of order q2. Proof: Toprove this, we needonlyshow that two distinct points are incident with a unique line and two lines either uniquely intersect or are disjoint. Let P and Q be distinct points of 7r and let P& denote the unique line of 7r containing P and Q, If P&is not a line of then P&is also a line of r(’D,). If P& is a line of there is a subplane TP,Q containing P and Q whose parallel classes are exactly those of Note that the infinite points together with P and Qgenerate a ‘unique’subplane containing P and Q. Hence, given two distinct points of T ,there is a unique line of containing the two points. The lines of are of two types; lines of and the subplanes of B. Two distinct lines must either be disjoint or share a unique point. By the above remark on the generation of subplanes, we need only consider the and a line t situation where there is a line of each type. A subplane T ~ , Q which does not lie in must share at least one common point by a simple counting argument and cannot share more than two points since otherwise t would not appear in the construction process.0
Definition 3.3 The afine plane is called the plane ‘derived’ fmm by the derivation of V m or merely the ‘plane derived from T
IT
We shall now establish that any AG(2,q2)is a derivable affine plane. We would like to determine the collineation group of the projective and affine geometries. It is now clear the subgroup which is induced from the
3.1. AFFINE FORMS.
19
general semilinear group is a collineation group. What is not completely clear is whether every collineation is of this form. This is, infact,true and the corresponding theorem is called the ‘fundamental theorem of projective geometry’. We postpone the proof until after we have considered coordinates. We note only the following at present.
Theorem 3.4 ( l ) The collineation group of PG(V,K ) contains the group P r q v ,K). (2) The collineation group of AG(V,K ) within PrL(V,K ) is PrL(V,K)p(w)where P ( W ) is the hyperplane deleted to form the afine geometry. (3) PrL(V, K ) is transitive on hyperplanes of PG(V,K ) . (4) Any two AG(V,K)’s are isomorphic. Proof It is clear that thecollineation group of AG(V,K ) is the subgroup of PG(V,K) that leaves P ( W ) invariant. Choose two hyperplanes W and S of V . Choose a basis Bw for W and a basis Bs for S and extend both to bases Bw and Bs respectively of V . Now the induced collineation which carries P ( W ) onto P ( S ) consists merely of a basis change from Bw to Bs which maps Bw onto 8s. Note that (3) now follows as the above shows that PYL(V,K ) acts transitively on hyperplanes. (4)now follows from (3).0
3.1
Affine Forms.
First consider PG(2,h);the projective plane over the finite field GF(h) of h elements. Let V3 denote the associated 3dimensional vector space over the field GF(h). Choose any basis and write the vectors in the form (z,y,z). Since any ldimensional vector space is a point of the projective plane over GF(h),we may identify (z, y , z ) , where not all z, y , z are 0, with (az,ay, az) for all a E GF(h) (0). This identification is called taking ‘homogeneous coordinates’ to represent the projective plane. Further, since the 2dimensional subspaces are thelines of the projective plane, we may choose the line which is deleted from the projective plane to
20
CHAPTER 3. BEGINNING DERIVATION
form the affine plane to be {(x,y, 0) ; x,y E K} and call this latter set loow We have noted previously that any two AG(2, h)’s are isomorphic. Considering homogeneous coordinates, we may take theremaining points to be represented in the form (x,y, 1) = (x,y) for all z,y E GF(h). Hyperplanes may represented as solutions to linear equations:
If z = 1 then the lines have the standard form y=xm+b,
x=d
for all m, b, d E GF(h).
Theorem 3.5 A n y a f i n e p l a n e AG(2,q2) is derivable.
Proof: It suffices to find the associated derivation set on loo.For this Dw,we choose {(m),(a); a E GF(q).} = PG(1,q). Choose any two points P and Q such that line joining them is in Voo. Since each of the following mappings (x,y) H(x a, y b) for a, b E GF(q2) is clearly a collineation of AG(2, q2) which fixes each parallel class, we may assume that P is (0,O). Hence, the point Q is on a line of the form y = x6 for S E GF(q) or x = 0. Now, we need to find a Baer subplane containing P and Q. Note that AG(2,q2) becomes a 2dimensional vector space over GF(q2) and GF(q2) is a 2dimensional vector space over GF(q) so that AG(2, q2) is a Cdimensional GF(q)vector space. Write vectors as ( x l , ~ , y 1 , y 2where ) xi,yi E GF(q) for i = 1 , 2 . We assert that the set T(a,b) = { ( m , ba,
[email protected],b p ) ; a,p E GF(q)} for a, b not both 0 is an &ne Baer subplane of AG(2,q2). To see this, note that by identifying (aa, ba,ap,bp)with (aa,ap) if a is not zero, we see that T(.,b) is isomorphic to an AG(2, q) and is hence an affine plane and thus an affine subplane whose lines incident with (0,O) are y = xp and x = 0 for all p E GF(q). Note that T(a,b)n (y = xp) is { (aa, ba,aap, bap) ; a E GF(q)}. It follows that T(a,b) is a subplane all of whose parallel classes are in Dw. First assume that Q = (c,CS) = (cl,c ~ , c ~ S , C Zthen S ) Q and P = (O,O, 0,O) are contained in the subplane T ( , ~ , , ~ )If. Q = (0, d ) = ((0, 0, d l , d2) then Q and P are contained in the subplane ~ ( d ~ , d ~ ) . O set
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3.2. PG(3,q) AND THE REGULUS.
21
Hence, every Desarguesian affine plane of order q2 is derivable and if GF(q2)is a field coordinatizing the plane then GF(q) U {W}, considered as a subset of the set of points on the line at infinity, is a derivation set.0
Definition 3.6 W e 'define' the 'Hall planes' as the b y the derivation of AG(2,q2).
afine planes obtained
Remark 3.7 Notethat, in theafineform above,we haverepresented AG(2, h) as a 2dimensional vector space over GF(h). Note that we started with a 3dimensional vector space over GF(h), formed the projective space PG(3, h) as the latticeof subspaces and then formedAG(2, h) by the deletion of a line which is also a hyperplane in this case. W e m a y also consider this more generally for any afine space. 3.1.1 Affine Space as a Vector Space.
Theorem 3.8 Consider the afine geometqAG(V, K)where V is an vector space over the skewfield K . Take any hyperplane W of V and choose a vector basis l3w for W and extend to a basis Bw of V containing Bw . Write V = W @(b) and represent vectors in the f o r m {((xi),a ) ; i E X} relative to the basis Bw so that (xi) f o r i E X are the elements of W f o r x ~ , a E K. Form the structure AW of points, and subspaces as follows: The points of AW are the vectors of W and the subspaces of AW are the additive cosets of vector subspaces of W , Two subspaces are said to be [parallel' if and only if the9 are in the same coset. Then AW is isomorphic to,AG(V, K). Proof The proof follows exactly as in the AG(2, h) case and is left to the reader to verify.0
3.2
PG(3,q) and the Regulus.
We shall consider the 'regulus' in a more general context later. For now, we are content to consider the finite case and P G ( 3 , q ) . In the previous example of a derivable plane AG(2, q 2 ) , we first realized AG(2, q2) as a 2dimensional GF(q2)vector space and then decomposed the
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C H A P T E R 3. BEGINNING DEWVATION
spaceintoa4dimensionalGF(q)vector space. From the4dimensional GF(q)vector space, we may then construct the associated lattice of subspaces in PG(3,q).
Definition 3.9 A ‘regulus’ R is a set of q + 1 lines in P G ( 3 , q) with the property that any line E which intersects R in at least three lines of R must then intersect each line of R. Any such line E is called a ‘transversal’ to R. The set of q + 1 of transversals to R form aregulus R o p p called the ‘opposite regulus Note that the opposite regulus to the opposite regulus of R is R.
3.2.1 The Standard Vector Space Form of a Regulus. Proposition 3.10 Let R be a regulus in P G ( 3 , q )and let VR denote the set of 2dimensional GF(q)subspacescorresponding to R. W e m a ychoose a basis for VR so that vectorsare represented in the form (XI,x2,yl,y2) where x = (x1,x2) and y = (91, y2) with scalar multiplication (Zl,Z2)6
= (216, S S ) .
Then the 2dimensional GF(q)subspacesof VR are P
= 0,y = 26 V 6 E G F ( q ) .
The opposite regulus in vector space form is the set of subplanes
where not both a and b are 0. Proof: Choose a basis so that three lines of R as 2dimensional GF(q)subspaces are represented as x = 0,y = 0,y = P where the vectors have the general form ( ~ 1 ~ 2y1, 2, y2). Now take any 2dimensional GF(q)subspace W which shares a ldimensional subspace with each of x = 0, y = 0,y = P. Note that this is the vector space version of the requirement for a transversal. Hence, W is ((O,O,g!, yz), (xi,x:, 0,O)) where in order to intersect y = P, we must have that (y:,yJ/2*) = (xi,x4)S for some S E G F ( q ) . Hence, W is n(,;,,;). That is, the transversals have the form maintained in the theorem.
DERIVATION. 3.3. INFINITE
23
Once we have R o p p represented in the form maintained, we may use the analogous argument showing that AG(2,q2) is derivable to show that R may be represented as maintained.0 Hence, in our example of a derivable affine plane, AG(2,q2),given any affine point 0, the set of lines incident with 0 and intersecting GF(q)U {m} = PG(1,q) projectively defines a regulus R in the associated threedimensional projective space PG(3,q). The set of subplanes of order q incident with 0 each of whose set of infinite points is GF(q)U {m} is the opposite regulus R o P P to R. As we mentioned previously, it is immediate that the subplanes of R are Desarguesian subplanes of order q.
3.2.2 The Regulus Question. We have seen that AG(2,q 2 ) is a derivable affine plane. In fact, Albert used this to construct the Hall planes by derivation. The derivation set used for the derivation corresponds to a regulus in the associated projective space PG(3,q). Perhaps the most important question dating from the is: Does every derivation set in a finite afRne plane correspond to a regulus? We have been motivated by two interesting types of planes. The Hall planes and the Hughes planes. We have begun with discussion of the Hall planes and will, in due course, discuss the ingenious Hughes planes. Appropriately, we shall end this book with a discussion on infinite analogues of the Hall planes which we call the ‘skewHall planes’ and which are derivable but can produce widely different planes.
3.3 InfiniteDerivation. It was noticed in Johnson [35]that the construction process of derivation does not depend upon finiteness but rather on the existence of a class of Baer subplanes. Definition 3.11 Let E be a projective plane, a subplane is a subset of points and lines which forms a projective plane. Let C, a subplane of C . C , shall be said to be a IpointBaer’ subplane if and only if every point of the plane is incident with a line of C,.
C H A P T E R 3. BEGINNING DERIVATION
24
C, shall be said to be a ‘lineBaer’ subplane if and only if every line of the plane is incident with a point of C,. C, shall be said to be a ‘Baer’ subplane if and only if the subplane is both pointBaer and lineBaer. A counting argument establishes that within a projective plane of order q2, any subplane of order q is both a pointBaer and a lineBaer subplane so that any finite plane of order q is a Baer subplane.
Definition 3.12 In the general case, an afine plane is ‘derivable’ if there is an associated derivation set D, which is a subset of the set of infinite points and such that for each two distinct points A, B whose join is in D, projectively then there is a Baer subplane TA,B containing A and B whose infinite points are exactly those of D,. The following is fundamental.
Proposition 3.13 A Baer subplane r0 of a projective plane T is maximal; if p is asubplane of T which properly contains a Baer subplane T , then p =T.
Proof Let p be a subplane which properly contains T,. Chooseany point P of p  T,. Then there exists a unique line l of T , incident with P. Choose any point Z of T  T , not on l and form PZ. There exists a unique point Q of r0incident with PZ. But, Q is a point of T , and P is a point of p so PQ = PZ is a line of p. There is a unique line m of T , (which is then a line of p ) incident with Z which is not l . Hence, m n PZ = Z is a point of p, So, all points of T not on l are points of p. Choose any point T of l and form TZ which intersects in a unique point R of r0.Then TZ = T R is a line of p so that T = l?n TR is a point of p.0 Ostrom’s theorem on finite derivable planes extends directly to infinite derivable affine planes:
Theorem 3.14 (See Theorem 3.2) Let T be a derivable afine plane with derivation set D,. Let B denote the associated set of Baer subplanes each of which has D, as its set of infinite points.
3.3. INFINITE DERIVATION.
25
Form thefollowingincidencestructure 7r(Dw): The points of 7r(Dm) are the points of 7r and the lines of 7r(Dw) are the lines of IT which do not intersect Dm in the projective extension and the subplanes of B. Then ~ ( 2 3 is~ an ) @ne plane. An interesting question is whether A G ( 2 , K ) is derivable when K is an infinite field. If K is a quadratic extension of a field, the answer is yes and the proof is almost identical to the proof that AG(2,q2)is derivable. The reader might try to consider the same question if K is an infinite skewfield which is not a field but which is a skewfield quadratic extension of a field.
Chapter 4
SPREADS In this chapter,we provide the ‘basics’ of spreads and translation planes due to Andre [2]. We begin with the definition of a translation plane.
4.1 Affine Planes with a Transitive Translation Group. We have previously defined pointline transitivities, we now consider the elements of such groups.
Definition 4.1 A ‘translation’ of an afine plane is a collineationwhich leaves each parallel class invariant and fixes each line of some parallel class. Hence the set of all translations of an afine plane form a subgroup of the full collineation group which we shall call the ‘translation group’. A ‘translation plane’ is an afine plane which admitsa group of translations which acts transitively on the points. Note that this implies that the translation group is regular on the set of points (transitive and no nonidentity element f i e s a n y ( a f i n e ) p o i n t ) . More generally, a ‘central collineation’ Q of a projective plane is a collineation which fixes a line t pointwise and a point P linewise. The line t i s called the ‘axis’ and the point P i s called the ‘center’ of Q. If P incident with t then U i s said t o be a n ‘elation’ and otherwise a ‘homology’. If Q i s a homology with axis t and center P and 0 is any pointof t then OP is called the ‘0coaxis of U ’ . 27
CHAPTER 4. SPREADS
28
Hence, a translation taken projectively is simply an elation with axis the line at infinity. The following result actually will be shown to be a generic way to represent translation planes.
Theorem 4.2 Let V be a vector space and let S be a set of mutually disjoint vector subspaces such that the s u m (direct sum) of each distinct pair is V and such thatUS = V . Also let C denote the set of all cosets of the members of S. .(V,S) := (V,C,I ) , whose Then the settheoretic incidence structure: (points’ are the vectors of V and whose ‘lines’ are the cosets in C , with settheoretic incidence I , is a translation plane and the full group T of translations consists of the group of translations of V regarded as a vector space:
T={TA:zHz+A;AEV}. Proof: We need to check that we obtain an affine plane and that there is a group of translations which acts transitively on the affine points. Let M be an element of S. First note that the group T willleave { M A ; A E V } invariant. Secondly, given any affine point P which is not the zero vector 0 of the vector space, we have that (0, P} is in a unique element, say OP, of S. Since the group T clearly acts transitively on the affine points (vectors), it follows that each two distinct points are incident with a unique line of the incidence structure so that we obtain an affine plane. The parallel classes are the cosets of elements of S and it is now obvious that T is a collineation group which leaves invariant each parallel class. It remains to show that each elementTAof T is actually a translation, that is, that TA fixes some parallel class linewise. But, every A is in some M of S so that M B I”) M B A = M B for each B E V . Hence, TA fixes exactly one parallel class linewise and is, therefore, a translation.0
+
+
+ +
+
Definition 4.3 W e shall call the set S a ‘spread’ of V and the elements of S the ‘components’ of the spread. Definition 4.4 Let G be a group. A (partition’ of G is a set N of nontrivial pairwise disjoint proper subgroups of G such that
4.1. P L A N E SW I T H
A T R A N S I T I V ET R A N S L A T I O NG R O U P ,
29
the members ofN are the ‘components of the partition’, and if all the components in N are normal in G,then N is a ‘normal partition’ of G.
Theorem 4.5 Let G be a group and N a normal partition of G such that
Then each of the following is valid: ( l ) G is a direct product of any two distinct subgroups of N ,
(2) each two distinct subgroups of N are isomorphic, and (3) G is Abelian. Proof The proof to (1) is immediate m the elements of disjoint normal subgroups commute. Consider (2). A groupcannotbe expressed as the disjoint union of two distinct subgroups. Hence N contains at least three members. So given distinct N I ,N2 E N , we may choose a thirdNo E N , and now N = NIONO and also N = NZ0 No. Hence
as required. (3) Since G is the direct sum of any two distinct members of N , we see that elements from distinct subgroups of N commute. So assume z, y E A E N and choose a nonidentity B E B E N  { A } ,and observe the following chain:
( q ) B = z(yB) and since yB not in A so = (yB)z = y(Bz) = y(zB) = (yz)B so that zy = yx.
This proves (3).0
Definition 4.6 A normal partition N of a group G is a ‘generating normal partition’ if G is generated by every pair of distinct components N I ,NZE N .
CHAPTER 4. SPREADS
30
Theorem 4.7 Let N be the components of a generating normal partition of a group G. Then the settheoretic incidence structure whose ‘points’ are the elements of G and whose ‘lines’ are the cosets of the elements of N i s an afine translation plane whose translation group consistsof the bijections of G, for every A E G of type: G + G 9
H
gA
Proof: As a spread of a vector space is a generating normal partition of the additive group, we see that the result in question is more general. However, one can merely notice that the proof for the vector space version can be applied for the proof.0 Now that we have a more general way to form a translation plane, the question becomes whether every generating normal partition of agroup forces the group to be a vector space over some skewfield. For this question, the answer is found in the following theorem, Note that, henceforth, we shall use ‘additive’ notation.
Theorem 4.8 Let G be a group and N a generating normal partition of G. Let Kenddenote the setof group endomorphisms which leave each component invariant. T h e n Kend is a skewfield and G is a vector space over Kend. The elements of the skewfield Kend are called the ‘kernel endomorphisms’ of the partition. The skewfield Kendis called the ‘kernel of the spread’. Furthermore, the nonzero endomorphism set which permutes the components is a subgroup of I’L(G,Kend),
Proof: Since G is Abelian by the previous result, the endomorphisms in Kendclearly form a ring. Hence it is sufficient to show that all the nonzero maps cj E Kendare bijective. Suppose A” = 0 for A # 0 in some component d.Let B be a component distinct from A and recall that G = A @ B. As 0 = A$ = ( A + B)$ (B)$, then as (d+B)4 is in a component C distinct from the previous two components, we have 0 = B$,whenever B is in any component B # A. Since the argument is symmetric, it follows that cj is the zero map. Thus, all nonzero mappings 4 E Kendare injective homomorphisms of G.
+
4.2. FUNDAMENTALTHEOREM OF TRANSLATIONPLANES.
31
Next we check the elements rj E Kendare surjective mappings, Let A be any nonzero element of G in a component d of N . Let B E B be a nonzero element in some other component of N , and define a third component 2 that contains B4  A. Consider (2 B ) n d. Since we have an underlying affine plane, this intersection defines a unique point of the plane which is a unique nonzero element W of G. Since r j fixes all components, it follows that A  W4 is in d as W is in d. But, A  W4 is in 2 if and only if A  W4 (B4  A) is in 2 which is valid if and only if ( B  W)d is in 2. Again, since r j fixes 2, the last statement is valid if and only if ( B  W ) is in 2 which we know is true as Wisin(Z+B)nd. So,AWd=OandhenceA=Wd. Hence, all nonzero mappings of Kendare surjective and thus, we have the proof to the theorem.0 So, we may refer to a generating normal partition as a ‘spread’ without confusion.
+
+
4.2
The Fundamental Theorem of Translation Planes.
In 1954, Andre provided the foundation for the theory of translation planes by proving that any translation plane may be identified with a generating normal partition of a group which actually turns out to be a vector space over a skewfield.
Theorem 4.9 The Fundamental theorem of Translation Planes. Let T be a translation plane with translation group T and let P denote the set of parallel classes of T . Let Ta denote the subgroup of T &ng all the lines of Q , for Q E P . ( l ) Then l? = U{Ta ;, Q E P ) is a spread on T and hence T is a vector space over the associated kernel Kend. (2) T is isomorphic to T T ~the , translation plane constructed from the spread of T . (3) The full collineation group G of T T is ~ TGo where Go is the full subgroup of the group I’L(T, Kend)that permutes the members of l? among themselves.
CHAPTER 4. SPREADS
32
Proof: (l)Ta is the subgroup of T fixing individually all the lines through a,hence it is normalized by T since T fixes each such class a. Since every translation in T has a unique center, T gets partitioned by its normal subgroups of type T,. It remains to show that T = Ta @ Tp whenever a and p are distinct parallel classes, Let t E T , and suppose t : A H B , where A is any affine point and B # A. Since Ta and Tp are normal and disjoint subgroups, it is sufficient to verify that t E (T,, Tp). Let a A n p = 2. Since Ta has as its nontrivial orbits all the affine points on each line through a, there is a g E Ta such that g : A H x and, similarly, there is an h E Tp such that h : x H B.Now clearly gh(A) = B. But the regularity of T now forces t = gh. Thus, T is generated by any two distinct Ta and Tp. This proves (1). (2) Fix an affine point 0 of T , and to each affine point A of T assign the translation T A E T that maps 0 onto A. Consider the bijection Q : A H T A , from the affine points of T onto the points of the vector space T. Consider the affine point A E e, where e is any affine line of T . Let be the unique line parallel to through 0 in the parallel class S. Clearly, T6 has e6 as its Oorbit, so Q(&) = T6. Next note that the points of e may be expressed as T A T ~ ( Oas) , the group T6 acts transitively on the affine points of each line of the class S. Now the image Q(TAT~(O)) = TAT^, i.e., a coset of T6. Thus we have shown the bijection Q maps the lines of T to cosets of the spreads associated with T , which means Q is a bijection from the affine plane T onto the affine plane associated with the spread on T that sends lines onto lines, so that Q is an isomorphism between the planes. This proves (2). The translation subgroup of the full collineation group G of TTTy may, of course, be identified with T itself. By the transitivity of T , G = TGo and, by its regularity, we further have T fl Go = (0). Let h E Go. Since T is normal in the translation plane TTTy, we have, for every T A E T , a unique TA/ E T such that hTA = T A )so~ ~ T A ( D=)T A , ~ ( D ) for all points D of TITy (elements of T ) . Hence,
+
h(D A) = A'
.
... .
+ h(D) for all points D in TTTy.
4.3. THE BASIS DECOMPOSITION THEOREM.
33
Letting D = 0,we observe that h(A) = A' and so the above identity yields
+
+
h(D A) = h(D) h(A) for all D,A points in 7rr. Hence, h is additive, and permutes the members of l?. Conversely, any map with these two properties also permutes the cosets of the components of F,and is thus a collineation of 7rTy. Kend(0) is a normal subgroup of Go. Hence, if h E Go, IC E Kend(0}, and A E T , then h(kA)= Ic"h(A) where Q is a multiplicative automorphism of Kend (0). For k,z E Kend {0}, we have (IC z)"h(A)= h((k z ) ( A ) )= h(kA+ zA) = h(kA) h(zA) = IC"h(A) z"h(A) which is equal to (C" + z')h(A). Thus, ((IC + z)"  IC"  z")h(A)= 0 for all A. Since h is bijective, it follows that (IC + 2)"  IC"  z" = 0 for all IC, z E Kend.Hence, Q is a field automorphism of Kendso that Go is a subgroup of l?L(T,Kend). This completes the proof of the theorem.0
+
+
+
+
Corollary 4.10 An afine plane 7r isisomorphictothe afine geometrg is a translation AG(2,K) for some skewfield K ifandonlyiftheplane plane whose associated vector space is 2dimensional over its kernel K . Proof: Note this follows directly from Theorem 3.8.0
4.3 TheBasisDecompositionTheorem. We now wish to make further use of the representation of a translation plane by use of a vector space. We shall prove a series of lemmas involving how to decompose a spread into a more usable model. Definition 4.11 Let S be a set of permutations of a set X containing the identity mapping. S is said to act 'regularly' on X if and only if for any pair A, B of elements of X , there exists a unique element QA,B of S mapping A to B. (More generally, the set acts 'semiregularly' if there is at most one such mapping.) Note that if X is finite then 1x1 = IS1 and it follows that for any element C in X then {Cc ; Q in S } = X . However, in the infinite case, thislast condition may not imply regularity.
CHAPTER 4. SPREADS
34
say that S has a ‘transitive (regular)
To distinguish this case,weshall orbit
Remark 4.12 A set acts regularly on X if and only if every element of X defines a transitive (regular) orbit. Remark 4.13 I n the following lemmas, we adopt the following conventions: Let T be a translation plane constructed .by a spread S of a vector space V . Let F denote any skewfield over which the components are Fsubspaces. W e f;3: the notation and specifv T as a left Fvector space. Let X , Y , 2 be any three distinct elements of S and form X @ Y and note that the componentsare Fisomorphic (isomorphic as Fvector spaces). The points of T are then elements (x,y) of X @ Y for all x of X and y of Y. In this representation, X is X @ 0 which we denote by y = 0 and Y is 0 @ Y which we denote b y x = 0. For each basis Bx for X and for each basis B y for Y ,take {Bx,B y } = B as a basis for T and write elements of T as tuples (xi,yi) f o r i in some index set X with respect to B where {(xi for i in X) ; xi in F } = X and {(vi for i in X) .;yi in F} = Y.Note that,such an indexing is possible as X and Y are Fisomorphic and therefore their bases have the same cardinality. Lemma 4.14 For each basis Bx for X , there exists abasis B y for Y so that taking {Bx,B y } as a basis for T then 2 has the representation { ( x i , x i ) for xi in F and i E X} which we denote by y = x. Proof: Since X and Y are Fisomorphic, it follows that for each x E X thereexists a y E Y such that x y = z isin 2. Define a mapping QZ : X : x I”) y if and only if x y E 2. It is claimed that uz is an Flinear mapping as a E F implies that ax ay = QZ so that QZ maps ax onto ay. It follows that the mapping is injective since x y = x y* if and only if y = g*. Now 2 @ X = V so given any y in Y ,there is a unique z E 2 and x E X such that y = z  x so that x y = z and the mappingis onto. Since (x x*) (y g * ) = (x y) (x*+v*), it follows that the mapping is a homomorphism so that 02 is an isomorphism. Then {Bx,~z(Bx)} is a basis for V such that 2 has equation y = x which proves the 1emma.O
+ +
+ + +
+ +
+
+
+
+
4.3.BASIS THE
DECOMPOSITION THEOREM.
35
Remark 4.15 Since X and Y are Fisomorphic, we now identify X and Y in this representation and take T = X @ X . Definition 4.16 Recall, by G L ( X ,F),we shall mean the group of all bijective Flinear endomorphisms of X . Furthermore, we point out that images ( X ~ Q W )= ((Cxiaij) for j in X) and observe that the elements of F due to the endormophism are written on the right of the elements xi whereas scalar multiplication is on the left. Theorem 4.17 {ow ; W E S } is called the set of ‘slope endomorphisms’ or ‘slope mappings’. We shall also call this set a ‘spread set’ for S . Given a spread set M ( S ) and any elementx, of X then {X,Q for all Q in M ( S ) }= X so that M ( S ) acts regularly on X . For each element W of S  { X , Y } , there is an element QW o f G L ( X ,F) such that W has the equation y = XQW. That is, W = {((xi,x p w ) for i in X) ; xi in F}. Proof Let W be any component so that for each x in X , there exists QW(X) is in W . Now if we identify X and Y then QW is an Fautomorphism of X . Hence, W has the equation y = XQW. This completes the proof of the 1emma.O QW such that x
+
Lemma 4.18 In the above representation,thefollowingset of endomorphisms of form a subskewfield Fend of the kernel of the spread:
Furthermore, there exists a skewfield K containing F such that each component is a left Ksubspace and such that {K, for all P in K } is equal to the kernel of T.

Proof Note that any element (x,9) (Px,Py) maps x = 0 to x = 0, y = 0 to y = 0 and (X,XQW) (Px,P(xa~) = ( P x ) Q ~so) that y = X Q maps to y = XQW. Hence, any such element is a kernel endomorphism. Let k* be a kernel endomorphism. Since k* fixes x = 0 and y = 0, define the k*image of (x,O) as (kx,O) and the k*image of (0,y) as (0, hy). Since k* is additive and fixes y = x it follows that k = h.
~
CHAPTER 4. SPREADS
36
We define the sum and product of the mappings Ic acting on X in the standard way and then note that K = {k ; k* in the kernel endomorphisms maps (x,y) to (Icx,Icy)} is a field isomorphic to the kernel. It follows that X is a Kvector space. This proves the lemma.0
Lemma 4.19 The full collineation group G of the translation plane is a semidirect product of the translation group T by the subgroup Go which fixes the zero vector 0. I n this representation, Go is a subgroup of r L ( n ,K) and r L ( T ,K ) is isomorphic to rL(.rr,Kend)where Kendis the kernel of n , Proof We know that the full collineation group of the translation plane is TGo where Go is a subgroup of FL(V,Kend)by Theorem 4.9 part (4). Note that Ic,p = K,Kp in the previous notation (using Tp(x, y) = (/?.,/?g)). Let u be a mapping of K and (T* a mapping of Kendsuch that u* : Icp
I")
Icp..
It is easy to verify that u* is an automorphism of Kendif and only if D is an automorphism of K. Let g E r L ( n ,Kend). Then g(Kp(x,y)) = Ic;'g(x,y) for some automorphism u* of Kendsince the kernel endomorphisms form a normal sub/?y) shows that then K;'g(x, y) = group. However, defining /?(x,y)= (/?x, Pog(x,y) = g ( / ? ( x y). , This shows that, in fact, g is also in l?L(n,K ) and furthermore that the two groups FL(n,K ) and r L ( n ,Kend)are isomorphic. This completes the proof.0 Lemma 4.20 Conversely, suppose X is a Kvector space and E is a 'set' of elements of G L ( X ,K ) which acts regularly o n X  (0). Then X @ X admits a spread x = 0, y = 0,y = xu for all u in E which defines a translation plane. Again, in this context, E is called a 'spread set'. Proof We merely need to show that theputative spread of E = 0, y = 0, y = xu for u E E actually is a spread of X @ X . In other words, we need to show 'that we have a cover and that any two components are Kisomorphic and generate the vector space. Let (x0,yo)be any nonzero vector which is not in either x = 0 or y = 0. Since E acts regularly on X , there is a unique (T E E such that yo = xou. Hence, we have a covering. As Q is in G L ( X , K ) ,obviously y = xu is Kisomorphic to y = O. It follows that any two components are Kisomorphic.0
4.3.BASIS THE
DECOMPOSITION THEOREM.
37
Remark 4.21 (1) Spread sets, spreads and translation planes are equivalent objects. (2) So, when representing a translation plane in this form, the set of points of the plane forms a Kvector space and the kernel of the spread is isomorphic to K . W e shall call K , the 'component kernel' of the spread. (3) For future reference, we shall refer to the contents of the previolls lemmas of this section as the 'Basis Decomposition Theorem'. The following corollary provides a particularly convenient construction method for finite translation planes.
Corollary 4.22 Let V = X @ X be a finite vector space over a field K isomorphic to G F W ) for some prime p . Assume that E G G L ( X ,K ) such that (i) IEl =p'  1 and (ii) for all 8 ,T E E U (0) then 8  T is also in G L ( X ,K ) U (0). Then E is a spread set and thus defines a translation plane. Proof: We must show that E acts regularly on X  (0). Note that for x nonzero, x8 = XT if and only if x(.  T) = 0 if and only if x = 0. Hence, E is injective and by the cardinality of E , it must be surjective. Hence, E acts regularly on X . 0
4.3.1 Some Examples. Although it is easy enough to provide examples without using the basis decomposition theorem, we elect to apply this theorem to illustrate the simplicity of its use.
Example 4.23 Desarguesian planes. Let K be any skewfield and form the %dimensional vectorspace K @ K and call its elements 'points'. A spread will be defined as the set of all 1dimensional Ksubspaces. If we identify three components as x = 0, y = 0 , y = x , the remaining components have the form y = X A f o r A in K and juxtaposition denotes skewfield multiplication. It is easy to verifg that a spread is indeed obtained in this way as clearly { A ; A E K  (0)) is a spread set. W e have seen this afine plane previously as AG(2,K ) .
CHAPTER 4. SPREADS
38
Example 4.24 Let K be a finite field isomorphic to GF(q) where q i s a n odd prime power. Form K @I K @I K K and call its elements 'points'. Write x = (XI,x2) and y = (VI,y2) where xi and yi are in K f o r i = 1,2. Define a set of 2dimensional Ksubspaces S as follows:
s={x=o,Y=o,
y=x
[y
'z]
Vu,t€K)
where c is an automorphism of K and y i s a .fixed nonsquare in K , T h e n S i s a spread for the associated vector space,
with derivable
i s a spread set and the associated afine plane is derivable partial spread (see definition below) forallaE~}.
{z=o,y=x[ U 0 Proof: Since there are q2  1 elements
it follows from previous results that if the set is transitive then a translation plane is obtained. Furthermore, given a nonzero vector z and spread set components M and N then z M # zN if and only if M  N is nonsingular. In this particular case, since the matrices are additive,we merely have to check that each is nonsingular or equivalently that each determinant is nonzero. Thedeterminant of
[ '1"
'r ]
nonsquare whereas both u2 and Notice that
{
IC
= 0,y = x
[
is u2 tU+lare
]
whichis
nonzero since y is
square.
for all U E K
l
is the vector version
of a regulus. Hence, we have another example of a finite derivable affine plane.0 We conclude this chapter with a basic isomorphism result.
4.3. THE BASIS DECOMPOSITION THEOREM.
39
Theorem 4.25 Let T be a translation plane with spread S, of X @ X = V where X i s a left Kvector space and let p be a translation plane with spread S, of Y @ Y = W where Y is a left Lvector space. Assume that K and L are the component kernels of T and p respectively. Let p and T be isomorphic by a bijective, incidence preserving mapping 4 f r o m p to T , ( l ) T h e n L and K are isomorphic skewfields and 4 m a y be considered a semilinear mapping from W onto V . (2) If T = p then 4 is an element of the group rL(V,K ) . Furthermore, the full automorphism group G of the translation plane T is a semidirect product of the translation group T by the subgroup Go of rL(V,K ) which permutes the components of the spread S . T h e subgroup Go of I’L(V,K) is called the ‘translation complement’ of G or T , firthermore, Go n GL(V,K ) is called the ‘linear translation complement.
Proof: We have seen (2) previously. We note that if g is in the kernel endomorphism skewfield Kendof T then gldg is in thekernel endomorphism skewfield Lendof p. Hence, K N KendN LendN L.0 Remark 4.26 The problem of connecting the various skewfields any one of which could conceivably be called the ‘kernel’ of the associated spread shall be takenupagain in thechapteronBaerextensions.Wepostponethis development until we have had a chance to look further at derivable afine planes and coordinatization.
4.3.2 Partial Spreads. Definition 4.27 A ‘partial spread’ of a vector space V = W @ W over a skewfield K i s a set of Kvector subspaces, also called ‘components’, which are Kisomorphic to W and which pairwise generate V . Given a partial spread, we may form a ‘translation net’ whose points are the vectors of V and whose lines are the translates of the partial spread components, Each component then defines a parallel class. W e shall also speak of a partial spread of PG(V,K).
CHAPTER 4. SPREADS
40 4.3.3
DerivablePartialSpread.
Definition 4.28 If a translation plane is derivable then the parallel classes of the derivable net define a partial spread which we shall call a 'derivable partial spread'. More generally, any partial spread is derivable if its associated translation net is derivable. We have seen that given a regulus in PG(3,q),there is an associated derivable partial spread and derivable net. The question is, given a derivable net, is there an associated derivable partial spread which is a regulus in some 3dimensional projective geometry? In the next chapter, we begin our analysis of the structure andrepresentation of derivable affine planes and derivable nets.
Chapter 5
DERIVABLE NETS It is possible to consider incidence structures called ‘nets’ which are analogous to the substructuresof a derivable affine plane defined by a derivation set. The definitions of pointBaer and lineBaer subplanes within a net instead of within a plane are the same as in the plane case with the condition on the lines restricted to the lines of the net. One may think of a net as an affine plane with some parallel classes of lines missing although not every net may necessarily be extended to an affine plane.
Definition 5.1 A ‘ n e t ’ N = (P,L,C,Z) is an incidence structure with a set P of ‘points’, a set L of ‘lines’, a set C of ‘parallel classes’ of lines and a set Z which is called the ‘incidence set’ such that the following properties hold: ( i )Every point is incident with exactly one line fromeach parallel class, each parallel class is a cover of the points and each line of C is incident with exactly one of the classes of C (parallelism is an equivalence relation and the equivalence classes are called ‘parallel classes 7 . (ii) Two distinct points are incident with exactly one line of L or are not incident. Definition 5.2 A ‘derivable net’N = (P,L,B,C,z> is an incidence structure with a set P of ‘points’, a set L of ‘lines’, a set B of ‘Baer subplanes’ of the net, a set C of ‘parallel classes’ of lines and a set Z which is called the ‘incidence set’ such that the following properties hold:
41
C H A P T E R NETS 5. DERIVABLE
42
(i) The structure is a net. (ii)If we refer to the setC as the setof ‘infinite points’ then the subplanes of B are afine planes with infinite points exactly those of the set C . (iii) Given any two distinct points a and b of? which are incident with a lane of C , there is a Baer subplane Ta,b of B containing (incident with) a and b. Remark 5.3 Given aderivable net N = (P,C,B,C,z>, we may definea parallelism relation on the set B of subplanes as follows: Two subplanes p and r of B are defined to be parallel if and only if they are disjoint on points. Define
N*= (P,B,C,C*,Z) = (P*, C*,B*,C*,Z*) as the incidence structure where Q* = P is the set of points, C* = B is the set of lines, B* = C is the set of Baer subplanes, C* is the parallelism on the set of lines C* defined on the setof Baer subplanes B and T is the incidence set Z. Then N*is alsoaderivable net which is called the ‘derivednet’. The transfer fromN to N*is called the ‘derivation’ of N .
5.1
Coordinatization.
Early attempts to recognize when a finite affine plane is derivable concentrated on the coordinate systems of the planes. We have not yet defined general coordinate systems but for Desarguesian affine planes, we have noticed that the coordinatizing field GF(q2)is a right 2dimensional vector space over GF(q). Coordinatization will play a major role in this book and it is a very important tool for geometers. In fact, many of the most interesting examples of affine and projective planes are and were constructed by distorting either the multiplication or addition of a field or skewfield. Such a modified system produces a ‘coordinate structure’ which can then be used to ‘coordinatize’ an affine plane conceivably quite different from the original. Note that when we wrote lines in the form y = zm+b, for m, b E GF(q2), the vector space was considered as a right vector space over GF(q). Whether a coordinate structure is a right or left .vector space is more important when we are not dealing with a field and we shall come back to this point in later chapters.
5.1. COORDINATIZATION.
43
There has been considerable research done by analysis of derivable affine planes using coordinate methods by Ostrom [62],Lunardon [57] in the finite cases and by Grundhofer [25] in the infinite case. Here we provide a short section on coordinates since these provide an easy way to view derivable nets within affine planes. Furthermore, it is possible to coordinatize arbitrary nets. In particular, coordinate based proofs will play a central role in various of our arguments on direct product nets.
5.1.1 Coordinatesof a Net. Let N be an arbitrary net with at least three parallel classes and let C denote a set which we call a set of ‘coordinates’ which may be put into 11 correspondence with the set of lines of a given parallel class. We set up a 11correspondence between the points of N and theelements of x Choose any parallel class and call the class (m). Using the 11 correspondence, if a line .l of (m) corresponds to c E C, assign the points of .l the coordinates (c, y) for all y E C and designate the line .l by z = c. Specify two particular elements of C as 0 and 1 and designate two other parallel classes relative to the 11correspondence as (0) and (1). Choose any line .l1 of (1) and assign the points (c, c) to this line by defining .l1 n (S = c) to be (c, c). Define the line of the parallel class (0) incident with (c, c) as y = c. Let P be any point of the net. Then, there is a unique line z = c of (m) and a unique line y = d of (0) incident with P. We designate the point with coordinates (c, d ) . Define CN = { m E C ; (0,O) and (1,m) are incident with a line of the net}. Since each parallel class contains a unique line joining (O,O), it follows that the set of parallel classes of the net may be indexed by C N ;the parallel classes are designated by (m) where m E CN together with the class (m). We now define the equations for the lines of the net.
c c.
5.1.2 The TernaryFunction. Given a line Mm of a parallel class (m) not equal to (m). This line intersects the line z = 0 uniquely in a point (0, b). Furthermore, the line intersects the
CHAPTER 5. DERTVABLE NETS
44

line z = zo uniquely in a point (zo,yo). Hence, there is a function T which depends upon m, b and z, that describes yo uniquely; T : C x CN x C C and T ( z ,m, b) = y if and only if (z,y) is incident with the line of (m) joining (0,b). We call T a ‘ternary function’ and designate the line Mm by y = T ( z ,m, b).
We point out that T ( z ,1 , O ) = z as (z,z) is incident with the line of (1) joining (0,O).
5.1.3
Addition.
+
We shall define a binary operation denoted by and called ‘addition’ as follows: There is a unique line incident with (0, b) of the parallel class (l),y = T ( z ,1,b) and there is a unique line z = a which is incident with (a, a)in the parallel class (m). We define (U, U
+ b) = (g = T ( z ,1,b ) ) n (Z=
U).
We note that thus, ( a , T ( a ,1 , b ) ) = (a,a + b) so that T ( a ,1,b) = a + b. Hence, we may denote the lines in (1) by the equations y = z+b for b E C.
5.1.4
Multiplication.
Choose the line y = T ( z ,m, 0) incident with (0,O) and in the parallel class (m) for (m) # (m) (also incident with (0,O) and (1,m)).We define a binary operation which we call ‘multiplication’ and denoted by * in terms of the intersection with x = a: (~,~*m)=(y=~(~,m,o))n(~=a).
Hence,T(a,m,O)=a*mandl*m=mforallmEC~. Note also that a * 0 = 0 for all a E C. We refer to our coordinate system by the notation (C, *,T) with lines z = c, y = T ( z ,m, b) for all c, m, b E C. Furthermore, we occasionally refer to m as the ‘slope’ of the line y = T(z,m,b).
+,
5.1. COORDINATIZATION.
5.1.5 Extension
45
to an Affine Plane andLinearity.
We note that our coordinate system canrefer only to a subnet or an abstract net or to an affine plane depending on the context. The coordinate structure is the coordinate structure for an affine plane if and only if CN = C in the notation developed above. Furthermore, in this case, we shall say that the coordinate structure is ‘linear’ if and only if T ( z ,m, b) = IC * m b. In this case, we denote the system more simply by (C, *) but call this a ‘leftcoordinate system’ with the variable IC written on the left of the ‘elements m.
+
+,
5.1.6 The Question of Linearity. Suppose we have an affine plane withcoordinatesystem (C,+,*,T). If there is a subset K of C such that (K, *) is a skewfield, we call K a subskewfield of C. Now assume that (C,+, *,T) is a coordinate system for a derivable net. Note that except for additionally defined elements in CN,if a derivable net is extended to an affine plane 7r then we may use the same basic underlying coordinate system with an extension of CN to C, and the corresponding extension of the function T from C x CN x C to C x C, x C = C x C x C. Most early analysis of derivable nets involved the coordinate approach and attempted to show that when the coordinate system for the affine plane is linear then the plane is derivable provided there is a subskewfield K such that C is a right 2dimensional vector space over K while writing lines in theformy=IC*m+b(s.ac=z*acforzECandacEKrepresentsscalar multiplication). Hence, we formulate the general question more generally for nets as opposed to affine planes containing them should they exist. Let D beaderivable net. Doesthereexistaleftcoordinate system C = (C,+, *) for D which is linear and a right 2dimensional vector space over a skewfield K? Shortly, we shall give examples of derivable affine planes with linearcoordinate structures and some with nonlinear coordinate structures. However, the subcoordinate systems coordinatizing the derivable net &re all linear with respect to the net.
+,
CHAPTER NETS 5. DERIVABLE
46
We now point out that nets which may be so coordinatized are, in fact, derivable.
Theorem 5.4 Let D be a net and let A denote a net containing D where
A m a y be D.
Suppose that there isa coordinate system (C,+, *,T)f o r A such that the lines of D have the fomn x = c, y = x I a + b for all c,b E C and for all (YECg=KGC. Assume thatK is a subskewfield of C such thatC is a right 2dimensional vector space over K. T h e n D is a derivable subnet of A and the Baer subplanesof the derivable net are Desarguesian subplanes. Proof: Consider the structures T(a,b)={(a*a,b*a,a*P,b*P)
;a,PEK)
with not both a and b zero. Note that since C is a right 2dimensional Kvector space, it follows directly that T(a,b) is a right 2dimensional Kvector space. Furthermore,
is a right ldimensional Kvector space and clearly r(a,b) E
{ (Y = x * 6)
T(a,b); 6 E
K) U {x = 01,
so we have a ‘spread’ in the terminology of the previous chapter. Thus, the setsT(a,b) are translation subplanes of the net D and the form indicates that the translation subplanes are Desarguesian. It also follows that the structures ~ ( ~ , b (c, ) d ) for all c, d E C are affine planes. Since the lines of the net D have the form = c, y = x * a b for all c, b E C and for all a E K, it follows that, given any two points P and Q, there is a Baer subplane of the net which contains P and Q and shares all parallel classes with D. Hence, this proves that the net is derivable.0 Note that this argument does not show that if A is an affine plane containing D then A is derivable but we shall come back to this in the chapter
+
+
5.2. COORDINATES FOR TRANSLATION PLANES.
47
on extension of a derivable net to an affine plane. However, the argument does show that the subplanes are Desarguesian in this case, As this was a question askedby early researchers in this area, we formally pose this question: Let D be a derivable net and x. any Baer subplane of the net, Is no always Desarguesian? This was shown to be true by Prohaska [67] for finite derivable nets in affine planes and for infinite derivable nets extendible to affine planes by Cofrnan [19]. However, Prohaska’s and Cofman’s ideas are completely different. In fact,it is Cofman’s ideas which provide the main insights for the solution of the structuretheory for derivable nets and we shall pay particular attention to this in the next chapter. We now show that there are awealth of affine planes containing derivable nets. In order to do this, we obtain coordinate structures for translation planes. We shall come back to thecoordinate approach when we discuss ‘pseudoregulus’ nets which are, in fact, derivable nets which may be coordinatized as above by skewfields. We also consider the socalled ‘direct product nets’ in a later chapter and show under certain circumstances that these become derivable nets. We shall use coordinates to show that the existence of certain Baer subplanes within an arbitrary net forces the net to be a direct product net. The exact description of these topics is postponed until we have a structure theory for derivable nets.
5.2
Coordinates for TranslationPlanesandTheir Duals.
Let 7r be a translation plane with spread S of the vector space
[email protected] a left vector space over a skewfield K. Hence, there is a set of left Ksubspaces of V such that each vector lies in exactly one of these subspaces. Choose any three components and choose coordinates as in the above setting for x and call these components z = 0, y = 0, y = z contained in the parallel classes (W),‘(O), (1) respectively and construct a leftcoordinate system (C,+, *, T) with lines z = c, y = T ( z ,m, b) for c, m, b E C.
CHAPTER NETS 5. DERIVABLE
48
Since T is a translation plane, we have a translation group G that acts transitively on the points.
Lemma 5.5 T ( z ,m,b) = z*m+b; coordinate systems for translationplanes are linear. (The reader might note that the following proof works more generally whenever there is a translation group with center (W) that acts transitively on the points (.fine) of a line incident with (W); when thereis an ((W),&)transitivity.) Proof The unique translation T ( o , ~of) 7r which maps (0,O)to (0, b) leaves invariant each parallel class and hence 'fixes' each slope. Note that since T ~  Jfixes ) z = 0, it must fix z = a for all a E C. Now this means that theline y = z is mapped onto the line of the parallel class (1) joining (0, b). Hence, y = z is mapped onto y = x + b. Since z = a is fixed,this implies that an arbitrary point (a, a)is mapped onto a point (a,a b): The line y = c is mapped to a line y = d such that (c, c) is mapped onto (c,c + b) so that d = c b. Thus, T(0,b) : (x,y) W (2,y b) for all 2,y E c. So, since y' = T(z,m,0) is mapped onto y = T ( z ,m,b), then it follows that (2, z * m) W (2, z * m b) E (y = T ( z ,m, b ) ) so that
+
+
+
+
T(x,m,b)= z * m + b . This completes the proof of the 1emma.O
Lemma 5.6 (C, +) is an Abelian group and (a+b)*c = a*c+b*c V a , b, c E C.

Proof Assume that T(,,O) is a translation of that maps (0,O) onto (c,0). An argument analogous to that of the previous lemma shows that T ( C , O ) : (z, 9) (+ .c, v)* Moreover, sincethe translations form a group, we have (z,y)~(,,o)~(d,o)= ((z + c) + d, y) and letting z = 0, it follows that T ( . , o ) T ( ~ , o= ) T(,+,.J,). Since the translations form an Abelian group, the first part of the lemma is proved. Now from the previous arguments, the translations may be represented in the form +r(a,O)T(O,b): (z, y) W (z a,y b) for dl a, b E
+
+
c.
5.2. COORDINATES FOR TRANSLATION PLANES.
Hence, y = x * m
49
+ e is mapped onto y = x * m + f such that
(x+a)*m+f=(x*m+e)+b. Choose x = 0 to realize that a * m Hence, we have
+ f = e + b.
(x+a)*m+(e+ba*m)=(x*m+e)+b so that
(x+a)*m+(a*m) =(x*m) which proves the second assertion of the 1emma.O
Definition 5.7 A ‘left quasifield’ is a spstem (C,+, *) with the following properties: ( l ) (C,+) is an Abelian group. (2) If a, b, c E C then ( a b) * c = a * c b * c. (8) a * 0 = 0 for all a E C. (4) For a , c E C with a # 0, thereexists a unique x E C such that a*x=c. (5) For a, b, c E C , and a # b, there exists exactly one x E C such that
+
+
x*ax*b=c, (6) there exists an element 1 E C all a E C .
 (0) such that 1* a = a = a * 1for
Thus, we have that a coordinate structure for a translation plane can be chosen to be linear and a leftquasifield.
Remark 5.8 Note that the conditions ( l )through (6) provide an alternative manner of defining translation planes as follows: ‘points’ are ordered pairs ( x ,y) ; x,y E C , and ‘lines’ are given by equationsx=c,y=x*m+b forc,m,bEC. The conditions ensure that an aflne translation plane is obtained. The reader might like to try to verify that this follows. Remark 5.9 In the following chapter, we shall be discussingtheHughes planes. These planes use a ‘right’ nearfield for their construction. A ‘left nearfield’ is a left quasifieield (C, *) such that (C  { 0 } ,*) is a group.
+,
CHAPTER NETS 5. DERIVABLE
50
5.2.1 Kernel Mappings and Kernel
of the Quasifield.
When we consider a translation plane as a left vector space W 63 W where W is a left Kvector space and K is a skewfield and select a leftquasifield as a coordinate structure, the kernel mappings imply algebraic properties of the coordinate system. Note that, in this case, we identify C with W as a left Kvector space. We assert that the following mappings define the ‘kernel mappings’:
for all c, m, b E C. These mappings fix eachparallel class and furthermoreany such mapping which fixes each parallel class has this form. To see the first assertion, note that
(0,k * b) and (l,m)
(0,b)

(IC, IC * m )
so that
Furthermore,
*
(c, c m
+ b)
(IC * c, IC * (c * m + b ) )
so that
Note that any collineation which fixes each component has the general form
where c is a additive mapping (i.e.’ aprime field linear transformation).
5.2. COORDINATES FOR TRANSLATION PLANES.
51
We want to show that a ( . ) = k * z for some k in C. If so, thena(1) = IC assuming there is an identify in the quasifield. The indicated mapping fixes each component y = z * m if and only if
a ( z * m ) = a(.)
*m
for all z and m E C. Hence, we obtain, in particular that
a(m)= a(1) * m V m E C. Now,
* *
*
a(1) (z y) = a(z y) = a ( . )
* y = (a(1) * z) * y
and a(1)* (z
+ y) = a(z + y) = a(.) + a(y) = a(1)* + a(1) * y.
This verifies that the kernel mappings areexactly given in the form maintained and, in fact, the kernel is isomorphic to the subset of k’s of C with the given properties. We define the ‘kernel’ of the leftquasifield as follows:
Definition 5.10 If (C,+, *) is a leftquasifield then the (kernel’ is {kEC;(k*c)*m+k*b=k*(c*m+b)}
f o r all c , m , b E C. The kernelof the leftquasifield, the component kernel and endomorphism kernel of the translation plane are all isomorphic.
5.2.2
Dual Affine Planes.
+,
Ifwe define a multiplication o by a o b = b * a where (C, *) is a left quasifield, then we obtain a ‘right’ quasifield and there is an affine plane coordinatized by such right quasifields.
Definition 5.11 A projectiveplane n isdualizedbytaking‘points’and (lines’ of the new plane to be the lines and points of the old plane respectively. The corresponding structure is a projective plane called the ‘dual’ of the plane T.
52
CHAPTER 5. DEHVABLE NETS
Hence, we may dualize the projective plane constructed from the projective extension of an affine translation plane. We may then delete any line and we have again an affine plane.
Remark 5.12 In the situation under consideration, we delete as a‘line’ from the dual plane the ‘point’ which we called (W) in the coordinate structure for the translation plane. If a coordinate left quasifield i s ‘dualized’ in the manner of interchange or reversal of the multiplication, then the plane obtained b y taking points as ordered pairs of the right quasifield is the plane obtained from theprojective translation plane b y deleting as a line the ‘point’ which was chosen as (W) in the coordinate structure for the aflne translation plane. The main point of our discussion is to see that this provides an easy way to find derivable nets within affine planes. Note that for spreads in PG(3,K), the points of the translation plane may be identified with a $dimensionalleft Kvector space. Ifwe take C = K x K, we may select a coordinate system (K x K,+,*) as a leftquasifield.
5.2.3
DualTranslation Planes and Derivation.
We now use a coordinatebased argument that dual translation planes obtained from spreads in PG(3, K) always contain derivable nets.
Theorem 5.13 Let T be an afSine translation planewith spread in PG(3,K)
where K is a skewfield. Let (K x K,+, *) be a leftquasifield for K , Then the associated rightquasifield (K x K, +,0 ) coordinatizes an aflne plane which contains a derivable net D. ThenetDhaslinesy=xoa+b,x=cforallb,cEKxKandforall ~ E wK herexoa=a*x.
+,
Proof It suffices to show that (K x K, 0 ) is a left coordinate system which is a right 2dimensional vector space over K as a subskewfield. However, this simply means that
5.2. COORDINATES FOR TRANSLATION PLANES.
53
and
When we consider a translation plane with spread in P G ( 3 , K ) and 7r as a left 4dimensional vector space, it follows that the kernel mappings contain
{(z,Y)(k*z,Ic*Y);kEK}. Note that the kernel could bea proper superset of this set as the plane could perhaps be Desarguesian. Hence, we have:
for all c, m, b E K x K and for all k E K. Now translate the subkernel and distributive properties in ( K x K , +, *) to ( K x K , +, 0) to see that the above required properties hold so there is a derivable subnet of this dual translation p1ane.O
Remark 5.14 W e have seen that certain afine duals of translation planes with spreads in P G ( 3 , K ) where K is a skewfield always contain derivable nets. What is not soclear is whether the dual translation plane isa derivable aflneplane.Thatis,thequestioniswhether,when deriving thenet, a corresponding afine plane is alwaysconstructed. Of course, this must be the case when the planes are finite. The main problem is whether the Baer subplanes of the derivable net remain 'Baer' when considered as subplanes of the afine plane. We shall see in the chapter on 'dual spreads' that this is not always the case.
5.2.4 ChainsandSemiTranslation
Planes.
Suppose that 7r1 is an affine translation plane with spread in P G ( 3 , q ) which contains a regulus net RI.Then a natural dual translationplane 7r2 is derivable. Let 7r3 denote the derived affine plane. Now extend to the projective extension of 7r3 and ask whether the dual plane has a natural affine restriction 7r4 which is derivable. If fact, amazingly enough, there is such a plane and one can see this by determining the coordinate structure for 7r4 and
CHAPTER 5. DERIVABLE NETS
54
verifying the right vector space property. Furthermore, there is a ‘chain of planes’ obtained in this way by alternately dualizing, restricting to an f i n e plane and then deriving. The chain is then: [TI
(dualize) H
T2
(derive) W
T3
(dualize) W
T4
(derive) H
**l.
Of course, since we are dealing with finite planes, any such chain must be finite, inthe sense that, eventually, we return to anaffine plane isomorphic to it is shown in Johnson [34] that the chains always have period 8, so that is isomorphic to Ti+8 for i = 1,2, ... . The proofgiven in [34] involves constructing the coordinate structures for the first nine planes in the chain each of which depends‘ultimately on the initial coordinate quasifield for the translation plane. The ninth coordinate structure may then be seen to be a quasifield isomorphic to the first. TI.Actually,
Definition 5.15 Let T be an aJgine plane. If there exists a group of translations T such that each orbit of T is a Baer subplane, we shall say that T is a ‘semitranslation’ plane. I n the finite case, T would have order n2 for some integer n and T would have order n2. Remark 5.16 Let V be a derivable net in a dual translation plane T such that derivation produces an afine plane T* and assume that V contains the center (CO) of the associated elation group (the dual of the translation group of the associated translation plane). Then T * is a semitranslation plane. Proof The associated translation subgroup of T with center (CO) (and axis the line at infinity) must leave the derivable net invariant &S the group fixes allinfinite points. Furthermore, the group acts transitivelyon the &ne points of all lines incident with (CO). These lines become Baer subplanes of the derived net which become Baer subplanes of the associated affine plane under our assumptions. The translation group of the dual translation plane becomes a translation group of the derived plane and it follows that this action forces the plane T * to be a semitranslation plane.0
Remark 5.17 Thus, we see that chains constructed as above involve translation planes, dual translation planes, semitranslation planes and their duals as well as the derived dual semitranslation planes. In Johnson [39],the
5.2. COORDINATES FOR TRANSLATION PLANES.
55
groups of derived dual translation planes are completely determined in the finite case. When the order is not 9 or 16, it turns out that the collineation group of any semitranslation plane derived from a dual translation plane must be thegroup inherited from the original dual translation plane. This result is important in tracing through the types of mutually distinct planes obtained in any chain. In the infinite case, basically nothing is known about the chains regarding the collineation groups of the planes or even if such chains universally exist. We shall return to these considerations in the chapter on dual spreads. We conclude this chapter by remarking that coordinates may be invoked to produce affine planes.
Theorem 5.18 Let C be a set containing elements designated as 0 and 1. Assume that there are binary operations + and * defined b y C. Let T be a function from C x C x C W C such that thefollowing properties are satisfied: (i) T(x,O,b) = T(O,m,b) = b for all x,m,b E C, (ii) T(m,1,0) = T(l,m,O)= m for all m E C, (iii) given xi, gi E C for i = 1 , 2 and x1 # x2, thereexists a unique ordered pair (m,b) E C x C such that g; = T(xi,m,b) for i = 1,2, (iv) given m,n,b, c E C for m # n, there exists a unique x E C such that T(x, m, b) = T ( x ,n, c ) . Then (C,+, *,T ) is a coordinate system for an aJgne plane whose ‘points’ are the elements of C x C and whose ‘lines’ are given by the following equations:
g=T(x,m,b), x=cVm,b,cEC Proof: The proof is moreorless immediate and is left for the reader to verify.0
Chapter 6
THE HUGHES PLANES As the existence of the Hughes planes and Hall planes provides something of the impetus for this writer's understanding of the derivation process, we provide, in this chapter, description of the important Hughes planes and show that these are, in fact, derivable. D.R.Hughes constructed these planes by appealing t o the construction of PG(2,q2) from a 3dimensional vector space over GF(q2). However, Hughes realized thatit was possible to construct an analogous projective plane by consideration of what might be called a 'vector space over a nearfield'.
Remark 6.1 E v e q PG(2,q) admits a cyclic subgroup G (called a 'Singer cycle ') of order q2+q+l acting regularly on the points (lines) of the projective plane. Proof Let GF(q3) be a 3dimensional vector space over GF(q) and note that there exists a cyclic group G of order q3  1of GF(q3)  (0). Hence, G acts on PG(2, q),is transitive and hence regular or order (q3  l)/(q  1) on the set of ldimensional GF(q)subspaces of GF(q3) which forms the points of PG(2, q).O
Lemma 6.2 Let K be a right nearfield of order q2 which contains a field N GF(q) in its center. Define a pointline geometry 7~sas follows:
F
57
CHAPTER 6. THE HUGHES PLANES
58
The ‘points’ of rTT+ are the sets
ya, za) and juxtaposition denotes nearfield where (x,y, z)a is defined as (xu, multiplication. (1) There are q4 + q2 1 points of r+. (2) Considering the 3dimensional vector space F3 . of vectors (x,y,z) over F of order q3, it is possible to extend the notation so there is a cyclic group G of order q2 + q + 1 acting on the pointsof r+. (5’) Let t E K  F or t = 1 and let L(t) denote the set of points whose ‘vectors’ (x,y, z ) satisfi x + y t z = 0. .Define the ‘lines’ of T+ as the sets of points of { L ( t ) g ; g E G}. Note that there are q2 1 ‘points o n each ‘line (4) Let r$ = {{(x,3, z)a ; a E K } ; ($,P,4 E F3  {(O,O,O)). T h e n r$ is Ginvariant, If ‘lines’ of r$ are defined as { L ( l ) g ; g E G } then r$ is a Desarguesian projective plane isomorphic to P G ( 2 , q) and is a subplane of the structure r+.
+
+
+
Proof Note that (1) and (4) are fairly immediate and are left to the reader to verify but note that the number of points is (q6  1)/(q2 1). If we consider the linear transformation group acting on F3 and assume ‘that F commutes with K, then define (x,y, z)ag = (x,y,z)ga for all a E K and for all g E G. This proves (2). For ( 3 ) , (x,y,z) satisfies x y t z = 0 if and only if (z,y,z) = (x, ( z x)tl, z ) . Hence, there are exactly (a4  1 ) / ( q 2  1 ) points per 1ine.O
+
+ +
Lemma 6.3 L ( t ) g k n L(s) is either a unique point or acting on the points.
S
= t and gk = 1
Proof: Let {(%,V,% ) U } E L(s) so that x + ys +,z = 0 and ( ~ , y , z ) g  ~E L(t) so there exist elements of F ,aij for i,j = 1 , 2 , 3 so that
+
(allx a129
+ a132) + (a213 + a229 + a23z)t + (a31x + a329 + a332) = 0.
Note that g is in GL(3,q) acting on P G ( 2 , q) extended to thesituation under consideration.
CHAPTER 6. THE HUGHES PLANES
59
Now solve the two equations substituting x = "z  sy and use the right nearfield properties together with the fact that F commutes with elements of K to obtain the following equations: Let
to obtain: yu
+ za + (yv + zb)t = 0.
Case I: b # 0. Then rewrite the above equation as
( p +2b)bb
+ y(u  vbla) + (yv + zb)t = 0.
Recalling that we have a 'right' nearfield, we obtain y(u  vb"a)
+ (yv + zb)(b'a + t ) = 0.
First assume that t = 1 to obtain that y(u + v) = z(a
+ b),
What this means is that we have y(a!
+ PS) = z6 for some a,P,6 E F.
If 6 # 0 then it follows that IC
Since X
= y(a*+ P * s ) for some a!*,@* E
F.
+ vs + z = 0, then we have an equation of the following form: $/(CY*
+ P * s ) + ys + y(a! + Ps)Sl
=0
[email protected]+
PIS).
SO, is nonzero and we have a unique solution of the form (E+us,l,p+ys)
for e,u,p,y E F.
CHAPTER 6. THE HUGHES PLANES
60 If t is in K
 F, let W = t + bla (yu
to obtain:
+ zb)w = y(u
 vbla)
which is equivalent to
y(v
+
(U
 wb'a)wl)
+ zb = 0.
By an argument similar to the above, it follows that there is a unique solution. Case 11: b = 0. Hence, we obtain
y(a +ut)
+ za = 0
and if a # 0, the previous argument applies. Hence, a = b = 0 so that
a13
+ a33 = all + agl and a23 = a2l.
Now consider (1,0, 1) and note that (1,0, l)gk = (alla13)(l, 0, 1). Thus, by regularity, gk = 1 which implies that L(t) and L ( s ) have exactly the point ( l , O , 1) in common when t # S. This proves the 1emma.O Hence, we have:
Theorem 6.4 Thepointlinegeometry 7r+ definedabove is aprojective plane of order q2 and is called the 'Hughes plane over the nearfield K' when K is not a field.
Remark 6.5 Note that it follows easily that, in fact, GL(3,g) acts on n+.
6.1 The AffineHughes Planes.
[i ! ]
[i ! ]}
= {(qv, x  y>a = {(S,y,O)a). 0 0 1 That is, the equation z = 0 represents a line which we designate . l , of n+. Form the affine plane 7r by deletion of .lm, Represent points of 7r by the sets of elements (z,y, l ) a for which we shall use the notation (qy)..
Let L(1)
0 0 1
6.1. THE AFFINEHUGHES PLANES.
61
Hence, lines of 7r have the general form:
alz + P l y
+ 61 + (a2x + p2y + 62)m = 0
for ai,&,& E F f o r i = 1 , 2 and m E K  F . Note that the lines L(1)g for g E G have the following form:
x = a, y = zs+p for all a,p, S E F. Then the lines of 7r have the following form: y = (")
[email protected], y = xS+b,
x =
C
fora,p,SEF,b,cEKandmEKF. Hence, we have the following:
Theorem 6.6 Let K be a right nearfield of order q2 which contains a field F of order q which is in the center of K . Then 'points' as elements of K x K and 'lines' as given by equations: y = (")
[email protected], y = x6 b,.
+
x = c for a,p,S E F , b,c E K and m E K q2
 F, forms an afine plane T
of order
Remark 6.7 ( l ) When K is a field, 7r is Desarguesian. When K is a nonfield nearfield, T is called the 'afine Hughes plane over K I, (2) Note that we have a coordinate system (K,+, ,T)with ternary function T defined as follows: Let { 1,S } be a basis for K over F and define T ( z ,m, b) = T ( z ,~ m +l m2,tbl + b2) = for m1 # 0, (x blm,') (sml+ m2) (b2  blm,'mz) and for m1 = 0, x. m2 (sbl + b2) for m;,bi E F , i = 1,2.
+
+
+
CHAPTER 6. THE HUGHES PLANES
62
Then the [points’ are the elements of K
X
K and the lines are given’ by
y = T(x, m, b), x = c f o r all m, b, c E K .
Remark 6.8 Thefinite nearfieldplanesarecompletelydeterminedby Zassenhaua [78]. W e areinterested in the finite right nearfields of order q2 whichcontainsubfields of order q in theircenters.Forthe socalled ‘regular nearfields’, these m a y be constructed as in the following lemma. Lemma 6.9 Let ( L ,+, be a field isomorphic to GF(q2)and F a subfield of L isomorphic to GF(q) and assume that q is odd. Define a multiplication * as follows: S)
x * m = xqm if m is a nonsquare in L and = xm if m is a square in L. T h e n (L,+, *) i s a left nearfield containing center.
(F,+, *) = (F,+,
e)
in its
Proof: It suffices to show that ( L  {0}, *) is a group and this is implied once associativity is proved. Let U,,,, = 1 or q if and only if m is square or nonsquare. Since x * (m * n) = ( x * m) * n if and only if x‘mbnmunn = xamunmunn, it is required to show that x‘mbn = x‘man for all x. This may be proved easily by taking the various cases of m and n square or nonsquare. These cases are left to the reader for verification. Also, note that when m is in F then m4 = m so that x * m = x m = m x = muxx= m * x so that F is in the center of the nearfield.0 Finally, we note:
Theorem 6.10 TheafineHughesplanes planes.
arederivablesemitranslation
Proof Note that the coordinate system K contains F and K is a right 2dimensional vector space over F . To see that the affine planes axe semitranslation planes, note that the following is more generally true of the projective plane n+: For every pointline pair (P,!) of the subplane n,f, there is a (P, @collineation group which is transitive on the nonfixed points of n$ that lie on lines incident with P. We call such a group a ‘(P,l,n$)transitivity’(0r ‘relative pointline transitivity’) .O
6.1. THE AFFINE HUGHES PLANES.
63
Remark 6.11 Obviously,muchmorecan be saidhereabout thederived planes from the afine Hughes planes. It turns out that although the Hughes planes admit the (P,!, .lr$)tmnsitivities mentioned previously, they admit no (P,!)transitivities. firthermore, the derived planes, called the 'OstromRosati planes' were the first examples of finite planes whose projective extensions admit exactly one incident pointline transitivity. The reader might like to read further about these planes in Hughes [,W], Ostrom [63] and Rosati P91.
Chapter 7
DESARGUESIAN PLANES In this chapter, we provide some basiccharacterizations of the Desarguesian planes. We have defined a Desarguesian projective plane as a P G ( 2 , K) where K is a skewfield.Similarly, a Desarguesianaffine plane is an A G ( 2 , K ) . There is a 'configuration' which occurs in a Desarguesian projective plane which provides its name and which, furthermore, characterizes Desarguesian planes. We present this here as it shall be required in a later chapter as a tool in the characterization of projective geometries. For this chapter, we shall allow the term 'Desarguesian' to refer to an arbitrary projective plane satisfying a certain configuration theorem.
Definition 7.1 Let C be a n arbitrary projective plane. Let ABC and A'B'C' be triangles defined by points A, B , C,A', B', C' with the property that the lines AA', BB', CC' are concurrent at a point 0. In thiscase,wesaythatthetrianglesare'perspective from a point'. Similarly, if the intersections of the corresponding sides, AC n A'C', AB n A'B' and B C n B'C' are collinear, we say that the triangles are 'perspective from a line'. A projective plane C is said to be 'Desarguesian' if and only if any pair of triangles which are perspective from a point are perspective from a line. 65
CHAPTER 7. DESARGUESIAN PLANES
66
AB n A'B'
B' We first show that any PG(2,K ) , for K any skewfield, has this configuration property.
Theorem 7.2 Any PG(2,K ) satisfies the Desarguesian configuration property. Proof: (Sketch) For every point line pair ( O , t ) , it is straightforward to show, and we take it as known, that every PG(2,K) admits a (0,t)central collineation group that acts transitively on the nonfixed points on lines incident with 0; the plane admits all possible pointline transitivities. Take a pair of triangles ABC, A'B'C' perspective from 0. Form the line (AB n A'B')(AC n A'C') = l. There is a central collineation g with center 0 and axis t which maps A to A'. It is not hard to show that C is mapped to C'. Moreover, since g fixes tpointwise, it also follows that B is mapped onto B' and '(BCnB'C') is fixed by g so must lie on t. Hence, the triangles are perspective from a line l.0
Theorem 7.3 (1) A projective plane C satisfies the Desarguesian configurationproperty if and only if for each pointlinepair ( O , t ) , there is a collineation group whichfixes each line incident with 0, fixes each point o f t and acts transitively on the remaining points on each line incident with 0. (2) A projective plane C satisfies the Desarguesian configuration property universally if and only if C is isomorphic t o PG(2,K ) for some skewfield K.
7.1. MOUFANG PLANES.
67
Proof (Sketch) First assume that C is a projective plane which admits a collineation group as in the statement. Then, select an affine restriction C A by the deletion of a line C,. This means that, with 0 varying over the line C,, and C = Cw, that the plane is a translation plane and hence, there is as associated spread of a vector space V = W @ W over the ‘kernel’ skewfield K . Now let 0 be any point of the translation plane which is not incident with C = Coo. Hence, it follows that the ‘kernel’ group is transitive on the affine points not equal to 0 on lines through 0. Hence, if the kernel is K then the translation plane is 2dimensional over K and the plane is an AG(2, K ) . From the previous theorem, we know that the extended plane PG(2,K) satisfies the Desarguesian configuration property. It remains to show that if a plane satisfies the Desarguesian configuration property then the plane satisfies the central collineation group property universally. Note that the proof uses almost exactly a reversal of the ideas of the proof that any PG(2, K) has the configuration property. To define a central collineation with center 0 and axis l , choose any two points A and A’ which are not 0 and not on l and AA’ is a line incident with 0 and define a mapping g which fixes each point on C, fixes 0 and every line incident with 0 and maps A to A‘. We need to extend the definition of g to include all points of C and to show that this mapping becomes a central collineation. Choose any remaining point B and form AB n C. Define g ( B ) as ( ( A B n l)A’)n BO. This mapping becomes a collineation provided the Desarguesian configuration holds and it is left to the reader to verify that this, in fact, is the case.0
7.1 Moufang Planes. In the previous section on Desarguesian projective planes, it was realized that such planes admit(0,C)transitivities for each pointline pair. If we restrict the class of transitivities to all ‘incident’ pointline pairs, the plane is called a ‘Moufang plane’. Note that this implies that anequivalent definition is that a Moufang plane is a projective plane such that any affine restriction is a translation plane. While every coordinate left quasifield of a Desarguesian plane then be
68
CHAPTER 7. DESARGUESIAN PLANES
comes a skewfield, the coordinate structures of Moufang planes are ‘alternative division rings’.
Proposition 7.4 Let 11 be a projectiveplaneandlet & be any line. Then II is a translation plane with respect to & (the associated a.#ine restriction is a translation plane) if and only if for any two points (m) and (0) on & the plane is (P,&)transitive for P E {(m),(0)). Proof: By the fundamental theorem on translation planes, it suffices to show that the translation group T generated by the two transitivities is transitive on the affine points (points of II e). However, we have seen that is validin the section on coordinates in translation planes when we considerd T(0,b) and T(a,O).U Proposition 7.5 Let II be a projective plane and (m) a point of II. Then II is a translation plane with respect to every line incident with (m) ifand only if there are two distinct lines &land e2 incident with (m) such that II is ((m),&i)transitive for i = 1,2 and also for points ( 0 ) E &l and P E l2 the plane is also ( ( 0 ) &l)transitive , and (P,&2)transitive. Proof By the previous proposition, it follows that II is a translation plane with respect to both lines &l and &. Since the transitivity (P,&) must move &l and fix (m) and act transitively on the lines not equal to &2 incident with (m), it follows that the plane is a translation plane with respect to each line incident with (m).O
Proposition 7.6 A projective plane is Moufang if and only if there is a triangZe ABC such that the plane is (A, AC)and ( A ,AB)transitive, ( B , A B ) and ( B ,BC)transitive and (C,BC) and (C,AC)transitive. Proof By the previous propositions, the plane is a translationplane with respect to every line incident with either A , B ,or C. It remains to show that the group generated by the transitivities actstransitively on the points of the plane. However, the translation group with axis AB is transitive on the points not incident with AB, the translation group with axis AC is transitive on the points not incident with AC and the translation group with axis BC is transitive on the points not incident with BC. It follows that the group acts transitively on the points so that, for every point D, the plane is a translation plane with respect to every line incident with this given point D.0
7.1. MOUFANG PLANES.
69
7.1.1 Coordinate Structures forMoufang Planes, Since a Moufang plane is a translation plane for any line, we may choose any line to designate as the line at infinity and obtain a left quasifield as a coordinate structure. The arguments for part of this section are adapted from Hall [26]. By the previous proposition, to accurately model a Moufang plane by a left quasifield, it suffices to assume, in addition, that the plane is ((m),x = 0), ((O,O),x= 0), ((O),y = 0) and ((O,O),y = 0)transitive.
Lemma 7.7 In a left quasifield (C,+,*),the existence of a ( ( m ) , x = 0)transitivity is equivalent to having the left distributive law: If a , b , c ~ C t h e n c * ( a + b ) = c * a + c * b . Proof: The indicated transitivity is transitive on the points of t,{(m)} so for each ( m )and m E C, there is a unique collineation fixing all points of x = 0, and alllines x = a and which maps (0) onto ( m ) . Assuming this is so, we then have ( 0 , b ) H (0,b) so y = b H y = x * m + b and Z=UHZ=U. Hence,(a,b)~(a,a*m+b). Then(l,k)~(l,m+lc) sothaty=z*k++y=x*(m+k). Since(a,a*k)~(a,a*m+a*k), we have (a,a * m a * k) E ( y = x * ( m IC)). Thus, we have
+
+
a*(m+k)=a*m+a*kforalla,m,kEC. Conversely, the proofalsoshows there is a left distributive law.0
that the collineations exist provided
Definition 7.8 A ‘distributive’leftquasifield is calleda ‘semifield’or a ‘nonassociativedivision ring’. Hence, a semifieldplaneis a translation plane that admits a group of affine elations with affine axis, say x = 0, and infinite center (m) which acts transitive on the remaining infinite points. We note that, in any left quasifield, there are ‘right multiplicative inverses’ and ‘left multiplicative inverses’ for a given a # 0 (that is, a * x = 1 = y * a for unique nonzero I and y ) but we don’t always know these two elements are identical. However,
CHAPTER 7. DESARGUESIAN PLANES
70
Lemma 7.9 A semifield plane (coordinatized b y a semifield) has identical right and left multiplicative inverses and the rule: a'
* ( a * b) = b for all a # 0 ,b E C
if and only if the plane is also ((0, 0), x = 0)transitive. Proof: The proof proceeds in a manner similar to the previous lemma. We assume that thetransitivity exists and find the multiplicative properties. Conversely, the multiplicative properties suffice to create the collineation. The lines incident with (0,O) are fixed so, for each a E C, there exists a unique collineation in the ((0, 0),x = 0)transitivity such that (0) is mapped onto (1  a, 0). Since (0,1+ a) is fixed, it follows that y=l+awy=x*m+b
+
which contains both ( 0 , l a) and (1  a, 0) which, in turn, implies that m=landb=l+aas(la)*m+(l+a)=O. Hence,
Similarly, since we have both distributive laws, it follows that g=b+a*bwy=x*m+b+a*b which contains both (0, b+a*b) and (1a,O) so that m = bas (l+a)*m = b+a*b=(l+a)*bandl+a#O. Since y = x * (1f a ) is fixed by the transitivity, then
if a # 0 (note if a = 0 then the image is the infinite point (1))such that Ic*(lfa)=IC+l+a. Note that IC = 0 if and only if a = 1.
7.1. MO UFANG PLANES.
71
In particular, since (00) is fixed, we have (z = 1) = (co)(l, 1+ a )
(0O)(IC,IC
+ 1+ a) = (z = IC).
+
Similarly, we must have: (1,b a * b) which is the intersection of three lines
(z=l)n(2,=b+a*b)n(y=z*(b+~*b))~ the intersection of the corresponding three lines
(z=k)n(y=z*b+b+a*b)n(y=a:*(b+a*b))= (k,k*b+b+U*b) (provided a is not 1) such that IC*(b+a*b)=IC*b+b+a*b.
+
Hence, we have IC * (a * b) = b a * b valid for any b and IC = 0 if and only if a = 1. Now let W 1 = IC to obtain:
+
w*(a*b)=b for all b. Now repeat the process for W so that there exists a t such that
t*(w*c)=c for all c. Letting c = a and using the fact that W
* a = 1, it follows that t * 1 =
t=c=a. Hence, we have w*a=a*w=lfora#O. This shows that left and right inverses are equal. We call W = a1 to obtain a'
* (a * b) = b for all a # 0 and b E C.
CHAPTER 7. DESARGUESIAN PLANES
72
Hence, we have shown that the transitivities assumed imply the multiplicative properties. Conversely, if we have a semifield with the multiplicative properties given then we need to show that there exists a ((O,O), x = 0)transitivity. Note that since we have a translation plane, the existence of such a transitivity will follow provided we have a nontrivial elation of this type since such a mapping will move the line at infinity. We define a mapping whichfixes x = 0 pointwise and fixes all lines incident with (0,O) with the f o l l o ~ n gadditional defining conditions: On points:
(4

( 1 9 4 , (1, m)

(4,
and (a,b) I$ ((a'
+ l)"',
+ 1)l * b)
(a
for a # 0 or 1. On lines:
L,I$x=l,x=lI$L,,x=k~z=(l+k
1 ) 1
for k # 0 or 1 and
In order to verify that this mapping is a collineation, it suffices to show that incidence is preserved. Hence, we need to show that (e,e
* m + b) H ((e' + I)',
This is valid if
+ I)'
+ I)' * ((e * m) + b ) )
+ b which is true if and only if * ( m b) + b = (e + I)' * ((e * m ) + b).
is incident with y = x * ( m  b) (e'
(e
7.1. MO UFANG PLANES.
73
and
(e'
+ I)' * (b) + b = (e +
* (b)
as we have both distributive laws. To verify the first equation, let e + 1 = a to obtain:
(e
+
(e * m) =
a'*(a*mm)
((U
 1) * m >=
= (a'r(a*m)al*m)

mal*m.
using the distributive laws and the multiplicative properties assumed. Thus, it remains to show that
mal*rn
= (la')*m=(l(e+l)l)*m = (e' + *m
which is valid if and only if
(E): m = ( ( e+ I)'
+ (e' + I)') * m.
We assert that (e'
+ l)' = e * (e +
To see this, note that (e'
+ 1) * (e * (e + I>'>
+
+
= ( e * (e e * (e = (e + + e * (e = ( e 1) * ( e 11l = 1.
+
+
+
+ I)'
Then (e
+
+ (e' + I)'
= (e
+ 11l
+ e * (e + 11l = (e
+ 1) * ( e + I)' = I .
The second equation can be easily transformed into E so the result is proved.0
CHAPTER 7. DESARGUESIAN PLANES
74
Lemma 7.10 A semifield plane (coordinatized by a semifield) has identical right and left multiplicative inverses and the rules:
* (U * b) = b and ( b * a) * al = b for all a # 0 , b E C
if and only if the plane is ((O,O), z = 0) and ((O,O),y = 0)transitive. Proof Consider the ((O,O),y = 0)elation which maps point (0, 1). Thus,
(W)
ontothe
(z = 1) = (oo)(l,O) I”) (0, l)(l,O)= (y = x  1).
Similarly,
(x = a) = ( W ) ( U , O )
I ” )
(y = 2 *.l
 1).
Also, (z=l)n(y=a:*((1~*b))=(1,1~*b)~ (y=zl)n(y=z*(la*b)=(e,f) such that el=f=e*(la*b) which implies that e*(a*b)=lande=(a*b)l, Thus,
(1,l a * b) c”t ( ( a* b y , ( a * b)1
 1).
so,
(0)(1,1 a * b) = (y = 1 a * b) W (O)((a* b y , (a * h)1
 1)
7.1, MOUFANG PLANES.
which is y = (a * b)l Now
75
 1.
* a'  1= h = g * (a'  b) which implies that g = bl and h = bl * a1  1. g
Thus, it follows that y=la*bHy=b'*all which is then equal to y = (a * b)l  1 and we have b  l * am1 = (a * b)l for all a, b nonzero in C.
To verify that ( b * a) * a1 = b notice that b = (b')l
= (a * (a'
b'))'
= (U'
* bl)l *
= ( b * a) * al.0
Definition 7.11 A semifield which has identical right and left multiplicative inverses and the rules: a1
* (a * b) = b
and
is said to be an 'alternative division ring'. Hence, we have proved the following theorem.
Theorem 7.12 A projective plane is a Moufang plane if and only if it may be coordinatized b y a n alternative division ring.
76
C H A P T E R 7. DESARGUESIAN PLANES
Remark 7.13 A finite Moufang plane is Desarguesian. Proof: Any Desarguesian plane is clearly Moufang. Now use the theorem of ArtinZorn mentioned in the prerequisites that any finite alternative division ring is a field.
Remark 7.14 If IT is a projective plane which is a translation plane with respect to a line tmthen either IT is Moufang or the line.lm is left invariant under the full collineation group of the plane. Proof: If this line is moved by a collineation of the plane then the plane is a translation plane with respect to all lines through a point (W)on .lm. If (W)can be moved then the plane is a Moufang plane by the preceding discussion. But, in any case, using the BruckKleinfeld/Skornakov/SanSoucie theorem (BK/S/S), the plane is Moufang if the plane is a translation plane with respect to all lines through a point.0 We now provide a characterization of Moufang planes.
Theorem 7.15 Let 7r be an afine translation plane. Then 7r is Moufang if and only if any leftquasifield coordinatizing 7r is a semifield. Proof The assumption is equivalent to saying that for any infinite point P there is (P,OP)transitivity in the plane. Hence, we have that the plane is a dual translation plane with respect to any infinite point. Thus, by the dual of the (BK/S/S) theorem, we have that the plane is a dual Moufang plane which is also then a Moufang plane due to the pointline transitivity properties.0
Definition 7.16 Let 7r be a translation plane and let S be a spread for 7r over the skewfield K . Hence, we may represent the components of 7r in the form x = 0,y = 0, v = xM where M is in GL(V,K ) and V represents the vector space over which 7r is defined. W e shall say that S is ‘additive’ if and only if {N ; y = XN E S } is an additive group. Theorem 7.17 A translation plane is a semifield plane if and only if the plane admits an additive spread.
7.1. MOUFANG PLANES.
77
Proof Assume first that the plane is a semifield plane. Hence, there is an elation group with finite axis, say z = 0, and infinite center, (CO) such that the group acts transitively on the remaining infinite points. Since the plane is also a dual translation plane, the elation group is Abelian by the chapter on translation planes (see Theorem 4.5). It is an easy matter to note that the elements of the elation group E may be written over G L ( V , K ) in the following form:
+
QN : (z, y) W (z, zN y) where if V = W @ W and W a Kspace then z,y E W and N E GL(W,K ) U (0).
Then, since E is Abelian, { N ; ITN E E } is also an Abelian subgroup. Nowwe assume within a spread set that we have the components x = 0,y = 0 and note that the elation group fixes z = 0 pointwise and maps y = 0 onto y = zN. Since E is transitive on the remaining components (not equal to z = 0), it follows that the spread is additive. Conversely, if there is an additive spread, it is a simple matter to construct an elation group using the spread set linear transformations exactly as above.0
Corollary 7.18 A translation plane isa Moufang plane if and only if every spread set is additive.
Chapter 8
PAPPIAN PLANES We shall have occasion to consider Desarguesian planes with spreads in PG(3,K) where K is an infinite field. That is, K is a skewfield with commutative multiplication. In this instance, such planes are said to be ‘ P a p pian’ for historical reasons. There is a configuration theorem called the configuration of Pappus from which the planes derive their designation. Let k‘ and k‘’ be distinct lines of a projective plane containing the points A, B , C and A’, B‘, C‘ respectively where A, B , C are mutually distinct and A’, B’, C‘ are mutually distinct. Form the points AB’n A‘B = N , AC‘n A‘C = M and BC’n B‘C = P. The configuration is said to be ‘Pappus’ if and only if N , M , and P are collinear. The projective plane is said to ‘Pappian’ if and only if for all such configurations of collinear points on lines, the points formed as above are collinear.
Remark 8.1 It is fairly direct that any projective plane satisfying universally the Pappus configuration theorem also universally,satisfies the Desarguesianconfigurationtheorem.However, acoordinateapproachalsowould showthatthe associatedskewfieldcoordinatizing system has commutative multiplication.Hence, a Desarguesianplane PG(2,K) is Pappian if and only if K is a field. 79
CHAPTER 8. PAPPIAN PLANES
80
A’
8.1
ReguliinPappianSpreads.
In this section, we consider the nature of Pappian planes with spreads in PG(3,K) for K a field and concentrate on the reguli within the spread. Later, we consider ‘parallelisms’ of PG(3,K) which are coverings of the lines by a set of mutually disjoint spreads (disjoint on lines) and utilize the analysis of Pappian spreads containing a given regulus for some constructions. We note that a ‘regulus’ here is a set of lines R completely covered by another set of transversal lines ROPP. A more general definition is taken up in chapter 14 (see definition 14.15) where a vector space form is also determined in Theorem 14.16.
8.1. REGULI INSPREADS. PAPPIAN
81
We begin with a fundamental under central collineations.
lemma on the invariance of subplanes
7 r : IT:
Theorem 8.2 Let 7rE denote a projective plane and an arbitrary projective subplane. Let a be a central collineation of 7rE. T h e n a leaves invariant if and only if the center and axis of a lie in and there exists some nonfixed point P of such that Pa i s also a point of
7r:
7r:
7r:.
Proof First assume that a leaves T: invariant. Let the center of a be denoted by C. If C is not a point of T: then assume there is a triangle of distinct points Pi for i = 1 , 2 , 3 of no two on a line incident with C.At least two of the two distinct lines Pic for i = 1 , 2 , 3 are not lines of say PIC and P2C. If P&' is not a line of 7rf then Pi is the unique point of incident with P&'. But, a fixes Pic and fixes T: so must fix Pi. Since C, PI,Pz are not collinear, the axis of a is P1P2.The argument shows that for any point Q of not incident with P1P2 then QC is a line of as a fixes Q otherwise. That is, all points of 7rf are on either PIP2 or P3C which is a contradiction. Since there are at least four points forming a quadrangle in T:, the choice of such a triangle is possible. Hence, the center and dually invariant. the axis of a must belong to 7rf if a is to leave Now assume that the center and axis of a are in and there exists a nonfixed point P of such that Pa is also a point of T:, Let Q be a which is not incident with PPa and not on the axis. Note that point of the center C of a lies on PPa. Form PQ and intersect this line with the axis !. of a in the point 2. Since a fixes 2,it follows that the line ZP is mapped by a to the line ZPa. Then, since QC is fixed by a,it follows that Q a = ZPa f l QC. But, P,Pa, Q,C are points of and t is a line of Hence, 2 is a point of and thus the intersection point Q a is also in as is a projective subplane. Hence, for any nonfixed point Q of which is not on PPa then Q a is in and, by reversing the roles of P and Q,it follows that is invariant under a.0 We shall also require several results regarding collineation groups of P a p pian spreads containing a given regulus.
7r:,
7r:, 7r:
7r:
7 r :
7 r :
7 r :
7r:
7 r :
7r:
7r:
7r:
7r:
7r:
7r:
7r:. 7r:
Theorem 8.3 Let V be a vector space and let C and IT denote two Pappian afine planes defined on the same points as V and coordinatized by fields L and F respectively.
CHAPTER PLANES 8. PAPPIAN
a2
( l ) Then, if the two planes share at least three components, there is a net defined by common Components which may be coordinatized by a subfield of L n F . (2) Consider a regulus net R whosepartialspread i s a regulus in PG(3,K ) forsomefield K . Thenany2dimensionalvector space T disjoint from the spread of R of the $dimensional vector space corresponding to the projective space may be embedded as a component into a unique Pappian plane containing R. (3) Assume that the two Pappian spreads may be embedded into the same projective space PG(3,K) where K is a field. If the two spreads contain a regulus and a component external to the regulus then they are, equal.
Proof Assume the conditions of (1).We choose coordinates so that the three common components are represented using equations z = 0, y = 0 , y = z and points are represented by (z,y) where 2,y are in a common reference subspace W such that W 8 W = V. Hence, we may arrange it so that the fields L and F share both identity elements 0 and 1. Choose coordinates in a common net, we obtain say a component g = x * m in one field which is equal to y = z o n in the other field. However, since the two components are equal and the fields share 1, it follows that m = n. From here, it is fairly direct that a common net must be coordinatizable by a common subfield of F and L. Now assume the conditions of (2). We have seen previously that a regulus in PG(3,K),for K a finite field, may be represented within the asso
{
ciated vector space in the standard form z = 0,y = z
[
U
0
]
; U E K}.
However, the same form is valid for infinite fields by Theorem 14.16. Now consider any 2dimensional subspace which is disjoint from the regulus net (standard net).Clearly, we may represent this subspace in the basic form y = z M where M is a nonsingular 2 x 2 matrix with elements in K. We assert that { M c Y I ~PI2 ; CY,P E K} is a quadratic extension field of K. Certainly, the set of matrices is additive as the difference of M and
+
6Iz is nonsingular by assumption. Let M = Ma12
+ PIz = [ "","
[
ba +P
]
and form
] C Y , P EK.
8.1. REGULI IN PAPPIAN SPRBADS.
83
Without loss of generality, we take d = 0. Suppose either c or b is equal to 0. Then, it is possible to obtain anelement of determinant zero contradicting the fact that the 2dimensional space is disjoint from the regulus net. Hence, let CCY= t denote an arbitrary element of K . Now letting P = U , we obtain the form for the set as:
{[
‘U lt
]
3
u,t E K } .
Clearly, this is a field of matrices which is a quadratic field extension of K . Moreover, it is the unique such field containing T and the regulus. Part (3) now follows immediately from part (2).0 We now turn to the consideration of common collineation groups.
Theorem 8.4 Let be a n affinePappianplanewhosespreadcontains a regulus R in PG(3, K ) where K is a field. Let G denote a collineation group W n g the zero vector of the regulw net corresponding to the regulus which is generated b y affine central collineations leaving R invariant. Then, G acts as a collineation group of T . Proof First of all, it is not completely clear what one might call a ‘central collineation group’ of a net. We simply mean by this a collineation of the net which fixes a line of the net pointwise and an infinite point of the net linewise. We may choose the socalled standard form for the regulus net corresponding to the regulus. If U is an affine central collineation which leaves the regulus net invariant and fixes an affine point then by Theorem 8.2, U leaves invariant each Baer subplane of the net incident with the zero vector since the infinite points are not all fixed but correspond to infinite points of the Baer subplanes. Choose the axis to be represented by z = 0. It is easily seen that we may always accomplish this without changing the basic structure of the regulus net. It follows easily that the group U may be represented as an element of the following group:
([
..
.
] ;o#O,PEK).
0 I aI PI
.
. . . .
.
,
.
.
I
84
CHAPTER PLANES 8. PAPPIAN
Now any Pappianplane that contains the regulus net represented in the standard manner has spread set in PG(3,K) of the form { M a PI ; Q, P E K} for some nonsingular 2 x 2 Kmatrix M . It follows immediately that any such group element c must leave the Pappian spread invariant and hence act as a collineation group on the plane. Hence, the group generatedby central collineations leaving the regulus net invariant must induce a collineation group on any Pappian plane containing the regulus.0
+
Chapter 9
CHARACTERIZATIONS OF GEOMETRIES In this chapter, we shall present the promised proof of the fundamental theorem. We also require recognition theorems for affine and projective spaces and these will be presented in the chapter.
9.1
The Fundamental Theorem of Projective Geometry.
Theorem 9.1 (The Fundamental Theorem) Let PG(V,K ) and PG(V', K') be an9 two projective spaces where V and V' are vector spaces each of dimension at least 3 over skewfields K and K' respectively. (1) If CT is an isomorphism of PG(V,K) onto PG(V', K') then K and K' are isomorphic and CT is induced from a semilinear mapping of V onto V'. (2) I n particular,thevector spaces havethesamedimension(Hamel dimension) and the skewfields are isomorphic. (3) The collineation group of PG(V,K ) is PrL(V,K ) . Proof We present the proof as a series of lemmas. 85
CHAPTER 9. CHARACTERTZATIONS OF GEOMETRIES
86
Lemma 9.2 Let {ea ; i = 1,2,3} be any set of linearly independent vectors of V. Let qei} = (ea ; i = 1,2,3). Then ( l )PG(V{ei},K). is a Desarguesian projective subplane of PG(V,’ ,K‘) so it is isomorphic to PG(2, F)for some skewfield F . (2) F is isomorphic to K’. Proof The first part of the lemma follows from Theorem 7.3. Since PI’L(V, K’) is transitive on Desarguesian subplanes, it follows that F must be isomorphic to K’.O
Lemma 9.8 K is isomorphic to K’ and Q restricted to PG(V{,,}, K ) is a semilinear mapping of V,,i>onto = V‘ei}. Hence, Y e i > and are isomorphic vector spaces.
ye{}
Proof Let {ei ; i = 1,2,3} be a basis ofV{ei} and let g be any isomorphism of PG(?,,), K ) onto PG(?,,,, K)a. Let V(ei} denote the subspace ofV’ giving rise to the projective subspace. Let g(ea) = fi and let g(e1 e2 es) = f4, Now considered projectively (ea) for i = 1,2,3 and (el e2 e3) are four points no three of which are collinear so that the points (fa) for i = 1,2,3,4 also have this property. It follows easily that {fa ; i = 1,2,3} is a basis for V[ei}. Then f4 = alf1 a& a3 f 3 for ai E K‘. If ai = 0 for i = 1,2 or 3 say a1 = 0 then {f4, f 2 , f3} is not a basis forViei}which clearly is a contradiction. Choose a basis {aifa = f! ; i = 1,2,3} so that f4 = f; fi +f$. Hence, there is an element h of GL(4, K’) such that gh maps ea to and maps el e2 e3 onto f; f;T f;. Write the points homogeneously as ( q y , z ) with respect to the basis {ea ; i = 1,2,3} and (X’, z’) with respect to the basis {ft ; i = 1,2,3}. Now let the 2dimensional vector subspace spanned by ( e l ,e2) be denoted by &, and consider the associated Desarguesian &ne plane A written in the form ( q v ) where x,y E K with lines in the form z = c, y = xm + b for c, m, b E K . Similarly, we let the 2dimensional vector subspace spanned by (f:, fi) be denoted by ! T , and A’ is the associated Desarguesian affine plane written in the form (X’, where x’,y‘ E K‘. We let context dictate which particular binary operation is actually used in the relevant skewfield.
+ + + +
+
+
+
+ +
+ +
fz
FUNDAMENTAL THEOREM. 9.1. THE
87
We note that now the isomorphism g acts as an isomorphism of A onto A' and we may assume that g maps x = 0, y = 0 , y = x onto x' = 0') y' = 0', y' = x' respectively and also maps the points (0,O) and (1,l)onto the points (O',O') and (l',1') respectively where 0' and 1' are the additive and multiplicative identities for K'. Hence we have that g : (x,y) H (x",y") where x = 0, x = 1, y = 0 and y = 1 are mapped into x' = 0', 2' = l', y' = 0' and y' = 1' respectively so that l" = 1' and 0" = 0'. Furthermore, (1,b) W (l',b") and (0,O) and (1)are mapped onto(0', 0') and (1') respectively so that y = x + b H y' = x' b". Hence, (c,c b) (c", (c b)") is incident with y' = x' b" so that we have
+
+
+
(C
+
+ b)" = c" + b".
Similarly, (1,m) H (l',m'") so that y = x m I+ y' = dma. Hence, it follows from (c, cm)H (c", (cm)")that
(cm)" = cum". In other words, in this context, g induces an isomorphism from K onto K' and therefore induces a semilinear mapping on threedimensional subspaces.0 We now may assume, without loss of generality, that V and V' are vector spaces over the same skewfield K .
Lemma 9.4 The Proof of the Fhdamental Theorem. Proof Wenow consider an arbitrary vector space V of dimension at least 3 over K. We note that the above lemmas state the following: K and K' are identified and if g is any isomorphism of PG(V,K ) onto PG(V', K) and C is any projective subplane of PG(V,K ) then glC is semilinear with respect to the 3dimensional vector subspace W of V corresponding to C. We wish to prove that given any two vectors v and U of V then
where CT is an automorphism of K (identifying K and K' under an isomorphism).
88
W
C H A P T E R 9. CHARACTERIZATIONS OF GEOMETRIES
Assuming that v and U are linearly independent, choose a third vector such that the set {v,U, W } is linearly independent. Let W = (v,u,w). By the previous lemmas, we must have g(Lyv) = d W g ( v ) and g(v
+
U)
+
= g(v) g(u)
where ow is an automorphism of K which is possibly dependent upon the three dimensional vector space W . However, by varying the vectors U , W , we see that the automorphism is independent of the subplane containing v. But, clearly then the automorphism is also not dependent upon the vector v. So, g is a semilinear mapping. It then follows that the vector spaces V over K and V‘ over K (or K’) are isomorphic and hence have the same dimension.0
9.2
Projective andAffineGeometries.
In this section, we give characterizations of both projective and affine geometries in terms of pointline properties. The proofs given are adaptations of proofs of similar results given in Cameron [18]. First, we need a few definitions.
Definition 9.5 Let ( P ,L,Z)= 7r be a triple of sets of ‘points’ P , ‘lines’ C and ‘incidence’ Z. Assume that there are at least two points incident with every line and that two distinct points are incident with a unique line. A ‘subspace’ (of points) is a subset P* of P such that if A, B are distinct points of P* then the points lying on the line AB also are points of P*. Hence,we may make P* andthecorrespondinglines defined b y pairs of points of P* into a system of points and lines. Assume further that there exists a set Aff of subspaces of 7~ which are, in themselves, afine planes and such that any three mutually distinct noncollinear points are contained in a unique afine plane of Aff. Let el and e2 be lines possibly equal. W e shall say that l!. is ‘parallel’ to &, written el 11 e2, if and only if they are parallel lines in some afSine plane of Aff. In the following,we shall be assuming that the ‘parallelism’ givenabove is a n equivalence relation. In this case, the equivalence classes of lines partition
9.2. PROJECTIVE AND AFFINE GEOMETRIES.
89
the points so there is a unique line of each class through each point. When a, is a subplane of Aff, let (A,E ) be a nonincident pointline pair of a,. Then, there is a unique line E A parallel to E and incident with A. The points of EA are included in the pointset of a, in this situation. Theorem 9.6 Let (P, L , Z ) be a triple of sets of ‘points’ P , ‘lines’ L and ‘incidence’Z with the following properties: (1) Any line is incident with at least three points. (2) Two distinct points are incident with a unique line. (3) Let A , B , Cbe distinct points such that C is not incident with the line AB. Call the points A , B , C‘vertex’ pointsof the triangle ABC fomned b y the lines AB, AC, BC. If a line E intersects the lines AB and BC in nonvertex points then E intersects AC nontrivially. (4) There exists at least four points no three of which are incident with a line (collinear). Then ( P ,L ) is theset of points and lines of a projectiveplaneor a projective geometry PG(V,K ) . Proof: We structure the proof by the use of a series of lemmas.
Lemma 9.7 Let P* be a subspace and let A be any pointwhich is not in P*.Denote b y (A,P * ) the set of points on lines AB where B is a point of P*. Then ( A ,P*)is a subspace. Proof Now let C,D be distinct points of (A,P*)and let Q be a point of the line CD. If C and D are collinear with A, there is nothingto show. Hence, assume that C A and DA are distinct lines. Clearly, C A and DA each contain a unique point of P*say C1 and Dl.If both C = C1 and D = D1 then Q is a point of P* and hence of (A,P*). Assume first thatboth C # C1 and D # Dl. Weshow that Q is in (A,P*)by two applications of property (3). Note that the line CD intersects the triangle ACID1 in two points C and D on the sides AC1 and AD1 respectively. Hence, CD intersects the side C1D1 in a point TI.Then the line AQ intersects the triangle CC1T1 in points Q and A on sides CT1
90
CHAPTER 9. CHARACTERIZATIONS OF GEOMETRIES
and CC1 respectively. Hence, AQ must intersect side C1T1 in a point El which is then a point of P* as P* is a subspace. But, then Q is a point of a line AE1 where E1 E P' so that, by definition, Q E (A,P*). Now assume without loss of generality that C = C1 but D # Dl. Since there are at least three points on AC, let C' be a point on AC not equal to A or C. Consider the triangle CC'D and note that AQ intersects two sides CD and CC' in Q and A respectively so AQ intersects the side C'D in a point say D'. Now take the triangle D1C'D and note that AQ intersects C'D in D' and intersects DID in A so must intersect the side C'D1 in a point El. Finally, consider the triangle CD1C' and note that AQ intersects CID1 in El, CC' in A so must intersect the side CD1 in the point TIand since P* is a subspace and C = C1 and D1 are points of P' then TI is a point of P*so, by definition, Q E ( A ,P*). Hence, we have shown that (A,P*) is a subspace.U
Lemma 9.8 If A is a point which is not incident with is a subspace which is a projective plane T,,.
a line !then (A,!)
Proof: Since a line !(set of points on a line) is trivially a subspace, the lemma follows.0 Thus, either the theorem is proved or there exists a pointD which is not in T,,. Thus, (D,7ra)is a subspace. Note that not all points are in the union of the two projective planes 7ra U (D,!) where !is a line of T,,. Hence, let D' be a point of (D, T,)  (D,e). Then, (D', l ) = 7r1 is also a projective plane distinct from T,, and T,, n 7r1 = !or otherwise they would be equal. Note also that D is not in either 7ro or 7r1. We consider the natural projection thru D of any triangle ABC of no ontoatriangle A'B'C' of 7r1, . If we think of two planes in real threedimensional space then the two planes intersecting in l force the two triangles which are perspective from D to be perspective from !. That is, we obtain a Desarguesian configuration in this way. We need to check that the projection is valid with condition (3). To see that AD intersects T I , use simply that (D,n,) = (D,nl) (the set of all points on lines thru D to points on 7ro is the set of all points on lines thru D to points on T I ) .
9.2. PROJECTIVE AFFINE GEOMETRIES. AND
91
Hence, by projection, we doobtain a triangle A'B'C'. Furthermore, since both no and n1 are projective planes, the corresponding sides can only intersect on the common line !provided they intersect at all. So, it remains to show that AB and A'B' intersect. However, this follows from property (3) by use of the triangle A'B'D and noting that the line AB intersects the sides A'D and B'D in A and B respectively so must nontrivially intersect the third side A'BI.0 The following argument is virtually identically to the argument showing that any PG(2,K ) satisfies the Desarguesian configuration theorem for any pair of pointperspective triangles.
Lemma 9.9 The projective plane no is Desarguesian and is isomorphic to PG(2,K ) for some skewfield K . More generally, the subspace (D,no) universally satisfies the Desarguesian configuration property. Proof: Now assume that we have two triangles in no say ABC and A'B'C' perspective from 0. Select any line of ( D ,no) no incident with 0 and choose two points of this line S and S' not in no. Note that (0,AS)= (0,A'S') since OS = OS' and OA = OA'. Hence, the lines AS and A'S' intersect in a point A*. Similarly, the lines BS and B'S' intersect in a point B* and the lines CS and C'S' intersect in a point C*. It directly follows that the triangles ABC and A*B*C*are perspective from S and the triangles A'B'C' and A*B*C*are perspective from S'. Let n1 denote the projective plane (A*, B*C*) and let != 7ro n T I . Hence, the triangles ABC and A*B*C*are perspective from !as are the triangles A'B'C' and A*B*C*by the previous argument. Note that
AB n A'B' = AB n A*B* = A'B' n A*B*, AC n A'C' = AC n A*C* = A'C' n A*B*, and
BC n B'C' = BC n B*C*= B'C' so the triangles
n B*C*,
ABC and A'B'C' are perspective from e.
92
CHAPTER 9. CHARACTERIZATIONS OF GEOMETRIES
So, in particular, any projective subplane satisfies the Desarguesian configuration and hence is isomorphic to PG(2, K ) for some skewfield K provided the system is not simply a projective plane. Furthermore, we have shown that (D,7r,,) is a subspace which also satisfies the Desarguesian configuration theorem.0 Now given any two planes within thissubspace,projection as above induces an isomorphism between the two planes. By Theorem 9.1, the two planes are ‘coordinatized’ by isomorphic skewfields K and K‘. However, since the two planes share a line, it follows that K = K’. Now let T,,and 7r1 be any two distinct projective subplanes. Let D,, and D1 be points of x,, 7r1 and 7r1 T,,respectively and let E,,, E1 be points of 7r,, and 7r1 respectively distinct from D,, and Dl. Then (E,,,D,D1) is a projective subplane sharing a line E,,D,, with 7ro and similarly, (El,D,,Dl) is a projective subplane sharing a line ElDl with TI.Since (D1,7ro)and (Do,7r1)are subspaces containing the line D,,D1, it follows that D,D1 may be coordinatized by a skewfield K such that both T,,and 7r1 are isomorphic to PG(2,K). Hence, every point A of the structure is a ldimensional Kvector subspace. Let V denote the union of the ldimensionalKvector subspaces which represent points. Let a and b be any vectors in points A and B respectively. Since A + B is the uniquely defined 2dimensional subspace given by AB, it follows that a b is uniquely defined. Furthermore, since scalar multiplication may be taken elementwise, it follows that V is a vector space over K whose ldimensional and 2dimensional vector spaces are the points and lines of the geometry. Since any subspace of the geometry is generated by its line joins to its points, it fol.lows that the geometry is isomorphic to PG(V,K).O We now prove an affine version of the previous theorem.
+
Theorem 9.10 Let (P, L,T,Aff) = 7r be a triple of sets of ‘points’P, ‘lines’ L and ‘incidence’z with a set of subspaces Aff which are
[email protected] planes and assume that Card Aff > 1. Assume that 7r has the following properties: (1) There are at least two points incident with e v e q line. (2) Two distinct points are incident with a unique line. (3) There are at least four noncoplanar points .
9.2. PROJECTIVE AND
AFFINE GEOMETRJES.
(4) Given three mutually distinct noncollinear points, there afine plane of Aff incident with them. (5) Parallelism is an equivalence relation. Then r is an afine geometry AG(V,K).
93 is a unique
Proof We break up the proof by various key lemmas. Note that by the definition of parallel lines, the relation is automatically reflexive and symmetric. Hence, we are further assuming that the relation is transitive. Let A, B , C,D be four distinct noncoplanar points. Then ( A ,B , C) = a1 shall denote the subspace generatedby the points on the line joins to A, B, C; the affine plane containing A , B and C. Note that given a parallel class ,O of this affine plane, there is a line of p of the affine plane incident with the point C. Thus, clearly, C D is not parallel to any line of the subplane but meets a1 in the point C. Similar to the proof to the projective version, we define a set which becomes a subspace. Since there are four noncoplanar points A, B , C,D then the affine plane ( A , B ,C) does not contain the point D . Form the line AD = k' so that k' intersects the plane ( A , B ,C ) uniquely in the point A. Let a denote the parallel class containing e. Then there is no line of ( A , B , C )in a because if so then the unique line k ' ~of the affine plane ( A ,B , C) incident with A and k' are not disjoint. If A, W and A, W' are points of k' and k ' respectively ~ is an affine plane with parallel lines k' and k ' which ~ are not then ( A ,W, W') disjoint .
Lemma 9.11 Let a, denote a plane of Aff. Let a be a parallel class of lines such that no line of a, is in a. Let (a,a,) denote the set of points on lines of a which meet a, in a point. Then (a,a,) is a subspace. Proof: Let A and B any two points of (a,a,). We first msume that both A and B are not in a,. Let A and B be on distinct lines k ' ~and of Q respectively which meet a, in points A' and B' respectively. If A and B are
k',
on the same line which meets a,, there is nothing to prove. Then A' and B' are distinct points as k ' and ~ are parallel. By property (4),there is an affine plane ( A ,A', B') of Aff and there is a unique line l?,,
94
CHAPTER 9. CHARACTERIZATIONS OF GEOMETRIES
of the subplane incident with B‘ and parallel to AA’ so that .~?BIis in Q. Thus, the points A, A’, B , B’ are coplanar. Now let C be any point of AB. The line tc parallel to AA‘ and incident with C is in the plane (A,A’,B’) as is A’B’ so &cn A’B’ is a point which is then common to a,. That is, any point C of AB is in (a, a,). It remains to consider the case when A E a, and B is not in a, and we may assume that t , does not intersect at A. However, the argument for this case is virtually identically to the previous so will be left for the reader to complete. Hence, (a, a,) is a subspace.0
Lemma 9.12 There is a unique subspace (a,a,) containing any set of four noncoplanar points and any four noncoplanar points within the subspace generate it. We shall use the t e r m ‘solid’ for such a subspace.
Proof: Now let A, B , C,D be four noncoplanar points. Then, as noted above, thereexistsasubspace of the form (a,a,) containingthese four points. Suppose there is an arbitrary subspace(p,c q ) containing A, B , C,D. Certainly, for one of the subspaces, say the first listed, we may assume that a, = ( A ,B , C). Thus, there exists a line of a incident with D which intersects a,. Since (p,al) is a subspace,it contains (A,B , C ) and D so if to is a line of a which intersects (A,B , C ) in a point E then all points of to are in (p,al). Since this argument is valid for any point D not in a,, it follows that (a, a,) C (p,a l ) . Since the argument is symmetric, the lemma is pr0ved.U Lemma 9.13 Two planes of a solid are either disjoint or share a line of points. Proof Let a1 be any subplane within (a,ao) which shares a common point A with a0 and let t i be a line of a1 and incident with A which is not a line of ao. Let B1 be any point of a1 which is not incident with l; and and incident with B1. Clearly, let L;, denote the line of a1 parallel to we may assume that is in a without loss of generality. The fact that is in (a,ao) implies that it intersects a0 in a point. Hence, a0 and a1 share two points and are both subspaces so they share a line of points.0 Now we examine when we might define two planes to be ‘parallel’.
.ti
ti
!Ll
9.2. PROJECTIVE AFFINE GEOMETRIES. AND
95
Lemma 9.14 Let a0 and a1 be distinct afine planes which have lines in at least two common parallel classes, say Q and p , Then a0 and a1 share all of their parallel classes. That is, every line of a0 is parallel to a line of a1 and conversely, and, in such a case, we would say that a0 is ‘parallel’ to al. Parallelism on afine planes of Aff is clearly an equivalence relation. Proof: To see this, let Ai be a point of ai for i = 0 , l and let t i , 6 for i = 0 , l and be lines of ai incident with A, and in the classes S = Q and p respectively. Note that if BO,6 # A0 arepoints of for 6 = Q or p then (h, & p , A I ) is a solid CI, which clearly contains ao. Also, (Ao,Bola,A1)is an affine plane and incident with A1 is parallel to incident with Ao. Hence,all points of el,, are within the solid Q and it further follows that a1 is in the solid as well. Assume that there is a parallel class y such that theline to,, of y incident with A0 is in a0 but the line el,, of y incident with A1 is not in a l . The plane (t~,,,tl,~) is a plane of the solid which shares a point A0 with a0 and a point A1 with a1 and hence shares the lines to,, and M1 with a0 and a1 respectively, Thus, M1 is not parallel to to,, as otherwise to,, would be el,, so M1 n to,, is a point common to a0 and a1 so that a0 n a1 is a common line &l. But, No1 cannotintersect a0 in a point offof to,, as otherwise a0 = (to,,,tl,,).Thus, the line No1 of a0 is parallel to the lines of &,a for S = Q, p which cannot occur. Hence, every line of a0 is parallel to a line of a1.0 The general idea of the proof is to extend this structure to a projective geometry within which the structure can be seen as an affine geometry obtained by deletion of a hyperplane. We therefore define a ‘projective point’ to be either a point of or an equivalence class of lines of Aff and a‘projective line’ to be either a line of C or an equivalence class of affine planes of Aff with incidence being inherited from the given structure. We need to verify the conditions of Theorem 9.6. We shall be showing that the set of projective points and projective lines form a projective space and the set of equivalence classes of lines form a hyperplane. Given any two projective points A and B. First assume they are both parallel classes of lines [a]and [b] respectively and let a and b be lines of the parallel classes [a]and [b] respectively which intersect. Clearly, (a,b) is
.
.. . ..
,
,
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CHAPTER 9. CHARACTERIZATIONS OF GEOMETRIES
an affine subplane of Aff and if a' and b' are lines of [a]and [b]respectively which intersect then (a', b') is an affine subplane which shares two parallel classes with ( a ,b) and hence the two affine subplanes are parallel. It follows that there is a unique equivalence class of affine planes and hence there is a unique projective line incident with [a]and [b]. The remaining situations are similar and are left to the reader for verification. Clearly, the nontriviality conditions are satisfied so it remains to verify condition (3) of Theorem 9.6. , be any triangle of projective points. Let k' be a projective Let A , BC line which intersects the triangle in nonvertex points on sides AB and AC. Assume that all points A , BC , are equivalence classes of lines [ a ] ,[b], and [c]respectively. Hence, k' intersects AB in the projective point [dab] and intersects AC in the projective point [clac] and all parallel classes indicated are distinct. Hence, we have distinct lines a , b,c,d,b, dac in the indicated parallel classes. Since we may choose any representative lines, we choose the lines so that they are concurrent with a common point C. Note that dab E ( a ,b) and d,, E (a, c) and ( a ,b,c) is clearly a solid. Hence, the configuration is within a solid and the planes (dab,clac) and (b,c) share a point C so share a line dbc. Then, the parallel class of a projective point [db,] is incident with the line .l so the 'third' side is nontrivially intersected. Finally, let ABC be a triangle where A is a point 'of P, If all are points of P then the points are contained within a unique affine plane of Aff . Similarly, if C = [c]but A and B are points of P then we trivially have a triangle and if c is a line incident with A then ( A B ,c) is an affine plane. If C = [c]and B = [b] and c and b are representive lines which intersect at A from the two parallel classes respectively,there is a unique affine plane of Aff containing A , c and b, and the unique projective plane extension contains the indicated triangle. In all of these cases, there is a unique extension to a projective subplane within our structure of projective points and projective lines so that condition (3) of Theorem 9.6 is automatically satisfied. Hence, this completes the proof to the the0rem.O
Chapter 10
DERIVABLE NETS AND GEOMETRIES Here we develop the main ideas of Cofman [l91 who associated an &ne space with a subnet of a derivable net. We shall consider what projective spaces can be associated with the derivable net itself. Later, we shall generalize these concepts in the study of ‘subplane covered nets’. We shall have occasion to use the collineation group of an arbitrary projective geometry. Thus, we shall utilize the theory presented in the previous chapter. In 1972, Foulser [23] completely determined thestructure of a finite derivable net D that lies in a vector space. Foulser showed that there exists a finite field K such that D corresponds to a regulus in PG(3, K).Foulser’s methods were largely linear algebraic and group theoretic. Also in 1972, Prohaska [67] showed, by a combinatorial argument, that when a finite derivable net is embedded into a finite &ne plane, the Baer subplanes involved in the derivation process are always Desarguesian. In this chapter, we are concerned with trying to obtain structuralinformation about a derivable net D without necessarily assuming the structure of an affine plane containing D. Note that if the derivable net is finite and embeddable in a vector space then questions of embeddability and extension are answered by Foulser [23] as a finite regulus type derivable net may at least be embedded in a Desarguesian affine plane.
97
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C H A P T E R 10. DERIVABLE NETS AND GEOMETRIES
In this chapter, we give a geometric characterization of a derivable net using projective geometries. In thenext chapter, this characterization is used to completely determine the structure of the derivable net. Our geometric characterization is obtained by extending both the ideas and framework of Cofman [ 191. Cofman studiedarbitrary derivable affine planes by developing a 3dimensional affine geometry associated with the net within the plane. The restrictions of the duals of the projective extensions of the Baer subplanes become &ne subplanes so that thesubplanes are always Desarguesian, thus extending Prohaska. The points of the affine space are the lines of a derivable net D which do not lie on a given parallel class (m). Hence, the affine space cannot adequately describe the structure of the derivable net. We show that the affine space determined by Cofman may be extended to a projective space in such a way so as to completely characterize the derivable net D whether D may be extended to an affine plane or not. We shall want to prove the results of Cofman along the way and, for this, we shall require various fundamental results on the intersection properties' of derivable nets. For convenience, we relist the definition of a derivable net.
Definition 10.1 A derivable net D = ( P ,C,B,C , Z) is an incidence structure with a set P of points, a set C of lines, a set B of Baer subplanes of the net, a set C of parallel classes of lines and a set Z which is called the incidence set such that the following properties hold: (i)Every point is incident with exactly one line from each parallel C h s , each parallel Class is a cover of the points and each line of C is incident with exactly one of the classes of C . (ii) Two distinct points are incident with exactly one line of C or are not incident. (iii)If we refer to the setC as the set of infinite points then the subplanes of B are afine planes with infinite pointsexactly those of the set C. (iv) Given any two distinct points a and b of P which are incident with a line of C , there is a Baer subplane Ta,b of B containing (incident with) a and b.
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99
Remark 10.2 In the following lemmas, we assume that D = ( P , L ,t3,C,Z) is a derivable net satisfying properties (i), (ii), (iii)and (iv). Recall that if p E C and Q a point of P,Q p shall denote the unique line of ,B incident with Q. Furthermore, QC shall denote the set of all net lines incident with Q.
Lemma 10.3 (1) Let A and B be distinct collinear points in the net. Then there is a unique Baer subplane of the net containing them. (2) Furthermore, two distinct Baer subplanes r0 and T I are either disjoint on points or intersect in a unique point Q # In the latter case, the two Baer subplanes share exactly the lines QC. (3') Also, there is a set of Baer subplanes that intersect in the point Q that completely cover the lines QC.
Proof: Let A and B be any distinct affine points which are collinear in the derivable net. By assumption, there is a Baer subplane TA,B containing these two points. Note that any Baer subplane containing A and B must contain the intersections of the lines AC to the lines BC. It follows easily that TA,B may be generated by these joins and intersections. Hence, this proves (1). Also, it is then clear that any two distinct Baer subplanes are either disjoint or intersect in a unique point Q. Clearly, the two subplanes share at least the lines QC. If there is another common line C then C must intersect each line of QC not in the same parallel class as C and hence, there must be at least two common points. Thus, the two subplanes share exactly the lines QC. This proves (2). Choose any line t of QC and choose any point A on l  {Q}. Then, there is a unique Baer subplane T Q , A . Since the point A may be chosen in an arbitrary manner, it follows that the subplanes generated by such a construction completely cover the lines QC and intersect exactlyin the point Q. This proves (3) and completes the proof of the 1emma.O A slight variation of the above lemma shows the following to be true.
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GEOMETRIES
Lemma 10.4 Choose any twolines say e, and !p from distinct parallel classes a and p. Then there is a unique set of Baer subplanes which share exactly the point e, r lt p = Q and the lines QC and otherwise partition the points on the lines QC. Lemma 10.5 If 7ro and 7r1 are distinct Baer subplanes which are disjoint on points then there is a unique parallel class p E C such that 7r, and 7r1 share all of their lines on this parallel class and share no other lines. Proof: Choose any point Q.of 7ro  711. Then there exists a unique line of 7r1 say of the parallel class p which is incident with Qo. Note that Qop is a line of 7ro and Qop = l1 so that the two planes have a common line. Now let e2 be a line of p and 7r1 which is not a line of no. Let Q1 be a point of 7r1 on & so Q1 is not a point of 7ro. Thus, there is a unique line t 3 of 7ro which is incident with Q1. Let Q16 = &3 for some parallel class 6. Since 4 is a line of 7ro but e2 is not, then 6 # p. Thus, t 3 and el are two common lines of 7ro and 7r1 and must intersect in acommon point which is a contradiction. Hence, the lines of 7ro of the parallel class p are exactly the lines of 7r1 of the parallel class p. In addition, there can be no further common 1ines.O
el
Lemma 10.6 If 7ro and 7r1 are distinct Baer subplanes which are disjoint on points and a third subplane 7r2 intersects 7ro in a point then 7r2 also intersects 7r1 in a point. Proof: By the previous lemma, there is a parallel class p on which the two subplanes 7ro and 7r1 share all of their lines. Assume that 7r2 and 7ro share a common point Q. Suppose that 7r2 and 7r1 are disjoint on points. Then, by the previous lemma, there is a parallel class a such that the subplanes 7r2 and 7r1 share all of their lines on a. Now first assume that a = p. Then the three subplanes would share all of their lines on a and if x, and 7r2 intersect in Q then they would be equal. Hence, a # p. Thus, Qa is a line common to 7r1 and 7r2. Furthermore, since Q is a point of 7ro, Qa is also a line of 7ro. Also, Qp is a line of 7r2 and of no and also of 7rl since 7r1 and 7ro share all of their lines on p. Thus, Qp and Qcr are lines common to 7r1 and 7ro which implies that Q is a common point of no and 7r1. Hence, this contradiction completes the proof.0
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Lemma 10.7 Choose any parallel class p E C and choose any two distinct lines E l and t 2 on p. Then there exists a set of mutually pointdisjoint Baer subplanes sharing E l , t 2 and all of their lines on p such that the union of the points on these subplanes partition the set of common lines on p. Proof Let A1 be any point of tl. Let a be any parallel class # p. Then Ala intersects l 2 in a point A2 distinct from Al. Hence, the points A1 and A2 are collinear in the net and there is a Baer subplane T A ~ , containing A ~ them and a previous lemma shows that this Baer subplane is unique. Choose another point B1 on E l and go thru the same construction as above to produce a subplane T B ~ , Bwhich ~ shares the lines E l and E2 with TA~,A~ Either , the two planes are identical or disjoint as if they share a point Q they share exactly the lines QC. Hence, by this construction, there is a set of planes which are disjoint and share the lines E l and 12. So, each of these planes shares the lines of T A , , A ~on p.0
Lemma 10.8 Let 7ro be any Baer subplane and let p(7ro) denote the set of lines of 7ro of the parallel class p. Choose any line E of p which is not in P(r0). Then, thereexists a uniqueclass of mutually disjoint Baer subplanes whichshare E and all of their @linessuch that these plines are disjoint from Proof Choose a parallel class CY # p and construct the class of mutually pointdisjoint subplanes that share the lines of 7ro on a and partition the points on these lines of a. There is a unique subplane 7r1 in this class which contains the line E. Since 7ro and 7r1 share all of their d i n e s , they share no other line. Hence, 7ro and 711 have disjoint sets of plines. Now assume that 7r1 and 7r2 are subplanes which share E and have both sets of &lines disjoint from p(7r0). Note that 712 and 7ro are disjoint for otherwise they would share a unique @line. But, we have seen that if two subplanes 7ro and 7r1 are disjoint and a third subplane 7r2 intersects one then the subplane intersects the other subplane. Hence, 7r1 and 7r2 are disjoint. Thus, there is a parallel class upon which the subplanes share all of their lines. So, and 7r2 share all of their lines on p.0
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10.1 TheCharacterizationTheorem. Theorem 10.9 Let V = ( P , C , B , C , Z ) be aderivable net where the indicated sets denote the sets of points P , lines C, Baer subplanes B , parallel classes C and incidence Z. (I) Then there exists a skewfield K and a 3dimensional projective geometry C isomorphic to PG(3, K ) together with a jixed line N of C such that (1) the points P of V are the lines of C which are skew to N , (2) the lines C of V are the points of C  N , (3) the parallel classes C of V are the planes of C which contain N , and (4) the subplanes B of V are the planes of C which do not contain N . (11) Conversely, if C N P G ( 3 ,K ) is a 3dimensional projective geometry over the skewfield K and N is any jixed line of C , define sets of points P, lines C , parallel classes C, subplanes B to agree with the correspondence above where incidence Z is relative incidence in C. Then V = (P, C,B , C,z> is a derivable net. Hence, derivable nets are equivalent to 3dimensional projective geometries. Proof: We begin with part (11). Let C denote the system described in the statement of the theorem with designated line N . For convenience, we shall refer to the structures P,C , B, C are ‘netpoints’, ‘netlines’, ‘netsubplanes’ and ‘netparallel classes’ respectively. Wewish to show that the combinatorial system of lines of C skew to N as the ‘net points’ P,the points of C  N as ‘net lines’ C, the planes (hyperplanes) of C containing N as ‘net parallel classes’ C and the planes of C that do not contain N as ‘net subplanes’ B form a derivable net. First consider property (i). Each netpoint is on exactly one netline from each netparallel class since each lineof C  N is incident with exactly on point from each plane containing N . Furthermore, each netline isin exactly one netparallel class as each point on CN is in a unique hyperplane containing N and each netparallel class partitions the netpoints as each subplane containing N intersects each line of C skew to N .
10.1. CHARACTERIZATION THE THEOREM.
103
Note further that two netlines in different netparallel classes must intersect uniquely since two points of C which do not lie on a hyperplane of C containing N are incident with a unique line skew to N , Since two distinct lines of C skew to N either uniquely intersect in a point of C  N or are skew, it follows that two distinct netpoints are incident with 0 or exactly one netline. This verifies property (ii). We now consider property (iii) and show that the elements of B are, in fact, affine subplanes of the incidence structure D = (P,L,C, B,T). To see this, let T,, be any plane (projective) of C which does not contain N. Let T: denote the dual projective plane. Notice that the points of are the lines of r0 and those skew to N are the netpoints. The lines of T: are the points of T,, and if we let (cm)denote the point T,, nN then the lines of T: not equal to (m) are netlines. Note that r0intersects each plane of C so, in particular, shares a unique line with each hyperplane of C which contains N. Translating this into the language of netpoints and netlines, this means that T:  (cm)is an &ne subplane of D whose parallel classes are exactly the parallel classes of C. We need to show that each subplane p of B is actually a Baer subplane of D. In order to prove this, we need to show that each point of D (of P) is incident with a line of p taken projectively and each line of D (of L) is incident with a line of p taken projectively. The latter propertyis immediate since the parallel classes of any subplane are the parallel classes of D (of C) and any line 1 of L therefore intersects the infinite line of the projective extension of p. We know that p is an affine plane and the projective extension p+ when dualized becomes a plane of C which intersects N in a point. A netpoint A is a line of C skew to N and a projective plane and a line must intersect and intersect in a point of C  N,This says, in the net language, that there is a netline of p incident with each given netpoint A. It remains to verify property (iv). Choose any two distinct netpoints A and B , we need to show that there is a subplane of B containing them. The netpoints are two distinct lines of C skew to N and as such generate a unique projective subplane containing them. The subplane cannot contain N for otherwise the given lines could not be skew to N. Since the planes of C which do not contain N correspond to the planes of B,we have a unique affine plane of B that contains A and B.
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Hence, we have shown that D is a derivable net. We now turn to the proof of part (I). Hence, assume that D = (?,C,B,C,z) is a derivable net. We wish to show that there is a corresponding projective space such that the. combinatorial structure listed above is the derivable net in question. It was mentioned above that Cofman [l91 developed a theory for derivable affine planes. It is not difficult to realize that Cofman’s arguments also are valid for arbitrary derivable nets. We begin by a proof of Cofman’s results stated in termsof nets. Cofman connects an affine geometry with each subnet of a derivable net obtained by deleting one parallel class. Once this is established, we show that there is a projective geometry connecting all of the component affine geometries which uniquely describes the derivable net.
10.1.1
Cofman’sTheorem(forDerivable
Nets).
For each a E C, we define a pointlineaffine subplane geometry SZ, as follows where the points, lines and subplanes of the geometry are referred to as affpoints, afflines and affsubplanes: The affpoints of
are the lines of C which belong to C  {a},
the afflines are the sets of lines of C which are of the following two types:
S ( i ) the set of lines of @
E
C  {a},for some given parallel class @
which lie on a subplane of
B or
P(ii) the set of lines of C E C  {a} incident with a point Q
of P. Two afflines shall be said to be ‘parallel’ if and only if they are disjoint or equal. We shall use the notation QC to denote such afflines.
To define the set of affsubplanes, let @ E C  {a}and let ap denote the following’pointline geometry of @points and @lines: The @points of ap are the lines of @ and the @linesare the lines of @ of a subplane of B (affline of type S(i)). Now let 7ro be a subplane of B. We define a pointline geometry an, where the points and lines are called spoints and slines as follows:
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105
The spoints are the common lines of C  {a}and 7ro (&points) and the slines are of two types; slines of type (a) are the sets of lines of 7ro which are incident with a point of 7ro (in P ) (af€lines of type P(ii))and the slines of type (b) are the setsof lines of p in 7ro where p E C  {a}(afflines of type S(i)). Hence, the affsubplanes are of two types: (i) ap for p E C  {a} and (ii) ar0 for no E B and the lines of an, are of two types (a) and (b).
We shall prove in due course that up and uno are affine planes and two affpoints are incident with at most one affline. If they are affine planes then we have a set Affof affine subplanes which are subpspaces by their definition. Recall that these two properties are the required affine subplane property and property (2) of Theorem 9.10. Note that thecardinality of Aff is > 1. If up and uno are affine planes and two affpoints are incident with at most one &line then we have a pointlinesubplane geometry A,. In order that & becomes an affine space, we would need only to verify properties (l),(3) and (4)of Theorem 9.10. Property (1) is that there are at least two affpoints on every affline. Since an affine subplane has at least two lines per parallel class and at least three parallel classes, property (1) follows immediately. Property (3) is that there are at least four noncoplanar points which is also trivially true. Property (4) is that given any three noncollinear affpoints A, B , 2, there is a unique affsubplane containing them. If the three affpoints areall lines of the same parallel class p E C  {a}, exactly ap contains these &points. If two affpoints A and B are lines of p but 2 is a line of y # p then the point intersections A n 2 and B n Z are on a unique subplane 7ro of B and the lines of 7ro not in a (affpoints) and the sets of lines of 7ro on parallel classes S E C  {a} (&lines of type S(i))and sets of lines of 7ro on affine points of 7ro (afflines of type P(ii))make up uno. Hence, A, B , 2 are affpoints on the unique affsubplane uro. If the three affpoints are lines in three distinct parallel classes say p, y and S E C  {a}then there are two points intersections since if there would
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be but one, the three afFpoints would be lines of some subplane of B and hence would be collinear. So, there is a unique subplane r1 of B containing the intersection points and the previous proof shows that there is a unique affsubplane a*l containing the three affpoints. Thus, it remains to show that up and a"0 are affine planes and that two affpoints are incident with at most one afFline. We first consider the structures ap. Choose two distinct affpoints t~ and t ~These . are lines of P and by a previous lemma, there is a set of mutually pointdisjoint Baer subplanes sharing these two lines and all of their lines on p. Hence, there is a unique affline incident with the two given affpoints. Now let a, be an &line and an affpoint not incident with a,. Then a, is the set of plines of a class of Baer subplanes and there exists aunique class of Baer subplanes sharing &A and all of their Plines all of which are disjoint from a,. Hence, there exists a unique affline containing t~ which is parallel to a,. Now to show that uno is an affine plane. This is essentially the affine restriction of the dual of the projective extension of ro but it is worth noting for later just exactly how this is so. If r, is extended to a projective plane and then dualized, the set of parallel classes of r, is a projective line which then becomes a set of points. The set of lines say incident with a (including the infinite line of r,), becomes the set of points at the line at infinity of the dual projective plane. The other points of the dual plane are the lines of r, which are not incident with (a); not in the parallel class a. Hence, by not taking the alines of r,, we obtain the corresponding set of affpoints and afflines to form the natural affine restriction of the dual of the projective extension of no. Now we need to show that two &lines intersect in at most one affpoint. First assume that the two afflines l o and 11 are both of type S(i). Then there are Baer subplanesr, and r1 such that p and y lines of T , and r1 are respectively t o and where p and y are parallel classes of C. If y = p and the two subplanes sharemore than one @linethen thesubplanes aredisjoint and share all of their @lines bythe previous lemmas. Hence, if y = p then the two &lines share at most one affpoint, Furthermore, If y # p then the afflines share no affpoint.
10.1. CHARACTERIZATION THE THEOREM.
107
Now assume that the two afflines are both of type P(ii). Hence, both afflines are of the form QC and QIC for some points Q and Q1. Clearly, there is at most one common line incident with Q and Q1 and hence at most one common affpoint , Finally, assume that the two afflines are of different types; one of type S(i)and one of type P(ii). Since one is a set of lines incident with a parallel class ,B and one is a set of lines incident with a point Q,it follows there is at most one common line. Thus, in all cases, there is at most one affpoint incident with two distinct afflines. Hence, we have proven Cofman’s theorem:
Theorem 10.10 Let D = (P,L,B,C,Z) be a derivable net and let a be any parallel class. Then the setof lines of La can be made into (points’ of a 3dimensional afJine geornetmJ A, by defining ‘lines’ of the afine geometry to be of two types: the sets of lines of a given Baer subplane incident witha given parallel class or setsof lines of a given Baer subplane incident witha given point and by defining the ‘planes’ of the afine spaceas two types: the natural afine restrictions relative to a of the projective extensions of Baer subplanes and the planes obtained from each given parallel class ,B by defining the ‘(points’ and ‘lines’ as above. Proof By Theorem 910, we have that this set forms an affine geometry. It remains to show that we have an AG(3,K). To see this, simply note that any two affplanes of type aToshare an affline of type P(ii) whenever they share a point and share an affline of type S(i) whenever they are disjoint.0
10.1.2 A Quilt of Affine Geometries. Now we may make specific the connections between derivable nets and projective geometries. Cofman’s theorem associates derivable nets minus a given parallel class with an affine geometry. Of course, we may extendany of these affine geometries to a projective geometry, but this geometry may not describe adequately the derivable net.
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Hence, we wish to sew together these affine geometries A,, for all Q E C, in such a way so as to realize the derivable net within the associated projective extension of one (of all) of these affine geometries; the projective geometry. We form a sort of ‘quilt’ of affine geometries. In order to do this, we define projective points and projective lines relative to the associated set of affine geometries and apply Theorem 9.6 to obtain a projective geometry which is the correct quilt. The parallel classes p form hyperplanes ‘Hp (ap in the previous notation) of the affine geometries A, for any p # Q. These hyperplanes turn out to extend projectively by adjunction of the same line N . Considered collec‘, union the line N forms a projective space tively, the ‘missing hyperplane’H which can be considered the hyperplane at infinity of A,. Hence, the main pattern for the quilt arises from the parallel classes as &ne hyperplanes.
Definition 10.11 Two Baer subplanes if and only if they are disjoint or equal.
7ro
and
7r1
are said t o be ‘parallel’
Lemma 10.12 Subplane parallelism is an equivalence relation and we shall denote by [rO] the class containing the Baer subplane 7ro. Proof: It suffices to note that if no and 7r1 are disjoint anda Baer subplane 7r2 intersects one of these subplanes then a previous lemma shows that the subplane intersects the remaining subplane. Hence, parallelism is transitive and is clearly symmetric and reflexive.0 Definition 10.13 A projective point, ‘propoint’, is defined as either an affpoint of the union of the afine geometriesA, f o r all a in C o r a nequivalence class [x,]where 7ro is a Baer subplane. Hence, the propoints are the lines of the net and the classes of disjoint Baer subplanes of the net. A projective line, ‘proline’, is of three types: (1) any affline of a n y a f i n e space A, with the adjunction of the propoint [7r0] in the case that the aflline is the set of lines o n a parallel class p of a Baer subplane 7ro, of type S(i), (2) any affline of type P(ii); of the form QC for .Qa point of the net, (3) U7r,Et3~~01‘
101. CHARACTERIZATION THE
THEOREM.
109
Note that any aflpoint is joined to every propoint of t h e f o r m [T,]as any line of p, f o r a n y p E C , is incident with some line of a Baer subplane Tl of [To]* To prove that this is a projective space, we apply Theorem 9.6 and verify the properties (1)thru (4),show that the structureis not a projective plane and then show that we obtain a 3dimensional projective space. Recall, the properties are: (1) Any proline is incident with at least three propoints. (2) Two distinct propoints are incident with a unique proline. (3) Let A , B2 , be distinct propoints such that 2 is not incident with the proline AB. If a line i? intersects the prolines AB and B2 in nonvertex propoints then i? intersects A 2 in a propoint. (4)There exists at least four propoints no three of which are incident with a proline (collinear). Hence, (l),(4)follow immediately and since the planes will turn out to be the duals of the projective extensions of Baer subplanes together with the extensions of the affine planes of type up, the structure obtained is not a projective plane. Thus, if the remaining properties are verified, we obtain a projective geometry which clearly has the extensions of affine planes of type up as hyperplanes and thus we obtain a 3dimensional projective geometry. So, we consider properties (2) and (3). Note that any affpoint A is joined to any propoint of the form [T,] by definition since such propoints are adjoined to the afflines. Moreover, given the line A of a parallel class p, there is a unique class of subplanes of [T,]one of whose &lines is A and such that all of their @linesare equal. That is, there is a unique affline joining A and [T,]. Furthermore, if A and B are two affpoints then, since there are at least three parallel classes, the affpoints are in some affine geometry SZ, so there is a unique affline joining them from which it follows there is a unique proline joining them. = Finally, if the two propoints are of the form [T,] and [ T I ] then UaOEB[~,] N is the unique proline joining them. It remains to verify property (3).
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CHAPTER 10. DERIVABLE NETS AND GEOMETRIES
Consider the triangle ABZ. If two of these are propoints in N then recall that up is an affine plane so that upUN is a projective plane containing the triangle. Hence, (3) is valid in this setting. Assume exactly two of the points A and B of the triangle are affpoints which are joined to the propoint [x,]. If A and B are in the same parallel class then, as above, the triangle lies in up U N.If A is in Q and B is parallel class p and a # p then A and B intersect in a point Q of the net and we may assume that x, is a subplane containing Q. Recall that in &, u n o is an affine plane and is an affine restriction of the dual of the projective extension of x,. If we include the lines of x , on Q and dualize T,, we obtain a 'punctured' projective plane (missing a point and all lines incident with it) lying in the structure of propoints and prolines. Adjunction of [x,] to u"0 U a(n,) U [x,] is a projective plane of the system of propoints and prolines containing A, B and [x,]. That is, the triangle A B 2 lies in a projective plane of the system so that (3) is valid. If A, B and 2 are all affpoints and if they are in the same parallel class p then since up U N is a projective plane within the system of propoints and prolines, (3) is valid. If not all three affpoints are in the same parallel class then since they form a triangle, there is a Baer subplane T , containing them and the triangle lies within u"0 U a ( x 0 )U [x,] so that (3) follows. Hence, (3) is valid for any configuration of propoints of the triangle so is generally valid. This completes the proof of the theorern.0
Remark 10.14 W e note that if we could ensure that the parallel classes ap are
[email protected] planes, we could avoid proving that the A, 'S are
[email protected] spaces. W e by have chosen this approach following the guidelines set down essentially Cofman and also for the insights provided. In the chapter on subplane covered nets, the afinespaces & are conceptually in mind but it is onlyrequired that the putative hyperplanes (planes in the derivable net case) corresponding to parallel classes are proven to be a f i n e spaces.
10.1. CHARACTERIZATION THE THEOREM.
111
The following diagram indicates the relationship between sets of points
P, lines L, Baer subplanes B, parallel classes C and incidence Z of the derivable net and the corresponding threedimensional projective space.
Chapter 11
STRUCTURE THEORY FOR DERIVABLE NETS In the previous chapter, we characterized a derivable net combinatorially within a threedimensional projective geometry. However, this representation does not provide algebraic or geometric information in the sense that it still does not answer the question of whether or not a derivable net is connected to a regulus net.Inthischapter, we use the projective space chacterization combined with the induced collineation group of the derivable net to more fully understand the nature of the net. Recall that a regulus in PG(3, q ) is a setof q + l lines which are covered by another set of q 1 lines. Furthermore, the corresponding net in the associated 4dimensional vector space is a derivable net. On the other hand, given a finite derivable net D, there is a 3dimensional projective space PG(3, K) where K N GF(q) and a designated line N , such that the derivable net D may be realized combinatorially within C. However, the regulus although giving rise to the derivable net, does not define the projective space within which it lives. The point is that, in this case, there are two 3dimensional projective spaces each isomorphic to PG(3,q). Now the question is, given a finite derivable net D and hence a 3dimensional projective geometry PG(3,K) where K N GF(q), is there a second 3dimensional projective geometry PG(3,q) such that the derivable net corresponds to a regulus in the second geometry? In other words: Is a finite derivable net a regulus net?
+
113
114 CHAPTER 11. STRUCTURE THEORY FOR DERIVABLE NETS We shall show in this chapter that, in fact, the answer is ‘yes’. In order to do this, we examine the nature of the collineation group as inherited from the collineation group of the combinatoriallyconnected 3dimensional projective geometry.
11.1 The Collineation Group of aDerivable Net. We shall adopt the notation of the previous chapter. Recall, that we have a derivable net D = (P,L,B,C,Z) withsets of points P , lines C, Baer subplanes B,parallel classes C and incidence Z. These sets are associated combinatorially with a threedimensional projective space C over a skewfield K with a designated line N such that the points P of the net are the lines of C skew to N, and the lines L are the points of C  N . The parallel classes of C extended by N, a$ are the hyperplanes(planes) containing N and the planes no of B which when extended projectively, dualized and properly interpreted are theplanes D(n,f)of C which intersect N in exactly one point. However, we should not push this connection further than what is given in the statement of the characterization theorem. That is, a skew line to N of C corresponds not directly to a point of P but directly to the set of net lines incident with P so this correspondence between points of the net and lines of the projective space skew to N should be kept in mind when dealing with collineations. Furthermore, when required, we let V denote the corresponding $dimensional Kvector space and fix the scalar multiplication so that V is a left Kspace (scalars are acting on the left of vectors). Let g be any collineation of the derivable net D. In order to consider g as a collineation g* of the projective space C,we must define g* to leave N invariant. Since g is bijective on the set of lines C, it follows that g* is bijective on the set of points of C  N as this statement is merely a reformulation of terms. A plane ro of B corresponds to a plane D(n,f)of C containing N. Hence, it follows that g*(D(r,f))= D(g(r,)+). Let W be any point of N and let W = D ( r , f )n N. Then, we define g*(D(r,f)n N) = D(g(ro)+)n N. Recall from the previous chapter that two distinct subplanes are saidto be parallel if and if they are disjoint. The set of ‘parallel classes’ of subplanes is thus automatically permuted by any collineation g of D.The dual structureof no is a punctured projective plane
11.1. THE COLLINEATION
GROUP OF A DERIVABLE NET.
115
and the same point [ T ~is] adjoined (and thus adjoined to N> for any plane parallel to T,. Hence, it follows that g* is a bijection on the set of points of C. Also, it follows that g* is a bijection on the planes of C that intersect N in a unique point as g* leaves N invariant. Similarly, g is bijective on the set of points P which, when properly translated combinatorially into the projectivegeometric structure, says that g* is bijective on the set of lines skew to N. If a line .t intersects N in a unique point W = [no]then there is a set of planes of C containing .t and we wish to understand how these are permuted. = Q[T,] where Q is any point of C  N. Let .l Now Q is, in fact directly, a line of C say in the parallel class Q. Realized back in the derivable net D,Q corresponds to a unique set of subplanes BQ of B which share the line Q and the setof lines of r0 incident with Recall that all of these subplanes are disjoint as they share all lines of a given parallel class a. Hence, there is a unique point [T,] adjoined to the dual punctured projective planes. Furthermore, these subplanes of B completely cover the set of net points of Q which means back in the projective geometry that the set of points of C N are completely covered by the corresponding projective planes D ( T , ~ ) . Thus, a line l of the projective space which intersects N in a unique point corresponds back in the derivable net to a unique set of disjoint Baer subplanes and such sets of Baer subplanes are clearly permuted by g. All of this simply says that the affine spaces (planes) ap for p E C are permuted by 9Hence, g* is a bijection on lines of C. Since incidence of the action on the projective geometry is inherited from that of the net, we have the first part of the following basic result.
a.
Theorem 11.1 Let D = ( P ,C , B,C,z> be a derivablenetandlet C N P G ( 3 , K ) denote the corresponding 3dimensional projective geometry with designated line N . Then the full collineation group Gz, of D is isomorphic to the stabilizer of the line N in the full collineation group of C. Hence, the full collineation group of D is isomorphic to PrL(4,K ) N . Proof: We have shown that any element of Gz, induces a collineation
of
c.
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C H A P T E R 11. S T R U C T U R E T H E O R Y F O R D E R I V A B L E N E T S
If U is any collineation of C which leaves N invariant then Q is an element of P r L ( 4 ,K ) and hence permutes the points of C  N,permutes the lines skew to N and the set of lines which intersect N in a unique point. Furthermore, B permutes the planes containing N and likewise permutes the planes which do not contain N. Translating these statements back into the net terminology, it implies immediately that Q induces a collineation of the derivable net D. We thus have a homomorphism of Gz, onto PrL(4,K ) N . It there is a collineation of G.o which induces the identity on C then all lines skew to N are fixed which implies that the collineation fixes all points of D and is hence the identity. This completes the proof.0
11.1.1 TheTranslationGroup
of a Derivable Net.
Definition 11.2 Let V be a vector space over a skewfield K. A ‘transvection’ r is an element of GL(V,K ) such that there exists a vector d and r(v)  v is in ( d ) f o rall v E V . In this case, r is said to be (in the direction of d W e shallalsousethe term ‘transvection’ torefertotheelement in PrL(V,K ) induced by a transvection in I’L(V, K ) . A transvection of P G ( 3 , K ) will fi. a plane pointwise. W e shall call the pointwise f i e d plane, the (axis of the transvection’. W e shall be considering transvections with axes planes containing the distinguished line N of the 3dimensional projective geometry C corresponding to the derivable net and in direction d a point of N . W e shall use the notation r* to denote the element in the quotient group corresponding b y an element r of FL(V,K ) . Definition 11.3 A collineation Q of the derivable net D shall be called a ‘translation’ if and only if c j k e s each parallel class in C and fixes some parallel class of lines linewise. The subgroup of D which isgenerated b y the set of translations of D shall be called the ‘translation group’ of D and denoted by T . Note that not every elementof a translation group is necessarily a translation.
11.1. THE COLLINEATION GROUP OF A DERIVABLE NET.
117
Lemma 11.4 Let T be a transvection of C with axis a plane a$ and in the direction d , a point of N . Then r* is a translation of the derivable net D , Proof: A transvection fixing a,f pointwise, translated into net language will fix all lines of the parallel class a. Let v be any vector in the associated 4dimensional Kvector space V . Then (v) is a point of C and assuming that v N,then v corresponds to a line of the derivable net V. So, ~ ( v ) v E ( d ) . This means that T* fixes each plane of C containing N which implies that r* leaves invariant every parallel class p E C of the derivable net V. Since T* fixes all lines of a,T * is a translation of D.0
Remark 11.5 We denote by T the translation group of the derivable net V
T : (r*; T* is a transvection with axis a; and some direction in N , V a E C) . We denote by Ta thesubgroup of transvections with f i e d axis a; with direction in N . Clearly, Ta is a normalsubgroup of T as T leaves each parallel class invariant. Note that any transvection with directiond will leave invariant any plane of C which contains d . Theorem 11.6 (1) The translation group T of a derivable net D is transitive on the points P of D. (2) The translation group T is normal in the full collineation group Gz, of D and G.0 = TGo where Go is the full collineation group which f i e s a given point 0 of P . Proof Let P1 and P2 be two incident points of P of the derivable net D. Let a denote the parallel class containing P1P2.There is a unique subplane 7rpl,h of B of the derivable net D containing the two indicated points. We consider the subgroup of transvections T,,[,,,,] with axis a$ and in the direction [7rpl,pJ. Note that any such transvection willfix D("&,&)and , ~a$) = J pointwise. induce a collineation fixing V ( T ; ~ n
118 CHAPTER 11. STRUCTURETHEORYFORDERIVABLENETS Since D ( T $ ~ ,is~ a) Desarguesian projective plane, the form of the group may be determined and we shall do this in due course. But, it is also clear that this induced translation group is transitive on the ‘affine’ points of D ( T & ~ i.e. ) , transitive on the points of D ( T & , ~ )  J .Any such translation group is also transitive on the set of ‘&ne lines’ which are incident withany given ‘infinite point’ of J. Taking the infinite point different from [ n ~ , p l ] then these affinelines correspond to lines which are skew to N . However, the lines skew to N incident with a fixed point of a$ are points of ~ a ,which p~ are incident with a given line of a. This says that thepoints of the subplane which are incident with a given line of a are in a transitive class under the induced translation group. Moreover, it says that there is a ‘translation’ of D which maps P1 onto P2. Thus, what we have proved is that for any two distinct points of the net D which are collinear, one may be mapped onto the other by a translation of the net D. Furthermore, the group induced is regular on the affine lines of D(x&,&)which says that if a element g of T maps P1 onto P2 then, in fact, g is a transvection. Now assume that P1 and P2 are distinctpoints of P which are not collinear. Let a and p be distinct parallel classes and form the lines P1a and P2p. Since these two lines are not parallel, there is a unique intersection p3 = p1a n 4 p . By the above argument, there is a translationof the netmapping P1 onto P3 and there is a translation of the net mapping P3 onto P2. Hence, there is an element of the generated group T which maps P1 onto P2. Moreover, we have shown that TaTpfor any fixed a,p and a # p E C is transitive on the points of the derivable net and hence, T is also transitive on the points of the derivable net. Since T is generated by normal subgroups of GD,it follows that T is also a normal subgroup. Let 0 be any point of D and let g E GD map 0 onto P1 and r E T map P1 onto 0. Hence, it follows that gr E Go so that g E TGo which is defined as T is a normal subgroup.0
11.1.2 A Representation for the TranslationGroup. Let 0 be a fixed point of the derivable net D and denote this point by the line MO skew to N in the projective geometry C . Then, in the corresponding bdimensional Kvector space V , MO becomes a 2dimensional Ksubspace as does N and we note that the two
11.1. THECOLLINEATION
GROUP OF A DERJVABLENET.
119
vector subspaces can only intersect in the zero vector of V . Choose a basis for V so that MO is represented by
and N is represented by
Furthermore, elements of V are represented by (21,$ 2 , z3,24) with scalar multiplication ~ ( 2 122, , 23,z4) = ( Q I C ~ az2, , az3,az4) for Q E K . Note that withthis agreement, matricesacton the right of vectors. Inparticular, elements of GL(V,K) acton the right when representing collineations of the corresponding derivable net.
Theorem 11.7 With the above choice of basis, the translation group of the derivable net V acting on the $dimensional vector spaceassociated with the 3dimensional projective space C combinatorially representing V has the following form:
Hence, T is Abelian and generated by any two transvection groupsT,,Tp f o r Q,P E C,a# P . Proof: Let ( q , z ~ , z 3 , z 4represent ) a vector (point) of V (of C ) and consider the matrix action by a 4 x 4 Kmatrix M be ( z l , z 2 , ~ 3 , z 4 ) M . We see that the indicated group in the statement of the theorem is the product of the following two transvection IZroups:
1
1 0
0 0 0 1 0 0 0 0 1 0 c3 cq 0 1
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CHAPTER 11. STRUCTURE THEORY FOR DERIVABLE NETS
which has axis the plane givenby the 3dimensional vector space whose equation is 2 4 = 0 in the representation of V and
[B H
0 c0; 0 "li;....14,\3,4 1
which has axis the plane given by equation
z3 = 0
in the representation of
V. Note that, T3T4 is Abelian and ~(c,,,,,0,0)~(0,0,c3,q)= T(q,CZ,cQ,q)' Now let T be any transvection with direction d = (yl,y2,0,0) E N for yi E K and i = 1 , 2 with axis a i . Since T fixes N pointwise and T is linear, must be represented in the general form
(Note that if the (1,l) and (2,2) entries are not 1, we may multiply by an appropriate scalar matrix by obtain the required form.) However, 7(21,2 2 , 23, 24)  (21,227 23, $4) = X(Yl, 030) for X E K. This means that
Since this isvalidfor any for i = 1,2,3,4, it follows that a = d = 1 and c = d = 0. This shows that any element of the transvection group may be written as a product of transvections from T3 and T4 respectively. Theactual transvectionswithin the transvectiongroup consist of r
9
r
the elements with corresponding X(,,,,,)(y1,y2) for all Z ~ , Z ~ , X ( , ~ ,E, ~K~ where the 4 s and yis are fixed and (yl, y2) # (0,O) and the subscript notation is meantto indicate that the X's depend on (23,24). In particular, if c3 = c4 = 0 then 2 3 ~ 1= X(,)yl
11.1. THECOLLINEATION GROUP OF A DERIVABLENET.
121
and x3C2 = X(,,,,,)y2 so yi = 0 if and only if q = 0 for i = 1,2. So, cj = 0 implies that A(,3,z4) = 2 3 ~ 4 4 : ~ where i # j (assuming the transvection is not the identify).Furthermore, if clc2 # 0 then we must have qy,’ = c2yZ1 = 1. So, in all cases, one may merely take (yl,y2,0,0) = (c1,c2,0,0) to produce the transvection T ( , ~ , ~ ~ , with o , ~ ) direction (cl, c2,0,0).Similarly, ~ ( 0 , 0 , ~ has ~ , ~ direction ) (O,O, CS, c 4 . 0
Corollary 11.8 ( l ) A n y line of C which is skew to N has a basis of the form (2) The element T
(
{(dl,d2,1,O),(d3,d4,0,1)}. , , ~ ~ ~ of, ~T ~maps , ~ ) the line with basis {(dl,d2,l,O),(d3,d4,0,1)}
onto the line with basis (3) The translation group T acts regularly on the points of the derivable net D. Proof (1) Any line W of C which is skew to N intersects the planes x4 = 0 and x3 = 0 in different points with uniquely determined representations (cl, c2,1,0)and (CS, c4,0,1) respectively. These constitute the unique basis of the form required in the statement of (1). Then (2) is immediate and (3) follows since the points of the derivable net are the lines of C skew to N . U
Remark 11.9 To distinguish between points of the net and skew lines with basis {(dl,d2,1,O),(d3,d4,0,1)}, we shall use the notation P(d1, d2, d3, d4) to represent points. I n this case, the element T ( ~ ~ , maps ~ ~ , the ~ ~point , ~P(d1, ) d2, d3, d4) onto the point P(& cl, d2 c2, d3 c3, d4 c4). Hence, T may be represented acting in D by the following:
+
+
+
+
{P(z1,22,23,24)  t ~ ( ~ l + C 1 , ~ 2 + C 2 , ~ 3 + C 3 , ~ 4 , C 4 ) } V q E K and i = 1,2,3,4.
122 CHAPTER 11. STRUCTURETHEORY FOR DEMVABLE NETS 11.1.3 The Group of QuasiSkewandSkew Perspectivities. We are trying to view a derivable net as sitting in a vector space over a skewfield F. We have the points P(&, d2, d3,d4)connected to a vector space. How do we make this set into a ‘left’ vector space, say over K? A natural way to do this is to let scalar multiplication aP(d1, d2, d3,d4) be either P(ad1, ad2, ad3, ad4) or P ( d l a , d p a , d 3 a , d 4 a We ) . shall show that both possibilities are consistent and allow the set of net points to be viewed as a left vector space over a skewfield. However, in the first case, the skewfieldis K and, in the second case, the skewfieldis K o p p . In order to see that such scalar multiplications are connected via the geometry of the derivable net within the 3dimensional projective geometry, we consider certain collineations of the net. We begin with linear bijections.
Lemma 11.10 Define g~p:(zl,~2,~3,~4)H(~1~r~2~,~3~,~4P)fOralla,PEK{0}.
Then (g,R,p ; % P E
is a group of linear bijections, i.e.
K)
in GL(4,K ) N which f i e s both MO and
N. If K is a field then g z p f i e s all ‘points’ of the two lines MO and N . Proof: Simply note that
maps to
Definition 11.11 A elementwhich f i e s all points of twoskewlines PG(3,K ) is called a ‘skew perspectivity’.
of
11.1. THE COLLINEATION GROUP OF A DERTVABLE NET.
123
W e shall call the group (s,R,p ;
 (0))
EK
7
the group of ‘quasiskew perspectivities’ fixing MO and N

Lemma 11.12 Define the following ‘semilinear’ bijection: L &,p : ( 2 1 ,2 2 , 2 3 , 2 4 )
( m ,QZ2, px3, P 2 4 ) f o r all a,p E K  (0) and alp E Z ( K ) (center of K).
Then (9:p ; a,P E K
is the group of skewperspectivities *ng subspaces. Proof In order that g&
 (0))
MO and N pointwise as projective
be a semilinear mapping, we must have that
a6al = 66 = psp1 where Q is an automorphism of K . Hence, it follows that a’@ must commute with S for all S E K so that alp E Z(K).O
Proposition 11.13 (l) If a o P(d1, d2, d3, d 4 ) = P(d1, d2, d3, d4)g& then the points P(d1, d2, d3, d4) f o r m a left vector space over K O P P , which is, of course, a natural ‘right’ vector space over K . (2) If a P ( d l , d z , d s , d4) = P(dl,d2,d3,d4)g&1 then the points P(d1, d2, d3, d4) f o r m a left vector space over K which is, of course, a natural ‘right’ vector space over K o p p . Proof We note that ( d l ,& , l ,0)
++
(&a,d2a, L O ) and ( 4 , &,O, 1)

(dsa, d 4 a , 0,1)
under g?, so that a o P(dl,dz,d3,d4) = P(dla,dza, d3a,d4a) for Q! E K (with t h i understanding that when Q is zero, the associated mapping is the ‘zero’ mapping).
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C H A P T E R 11. S T R U C T U R E T H E O R Y F O R DERIVABLE N E T S
Note that g 2 1 g i l = we have
so that with K o p p multiplicationdenoted
by
‘0’
Note that we may make the group
into a field with the inclusion of the zero mapping in the standard way so that it follows that we obtain a right vector space over K o p p , This proves (1).The proof of (2) is analogous to theproof of (1) and isleft for the reader  gllalgl,6l R R .U to complete but notesimply that gc(06)1 
Theorem 11.14 (1) If the points set of the derivable net is made into left vector space over K o p p then the lines thru the origin have the form
a
x = 0, y = Sa: f o r all S E K
and are left KOppsubspaces whereas the subplanes incident with the origin have the form
and are not all necessarily left KOppsubspaces. (2) If the points set of the derivable net is made into a left vector space over K then the subplanes incident with the originare of the form ra11a2 =
((aa1,aa2,pal,pa2) ; a,P E K }
and are left Ksubspaces whereas the lines thru the origin have the form x=O,y=Sx f o r a l l S E K
and are not all necessarily left Ksubspaces. Proof Inthenotation of the above proposition, we have identified p(dl, d2,d3 d4) with (dl 7 d2 d3 d4) The lines of the net incident with 0 are thepoints of C incident with the line MO skew to N . That is, they are the ldimensional left Ksubspaces ((O,O, T , S)) for fixed r, S E K not both zero.
11.1. THE COLLINEATION GROUP OF A DERIVABLE NET.
125
However, we want to use our Pnotation. What this means is that a line incident with 0 is a set of Ppoints which, when represented in the projective geometry C, is a set of lines skew to N and each represented by a unique basis of the form {(cl, C Z , ~0), , (c3, c4,0, l)}. In order that the Ppoints lie on a line incident with 0 in the derivable net, we must have the corresponding lines skew to N intersecting MO in a common point say (O,O, a,@)for a,P E K and not both zero. What this means is that
so that we must have a(c1, c2) = P(c3, c4).
If P p'a.
= 0 then c1 = c2 = 0. Otherwise, (c3,cq) = p(c1,cz) where p =
Now translating this back into the Ppoint notation, we have that the lines of the derivable net incident with 0 have the form
Now we consider the Baer subplanes of the derivable net which contain the point 0. These correspond in C to the planes of C which intersect N in a point and contain MO. Using the notation established above, let MO CB ((dl, d2,0,0))be an arbitrary plane of C which contains the line MO. We now want to suppress the P notation. We assert the following: (1) Then the set of lines of this plane which are skew to N have the following for their bases:
(2) The Baer subplanes of the derivable net which are incident with 0 = (O,O, 0,O) have the following form: rd,,da
= {(adl,ad2,Pdl,PdZ) ; a,P E K }
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CHAPTER 11. STRUCTURETHEORYFORDERIVABLENETS
for ( d l ,d2) fixed and not (0,O). Since we have seen that (1) is true, we consider the validity of (2). The plane as a vector subspace is {(adl, adz,p, y) ; a,p, y E K} which is generated as a vector subspace by the vector ( d l ,d2,0,0) of N and the subspace MO where (dl, d2) # (0,O). The points of intersectionwith x4 = 0 and x3 = 0 are (a&,ad2,1,0) and (pdz, pd2,0,1) respectively. Now translating into the Ppoint notation shows that the subplanes have the required form. Now simply use the two scalar multiplications defined above to verify the remaining parts of the theorem. This completes the proof.0 In the following, we shall use the second possible definition of scalar multiplication so that we have a left vector space over K where K is the corresponding skewfield of the projective space associated with the derivable net.
11.2
TheStructureTheoremforDerivable
Nets.
Theorem 11.15 Let V = (P,,C,B,C,z>be a derivable net. ( l ) Then there exists a skewfield K such that thepoints of V may be considered the vectors of a left Ksubspace representedin the form (XI, 22,yl,y2) and let = (x1,52), = ( 3 1 , ~ 2 ) . (2) The lines of the net which are incident with the zero vector 0 are left Z(K)subspaces which may be represented by the following equations: X
= 0, y = 6 x V 6 E K.
(3) The Baer subplanes of the net which are incident with the zero vector are left Ksubspaces which may be represented by the following equations: rdl,d2
= { (ad17 ad2 7 @l
pd2) ; a,P E K }
where (dl 7 d2) # (0,O). (4) The remaining lines and Baer subplanes are translates of the lines and Baer subplanes incident with the zero vector. Proof: Note that in the statement of the theorem, we are suppressing the Fnotation. We also note that the r d l , d 2 are clearly left Ksubspaces and the mappings g&l : (21, Q, y1, y2) (6x~,S22,6y1,6y2)for 6 E K leave each such Baer subplane invariant (note here we are acting on Fpoints not the vectors of the vector space associated with the projective space).
11.3. T HAEN S W E R
TO THE REGULUS QUESTION.
127
11.3 The Answer to the Regulus Question. Now we may answer the ‘regulus’ question for finite derivable nets.
Corollary 11.16 Every finite derivable net corresponds to a regulus in some 3dimensional finite projective geometry. Proof There is a finite field K 31 GF(q) such that the lines incident with the origin have the basic form x = 0 , y = 6z. Since Z ( K ) = K in this case, we see that both the lines and the Baer subplanes incident with the zero vector are 2dimensional Ksubspaces. Hence, there is a corresponding regulus in P G ( 3 , K).
11.4 The Solution to the CoordinatesProblem. It was mentioned previously that the early researchers in the area of deriv& tion asked if a necessary and sufficient condition that an affine plane be derivable is whether the coordinate structure is a right 2dimensional vector space while writing the slopes for the derivable net on the right. More generally, one could ask if this is true for the coordinate structure for any net containing a derivable net. We now show that the answer is, in fact, ‘yes’. Theorem 11.17 Let V be a derivable net and let A be any net containing V in the sense that the pointsof A are the points of D and all parallel classes of V are parallel classes of A. Then there exists a coordinate svstem (Q, f, for A such that lines of D may be represented in the form x = c, y = x a b where b E Q and a E L for L a subskewfield of Q . Furthermore, Q is a right 2dimensional vector space over L. S)
+
Proof In the representation of the derivable net in the Structure Theor e m l i s t e d a b o v e , l e t Q = K @ K . LetD=(O,O),i=(l,O). Furthermore, let (x = 0) = {(O,O,Yl,Y2) ; Yi E K,i = 1,217 (y =D) = ( I C ~ , X ~ , O , O; E K,i= 1,2}
128 CHAPTER 11. STRUCTURE THEORY FOR DERIVABLE NETS and
We now form the coordinate system for the net on

~t: = 0,
y = U, y = X, @,U) and
(i, 7).
+
To form a b, for a, b E Q, one forms the line parallel to y = z thru (U,b) on z = U. Form the line thru the point (a,a) parallel to x = 0 and determine the intersection point of these latter two lines to obtain thepoint (a,a b). We know that the lines of the net A are translates of the lines thru the origin since there is an associated translation group. Thus, for b = ( b l , b 2 ) for bi E K ,i = 1 , 2 , the line e b defined by
+
in the original representation must be the line y = x For y = (91, y 2 ) and z = (21,Q ) , we obtain
+ b.
so that (Q, +) in the coordinate system is vector addition on K @ K. We now determine a a for a E Q and a E K @ (0). First determine (U, a)on z = 'ii and form the line y = a parallelto y = 'ii thru @,a).Then determine (i,b) on y = 'ii and form the line z = 1 parallel to z = U thru (i,'ii). Now intersect these latter lines to obtain the point (i, a). Finally, form the line y = z a as the join of the points @,U) and

e
(Tal
It is clear that the line of the original representation ((z1,z2,az1,az2}; xi E
K,i= 1,2}
contains (O,O,0,O)= @,U) and ( l , O , a,0) = (i, a) (using a in two ways). Hence,y=z~ais((z~,z~,az~,az~};z~~K,i=1,2}. Now to obtain the point (a,a a),one finds (a,a) on y = z, and forms x = a parallel to z = 'ii thru (a,a). Then intersect y = z . a and z = a t o
11.4. THESOLUTION
TOTHECOORDINATESPROBLEM.
129
obtain (a,a a)=
= W} n { ( ~ ~ 1 , 9; 2 E)
{(WWW,QQ) ; xi E from which we have
a ' a = (a1,Q
= 1,2}
) ' a = (aal,aa2).
Let L denote the dual skewfield with multiplication
'0' defined
by
aob=ba where justaposition denotes skewfield multiplication in K . It remains to show that (Q, f, is a right vector space over L. Note that we have three multiplications the notation of which is suppressed for one (that of K ) . That is, by a a in the coordinate system and a also in L then a  a = a o a . Tosee thisnote that asp = (a,O) = (Pa,O) = (1,o)  (pa)= i (pa) = (Pa)= (a0 p). As above, we have S)
a . a = (aa1,aaz)= (a1 0 a,a2 0 a).
Then (U '
a)' p = ((a10 a,a2 0 a))' p
= (@(a10 a),P(a2 0 4 ) = ((a10 a)0 P, (a2 0 a)0 P) =(alo(aoP),a2O(~oP)) =a.(aOp)=a.(a*p).
Furthermore, (a
+ b)
'
+
a = ((aha2) ( h ,b 2 ) ) * a = ((a1 h , a 2 b2)) ' a = (+l b1),a(a2 b 2 ) ) = (0% aa2) ( a h ,ab21 =al.a+b*a.
+ +
+
+
+
+ P) = a a + a
Similarly, it follows that a (a all a,P E L . 0
p for all a E Q and for
Chapter 12
DUAL SPREADS AND BAER SUBPLANES Previously, we have discussed spreads in PG(3,K ) which are coverings of points by lines. The dual concept and the subject for this chapter is the nature of coverings of planes bylines. Furthermore, we generalize these notions to arbitrary projective spaces.
Definition 12.1 Let V = W @ W be a vector space over a skewfield K andlet S be a spread.Hence, S i s a set of mutually disjoint vector subspaces which are Kisomorphic to W , whose union is V and which pairwise generate V . Thus, a spread has the property that each nonzero vectorof V is incident .with exactly one component of S . A ‘dualspread’ S* of V is a set of mutually disjoint vector subspaces which are Kisomorphic to W which pairnoise generate V but has the property that each hyperplane of V contains exactly one component of S. Note that we are not claiming that a dual spread is a spread but it is, of course, a partial spread. At this stage, it is not totally obvious that there exist duals spreads in general. However, at least when K is a field and V is finite dimensional over K, we may construct dual spreads from spreads as follows:
Proposition 12.2 Let S be a spread of V where V is a n (n = 2m)dimensional vector space over a field K where n < 00. Let V* denote the dual
131
132
CHAPTER 12. DUAL SPREADS AND BAER SUBPLANES
space of linear functionals of V into K . If W is a subspace of V , denote by W* the dual space of W within V * . Then S* = {W* ; W E S ) is a dual spread of V*. Proof: In this situation, n = 2m and the components (elements) of S are mdimensional vector subspaces of V . The duality mapping is inclusion reversing in the sense that for subspaces W E 2 we have 2 ' S W*. Thus, we see that the dimension of W* is 2mdim W and, in particular, if W is mdimensional then so is W*. Given any nonzero vector vo, there is a unique component W of S containing v0 so that (vo)* 2 W*. Since V* is isomorphic to V , givenany hyperplane 31, there is a nonzerovector v0 such that (vo)* = 31. Hence, given any hyperplane of V * ,there is a unique component of S* which is contained in the hyperp1ane.O
Remark 12.3 When K is finite anyspread is a dual spread and conversely.
+
Proof: We give the proof in PG(3,q) so that a spread consists of q2 1 lines. Each linelieson q 1 hyperplanes (planes) and there are exactly (q4  l ) / ( q  1) = q3 q2 q 1 = (a2 l ) ( q 1) planes. Thus, the set of planes containing lines of the spread constitutes all possible planes. Thus, each spread is a dual spread. The proof that each dual spread is a spread is virtually identical to the above.0
+ + + +
+
+
Remark 12.4 (1) Suppose aspread S is also a dual spread then the dual spread S* is also a spread. In this case, if T S is the translation plane corresponding to the spread S and TSI is the translation plane corresponding to the spread S*, the question becomes: When is T S isomorphic to rs*? (2) Assume that both T S and TS* are translation planes and assume that K is a field. By theBasisDecompositionTheorem,'choosethe spread S represented in the form x = 0 ,y = 0,y = xM for M in spread set M . Assume that the dimensions of the associated vector spaces are finite 2m so that the M's may be considered m x m matrices with elements from a field K . Then x = 0 , y = 0, y = x M t where t denote the transpose of M for all M E M is a spread St. The spread St is called the 'transposed spread'. Fhrthermore, it is moreorless straightforward to show that St is S*.
CHAPTER 12. DUAL SPREADS
ANDBAER SUBPLANES
133
Proof: (Sketch) We shall show that St is a spread in the finite case and leave the fact that St is S*and the more general case to theinterested reader to complete. By the Basis Decomposition Theorem, since the cardinality is correct, it remains to show that the differences of the transposes of distinct matrices of M are always nonsingular. Since ( M  N ) t = M t  N t , this latter condition is immediate.0 We have defined spreads of vector spaces which then provide covers of the points of the associated projective geometries. We shall also refer to thesesets as 'spreads'. For example, a 'spread' of PG(3,K) is a set of mutually skew lines which cover all points. As such projective spreads are clearly equivalent to vector space spreads, we shall not distinguish between then and allow context to dictate which variety we are considering. The basic question is whether therearespreads which arenotdual spreads and we have seen that for such to exist we must look to the infinite fields or skewfields. In fact, for any infinite skewfield, there are spreads which are not dual spreads and dual spreads which are not spreads. The proof to this general result depends on certain aspects of mathematical logic and, in particular, transfinite induction. However, we shall give a elementary proof for countable skewfields. There are examples of spreads which are not dual spreads in PG(3,K ) and the reader is referred to Cameron [l81 (p.41,4.1.6) for more details. The spreads mentioned in Cameron are generalizations of those given in Bruen and Fisher [l71 for spreads over countable fields. The reader is also referred to Barlotti [4] and Bernardi [7] for additional background and details.
Theorem 12.5 Let K be any skewfield. Then there are spreads in PG(3,K). If K is an infinite skewfield then there are spreads in PG(3,K ) that are not dual spreads. Proof We may assume that K is infinite since for GF(q),there is always a quadratic extension field which implies that there arealways Desarguesian spreads in PG(3,q). The proof in general relies on transfinite induction. However, we shall give a sketch of the proof for countable skewfields which is adapted from
134
CHAPTER 12. DUAL SPREADS AND BAER SUBPLANES
a proof when K is a field by Bruen and Fisher [l71 who further construct regulusfree spreads. Let 7ro be any plane of PG(3,K). Enumerate the points Pi of 7ro and enumerate the points Qi of PG(3, K )  no for i = 1,2,3, ... . We shall show that there exists a spread S such 7ro does not contain a line of S. We note that no PG(2,K) can be covered by a finite number of points and lines since there are infinitely many points and lines in the projective plane. Form the line P1Q1 = P1 and note that P2 cannot be incident with el. Let 2 1 = Q1 and Qn2 = 2 2 be the first listed point not in 7ro which is not in the plane (el, P2). We see that the line P222 = t2 is skew to t l as otherwise, Qnz would be in the plane (el, P2). Let 2 3 be the first listed point Qj which is not a point of { t i ; i = 1,2}. Let Wi = Pi for i = 1,2. Now take the first listed point Pj = W 3 which is not in union of the planes ( t i , 2 3 ) for i = 1,2,3. In order to see that there is such a point, consider the quotient space V / ( 2 3 ) in the projective sense. This is a projective plane which cannot be covered by finitely many points and lines as K is infinite. Note the lines of this projective plane are the planes (e, 2 3 ) for lines t not incident with 2 3 . Hence, there is a point of the quotient space not incident any of the ‘lines’ ( t i , 2 3 ) for i = 1,2,3 and so there is a line incident with 2 3 which is skew to {ti ; i = 1,2}. Thus, there is a point of 7ro incident with a line thru 2 3 which is skew to the previous lines and there is a first listed point Pj = W 3 . Form the line W 3 2 3 = t 3 . Let W4 be the first listed point P k which is not a point of {ti ; i = 1,2,3} and repeat the procedure above. Since K is infinite, an easy induction shows that the above procedure produces an infinite set of lines which covers the points and such that no line of the set belongs to the plane T,,. Hence, the spread constructed is not a dual spread.0 We shall see that there is an intimate connection between spreads which are dual spreads and Baer subplanes. We have previously defined pointBaer and lineBaer sublanes of a projective plane and noted that when the projective plane is finite then every pointBaer subplane is also a lineBaer subplane. However, in the infinite case, this is no longer the case as is shown by Barlotti [4].
CHAPTER 12. DUALSPREADSANDBAERSUBPLANES
135
There are certain critical situations when it can always be guaranteed that a given subplane is a lineBaer subplane. Recall that we refer to an affine subplane ro of an affine plane n as pointBaer or lineBaer if the projective extension subplane nr is pointBaer or lineBaer respectively in the projective extension rE of the affine plane.
Theorem 12.6 Let S be a spread in PG(2k  1,K) for k finite. Let ns denote the associated afine translation plane obtained b y taking vectors in the 2kdimensional vector space over K as points and translates of the elements of S (taken as vector subspaces) as lines. Let no be a kdimensional Kvector subspace of T S . Assume that k is even and assume that nontrivial intersections with the spread elements with no are always k12dimensional Ksubspaces of ro. ( l ) Then no is an afine translation planewhich is a lineBaer subplane. (2) If k = 2 then a spread S is a dual spread if and only if any 2dimensional Ksubspace which is not a line of rs is a pointBaer subplane of r s and hence a Baer subplane of rs. Proof Note that the intersections of no with elements of S determine a spread in P G ( k  1,K) as the lattice of Ksubspaces of no. Clearly, this forces r0 to be an affine translafion subplane of ns. Let N,, denote the subnet of ns which is defined by the elements of S which nontrivially intersect no. Let L be any line of rs. If L is a line of N,, then the projective extension of L intersects the projective extension of no. Thus, assume that L is not a line of Nno. By the construction of ns, L = V, c where V, is in S (as a vector subspace) and c is a vector of rs. Thus, V, and no are mutually disjoint kdimensional Ksubspaces and since k is finite, V, @ no = ns as a Kvector space. In order to show that L contains a point of no,it suffices to show that c may be taken in no. Since c = W c* where v in V, and c* in no,then L = V, + c = V, + W c* = V, +c*. This proves (1). To prove ( 2 ) , assume that k = 2 and note similarly as above that any 2dimensional Ksubspace which is not a line has a spread of ldimensional subspaces automatically induced upon it so there is an induced affine translation subplane.
+
+
+
136
C H A P T E R 12. DUALSPREADSANDBAERSUBPLANES.
By (l),the subplane constructed from any 2dimensional Ksubspace is a lineBaer subplane of the projective extension. Assume that S is a dual spread. Suppose that 7ro is an affine plane which is a 2dimensional vector subspace with spread induced from S as above. Let b be any point of the projective extension of 7rs. Since the line at infinity is a line of the projective extension of the subplane 7ro, b may be assumed to be in 7rs. Let (b,7ro) denote the 3dimensional subspace generated by b and 7ro. Then, by assumption, there is a spread line V, of S (asa 2dimensional vector subspace) contained in (b, 7ro). Hence, V, n 7ro is a ldimensional Ksubspace. So, for d a point of 7ro  V, nno, then (b, 7ro) = V, @ (d). Thus, b is either in V, or V, d both of which are lines of no. Therefore, 7ro is a pointBaer subplane and, as observed above, also a lineBaer subplane so is a Baer subplane. Conversely, assume that every 2dimensional Ksubspace which is not a line defines a pointBaer subplane and hence a Baer subplane. Let C be any plane of the projective space PG(3,K) and consider C as a 3dimensional vector subspace of 7rs. Let 7ro be any 2dimensional Ksubspace of C. If 7ro is a line of S then certainly C contains a line of the spread. Thus, assume that 7ro is not a line of S and so defines a pointBaer subplane by assumption. If b is a point of Cno then C = (b,T,,).Since 7ro is a pointBaer subplane, there is a line of 7ro, V, d for d in 7ro, which nontrivially intersects b. However, 7ro n V, # 0, so it follows that V, = 7ro n V, @ ( b  d) c C . We have thus shown that any plane of the threedimensional projective space contains a line of the spread so that S is a dual spread. This completes the proof of (2) s i We now obtain the following corollary:
+
+
Corollary 12.7 Let S be any spread of PG(3, K) for K a skewfield.
Let
7rs denote the corresponding translation plane. Let x$ denote the projective extension and let V ( n $ )denote the dual projective plane. Then any 2dimensional Kvector subspace of 7rs which is not aline defines a pointBaer subplane in V(#.
Proof Note that any 2dimensional Kvector subspace defines a line
Baer subplane which when taken projectively and dualized becomes a pointBaer subplane of the dual translation p1ane.U
12.1. COORDINATEFREE DERIVATION.
12.1
CoordinateFreeDerivationand sion Question.
137
the Exten
We have seen that any finite derivable net corresponds to a regulus in some threedimensional projective space over a field K isomorphic t o GF(q).Furthermore, any regulus with q l lines may be 'coordinatized' by GF(q). Since any field of order q has quadratic extensions to fields of order q2, there are always Desarguesian affine planes which extend any finite derivable net. We have also seen that any derivable net may not be a regulus but is at least a pseudoregulus (see also the chapter on embedding) so it might be possible that there is a Desarguesian extension of any derivable net. We pose this formally: Given any derivable net, does there exist an affine plane which extends the net? In thissection, we provide a constructionwhich produces a nonderivable or derivable affine plane containing a derivable net depending upon the nature of a given spread. We shall state the theorem only when the spread is not a dual spreadwhereas a more general construction procedure is possible and is, basically equivalent to the coordinate approach showing that any rightquasifield of dimension two over a skewfield K is always derivable. We have seen in the chapter on coordinate structures that any dual translation %orK a skewfield, always contains a plane arising from a spreadin PG(3,K), derivable net. What is not known is whether the corresponding plane is also derivable and this depends on the nature of the subplanes of the derivable net having the Baer property in the affine plane which the above suggests hassomething to dospreads which are dual spreads. We shall consider these problems without the use of coordinates in this chapter. Moreover, this 'coordinate' free approach is both interesting and more suitable to the determination of specific derivable nets.
+
Theorem 12.8 Let S be a spread of PG(3,K) which is not a dual spread where K is an infinite skewfield. Choose a plane C of PG(3,K) which does not contain a line of S . Let 7ro be a line of C and write no = X , CBYo where X , and Yo are ldimensional vector Ksubspaces. ( 1 ) Then 7ro is a lineBaer subplane of the associated translation plane TS
Let 7rf denote the projective extension of ns and V ( r f )the dual of 7 r f .
138
CHAPTER 12. DUAL SPREADS AND BAER SUBPLANES.
Denote the parallel classes of TS containing the lines of S containing X , and Yorespectively by (m) and (0). Form the lines (m)o for all o in Yo. Dualize the tmnslation plane, sending (m) to the line at infinity of D ( T ~ ) and the line at infinity of the translation plane to (W)*. (2) The netR with infinite points(m)*,(m). for all v in Yois a derivable net in the afine dualtranslationplane Ss obtained by deleting (W) from D(T3 (3) The afine dual tmnslation plane Ss containing R is not derivable by the derivation of R. Proof: (1) is immediate from the previous section as C does not contain a line of S. In order to prove (2), we need to show that given any two affine points a, b of the affine dual translation plane such that the join ab is a line of a parallel class either (m)* or (m). for some v in Yothen there is a Baer subplane T a , b of the net R which contains a and b. Note that this simply says that the subplane has the same parallel classes as R. For an affine subplane with the same parallel classes as R to be a Baer subplane of the net, it is required only that the subplane be a pointBaer subplane as it is automatically a lineBaer subplane with respect to the net R since every line of the net contains one of the infinite points of the affine subplane. Considered projectively, it is required that the affine restriction of the dual subplane of the corresponding projective extension translation subplane be a lineBaer subplane in the translation plane. We have seen that any subplane of the translation plane which originates as a 2dimensional vector subspace is always a lineBaer subplane. Thus, to show that the net R is a derivable net, it suffices to show that there is a set of lineBaer subplanes of the translation plane which projectively contain the lines (m)v forall v in Yoand which dualize to appropriate subplanes of R. Given points a and b of the affine dual translation plane such that the join ab is a line of R, there is a Baer subplane Ta,b of R containing a and b if and only if there is a projective subplane of T: which contains the lines (m)v for all v of Yo, contains the line at infinity I , of T S and contains the lines a and b of T:.
12.1. COORDINATEFREE DERIVATION.
139
Suppose that a and b are (necessarily affine) lines of 7rs which are concurrent either on the line at infinity of 7rs or at a line (w)w,for some W,. Hence, it is required to construct a lineBaer affine subplane of 7rs which contains the lines a and b as well as the lines (m)w for all W in Yo. Note that thetranslation group withcenter (W), T(m), of the translation plane acts as a translation group of the dual affine translation plane and so fixes each parallel class of R. Moreover, the group T(m)acts regularly on the lines of each parallel class of 7rs distinct from (m). Since the lines a and b are concurrent at (W)if and only if they are infinite points of the corresponding affine dual translation plane,it may be assumed that there is a translation with center (m) which maps b onto a line which intersects the zero vector of the line (w)O. There are two cases depending on whether a and b are concurrent at an affine lifie or at the line at infinity of 7rs. The arguments are similar but we shall take these separately. If a and b are concurrent at the line at infinity, they belong to the same parallel class say 6. Assuming that b is 0 6 , take any point W on bn (w)wo for w o # 0 in Yo.There is a kernel homology group H of 7rs which acts transitively on the nonzero vectors of any ldimensional Kvector subspace and fixes each infinite point of 7rs. It follows that H fixes (m)O and acts regularly on the set of lines (m). for all W in Yo ( 0 ) .Take a n (w)O = A and form the 2dimensional Ksubspace (A,W )and note that this subspace becomes a lineBaer affine subplane 7r1 of 7rs. Moreover, the above remark shows that the subplane 7r1 contains the lines (m). for all W in Yoand as a projective plane contains the line at infinity. Therefore, the plane contains 0 ,W and A as points and OW = b projectively intersects the line at infinity at 6. Hence, it follows that 7r1 contains the line A6 = a , Thus, in the dual translation plane, there is a pointBaer affine plane of the net R which contains the points a and b. Now assume that the lines a and b are concurrent on a line ( w ) ~ ,First . # 0 of Yo. assume that W, Then, a n b = W is a nonzero vector and assuming again that bn(m)O = 0 and letting a n (w)O = A, again it follows that (A,W ) is a lineBaer subplane 7r1 which contains the points 0,W,A and hence the lines OA = a and O W = b . Now assume that thelines a and b are concurrent on (m)O and, without
140
CHAPTER 12. DUAL SPREADS AND BAER SUBPLANES.
loss of generality, assume that a n b is 0. Let W = b n ( o o ) ~A, = a t lW)..( for some W # 0 of Yo.Form the 2dimensional K space ( A ,W ) . Since the group H is regular on the lines (W). for all v # 0 of Yo, it follows that as a lineBaer affine subplane, ( A ,W ) contains the lines (m). for all U of Yo. Moreover, ( A ,W ) contains the lines O A = a and OW = b. Hence, in all cases, given affine points a and b of the affine dual translation plane such that the join ab is a line of R then there is a pointBaer subplane of the associated dual translation plane containing a and b. Since any pointBaer subplane of the net R is also lineBaer with respect to the net, it now follows that R is a derivable net. This proves (2). To prove (3), we need to show that the subplanes of R are not all Baer subplanes of the affine dual translation plane. This is equivalent to showing that not all subplanes of R are lineBaer in the dual translation plane. To construct the set of lines (oo)~, we used a 2dimensional Ksubspace ro. When extended and dualized to say D(T$),this subplane becomes a pointBaer subplane of the net R considered projectively. We assert that D(T$)is not a lineBaer subplane of the dual translation plane. Hence, this assertation is equivalent to the assertation that T~ is not a pointBaer subplane of the translation plane. So, suppose that T~ is a pointBaer subplane of T S . Recall that we obtained ro from a 3dimensional vector subspace C which did not contain a,line of the spread (asa set of 2dimensional Ksubspaces). Take any point A of C  T ~ If. there is a line of T~ as a subplane which contains the point A then the previous arguments show that there is a line of the spread contained in C. Hence, rIT0 is not a pointBaer subplane and thus, D(T$)is not a lineBaer subplane. So, although R is a derivable net which is contained in the affine dual translation plane, the &ne dual translation plane cannot be derivable by derivation of R. Note that an attempt at derivation would result in a structure whose lines are the lines of the affine dual translation plane which are not in R together with the Baer subplanes of R. However, since D(T$)is not lineBaer, there exists a line of the affine dual translation plane which does not intersect D(T$)and would.,not be considered in the same parallel class as D(T$)so that the derived structure would fail to be an affine plane. This completes the proof of the theorem.0
Remark 12.9 The general construction above also is valid for spreads which are dual spreads producing derivable dual translation planes.
12.1. COORDINATEFREE DERIVATION.
12.1.1
141
CoordinateFree Derivation.
Here we have provide a coordinatefree view of the derivation of a dual translation plane based merely on the definition of a derivable net. The previous method verifies that when there is a spread in PG(3,K ) for K a skewfield, there is an associated derivable net in the dual translation plane. What is not so completely clear ishow to consider this in that there are many and various ways to construct such a derivable net. Of course, many different coordinate leftquasifields are possible and hence many different derivable nets in various associated dual translation planes. The coordinatefree approach allows a more complete focus on the derivable nets obtained rather than the choice of coordinates although basically the two methods are equivalent.
Remark 12.10 W e have constructed derivable nets in nonderivable dual translation planes. Is it possible to construct derivable nets in nonderivable translation planes? Also, we have not determined conditions on spreads which force them to be dual spreads. W e shall see in the section on conical and ruled spreads over fields that there isa wide class of spreads which are dual spreads so that the derivation of dual translation planes with spreads in PG(3,K )produces afine planes in a variety of cases.
Chapter 13
DERIVATION AS A GEOMETRIC PROCESS We have raised the question whether the process of derivation is geometric. We have associated with a derivable net a 3dimensional projective space over a skewfield K . Since the derived net is also a derivable net, there is a corresponding 3dimensional projective space over a skewfield K D .In the present context, we might ask how the two combinatorial structures relate within the threedimensional projective spaces corresponding to a derivable net and its derived net. We recall that if ( K ,f , is a skewfield, the 'opposite skewfield' K o p p is defined by K o p p = ( K ,f , 0) where a o b = b . a. S)
13.1 The Annihilator Mapping. Let V denote a left vector space over a skewfield K E (K, +, such that the lattice of left vector subspaces defines the projective space II. Let V* denote the dual space of V (the space of linear functionals). We note that V* is a right vector space over K by defining scalar multiplication as follows: If f is in V* define fa by f a ( z ) = ~(II:). for II: in V . Clearly, V* is a left KoPP ( K ,+, 0)space where ab = a b = b o a. Now consider the annihilator mapping Idefined from V onto V* where if W is a left Ksubspace of V then WL = (f E v,* ; f(w)= 0 'dw in W ) is a left Koppsubspace of V*. S)
143
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CHAPTER 13. DERIVATION AS A GEOMETRICPROCESS
Let I I ( P G ( V ,K ) )and II*(PG(V*,KOPP) be theprojective spaces defined by the lattices of left vector Ksubspaces of V and left KoPPsubspaces of V* respectively. Note that theannihilator mapping is dimension inverting and when considered on the associated 3dimensional projective spaces ll and II* maps lines of II onto lines of II*. Furthermore, points and planes of II are interchanged with planes and points of II* respectively &S W E 2 for W,2 left Ksubspaces of V if and only if ZLC WL for ZL, W* left KoPPsubspaces of We shall be interested in ‘Correlations’of projective geometries but we shall see that these are all basically given by semilinear mappings in the case at least when K is a field.
v*.
Definition 13.1 A ‘correlation’ a of a projectivegeometryis a bijective mapping on the set of projective subspaces which reverses inclusion. A ‘polarity’ is a correlation of order 2. In particular, a polarity of P G ( 3 , q ) will map points to planes and lines to lines. Theorem 13.2 Let V be a finite dimensional vector space over a field K and let V* denote the dual space of V . Let A denote the annihilator mapping of V onto V* and use the same notation for the mappingof PG(V,K)into PG(V*,K) in the sense that points map to hyperplanes, etc. Let a be any correlation of PG(V,K ) . Then a d i s induced from a semilinear bijection from V onto V*. Proof Note that PG(V,K)and PG(V*,K ) are isomorphic since both V and V* are isomorphic vector spaces over K of the same finite dimension. We now may apply the Fundamental Theorem of Projective Geometry.0
Definition 13.3 Let r be any semilinear bijection of V onto V* and let K be afield. Define a mapping from V x V into K as (v,u) c,v(ur) (w,u). Such a mapping is called a ‘sesquilinear’ f o r m and there is an theory of such forms that characterize the possible correlations. Then, define W l o = {W E V ; (v,u)= 0 f o r all U E W } . It follows that the mapping W c,W L o is inclusion reversing. Hence, depending on the semilinear bijection, it is possible to construct correlations and/or polarities. Define a (nonsingular) sesquilinear form abstractly as a mapping from V x V into K satishing the following properties:
13.2. A 'GEOMETNC' I N T E R P R E T A T I O N .
(i)
(W,U + W ) =
(v, U )
145
+ ( v , ~ and )
= apU(v,u) f o r all v,u,w E V and for all a,P E K where (ii) (av,pu) is an automorphism of K .
U
(iii) h r t h e r , (v,u) = 0 f o r all U E V implies v = 0 andanalogously, (v,u) = 0 f o r all W E V implies that U = 0. Remark 13.4 Correlations on finite dimensional projective spaces over fields are equivalent to sesquilinear forms. Proof Simply note that every sesquilinear form defines a semilinear mapping from V* onto V . 0
13.2
A 'Geometric'Interpretation.
Let II and 11* be the projective spaces associated with the fourdimensional vector space V4 over the skewfield K and its dual space V$ over K o p P both taken as left spaces. Let V denotea derivable netand assume that II corresponds combinatorially to V. Let N denote the line of II corresponding to the construction of the derivable net V. Let N* = Nl. Then, clearly points of II  N map to planes of II* which do not contain N*. Similarly, planes containing N are mapped to points of 11* incident with N' and points of N are mapped to planes of ll* containing N*. Hence, we have the following theorem:
Theorem 13.5 Let V be a derivable net and let V4 denote the left vector space over a skewfield K such that the 3dimensional projective geometry PG(3,K ) combinatorially corresponds to V . Then, the annihilator mapping defines a derivable net D* which corresponds naturally to the 4dimensional left KOPPvector space V,. and to a 3dimensional projective geometvPG(3,K O P P ) . If we identify the lines of V4 and V$ via the annihilator mapping then we have made an identification between the points of V and V* which provides an interpretation of 'derivation'.
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CHAPTER 13. DERJVATION AS A GEOMETRIC PROCESS
We have interpreted aderivable net within athreedimensional projective space I'I isomorphic to PG(3,K) and its derived net similarly interpreted within a threedimensional projective space I T 'isomorphic to PG(3,KOPP). The interpretation may be considered 'geometric' in that it says something about projective geometries. However, we can do better than connecting with dual spaces when K is a field by the consideration of correlations or polarities. Suppose that D is a finite derivable net of order q2 and degree 1 q and let C N PG(3,q) with distinguished line N correspond combinatorially to D.The q4 points of D correspond to theq4 lines skew to N and the q 2 ( l + q ) lines correspond to the (q4  l)/(q  1)  ( q 1) points of PG(3,q) which do not lie on N. Of course, the q2(1 q ) Baer subplanes of D correspond to the q2(1 q) planes of C which do not contain N. Note that there are the same number planes as points in C and the q 1 parallel classes of D correspond to the q 1 planes of C that contain N. Now consider any polarity cr of C which leaves the line N invariant. The points and hyperplanes are interchanged by Q whereas the set of lines is left invariant. The set of points which are on C N is interchanged with the set of planes that do not c0ntain.N. Translating the mapping conditions back into net language, we see that the set of points of the net is left invariant and the lines and Baer subplanes of the net are interchanged. Everything is the same if the associated skewfield K is a field so we assume this in the following. We simply need to produce the desired polarity. We do this by appealing to sesquilinear forms. 0 0 0 1
+
+
+
+
+
Let ( v , ~ )= v
+
1
[H
ut when considing v and
U
as $vectors
over a field K and ut denotes the transpose of U. Let v = (211,212,213,214) and = (U1,U2,U3,U4) so that
U
+
(v,u> = v4u1 f U3u2 v2u3 f v1u4 = = U4211 uQv2 U2213 f U1214 = (v,U) since K is a field.
+
+
Hence, (v,u) = 0 if and only if ( v , ~ = ) 0. It is now essentially immediate
13.2. A ‘ G E OI NMTEETRRPI RCE’ T A T I O N .
147
that the induced mapping I, is a polarity of the associated projective space. We take the basis setup exactly as in the structure theory chapter and let h = N @ MO and note that
Hence, we have the following geometric interpretation of the derivation process.
Theorem 13.6 Let V be a finite derivable net or a derivable net written over a fieldandlet C denotethecorresponding3dimensionalprojective space with distinguished line N . If Q is a polarity of C which leaves invariant the line N then the derivation process maybe realized by the action of the polarity; the set of points of C  N is interchanged with the set of planes not containing N,and the set of lines skew to N is left invariant; the sets of lines and Baer subplanes of the net V are interchanged and the set of points of the net is left invariant. Remark 13.7 So, we have connected the derivation process, at least when the associated projective space is overa field, by a purely ‘geometry’ concept, that of a polarity of a projective space. Hence, we m a y abuse notation slightly and say that ‘derivation is a polarity’.
Chapter 14
EMBEDDING We have seen that any derivable net may be embedded into a projective geometry in the sense that the set of points of the net become lines of the geometry and the lines of the net become points. In this chapter, we determine the nets which can be embedded into projective geometries in this fashion: the pointsand lines of the net become lines and points of the geometry. In point of fact, part of the net is embedded into an affine geometry so wewill be concerned with nets which can be embedded into either affine or projective geometries. Our main chapter isconcerned with the socalled 'subplane covered nets' and these become relevant here.
Definition 14.1 Let R = ( P ,C , B,C,Z) be a net with setsof points P , lines C , subplanes B, parallelclasses C and incidence Z. W e shall say that the net is 'subplane covered' if and only if given any two distinct points a and b which are incident with a line then there is an afine subplane Ta,b containing a and b whose parallel class set is exactly C . Definition 14.2 W e shall say that a net (P,,C,z> may be 'embedded' into an afine or projective geometrg Q if and only if each point of P becomes a line of Q and each line of C becomes a point of Q and incidence within Z implies incidence in the corresponding embedding. Sometimes, it is said that the 'dual net' is embedded in the geometry Q. Remark 14.3 W e notethat,
of course, any derivable net is a 'subplane
149
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CHAPTER 14. EMBEDDING
covered net’ which may be embedded into a projectivegeometry.Furthermore, we have proved Cofman’s theorem which states that certain subnets of a derivable net may be embedded in afine geometries. In the following, we give examples of subplane covered nets which ultimatelyare canonical. For this discussion, we consider possibly infinite dimensional projective geometries. We first consider nets which arise from projective geometries.
Theorem 14.4 Let V be any Kvector space where K is a skewfield and let C denote PG(V,K ) . Let N be any codimensiontwo subspace. ( l ) W e define the following set R = ( P , L ,B,C,z>of ‘points’ P , ‘lines’ C , ‘subplanes’ B, ‘parallel classes’ C and ‘incidence’ Z: Q is the set of lines of C skew to N , L is the set of points of C  N , B is the set of planes of C that intersect N in a point, C is the set of hyperplanes of C that contain N , and Z is the incidence set induced from the incidenceof C . T h e n R is a subplane covered net. (2) Now consider the afine geometry AG(V,K ) obtained by the deletion of any hyperplane A which contains N . Let R denote the structure defined by the deletion of the hyperplane A from C and the required point and line deletions. W e denote this structure by R = (P, C,C,Z), so that Q is the set of lines of C which are not in A and which are skew to N and hence equal to P , L is the set of points of C  A, C = C  A, and Z is incidence of Z restricted to the indicated sets. T h e n R is a net which may be obtained f r o m R bythedeletion of a parallelclass. We note that the points of R and the points of R are identical. Proof: Clearly, if R is a (subplane covered) net then R is a net as well. Let A be any point of R. We need to show that there is a unique line of each parallel class incident with A. However, since a point A is a line of C skew to N, it follows that A intersects each hyperplane of C containing N in a unique point of C N.
14.1. EMBEDDING NETS INTO AFFINE SPACE.
151
If A and B are points of R then they are lines of C skew to N and hence intersect in at most one point of C  N which is then a line of R. Hence, R is a net. If A and B are collinear points of R then there is a unique point of C which is their intersection as lines of C. It follows that (A,B ) is a projective subplane of C which intersects N in a unique point [7r0]. Deleting this point and dualizing this structure, we obtain a unique subplane 7r0 which shares all of the parallel classes of C. Hence, R is a subplane covered net.0 We now show that any net which may be embedded into an affine space is isomorphic to a net of type R and any net which may be embedded into a projective space is isomorphic to a net of type R.
14.1 Embedding Nets into Affine Space. Definition 14.5 W e define a net which may be embedded into a projective space C , as in the previous theorem part (l), a ‘codimensiontwo net’. A net R which may be embedded into an afine space as in part (2) is called a ‘lclass retraction’ of a codimensiontwo net R. The codimensiontwo net R is also said to be a lclass extension’ of the net R. Theorem 14.6 Any net which may be embedded into an afine space is a lclass retraction of a codimensiontwo net.
Proof Let R = (P, C,C,Z) denote the net. We shall give a proof as a series of lemmas. Assume that II is an affine space such that P is a subset of lines of II and C is a subset of points of II. Remark 14.7 We adopt,aswe did previously,the usage of ‘netpoint’, ‘netline’ and ‘affline’) ‘affpoint’ to denote the net points and lines which are to be taken as lines and points in the associated afine space. However, there are affpoints and afflines which may not correspond to netlines and netpoints and we need to distinguish between these. Lemma 14.8 Let Q and p be distinct parallel classes of C“ and let a and b be netlines of Q and p respectively so are then affpoints of Let ab denote the a#line in I I joining a and b. W e note that ab is the netpoint a f l b in the net.
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CHAPTER 14. EMBEDDING
The affpoints in II of ab intersect each set y of C (as a set of affpoints in I I )in a unique a#point and ab C UpGCp . Proof Since a n b is a netpoint of the net which is incident with a netline of each parallel class, the affpoints on ab in II are all netlines of the net so the lemma follows immediately.0
Lemma 14.9 Let Q be any parallel class of C as a set of a#points of II. Let P, and Q, be two netlines of Q. Form the a#line P,&, of II joining these affpoints of II. Then all affpoints of II on P,Q,are netlines of a in the net. Proof Let T be any affpoint of P,Qa and assume that T is not in a; that is, is not a netline of the net. Let S be any parallel class distinct from Q and let P6 be any netline of the net in S. Form the affine plane 7ro in II generated by the affpoints P,, Q, and PS. Let UpECp = II*. If the affline P6T intersects II* in an &point of II*  (6) then it follows from Lemma 14.8 that Tis in U*.If T is in II* and not in CY then P,T = P,&, is incident with a unique &point of a (asa set of af€points) by Lemma 14.8which is clearly a contradiction. Thus, we may assume that P6T n II* C S. Let Pp = Pap6 n p and &p = &,P6 np for p # a or S. The affline PpT cannot be parallel to both P,Qp and Q,Qp = Q,P6 unless all of these afflines are equal so that P6T contains an affpoint of p (asa set of affpoints), contrary to the above. Hence, PpT intersects one of these afflines P,Qp or &,&p in an affpoint of II* and this affpoint must be in p. Thus, by uniqueness PpT intersects both of these afflines at &p. Now let U be the affline of 7ro which is incident with the affpoint P, and parallel to affline PpT. Hence, U must intersect QaQp = &,P6 in an affpoint or is affparallel to &,&p. In the latter case, PpT = &,&p since T ,Pp, &p are affcollinear. However, this implies that T is a netline of Cby Lemma 14.8. Since U must intersect &,&p and all such affpoints are in II*, assume that the intersection is in Q (as a set of affpoints), so it must be Q, by uniqueness. However, this would mean that T is in U but U is &parallel to PpT and distinct from it.
14.1. EMBEDDING NETS INTO AFFINE SPACE,
153
Thus, U intersects &,&p in anaffpoint of some parallel netclass distinct from a and hence U intersects each parallel netclass of C in affpoints of IT*. In particular, U contains a unique affpoint R8 of 6 (as a set of affpoints). Form Rp = (RsP, = U ) np. Since the affpoints Rs, P,, Q, also generate rT0, we see that the affline RpT must intersect one of the afflines P,Qp and QaQp in an affpoint of p and hence intersects these afflines at &p. Thus, the affpoints RP,Pp, &p, T are all collinear. However, Rp is incident with the affline U which is parallel to PDT.This contradiction completes the proof of the 1emma.O
Lemma 14.10 (i) For any parallel class a E C of the net, a is a n a f i n e subspace of the afine space IT. (ai) UpECp = 11* is an a f i n e subspace of IT. Proof To prove (i), we note that since given any two affpoints p and q of a,the affline joining these points pq contains only affpoints of CL! by Lemma 14.9 so that a is closed under joining by affpoints and hence a is an affine subspace of II. Let p and r be affpoints of IT*. If both are in some parallel class (as lines of the net) p then (i) applies to show that the affpoints on pr are in TI*. If both are not in the same parallel class as netlines then since they are each in some parallel netclass, it follows from Lemma 14.8that the affpoints on p r are in TI*. Hence, 11* is an affine space.0
Lemma 14.11 In t h e a f i n e space IT*, the afine space a is parallel to the a f i n e space p f o r a n y CL!,^ E C. Proof: Let P, and Q, be distinct affpoints on an affline U in a where a is considered an affine subspace of II*. Since this argument will be symmetric, it suffices to show that there is an affline v of p in 11* which is parallel to U. Let Pa be an affpoint of S as a subspace of IT* where S is distinct from (Y and p. Recall that Paps corresponds to a netpoint of the net and there is a netline of the net Pp and hence an &point of p as a subspace of IT* incident with P&. Similarly, there is an &point &p of p incident with &,P6 and note that &p # Pp. Form the affine plane (Pa,Q,,P,) (of affpoints) and note that P,&, and PpQp are disjoint since a and p are disjoint on &points. Hence, it follows that P,&, and PpQp are parallel.
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CHAPTER 14. EMBEDDING
So, given any affline U of a,there is an affline W of p such that U is parallel to W. Thus, a is parallel to p. 0 Now let C* denote the projective extension of the affine subspace II* and let H denote the hyperplane at infinity extending 11* to C*. Extend a to a projective subspace a+ of C* and let N“ denote the hyperplane at infinity of a+ in H . Note that N”= Np = N for all a,P E C as a is parallel to p. Further denote H by (W).Note that every affline of II* is either parallel to an af%line of a or intersects a nontrivially. Hence, a+ is a hyperplane of C* and Af is a projective subspace of codimension two (a hyperplane of a hyperplane of C*). Thus, {a+,(W) ; a E C} is the set of hyperplanes of C* which contain the codimensiontwo subspace N. Nowwe have seen previously that the combinatorial structure R = (P,C,C,B,T) of the set of ‘lines’ C as the set of points of C*  N, the set of ‘points’ P as the set of lines of C* skew to N, the set of ‘parallel classes’ C as the set of hyperplanes of C* containing N, and the set of ‘subplanes’ B as the set of planes of C* that intersect N in a point, with naturally induced incidence is a subplane covered net and is a codimensiontwo net. Clearly, R is a lclass retraction of R. Hence, we have completed the proof.0
14.2 Embedding Nets into Projective Space. Remark 14.12 W e shall now consider embedding nets into projective space. W e shall use the notation of ‘netlines’and‘netpoints’,‘propoints’and ‘prolines’ to incidate whether the points or lines are considered within the net or the projective space. Theorem 14.13 A n y n e t R = (P,C,C, B,z> which may be embedded into a projective space is a codimensiontwo net. Proof Let Pa!and Q, E a as propoints (lines of the net and points of the projective space C*). Let Pa E 6 # a for a,S E C. Let T be any propoint of the projective space on the proline P,&, which is not in a. We may use exactly the same argument as in the proof of the embedding of a net into affine space to see that PaT n (UpEC p = II*) is contained in S. We see that there is a direct analogue of Lemma 14.8 of the proof of the &ne
14.2. EMBEDDING NETS INTO PROJECTIVE
SPACE.
155
space embedding in this case, Hence, let Pp and Qp be in p and incident with Pap6 and Q& respectively. Similarly, PpTn 11* C p. Hence, PpTn QaQp must be Qp so that Pp, T and Qp are collinear. Suppose there is a propoint S of PQQa a U {T}. Form PgS and argue as above to note that Pp, S and Qp must be collinear so that ST = PaQa = PpQp which is acontradiction since when there are propoints of different parallel netclasses of the net on a line of the projective space, there is a unique propoint from each parallel netclass on this proline. Thus, there exists at most one propoint of POQa which is not in a (as a set of propoints). Note that PpQp f l P,&,, is a propoint R which cannot be in p or a since if it is in a then RPp contains a unique propoint on a. Furthermore, it follows that this propoint is in the intersection of prolines PpQp for all p E C for certain propoints Pp,Qp of p. Let (a)denote the projective subspace generated by a. Note that we have proved that every proline of (a)contains exactly one propoint not in a and this propoint is in (a)= N, We assert that any plane 7ro as above, in the argument for the affine embedding, is ‘punctured’ by N. To see this, we note that PpQp is the only proline of 7ro in (p) since otherwise, then no would be a subspace of (p) and we have seen that prolines of (p)contain at most one propoint which is not in p and P6Qa contains propoints of the space of each parallel netclass of the net. Two prolines of 7ro intersect within 11* unless one is the unique such proline in say (a) and the other is in a different parallel class ,&space (p) (as a proline) and hence there is a unique propoint of intersection of all such prolines. Thus, 7ro is punctured by N. So, we have shown that any proline joining two propoints of a intersects N and any proline joining two propoints of N lies in N so that N is a hyperplane of (a).We note that a n N is empty since any propoint of N may be obtained as the intersection of prolines say PaQa and PpQp. Clearly, this intersection propoint cannot be in a. Thus, we have shown that (a)= a U N. Now we form UQEC (a)= UaEC(a UN) = II* U N.We assert that this is a projective subspace. We note that given two propoints p and q of this set if p and q are both in 11* then the preceding shows that the propoints of pq are in II* U N. If p is in a (as a set of propoints) and q is in N then
naEC
156
C H A P T E R 14. EMBEDDING
pq is a proline of (a)= a U N . Hence, Il* UN is a projective space and note that (a)is a hyperplane of the projective space since every proline intersects (a)nontrivially. Furthermore, N is a codimensiontwo subspace. Thus, we have shown that the net R is a codimensiontwo net.0
14.3
CodimensionTwo and PseudoRegulus Nets.
We have seen that there is a natural algebraic representation of a derivable net obtained by the embedding of the derivable net into a projective space. There is a sirnil&algebraic representation of a subplane covered net.
Theorem 14.14 Let W be a. left vector space over a skewfield K. Let V =
[email protected] W Let.
and
(y=6z)={(z,6~);z~WforafiedS~K}. Define R = ( P ,C ,C , B,z> as follows: The set of (points’ P is V , the set of ‘lines’ L is the set of translates of x = 0 and y = 6 x V 6 E K and theselines(incidentwith (0,O)) are representatives for the set of parallel classes C . The set of subplanes is the set of translates of the following set of subplanes 7rW = { (020, /?W) ; a,p E K} f o r f i e dW E W  { (0, 0)) and incidence Z is the natural, induced incidence. Then R is a subplane covered net which is called a ‘pseudoregulus net’. When theaisociated’skewfield K defining the pseudoregulw isa field K, the pseudoregulus is called a ‘Kregulus’ or merelya ‘regulus’ if K is understood from context. Proof: We note that 7rw is clearly isomorphic to the Desarguesian affine plane coordinatized by K with affine points (.,p) for all a,@ E K and lines x = y, y = 6z+p for y, a,,8 E K. It is a simple matter to check that 7rw is a subplane of the net butthe reader is cautioned to note that theintersections of subplanes with lines y = 6z of the net are not always ldimensional left Ksubspaces.0
14.3. CODIMENSIONTWO AND
PSEUDOREGULUS NETS.
157
Previously, we have realized a vector space form of a regulus in PG(3,K ) where K is a field. We want to generalize this notion.
Definition 14.15 Consider PG(2n1, g ) . A set of g+l (n1)dimensional projective subspaces of P G ( 2 n  1,g ) which is covered by a set of lines with the property that if a line intersects at least three members of the set then the line intersects all members and the set is said to be a n '(n 1)regulus'. W e call such covering lines 'transversal lines'. More generally, af K is a field, an '(n  1)regulus in P G ( 2 n  1,K ) ' is a set of (n  1)dimensional projective subspaces which is covered bya set of transversal lines (with the definition given above) and such that the transversal lines are themselves coveredby the set of ( n  1)dimensional subspaces. Still more generally, for a vector space V = W @ W over a skewfield K , it is possible t o speak of Wreguli in PG(V,K ) . A 'Wregulus' is a set of projective subspaces isomorphic to the projective version of W which are covered by transversal lines and such that the transversal lines are covered by the set of subspaces. When there is no chance for ambiguity, we shall call a Wregulus simply a 'regulus' or a 'Kregulus' when we want to place emphasis on the associated field. Theorem 14.16 Let V = W @ W be a vector space over a skewfield K of dimension at least four and let W be a Wregulus. Then ( l ) K is a field and (2) W m a y be represented by
x = 0, y = Sx f o r all 6 E K where x = (xi)and y = (yi) are vectors in W with respect to some basis for W for xi, yi E K and Sx = (Sxi) for S E K and for all i E X where X i s a n index set for the basis. Proof Choose a basis so the three distinct members of the regulus are
x = 0, y = 0, y = x (in vector form) where the associated vector space is a direct sum of the first two and x = (xi)and y = (yi) are vectors in W with respect to some basis for W where zi,yi for i E X are in K .
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CHAPTER 14. EMBEDDING
Let L denote a line of the net intersecting the three indicatedsubspaces. Hence, we may write L as (((ad),( O j ) ) , ((Ok,), ( b i ) ) ) as a 2dimensional vector subspace where ai, bi are in K for all i E X and (0,) for z E X represents the Ovector of W . Since L intersects y = z in a ldimensional subspace, it follows that (bi) = b(ai) for some nonzero element 6 of K. Thus, it follows that L = ((a(ai),P(G(ai))) ; a,P E K and i E X}. Since L is a Kvector subspace, it follows that a(&) = S(az) for all a E K. Now vary ai and bi to realize that any such vector subspace of the regulus representing a line of the net must have the form y = pz for some p in K. However, this implies that K is commutative. Thus, we have proved both (1)and (2).0 Hence, we have:
Corollary 14.17 The vector f o r m of a regulus is also a pseudoregulus. A pseudoregulus corresponds to a regulus if and only if the associated skewfield is a field. Proof: Apply the previous definition and theorem.0 We now concentrate on codimensiontwo nets. Theorem 14.18 Any codimensiontwo net is a pseudoregulus net. Proof: Let C denote the projective space into which the codimensiontwo net may be embedded and let N denote the projective subspace of C of codimension two so that the points P, lines C, parallel classes C and subplanes l3 of the net R = (P,C, C, B,z) are thelines of C skew to N, points of C  N , hyperplanes of C containing N and planes of C that intersect N in a point respectively. Let K denote a skewfield such that C is the latticeof left vector subspaces of a vector space V over K. Since N is of codimension two, there is a basis W for N which extends to a basis W U {el, ez} for V so that representing V with respect to this basis, we have
V = ~ ~ ~ N = { ( ( z 1 , z z ) , ( y i ) ) } V z 1 , z z ~ K a n d V3 y i ~ K N = {(vi) ; yi E K} and i E X for X an index set for the basis W.
14.3. CODIMENSIONTWOAND PSEUDOREGULUS N E T S .
159
Note that for a vector v and scalar a,and v = ((zl, 4 , (pi)) then crv is simply given by ((aq,az2),( a y i ) ) . The proof of this result follows along similarly as the algebraic description of a derivable net. We shall give slightly abridged versions of the proofs. The reader should be able to fill in the details afterlooking back at the corresponding proofs in the derivable net case.
Lemma 14.19 Any line of C which is skew toN has a unique basis of the following form:
+ v2)
{el
(1,o)), ((d2,i), (091))) f o r vectors vj = (dj,i) of N , j = 1,2, i E X.
E
Proof: Suppose the line M as a 2dimensional vector subspace has basis {ael+ be2 w1, cel de2 w2) where a, b, c, d are in K and w1 and w2 are vectors in N . Note that none of the pairs (a,b), (a,c), (c,d) or (b,d) can be (0,O)since M is skew to N . Now M must intersect the subspace N (el) in anonzero vector say a(ael+be2+wl)P(cel+de2+w2) = (aapc)el+(abpd)e2+(awlpw2) for some nonzero elements a and p of K so that a b = pd. Now it may be noted that aa  p c # 0. Hence, there is a point of the form e2 + W ; in M where E N and similarly, there is a point of the form el +w;C in M such that the two vectors are linearly independent. To show uniqueness, let (el w1, e2 w2) = (el + V I , e2 + v2) where wi,vj E N for i , j = 1,2. Let el 201 = p(e1 V I ) y(e2 v2) where p , y ~ K T. h e n ( p  l ) e ~ + y e ~ E N s o t h a t i t m u s t b e t h a t p = l a n d y = O which implies that w1 = v1 and similarly we have w2 = v2.0
+
+ +
+
+
+
+
+
+
Remark 14.20 If M is a line of C which is skew to N with unique basis of the f o r m {el + V I ,e2 v2) f o r vectors V I ,v2 of N , we denote M b y the tuple (VI, 212) when we consider M as a point of the corresponding net.
+
We now find a suitable vector space associated with such tuples.
Theorem 14.21 Any codimensiontwo netR corresponding to a projective space C over a skewfield K with associated vectorspace V and codimensiontwo subspace N admits PrL(V,K ) N as a collineation group.
160
CHAPTER 14. EMBEDDING
Proof: Simply note that the stabilizer of N leaves invariant the set of lines skew to N , the set of parallel classes which contain N , the set of points of C  N , and the set of planes which intersect N in a point.0 We have previously defined transvections.
Proposition 14.22 The subgroup T of G L ( V , K ) Nwhich is generated b y the transvections with axis a hyperplane containing N and in the directions d in N may be represented as follows:
Proof First consider the transvection T with axis N @ ( e l )with direction If T maps e2 onto e2 a(di) for a E K then we must have 7 ( 2 1 , 2 2 , (pi)) = (z1,22,(pi) +z2a(di)). Similarly, the transvection p with axis N @ (e2) and direction (si)has the form p(z1,~2, (pi)) = (z1,52,(pi) zlb(gi))for b E K . Let T I ,T2 denote the two groups of transvections listed above with axes N @ (el) and N @ ( e 4 respectively. We shall show that the full transvection group is TlT2. Let 7 be any transvection with axis a hyperplane H = N @ (ael+ be2) containing N and direction ( d i ) and suppose 7 maps el onto el + a(di) and (~i)+zl(adi)+sz(bdi)) , maps e2 onto ez+b(di). Theny(zl,zz, (pi)) = ( z l , ~ and y is clearly in T1T2.0
+
(ai) on N .
+
Remark 14.23 We note that not every element in T is a tmnsvection. In particular, in orderthat T be a transvection, it must be that ( d i ) and (si) generate the same ldimensional Ksubspace. Also, T induces an isomorphic subgroup in PGL(V, K)N as there are no nontrivial elements in T which fi all ldimensional Ksubspaces. We now have: Corollary 14.24 If M is a line of C skew to N with basis {el +v1, e2 +v2} then the group T maps M to the set of all lines with bases of the form { e 1 + v l + w l , e 2 + v 2 + ~ 2 }f o r a l l w 1 , w 2 E N .
14.3. CODIMENSIONTWO AND PSEUDOREGULUS NETS.
161
Thus, what we have shown is that we may identify the points of the net as ordered pairs of elements of N. This set admits a ‘translation group’ T which is the group generated by the transvections with axes the hyperplanes of C containing N and in the directions in N. Note that T acts on N @ N in the standard manner (2, y) H (z a, y b) for all a, b E N. It is easy to check that the scalar mappings z H kz for k E K are semilinear mappings of I’L(V, K) and are exactly those semilinear mappings that fix each ldimensional Ksubspace. Hence, any semilinear mapping p which fmes MO = (el,e2) pointwise as a projective space must have the form z I$ kz for some k E K acting on MO. Similarly, if p fixes N pointwise as a projective space then p acting on N is of the form z H sz for some S E K. Hence, p(z,y) = ( k x ,sy). In order that p be semilinear, we require that p(az, cry) = awp(z,y) where W is an automorphism of K. Hence, we must have kakl = aw = S C Y S  ~if and only if kslis in the center of K.
+
+
Proposition 14.25 The subgroup of PI’L(V, K)N which fixes the line MO andthe codimensiontwo subspace N pointwise(asprojectivespaces) is isomorphic to the group acting on N @ N represented as follows (76
; 76(2,W) = (62, SW) v6 E Z(K)  (0))
.
Proof: Suppose we represent an element T in the form ~ ( zy), = (kz,sy) on MO @ N and x E MO , y E N where kls = 6 E Z(K) the center of K. Note that a line of C skew to N with vector basis {el w1, e2 w2) is mapped to {kel S W ~ kea , S W ~ } .It follows that the tuple ( ~ 1w2) , is mapped to ( 6 ~,6w2) 1 .U
+
+
+
+
Proposition 14.26 The set of lines of C skew to N can be made into a vectorspaceoverthecenter of K, Z(K). firthemnore, if Q is any .fixed point of MO then the set of lines of C skew to N and containing Q is a Z(K)subspace.
+
Proof: Let Q = (re1 se2) for some fixed r, S E K. A line with basis {el+wl, e2+~2} that contains Q must satisfy rw1+9~2= 0 since rel+sez = a(e1 w l ) @(e2+ w2) forces r = a,@ = S and awl ,Ow2 = 0. But, rw1 + sw2 = 0 if and only if 6(rwl + S W ~ )= r6wl + s6w2 = 0 where 6 E Z(K).O
+
+
+
CHAPTER 14. EMBEDDING
162
We have a similar situation as in the derivable net case when we would like to define the points of the net as vectors over a skewfield. When K is a field, the above mappings will allow a natural definition of netpoints as left vectors over K. However, if we define the mappings SZp : (
(..,
[email protected])
for all a,P E K
W )
 (01,
we note that (S:@ ; % P E
K  (0))
is a group within GL(V,K),v which also fixes MO. We have the netpoints represented by
{el + W ,e2
+ ~ 2 =) { ( ( d l l i )(,L O ) ) , ((d2,iL
(0,1))) for vectors vj = (dj,i) of N,j = 1 , 2 , i E X.
We use the notation P(d1,i,dz,i)to represent the netpoints similarly as in the derivable net case where X has cardinality two. Note that

which means that
S$l : P(dlli,d2,i>
W d l , i ,Sd2,i).
Hence, just as in the derivable net case, we may consider the points of the net as the vectors of a left vector space over K by defining scalar multiplication by the image of g&1 and extending the mappings to a field isomorphic to K. We now may complete the proof of the theorem providing the algebraic characterization of codimensiontwo nets. Let MO be the zero vector in question identified with a particular point of the net. In the above proposition with Q = (re1 + se2) let (T, S) = ( 1 , O ) . Then the lines with bases {el w1, e2 z u 2 ) containing Q satisfy w1 = 0.
+
+
14.3. CODIMENSIONTWO AND PSEUDOREGULUS NETS.
+
163
+
If r = S and S = 1 then the lines with bases {el w1,e2 w2} containing Q satisfy w2 = S q . Using the notation established above, we have the representation for the lines of the net which are incident with the zero vector. Since the net is a translation net, all lines are translates of the lines incident with the zero vector. The subplanes of the net are the planes of the projective space which intersect N in a point. Any such plane MO @ (w1) which contains MO (the zero vector as a point of the net) must have a basis of the form {el,e2,w1} forsome w1 E N. The intersections of this plane with the hyperplanes N @ ( e l ) and N @ (e2) are the subspaces ( e l ,w1)and (e2, w2)respectively. The lines joining a point from each of the two subspaces have bases of the form (ael Pw1, Se2 ywl) for all a ,P, y, S E K . In order that such a line be skew to N,we require that a6 # 0. Hence, the set of all such lines skew to N have bases {el awl, e2 Pwl} for all a,p E K. This translates to the set of points of the net having the form
+
+ +
+
{(QWl,PWl)
; a,P E K ) .
Note that clearly this set is a left Ksubspace whereas the lines incident with the zero vector are not necessarily left Ksubspaces.
Remark 14.27 Note that analogous to the derivable net situation, we may define a left vector space over K o p p so that the lines of the net incident with the zero vector are left KOpPsubspaces whereas the subplanes incident with the zero vector are not necessarily left KOPpsubspaces.O
14.3.1 Embedded Nets Are PseudoRegulus Nets. By combining our previous results, we have the following theorem:
Theorem 14.28 Any net which may be embedded in a projective space is a pseudoregulus net. Corollary 14.29 Any finite net which may space is a regulus net.
be embedded in a projective
Proof A finite skewfield is a field so a finite pseudoregulus (in vector form) is a regu1us.O
Chapter 15
CLASSIFICATION OF SUBPLANE COVERED NETS Just as there are many examples of affine planes containing derivable nets, there are similarly various examples of affine planes containing subplane covered but not derivable nets. Separating the net from the plane, we are interested in the structure of any subplane covered net. In this our main chapter, we provide a complete characterization of the nets which are covered by subplanes. We have previously associated a projective space with a derivable net and shown it is possible to completely characterize derivable nets by this embedding. We have also discussed arbitrary nets which may be embedded into projective space and have shown that such nets are codimensiontwo nets which are, in turn, pseudoregulus nets. What we have, of course, not shown is that any subplane covered net can be embedded into a projective space. We point out that a ‘subplane covered net’ has essentially the same properties of a derivable net with the term ‘Baer subplane’ replaced by the term ‘subplane’. Recall that a Baer subplane in an arbitrary net is .a subplane such that every point lies on a line of the net and every line contains a point of the subplane (in the projective setting). Thisproperty of Baer subplanes is vital
165
166
CHAPTER 15. CLASSIFICATION
to the construction of an associated threedimensional projective space and becomes the main obstacle in considering the problem in the arbitrary case where ‘subplane’ replaces ‘Baer subplane’. However, this obstacle may be overcome once it is realized that within any subplane covered net, there is always a derivable ‘subnet’ withinwhich the subplanes areBaer. We are then able to use our derivable net embedding theorem as a tool inthe embedding of an arbitrary subplanecovered net into a projective space. To be clear, we recall our notation and assumptions.
Remark 15.1 Let R = ( P ,L, 8,C,Z) be a subplane covered net where the sets P , L,B,c, Z denote the sets of points, lines, subplanes,parallel classes, and incidence respectively. Note it is assumed implicitly that there is more than one subplane for otherwise any afine plane would be asubplane covered net. Furthermore, occasionally we shall refer to the setof parallel classes C as the set of infinite points of the net. If P is an afine point and a is aparallelclass, Pa shall denote the unique line of a which is incident with P. Also, notethatgiven a pair of distinctpoints P, and Q which are collinear in N then there is a subplane n p , ~which contains P and Q and which has C as its set of infinite points. Proposition 15.2 Under the above assumptions, the subplane unique subplane of B which contains P and Q.
T ~ , Q is the
Proof: Let R be any point of the subplane which is not on the line P&. Then RP and RQ are lines of distinct parallel classes say a and @ respectively. Then RP = Pa and RQ =
[email protected] and R = Pan
[email protected], any ~ is not on the line P& may be obtained as the intersection point of n p , which of the lines in {PS ; S E C} and in {Qp ; p E C}. Similarly, any point of P& may be obtained as the intersection of lines Ra and Pp for a particular point R (of intersection as above) for certain a, p in (7.0 Our first main section is devoted to a theorem which states that once two distinct subplanes of 8 share at least two distinct lines then the lines are parallel and the two subplanes share all of their lines of this parallel class.
15.1. SKETCH
OF THE EMBEDDING.
167
15.1 Sketch of the Embedding. We have seen previously that any subplane covered net which may be embedded into a projective space is a codimensiontwo net. Keeping in mind that it is our intention that we have such an embedding, the following is a sketch of how this comes about. If R is a subplane covered net which may be embedded then the ‘point’ set of the projective space contains the line set of the net and the ‘line’ set of the projective space contains the ‘point’ set of the net. The reader might study the diagram at the end of the chapter which provides a mental picture of the net embedded into the projective space. Furthermore, several ‘net diagrams’ are given that are indispensable for the understanding of the theorem. The main parts of the theorem are as follows: I. Any two planes of the net that share two lines of a parallel class share all of their lines. (Two distinct ‘points’ of the projective space are incident with a unique ‘line’,) 11. There are subnets which are derivable netsin the sense that the subplanes of the subnets are Baer within this structure. (Two skew ‘lines’ in a projective space generate a unique three dimensional projective space.) 111. If a is a parallel class, the set of lines of the net not in a can be made into an affine space A, as follows: The ‘points’ are the lines of the net not in a,The ‘lines’ are of two types: A line of the first type is obtained by defining a ‘line’ to be the set of lines on a parallel class p # a of a subplane of the net. A line of the second type is the set of lines of the net not of a incident with a point of the net. The ‘affine planes’ are also of two types; If D is a derivable subnet and p a parallel class let Dp denote the ‘affine plane’ induced from the derivable subnet. The second type is defined by the induced structure from any subplane. Each parallel class of lines p # a becomes an affine hyperplane ‘Hp of
&. This approach would parallel the framework set up for the classification of derivable nets. However, it is possible to bypass the proof that the d a ’ s are affine spaces and prove only that the llp’s are affine spaces. IV. Parallelisms are defined on the ‘lines’ and ‘planes’ of the Xa’s so that the projective space extensions Np of the hyperplanes llp of the affine
168
CHAPTER 15. CLASSIFICATION
spaces A, are equal; N" = Np for all a,,f?E C. Notice that the hyperplane ' H p is not in d p . V. So, one visualizes the set of parallel classes of the subplane covered net as a set of 'affine' hyperplanes of a putative projective geometry such that the deletion of any one of these is an affine geometry. Tying together the affine spaces A, with the binding N = N" or rather tying the hyperplanes together for all a E C produces a projective space C.R where N is a hyperplane ofH ' ,andH ' ,UN is a hyperplane of ER;N is a codimensiontwo projective subspace where the set P of points of R is the set of lines of ER skew to N, the set L: of lines of R is the set of points of ER N,the set B of subplanes of R is the set of planes of E.R that intersect N in a single point and the set C of parallel classes of R is the set of hyperplanes of that contain N. Note that, although we are basically extending sl, to a projective geometry, we choose not to consider the construction in this manner since the ' ,is a hyperplane not in sl, but in any parallel class a as a hyperplaneH d p for ,t?# a since strictly speaking the projective extension of A, need not contain the projective extension of H ' ,. So, the proper interpretation of this construction is the quilting together of a set of 'affine' hyperplanes by the common binding of their 'common' projective extensions to construct a projective geometry. Thus, the mental picture is of a tying together by a codimensiontwo projective subgeometry of a certain set of hyperplanes of the putative projective geometry.
15.2
The ShareTwoTheorem.
Theorem 15.3 If T I and are subplanes of B that sharetwolines parallel class a in C then the subplanes share all of their lines on a.
of a
Proof We shall give the proof as a series of lemmas. Note that is possible that there are infinitely many parallel classes. We begin the proof assuming that there are at least five. But, even ifwe have finitely many parallel classes, it is still possible to have infinitely many subplanes. The reader is encouraged to work thru the proof using the figures.
15.2. THE SHARE TWO THEOREM,
169
a
P
Lemma 15.4 Let the cardinality of C be c 2 5 , Then the subplanes T I and 1r2 have at least c  2 common lines of the parallel class a. Proof Let z and y be common lines to T I and 7r2 in the parallel class a. Let z1 and 22 be lines of parallel classes p and S respectively where a, P , S are mutually distinct and lines of T I ,~2 respectively. Let L1,M1 be z1 n z, z1 n y respectively so that T I = T L ~ , M ~Similarly, . let L2, M2 be 22 n z, 22 n y respectively so that 7r2 = TL,,M,. Note that (L1, M I } and (L2,Mz} must be disjoint in order for the subplanes x1 and ~2 to be distinct. Let W = z1 n 22.
170
CHAPTER 15. CLASSIFICATION
Also, note that if G is a point of a subplane no then any line G6 for 6 E C is a line of no; the lines thru G are lines of no. So, it follows that W is a point of the subplanes ~ L ~ and , L~ ~M ~ as, , Mfor~ example, z1 and z2 are lines thru L1 and L2 and thus lines of the subplane T L , , L ~ . We point out that such subplanes exist since L1, L2 are collinear with x and thus, the intersection point W is a point of the subplanes. It is pointed out that W a must be distinct from L l a = z and from M1a = y since n1 and 7r2 are distinct. (If W a = z then L1 = L2 = W so that n1 = ~ L ~ and , M n2 ~ = ~ L ~ share , M ~a point W and two lines z and y which implies the two subplanes are equal.) Choose any point U on W a distinct from W and in ~ L ~ , Hence, L ~ . U and L1, and U and L2, are collinear. Choose any line T I not equal to y thru M1 and intersect W a in R I . Since W a and T I then become lines of T M ~ , M ~ , it follows that R1 and M I , and R1 and M2, are collinear. Hence, T I = R1M1 and there is a line R1 M2. Thus, we have the lines UL1, UL2,R1M1, and R1M2. Note that at this point, it is not clear that the intersection points are affine; various of the lines could belong to the same parallel class. Extend the notation so that two parallel lines “intersect” in the infinite point p if and only if they belong to the parallel class p. Form UL1nR1M1 = S and ULznR1M2 = T.We may choose T I = R1M1 to be not parallel to UL1 but it is still possible that RlM2 is parallel to UL2. Let UL1 = LIP, and UL2 = L2p2 where /l1 and p 2 E C, A different choice of T I produces a different intersection point R1 on W a and all of these intersection points are collinear with M2 so the lines formed belong to different parallel classes. Hence, there is at most one line T I which will produce an intersection point R1 so that RlM2 is parallel to UL2. So, choose T I different from y, different from z l , not on &(i.e. not parallel to UL1) and distinct from a line (at most one) which produces the intersection point R1 such that &P2 = R1M2. This choice is possible since c 1 5. Then the intersection points S and T where S = UL1 n R1M1 and T = UL2 n RlM2 are both affine. Note that U and R1 are collinear (they are both on W a )and distinct for otherwise, R1M1 = UM1 and z1 would be lines of TL,,L, which intersect in M1 so that M I , L1 and L2 are points of
SHARE 15.2. THE
TWO THEOREM.
the same subplane which cannot occur if distinct subplanes.
171 7r1
=7
r ~ ,and , ~7r2~ = 7
r
~ are~
,
So, there is a subplane 7ruI~,.Allof the indicated lines are lines thru either U or R1 so that the intersection points S and T are in T U , R ~ . Furthermore, the point S is in = 7r1 as it is the intersection of two lines of this subplane, and similarly T is a point of T L ~ , M=~ 7 r 2 . Hence, ST is a line which must be common to both subplanes T I and 7 r 2 . However, if the subplanes are distinct then ST = Sa = Ta since otherwise, ST intersects z and y in distinct affine points which, by the remark on uniqueness above, forces the two subplanes to be identical. Thus, ST = Sa = T a is a line of a which is common to both subplanes. If ST = z then S = L1 and T I = z1. Similarly, ST = y forces S = M1 and T = M2 so that
TI
= y.
Hence, we have shown that with the exception of at most four lines thru M I , any such line T I produces a line of a common to both subplanes.
If Sa = S2a then SS2 = Sa = S2a = UL1 as S and S2 are both on UL1 which is a contradiction since UL1 cannot be in the parallel class a as U is a point of W a # &cr. Hence, each such line T I produces a distinct common line of 7r1 and ~ 2 Therefore, there are at least (c  4) 2 = c  2 common lines each of which must be lines of the parallel class a.0
+
.
~
~
172
CHAPTER 15. CLASSIFICATION
15.2. THE SHARE TWO THEOREM.
Lemma 15.5 Assume that c 1 6. Then 7 r 1 and of the parallel class a.
173 7r2
share all of their lines
Proof Now assume that 71 and 7r2 do not share all of their lines on a but share at least two so they share at least c  2. Let y1 be a line of a of 712 which is not a line of 7 r 1 , Let z1 n y1 Nl and ~2 n ~1 = N Z . Form the subplane T L ~ , N=~ 7r3 (note that L1 and NI are distinct points of 21). Furthermore, 7r2 = TL,,M, = XL,,N, because yl is a line of 7r2 and note that W is a point of T N , , N ~ as well as a point of TL,,L, and TM,,M,. Note that W a cannot be a common line of 7 r 1 and 7r2 for if so then W is a common point which cannot occur. Let v be a common line of 7r1 and 7r2 on a and distinct from z or y, Let A be a point (affine) of v n 7r2 which is not on z2. Since A is a point of 7 r 2 , A and NZ are collinear. Form AN2. Recall that W a is a line of TN,,N, as is AN2,so the intersection W a n AN2 = R2 is a point of T N ~ , N ,. The point R2 is affine since otherwise AN2 would be in the parallel class a and A would be on 91 which cannot be since 91 is not a line of T I . Since A and L2 are distinct points of 7 r 2 , form AL2 n W a = U1 so that U1 is an affine point (similarly AL2 is not parallel to W a for otherwise, A and L2 would be on z and A a = v would then be z). Thus, U1 is a point of TL,,L, as W a and L2A are lines of TL,,L, and hence U1 and L1 are collinear. Form R2N1 which is possible since the joining points are in the same subplane TN~,N,. Now U1L2 n R2N2 = A , R2 = AN2 n W a and AL2 f l W a = U1 so that U1L2 = AL2 and R2N2 = AN2 and is, of course, in 7 r 2 . Similarly, U1L1 nR2N1 = S1 is in TL,,N, = 7r3. Note that R2 and U1 are both on W a and if distinct determine a unique subplane TU,,R, (also the indicated intersection S1 is defined if R2 is not VI). Similar to theabove argument, if R2 = U1 then R2N1 and z1 are common lines of T L , , L ~so that L1, L2, and NI are in the same subplane asz1 is L1N1. However, TL,,N, = 7r3 and T L ~ , M , = 7r2 so that 7r3 and 7r2 share a common point (namely L2) and two common lines z and yl on a which
CHAPTER 15. CLASSIFICATION
174
forces these two subplanes to be equal. But, in this case, 7r3 contains L1 but 7r2 cannot contain L1. Thus, S1 and A are points which are common to T U ~ , R ~ . However, we don't know yet know that S1 is an affine point. We know from the Lemma 15.4 that there are at least c  2 lines on a which are common to and 7r2 and since c > 5, there are at least 4 such lines. Let v1 be a line on a common to Form AN2 n v1 = TI. Then TIis a point of
7r2
7r1
and
7r2
and distinct from x,y,or v.
distinct from A or N Z .
Form TIL2 nW a = U2 and notethat TIN2 nWa = AN2 nW a = R2 and since U2 is a point of 7 r ~ =, 7 ~r ~~ ~then , ~ we ~ may , also form the intersection S2 = U ~ L ~ ~ .R(Note ~ Nthat I U2Ll and R2N1 are distinct for otherwise the line is 21 which implies that R2 = W = U2 so that U1L1 n (R2N2 = 22) = A which forces A on z2. Hence, in particular, U2 # R2.) Note that since U1L1 and U2Ll intersect in L1 then both cannot be parallel to R2N1. Also, U2 # U1 since otherwise T would be on z2. Now both S1 and S2 are points of ~ L , , N ,= 7r3, T and S1 are points of and TI and S2 are points of TU,,R, .
TU,,R,
Without loss of generality, we may assume that S1 is an affine point. (Note that both points S1 and S2 arepoints of R2N1 so either the two points are equalor one is affine but they cannot beequal. That is, if so then S1 on U1L1 and S2 on U2L1 and both points are on R2N1 since U1 # U2 and U1 # R2. So, the three lines are concurrent at L1 which implies' that L1, R2, NI are collinear. Therefore, R2 is on 21 which implies that R2 = Q so that T l A = R2N2 = WN2 = 22 so that A is on z2, a contradiction.) 7r3
Since S1 and A are collinear it follows that & A is a line common to and to 7r2 but since 7r2 and 7r3 share x and VI, itthen follows that
S1A =
=Aa =V .
Hence, 7r3 and 7r1 share a point L1 and two common lines x and v which implies that 7r1 and 7r3 are identical which cannot be the case as y1 is a line of 7r3 but not ~ 1 . 0
15.2. THE SHARE TWO THEOREM.
175
l' Lemma 15.6 Assume c = 4. Then 7r1 and
7r2
share all of their lines on a.
Proof: We need to show that 7r1 and 7r2 share three lines given they share two. With the structure of the proof set up exactly as above in the previous two lemmas since there are exactly four parallel classes, there are exactly four affine lines thru MI, namely y, z1 and say TI and r2, Let R1 = T I n W a and R2 = r2 r l W a . There are three affine points of 7 r ~ , on , ~W~ a , namely W and say U1,U2. Note that neither R1 nor R2 can be in 7 r ~ since ~ , if~ so, ~ for example if R1 is a point of TL,,L, then T I and z1 would be lines of this subplane which forces T I n z1 = M1 to be a point of T L , , L ~which cannot occur as
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CHAPTER 15. CLASSIFICATION
= 7rL1,Mlwould equal 7 r ~ is ~this , instance ~ ~ implying that 7r2 = T L ~ , M ~ and 7 r 1 share L2, Now consider U1L1 and U2L1. At least one of these twolinesis not parallel to RIM1 and at least one is not parallel to &MI. Without loss of generality, assume that U1L1 is not parallel to R1M1. Note that R1 and R2 are in 7 r w ~ =~ 7 r ~ , ,so ~ ,R1 and M2 are collinear. , collinear. ~ ~ Now form R1M2 and U1L2. Also, U1 and L2 are in 7 r ~ so~ are If these latter two lines are not parallel, then we may find a common line on a of 7 r 1 and 712 distinct from z and 9 by the above argument taking U1 as U in the argument of Lemma 15.4. Hence, assume that RlM2 is parallel to UlL2. If U1L1 is also not parallel to R2M1 then forming U1L2 and RzM2 and noting that R1M2 is parallel to U1L2 shows that U1L2 cannot be parallel to R2M2 which means again we apply the argument to Lemma 15.4 to produce a common line of 7r1 and 7r2 on a distinct from z and.9. Hence, it must be that U2Ll is not parallel to R2M1. Forming U2L2 and R2M2, we must have these two lines parallel or we are finished. Summarizing, we are forced into the following situation: U1L2 is parallel to R1M2 (so is not parallel to RzMz), (so is not parallel to R1M2), and U2L2 is parallel to U1L1 is parallel to R2M1 (since U1L2 is not parallel to &L1 is parallel to R1M1 (since U2L2 is not parallel to R1M2). We have exactly four parallel classes say a,p, 6, y. We have that U1L2 is parallel to RlM2 so these lines liesay in p (asthey can’t lie in CY). Similarly, U2L2 is parallel to R2M2 but U2L2 cannot liein a or p so these lines lie say in S. Also, U1L1 is parallel to R2M1 but U1L2 cannot lie in p as U1L2 does and R2M1 cannot lie in 6 as R2M2 does so that these two lines lie in y. We have U2L1 is parallel to R1M1 but U2Ll cannot lie in S or y as U2L2 lies in 6 and U1L2 lies in y and since R1M1 cannot lie in p since R1M2 does, we obtain that U2Ll and R1M1 are forced to lie in a which is a contradiction.0 7r1
Lemma 15.7 A n y subplane covered net has at least 3 components and if c = 3 t h e n 7r1 and 7r2 share all of their lines on CY.
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Proof A subplane has order at least 2. If c = 3 and 7r1 and 712 share at least two lines on a then they automatically share all their alines.
Lemma 15.8 If c = 5 then 7r1 and
7r2
share all of their alines.
Proof: We know that there are at least c  2 = 3 common lines. Let 7r1 and 7r2 share lines x, y and say v on a. Let v1 be the fourth line,of 7r2 on a. Form the subplane 7r3 which contains L1 and v1 (that is, 7r3 = 7rh1,zlnvl). Then 7r3 shares x, v1 with 7 ~ 2and since they share then at least c  2 = 3 lines also shares either y or v . In either of the above cases, 7r3 and 7r1 must share L1 and two distinct lines on a. Hence, 7r1 = 7r3. This shows that 7r2 and ~1 share all four of their lines on a.0 This completes the proof of the share two theorem.0 In the next part of the analysis of subplane covered nets, we determine derivable subnets within any subplane covered net.
15.3 The Derivable Net Substructures. Remark 15.9 Let L and N be any two afine points of the net which are notcollinear.Let x be anyline incident with N . Formthe intersection LP n x if x does not lie in P E C and determine the subplane T]rLILpnz. This subplane contains all of the points L6nx so that by remark 15.2 any such intersection point together with L uniquely determines the subplane. We shall use the notation 7rhIz for this subplane. We shall also use the A ,denote ~ a plane defined b y a line thru a point A and a line z notation T . T Tto incident with a point B where A and B are noncollinear points. The point B incident with z shall be noted b y context. Definition 15.10 Let L and N be any two noncollinear points of the net. We define the pointline geometric structure Sf as follows: The Lpoints'are the points of U N 7 r ~ , = where x varies over the set of lines incident with N .
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The ‘lines’ of SF are the lines of a subplane T L , ~whereas the points of Sf are defined as intersections of nonparallel lines of the subplanes T L , ~ for various lines x.
Remark 15.11 Note that it is possible that there are other subplanes within SF which are not of the type T L , ~ . In the following theorems and lemmas, we shall describe the properties of the structures Sf.
Theorem 15.12 (i) Let P be an afine point of Sf. Then every line of the net incident with P is a line of Sf. (ii) Let Q be any afine point of SF which is not collinear to L. Then UQT L , = ~ S ! = Sf. Proof Note that (ii) implies (i) since if y is a line incident with P and P is incident with L then y is a line of any subplane T L , ~for any line z incident with N and if P is not incident with L then y in T L , and ~ Sf = Sf implies that y is in SF. Hence, it remains to prove (ii). We shall break up the proof into four lemmas. Since Q arises as an intersection of two lines of Sf, there is a line z incident with Q such that z is in T L , for ~ some line 2 incident with N . Consider z is in the parallel class a and form La.
Lemma 15.13 If z is parallel to x and N and Q are collinear (but N and Q are both noncollinear with L) then the theorem follows. Proof: Then z , x and La are all lines of the subplane T L ; and ~ since Q and N are collinear, we may assume that z and 2 are distinct. Note that since the points of TL,, on x are collinear to L,it follows that if Q E T L , then Q is not on z. La is distinct from z and from z as otherwise L would be collinear to Q or N . Since Q and N are collinear, we may form the subplane TQ,N and note that this subplane has E(: and z as lines on Q. Thus, TQ,N shares 2 and z with T L , ~and by the Share Two Theorem must share all lines with X L , + on a.
~
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Thus, La is a line of X Q , N . Now take any line x1 incident with N and not in a and intersect La say in P. Since L and N are not collinear then P is distinct from L. Hence, P is a point of X Q J . So, P and Q are collinear so form PQ = z1. Now form the subplanes X L , and ~ ~ X L , , ~and note that both subplanes contain L and P since La n q = P = La n z1 so that by remark 15.2, we must have X L , ~ ,= XL,,,. Hence, for each line z1 incident with N , there is a line z1 incident with Q such that X L , , ~= X L , ~ , .Note that X L , = ~ XL,~. Suppose that z1 = z2 and XL,,, = XL,,, and X L , , ~= X L , where ~ ~ z1 is a line incident with Q and x1 and x2 are lines incident with N . Then this forces x1 and 22 to be lines of the same subplane so that 21 nz2 = N (assuming x1 and x2 distinct) which is a contradiction as this would imply N and L are collinear. Hence, in the case where z and x are parallel, we obtain ( U N r ~ , , )
c (UQ X L , y ) m
Conversely, the previous argument may be seen to be symmetric. Let z1 be anyline distinct from z and incident with Q and form z1nL a = K so that K is a point of X Q J as z1 incident with Q forces z1 to be a line of XQ,N (see remark 15.2). Hence, K and N are collinear so form K N = x1. Form the subplanes T L and ~ XL,,, ~ ~ and note that both contain K and L so are equal. ) (UNX L , ~ so ) that Sf = SF in the case This proves that (UQX L , ~ E that z and x are parallel and Q and N are collinear.0
Lemma 15.14 If z i s parallel t o x and Q and N are not collinear, the theorem is proved.
Proof: Consider any line W incident with N and any line U incident with Q and if W and U are not parallel form the intersections W n U . Let W lie in the parallel class ,B and let U and v be lines incident with Q and in parallel classes distinct from ,B. As the degree of the net is at least 3, we may select lines as above. Form XwnocIwn2). Assume both intersection points W n U and W n U are collinear with L. Then we have Q and L points of the same subplane which
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implies that Q and L are collinear
CHAPTER 15. CLASSIFICATION
(as Q = U n v and U and v are lines of
rwnu,wnv)
Now Q occurs as the intersection of two lines u,v of SF. Take a line W incident with N and not parallel to U or v and note that all lines on N are in SF. Without loss of generality, we may assume that E = w n v is not collinear with L. Hence, it follows that there is a point E of Sf which is collinear to both Q and N but which is not collinear to L. The lemma below considers the case where the points Q and N are collinear, but the lines z and z are not collinear, in a general or generic sense. Hence, in either situation for E in r l ~ 1 ; and , ~ ~ ZE parallel to XE or ZE not parallel to X E for E , N collinear, we have SF = Sf = Sz.0
Lemma 15.15 If z is not parallel to x and Q is collinear'to N , the theorem holds. Proof: Again note that Q is not on z. Let z1 be any line incident with Q and distinct from z , Consider z n z = P. Since z and z are lines of T L , by ~ assumption, we have that P and L are collinear. First assume that z1 is not parallel to PL, let T = z1 nPL and note that T is distinct from L as Q and L are not collinear and z1 is a line incident with Q. and note that N = N Q n (x = P N ) and T = Form the subplane T ~ , Q PL n (21 = TQ) so that both N and T are points of the subplane T ~ , Q . Hence, N and T are collinear so form NT = z1. Note that the subplanes 7 r ~ and , ~ 7~ r ~ both , ~ ~contain the points T and L so are identical by remark 15.2 ( T L , is ~ ~relative to Q). Now suppose z1 is parallel to PL. Note that 21, z , and PL are lines of TQ,N ( P is a point of the subplane and PL is a line incident with P ) . Assume that z1 and PL belong to the parallel class S so that NS is also a line of T Q , N . Form 7rL,N6 and note that this subplane shares two lines PL and NS on S with TQ,N so, by the Share Two Theorem, the two subplanes share all of their lines on S. Hence, 21 is a line of T L , N so ~ that r l ~ = ~ 7rL,N6 , ~ ~ as z1 is not incident with L.
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Thus, for each line z1 incident with Q, there is a line x1 incident with N such that X L , ~ = , XL,,~, Conversely, let x1 be a line incident with N and first assume that x1 is not parallel to PL. Let T = x1 n PL. Form X N , and ~ notice that P L and x1 are lines of this subplane as are z and Q N . Recall Q = z fl Q N , so that Q is in x ~ , N . Note also that T = x1 fl P L so that T and Q are collinear. Hence, let T Q = z1 and observe that X L , ~ and , X L , ~both ~ contain the points T and L so are identical. If x1 is parallel to P L and both lines are in the parallel class 6, note that x1 and P L are both in TQ,N (x1 is incident with N , P is a point of X Q , N and P L is a line incident with P). Form nL,Q6 and note that P L and Q6 are also lines of XQ,N so that, by the Share Two Theorem, x1 is also a line of 7rL,Q6 which implies that TL,xl
=TL,Q~.
Hence, the previous arguments show that Sf.0
U Q ~ ~ =, 2Sf/
= UN XL,,

Lemma 15.16 If z and x are not parallel and Q and N are not collinear then the theorem follows. Proof: Then there is a point E = z n x of SF which is not collinear to L but is collinear to Q and to N . (If E is collinear to L then E on X L , , and on X+ and E # L implies that X L , , = X L , ~ . ) Hence, SF = Sf = Sf.0 This completes the proof of our theorem in all cases z parallel to x and z not parallel to z.0
Remark 15.17 In the following, let SL = Sf = Sf for all points Q of SF which are noncollinear with L. Furthermore, N is in SF.
Proof Let x( be two distinct lines on N for i = l, 2 and form XL,,( and let LP n xi = Ti for ,L? not a in C. Note that X L , , ~is not X L , , ~as N and L are not collinear. Let TISn Tia = Q for a,p, 6 distinct parallel classes and ~ noting that Q note that N = NT1 n NT2 = x1 n x2 so N , Q fiX T ~ , Tand and L are not collinear (asotherwise T1T2 n LQ = L is a point of X T , , T ~ ) , N is the intersection of lines in XL,,, and X L , ~ ~ . O
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Theorem 15.18 Let A, B be points of SL where A i s n o t collinear to B and B is not collinear to L. Then TA,= ~ S$ = SL.
UB
Proof First assume that A and L are collinear and form TA,L. Since A is in SL, every line incident with A is a line of SL and as such is in some subplane T L , where ~ z is a line incident with N . It then follows that TA,L is one of the basic subplanes T L , ~ . Let B be any point of SL which is not collinear to L. This subplane is equal to a subplane T L , where ~ W is a line of SL incident with B by Theorem 15.12. Hence, TA,L = T A , for ~ some line W incident with B. Any line z thru B is a line of SL by Theorem 15.12. Any line thru L is a line of T A J . Form X L , % . So z is a line of U BTA,% and if L6 is a line of TA,L = TA,,,, then z , L6 and z t l L6 are in U BTA+. If P is such a point then P6 for all 6 E C is a line of the subplane. Since all lines thru L are lines of T A J = TA+ and all lines thru B are lines of U BT A , ~it, follows that all these initial intersection points arealso points Of
U BTA,~
Since the remaining points of T L , are ~ generated by these initial intersection points, it follows that the points of each of the subplanes T L , for ~ z incident with B are points of TA,~. By applying part (i) of Theorem 15.12 to T A , ~it, follows that on any point Q of T A , ~ all , lines on Q are also lines of TA,~. Moreover, since the lines of the subplanes T L , for ~ z incident with B may be obtained by taking the points P and forming Pa for all a E C and since P is also a point of T A , then ~ such lines also become lines of TA,~. Hence, all lines of the subplanes T L , for ~ all lines z incident with B are so that all subsequent points of SL are also points of also lines of
UB
UB
UB
UB
UB
UB
UB U BTA,~* Thus, SL c U BTA,y
Since A and B are points of SL, all lines incident with A and all lines incident with B are lines of SL by part (i) of Theorem 15.12 and hence all subsequent points and lines generated within T A , are ~ likewise in SL, as the previous argument is symmetric, so that TA,C ~ SL. Hence, we have shown that if A and B are points of SL which are not collinear and A and L are collinear but B and L are not collinear then
UB UB
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183
U Br d , y = SL. Now assume that there is a point C in SL such that A is collinear with C, C is collinear with L and A, C, L are each not collinear with B. Then U B7 r q W = U BTL,= = SL and since A is then in U Bre,+,, it follows from the above argument that U BT A , ~= U BT C , = ~ SL. If A and L are not collinear take any two lines U and v thru A. These lines U and v are lines of SL by the previous lemma. Take any line W thru
L which is not parallel to either U or v. Suppose both intersection points U n W and v n W are collinear with B. Then since A is collinear with both intersection points ( A is u n v ) , it follows that A and B are points of the subplane runw,vnw which forces A and B to be collinear. Since u,v and W are lines of SL, the intersection points are also in SL and one of these, say C, is not collinear with B but is collinear to both A and L. Hence, it follows that UB rdly= UB r ~ = SL.O , ~
15.3.1
The Subnet Theorem.
Definition 15.19 W e define a ‘subnet’ as a triple of subsets of the sets of points, lines, and parallel classes as follows: The ‘lines’ of the subnet will be the lines of thesubplanes r ~for, x ~ incident with N where L and N are not collinear. The ‘points’ of the subnet shall be the intersections of lines of the subplanes indicated. The subset of lines of each parallel class Q E C is the union of the sets of lines belonging to the subplanes r ~which , lie ~ in Q. Theorem 15.20 Thestructures SL are derivable subnets;thestructures SL are subnets with parallel class C and the subplanes contained within the structures are Baer subplanes of SL. Proof Note that each line on each point of SL is a line of SL by Theorem 15.12 so that each point P is on exactly one line of each parallel class. Hence, it easily follows that we have a subnet. It remains to show that given any pair of distinct collinear points P and Q of SL then the subplane TP,Q is a subplane of SL and to show that the subplane is Baer within SL.
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184
As each line incident with P or Q are lines of the subnet and the points of T ~ , Q are obtained via intersections of Pa and Q p for all a,P E C then all points are then back in SL as are all subsequent lines by applications of Theorem 15.12. This shows that T ~ , Q is a subplane of SL. It remains to show that the subplane is Baer within the subnet. Take any subplane 7r1 of the net which is within SL and let A be any point of SL. Recall to show that 7r1 is Baer within SL, we must show that every line of the net contains a point of the projective extension of T I , and that every point of the net is incident with a line of T I . The first Baer condition is trivial since each line projectively contains an infinite point (point of C) of T I . To show the second condition, we first show that 7r1 is of the form T Q , ~ where x is a line incident with a point B which is not collinear to Q and Q and B are points of SL. Let T I = TP,Q where P and Q are any two distinct affine points of the subplane and note that P and Q must be in SL. Take any line U of .x1 incident with P and not PQ. It must be that U is a line of SL. If U containsa point B in SL whichis not in 7r1 then B cannot be collinear with Q for otherwise B would lie on two lines of,.rrl and hence be a point of 7r1. But U is in some subplane T L , where ~ x is a line incident with N and as such U contains at least two affine points of X L , in ~ SL. If both of these points are in TP,Q then T ~ , = Q T L , by ~ remark 15.2. If one of these points say B on U in 7 r ~ is, ~not in T ~ , Qthen B and Q are not collinear and T ~ , Q = T Q , ~ Now . to show that there is a line of TQ+ incident with A. If A and Q are collinear, clearly AQ is a line of T Q , ~ incident with A. First assume that T Q , is ~ a subplane of the type T L , for ~ ~some line 21 incident with N . We may assume that A and L are not collinear as otherwise La is of line of 7 r ~ on , ~A. Then, UAT L , ~= TL,~ andfurthermore,there is a 11 and onto correspondence x + z of lines z incident with N and lines z incident with
UN
15.3.DERIVABLE. THESUBSTRUCTURES. NET
A such that
185
TL,= ~ 7 r ~ , ~ .
Hence, there exists a line z1 thru A such that T L , contains ~ ~ this line; 7r1 contains a line incident with A. Now assume that T Q , is~ not a subplane of the type T L , but ~ note that U is a line of T L , ~ for , some line 2, incident with N . We want to show that A is in T Q , where ~ D is a point of SL on U. However we know that A is in SL and U DT Q , C~ SL. On any line t thru Q of TP,Q = T I ,assume two points of t in TP,Q are incident with L , then L must be in x p , ~ . Hence, if TP,Q is not of the form T L , for ~ some line 2 incident with Q (which is then also a T L , ~ then ) at most one point of t in TP,Q is incident with L. If the number of parallel classes is > 3, we may assume without loss of generality that neither P or Q are incident with L and we have that A and Q are not collinear. Let B be a point of T L , ~ on , U which is not in T Q , ~ . Form the subplane T B , (note ~ U = B P ) and note that this subplane ~ T L , ~ since , if T B , is~ T L , ~ then , P and must be distinct from either T Q , or L are collinear. We have established that 7r/~g,pis a subplane of SL assuming that the number of parallel classes is > 3. Hence, any point D on U of T B , distinct ~ from B or P is not in either plane T Q , or ~ T L , ~ ,(if D is in T Q , then ~ T B ,= ~ T D ,= ~TQ,~). Then D is not collinear to L or Q so that = = SL (note if D is collinear to L then T L , ~ ,= ‘IT,+ implies D in 7 r ~ so , ~that , T B , = ~ T B , D = T L , ~ , , a contradiction). Hence, A must be in U DT Q , so ~ that we may apply the previous results to show that UAT Q , = ~ TQ,~. Moreover, there is a 1  1 and onto correspondence W + y of lines W incident with D and lines y incident with A such that the T Q , = ~ TQ, This implies that for the line U there is a line z incident with A such that T Q ,= ~ T Q , so ~ that the subplane T I = TQ+ contains a line incident with A. Thus, itremains to show that when the degree is exactly 3, the subplanes contained in SL are Baer. Note again that, in this case, we are not necessarily assuming that the net is finite. However, there are exactly threelines of SL incident with N and on each
UD
u D ~ ~ u, Dw ~ ~ , y UD
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C H A P T E R 15. CLASSIFICATION
such line there is a unique point incident with L so there are exactly 4 3 lines of SL and it follows that on each line there are exactly 4 points of SL. That is, SL is a subnet of degree 1 2 = 3 and order 22. Since the subplanes contained in SL now haveorder 2, it follows that such subplanes are Baer within SL.This completes the proof of the the0rem.O From the classification of derivable nets, we obtain:
+
Corollary 15.21 Considerany of thesubnets SL of points,lines, subplanes, parallel classes, and incidence. Then there isa 3dimensional projective space C and a line N of C such that the lines of C skew to N are the points of SL,the points of C  N are the lines of SL,the planes of C which intersect N in a point are the subplanes of SL and the planes of C which containN are the parallel classes of SL. Wenow show that we can construct an parallel classes.
affine space structure on the
15.4 The ParallelClassesAreAffine
Spaces.
Definition 15.22 Let a be any parallelclass. Define the structure H ‘ ,as follows: The ‘points’ ofH ‘ ,are the lines of the net on a. The ‘lines’ ofH ‘ ,are the sets of lines of subplanes T ~ , Qwhich lie on a. The ‘planes’ of 3 1, are defined via the sets SL(derivable subnets) and are denoted by SL,,. By this we mean that the restriction of any SL toH ‘ ,is an afine plane as seen in the previous corollary. This set of afine planes defines the set Aff, for 31,. The boints’ of SL,,are the lines on a of the set of subplanes of SL. A ‘line’ of SL,,is, of course, the set of lines o n CY of a subplane of SL. Recall that we used the notation a, to representH ‘ ,in the derivable net classification. Definition 15.23 W e shalldefine two lines ofH ‘ ,to be ‘parallel’ if and only if the two lines correspond to subplanes which belong to some SL and their lines on a are disjoint or equal.
15.4. THE PARALLEL CLASSES ARE
AFFINE SPACES.
187
Note that it is clear that the relation of being parallel is symmetric and reflexive. The previous result that there arederivable subnets is vital for the results in this section. l, are interconnected to the net,we shall Furthermore, as the structures" require net properties to show that theH ' ,are affine spaces. Ultimately, we shall use the characterization of affine spaces, Theorem 9.10, to prove this. Hence, we need to show A(0) Card Aff, > 1, A(l) there are at least two points incident with every line, A(2) two distinct points are incident with a unique line, A(3) there are at least four noncoplanar points, A(4) given three mutually distinct noncollinear points, there is a unique affine plane of Aff, incident with them and A(5) parallelism is an equivalence relation. We are assuming that there are at least two subplanes inthe net,so A(0) is automatic. Since there are at least three parallel classes A ( l ) is clear. It remains to prove the remaining conditions. In order to prove that each of these structures is an affine space, we need to consider all such structures globally. Hence, we define parallelism of 'lines' of different Z & s .
Definition 15.24 We define two lines a and b of the structures 'Ha,' H p for Q # ,f3 E C to be parallel, written a 11 b if and only if these sets are the sets of lines on Q , ,f3 respectively of a subplane ro. Again, it is clear that this relation is symmetric. Lemma 15.25 Given any subplane ro and any line U of the net which is not a line of 7ro, there is a unique derivable subnet (no,u)containing ro and U.
Proof Take any line v in ro which is not parallel to U . Let N = U n t~ and let L be a point of ro which is not collinear with N . Note that N cannot be a point of 7ro Form UN7r~,z = S2 and note that this derivable subnet contains 7ro (simply take 2 to be v) and U (take 2 to be U ) .
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Note that any derivable net containing IT,, and U must contain the intersection point N as a point and hence, must contain the set of lines incident with N . Thus, any such derivable net contains Sf so existence and uniqueness is verified.0
Lemma 15.26 Any two distinct subplanes T,,and IT^ which share a parallel class of lines are in some unique derivable subnet (IT,,,TI). Proof Let U be any line of IT^ whichis not a line of x,,. Form the derivable subnet (T,,,U ) . Assume that theindicated subplanes share all of their lines on the parallel class a E c. Since (T,,,u) is a derivable net containing U , there is a subplane IT;of this derivable subnet which contains U and which shares the lines of IT,, on a by the classification theorem for derivable nets. Since T I and T; share U and share all of their lines on the parallel class on a,it must be that T I and IT;are identical.0
Lemma 15.27 Let a, b, c be lines of various of the structures ' H 6 for S E C . Ifa (1 b and b 11 c then a 11 c. Proof: We consider the following cases: respecCase (1): the lines a, b, c belong to the structures 'Ha,' H p , tively where Q, p,y are mutually distinct. In this case, there are subplanes IT,, and T Isuch that a and b are the sets of lines of T,,on a and p respectively and b and c are the sets of lines of T I on p and y respectively to T I and IT,, share b. Form the derivable subnet (IT,,,T I )and note that a, b, and c are lines of this subnet. Then, within this derivable subnet, there is a subplane x2 such that a and c are the sets of lines of ~2 on a and y respectively by the classification theorem for derivable nets. Hence, a 11 c. Case (2): a and b belong toH ' ,but c belongs to for Q # y, By assumption, there is a derivable subnet ( T , , , T ~ )and hence a unique such net such that a and b are the setsof lines ona of T,,, and T Irespectively. Within this derivable subnet, there is a subplane which contains a and say d not on a or ,t3 (a set of lines of this subplane which does not belong to
H ',
H ',
15.4. THE PARALLEL CLASSES ARE AFFINE
SPACES.
189
either parallel class) and a subplane which contains b and d (since a and b are sets of lines of a parallel class of subplanes of the derivable net), That is, a 11 d and b 11 d. Hence, c 11 b 11 d and all three lines a, b, c are in distinct substructures X,, for various values p E C, it follows from case (1) that c 11 d. Hence, c 11 d 11 a so that another application of case (1) shows that c 11 a. Case ( 3 ) : a and c are in 3 1 ,and b is in 3  1 ~for a # p. Let 7ro be a subplane whose sets of lines on a and p are c and b respectively and let 7r1 be a subplane whose sets of lines on Q and p are a and b respectively. We may form the derivable subnet (no,T I )since the subplanes share b. Then a, b and c are lines of a derivable subnet and a I( b 11 c so that a automatically becomes parallel to c. If b and c are in 3Ip and a is in 3 1 ,then c 11 b and b )I a implies c 11 a by case 2 . Case (4): a r b ,and c are in 31,. Since a is parallel to b, there is a unique derivable subnet (7rITo,7r1)such that the lines on a of 7ro and 7r1 are a and b respectively. Similarly, there is a unique derivable subnet (7r2,7r3) such that the lines of 7r2 and 7r3 are b and c respectively. Take any set of lines d of 7r1 on a parallel class p distinct from a. Then a 11 b 11 d implies a 11 d from case ( 2 ) and d 11 b 11 c implies d I( c (i.e. c 11 b 11 d ) again from case ( 2 ) . Then a 11 d 11 c implies that a 11 c from case (3).0
Theorem 15.28
X,
is a n a f l n e space for each parallel class a E C .
Proof First take two distinct points A and B of 31,. Recall that A and B are lines on a. Take any line U of the net which is not in a. Then the intersections of U with A and B produce a subplane 7ro such that any other subplane which shares A and B with 7ro must share all of the lines on Q with 7ro by the Share Two Theorem. That is, given two distinct points of 3 1 ,there is a unique line joining them. This proves A(2). Since there are derivable subnets, thereare at least four nonAff,coplanar points which proves A(3).
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190
Now take three distinct points of ? ' l, A , B ,C not all collinear. Then there is a unique 'plane' of Aff,, ( A , BC), , containing these points. To see this, let U be any line of the net which is not in a. Form the intersection of U with A and B and thecorresponding subplane TO.
, are not collinear so C is not a line of 7ro. By assumption, A , BC Form the intersections of U with B and C and construct the corresponding subplane 7r1. Let P = U n B so that P is a common point of 7ro and T I . Take any line IC of 7ro which is not on P and take any line z on 7r1 which is not on P and not parallel to IC. Let N = IC n z. If P and N are collinear then P N intersects II: in N so that N is a point of 7ro and similarly also a point of 7r1 which forces 7ro to be 7r1 ~ 1 sthen the two subplanes both share N and P. Hence, P and N are not collinear. Form which contains 7r1 = 7rp,= and 7ro = ~ p , ~ . Therefore, there is a derivable subnet containing 7ro and 7r1 so that there is a plane of Aff, of '?la containing A , B , C. Let D be any derivable net containing A , B and C.Then the set of lines of the derivable net on cr form a plane of 7 1 containing , A , BC , . Since any plane is generated by any of its triangles, it follows that the plane is unique. Now assume that there are two derivable subnets that share the lines
UN
A, B.
We assert that if two planes of 7 1 share , two distinct points A and B then they share all points on the line AB. To prove this, note that the two planes are defined by two derivable nets D1 and D2. Within Dl,there is a subplane 7ro which contains the lines A and B . Any other subplane which contains the lines a and B contains as lines all of the lines of 7ro on a by the Share Two Theorem. Hence, any subplane on D2 which contains A and B must contain the lines of 7ro on cr and thus each plane of 7 1 containing , A and B contains all of the points on the line AB. Thus, we have shown that A(4) is valid. Now Lemma 15.27 shows that parallelism is an equivalence relation so that 4 5 ) holds.0
SPACE. PROJECTIVE 15.5. THE
191
It now follows that the structures
3 1 ,are affine spaces.0
15.5 The Projective Space. In this section, we show that we may weave the &ne spaces 3 1, together to form a projective space that completely characterizes the subplane covered net.
Remark 15.29 ( I ) By the classification theorem for derivable nets, we may assume that the net is not a derivable net. (2) Since derivable nets induce planes in N,, it follows that we may assume that the structures are afine spaces of dimension 2 3. (3) Let D and R be derivable subnets which share three lines of the same parallel class CY E C not all in the same subplane. Then the derivable subnets share all of their lines o n a and we denote this by D, = R, as the 31s; are afine spaces. (4) The reader will need to distinguish between lines of the net or subnet and lines of the afine spaces 3 1,or D, since a line of a derivable subnet D, is the set of net lines on CY of a subplane of D.
31,
We now define the 'infinite points':
Definition 15.30 W e consider the projective extensions of the a p n e spaces
31,.
31,
Let N" denotethehyperplane of at infinity obtained by defining 'infinite points' to be parallel classes of lines of 3 1,and 'infinite lines' to be parallel classes of planes of 31,. W e want to show that N a = Np for all a ,p E C. What this basically implies is that there is a projective space CR such that the parallel classes when properly extended become hyperplanes in CR that contain a common codimensiontwo subspace N. In order to do this, we need to define what it means for two planes of diflerent afSine spaces 3 1, and 31p to be parallel for possibly diflerent parallel classes CY and p.
31,
Definition 15.31 Let TI,, TIp be planes of and 31p respectively. W e shall say that T I ,is 'parallel' to l$, written I I ,11 IIp if and only if each line of T I ,is parallel to some line of TIp.
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CHAPTER 15. CLASSIFICATION
Before provingthat the relation defined above isan equivalence relation, we provide some lemmas on derivable subnets.
Lemma 15.32 Let D be a derivable subnet. (1) Let a be a parallel class of the net and let x be a line which is not in a. There there is a unique derivable subnet generated b y x and'D, which we denote b y (x,D,). (2) Let ro and r1 be any two distinct subplanes that share a point. Then there is a unique derivable subnet containing ro and rl. Proof (1) Take any three lines u,v, W of D on a not all in the same parallel class of lines of a subplane of D. Form the intersections U n x = P,v f l x = Q, and W n x = R and form ~ TQ,R. Let L and R be noncollinear points of r p , ~ the subplanes r p , and and TQ,R respectively with L on U (x is common and incident with R if another point on U of r p , is ~ collinear with R then LR n x is common to both subplanes implying R is common, a contradiction). Form the derivable net S,". So x is a line on R implies that x is a line of Sf. So, L a = U , R a = W are lines of Sf and L is collinear with Q as is R so that LQ n RQ is a point of Sf and hence v is a line of S?. Thus, (S,"), = D, asH ' ,is an affine geometry. This implies that there is a unique derivable net R containing these subplanes and clearly R, = D,. Since R contains x, R = (x,D,). This proves (l). Now assume that ro and rl are two subplanes that share a point Q. Take the common line v = Qa. Let x = QP in a parallel class p # a. Let U and W be lines of ro  r1 and r1 ro respectively on a. letting U n x = P, and w n x = R and we have v n x = Q,then no = r p , and ~ r1 = ~ Q , R Since . F 'I , is an affine geometry, there is a unique affine plane ofF 'I ,incident with U, v, W. Hence, there is a subplane r2 of some derivable subnet that shares the lines U and v. By the Share Two Theorem, the lines on a of r2 and ro are identical. Using the above argument, it follows that ro and 7rl are in Sf defined exactly as in part (1) and uniqueness is implied by the previous analysis of the derivable subnets.0 Weknow that planes ofF ', Imust fall into parallel classes asF ', Iis an affine space. What we don't know is how the derivable subnets that define these planes are related.
15.5. THE PROJECTIVE SPACE.
193
Lemma 15.33 Let D be a derivable subnet so that D, i s a plane of 3.1,. Let x be a line of a which is not in D,. Then the unique plane of 31, incident with x and parallel to D, m a y be constructed as follows: Take any line z of D not in Q . Then there exists a unique derivable net R containing x and z with the property that R, is parallel to D,. Any other derivable netB so constructed from any derivable netT where T, = D, and containing x has the property that B, = h. Proof Let b be a line of D, in H, (a set of lines on a of a subplane). Let z be a line of D in p distinct from a. Then z intersects b in a uniquely defined subplane 7ro which does not contain x by the Share Two Theorem. Further, there is a unique derivable net containing 7ro and x, (x,7 r o ) , Note that since (x,no),is an affine plane in X,, it follows that there is a unique line &,x of (x,ro), parallel to b thru x. Recall that this line on 8, is the set of lines on Q of some subplane. In (x,ro), there is a unique subplane 7r1 which has Lb,x as its lines on Q and which contains z. Let Lb,z denote the line of H p which is the set of lines of 7r1 on p. Note that b 11 Lb,z 11 Lb,z so that b 11 Lb,z So, there is a unique subplane 7 ~ 2containing b and Lb,z as its sets of lines on Q and p respectively and since n2 contains b and z , it follows that 7r2 = TO.. Hence, U{Lb,z ; b is a line of D,} = Dp. Note that (x,Dp) is a derivable net by Lemma 15.32 and there is a unique subplane containing Lb,z and x and this is a subplane 7r1 containing Lb,z and Lb,z SO that E (x,Dp). Hence, U{LbIx; b is a line of D,} = (x,Dp),. So, we have produced a derivable net R containing x such that every line of D, is parallel to some line of h. Let c and b be any two lines of D, then since F ‘I , is an affine space, the plane generated by c and x is unique and hence the line parallel t o c thru x is unique. A similar statement is valid for b and x. Hence, let B be any derivable net which contains x and contains the lines on x parallel to c and b. Then B, is uniquely determined.
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CHAPTER 15. CLASSIFICATION
It follows that RQI and D, are mutually parallel (since they are planes of an affine space and one is parallel to the other). Furthermore, since each line of R, is parallel to a line thru IC and parallelism on lines of the affine spaces 7iks is an equivalence relation, it follows that each line of R, is parallel to a line on z of Dp and conversely each line of Dp is parallel to a line of & containing IC. Hence, it follows that & and Do are parallel planes.0
Lemma 15.34 Parallelism on planes of the aflne spaces 'Hks is an equivalence relation. Proof Note that if D, 11 Rp where D and R are derivable nets and a and ,L? are distinct and if z is any line of Rp then there is a derivable net (2, D,). Take any line b of D, and note there is a unique subplane r 0 of (z, D,) containing z and with b as its set of lines on a. Since Rp is an affine plane, every line of Rp is parallel to a line which contains z. Hence, since b is parallel to some line of Rp, it follows that b is parallel to a line c which contains z and this line c must be exactly the sets of lines of r0 on p. It follows that ( 2 ,D,)p = Rp, So, any line of Rp is parallel to some line of D,. Toprove transitivity, simply note that if three planes D, 11 Rp 11 B, where D , R, B are derivable subnets then every line b of D, is parallel to some line c of Rp and every such line c is parallel to some line d of B, . Since parallelism on lines is an equivalence relation, it follows that b is parallel to d and hence, every line of D, is parallel to some line of B, so that D, 1) B,. This proves the lemma.0
Lemma 15.35 If D and R are derivable nets and for some parallel class a, D, 11 R, then for all parallel classes p , Dp (1 Rp. Proof Clearly for any derivable net B and any parallel classes y and p, it follows that B, 11 Bp. Hence, D, 11 & 11 Rp implies that D, 11 Rp and Dp 11 D, 11 Rp implies that Dp 11 Rp.0
15.5. THE PROJECTIVE SPACE.
195
Lemma 15.36 Let 3 1 , be any afine space f o r a E C and let N f f denote the hyperplane at infinity of the projective extension 312 of 31,. T h e n N f f= Np which we denote universally by N f o r all a,P E C . Proof In order to construct N f f ,we define the points of N f fto be the equivalence classes of lines of 3 1 ,and the lines of N f f as the equivalence classes of the planes of 31., Recall that any plane of 3 1 ,is defined by a derivable net D as D, and 1 ,is defined by a subplane of D. any line of 3 Since a parallel class of lines of 3 1 has , a representative in any ‘Hp and 1 ,has a representative in any ‘Hp it follows any parallel class of planes of 3 that N f f= Np = N.0
15.5.1 The Classification Theorem. Definition 15.37 Let R = (P,,C,B,C,z>be any subplane covered net. Define the pointline geometry CR as follows: Call the lines of a given parallel class of a subplane ‘class lines’ and call the lines of a given parallel class of a derivable subnet ‘class subplanes’. Note that there are equivalencerelations o n both the set of class lines and on the set of class subplanes. Calltheequivalenceclasses of the class lines (infinite points’ and the equivalence classes of the class subplanes ‘infinite lines’. Also, note that the infinite points and infinite lines form a projective subspace N . T h e ‘points’ of CR are of two types:
(a) the lines
,C
of the net and
(ii) the infinite points defined above. T h e ‘lines’ of ER are of three types: (i) the sets of lines on an afine point (identified with the set P ) , (ii) the class lines extended by the infinite point containing the class line, and (iii) the lines of the projective subspace N .
CHAPTER 15. CLASSIFICATION
196
The 'planes' of C R are of three types:
(i) subplanes of B extended b y the infinite point on the equivalence class lines of each particular subplane where the points and lines of the subplane are now considered as above (actually the dual of the subplane extended), (ii) the afine planes whose points are the lines of a net parallel class and lines the class lines of a derivable subnet of the net parallel class extended b y the infinite points and infinite line, and
(iii) the projective planes of the projective space N = N" for all
a!
E C.
The hyperplanes of CR that contain N are the parallel classes C extended b y the infinite points and infinite lines of N. Note that N becomes (or will become) a codimensiontwo subspace of CR.
Theorem 15.38 C R is a projective space. Proof: To show that C R is a projective space, we shall use Theorem 9.6. Recall that we must show that the following conditions hold:
P(l) any line is incident with at least three points, P(2) two distinct points are incident with a unique line, P(3) if ABC is a triangle and .t is a line which intersects two of the three sides in nonvertex points then .l intersects the third side in a point, P(4) there exists at least four points no three of which are collinear and
P(5) C R is not a projective plane. The conditions P ( l ) and P(2) follow from previous lemmas and P(4) is obviously satisfied. Hence, it suffices to prove that P(3) is valid. Since we have more than one projective plane within ER, then P(5) is valid and to complete the proof, we need only show that any three distinct points A , BC, not all collinear generate a unique projective subplane.
SPACE. PROJECTIVE 15.5. THE
197
If the points are all infinite points then since N is a projective subspace, the result is clear. Assume that A, B and C are all lines of the net, If all are pointsof the same 31, then since 3 1 , is an &ne space, the points will generate an affine plane which then uniquely extends to a projective plane in 3 1, U N. If A and B are in A, and C is in 3  1 ~where a and p are distinct then by taking intersection points of the lines, there is a unique subplane of the net containing A, B and C. By extending the subplane with the infinite point corresponding to the class points, it follows that there is a unique projective plane interpreted in the notation in the statement of the theorem generated by these points A, B and C. Similarly if A, B and C are all in mutually distinct affine spaces X,,'Flp, 7i7,there is a unique subplane of the net containing A , B and C and the previous argument applies. Suppose that A and B are infinite points and C is a line of the net. Let C be in the parallel class a. Since A is an infinite point, there is a unique representative class line A1C1 which contains C (as a line of the net). Similarly, there is a unique representative class line BlCl in Q of B (thought of as an equivalence class) which contains C. Note that A1Cl and B1C1 extended are lines of the structure CR. Now the two class lines contain C and thus, by the proof to Lemma 15.32,there is a derivable subnet D which contains these two class lines and any other derivable subnet containing these class lines agrees on the parallel class a with D ,The set D, is a plane of 3 1 ,which when extended becomes the unique projective subplane generated by A , B , and C. Assume that A and B are lines of the net and C is an infinite point. If A and B are in the sameparallel class a,consider the set of subplanes which contain A and B.' Recall that the line of 31, AB, , is uniquely determined as the set of lines of any subplane containing A and B. Now as A, B and C are not collinear then C is not a equivalence class of any subplane that contains A and B . Hence, there is a unique representative class line on a which contains A but not B . Take any line x not in a and intersect the lines of the class
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CHAPTER 15. CLASSIFICATION
point and B. Then there is a unique derivable net D containing 2 and these intersection points by the proof to Lemma 15.32. Furthermore, any other derivable net containing the class line and B shares the lines on a with D. Hence, there is a unique affine plane D, of 3 1 which , when extended is the unique projective plane generated by A, B, and C. Finally, assume that A and B are lines in different parallel classes of the net and C is an infinite point. Let P = A n B . The set of lines of the net incident with P is a line of the structure which does not intersect the projective subspace N so that A,B and C are intrinsically noncollinear in this case. .. Take a representative class line c on the parallel class a of the net containing A. Form the intersection points of this class line (which is a set of lines of a subplane) with B and note that there is a unique subplane x,., of the net generated. This subplane contains A, B and when extended by C is the unique projective plane containing A,B and C when interpreted in the notation of the theorem. Note that if the class line b is on B then b 11 c so that r0 may be taken as the unique subplane with these properties. This completes the proof of theorem.0 Clearly, we have:
Corollary 15.39 Any subplane covered net is a codimensiontwo net. We have discussed pseudoregulus nets previously and have shown that any codimensiontwo net is a pseudoregulus net. Hence, since any subplanecovered net is isomorphic to the structureER, we have the following characterization of subplane covered nets.
15.5.2
TheTheorem on PseudoRegulus Nets.
Theorem 15.40 A subplane covered net is a pseudoregulus net. We also obtain the following result first proved by De Clerck and Johnson using results of partial and semipartialgeometries which are necessarily finite. These geometries are not considered in this monograph but the interested reader is referred to the papers of F. De Clerck and J.A. Thas for further detailed information and results.
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199
Theorem 15.41 (De Clerclc and Johnson [20]). A finite subplane covered net is a regulus net (corresponds to a regulus in some projective space). Remark 15.42 It is important to note that if the net has order g" and degree q 1 (number of parallel classes), the projective space corresponding to the net is isomorphic toPG(n+ 1,g) whereas the projective space wherein lives the associated regulus is isomorphic to PG(2n 1,g). Thus, only in the case of derivable nets do subplane covered nets correspond to two different threedimensional projective spaces. For, in the derivable net situation, when n = 2, the projective space corresponding to the net is isomorphic to P G ( 3 , g) and the net is a regulus net and corresponds to a regulus in another threedimensional projective space P G ( 3 , g ) .
+
a
.
.
Chapter 16
SUBPLANE COVERED AFFINE PLANES We have previously discussed pseudoreguli and reguli and realized in the latter case that they may only be embedded in projective spaces over fields. We shall be interestedin spreads which are covered byreguli in various ways. We begin by the consideration of spreads which contain as many reguli as possible.
16.1 Regular Spreads. Definition 16.1 a 'regular spread' is a spread S over a field K such that for each triple of distinct components L, M , N of S, the unique Kregulus containing L , M , N also belongs to the spread S.

Remark 16.2 Every regular spread over K admits the collineation group
((x,v>
(02,P P )
; 0,P E K )
Proof Choose two components to be represented by x = 0, y = 0. Now choose a third component which must then have the form y = z A where A is an appropriateKlineartransformation. For example, if the vector space V is W CBW then A is in GL(W,K ) . Consider the sets y = zyA where y E K  (0). Change bases by the mapping (x,g) I") (z, yAI) t o realize that we have the regulus z = 0, y = 201
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C H A P T E R 16. SUBPLANE COVERED AFFINE PLANES
0,y = z7 for y E K(0) which are componentsof the spreadby assumption. Hence, y = zyA is a component whenever y = z A is a component. Since K is a field and A is in GL(W,K ) then yA = AyI so we have that the above mappings are collineations of the corresponding translation p1ane.O Now assume that y = z A and y = zB are distinct components. Choose a new basis by the mapping (z,y) H (z,  z A p) which maps y = z A onto y = 0, y = zB onto y = z ( B A ) and leaves z = 0 invariant. We know that ( B  A) E GL(W,K ) since the components are disjoint. Now change bases again by the mapping (z,y) I”) (z, y(B  A )  l ) . We now have z = 0, y = 0, y = x y l within the spread as we have x = 0,y = 0, y = x in the spread. Wenowwork out the preimages of the components y = zy1. If y = zT maps to y = zyI it must be that
+
(T  A)(B A)l = yI which implies that
T=(BA)yI+A=yB+(ly)A. Notice that we could have started with wA for W E K. Hence, we have = z(yB
+ (1 y)wA)
is a component for every y, W E K . Now assume that K is not isomorphic to G F ( 2 ) . Then, there exists a nonzero y such that (1 y) is nonzero and we may choose W = (1  y)ly which implies that we have y = zy(B
+ A) and hence y = z ( B + A )
is a component. That is, we have an additive spread. Since we could have originally chosen any component to be represented as z = 0, we have that every spread set is additive. By previous results in the chapter on coordinates, the plane is a Moufang plane. Hence, this proves the following theorem:
Theorem 16.3 (BoseBruck [13]). If S is aregular spread over afield K # G F ( 2 ) t h e n S defines a Moufang plane. If S is a finite regular spread over a field not equal to G F ( 2 ) t h e n the plane i s Desarguesian.
SPREADS. 16.1. REGULAR
203
Proof: We have noted previously that any finite alternative division ring is a fie1d.O If we define a subplane covered affine plane in the obvious manner (also see below), it may be possible to prove that the affine plane is a translation plane. Recall that we are then looking for a vector space whose vectors are the points of the affine plane. The main problem is that although there are projective spaces and hence vector spaces corresponding to each subplane covered net, there may not be a vector space which corresponds generally to the affine plane. In particular, we consider the possibility that it may be possible to discuss the translation planes which are defined by regular spreads in terms of subplane covered nets.
Definition 16.4 Let T be an afine plane (finite or infinite). W e shall say that x i s a ‘subplane covered afine plane’ provided there exists a set B of proper afine subplanes of x’ possibly of diferent cardinalities, whose union covers the plane with the property that givena subplane r0 E B, the net Nro defined by the parallel classes of x. is subplane covered. In various situations which we shall consider in a forthcoming chapter, the affine plane is a translation plane and the subplane covered nets are all of the same type (derivable nets). In these cases, the nets all share one or more parallel classes. We normally consider that a nontrivial net must have at least three parallel classes and it is possible that the elements of a set of subplane covered nets share a nontrivial net.
Definition 16.5 Let x be an afine plane (finite or infinite). Let C be a fixed set of parallel classes of x. W e shall say that x is a ‘subplane covered afine plane with embedded net R’ defined by set of parallel classes C if and only if the plane is subplane covered and for each subplaneof B , the set of parallel classes of the subplane contains C . W e shall say that the embedded net R is ‘nontrivial’ if R has at least three parallel classes. Definition 16.6 Let x be an afine plane (finite or infinite). A subplane covered afine plane shall be said to be ‘subplane regular’ if and only if f o r
CHAPTER 16. SUBPLANE COVERED AFFINE PLANES
204
any three distinct concurrent lines L , M , N , there is a subplane covered net of T which contains L, M and N . Note that there are many embedded nets but not necessarily any which are nontrivial. Definition 16.7 A projective plane shall be said to be ‘subplane covered’ (‘subplane regular’) if and only if every afine restriction is subplane covered (subplane regular). A projective plane shall be said to be ‘subplane covered with embedded net of class a’if and only if for every afine restriction A, there is a net RA with a parallel classes so that A is a subplane covered afine plane with embedded net RA. To illustrate some examples, we note subsequently that any translation plane of characteristic two is a subplane covered affine plane. In a forthcoming chapter on direct products, we shall construct translation planes of order q4 with embedded nets with q 1 parallel classes which admit a collineation group isomorphic to SL(2,q). There is a set of 1+q+ q2 derivable nets which contain a given subplane covered net of degree 1 q. Since derivable nets are subplane covered nets where the subplanes are all Baer subplanes,such planes become subplane coveredaffine planes. Actually, such planes are interesting for other reasons as these are connected to packings or parallelisms of projective spaces and will be discussed further in the chapter on parallelisms. These subplane covered affine planes of order q4 with embedded net of degree 1 q are translation planes. Conversely, we are able to show that for q # 2 any such subplane covered net is a translation plane. So, there is a possible wayof discussing regularspreadsinterms of subplane regular affine planes or subplane regular translation planes,. The, main questions then become: (1) Is a subplane regular translation plane Desarguesian? More generally, (2) Is every subplane regular affine plane a translation plane? Note that a subplane regular affine plane is also a subplane covered affine plane with embedded net. We have noted above that when the embedded net is nontrivial, we may show that the plane is a translation plane
+
+
+
Definition 16.8 A ‘Fano plane’ is a projective plane with the property that every quadrangle generates a projective subplane of order 2.
16.1. REGULAR SPREADS.
205
We restate the relevant results and related theorems given in the prerequisites.
Theorem 16.9 (Gleason [g,$]) A finite Fano plane is Desarguesian. Actually, it is shown that a Fano plane satisfies the little Reidemeister condition (see below) and more generally, it followsfrom the work of Gleason, Luneburg and Kegel [55] and Luneburg [60] that such finite planes satisfying this condition are always Desarguesian.
Theorem 16.10 (Gleason 1.41, Liineburg [60], KegelandLiineburg [SS]) A finite projective plane is Desarguesian if and only if it satisfies the little Reidemeister condition. We now consider the definition of an &ne Fano plane.
Definition 16.11 An afine plane shall be said to be an 'afine Fano plane' if and only if given any quadrangle (P,Q , R, S ) such that that P& 11 RS and P R 11 &S then PS 11 QR. Remark 16.12 Given aline L in a Fanoplane, clearly the afine plane obtained b y deleting this line L is afine Fano. Furthermore, any translation plane of characteristic two is afine Fano. Proof Let C be any Fano plane and let L be a line of C. Form the affine plane A by deleting L. Let (P,Q, R, S) be any quadrangle so that P& n RS and P R n &S are both points of L. Then it follows that PS n&R is a point of L as there is a generated projective subplane of order 2. Hence, A is affine Fano. Let be a translation plane of characteristic two. Let (P,Q, R, S) be any quadrangle such that P& 11 RS and PR 11 &S. There exists a unique translation Q with center the class, say (l),containing P& and mapping P onto Q. Since Q fixes all parallel classes, Q fixes the classes containing P R and PS, say (0) and (W) respectively. Hence, the line (0)P = P R is mapped onto the line (0)Q = &S. Since RS is fixed, then R = RS n RP is interchanged with RS n SQ = S as Q has order 2. Hence, (w)P= PS is mapped to (w)Q= QR so that PS 11 QR.0 This brings up the following question:
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CHAPTER 16. SUBPLANE COVERED AFFINE PLANES
(3) Whatis the connectionbetweensubplanecoveredaffine planes which are covered by subplanes of even orderand afthe Fano planes? Finally, we ask (4) Is every subplane covered projective plane Moufang? Our basic tool for this chapter is the characterization of subplane covered nets given in the previous chapter.
16.2 The Affine Little Reidemeister Condition. Definition 16.13 Let II be a projective plane and let L be a line and let c,u,v denote three distinct points of L. Let p1,pz be two distinct points not equal to c o n a line M # L and incident with c and not incident with U or v. Let Z , W be distinct lines diflerent from M incident with c and define zi = Z n (piu) and wi = W n (piu) where i = 1,2. h r t h e r m o r e , form mi= ( q v )n (w~u), i = 1,2. If the points c, m l , m2 are collinear then the configuration satisfies the ‘Reidemeister condition with respect to the ordered triple c, U ,v
V
16.2. THE AFFINELITTLEREIDEMEISTERCONDITION.
2 07
Definition 16.14 W e shall say that a projective plane satisfies the 'little Reidemeister condition' provided the little Reidemeister condition is satisfied with respect t o all collinear ordered triples c, U , v. An afine plane satisfies the lafine little Reidemeister condition' if and only if the little Reidemeister condition is satisfied with respect to all triples c, U , v on the line at infinity. Definition 16.15 Let TT be a projective plane. Let J and I be two lines and A any point not incident with either line. The 'projection of I onto J thru A afJ is the mapping taking the points P E I onto AP n J. Let AB = A(I n J) where B is not I n J. Then afJa7I is a permutation of the points of I . Let T be an afine plane and let c, U , v be three distinct 'infinite points' and I be a fixed afine line incident with c. f o r all a f i n e h e 8J incident with c} is a 'set' of permuThen {ayJa"J~; tations which acts regularly on the points of I . The verification of this is left for the reader. We also require the following result of Gleason [24].
Theorem 16.16 (Gleason [24, Lemma (2.2)]) Let ll be a projective plane and let c,u, v be collinear points incident with the line L. Let I # L denote any line of ll incident with c. If w is any point of L , define the projection of I onto line J thru w by ayJ;.Let = {ayJa:I ; J is a line # L incident a group f o r alllines I # L incident with c if and with c}. T h e ni s only if the little Reidemeister condition with respect to c, U , v is satisfied. rf"l"
We first note the following connections with subplane regular affine planes and affine Fano planes.
Theorem 16.17 An afine plane is a subplaneregular afine plane with covering subplanes of even order if and only if the plane is afine Fano. Proof: Let T be a subplane regular affine plane. Let Q,R,S, T be a quadrangle such that &R 11 ST and &S 11 RT. Form the lines SQ,SR,ST. Since the plane is subplane regular, there exists a subplane covered net M containing the indicated three lines. Moreover,the net M is a pseudoregulus
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CHAPTER 16. SUBPLANE COVERED AFFINE PLANES
net by a previous chapter. Hence, thereexists a unique subplane TQ,S containing Q and S. Assume that theparallel classes containing S Q , SR, S T are a,,B,7 respectively. Form the point Q r n S p = R and thepoint I3anSy = T to observe that the points S, Q , R , T are points of the subplane TQ,S. However, the plane TQ,S is Desarguesian of even order so that thequadrangle generates a Desarguesian subplane of order 2; in other words, we may use the quadrangle to coordinatize the Desarguesian plane. Hence, the plane is affine Fano. Now assume that the plane is affine Fano. Let L, M, N be three distinct lines in parallel classes a,@,? respectively and incident with a point S. Choose any point Q # S on L and form Q7 n Sp = R and Ro n ST. Then there is an affine subplane containing Q and S and containing the lines L, M , N . Since we may do this for arbitrary points Q # S on L, we have that the points on the lines of S are covered by subplanes of order 2 and with the same parallel classes. Since every affine point S is incident with a line from each parallel class, it follows that the net defined by the parallel classes a , p , y is subplane covered. Hence, any affine Fano plane is a subplane regular affine plane. Thus, we have the proof of the theorem.0
Corollary 16.18 A finite afine plane of even order is subplane regular and only if the plane is afine Fano.
if
Proof To prove the corollary, we need to show that all of the subplanes of the cover are of even order. To see this, we simply note that a finite subplane covered net covers lines by subplanes of the same prime power order pr and the cardinality of any line is also p". In other words, the order of any finite subplane covered affine plane is always a power of a prime. If the order of the affine plane is even then all subplanes have even order and the corollary follows.0 Now we connect subplane regular affine planes to the affine little Reidemeister condition. Theorem 16.19 A subplane regular afine plane satisfies the afine little Reidemeister condition. Proof: We shall show that the setsrc1,u3" defined above are always groups for all triples of points c, U , v on the line at infinity and all affine lines incident
16.2. THE AFFINE LITTLE REIDEMEISTER CONDITION.
209
with c and then invoke Gleason's lemma. Consider three lines incident with a point S in the parallel classes c, U , v (points) on the line at infinity. Since the affine plane is subplane regular, there is a subplane containing S and having c, U , v as parallel classes. Furthermore, the net N defined by this subplane is a subplane covered net and hence is a pseudoregulus net. Thus, there is a translation group T which acts on the net N and is regular on the affine points. T is Abelian and generated by anytwo of the subgroups with fixed centers. That is, if d is a parallel class of the net N , there is a subgroup Td of T which fixes allparallel classes of N , fixes and acts regularly on the affine points of any line in the class d (incident with d i n the projective extension). Since TUTU= T , it follows that each element gc of Tc can be written as a unique product of elements gu E Tuand gv E TVso that gc = gprgu. We assert that TcII, for I any affine line of the parallel class c, is the set
r;ulv*
To see this, let P be a point on I and let g c ( P ) = Q.Suppose g u ( P ) = R and let J = RC. Since gu is a central collineation, this must mean that gv maps I to J. Furthermore, since gu fixes each line of U , it follows that U ,P and R are collinear. It is then clear that gu induces the mapping ayJ from I to J . Since gc leaves I invariant and gc = gugv, it follows that gv must map J to I . Similarly, it follows that gv induces the mapping from J to I and hence the product gc = gvgu induces the mapping ayJdjI (note the notations are inverted by our convention). Since, and Tc areboth regular on the affine points of I , it follows that TclIis l?;u1v. Hence, is a group for all ordered triples c , U , v of points on the line at infinity and for all lines I incident with c. Thus, the plane satisfies the affine little Reidemeister condition.0
I'Y3'
Theorem 16.20 A subplane regular projective plane satisfies the little Reidemeister condition. Proof:A subplane regular projective planeis subplane regular with respect to any afine restriction so that the little Reidemeister condition holds for any collinear triple of points.0 Corollary 16.21 A finite subplane regular projective plane is Desarguesian.
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CHAPTER 16. SUBPLANE COVERED AFFINE PLANES
Proof Since the plane satisfies the little Reidemeister condition, the plane is Desarguesian by the results of Gleason, Liineburg and Kege1.O
16.3 Embedded Nets. Let 7r be a subplane covered affine plane with an embedded net R so that each subplane of the cover contains the parallel classes of R as a subset of its parallel classes. We note that subplane regular affine planes may be considered subplane covered affine planes with embedded nets of degree 2. If the embedded net has degree 2 3, i.e. nontrivial, then we can show that the affine plane is a translation plane.
Theorem 16.22 Let 7r be a subplane covered
[email protected] plane with nontrivial embedded net R. T h e n 7r is a translation plane. Furthermore,. any maximal embedded n e t is a subplane covered net and hence is a pseudoregulus net. In addition, any finite projective subplane covered net with embedded net of class at least three is Desarguesian. Proof: We shall show that T admitsatranslation group which acts regularly on the affine points. Let L be any line not in the embedded net R. Then there exists a subplane covered net M containing the line L and containing R by assumption. Hence, there exists a translation group T which acts on the affine points and acts as a translation group of the net M . Let J be any other line not in the net M and let W denote a subplane covered net which contains J and R. Similarly, there is a translation group G of the net W. Let c, U ,U be distinct parallel classes of the net R.The translation group T is Abelian and generated by any two of the subgroups with fixed centers. Letting Td denote the subgroup of T which fixes each line of the parallel class of d in R,we note that T = TuTv.Similarly, G = GuG,. For t, E T,, let t,,t, E Tu, TVrespectively so that t, = tutv and for g, E G,, let gu,gv E Gu,Gv respectively so that g, = gugu and note that these representations are unique as the individual groups are Abelian and mutually disjoint, We assert that T, = G,. Suppose not! Then there exists an element of the permutation group on the affine points and on the net R which is
16.3. EMBEDDED NETS.
211
generated by T, and G,, (T,,G,) which contains an element h that fixes all lines of c, all parallel classes of R (or of M fl W ) and fixes an affine point P , as each of the groups T, and G, act regularly on the affine points on any line of c. Since h fixes U and v, it follows that h leaves invariant the line Pu. However, since h fixes all lines of c, it is immediate that h leaves the line Pu fixedpointwise. But,this implies that h leaveseachlineof v fixed. Furthermore, h leaves Pu fixed so that h also fixes each line ofU. Any affine point Q is incident with exactly one line from each parallel class so that h must fix Q. Thus, h acts trivially on the affine points. This says that T, and G, act identically on the net R. Similarly, we have TV= G, as acting on the net R. Since T, and G, act regularly on the affine points of lines of c, it follows that we may assumethat h = t,g, fixes the point P for t,, g, E T,,G, respectively. This says that t, = gF1 when acting on the &ne points. But, this also states that thegroup T,acts on the union of the netsMUW.Hence, it is clear that T,= G,. Similarly, TV= G, so that T = TcTv= G,G, = G. Thus, T is a translation group of the union of the nets M and W which acts transitively on the affine points. Since the net W was chosen arbitrarily, it follows that the affine plane admits T as a translation group. That is, 7r is a translation plane. It remains to show that the embedded net R is a subplane covered net. By assumption, R is not the entire affine plane nor can but a single subplane covered net contain R, for otherwise, there would be a single subplane. Hence, there exist at least two subplane covered nets, say M and W , which properly contain R. Furthermore, assuming that R is maximal, we may assume that M and W share precisely the net R (or more properly the intersection of the subplane covered nets is exactly R). Let P and Q denote any two distinct affine points such that PQ is a line of the net R. Since M is subplane covered, it follows that there is a unique subplane of M , 7 r which ~ contains the points P and Q. Similarly, it follows that there is a unique subplane of W , 7rw which contains the points P and Q. Since 7 r and ~ 7rw share at least three parallel classes, it follows easilythat T M and 7rw share a quadrangle of affine points. Hence, 7 r and ~ 7rw share a subplane which has exactly the parallel classes of R as its set of parallel classes. Since we can vary the points P and
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CHAPTER 16. SUBPLANE COVERED AFFINE PLANES
Q, it follows that R is a subplane covered net and hence a pseudoregulus net (note that the same proof works for the intersection of the subplane covered nets and intersection of the subplanes T M for all subplane covered nets M ) . This proves the theorem.0 Theorem 16.23 Let T be a finite subplane covered afine plane with maximal nontrivial embedded net R. T h e n (1) T i s a translation plane of order q", the kernel of T contains a field K isomorphic to GF(q) and R i s a Kregulus net of degree 1 + q. (2) Furthermore, T admits a collineation group isomorphic to rL(2,q) such that the linear pelements are elations, where p' = q f o r pa prime. More generally, the full collineation group of the net R which leaves each subplane covered net invariant is a collineation group of T . Proof: If the plane is finite then any subplane covered net is a regulus net by previous results. Hence, the net R has order qn and degree 1 q for some prime power q. To see that the kernel contains a subfield isomorphic to GF(q),let K E GF(q)coordinatize the net R. Clearly, there is a Desarguesian plane of order q" which contains R and admits the subhomology group determined by elements of K* and note that any Desarguesian plane of order q" will admit the subhomology group determined by elements of K* = K  (0) as the larger Desarguesian planes are coordinatized by field extensions of K*. Let L be any component and let M be a subplane covered net containing L and R. Then M is a regulus net and there is a Desarguesian plane which contains M . The field coordinatizing the Desarguesian plane containing M is a field extension of K so that the homology group determined by elements of K* leaves each component of the affine plane invariant. That is, the kernel of the plane contains K zi GF(q). Furthermore, any collineation group of the net R which leaves invariant each of the subplane covered nets M containing R will then act . a s a collineation group of the plane T . Consider any net M containing R. Since both are subplane covered nets and hence both regulus nets, there existsa Desarguesian plane C M containing M . Hence, there are fields F 2 L 2 K such that K coordinatizes R, L coordinatizes M and F coordinatizes C M . Clearly, there is a collineation group of all C M isomorphic to rL(2,q) where the pelementsof GL.(2,q) are elations which leaves R,M invariant. Since M is arbitrary, it follows that
+
16.4. SUBPLANE COVERED TRANSLATION PLANES.
213
there is a collineation group of the plane isomorphic to rL(2,q)where the linear pelements are e1ations.U
Remark 16.24 In the chapter on parallelisms, we shall show a connection with the described planes in the previous theorem when n = 4 andparallelisms in projective spaces. Definition 16.25 A subplane covered afine plane shall be said to be ‘normal’ if and only if the subplane covered nets of the couer are all isomorphic.
16.4 SubplaneCoveredTranslationPlanes. In this section, we show that the concepts of subplane regular and regular are equivalent for translation planes. Recall that a ‘translation net’is a net which admits a collineation group called a translation group that acts regularly on the points and which leaves each parallel class invariant. An ‘Abelian translation’ net is a translation net with Abelian translation group. Note that it is not clear that anAbelian translation net has a unique Abelian translation group.
Proposition 16.26 Let R and M be Abelian translation nets on the same points and assume that the degree of R is at least three. If the set of parallel classes of R is a subset of the set of parallel classes of M then the translation groups of R and M are identical. Proof: Let the translation group for R be T and the translation group for M be G. Choose any commonparallel class Q and let P and Q be points of a line incident with a. Let 7 E G map P to Q and p E T map Q to P. Since both translation groups are Abelian, it follows that p and 7 fix all lines of the parallel class Q and fix all parallel classes of R. Furthermore, p7 fixes the point P. It followseasilysince there are at least three fixed parallel classes that p7 fixes each affine point and hence is the identity. If P and Q are not collinear, choose a point R which is collinear to both P and Q. The translations mapping P to R and R to Q are then common to T n G. Hence, it follows that T = G.U
Theorem 16.27 A translation plane is regular (over the prime kernel subfield) if and only if it is subplane regular.
2 14
CHAPTER 16. SUBPLANE COVERED AFFINE PLANES
Proof First 8ssume that the plane is regular. That is, there is a subfield
K of the kernel of the plane such that every three distinct components are contained in a Kregulus. Since every Kregulus is subplane covered, it follows that the plane is subplane regular. Now let the plane be subplane regular and choose three distinct components. There there is a subplane covered net R containing thesecomponents. Since the net R and the plane (net) n are Abelian translation nets and R is a subnet of n of degree 2 3, it follows from Proposition 16.26 that R is a translation net with translation groupT equal to the translation group of the plane T. R is a pseudoregulus net since it is subplane covered and hence may be coordinatized by a skewfield K and hence contains a Z(K)regulus net where Z ( K )denotes the center of the skewfield. Let P denote the prime field contained in Z ( K ) so that, in turn, there is a Pregulus net also admitting the translation group T of the plane. We need to show that there is a subfield J of the kernel of the translation plane so that the Pregulus net is also a Jregulus net for this will imply that the plane is regular. We shall show that we may choose P as a subkernel field. Let J denote the prime field which is in the kernel of the translation plane. Let L be a component of T , Then there is a translation group TLwhich +g, acts regularly on the points of L. Let g E TL,Then n g = g +g +g n times for n a positive integer, means the same relative to P or to J. That is, this element is in the ldimensional Pspace or the ldimensional Jspace. Hence, it follows that we may assume that elements 1 n are in P n J for all integers n. Since both are prime fields, it follows that P and J may be identified. Hence, each Pregulus net is a Jregulus net and hence the plane is regular. This completes the proof of the the0rem.D
+
a
Corollary 16.28 A subplane regular translation plane is either afine Fano or Moufang. Proof We have seen that every regular translation plane is either Moufang or the regulus netsare considered over GF(2). Since even order translation planes are affine Fano, we have the proof to the corollary.0
Chapter 17
DIRECT PRODUCTS In this chapter,we shall provide sufficient theory of direct products of affine planes so as to consider ‘parallelisms’in the next chapter and the structure of nets containing at least one Baer subplane in the chapter after that. We shall consider the direct product of twoaffineplanes. This may be generalized to more general products of nets and other structures. The reader is referred to Johnson and Ostrom [51]for more complete details.
Definition 17.1 Let 7r1 and 7r2 beaJgine planes equal to (Pi,L;,Ci,X;)for i = 1’2 respectively where P, denotes the set of points, L, denotes the set of lines, C; denotes the set of parallel classes and Z; denotes the incidence relation of ~i for i = 1,2. Assume that there exists a 1  1 correspondence U from the set of parallel classes of T I onto the set of parallel classes of 7 r 2 . W e denote the following incidence structure ( P 1 x P 2 , L1 x, &,C,,,Z,,) b y 7r1 x. T2, and called the ‘direct product of T I and 7 ~ 2 defined ’ as follows: The setof ‘points’ of the incidence structure is the ordinary direct product of the two point sets P 1 x P2. The ‘line set’ of the incidence structure L1 x, L2 is defined as follows: Let el be any line of 7 r 1 and let el be in the parallel class CY.Choose any line e 2 of nz of the parallel class CYUand form the cross product el x e 2 of the set of points of the two lines. The ‘lines’ of L1 x,, L2 are the sets of the type
el X e2.
The set of ‘parallel classes’ C,, is defined as follows: Two lines
215
El
x
e2
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CHAPTERPRODUCTS 17. DIRECT
and el, x el, are parallel if and only ifl!. is parallel to .!l,. Note that this implies that e 2 is parallel to .!l, since both lines are in the parallel class au. Parallelism is an equivalence relation and the set of classes is C,,. Incidence Z, is defined by the incidence induced in the above defined sets of points, lines, and parallel classes. Definition 17.2 A net is a ‘vector space net’ if and only if there is a vector space over a skewfield K such that the points of the net are the vectors and the vector translation group makes the net into an Abelian translation net. Theorem 17.3 (1) If 7r1 and 7r2 are afine planes and U is a 1 1 correspondence from the parallel classesof 7r1 onto the parallel classes of 7r2 then 7r1 x,, 7r2 is a net whose parallel class set is in 1 1 correspondence with the parallel class set of either afine plane. Fbrtherrnore, this net admits Baer subplanesisomorphicto 7r1 and 612 anygivenpoint. (2) If 7r1 and 7r2 both have order n then the order of 7r1 x , 7r2 is n2 and the degree (number of parallel classes) is n f 1. (3) If 7r1 and 7r2 are both translation planes with translation groups Z f o r i = 1 , 2 respectivelythen x , Ta isanAbeliantranslationnetwith translation group TI x T2.
Proof It follows easily that since parallelism is an equivalence relation on the line set of 7ri for i = 1,2, then parallelism is an equivalence relation in the line set of nl x,, 7 r 2 . Hence, if cy is a parallel class of T I ,we may denote by a x cyu or (a, ).a the corresponding parallel class of 7r1 x07r2. It is then clear that thismapping is a 1 1 correspondence from the parallel classes of 7r1 onto the parallel classes of 7r1 x,, 7r2. Given any point ( 4 , P 2 ) of 7r1 x,, 7r2 where Pi E Pi for i = 1,2, choose any parallel class (a, au)of C, . Then exists a unique line t a , p l , &,,h of for i = 1 , 2 respectively that contains PI, P2 respectively. Thus, there is a unique line &,pl x ea,,,& of ,C1 x , C2 containing (P1,Pz). This proves that the incidence structure is a net and furthermore proves (1) and (2) since it follows that the subplanes themselves become Baer subplanes of the net. To prove (3), let (P1,Pz) denote points of the direct product for P,E Pi, i = 1,2. Then, it is easy to verify that the mapping 7 : ( 7 1 , ~ for ) 7, E Ti , i = 1 , 2 defined by (PI, P 2 ) 7 = (PITI, P 2 7 2 ) is a collineation which fixes each
C ,,
17.1. THE RECONSTRUCTION THEOREM.
217
parallel class of the direct product. Furthermore, since Ti acts regularly on the points of Pi for i = 1,2, it follows that TI x TZacts regularly on the points of the direct product. Hence, the direct product net is a translation net and since Ti are both Abelian, the direct product net is an Abelian translation net. This completes the proof of (3).0
Remark 17.4 In part (3) of the previous theorem, if the two translation planes are defined over the same prime field then the direct product net is a vector spacenet which is an elementary Abeliantranslation net. On the other hand, onecannotguarantee this with the existence of two Baer subplanes within a net. However, when there are three Baer subplanes incident with a given point of an Abelian translation net, one always obtains an elementary Abelian translation net as seen below. Definition 17.5 A direct product net where CT is induced from an isomorphism of 7r1 onto 7r2 is called a ‘regular direct product’ b y T I . Identiking n1 and n2, we also refer t o n1 x 7r1 as a ‘2fold product’.
Corollary 17.6 A regular direct product net contains points incident m*th at least three mutually isomorphic Baer subplanes. Proof: Note that as we have the regular direct product 7r x n,we also obtain the Baer subplane defined by {(P,P ) ; P a point of n} as well as n x 0 and 0 x n.0
17.1 The ReconstructionTheorem. In the forthcoming chapters, we shall be considering how the existence of Baer subplanes within nets (Baer nets) effects the structure of the net. We have considered derivable nets where the net is covered by Baer subplanes and showed there is a complete characterization theory. In this section, we shall showhow the use of direct products of affine planes facilitates the theory of Baer nets. Since we shall be restricting ourconsiderations to nets which lie in translation planes, we focus on Abelian translation nets. In the following, we shall often use the term ‘pointBaer’ to emphasize the fact that the subplane is Baer in the net but perhaps not Baer in a net
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CHAPTER 17. DIRECT PRODUCTS
or translation plane extending the given net whereas the subplane remains a pointBaer subplane of the extension.
Theorem 17.7 (The Reconstruction Theorem) Let M be an Abelian translation net. If M contains two distinct pointBaer subplanes incident with a point whose infinite points arethe infinite points of M then M is a direct product net. Proof First, we prove that any such pointBaer subplane p. is a translation plane. Let P be a point of po. Assume that there is an element g of the assumed translation group T such that Pg is a point of p. and P and Pg are incident with a line PPg of M in the parallel class a. It follows easily that g must fix each line of the parallel class Q. Now let Q be a point of p. which is not incident with PPg and note that Q and P are incident. Hence, the line &a is fixed by g so that &g is incident with Q. Note that P& is in a parallel class p # a of M and g fixes each parallel class. So, &g is incident with (PQ)gand with &a both of which are lines of the subplane po. Therefore, &g is a point of po. Now let S be any point of p. incident with PPS. Let a,p, S be mutually distinct parallel classes of M . Let R = S,8 fl PS so that R is a point of p. and the previous argument shows that Rg is also a point of po. Now use the above argument with R in place of P to show that Sg is a point of po. Hence, g leaves p,, invariant. This is to say that once any point image Pg of a point P in p. is back in p. then g leaves p. invariant. Since the group is transitive on the affine points, this implies that there is a subgroup Tpoof T which fixes p. and is a translation group acting transitively on po; p,, is a translation plane. Now let p1 and p2 be distinct subplanes which share the point P and let Ti for i = 1,2 denote the translation subgroups acting regularly on pi for i = 1,2 respectively. It follows that the group T1T2 is isomorphic to the external direct product (sum) of the two groups. For each parallel class a of the net M , there is a corresponding parallel class ai of pi respectively such that the lines of a;are lines of a. Isolate on p1 and define the mapping o by a(a1)= 0 2 . Form the direct product net p1 x,p2 and note that this net is an Abelian translation net. We will show that this direct product net is isomorphic to the net M .
RECONSTRUCTION 17.1 THETHEOREM.
219
Our method of proof is to consider the four coordinate systemsfor pl, p2, p1 xUp2 and M . All of these coordinate systems have linear equations representing lines, so we suppress the notation for the various ternary functions. We shall use the notation (Qi, +i, *i) to represent the coordinate quasifields for pi for i = 1,2. We use the notations (Q, +, *) and (Q1 x,, 92,e,0 ) to denote the coordinate systems for M and p1 xUp2respectively. Furthermore, we use the notations for infinite points (mi),(Oi) and (li)for a particular set of three infinite points of pi for i = 1,2. We choose a coordinate system for M such that P = (0,O) and infinite points (m), (0) and (1) so that (mi) = (m)i,(O= i)(0)i and ( I d ) = (1)i for i = 1,2 to connect the parallel classes of the direct product net to the net M . Note that this does not imply that 11 = 12. However, since P is a point common to both subplanes pi, we may agree that 01 = 02 = 0. The points of pi have the form (z:i,yi) for i = 1,2 and si,yi E Qi.Since a point of the direct product net is of the form (P1,Pz) for fi points of pi and i = 1,2, we may take the form for points of the direct product net to be (21,y1,22,92) for zi,yi E Qi for i = 1,2. Note that the group TIx T2 acts on the direct product net and T1T2 acts on the net M . Furthermore, the action of Ti on the points of pi is as follows: Let Ti = (g,&) such that gai,bi
(pa,. pa.)
(G
+i
ai,yi +a bi) ; ai, bi E Qi
for i = 1,2. Letting T denote the translation group of the net M , the action of T on the points of M is as follows: Let T = (Ta,b) where 7a,b:(2,Y)~(Z+a,y+f);a,bEQ.
x
As also acts on M , it follows that for each (ai,bi) in Qi x Qi, there exists a unique (a,b) for a, b E Q such that gairbi = Ta,b. Since cai,bi maps (Oi,Oi) onto (ai,bi) and Ta,b maps (0,O) onto (a, 6) and Oi = 0, it follows that ai = a and b, = b. So, it follows that Q i is a subquasifield of Q for i = 1,2. Now T1 x T2 acting on p1 x, p2 has the following form: ( ~ 1 , ~ 1 , 2 2 , ~+ 1 2 )(XI+I a l , +l~ b1,21+2 ~
~
2 +2 ~b2) ;~ai,bi 2 E Qi
for i = 1,2.
CHAPTER 17. DIRECT PRODUCTS
220
We note that (01,01,02,02) maps to ( a l ,b l , a2, b2) under the action of Tl x T2 and (0,O) maps to (a1+a2, bl +b2) under the corresponding elements of T1T2. Define M  as the subset of M with points Q1 X Q2. We assert that M  is a subnet of M containing pi for i = 1 , 2 which is isomorphic to p1 x u p2 under the mapping $ : (a&,a2,b2)

(a1 + a 2 , h +b2).
We have the lines
xi = Q , yi = xi *i mi +I bi for %,mi,bi E Qi. Moreover, by our correspondence CT where mi)^^ = (mz),and points of the direct product are in the form (z1,91,22,y z ) , we have a line of the direct product represented as a point set by
{ (z1,z1 * I
m1 +I
b l , z2,z2 * 2 m2 +2 b2) ; zi E Qi for i = 1,2}
where mi, bi E Qi for i = 1,2 and ((cl,yl,c2,92) ;pa E
Qi for i = 1,2} where Q E Qa, i = 1,2.
The images under $ of the points respectively are:
{ ( z l +z2, (x1 * m1 +I h )
+ (z2 * m2 +2
b2)) ; xi E Qi, i = 1,2}
and {(~1+~2,91+92);9i€Q9i,i=1,2}.
Now note that a line of pi containing (Oi,bi) with slope (m)i = (mi) is the line = zi si mi +i bi which is the same line as represented by 9 = x * m (bl b2). Hence,
+ +
{z1+ 32, ($1
* m1 +I
h ) + ( m * m2 +2 h ) ) ; xi E Qi, i = 1,2}
+ (bl + b z ) , or rather as a point set ‘contains’ ( ( 2 1 + 2 2 , ( m * m+ b l ) + (x2 * m + b2)) ; zi E Q*,i = 1,2}
is the line y = z * m
17.1. RECONSTRUCTION THETHEOREM,
221
and note that
Similarly, the line z = c = (cl, cp) of the direct product maps to the line maps to the line z = (c1 c2) of M . Hence, y = z 0 m @ b maps to y = z * m b and z = c maps to z = (c1 c2) which proves that M  is isomorphic to p1 xu p2 and note that M  is a subnet of M ; a subset of points and lines which it itself a net. It remains to show that M = M  . Assume that M is not equal to M and let Z be a point of M  M  . Then there is a unique line Li of pi for i = 1,2 which is incident with Z since the subplanes areBaer subplanes of M and hence, are pointBaer subplanes. First suppose that L1 is not parallel to La. Then Z is the intersection of these two lines which shows that Z is in M. Hence, L1 is parallel to L2 and thus L1 = L2. Thus, the only possible points of M which are not in M  lie on common lines of the two Baer subplanes pi for i = 1,2. Choose any parallel class P distinct from the parallel class containing L1 and form ZP. Let B be any point of ZP which is not on a common line of both Baer subplanes. Then, B must be a point of M  so that ZP = BP which implies that 2 is the intersection of two lines of M  , namely ZP and L1 and since M  is a net this forces 2 to be a point of M. Therefore, no point of ZP can be a point of M  and all points on ZP are points on common lines of pi for i = 1,2. It also follows that all lines incident with Z and not equal to L1 have this property. Note that the two Baer subplanes share aunique affine point P by assumption and it therefore follows that L1 = P Z . Take any point E of ZP and note that there is a common line of the two subplanes incident with E which forces the common line to be PE. So, every point on ZP is incident with a common line ofp1 and p2 incident with the common point P or we are finished. Hence, there are card(2P) parallel classes of M . If the net is finite, we see that we have a contradiction. So, we may assume that there are infinitely many parallel classes. Let D be a point of p1 and let 6 be a parallel class distinct from P and the parallel class containing P2 and we may assume that 6 does not contain PD since there are more than 3 parallel classes by above. Then D6 intersects Zp in a point F since they are lines from distinct parallel classes and M is a net. However, the point F as it is incident with ZP ison a common line of the two subplanes incident with P. That is, F = FP n D6
+
+
+
222
C H A P T E RPRODUCTS 17. DIRECT
which means that F is a point common to two distinct lines of M  so that F is in M  contrary to our assumptions. Therefore, M  = M and we have completed the proof to our theorem.0
Corollary 17.8 Let M be an Abelian translation net. If M contains three distinct pointBaer subplanes incident with a point whose infinite points are the infinite points of M then M is a regular direct product net and each pair of the planes are isomorphic. firthermore, M is then a vector space net over a field L and the pointBaer subplanes may be considered Lsubspaces. If one of the subplanes ro has kernel KOand M is isomorphic to ro x r 0 then M is a KOvector space net. At least three of the pointBaer subplanesof the net whichshare an afine point and all of their parallel classes are Kosubspaces but not all pointBaer subplanes are necessarily K,subspaces. Proof First of all, from the previous result, we note that any two pointBaer subplanes direct productto thenet as a set of points. If we identify the points of the net as (21,y1,22, yz) where si, are in two of the subplanes pa, i = 1,2, then, since the net is an Abelian translation net, we may regard the direct product of points (21,91) x ( 2 2 , yz) unambiguously as ( 2 1 , y1,22,y2). We identify the zero elements in the two quasifields corresponding to thetwo subplanes and notethat, in particular,we are identifying the point (0, O,O, 0) = P as common to the two subplanes. Let T denote the translation group of M . Then we may identify the points of M with the elements of T. Furthermore, given any Baer subplane p incident with a given point P = (0, O,O, 0), there is a subgroup Tp of T which acts transitively on p so that p is a translation plane with translation group Tp. Furthermore, the ReconstructionTheorem shows that for any two Baer subplanes pi for i = 1,2, the two translation subgroups TpI and Tp2generate T. Hence, T = TP1TP2. Assume that there is a third subplaneincident with thecommon point P and denote the subplanes by pi for i = 1,2,3. Hence, there is a translation subgroup Tp3such that T = Tp3Tp2= Tp3Tp1. Identify a point W of M by the unique element TW of T which maps the given common point P onto W .
RECONSTRUCTION 17.1. THETHEOREM,
223
We define an ‘addition’ on M by requiring that W + E = C if and only
if
TWTE
= TC.
Now it follows that the ‘sum’ of p3 and p2 is M Represent the net as a direct product net p1 x,p2 and consider the set of points C3 = { ( Q , O ) ; (Q, M ) is a point of p 3 } . If there exists a point of p l , say (Q, 0) which is not in C3 then since p2 x p3 = M as a setof points, then, in order to write the point (Q, 0) as a sum of elements of (0, S) (2,N ) for 2 E p2 and ( 2 , N ) E p3, it follows that Q = 2. Now assume that there are two points of p3 of the form (Q, N ) and (Q, U) then clearly N = U as otherwise p3 would intersect p2 in a point other than (O,O,0,O) = P which would force p3 = p2. Hence, we may define a mapping r from the points of p1 onto the points of p2 in the following manner: If ( Q , N ) is a point of p3, let .(Q) = N . It follows that r is bijective on points by repeating the above argument with p2 and p1 interchanged. Regard M as the direct product of p1 x, p2. Consider any line L of M incident with P. This line arises as the directproduct of lines L; of pi incident with ( O i , O i ) so that L is L1 x La. Taken in the context of the addition defined on points of M , it follows that any line incident with P is the sum of its intersections with any two distinct Baer subplanes containing the point P. Hence, L contains points # P of p3. Thus, if (Q, N ) is a point of p3 on L not equal to ((0, 0), (0,O)) then Q is a point of p1 collinear with (01,Ol) and hence lies on L1 and similarly N is a point of p2 collinear with (02,02) and therefore on La. So, if we extend the mapping r on lines L1 of p1 incident with the origin (01,Ol) to map to the lines L2 of p2 incident with the origin (02,02) where L = L1 x L2 is a line incident with (O,O, 0, 0), we have extended the definition of r and note that it is clear that acting on the parallel classes T = CT.Since the two planes are translation planes, it follows that r is a isomorphism of p1 onto p2. By the Reconstruction Theorem, we may interchange p3 with p1 to conclude that all three planes are isomorphic and that we have a regular direct product net. Since the subplanes are isomorphic translation planes, let planes pi have prime fields . F d for i = 1,2,3. Let g be an isomorphism from F1 into F2. The planes pi admit a kernel homology group corresponding to Fi  (0) whichfixes each infinite point of pi and fixes (Oi,Oi). It follows directly
+
CHAPTERPRODUCTS 17. DIRECT
224
that the direct product net p 1 x p2 admits as a collineation group the group ( 3 1 (0)) x (F2 (0)) that fixes each parallel class of the direct product net and fixes the point (01,01,02,02) = (O,O,O, 0). Hence, the net admits the field mappings (h,h g ) for all h E 3 1  (0) as collineations. Since the groups 3; (0) normalize the corresponding translation groups of pi, it follows that the indicated mappings normalize the translation group of the net. Hence, we have shown that the net is a vector space net over the field ( ( h , h g ) ; h E FI},Similarly, if one of the subplanes x, has kernel K,, it follows as above that the net is isomorphic to x, x x, and is a vector space over a field isomorphic to K,. If the three subplanes in question are represented as ((Q,O) ; Q E T O ) , { ( O f & ) ; Q E ro) and ((999) ;Q E x,} then identifications may be made so that the three subplanes are K,subplanes, but it is not necessarily the case that all Baer subplanes of this net are, in fact, K,subplanes. We shall return to this situation laterin the text. This proves the result.0
Theorem 17.9 Let M be a vector space net over a skewfield K where M i s a regular direct product net of two isomorphic pointBaer subplanes with kernel K,. T h e n M admits N GL(2,K,) as a collineationgroupthat f i e s a n aflne point and fieseach parallel class. firthemnore, l? is generated bv the groups whichfipointBaer subplanes pointwise. If M = T, X K, and K, is the kernel of x, as a left K,subspace then the
action of a n invertibleelement
:]
[
T(a,b,c,d) : (Po,P 1 )
onMis
++
(ape + cpl, bp,
f
dpl)
f o r a b, , c, d in K, and p,, p1points of x,, Proof: Represent the direct product net as noted by x, x T,, let K, denote the kernel of x, and consider x, a left K,vector space. The definition of the mappings above clearly implies that these induce collineations of the direct product net. Note that (p,,O) W (ap,, bp,) so the mappings when a = 1 and b = 0 are ‘Baer collineations’ of the direct product net in the
17.1. RECONSTRUCTION THETHEOREM.
225
sense that these fix the subplane 7ro x 0 pointwise. Notice that T(l,O,c,d)T(a,b,l,O)= 7(a,b,ac,c+d)
and since a and c are not both zero, it follows that Baer collineations generate the group which is clearly isomorphic to GL(2,K,).U We now turn to
17.1.1 Desarguesian Products. We consider the nature of the lines of 2fold products (regular direct products) of Desarguesian affine planes 7ri, i = 1,2. We identify 7r1 and 7r2 and consider the corresponding regular direct product or 2fold product. In the proof of the Reconstruction Theorem, we considered a coordinate setting. Here we repeat this setting where the subplanes are Desarguesian. However, we need not worry about the two additions and quasifields. Hence, the addition that we use is the addition in any coordinate skewfield for a Desarguesian plane TI. Let TI have points ( z 1 , y l ) where z 1 , y l are in a skewfield KT1coordinatizing T I . Hence, the points of 7r1 x 7r1 are ( ( q , y 1 ) , ( ~ 2 , y 2 ) ) which we rewrite as ( q , y 1 , ~ , y 2 )for all zi,yi E KT1and i = 1 , 2 . We consider the lines of the cross product incident with (0, 0, 0,O). Now assume that the associated vector space is a left K,,vector space. Recall for Desarguesian planes, the 'slopes' are then written on the right to ensure that the components are left K,,subspaces. Choose a line y1 = z 1 a for a E ITrl. The parallel class of 7r2 associated with a is then a and the unique line incident with (0,O) of 7r2 is 9 2 = 2 2 a . Hence, the line of the cross product is ( ( z 1 , ~ 1 a , 1 ~ 2 , 5 2 a;)xi
E KT1fori = 1,2}.
Similarly, the line of the cross product incident with (O,O, 0,O) in the parallel class ((W), (co)a)is
Since the plane 7ri admits the translations (zi, yi) all a, b E K T l ,i = 1,2, it follows that the mappings (51,91,22,Y2)

(21 +%Ill

+b,Z2+C,YZ
( q + a , yi + b) for
+4
226
C H A P T E RPRODUCTS 17. DIRECT
are collineations for each a, b, c, d of K,, which fix each parallel class of the net. Hence, the net admits a group of translations and it follows to specify the lines, we need only specify the lines incident with the zero vector. In thechapter on the structureof derivable nets embeddable in projective spaces PG(3,K ) for K a skewfield, we were able to consider the underlying vector space in such a way so that we may assume that the vector space is a left space over K where the components of the space have the basic form z = 0, y = Sz for all S E K and the Baer subplanes have the basic form Ta,b = ((aa,crb,pa,pb) ; a,P E K} where (a,b) E K x K  {(O,O)}. The components are not alwaysleft Ksubspaces whereas the Baer subplanes incident with the zero vector are left Ksubspaces. Now identify K and K,,. Note that
and {(O’Yl,O,Y2)
;vi
E K,, f o r i = V } = TOJ.
Thus, the components of the direct product net are Baer subplanes of a corresponding derivable net and hence a 2fold product is a derivable net. Also, note that while the components are left K,,subspaces, the Baer subplanes of the direct product net (the original y = and z = 0) are natural ‘right’ K,,subspaces. Now suppose that x is given by a left tspread over a skewfield K . We formalize this as follows:
Theorem 17.10 Let x be a Desarguesian afine plane defined b y a spread of left tdimensional vectorsubspaces of a left 2tdimensional Kvectorspace where K is a skewfield. (1) Then, corresponding to 7r is a dimension t extension skewfield KT of K considered as a left Kvector space. (2) A 2foldproduct x IT producesaderivable net whosecomponents (line incident with the zero vector) are left 2dimensional K,subspaces and 2tdimensional left Ksubspaces of a left &dimensional left vector space V& over K .
17.1. RECONSTRUCTION THE THEOREM.
227
(3) There is a natural definition of the set of vectors of T x T as a right 4dimensional K,subspace and as a right 4tdimensional Ksubspace. firthermore, the Baer subplanes incident with the zero vector are right 2dimensional K,subspaces and right 2tdimensional Ksubspaces. Proof The proof to the previous result shows that (1) is valid. Furthermore, if y1 = x1a for a in K , is a component of T then
is a component of T x T . It follows that the latter is a left 2dimensional K,subspace provided g1 = z1a is a left ldimensional K,subspace. In this representation, the Baer subplanes are T x 0, and
& = {(P, aP); P is a point of T ; for a fixed in K,}. Representing points inthe form (21,y1,22,yz) for xi,yi in K,, we may define a right K,vector space by defining (z1,y1,22,y2)a = (21a,yla,22a,y2a) for all a in K,. Note that this makes T x 0,O x T and into both right and left K,subspaces but l , is a right K,subspace which is left if and only if a commutes with K,; if and only if a is in the center of K,. Similarly, if we decompose the left tvectors zi, gi as ( q i , z2i,...,xti) and (yli,y2i,...,yti), respectively, for i = 1,2, there is a natural way to make T x 7r into a right 4tdimensional Ksubspace. This proves (2) and (3).0
Definition 17.11 W e define ‘left’ and ‘right’projective spaces as follows. If Vzt is a left 2tdimensional Kvectorspace and Vqt is a right 4tdimensional Kvector space, denote the corresponding projective spaces b y PG(2t1, K)L: and P G ( 4 t  1,K ) R respectively. Definition 17.12 Let V be a left vector space over a skewfield K . A ‘lefttspread’ S2 is acover of thenonzerovectors of V b y left tdimensional Ksubspaces. W e shall say that a left tspread is ‘Desarguesian’ if and only if there is a extension skewfield K’ of K of degree t such that elements of the left tspread are left ldimensional K’subspaces.
228
C H A P T E RPRODUCTS 17. DIRECT
Note that if K has a tdimensional extension skewfield K' then there isa natural lspread over K' (covering of V b y the ldimensional K' subspaces) and since each such lspace overK' is a tspread over K , it follows that Desarguesian left tspreads are equivalent to tdimensionalskewfield extensions of K (considered as left Kspaces). Analogously, we may consider right tspreads. Remark 17.13 Furthermore,partial spreads of left subspacesare called 'left partial spreads'. This is somewhat problematic terminology for pseudoregulus nets as a net of the form y = Sx, x = 0 f o r all S E F and F a skewfield has right components and therefore is termed a 'right' pseudoregulus. Remark 17.14 (1) Let NK denotethenet defined by the 'subplanes' of x T,IT x 0,Ox IT and k ' ,for all cy in K . Then NK is a 'right' pseudoregulus net which may be coordinatized by a 'right' 4tdimensional Kvector space whose components (the subplanes) are 'right' 2tdimensional Ksubspaces. In this representation, NK has components x = 0,g = ax for all cy E K . Analogously, derivation of a leftnet over a skewfield K produces a leftnet over the skewfield K o p p which is then a rightnet over K . (2) Notice that the set of Baer subplanes of the derived net (IT x T)* intersects the original Desarguesian afine planeIT in the original spread for
IT
n.. Definition 17.15 Any pseudoregulus net is a subplane covered net and b y the classification theorem, every subplane covered net is an Fpseudoregulus net for someskewfield F . If a left (respectively, right) Fpseudoregulus net contains a left (respectively, right) Kpseudoregulus net, where F is a skewfield extension of K , we shall say that the Fnet is a 'left (respectively, 'right') extension of the Knet'.
17.2
LeftSpreads Produce Right Partial Spreads.
Corollary 17.16 A Desarguesian spread in P G ( 2 t  1, K ) & producesa derivable partial spread of (4tl)dimensional right Ksubspaces in PG(4t1, K)R.
17.3. LEFTSPREADSPRODUCELEFTPARTIALSPREADS.
229
( l )If p and T are two Desarguesian afine planes defined by left tspreads over a skewfield K then both p x p and IT x IT may be regarded as the same
right 4tdimensional Ksubspace whose corresponding Baer subplanes may be considered right 2tdimensional Ksubspaces. (2) Define the subnet NK given by p x 0 = IT x 0 as y = 0, 0 x p = 0 x IT as x = 0, and {(P,"P); P a vector of p (= T ) ) as y = a x for all (Y E K . Then the derived nets of p x p and T x IT both contain NK.
Proof We may identify the vectors V of the two associated Desarguesian affine planes. Choose any left basis B for V and define a right Kspace via this basis using the representation (XI, 22, ...,x2t)a = ( X ~ C Y...., , x2ta). It follows that the Baersubplanes of the common set NK are right 2tdimensional Ksubspaces. If p is Desarguesian then there exists a skewfield extension Kp of K such that p is a 2dimensional left Kpvector space. Hence, there are Baer subplanes of p x p, { ( P ,UP); P E p } for a in Kp and hence of the basic form y = ax where x is a left 2tvector over K. In this form, a is a 2t x 2t matrix with entries in K . Thus, it follows that (ax)a = a(xa) with respect to the basis B. Hence, all Baer subplanes of either of the two direct product nets may be regarded as right Ksubspaces. U
17.3
Left Spreads Produce Left Partial Spreads.
Now we shall want to consider some 'left' partial spreads in PG(4t  1,K)L: defined by leftspreads inPG(2t  1,K)r. whether the spread is Desarguesian or not. When t = 2 in the previous section, we may construct some left partial spreads in PG(7, K) which may not come from derivable nets as follows: Let V, U, W be any three mutually disjoint 4dimensional left subspaces of a 8dimensional left Kvector space h.(Think of V,U, W as IT x 0,O x IT and .l1 of the previous discussion.) Decompose V8 relative to V and W so that writing vectors of v8 in the form (x, y), we assume that V, W and U have the equations x = 0, y = 0, y = x.
230
C H A P T E R 17. DIRECT PRODUCTS
Lemma 17.17 Consider any 2dimensional left Ksubspace X of V . Let X" denote the image under the mapping (x,y) H (y, x) where x,y are 4vectors over K . Then X @ X" is a 4dimensional left Ksubspace of V8 which intersects x = 0, y = 0 , y = x in 2dimensional left Ksubspaces. Conversely, any 4dimensional left Ksubspace that intersects x = 0, y = 0, y = x in 2dimensional left Ksubspaces is of this form.
Proof Let P be any point of a 4dimensional left Ksubspace with the indicated properties and which is not incident with x = 0,y = yz for any y E K. By uniqueness of representation, there exist unique nonzero vectors on x = 0 and y = 0 such that P=
+
( ~ , ~ , ~ , ~ , ~ 1 , 2 / 2 , ~ 3( ,~~ 14 /, 4~) 2 , ~ 3 , ~ 4 , ~ , ~ , ~ , ~ ) ~
Inparticular, for any vector (O,O,O,O,yl,y2,y3,y4), thereisa vector ( x ~ , x ~ , x ~ , x ~ , O,O, O , 0) which is not linear a combination of (yl,y2, y3, y4,0,0,0,0)as the intersection on y = 0 is 2dimensional. Similarly, there exist unique nonzero vectors on x = 0, M = x such that
and there exist unique nonzero vectors on y = z,y = 0 such that
It follows that
and (~1,~/2,~3,~4)=(~1+~1,~2f~2,Z3,f~3,Z4+~4)=(~1,S2,S3,S4).
Thus, the left 4dimensional space generated by these vectors is generated bY
17.3. LEFTSPREADSPRODUCELEFTPARTIALSPREADS.
231
and is of the form X @ X" for X the left 2dimensional Kspace ((0,0,91,Y2,2/3,y4),(o,o,Xl,x2,~3,~4))
Note that we know that (yl,y2,y3,y4)# y(x1,x2,x3,x4)for y E K as otherwise P would be on y = yx contrary to assumption. Hence, the subspace must be generated by the four linearly independent vectors. Since this space is of the form X @ X",we have completed the pro0f.U
Lemma 17.18 Define a 'right' &dimensional Kspace relative to V @ W so that V,U, W are both right and left Ksubspaces. Let NK denotethe right pseudoregulus netwithcomponents x = 0, y = ax for all a E K . Note that the set of components contains right 4dimensional Ksubspaces and contains V,U,W . Let P(NK)denote the set of vectors which lie on the components of N K 8 If P is any vector notin P(NK)then there existsa unique 2dimensional left Ksubspace of V , X , such that P is incident with X @ Xu;there is a unique 4dimensional left Ksubspace containing P that intersects V,U,W in 2dimensional left Ksubspaces. Proof: The previous lemma shows that there is a unique such 4dimensional left subspace containing P that intersects V,U, W in 2dimensional left Ksubspaces and it has the form X @ X'.U
17.3.1 2fold Products. Theorem 17.19 Given a left spread S of V of 2dimensional left Kvector subspaces, { X @X" ; X E S } = PS is a partial spread of 4dimensional left Ksubspaces which contains the pseudoregulus partial spread defining NK as a set of Baer subplanes. Let 71s denote the afine translation plane determined b y the spread S. Considering (T as an isomorphism, we may form the regular direct product 71s X" rs. Then PS is defined by 71s x, 71s where 71s. Proof Let p denote the affine translation plane determined by the spread
S. We note that p has kernel containing K and may be considered a left Cdimensional subspace. Form p x p to obtain a net. But, note that this is
CHAPTERPRODUCTS 17. DIRECT
232
not necessarily a derivable net as p is not necessarily a Desarguesian affine plane. Now clearly, p x 0,O x p and {(P,QCP) ; a E K} are Baer subplanes of the net which are isomorphic to p. We can make the space into a right 8dimensional Ksubspace so that these Baer subplanes are in the net and are 4dimensional right Ksubspaces. The lines of the net incident with the zero vector are the lines of the regular direct product net and are 4dimensional le$ Ksubspaces of the form X $ X" for all X E K by the previous uniqueness result. This proves the theorem.0
Theorem 17.20 U{X $ X"  P(NK); V2dimensional left subspaces X of S } is a disjoint cover of the vectors of V8 which are not in P(NK). Proof Note that if two left spaces X $Xa and Y $ Y" have a point P in common outside of NK then the previous argument shows that X = Y. Hence, we have a disjoint cover as maintained. Alternatively, simply note that we obtain a 2fold product net.U
Corollary 17.21 The net p x p containing the pseudoregulus netNK (as a set of Baer subplanes) admitsa collineation groupG isomorphic to GL(2, K) defined as follows:
(the elements of K act on the left b y scalar multiplication). firthermore, G leaves invariant each 4dimensional KsubspaceX $X" f o r X a 2dimensional Ksubspace of V and acts on the pointsof X $Xu 
~(NK). Proof Represent
NK in the form x = 0, y = s a 4 for all QC E K.
+
Suppose (x,y) = (x,0) (0,y) is in X $X" so that both s and y are in X. Then QCZ Py, 6s yy are both in X . Hence
+
is i n X $ X '
+
for all a,P,6,'y E K.0
Chapter 18
PARALLELISMS A ‘parallelism’ in PG(3,q ) is a set of 1 + q + q2 spreads which cover the lines. Hence, each line lies in exactly one of the spreads. More generally, a ‘parallelism in PG(3,K)’ where K is a skewfield, is a set of line disjoint spreads which cover the line set of the projective geometry. Recall, when K is a field, a spread S is said to be ‘regular’ exactly when the regulus generated by any threedistinct lines of S is contained in S. A ‘regular parallelism’ is a parallelism where the spreads are all regular. In this chapter, we shall use the direct product constructions of the previous chapter to construct translation planes or rather spreads in PG(7, K) from what we call ‘Desarguesian parallelisms’.
18.1 Desarguesian Parallelisms. Definition 18.1 A set of left tspreads of a left vector space V over K is a ‘partial left tparallelism’ if and only if the tspreads contain no common tdimensional subspace. A ‘lefttparallelism’ is a partial left tparallelism such that every left tdimensional Kspace of V is contained in exactly one of the left tspreads. A partial left tparallelism shall be said to be a ‘Desarguesian partial left tparallelism’ if and only if each left tspread in the partial left tparallelism is Desarguesian. Similarly, we have the natural definition of a ‘Desarguesian left tparallelism 233
234
CHAPTER 18. PARALLELISMS Analogously, we may consider the ‘right’ structures.
We first note that from right partial spreads of pseudoregulus tdimensional extensions of Kpseudoregulus nets, we obtain Desarguesian left partial parallelisms.
Theorem 18.2 Let S be a partial spread in PG(2tk  1,K)R defined by a set of right (tk  1)dimensional Ksubspaces where K is a skewfield which consists of a set X of pseudoregulus right tdimensional extensions of the right Kregulus net NK . Then the set of subplanes of each pseudoregulus net in X induces a Desarguesian left tspread on any component of NK and the union of the spreads is a Desarguesian left partial tparallelism of PG(tk  1,K).
Proof: The subplanes of a right Fpseudoregulus net areF2dimensional Desarguesian affine planes and are left Fsubspaces in the manner described in a previous lemma. If F is a right tdimensional skewfield extension of K then each Fpseudoregulus gives rise to a spread in l& over K of left tdimension Ksubspaces which are ldimensional left Fsubspaces. That is, we have a Desarguesian left tspread of l&. It remains to show that we obtain a Desarguesian left partial parallelism for the set of tdimensional extensions. Assume that there is a common tspace X in the two induced Desarguesian spreads. It is always possible to coordinatize so that V2tk is isomorphic to l& @ l& and that the subplanes of the Fpseudoregulus nets are of the basic form Y CBY for some tdimensional left Kspace Y. Hence, if there is a common tspace X ,then there is a common subplane of the two pseudoregulus nets within the net defined by the partial spread. Since any common subplane may be coordinatized by a skewfield identical to the skewfield coordinatizing the pseudoregulus net, it follows that the two nets are identica1.O
Remark 18.3 We have seen that spreads consisting of tdimensional extension pseudoregulus nets induce Desarguesian partial tparallelisms. over K What is not clear is if Desarguesian partial tparallelisms in produce spreads consisting of tdimensional extension pseudoregulus nets of a Kpseudoregulus net. For simplicity, we consider the situation when t = k = 2.
18.2. WHEN THE VECTOR SPACE IS 4DIMENSIONAL.
235
18.2 When the VectorSpace Is 4dimensional. We now specialize to the case when V4 over a skewfield is 4dimensional so that a left 2spread is a spread of left 2dimensional subspaces of V . In this case, we call partial left 2parallelisms merely ‘left partial parallelisms’ since a 2spread S determines an affine translation plane T S . Furthermore, we also say that theset of left spreads as setsof ldimensional projective subspaces in P G ( 3 , K)c is a partial parallelism of the projective space. But, usually, we shall not make a distinction between partial 2parallelisms of the vector space and the parallelisms of the associated projective space.
Remark 18.4 Note that a Desarguesian left 2spreadis nowsimp19 a spread which produces a Desarguesian affine plane. We now show how to obtaina translation plane with spread inPG(7,K ) from a parallelism in P G ( 3 , K ) . We repeat some of the results on direct products nets to establish the pertinent situation.
Theorem 18.5 Let S2 be a Desarguesian left partial parallelism of a left 4dimensional left vector space V over a skewfield K . Form V @ Vas a 8dimensional left Kvector space and choose three mutually disjoint 4dimensional vector Ksubspaces,O @ V, V e0 and { (v,v) ;E V } and refer to these as x = 0 , = 0 , y = x respectively. For S E S2,let T S denote the affine Desaryuesian plane constructedfrom the Desaryuesian spread. ( l ) Then T S x T S is a derivable net Vs whose components are 3dimensional projective left Ksubspaces in PG(7,K)&. Let V; denote the derived net represented in the form x = 0 , y = a x for all a in K [ q where K [ S ]is the quadratic extension skewfield of K coordinatizing T S = ( x l , y l ) and note that now x = 0 is a representation of T S , Note that the components of the derived net are ‘right’ 4dimensional Kvector subspaces. Let NK denote the Kpseudoregulus net defined by x = 0 , y = ax for all a in K so that NK is a subnet of V; for all S E S2. (2) For S, S’ E S2, { T S x T S } * U {nst x rst}* defines a net with components in PG(7,K ) R such that {TS x T S } * n {TSI x Tst}* = NK.
236
CHAPTER 18. PARALLELISMS
(3) U s E s z { ~xsT S } * = TS, defines a translation net defined by components within PG(7,K)R consisting of derivable nets which mutually share
NK (4) nsz admits a collineation group isomorphic to GL(2, K) which is defined as follows: Assume that the ambient vector space is a left space over K. Then define the elements of GL(2, K) as follows on points ( P I ,P2) of the direct product: ( P I ,P2) W (aP1 +pP2,SP1 +yPz) for all a,p, S,y of K where
[ { ] is invertible and Pi is a point
of some fixed 7rs fo,r i = 1,2.
Further, GL(2, K) leaves invariant eachderivable net containing NK of the corresponding set of derivable nets and acts o n N K canonically as GL(2, K). Note that SL(2, K) is generated by elation groups of NK. Proof: (1)follows from the section on regular direct products. Note that we consider a fixed left basis and define the right space via this basis. In this context, for az, the elements a are matrices with respect t o this basis. Hence, we may regard the components as right Ksubspaces. Note that each derivable net is coordinatized by a quadratic skewfield extension of K. If the structure forms a net and there is a common subnet properly then the common subnet may be coordinatized by a subcontaining NK, skewfield properly containing K which cannot be the case. Hence, if the structure is a net then the result follows. Thus, assume that (2) is not true so that the structureis not a net. This means that the components outside of the components of N must not all be disjoint. Let P be a point in common in {TS x T T ~ }n * { T S I x 7rst}* P(NK).The Baer subplanes incident with the zero vector cover the points on lines thru the zero vector of each net. Hence, there are unique Baer subplanes P ~ , ~ , p ~ tof, the ~ derivable nets {TS x7rs}*, (7rst x r s t } * respectively which contain 0 and P. These Baer subplanes are 4dimensional left Ksubspaces which intersect x = 0,y = 0,y = z in 2dimensional Ksubspaces. For example, a subplane ps,p must have the general form {(ae, be) ; a, b E K[S]}so, for example, the intersection with z = 0 is {(ae, 0) ; a E K[SI}. Since K[SI is a 2dimensional left Ksubspace, the intersection on z = 0 is a 2dimensional
18.3. DESARGUESIAN PARALLELISMS. LEFT
237
left Ksubspace. Similarly, the intersections on y = 0 and y = z are 2dimensional left Ksubspaces.(Note that it is not necessarily true that the Baer subplanes intersect each component of NK in a 2dimensional Ksubspace if K is not commutative.) Since there is a unique 4dimensional left Ksubspace with this property, it follows that ps,p = psj,p. But, then the intersection of these subspaces on say x = 0, produces the same 2dimensional Ksubspace X . But, this means that X is in the two spreads S and S’ contrary to assumption. Wenow consider part (4). We may identify the points of the union of the derivable nets with the points of any direct product net(of any derivable net). We have seen a variation of this theorem in the previous chapter. For each derivable net where the subplanes incident with the zero vector are 2dimensional left Fsubspaces, there is a collineation group defined as in the statement of the theorem except that the elements a,p, S,y are in F. Since the subplanes may be coordinatized by a 2dimensional skewfield extension of K, it follows that each derivable net admits the indicated group so that the union admits this groupas well. This proves (4) and completes the proof of the theorem.0
18.3 Desarguesian Left Parallelisms Construct Right Spreads. Theorem 18.6 Let P be a left Desarguesianparallelism ofPG(3,K)L:where K is a skewfield, Then there is a corresponding right spread r p in PG(7,K)R consisting of derivable right partialspreadscontaining therightpseudoregulus defined by K . Conversely, a right spread in PG(7,K)R which is a union of derivable right partial spreads containing the right Kpseudoregulus produces a Desarguesian left parallelism of PG(3,K)L.
Proof: Given a left parallelism P of PG(3,K)L,there is a corresponding that’ the translationnet np. We assert that r p is atranslationplane; derivable right partial spreads cover the vectors. If not, there exists a point P outside of the points of NK so there is a unique 4dimensional left vector space X CBXu containing P which intersects x = 0, 9 = 0, y = x in 2dimensional left Ksubspaces by Lemma 17.17.
238
C H A P T E R 18. PARALLELISMS
Since P is a left parallelism, there is a Desarguesian spread S containing X and let n denote the Desarguesian affine plane determined by S. Then, in T x n, there is a unique Baer subplane of the net containing P. But, then the derived net of n x 7r is within the structure np. Now assume that C is a spread in PG(7,K)Rconsisting of derivable right partial spreads containing the Kpseudoregulus. Let V be a component which is also a left 4dimensional subspace. Then, the set of Baer subplanes of each derivable net defines a left 2spread on V. Since no two derivable nets can share a left 2space on V , it follows that there is a corresponding Desarguesian left partial parallelism Pc. Assume that there is a left 2space X over K which is not covered by Pc. Then X @X" is a 4dimensional left Ksubspace. If P is a point of X @ X" which is not a point of a component of the Kpseudoregulus, then there exists a unique derivable net containing P. The Baer subplane containing of the derivable net is a 4dimensional left Ksubspace containing P and by uniqueness must be X @ X" which is a contradiction. Hence, Pc is a Desarguesian left parallelism.0
Remark 18.7 Essentially all previous research o n regular or Desarguesian parallelisms in P G ( 3 ,K) has been done over finite fieldsK N G F ( q ) . However, although parallelisms over infinite fields are possible (see Beutelspacher [g]), it is an open question whether there are infinite Desarguesian parallelisms where K is an infinite field or a skewfield. W e have presented this theory basically free of the assumption of finiteness or commutativity because of our general theme of derivable nets and their applications. But, clearly, there is much to be done in this direction. In the following, we present a class of parallelisms in P G ( 3 ,K ) for arbitraw fields with mild restriction. firthermore, although the parallelisms constructed are not Desarguesian, they are interest to us here as they depend on derivation for their construction. In the finite case, there is a combinatorial proof given in Beutelspacher [8] using properties of an embedding of P G ( 3 , q ) i n t o P G ( 3 , q 2 ) which is valid for q > 2.
18.4. AN INFINITECLASS OF ‘DERIVED’ PARALLELISMS.
239
18.4 An InfiniteClass of ‘Derived’ Parallelisms. We assume that K is an arbitrary field, finite or infinite. In the following, we shall let R be a regulus in PG(3,K)where K is a field.
Remark 18.8 If K is any field which has a quadratic extension field K’, then every element e of K’  K defines an irreducible quadratic x2  xf  g exactly when e2= e f + g. If K is not GF(2), there exist two fields of matrices in GL(2,K ) which share exactly the matrix
field
{ [ i :]
; U E K } . Hence, there exist two
distinct Pappian spreads containing a regulzls.
Proof If K is not GF(2),then there exist at least two elements e and e’ in K’  K that satisfy distinct irreducible quadratic equations over K . Let e2 = ef g and el2 = e’f’ g’. Then the two matrix fields are
+
+
{ [ ;  t f :]
;u,tSK}
and
{ [ t  tf‘ tg’ ] U
; u,t E K } .
These fields define Pappian spreads containing the regulus given by
Lemma 18.9 Assume that there are two Pappian spreads C and C‘ which contain a regulus R. Let L be any line of R and let G denote the central collineation group of the Pappian afine plane associated with C which f i e s L pointwise. Then G acts semiregularly on the set of all lines of PG(3,K ) which are skew to L and not in C. Proof: We consider the associated Desarguesian affine plane A @ ) . Let L’ be any line of PG(3,K )which is not in C and which is skew to L. Then
... .
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CHAPTER 18. PARALLELISMS
L' is a 2dimensional Kvector subspace. The intersection subspaces with L' and the spread lines of C define a spread for L'. Hence, L' becomes a Baer subplane of C incident with the zero vector which is skew to L. Let g be in the central collineation group G of A ( C ) with axis L. Then, L'g # L' by Theorem 8.2 since otherwise, the center must be in the Baer subplane whereas L' and L are disjoint.0 Hence, the group G is semiregular.0 Lemma 18.10 G R fixes any Pappian spread I? containing R and acts as a collineation group of the associated Pappian afine plane A(r)which acts regularly on the partial spread  R. Proof By Theorem 8.4, any Pappian spread that contains R admits as a collineation group any groupwhich fixes the opposite regulus linewise and which is generated by a central collineation group of the regulus net. But, any centralcollineation of the ambient Pappian spreadwhich fixes R and has center and axis in R projectively must induce a central collineation group of the net. Moreover, the proof of Theorem 8.4 shows that the subgroup GR which leaves R invariant acts transitively on the components of the spread which are not in R.0
Lemma 18.11 C'g and C'h, f o r g , h E G,share a line skew to L and not in C if and only if C'g = E h . Proof Since GR also is semiregular on the images of L', it follows that the spread for C' = R U {L'g ; g E GR}.Assume that C'g and C'h for g, h E G share a line skew to L and not in C. Then C'ghl and C' share a line M' which is not in C and skew to L. But, all lines not in R of C' are images of L' under the group GR.Hence, M' = L'w for some element W E GR,Moreover, M' in C'ghl is the image under ghl of a line N' of C'. If N' is in R then N'ghl is a line of C which implies that ghl is in GR.If N' is not in R then M' = L'sgh1 where S is in GR and N' = L's. Thus, L'sghl = L'w but the group acts semiregularly so this implies that sghl = W . Since S and W are in GR,this, in turn, implies that ghl is in GR which implies that C'ghl = C' or equivalently that C'g = C'h. Hence, if two images share a line skew to L and not in C then there are equa1.O
18.4. AN INFINITECLASSOF'DERIVED'PARALLELISMS.
241
Theorem 18.12 Let C be any Pappian spread in PG(3, K) where K is a field. Let R be any regulus in the spread and L any line of R. Assume there exists a second Pappian spread C' in PG(3, K) containing R (i.e. K not GF(2)). Let L' denote a line of C' which is not in the spread of C. Let G denote the full central collineation group of C with axis L. Then, {C'g ; g E G } is a set of Pappian partial spreads which cover the lines which are skew to L and not in C . Proof It remains to show that every line M' which is skew to L and not in C is in the same Gorbit. Represent C in the form x = 0, y = x pit y l t for all U , t E K
[ z"
+
U
+
]
where K is a field and u2 uplt  y l t 2 = 0 if and only if U = t = 0. Let M' be a line skew to L. This means that as a 2dimensional Ksubspace, we may write M' in the form y = X A for A any 2 x 2 Kmatrix,
A=
[ ," 51 1.
Note, for example, a basis for M' may always be taken in
the form: { ( 1 , 0 , 0 ) , (0,1,c,4}
so the Kspace M' may be written as y = x We consider the image of y = x
[ ]
[:511, first under the affinehomology
group elements
for all
U , t , (U,t )
# (0,O)
to obtain 2subspaces of the following form:
y=x for all m, s E K  {(O,O)} elation group elements:
m
0 0
and then follow these images under the affine
242
CHAPTER 18. PARALLELISMS
for all U ,t E K . Hence, we need to show that
has a unique solution (m,s,u, t ) where (m,S)
#
(0,O) if y = z
1:
: l Ai s
L
not a component of C. Thus, we obtain the equivalent system of equations:
Now ( m , s ) = (0,O) if and only if
[ ac bd ] = [
'l" t lt
U
]
if and
only if M' E C . 0
Theorem 18.13 Let C be a Pappian spread in PG(3,K ) for K a field. Assumethatthereexists aregulus R whichiscontained in atleasttwo distinct Pappian spreads C and C'. Let e be a jixed component of C and let G denote the full group of central collineations of the afine translation plane A associated with C with axis e. Consider the set of spreads {C'g ; g E G } and form the Hall spreads c'g b y derivation of each Rg. ( l )c'g = Eg;there is a group of C acting transitively on the setof Hall spreads. (2) C U { c'g ; g E G } is a parallelism consisting of one Pappian spread and the remaining spreads are Hall spreads. Proof We have pointed out that the infinite Hall planes may be constructed by derivation of Pappian planes.
18.5. THE PARALLELISMS IN PG(3,2).
243
It remains only to show that the lines are all uniquely covered by the indicated spreads. A line skew to L and in C is, of course, covered by C . The lines of C‘g which are not in Rg are not in C and the set of such lines is covered by the set of lines of C‘g’s which are not in any of the Rg’s. A Baer subplane of Rg cannot be a Baer subplane of any Rh unless Rg = Rh since a Baer subplane within a Desarguesian affine plane completely determines the regulus net containing it as a Baer subplane. Now take any line .t which does intersect z = 0. Then there exists a unique regulus net R1 defined by .t which lies in d.The question is whether any regulus net R1 is the image of R under a collineation g on G. Take a regulus R1 containing ~t:= 0. Choose two distinct lines L1 and L2 of R1 not equal to z = 0. Since the group of central collineations is doubly transitive on the components of C  { L } , it follows that there is a central collineation g with axis z = 0 which maps R onto R I . Hence, we have a parallelism.0 Using the previous construction, we may obtain another parallelism by the derivation of C and
z.
Theorem 18.14 Under the assumptions of the previous theorem, let E denote the Hall spread obtained b y the derivation of R and let P denote the previous parallelism. Then E U C’ U {P  { & g }is}a parallelism of PG(3,K ) . Proof Note that R is in C and C’ so R U R‘ is a set of lines of PG(3,K) which is covered by C U F. Hence, this set is covered also by 5 U C’.O Thus, there areparallelisms in PG(3,K)for every field K # GF(2)that admits quadratic extensions. We end this chapter with a shortdiscussion of the parallelisms over fields of two elements.
18.5 The Parallelisms in PG(3,2). There are exactly two parallelisms in PG(3,2) and since any affine plane of order 4 is Desarguesian, the parallelisms are regular. As mentioned previously, these are particularly interesting in that the associated translation
CHAPTER 18. PARALLELISMS
244
planes of order 16 with spread in PG(7,2) have collineation groups which fix a Baer subplane no and act transitively on the lines of the plane which are tangent to the plane x. at a particular affine point. The two planes are the socalled LorimerRahilly and JohnsonWalker planes of order 16. The naming of two planes by four individuals refers to the fact that there are various ways of going about the construction of these two planes. We have noticed that theplanes constructed from parallelisms in PG(3, q) are always composed of a set of 1 q q2 derivable nets. The derivation of any of these nets thus produces a corresponding translation plane. In the case of the LorimerRahilly and JohnsonWalker planes, any such derived plane is a semifield plane (a plane admitting an affine elation group of order 16 transitive on the components of the plane not equal to the axis of the group). The two semifield planes used in the derivation process are ‘transposes’ of each in the sense that their spreads are connected via a dualization (i.e. a polarity). It turns out that the construction processes of ‘transpose’ and ‘derivation’commute which makes the LorimerRahilly and JohnsonWalker planes transposes of each. Moreover, the derived planes are not isomorphic as the two semifield planes when transposed have different affine homology groups. So, Letting TLR and TJW denote the LorimerRahilly and JohnsonWalker and x i F and x;W denote the derived planes, we obtain the following lovely pattern:
+ +
[ 2:
J: (derive)
(transpose) x;W 5 (derive) ct (transpose) x ~ w ct
1
.
The reader interested in the particulars of these planes is referred to Johnson [36] and Walker [77].
Remark 18.15 Thesetwoplanescorresponding to theparallelisms in PG(3,2) admit collineation groups which act transitive on the 1 2 + 22 derivable nets which, in turn, implies that thegroup acting on theparallelism is twotransitive on the spreads.
+
Chapter 19
PARTIAL PARALLELISMS WITH DEFICIENCY In this chapter, we consider partial parallelisms and their possible extensions.
Remark 19.1 Recall that for any nonzem vectorv0 of a &dimensional vector space V over a field K isomorphic to GF(q), thereare exactly 1+ g +g2 2dimensional K subspaces containing vo. Proof: Simply count the number of ldimensional vector spaces of V/ (vo).O Definition 19.2 A partial left parallelism in P G ( 3 ,K)L:has 'deficiency t ' < 00 if and only iJ for each point P, there are exactly t 2dimensional left Ksubspaces containing P which are not in a left spread of the partial left parallelism or equivalently, there are exactly t lines not incident with P. Thus, in the finite case, a partial left parallelism of deficiency t if and only if the number of spreads is 1 q q2  t .
+ +
Now we note,
Theorem 19.3 Let P, denote a partial left parallelism of deficiency one in P G ( 3 ,K)r. for K a skewfield. Then, there isa unique extensionof P, to a left parallelismof P G ( 3 , K ) L . 245
246
CHAPTER 19. PARTIALPARALLELISMSWITHDEFICIENCY
Proof: Note that for each nonzerovector, there is a unique 2dimensional subspace which covers it and which does not appear in the parallel parallelism. Hence, the remaining 2spaces must form a spread S which extends P, as otherwise there is some vector which liesin two of these 2subspaces.0 Wenow consider Desarguesian partial parallelisms whichhave extensions.
Theorem 19.4 Let St be a Desarguesian partial left parallelism of deficiency t < 03 in PG(3, K ) , f o r K a skewfield. Let TS, denote the corresponding translation net defined by a partial spread in PG(7, K ) R consisting of derivable right partial spreads sharing the Kpseudoregulus net NK. ( l )If St can be extended to a partial left parallelism of deficiency t  1 by the adjunction of a leftspread S , let T S denote the translation plane determined by the spread. T h e n a partial spread C in P G ( h , Z ( K ) ) m a y be defined as follows:
C = ( T S  ~ N K )U (TS X .S). where the notation is intended to indicateright the 4dimensional Ksubspace components of the net TS, which are not in NK union the 4dimensional left Ksubspace components of T S x T S . (2) C admits GL(2, K ) as a collineationgroupwherethesubgroup SL(2, K ) i s generated by collineations that f;a: Baer subplanes pointwise. Proof We have seen previouslythat TS X T S is a partial spread containing
NK as defining Baer subplanes. If a component l? of TS, not in NK and T S x T S share a common point P then there is a unique Baer snbplane of one of the derivable nets incident with .t which contains P. Since the Baer subplanes incident with the zero vector are 4dimensional left Ksubspaces, and intersect three common components of NK in 2dimensional left Ksubspaces, it follows by uniqueness of representation that the two 4dimensional subspaces are equal which is a contradiction that S is a spread which extends the parallelism. This proves (l). To prove (2), note that we now are using a partial spread X CBX" for X E S as components of the new translation net, The components are
ONE. 19.1, DEFICIENCY
247
each invariant under GL(2,K ) . The union of these partial spreads X Xu for X E S cover the components of the regulus net NK and the elation groups fix the components of the regulus net pointwise. In the new net, the components of the regulus net now appear as subplanes of the union of the set of partial spreads X @ Xu for X E S and hence are Baer subplanes of the new translation net. This completes the proof of ( 2 ) and the proof of the theorem.0
19.1 Deficiency One. Theorem 19.5 Let S1 denoteaDesarguesianleftpartialparallelism of deficiencyone in P G ( 3 , K ) c where K is askewfield. Let S denotethe spread extending P, and let 7rs denote the afine plane given b y the spread S. Then there is acorresponding translationplane with spread C in PG(Vs,Z(K))defined as follows:
C = (TITS, N K )U ( ~ Xs~ s ) . where the notation is intended to indicate the right 4dimensional Ksubspace components of the net TS, which are not in NK union the 4dimensional left Ksubspace components of 7rs x 7rs. (2) Conversely, a spread with right and left$spaces definedas above produces a Desarguesian left partial parallelism of deficiency one. (8) The unique extension of a Desarguesian left partial parallelism of deficiency one to a left parallelism is Desarguesian if and only if the net 7rs x ITS defined b y the extending left spread is derivable. Proof: Let So = S1 U {S} denote the unique parallelism extending S1. If C is not a spread then there is a point P which is not covered by the components. Hence, there is a unique left 4dimensional space X @ X" containing P which intersects x = 0, y = 0 , y = x in 2dimensional left Ksubspaces. But, X is in a unique spread S' of SOso is either in S1 or is
S. The converse is now essentially immediate. If the extension parallelismis Desarguesian, then 7rs x ITS = D is a derivable net and derivation produces the type of translation plane associated initially with a Desarguesian parallelism.
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CHAPTER 19. PARTIALPARALLELISMS WITH DEFICIENCY
Now assume that the net D is derivable. Then there is a skewfield F so that the Baer subplanes of the net are all 2dimensional right Fsubspaces. Since all of the Baer subplanes are isomorphic Desarguesian subplanes coordinatixable by F, it follows that F must be an extension of K. Furthermore, the components are 2dimensionalleft Fsubspaces. But,the Baer subplanes of NK are 4dimensional right Ksubspaces so it must be that F is a quadratic extension of K and that all Baer subplanes of D are 4dimensional right Ksubspaces. Note for ‘derivation’ in the infinite case to produce an affine plane (a translation plane) the subplanes which are Baer in D must be Baer within the containing plane. That is, the subplane is automatically ‘pointBaer’ as the points of the net are the points of the translation plane. The question is then whether the subplane is also a ‘lineBaer’ subplane of the translation plane. In this case, the Baer subplanes are right 4dimensional Ksubspaces and the components not in the net D are right 4dimensional Ksubspaces. We shall see in the next chapter that, if any pointBaer subplane and any disjoint component direct sum to the ambient vector space then the subplane is also pointBaer. Hence, derivation produces a translation plane whose spread iscoveredby derivable right partial spreads containing the Kpseudoregulus partial spread which, in turn, produces a Desarguesian left parallelism extending P1.0 In the finite case, we may obtain a characterization by a collineation group:
Theorem 19.6 ( l ) Given a Desarguesian partial parallelism P, of deficiency one in PG(3, q ) , there is a corresponding translation plane C with spread in PG(7, q). (2) Thetranslationplaneadmits a collineationgroupisomorphicto GL(2, q ) which contains asubgroup isomorphic to SL(2,q) that is generated b y Baer collineations. Furthermore, (l!)(.) there are exactly (1 q) Baer axes which belong to a common net N* of degree ( 1 + q 2 ) . (2)(b) SL(2, q ) has exactly q(q + 1) component orbits of length q(q  1). These orbits union the (1+ q) Baer axes define derivable nets.
+
PARALLELISMS. 19.2. DUAL PARTIAL
249
+
The q(q 1) derivable nets induce a Desarguesian parallel parallelism of deficiency one on any Baer axis considered as PG(3,q). @)(c) The Desarguesian partial parallelism maybe extended to a Desarguesian parallelism if and only if the net N* is derivable. (3) Conversely, let x be any translation plane with spread in PG(7,q) that admits SL(2,q) as a collineation group generated by Baer pelements where q = p' and p is a prime. If the component orbits of SL(2,q) have lengths either 1 or q2  q then there is a corresponding Desarguesian partial parallelism of deficiency one induced on any Baer axis. Proof Except for ( 3 ) , the proof follows the previous theorem. Assume the conditions of ( 3 ) . Then, it follows that each orbit of length > 1union the set of Baer subplanes fixed pointwise by Baer pelements defines a derivable net. If a Baer pelement g fixes a Baer subplanes pointwise, then by the order of the group, each component of this Baer subplanes is in an orbit of length 1 under the group SL(2,q). Hence, each Baer pelement fixes a Baer subplane pointwise that resides in a net of degree q 1. Therefore, there are exactly q2 q orbits of length q(q  1) defining the same number of derivable nets. Each derivable net induces on any Baer subplane a Desarguesian spread and the set of q2+q component orbits induces a Desarguesian partial parallelism of deficiency one.0
+
19.2
+
PartialDual Parallelisms.
Definition 19.7 Let K be a skewfield. A set of dual spreads in P G ( 3 , K ) R (or PG(3,K ) L ) which are mutually disjoint on lines is said to be a 'partial dual parallelism'. Remark 19.8 Any partial dual parallelism in P G ( 3 ,K ) R ( o r P G ( 3 ,K ) L ) is equivalent to a partialparallelism in P G ( 3 ,K ) L (or PG(3, K ) R ) . I n particular, any finite parallelism produces a related 'dual' parallelism. Proof Apply the annihilator mapping to the associated 4dimensional right or left space. It follows that spreads map to dual spreads.0
250
CHAPTER 19. PARTIALPARALLELISMS W T H DEFICIENCY
Theorem 19.9 Let P, be any partialparallelism of deficiencyone in PG(3,K)c such that all spreads of P, are dual spreads. (1) Then there isa corresponding deficiency one partial parallelism Plin PG(3,K ) R called the ‘dual’ partial parallelism of deficiency one. (2) Let P denotetheuniqueextension of P, to a parallelism in PG(3,K ) c then there is a uniqueextension of P!.l? say P*, which is a parallelism in PG(3,K ) R . Remark 19.10 If P is a Pappian partial parallelism in PG(3,K) then all spreads are duals spreads and there is a corresponding Pappian ‘dual’ partial parallelism. Proof: See Theorem 20.7 and its corollary.0 Hence, we obtain:
Corollary 19.11 Let T be a translation plane corresponding to a deficiency one Pappian partial parallelism. Then, there is a corresponding translation plane T* obtained from the dual partial parallelism. Proof All but one of the spreads are Desarguesian and hence all such are dual spreads. Thus, the dual partial parallelism is a set of dual spreads which are also spreadsand since these spreads are still Desarguesian, it follows that we now obtain a deficiency one ‘right’ partial parallelism and hence a corresponding translation plane.0
Remark 19.12 In the next chapter, we will again deal with left and right pmjective spaces. However, we shallchoosea diferent framework for the projective spaces. Note that a right space over a skewfield K becomes a naturalleft space over K O P P . Hence, we could have considered theprojective spaces PG(7,K ) R as PG(7,KoPP)c. Moreover, spreads in PG(3,K ) c produce dual spreads in PG(3,K0PP)e i n a natural manner. We shall see that there are many variations of such projective spaces.
Chapter 20
BAER EXTENSIONS In aprevious chapter, we completed the analysis of nets which are completely covered by subplanes. In this chapter, we consider what occurs when it is not necessarily the case that the net is subplane covered but merely that there are subplanes within the net sharing all of its parallel classes. We have discussed pointBaer and lineBaer subplanes previously. Furthermore, we have discussed collineations of derivable nets which are generated by net collineations which fix a Baer subplane pointwise. We have noticed that there is a problem for infinite derivation in that a Baer subplane of a net need not determine aBaer subplane of an &ne plane extending the net. We are therefore interested in whether pointBaer subplanes admitting certain collineations are then Baer subplanes.
20.1
PointBaerCollineations.
Definition 20.1 A collineation U of an afine plane which &es a pointBaer subplane pointwise is said tobe a ‘pointBaer perspectivity if and only if the collineation fixes each subplane of a set C of pointBaer subplanes which f o r m a cover of the points of the afine plane. The collineation U is a ‘pointBaer elation’ if andonly if Fixa is in C. Otherwise, a is a ‘pointBaerhomology’.Furthermore, C is called the ‘center’ of the collineation, the elements of C are called the ‘central planes’ and Fix a is the ‘axis 25 1
252
CHAPTER EXTENSIONS 20. BAER
If a collineation fixes a pointBaer subplane pointwise then, conceivably, it is not a pointBaer perspectivity. However, the structure of pointBaer collineations is essentially completely determined for translation planes. Definition 20.2 An ‘axialBaer perspectivity a is a pointBaer perspectivity such that Fixa projectively nontrivially intersects each pointBaer subplane of the center. (The set of points of the projective extension nontrivially intersects each of the point sets of the projective extensions of the central subplanes.) Theorem 20.3 Let T be a translationplane with spread andlet a be a collineation which @es a pointBaer subplane pointwise. Then a is either a pointBaer homology (and hence an axialBaer homology) or a is a n axialBaer elation and, in this case, all the planes of the center are Baer subplanes. In particular, in all cases,theaxis Fixa is a Baer subplane of 7~ and a has a unique center. Proof: Assume that Fixa is lineBaer. Let P be any point of the plane which is not incident with the subplane Fixa. Then P is distinct from Pa and we consider the line PPa. This line meets Fixa in a point A since the subplane is lineBaer. Clearly, these three points are mutually distinct. However, this implies that the line is ainvariant as a maps P to Pa and fixes A. This means that any point of the plane is incident with a line of Fixa so that Fixa is pointBaer. Now extend the affine plane to a projective plane and apply the dual argument above to conclude that if Fixa is pointBaer then it is also lineBaer , We let T Odenote the subspace of the vector space T associated with the points of TO= Fixa, and let 7 ~ 1= .(a  1) where we may assume that a is in GL(r,P) for P the prime subfield of the kernel of T , We shall show that T I is a translation subplane. Thus, we want to first establish that T I is equipped with a spread with component set rl = {Fn T I ; F E ro}, where F0 E l? denotes the components of TO,i.e., all the components 7 E that meet the subspace T O nontrivially. Now consider any affine pointT not in T O . By our pointBaer hypothesis, T lies on a line t of T O ,and by the ainvariance o f t we clearly have t = P T
20.1. POINTBAER COLLINEATIONS.
253
(here we denote the aimage of T by Tu). Since TOT is a line of T O ,it must be a translate of a component L of T O . That is, T and Tu are both points of L c for some vector c of 7ro. Hence, the difference Tu T = T ( a  1) is in a component L of T O . Thus, the vector subspace nl=.rr(a 1) of 7r is contained in the union of the set of points of 7r that lie on components of T O . That is, T I is covered by the partial spreadro defined by the components of T O . In fact, every component of r o meets 7r1 nontrivially because if F E ro then F(.  1) is clearly a nontrivial subspace of F. Now T I is partitioned by l?. So, to demonstrate that thisdefines a spread on T I , it is sufficient to establish that if H and F are arbitrary but distinct members of F0 then H1 @ F1 = T I , where H1 = H n7r1 and F1 = F n 711. This holds because since T = H @ F and a leaves H and F invariant, we have
+
.rrl=n(a")=H(al)$F(al)~Hl~FlE.Irl, yielding the desired result. Thus, we have established that .rrl=r(a 1) is an &ne translation subplane of that has precisely the same components as the pointBaer subplane T O . Hence either T O = q or the only affine vector they share is the zero vector. We assert that in either case a is an axialBaer perspectivity. First assume that the two subplanes are identical. Then m ~ ( a 1) = .(a is the zero space so that (a I 0. We assert that a is an axialBaer elation. Note that a leaves each translate of 7ro invariant and the translates of T O are clearly pointBaer subplanes which partition the affine points of n and share the same slopes as T O . This is equivalent to the collineation a acting trivially on the quotient vector space T / T O which is implied by the fact that ( a I 0. Hence we may assume that thesubplanes 7ro and 7r1 are distinct andhave trivial intersection. So, we have two distinct pointBaer subplanes sharing the same infinite points and a common affine point. By the Reconstruction Theorem 17.7, the net is a direct product net so that 7~ = ~1 @ T O as the subplanes are also subspaces. Now let C be the set of subplanes of T that are translatesof T I . Because T I @TO= x , every plane Q E C is of the form +PO,where p0 E Q, and so Q is ainvariant. Thus, C is a class of ainvariant subplanes that are translates
CHAPTER 20. BAER EXTENSIONS
254
of T I and since distinct cosets of 7 ~ 1are pairwise disjoint planes that share precisely the same slopes as T I ,we see that the planes ‘1c, partition the affine points of the plane T. Furthermore, all the (affine) planes of C have precisely the same set of infinite points, namely, those of the pointBaer subplane TO. To show that Q is a pointBaer homology (and hence, of course, an axialBaer homology), it only remains to check that theplanes in C are pointBaer, or equivalently, that T I is pointBaer. Thus, we must check that if T is any affine point of 7~ outside T I then T lies on a(unique) line of TI. Since we have already seen that the components of T O and T I coincide, we may assume without loss of generality that T is not in TOU TI. Now since T = T
[email protected] T I , the coset T O+T must meet the subspace ?rl at a unique point S (T = R+S where R in T O and S in T Iso that TO+T= rr,+S). But since the slopeset of T O T is the same as the slopeset of T O and of T I ,ST meets the line at infinity at a point S of the affine plane T I ,and ST shares two points with the projective extension of T I . Hence the line ST E T I and certainly meets T . It follows that T I ,and hence every member of C is a pointBaer subplane of T. Thus, in all cases, g is an axialBaer perspectivity, so TO is a Baer subplane of T. Furthermore, since the center is the set of translates of T I , we further have that all the central planes are Baer when T I = T O . This completes the proof of the theorem.0 We now give a necessary and sufficient condition for a pointBaer subplane of an affine translation plane to be a Baer subplane.
+
Theorem 20.4 Let 7~ be a translation plane, associated with a spread l? o n a vector space V and let T O be a pointBaer subplane of 7~ incident with the zerovector 0 E V . T h e n T O is a proper Baer subplane if and only if
Proof: Assume the directsum hypothesis. To show that T O is a proper Baer subplane of T ,it is clearly sufficient to show that every affine line of 7~ meets TO at least once. So we need only consider lines of type t T , where t is a component of the spread that trivially intersects TO. Hence, by the directsum hypothesis, T E [@no, say T = Q + P where Q E t and P E 7 ~ ~ . Then the line T t = P t meets TO. Thus T O is a Baer subplane of 7 ~ . The converse follows by reversing the argument.0
+
+
+
20.1. COLLINEATIONS. POINTBAER
255
There is another convenient directsum type theorem for pairs of planes which allows us to decide that a subplane of a spread is pointBaer.
Theorem 20.5 Let 7ro and x1 denote two subplanes of a translation plane7r with the same slopeset and sharing an afine point0 such that IT = TO$ T I , when 7r is viewed as a vector space and T O and n1 are subspaces. Then no and 7r1 are both pointBaer subplanes of 7r. Proof It is required to show that any &ne point T E lies on at least one line of each of the subplanes T O and T I . So assume T does not lie in no or T I . Since T E TO $ n1, the coset no T must meet the subspace 7r1 at a unique point S, and S # T as otherwise T = S = 0, contradicting the fact that T is not a point of TO.But since the slopeset of no + T is the same as the common slopeset of 7ro and T I , the line ST meets the line at infinity at a point of the affine plane 7r1, and ST shares a second point with 7r1, S. Hence the line ST E 7r1 and certainly meets T. It follows that 7r1 is pointBaer. Interchanging the roles of 7r1 and 7ro above shows that TO is also pointBaer.0
+
Remark 20.6 W e have already noted that a partial converse of the above propositionistrue,Thatis, if two subplanes TO and of a spread (V,I’) share the same components and directsum to V , then both the subplanes are ‘pointBaer’ subplanes; the complete converse requires the subplanes to be Baer subplanes of thegiven spread. What is not known in general is whether the planes of the center are Baer when a is a pointBaer homology. This can be overcome in certain situations. Since we have been interested in spreads in PG(3,K ) , we consider this in the following. Theorem 20.7 Let S be a spread in PG(3,K ) , for K a skewfield. Let 7r denote the corresponding translation plane. Let a be a collineation of n which fixes pointwise a 2dimensional K subspace which is not in S . Let no = Fixa. (1) Then a is in GL(4, K ) and .(U  1) is a 2dimensional Ksubspace such that 7ro and n(a  1) are lineBaersubplanesthat shareallparallel classes. (2) Furthermore, no and n(a  1) are pointBaer subplanes so are both Baer subplanes.
256
C H A P TEEXRT20. E N SBI O A ENRS
Proof: First recall that any %dimensionalKsubspace which is not a component of the translationplane defines a lineBaer subplane by the chapter on dual spreads. Since Q is a collineation of n,we may assume that CT is in rL(4, K). However, writing the elements of n in the form (z1,z2,ylry2) and the equation for no as zl = y1 = 0, then since no is a Ksubspace, we have Q represented in the form
where A is an element of G L ( 4 ,K) and p an automorphism of K. It follows that zf = z1 for all 21 E K so that CT is linear and hence T(Q  1) is a left Ksubspace. Furthermore, since n/nois now Kisomorphic to n(a  1) then n(a  1) is a 2dimensional left Ksubspace. The argument given in Theorem 20.3 shows that n1 and no share the same parallel classes. So, by Theorem 20.3 we are finished or no # n(a  1). Hence, n = no 63 T ( Q  1). We have seen above that this implies that both subplanes are pointBaer and hence both are Baer subplanes. This proves the resu1t.O
Corollary 20.8 ( l ) Let S be any spread in P G ( 3 , K ) and n the corresponding translation plane. If any 2dimensional Ksubspace which is not a component is f i e d pointwise b y a collineation of n then S is a dual spread. (2) Any Pappian spread in P G ( 3 , K) is a dual spread.
Remark 20.9 Thus, for spreads in P G ( 3 ,K), for K a skewfield, any collineation which fixes a 2dimensional Ksubspace which is not in S becomes a Baer collineation with a unique center and axis. However, we still don't know anything of the nature of the 'Baer group' itself. In order to regard this more generally, we need to consider arbitraw spreads over skewfields.
20.2
Notes on the NonCommutativeCase.
When K is a skewfield which is not a field, we have seen that there are some important differences between the commutative and noncommutative cases with the consideration of the group G L ( 2 ,K).
20.2. NOTES ON THE NONCOMMUTATIVE CASE.
257
We include this section to somewhat ease into thecomplications of what comes later  the analysis of Baer groups within translation planes. We first consider Desarguesian spreads over skewfields and worry about commutativity or the lack thereof in K. We then consider arbitrary translation planes 7r with kernel K and consider the kernel mappings acting on the left whereas elements of the group GL(n,K) necessarily act on the right when 7r is a left Ksubspace. All of this has been covered in previous chap ters. We complicate the issue by considering a pointBaer translation subplane 7ro with its own kernel K, and ask when it is possible that there is a direct product net 7ro x 7ro sitting in the tr.anslation plane 7r. We have seen that there is a natural group GL(2,K,) which is defined via scalar mappings of 7ro so this group must act on the ‘left’. This group is generated by certain collineations which fix Baer subplanes pointwise and, in particular, certain collineations of the translation planes could fix 7ro pointwise and then simultaneously be in GL(7r,K)and GL(2,K,)posing a potential problem as one group acts on the right while the other group acts on the left. Such situations seem to imply some sort of commutativity of K in the case when no is a Ksubspace of 7r. Actually, the use of the notation is a bit problematic as the elements of the group are not necessarily Klinear mappings in the traditional sense. Although, in the chapter on coordinates we have considered various different fields for which the designation ‘kernel’ is appropriate, we revisit some of these ideas in this chapter to fucther cement the concepts. Consider a Desarguesian affine plane (2,y) considered as a2dimensional left vector space over a skewfield (K, +, Since we may also consider the affine plane as a 2dimensional right space over K, we take components to have the form y = za for Q in K a n d z = 0 and note that y = ZQ and z = 0 are ldimensional left Ksubspaces. We may consider the mappings called the kernel mappings : (z,y) $ (Pz, @g). It follows easily that {Tp for all P in K} forms a field isomorphic to K S (K,+, and fixes each component of the Desarguesian plane. S).
2’’
e)
Define the determinant of
1 1
as acld
. I
 b if c # 0 and ad other
+ yb, zc + yd) such that the
wise. Now consider the mappihgs (z, p)
+ (za
corresponding determinant of
is nonzero. Then it follows easily
[ 1]
258
CHAPTER 20. BAER EXTENSIONS
that each mapping is a (2'')linear mapping. Hence, we may justify the designation GL(2,K ) . Previously, we have defined the kernel of a translation plane as the set of endomorphisms which leave each component of the plane invariant. Hence, {Tp ; p E K } = Kendis the kernel of the Desarguesian plane K . Furthermore, the full collineation group of K which fixes the zero vector (the Since the use of K or translation complement) is I'L(2,K) or is I'L(2,Kend), Ked N K is merely in the distinction between K and the associated kernel mappings, we also refer to K as the kernel of the plane. So, considering the translation complement of K as I'L(2,K ) then Kend*= Kend {To}is a group of semilinear mappings and as a collineation Tp is in I'L(2, K ) but is in G L ( 2 , K ) if and only if p is in Z ( K ) . That is to say that the elements of GL(2,K ) are the elements of I'L(2,K ) which commute with Kendand Tp commutes with Kmd if and only if p is in Z ( K ) . The notation can be particularly tricky if one considers ,Ox as a linear mapping over the prime field P of K . For example, PICis normally written xp when considering p as a Plinear endomorphism. Then considering an element U in K as a Plinear endomorphism, it follows then that notationally pu = up when considering the elements as linear endomorphisms whereas it is not necessarily the case that Pu = up when the operation juxtaposition is considered as skewfield multiplication. Note that elements of G L ( 2 , K ) act on the elements (IC,y) on the right ) the left. whereas Tp acts on the elements ( I C , ~on We now generalize to an arbitraryaffine translation plane with kernel K as follows: Let X be a left Ksubspace and form V = X @ X. We denote points by (2, y) for IC,y in X. When we have an affine translation plane C with kernel K , we similarly consider the lines through (0,O) (components) in the form IC = 0, y = xM where M is a Klinear transformation. The kernel K then gives rise to a set of kernel mappings { T p : (z,y) + (@z,py) ; Vp E K } = Kend. In the finite dimensional case, we may take M as a matrix with entries in K say as [aaj] and define X M = (IC~,ICZ, ...zn)M as (Czaail, ...,Cziuin). It follows that M becomes a left Klinear mapping with scalar multiplication defined by PIC = (,f?51,@22,...,PICn) and furthermore, {(IC,z M ) } is a left Klinear subspace. In this case Kend is a skewfield isomorphic to K and as a collineation group of C, Ked* is a semilinear Kgroup. Similar to the
20.2. NOTES ON THE NONCOMMUTATIVE CASE.
259
Desarguesian case, one may consider the left scalar multiplication as a linear endomorphism over the prime field P of K . When we do this, we shall use the notation KL. Hence, the M‘s now commute with the elements of KL. To be clear, we nowhave three different uses of the term ‘kernel’ of a translation plane. We always consider the translation plane as X @ X where X is a left vector space over a skewfield K , the kernel mappings are denoted by Kmd and the component kernel mappings thought of as prime field endomorphisms are denoted by KL. All three skewfields are isomorphic and each is called the kernel of the translation plane where context usually determines which skewfield we are actually employing. Of course, there is a fourth usage of ‘kernel’ ifwe are to include the kernel of an associated quasifield.
20.2.1 LeftandRightinDirectProduct
Nets.
We now consider a coordinate set for a regular direct product net. Recall in the Reconstruction Theorem 17.7, it is noted that for Abelian nets, the direct product of two pointBaer subplanes that share the same infinite points is the entirevector space and furthermorethe (direct) product of their intersections on any line concurrent with the common point is that line. We may identify any pointBaer subplane as T , within the direct product so that the points of the net have the general form ( p l , p 2 ) for p1 and p2 in rITO and the lines have the form L1 x L2 for L1 and L2 parallel lines of T,. It follows that the net M is T , x T , with the identity mapping defined on the set of parallel classes. Considering the translation plane T , with kernel K,, we specify two lines incident with the zero vector as z, = 0 and yo = 0. Recall, that we have a group GL(2,K,) acting on the direct product net with elements from K, acting on the ‘left’. We further decompose in terms of these two subspaces and write the elements of rITO as (x,,yo) where x,, yo are in a common &subspace W,. We may take g,, = zoas the equation of a line of T , incident with the zero vector so that the remaining lines are of the general form yo = z,M where M is a &linear transformation of W, for M in a set ll,. The points of the net now have the general form (z,,y,, zl,y1) where
CHAPTER EXTENSIONS 20. BAER
260
x,, yo, x1,yl are in W,. The lines of the net are as follows: (Yo = %M
+ c,)
x
(yo
=XOM
+ Cl)
for all M in I T , containing I and 0 and (xco= c,) x (x, = Cl).

Note change bases by the mapping
x:
(ZO,YO,%Yl)
(~o,%Yo,Yl).
Finally, we write ( x o , x l ) = x and (y,,yl) = y when (zo,yo,xl,yl) is a is a point after the basis change. original point of the net or (zo,x~,yo,y~) Note that, before the basis change x, the lines of the net are sets of points
{(x,, x,M
+ c,,
21,z1M
+ cl) for all x,, x1 in W,}
for fixed c, and c1 in W, and {(c,,y~,cl,yy~) for all y1,y2 in W,}, for fixed c, and c1 in W,.
Hence, after the basis change, the lines of the net have the basic form
1 O M 1+
x = (co,cl)and y = x M O
(co,cl).
Before the basis chinge x,the pointBaer subplanes incident with the zero vector which are in a GL(2,K,) orbit of T , have the following form: Po0 = {(O,O,~l,Yl)tr'Z1,Vl E
W O }
and
Definition 20.10 W e shall call these subplanes poor or pa! the 'base subplanes'.
20.2. NOTES ON THE NONCOMMUTATIVE CASE.
Remark 20.11 We nowobservethatthegroup left is represented b y mappings of the f o r m

(50,~1,Yl,92)
( W +bz2,m +dz2,ay1
261
GL(2,K,) acting on the
+bY2,Cyl+~Y2).
Wenow connect the ideas of coordinates of a translation plane which contains a regular direct product net and coordinates of the assumed pointBaer subplanes. Now assume that we have a translation plane C with kernel K and there are at least three pointBaer subplanes as above with kernel K, which are left invariant under the mappings Kendoorequivalently are Ksubspaces. Then there is a regular direct product net N isomorphic to T,x T, embedded in C. The translation complement of C is a subgroup of l?L(C,K ) with the linear elements acting on the right. Furthermore, there is a group of the direct product net N which is isomorphic to GL(2,K,) and naturally embedded in GL(4,K) with the elements acting on the left. It is easy to see that if a collineation g of C fixes a Ksubspace T, pointwise then g is in G L ( C , K ) and hence commutes with the mappings Tp. Now any kernel homology group Kend*induces a faithful kernel group on any invariant pointBaer subplane so K may be considered a subskewfield of K,. 20.2.2
When K Is K,.
We specialize to the case that K = K,. Then C is a K = &vector space and the above decomposition preserves the structureof the translationplane. In order that g be in GL(C,K ) and GL(2,K,), it must be in the product of the following two groups (Diag (Dag
[ ] [6 0]
such that X E K,
)
such that 6 E K,
 IO})
acting on the left. On the other hand, each such mapping is in G L ( C , K ) acting on the right. For example, assuming that 6 = 1, we would have g : (z1,z2,y1,y2)

+z2,91,Xy1 + Y 2 )
262
CHAPTER EXTENSIONS 20. BAER
is in GL(C,K ) . In order that the mapping commutes with each Tp it must be that
so that
X must be in Z ( K ) . Similarly, if Diag
[ ]
= g then
6 isin
Z ( K ) . Alternatively, the above mapping is Klinear for X in K if and only if the mapping x1 W X q is Klinear if and only if X is in Z ( K ) . We note that although we are assuming that K = K,, there is a distinction between KZendand Kendas they are the associated kernel mappings on r0 and the translation plane C respectively and Kendlr,= KZnd, So, we have noted that when there are at least three pointBaer subplanes sharing an affine point and all of their infinite points which are left invariant by the kernel K of the supertranslation plane then any pointBaer perspectivity induces a degree of commutativity in K in the case when K is (i.e. isomorphic to) the kernel of the subplane. Of course, it is not necessarily the case that any given pointBaer subplane which is pointwise fixed by a collineation of a translation plane with kernel K is automatically necessarily a'Ksubspace. Thus, part of our analysis in the next section details situations when such pointBaer subplanes are, in fact, kernel subspaces. Moreover, we shall expect that certain large pointBaer collineation groups in translation planes with kernel K could induce a certain degree of commutativity in K.
20.3
PointBaer Subplanes in VectorSpace Nets.
In the following three subsections, we consider what is implied of a translation plane due to the existence of a Baer subplane, Since the points of a translation plane form a vector space, we relax this condition and assume only that there is a vector space net admitting a pointBaer subplane. The theory decomposes naturally into three separate cases: When there are at least three pointBaer subplanes incident with the zero vector and sharing all parallel classes, when there are exactly two such pointBaer subplanes and when there is but one.
20.3. POINTBAER SUBPLANES IN VECTOR SPACE NETS.
263
Apart from a general theory, our goal is to apply these resultsto Baer or pointBaer subplanes within translation planes and particularly those with spreads in PG(3,K) for K a skewfield.
Definition 20.12 Any net whose parallel classes are those of a Baer subplane is said to be a ‘Baer net’. 20.3.1
ThreePointBaerSubplanes.
In this subsection, we show that if there are at least three pointBaer subplanes incident with the zero vector and sharing all parallel classes with a vector space net then the ‘base’ subplanes form the entire set of pointBaer subplanes incident with the zero vector and in the net (sharing all parallel classes with the net).
Remark 20.13 We adopt the notation of the previous preliminary section. In particular, we acceptthe notation x = (xo,x1) and y = (yo,yl) if and only if (xo,yo,x1,91) is a point originating from the direct product notation. We shall use the notation (m) to denote the parallel class containing the line x = 0 and (0) to denote the parallel class containing the line y = 0. Note that, in our notation, the points on x = 0 are the points (0, yo, 0,yl) and the points on y = 0 are the points (xo,O,x1,0). Similarly, points on y = x have the representation (xo,xo,x1,xl).We shall use both the original direct product point notation and the notation after the basis change x moreorless simultaneously. After our main structure theorem, we shall use the representation after the basis change exclusively.
Lemma 20.14 Let p be any pointBaer subplane incident with the zero vector and sharing all parallel classeswiththe net. Then (O,xo,O,xl)is in p n ( x = O ) ifandonlyif(xo,O,zl,O) isinpn(y=O). Proof Let the infinite points of x = 0, y = x (m) and ( M ) respectively.
[ L]
be denoted by
Let (xo,O,xl,O)be apoint of p n ( y = 0). Form the line (m)(z0,0,x1,0) (x = (z,,zl)) and intersect the line y = x to obtain (xo,xo,zl,~l). Since all such lines are lines of p, the intersection is a point of p. NOWform the
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CHAPTER EXTENSIONS 20. BAER
line of p, (O)(z,, x0,x1,q) and intersect x = 0 to obtain (O,x,,O, z1) in p n (x = o).O
Lemma 20.15 Now assume the subplane p is not a base subplane. For (0, x,,0, X I ) in p n (x = 0 ) define a mapping X o n W, which maps x, to 51 (Ax, = XI). T h e n X is a 1 1 and onto additive transformation of W,. Furthe,nnore, p = { (x,,y,, Ax,, Ay,) f o r all x,,yo in W,}. Proof: We noted previously that no two distinct pointBaer subplanes incident with a common affine point and sharing all of their parallel classes can share two distinct affine points. Hence, z, = 0 if and only if x1 = 0 when (0, x,, 0, q ) is a point of p and p is not the base subplane pm or p,. It follows that the subplane p is a translation affine subplane and hence a subspace of the underlying vector space taken over at least the prime field. Hence, it follows that X is 1  1 since the intersections with any of the base subplanes contain exactly the zero vector and it is also now clear that X is additive as p is additive. It remains only to show that X is an onto mapping. From the above remarks, any two distinct pointBaer subplanes sharing a common affine point and their infinite points sum to the vector space and their intersections with a line incident with the common point sum to the line. Hence, given any element x; of W,, we consider the vector (0, 0, 0, x;). There exist vectors (0, x:, 0,O ) in p o n ( x = 0) and (0, x,, 0, x1) in p n ( x = 0) such that (O,O, 0, xi) = (0, z :, 0,O) (0, x,, 0, x1). It follows that x1 = SO there exists a vector (0, zo, 0, x!) in p. Hence, the mapping X : x, + x1 is onto. If (0, x,, 0, Xz, 0) is in p n (x = 0) then (xo,0, Ax,, 0) is in p n (y = 0) by the previous lemma so that (xo,yo,Ax,, Ay,) is in p for all x,, yo in W, as p is the direct sum of any two components. Let (x:, p:, xi, y:) be any point of p then it follows that p also contains (O,O, Ax:  x;,Ay:  PT) and since p n pm = (O,O, 0,O) this forces x; = Ax: and g: = Ay:. This completes the proof of the 1emma.O To see that it is not possible that p is not a base subplane, we show that, in fact, X is in K,.
+
CC;
20.3. POXNTBAERSUBPLANESIN Lemma 20.16 For p then
=
VECTOR SPACE NETS. a line of the net and
265
(x,,0 , X I ,0 ) in
(x,,z o M ,5 1 , z l M ) is also in p.
Proof: We have seen this previously in the preliminary section. We form (z,, O , q , O)(oo)
= (z = (z,,
21))
and intersecty = z
[ Y h]
to obtain the point (z,, x o M , z l , z l M ) . Since all of the points and lines are points and lines of C, it follows that the intersection point is also in p , The previous lemma shows that if (z,, O,z1,0) is in p then so must be (z,, %,M, q , q M ) which, in turn, implies that (O,z,M, 0 , z l M ) is in p. However, also we have that z1 = Xz, and we know that (0, z o M ,0, X(z,M)) is in p. Subtracting, since p is additive, we have that (O,O, 0, (Xz,)M X(z,M)) is in p for all z,. Since pnpw = (0, 0, O,O), it follows that (Xz,)M = X(Xc".
Let L, be any skewfield such that { M for M in E,} is a set of L,1']near transformations. Then it follows that L, must be contained in the kernel K, of T , = pw. Hence, X is in L, G K,.O Therefore, we have proved the following result:
Theorem 20.17 Let M be any Abelian Baer net which contains three pointBaer subplanes that share the same aflne point. Then there is a skewfield K, such that M is a K,vector space net and there is a K,space W, such that the points of M m a y be identified with W, @ W, @ W, @ W,. The set of all pointBaer subplanes of M that share the zero vector is isomorphic to the set whose subplanes are
Furthermore, there is a collineationgroup l? of the net isomorphic to GL(2,K,) which $xes (0, 0, 0,O) and all parallel classes and acts triply transitively on the setof all pointBaer subplanes incident with( 0 ,O,O, 0 ) . Moreover, if B denotes the set of all pointBaer subplanes of M and l? is the pointwisestabilizer of a subplane 7ro of B then l? = ; T, E B\,
CHAPTER 20. BAER EXTENSIONS
266
Corollary 20.18 Let M be anyAbelianBaer net whichcontains three pointBaer subplanes that share the same afine point P . If one of the pointBaer subplanes has kernel KOthen the set of all pointBaer subplanes of M incident with P is isomorphic to PGL(1,KO). Proof We consider the above representation after the basis change The product of the two groups
x.
[ ] such that E KO) (Diag [ 6 0 ] such that S KO IO}) (Diag
X
E
fixes r, = pm pointwise and actsdoubly transitively on the remaining pointBaer subplanes.0 Below, we completely determine the collineation group of a net of type in the statement of the above theorem. We first note the following:
Remark 20.19 Let R be any Abelian Baer net which contains three pointBaer subplanes that share the same afine point A . Let ro be any pointBaer subplane incident with A . Then ro is an afine translation plane with kernel K,. Let G,, denote the full translation complement of ro. Then there is a collineation group of R isomorphic to GAowhich leaves ro invariant. Proof: We have noted that R is a regular directproduct net. The result is then not difficult and is left for the reader to verify.0
Remark 20.20 In the net R defined as a regular direct product of a translation plane as a left vector space, the components of the net R become ‘left’ subspaces whereas the Baer subplanes incident with the zero vector become natural‘right’subspaces. Of course,theoriginal translation plane may be considered both a left and a right space. Furthermore, any three given Baer subplanes may be chosen so that the three subplanes are both left and right spaces.
20.3. POINTBAERSUBPLANES IN VECTOR SPACE NETS.
267
Theorem 20.21 Let R be any Abelian Baer net which contains three pointBaer subplanes that share the same afine pointP. Let x, be any pointBaer subplane incident with P. T h e n 7ro is an afine translation plane with kernel K,. Let G=, denote the full translation complement of 7ro obtained as a collineation group of R which leaves 7ro invariant. Then the full collineation group of R which fixes P is isomorphic to the product of G=, bp GL(2,K,) . The two groups intersect in the group kernel of T, naturally extended to a collineation group of R. Proof The group GL(2,K0)acts 3transitively on the pointBaer subplanes of the net R and fixes R componentwise. Hence, we may assume that a collineation fixes the zero vector and permutes the pointBaer subplanes 7rW = ( ( 0 , p ) such that p E T,},TA = { ( p , Xp) such that p E 7ro and X in the kernel of 7 r o } (when X = 0 the subplane 7ro is identified with 7ro x 0). So, if g is a collineation of R which fixes the zero vector then we may assume that g leaves rm,no,and 7r1 invariant. Hence, g is in G=, as it acts faithfully on K , . Wenow consider intersections. Since GL(2,K,) fixes R componentwise, assume g fixes R componentwise. Then g induces the kernel mappings on T , and on 7r1 and is fixedpointfree as it also leaves rooinvariant. Thus, the faithful stabilizer of 7ro in GL(2,K,) which fixes 7roo,7ro, and 7r1
( [ PO P ]
in thisrepresentation. It then follows that ) the collineation group of R is the product as maintained.0 is
such that ,B E K,
Corollary 20.22 Let N be a vector space Baer net which contains three pointBaer subplanes which have a common afine point. If one of the subplanes is Desarguesian then the net N is derivable. Proof: Under the assumptions, it follows that if KOis the kernel of one of the subplanes ro then 7ro is a left2dimensional K,subspace. In the direct product net, generally, Baer subplanesare right spaces where the components are left spaces and note the direct product net becomes a left 4dimensional K,subspace. However, by threetransitivity, any three pointBaer subplanes may be considered as left K,subspaces in thesense that the components of the net are left Ktndsubspaces (under what we have been
268
CHAPTER EXTENSIONS 20. BAER
calling {Tp forall p in K,}). Take anycomponent T of the net. Since the intersections of T with any two of the pointBaer subplanes sum (direct sum or product) to the component and there are two of the pointBaer subplanes that are left ldimensional K,subspaces, it follows as before that each component may be considered a 2dimensional left K,subspace. Since GL(2, K,) acts transitively on any left 2dimensional K,subspace which it leaves invariant, it follows that thenet N is coveredby pointBaer subplanes. To see this in coordinate notation, after the basis change x, we have points of the net (x0,xl,y0,y1)where (s0,yo) denote the points of T,. The group GL(2,K,) acts on the left mapping (xo,yo,z1,y1)onto
Now x = 0 is represented after the basis change by the points (O,O,yo,91) which is mapped by GL(2, K,) onto the points (0, 0, ay0 Syl, pyo 791). Since r0 is 2dimensional, we may take zi, yyi E K,. In particular, (O,O, 0 , l ) maps to (O,O, S,7) for all S, y. Hence, the net is derivab1e.O
+
+
20.3.2 When There Are Two PointBaer Subplanes. In this section, we consider the situation where there are exactly two pointBaer subplanes that share the same parallel classes and an affine point. If we consider that the net involved is a vector space net then it follows from the Reconstruction Theorem17.7 that thenet is a direct product netby the two pointBaer subplanes. However, the net is not necessarily a regular direct product net. Still, all of the above lemmas involving decomposition of the space viathe pointBaer subplanes remain valid. We may write the space relative to a vector space WOas WO63 W, 63 W, 63 W,. We choose coordinates so that (0,52,0, y2) defines a pointBaer subplane T , and ( q ,0, y1,O) defines a pointBaer subplane T I . Since the net is a direct product net, we may set up a choice of coordinates for the pointBaer subplanes so that the z = O's, y = O's, and y = x's correspond under the correspondence between parallel classes. Note that we have a subtle distinction between having a net that contains two pointBaer subplanes sharing a point and all parallel classes and realizing the net as a direct product net r0x, T I where Q is a bijection from the parallel classes of T , onto the parallel classes of TI. In the latter
20.3. POINTBAERSUBPLANESINVECTOR
SPACE NETS.
269
case, we choose coordinates for the spreads as noted directly above. If the components for 7ro are say 21 = 0,yl = q M 0 and the components for 7r1 are 2 2 = 0, 32 = z2M1 then we may agree that Q maps 21 = 0 onto 2 2 = 0, yl = 0 onto y2 = 0 and yl = 21 onto y2 = 5 2 . In this way,we may then rewrite the equations for 7ro and 7r1 so that allowing that { (21,22,31,32)) now represents points of the net then 7ro has equation 21 = 0 = y1 and 7r1 has equation 2 2 = 0 = 32. If one considers how components of the direct product net are formulated with the switch of coordinates mentioned above, it then follows that components of the net have the general form
Theorem 20.23 If a vector space Baer net contains two pointBaer subplanes sharing an afine point then the net is a direct product net by pointBaer subplanes. Let the two pointBaer subplanes be denoted by 7ro and 7r1 and let the kernels of no and 7r1 be denoted b y KO and K1 respectively and assume that both planes are left vector spaces over their respective kernels. If the net does not contain three pointBaer subplanes sharing an afine point and allparallel classes then the collineation group of the net which @es 7r0 pointwise induces the full kernel homology group K1 o n 7r1 and may be represented in the form
Proof Clearly, if there are not three pointBaer subplanes, the group fixing 7ro pointwise must also leave 7r1 invariant so that the group has the general form and certainly induces kernel homologies on the subplane From the general set up of the direct product net, it is clear that any kernel homology of can be used to define a collineation of the net whichfixes 7ro pointwise. We consider the mappings acting on the 1eft.O
m.
270
20.3.3
C H A P TEEXRT20. E N SBIAOENRS
The Space Is Decomposable by a PointBaer Subplane.
We assume that there is but one pointBaer subplane withinthe vector space net. In this situation, we must make an assumption which is automatically valid in the finitedimensional case. We assume that it is possible to decompose the space as (z1,zz,y1,y~) for zi,ya in W, for i = 1 , 2 such that a pointBaer subplane r, is given by z1 = y1 = 0 while the components of the net have the form z=o, y = o , y = z
[t
:0]
where MO defines a component of no and the elements A , B ,MO are considered as prime field linear mappings of WO.Furthermore, suppose y is a prime field linear endomorphism which commutes with all such M$. Then the mapping (zo,yo)W (zoy,yoy) will fix each component zo = 0, yo = 0, yo = soMOof no. Hence, defining left scalar multiplication by yz, = z,y (recall that the right hand notation considers the image of zo as a prime field endomorphism) then the mapping is T, as previously discussed so that y is considered within the kernel skewfield KOof ro. In this situation, we say that the space is ‘decomposable by a pointBaer subplane’. Note in the following, there is a natural extension of the definition of ‘central collineation’ of an affine plane to a net.
Theorem 20.24 Let N be a vector space Baer net which is decomposable b y a pointBaer subplane ro. Recall that we denote b y ( K o )the ~ kernel of ro taken as the centralizer of the set {M,} such that yo = xoMo defines a component of ro and ro is a left KOvector space. Assume that N contains exactly one pointBaer subplane incident with a given point and sharing its parallel classes. If G is thefull pointBaer central collineation group of the net which jhes ro pointwise and the central planes are the translates of no then G may be
20.3. POINTBAERSUBPLANESINVECTOR
SPACE NETS.
271
represented in the form
acting on the left. O O O I Proof: Under the given initial set up, anycollineation fixing T, pointwise
ra
may be represented in the form
S o
I i
01
1
where a,S,0, I are consid
L o 0 0 IJ ered as linear endomorphisms over the prime field written on the right. Note that we will showing that S is in (&)L. By assumption, each element of the group leaves invariant each translate of T,. This implies that for any elements c,d of W, then { ( ~ , 1 1 : 2 , d , g 2}) is left invariant. Hence, it follows that a = I in all cases. Now represent the components of T, in the form II: = 0 = ( 2 1 , ~ )and y = II:
[ A cB ]
where A, B,C are considered as linear endomorphisms over
the prime field. Since, the indicated collineation fixes each component, it follows that
=
K
B  6 CC+ A 6
1
so it follows that SC = AS. Now, suppose that S is singular. Then the collineation fixes points outside the pointBaer subplane.Thenthere is a translate of T, say rc = {(c, Q,c, y2) ; 11:2,g2 E W,} (such that CS = 0) which is fixed pointwise by the collineation. Let J be any point. Then, since both r0 and rc are pointBaer subplanes, there exist lines ! ,and !,of T, and 7rc respectively which are incident with J and must be fixed by the collineation. However, this says that J is fixed and as J is arbitrary, we have a contradiction. Thus, S is nonsingular as a prime field Plinear endomorphism.
CHAPTER EXTENSIONS 20. BAER
272
For some fixed collineation, change bases over the prime field P by
T O O 0 O O O T where T is a Plinear automorphism so that the group
;l)
I P O O
([:;;
O O O I
becomes
I T  ~ P To
([: O
I0
O
o
OI T.],'
O
I
Choose a particular 6 so that Tl6T = I. The components of r0 now have the form
which is
T~AT T~BT A* B* 0 TlCT]=[ 0 C*]'
Since we now have an element of the form
8 8 1 [ O O O I
A* = C*. Following the previous argument, we have that if
, it follows that
20.3. POINTBAERSUBPLANESINVECTOR
SPACE NETS.
273
is an element of the groupafter the basis change then y is nonsingular and ?A* = yC* = C*y. We are now thinking of the elements y as linear endomorphisms over the prime field. The points of the subplane x, have the form (zo,yo) and components have the general form z, = 0, y = z,C* where C* is a Plinear automorphism of W,. Hence, y is in the kernel of no considered BS an element of (&)L. Defining left scalar multiplication as in the preliminary section, we may consider the above matrix group acting on the left. This proves the theorem.0
Chapter 21
TRANSLATION PLANES ADMITTING BAER GROUPS 21.1 General Spreads with Baer Groups. In this section, we considerwhen an Abelian net 'with three pointBaer subplanes can be embedded in a translation plane with kernel K in such a way so that K may be considered to be contained in the kernel KO of a given pointBaer subplane.
Definition 21.1 Let D be a vector space net. Let ro be a translation pointBaer subplane with kernel KO. A collineation group of D W n g no pointwise and which may be identified with
is called a (full K,pointBaer elation group'. A collineation group of D jibing ro pointwise which may with
275
be identified
276
CHAPTER 21. BAER GROUPS
is called a %l1 K,pointBaer homology group'. The following somewhat omnibus theorem tries to detail various situations in which a pointBaer subplane of a translation plane with kernel K is left invariant under the kernel mappings Kend.The proofs to all parts of the theorem will be given collectively. Theorem 21.2 Let n be a translation plane with kernel K . Assume that CT is a pointBaer collineation and Fixa = no. Let K, denote the kernel of no. Let G denote a collineation group of n in the translation complement which f i e s no pointwise. Let h0denote the Baer net defined b y the components of no. Then no is a Ksubspace in an9 of the following situations. (i)K is a field and [GI> 2. (ii) The characteristic is not 2 and c is a pointBaer elation. (iii)G is a full K,pointBaer elation group and [K,[ > 2 (iv) L, is a subskewfield of K,, \Lo/> 3 and G is a full L,pointBaer homology group. (iv)' More precisely, If the collineation group of n which f i e s no pointwise contains a pointBaer homology group isomorphic to the multiplicativegroup of a subskewfield of K, and has order > 2 then no is a Ksubspace and there exists another pointBaer subplane of &o that shares an aBne point withno and is invariant under the pointBaer group and which is also a Ksubspace. Proof Either no is a Ksubspace and we are finished or not every kernel homology Tp leaves no invariant. First assume that there are exactly two pointBaer subplanes no, n1 incident with the zero vector that share the infinite points of no and g in the kernel of n interchanges no and n1. Let H denote the subgroup of the multiplicative group fixing no. Then H U gH = K* = K  (0). We assert that HU(0) is a skewfield. To see this, let h, IC be in H.Then h f k is either in H or gH.However, if a E T, then a(h+ k ) = a h + a k and since both ah and a k are in no and no is a subspace over the prime field, it follows that h + IC is in H U (0). It is immediate that 1 in K must leave every subspace over the prime field invariant so that it follows that h in H implies that h is in H . Hence, H U (0) is an additive group, H is a multiplicative group
21.1. GENERAL SPREADS
WITH BAER GROUPS.
2 77
and it follows that H must be a subskewfield as the remaining skewfield properties are inherited from those of K. It then follows that gH U {0} is additive. Now consider gh h = ( g + 1)h for h in H and note that this element is either in H or gH. If (g + 1)h is in H then g 1 is in H but 1 is in H and H is additive so that g is in H. If ( g + 1)h is in gH then, since gH is additive, this forces h to be in gH. Thus, we obtain a contradiction in either situation. Hence, we may assume that there are at least three pointBaer subplanes incident with the zero vector that share the infinite points of T,. By the previous chapter, we have a complete determination of & o . Recall that we obtain the representation of the points of the net in the form (s1,y1,22, yz) where T , is given by s 1 = y1 = 0. Change bases by x as before so as to represent the points in the form (z1,x2,y1, y2) for all si,yi in W, for i = 1,2 and W, a left K,subspace now considered as a Psubspace where P is the prime field. Nowwe may allow that the equations
+
x=O,y=O,y=xMforM=
+
[ :: ]
define the components of the plane where the entries mi are Plinear transformations of W,. We may also assume that y = s is one of the lines of the plane. The pointBaer subplanes of &o incident with the zero vector are now T, and TA
= { (21, As1,yl ,Xlyl)} for all z1, y1 in W, for each X of K,}.
Furthermore, the components of Goare s = 0, y = s
[
1,
for a set
of Plinear automorphisms {W}, such that the components of T , are z 1 = 0 and = Z ~ W . Now what has occurred is that we have changed bases with respect to the prime field P to obtain this representation. Let T denote the basis change. Hence, there is an isomorphic translation plane TT and subplane T,T. The original kernel mappings Tp are represented as Plinear mappings, so let ,&c = sB over the prime field where B commutes with M when y = z M
CHAPTER 21. BAER GROUPS
278
is a component. Since we are free to choose x = 0, 9 = 0, 9 = x without destroying the original kernel K, we now have a kernel of the form
We further denote the new kernel by S”KS. Hence, there exists afield of Plinear automorphisms, KS, isomorphic to K which fixes all components of &,r. We will show that this kernel for TT leaves T,,T invariant. Hence, this implies that the original subplane T,, is a Kspace. We have determined the full collineation group GT, GL(2, K,,) of the net. Note that the subgroup of Gn, which fixes each component is actually in GL(2,KO),Suppose that cr is an element of the kernel of the translation plane. Since cr in Gs, GL(2, K,,), it follows that we may assume that the kernel is in GL(2,K,,) in this representation. Notice that the group GL(2,K,) of the net h,in this representation is (Diag
c
[ 06 ]
for all a,& 6 ,in~K,,
)
such that the mappings are nonsingular and where the entry elements p of K, are writtenas left K,,scalar mappings and then,in turn, as Flinear mappings acting on W,. In other words, the action of the element Diag on a point
( 2 1 , 2 2 , 91,92) is
Furthermore, such an element leaves T,, (x1= 91 = 0) invariant if and only if 6 = 0. Hence, each element B* = a p for some a,p, S, y of K,,. w e note 71 that the set is an additive set of matrices with entry elements thought of as Plinear automorphisms. Each collineation of the plane which fixes the zero vector is an element of rL(x,Ks) (a semilinear automorphism over the kernel). Each collineation
L
WITH BAER GROUPS.
21.1. GENERAL SPREADS
2 79
which fkes T , pointwise in this representation is in the group
[ (Diag [
:]
(Diag
P O
).
such that X E KO
] such that p
E KO IO})
.
Furthermore, each such collineation normalizes KSwhich is a subskewfield of (Diag
[
f]
forall a , p , S , r i n K,
)
such that the mappings are nonsingular. We point out that all of the operations are considered on the left. First assume that a pointBaer collineationa of the plane is Diag Suppose that k = Diag
[
[::l
f ] is in the kernel of the translation plane.
Then consider the upper 2 x 2 submatrix of alcal k(note that alc0l k must be in the kernel of the translation plane):
[;:][ S [;;"I[ a P
a P
S
[y
6X (a+XS)X+Xy
I.
Apply the conjugate subtraction construction again to obtain the element
so that for X # 0, it must be that S = 0 or the characteristic is two. Again note that if the characteristic is not two then since the subplane T , has the representation {(O,z2,O,p2)}, it follows that any kernel element
f ] must leave invariant. [ Hence, when the characteristic is not two and there is a pointBaer ela
represented on the net in the form Diag
To
tion with fixed point set 7ro, it follows that the kernel must leave T , invariant and hence rIT0 is a Ksubspace. This proves part (ii).
CHAPTER 21. BAER GROUPS
280
Now assume that K is a field. Note that we do not necessarily then have that KOis a field. First assume that there is a pointBaer elation. We may assume that the characteristic is twoby the above argument. Also,by the previous argument, we consider the following element:
X6 (a+X6)X+Xy 6X
(a+X6)X+Xy 6X
Then (a+XS)X+Xy
[ + +6X 6x6 =[ and [; /3] [ ^o"  ( a : + W X + h X6a
((a X6)X
+ Xy)6
Asp + ((a+ X6)X + Xy)y
1
+ + Ay) + psx + + Ay) + y6X Equating the (2,2)entries, we obtain 6((a+ X6)X + Ay) + y6X = 6x7 so that 6((a+ X6)X) = ySX which implies that 6(a+ AS) = 76. This is to say 6X
l=[
1
ax6 .((a X6)X 6x6 6((a X6)X
that whenever there is a pointBaer elation of the form Diag
[
*
] then
S(a + M ) = y6. So, if there are two pointBaer elations then there exists another element p of KO {X} such that 6(a+ p&) = y6 = 6(a+ M ) which implies that 6p6 = 6x6. Hence, either 6 = 0 or p = X. So, if the pointBaer elation group has order > 2 and K is a field, we have a contradiction. Thus, assume that K is a field and there is a pointBaer homology. We may assume that there is a pointBaer homologywhichfixes 7ro pointwise of the followingform:Diag
[ y ] for A Z O .
Form
[ t ;][6
a
p y][
xl 0
0 a P XaX"1  a (X  1)p l ]  [ & r]=[a(x"l) 0
I.
Since K is a field then KS is a field. (Note again that we are not assuming that KOis a field.)
21.1. GENERAL SPREADS WITH BAER GROUPS.
281
Then
Hence, we have
[ (XaX16
+ (X  1)p6  1)a
(XaX1
a)a
(x1
+
XaXl a) p6 (X1 XaX"a)+yS(Xll)
 a)p + (X  1)p7
6 (x1  1)p
 1)
a (X  1)p
S(Xl)P
1
1
*
Equating the (2,2)entries, we have S (X  1)p = S (Xl  1) p. If
[email protected] # 0 then X = X' so that the group has order 2. If Sp = 0, assume p = 0. Then the element Diag
] is in K S which implies 6 0 as [ Hence, $;I1lIG p 0 S 0 which implies that is left =
X' # 1 and XaXl = a. = = = 7ro invariant by the kernel mappings of the translation plane. So, if there is a pointBaer homology, K is afield and the fixedpoint subspace is not a Ksubspace then the only pointBaer homology is Diag
Thus, there must be pointBaer elations if the group has order larger than 2. However, by previous arguments, the characteristic must be 2 and the condition X' = X provides a contradiction unless 6 = 0. Hence, in the case that K is a field, and the group fixing no pointwise has order > 2 then 7ro is a Ksubspace. This proves part (i). We now assume that there is a full pointBaer elation group
The previous arguments show that the following elements
[
(a+XS)X+Xy 6X
I
for all X in KOare obtained so that Diag
[ ^o"
(a+XS)X+Xy
6X
1
CHAPTER 21. BAER GROUPS
282
is in the kernel. Moreover, we may assume that the characteristic is 2. We have
[y
(a+XS)X+Xy SA
=[
(X +PIS 0
(a
] [;
(a+pO+m SP
+
1
+ (X +P)S)(X + P ) + (X + P)/ 6(X
+ P)
1
which implies (using the (1,2) elements) that p6X = XSp for all p, X in K,.
Hence, K, is a field or 6 = 0. Assume the former. Then
[ ^o”
(a+M)X+Xy 6X
1
6P
On the other hand,we already would have an element constructed from a suitable pointBaer elation of the form
[
(XPW
(a + (XPS)S)(XPS) + ( X P S h
o
WPS)
1
This implies that
+
+
+
+
XS((a PS)P m ) ) ((a W X = (a (Xp6)6)(XpS) (XpS)y.
+
+
+ WW
Since KOis a field, we are reduced to XS2p2
+ X2S2p = aXpS + X2p2fi3 + XpSy for all X, p in KO.
Keep X fixed and nonzero then p satisfies a quadratic in KO for all p in K,. By our assumptions, this implies that AS2 = X2S3 for all X in KOso that S = 0. Hence, T, is a Ksubspace in this situation. This proves part (iii). Now assume that K is not necessarily a field and assume the conditions of part (iv) so that every element fixing T , pointwise has the form
21.1. GENERAL SPREADS
WITH BAER GROUPS.
283
[ y]
for all X # 0 in a subskewfield L, of K,. So, there are at least Diag two pointBaer subplanes which share the same infinite points and exactly one affinepoint. Either both areKsubspaces or we have at least three pointBaer subplanes that share an affine point and all parallel classes. Hence, we may assume the latter case. Conjugate the kernel element Diag and subtract to obtain Diag
[
XaX'
a
S(X1
1)
(X
 1)p 0
1
is in the kernel for all elements X # 0 in L, and fixed p, 6,y. Assume that PS # 0. Fix X as X, # 1 and take the inverse of this element and note it is 0 ((X  1)p>'
Diag = Diag
(6 (X1  l))1 ((X  1) p p (  (XaX1  a))(S
(X1
1
 l))1
[O f ]. g e
Apply the conjugationsubtraction method again using a general element Diag
[ y ] as above to obtain an elementin the kernel of the form [
0
Diag g(p1 where g = ((X,
 l)p)'
is in the kernel, we have
 1) = d
and f = (6 X(';
(p
0 =
1
 l))'. Assume bd # 0.
Since
CHAPTER 21. BAER GROUPS
284
is in the kernel for all elements X in L:. Since the kernel is fixedpointfree, it follows that
Hence, we have Xlp' = plX' so that L, is a field provided bd # 0. Thus there is a function U : L, $ K, such that X" = dXldl for X # 0 and where we may assume that 0" = 0. Since U is 11 and onto and preserves addition and multiplication, it follows that U is an isomorphism of L, into K,. For fixed d, and r E K,, the mapping P : K, "t K, such that r p = d7dl is an inner automorphism of K,. Hence, Xp = X" so that X' = Note that L! = LE. Hence, q 3  l = 8 is an automorphism of L, such that X' = X" for all X # 0 of L, and if bd # 0, L, is a field. In this latter case, it followseasily that L, E GF(2), GF(3) or GF(4). However, the group is assumed to have order > 2, so we have a contradiction unless we have GF(4). If L, is isomorphic to GF(4), let the elements be denoted by 0,1,8 and e2 = e + 1.
[ dXldl O I , we apply the argument say d*. It then follows that since the
Since we have the elements Diag
again for a possibly different d d = d* kernel is additive. Recall that d = g(pl  1) = ((X,  l)P)l(pl  1). Fix p for this particular d as p,. We could have obtained an element d* for fixing say X1 initially and then isolating on pl. Hence, it follows that
and hence, for p nonzero we have
(X,  1)l(p;l
 1) = (X1  l)l(pTl  1)
for all X,, X1,p,,pl in GF(4) not equal to 0 or 1.
21.2. T R A N S L A T I O N P L A N E S W I T H SPREADS IN PG(3,K).
285
Since 7l = 72for 7 in GF(4) {0}, we obtain ((X,
+ l)(p, +
=(
( h + u p 1+
+
Let X, = p, = 8 and X1 = p1 = 8 1 toobtain O4 = (8 contradiction. Hence, the situation GF(4) does not occur, Hence bd = 0 and note b = 0 if and only if d = 0. Thus, ((X,
+
a
 1) p>l (p1  1) = 0 = ( p  1) (6 (Xi1 
for all such nonzero elements of L,. This implies 6 = 0 = p. Thus, the kernel K must leave T, invariant. This completes the proof to part (iv). So, this completes the proofs to all parts of the theorem.0
21.2
Translation Planes with Spreads in PG(3,K ) .
In this section, we shall show that spreads in PG(3,K), for K a skewfield, that admit full K,Baer elation or homology groups can only exist if the skewfield K is a field. More precisely,
Theorem 21.3 Let T be a translation plane with spread in PG(3,K) for K a skewfield. Assume that T , is a pointBaer subplane with kernel K,. If T admits a full K,pointBaer elation group of order > 2 or a full K,pointBaer homology group of order > 2 j h n g T , pointwise then (1) K, is isomorphic to K and (2) K is a field. Proof By the previous section, T, is a Ksubspace. Hence, it follows that is a 2dimensional Ksubspace. The components of no are 2dimensional Ksubspaces so that there is an induced Desarguesian spread which forces T , to be a Desarguesian affine subplane. Hence, the kernel of no is isomorphic to K. Choose coordinates so that T is T,
where scalar multiplication is on the left
CHAPTER 21. BAER GROUPS
286
and 7ro is { (0, x2,0,92) for x2,yz in K} and where components of the general form
7r
have
and f , g functions : K x K into K. It is easy to see that the components of are
7ro
Let f(u, 0) = f(u)and g(u,0) = g(u). Assume that 7r admits a full KOpointBaer elation group B with axis 7ro and center C the set of translates of no. Then B fixes 7ro pointwise and since 7ro is a Ksubspace, it follows that B is a linear group representable by a 4 x 4 matrix with elements in K and acting on the right. Furthermore, B leaves invariant every subplane of C. The group which fixes 7ro pointwise is Klinear and has the matrix form
But, each pointBaer elation fixes each translate of 7ro so that it follows that a = 1 in all cases. We emphasize that this matrix group is acting on the right whereas previously the action was defined on the left when we were considering the group acting on the net and elements defined via the kernel of the subplane KO.It is this fact which we shall see will force the commutativity of K. The pointBaer elation groupfixes each component y = x for U in K. Hence, we must have 0
1
o
u
0 1
1
21.2. TRANSLATIONPLANESWITHSPREADSINPG(3,
K).
287
Thus, bu = f(u)b for all b,u in K. So f(u) = U for all U in K and bu = ub for all b,u in K. Hence, K is commutative. Now assume that there is a full K,pointBaer homology group B fixing 7ro pointwise. By the previous section, 7ro is a Ksubspace. By results of the previous chapter, we see that there aretwo pointBaer subplanes 7ro and T ( Q  1) sharing the same parallel class both of which are Ksubspaces for Q in B  {l}.Note that T ( Q 1) is clearly ainvariant and distinct from 7ro. Hence, we may choose bases as above but now 7r(a  1) is represented by {(z1,0, y1,O) for all 51,y1 in K}. It follows that the collineation group B is Klinear and has the matrix form
acting on the right. It is immediate that, in this case, b = 0 for all elements. The components of all U in K. Thus,
7ro
now have the form y = z
!][ f ( 4
0
0
[ ft' g ] , z = o f o r
!]
so, .l
f (U). = f (U)
for all a # 0, U in K. Since { f ( u )for all U in K} = K, it follows that av = vu for all a,v in K so that K is commutative. This proves the the0rem.O Wenow prove that a pointBaer homology group becomes a Baer homology group under certain circumstances. Theorem 21.4 Let 7r be a translation plane with spread in PG(3,K),for K afield. Then, any pointBaer homology group of order > 2 is a Baer homology P U P .
CHAPTER 21. BAER GROUPS
288
Proof Let B denote the pointBaer homology group and let no = FixB. Then, by convention, n(a  1) = 7r1 = T(T  1) for all 6 ,in ~B. We know from the previous results that no is a Kspace. Moreover, the previous proof shows that the unique 0coaxis of B (unique such pointBaer subplane # no incident with the zero vector) is also left invariant by K . The action of B on n(6  1) induces a kernel homology subgroup on the subplane. On the other hand, n(a  1) is a Pappian plane with kernel K. Hence, there exists a subgroup of the kernel homology group which induces the same action as does B on n(a 1). So n(a 1) is fixed pointwise by a nontrivial collineation group which forces the subplane to be Baer. This completes the proof of the theorem.0
Corollary 21.5 Let n be a translation plane with spread in PG(3,K ) , for K a skewfield. Let no be a pointBaer subplane with kernel KOwhich is f i e d pointwise by a full K,pointBaer group B of order > 2 (either a pointBaer elation or a pointBaer homology group) Then K is a field and B is a Baer elation or a Baer homology group.
Proof: We may assume that B is a full K,pointBaer homology group. Since K is now a field by the above remarks and no is a Ksubspace, it follows from the above theorem that B is a Baer homology group.0 Theorem 21.6 Let n be a translation plane with spread in PG(3,K ) for K a skewfield. Let 7ro be a pointBaer subplane with kernel KO. If B is a full K,pointBaer elation group of order > 2 then K is a field and one of the following three situations occur: (i) The net defined by the components of no is a regulus net relative to PG(3,K ) . (ii) The net defined by the components of no contains exactly one pointBaer subplane incident with the zero vector, the net is not derivable and the full collineation group which f i e s no pointwise is
([ i
H1
l a 0 0
foralZuinK).
21.2. T R A N S L A T I O N P L A N E S W T H SPREADSIN PG(3,K).
289
(iii)The net defined b y the components of r, is a derivable net and there is exactly one pointBaer subplane incident with the zero vector which is a Ksubspace. I n this case, the components of the net have the following form: x=o,
1
y=x
for all U in K and where A is a linear transformation considered over the prime field P of K such that Au  uA is in K for all U in K. Proof: We see that K is a field isomorphic to K,. Furthermore, r, is a Kspace so is 2dimensional and hence we have the decomposable situation described in the previous chapter. Either there exists another pointBaer subplane incident with the zero vector and sharing the same parallel classes or we may apply the results of the previous chapter. Since B acts regularly on the nonfixed ldimensional Ksubspaces that are on components of x,, it follows that BK* acts transitively on the set of points not in ro on components of r,. If there exists another pointBaer subplane, there is a covering by pointBaer subplanes and the net is derivable. Hence, it remains to consider the situation when the net is derivable but not a regulus net. The derivable net must have the basic components
x=o, y=x
[8
g?)]
forall U in K where g is a function on K. And, we note from previous results that K is a field. If the net is a derivable net then there exists a Baer group which fixes r, and any particular vector on say x = 0. We assume that considered over the prime field P of K a basis change leaves r , invariant as well as x = 0,y = 0,y = x and is thus represented as
Ti ' T 0
0
0
0
OTq
C H AGROUPS P T E R 21. BAER
290
where Ti represent nonsingular linear mappings over the prime field P. Furthermore, we assume that the standard representation of the derivable net is now obtained. That is, we have:
Hence, it follows after a bit of calculation that
g(u) = T2TC1u  uT2TL1
[%a;I]
for all U in K.Letting A = T2Tc1 shows that case (iii) is the only remaining possibility. Conversely, any setof matrices of the form in case (iii) determine a derivable net by the basis change O O I A '
This completes the proof of the theorem.0
Corollary 21.7 Under the assumptions of the above theorem, if K is finite then Au  uA = 0 for all U in K and case (iii) does not occur.
+
Proof g(u) = Au  uA implies that vs(.) us(.) = g(uv). Hence, 2ug(u) = Au2  u2A. In particular, for characteristic two perfect fields, it follows that g(w) = 0 for all W in K. An easy induction argument shows for all integers that g acts like a formal derivative so that g(u') = rurplg(u) r. Hence, if K is GF(q) then g(u) = g(uq)= qzP1g(u)= 0 for all U in K. Example 21.8 Let L be a field of characteristic two which is not perfect.
Let 7 denote anonsquare
in L. Then {
quadratic extension field K of L.
1 1 L
for u,t in L } definesa
.l
Furthermore, for any 2 x 2 matrix A over L , x = 0 , y = x for all W in K defines a derivable net which is a regulus n e i i n PG(3,K ) if' and only if A is in K.
21.2. TRANSLATIONPLANES WITH SPREADS IN PG(3,K ) .
291
Proof: It suffices to show that for any 2 x 2 matrix A over L then Aw wA is in K . Let A = Then, for W = and U , t in L , we obtain:
[
x: 1.
AwwA=
[p ]
(a2
+V 3 ) t
(a4 + a1)t
++V a1)t 3 ) t
Y(a4
(a2
1
which is always in K . This expression is identically zero if and only if a2 = ya3 and a4 = al which is valid if and only if A is in K .
Corollary 21.9 Let K be any field which admits a nontrivial derivation 6. Then there is a derivable partial spread in PG(3,K ) which is not a regulus and whose corresponding derivable net contains exactly one Ksubspace pointBaer subplane. Theorem 21.10 Let r be a translation plane with spread in PG(3,K ) for K a skewfield. If r admits a pointBaer subplane ro with kernel KO and a collineation group of order > 2 an the translation complement which is a full KOpointBaer homology group then one of the following occurs: (i)There are exactly two pointBaer subplanes incident with the zero vector and sharing their parallel classes, the net defined by the Baer subplanes is not derivable and the f i l l collineation group which fixes r0 pointwise is
0 0 0 1
(ii) There are exactly two pointBaer subplanes incident with the zero vector which are also Ksubspaces but the net defined by no is a derivable net. The components maybe represented in the formx = 0 , y = x
[T e]
for all U in K and Q in AutK. (iii) There are at least three pointBaer subplanes incident with the zero vectorwhich are Ksubspaces and the net defined by ro is aregulus net (corresponds to a regulus in PG(3,K)).
292
CHAPTER 21. BAER GROUPS
Proof By previous results, there are at least two pointBaer subplanes which are Ksubspaces, K is a field and K1 is isomorphic to K where K1 denotes the kernel of T(Q  1) = 7r1 and where Q is a pointBaer collineation. The group B mentioned inpart (i) is clearly a collineation group of the plane. Moreover, the group BK" acts regularly on the points on component of 7ro which are not in 7ro U TI.Hence, if there exists a third pointBaer subplane, the net is derivable. If there exists a third pointBaer subplane which is a Ksubspace, the net is a regulus net. Finally, when there are exactly two Ksubspace pointBaer subplanes, one can diagonalize the space and when the net is derivable, the associated matrices of the spread components form a field. Hence, (ii) follows immediately.0
Chapter 22
SPREADS COVERED BY PSEUDOREGULI. In thischapter, we consider spreads in P G ( 3 , K ) which are coveredby pseudoreguli that share a given line that we call ‘conical spreads’ and we consider spreads which are covered by pseudoreguli that share two given lines which we call ‘ruled spreads’ and formulate the corresponding theory. Although it seems natural enough to consider this study in the context of derivable nets, this material originated not with derivable nets but with the consideration of flocks of quadric sets. In the next chapter, we sketch part of the theory interconnecting coverings of quadrics by planes to the analysis of spreads covered by reguli. However, we choose to work from the general to the specific in this instance.
22.1
PseudoReguli.
There are some technical problems forming unions of pseudoreguli that can occur due to the possible noncommutativity of multiplication of K so we consider what are called ‘normal sets’ of pseudoreguli. We have discussed a geometric version of a ‘regulus’ in PG(3,q) and a pseudoregulus is only defined algebraically. Furthermore, we have defined a Kregulus as a pseudoregulus over a field and have given a geometric definition in definition 14.15
293
294
C H A P T E R 22. SPREADS COVERED
BY PSEUDOREGULI
Wenow consider a possible geometric definition of a pseudoregulus. This is not completely satisfactory but it gives some insight as to the difficulties in working with noncommutative geometries. It has been shown that given a derivable net N algebraically represented as a pseudoregulus net with reference to a skewfield K then thederived net N* may be algebraically represented with reference to the skewfield K o p p where multiplication o in K o p p is defined by a o b = ba where juxtaposition denotes multiplication in K. We recall that if the vector space V of points is a left vector space over the skewfield K then the lines of the derivable net incident with the zero vector are not necessarily always 2dimensional left Kvector subspaces although they are natural2dimensional right Kvector subspaces. However, the Baer subplanes incident with the zero vector are 2dimensional left Kvector spaces. For the derived net, the situationis reversed. The lines of the derived net incident with the zero vector are not always 2dimensional left KoPPvector subspaces but they are natural 2dimensional right KoPPvector subspaces as they are always 2dimensional left Ksubspaces. Similarly, the Baer subplanes incident with the zero vector of the derived net are 2dimensional left KoPPvector spaces as they are thelines of the original net incident with the zero vector which are 2dimensional left Kvector subspaces. Since we would like to represent the lines of our derivable net within the lattice of left subspaces of a 4dimensional left vector space, we dualize everything and note the following:
Remark 22.1 Let N denote aderivable net with lines incident with the zero vector represented in the form = x6 for all 6 in a skewfield J where the associated vector space V is a $dimensional left Jspace and the lines indicated are 2dimensional left Jsubspaces. So, the lines of N incident with the zero vector become lines in the projective space C isomorphic to PG(3,J ) & defined as the lattice of left vector Jsubspaces. In terms of a given basis, V may also be defined as a 4dimensional right or left JOppvector space V* and the lines of N* incident with thezero vector become lines in the projective space C* isomorphic to PG(3,J0PP)& defined as the lattice of left vector JOPPsubspaces.
22.1. PSEUDOREGULI.
295
Choose a left Kbasis B = {ei f o r i in X}. For a vector Cxiei, xi in K f o r i = 1,2,3,4, a left space over K o p p may be defined as follows: uoCxiei = Cxiuei = C(uoxi)ei for U E K . So, there are many ways toform a projective space PG(3,K O P P ) if K is a noncommutative skewfield. Remark 22.2 In the following, we shall be considering always ‘left’ vector spaces over either K or K o p p , so we shall use simply PG(3,K ) to denote PG(3,K ) L . Note that a right ldimensional Ksubspace may notbe associated with a ‘point’ of PG(3,K ) but it would be a ‘point’ of PG(3,K O P P ) . If the associated vector space is considered over Z ( K ) ,we call ldimensional Z(K)subspaces, ‘Z(K )projective points Definition 22.3 Let S be any set of mutually skew lines of PG(3,K ) . A ‘vectortransversal’L to S is a line of some PG(3,K O P P ) of Z(K)projective points with the property that L as a left Z(K)subspace has a nontrivial vector intersection with each line of S as a left Z(K)subspace such that the direct sum of any two such intersections is L. A ‘pointtransversal’ to S is a line of PG(3,K ) which is also a vectortransversal. A ‘projective pseudoregulus’R = R!Z”::>\ in PG(3,K)is a set of lines (as left 2dimensional Kvector spaces) containing { L , M , N ) with a set of points {U, V , W of } L such that any line T which intersects L in either U,V, or W and also intersects M and N intersects each line of R and T is contained in the set of these intersections. So, any such line T becomes a pointtransversal. Remark 22.4 W e have previously defined the pseudoregulus net. Here we consider the corresponding net defined by aprojectivepseudoregulusalso called a pseudoregulus net. We shall show that there is no distinction between the two nets and thus the terminology is justified. firthermore, when it does not present problems, we shall use the same notation for the pseudoregulus and the corresponding net and allow context to dictate which is under consideration. Theorem 22.5 Choose any three mutually skew lines L , M , N of PG(3,K) and let U,V,W be any three distinct points o n L.
296
CHAPTER 22. SPREADS COVERED
BY PSEUDOREGULI
(1) Then there exists a unique projective pseudoregulw RCL,M,N} {SV,W in PG(3,K) which contains L, M , N and which has pointtransversals intersecting L in U,V, and W . Furthermore, there is a unique basis such that L , M , N may be repre(V I) sented in the form y = x, x = 0, y = 0, respectively where L = ( U ) @ andW=(U+V). The pseudoregulus then has the form x=o,y=x
[;
:]
for all U in K. (2) Choose any set {L*,M * ,N * } of three mutually skewlines of the pseudoregulus R:;:; , Then .thereexist points U*, V*, W*on L* such
*,v*,w*)  R{u,v,w)
that R I i * l M * , N * }  {L,M,N}' (3) Any pseudoregulw net is a derivable net (is a standardpseudoregulus net). Proof: Consider the associated 4dimensionalleft Kvector space V . Choose a basis for L as (U, V ). Then W = aU PV for a,p in K. Hence, projectively, we may assume without loss of generality that W = U V . Represent V4 = M CBN , then there exist unique elements mu, m, of M and nu,n, of N such that U = mu nu and V = n, nu. Then
+
+
It is immediate that N. Now form
{h, m,}
+
+
is a basis for M and {nu,n,} is a basis for
Clearly, Tu n L = ( U ) , TV n L = ( V )and TW n L = ( W ) . Choose a basis {m,m,, nu,n,} for V4. In terms of this basis, Vq={(~1,~2,y1,~2);~k,~iEK,i=1,2}.
22.1. PSE UDOREG ULI.
297
Let = ( 2 1 , ~and ~ ) y = (yl,y2). Then L, M ,N are y = z, y = 0, z = 0 respectively. Furthermore,
TU = {(~1,0,1/1,0); za E K , i = L2}, T V = {(O,y1,0,!/2) ; yi E K , i = V } , Tw = { ( ~ l , ~ l , ? / l , Y; Z1,Yl l) E K}. It follows that any component of the pseudoregulus isof the form y = z
[ 1] for
a, b, c, d in K and the intersection with
[ i :] for
TU,TVand TWshows
that a = d = U and b = c = 0. Let the pseudoregulus R be represented in the form z = 0,y = 0,y = z
U
E X E K. Now in order that TU
is contained in {TUn 2 ; z E R}, we have { ( ~ 1 , 0Z, ~ U 0) , ; ~1 E K } = TUV U E X.
It clearly follows that this forces X = K so that the pseudoregulus has the required form. It follows immediately that any pseudoregulus is a derivable partial spread. The derivable net R defined by 5=O,y=a:
[ i 81 V U E K
has Baer subplanes
and (a,b) # (0,O)and by such a choice of basis for the vector space, we see that there are at least three 2dimensional KoPPsubspaces whichare vectortransversals that arealso pointtransversals (lines of PG(3,K ) ) ,namely po,l, pl,o and pl,l . Note that if Z ( K ) is isomorphic to GF(2), there are exactly three lines of PG(3,K) which are pointtransversals to this net. Now assume that there is another pseudoregulus satisfying these conditions. Assume that exist three pointtransversals yo,yl ,y2 (2dimensional left Ksubspaces) such that yon L = U,yl n L = V and y2 f l L = W . The
298
CHAPTER 22. SPREADS COVERED BY PSEUDOREGULI
question is whether yo actually turns out to be Tu above. Since it intersects L in the same subspace, we may assume that yo contains (1,0,1,0). Let (z1,z2,y1,y~) and ( l , O , 1 , O ) generate yo. Since yo intersects z = 0 and y = 0 in a ldimensional Ksubspace, it follows that the only way to manage this is for z2 = 0 and for y2 = 0. The two indicated vectors are linearly independent so it is clear that yo = ( ( l , O , O , O ) , (O,O, 1 , O ) ) = Tu.Similarly, 71 = TV and 72 = Tu. In other words, any two sets of pointtransversals of three elements to {L, M , N } which intersect L in the same set of points are identical. Since thesepointtransversalsdetermine a unique representation, we have a uniquely defined pseudoregulus. We have noted that there are at least three lines of PG(3,K) which are pointtransversals to the pseudoregulus R{L;$,Nl. {U VWl By the structure theory for derivable nets previously determined, there exists a collineation group of the pseudoregulus net which is triply transitive on the components of the pseudoregulus and fixes each Baer subplane incident with the zero vector and hence fixes every vectortransversal . Thus,thereexists a collineation a of the net which carries {L, M , N } onto {L*,M*,N * } as ordered sets. Choose U*, V * , W* as Ua,Va, Wa respectively. Then the above construction is merely a basis change so that the two pseudoregulus nets are identical.0
Remark 22.6 Any pseudoregulus in PG(3,K ) has a set of transversal lines in 1 1 correspondence with a set of cardinality Z(K)+ 1. We note that since the vectortransversals define Baer subplanes of the net incident with the zero vector, it is not necessarily true that every Baer subplane incident with the zero vector intersects each component in a 1dimensional left Ksubspace (point of PG(3,K ) ) . In fact, thepointtransversals are determined by any one intersection.
Corollary 22.7 Let R be angpseudoregulus in PG(3,K ) . If avectortransversal intersects some line of R in a point (a ldimensional left K space) then the vectortransversal is a pointtransversal (line). Proof: We may represent R in the standard form x=o,y=x
[
e]
VuEK.
22.1.
PSE UDOREG ULI.
299
The vectortransversals are exactly the Baer subplanes Pa$ . Suppose pa,b intersects ~t:=.O in a ldimensional left Kspace. Then, it follows that the intersection is {(O,O,aa, ba) V a in K } . However, this is a ldimensional left Kspace if and only if a and b are in Z ( K )which implies that theintersection
[
with y = ~t:
]
whichis
{ (aa,ba,aau, bau) V a E K} and is a ldimensional left Kspace. Similarly, if any such intersection
{ (aa,ba,aau,bau) V a E K} is a ldimensional left Kspace then a and b are in Z(K) so that all intersections are ldimensional left Kspaces.O This also proves the following wellknown corollary which we repeat in order to emphasize the ensuing remarks.
Corollary 22.8 If K is a field then there is a unique regulus containing any three mutually skew lines L, M , and N . Proof: Choose any three points on L and construct the corresponding regulus net. Any vectortransversal is a pointtransversal and corresponds to a Baer subplane of the net. As any three points on L correspond to unique Baer subplanes of this regulus net, it follows that the choice of three points on L is arbitrary.O
Remark 22.9 Toillustratethattheprevious corollay(l is notnecessarily valid for pseudoreguli, suppose K is a skewfieldsuchthat K # Z ( K ) N GF(2). Then, the above resultshowsthat for any three distinct points U,V,W on a line L of PG(&K), there is a unique pseudoregulus R);Z>\ containing { L ,M ,N } . Furthermore, thereare exactly three Baertransversals whichare pointtransversals to this pseudoregulus. So, take any threedistinct points U*) V', W*on L such that {U,V,W } # {U*, V * ,W " } . Then, it {V*,V*,W*} is not possible that RIL:n;,N) 1' is equal to . Hence, there exist skewfieldsK such that thereare infinitely many pseudoreguli that share any three mutually skew lines in PG(3,K ) .
300
22.2
C H A P T E R 22. SPREADS COVERED
ConicalandRuled
BY PSEUDOREGULI
Spreads over Skewfields.
Now suppose that D is a pseudoregulus net represented in standard form. Suppose that a subplane Pa$ intersects x = 0 in {(O,O,ap,bp)Vp E K}. Assume that a and b are not both in Z(K). Choose the vector (O,O,a,b) and let ((0, 0, a, b ) ) = U denote the left ldimensional Ksubspace generated by (O,O,a,b). Assume that {x = 0, M , N } is a set of skew lines in P G ( 3 , K ) which is not in D. Choose any other points V, W of x = 0 and form R{2=O,M,N {v’’w) } = R. Then R and D are pseudoreguli which share at least one line but do not have the property that the vectortransversals to the two pseudoreguli partition x = 0 in the same set of sublines. We shall be interested in situations where there is such a partition and, to this end, we formulate the following definition.
Definition 22.10 Let D1 and D2 be any two pseudoregulus nets whose union defines a partial spread in P G ( 3 ,K) that share either one or more components. Assume that on one of the common components L there exists two points of P G ( 3 , K) (ldimensional left Ksubspaces) each of which are in pointtransversals to ~i for i = 1,2 respectively. Any two pseudoreguli sharing one or more two lines whose nets satisfy the above property shall be said to be ‘normalizing’. Furthermore, any set of pseudoreguli sharing one or more lineseach pair of which satisfies the above property with respect to the same two points shall be said to be ‘normal set’. Any such point shall be said to be a ‘normal point’. Remark 22.11 If K is afield, any tworeguli satisfying the above conditions are normalizing pseudoreguli. Theorem 22.12 (1) Two distinct normalizing pseudoreguli in P G ( 3 ,K) share one or two lines. (2) If twonormalizingpseudoregulishareexactlyoneline then there is an elation group with axis the common line which acts regularly on the remaining lines of each pseudoregulus. Furthermore, the vectortransversals to each pseudoreguhs net induce the same partition on this common line. ( 3 ) If two normalizing pseudoreguli share at least two lines then there is a homology group with axis and coads the two common lines which acts regularly ontheremaininglines of each pseudoreguhs. Furthermore, on
22.2. CONICALANDRULEDSPREADSOVERSKEWFIELDS.
301
one of the common lines, the vectortransversalsto the various pseudoreguli induce the same partition on this line. Proof: Assume the hypothesis of (2). By appropriate choice of coordinates, any given pseudoregulus may be brought into the form x = 0,y =
[
]
x V u E K . Let D1 have thisstandard form. We assume that x = 0 is the common line. The Baer subplanes incident with the zero vector are
Note the Baer subplanes which are transversal lines are r a , b where both a and b are in Z ( K ) .Furthermore, we may assume that the two ldimensional left Ksubspaces on x = 0 which belong to transversal lines of each of the two pseudoreguli have the general form ((0, 0,1,0)) and ((0, O,O, 1)). Note that the Baer subplanes of D1 containing the indicated ldimensional left Ksubspaces are r1,o and ro,l respectively. Assume that y = x q for i in X and II: = 0 are the lines of the second pseudoregulus net D2. Select two distinct values c, d of X and change bases by the mapping
This mapping fixes x = 0 pointwise and carries y = ST,onto y = 0. Now change bases again by the mapping
It follows, after the basis change, that thecomponents of the second pseudoregulus net are
x = 0,y = 2(Td  Tc)l(z  T,). In particular,x = 0, y = x,y = 0 are components of the second net D2. Since x = 0 is fixed pointwise by both of the above basis changes, it follows that there are two transversal lines L, M to D2 intersecting x = 0 in ((0, 0,0 , l ) ) and ((O,O, 1 , O ) ) respectively such that L n (y = 0) = ((rn,n,O,0)) and
302
CHAPTER 22. SPREADS COVERED
BY PSEUDOREGULI
M n (y = 0) = ((S, t , 0,O)). Since L and M both intersect y = x in a 1dimensional left Ksubspace, it follows that m = 0 = t. Hence, L and M are T O J and T ~ , O respectively. It then easily follows since D2 is a derivable net with partial spread in PG(3,K) such that the components of the net have the general form x=O, y = x [
:]
VUEK
and CTan automorphism of K. However, there is at least a third transversal line to the derivable net. Since the Baer subplanes of the net in question now have the form Pa& = {(aaa,ba,apab, p)} then there exist a, b such that ab # 0 and a and b are in Z(K) so it must be that U = 1. Hence, the set of images X
= 0,
= 0,
= x(Td  Tc) 1( T i  Tc)
is equal to
x=o, y=o, y = x
[ t g] V U E K .
Thus, = Tc
+ (Td  TC)UIV U E K.
To prove part (2), it suffices to show that Td  Tc = v,I for some v, in K. Since z = 0 is fixed pointwise, it would then also follow that the vectortransversals to both pseudoregulus nets share the samepoints on x = 0. Hence, we have a partial spread
Let Td =
[ : ii ] and Tc [ z i 2 3. =
'
It then follows that all
matrix differences are nonsingular or zero so that we must have
22.2. CONICAL AND RULEDSPREADSOVERSKEWFIELDS.
303
is nonsingular V u ,W E K . Assume that (bl  b2) # 0. Then choose v so that b2
+ (bl  b2)W = 0.
Then there exists a U E K such that a2
+ (a1  a2)v  U = 0.
This is a contradiction so bl = b2 and similarly c1 = q. Note that if c2 = 0 then choosing U = a2 shows the matrix difference to be singular. Hence, c2d2 # 0. Also, since c1 = c2 and Td  Tcis nonsingular, it follows that a1  a2 # 0. Then, for each U in K , we may choose v in K so that a2 + (a1  a2)v  U = 1. This leaves us with the matrix
Then, in order that the matrix be nonsingular, we must have b2
 %'(l a2 + d2 + ((dl  d2)  (a1  a2))v)# 0 V v E K.
Clearly, this forces ( d l  d2)  (al  a2) = 0. . Thus,there is an associated Hence, Td  Tc= a1  a2 0 a1 O a2 I elation group E of the form
Assume the conditions of (3). Let D1 be represented in the form
.
..
.
..
..
.
304
CHAPTER 22. SPREADS COVERED
BY PSEUDOREGULI
and choose the two common components to be z = 0,y = 0. Let 0 2 have components y = zTa and x = 0,y = 0. Further, assume the two 1dimensional left K subspaces lie on z = 0 and are ((0, 0,1,0)) and ((O,O, 0 , l ) ) . Choose a new basis by (z,y) + (xTd,y).Then there are Baer subplanes in the image of 0 2 which are 2dimensional left Ksubspaces that share z = 0, y = 0, y = z and such that the subplanesintersect z = 0 in ((O,O, 1,0)), and ((O,O, 0,l)). It follows, similarly as in the previous argument, that the two Baer subplanes have the form ((1,0,0,0), (O,O, 1,O)) and ((0, 1,0,0), (0, O,O, 1)).Hence, as before, 0 2 is now represented as
Thus, Ti = T d U I . Hence, there is an associated homology group H of the form
([B ;::]
V.EK{O)).
o o o u
Since y = 0 is pointwise fixed by the basis change above, this proves (3). Furthermore, if 0 2 shares at least three components with 0 1 then this will force Td = v,l so that the two pseudoregulus nets are identical. This proves ( W Thus, we have the following result:
Theorem 22.13 (1) Let 7~ be a translation plane with spread in P G ( 3 ,K), for K a skewfield, which is a union of a normal set of pseudoreguli that share exactly one line L. Then there is an elation group E with axis L of T which acts regularly on lines # L of each pseudoregulus. (2) Let T be a translation plane withspread in P G ( 3 , K )which is a union of a normal set of pseudoreguli that share exactly two lines L and M . Then there is a homology group H of 7~ with axis and coaxis L and M which acts regularly on lines # L or M of each pseudoregulus . In the last chapter, we discuss what are called “skewHall” planes which are defined by deriving translation planes which, although are notnecessarily
22.2. C O N I C A LA N DR U L E DS P R E A D SO V E RS K E W F I E L D S .
305
always Desarguesian, are related to translation planes admitting families of reguli in their spreads. Also, because of the connection with flocks of quadratic cones and flocks of hyperbolic quadrics in P G ( 3 ,K),for K a field, we formulate the following definitions.
Definition 22.14 Let T be a translation plane withspread in P G ( 3 , K ) ,for K a skewfield. If the spread for x is a normal set of pseudoreguli sharing exactly one line L, we shall call T a‘conical translation plane’ and the corresponding spread is said to be a ‘conical spread’. If the spread for x is a normal set of pseudoreguli sharing exactly two lines M and N , we shall call x a ‘ruled translation plane’ or a ‘hyperbolic translation plane’ and the corresponding spread is said to be a ‘hyperbolic’ or ‘ruled’ spread. Theorem 22.15 (1) If T is a conical translation plane then the spread for x may be represented in the form
and for g, f functions on K . (2) f and g are functions on K such that x2t x g ( t )  f ( t )= &(t) is bijective V x in K if and only if the functions define a spread of the form in (1). ( 3 ) If T is a ruled translation plane then the spread for x may be represented in the form
+
x = o , y=o, y = x
[o v],
= X
[
tu
f ( t ) u ]Vt,V,U,ut U
# 0 EK
where g , f are functions on K . (4)If x is a conical translation plane with line L and also a ruled translation plane with linesL and M such that two normal points (ldimensional left Ksubspaces) lie on L then the spread for T may be represented in the form x=O,y=x[ where g and f are constants in K .
t]
Vt,uEK
306
CHAPTER 22. SPREADS COVERED
BY PSEUDOREGULI
Proof If T is a conical translation plane, choose coordinates so that the common component is x = 0, and one of the pseudoreguli has the form
x = 0,y = x
[::]V u E K .
We shall refer to this as the ‘standard form’. It follows fromthe above results that each remaining pseudoregulus has the form z = 0, y = x ( S uI ) for a set {S}of 2 x 2 Kmatrices. Recall that the set of components must be of the general form
+
and for functions G and F from K x K to K. Since the group E exists as a collineation group, the result (1) now directly follows. If T is a ruled translation plane, choose coordinates so that the two common components are x = 0, y = 0 and that the two ldimensional left Ksubspaces referred to in the statement lie on z = 0. Choose one pseudoregulus to have the standard form. Since the plane now admits the group H listed previously, the result (3) now follows immediately. If T is a ruled translation plane of the type listed in statement (4),use the form of (1) and apply the group H to obtain the conclusion that
Hence, letting g(l) = g and f(1) = f, (4) is now clear. Assume the conditions of (2). We have that z = 0,y
=x
[
f(t) U
] Vt,u
EK
and for g, f functions on K defines a spread if and only if, for each nonzero vector (a,b, c, d ) such that not both a and b zero, there exists a unique pair (u,t)such that
a(u + g ( t ) ) + bt = c and a f ( t ) + bu = d.
22.2. CONICALANDRULEDSPREADSOVERSKEWFIELDS.
307
If a is zerothen thereis a unique such pair, namely (bld,blc). If b = 0 then since f(t) is bijective, the unique pair is (fl(ald),ulc  g(fl(ald)). If ab # 0 then
 g(t)  albt z2t + zg(t)  f ( t )
blaf(t)
= bld = ald
 a'c if and only if  albalc for z = alb.
Hence, if & is bijective then, for any given d,c,there is a unique t in K that satisfies the above equation and defining U
= a%
there is a unique pair
(U,
 a%  g ( t ) = bld
+blaf(t),
t ) which satisfies the first system of equations.0
Chapter 23
CONICAL AND RULED PLANES OVER FIELDS We have considered &ne planes which are covered by subplane covered nets and, in the last chapter, we studied translation planes covered by pseudoreguli in two ways. In this chapter,we restrict the skewfield K to be a field and formulate some of the related theory of flocks of quadric sets.
23.1
Conical Flocks.
Definition 23.1 Represent PG(n  1, K ) , where K is a field, b y ldimensional vector subspaces of a ndimensional vector space V n over K where n is finite. W e m a yrepresent points by nonzero vectors (xo,X I ,...,xn1) if we agree to identify vectors in the same ldimensional vector subspace. A (symmetric bilinear form'is a mapping f from Vn X Vn into K such f(
[email protected],U ) = a f ( v ,u)
[email protected](w,U ) and f (v,U ) = f (U, v ) f o r all a,P E K and for all U , V , W E V,. A 'quadratic form' Q is a mapping of V n X Vn into K with the following properties: ( l )Q(av)= a2Q(v)and (2) &(v + W ) = &(v) + &(W) f(v, W ) where f is a symmetric bilinear f o r m f o r all a! E K and for all v,w E Vn.
+
309
310
C H A P T E R 23. CONICAL AND RULED PLANES OVER FIELDS
A ‘quadric’ is the set of points v E PG(n  1,K) such that &(v) = 0 for some quadratic form. A ‘nondegenerate quadric’ is a quadric such that &(v) = 0 and f(v,u) = 0 for all U E Vn implies that v = 0. A ‘conic’ is a quadric in P G ( 2 ,K). If K is finite and isomorphic to GF(q), the number of points of a nondegenerate conic is q + 1. Definition 23.2 Let C be a 3dimensional projective geometry isomorphic to PG(3, K ) where K is a field.Choose a plane ro (so isomorphicto P G ( 2 , K))and a point v0 of C  {ro} and let C be a nondegenerate conic of T o . W e define the ‘quadratic cone’ defined by C and v0 as the set of points which lie o n lines voc for all c E C. W e call v0 the ‘vertex’ of the cone. A fflock’ of a quadratic cone is a set of mutually disjoint conics in planes of C whose union is C  {vo}. For convenience, we represent the points of C by (xo,x1,x2,x3) not all xi = 0 for i = 1,2,3,4 and the points of 7ro by (50, X I ,x2,O). Then, we may choose WO as (O,O, 0 , l ) .
It turns out that there are conical forms for conics in P G ( 2 , K) and quadrics in P G ( 3 , K)but we shall not be overly concerned with such forms. What we are trying to do is to show that conical translation planes when K is a field and flocks of quadratic cones are in 11correspondence. Lemma 23.3 Given PG(2,K),and points (zo, ~ 1 , 3 2 2 )then x051
istheequation
Q :Q(20,21,22)
= 2 22
of thenondegenerateconicwithdefiningquadratic = ~ 0 x1 xi.
form
ProoE The associated bilinear form is
If Q(xo,x1,x2) = 0 then letting (55,x;,x4) equal (1,0,0), (0,1,0) and (0, 0 , l ) forces x0 = $1 = x2 = 0 so that the conic is nondegenerate.0
23.1. CONICAL FLOCKS.
311
Theorem 23.4 Represent the points of PG(3,K ) by homogeneous coordinates ( x o , x ~x2,x3). , A flock of a quadratic cone in PG(3,K ) with conic 20x1 = x i in the plane whose points are given by the equation x3 = 0 and with vertex v0 = (0, O,O,1) has planes represented in the following form:
txo
+ G(t)xl+F ( t ) ~+2 x3 = 0
for all t E K and for functions G,F o n K . Proof: Every plane which does not go thru (0, 0, 0 , l ) has the basic form axo+bxl+cx2+~3=Ofora,b,cEK. If there is a flock and two planes have the same a's then their intersection is of the general form (bb*)x~+(cc*)x~= 0. In addition, their intersection on the quadratic cone whose equation is given by 20x1 = xf contains the point (0, (c  c*)/(b  b"), 1,0) unless b = b*. Similarly, c = c* and the two planes are identical. Hence, the U'S must vary over K . It is equally straightforward to see that the b's and c's are functions of a.0 Wenowshow the equivalence betweenflocks of quadratic cones and conical translation planes.
Theorem 23.5 For every translation plane with spread
spread in PG(3,K ) with
thereisan
associated flock of aquadraticconedefinedas follows: Let denote homogeneous coordinates for PG(3,K ) , and consider the conic C with equation 20x1 = xf in the plane x3 = 0. Let v0 = (O,O,0 , l ) and form the cone voP where P E C. Then the following planes define, b y intersection, a flock of the cone: (50,X I , 2 2 , x3)
7rt
: txo
 f ( t ) X l + g(t)x2 + x3 = 0.
Conversely, a flock of a cone in PG(3,K ) with equation 50x1 = x i and vertex (O,O,0 , l ) has plane equations of the form given above and defines a translation plane withspread in PG(3,K ) with components defined as above.
312
CHAPTER 23. CONICALAND RULED PLANESOVER FIELDS
Proof: We shall first show the converse. Suppose that {rt ; t E K} with the given form provides a disjoint cover of the cone of points # v0 (vertex). This means that for any zo,z1, z2 such that zozl = z# and 6 E K then there is exactly one plane rt which contains the point (zo,z1,z2,6)of the line ((zo,z1,z2,0), (O,O, 0,l)). That is, we have the following: For any point (zo, z1,z2,6) such that zozl = z# where S E K, there is a unique t E K such that tzo  f ( t ) z l +g ( t ) z 2 S = 0. We first note that f is 11and onto, Consider the point (0,1,0,6). Hence, there exists a unique t such that  f ( t ) + S = 0. Now assert that u2 ug(t) t f ( t )# 0 unless U = t = 0. To see this, consider the point (z,",1,z2,0). There is a unique value for t , namely 0, such that tz#  f ( t ) g ( t ) z 2 = 0. Note that if t is not zero, the equation can be transformed into the equation above by multiplying the equation by t and letting U = tz2 to obtain u2  t f ( t )+ g(t)u= 0. Now in order that
+
+
+
z=O,y=x
[
f ( t ) ] Vu,t E K U
is a spread in PG(3, K), we need only to show that for any nonzero vector ( V I , 212, w1, w2) there is a unique pair (U, t ) such that thevector is on the indicated component (note that the sets above are 2dimensional Ksubspaces). That is, we must show that when 211, 212, w1,w2 are fixed and not all zero then the following system has a unique solution for U and t : W1
=
W2
=
+ g @ ) ) + 2126 V l f ( t ) + 212% Ul(U
+
or 211 = 212 = 0. Note that if 211 = 212 = 0 then since u2 ug(t)  t f ( t )# 0 for (U, t ) # (O,O), it follows that z = 0 contains the indicated vector and is the only component which does. Assume that 01 = 0 but 212 # 0. Then t = W ~ / V Zand U = w2/212. If 212 = 0 but 211 # 0 then w2/211 = f ( t ) and since f is 11 and onto, this uniquely specifies t. Then U = w 1 / q  u l g ( t ) . Hence, assume that 211212 # 0. Then W1212
 W2211 = v1212U + vl'U29(t)+ 'U&  ('U:f(t)+ V l V z U )
23.2. HYPERBOLIC FLOCKS.
313
which is equal to
+
v1v2g(t) v;t Now call

( ~ 1 ~ w2v1)/v; 2 =
 v: f (t).
S and v1/v2 = W to obtain the equation
t  f ( t ) W 2 + g ( t ) w + S = 0. By the above, for given satisfies this equation. Note that (W1
W
and 6 in K, there is a unique t E K which
 W ( t ) v2t)/v1 = (W2  ' U l f ( t ) ) / ' u 2
then uniquely defines U . Hence, we obtain a spread from a flock. To show that a spread of the form listed above produces a flock, it is possible to essentially reread the proof to show that a spread produces a cover of the associated cone and such details are left to theinterested reader to verify.0
Remark 23.6 GeneralizedQuadrangles. There have been a tremendous number of developments which interrelate the two types of geometric incidence structures of the previous theorem. f i r thermore, there are also connections to 'generalized quadrangles' and other combinatorialgeometries.Howeverthisdirectiontakes us away from the ideas of subplanes of nets and the subsequent immediate consequences. The interested reader is referred to the survey article %locks of Laguerre Planes and Associated Geometries ' b y the author and S.E. Payne [52] f o r examples of planes, flocks and generalized quadrangles and more complete information and theow both in the finite and infinite cases. Furthermore,thisconnectionwith generalized quadrangles,flocksand spreads was first noticed by J * A . Thas in [78] and the reader is also referred to the article b y Thas.
23.2
Hyperbolic Flocks.
Definition 23.7 Consider C isomorphic to PG(3, K ) and represent points in the form (20,21,22,23) f o r zi E K ,i.= 0 , 1 , 2 , 3 .
314
C H A P T E R 23. CONICAL AND RULED PLANES OVER FIELDS
Then 20x3 = 21x2 represents a nondeneratequadric 31 in P G ( 3 ,K ) which we call a ‘hyperbolic quadric’Q : Q ( x ~ , x ~ , x ~=,~x0~x)3 21x2. A ‘flock of a hyperbolic quadric’ is a set of mutually disjoint nondegenerate conics in planes of C whose union is 31. W e are interested in the ‘ruling’ lines of the hyperbolic quadric. These are the lines that lie within 31 , There are two designated classes and it is not dificult to show that the classes are reguli which are mutually opposite. Thus, a hyperbolic or ‘ruled’ quadric is the set of points of a regulus and the lines of the quadric are of a regulus and its opposite regulus. In the finite case when K is isomorphic to GF(q),31 has (g+ points and there are q 1 conics in a flock as each regulus has q + 1 lines. Also, a finite hyperbolic quadric has 2(q + 1) lines contained in it.
+
We nowshow that there are connections with hyperbolic translation planes and hyperbolic flocks in a manner similar to thatof conical translation planes and conical flocks. Theorem 23.8 (1) Let F be a flock of the hyperbolic quadric ~ 1 x = 4 ~2x3 in P G ( 3 , K ) whosepoints are represented by homogeneouscoordinates ( 2 1 , x 2 , 2 3 , 2 4 ) where K is a field. Then the set of planes which contain the conics in F may be represented as follows: p : x2 = x 3 , 7rt
: x1  tx2
+ f (t)x3 g(t)24 = 0
for all t in K where f and g are functions of K such that f is bijective. (2) Corresponding to the flock F is a translation plane TF with spread in P G ( 3 , K ) written over the corresponding 4dimensional vector space over K as follows: Let V4 = (x,y) where x and y are 2vectors over K . Then the spread may be represented as follows:
23.2. HYPERBOLIC FLOCKS,
315
Define
Rt
R,
= { y = x [ f (Ut ) u
tu
],
X
= 0 ;U E K }
,
= {y=x[
Then {Rt,L }is a set of reguli that share two lines (components x = 0,y = 0).
The translation plane admits the collineation group
whichcontainstwo afine homologygroupswhosecomponentorbits union the m ' s and coaxis define the reguli (regulw nets). (3) A translation plane with spread in PG(3,K ) which is the union of reguli that share two components may be represented in the form (2). Equivalently, a translation plane with spread in PG(3,K ) which admits a homology group one of whose component orbits union the axis and coaxis is a regulus may be represented in the form (2). In either case, such a translation plane produces a flock of a hyperbolic quadric in PG(3,K). Proof: Suppose that a translation plane with spread in PG(3,K ) admits an &ne homology group one of whose component orbits union the axis and coaxis is a regulus R in PG(3,K ) . Choose a representation so that the axis is y = 0, the coaxis x = 0 and y = x is a component (line) of the regulus R. Then R is represented by thepartial spread x = 0,y = x W E
[
K ,Moreover, the homology group takes the matrix form:
([
! o o o u
;urK{O}).
3
forall
316
CHAPTER 23. CONICAL AND RULED PLANES OVER FIELDS
There are functions f and g on K and components of the following matrix form:
Note that, in particular, this says that the function f is 1 1 as otherwise differences of certain corresponding matrices are singular and nonzero contrary to the assumption that .the components form a unique cover of the vector space. The homology group maps these components into
for all nonzero U in K. Hence, the regulus R and these components for all W , t , U # 0 in K define the spread in PG(3,K). Take any value a in K and consider the vector (1,a, 0,l). Since this vector is not on x = 0 or y = xw1 and we are assuming a 'cover', there is a unique pair (u,t) with U nonzero such that (1, a, 0 , l ) is incident with the component a, = x
1 ft)u 1. U
Hence, we have f ( t ) u au = 0 and
g(t)u atu = 1. In pkrticular, sinceL is nonzero, we must have f ( t ) = a. Hence, f is 'onto'. In order to see that the planes listed in the theorem intersected with the hyperbolic quadric in PG(3, K) form a unique cover of the hyperbolic quadric and hence define a hyperbolic flock, we must show that for all points (a,b, c, d ) for b # c and ad = bc, there is a unique t in K such that the point is on the plane nt. Since we have a cover of the 4dimensional vector space, we know that for a vector (e,h, m, n) where not both e and h are zero and ((m,n)) is not in ( ( e ,h ) ) ,there is a unique ordered pair ( t ,u ) such that
(e,h, m, n) is on the component y = x
tu To distinguish between points of P b ( 3 , K ) t.hat rglate to the flock and vectors of V4 which relate to the translation plane, we shall use the terms 'points' and 'vectors' respectively. That is, for all e, h, m, n such that not both e and h are zero and the vector (m,n) is not in the lspace generated by (e, h), there is a unique ordered pair ( t , u ) such that
+
+
ef(t)u hu = m and eg(t)u htu = n.
23.2. HYPERBOLIC FLOCKS.
317
+
The point (a,b, c, d ) is on 7rt if and only if a  bt f(t)c  g ( t ) d = 0. First assume that bc # 0. Then, without loss of generality, we may take b = 1 so that ad = c (recall that the point is considered homogeneously). Then the above equation becomes
ad
 dt + f(t)cd  g ( t ) d 2 = 0 = C  d t + f(t)cd  g ( t ) d 2 .
(*l
We may multiply (*) by an element r in K and subtract to obtain (m  nr
+ hu)  t(Thu)+ f(t)(eu)  g ( t ) ( r e u ) = 0.
Hence, clearly we may solve the equation
+ f(t)cd  g ( t ) d 2 = 0 for some t = to E K. First assume that f(t,)d + 1 = 0 = z . Then g(t,)d + to = 0 = W c dt
the vector ( l , d  l , O , 0) is on the component y = z
[ f(t'J)d d
g(to)d t,d
]
and and
y = 0, which is a contradiction. Hence, zw # 0. So, W = zed" and the vector ( l , d  l , ~ , ~ c d  l is ) on the component
Now assume that there exists another element c  ds,
S,
such that
+ f(s,)cd  g(so)d2 = 0.
Then f ( s , ) d + 1 = z* # 0 and there exists an element v in K such that = 2. So, the vector ( 1 , d  l ,z , zcdl) = (1, d  l , z*v, z*vcd') is also on
2 ' 2 1
By uniqueness of the vector space cover, it follows that (to,d ) = (s,,dv) (as components). Hence, there is a unique plane 7rt containing the point (a,b, c, d ) such that b # c and ad = bc where bc is nonzero. Now assume that bc = 0. If b = 0 and d = 0 then, without loss of generality, we may take c = 1 so we are considering the point ( a ,0,1,0). We
318
CHAPTER 23. CONICAL AND RULED PLANES OVER FIELDS
+
need to determine a t in K such that a f ( t ) = 0. Since f is 1 1 and onto as noted above, there exists a unique value t which solves this equation and hence a unique plane 7rt containing the point (a,0,1,0). If b = 0 and a = 0 and c = 1, it is required to uniquely cover the point (0, 0,1, d ) by a plane so we require a unique solution to the equation
f ( t ) g ( t ) d = 0. By (*), certainly there is a solution tl. Moreover, f ( t l )= z1 # 0 as otherwise the spread would contain y = x ony=z
[
f(ltl)
1,
[ t", ] .
Hence, the vector (1,O,z1,z l / d ) is
If there exists anothersolution
s1 then
f(s1) = zi
and there existsan element W of K such that zrw = z1. Hence, the previous vector also belongs to y = IC
[
W
S1W
]
which, by uniqueness of
the vector space cover, implies that (tl,1) = (s1,w) as components. If c = 0 then a = 0 or d = 0 and b = 1 without loss of generality. We are trying to show that there is a unique solution to a  t g ( t ) d = 0. If d = 0 this is trivial. Thus, assume that a = 0. By (*), there is a solution t 2 to t+g(t)d = 0. Let f ( t 2 ) d  l = z2 . Then
+
the vector (1, dl,
z2,O)
is on the component y = z
d clearly z2 # 0. If there is another solution 92 then let f ( s 2 ) d  1 = zz so ;hat there exists an element W such that z,*w = z2. Hence, the previous point is also on the component y = x
1
1
f(s2)dw so by uniqueness, dw S Z ~ W we must have ( t 2 , d ) = ( s 2 , d w ) as components. Thus, a translation plane with spread in PG(3, K )which admits an affine homologygroup of the type listed above produces a flock of a hyperbolic quadric. To complete the proof of part (3), we must show that if a translation plane has its spread in PG(3, K)and the spread is a union of reguli sharing two components, then there is a homology group of the type mentioned above. However, this follows from the relevant theorem in the previous chapter as the reguli are normalizing. To prove (l),we may choose a basis so that a given plane of the flock has equation x2 = 23. From here, it is fairly direct that we may represent L
J
the flock in the form given. The function f ( t ) is 1 1 to avoid intersections and must be onto in order to ensure a cover. The proof of (2) follows along the lines of the proof (3) and is left to the reader.0
Remark 23.9 Minkowski Planes. There are some nice connections with other geometries and translation generplanes admitting spreads covered by derivable partial spreads which alize the notions in this section in the sense that the spreads need not lie within athreedimensionalprojective space. We haveseenthatthere are connections with the classical hyperbolic quadric and spreads covered by reguli in a particular manner. The underlying point/section geometry is thatof a Minkowski plane which is classical in the hyperbolic case. The interested reader is referred to Knarr [56] and Johnson 1451 and [46]. Furthermore, there are various fairly complete theories in the finite case. I n particular,the beautiful theorem of Thas,BaderLunardonprovides a complete determination of finite hyperbolic flocks (see Thas [74], Bader and Lunardon [3’]). The reader interested in more details on hyperbolic flocks and hyperbolic translation planes is referred to the articles by the author 1421 and [48] and the references therein.
Chapter 24
SPREADS WHICH ARE DUAL SPREADS In the previous chapter, we have seen that there exist derivable nets in nonderivable affine planes. The afFine planes in question are dual translation planes. Could such a plane be a translation plane? Note that when K is a field in the above context then a derivable net corresponds to a regulus in some projective space isomorphic to PG(3,K ) . So, we have a nonderivable extension plane of a regulus net. However, the plane does not necessarily correspond to a spread of PG(3,K). Thus, we arrive at the following question: Given a regulus R in PG(3,K),for K a field, assume that there isaspreadin PG(3,K) whichcontains R. Is the corresponding translation plane derivable? Another related question is: If aspreadin PG(3,K),for K a field, is a dualspreadand contains a regulus and if the translation plane is derivable, is the derived spread also a dual spread? In Bruen and Fisher [17], it isshown that any subregular spread in PG(3,K ) , for K a field, is a dual spread where is a ‘subregular spread’ is one which is obtained by a sequence of multiply deriving translation planes originating from a Pappian plane. More precisely,the proof taken in the context of the results on Baer subplanes and duals spreads allows the following extension. 321
322
C H A P T E R 24. SPREADS WHICH
ARE DUAL SPREADS
Theorem 24.1 Let S be any spread in PG(3, K) which contains a regulus net R. Then the tmnslation plane is derivable with derivable net NR and the spread obtained by derivation of the regulus net is a dual spread if and only if the original spread is a dual spread. Proof: First of all, we see that any regulus net defines a derivable net so that any 2dimensional Ksubspace which is a subplane of the net automatically becomes a pointBaer subplane of the net and hence also of the plane since the points of the net and the points of the plane are the same. But, we have also noted that such subplanes are also lineBaer. Hence, any such affine plane incident with the zero vector becomesa Baer subplane. Clearly, any translate of a Baer subplane is also Baer. Hence, the translation plane is derivable. Now derive the plane to 7r* using the regulus net N R . We need to show that every 3dimensional Ksubspace C contains a line of the new spread. We give a slightly modified version of the proof of [l71 (Theorem 2 , p. 802). Let 7ro be a subplane of the net NR which is in C and let 7r1 and 7r2 be two other distinct subplanes of the net which are incident with the zero vector. Then C m i = (Pi)# 0 for i = 1,2. Thus, (P1,Pz) is a 2space of C (line of the plane C) and this 2space nontrivially intersects 7ro as both are 2spaces within a 3space. Hence, the space ( P I P2) , is a line of the opposite regulus to R. Since this is a component by assumption, then C contains a line of the new spread.0 We state the equivalency as a corollary.
Corollary 24.2 Let S be a spread in PG(3,K), for K a field, and let R be a regulus of S . Then ana, plane of PG(3, K) which contains a line of the opposite regulus to R contains a line of the spread. We may use the above results to show that any conical or hyperbolic spread which corresponds to a flock of a quadratic cone or to a flock of a hyperbolic quadric in PG(3, K) must be a dual spread. Our proof depends on the structure of the spread based on functions acting on K. Although our proof seems to depend on the assumption that K is a field, this is probably not necessary. As we mentioned in the last chapter, we shall provide
CHAPTER 24. SPREADS WHICH ARE DUAL SPREADS
323
examples of spreads which are unions of pseudoreguli in PG(3, K ) when K is a noncommutative skewfield. However, it is an open question when such spreads are dual spreads.
Theorem 24.3 Let S be a spread in P G ( 3 , K ) , for K a field. Then S is a dual spread in either of the following two situations: (1) S is a union of reguli that share exactly one line. ( 2 ) S is a union of reguli that mutually share two lines Proof In either of the two situations, let C be any 3dimensional Ksubspace. It is asserted that either C contains a line of the standard regulus net x = 0, y = x
[
]
for all U E K or coordinates may be chosen so that
Cn(x = 0) = ((O,O, 0,l)) and Cn(y = 0) = ((1,0,0,O)), To see this, we note from above that if C contains a Baer subplaneincident with the zero vector of the standard regulus net N then C contains a line of N . So, assume that C does not contain aline of N . Then C intersects x = 0 and y = 0 in distinct Baer subplanes of N . We may choose coordinates so that N retains the form and the two Baer subplanes are 7ro,1 and 7r1,o in the standard notation for regulus nets. Now since N retains its form.andsince we either have an elation group or homology group whose line or component orbits union the axis or axis and coaxis respectively determine the base regulus nets, this means that the forms of the elation group and/or homology group have not changed. Furthermore, although we allow that perhaps the functions describing the spread have changed, the basic properties listed in the theorems 23.5 and 23.8 above are still valid for the possibly new functions. Let C n (g = x) = ((1,a,,1,a,)),We note that a, # 0 since otherwise, C would contain the line x = 0. It is noted that the choice of the lspace is made possible by the fact that C does not contain the line y = 0 by assumption. Hence, we may assume that C = ((0, 0, 0, l),(1,0,0,0), (1,a,,1,ao)). Now assume the conditions of (1). It suffices to show that there is a Baer subplane of one of the regulus nets which is also in C. Since C does not contain2 = 0 but intersects C at ((0, 0, 0, l)),it follows that theonly way that C could contain one of the Baer subplanes in question is if it contains
324
CHAPTER 24. SPREADS WHICH ARE DUAL SPREADS
an image of the Baer subplane ((0, z2, 0,yz)) = T O Junder a mapping
L o o
o
1 J
The reader is directed back to the statement of Theorem 23.5 for the significance and properties of the functions g and f . Thus, the question now becomes whether C contains a Baer subplane of the form ((O,z2,zzt,yz)) = Rt.
+
Notice that (y2  m ) ( O , O,O,1)  (z2/ao)(l,O,O,0) (z2/ao)(l,ao,1,a,) are vectors in both C and ~ 1 / Furthermore, ~ ~ . we see that, for each t in K, the regulus nets St are given by the partial spreads:
Hence, C contains a Baer subplane of one of the regulus nets, namely S1,/ and so contains a line of one of the reguli and hence contains a line of the spread. Now assume the conditions of (2). In this case, we show that
((O,O,O,11, (1,0,070)) is actually a Baer subplane of one of the regulus nets that share both z = 0 and y = 0. We consider the image of 7ro = ((1,0,0,0), (O,O,1,O)) under mappings
Again, we direct the reader to Theorem 23.8 for the significance and properties of the functions f and g but it is pointed out, in particular, that f is bijective on K. The subplane 7ro maps under ht to (( 1,0,0,0), (0, 0, f(t),g ( t ) ) ) . Since f is bijective, there exists an element to such that f (to)= 0.
C H A P T E R 24. SPREADS WHICH A RD E U ASLP R E A D S Furthermore, since
[
(to)
=
1
g(to)
t0
325
] is nonsingular, it follows that
g (to)# 0. Note that, for each t in K , the regulus nets & other than N are
So, C contains a Baer subplane of &, and hence a line of Rt, and thus contains a line of the spread. This completes the proof of the theorem.0 Corollary 24.4 Any spread in P G ( 3 ,K ) that corresponds to a flock of a quadratic cone in P G ( 3 , K ) or to a flock of an elliptic or hyperbolic quadric in P G ( 3 ,K ) is a dual spread. Corollary 24.5 Let S denote any spread corresponding to a quadric set in P G ( 3 ,K ) where K is a field. Let N be any regulus in the spread and let N* denote the opposite regulus. Then ( S  N ) U N* is a spread which is a dual spread. Remark 24.6 A s noted in Bruen and Fisher ([l71 Corollary to Theorem 3)) it is proved that any subregular spread is a dual spread. There are essentially three types of quadric sets in P G ( 3 , K),for K a field;conical,hyperbolicandelliptic. A ffEock' of such aquadricset is a covering of all but at most two points b y plane intersections. I n Biliotti and Johnson [l01,it is pointed out that any flock of an elliptic quadric in P G ( 3 ,K ) corresponds to a spread which may be obtained from a Pappian spread b y multiple derivation. Hence, any such spread is a dual spread. W e have avoided use of the Klein quadric in P G ( 5 ,K ) when K is a field. It is possible to produce proofs of the results o n dual spreads of conical and hyperbolic spreads over fields by the w e of the Klein quadric, However, as we mentioned previously, it is more in keeping with our general analysis to present our proofs independent of this theory. The interested reader might also like to look at the corresponding theory of 'topological flocks' in the work of Riesinger [68].
326
C H A P T E R 24. SPREADS WHICH
ARE DUAL SPREADS
Corollary 24.7 A n y semifield spread S in P G ( 3 ,K )that contains a regulus N sharing an axis L of a nontrivial elation group of S is a dual spread. Furthermore, ( S  N ) U N * is a spread which is a dual spread. Any nearfield spread S in P G ( 3 ,K ) that contains a regulus sharing an axisand coaxis of a nontrivialhomology group for the spread is adual spread. The derived structures are spreads which are dual spreads, Proof: Apply the groups to obtain the spread as a union of reguli that share L or as a union of reguli that share L and M . Now apply the above theorem.0
Theorem 24.8 Let S be a spread in P G ( 3 ,K ) which is a union of reguli sharing one line or a union of reguli sharing two lines. Let X , be any 1dimensional K subspace and let L and M be components of the corresponding translation plane such that L contains Yo. Let (m) denote the parallel class containing M and form (00)afor all a in Yo. Then { ( m ) afor all a in Yo} is a derivation set for the associated dual translation plane which is then derivable. Proof Previously, the net defined by the set in question was shownto be a derivable net. Since the spread is now also a dual spread, this forces the dual translation plane containing the net to be derivable as the subplanes in question now must be Baer in the translation plane and so are Baer in the dual translation p1ane.O
24.1
ChainsandSemitranslation ited.
Planes Revis
We have considered the possibility of considering ‘chains’ of planes starting with a translation plane then asking if a suitable affine restriction of the projective extension is derivable. We have seen that there is a wide variety of infinite translation planes whose duals are derivable. It is still, however, not completely clear that a chain can be built on any translation plane with spread in P G ( 3 ,K),for K a skewfield, whichis also a dual spread,although at least one can derive the dual translation plane. It also follows that the
24.1. CHAINS AND SEMITRANSLATION PLANES REVISITED.
327
derived dual translation plane is a semitranslation plane. But, as little is known of the nature of the full collineation group of the semitranslation plane, the possibility of chains and the corresponding isomorphism classes of the planes of the chain is an open problem. On the other hand, Barlotti and Bose [S] have extended the results on finite chains mentioned based on translation planes with spreads in PG(3,K) that contain reguli to the infinite case. Here, the chain is periodic of period 8 exactly as in the finite case.
Chapter 25
PARTIAL FLOCKS OF DEFICIENCY ONE In this chapter, we consider structures which are essentially flocks‘missing’ one conic; partial flocks of deficiency one. We have considered partial parallelisms of deficiency one previously. The general idea for the study of deficiency one comes about when deriving an affine plane covered by derivable nets. The derived plane, by the derivation of one of these nets, now is not covered by derivable nets but a good share of its parallel classes still are connected to derivable nets although it is apparent that the remaining lines of a given derivable net are now Baer subplanes of the derived plane. If one forgets that the derived net is, in fact, derivable, it is natural to consider such planes abstractly and ask what geometric pointline geometries are associated and what is implied of such structures when the corresponding net does turn out to be derivable. We first consider the import of the theory of Baer groups for spreads in PG(3,K ) .
25.1
PointBaer Subplanes in Spreads of PG(3,K ) .
Theorem 25.1 Let be a translation plane with spread in PG(3,K),for K a skewfield. Note that the following are considered the kernel mappings:
329
330
CHAPTER 25. PARTIAL FLOCKS OF DEFICIENCY
ONE
whereas linear collineations may be represented b y 4 x 4 matrices acting on the right and components have the general f o r m x = 0,y = X M where M is a 2 x 2 matrix acting on the right. (1) Let r admit a nontrivial pointBaer elation group B which f i e s the pointBaer subplane ro pointwise. Let the kernel of ro be KO. If B is a full K,pointBaer elation group of order > 2 then ro is a Kspace, K is a field isomorphic to KO and B may be represented in the form
([
i i]
VJEK).
0 0 0 1
( 2 ) If r admits a nontrivial pointBaer homology group C then let FixC = ro have kernel K,. If C acts as the full K,pointBaer homology group of the net containing FixC and has order > 2 then K is a field isomorphic to K , and C may be represented in the form
L
J .
Note, in this case, there is another pointBaer subplane sharing its infinite pointswith FixC and invariantunder C. This second pointBaer subplane is called coFix C. (3)If L is any 2dimensionalleft Ksubspace whichis disjoint from Fix B then the orbit of L under B union Fix B is a regulus net in PG(3,K ) . (4)If L is any 2dimensionalleft Ksubspace whichis disjoint from Fix C or coFix C then theorbit of L under C union Fix C and coFix C is a regulus net in PG(3,K ) . Proof: (1) and ( 2 ) follow from the chapter on Baer extensions. Now assume the conditions of (3). Change coordinates so that FixB is represented by x = 0. Clearly, B
25.2. DEFICIENCYONE PARTIAL CONICAL FLOCKS.
331
now has the following form
([B
o u o
0
VUEK),
0 1 0 0 0 1
The set of B images of y = o is y = z
:]
[
v u E K. w e may assume
that a 2dimensional subspace disjoint from z = 0 has the general form g = zN where N is a 2 x 2 matrix (possibly singular). By a coordinate change of the form
[6
]: ,
we may take the 2dimensional subspace as
y = 0 without changing the new form of the group B. This then proves (3). Assume the conditions of (4). Change bases over the prime field so that the fixed subplanes have the form x = 0 and y = 0. The group now has the form
1 1 U 0 0 0 o u o o
0 0 1 0 0 0 0 1
v),.
Any 2dimensional Ksubspace disjoint from z = 0 and y = 0 has the form y = zN where N is a nonsingular 2 x 2 matrix. Change bases by
1.
[
This basis change leaves the new form of thegroup invariant
and the images of the 2dimensional subspace are y = x
[::]V U E K * .
Hence, we have the proof to (4).0
25.2
DeficiencyOnePartialConicalFlocks.
Definition 25.2 Let C, be a nondegenerate conic in a plane of PG(3, K), for K a field. Let U, be a point exterior to C, and form the quadratic cone.
A ‘partial flock of the quadratic cone’ is a set of mutually disjoint conics which lie in the CO {U,}.
332
CHAPTER 25. PARTIAL FLOCKS OF DEFICIENCY ONE
A ‘partial flock of deficiency one’ is a partial flock such that for each line of the cone, the union of the conics cover all but exactly one of the nonvertex points. Remark 25.3 W e have seen previously how to connect aflockof a quadratic cone with conical spread in PG(3,K ) . Each plane of the flock corresponds to a conic of the cone which, in turn, corresponds to a regulus net of the conical spread. Two conics of the flock correspond to two regulus nets but the points of the conics do not correspond to the partial spread components which share a line but to the set of Baer subplanes incident with the zero vector of the regulus nets. Under this sort of correspondence of points on the cone with Baer subplanes of the regulus nets, the cone points on a line correspond to the set of Baer subplanes which share a given ldimensional Ksubspace on the common line. When we have a partial flock, we have the same sort of connections with cone points and Baer subplanes. Note, if the planes of a partial flock cover all but one of the cone points on a line then there would be a set of regulus nets containing a set of Baer subplanes that intersecta given ldimensional Kspace on the common line and, asa set of 2dimensional Ksubspaces would be one short of containing all such 2dimensional Ksubspaces with a given mutual lspace intersection. Theorem 25.4 (1) The setof partialflocks of a quadratic cone of deficiency one in PG(3,K ) , for K a field, is equivalent to the set of translation planes with spreads in PG(3,K ) that admit a pointBaer elation group B which acts transitively on nonfixed ldimensional Ksubspaces of components of the &ed point subplane. ( 2 ) A partial flock of a quadratic cone of deficiency one in PG(3,K ) , f o r K a field, may be extended to a flock if and only if, in the corresponding translation plane which admits a pointBaer elation group B as in ( l ) , the net defined b y the pointBaer afine plane FixB defines aregulus in PG(3,K). Proof: Assume that such a translation plane T exists. We point out that when we use the term Baer subplane of a regulus net we are not asserting that the subplane indicated is a Baer subplane of any other net containing the regulus net as we have seen that this is not always the case.
25.2. DEFICIENCYONE PARTIAL CONICAL
FLOCKS.
333
Since K is a field, the pointBaer subplane isalways a 2dimensional Ksubspace. Hence, the group has the following representation:
H4
([
VPEK).
0 0 0 1
The components of the net N containing r0 have the basic form x=o,*=x
[:
VuEK
and b a function on K . Since the kernel may now be given unambiguously by the mappings
rP o o 01
o o o p then the components of the spread which are not in the net N have the following general form:
where G and F are functions of K x K. Change coordinates by (21, x2,g1, y2) has the following form:
+
(XI,
y1,x2,92) so that the group
The components which are not in the net have the form: *=x[
G(t, u)t' tl
F ( t ,U )  G(t,2L)th tlu
] '#t#O,uEK.
CHAPTER 25. PARTIAL FLOCKS
334
OF DEFICIENCY ONE
Hence, the components not in N have the general form
for functions G* and F* on K x K . Since the plane admits the group listed as above, it follows that G*(t,u)= g ( t ) + U and G*(t,u)= f ( t ) for functions g, f on K. It is now clear that there is a set of regulus nets Rt with partial spread x = 0 (which is the new equation for T,) and
[
y =x
U
] Vu
E
K and t fixed in K  (0).
We are not claiming that the matrices involved are nonsingular. However, we may select one of the matrices and change bases without changing x = 0 so that this matrix is zero. The resulting partial spread has the form: Rt with partial spread x = 0 (which is the new equation for T,) and &/=X
[
t
U
]
Vu E K and
t fixed in K  {to}, for to # 0
where the matrices and differences are either zero or nonsingular. Now given any ldimensional Ksubspace X.If X is in the original net N then the Borbit of X generates a 2dimensional Ksubspace which is a component of the net and which intersects T, in a ldimensional subspace. If X is not in the original net N then the Borbit of X generates a 2dimensional Ksubspace which again intersects T, in a ldimensional K subspace. Take the set of 2dimensional Ksubspaces each of which is invariant under B and each of which intersects r0 in the sameldimensional K subspace Po. Then, there is exactly one which defines a component of the net N . It then follows directly that any other such 2dimensional subspace is a Baer subplane of one of the regulus nets Rt. We consider the following partial flock of a quadratic cone in PG(3,K ) defined by equation 20x1 = xi in the plane x3 = 0 given by homogeneous ~2~ 23) with vertex (0, 0, 0 , l ) : coordinates (xo,q,
25.2. DEFICIENCYONE PARTIAL CONICAL FLOCKS.
335
The planes containing the conics of intersection are: pt : zot  Z l f ( t ) zzg(t) 23 = 0. It follows easily that { p t V t # 0) defines a partial flock of the quadratic cone. Recall that the points on the linesof the cone correspond to Baer subplanes incident with the zero vector of the regulus nets and each of these is a 2dimensional Ksubspace whichis invariant under B. Furthermore, for each ldimensional K subspace on r0,there is exactly one component containing this ldimensional Ksubspace. Thus,it follows that thereis exactly one point on each line of the cone which is not covered by the partial flock. This shows that the deficiency is one. Now assume that there is a partial flock F of a quadratic cone of deficiency one in PG(3,K). Again, note that there is a corresponding partial spread PF in PG(3,K ) which is the union of a setof reguli that share acommon component and admit a groupB. The points on a line of the cone which are covered by the partial flock correspond to Baer subplanes of the regulus nets of the translation net corresponding to the partial spread. Hence, for each line of the cone, there is exactly one point which then corresponds to a 2dimensional Ksubspace that is Binvariant in the corresponding vector space (which is the ambient space of the translation plane) and which is not a Baer subplane of one of the regulus nets. The points on the line of the cone correspond to the set of all Binvariant 2dimensional Kspaces which intersect the common component L in a fixed ldimensional Ksubspace. Hence, for each ldimensional Ksubspace 2 of L, there is a unique 2dimensional Ksubspace 7rz containing 2 which is Binvariant which does not lie as a Baer subplane in one of the regulus nets of the partial spread.So, there is a set R = { r z ; 2 is a ldimensional Ksubspace of L} of Borbits which then must consist of mutually disjoint 2dimensional Ksubspaces. It also clearly follows that these subspaces are disjoint from the partial spread excluding the common component L. Clearly, RU (PFL) covers the set of all ldimensional Ksubspaces and is a set of 2dimensional Ksubspaces. Hence, we have a translation plane with spread in PG(3,K) and which admits B as a collineation group that fixes L pointwise. Since L is a 2dimensional Ksubspace, L becomes an affine plane of the new translation plane C. We note that L defines a lineBaer subplane of C which admits a collineation group B fixing it pointwise. We have also noted that this implies that the subplane is Baer and hence
+
+
336
C H A P T E R 25. PARTIAL FLOCKS
OF DEFICIENCY ONE
pointBaer. This completes the proof of part (1)of the theorem. If the partial flock can be extended to a flock then there is a translation plane T+ corresponding to theflock which admits B as an elation group. By (l),there is a translation plane T corresponding to the partial flock which admits B as a pointBaer group and which shares all components with 7r+ which do not lie in the net N defined by the components of FixB. Let N + denote the subnet of 7r+ which replaces N . We notice that the components of N are generated by Borbits. Moreover, it follows that N + is a regulus net so that Borbits of ldimensional Ksubspaces within N f generate the Baer subplanes of N+ which are the components of N . Hence, N is the opposite regulus net of Nf. Now assume that the plane T is derivable with net N . Then the net contains at least three pointBaer subplanes so it follows from the preceding section that each of the pointBaer subplanes is a Ksubspace. This means the net N is a regulus net. We have seen that the derivation of this net determines a translation plane with spread in PG(3,K) which admits an elation group B such that any component orbit union the axis of the group forms a regulus net. It follows that thederived translation plane corresponds to a flock of a quadratic cone which extends the original partial flock. This completes the proof of the theorem.0
25.3
DeficiencyOnePartial Hyperbolic Flocks.
Definition 25.5 Let H be a hype’rbolic or ruled quadric of the form 2124 = in PG(3,K ) represented b y homogeneous coordinates (XI,x2,23,24). A partial flock of H is a set of mutually disjoint conics which lie in H . A ‘partial flock of deficiency one’ is a partial flock such that on any line of either ruling, there is exactly one point which is not covered b y the conics of the partial flock.
52x3
Remark 25.6 W e havementionedthatgiven ahyperbolicquadric in PG(3,K), there are two ‘ruling’classes of lineswhich are reguli. When we considered flocks of hyperbolic quadrics, we found that there are associated hyperbolic spreads in PG(3,K). The conics are associated with regulus nets and the points on the conics are associated with the Baer subplanes incident with the zem vectorof the regulus nets. Hence, we have connected the
25.3. DEFICIENCYONE PARTIAL HYPERBOLIC
FLOCKS.
337
points of a hyperbolic quadric with the Baer subplanes incident with the zero vector of a set of regulus nets sharing two common components. Each lineof a given ruling class (one of the reguli of the hyperbolic quadric) corresponds to the set of Baer subplanes that are incident witha given ldimensional K subspace on one of the common lines of the regulus nets. A line of the other ruling class (the other of the reguli of the hyperbolic quadric) corresponds a given ldimensional to the set of Baer subplanes that are incident with Ksubspace on the other of the common lines of the regulus nets. Now these same connections hold for partial flocks. In particular, if the planes of the partial flock cover all but one of the points on a line of one of the ruling classes, there is a corresponding set of Baer subplanes which as 2dimensional Ksubspaces share a common lspace such that as a subset of the set of all 2dimensional Ksubspaces sharing that common lspace, there is exactly one missing 2dimensional Kspace.
Theorem 25.7 (1) The set of partial flocks of ruled quadrics in PG(3,K ) , f o r K a field, of deficiency one is equivalent to the set of translation planes with spreads in PG(3,K ) which admit a pointBaer homology groupB which is transitive on the nonfied ldimensional Ksubspaces on any component of Fix B. ( 2 ) A partial flock of deficiency one of a ruled quadric m a y be extended to a flock if and only if the netdefined by Fix B of the corresponding translation plane is a regulus net. Proof A partial flock of a ruled quadric in PG(3,K ) givesrise to a partial spread in PG(3,K ) which is the union of a set of reguli that share two lines. Furthermore, a flock corresponds to a translation plane whose spread has the same property. In addition, the translation plane admits an affine homology group B each of whose components orbits union the axis and coaxis is a regulus net. This isalso true of any such corresponding partial spread. In general, the Baer subplanes of the regulus nets correspond to points on the ruling lines. Again, we are not necessarily assuming that the subplanes are Baer in any net other than theregulus net in question. Furthermore, the set of points on a given ruling line correspond to the set of Baer subplanes that share a given ldimensional Kspace on a given common component of the partial spread. The corresponding Baer subplanes are 2dimensional
338
CHAPTER 25. PARTIAL FLOCKS
OF DEFICIENCY ONE
Ksubspaces generated by point orbits under B. Let the two sets of ruling lines be denoted by R+ and R. Let J+ be a line of R+ The points on J+ correspond to the 2dimensional Ksubspaces which are generated by Borbits of ldimensional Ksubspaces which are not on the common components L and M and which intersect on one of the common components, say L, in a particular ldimensional Ksubspace. First, let 7r be a translation plane with spread in PG(3,K) that admits a pointBaer homology group so there are at least two pointBaer subplanes incident with the zero vector which share the same infinite points. By the theory of the previous chapters, since K is a field, each pointBaer subplane is a Ksubspace and the group B may be represented in the form
.
x 0 0 0
([; ;;;]W ’ ) . 0 0 0 1
It follows that each component orbit disjoint from the two subplanes 7ro and T I under B union 7ro and 7r1 defines a Kregulus in PG(3,K). Let N denote the net containing the two pointBaer subplanes. It then follows that corresponding to the partial spread { 7 r ” N } U { L , M } is a partial flock since {T  N } U { L ,M } admits the appropriate‘homologygroup’ with axis r0 and coaxis T I if FixB = r0 and { L , M } = {r0,m}. The 2dimensional Ksubspaces which are generated by point orbits under B intersect r0 and T I in ldimensional Ksubspaces. Each such pointorbit of a point which is not on a component of N lies in a Baer subplane of a Kregulus net which corresponds to a conic. For a given ldimensional Ksubspace of r0,there is a unique Borbit 2dimensional Ksubspace which is not in one of the regulus nets. This unique Borbit is the component of N containing the subspace in question. This component also nontrivially intersects T I . Hence, it follows that, for eachline of either ruling of the ruled quadric, there is a unique point of the line which is not covered by conics corresponding to the regulus nets of {r  N } U { L , M } . Thus, the partial spread has deficiency one. Now assume that a partial spread has deficiency one. Then, there is a partial spreadof the form P U { L ,M } admitting a homology group B which
25.3. DEFICIENCYONE PARTIAL HYPERBOLIC FLOCKS.
339
may be represented in the form
x o o o
([:;;:]v.=*) O O O I
and such that each component orbit of P union {L, M } defines a regulus in PG(3,K). Similarly as above, the points of the ruling lines correspond to the 2dimensional Ksubspaces which are Binvariant and not equal to L or M . The deficiency one assumption implies that for each ldimensional Ksubspace of L, there is a unique 2dimensional Ksubspace invariant under B which is not in a regulus net of the partial spread. The union of the set S of these 2dimensional subspaces cover both L and M and are mutually disjoint. Form P U S. Clearly, this partial spread completely covers all 1dimensional Ksubspaces and consists of 2dimensional Ksubspaces so that a translation plane with spread in PG(3, K) is obtained. Since the components are defined via Borbits, B is a collineation group of the constructed translation plane. However, now L and M are 2dimensional Ksubspaces which then define lineBaer affine planes. Since
is a collineation group of the translation plane that fixes L pointwise and
(I:;::l x o o o
o o o x
vx*)
defines the kernel homology group, it follows that
340
C H A P T E R 25. PARTIAL FLOCKS OF DEFICIENCY ONE
is also a collineation group of the translation plane that fixes M pointwise. We have noted in the chapter on pointBaer collineations that any pointBaer or lineBaer subplane which is pointwise fixed by a collineation is a Baer subplane. This completes the proof of part (1). Now assume that there is a partial flockof deficiency one which may be extended to a flock. Let 7r denote the translation plane admitting the pointBaer group B and let C denote the translation plane corresponding to the flock. Let G denote the regulusinducing homology group of C. We note that the subplanes of C which are the Baer subplanes of the regulus nets sharing two components L and M correspond to the 2dimensional K spaces which are Ginvariant. Hence, all of the components of the net N of 7r are now various of these subplanes of C. Hence, G = B. In other words, N is a regulus net and 7r and C are derivates of each other by replacement of N . Conversely, if N is a regulus net in n then, it must be that derivation by N produces a translation plane C and B becomes a homology group in C of the correct form to produce a flock of a hyperbolic quadric.0
Definition 25.8 The translation planes corresponding to partial flocks of deficiency one of either quadratic cones or hyperbolic quadrics shall be called ‘deficiency one translation planes Remark 25.9 Just as with Desarguesian partial parallelisms of deficiency one and Desarguesian parallelisms, little is known about deficiency one partial flocks in the infinite case. There are examples of deficiency one partial hypergolic flocks in PG(3, g) for certain values of q (see Johnson [@], Johnson and Pomareda [53] and Biliotti and Johnson (111). However, there cannot be proper finite partialconical flocks of deficiency one in PG(3,g ) b y results of Payne and Thas [66].
Chapter 26
SKEWHALL PLANES As, in some sense, I was introducted to the concept of derivation by trying to understand Albert's proof that the finite Hall planes are those obtained by derivation of AG(2,q 2 ) , it seems appropriate to conclude this monograph with a chapter on variations of the famous Hall planes. The finite Hall planes of order q2 originally constructed by changing multiplication in a finite field so as to construct an associated quasifield are exactly those translation planes obtained by the derivation of a regulus net in a Desarguesian spread in PG(3,GF(q)). If K is a fieldwhich admits a quadratic extension K[8]then there are analogous infinite Hall planes constructed via Pappian spreads in PG(3,K ) . A Pappian plane defines flocks of both quadratic cones and ruled quadrics in PG(3,K ) . We have seen that flocks of quadratic cones or ruled quadrics are equivalent to translation planes with spreads in PG(3,K ) which are unions' of reguli either sharing one or two lines respectively. A regulus cannot exist in P G ( 3 , K ) ,for K a skewfield, unless K is a field. However, pseudoreguli do exist in P G ( 3 , K ) as we have seen. In this chapter, we againconsider translation planes with spreads in PG(3,K ) which are unions of pseudoreguli sharing either one or two lines. We construct a class of translation planes which may be derived from a translation plane which has these properties. Because of the similarity to Hall planes, we call these 'skewHall planes'. Note thatthe translation planes deriving the skewHall planes are both conical and ruled. We begin with a technical lemma.
34 1
CHAPTER 26. PLANES SKEWHALL
342
Lemma 26.1 Let D be apseudoregulus netwithcomponentslines of PG(3, K ) , for K askewfield. Letdenotethe associatedleft Kvector space. If T is a line of PG(3, K ) which as a vector subspace of V4 is contained in the union of the components of D then T intersects each component of D and defines a Baer subplane of the net whose infinite points are exactly those of D. Proof Represent the components of D in standard form
x=o, y=x
[::]
VUEK.
Any line T is a 2dimensional left vector space. Certainly, if K is finite then T intersects each component of D. Hence, assume that K is infinite. T intersects each component in a Odimensional or ldimensional Kvector subspace. Hence, there are infinitely many components which T intersects in ldimensional Kvector subspaces. Choose any three of these m x = 0, y = 0, y = x and choose a new basis, if necessary, to preserve the standard form. Hence, relative to this basis, T = ( ( s , t ,O,O), (O,O, S,t ) ) where not both S and t of K are zero. Since T is a left 2dimensional Ksubspace, the ldimensional subspace Tu= ((S,t ,us, ut))for U E K must be contained in some component of D. Clearly, this component is y = x '  l u s Thus, slus = 0 tlut tlut = u' V u E K . So, T must intersect each component y = x
[
V u E K and since
1.
[T
:U]
is an automorphism of K , it follows that T must intersect each component of D in a ldimensional vector subspace so there is a spread induced on T w h h forces T to be a Desarguesian'affine Baer subplane of D whose infinite points are exactly those of D.0 0
Theorem 26.2 Let K be a skewfield and let T denote a 4dimensional left Kvector space. (1) Then the set of left 2dimensional Ksubspaces
CHAPTER PLANES WHALL 26. SKE
343
+
is a spread in PG(3,K ) if and only if z2 zp  y # 0.Vz E K where x and y are 2vectors. Let rP,? denote the corresponding translation plane. (2) rP,? is a Desarguesian plane if and only if both p and y are in the center of K. (3) rP,? is a semifield plane which admits the elation group
The component orbits union theaxis x = 0 of Eu define a set Cu of pseudoregulus nets that share exactly the component x = 0 and each of these nets may be derived to produce an associated translation plane. (4) rP,?also admits the elation group
:]
1 0 pt yt
Z=([:
i
0 0 0
VtEK).
1
The componentorbitsunion the axis x = 0 of Eu defineaset Ct of pseudoregulusnetsthat shareexactlythecomponent x = 0 , andeach of these nets may be derived to produce an associated translation plane. ( 5 ) rP,? also admits the homologp group
: i :] 1 0 0 0
H=([
VuEK*).
o o o u
The Componentorbits uniontheaxis (y = 0 ) and the coaxis (x = 0 ) define a set R of pseudoregulus nets that share two components x = 0,y = 0 and each of these nets may be derived to produce an associated translation plane. ( 6 ) For each derived translation plane, the kernel of the plane is Z ( K ) .
.
. . .
.
344
CHAPTER 26. PLANES SKEWHALL
In particular, if K is not a field then the plane derived from a Desarguesian plane does not have its spread in PG(3,K ) and hence cannot be considered a Hall plane even in this case. Also, note that if K is infinitedimensional over Z ( K ) then the derived planes are infinitedimensional over their kernels. Proof: By previous results, we obtain a spread in and only if the equation z2t zpt  7t = ( t )is bijective for each z in K . Hence, we obtain a spread if and only if z2 + zp  7 is always nonzero for z in K . This proves that a spread is obtained under the stated conditions. Direct computation shows that the spread is multiplicative if and only if p and 7 E Z ( K ) and furthermore multiplicative inverses exist provided z2 zp  y # 0 for z in K . We have noted that
+
+
x=o, y=x
[ x :]
VUEK
is a derivable net and also a pseudoregulus net in PG(3,K ) . Clearly, any nontrivial image of a component by Eu union the axis of Eu is then a pseudoregulus net and similarly, any nontrivial image of a component by H union the axis and coaxis of H is a pseudoregulus net. It remains to show that any nontrivial image of a component by Tt union the axis of is a pseudoregulus net in PG(3,K ) . Change bases by the matrix
This turns Tt into Eu. It remains to show that when
x=O,y=x[x
:]
VUEK
is derived, there is a constructed translation plane of kernel Z ( K ) .
CHAPTER 26. PLANES SKEWHALL
345
In order that the constructed structure is, in fact, an affine plane, it must be that the subplanes of this derivable net are Baer subplanes of the translation plane. Since the net is derivable, the subplanes must be pointBaer subplanes. It remains to show that they are also lineBaer subplanes. The affine subplanes sharing the parallel classes incident with the zero vector are pa,b = { ( m , ba, up, b p ) V a ,p E K} and a and b not both zero. We note from the chapter on dual spreads that any left 2dimensional Ksubspace defines a lineBaer subplane. If either a or b = 0 then pa,b is a left 2dimensional Ksubspace. Hence, assume ab # 0. Take any line y = IC
pt .L
.l +
yt
(c, d). It must be shown that this
line intersects Pa,b projectively. Clearly, we mayassume that t # 0 as if t = 0 then a line of the net containing the subplane is obtained and the line projectively intersects the subplane. Hence, it remains to show that there exist a,P E K such that
(aa,ba,aa(u + p t ) + bat + c, aayt + bau + d ) = (aa,ba,up, bp). Since ab # 0, we obtain the following equations:
+
+ + + +
a  y a a p ba)t a% au = blaayt au bld =
p,
p.
If a = 0 then there is an intersection with z = 0 and the givenline at (O,O,c,d). Hence, if (c,d) = (up,b p ) , there is an intersection with otherwise not. Thus, assume that (c, d ) # (up,bp). Hence, there exists such a ,p if and only if there is a solution to ("p
+ a%a  blaay)t = (alc  b"1d).
Note that (a"c
 bld)
= 0 if and only if (c, d ) = (up,bp) for some p E K.
Since we have assumed t
# 0, there is a solution if and only if
("p
+ alba
 blaay) # 0.
Since we can only have that CY # 0, multiply by alon the left to obtain (p+alalbaalblaay). Now multiply by alalba = z on the left to
CHAPTER 26. PLANES SKEWHALL
346
obtain the expression zp+ z2 y and since this is # 0, we obtain a solution. Hence, each subplane pa,b is lineBaer and hence a Baer subplane. It remains to show that the kernel of any derived plane II is "(K). We may isolate on the net
x=o,y=x
[:0 1 V U E K
as all other nets may be transformed into this net eitherby a collineation of the plane or by a Kbasis change. We note that the Baer subplanes pa,b are components of the derived translation plane. As these are right Kspaces but are left invariant only under Z(K) since a and b may vary over K, this shows that Z(K) is a subfield of the kernel. If there exists an element g such that g is not in K and in the kernel of the plane then the skewfield generated by K and g , (K,g ) , must fix all components of the translation planewhich are not in the derivable net. Since all such components are 2dimensional Ksubspaces, this forces the components to beldimensional (K,g) subspaces. This implies that these components may be embedded into a Desarguesian affine plane C coordinatized by (K,g) . Let D denote the derivable net of the original translation plane and let M denote the complementary net. Then 7r = 7rP,, = D U M and C = N U M for some net N . It follows that D and N are replacements for each other and note that all components are at least Ksubspaces. Assume that D and N are not equal. Then each component T of N which is not a component of D is a 2dimensional Ksubspace which intersects each component of D either trivially or in a ldimensional Ksubspace and is contained in the net D. By the above lemma, it follows that T is an affine Baer subplane of D sharing all infinite points with D. Hence, a subnet N* of N is defined by the set of Baer subplanes of D so that 7r and C are derivates of each other. Since the Baer subplanes of D are left 2dimensional subspaces, it follows that K is a field. Since N and N* cover the same points, N* = N . Thus, it follows that D and N are either equal or one is the opposite regulus of the otherand, in the latter case, K mustbea field. In this situation, the original plane is Pappian and the derived plane is Pappian. However, the derived plane will now admit a full pointBaer homology group or a full pointBaer elation which acts regularly on thecomponents of
26.1. DOUBLECOVERS.
347
the pseudoregulus net other than the axis and coaxis or axis respectively, It follows that the kernel of one of the Baer subplanes is K and we have seen that this would force the axis Baer subplane to be an Lspace where L is the kernel of the plane. In other words, the kernel can’t be as large as a skewfield coordinatizing the entire space since the superplane contains a kernel group acting transitively on the nonzero points of any component. Hence, the kernel of the derived plane is Z ( K ) = K in this situation. The other situationis when T is actually Desarguesian and coordinatized by L = (K,g)and derivable and there is an element g not in K which fixes each Baer subplane of the derivable net D.Hence, g induces an element of the kernel on each Baer subplane. Since the kernel of each Baer subplane is KoPP it follows that g must be in Z ( K ) acting as a kernel homology of the derived plane which is a contradiction to our assumptions. This completes the proof of the theorem.0
Definition 26.3 Any translationplane T ~ shall , ~be called a ‘skewDesarguesian plane’ and any plane derived as above from rP,,.shall be called a ‘skewHall plane’. Remark 26.4 W e noted previously when there is a full K,pointBaer homology or full K,pointBaer elation group of order > 2 acting o n a translation plane with spread in PG(3,K ) then this forces K to be a field. In the skewHall planes, thereare both full KpointBaer homologygroups and elation groups for subplanes with kernel K = K, and K is not necessarily commutative. On the other hand, the kernel of the associated translation plane is Z ( K ) . It wouldappear thattheexistence of fullK,pointBaer elationorhomology groups forcesthekernel of acorrespondingtranslationplaneto be a field.However,thishas been proved only for spreads in PG(3,K ) .
26.1 Doublecovers. In BiliottiJohnson [lo], the concept of a doublecoverisdevelopedfor spreads in PG(3,K ) , where K is a field. Here, we consider this in a more general setting.
CHAPTER 26. SKEWHALL PLANES
348
Definition 26.5 Let x denote aconical translation plane with spread in PG(3,K).Let C denote a normal set of pseudoreguli which sharea common line L and whose union is the spread for xmIf there exists a second normal set of pseudoreguli C* # C which share a common line L* whose union is the spread for x, we shall call {C,C*) a ‘doublecover’ for x. If L* = L and {C,C*) is a normal set of pseudoreguli then we call {C,C*) a ‘normal, doublecover with common line’. So, we see that the skewDesarguesian planes admitnormal doublecovers witha common line. The converseis also true as wenow point out.
Theorem 26.6 Let x be any conical translationplanewith spread in PG(3,K),for K a skewfield. If x admits a normal, doublecover with comm o n line then T is a skewDesarguesian plane. Proof: Represent x in the .form
with respect to the cover C and choose the common line to be z = 0. Let D be any pseudoregulus net in C*. By our assumptions, ifwe choose the same basis for x = 0 with any basis for the entire space then a choice of lines of D  {x = 0) as y = 0,y = z forces D to be in standard form. Moreover, D can share at most two components with any net of C. Let X denote the subset of K so that D shares two components with the pseudoregulus net
{x = 0, y = z
[
U
By previous results, we may assume that 0 is in X. That is, we may assume that D shares z = 0 and y = 0 with the standard net in C. Hence, there exists a subset X of K such that the components of D have the form
26.1. DOUBLE COVERS.
349
and for each t , ut is in K . Choose a basis change which fixes x = 0 pointwise and leaves y = 0 invariant and maps
onto y = x for tl E K
 (0).
Let a =
and change bases applying
the mapping a. Then, it follows that
Since we obtain the standard pseudoregulus net, it follows that
{h(t); t E X} = K. In other words, the mapping t H h(t) is surjective from X to K . Since, differences of matrices of D are nonsingular, it follows that the indicated mapping is also injective. Hence, X = K and
for MD a nonsingular matrix depending only on D. Another way to see this is to apply a previous result which states that if two normalizing pseudoreguli share two components then they share a homology group of the form
([ i ;
r;
V.EK{O}).
o o o u
Furthermore, when we choose D to be represented in the standard form then all pseudoregulus nets in C* have the form
for functions g* and f*.
CHAPTER 26.PLANES SKEWHALL
350
Notice that we have shown that any pseudoregulus net of C* must share two components with each pseudoregulus net of C. There are also two elationgroups Ec and where
It follows that
Since the pseudoreguli from C share two components withpseudoreguli from C*, it follows that Ec n E p = (l),Let MD = a component of the form y = x
[ 1]
then there is [ in the] standard net.
whichis not Hence, then clearly neither c nor b can be zero. Taking the image of y = 0 under E p and then Ec , in turn, we obtain y = x
ct
d tbt+ u
1
,
of the form y = x
Let d t + U = v and
S
[ 2 ] andthen
= ct to obtain components
where p = (a  d)cl and y = bcl.This
completes the proof of the theorem.0
26.2
Postscript.
In this monograph, we have been concerned with what could be said about derivable nets and about nets which are covered by subplanes or those merely containing at least one Baer subplane. Natural developments have led to the realization of derivable nets in3dimensional projective space andto the concept of a pseudoregulus net. As pseudoregulus nets are direct products of any two Desarguesian Baer subplanes incident with a given point, we have considered families of such direct products and the consequent theory of Desarguesian parellelisms in PG(3,K ) for K a skewfield.
26.2. POSTSCRSPT.
35 1
Following the general theme of studying ‘coverings’ or at least partial coverings, the concepts of affine and projective planes covered by subplane covered nets were developed as well as the theory of spreads covered by reguli or pseudoreguli and considered in a variety of ways. Furthermore, we have seen some applications and connections to other geometries such as flocks of quadric sets. In addition, the theory of structures analogous to those of derived versions of such affine planes leads to the theory of Baer nets, Baer groups, partial flocks and partial parallelisms (of deficiency one). We have been able to provide a fairly complete theory without restriction to either the finite case or by imposing commutativity conditions. However, examples of the possibilities are not complete known nor, in fact, are there examples in various situations. As I mentioned in the preface, many of the ideas encountered here each wearing the face of nets and or disguised within translation planes were originally developed for other combinatorial incidence structures. Thus, it ismy hope that this monograph both provides a‘theory of derivation’ as well as an invitation to the reader to investigate further the myriad interconnections that may be found with other varied combinatorial geometries and the simple and beautiful ‘subplane covered net’.
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INDEX (P,L)transitivity, 16 2fold product, 217
Central Collineation, 27 Central Collineation Group of a net, 83
Addition in coordinate systems, 44 Additive Spread, 76 Affine Fano plane, 205 Affine Geometry, 11 Affine Hughes Plane over K, 61 Affine Little Reideimeister condition, 207 Affine Plane, 15 Affine Plane over K, 12 Alternative Division Ring, 75 Annihilator, 143 AxialBaer perspectivity, 252 Axis of a central collineation, 27 Axis of a pointBaer perspectivity,
Central plane, 251 Codimension, 10 COdimensiontwo net, 151 Coaxis of a Homology, 27 Collineation, 15 Component Kernel, 37 Components of a Partial Spread, 39
Components of a Partition, 29 Components of a Spread, 28 Conic, 310 Conical Flock, 310 Conical Translation Plane, 305 Coordinates of a net, 43 Coordinates question, 45 Correlation, 144
251
Axis of a transvection, 116 Baer subplane, 24 Base subplanes, 260
Deficiency of a partial parallelism,
Center of a central collineation, 27
Deficiency one partial flock, 332
245
359
INDEX
360 Deficiency one Translation Plane, 340 Derivable Affine Plane, 18, 24 Derivable Chain, 54 Derivable Net, 41 Derivable Partial Spread, 40 Derivation set, 17 Derived Net, 42 Derived Plane, 18 Desarguesian Affine Plane over K, 12 Desarguesian Configuration, 65 Desarguesian configuration, 65 Desarguesian Parallelism, 233 Desarguesian partial Parallelism, 233 Desarguesian Plane, 65 Desarguesian Products, 225 Desarguesian Projective Planeover K, 9 Direct Product of Affine Planes, 215 Distributive quasifield, 69 Doublecover, 348 Dual Net, 149 Dual Planes, 51 Dual Spread, 131 Elation, 27 Embeddable Net, 149 Embedded net, 203 Extension of a net, 151 Extension question, 137 Fano Plane, 204 Finite Affine Plane, 15 Finite Affine Plane of Order n, 15 Finite Projective Plane, 15
Finite Projective Planeof order n, 15 Full pointBaer elation group, 275 Full pointBaer homology group, 276 Fundamental TheoremTranslation Planes, 31 Generalized Quadrangle, 313 Generating Normal Partition, 29 Hall Plane, 21 Homogeneous Coordinates, 19 Homology, 27 Hughes Planes, 57 Hughes Planes over a nearfield, 60 Hyperbolic Flock, 314 Hyperbolic quadric, 314 Hyperbolic translation plane, 305 Hyperplane, 10 Infinite Derivation, 24 JohnsonWalker plane, 244 Kregulus, 156 kRegulus, 157 Kernel Endomorphism, 30 Kernel mappings, 50 Kernel of a quasifield, 51 Kernel of a Spread, 30 Left and Right Projective Spaces, 227 Left and Right Spreads, 227 Left Extension of a Knet, 228 Left Partial Spread, 228 LineBaer subplane, 24
INDEX Linear Coordinate systems, 45 Linear Group, 12 Linear Translation Complement, 39 LorimerRahilly plane, 244 Minkowski plane, 319 Moufang Plane, 67 Multiplication in coordinate systems, 44 Nearfield, 49 Nearfield Plane, 62 Net, 41 NonAssociative Division Ring, 69 Normal Double Cover, 348 Normal Partition, 29 Normal set of reguli, 300 Normalizing pseudoreguli, 300 Opposite Regulus, 22 OstromRosati Plane, 63 Pappian Plane, 79 Pappus Configuration, 79 Parallelism, 233 Parallelism in AG(V,K), 12 Parallelism of Baer subplanes, 108 Partial Dual Parallelism, 249 Partial Parallelism, 233 Partial Spread, 39 Partition of a Group, 28 Perspective from a line, 65 Perspective from a point, 65 Point Transversal, 295 PointBaer elation, 251 PointBaer homology, 251 PointBaer perspectivity, 251 PointBaer subplane, 23
361 Polarity, 144 Projective General Linear Group, 14 Projective Geometry, 9 Projective Line over K,9 Projective Plane, 14 Projective Plane over K, 9 Projective PseudoRegulus, 295 Projective Semilinear Group, 13 Projective Special Linear Group, 14 Projective Spread, 133 Pseudoregulus net, 156 Quadratic cone, 310 Quadratic Form, 309 Quadric, 310 Quasiskew Perspectivity, 123 Quasifield, 49 Regular action, 34 Regular Direct Product net, 217 Regular Nearfield, 62 Regular Parallelism, 233 Regular Spread, 201 Regulus, 22 Regulus Question, 23 Reidemeister condition, 206 Relative pointline transitivity, 62 Retraction of a net, 151 Right and Left Parallelisms, 233 Right and Left Pseudoregulus, 228 Ruled Translation Plane, 305 Semilinear group, 12 Semiregular, 34 SemiTranslation Plane, 54
INDEX
362 Semifield, 69 Sesquilinear form, 144 Singer Cycle, 57 Skew Perspectivity, 122 SkewDesarguesian Plane, 347 SkewHall Plane, 347 Slope Endomorphisms, 35 Slope mappings, 35 Solid, 94 Special Linear Group, 13 Spread, 28 Spread Set, 36 Subnet, 183 Subplane covered affine plane, 203 Subplane Covered net, 149 Subplane covered net, 3 Subplane covered projective plane,
204 Subplane Regular, 203 Subspace of a Geometry, 88 Ternary Function, 43 Transitive Group, 16 Translation, 27 Translation Complement, 39 Translation Group, 27 Translation Net, 39 Translation of a net, 116 Transposed Spread, 132, 244 Transvection, 116 Transversal, 22 Vector Space Net, 216 Vector Transversal, 295 WRegulus, 157 Z(K)projective point, 295