Rudiments of Plane Affine Geometry: Mathematical Expositions No. 20 9781487575649

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RUDIMENTS OF PLANE AFFINE GEOMETRY P. SCHERK AND R. LINGENBERG

P. SCHERK is Professor of Mathematics at the University of Toronto. R. LINGENBERG is Professor of Mathematics at the Universitli.t Karlsruhe, Germany, and the author of three books on linear algebra and geometry.

In recent decades research into the foundations of geometry has completely transformed the field and created a need for new methods of presenting geometry at the university

level. Rudiments of Plane Affine Geometry develops rigorously and clearly one geometric theory accessible to the reader with no previous experience. It reflects the spirit and displays some of the basic ideas of modern geometric axiomatics. The volume is intended for undergraduates with a modest knowledge of algebra and linear algebra and provides a sound introduction to deductive geometry.

MATHEMATICAL EXPOSITIONS Editorial Board

H.S.M. COXETER, G.F.D. DUFF, D .A.S. FRASER, G. de B. ROBINSON (Secretary), P.G. ROONEY Volumes Published

1 The Foundations of Geometry

G. DE B. ROBINSON

2 Non-Euclidean Geometry H.S.M. COXETER 3 The Theory of Potential and Spherical Harmonics w.J. STERNBERG and T.L. SMITH 4 The Variational Principles of Mechanics CORNELIUS LANCZOS

5 Tensor Calculus J.L. SYNGE and A.E. SCHILD 6 The Theory of Functions of a Real Variable R.L. JEFFERY 7 General Topology WACLAW SIERPINSKI (translated by c. CECILIA KRIEGER) (out of print) 8 Bernstein Polynomials G.G. LORENTZ (out of print) 9 Partial Differential Equations G .F.D. DUFF 10 Variational Methods for Eigenvalue Problems s .H. GOULD 11 Differential Geometry ERWIN KREYSZIG (out of print) 12 Representation Theory of the Symmetric Group G . DE B. ROBINSON 13 Geometry of Complex Numbers HANS SCHWERDTFEGER 14 Rings and Radicals N.J. DIVINSKY 15 Connectivity in Graphs w.T. TUTTE 16 Introduction to Differential Geometry and Riemannian Geometry ERWIN KREYSZIG 17 Mathematical Theory of Dislocations and Fracture R.w. LARDNER 18 n-gons FRIEDRICH BACHMANN and ECKART SCHMIDT (translated by CYRIL w.L. GARNER) 19 Weighing Evidence in Language and Literature: A Statistical Approach BARRON BRAINERD

20 Rudiments of Plane Affine Geometry

P. SCHERK

and

R. LINGENBERG

MATHEMATICAL EXPOSITIONS NO. 20

Rudiments of plane affine geometry P. SCHERK and R. LINGENBERG

UNIVERSITY OF TORONTO PRESS Toronto and Buffalo

© University of Toronto Press 1975 Toronto and Buffalo Printed in Great Britain

Reprinted in 2018 ISBN 978-1-4875-7274-7 (paper) LIBRARY OF CONGRESS CATALOGING IN PUBLICATION DATA

Scherk, Peter Rudiments of plane affine geometry (Mathematical expositions; no. 20) Bibliography: p. Includes index. 1. Geometry, Affine. I. Lingenberg, Rolf, 1929- joint author. 11. Title. III. Series. QA477.S33 516'.4 75-11705 ISBN 0-8020-2151-4 CN ISSN 0076-5333 AMS 1970 subject classifications 50-A99, 50-o05

IN MEMORY OF HANS HEILBRONN

Contents

PREFACE

ix

1 AFFINE PLANES

1.1 The axioms 1.2 Examples 1.3 A co-ordinate plane 1.4 Finite planes Exercises

3 3 7 8 10 13

2 COLLINEATIONS

2.1 2.2 2.3 2.4

Bijections Collineations Fixed elements Homotheties 2.5 Translations 2.6 Dilatations . 2. 7 Axial affinities Exercises

14 14 15 19 21 23 26 28 30

3 TRANSLATION PLANF.S

3.1 Linear transitivity 3.2 The configurations of translation planes 3.3 The prime kernel of a translation plane Exercises 4

32 32 32 44 49

DESARGUESIAN PLANES

52

4.1 The dilatation groups D( 0) 4.2 The shear theorem

52 57

viii 4.3 The linear transitivity of the groups A(a) Exercises

5

CONTENTS

61 64

PAPPUS PLANES

66

Appendix Exercises

69 71

6 CO-ORDINATES IN DESARGUESIAN PLANES

72

6.1 Co-ordinate planes 6.2 Co-ordinates in desarguesian planes 6.3 The fundamental theorem of affine geometry Exercises

72 74

80 84

7 THE PROJECTIVE CLOSURE OF AN AFFINE PLANE

85

7.1 Motivation 7.2 The axioms of a projective plane 7.3 The duality principle 7.4 Collineations 7.5 The canonical duality 7.6 Affine restrictions of collineations of projective planes Exercises

85 86

88 90 95

96 99

8 DESARGUESIAN PROJECTIVE PLANES

IOI

8.1 Projective and affine desarguesian planes 8.2 The projective theorem of Desargues 8.3 Projective Pappus planes Exercises

IOI 102 104

APPENDIX

107

A.I Groups A.2 Skew fields A.3 Right vector spaces

107 108

REFERENCES

111

INDEX

113

105

108

Preface

There is fairly general agreement that the traditional presentation of euclidean geometry at the colleges (and high schools) on this continent is an anachronism. We have to look for alternatives if we wish to keep geometry alive. Affine geometry is both modern and simple. It is closely related to euclidean geometry. We can introduce it by stripping euclidean geometry of its topological and metric features. The beginnings of the resulting theory are simple enough. It is enriched gradually until the transition to affine (and euclidean) analytic geometry, which is more or less hidden in today's linear algebra, can readily be made. This book is suitable for a one-term course for undergraduates with a modest knowledge of algebra and linear algebra. Its detailed presentation is the result of our intention to make it suitable for the beginner and the non-specialist. Some of the more difficult sections and problems are marked by asterisks and can be omitted if necessary. Some instructors may also wish to skip the proof of Theorem 3 of Chapter 2. For the convenience of the reader some algebraic concepts are collected in an appendix. It will not be referred to in the text. The literature is extensive and recent. Very little in this book could have been written thirty years ago. The interested reader will find references at the end of the text. In conclusion the authors wish to thank Dr E. Ellers for his careful reading of their manuscript and his criticisms. and R. LINGENBERG Toronto and Karlsruhe March 1974

P. SCHERK

RUDIMENTS OF PLANE AFFINE GEOMETRY

1

Affine planes

1.1 THE AXIOMS

Euclidean geometry as it is taught in our high schools is a highly sophisticated theory. It combines three classes of concepts: (i) linear concepts such as 'point,' 'straight line,' 'incidence,' 'parallelism,' etc. ; (ii) metric concepts such as 'length' and 'angle' as well as 'orthogonality' and 'congruence'; (iii) topological concepts such as 'betweenness.' We arrive at affine geometry by dropping both the metric and the topological concepts. Thus the purpose of this book is the discussion of those parts of euclidean geometry which deal with linear concepts only. Our treatment will have to avoid metric or topological reasoning. We shall start with some natural assumptions which are satisfied in euclidean geometry. They will be our axioms. (Every mathematical theory needs, in addition to some basic undefined concepts and relations, some assumptions ('axioms'] on them which we agree to accept. A 'proof' reduces other statements ['theorems'] to these axioms.*) Let P and L denote two non-void disjoint sets. We call the elements of P points and those of L (straight) finest and denote the former by capital letters P, Q, A, B, ... and the latter by small ones g, h, /, .... In addition, we are given a relation I called incidence between certain points and lines. Thus I is a subset of the set of all the pairs (P, /) of points and lines. If the pair (P, /) belongs to this subset, we write

PI I or II P and say 'P and/ are incident' or 'Plies on/' or'/ passes through P.' If P and I are not incident, we write Pl I. • The reader should be warned though that some of the 'planes' which satisfy these axioms will look rather different from the euclidean plane. f In this text, 'line' will be synonymous with 'straight line.' Note that 'line' means 'curve' in differential geometzy.

4

AFFINE PLANES

If I is incident with two points P and Q, i.e. if

/IP,

/IQ,

(1)

we say I connects P and Q. If the point P is incident with the lines / and h, i.e. if

Pih,

PI/,

(2)

we say I and h have the point P in common. The lines I and h are said to be parallel,

I

II h,

if either I = h or I and h have no point in common. Thus two distinct lines / and h are parallel if (2) has no solution P. If I and h are not parallel, we write

/,Ith. We state AXIOM

Al If P I: Q there exists one and only one line connecting P and Q.

We denote this line by [P, Q]. Thus/ P-:/: Q. THEOREM

= [P, Q] is equivalent to formula (I) if

I Two non-parallel lines I and h have exactly one point in common.

PROOF We have/ ,Ith if and only if/-:/: hand/ and h have at least one point in common. Thus we have to prove that / and h have not more than one point in common. Suppose / and h have the distinct points P and Q in common. Then P and Q would be connected by the two distinct lines I and h. This contradicts Axiom

Al. □

If / ,It h, we say that / and h intersect and call the common point the intersection of / and h. It will be denoted by

[/, h] = [h, /]. Our next assumption is the parallel axiom of euclidean geometry. AXIOM A2 Given I and P, P l I, there exists one and only one line through P parallel to I.

If PI /, there is at least one line through P parallel to/, viz. I itself. If h I: I is another line through P, then h and / are both incident with P. Hence h ,II' /. Thus we can improve Axiom A2 slightly:

5

1. I THE AXIOMS

THEOREM 2 To every line I and every point P there is exactly one line through P parallel to I. THEOREM

3 Parallelism is an equivalence relation.

We remember that a relation is called an equivalence relation if it is reflexive, symmetric, and transitive. (i) Reflexivity We have to show that every line is parallel to itself. This follows from our definition. (ii) Symmetry To show that / II h implies h II/. This is true if/= h. Let I =f h; I II h. Then / and h have no point in common. Thus h 111. (Or more briefly: our definition of/ II his symmetric in/ and h.) (iii) Transitivity Let / II g, g II h. We have to show that / II h. This is certainly true if / and h have no common point. Suppose then that / and h are both incident with P. By our assumptions and by part (ii) of our proof we have

/ II g, II P and h II g, hI P. Thus both lines / and h pass through P and are parallel to g. Hence by Axiom A2, I = h and therefore / II h. □ Let Il I denote the set of the lines parallel to /. Thus

n, = {h I h 11 /} and le Il 1•

(3)

We call Tl 1 a parallel pencil. Our next goal is THEOREM 4 (i) Every line belongs to one and only one parallel pencil. (ii) Two lines I and h are parallel if and only if they belong to the same parallel pencil. PROOF

(i) Let / be any line. By (3) I belongs to the parallel pencil Il 1• Suppose the parallel pencil nh also contains/. We have to show that nh = Il 1• Let g e nh· Thus g II h. Since / e nh, we have / II h. Hence by Theorem 3, g II l or g e Il 1• Since this applies to every g e nh, we obtain (4)

Conversely, assume g e Il 1• Thus g II I. Since / II h, this yields g II h or g e Ilh and therefore

n, c nh. Combining (4) and (5), we obtain

(5)

nh =

Il 1•

6

AFFINE PLANES

(ii) If n II/, then h and / both belong to I1 1• Conversely, if h and / both lie in rr,, then h II g and / II g, and therefore h 111. □ THEOREM

5 Let 11 1112 , 11 ,j/' h. Then 12 ,j/' h. The points [/1, hJ and [/2, hJ are equal

if and on~y if 11 = 12 •

PROOF Suppose that 12 II h. Then /1 II 12 implies /1 II h by Theorem 3. Thus /2 ,j/' h and the points [/1, hJ and (/2 , hJ exist by Theorem 1. Obviously, /1 = /2 implies [/., hJ = (12 , hJ. Conversely, if / 1 =/: 12 , the points [/1, hJ and (12 , hJ cannot be equal since they are incident with distinct parallel

lines. □

We call three points collinear if there is a line that all three of them are incident with. One third assumption reads: AXIOM

A3 There are three non-collinear points.

A triplet

~( = (P, L, I) satisfying the axioms Al-A3 is called an affine plane. Note that P is not void by Axiom A3 and that Axiom Al then ensures the existence of lines. We call a triplet A, B, C of non-collinear points a triangle ABC and the lines

a= [B, CJ,

b = [C, AJ,

C

= [A, BJ

(6)

the sides of the triangle. THEOREM

PROOF

6 No two sides of a triangle are parallel.

See Exercise 6.

The set of all the lines incident with a given point P is called the pencil of the

lines through P.

THEOREM 7 (i) Every parallel pencil contains at least two lines. (ii) Every line is incident with at least two points. (iii) For every P, the pencil of the lines through P contains at least three lines.

PROOF By Axiom A3, there exists a triangle ABC with the sides a, b, c. (i) Let rr, denote a parallel pencil. Then at least one of the points A, B, C, say A, is not incident with 1. The line through A parallel to / belongs to I1 1 and is distinct from I. (ii) Given the line /, Theorems 3 and 6 imply that at least one of the sides of ABC, say a, is not parallel to /. We have already shown that I1 0 contains a line

1.2

7

EXAMPLES

a' =/: a. Since a' II a and a ,It/, Theorem 5 yields a' ,ll' I and [a,/] =/: [a',/]. Thus / is incident with the distinct points [a,/] and [a',/]. (iii) Let P be arbitrary. Let a', b', c' denote the lines through P parallel to a, b, c, respectively. By Theorems 6 and 4, the three parallel pencils II 0 , II,,, Ile are mutually disjoint. Hence the three lines a', b', c' through Pare distinct. □ THEOREM

common.

8 Two lines are non-parallel if and only if they have exactly one point in

PROOF By Theorem I two non-parallel lines / and h have exactly one point in common. Suppose then / II h. If / =/: h, I and h have no point in common. If I= h, there are, by Theorem 7(ii), at least two points on this line. □

1.2 EXAMPLES

To familiarize ourselves with the three axioms, we first mention three somewhat intuitive examples from euclidean geometry. (i) The plane euclidean geometry satisfies our three axioms. (ii) The geometry on the euclidean line satisfies Al and A2 but not A3. (iii) P consists of the points of euclidean three-space, L of its lines. 'Incidence' means ordinary incidence. Thus two skew lines are 'parallel' and A2 does not hold. However, Al and A3 remain in force. The next examples are rigorous. They will demonstrate the independence of the three axioms Al-A3; cf. Exercises 4-6 and Section 1.4. (iv) Let P be any non-void set and L a set consisting of one element / ¢ P. The relation I consists of all the pairs (P, /) with P e P. Thus P I / for all P e P. Then the triplet (P, L, I) satisfies Al and A2 but not A3; cf. Example (ii). (v) Let P = {A, B, C} and L = {a, b, c} denote any two disjoint sets of three elements each. The incidence relation consists of the pairs. (A, b),

(A, c),

(B, a),

(B, c),

(C, a),

(C, b).

Thus

BI a, c, Cia,b; . cf. Figure I.I. Then the triplet (P, L, I) satisfies Al and A3 but not A2 because AI b, c,

no two distinct Jines are parallel. (vi) Let P = {A, B, C} be an arbitrary set of three elements and Jet L = {a, d} be any set of two other elements. Furthermore the relation I consists of the pairs (A, d), (B, a), (C, a). Thus A Id,

BI a,

CI a;

cf. Figure 1.2. Then the triplet (P, L, I) satisfies A2 and A3 but not Al.

8

AFFINE PLANES

C

Figure 1.1

In the next two sections we shall discuss some triplets (P, L, I) which satisfy the axioms Al-A3, thus obtaining rigorous examples of affine planes.

1.3 A CO-ORDINATE PLANE

The following example illustrates an important · method of constructing affine planes. Let IR denote the field of the real numbers. Put

P

= IR

x IR

= {(x, y)

Ix, ye IR}.

Thus the 'points' are pairs (x, y) of real numbers. The 'lines,' i.e. the elements of L, are certain point sets, viz.: (i) for every c e IR we take the point set I

=

l(c)

= {(x, y) I x =

c }.

(7)

(ii) for every pair of elements m, d of IR, the point set I = l(m, d) = {(x, y) I y = mx+d} shall be a line. A ---0----------d

----------c--a C B

Figure 1.2

(8)

1.3 A

9

CO-ORDINATE PLANE

The incidence relation I consists of all the pairs (P, /) such that the point P is an element of the point set /. Thus (9)

Before we prove that the triplet ~ = (P, L, I) is an affine plane we investigate parallelism. By definition two lines are parallel if they are either identical or disjoint. (cc) Obviously, any two lines l(c) and l(c)' are parallel. (ft) No two lines I = l(m, d) and /' = /(c) are parallel. Evidently/ :p I' and the point (c, mc+d) lies on both/ and/'. (y) J(m, d) II J(m', d') if and only if m = m'. Let I= J(m, d) and/' = l(m', d'). Assume first m = m'. If d = d', then I= I'; thus 111 l'. If d :pd', I and /' have no common point (x, y) ; for it would satisfy

= y = m'x+d' = mx+d'. Thus d = d', contrary to our assumptions. mx+d

Conversely, let/ II/'. Ifm :pm', the equation mx+d = m'x+d' has a solution x and the point (x, mx+d) would be incident with both I and/'. Since I !I I', this implies I = I' and m = m'; contradiction. We now verify the axioms Al-A3.

To Axiom Al Let P = (a, b) and P' = (a', b') denote two distinct points in ~- If a = a', then l(a) connects P and P'. Suppose there is another line connecting P and P'. Then it must be a line l(m, d). Thus b = ma+d = ma' +d = b' or P = P'; contradiction. Foe a :pa' there is one and only one m such that m(a-a') = b-b'. Put d = b- ma. Then both P and P' lie on J(m, d). Any other line through P and P' must be a line l(m', d'). Hence m'a+d' = b, m'a' +d' = b', and m'(a-a') = b-b'. This yields m = m' and d' = b-m'a = b-ma = d. Thus l(m, d) = /(m', d'). In either case we have one and only one line connecting P and P'. To Axiom A2 Let P = (a, b) be any point and/ a line. If/= l(c), the line /(a) is incident with P. By (cc) it is parallel to/. By (/3), l(a) is the only line through P parallel to /. If/= l(m, d), the line /' = l(m, b-ma) through P is parallel to I by (y). Conversely, any line through P parallel to I is a line l(m', d') by (ft) . By (y), l(m', d') 11 l(m, d) implies m' = m. Since Pe l(m', d'), we obtain d' = b-m'a = b-ma

=d

and l(m', d')

= l(m, d).

To verify Axiom A3, choose the points (0, 0), (0, l), and (I, 0). They are readily shown to be non-collinear.

10

AmNEPLANF.S

1.4 FINITE PLANES

In the examples (iv)-(vi), we studied triplets of.finite point and line sets with an incidence relation. But we did not construct any affine planes with a finite number of points and lines. Such a plane is obtained if we try to construct an affine plane m: = (P, L, I) with a minimum number of points. By Axiom A3, the set P must contain a triangle ABC. Thus L contains its sides (6); cf. Axiom Al.

Applying Axiom A2, construct the Jines d through C parallel to c and e through B parallel to b. Thus

d II c ,It b II e

(i.e. d

II c, c ,It b, b II e).

Hence d ,It e and these lines intersect at a point D. The point Dis distinct from A, B, C: since CI c, we have d =/: c. Hence d II c implies D =/: A, B, and similarly D =/: C. Let f = [A, D]. The six lines a, b, c, d, e, fare mutually distinct. This is obvious for a, b, c. Furthermore, d =/: a, b, c. Otherwise either a = d II c or b = d II c (contradicting Theorem 6) or CI c = d. Symmetrically, e =/: a, b, c. Since c ,It b II e, ABD is a triangle. Hence B If. As B is incident with a, c, e, these Jines must be distinct from/. Similarly,/=/: b, d. Put

P = {A, B, C, D},

L = {a, b, c, d, e,f}.

Thus P and L are sets of four mutually distinct points and of six mutually distinct lines, respectively. The incidence relation is the following set of pairs

I

= {(A, b), (A,

c), (A,f), (B, a), (B, c), (B, e),

(C, a), (C, b), (C, d), (D, d) (D, e), (D,f)}.

It is indicated in Figure 1.3 and in the following incidence table, the asterisks

Ia I b I c A

B

C D Figure 1.3

•

Id

I• I• I I • I

·1· I

e

• •

•

• • •

1.4 FINITE PLANES

11

denoting incidence. It shows that Axiom Al is satisfied. Also we have the following pairs of parallel lines

a Ill,

b

II e,

c II d.

This implies A2. Finally A, B, C are not collinear. Thus ~2

= (P, L,I)

is an affine plane. In ~ 2 , every line is incident with exactly two points. We call ~ 2 the affine plane of order two. Note that every point is incident with exactly three lines. We wish to generalize our last remarks and to study the simplest properties of finite planes, i.e. of planes with only a finite number of points. We prepare our discussion by LEMMA 9.1

If one line is incident with at least n points, then every line is.

Suppose the line a is incident with then mutually distinct points A 1 , • • • , An. Let b =I= a. By Theorem 7(iii) and Theorem 2 there is a line /1 through A1 which is parallel to neither a nor b. Let lk denote the line through Ak parallel to / 1 • Thus the lines /1 , • • • , In are parallel. We have lk ,fl' a and Ak = [lk, a]. Since

PROOF

fork =I= j,

Theorem 5 implies lk =I= 11. Thus the n parallel lines /1 , ••• , In are mutually distinct. Furthermore lk II /1 ,fl' b. Hence lk ,fl' b and Theorem 5 implies that the n points

(k = I, ... , n) on b are mutually distinct. □

If the affine plane ~ contains one line which is incident with a finite number of points only, then there exists an integer n ~ 2 such that every line is incident with exactly n points.

THEOREM 9

PROOF Assume the line / is incident with exactly n points. Let h denote a second line. By our lemma, h is incident with at least n points. Suppose h is incident with not less than m points. Then our lemma implies that / also is incident with at least m points. Thus h is incident with exactly n points. D

A plane has the order n if every line is incident with exactly n points. THEOREM 10

n+ I lines.

Every point in an affine plane of order n is incident with exactly

PROOF Given a point P in our plane, there exists, by Axiom A3, a line / not incident with P. Let A1 , • •• , An denote the points on/. Then [P, A1], • •• , [P, An] and the line through P parallel to / are n + I mutually distinct lines.

12

AFFINE PLANES

Conversely, any line through P intersects I at some point A" unless it is parallel to /. Thus there are no other lines through P. THEOREM 11

Every parallel pencil in a plane of order n contains exactly n lines.

PROOF Let II I denote any parallel pencil. By Theorem 6, there exists a line h ,II' I. Through each point P of h there is exactly one line Ip of II 1• Conversely, each line of II 1 intersects h. Hence it is equal to one of the lines Ip where PI h. Thus these are all the lines of II 1• If P and Q are distinct points on h, Theorem 5 yields Ip =I= la. Hence the number of lines in II I is equal to the number n of points on h. □

THEOREM 12

An affine plane of order n contains exactly n+ 1 parallel pencils.

PROOF Choose any point P. Let ai, ... , an+i denote the lines through P. The n + I parallel pencils

(10)

are mutually distinct by Theorem 4. Hence there are at least n + 1 parallel pencils. Conversely, any parallel pencil II contains a line a through P by Axiom A2. Hence II = II 0 is equal to one of the pencils (10). D THEOREM 13 ·An

lines.

affine plane of order n contains exactly n 2 points and n(n+ 1)

PROOF

(i) Let II 1 = {/i, ... , ln} denote a parallel pencil. By Axiom A2, each point is incident with exactly one line parallel to I, i.e. with precisely one of then Jines Ii, • .. , ln. By Theorem 9, each of them is incident with exactly n points. Thus the plane contains exactly nxn points. (ii) By Theorem 12 there are precisely n+ I parallel pencils. By Theorem 10 each of them contains exactly n lines. As every line belongs to one and only one parallel pencil [cf. Theorem 4), we have altogether exactly n(n+ 1) lines. □ Note that by Theorem 9 an affine plane is finite if and only if it has order n for some n. The question whether there are affine planes of a given order n is only partially solved. We know that there are such planes if n is the power of a prime number, and we do not know any planes whose order is not a power of a prime number. The famous Bruck-Ryser Theorem states: Let n = 1 or 2 (mod 4) and suppose n is neither a prime power nor the sum of two squares. Then there is no affine plane of order n.

13

EXERCISES

We thus know: (i) There are planes of the orders 2, 3, 4, 5, 7, 8, 9, 11, 13, 16, .•• (ii) There are no planes of the orders 6, 14, 21, 22, ••• But we do not know whether or not there are planes of the orders 10, 12, 15, •••

EXERCISES 1 Suppose the triplet P, L, I, satisfies the following axiom and no other: Two distinct lines have not more than one point in common. {i) Show that two distinct points have not more than one connecting line. (ii) Show by an example that two distinct points need not have a connecting line. 2 Which sections of Theorems 2 and 3 would remain valid if we call two lines parallel if and only if they have no point in common. 3 Suppose the set P consists of the four vertices A, B, C, D of a regular tetrahedron in euclidean three-space, and L consists of its six edges

a

= [B, CJ,

b

= [C, AJ,

c

= [A, BJ,

d

= [C, DJ,

e

= [B, DJ, f = [A, DJ.

=

•incidence' means ordinary incidence. Show that ~ (P, L, I) is an affine plane of order two [interpret Figure 1.3 as the drawing of a tetrahedron !J. 4 Using Exercise 3, construct a triplet (P, L, I) satisfying the axioms Al and A3 but with a point A and a line l such that more than one line through A is parallel to 1. 5 Let O be a point in euclidean three-space. The •points' of Pare the lines through 0. The •Jines' of L are the planes through 0. A 'point' is •incident' with a •Jine' if it is contained in the latter. Verify that the triplet (P, L, I) satisfies the axioms Al and A3 but that any two •Jines' always •intersect' in exactly one •point.' Check that these results are not affected by interchanging the definitions of •points' and •Jines' if the definition of •incidence' is suitably modified.

6 (i) Prove Theorem 6. (ii) Show that there is no point incident with all the sides of a triangle. 7 Suppose the affine plane ~ contains a line with at least n points. Show that (i) every parallel pencil contains at least n lines; (ii) each point is incident with at least n + I lines. 8 Given a finite affine plane of odd order n, show that any set of n +2 points contains three (mutually distinct) collinear ones. 9 Repeat the construction of a co-ordinate plane replacing IR by any skew-field; cf. p. 108.

2 Collineations

2.1 BIJECTIONS

A bijection ex of a set S onto a set T maps each element A of S onto an element

B

= Aex

(I)

of T such that (i) Aex = A'ex implies A = A' if A, A' lie in S, (ii) to any B e T there is an A e S which solves (I). By (i), A is uniquely determined by B. We call A the inverse image of B. By associating with each element B of T its inverse image A, we obtain a welldefined mapping ex - i from T into S, the inverse of ex. Thus Bex- 1

= A-Aex = B

for all BeT.

Obviously, ex- 1 also is a bijection. It satisfies (ex-1)-1

= ex.

(2)

A trivial but important example of a bijection is the identical map ,5 from S onto itself defined through

A,5

=A

for all A e S.

If ex and fJ are bijections of S onto T and of Tonto the set U, respectively, the mapping exfJ of S onto U is defined through

AexfJ = (Aex)fJ

for all A e S.

Then a.fJ is readily seen to be a bijection of S onto U with the inverse (exfJ)- 1

= p- 1ex- 1 •

(3)

We call a./J the product of ex by /J. Note that (4)

and a.ex- 1

= ,8

and ex -1 ex = 'T•

(5)

2.2 COLLINEATIONS

15

Our composition is associative. If y is a bijection of U onto a set V, then (a./J)y = a.(/3y)

(cf. Exercise I).

(6)

2.2 COLLINEATIONS

Let m: = (P, L, I) and m:• = (P', L', I') denote two affine planes. We call a pair ex of bijections a. 1 of P onto P' and a. 2 of L onto L' a col/ineation of m: onto ~(' if

PI/Pex 1 I'la. 2 •

It is customary to drop the indices and to use the same letter ex not only for the pair but also for the individual bijections. Furthermore, as a rule, the same symbol I is used to denote incidence in both m: and m:'. Thus we usually write the preceding formula in the form \ PI I Pa. I /ex.

I

(7)

This notation will be justified by Theorem 3. The identical bijections 'P and 'L induce a collineation of m: onto itself, the identity , = '!!I· Another obvious but important example is the following: Given any affine plane~(= (P, L, I), define

I' = {P I PI /}

for each / e L.

Thus P e I' if and only if PI I. Put L'

= {/' 11 EL}

and define

m:• = (P, L', e). Thus our incidence relation in m:• consists of all the pairs (P, /') such that Pe/'. We readily verify that m:• is an affine plane and that

Pex = P, la. = I'

for all P e P and all l e L

defines a collineation ex of m: onto m:•. Two affine planes m: and m:• are called isomorphic if there is a collineation of m: onto m·. THEOREM

PROOF

1 Isomorphism of affine planes is an equivalence relation.

(i) Reflexivity The identity is a collineation of m: onto itself. (ii) Symmetry Suppose the affine plane m: is isomorphic to the affine plane m:'.

16

COLLINEATIONS

Then there is a collineation a. ofm: onto m:'. It consists of two bijections, a. 1 and By (7),

°'2·

a. -1

=

(

°'1

-1

, °'2

-1)

(8)

is a collineation of m:' onto m:. Thus m:' is isomorphic to m:. (iii) Transitivity Assume there are collineations a. of m: onto onto m:". Let a. = (a. 1 , a.2 ) and /3 = (/3 1 , /32 ). Then

m:' and /3 of m:' (9)

is a pair of bijections. If P is a point and / a line in equivalent to

m:, then PI I is

by (7)

Pa. I la. and Pa.{3 I la.{3.

Thus a.{3 is a collineation from m: tom:". □ We call the collineation a. - 1 the inverse of a. and a./3 the product of the collineations a. and {3; cf. (8) and (9).

THEOREM 2 Let a. be a col/ineation of an affine plane 1, h be lines of m:. Then (i) P =/: Q - Pa. =/: Qa.. (ii) If P =/: Q then [P, QJa. = [Pa., Qa.J. (iii) I II h - la. 11 ha. and I ,It h - la. ,It ha.. (iv) If I ,Ith then [/, h]a. = [/oi:, ha.].

m:. Let P,

Q be points and

PROOF

(i) is trivial. (ii) Let 1 = [P, QJ. Then /I P, Q. Hence by (7), la. I Pa., Qa.. Thus by (i), la. = [Pa., Qa.]. (iii) It is sufficient to prove the second statement. Let /,Ith. Thus there is exactly one point PI 1, h. By (7), Pa. I la., ha.. If P' I la., ha., then (7) yields P'a.- 11 l, hand hence P'oi:- 1 = P or P' = Pa.. Thus la. ,It ha.. Conversely, let la. ,It ha.. Then from the above I

= (la.)a. - 1 ,It (ha.)a. - 1 = h.

(iv) follows from

(iii). □

Theorem 2(iii) implies COROLLARY 2.1 A col/ineation a. of an affine plane maps a parallel pencil onto a parallel pencil; more precisely

Il 1 oi: = IT,.,. Since a. maps non-parallel lines onto non-parallel lines, we obtain from Corollary 2.1 :

17

2.2 COLLINEATIONS

2.2 A collineation of the affine plane monto the affine plane ~{' maps the set of the parallel pencils ofm bijectively onto that ofm'.

COROLLARY

Let a:= (a:., a: 2) denote a collineation ofm: = (P, L, I) onto m:' = (P', L', I). Suppose the points A, B, C e P are collinear. Thus there is a line 1 incident with all three of them. By (7) this implies la: 2 I Aa: 1 , Ba:., Ca: 1 ; i.e., the three image points are collinear. We say that collineations preserve collinearity. We wish to prove a converse of this remark. 3 Given two affine planes m: = (P, L, I) and m:' = (P', L', I) let a: 1 be a bijection of Ponto P' which preserves collinearity. Then there exists one and only one bijection a: 2 ofL onto L' such that a: = (a:i, a: 2) is a collineation.

THEOREM

PROOF We first verify the uniqueness of a: 2. Let /32 also be a bijection of L onto L' such that /3 = (a: 1, /32) is a collineation. Put

r=

a:/J- 1 = (a:1 a:1 -

1,

a:2/32- 1) (cf. (8) and (9)).

Thus r = (r., y2) is a collineation ofm: onto itself. Here y1 = a: 1a: 1- l = 'Pis the identity mapping of P onto itself and y2 = a: 2/32- l. If 1 e L choose any two distinct points P, Q incident with 1. Thus I = [P, Q]. By Theorem 2 (ii), ly2 = [Pr1, Qy1] = [P, Q] = 1.

Hence r 2 maps every line of 1 onto itself, i.e. y2 = 'L· This yields /32 = y2/32 = a:2/32 -•p2 = a:2. It remains to construct a bijection a: 2 of L onto L' such that a: = (a:., a: 2) is a collineation. Let I= [A, B] = [C, D]. Without loss of generality assume C =fa A, B. Thus A, B, Care mutually distinct collinear points and so are Aa: 1 , Ba: 1 , Ca: 1 • With A, C, D, the points Aa: 1 , Ca: 1, Da:1 are collinear. Hence [Aa: 1 , Ba: 1] = [Aa:i, Ca: 1] = [Ca: 1 , Doei]. We may therefore define a map a: 2 of L into L' through [A, B]a:2 = [Aa:i, Bo:1]-

(IO)

It remains to prove that a: = (o: 1, a: 2) is a collineation, i.e. that oe satisfies (7) and

that o: 2 is bijective. We first verify part of (7), viz. that a: preserves incidence: let PI/; choose QI/, Q =fa P. Then I= [P, Q] and by (10), la:2

= [P, Q]a:2 = [Poei, Qa:i];

in particular Pa: 1 I /a: 2. Next a: 2 is surjective, i.e. every line of L' is an image: ~et I' = [A', B'] e L'; thus A' =fa B'. Since o: 1 is bijective, the points A'a: 1- l and B'o: 1 - l of Pare distinct. By (10) and (5), we have

18

COLLINEATIONS

[A ,°'1 - 1 , B' °'1 - 11°'2

= [A'°'1 - 1°'1• B'°'1 -t °'1 1= [A' , B'1 = I' •

Thus /' is the image of the line [A' oc 1 - 1 , B' oc 1 - 11. In the next steps we show that oc 2 is injective, i.e. that different lines of L have distinct images. Since ex preserves incidence, / ,It' h implies that /oc 2 and hoc2 have points in common. Thus if /oc 2 and hoc2 are distinct parallel lines, / and h must be parallel. Assume first that the lines / 1 and /2 of L are not parallel. Choose h' II /1 oc 2 , h' # /1oc 2 • Since oc 2 is surjective, there is a line he L such that h' = hoc2 • From the above, we have h 1111 and hence h ,It' 12 • Therefore h' and /2 oc2 have points in common. As h' and /1oc 2 have no points in common, this implies /1oc 2 # /2 oc 2 • Consider now two parallel lines / 1 and /2 • Choose h ,It' 11 • Then the points Pk = [h, lk] exist and are distinct; thus h = [Pi, P 2 ]. Since lk ,It' h, we have

k = I, 2. The point P 1oc 1 being incident with /1oc2 , we obtain

P2oc1 I

/1°'2•

As P 2 oc 1 I /2 oc2 , this yields /1oc 2 # /2 rx 2 • We have now proved that distinct lines of L have distinct images, and the proof that oc 2 is bijective is complete. We still have to verify the second half of (7), viz. that Poc 1 I /oc 2 implies PI /. Suppose PI/. Choose a point QI /; thus P # Q. Put h = [P, Q1. Then h¥,/and

hrx2

= [Paci, Qoc11 #

loc2•

Since P # Q = [h, 11, we have Poc 1 # Qoc 1 = (hoc 2 , /oc 2]. As Poc 1 I hoc 2, this implies Poc 1 I loc2. □ The last part of our proof yields COROLLARY 3.1 Given two affine planes & = (P, L, I) and&' = (P', L', I), a pair of bijections of P onto P' and of L onto L' which preserves incidence is a collineation.

From now on we study the collineations of an affine plane onto itself. THEOREM

element,.

4. The col/ineations oc of an affine plane form a group with the unit

The product of two collineations was defined in (9). The associativity of the composition oi bijections induces that of the multiplication of collineations. By (4) and (9), we have ,ex = ex, = ex. Finally by (8), ococ- 1 = oc- 1oc = ,. □

PROOF

2.3 FIXED ELEMENTS

19

2.3 FIXED ELEMENTS

We call the point P a fixed point of the collineation a if Pa = P. Similarly if la = /, I is called a fixed line of a. The fixed set F(a) of a consists of all the fixed points and lines. Thus

F(,)

= Pu L,

(11)

F(a- 1) = F{a}, F{a}

c

F{an)

(12) for n

= 2, 3, ...

(13)

We collect three basic properties of F(a). THEOREM 5 (i) If P and Q are two distinct fixed points of a, then [P, Q] is a fixed line of a. (ii) If I and hare non-parallel.fixed lines of a, then [I, h] is a.fixed point of a. (iii) If P is a fixed point and I is a fixed line of a, then the line through P parallel to 1 is a fixed line of a, PROOF Suppose Pa. = P, Qa. = Q, and P i= Q. Then by Theorem 2(ii), [P, Q]a = [Pa, Qo:] = [P, Q]. Part (ii) follows similarly. Finally, let h denote the line through P parallel to 1. By Theorem 2, a preserves not only incidence but also parallelism. Hence ha is the line through Pa. = P parallel to la. = 1. Thus ha. = h. □

COROLLARY

S.l The triplet

(F(a.) r. P, F(a.) r. L, I')

(14)

satisfies the axioms Al and A2. Here I' is the restriction of! to (F(a.) r. P) x (F(o:) r. L}. Note that, e.g., F{a.) r. Pis the set of the fixed points of a.. Axiom Al for the triplet (14) follows from Theorem S(i). Similarly, Axiom A2 for (14) is obtained from Axiom A2 for the whole plane and Theorem 5(iii). □ If P' c P, L' c L, and if~, = (P', L', I') satisfies the axioms Al-A3, ~, is called a subplane of~(. Corollary 5.1 implies COROLLARY

5.2 /fF(a.) contains three non-collinear points, (14) is a subplane of~.

We next verify a simple condition for two collineations i.e. for a{J = {Ja.. THEOREM 6

onto itself. PROOF

a

and fJ to commute,

If the collineations a. and fJ commute, a must map the fixed set of fJ

Let a.{J

= {Ja.. Suppose, e.g., that Pis a fixed point of fJ. Then

20

COLUNEATIONS

Prx. = (P{3)rx. = P(f3rx.) = P(rx./3) = (Prx.){3. Thus Prx. e F(/3) and rx. maps F(/3) into itself. Since rx.{3 = {3rx. implies

fX.-113 = fX.-1({3rx.)rx.-1 = fX.-1(rx.{3)rx.-1 = /3rx.-1, rx. - I and f3 also commute and rx. - 1 maps F(/3) into itself. Hence F(/3) is mapped by rx. onto itself. □ An axis of the collineation rx. is a line, all the points of which are fixed. If every line through a point is fixed, the point is called a centre of rx.. Thus every axis is a fixed line and every centre is a fixed point; cf. Theorem 5. If rx. = ,, every line is an axis and every point is a centre. We next show that a collineation =/a, cannot have too many axes and centres. THEOREM 7 Let rx. be a co/lineation, rx. =fa ,. (i) If rx. has a centre 0, then it has no other fixed point. (ii) If rx. has a centre 0, then any fixed line is incident with 0. (iii) If rx. has an axis a, then any fixed point is incident with a. PROOF

(i) Suppose rx. also has the fixed point F =/a 0. We first verify that Fis a centre: let / I F. The line through 0 parallel to I being fixed, / itself is fixed by Theorem 5(iii). Thus every line through Fis fixed. Next let P be any point. If PI (0, F], Pis the intersection of the fixed lines (0, P] and [F, P] and hence fixed. From the above, Pis even a centre. Finally, if PI [0, F] but P =fa 0, F, choose any QI (0, F] and repeat the preceding construction replacing F by Q. Thus every point is a centre and every line is fixed. Hence rx. = ,. (ii) Suppose there is a fixed line/ I 0 . Let PI/. Then Pis the intersection of the fixed lines (0, P] andf. By Theorem 5(ii), Pis a fixed point of rx.. This is excluded by the first part of our theorem. (iii) Suppose there is a fixed point 0 l a of rx.. By Theorem 5(iii), the line through 0 parallel to a is fixed. Let /I 0, I ,JI' a. Since 0 and [/, a] are fixed points, I = [0. [/, a]] is fixed by Theorem 5(i). Thus every line through O is fixed and 0 is a centre. Since a I 0, this is excluded by the second part of our theorem. □ COROLLARY 7.1 The total number of centres and axes of a collineation =/a, is not greater than one.

Let rx. =/a ,. If rx. has a centre, Theorem 7(i) implies that rx. cannot have an axis or another centre. By Theorem 7(iii), rx. cannot have two axes.

PROOF

We shall see later that a collineation =fa, may have other fixed lines in addition to an axis without being the identity.

REMARK

2.4 HOMOTHETIES

21

2.4 HOMOTHETIES

The collineation 'T/ is a homothety if

l'TJ II I

for all I e L.

(15)

If 'T/ = ('TJ., 'f/2) is a homothety, our definition implies [P, Q] II [P'T/ 1 , Q'T/ 1)

for every pair P ::/: Q.

(16)

We show that (16) characterizes the homotheties : 8 Given an affine plane~ = (P, L, I), let 'T/t denote a bijection of Ponto itself which satisfies (16). Then there exists one and only one bijection 'f/ 2 ofL onto itself such that 'TJ = ('T/i, 'T/ 2) is a collineation. 'TJ then is a homothety.

THEOREM

PROOF If we can show that 'T/i preserves collinearity, Theorem 3 implies the existence of a unique 'T/2 such that 'T/ = ('T/i, 'T/2) is a collineation. If I = [P, Q] is any line, then

l'T/2

= [P, Q)'T/2 = [P11i,

Q111J II [P, Q)

= /.

Thus 11 will be a homothety. Assume then that P, Q, Rare collinear. We may also assume that they are mutually distinct. By (16)

[P11i, Q11iJ II [P, Q) = [P, R] II [P111, R11iJ. Hence [P11 1 , Q11 1J = [P11i, R111 ], and P11 1 , Q11i, R11 1 are collinear. Thus 111 preserves collinearity. □

The homotheties of the affine pla,te the group of the collineations of~.

THEOREM 9

~

form a normal subgroup H of

PROOF If 11 and 11' satisfy (15), then (/7J}r,' II /'T/ II/ for all/ e L. Since the collineation 11 - l preserves parallelism, /11 II / implies / = (/11)11- 1 II /11- 1 for all / e L. Thus His a subgroup. Next, let a: denote any collineation. Then by (15)

(la -1 }r,

II la -1

and hence by Theorem 2(iii), ((/oc- 1)11)oc II (/oc- 1 )oc or loc- 111oc !I I

for all / e L. Thus oc - 111oc also is a homothety. D THEOREM

l O A fixed point of a homothety is a centre.

22

COLLINEATIONS

PROOF

ITJ

Let 07]

= I. □

=

0; I I 0. Then ITJ I 07]

=

0 and by (15), ITJ

II I. Hence

Theorem IO and Corollary 7.1 imply COROLLARY 10.1 A homothety ,f=, has not more than one.fixed point. In particular, it has no axis.

If ct is any collineation and if P :/= Pct, the line [P, Pct] is called a trace of ct. Every fixed line is either an axis or a trace. For if the fixed line f is incident with a point P I= Pct, then Pct I fct = f and therefore f = [P, Pct]. Let 7/ again denote a homothety. By Corollary 10.1 every fixed line of 7/ is a trace. Conversely, let I = [P, PTJ] be a trace of 7/· Then /IP implies ITJ I P7J. Thus by (15), ITJ is the line through PTJ parallel to /. Since /I P7J, this implies ITJ = I. We thus have THEOREM

11 A line is the trace of a homothety =/:,

if and only if it is fixed.

We cannot deduce the existence of homotheties other than the identity from our axioms. On the other hand, it is easy to show that there cannot be too many homotheties. 12 Let A :/= B. Then there is not more than one homothety mapping A onto C and B onto D.

THEOREM

PROOF

Suppose the homotheties 7/ and 7J' satisfy

BTJ = BTJ' = D. By Theorem 7, 7/ '7/ - 1 is a homothety. It has the fixed points A and B. Hence by Corollary 10.1, 7/,7/- 1 =, and 7J' = TJ. □ Concluding this section we study the set H(a) of all the homotheties with the given fixed line a. THEOREM

Then 7/ 1-+

13 For any line a, H(a) is a subgroup of H. Let ct be any collineation.

TJ"'( = ct- 1TJct),

7J

e

H(a),

(17)

defines an isomorphism of H(a) onto H(act). PROOF Let 7/ ', 7/ e H(a). Since a is a fixed line of 7/' and of 7/ TJ,TJ- 1 • Thus 7/,7J- 1 e H(a). Next consider the map (17). Since

it is a bijection of H(a) onto H(act). Also ( 7/7/ ')"' =

ct -17]7], ct = ct -17/ct. ct -17], er. = TJ"'TJ '"'.

t,

it is one of

2.5

23

TRANSLATIONS

Thus (17) defines an isomorphism of H(a) onto H(aa.). □ THEOREM 14 Given a line a and two points A and B not on a, there is not more than one homothety in H(a) which maps A onto B.

Let TJ, TJ' e H(a); AT/= ATJ' = B. By Theorem 13, TJ 1TJ-i e H(a). Since ATJ'TJ-i = 4, Theorem 10 implies that A is a centre of the homothety T/,T/- 1 • Since a is a fixed line of T/ 'T/ - 1 and A l a, Theorem 7(ii) implies T/ 'T/ - i = ,. Thus TJ' = TJ- □

PROOF

2.5 TRANSLATIONS

By Corollary 10.1, there are two types of homotheties ,fa,, those with one fixed point and those with none. We first study the latter, defining: a translation is,a homothety which either has no fixed points or is the identity. If the translation -r is not the identity, the fixed set F(-r) consists of the traces of-r; cf. Theorem 11. THEOREM

15 If the translation

-r

is not the identity, F(-r) is a parallel pencil IT.

We call -r a translation in the direction of IT. Through each point P there is a trace [P, P-r], i.e. a fixed line; cf. Theorem 11. Since -r has no fixed points, Theorem 5(ii) implies that any two fixed lines are parallel. Hence the set of fixed lines is a parallel pencil. D PROOF

We next verify a uniqueness theorem for translations. 16 For every pair of points A, B there exists at most one trc. .,uiution which maps A onto B.

THEOREM

-r

PROOF Let-rand -r' denote two translations; A-r = A-r' = B. If B = A, we have -r = -r' = ,. Let B ,fa A. Then a = [A, B] is a trace and hence a fixed line of -r and -r'; cf. Theorem 11. Thus -r and -r' belong to H(a) and our assertion follows by applying Theorem 14 to A, B, and a line a' II a, a' # a. □

Alternatively, we can prove Theorem 16 by a construction which will become important later. Let -r be a translation, A # B = A-r. If Cl [A, B], then [C, C-r] II [A, B] by Theorem 15. Furthermore, by Theorem 2(ii) and (15), [B, C-r] = [A-r, C-r] = [A, C]-r 11 [A, C]. Since [A, C] ,It [A, B], this determines the point C-r: it is the intersection of the lines through C parallel to [A, B] and through B parallel to [A, C]. If CI [A, B], we repeat our construction, replacing A by some point D B

24

COLUNEATIONS

DT

D

A

C

CT

B

Figure 2.1

outside [A, B] and B by Ih. Thus the image of each point is uniquely determined. The same will therefore apply to the lines; cf. Figure 2.1. THEOREM

17 The translations form a normal subgroup T of the group of all the

col/ineations.

Let-rand -r' be translations; a is any collineation. We have to verify that -r- 1 , -r'-r, and a- 1-ra are translations. By Theorem 9, they are homotheties. As F(-r- 1) = F(-r), -r- 1 has no fixed points unless -r = ,. Suppose P-r'-r = P. Then P-r' = P-r - l. Since -r - l also is a translation, Theorem 16 implies -r' = -r- 1 or -r'-r = ,. Thus either -r'-r has no fixed point or PROOF

,

TT='•

Finally, Pa- 1-ra = P implies (Pa- 1)-r = Pa-•. Thus -r has the fixed point Pa-• and -r = ,. Hence a- 1 -ra = ,. □ Let T(II) denote the set of the translations in the direction of II. Thus T(II)

=

{-re TI F(-r)

=

II or-r

= , }.

Our definition implies

if II =/: II'.

(18)

THEOREM 18 For any parallel pencil II, T(II) is a normal subgroup of H, in particular ofT. For every co//ineation a,

-r 1-+ -r°'

= a - 1-ra

for all -r e T(II)

defines an isomorphism of T(II) onto T(IIa). PROOF

If both-rand -r' belong to T(II), then by Theorem 17, -r'-r e T. Further-

more,

1-r'-r

= 1-r = l

for every le II.

Hence -r'-r e T(II). Since F(-r) is a subgroup of T.

= F(-r- 1), -re T(II) implies -r- 1 e T(II). Thus T(II)

2.5 TRANSLATIONS

25

Let a be any collineation. By Theorem 17 a - 1-ra is a translation. If I (la)(a- 1-ra) = 1-ra = la. Hence

e

II, then

a- 1 -ra E T(Ila); cf. Corollary 2.1 . For any homothety a, Ila = II. Thus T(Il) is a normal subgroup of H. The second statement is obvious; cf. the proof of Theorem 13. □ A fairly weak additional assumption yields some interesting results on the group T. THEOREM 19

If there are

translations t,, in distinct directions, Tis abelian.

PROOF

(i) Let -r and -r' denote two translations =,=, in distinct directions, say -r e T(Il), -r' e T(II '), II t, II'. By Theorem 18, -r'- 1 and -r- 1 -r'-r both lie in T(Il'). Hence -r'- 1(-r- 1-r'-r)

E

(19)

T(ll').

On the other hand, T(II) contains -r' - 1-r - 1-r' and -r and hence (-r' - 1-r - 1-r')-r. Thus -r' -

l'T - 1

-r''T

E

T(II) () T(ll ') = {'}

or (TT')- 1(-r'-r) =, or -r'-r = TT';

cf. (18). (ii) Let-r, -r' denote two translations in the same direction II. By our assumption there is a -r" ¢ T(Il). Since -r'-r" e T(ll) would imply -r" = -r'- 1 (-r'-r") e T(ll), we have -r'-r" ¢ T(II).

Hence by (i),

Suppose there are translations :f,, in distinct directions. If there is one translation :f,, offinite order, then every translation has finite order and there is a prime number p such that the order of every translation :f,, is equal top.

THEOREM 20

Let -r0 :f, , be a translation of finite order n and let p be a prime divisor of n ; thus n = pm. Then the order of -r = -r0 m is equal to p. Let-rand -r' satisfy (19), -r' :f, ,. Thus

PROOF

II"= F(TT') :f, II' and

II"

c

F((TT')P).

26

COLLINEATIONS

By Theorem 19, Hence F(r'P) contains the distinct parallel pencils TT' and TT". This yields T ' P = ,. Thus every translation T 1 e T(TT) is of order p. Repeating our argument starting from T 1 , we obtain that every translation not in T(TT '), in particular every element =ft, of T(TT) also has the order p. □

2.6 DILATATIONS

We call a homothety with a fixed point a dilatation 8. If 8 =ft ,, we obtain from Theorem 10 and Corollary 10.1 that 8 has exactly one fixed point 0, and 0 is a centre of 8. By Theorem 7, 8 has no fixed lines other than those through 0 . Thus F(8)

=

{0} f"'I {/III 0}.

By Theorem 11, the traces of 8 are its fixed lines. THEOREM 21

a group.

For any point 0, the set D(0) of the dilatations with the centre 0 is

If oc is any collineation, the mapping 8 e D(0),

defines an isomorphism of D(0) onto D(0oc). The proof is obvious. From Theorem

12

we obtain :

If A and B are distinct from 0, there is not more than one dilatation with the centre 0 which maps A onto B. If there is one, the points 0, A, B must be collinear.

THEOREM 22

By Theorem 9, the product rr'l' of two homotheties 11, 11' is a homothety. A homothety was either a translation or a dilatation. We obtain special cases by prescribing for each of the homotheties 11, 11 ', 1111' whether it shall be a translation or a dilatation. However, not all of these cases are possible. THEOREM23

(i) The product of two translations is a translation. (ii) The product of a translation T and oc dilatation 8 =ft , is a dilatation. (iii) The product of two dilatations with the same centre is a dilatation. (iv) If the product of the two dilatations 8, 8' =ft , with distinct centres 0, 0', respectively, is a translation, then 88" is a dilatation for every 8" =ft 8' in D(0'). REMARK Suppose the groups D(0) and D(0') are linearly transitive, 0 =ft 0'; cf. Chapter 3. Then there exists to every 8 e D(0) a dilatation 8' e D(0') such that 88' is a translation; cf. Chapter 4, Exercise 3.

2.6 DILATATIONS

27

23 The assertions (i) and (iii) have been proved in Theorems 17and21. To (ii). Assume T8 is a translation. Then 8.= T- 1{T8) would be the product of two translations and therefore a translation; contradiction. To (iv). Put 88' :z: T . Assume 88" = T is a translation for some 8" in D(O'). Then PROOF OF THEOREM

1

is a translation. Since it has the fixed point O', it must be the identity. Hence 8' = 8". □ A collineation ex is an involution [is involutory] if ex2

= ,,

ex

"F ,.

A reflection a in the point O is an involutory dilatation with the centre O; thus a e D(O), THEOREM

a2

= ,,

24

(i) The product of a translation with a reflection in a point is a reflection in a point.

(ii) The product of two reflections in distinct points is a translation. (iii) The product of three reflections in points is again a reflection in a point, unless all the three points coincide. (iv) There is at most one reflection in the point O if there is a collineation ex such that Oex =I: 0. (v) If there are involuntary translations, there are no reflections in points. PROOF To (i). a- 1Ta is a

T,

Let T be a translation and let a denote a reflection in a point. With translation ; cf. Theorem 17. Hence

{Ta) 2 = T(aTa) = T{a- 1Ta)

is a translation. By Theorem 23(ii), Ta is a dilatation. Thus by Theorem 21, (Ta) 2 also is a dilatation. This yields (Ta) 2 = ,. lfTa = ,, a= T- 1 would be a translation =I:,. Thus w =I: , and Ta is an involution. To (ii). Let a, a' be two reflections in distinct points. Thus a =I: a' and aa' = a - 1a' =I: ,. We have to show that the homothety aa' is not a dilatation. Suppose aa' has the fixed point 0. Then Oaa' = 0

or Oa = Oa' = Q, say.

Then Qa' = Oa' 2

=0

(20)

and

Qaa' = Oa 2a' = Oa' = Q.

•

Thus aa' would have the two fixed points O and Q. Since aa' =I: ,, this implies 0 = Q; cf. Corollary. I 0.1. By (20), 0 would be a common fixed point of a and a'; contradiction.

28

COLLINEATIONS

To (iii). Let u, u', u" denote reflections in 0, 0', 0", respectively. Thus either 0 =/: 0' or 0' =/: 0". In the first [second] case, uu' [u'u") is a translation by (ii) Hence by (i), uu' u" is a reflection in a point. To (iv). Let u and u' denote two reflections in 0. By Theorem 21, u" = ex- 1uex is a reflection in 0ex =/: 0. By (ii), uu" and u"u' are translations and so is their product uu' = uu"·u"u'. As ua' has the fixed point 0, it is the identity. Hence

u' = u- 1 = u.

To (v). Assume we had both an involuntary translation T and a reflection u in a point. By (i), Tu would be a reflection in some point. Thus, = (Tu) 2 or Hence u = TUT = T- 1 Bk> Ck be incident with lk; k = 1, 2. Assume

[A1, B2] II [B1, A2], Then

(5)

[A., C2] II [C1, A2],

(7)

We repeat: THEOREM

2 Statement (p) holds in translation planes.

Our argument has also yielded COROLLARY 2.1 IJT(IT) is linearly transitive and abelian in an affine plane~( then (P)n holds true in~- [We obtain (P)n from (p) by adding the assumption that / 1 and 12 belong to IT.] A special case of Statement (p) applies to any affine plane. REMARK 2.2 Statement (p) always holds if A 1 , B 1, C 1 or A 2 , B 2 , C2 are not mutually distinct.

The proof is analogous to that of Remark 1.1. THEOREM

3 (d) implies (p).

PROOF (cf. Figure 3.3) On account of Remark 2.2 we may assume that the three points Ak> Bk, Ck are mutually distinct; k = 1, 2. Thus A1 =I= C1 and the lines [A 1, B 2 J and [C1, B 2 ] intersect. Hence the lines through C1 parallel to [A 1, B 2J and through A 1 parallel to [C1 , B2 ] also intersect, say at D. Since [D, Ad II [C1 , B2 ] II [B1 , C2 ], we may apply Statement (d) to the triplets Ai, B 2 , C2 and D, C1 , B1 • Thus

[A 1, B2J II [D, Cd and [B2, C2] I! [C1, Bi] implies [Ai, C2J II [D, B1]. • Pappus (approximately 320 AD) lectured at the Museum in Alexandria on mathematics, astronomy, geography, and the interpretation of dreams. His main work, the Synagogue [ =collection], reports on and complements classical mathematics. He seems to have anticipated Desargues' Theorem.

36

TRANSLATION PLANES

D

--Go---.....a---------,_____ Lz A2

B2

Cz

Figure 3.3

Symmetrically, we obtain

II [D, Bil• Hence [A 1, C2) II [D, B.J II [C1, A2J- □

[C1, A2J

Combining Theorems I and 3, we once more obtain Theorem 2. We require a third statement. To motivate it we first assume that~ is a translation plane. Let -r be a translation #, and let P1 be any point. Using Q1 = P1-r we construct the point Qi' = Pi'-r for a given point P1' on / 1 = [P1, Qtl in two different ways: choose /2 II /1, / 2 # 11 • Let P2 be any point on /2 • Construct the intersection Q2 of /2 with the line through Q1 parallel to [P1 , P 2 ]; thus Q 2 = P2 -r and

[P1', P2J II [P1 '-r, P2-rJ = [Qt', Q2l• Replacing P2 by any point Pz' on /2 and repeating our construction, we arrive at a point Qz' = Pz'-r and

[Pi', Pz'J II [Qi', Qz'J; cf. Figure 3.4. It is now easy to verify that the following theorem holds in a translation plane. (s) ['Little Shear Theorem']: Let 11 1112; 11 # 12. Let Pk, Pk', Qk, Q/ be incident with 111 (k = I, 2) and let

STATEMENT

[P1, P2J II CQ1, Q2J,

(8)

Then (9)

3.2

37

THE CONFIGURATIONS OF TRANSLATION PLANES

Figure 3.4

The preceding statement is symmetric both in the P's and Q's and in the indices I and 2. If, for example, P 1 = Q 1 (8) implies P2 = Q 2 , P 2 ' = Qz', Pi' = Qi' and hence (9). Similarly, (9) follows from (8) if Pi' = Qi' or if P 1 = P/. This yields REMARK

pk

-::/=

4.1 Statement (s) is valid in any affine plane unless

Qk,

P,/

-::/=

Q,/,

pk

-::/=

P,/,

Qk

-::/=

Q,/,

k = 1, 2.

(IO)

We next prove THEOREM

PROOF

4 (p) implies (s).

(cf. Figure 3.5) Construct Pz'' I /2 and

Qi" I

/1 such that (11)

Since [P1, P2] II [Q1, Q2] and [P1, Pz"] II [Qi", Q2], Statement (p) yields [Qi, P2"] II [Qi", P2],

Figure 3.5

38

TRANSLATION PLANES

Furthermore [Pi', Pz']

II [Q 1", Q2] and [Pi', P2] II CQ1', Q2]

imply on account of (p) that [Qi", P2] II [Qi', Pz']. Hence

[Qi, P/] II [Qi', Pz']. Also [P1, P 2 '] II [Qi, Qz']. Therefore by (p), [P1, Pz"] II [Qi', Qz']. Thus ( 11) implies (9). □ We obtain from our proof COROLLARY

4.2 (P)u implies (s)n-

[The statement (s)rr is obtained from (s) by adding the assumption that {/1,/2} C Il.] By Theorems 3 and 4, (s) follows from (d). Since we could not deduce (P)u from (d)u, the preceding results are not sufficient to prove REMARK

4.3 (d)rr implies (s)n.

We give a direct proof assuming (10). If [P1, P2] II lP1', Pz'] and CQ1, Q2] II CQ1', Q2'], (s)rr follows from [P1 , P2 ] II [Q 1 , Q 2 ]. Thus we may assume that, e.g., [Pi, P2 ]

~

[Pi', P2 '];

cf. Figure 3.6. Let P denote the intersections of these lines. Thus P is incident neither with / 1 nor with /2 • The line/ E Il through P intersects [Q 1 , Q 2 ] at a point Q. Since ·

Figure 3.6

3.2 THE CONFIGURATIONS OF TRANSLATION PLANES

39

[Pi, P] 11 [Q 1, Q] and [P1 , Pz'] 11 [Q 1 , Qz'], (d)n implies (Pz', P] 11 [Qz', Q].

Since [P2, P] II (Qz , Q] and [Pi', P2] II [Qi', Q2], (d)n yields [Pi', P] II [Qi', Q].

(12)

Thus [Qi', Q] II [Pi', P]

= [Pz', P] II (Qz',

Q]

and hence [Qi', Q] = [Qz', Q]. The points Qi', Q 2 ', Qare therefore collinear and (12) implies CQ1', Qz'] = [Qi', Q] ll[Pt',P] =

[Pt',Pz'] . □

The main result of this section is the following 5 An affine plane is a translation plane theorem (d) of Desargues holds true.

THEOREM

if and only

if the little affine

Because of Theorem 1 we may assume that the plane~ = (P, L, I) satisfies (d). Let P0 , Q 0 be two distinct points. We have to construct a translation T such that (13) (i) We first construct a bijection T of Ponto itself which satisfies (13). Put 10 = [P0 , Q0 ] . If P l 10 , let Q = PT be the intersection of the lines parallel to 10 through P and parallel to [P0 , P] through Q0 ; cf. Figure 3.7.

Figure 3.7

40

TRANSLATION PLANES

Figure 3.8

Before defining Q = P-r for points Pon 10 , we choose any point Pi outside 10 and put Qi = Pi-r, Then we repeat our construction for a point PI 10 replacing P 0 and Q0 by Pi and Qi, respectively. This defines a map -r of Pinto itself, which, obviously, satisfies (13); cf. Figure 3.8. In order to show that -r is a bijection, construct a second map -r' repeating our construction but interchanging the P's and Q's. Thus

P-r

= Q Q-r' = P.

(14)

Hence -r is a bijection with the inverse .,. - i = -r'. (ii) We next prove

[A, B] II [A-r, B-r]

(15)

for all the pairs of distinct points A, B. As this is trivial if [A, B] 1110 , we may assume [A, B] ,II' 10 •

First case (cf. Figure 3.9): A, Bl 10 • By our construction

[A, A-r] II [B, B-r] II lo, [P0 , A] II [Q0 , A-r], and [Po, B] II [Qo, B-r].

Figure 3.9

3.2

THE CONFIGURATIONS OF TRANSLATION .PLANES

41

Figure 3.10

Thus (d) implies (15).

Second case (cf. Figure 3.10): A J'. 10 , A J'. [Pi, Q1], BI 10 • Then [A, A-r] II [P1, Qi] II lo,

B-r I /0 , and

[P0 , A] II [Q 0 , A-r],

[P0 , Pi] II [Qo, Qi],

Hence (d) applied to the triangles AP0 P1 and (A-r)Q 0 Q 1 implies

[A, Pi]

11

[A-r, Qi].

Applying (d) once more to the triangles P1 AB and Q1 (A-r)(B-r), we now obtain (15).

Third case: A I [P1 , Qi] and BI 10 • Then A-r I [Pi, Qi] and B-r I 10 , [P1 , Q 1] II 10 , and

[P0 , Pi] II [Qo, Qi],

[P0 , A] II [Q 0 , A-r],

[B, Pi] II [B-r, Qi].

Thus (15) follows from (s), while (s) is a consequence of (d). (iii) By Chapter 2, Theorem 8, (l 5) implies that the bijection -r of P onto itself can be completed to a homothety -r of 21 in one and only one way. By our construction, the point mapping had no fixed points. Thus -r is a translation. It satisfied (l 3). D By Remark 4.3, (s)rr follows from (d)rr. Thus we obtain COROLLARY

5.1 Statement (d)rr holds true if and only if'T(IT) is linearly transitive.

We collect the preceding results in two diagrams. It is an unsolved problem whether (d) and (p) or (p) and (s) are equivalent. It can be shown that drr does not imply (P)rr.

42

TRANSLATION PLANES

(d)

1l

( p)

!

(s )

>

(d

>,r

>

( p

>,r

==>

ll

(s

>,r

Diagranu

~

translation plane

( d)

T(II) linearly transitive (d)rr T(II) Jin. trans. and abelian (d)rr and (p)n.

Diagramo

In order to show that Statement (d) is an independent axiom which is not implied by Axioms Al-A3, we give an example of an affine plane which is not a translation plane. If a, b are real numbers, define 0

0

b = {2ab if b > 0, ab otherwise.

The points of our plane are the pairs of real numbers (a, b). The lines are the point sets /(c) = {(x, y) Ix= c} and l(m, d) = {(x, y) I y =mo x+d}.

(16)

A point is incident with a line / if it belongs to the point set /. Thus this plane is obtained from the real euclidean plane by replacing the lines of positive slope by lines broken along the y-axis; cf. Figure 3.11 . The reader will verify that the axioms Al-A3 are satisfied and that (x, y) H (x, y+a)

defines a translation parallel to the y-axis. However if II denotes the pencil of the lines parallel to the x-axis, there are no translations =/:, parallel to II. [Also (s)rr does not hold; cf. the diagrams on this page and Figure 3.12.]

3.2 THE

43

CONFIGURATIONS OF TRANSLATION PLANES

Figure 3.11

Y-axis

l,

(x-axis)

Figure 3.12

44

TRANSLATION PLANES

3.3* THE PRIME KERNEL OF A TRANSLATION PLANE

In the second half of this chapter we study the group of the homotheties of a translation plane m:. By Chapter 2, Theorem 20, either every translation '#, has infinite order or there is a prime number p such that every translation '#, has the order p. In the first case, put p = 0. In either case, we call p =charm:

the characteristic of m:. Thus a translation plane of characteristic zero has no translations of finite order other than ,. We remember that the prime.field FP of characteristic pis the rational field if p = 0 and the field with p elements if p > 0. We now associate with each translation plane m: of characteristic p the prime field FP of the same characteristic. FP is called the prime kernel ofm:. Our results wi11 be collected in the next three theorems. m: wi11 always denote a translation plane; FP will be its prime kernel. THEOREM 6

(i) Given any parallel pencil IT ofm:, T(IT) contains a subgroup isomorphic to the additive group Fp+ of Fr (ii) For every point O of~{, D(0) contains a subgroup isomorphic to the multiplicative group FP * of Fr By Theorem 6(ii), every translation plane of characteristic '# 2 contains nontrivial dilatations. The two parts of this theorem i11ustrate the close connection between the geometric structure of a translation plane and its prime kernel. This connection is elaborated in the following theorems. THEOREM

7

(i) There exists exactly one mapping (r, r) i--. Tr of Tx FP onto T satisfying T 1 = Tand (17) for every T e T and every r, s e Fr We call Tr the rth power of T . (ii) Given To E T(IT), To '# ,, the formula for all re FP

(18)

defines a monomorphism [i.e. an injective homomorphism] of FP + into T(IT). THEOREM 8 (i) Choose a point O ofm:. For every point P let Tp denote the translation which maps O onto P. Let r be any element of the multiplicative group FP * of FP. Then

Por

=

(OTp')

for all P

defines a dilatation or with the centre O.

(19)

3.3 THE PRIME KERNEL OF A TRANSLATION PLANE

45

(ii) Through r 1-+ 8,

for all r e FP *,

(20)

a monomorphism of FP * into D(0) is defined. In particular

8,. = 8,8. = 8.8,

for all r, s

e

FP *

(21)

and

8,-, = 8, - i

for all re F/.

(22)

Theorem 6 being a corollary of Theorems 7(ii) and 8(ii), only Theorems 7 and 8 need to be proved. This proof will be prepared by two lemmas dealing with integral powers of translations. Any translation T generates the cyclic subgroup of T which consists of the integral powers of T. Like every cyclic group it satisfies and ~m = (~)m

Tn+m = TnTm

(23)

for any integers n, m. Also, the group T being abelian, we have Tt"T2n

=

(24)

(T1T2)"

for every integer n and any translations From now on, let char ~ = p. Then

T1

and

T2 •

(25)

Tn=,pln,

if T "# , [a I b means a divides b. Thus O I n n = OJ. LEMMA 7.1 Let Ti, T2 denote two translations

-1=, and in distinct directions. Then

[PTt", PT/] Ii [PTi, PT2]

(26)

for every point P and every integer n such that p { n. PROOF By our assumptions T1 - 1T2 "# , and thus (T1 - 1T2)" (24) and (25). We have

PT1(T1 -IT2)

= T1 -"Ti" -I=

,; cf.

= PT2,

Hence [PT 1, PT2] is a trace of T1 - 1T2• Thus it also is a trace of (T1 - 1T2)"=T1 -"T2". Similarly (PTt")(T1 -nT/)

= PT2"

implies that [PT1", PT/] is a trace of T1 -nT2". Any two traces of this translation being parallel, we obtain (26). O LEMMA

7.2 Let Toe T, p

,r n. Then the equation (27)

has one and only one solution

T

e T.

46

TRANSLATION PLANES

Figure 3.13 PROOF (cf. Figure 3.13) Choose any point P and any translation T 1 =/; , which is not parallel to T0 • The line through P.r 1 parallel to [P-r 1n, Pr0 ] intersects [P, PT0 ] at a point Q. Construct the translation T which maps Ponto Q. By Lemma 7.1,

[Prt, PTn] II [PT1, PT] = [Pr1, Q] II [PTt, PTo]. Thus

[Pr/, P~] = [PT/, PT0 ]. Since Prn and Pr0 are both incident with [P, Q] = [P, PT0 ], this yields PTn = PT0 • As there is only one translation which maps P onto PT0 , we obtain Tn = To. If rn = rm, (24) implies Since p ,r n, this yields solution r. □

T- 1r'

= , or T' = r. Thus equation (27) has exactly one

PROOF OF THEOREM 7 To (i). We first define r' for every re FP. If p > 0, the prime field FP can be identified with the field of the residue classes of the integers modulo p. Put

fl= {m Im= n (modp)}.

By (25), T" is uniquely defined for every integer n by (28) This defines r' for every r e FP and every r e T. In this case, (17) follows at once from (28) and (23). Let p = 0. Thus FP is the rational field. Let r = m/n denote any rational number. By Lemma 7.2, there is to every r one and only one r 0 such that r 0 n = r'". We therefore try to define (29)

This definition is justified if it is independent of the representation of m/n as a quotient of two integers.

3.3

47

THE PRIME KERNEL OF A TRANSLATION PLANE

= m'/n', Then mn' = m'n and = Ton 'Tmm' = To nm' = Tomn' Tm' = 'Ton' .

Let m/n Tm

This equivalence now permits us the definition (29). Thus T' is defined for every r e F0 and every T e T. It remains to verify (I 7) for p = 0. Let r = m/n and s = m' /n' denote rational numbers. Put Thus _mn' +m 'n _

'T

-

'To

nn'

'T

'

m'

= 'T2 n' •

Hence

and therefore, by Lemma 7.2,

T 1T 2

= T 0 • Thus for any

T

e T.

For the proof of the second half of (17), write

Thus

Hence 'T2n'n

=

('T 2 n')n

and so T 2 =

=

To

(Tim')n

=

('Tin)m'

= (Tm)m' = 'T0 nn' = 'Ton'n,

and

for all 'T e T. This proves the first part of Theorem 7. To (ii). By (17), (18) defines a homomorphism of FP + into the group T(TT) of the translations parallel to T 0 • We have to verify that it is injective. Suppose r lies in the kernel of this homomorphism. If p '# 0, then r = n for some integer n, and implies p In or r = 0; cf. (27). If p = 0, r = m/n, then T' = T"' 1n = i implies by (29) that T"' = ,n = i and hence m = 0 and again r = 0. Thus the kernel of the homomorphism (18) is always equal to {0} . □ Before proving Theorem 8, we extend Lemma 7.1.

48

TRANSLATION PLANES

LEMMA

Then

8.1 Suppose the translations T 1, Tz, , are mutually distinct. Let re FP •.

for all P. PROOF Our assertion being trivial if T 1 and Tz have the same direction, we may assume that their directions are distinct. (i) Let p > 0. Since every r e FP * is equal to a residue class n with p { n, our statement foJJows from (28) and Lemma 7.1. (ii) Assume p = 0. Let r = m/n denote any rational number. Then there are translations -ri', T2' such that -r1 m = T 1'" and Tzm = Tz'". Hence by Lemma 7.1,

[PT{, P-r21 = [P-r/, P-rz'] II [P-r/n, P-rz'"] = [PT1m, P-r2m] II [PTi, P-r2] foraJJP. □ PROOF OF THEOREM

8

m:

To (i). Let re FP *. Obviously, 8, is a map of the set of the points of into itself. We prove that 8, is bijective by verifying that it has the inverse 8,-1, i.e. that (22) holds true: We have

Q

l:> = Po, = 0 'Tp r Ta = Tp r 'Tp = Ta ,-1 p = 0Tp = 0-rar-l = Q8,-1.

This yields

Q

= P8, P = Q8,-1. By Lemma 8.1,

[P8,, Q8,]

= [0Tp',

0-ra1 II [0-rp, 0-ra]

= [P,

QJ.

Hence by Chapter 2, Theorem 8, the bijection 8, of P can be extended to a homothety 8, of~- As

08, = 0-r0 ' = 0,' = 0, this homothety is a dilatation wittt the centre 0 . This proves (i). To (ii). Obviously, (20) defines a map of Fp* into the group D(0). By (19),

Hence by (17),

P8,s = 0-r/s = O(T/)5 = 0-rn/ = (P8,)8s = P8,8s for every P. This yields (21). In particular, our map (20) is a homomorphism of FP * into D( 0). It remains to prove that (20) is injective. Let r lie in its kernel. Thus 8, = , and

0Tp = P = P8, = 0-r/.

49

EXERCISES

Hence Tp = T/ and T/- 1 = T/Tp - 1 = i for all P. Thus r-1 Theorem 7(ii), and (20) is a monomorphism. D

= 0 or r = 1, by

COROLLARY 8.2 /f'iJl is a translation plane of characteristic p -:/= 2, there exists to every point 0 a reflection in 0.

By Theorem 8 (22), we have - 1 -:/= 1 and we obtain

PROOF

o_1 - 1 = o_1 or o_.2 = ,.

If p -:/= 2, then

if P-:/= 0. Hence

o_ 1

-:/=

i.

Thus

o_ 1 is an involution. This proves our corollary. □

EXERCISES 1 Prove that Statement (p) is equivalent to the following Statement (p'): Let /1 and / 2 be parallel and distinct; Pt, Pt', Qt, Qt'Ilt fork = l, 2; Then (P1 ', P2'] II [Qi, Qz] ; cf. Figure 3.14. 2 Let Ill: be any affine plane such that the diagonals of every parallelogram are parallel. Show that (p) holds true in Ill:. If Ill: admits a translation -r t, i, what is the order of -r?

3 Show that (d) implies the following Statement (s'): Let lo, Ii, /2 be parallel and mutuallydistinct;Po,Po', Qo Qo'Ilo;Pi, Q,Ili ;Pz, Qzl/2. Assume[Po, Pt] II [Qo, Q,], [Po', P1] II [Qo', Qi], [Po, P2) II [Qo, Q2). Then [Po', P2) II [Qo ', Qz) ; cf. Figure 3.15.

4• Prove that (d) is equivalent to Statement (r) : Let /1 = [Pi, Qi) II /2 = [P2, Q2]. Assume [Pi, P2) II [Qi, Qz], [Pi, P3) II [Q2, QJ), [P2, P3) II [Qi, Q3). Then [Qi, P3) 11 [P2, Q3); cf. Figure 3.16.

Figure 3.14

50

TRANSLATION PLANES

Figure 3.15

Figure 3.16

5 A set T of bijections of a set S onto itself is called [sharply simply] transitive if to every pair A, Bin S there exists [exactly] one-re Tsuch that A-r = B. Prove: Suppose the set Tis transitive and abelian. Then every bijection of S onto itself which commutes with every -r e T belongs to T, and Tis a sharply simply transitive group. 6 Let A be a translation plane. Prove that a collineation which commutes with every translation is itself a translation. 7 IT (p)rr holds in an affine plane, T(II) is abelian. 8 Elaborate the example on p. 42. 9 Let F denote the field of order pk; p > 2, k > l. Let p+ be the normal subgroup of F* consisting of the squ~res of all the elements '# 0. Define xoy

xy = { x"y

ifyeF+, otherwise.

51

EXERCISES

The points of 21 are the ordered pairs (x, y) of elements of F; the lines are the point sets (16). (i) Let n e F"---F+. Show that xP - nx yP - ny implies x = y and deduce that the equation xP - nx = c has, for any c e F, one and only one solution x. (ii) Prove that 21 satisfies the axioms Al-A3. (iii) The group of the translations parallel to the y-axis is linearly transitive and abelian. (iv) There are no translations -:/: , parallel to the x-axis. (v) Deduce that there are at least two non-isomorphic planes of order pk if p > 2, k > 1; cf. Chapter 1, Exercise 9 and Chapter 4, Exercise 7.

=

10 If an affine plane has reflections in each point it is a translation plane of characteristic -:/: 2.

11 • Prove Corollary 8.2 directly. 12 Let re Fp*• (i) Verify (T1T2)' = T1' Tz' and show that the mapping Tl-+T' defines an automorphism of T. (ii) Show that T' = 8, - 1 TS,. 13 (i) Show that the group T of the translations of a translation plane 21 is isomorphic to the additive group of a vector space over a prime field. (ii) Prove that this isomorphism induces a homomorphism of the group of the collineations of 21 into the group of the linear transformations of the vector space; cf. Exercise 12(ii). Determine the kernel of this homomorphism. (iii) Let 21 be finite. Show that the order of 21 is a power of char 21.

=

14 Let char ~ -:/: 2. If Q PT, the point PT½ is called the mid-point between P and Q. (i) Show that this concept is symmetric in P and Q. (ii) Let / II h. Prove that the locus of the mid-points of all the pairs P, Q with PI / and QI h is a line parallel to both.

4 Desarguesian planes

4.1 THE DILATATION GROUPS

An affine plane

~

D(O)

is called desarguesian if the following statement holds true.

['Affine Theorem of Desargues']. Let /0 , Ii, 12 denote three mutually distinct lines with a common intersection 0. The points Pk> Qk are incident with lk and distinct from O; k = 0, l, 2. Assume

STATEMENT (D)

[Po, Pi]

II [Qo,

Q1]

II W1,

Q2J;

and [Po, P2J

II [Qo,

Q2l•

(l)

Then [P1, P2J

cf. Figure 4.1.

Figure 4.1

(2)

4.1

THE DILATATION GROUPS

53

D(0)

1.1 Statement (D) holds true in any affine plane Q0 , Qi, Q 2 are collinear or if Pk = Qkfor some k.

REMARK

PROOF

if either P0 , Pi, P 2 or

See Chapter 3, Remark 1.1.

1 If the dilatation group D(0) is linearly transitive for every choice of 0, the plane is desarguesian,· cf. Chapter 3, Theorem 1.

THEOREM

PROOF We use the above notation. By our assumption there exists a dilatation 8 in D(0) such that P0 8 = Q0 • Since h8 !I h for every hand the traces of 8 are the lines through 0, P 18 is the intersection of [0, Pi] with the line through P0 8 = Q0 parallel to [P0 , Pi]. Thus P 18 = Q1 • Similarly P 2 8 = Q 2 and thus [Pi, P 2 ] 11 [Pi, P 2 ]8 = [P18, P2 8] = lQi, Q 2 ]. □

Let (0) 0 denote Statement (D) with the additional assumption that the lines 10 , Ii, 12 are incident with the given point 0. Our proof of Theorem 1 shows COROLLARY

1.2

If the group D(0) is linearly transitive Statement (0)0 will hold.

Before studying the converse of Theorem 1, we prove THEOREM

2 (D) implies (d).

COROLLARY

2.1 Every desarguesian affine plane is a translation plane.

2 Let 10 , Ii, 12 denote three mutually distinct parallel lines Suppose the points Pk and Qk are incident with lk (k = 0, 1, 2) and satisfy

PROOF OF THEOREM

[Po, P1J

II [Qo,

(1)

Q1J,

We have to prove [Pi, P2J

II [Qi,

Q2l•

In our proof we may assume that Pk#: Qi,; for k P0 , P1 , P2 nor Q0 , Q 1 , Q 2 are collinear. Assume

= 0,

1, 2 and that neither

(2') We first construct a configuration to which we shall later apply (D): Leth be the line through Q1 parallel to [Pi, P 2 ] and Qz' the intersection of hand [Q0 , Q2]. Put lz' = [P2 , Qz'), 0 = [lz', /1), and /0 ' = [0, P 0 ]. Finally, let Q 0 ' denote the intersection of /0 ' and [Q0 , Q 2 ]; cf. Figure 4.2. By our construction [P1, P2J

II CQ1,

Qz')

and [Po, P2J

II [Qo',

Qz'].

Thus we may apply (D) to the triangles P0 P1P2 and Q 0 'Q 1Qz' obtaining [Po, Pi] II [Qo', Q1l• Hence [Qo', Q1J II CQo, Q1J and Q1 I CQo', Q1J, [Qo, Qi).

54

DESARGUESIAN PLANES

Figure 4.2

Thus [Q 0 ', Qi]= [Q 0 , Qi]. This implies Q0 ' = Q0 and /0 ' = [P0 , Q0 ] = lo, Since O I /0 , / 1 , and /0 II /1 , we obtain /0 = /1 ; contradiction. □ This proof is not complete : we have not discussed the existence and uniqueness of the elements of our construction and we have not made sure that the assumptions of (D) are satisfied. The details of such verifications are easy but cumbersome. It may be instructive to go through them at least once. In later proofs we shall tend to omit the corresponding discussions. By our assumptions,

I; # lk pk =I- Qk,

if i =I- k; i, k = 0, 1, 2 ; k

= 0, 1,2;

I; =I- [Pk> Ph], [Qk> Qh], [P;, Pk] =I [P;, Ph],

(3) (4)

i, k, h = 0, 1, 2; k =I- h;

[Qi, Qk] =I- [Q;, Qh]

(5)

(6)

if i, k, h = 0, 1, 2; i, k, h mutually distinct. 1

We first prove the existence of the elements of our construction:

(i) Qi' exists: since [Q 0 , Q 2] II [P0 , P 2 ] ,It [P1 , P 2 ] II h, the lines [Q 0 , Q 2] and h must intersect.

4.1

THE DILATATION GROUPS

D(0)

55

(ii) Ii' exists: suppose P2 = Q/. Then P2 , Q2 I /2, [Q0, Q2], and /2 :/: [Q0, Q2J would imply P 2 = Q 2 ; cf. (5); contradiction. (iii) Ii' ,I!' /1 and thus 0 exists: suppose /2' II /1. Then Ii' 1112. Since P2 I//, /2, this would imply// = /2. As Q2 , Qi' I /2, [Q0, Q2], and /2 ¥= [Q0, Q2], we would obtain Q2 = Qi' and hence [P., P 2] II [Q., Qi') = [Q1, Q2], This contradicts (2'). (iv) /0 ' exists: since /0 and /1 are distinct parallel lines, P0 I /0 and 0 I /1 implies P0 ¥= 0. (v) Finally, Q0 ' exists: suppose /0 ' II [Q0 , Q2]. Then /0 ' II [P0 , P 2 ]. Since P 0 I lo', [P0 , P 2], this yields /0 ' = [P0 , P 2 ]. Thus O = [/1 , [P0 , P 2 ]]. Since /1 and /2 are distinct parallel lines and 0 I I., P 2 I /2, we have 0 ¥= P2 • Hence /2' = [0, P 2] = [P0 , P2]. The lines [P0 , P 2 ] and [Q0 , Q 2] being parallel and having the point Q 2 ' in common, they would have to be identical. This is excluded by (4). In order to verify that P0 P1P2 and Q0'Q 1Q/ are triangles which satisfy the assumptions of (D), we have to show: II

III

The lines /0 ', / 1 ,

/ 2'

are mutually distinct.

The points P0 , P., P 2 , Q0 ', Q1, Q 2 ' are distinct from 0.

To II (vi) Since P0 I /0 ' and P0 l / 1 , we have /0 ' :/: 11 • Similarly Ii' ¥= 11 • (vii) /0 ' :/: 12 ': suppose /0 ' = //. As P0 and P2 lie on the distinct parallel lines / 0 and /2 , we have P0 ¥= P 2 • The lines [P0 , P 2] and lo' = Ii' having the points P0 and P2 in common, we obtain // = [P0 , P 2 ]. Repeating the argument of (v), we arrive at a contradiction. To III (viii) By (iv), we have 0 ¥= P0 ; (v) yielded 0 ¥= P 2 • (ix) Suppose 0 = P 1 • Then 0 ¥= P 2 implies // = [0, P 2 ] = [P., P 2] II h. As Q2 ' I //, h, this would yield /2 ' = h. Hence P 1 , Q 1 I /1 //. Since /1 ¥= Ii', we would obtain P 1 = Q 1 • This contradicts (4). (x) Suppose 0 = Q 1 • Then//, hI Q., Q2 '. Thus either Q 1 = Qi' or /2 ' = h. If Q 1 = Qi', this point would be incident with [Q0, Q2]. Thus [Q0, Qi] = [Q0 , Q 2], contradicting (6). Suppose then// = h. Then [P1 , P 2 ] II h = 12' and P 2 I /2 ', [P1 , P 2 ] would yield [P1 , P2 ] =//and O = [I/, /1 ) = P 1 • Hence P 1 = Q., contradicting (4). (xi) 0 = Q 2' would imply 0, Q 1 I h, /1 • Since O :/: Q 1 by (x), this would yield / 1 = h II [P., P 2 ]. As P 1 I /1 , [P1 , P 2 ], we would obtain [P., P 2 ] = / 1 , contradicting (v). (xii) _Suppose 0 = Q0 '. Since 0 ¥= Qi', this would imply Ii' = [0 1 Q2 ') = CQo', Qi'] = CQo, Q2] II [P0, P 2]. The lines Ii' and [P0, P 2] having P2 in common, we would obtain [P0, P 2] = Ii' = [Q0, Q2]; in particular, P0 = [/0, [P0, P 2]] = [lo, [Q0 , Q 2]] = Q0 • This contradicts (4). IV In conclusion, we elaborate the application of (D) in our proof: applying (D) to P0 P1P2 and Q0'Q1Q/ we obtained [Q0', Q 1] = [Q0, Qi]. Hence Q0 , Q0 ' I

56

DESARGUESIAN PLANES

CQo, Q1], [Qo, Q2l• By (6), [Qo, Q1] =/: [Qo, Q 2]; therefore Q 0 = Q 0 '. Also Q0 , P 0 I 10 , / 0 ' and Q0 =/: P 0 , by (4). Hence 10 = /0 '. Thus O I 10 , 11 while 10 and 11

are distinct parallel lines by (3); contradiction.

We can characterize the desarguesian planes : 3 An affine plane is desarguesian if and only D(0) is linearly transitive for every choice of 0 .

THEOREM

if the dilatation group

On account of Theorem I, we have to prove: suppose the affine plane ~ = (P, L, I) is desarguesian. Then each group D(0) is linearly transitive. Suppose the points 0, P 0 , Q 0 are collinear and mutually distinct. We try to construct a dilatation 8 with the centre O such that (7)

cf. Figure 4.3. Let /0 = [0, P 0 ] = [O, Q 0 ] . Choose a point P 1 not incident with 10 and put 11 = [O, Pi]. We now define the mapping 8: P---+ Pas follows: (i) If PI /0 , intersect I= [O, P] with the line through Q 0 parallel to [P0 , P]. If Q is the intersection, put PS = Q. In particular, Q 1 = P 18 is the intersection of 11 with the line through Q 0 parallel to [P0 , P1] . (ii) If P 0 ' I /0 , let Q 0 ' = P 0 '8 be the intersection of /0 with the line through Q 1 parallel to [P1 , P 0 ']: in particular 08 = 0. We have now defined a map 8 of P into itself with the one fixed point O and

Figure 4.3

4.2

THE SHEAR THEOREM

57

which satisfies (7). Interchanging the P's and Q's we obtain another map 8' of P into itself such that

P8

=

Q ¢> Q8'

= P.

Thus 8 is a bijection with the inverse 8- 1 = 8'. Continuing our proof we next show [P, P'] 11 [P8, P'8J

for all P =IP'.

This is obvious if O I [P, P']. From now on we may assume that 0, P, P' are not collinear. Put P8 P'8 = Q'.

(8)

= Q,

Case (i) Neither P nor P' is incident with 10 • Then

[P0 , PJ II (Q0 , Q] and [P0 , P'J II (Q0 , Q']. Thus (8) follows from Statement (D). Case (ii) From now on we may assume that, for example, P' = P 0 ' is incident with /0 • Thus PI /0 • Put Q0 ' = P0 '8. (1X) PI / 1 • By Case (i), [P1 , PJ II (Q 1 , Q]; and by our construction [P0 ', Pi] II (Q0 ', Q1]. Hence (D) yields

[Po', PJ II (Qo', QJ. (P) The point P

= Pi' is incident with /1 • By Chapter I, Theorem 7, there is a line /2 through 0 distinct from /0 and /1 and a point P2 on 12 distinct from 0. Put Qi' = P/8 and Q2 = P28. By the preceding cases,

[Pi', P2J II CQ1', Q2J and [Po', P2J II [Qo', Q2l• Hence by (D) [Po', Pi'] II (Qo', Qi']. This proves (8). We shall return to Case (ii) (P). By Chapter 2, Theorem 8, (8) implies that our bijection 8 can be completed to a homothety 8 of~(. As it has the fixed point 0, this homothety belongs to D(0). It satisfied (7).0 Our proof and Corollary 1.2 imply COROLLARY

3.1 D(0) is linearly transitive if and only if Statement (D) 0 holds true.

4.2 THE SHEAR THEOREM

We formulate STATEMENT

(S) ['Shear Theorem']. Let 11 =I 12 • Suppose P 1 , Pi', Q1, Qi' are

58

DESARGUESIAN PLANES

incident with 11 but not with 12 , and P2 , Pz', {2 2 , Qz' are incident with 12 but not with 11 • Assume [Pi, P2]

II CQ1, Q2],

Then [Pi', Pz')

11

[Qi', Qz'];

cf. Figure 4.4. REMARK

4.1 Statement (S) holds in any office plane unless

pk =f, P/,

Qk =f, Q/, pk =f, Qk, P/ =f, Qt'

cf. Chapter 3, Remark 4.1 .

fior k = l ' 2·'

(9)

REMARK 4.2 For 11 1112 we obtain the Little Shear Theorem (s). By Chapter 3, Theorems 3 and 4, it follows from (d). Hence (D) implies (s) by Theorem 2.

If Ii and /2 are not parallel, we come back to case (ii)(~) of the proof of (8). This yields THEOREM

4 (D) implies (S).

Let (S)0 denote Statement (S) with the additional assumption that /1 and /2 are incident with the given point 0. Then the proof of (8) yields COROLLARY

4.3 (0)0 implies (S)0 •

We next wish to prove the rather surprising result that the shear theorem (S) implies the affine theorem of Desargues (D). Assume then that (S) holds true in our affine plane. The lines and points lk, 0, Pk> Qk satisfy the assumptions of Statement (D). We wish to prove formula (2). We may assume that Pk =f, Qk and that neither P0 , P 1 , P2 nor Q0 , Q 1 Q 2 are collinear; k = 0, 1, 2. (i) We first prove that, if (2) is false, the intersection R 2 of[P1 , P 2] and [Q 1 , Q 2] is incident with /0 ; cf. Figure 4.5.

Figure4.4

4.2

59

THE SHEAR THEOREM

Figure 4.5

Assume R 2 l 10 • By its definition, R 2 is not incident with / 1 or 12 • The line through R 2 parallel to / 2 intersects /0 at some point O' and the line through R 2 parallel to [P0 , P 2 ] and [Q0 , Q 2 ] intersects /0 at a point Ro, Finally, let Pi' and Q 1' denote the intersections of [P1, P 2] and [Q1, Q 2], respectively, with the line through O parallel to /1 • Since

(0, P1] II [O', P1'],

[Po, P2]

II [Ro,

R2],

Statement (S) yields [Po, P1J

II [Ro, P/J.

Symmetrically, we obtain [Q0 , Q 1] [Ro, P1'J

II [Po, P1] II lQo, Qi] II [Ro,

II [R0 , Q 1 ']. Hence Qi']

and

= [Ro, Qi']. As the points Pi' and Qi' are the intersections of this line with the line through [Ro, Pi']

O' parallel to Ii, this yields P 1' = Q 1' . This point is incident with both [Pi, P 2] and [Qi, Q 2]. Hence Pi' = Qi' = R2 • Thus we would obtain

60

DESARGUESIAN PLANES

Figure4.6

Hence [P0 , P 2] = [P0 , P 1] and the points P0 , P 1 , P2 would be collinear; contradiction. (ii) We can now prove that (S) implies (D); cf. Figure 4.6. Assume that (2) is false. Then by (i), the lines [P1 , P2 ] and [Q 1 , Q 2 ] intersect at a point R 0 incident with 10 • The line through R 0 parallel to [P0 , Pk] intersects lk at a point Rk. Thus k

= 1, 2.

Since P0 , Pi, P2 are not collinear, R0 , Ri, R 2 cannot be collinear either. Thus [R 1 , R 2 ]. If [R 1 , R 2] and [Pi, P2 ] were not parallel, their intersection would be incident with 10 by (i). Thus it would be equal to R0 • As this has been excluded, we obtain

Ro l

and symmetrically

Thus we arrive at formula (2) after all;

contradiction. □

Combining our result with Theorem 4, we obtain 5 The affine theorem of Desargues (D) and the Shear Theorem (S) are equivalent.

THEOREM

4.3

THE LINEAR TRANSITIVITY OF THE GROUPS

4.3 THE LINEAR TRANSITIVITY OF THE GROUPS

A(a)

61

A(o)

Our goal is the following result. 6 The affine plane mis desarguesian linearly transitive for every choice of the axis a.

THEOREM

if and only if the group A(a) is

On account of Theorem 3, this assertion is a corollary of the following theorem. THEOREM 7 Let m= (P, L, I) be any affine plane. Let a be any line in m. Then the following statements are equivalent: (i) For every O I a, the group D(O) is linearly transitive. (ii) The group H(a) of the homotheties with the fixed line a is linearly transitive. (iii) The group A(a) of the affinities with the axis a is linearly transitive; cf. p. 32.

7 We exclude the trivial case of the plane of order two. (i) implies (ii): Let P, Q be two points not incident with a. Choose any point 0 I a. Then there exists a point O' =I Oona such that [O, P] ,It [O', Q]. Let R be the intersection of [O, P] and [O', Q]. Then R J'. a and there are dilatations Se D(O) and S' e D(O') such that PS = Rand RS' = Q. Hence 1J = SS' e H(a) andP11 = Q. If P, QI a, choose any point O on a distinct from P and Q. Then there is a Se D(O) such that PS = Q. (ii) implies (iii): Given two points P 0 and Q0 not incident with a, we have to construct an affinity oc with the axis a such that PROOF OF THEOREM

(10)

cf. Figure 4.7. By our assumption (ii) and by Chapter 2, Theorem 14, there is to any pair of points P, Q not incident with a, one and only one homothety in H(a), say 1JPQ which maps P onto Q. Thus

Figure4.7

62

DESARGUESIAN PLANES

TJPQ

e

H(a).

Define the map oc of P into itself by Poc

= QoT/PoP

if Pl a,

Poc

=P

if PI a.

(1 I)

Interchanging the P's and Q's, we obtain a second map oc' of P into itself. By our definition Poc = Q Qoc'

=P

for all P.

Thus"' is a bijection of Ponto itself with the inverse «- 1 Before proving that oc preserves collinearity, we show

= oc'.

(*) If P f= Q; P, Q l a, then the lines [P, Q], [Poe, Qa.], a

belong to the same pencil.

As the homothety T/PoP - i T/PoQ Hes in H(a) and maps Ponto Q, we have TJ PoP

-1

TJ PoQ = TJ PQ•

Hence (Poe)TJPQ = (QoTJp0 p)(TJPoP -l

T/P0 Q)

= QoT/PoQ = Qoe.

Thus the lines [P, Q], [Poe, Qa.] and the fixed line a are traces of the same homothety TJ PQ· In particular they belong to the same pencil. This proves (*). Let P, Q, R be three mutually distinct collinear points. We wish to show that Pa., Qa., Roe are collinear. We may assume that not more than one of the points P, Q, R is incident with a. If RI a, then R = [a, [P, Q]]. Hence by (*), RI [Poe, Qa.]. Thus Poc, Qoe, R = Ra. are collinear. If none of the three points P, Q, R is incident with a, then (*) implies that [P, Q], [Poc, Qa.], a and similarly [P, R], [Poe, Roe], a belong to the same pencils. As [P, Q] = [P, R] f= a, these two pencils are identical and [Poe, Qa.] = [Poc, Roe]. Thus Poe, Qoe, Ra. are again collinear. By Chapter 2, Theorem 3, our bijection oe can be completed to a collineation "'· By (11) it has the axis a and maps P0 onto Q0 • (iii) implies (i): Assume (iii). On account of Corollary 3.1, it is sufficient to prove Statement (D) 0 for every O I a. Let 10 , / 1 , / 2 denote three mutually distinct lines through 0 . We assume first that they are distinct from a. Suppose the points Pk, Qk satisfy the assumptions of Statement (D)0 ; k = 0, 1, 2; cf. Figure 4.1. There are two affinities oe and f1 with the axis a such that Po«= Pi,

4.3

THE LINEAR TRANSITIVITY OF THE GROUPS

A(a)

63

If we can show that Qi/3

= Qz,

(12)

both [Pi, P 2] and [Q 1, Q2] will be traces of /3; thus they will be parallel. We have

l0 ex.

= [O, P 0 ]ex. = [Oex., P0 ex.] = tO, Pi] = /1 •

Thus Q0 I /0 implies Q0 ex. I /1 • The traces of ex. being parallel, Q0 ex. must be the intersection of 11 with the line [Q 0 , Qi] through Q0 parallel to [P0 , P 1], i.e.

Qi

= Qoex..

Since a./3 also is an affinity with the axis a, we obtain in the same way Qz = Q0 ex.f3. This yields (12). If a= /0 , apply only /3. The points P0 and Q0 now being fixed, (1) implies

[Qo, Q2] II [Po, P2]

= [Po, Pi]/3 II [Qo, Qil/3 = [Qo, Q1/J]. Since Q 1/3I 11/3 = /2 , we arrive again at (12). If a = 12 , we use only a.. Since P2 and Q 2 are fixed, we have [Qi, Q2] = [Qoor, Q2a.] = [Qo, Q2]a. II [Po, P2Jex. = [P1, P2]. The case a

= 11 is symmetric to the case a = 12 • □

We have not defined the linear transitivity of the group of all the collineations. The following theorem contains a first result. A more complete treatment will be given in Chapter 6.

8 Let P1P2 P 3 and Q 1 Q2 Q3 denote two triangles in a desarguesian plane. Then there are three axial affinities a. 1 , ex. 2 , ex. 3 such that

THEOREM

(13) Thus any given triangle can be mapped by a suitable collineation onto any other triangle. Choose any line a 1 incident neither with P 1 nor with Q 1. Since the group A(a 1) is linearly transitive by Theorem 7, there exists ex. 1 e A(a 1) such that P 1ex. 1 = Q1. The image points Q1, P2 a.i, P 3 ex. 1 of the non-collinear points Pi, P2 , P 3 are again non-collinear. Next choose a line a2 through Q 1 which is incident neither with P 2 ex. 1 nor with Q 2• Then there is a. 2 e A(a2 ) which maps P2 ex. 1 onto Q 2 • Since Q 1 I a2 , we have PROOF

The images Qi, Q2 , P 3 ex. 1a. 2 of Qi, P2 ex.i, P 3 ex.i, respectively, are again noncollinear. Finally put a 3 = [Q 1, Q 2]. Then P 3a. 1ex. 2 and Q 3 are not incident with a 3, and

64

DESARGUESIAN PLANES

there is an affinity oc 3 in A(a3 ) which maps P 3 oc 1 oc 2 onto Q3 • As Q 1 and Q 2 are fixed points of oc3 , oc 1 oc 2 oc 3 satisfies (13). □

EXERCISES 1 The group H of the homotheties of the plane 2l is said to be linearly transitive if there is a homothety mapping P1 onto Q1 and P2 onto Q2 whenever [Pi, P 2 ] and [Qi, Qz] are parallel; cf. Chapter 2, Theorem 12. Show that an affine plane is desarguesian if and only if the group of its homotheties is linearly transitive. 2 The plane 2l is desarguesian if and only if it is a translation plane and there is a point 0 such that D(0) is linearly transitive. 3 Suppose that D( 0) is linearly transitive and that there is a collineation °' such that 0 f:. Ooc. Put a = [0, 0oc]. Then H(a) is linearly transitive. [Thus Theorem 2 can also be deduced from Theorem 3.] 4 Suppose D(0) is linearly transitive and there are two collineations °'• f3 such that 0, Otx, 0{3 are not collinear. Then 2l is desarguesian. 5 Suppose that A(a) is linearly transitive and that there is a collineation °' such that a°' f:. a. Then 21 is desarguesian.

6 A shear is an axial affinity whose fixed lines are parallel to the axis. Let 21 be desarguesian. Prove (i) The product of two shears with axes in the parallel pencil Il is either a shear with an axis in Il or a translation in the direction of n. (ii) Every translation in the direction of Il is the product of two shears with axes in Il. (iii) The union of the translation group T{Il) with the set of the shears with axes in n is a group. (iv) This group is abelian. 7 The field F, the group p+, and the operation x o y are defined in Chapter 3, Exercise 9. The points of 2l are again the ordered pairs (x, y) of elements of F. However, the lines are the sets {(x,y)lx=c}

and

{(x,y)ly=x

0

m+d} .

(i) Check that ~ is an affine plane. (ii) Show that 21 is a translation plane. (iii) Suppose r lies in F but not in the prime field of F; thus rP f:. r. Prove that there is no dilatation with the centre O which maps the point (1, 0) onto the point (r, 0). Thus 21 is a non-desarguesian translation plane.

EXERCISES

65

8* Show that (D) is equivalent to the following Statement (D): Let I., lz, /3 denote three mutually distinct parallel lines. Let P,PzP3 and Q,QzQ3 be two triangles such that P,, Q,I I,; i = 1, 2, 3. Then either corresponding sides of the two triangles are parallel or there is a line a such that any pair of corresponding sides of the triangles together with a belong to the same pencil.

5 Pappus planes

We have narrowed the class of all the affine planes first to that of the translation planes, then further to that of the desarguesian planes. In this chapter we introduce a subclass of the latter. (P) ['Affine Theorem of Pappus']. Suppose the lines 11 and 12 intersect, say at 0. The points Ak, Bk, Ck are incident with lk and distinct from 0; k = l, 2. Let

STATEMENT

[A1, B2] II [B1, A2],

(l)

Then

[Bi, C2] II [C1, B2];

(2)

cf. Figure 5.1. An affine plane is called a Pappus plane if the affine theorem (P) of Pappus is satisfied. We note: I.I Statement (P) always holds if A1, Bi, C1 or A 2 , B 2 , C2 are not mutually distinct; cf. Chapter 3, Remark 2.2.

REMARK

THEOREM PROOF

I Every affine Pappus plane is desarguesian.

By Chapter 4, Theorem 5, it is sufficient to deduce Statement (S) from

Figure 5.1

67

PAPPUS PLANES

(P). This can be done by repeating the proof of Chapter 3, Theorem 4, merely replacing the parallel lines 11 and 12 by intersecting lines and Statement (p) by (P). We can now reformulate the definition of Pappus planes.

(P ') Let 11 and 12 denote two distinct lines. Suppose the points Ak, Bk. Ck are incident with lk but not with the other line; k = l, 2. Then (1) implies (2). If 11 ,lt 12 , (P') becomes (P) while for 11 1112 we obtain (p).

STATEMENT

COROLLARY

ment (P').

1.2 An affine plane is a Pappus plane

if and only if it satisfies State-

Obviously, (P') implies that the given plane is a Pappus plane. Conversely, being desarguesian, a Pappus plane is a translation plane, by Chapter 4 Corollary 2.1. Hence by Chapter 3, Theorem 2, it satisfies not only (P) but also (p) and therefore (P'). □

PROOF

We wish to describe Pappus planes in terms of their dilatation groups. Given a point 0, (P)0 will denote Statement (P) if the lines 11 and 12 are required to intersect at 0. 2 If the group D(0) is linearly transitive and abelian, (P)0 holds true. Conversely (P)0 implies that D(0) is abelian.

THEOREM

PROOF

(i) Let D(0) be linearly transitive and abelian. The lines lk and the points At, Bk, Ck shall satisfy the assumptions of Statement (P)0 ; k = l, 2. In particular, we may assume (1). Since D(0) is linearly transitive, there are dilatations o, o' with centre Osuch that Then by our assumptions, A 2 o = B 2 and symmetrically C2o' = A 2 ; cf. the proof of Chapter 4, Theorem 1. Hence C2 oo' = C2 o'o = A 2 o = B 2 • As B 1 oo' = A 1o' = C1 and oo' e D(0), we obtain [B1, C2 ]

II [Bi,

C2]oo' = [B1oo', C2 oo'] = [C1, B 2 ].

This yields (2) and proves (P)0 • (ii) Conversely, assume (P)0 • Leto and o' be dilatations with the centre 0. We have to show oo'

= o'o.

Let 0B1 C2 be a triangle, 11 = [O, B 1], 12 = (0, C2 ]. Put

(3)

68

PAPPUS PLANES

Thus Ak, Bk, Ck are incident with lk and distinct from O; k Since 8 and 8' map a line onto a parallel line, we have

=

l, 2.

[B1, A2] II [B1, A2 ]8 = [Ai, B2] and [A1, C2] II [Ai, C2]8' = [A2, C1], Hence by Statement (P)0 ,

[Bi, C2] II [B2, C1l•

(4)

Since 88' and 8'8 map the line [B1 , C2 ] onto the parallel line through B 1 88' A 18' = C 1 and C 2 8'8 = A 2 8 = B 2 , respectively, (4) implies

=

[B1, C2]88' = [Bi, C2]8'8 = [B2, C1]. In particular, B 18'8 is the intersection of this line with Ii, i.e.

By Chapter 2, Theorem 22, there is only one dilatation with the centre 0 which maps B1 onto C1 • Hence the last relation implies (3). O Theorems 2 and 1 yield the group theoretic description of Pappus planes for which we have been looking: COROLLARY 2.1 An affine plane is a Pappus plane if and only if all the groups D(0) are linearly transitive and abe/ian.

We can now complete the diagrams of p. 42 to a diagram which contains the principal results of the last three sections:

Pappus planes:

Desarguesian planes: { Translation planes:

Every D(0) lin. trans. -(P) and abelian Every D(0) lin. trans. Every A(a) lin. trans.

T linearly transitive

l

)

(D)-(S)

-

(d) ➔ (p) ➔ (s)

l

Diagramm

In Chapter 3, Exercise 9, finite planes were discussed which were not translation planes. In Chapter 4, Exercise 7, we had examples of finite nondesarguesian translation planes. In Chapter 6, we shall meet desarguesian planes which do not satisfy Statement (P). However, such planes are necessarily infinite.

69

APPENDIX APPENDIX

By Theorem I, we have THEOREM

l' Statement (P') implies Statement (D).

We wish to give a direct proof, cf. Figure 5.2. Suppose the lines lk and the points, 0, Pk, Qk satisfy the assumptions of Statement (D). Thus in particular

[Po, P1] II [Qo, Q1] and [Po, P2] II [Qo, Qz].

(5)

We have to show that (6)

By Chapter 4, Remark l.l we may assume that P0 , P1 , P 2 and Q0 , Qi, Q 2 are not collinear and that Pk ::f, Qk fork = 0, l, 2. If both [P1 , P 2 ] and [Q 1 , Q 2 ] are parallel to 10 , (6) is trivial. Thus we may assume that, e.g., (7)

The following construction will enable us to apply (P'). We shall not discuss the existence and non-incidence of the various points and lines; cf. the remark in the proof of Chapter 4, Theorem 2. Let S denote the intersection of [P0 , P 2 ] with the line through O parallel to [P1, P2 ]. Let R be the intersection of / 1 with the line through P2 parallel to 10 • Then

s

Figure 5.2

70

PAPPUS PLANES

[O, S]

II [P1, P2] and

[O, P 0]

II [R, P 2].

Hence by (P')

[R, S] II [Pi, P 0]. The line h

(8)

= [S, Q0 ] intersects [P2 , R] at some point T. Then

II lQo, Q2] and

[S, P2]

lQo, O]

II [T, P 2].

Thus (P') implies [O, S]

II [Q2,

(9)

T].

If Q 1 I h, (8) implies h = [Qo, Qi]

II [Po, P1] II [S, R].

Hence h = [S, R]. As this line intersects [P2 , T] both at Rand at T, this yields R = T. Since h intersects /1 both at R and at Q 1, we obtain R = T = Q 1• Hence by (9), [Pi, P2]

II [O, S] II CQ2, TJ = CQ2, Qi],

This yields (6). From now on we may assume Q 1 l h. Then [O, Q0 ]

II [R,

T] and

[R, S]

II [Qi,

Q 0 ],

Statement (P') now yields [O, S]

11

[Qi,

TJ.

(10)

By (9) and (IO), [T, Q1] [Qi, Q2J

=

[T, Q1J

11

[T, Q2]. Hence

II [O, S] II [Pi, P2J,

This completes the proof of (6). □

Figure 5.3

EXERCISES

71

EXERCISES 1 Deduce (p) directly from (P); cf. Figure 5.3. 2 Show that every dilatation in a Pappus plane is the product of two suitable axial affinities. 3 Every translation plane contains Pappus subplanes of finite or countable order.

6

Co-ordinates in desarguesian planes

6.1 CO-ORDINATE PLANES

Let V be a two-dimensional right vector space over the skew field F and let 0, A, B denote elements of V. By means of V we can construct a desarguesian plane. Our points will be the vectors of V. A line is a point set I= A+(D) = {A+Dx Ix eF},

{I)

D corresponds to a 'direction vector' of/. Two linesareequa/ if they are identical point sets. Let L denote the set of the lines. The point Pis said to be incident with the line / if P e /. Thus two lines are parallel if, considered as point sets, they are either identical or disjoint. We have now defined a triplet ~

= (V, L, e).

In this plane, lines have been defined as point sets. Nevertheless, it was important that we did not do so in the first chapter. We shall later complete every affine plane to a projective plane, and there will be complete symmetry between points and lines in such a plane : If its points are reinterpreted as lines and its lines as points, another projective plane is obtained. We prepare our discussion by the following remark. LEMMA

I.I Let

I= A+(D), I'= A'+(D').

Then (i) (D) = (D') implies 1111'; (ii) if (D) =/: (D'), then I and I' have exactly one point in common. [Here (D) = {Dx Ix e F}.] PROOF

(i) Let (D) (D'); thus /' = A'+ (D). Obviously, the lines / and /' are either disjoint or identical. Thus / II/'. ::ic

6.1 CO-ORDINATE PLANES

73

(ii) Let (D) =/: (D'). Then D and D' are linearly independent. Thus they form a base of V. The equation

A+Dx

= A'+D'x'

(2)

in the unknowns x and x' is equivalent to

A-A'= D(-x)+D'x'. Since A -A' can be written as a linear combination of D and D' in one and only one way, the last equation has one and only one solution x, x'. The same applies to (2). Thus / and /' have exactly one point in common. In particular, they are not parallel. D LEMMA

1.2

m= (V, L, e) is an affine plane.

To Axiom AI: Let P and Q be two distinct points. Then Q-P =/: 0 and the line P+ (Q-P) passes through both P and Q. By Lemma 1.1, there is not more than one line incident with both P and Q. To Axiom A2: Given the point P and the line / = A+ (D), Lemma 1.1 implies that the line P+ (D) is parallel to/ and passes through P and that it is the only line with both of these properties. To Axiom A3: If the vectors A and B are linearly independent, the three points 0, A, Bare not collinear. □ PROOF

THEOREM l m= (V, L, e) is a desarguesian affine plane. We call it the co-ordinate plane over the skew field F.

We have to show that the group D(C) is linearly transitive for any choice of CinV. Suppose C, P, Qare collinear and mutually distinct. We have to construct a dilatation 8 with the centre C such that P/5 = Q. Since Q lies on the line {C + (P- C)x I x e F} through C and P and since Q =/: C, there exists d =/: 0 in F such that

Q = C+(P-C)d. Define the maps 8 and 8' of V into itself through

xa =

C+(X-C)d,

Y/5' = C+(Y-C)d- 1 •

Then

xa =

Y ¢> Y-C

or

xa =

= (X-C)d¢> X-C = (Y-C)d- 1

Y/5' = x. Thus 8:V-+ Vis bijective and has the inverse a- 1 = 8'. If X=/: X', then [X, X'] = X+(X'-X) and [X/5, X'/5] = X8+(X'8-~). Y-

74

CO-ORDINATES IN DESARGUESIAN PLANES

Here

X'S-n

= (C+(X'-C)d)-(C+(X-C)d) = (X'-X)d.

Hence by (2), (X'S-n) = (X'-X) and by Lemma l.1 en, X'S] II [X, X']. By Chapter 2, Theorem 8, our bijection S can be extended to a homothety S of ~- As it has the fixed point C, it belongs to D(C). Finally PS = Q. □ THEOREM PROOF

2 ~

= (V, L, e) is a Pappus plane if and only if Fis afield.

Let Sand S' be two dilatations with the centre C; say,

n = C+(X-C)d,

n' = C+(X-C)d'.

Then ns' = C+(n-C)d' = C+(X-C)dd' and n·s

= C+(X-C)d'd.

Thus SS' = S'S if and only if dd' =

d'd. □

6.2 CO-ORDINATES IN DESARGUESIAN PLANES

Our next goal is the theorem that every desarguesian plane is isomorphic to a plane (V, L, e) over some skew field F. Let~= (P, L, I) be a desarguesian affine plane. Choose two distinct points 0 and E of~; I= (0, E]. Put F= {PIPI /}

and let A, B, C, .. . denote elements of F. Since ~ is, in particular, a translation plane, there is to each A exactly one translation which maps O onto A. We denote it by -rA.• Thus O-r,4.

=A

for all A eF.

(3)

Similarly if A ,f, 0 there is one and only one dilatations.= S,4. with the centre 0 which maps E onto A: £8"'

=A

for all A

e F"'-. {O}.

(4)

Define (5)

6.2

75

CO-ORDINATES IN DESARGUESIAN PLANES

= E8,.8B if A:/: A·O = O·A = 0 . A·B

0, B :/:

o,

(6) (7)

By (5) and (3),

A+B = ATs,

(8)

Furthermore, OT,.+s = A+B = OT,.Ts. Since T"Ts is a translation, Chapter 2, Theorem 16 implies (9)

By (6), (7), and (4), if B :/: 0.

(10)

If A and B are distinct from 0, (6) implies A· B :/: 0 . By (4) and (6), ES,..a = A · B = ES,.s a• Hence by Chapter 2, Theorems 21 and 22, if A :/: 0, B :/: 0 . LEMMA

(11)

3.1 F is a skew field.

PROOF

(i) By (9), the mapping of each T,. onto the corresponding point A is an isomorphism of the abelian group of the translations in the direction of I onto F. Thus Fis an abelian group under addition. For the convenience of the reader, we elaborate this argument. By (9) and Chapter 2, Theorem 19, Hence by (3),

A+B = OTA+B

= OTB+A = B+A.

Similarly by (9), T(,i+B)+C

=

T,i+BTc

= TATBTc

=

TATB+C =

TA+(B+C)

and thus

(A+B)+C = A+(B+C). Since To

= ,, (9) implies for all A

Given A, put B

e F.

= OT,. - 1 • Then T8 = T" - 1 and

(ii) The analogous arguments show that F satisfies the multiplicative laws of a skew field.

76

CO-ORDINATES IN DESARGUESIAN PLANES

(iii) It remains to prove the distributive laws

(A+B)·C

= A·C+B·C

(12)

= C·A+C·B.

(13)

and

C ·(A+B)

We may again assume that C is distinct from O.

Proof of (12). By Chapter 2, Theorem 17, 8B - 1-rA8B is a translation. By (3) and (10),

oaB- 1-r...aB = o-r...aB = A8B = A ·B = o-rA•B• Hence 8B - 1 -rA8B = -rA•B and = 8c - 17..t+a3c

7 (..t+B)C

= 8c - 17..t7 B8c =

8c - 17..t8c 8c - 17B8c

This yields (12). For the proof of (13), we choose a line a through O distinct from /. By Chapter 4, Theorem 6, the group A(a) of the affinities with the axis a is linearly transitive. Thus there is to every Ae F, A =/: 0 one and only one a e A(a) such that Ea = A. Write a = a..4., Thus

Ea..t = A

for all A e F"-._ {O}.

By Chapter 2, Theorem 29, we have aB8A AaB

= E:8..4.as =

Eas8A = B8A

(14)

= 8AaB. Hence

= B·A.

This formula yields since as - 1-r..ta s is a translation which maps O onto

Oas - 1-rAas = 0-rAas = Aas = B· A = 0-rs•..t· We now obtain 7

C•(A+S)

= ac -1 7 ..t+sac = ( ac -1 T..4.ac)(ac -1 Taac)

This completes the proof of our lemma. TIIEOREM 3 PROOF

Every desarguesian plane is isomorphic to a plane over a skew field.

Given the desarguesian plane A = (P, L, I), let

I= [O,E]eL, F= {PIPI/}.

6.2

CO-ORDINATES IN DESARGUESIAN PLANES

77

Define addition and multiplication in F by (5)-(7). Thus F becomes a skew field. The ordered pairs

A=

(Ai, A2)

of elements of F form a two-dimensional right vector space V over F if we define

A1, A2 )+(B1, B2)

= (A 1 +Bi, A2 +B2 )

(and

(A 1 , Az)C

= (A 1 ·C, Az·C).

By Theorem 2, V determines a desarguesian plane

9I = (V, L, e) where the elements of L are the point sets A+

= {A+B·CI CeF}.

We wish to construct a collineation of fil onto 9I. (i) Choose two lines a 1 and Oz through O distinct from one another and from /. Given any point A in fil, the lines through A parallel to Oz and a 1 intersect/ at two points A 1 and Az, respectively. Then A