Contributions to the theory of distance regular graphs

Citation for published version (APA):Lambeck, E. W. (1990). Contributions to the theory of distance regular graphs. Technische UniversiteitEindhoven. https://doi.org/10.6100/IR340078

DOI:10.6100/IR340078

Document status and date:Published: 01/01/1990

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CONTRIBUTIONS TO THE THEORY

OF. DISTANCE REGULAR GRAPHS

E.W. LAMBECK

CONTRIBUTIONS TO THE THEORY OF DISTANCE REGULAR GRAPHS

CONTRffiUTIONS TO THE THEORY

OF DISTANCE REGULAR GRAPHS

PROEFSCHRIFf

ter verkrijging van de graad van doctor aan de Technische Universiteit Eindhoven, op gezag van

de Rector Magnificos, prof. ir. M. Tels, voor een commissie aangewezen door het College van Dekanen in het openbaar te verdedigen op dinsdag 6 november 1990 te 16.00 uur

door

ERNST WILLEM LAMBECK geboren te Siddeburen (gemeente Slochteren)

druk: wibro dissertatiedrukkerij, helmond.

Dit proefschrift is goedgekeurd door de promotor prof. dr. A.E. Brouwer

Aan mien olders

ACKNOWLEDGEMENTS

At this place I would like to thank several persons for their support in one way or another. First of all I would like to thank my thesis supervi­sor Andries Brouwer, he introduced me to distance regular graphs and gave me all the help and inspiration I needed, it was a privilege working with him. Also I would like to thank Prof. dr. J.H. van Lint and Dr. ir. H.C.A. van Tilborg for giving me the chance to work in Eindhoven. I want to thank the Department of Mathematics and Computing Science for giving me the possibility to teach and for giving me one extra year to finish this thesis. The first version of the computer program used in Chapter 4 of this thesis was written by F.C. Bussemaker, for which I want to thank him. I wish to thank my former teachers Drs. H.J.C. Vlogtman and Prof. dr. M. van der Put from who I learned to love mathematics. To the people of the Discrete Mathematics group and the other colleagues of the Eindhoven University: I want to thank you for the pleasant working atmosphere, it was really nice working in your midst. I want to thank all my roommates at the University the last five years, especially Aart Blokhuis, for every mathematician it should be a privilege sharing an office with him, and Vladimir Tonchev, for his permission to add our theorem to the list of "stellingen". I want to thank my friends in Eindhoven and Veldhoven, they made my stay in Eindhoven a pleasant one. Finally I want to thank my family and friends from Groningen for their part in my diversion the last five years.

CONTENTS

1. INTRODUCTION . . . . . . . . 1

2. ON THE INTERSECTION ARRAY 8

2.1 Distance regular graphs with c; = b1 8

2.2 An application of circuit chasing . . 9

2.3 An inequality on the parameters . . 14

2.4 Distance regular graphs with ~ = 1 15

2.5 Graphs with J.L = 2, c3 = 3, a2 = 2A and A =f. 2 17

3. CHARACTERIZATION BY PARAMETERS . . 21

3.1 Distance regular graphs with d = 3i - 1, b; = 1, k > 2 21

3.2 Uniqueness of a graph on 506 vertices 22

3.3 Uniqueness of a graph on 280 vertices 27

3.3.1 On the structure of the graph 28

3.3.2 Affine subplanes in a projective plane 41

3.3.3 An isomorphism

4. THE HYPERPLANES OF THE REGULAR NEAR HEXAGON ON 759 POINTS

4.1 llyperplanes and embeddings .

4.2 A small example . . . . . . .

4.3 The near hexagon on 759 points

4.4 The hyperplanes generated by two points

4.5 The hyperplanes generated by three points

4.6 The hyperplanes generated by four points .

4. 7 The hyperplanes of :F0 • • • • • • • • •

4.8 The hyperplanes of 1-{ • • • • • • • • •

5. CONVEX SUBGRAPHS OF CLASSICAL GRAPHS

5.1 Introduction

5.2 When 2-convexity implies convexity

5.3 Subspaces

5.4 Johnson graphs .

5.5 Hamming graphs

5.6 Grassmann graphs

44

49

49

51 52

56

57

61

65

66

70

70

70

73 73 75

76

5. 7 Dual polar graphs ... 77

5.8 Bilinear forms graphs 80

5.9 Alternating forms graphs 82

5.10 Hermitean forms graphs 84

5.11 Quadratic forms graphs 87

APPENDIX . 95

A.l Graphs 95

A.2 Steiner systems 96

A.3 Near polygons . 97

REFERENCES 98

SAMENVATTING . . 101

CURRICULUM VITAE . 103

CHAPTER 1: INTRODUCTION

In this chapter distance regular graphs are introduced and some exam­ples are given. Furthermore we will give a brief summary of the contents of each chapter in this introductory chapter. Usually the definitions of unfamiliar objects and terminology can be found in the Appendix.

Before giving the definition of a distance regular graph, let us introduce some notation. If r is a graph and "Y is a vertex of r, then let us write ri("Y) for the set of all vertices of r at distance i from ,, and r( "'() = r 1 ( "'() for the set of all neighbours of "'( in r.

DEFINITION. A connected graph r is called distance regular with diameterd and intersection array {bo, ... , bd-ti c1, ... , cd} if for any two vertices"'(, li at distance i (0 $ i $d) we have 1rH1 ("Y)n r(li)l = bi and

lri-t("Y) n r(li)l Cj.

Clearly bd = Co = 0 and Ct = 1. Also, a distance regular graph r is regular of valency k = b0 , and, if we put ai = k - bi - Ci, then lri("Y)nr(li)l ai whenever d("'f,li) = i. We shall also use the notation ki = lri("Y)I (this is independent of the vertex "f), .X = a 1 and J-t = c2 •

Furthermore we will write "Y,..., li if {"'1,8} is an edge of r. Let us first give some examples of distance regular graphs:

(1) The polygons are distance regular graphs, they have intersec­tion array {2, 1, ... , 1; 1, ... , 1, cd}, where Cd = 2 for the 2d-gon and cd 1 for the (2d + 1 )-gon.

(2) The vertices and edges of the five Platonic solids are distance regular graphs. Their intersection arrays are { 3; 1} (the tetra­hedron), {4, 1; 1, 4} (the octahedron), {3, 2, 1; 1, 2, 3} (the cube), {5,2,1;1,2,5} (the icosahedron) and {3,2,1,1,1;1,1,1,2,3} (the dodecahedron).

(3) Let X be a finite set of cardinality q 2:: 2. Let r be the graph with as vertex set Xd, the Cartesian product of d copies of X, two vertices being adjacent whenever they differ in precisely one coordinate. Then r is distance regular with intersection array given by bj = (d- j)(q-1) and Cj = j for 0 $ j$ d. r is called the Hamming graph with diameter don X and is denoted by Xd, qd or H(d,q).

( 4) The Petersen graph is a distance regular graph with intersection array {3, 2; 1, 1}, which one obtains by identifying the antipodal points of the dodecahedron. We will meet the Petersen graph a number of times in this thesis.

Fig.1 The Petersen graph

Distance regular graphs were introduced by Biggs as a generalization of distance transitive graphs about 20 years ago, see e.g. [Bil] and [Bi2]. On the other hand, distance regular graphs are just P-polynomial asso· ciation schemes. From this point of view, Delsarte studied distance reg­ular graphs, also about 20 yeats ago, although he called them metrically regular, see e.g. [De]. A book by Bannai and Ito on association schemes, [BI], appeared in 1984. In this book one can also find a chapter on dis­tance regular graphs. About one year ago a book on distance regular graphs appeared, written by Brouwer, Cohen & Neumaier. This book, [BCN], contained almost all information on distance regular graphs known at that moment.

The central problem in the theory of distance regular graphs is their classification, which seems to be very hard. Up to now only the dis­tance regular graphs of valency 3 have been classified, viz. by Biggs, Boshier & Shawe-Taylor, see (BBS]. Questions related to this central problem are questions like: What can one say about the intersection ar­ray of a distance regular graph'! or Is a distance regular graph uniquely determined by its intersection array'!

The book of Brouwer, Cohen & Neumaier, [BCN, Ch.4,5], contains several restrictions on the intersection array of a distance regular graph.

· Some of these restrictions will be given below, most of them will be used

2

later on. More restrictions on the intersection array will be obtained in Chapter 2 of this thesis.

Let r be a distance regular graph of diameter d with v vertices and intersection array { bo, ... , bd-1; Ct, ..• , cd}. Define, for i = 0, ... , d, v x v-matrices Ai (where rows and columns are indexed by the vertices of r) by

{ 1 if d(i,6) = i,

(Ai),.6 = . 0 otherwise .

. Then we have

(i = 0, ... ,d).

It follows that the matrices Ai can be written as polynomials in A= A1

of degree i, so that we find

d

AiAj = LP~jAl l=O

for certain numbers P~j· In fact, the number P~j is just the number of vertices at distance i from the vertex 1 and at distance j from the vertex 6, where(!, 6) is any pair of vertices with d(/,6) = l. One can calculate these numbers from the intersection array of r. The adjacency matrix A of r has exactly d + 1 distinct eigenvalues, each of them with a certain multiplicity /j, 0 ~ j ~ d, which can also be computed from the intersection array of r. Now we can state the first restrictions on the intersection array:

PROPOSITION (1.1). For a distance regular graph, the following restrictions on the intersection array hold.

(i) k = bo > b1 2:: b2 2:: · • • 2:: bd-1 > bd = 0.

(ii) 1 = Ct :5 C2 :5 · · · :5 Cd ~ k.

(iii) If i + j ~ d, then Ci ~ bj.

(iv) All parameters P~j (including the ki = P?i) are nonnegative integers.

( v) All multiplicities /j are integers.

An intersection array satisfying the conditions above is called feasible in [BCN]. This definition is slightly different from the one of Biggs in [Bi2]. There are a few restrictions involving v and the ki:

3

LEMMA (1.2). For the parameters of a distance regular graph we have the following divisibility conditions: vki = kiai = 0 mod 2 (1 ~ i ~ d) and vk>. = 0 mod 3.

PROPOSITION (1.3). Let r be a distance regular graph with d;::: 3.

(i) 1 < kt < · · · < kh = · · · = k, > · · · > kd for some h and l with 1 ~ h ~ l ~ d, i.e., the sequence 1 = k0 , k1 , .•• , kd is unimodal.

(ii) ki ~ k; (0 ~ i ~ j, i + j ~d). (iii) If ki = k; (0 ~ i < j, i + j ~d), then bi = c;,bi+l = Cj-b· •• ,

b;-1 = Ci+l, and in particular ki+l = k;-t·

Many more restrictions can be found in [BCN], but let us give just one more restriction, concerning the parameters ai. It is Proposition (5.5.1) from [BCN], but note that in the original statement the condition in (ii) was erroneously omitted.

PROPOSITION (1.4). Let r be a distance regular graph of diame­ter d with A> 0. Then, for 1 ~ i ~ d -1:

(i) 2ai ;::: A + 1;

(ii) ai + ai+l ;::: A+ 1, unless i + 1 = d and ad = 0;

(iii) bi + Ci+J ;::: A + 2.

A distance regular graph is not necessarily uniquely determined! by its intersection array. Let us give an example.

Let q be some prime power and d some positive integer. Let V be the vector space GF(q)2d equipped with a nondegenerate symplectic form and let W be the vector space GF(q)2d+1 equipped with a nondegener­ate quadratic form. Subspaces of V or W are called isotropic if the form vanishes completely on the subspace. In both spaces maximal isotropic subspaces have dimension d. The dual polar graph on V, respectively W, is obtained by taking as vertices the maximal isotropic subspaces, two subspaces are adjacent whenever their intersection has dimension d-1. The dual polar graphs Cd(q) (from V) and Bd(q) (from W) have the intersection array given by

"+I [d- j] [j] bj = qJ 1 ' Cj = 1 (0 ~ j ~d),

where [ ~] is the q-ary Gaussian binomial coefficient:

[n] = [n] = (qn-l) ... (qn-m+l_l) m · m q ( qm - 1) ... ( q - 1)

4

Hence they have the same intersection array, but it turns out that they are isomorphic if and only if q is even, see [BCN, §9.4].

Nevertheless, a lot of distance regular graphs are uniquely determined by their intersection arrays. Let us give some examples:

(1) Let r be the graph with as vertices the blocks of the Steiner system S(5, 8, 24), two blocks being adjacent whenever they are disjoint. Then r is a distance regular graph with intersec­tion array {30, 28, 24; 1, 3, 15}, as. can be easily checked from the properties of the Steiner system. Brouwer [Br 1] showed uniqueness of this graph.

(2) Let~ be the graph obtained from r by deleting the vertices that are blocks containing some fixed element from the 24-set. Then ~is distance regular with intersection array {15, 14, 12; 1, 1, 9}. ~was shown to be unique, see [BL] or Chapter 3 of this thesis.

(3) Let :E be the graph obtained from r by deleting the vertices that are blocks containing exactly one of two fixed symbols from the 24-set. Then :E is distance regular with intersection array {7,6,4,4;1,1,1,6}. :E was shown to be unique in [Br3] by Brouwer.

(4) Let C be a double truncation of the binary Golay code (For undefined terminology concerning coding theory, see e.g. [MS] or [Li]). Then the graph r, obtained by taking the cosets of C in V = GF(2)21 as vertices and joining two vertices if the cor­responding cosets have representatives at Hamming distance 1, is distance regular with intersection array {21,20, 16; 1,2, 12}. r was shown to be unique by Ivanov & Shpectorov [IS1].

(5) For 1 E r of (4), the subgraph ~ = r3(1) of r is distance reg­ular with intersection array {9, 8, 6, 3; 1, 1, 3, 8}. ~ is uniquely determined by its intersection array, as we will see in Chapter 3 of this thesis.

Chapter 3 deals with characterization of distance regular graphs by their parameters. Apart from the graphs mentioned in (2) and (5) above, we will characterize the dodecahedron as a distance regular graph of diameter 3i - 1 satisfying bi = 1, i ~ 2 and k > 2.

The last two chapters of this thesis deal with substructures of a distance regular graph. In fact, we already will show the existence of Petersen subgraphs in certain distance regular graphs in Chapter 3 and use their

5

existence in some of the uniqueness proofs given in the same chapter.

Another kind of substructures are the singular lines. In order to intro­duce them, we need some notation. For a graph r, a vertex 'Y E r and a subset X of r, we define

'Y..L = {'Y} u r('Y) and x..L = n 'Y..L· ;EX

If 'Y "' S, then {;, 6}..L..L is called a singular line. Observe that u E {;, S} ..L..L if and only if u is 'Y, u is S or u is a common neigh­bour of 'Y and S adjacent to all other common neighbours of 'Y and 6.

The graph r of (1) above together with its singular lines is a regular near hexagon. In this sense we can ask for its hyperplanes, i.e. we ask for subsets of r such that each singular line is either contained in it or has exactly one point in it. This question is the subject of Chapter 4 of this thesis.

Let r be a graph of diameter d and C a sub graph of r. Let t be an integer, t :5 d. Suppose that for all j :5 t and that for any two vertices 'Y and S of C with distance j in r the following holds: r i( 'Y) n r i-i( S) c C for all appropriate i, i.e., 0 ~ i ~ j. In this case we will call C t-conve:c. C is called convex or geodetically closed when it is t-convex for all t ~ d. In Chapter 5 we will determine the convex subgraphs of some infinite families of distance regular graphs. We will also pay attention there to the question whether a connected 2-convex subgraph is convex or not.

Let us finish this chapter by introducing distance distribution diagrams for distance regular graphs. If r is a distance regular graph, 'Y a vertex of r and { bo, .. . , bd-1 j Ct, .•• , Cd} the intersection array of r, then the distance distribution with respect to 'Y can be given in the following diagram:

The interpretation of this figure is obvious. One can also give a distance distribution diagram with respect to an edge a "' (3 of r. It will be given in a form as shown in Figure 2, following Boshier & Nomura

. (BN], where Dj = r i( a) n r i((3).

6

A line between D~ and Df indicates that there might be edges between them. At the place of Dj there will be a balloon with the number IDjl

·inside. If some Dj = 0, then we will delete this set from the diagram (together with the lines joined to it). If the number of edges joining a given vertex 1 in Dj to vertices of Df does not depend on 1 but only on i, j, k, l, then this number of edges is put on the line between the corresponding balloons, just outside the balloon corresponding to Dj. If there are no edges between Dj and Df, then the line between the two corresponding balloons will be deleted.

7

CHAPTER 2: ON THE INTERSECTION ARRAY

2.1 Distance regular graphs with c; = b1 •

In [BCN, §5.4] Brouwer, Cohen & Neumaier ask whether c; = b1 (for some i < d) implies b; = 1. In this section we will prove the following theorem, which gives an affirmative answer to this question.

THEOREM (2.1). Let r be a distance regular graph. Suppose for some i > 0 we have c; = b1• Then we must have b; ~ 1.

Proof. Assume that r satisfies Ci = bl and b; > 1 for some i > 0. Then A > 0 because 1 < a; + bi = k - Ci = k - b1 = A + 1. In that case Proposition (1.4) yields ai ~ (A+ 1)/2, so 1 < b; ~ (.A+ 1)/2, from which we get A ~ 3, ai ~ 2. Take a vertex a of r, let (3 E r i( a) and 8 E rHt(a) n r((3). The A common neighbours of (3 and 8 must be in the set

in which (3 has exactly ai+bi-1 =A. neighbours distinct from 6. Hence:

(1) if (3 E r;(a), 6 E rHt(a) n r((3), then 6 "' (j for all ~

(j E A( a) n r((3) \ {6}.

This gives ai+1 ~ bi- 1 > 0 and thus a;+ ai+I ~ A.+ 1 = a;+ b;, i.e, ai+1 ~ b;, by Proposition (1.4). So we can find p E r(6) n ri+I(a:) with p not adjacent to (3. If p and (3 would have a common neighbour T E r;(a), then (1) applied to the edge T,..., p gives p,..., (3, a contradic­tion. Thus we must have (r(p) n r((3)) ~ (r i+I (a) n r((3) ), hence:

(2) p ~ bj.

Let (j and T be two distinct neighbours of (3 in ri(o:). Then by (1) they have at least 1 + b; common neighbours, viz. the vertices of {(3} U (rHI(o:) n r((3)). Therefore they must be adjacent by (2). It follows that A( a:) n (3.!. is a (A.+ 2)-clique. In particular we have:

(3) for any vertex o: and any edge (3 ,..., 1 with (3, 1 E r;(o:), the common neighbours of (3 and 1 have distance i or i + 1 to a:.

Now let us take any two neighbours 6 and p of (3 in rHI(a:), then 6 ,..., p by (1). But for any vertex e E r(a:) n ri-t(f3) we have d(e,6) = d(e,p) = i, d(e,(3) = i- 1, which contradicts (3). Hence our initial assumption was false, which shows the theorem. 0

8

2.2 An application of circuit chasing.

Circuit chasing is a technique which was developed by Biggs, Boshier and Shawe--Taylor in their paper [BBS] on the classification of dis­tance regular graphs of valency 3. The same technique was also used by Boshier and Nomura in [BN], where they studied intersection ar­rays with the property (c.;, a.;, b.;) = (1, 0, k - 1) for 1 ::; i S r and (c.;, a.;, bi) = (1, 1, k- 2) for r + 1 ::; i ::; r + s. We will use this technique to give an upper bound on the valency k when the intersection array has certain properties, one of them similar to the Boshier & Nomura

· case. This upper bound implies that the dodecahedron is the only dis­tance regular graph of diameter 3i- 1 with bi = 1, ~2:: 2 and valency k > 2, as we will see in Section 3.1.

We will first introduce some notation. Let r be a distance regular graph. For subsets A and B of the vertex set of r let e(A, B) denote the number of edges between A and B. If a is a vertex of r, then we put e( a, B) : = e( {a}, B). Finally for vertices a and {3 we define Dj(a,/3) := ri(a) nri(/3). From [BN] we quote the following lemma:

LEMMA (2.2). Let a"' {3 be an edge ofr and let Dj = Dj(a,/3). Let 'Y E D~+l, i 2:: 0, and 8 E D1, j > 0. Then we have:

(i) b1 = e('Y,D1+1 ) + e(!',D!tD + e(r,D!ti);

(ii) bi+l = e(,,D~ii);

(iii) Ci = e(r,Di-1);

(iv) Cj+J = e('Y,Dt_l) + e(r,nn + e(,,Di+l); . '+1 '41 '41

(v) bj = e(8,D1+1) + e(8,D~+1) = e(8,DJ ) + e(8,D]+t);

(vi) Cj = e( 8, n;-1) + e(8, nJ:D = e( 8, n;_1) + e(8, n;:i);

(vii) b; = bi+t if and only if e( D!+l, D!t}) = e( Di+l, n;+t) = 0;

(viii) c1 = Ci+J if and only ife(Df+I,DD = e(n;+1,Di+1) = 0.

Proof. Straightforward. 0

Now let us assume that r is a distance regular graph of diameter d > 2i, i 2:: 2, such that (ci-bai-I.bi-d = (cha1,bt) and (c2i-t.a2i-I.b2i-d = (ci,a;,bi)· Assume moreover that a1 = 0, Ci = 1, ai > 0 and a2i < ai. We will show that the valency k is at most 2bi + 1 (Theorem (2.4)), but we need another lemma:

9

LEMMA (2.3). Let a"" f3 be an edge ofr and let Dj = Dj(a,/3).

(i) If 1 E D1+I, 1 ~ j ~ i- 2, then e('Y,D1_1 ) = 1 and '+2

e( 1, D~+I) = k .- 1.

(ii) If 1 E DLH then e(I,Di:~) = 1, e(1,Di) = a; and e(I,Di+1) = b;.

(iii) If 1 ED!, then e('Y,Di_1 ) = e(I,Di-1 ) = 1, e(I,DiiD = bi and e('Y, DD = a; - 1.

(iv) If1 E D;+1, i ~ j ~ 2i-2, then e('Y,D;_1 ) = 1, e('Y,D1!~) = bi '+1

and e(I,D~ ) = ai.

(v) If1 ED;, i+1 ~ j ~ 2i-2, thene(I,D1::) = 1, e(1,D1) = ai '+1 and e('Y, D~+l) = b;.

In each of the five cases above e( 1, Di) = 0, unless stated otherwise.

(vi) If 1 E D~;_ 1 then e(I,Dii:i) = 1, e(/,D~i+l) = ~i and e( [, D~i=~) > 0.

(vii) If[ E Di!:L then e('Y,D~~:i) = 1 and e('Y,Di!:D ~ k-1-2bi.

Proof. We have bj ~ bi+l hence:

(cj,aj,bj) = (1,0,k -1) (cj,aj,bj) = (1,a;,bi)

and Cj ~ c;+1 by Proposition (1.1),

if 1 ~ j ~ i -1; ifi~j~2i-1.

Moreover, we have D1 = 0 if j < i, for ai = 0 in that case. Now everything follows immediately from Lemma (2.2). Let us show (vii); the other claims follow in a similar way. By Lemma (2.2) (v), (vi) and (vii) (using ~i-2 = b2i-l) we have for any vertex 1 E Di!:t:

e('Y, D~Lt) + e('y, D~i) = ~i-1 = bii

e(I,D~~-l) + e([,Dit) = b2i-1 = b;;

( D2i-2) 1 e [, 2i-2 = c2i-l = ·

Since e(Di!:LD~i:D = e(Di!:LDii:D = 0 we get e('Y,Dii:D = k- e(,,Dii:i)- e('Y,Dij-1)- e('Y,D~L1 )- e(1·,Di!) ~ k -1- 2bj. D

Let /O "' /I "' . . . - lr "' /0 be a circuit of r and suppose 11 E D;:('Yo,[I). Then the profile of the circuit with respect to the edge 'Yo ""/1 is the series of ordered pairs (io,jo),{it,it), ... ,(ir,ir).

THEOREM (2.4). Let r be a distance regular graph of diame­. ter d > 2i, i 2: 2, such that (c;-t,a;-t,bi-1) = (ct,at,bt) and

10

(c2i-t. a2i-l, b2i-l) = ( Cj, ail b;). Suppose moreover that a1 = 0, Ci = 1, ai > 0 and a2; < a;. Then the valency k of r must satisfy k :5 1 + 2bi.

Gj

k-1 1

1

1

k-1 a;

Fig.9 The distance distribution around an edge in Theorem (2.4)

Proof. Suppose k > 1 + 2b;. Then Lemma (2.3)(vii) implies e( T• Di~=l) > 0 for all T E D~~:f. The distance distribution around an edge is as shown in Figure 3. From this figure we see that there exists a circuit TO rv T1 rv • • • "" T4i "' TO with profile (with respect to the edge

To ""'Tl)

(0, 1),(1,0), ... ,(i,i -1),(i +1,i), ... ,(2i,2i -1),(2i -1,2i -1),

(2i 1,2i -1),(2i -2,2i- 2), ... ,(i,i),(i -l,i), ... ,(1,2).

We can determine the profiles with respect to the edges Tj "'/j+I for j = 1, 2, ... , 2i successively, using Lemma (2.?) and our knowledge of

the distances d(TO•Ti+I),d(Tt,Tj+t), d(Tj,TI) for 0 :5 l :5 4i. With respect to Tl ""' T2 the profile is

(0,1),(1,0), ... ,(i -1,i- 2),(i,i -1),

( i + 1, i), ... '(2i- 1, 2i - 2), (2i 1, 2i 2), (2i - 1, 2i - 2),

(2i -2,2i -3), ... ,(i,i l),(i,i),(i -1,i), ... ,(1,2).

With respect to the edge Ti-l "'li we have TO E D~-l, Tl E D1:r and the profile is

(O,l),(l,O), ... ,(i,i -1),(i + l,i),(i + 1,i),

(i + 1, i), (i, i- 1), (i -1, i 2), ... '(3, 2), (2, 1),

(3, 2), ... ' ( i, i- 1 ), ( i, i), ( i - 1, i), ... '(1, 2).

11

With respect to the edge "Yi "' "YiH we have 'Yo E D!+t• "Yt E Di-1 and the profile is

(0, 1),(1,0), ... ,(i -1, i- 2), (i, i -1),(i, i), (i, i),

( i- 1, i),{ i- 2, i- 1 ), ... '(2, 3), (1, 2), (2, 3), ... ' ( i - 1, i),

(i~ i + 1), (i,i + 1), (i- 1, i), ... ,(1, 2).

With respect to the edge "Y2i-2 ,..., /2i-t we have 'Yo" E D~!:r, "'t E D~~=~ and the profile is

(0, 1),(1,0), ... ,(i,i -l),(i,i),(i,i),

(i -l,i), (i,i + 1), ... ,(2i- 3,2i 2),(2i- 2,2i -1),

(2i- 2,2i -1),(2i- 3,2i- 2), ... '(i -1,i), ... ,(1,2).

With respect to the edge /2i-1 "' "'(zi we have 'Yo E Dil-1, 'Yt E Di!:i and the profile is

(0, 1), (1, 0), ... '(i, i- 1), (i, i), (i, i),

(i + 1,i + 1), ... ,(2i -1,2i -1),(2i -1,2i),

(2i- 2,2i -1), ... ,(1,2).

With respect to the edge 'Y2i "' {Zi+l we have 'Yo E Dii-t, 'Yt E Di!:l and the profile is

(0, 1), (1,0), ... '(i, i -1),(i, i), (i + 1, i + 1), ... , (2i- 1,2i- 1),

(2i, 2i - 1 ), (2i - 1, 2i - 1 ), ... ' ( i, i), ( i - 1, i), ... '(1, 2).

Using d('Yzi+ll 'YI) for 0 $ l $ 4i, as given by the previous profile, and Lemma (2.3) we would find with respect to the edge /2i+I "" /2i+2 the profile

(0,1),(1,0), ... ,(i,i -1),

( i + 1, i), ... '(2i - 1, 2i - 2), (2i - 1, 2i - 2),

(2i - 1, 2i- 2), ... ' ( i, i - 1 ), ( i, i), ( i - 1, i), ... '(1, 2).

On the other hand, we had {o, 'Yt E Dii:H "Y2i+h 'Yzi+z), a contra­diction. Hence, our initial assumption was false, i.e. we must have k $ 1 + 2bi. D

Looking at the proof just given, we observe that we found a contradic­tion by forcing an edge that cannot be there, viz. an edge between the

sets Dii:H"Y2i+b"Y2i+2) and Di~:i('Yzi+t,/2i+2)· Boshier & Nomura ([BN]) obtained divisibility criteria by looking when an initial profile comes back for the first time. There is another way to use the technique

· of circuit chasing, which is used to prove the following lemma.

12

LEMMA (2.5). Let r be a distance regular graph of diameter 6 with intersection azTay {15,14,12,6,1,1;1,1,3,12,14,15}, v = 1518, and distance distribution

2 6 2

Then e( 7, Di) > 1 for some edge a "' f3 and some vertex 7 € Df.

2 2

Fig.4 The distance distribution around an edge in Lemma (2.5)

Proof. Note that r must be an antipodal 3-cover of the distance regular graph on 506 vertices of Section 3.2. For the folded graph it is easy to determine the distance distribution diagram with respect to an edge, using Lemma (2.2) and the information obtained in Section 3.2. Using this, for r all numbers e( 1, D{) with 1 1/:. D~ can be determined. The number of edges between Di and D~ is' equal to the number of vertices in Di. Now suppose the claim is false. Then e(!, DD = 1 for all1 € D~ and we know all e("', D{ ). The distance distribution diagram around an edge in this case is given in Figure 4.

Starting with a circuit /o ,...., "11 "' · · · "' /6 "''Yo of profile

(0, 1), (1, 0), (2, 1), (2, 2), (3, 3), (2, 3), (1, 2),

one does not find this profile with respect to another edge of this circuit, and with respect to the edges /l ""72, 14 ,...., /5 and /5 "'/6 one finds the profile

(0, 1), (1, 0), (2, 1), (3, 2), (3,3), (2, 3), (1, 2).

13

Thus, if we would count pairs ( /o i/o ,..., /1 ,..., · · · "" /6 ,..., /o ), then 3n1 :5 n2 , where n1 is the number of pairs with our initial profile and n2 is the number of pairs with the other mentioned profile. But we can count these pairs. For ·n1 we get n1 = 1518 · 15 · 14 · 2 ·12 · 2, but for n2 we get n2 = 1518 · 15 · 14 · 12 · 4 · 2 = 2nt, a contradiction. 0

Although the intersection array of r in the previous lemma is feasible, r does not exist. This result was obtained by Ivanov and Shpectorov ([IS2]) using group amalgams. Lemma (2.5), however, may be the starting point for a combinatorial proof of the nonexistence of r.

2.3 An inequality on the parameters.

In this section we will derive an inequality for the parameters of a distance regular graph. The geometric information obtained from the proof of this inequality in the case of equality will be used in Chapter 3 to prove uniqueness of two distance regular graphs, viz. one with 506 vertices and one with 280 vertices. The contents of this section is essentially the first part of [BL]. Note however that the case of equality in (ii) and (iii) of the next theorem was formulated incorrectly in [BL] and in [BCN, Prop.(5.5.4)].

THEOREM (2.6). Suppose r is a distance regular graph and ai f:: 0 for some i, 1 :5 i :5 d. Put ad+l = 0. Then we have:

(i)

(ii)

(iii)

Equality holds if and only if the following is true. For any four vertices a, {3, 1 and S ofr such that a,..., {3, 1 ""S and d(a,1) = i, the three distances d(a,S), d(/3,1) and d(f3,S) are not all equal. In particular~ equality implies ai+l :5 a; and ai-l :5 ai.

Equality implies a;+l :5 ai :5 ai-l·

Equality implies a;-1 :5 a;, and, if i < d, ai :5 ai+l·

14

Proof. Count 4-tuples (a,/3,1,8) with a"' /3, 1""' 6, d(IJ,a)-:/: i, d(!,a) = d(i,/3) = i. Clearly, there are vkiaa(b& + ci) such 4-tuples. We find the inequality

by covering this collection of 4-tuples with the three sets

{(a,/3,1,6)la""' /3,1 "'6,d(l,a) = d(/,/3) = i,d(6,/3) = i},

{(a,/3,1,6)la "'/3,1 ""'6,d(8,a) = d(6,/3) = i + 1,d(!,a) = i},

{(a,/3,1,6)la ""{3,1 ""'6,d(6,a) = d(6,/3) = i -1,d(!,a) = i}.

For example, the first set has cardinality vkiai2 , as can be seen by choosing a, 1, /3,6 in this order. This proves the inequality in (i) (note that kjbj = ki+l c;+I); those in (ii), (iii) follow similarly by restricting d(a, 6) to be i + 1 and i- 1, respectively. Equality in (i) is equivalent with: if a"' /3,1 "'8,d(a,1) = i, then the three distances d(a,6), d(/3, 1) and d(/3, 8) are not all equal. With d( 1, /3) = d( 8, /3) = i ± 1 this implies ai±1 .$ ai (varying 8). In the same way we find ai+I .$ ai

and ai-l .$ ai in case of equality in (ii) and (iii), respectively. Using (i) we find the second inequalities in case of equality in (ii) and (iii) respectively. 0

Clearly, some more geometric information is available in case of equality in (ii) or (iii). A special case of this inequality has already been given by Faradjev, Ivanov & Ivanov [FII]. Also Nomura has independently obtained a slightly weaker form of this result, see [Nol].

2.4 Distance regular graphs with ~ = 1.

In [BCN, §5.7] the authors ask whether b2 = 1 forces r to be antipodal. Of course one has to take k > 2, since for k = 2 we have the polygons as an example and not all of them are antipodal. In this section we will give a partial answer to this question (Proposition (2.8)). We will need a lemma from [BCN] concerning restrictions on the intersection array whenever bi = 1 for some i ~ 1.

15

LEMMA (2.7). Let r be a distance regular graph with k > 2 and b; = 1 for some i ;;,:: 1. Then r has diameter d < 3i and we have

(i) Cj = 1 and Ci+j > ai if d;;,:: i + j; (ii) C2i > 1 if d ;;,:: 2i;

(iii) aj=Oifd;;,::2i+j;

(iv) a;;;,:: (J.L -1)ai+I +.A.

Proof. See [BCN, Lemma (5.3.1)] 0

PROPOSITION (2.8). Let r be a distance regular graph with b2 = 1 and k ;;,:: 3. Then exactly one of the following holds.

(i) r is an antipodal graph.

(ii) r has diameter 3 and satisfies C3 > max(p,A + 1) and .A + 1 :::;; a3 < b1. Furthermore we have: Jl divides b1 prop­erly and either Jl :::;; 2 or C3 ;;,:: Jl + 2.

Proof. By Lemma (2.7) we have 3:::;; d:::;; 5. If d = 5, then r must be the dodecahedron, see [BCN] after Corollary (4.3.12) or Theorem (3.1) of this thesis. If d = 4, then by Lemma (2.7) we have c2 = 1 and c4 > a2 = k- 2, hence c4 = k - 1 or c4 = k. Suppose C4 = k. Then we have a4 = 0, k2 = kbl and k4 = ~· If r is not antipodal, then [BCN, Prop.(5.6.1)] gives kb1 :::;; ~(~ -1), which yields kbt :::;; b1(bt- 1), a contradiction. Suppose c4 = k - 1. Then we have k4 = caN!:..t) and (k - 1)lbt follows, since k > 2. Hence bt = k - 1 and c3lk. On the other hand, p~4 = k;;/. It follows that c3 = 1, r has intersection array {k, k- 1, 1, 1; 1, 1, 1, k - 1} and k > 3 ( k = 3 is impossible by [BBS]).

Let d('yo, 74) = 4 and take a path /o "' it "' /2 "' /3 "' /4. Let a E r('y4) n r4(io). Then d(72,a) = 3 (~ = 1,.A = 0). We can find two vertices (3, 6 E r('y4) n r3(io) \ {13}. For these vertices we have d('y2 ,{3) = d( 12, 6) = 2. With b2 = 1 and .A = 0 it follows that there exist vertices (3',6' E r2("Yo) n r('y2) and (3' "'(3, 6' "'6. Now consider the path 'Yo ,...., (3" "'(3' "'(3 "'"'f4, where (3" E r( 'Yo) n r((J'). The same arguments as before give d((3', 6) = 2 and li' "" (3'. We have found a triangle: /2 "" (3' "" li' "' 12 , which contradicts .A = 0. Suppose d = 3. If a 3 = 0, then k3 = ~ and k2 = ~. If r is not antipo­

dal, then [BCN, Prop.(5.6.1)] yields~:::;;~(~ -1), a contradiction.

16

Suppose d = 3 and a3 > 0. If c3 = p, then c3 = f-L = 1 by [BCN, Th.(5.4.1)] and A = 0 by Lemma. (2.7). r has interse~ti~n array { k, k - 1, 1; 1, 1, 1}, which is impossible by [BCN, Cor. 4.3.12] and [BBS]. Thus if r is not antipodal, then r has diameter 3, a3 > 0 and CJ > f-L· Considering an edge in the set r 3 ( 'Y) for some vertex 'Y E r we see a3 2:: A+ 1. Corollary {5.5.3) from [BCN] yields the bounds on both aa and ca. Also we have Ph = ~' giving the divisibility criterion in the statement. The last statement is an immediate consequence of

· Corollary (5.4.2) from [BCN]. 0

Unfortunately the restrictions in (ii) are not strong enough to kill this case. Nevertheless, there are no feasible intersection arrays with v :$; 107 and k :$; 100.

2.5 Graphs with f-L = 2, ca = 3, a2 = 2A and A #= 2.

Instead of distance regular graphs we will consider a larger family of graphs in this section. We will consider regular graphs (undirected, without loops or multiple edges) with the following properties:

(1) there is a number A such that any two adjacent vertices have exactly A common neighbours;

(2) there is a number f-L such that any two vertices at distance 2 have exactly p common neighbours;

(3) there is a number c3 such that for any two vertices a and (3 at distance 3, a has exactly c3 neighbours at distance 2 from (3;

( 4) there is a number a2 such that for any two vertices a and (3 at

distance 2, a has exactly a2 neighbours at distance 2 from (3.

We will prove that such graphs with f-L = 2, c3 = 3, a2 = 2A and A #= 2 are quotients of Hamming graphs, generalizing results by Brou­wer ([Br2]) and by Egawa. ((Eg]). We will use a result obtained by Nomura ([No2]). He stated the following result for distance regular graphs, but actually he proved:

THEOREM (2.9). Let r be a regular graph with c2 = 2, ca = 3, a2 :5 2A and A#= 2. Then every edge off is in a unique (A+ 2)-clique, i.e. all singular lines of r have size .:\ + 2.

Proof. See [No2]. 0

17

By [BCN, Prop.(4.3.3)] we have (..\+l)lk and a2 ~ 2..\ as a consequence. Let us first prove the following lemma.

LEMMA (2.10). If r satisfies C2 = 2, Ca = 3, a2 = 2..\ and ,\ :f. 2, then any 3--claw (i.e. an induced subgraph K 1 3 ) of r detennines a

• I

unique geodetically dosed 3-cube (i.e. a Hamming cube 23 ).

Proo£ Fix a vertex a of r. Each vertex f3 at distance 3 from a determines a unique 3-cube (for c2 = 2 and c3 = 3), which contains a 3-claw with "central" vertex a. Obviously, such a 3-claw can be in at most one 3-cube. Now from Theorem (2.9) we find that the number of 3-claws with "central" vertex a is

k(k- ,\- 1)(k- 2..\- 2) kbt b2 =--, 3! CtC2C3

which is the number of vertices at distance 3 from a. This proves our lemma. D

THEOREM (2.11). Let r be a regular graph with c2 = 2, ca = 3, a2 = 2..\ and,\ :f. 2. Then s := .x!t is an integer and there is a map II : H(s, ,\ + 2) - r preserving distances :$ 2, where H(s, ,\ + 2) is a Hamming graph.

Proof. We will proceed by induction on the weight of the vectors of the Hamming graph(..\+ 2)8

• Fix a vertex 'Yo of rand let 1r{O) ='YO· Map the vectors of weight 1 to the neighbours of 'Yo in such a way that d( u, v) = d( 1r( u ), 1r( v)) for all vectors u and v of weight :$ 1. The fact that we have(..\+ 2)-cliques assures that this indeed can be done. Let w ~ 2. Assume 1r(u) is defined for all vectors u of weight :$ w- 1 and that for any two vectors u and v of weight :$ w - 1 we have d(u, v) = d(1t'(u),1r(v)) whenever d(u, v) :$ 2. We now will define 1r(u) for all vectors u of weight w and afterwards show that for any two vectors u and v of weight :$ w we have d( u, v) = d( 1r( u ), 1r( v)) whenever d{ u, v) :$ 2. Let u = ( Ut, ... , u8 ) be a vector of weight w. Choose i and j such that ui :f. 0 :f. Uj (i :f. j), and define 1r(u) to be the common neighbour of 1r(u- Uiei) and 1r(u- Ujej) distinct from 1r(u- Uiei- Ujej). 1r(u) is well-defined since any 3-claw determines a unique 3-cube. Note that by the induction hypothesis we have d(1r(u uiei),1r(u- Ujej)) = 2

· and they have 1r(u- Uiei- Ujej) as a common neighbour.

18

Now suppose d(u, v) :5 2, wt(u), wt(v) :5 w. We want to show d('7r(u),7r(v)) = d(u,v). Take u = (ub···,u,.), v = (v1 , ••• ,v,.). We have to consider the following cases:

(a) d(u, v) = 1, wt(u), wt(v) :5 w -1;

(b) d(u, v) = 1, wt(u) = w -1, wt(v) = w;

(c) d(u, v) = 1, wt(u) = wt(v) = w;

(d) d{u, v) = 2, wt(u), wt(v) :5 w- 1;

(e) d(u, v) = 2, wt(u) = w- 2, wt(v) = w;

(f) d(u, v) = 2, wt(u) = w -1, wt(v) = w;

(g) d(u, v) = 2, wt(u) = wt(v) = w.

In the cases (a) and (d) we are done by the induction hypothesis and in the cases (b) and (e) we are done by the construction of 7r( v). Case (c). Now we have some i 0 such that 0 f:. Ui 0 f:. Vj0 f:. 0, Ui =Vi

for all i f:. io and there is a j f:. io with Uj = Vj f:. 0. Obviously, 7r(u) and 7r(v) have a common neighbour, viz. 7r(u- Ui0 ei0 ). Suppose d(7r(u), 7r(v)) = 2. Consider the four vertices 7r(u), 7r(v), 7r(u- Ui0 ei0 )

and 7r( u - Uio ei0 - u jej ). By (e) they form a 3-claw, so they determine a geodetically closed 3-cube, giving d(7r(u Ujej), 7r(V- Vjej)) = 2, a contradiction to the induction hypothesis; If 7r(u) = 7r(v), then both 7r(u- ujej) and 7r(v- Vjej) are adjacent to both of 7r(u) and 7r(u- Ujej- Uj0 ei0 ), hence by Nomura's theorem we have d(7r(u),7r(u- Ujej- Ui0 ei0 )) :5 1, contradicting case (e). Thus d(7r(u), 7r(v)) = 1. Note that now we have d(7r(u),7r(v)) :5 d(u,v) and we only have to show d(7r(u), 1r(v));:::: 2 in the cases (f) and (g). Case (f). We have j and k such that Uj = 0 f:. Vj, 0 f:. Uk f:. Vk f:. 0 and Ui = Vi if j f:. i f:. k. If 7r(u) = 7r(v) then 1r(v) ,.... 7r(u- Ukek), contradicting case (e). If 7r(u) "' 7r(v), then, since they are both adjacent to 7r(v- Vjej) and 7r(u), 1r(v- Vjej) are both adjacent to 1r( u- ukek), Nomura's theorem yields d( 7r(v ), 7r( u- ukek)) :5 1, con­tradiction to case (e). Case (g). We have to consider two possibilities:

(g1) there are j and k such that Uj = 0 f:. Vj, uk f:. 0 = Vk,

Uj = Vj if j f:. i f:. k;

(g2) there are j and k such that 0 f:. Uj f:. Vj f:. 0, 0 f:. uk f:. Vk f:. 0, Uj = Vi if j f:. i f:. k.

19

Case (g1 ). There is a l such that j =f:. l =f:. k and ul = Vt =f:. 0. By the induction hypothesis 1r(u- u~:ek- u,ei), 1r(u- u~:ek), 1r(u- u,ea) and 1r(v- Vtea) form a 3-claw, hence they determine a geodetically closed 3-cube and we'll find d(1r(u), 1r(v)) = 2. Case (g2). If 1r(u) = 1r(v), then we would have the following ad­jacencies: 1r(u) "" 1r(u- (uk - v~:)ek) "' 1r(u - u~:ek) "' 1r(u) "" 1r(v- Vkek) "" 1r(u- u~:ek). By Nomura's theorem we would have 1r(v- v~:ek)"" 1r(u- (uk- v~:)ek), contradicting case (f). If 1r(u) "" 1r(v) then we would have the following adjacencies: 1r( V) "' 7r( U- ( Uj - Vj )ej) "'1r( u) "-' 1r( U- Ujej) "" 1r( U ~ ( Uj - Vj )ej), hence by Nomura's theorem 1r(v) "'1r(u- Ujej), contradicting case (f) again. D

COROLLARY (2.12). Let r be a regular graph with c2 = 2, ca = 3, a2 = 2..\ and ..\ =f:. 2. Then vi(..\+ 2)8

, where v is the number of vertices

ofr and s := >.!t• Proof. Observe l1r-1(a)l = l1r-1(P)I for any two adjacent vertices

a and p of r. D

In [BCN, §11.1] regular, uniformly regular and completely regular par­titions of distance regular graphs are defined. As a first remark we give the following consequence of Theorem (2.11 ):

COROLLARY (2.13). The partition II = {Gala E r} of the Ham­ming graph ( ..\ + 2 )8

, where C a = 1r -l (a), is a uniformly regular parti­tion and r ~ (..\+2)8 /II. Ifr is distance regular, then II is a completely regular partition.

Proof. It is straightforward to check that II is a uniformly regu­lar partition. By [BCN, Th. (11.1.6)] II is completely regular if r is distance regular, since obviously we haver~ (..\ + 2)8 /II. D

REMARK. In fact we have shown that r is a quotient of a Hamming graph whenever c2 = 2, all singular lines have size ..\ + 2 and each 3-claw determines a 3-cube as an induced subgraph. However, if..\ = 1, then Egawa ( [Eg]) showed that there are distance regular graphs with c2 = 2, c3 = 3, a2 = 2, ..\ = 1 which cannot be quotients of Hamming graphs.

20

CHAPTER 3: CHARACTERIZATION BY PARAMETERS

3.1 Distance regular graphs with d = 3i - 1, bi = 1, k > 2.

As an application of Theorem (2.4) we will determine in this section all distance regular graphs with b; = 1 and k > 2 which meet the diameter bound of Lemma (2.7), i.e., which have diameter d = 3i-l. This settles an open problem mentioned in a preliminary version of [BCN, §5.3] and generalizes a result by Godsil ([Go, Cor.(4.3)]). A matching of a com­plete graph K2n with 2n vertices is a collection of n edges such that every vertex is in exactly one of those edges.

THEOREM (3.1). Let r be a distance regular graph of diameter d = 3i - 1, i 2 1, with bi = 1 and k > 2. Then either i = 1 or i = 2. Moreover, if i 2, then r is the dodecahedron, and, if i = 1, then k is even and r is a graph obtained from the complete graph K~;+2 by deleting a matching. In both cases r is an antipodal 2~cover.

Proof. From Lemma (2.7) we get the following information about the intersection array of r:

(c;,a;,b;) = (O,O,k) (c;,a;,b;) = (1,0,k -1) (c;,a;,b;) = (l,k- 2, 1) (c;,a;,b;) = (c;,a;,1) (c;,a;,b;) (cd,ad,O)

if j = 0; if 1 ~ j ~ i -1;

ifi ~ j ~ 2i - 1; if 2i ~ j ~ d - 1; if j =d.

Furthermore we have c2i > 1, thus a2i = k- 1- c2i < k- 2 = ai and Theorem (2.4) gives: i = 1 or k ~ 3 (hence k = 3). Suppose i > 1, i.e. k = 3. As already mentioned in Chapter 1, the classification of the distance regular graphs of valency k = 3 was completed by Biggs, Boshier & Shawe-Taylor in [BBS]. Consulting their list we see that r must be the dodecahedron and i = 2. If i = 1, then we have d = 2, b1 = c1 = 1 and a1 = k - 2. Take a vertex a of r. A neighbour fJ of a has k - 2 neighbours in common with a, a has exactly k neighbours, which means that there is exactly one neighbour 'Y of a not adjacent to fJ. We see that the neighbours of a come in pairs and k must be even. On the other hand, fJ and 'Y have at least k- 1 common neighbours, viz. a and all the neighbours of a distinct from fJ and/, giving J.t 2 k-1. However, k2 =~=~'which must be an integer. This means J.l. = k, k2 = 1 and a is adjacent to all vertices of r, except one, and r is as claimed. 0

21

In the mean time Suzuki ([Su]) showed the following. Let r be a dis­tance regular graph with diameter d > 2 and valency k > 2. Suppose bi = 1 for some i with 2d 2 5i - 7 and d 2 2i + 1. Then r is an antipodal 2-cover.

3.2 Uniqueness of a graph on 506 vertices.

In this section we will show the uniqueness of a distance regular graph on 506 vertices, the contents of this section is essentially the last part of [BL]. Let us start by defining two distance regular graphs, E and r, respectively. Let E be the graph with as vertices the 759 blocks of the Steiner system 8(5,8,24), where two vertices are adjacent if the corresponding blocks are disjoint. Let r be the graph obtained from E by deleting all vertices for which the corresponding blocks contain a fixed symbol. E and r are distance regular graphs with intersection arrays {30, 28, 24; 1, 3, 15} and {15, 14, 12; 1, 1, 9} respectively. Their distance distribution diagrams are:

3 IS

and

In both cases, two vertices are at distance 0, 1, 2, 3 if the corresponding blocks meet in 8, 0, 4, 2 symbols, respectively. In fact, both graphs are distance transitive, E under the action of M24 and r under the action of M23 • Brouwer ([Br1]) showed the uniqueness of the graph E. We will show the uniqueness of r by embedding it in E. First we need some lemmas.

LEMMA (3.2). Let .6 be a graph with .A 0, p = 1, a2 = 2, and suppose that each induced hexagon with at least one pair of vertices at

distance 3 contains three such pairs. Then for each vertex 'Y of .6, the subgraph .62 ('Y) is the disjoint union of ik2 hexagons, and hence any two vertices of .6 at distance 2 are contained in a (unique) Petersen subgraph.

Proof. Fix oo E .6 and let for a E .62( oo ), a' be the unique common neighbour of oo and a. From a2 = 2 it follows that .62 ( oo) is a union

· of polygons; let ao "' · · · ""' ag-1 ""' ao be such a polygon; clearly g 2 5.

22

Since d(ai,a~+I) = 2, the vertex a~+l has two neighbours in .6.2(a;), say oo and f3i· Let /i be the common neighbour of a; and {3;. Since d(a;,ah1 ) = 2, it follows-'from the hypothesis applied to the hexagon

oo "' a~ "' a; "' li "" {3; - a~+t - oo that also d( oo, li) = 2. But then /i E {a;-t,ai+I} and {3; E {a;-z,ai+2}· If {3; = ai+z then we have the triangle a~+t IV ai+I "' ai+2 IV ai+I, contradicting ,\ = 0. Thus {3; = ai-2, so that a~+I = a~_2 (i E Z9). Now a~ "'aaj "'aaj+I ""ai for all j E Z 9 , so that there are only two points aaj, and g = 6. 0

For the remainder of this section we assume that r is a distance reg· . ular graph with intersection array {15, 14, 12; 1, 1, 9}. Then ca = 9, a2 = 2, aa = 6 and we have equality in (iii) and (i) of Theorem (2.6). The information about the case of equality stated there immediately implies that the hypothesis of Lemma (3.2) about hexagons is satisfied, and hence the conclusion of Lemma (3.2) holds: any two vertices of r at distance two are contained in a unique Petersen subgraph of r. For a Petersen subgraph P of r we will write fj{P) for the set of vertices of rat distance j from P, and f(P) = f 1(P). Let us define a relation II on the edges of r as follows:

e1 II ez if and only if there is a Petersen subgraph P containing them such that e1 U ez induces a subgraph isomorphic to 2K2

(i.e., each vertex of e1 has distance 2 to each vertex of e2 ).

Observe that II restricted to a Petersen graph is an equivalence relation with equivalence classes of size 3. We will show that II is an equivalence relation on the edges of r. The equivalence classes will be the missing points from E.

LEMMA (3.3). For r, p and II we have the following.

(i) For every/ E f(P) we have lf(l)nPI = 1 and lf(l)nf(P)I = 0.

(ii) If 1 E f2(P), then fz(l) n P does not contain 2-claws.

(iii) If1 E f 2 (P), then fz(l) n Pis either a codique or it induces 3K2 as a subgraph ofP.

(iv) The relation II is an equivalence relation.

( v) r has precisely 253 equivalence classes with 15 edges in each equivalence class.

Proof. (i) There are no triangles or 4-gons in r, every 2-claw is in a unique Petersen subgraph and P is geodetically closed.

23

(ii) Suppose lt7/2{Y3 E r2(1) n 'P such that It "' 1'2 "' 1'3· Then 1 and 12 determine a unique Petersen subgraph P, which contains r(l2) n r2(1). But now ihl'2,[3 E P n P', a contradiction. (iii) By (ii) there are only 10 possibilities left for the isomorphism type of the subgraph of P induced by r 2(1) n 'P, namely (a) 3K2 ,

(b) 2Kz + Kb (c) 2Kz, (d) K2 + 2K11 (e) K2 + Kll (f) Kz, (g) 4K1l (h) 3Kt, (i) 2K1 and (j) Kt. (Note that in case (h) there are two es­sentially distinct embeddings in P.) Write Q = P\(Pnr2(1)) = Pnr3(1). For each vertex 8 of Q we find C3 = 9 points in r(l) n r2(6). If we put e6 = lr(6) n (P \ Q)l, then e6

of these points lie in r(P), and the remaining 9 e6 are in r 2(P). Put A(8) = r(l) n r2(6) n rz(P). Now suppose Q contains a path 61 "" 62 "" 63 ; then by (ii) the set A(61)nA(62 )nA(63) must be empty. In particular, if we put a= IP\QI, so that lr(l) n r 2(P)I s 15- a, then the following inequality holds: 27- e61 - e6,- e63 S 30- 2a, or, equivalently, 3 + e61 + e62 + e63 ~ 2a. This observation will kill most of our ten possibilities. Indeed, suppose we have situation (b). Then we can find a path

61 "" 62 "' 63 in Q such that ~ = ( e6u e62 , e63 ) = (2, 1, 2), but a = 5, contradicting the above inequality. Similarly, in cases (c), (d) .and (e) we find a= 4, ~ = (2,0,0) and a = 4, ~ = (1, 1, 1) and a = 3, ~ = (1, 0, 0), respectively, a contra­diction in each case. In case (f) this argument is not strong enough, but we can find a Kt,3 in Q, say with vertices 6o,6t,6z,63, where So is the vertex of degree 3, and there must be a pair iJ in 1,2,3 such that A(60 )nA(6i)nA(6j) #- 0, again a contradiction. The cases (h), (i) and (j) can be ruled out by similar arguments, but since we shall not need the absence of these, we do not give the details. (iv) Suppose e1 II ez and e2 II e3. If e~, e2 and e3 are in one Petersen subgraph, then, as we observed above, e1 II e3 • So assume et,e2 C P1

and ez,e3 C 'P2, and put e1 = {ft,[2}· Now it follows from (i) that d('Y;, P2) = 2 (i = 1,2) and e2 c r2(f;)n'P2 (i = 1,2). By (iii) we have e3 c r2(l'i) n p2 (i = 1, 2) and therefore el and e3 determine a unique Petersen subgraph P3 containing them and e1 U e3 induces 2K2 as a subgraph of 'P3. This means that e1 II e3. (v) Each edge is in seven Petersen subgraphs and hence equivalent to 14 other edges. 0

· Let us define the following sets:

24

X 0 , the vertex set of r;

X 1 , the set of equivalence classes of edges of r;

X= Xo UX1 and

c = { {;, tS,E}I{;, 8} is an edge of rand E is the equivalence class of{;, 8} }.

LEMMA (3.4). (X,C) is a regular near hexagon with parameters (s, t2, t) = (2, 2, 14).

Proof. First of all, (X, C) is a partial linear space: an equivalence class E of edges and a collinear vertex 1 determine uniquely an edge e = {'y, 8} E E. Next, the partial linear space is connected, since r is connected. Note that X 1 is a coclique and that the collinearity graph of (X, C) induces ron X 0 •

Next consider the parameters. We have IXo I = 506, IXtl = 253 and hence lXI = 759. Each point is on t + 1 = 15 lines, each line contains s + 1 = 3 points. In order to show that t2 + 1 = 3, we must show that any two points e, 11 at distance 2 have three common neighbours. If e, 11 E X 0 , then they determine a Petersen subgraph P, and the three common neighbours. are their common neighbour in r and the two equivalence classes containing edges on them. If e E X 0 , 11 E Xb then 11 contains an edge {a,p}, where e ""' a. Now a, p and { are contained in a Petersen subgraph P, and any common neighbour of e and 11 must be in Xo (since X 1 is a coclique), and therefore must be one of the three neighbours of e on the three edges that 11 has in p. Finally, if e, 11 E X}, then e contains an edge {a, P} and 11 contains an edge {a,'{}; again a, p, 1 are contained in a Petersen subgraph P, and any common neighbour of e and 17, distinct from a, must be a vertex 8 at distance two from each of a:, {3 and 1, and therefore, by Lemma (3.3)(ii), in P; but in P we see three common neighbours of e and 11 (including a). This proves that any two points at distance 2 have precisely three common neighbours, i.e., t2 = 2. Now, let us check the near polygon axiom. Let L be a line, say L = {1,6, E} with E E Xt, and let e be a point with d(e,L) = 1. If e "" E, then e, 1 and 8 are contained in a Petersen subgraph P of r and {has distance two to the edge {;, 8}. Otherwise, if e E Xo, then e is adjacent to either 1 or 8 but not both, since r does not contain triangles, while if e E Xt and { "" 'f, 8 then e = E. Thus, the near polygon axiom is satisfied in this case.

25

Suppose d(e,L) = 2. He E Xo and d(e,7) = d(e,o) = 2, then e, 7 and o are contained in a Petersen subgraph P, and e is on an edge parallel to { 7, o}, so that e is collinear with E, contrary to our assump­tion. If e E Xo and d( e, 7) = d( e, E) = 2, then E contains an edge {a:, I'}' where e "' a:, but now a:, ,, i' 0 and e are all contained in one Petersen subgraph P of r, and we find e "' o, a contradiction. If e EXt and d(e,i) = d(e,o) = 2, then e contains two edges {a:,f'} and { (, 71}, where 7 ,...., a: and o ,...., (. But these two edges are contained in a Petersen subgraph 'P, and this situation contradicts Lemma (3.3)(i). Finally, if e E X 1 and d(e,7) = d(e,E) = 2, then e contains two edges {a:, I'} and{(, 77}, where a:"' 7 and, for some 9 E X 0 , {77, 9, E} is a line. But then the edge {a:,7} is contained in r 2(77), so there is a Petersen subgraph P of r containing 7], a: and 7, and therefore also f', (, o, 6. It follows that d(e, o) ~ 2, a contradiction. Thus, the near polygon axiom is also satisfied in case d( e, L) = 2. Given a line L, there are three points on L, 3 · 2 · ( 15 - 1) = 84 points at distance one to L, and 84 · 2 · (15- 3)/3 = 672 points at distance two from L. But 3+84+672 = 759, so (X, C) has diameter 3 and each point is at distance at most 2 from any line. 0

Brouwer ([Br1] showed there is a unique such near polygon, namely that with as points the blocks of the Steiner system S( 5, 8, 24 ), and as lines the triples of pairwise disjoint blocks. We observed that our set Xt is a coclique of size 253 in this near polygon. However, we know all cocliques of this size, as is stated in the following lemma.

LEMMA (3.5). (X, C) contains precisely 24 cocliques of size 253, namely tbe 24 sets of blocks of S( 5, 8, 24) containing any given symbol.

Proof. Let C be such a coclique. Since there are 3795 lines and each point is on 15 lines, it follows that C meets each line in precisely one point. In particular, if Q is any quad, then C n Q is an ovoid in Q, that is, the set of :five blocks containing a :fixed tetrad T. Each block in C meets T, for blocks disjoint from T are disjoint from some block containing T. If Q' is another quad, we :find another tetrad T'. If TnT' = 0, then there is a block on T disjoint from T', a contradiction. Thus, the set of 1771 = (2;) tetrads found by varying Q is a linked sys­tem and by a theorem of Erd&, Ko & Rado (see [EKR]) this collection of tetrads consists of all tetrads on a :fixed symbol. It follows that all

· blocks in C contain this symbol. 0

26

REMARK. At the end of Chapter 4 we will give an alternative proof of Lemma (3.5) using the results obtained in that chapter.

THEOREM (3.6). Up to isomorphism, there is a unique distance regular graph with intersection array {15, 14, 12; 1, 1, 9}.

Proof. Starting from an arbitrary graph r with the given inter­section array we find r as the complement of a 253-coclique in the collinearity graph of the unique regular near hexagon on 759 points. By Lemma (3.5) it follows that r is uniquely determined up to isomor­phism. 0

3.3 Uniqueness of a graph on 280 vertices.

In this section we give a rather tedious proof of the uniqueness of a distance regular graph with intersection array {9, 8, 6, 3; 1, 1, 3, 8} and distance distribution diagram

1 9 9 3 144 3

2 3 In [BCN, §11.3. 7] a few descriptions of such a graph r are given:

(i) the vertices of rare the rank 3 Hermitean 3 x 3-matrices over GF(4), where matrices are adjacent if they differ by a rank 1 matrix;

(ii) the vertices of r are the subplanes AG(2, 3) of PG(2, 4 ), where subplanes are adjacent whenever they have precisely one point in common and have a common tangent at that point;

(iii) the vertices of rare the words of weight 9 in a double trunca­tion of the binary Golay code, where adjacent codewords have Hamming distance 16;

(iv) r = .63(6), where {J is a vertex of .6, the unique distance regular graph with intersection array {21, 20, 16; 1, 2, 12}, cf. Chapter 1.

r was first described by Biggs & Gardiner in [BG] and also studied by Tchuda in [Tc]. The uniqueness of r will be shown by using description (ii), so let us finish this introduction by giving the other relations in this description, cf. [BCN, §11.3. 7]:

(*) two affine subplanes have distance 3 whenever they have pre­cisely three (non collinear) points in common, and distance 2 (respectively 4) whenever they have five points in common, car­rying one (respectively two) collinear triples.

27

3.3.1 On the structure of the graph.

For the remainder of this section r will be a distance regular graph with intersection array {9, 8, 6, 3; 1, 1, 3, 8}. First we will determine the distance distribution around an edge.

LEMMA (3.7). Let a"' f3 be an edge and let Dj = Dj(a,/3).

(i) There are no edges between Di and D~; equivalently, any in­duced hexagon in r with at least one pair of vertices at distance 3 contains three such pairs.

(ii) There are no edges contained in D~.

(iii) There are no edges between D! and D!.

Proof. Fori= 3 we have equality in (i) of Theorem (2.6). Hence, if 1 "' 6 and d( 1, a) = 3, then the three distances d( a, 6), d(/3, 1) and d(/3, 6) are not all equal. Thus, for j = 2, 3, 4, there are no edges between Dj and D~. 0

LEMMA (3.8). Any two vertices a and f3 of r at distance 2 are contained in a unique Petersen subgraph P = P(a,/3) ofr.

Proof. Immediate consequence of Lemmas (3.2) and (3.7). 0

Let P be the collection of Petersen subgraphs of r.

LEMMA (3.9) .. Let P E P, a rJ. P and a"' f3 E P. Then we have d(a,1) = 1 + d(/3,1) for all 1 ~ P.

Proof. This follows since both P and r satisfy .A = 0, I' = 1 and a2 = 2. 0

LEMMA (3.10). Let d(/,6) = 3 and let r2('Y)nr(6) = {c:be:2,e:3}. Then the three Petersen subgraphs Pi = P('Y,e:i) (i = 1,2,3) each contain one vertex of r(l) n r2(6) and two vertices of r('Y) n r3(6). Conversely, ifa,/3 E r('Y)nr3(6), then P(a,{3)nr(6) f= 0, moreover, each a E r( 1) n r 3 ( 6) is contained in precisely two of the Pi.

Proof. First of all, Pi f= Pj, otherwise 6 E P('Y,c:i)· Lemma (3.9) shows that Pi contains one vertex of r('Y) n r 2(6) and two vertices of r(l) n r 3(6). These three Petersen subgraphs thus coincide with the three Petersen subgraphs determined by the three pairs of vertices in

. r('Y) n r3(6). o

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LEMMA (3.11). r .bas distance distribution with respect to any edge a "' {3 as shown in Figure 5.

- 2 1

- 2 1

Fig. 5 Distance distribution of r with respect to an edge

Proof. Let 1 E Df. By Lemma (3.10), two of the three Petersen subgraphs on a containing a neighbour of 1 in r 2(a) do contain {3.

Thus, 1 has two neighbours in D~. Now we can uniquely complete the diagram, using the intersection array of rand Lemma (3.7). D

We will also need the distribution of the distances of the vertices of a Petersen sub graph P to a point 1 outside. If 1 has a neighbour in P, this was already done in Lemma (3.9).

LEMMA (3.12). Let P E P and d(1, P) ~ 2. Then d(T, P) = 2. Furthermore, we .have one of the following:

(i) P n r 3(1) is a 4-coclique and P n (r2(1) U r4(1)) is a 3K2

subgrap.b ofP, where there is exactly one edge in Pat distance 4 from 1;

(ii) Pn(r2(1)ur4(1)) is a4-codique, Pnra(1) is a 3K2 subgraph of P and there is a unique vertex in P at distance 2 from 1·

29

Proof. (1) Suppose P has an edge a . "' j3 in r 4(; ). Let r4(a) n r4(j3) n r("Y) = {6}. Since all distances between Di(;,6) and D1("'Y, 6) are 2, the four Petersen subgraphs on a "' j3 must together cover the 16 points of·Di. Thus, each contains 4 of them and is of type (i). (2) Suppose P contains a pentagon a "' j3 "' YJ "' 6 "' e "' a with a, j3 E ra("'Y). Then; E Di, 6 ED~, so d(J', 6) f= 4: ; has a, neighbour Tin Dj, T has distance 2 to 6. If d("Y,6) = 2, then by Lemma (3.11) we must have ; "' c ,...., 6 with c E Di, i.e. c E P and d("Y, P) = 1, a contradiction. Thus, all vertices of P not adjacent to a and j3 are in r a (I'). (3) Suppose P has an edge a "' j3 in r 2("Y). Then P is one of the three Petersen subgraphs on a ,...., {3 distinct from P(;, a). We have p n (r(a) u r({3)) c ra("Y) and, by (2), the remaining vertices in r2(;) u r4("Y). By (1), P has either three edges in r 2("Y) or two edges in r 2(T) and one edge in r 4("Y). But; has distance 2 to six vertices of Di \ P(a,;), i.e. on the average two for each choice of'P on a"' j3, so

the former case cannot occur and Pis of type (i). (4) Now we have seen 12 + 72 + 108 Petersen subgraphs (12 on ;, 72 with d(;, P) = 1 and 108 with an edge in r 4("Y)) and by counting it follows that the remaining 144 Petersen subgraphs have on the average one point in r 2(T), six points in r 3(;) and three points in r 4(T). But by the foregoing we know that each has at least six, and hence precisely six points in ra(;). For a E r 3(;) we have already seen the Petersen subgraphs on a not meeting r(a) n r 4(;), so all remaining Petersen subgraphs on a do meet r(a) n r4(T), and hence have at least three vertices in r 4 (T ). Consequently these must be of type (ii ). 0

If a E P E P, then the set r( a) n P will be called an a -line.

LEMMA (3.13). The a-lines form an affine plane AG(2,3). More­over, if i E ra(a), then the vertices of r(a) have distances to ; as indicated in Figure 6. Thus, we have three a-lines meeting r 3 (;) twice and r2(i) once, three a-lines meeting r4(;) twice and ra("Y) once, three a-lines meeting r2( ;) twice and r 4( ;) once and three a-lines meeting r2(;), ra(;) and r4( ;) all once.

Proof. Each Petersen subgraph containing a determines an a-line. For any pair of vertices 61 , 62 E r( o:) there is a unique Petersen sub­

. graph P = P(6t,62) containing a. Hence the a-lines form a Steiner

30

3 3 2

3 4 4

2 4 2

Fig.6 The distances from 1 to r(a) when d(1,a) = 3

triple system on nine points, i.e., an affine plane AG(2, 3). The dis­tance distribution on r(a) with respect to 1 E r3(a) is an immediate consequence of the Lemmas (3.10) and (3.12). 0

The aim of this subsection is to show that for any vertex oo of r the induced subgraph r 3 ( oo) is bipartite and has eight connected compo­nents. Furthermore we want to show that each vertex a of r 4 ( oo) has exactly one neighbour in each of the 8 connected components of r 3 ( oo ). In order to achieve this, we need a graph G on 72 vertices.

LEMMA (3.14). Let the graph G be defined as follows: the vertices of G are the noncollinear triples of points of AG(2, 3), where two triples

{x, y, z} and {a, b, c} are adjacent if and only if the set {a, b, c, x, y, z} is the set of six points on two parallel lines of AG(2, 3). Then G satisfies the following properties.

(i) G has four connected components.

(ii) G is bipartite.

(iii) Each connected component of G is an antipodal 3-cover of the complete bipartite graph K3,3 and has intersection array {3, 2, 2, 1; 1, 1, 2, 3}, i.e. each connected component of G is iso­morphic to the Pappus graph.

Proof. Let {x,y,z} be a noncollinear triple of AG(2,3). Then the three joining lines xy, xz and yz belong to three different parallel classes of AG(2,3). If {x,y,z}"' {a,b,c}, then the joining lines ab, ac and be belong to the same three parallelclasses, so we have at least four connected components. Now everything is straightforward. 0

31

LEMMA (3.15).

(i) r 3 (oo) is a 2--cover of G with quotient map 1r : r 3(oo) ....... G defined by 1ra = r(oo) n ra(a) for every a E ra(oo).

(ii) H {3 E ra(oo), then there is precisely one "Y E ra(oo) \ {{3} with 1r"'f = 1r{3. Moreover, r(oo) n r2(f3) = r(oo) n r2("Y) and each P E P with oo E 'P and d(/3, 'P) = 1 satisfies d("Y, P) = 1. In that case 61 "' e """ 62, where 61 E 'P n r(j3), 62 E 'P n r( "Y ),

61 'f= bz and e E 'P n r( oo).

Proof. In order to prove (i), we must show: a""" f3 implies 1ra,..., 1rj3 for a,/3 E ra(oo); for every a E ra(oo) and every X E G with X,..., 1t"a there is a f3 E ra(oo)nr(a) with 1r{3 = x; lra(oo)nr(a)l = IG(1ra)l for all a E r 3 (oo); lr3(oo)l = 2IGI, where IGI is the number of vertices of G. We will do this in six steps, the last step will also prove (ii). (1) 1ra is a vertex of G for every a E r 3(oo) by Lemma (3.13). (2) If a"' f3 is an edge contained in ra(oo), then oo E D:(a,/3) and Lemma (3.11) gives:

(1) lr(oo)nra(a)nr2(f3)1 = 1;

(2) !r(oo)nr2(a)nra(f3)1 = 1;

(3) lr(oo)nr4(a)nr4(f3)1 = 1;

( 4) 1rc oo) n r a (a) n r 4 (/3) 1 = 2;

(5) lr(oo) n r4(a) n ra(/3)1 = 2;

(6) jr(oo) n r2(a) n rz(/3)1 = 2.

Thus we must have 1ra "'11'{3 with Lemma (3.13). (3) lra(oo) n r(a)l = IG(11'a)i = 3 for every vertex a E ra(oo), for G has valency 3 by Lemma (3.14) and aa = 3. (4) Let f3 "' a ""' "'(, j3,a,"'( E ra(oo). Suppose 1rj3 = 11'"'(. By Lemma (3.13) we haver( oo) n r 2(.8) = r( oo) n r 2("Y). Let us take 6 in r(oo)nr2 ({3)nr3(a) = r(oo)nr2 ("Y)nr3 (a). (These sets have cardinal­ity 1 by Lemma (3.11): oo ED~ with respect to both the edges a""' f3 and a"' "'f.) Thus we have {3,"'( E r2(6) n ra(oo), a E ra(oo) n ra(6). By Lemma (3.11), with respect to the edge oo ,..., 6, we have f3 = "Y·

Thus 11'{3 i= 11'"'( or f3 = "Y·

(5) Ira( oo )I = 144, IGI 72. (6) Let x be a vertex of G. We will show that there at most two ver­tices in r 3 ( oo) which are mapped onto x by 11'. Suppose 1ra = 1rj3 = x, a,/3 E ra(oo). Let X= {et,e2,ea}, 'Pt = 'P(et,e2), 'Pz = P(et,ea)

32

a7 aa au Fig. 7 Two parallel cla.'lses of oo-lines

and "P3 = P(e2,e3). Then "Pin r(a) =f. 0 =f. "Pin r(t#), fori= 1,2,3 by Lemma (3.10). Let hi} = "Pin r(a), {6i} r('Yi) n r(oo) and {ai} = (r(6i) n r2(oo) n "Pi)\ {'yi}. Now "Pin r(/.:1) ~ {'Yi,ai}, because r(oo) n r 2(,8) = r(oo) n r 2(a), by Lemma {3.13). We want to show "Pi nr(,B) = {ai}, which proves both (ii) and our claim above. Suppose for some io, "Pion r(,B) = hio}· Then Pin r(,B) = {aj} for j =f. io, otherwise a and ,8 would have at least two common neighbours, contra­dicting p = 1. So we have an hexagon a "' 'Yio "' ,8 "' a j "' 6j "" 'Yi "" a (io =f. j), where d(,B,-y;) = 3 by Lemma (3.9). Hence, by Lemma (3.7)(i), d(a,aj) = d('Yi0 ,6j) = 3. But oo, 'Yi and O"j are in r2{'Yi0 ) and oo, 'Y;, O"j E "Pj n r( Dj ), so the 6;-line { oo, 'Yj, a;} is contained in r2{'Yi0 )

contradicting Lemma (3.13). Hence, "Pin r(,B) = {ai} for all i. 0

LEMMA (3.16).

(i) Let a E r4(oo) and B := {xlx E tr,B for all ,8 E r3(oo) n r(a)}. Then B =f. 0. In particular 6 E B, where {6} r(oo) n r 4(a).

(ii) Let a E ra(oo), ,8,-y E r4(oo) n r(a), ,8 =f. 'Y· Then r4(,8) n r4('Y) n r(oo) = 0.

Proof (i) Take ,8 E ra(oo) n r(a), 6 E r(oo) n r4(a). Now a E Dj(oo,6) and, since ,8 E r3(oo), ,8 E D~(oo,6) by Lemma (3.11). Thus 6 E r( oo) n r3(,8), i.e., 6 E tr,B. (ii) Suppose on the contrary there is a vertex 6 E r 4 (,B) n r 4 ( 'Y) n r( oo), then we have ,8,-y E Dl(oo, 6). By Lemma.(3.11) we have a E Df(oo, 6), but the same lemma tells us a has only one neighbour in Dj giving ,8 = 'Y, a contradiction. 0

33

Now let us fix a vertex oo of r. Let r( oo) = { O't, u2 , ••• , u9 } and let the oo-lines be as in Figure 7, i.e., we have the following oo-lines (with lines from one parallel class on one line):

{u1,u2,u3}, {u4,us,u6}, {ur,us,ug};

{ullu4,0'7 }, {u2,0's,us}, {ua,u6,u9};

{ut,O's,u9}, {u2,u6,0'7 }, {ua,u4,us};

{ubu6,us}, {u2,0'4,u9}, {ua,us,ur }.

Figure 8 shows one connected exponent of G explicitly, the other three components are isomorphic to this one. Each balloon is a vertex of G, where the ij k inside stands for { O'i, 0' i, O'k}. We will use this component to achieve the wanted information on r 3 ( oo ).

Fig.8 A connected component of G

34

Furthermore let us define certain types of Petersen subgraphs. Let 'P E P, then we call 'P of type:

(1) if d(oo, 'P) = 1;

(2) if d( oo, 'P) = 2 and 'P satisfies (i) of Lemma (3.12);

(3) if d(oo, 'P) = 2 and 'P satisfies (ii) of Lemma (3.12).

LEMMA (3.17). Let 'P E P be of type (3). If {3 E 'P n f4(oo) and r(,B) n 'P = { O:t, a2, a a}' then 7r<l~l' 11"0:2 and 7l"O:a belong to three different components of G.

fJ

«t

Fig.9 The vertices of 'P in the proof of Lemma (3.17)

Proof Let r be the unique vertex at distance 2 from oo in 'P, and name the vertices of 'P as is shown in Figure 9, then we have a:baz,a:a,6t,6a,6s E fa(oo) and {3,6z,64 E f4(oo). We have 1l"O:t "' 11"61, 1ra:2 ,...., 1rlia and 11"0:3 "' 7r6s. Suppose 1ra2 and 7l"O:a belong to the connected component of G given in Figure 8, then 1r8a and 1r65 are also in this component. If 'P' = 'P( oo, 1 ), then 'P' n r 2 ( r) n f( oo) is con­tained in fa( at) n r3(6a) n fa( lis), thus, j1ra:t n 1r6a n 7r6sl2 2. Lemma (3.16)(i) shows that each of the sets 7rO:t n 71"0:z n ?ra:a, 7r6t n 7r63 n 7rO:a and 1r61 n 1r65 n 1ra:2 is nonempty. If 1r83 = 1r85 , then 1ra:3 n 1r63 = 0, impossible, so l1r63 n 7r8sl 2. We may assume 1r63 = { O't, a2, u4} and

35

1rlis = {ut.u2,us}. Then we get 1ra1 = {ut,u2.,ua}, 1rlii = {u4,ug,ug}, 1raa = {u4,ua,ug}, 1ra2 = {ua,us,ug} or 7rO:t = {u1,u2,us}, 1rlit = {u4,ur,us}, 1raa = {u4,us,ur}, 1ra2 = {us,ur,us}. Let us assume we are in the first case. Now Ut.U2 E P', so d(1,u3 ) = 1, d{l,ut) = d{l,u2) = 2 and d(/,ui) = 3 (i 2:: 4), since 1J.z = 6. Let P" = P(ua,ug). Then ua E P", for {ua,ua,u9} is a oo-line. We have d(a2, P") = d(aa, P") = 1 by Lemma (3.10), so let us take ct, respec­tively c:2, to be the neighbour of a 2, respectively a 3 in P". Lemma (3.10) gives C:t ,..,., ua ,.., c:2. Now ct =I c2 (for p, = 1 and a2 ,..,., (3"' aa), ct,€2 f/_ p (for r2(oo) n p = {t}) and d(l,c:t) = d(/,£2) = 3 (for ci and e2 have distance one to P). But 1 "" ua ""' C:t. e2: a contradiction. Hence 1ra2 and 1ra3 do not belong together to the given component of G. Similarly one can handle the other cases. D

LEMMA (3.18). Let a1 "' a2 ,.., a3 "' a4 be a path in ra(oo). Then:

(i) !1rat n 1ra4l = 1 (respectively 2) if and only if d(ab a4) = 3 (respectively 2 );

(ii) if d(a1.a4) = 2, then there are a 5 ,a6 E r 3(oo) such that

a1 "' a2 "' aa "' a4 "' as "' aa "' O:t is a 6-gon;

(iii) if d(at,a4) = 3, a1 "'f3t "'(32 "'a4 and a1·"" /1 "'/2 "'a4, then d(f3t, oo) = d(f32, oo) and d{II, oo) = d(12, oo );

(iv) there is a 6-gon contained in r 3 ( oo ), which contains all the a;, 1 :5 i :5 4.

Proof. (i) Let P = 'P( ai, a a), then P is of type ( 1) and hence

lr(aa) n ra(oo) n PI = 2. Since lr(aa) n ra(oo)l = 3, aa has one neighbour in ra(oo) n r2(at) and one in r3(oo) n ra(at), say /I and 72, respectively. (Note that a 4 must be one of those two;) We may assume 1ra1 = {ut.u2,u4}, 1ra2 = {ua,us,ua} and 1ra3 = {ut.ur,us}. (The other cases can be handled similar.) We have /I E P and hence r(ai) n r(;t) c r2(oo), thus 17rat n 'lr/tl 2:: 2 by Lemma (3.9). If 1ra1 = 7r/t, then 7rO:t "' 1ra2 "' 1raa ""' 7r/t = 1ra1 is a triangle in the bipartite graph G, which is impossible. Hence l1ra1 n 1r11l = 2, 'lr/1 = {o-2,174 1 175} and 'lr/2 = {u2,u3,U9}, i.e. l1ra1 n 'lr/21 = 1. (ii) In this case we have a4 E P = P(at,aa), which is of type (1). Now p n r 3 ( 00) is the desired 6-gon. (iii) Suppose (31 E r2( oo ), (32 E f3( oo ). Then, with P = P( oo, f3t), we

36

Fig.JO The vertices of'P in the proof of Lemma (9.18}

have d(f3z, 'P) = d(it1 , 'P) = 1. By Lemma (3.9) we have 111"it1 n11'/3zl ~ 2. On the other hand, d( 11'iti, 11'it4) = 3, for G is bipartite and contains no quadrangles, cf. Lemma (3.14). However, 1r/32 ""' 11'it4 in G and thus l1r/32 n 11'ittl :5 1, a contradiction. Suppose f3t E r4(oo), f3z E r3(oo), then the previous lines show l1r/3z n 11'it1 1 :5 1 and, by Lemma (3.16), hence l1r/32 n 11'itd 1. Let 'P = 'P( it}, f3z) and name the vertices of 'Pas shown in Figure 10. Since it}""' it2 "'it3 "'it4,..., !32 is a path in r3(oo), 11'itt and 11'/3z must be in one connected component of G and have distance 2 or 4 (recall d(11'itt,11'it4) = 3). By Lemmas (3.12) and (3.17) 'P must be of type (2), giving /31>61 E r4(oo), itl>/32,64,65 E r3(oo) and 62,63,66,67 E r2(oo). Now let us take 11'it1 = {ut,O"z,u4}, 11'it4 = {uz,u3,u9} (the other cases can be treated similar again), then 1r/32 = {u1,u7,u8 } without loss of generality. Let 'P' = 'P( 62, oo ), then 63 E 'P'. By definition of 11'it1 and our knowledge of the oo-lines, · 'P' corresponds to one of the following oo-lines: {u11u2,u3}, {u11u4,u7} or {u2,0'4 1 0'g}. In the same way, considering 11'/3z, 'P' corresponds to one of the following oo­lines: { u1, 0' 4, u7}, { ub <76, O"s} or { u7, us, O'g }. Thus 'P' corresponds to the oo-line {ut,0'4,u7}, i.e. 'P' 'P(u1,u4). In the same way we see

37

P" = 'P{oo,66) = P(u11u4), i.e. P' = P", soP'= P(6z,66) = P and oo E P, a contradiction. (iv) If d(at, a 4) = 2, then we are done by (ii). So we are left with the case d(a11a4) = 3 and l1rat n 1ra4l = 1. Let f3tJ3z,"'fb'Y2 be as in (iii). It satisfies to show {/3t,'Yt} n ra(oo) =I= 0, since then our claim follows by (iii). Let us take 7r<:¥t = {ut,O'z,u4}, 1raz = {ua,us,u6}, 7raa = { O'J, 0'7, us} and 1ra4 = { uz, O'a, 0'9} (the other cases can be treated similar again). Suppose {f31l'Yd n r 3 (oo) = 0. Then we are in one of the following three cases:

(a) /3t,"'(I E r4(oo);

(b) f3t E rz(oo), 'Yt E r4(oo);

(c) f31!'Yl E rz(oo).

If (a) holds, then, by Lemma {3.16)(ii), r 4(/3t) n r 4('Yt) n r(oo) = 0, r 4(/3z) n r 4('Yz) n r(oo) = 0, and, by Lemma (3.11) we have lr4(f3t) n r4(/3z) n r(oo)l = jr4('Yt) n r4('Yz) n r(oo)l = 1 (note that oo E D1 with respect to both the edges f3t ,..., f3z and 'Yt ,..., 'Yz). Now l1rat n 1ra4l ~ 2, by Lemma (3.16)(i), which contradicts (i). If (b) holds, then, by Lemma(3.16)(i), r 4('Yt) n r(oo) C {ullu2 ,u4} and r 4('Y2 ) n r(oo) c {u2 ,u3 ,u9}. As in case (a) above, we have

lr4('Yt)nr4('Yz)nr(oo)l = 1, hence {uz} = r4('yt)nr4('Yz)nr(oo). Now we have a 6-gon a1 "" f3t "' f3z "' a4 "' 'Yz "" 'Yt "' a 1 with d( a11' a 4) = 3 and hence d(f3t,"'fz) = 3 by Lemma (3.7)(i). Let P = P(oo,/31), then f3z E P and by definition of 1ra1 , 1ra4 P corresponds to one of the fol­lowing oo-lines: { O't, O'z, ua} or { O'z, 0' 4, 0'9} and f3t "' ua or f3t "". 0'9

by Lemma (3.9). Hence P contains an edge oo '"" u2 in r 4('Y2 ). By Lemma (3.12) and the fact d(f31!'Y2) = 3 we get f3t"" oo or f3t "'O'z, a contradiction. If (c) holds, then take P = P( oo, /31) and P' = P( oo, 'Yt ). Observe P =/= P', for otherwise the common neighbour 0:1 of f3t and 'Yt would be in 'P, which is obviously ridiculous. By definition of 7re¥t, 1ra4 we may assume P corresponds to the oo-line { 0'1, O'z, ua} and P' to the oo-line {uz,0'4 1 0'g}. Recall our assumptions on 7I"O:t and 1ra4; by Lemma (3.9), observing /32 E 'P, we have f3z "' u1 and d(/32 , ui) 3 for 4 ~ i ~ 9. Furthermore, as before we have d(/3z,'Yt) = 3. Now we have an edge u9 ,..., 'Yt in P' nr3 (/32 ) and an edge oo,..., u2 in P' nr2 (/32), which is im­possible by Lemmas (3.9) and (3.12). This finally proves the lemma. D

38

LEMMA (3.19). f 3 (oo) has eight connected components, all of them isomorphic with a connected component of G.

Proof. Lemma (3.18) shows that every path of length 3 in f 3(oo) is contained in a 6-gon in ra( oo ). In G every path of length 3 is in a unique 6-gon. Now our claim is straightforward. 0

LEMMA (3.20). Let a E r 4 ( oo ). Then a has exactly one neighbour in each of the 8 connected components of r 3 ( oo ).

Proof. It suffices to show IBn r( a) I ~ 1 for each connected compo­nent B of r 3(oo) (for c4 = 8). If a E PEP and Pis of type (3), then Lemma (3.17) shows IPnr(a)nBI ~ 1 for each connected component B of ra(oo). Let us suppose fJ,"' E r( a) n B with (3 =J. "' for some connected component B of r 3(oo). Then P = P(f3,"f) is of type (2). Lemma (3.16)(i) gives 1r(3 n 1r"f =J. 0, hence in G we have d(1r(3,1r'Y) E {2,3}, thus, by Lemma(3.19), d(f3,7) E {2,3} in the subgraph ra(oo). Thus d(f3,"f) = 3 in r 3 (oo) (for p = 1). But the paths of length 3 join­ing (3 and"' must be in P and P does not contain edges of r 3(oo): a contradiction. 0

The last lemma of this section will be useful in section 3.3.3, but before we will prove this lemma, we will need one other lemma.

LEMMA (3.21). Let a E r4(oo), (3 E r2(oo) n ra(a). Then (r(a)nr2((3))u(r2(a)nr((3)) ct. r 3(oo).

Proof. First of all, we have lr 2 ( 00) n r 3 (a) I = P~3 = 32. Let {at}= r(oo) n r 4 (a). By Lemma (3.12) the four Petersen subgraphs containing oo"' a 1 all contain two elements ofr3(a)nr2(oo), but they have a neighbour at distance 2 from both oo and a. The other eight Petersen subgraphs containing oo, but not a 1 , all have three elements of ra(a) n r2(oo), but in each such Petersen subgraph two of those three elements have again a neighbour at distance 2 from both oo and a. Hence, there are eight elements, say 'Yb····"fs, of ra(a) n r2(oo) with the property r2(a) n r(-Yi) c ra(oo). FUrthermore, there are 8 paths a "' Cl "' £2i "' C3i with Cl E r 4( 00 ), C2i E fa(oo) and £ai E r2(oo)nra(a) (i = 1, ... ,8). Let a"' £t "'£2 "'£3 be such a path and suppose £3 "'T E r2(a) n r2(oo) and T "' 6 E r(a). By Lemma (3.7) we have d(6,c2) = 3 and by

39

Lemma (3.11) we have ct E Dl(oo,O't) and c,e2 E D~(oo,O't), i.e., O't E 7rC n 1re2. On the other hand, r,ea fl. Di(a,et) hence, by Lemma (3.11), O't f/. 'P(oo,r). Suppoae 'P(oo,r) = 1'(0'4,0's), T "" 0'4, ea "' O's, then we get 1r8 = {O't,O's,0'6} and 1re2 = {O'i,0'4,0'6}· This gives 1r8"' 8' and 1re2 ""e' with 8',e' E 'P(O'b0'6) n r(O's) and 8' ::/:: e'. Now d( r, 8') = d(e3 ,e') = 2, but by Lemma (3.12) we must have d( e3 , c') = 4, ridiculous. We conclude that, after suitable renumbering of the "fi, 'Yi = eai (i = 1, ... ,8), which proves our lemma. 0

LEMMA (3.22). Let 1'11 1'2 and 1'3 be three distinct Petersen subgraphs containing oo. Then the distances d( a, {3), with a E 1'1 ,

f3 E 1'2 U 'Pa, determine uniquely the distances d('Y, c), with 'Y E 1'2 and 8 E 'Pa.

Proof. The strategy is to show that d( a, /3) is determined for some a E 1'2, f3 E 'Pa, with a at distance 2 from 'Pa, a,/3 E r2(oo). Then all distances are determined by the Lemmas (3.9) and (3.12), knowing d( 0' h a) for all i. We have to consider four cases:

(a) 1'1 II 1'2 II 'Pa;

(b) 1'1 II 1'2, I'Pa n 'Ptl = I'Pa n 'P2I = 2;

(c) I'Pt n 1'2 n 'Pal= 2;

(d) I'Pt n 1'2 n 'Pal = 1, no two of the 3 Petersen subgraphs are parallel.

Furthermore, if oo E 'Pt E P and 1'1 n r(oo) = {O'i,O'j,O'k}, then let au -ali "" alk "" bu "" btj ,...., btk "' ali be the 6-cycle p, n r2( 00 ), where au "" 0' i, etc. (a) Let O't,O'z,O'a E 1'1. 0'4,0's,0'6 E 'Pz and 0'7,0's,0'9 E 'Pa. Using Lemma (3.12) we may assume d(bn, bz4) = d(b13, bag) = 3. We will show d( b24, bag) = 3. If this is not the case, then d( b24, bag) = 4. We would have b13 "' cp "' T "" bz4 and b1a ""' cp "" e "' bag with 1rr.p = {O't,O'z,O'.t}, 7rT = {O'a,O's,0'6} and 1re = {O's,0'7,0's}. We get d(r.p,0'8 ) = d(r,0'8 ) = 4 by Lemma (3.13), but d(b38 ,cp) s; 3 and d(bas,r) = 4 (by Lemmas (3.12) and (3.13), for d(r,bag) = 3), a con­tradiction with Lemma (3.11) (b) Let O't, 0'2, 0'3 E 'Pt. 0'4, O's, 0'6 E 'Pz and O't, 0'4, 0'7 E 'Pa. Using Lemma (3.12) we may assume d(b13, azs) = 3 and d(b13, aa7) = 2. Then d(a2s,aa7) E {2,4}. Suppose d(azs,aa7) = 4. Let 'P' = 'P(aa7,b13),

'then 'P' is of type (3) with respect to the vertex azs. Hence, if

40

"Y E r(bta) n r2(a2s) n r2(aa7), then "Y is a neighbour of aa4 and bu, i.e., "Y E ra( oo) with tr"'( = { u~, u2, u1 }. On the other hand, 6 E r(bta) n r2(a2s) implies rr6 \ {u1,u2} C {us,us,ug}. Thus, by Lemma (3.21), {a26} = r4(aa7) n r(a2s) c r2(bta), a contradiction. (c) Let O't,0'2,ua E P1, O't,0'4,0'7 E 'P2 and O't,0'6 1 0's E 'Pa. We may assume d(au,a27) = d(au,aas) = 2. Also we have d(a27,aas) E {2,4}. Suppose d( a21, aas) = 2, then there is a 6-cycle a13 "" "'( "" aas "" 6 "" a21 "" c ,....., a13 with "'(,6,e E ra(oo) and tr"'( = {ubu2,u6}, rr6 = {ut,u4 , u6} and Tre = {ut, u2, u4}. By Lemmas (3.12) and (3.13) it follows that d(a1a,6) = 3 and, by Lemma (3.7), d(a21,"'() = 3. Con­sider 'P' = 'P( 0'7, e); it contains 6 and a27 . But we have a27 E r 3 ('1) and 6,u7,e E r2("Y), which contradicts Lemma (3.12). (d) Let 0'1!0'2,0'3 E 'P11 u1,u4,0'7 E 'P2 and u2,u4,0'9 E 'Pa. We may assume d(au,a27) = d(a13,aa9) = 2. Then the neighbour "Y E ra(oo) of a13, with tr"'( = {ut,0"2,u4} is adjacent to both a27 and aag, which therefore have distance 2. 0

3.3.2 Affine subplanes in a projective plane.

In this subsection we construct the projective plane PG(2,4) using our graph r and our fixed point oo. After that we will study affine subplanes AG(2, 3) of PG(2, 4) containing a given collection of points.

Let { L1 , L2, ... , L12} be the collection of oo-lines.

LEMMA (3.23). Define a 21 x 21-matrix M with entries from {0, 1} and both rows and columns indexed by ul! ... , u9 , Lb ... , L 12 as fol­lows:

1 ifi =i = O'k, 1::5 k ::59;

1 ifi = O'k, j = Lt (or conversely) and O'k E L,,

M;; = 1 ::5 k ::5 9, 1 ::5 l ::5 12;

1 if i =Lt., j = Lt and Lk n Lt = 0, 1 ::5 k, 1 ::5 12;

0 otherwise.

Then M is the incidence matrix of the projective plane PG(2, 4).

Proof. After observing the fact that M is symmetric it suffices to show that each row contains exactly five ones and that each pair of distinct rows has exactly one 1 in common in order to prove M to be the incidence matrix of the projective plane PG(2,4). These two claims, however, can be checked straightforwardly. 0

41

Observe that, since M is symmetric, interchanging points and lines gives us an Hermitean polarity of PG(2,4) with nine fixed points, viz. O"t, ... , u9, that form an affine subplane AG(2, 3). Let us denote U00 = {ut, ... ,u9 } and let us call the affine subplanes AG(2,3) of PG(2, 4) unitals.

a;:s '----

Fig.11 The projective plane PG(2, 4) with U 00

Figure 11 shows us PG(2,4), the quadrangle with 9 points, shown in the middle, is U00 • There are four triples of points outside, each triple corresponds to a parallel class of U 00 : if { £1, £2, £3} is such a parallel class, then we have a triple of points, also denoted by { Ll! L 2 , L 3 },

such that Li E Lj if and only if i of: j. Except the 12 lines, which are extensions of Lt, ... , £12, there are 9 more lines. Each of these lines, indexed by ul! ... ,u9 , contains one point of U00 and one from each of the four triples: O"i contains O"i and the points Lj with O"i E Lj. The line O"i is a tangent of U00 through the point O"i· We will use this figure in the proof of the next lemma.

LEMMA (3.24). Let {x, y, z} be a given noncollineartriple of points , ofU00 • Then there are two unitals U, such that Un Uoo = {x,y,z}.

42

If U A and U B are these unitals, then U A contains all three points Li, L;, Lk of the parallel classes not containing the joining lines xy, xz and yz and from each of the other triples U A contains exactly one point, viz. the intersection point of a joining line and the line parallel to it containing none of x, y and z. U B contains the six points outside U 00 which are not contained in U A.

Proof. Let U be a unital such that U n U 00 = { x, y, z}. Each line of U contains exactly 3 points, so the lines xy, xz and yz contain exactly one point outside U00 • Hence: b1 or b2 in U, a1 or aa in U, c2 orca in U. If b2 E U, then the line b2z has another point in U, hence b3 E U. Similarly a3 E U implies a2 E U and c2 E U implies c1 E U. If b1 E U, then ba ft U, for otherwise the line b1 ba would have a third point in U, but this has to be a point of U 00 \ { x, y, z}. Hence, either b1 E U or b2,b3 E U, and in the same way either a1 E U or a2,a3 E U and either c3 E U or cllc2 E U. Furthermore, if some di E U, then all dj are in U. Now U contains nine points, so if U does not contain any di, then

we must have a2, a a, b2, ba, c1 , c2 E U and if U contains all the d j, then all bb c3 E U. The last possibility gives us U A, the first one U B· 0

Let us call a unital U with IU n Uool = 3 of type A (respectively of type B) if U is like UA (respectively like UB) in Lemma (3.24).

LEMMA (3.25). LetS= {u,v,x,y,z} be a set offive points ofU00

and suppose that S satisfies one of the following:

(i) Sis the set of points on two intersecting lines of U00 ;

(ii) Sis a set offive points on two parallel lines ofU00 •

Then there is exactly one unital U with U n U00 = S.

Proof. Similar to the proof of Lemma (3.24). 0

LEMMA (3.26). For each Ui there is exactly one unital Uu; which intersects U00 only in Ui.

Proof. First observe that U can contain at most two points of each

of the four triples { a1, a2, a a}, ... , { d1, d2, d3} outside U. Also U must contain nine points, hence each triple must have exactly two points in U. Each triple corresponds to a parallel class of lines of U 00 , one of those lines meets Uij the two points must be the points on this line outside U00 • D

43

3.3.3 An isomorphism.

In this final subsection we are going to construct an isomorphism u : r - r*' where r· is the graph of description (ii) in the intro­duction of this section: In other words, we will construct a bijection U between the vertices of r and the vertices of r•' which preserves adja­cency: a "' {3 if and only if U a "' U {3.

Let us start by defining Uoo := U00 and Uui := Uap 1 ::::; i::::; 9, then we already have Uoo"' Uui in r• by Lemma (3.26).

Let a E r 2( oo ), 'P = 'P( a, oo) corresponding to the oo-line L 1 and let L2, L 3 he the two oo-lines parallel to L 1 . Looking at r•, we see that there are two possible definitions for U a: U a = U, where U n U oo = ( Li U Lt) \ r( a), for i = 2 or 3.

Let {3 E r 3(oo), then, looking at r• again, we see that there are two possible definitions for U{j: either U(3 n Uoo = r 2((3) n r(oo) and U(3 is of type A or U{j n Uoo = r 4((3) n r(oo) and U(3 is of type B.

LEMMA (3.27). Let a"' {3, a,{j E r2(oo) U f3(oo), Ut, U2 the possible choices for U a and U3, U4 the possible choices for U {3. Then, maybe after interchanging u3, u4, we have the following intersection sizes in the following ca8es:

(i) if a,(3 E r2(oo), then lUI n U31 = IU2 n U41 = 5 and IU1 nU41 = IU2 nU31 = 1;

(ii) if a E r2(oo) and {3 E r3(oo), then lUI n U31 = IU2 n U41 = 1 and IUt n U41 = IU2 n U31 = 3 and Ua, u4 are of distinct types;

(iii) if a,(3 E ra(oo), then IUt n Ual = IU2 n U41 = 5 and lUI n U41 = IU2 n Ual = 1 and Ut, u3 are of one type and u2, u4 are of the other type.

Proof. (i) If P = 'P(oo,a) corresponds to the oo-line {ut,0'2,u3 }

and a "" u1 , (3 "' 0'3, then we have:

UI n Uoo ={u2, ua, 0'4, Us, C!6},

U2 nUoo ={u2,ua,u7,us,u9},

Ua n Uoo ={ut,u2,u4,us,u6},

U4 nUoo ={ubu2,u7,us,u9}.

Now our claim follows by an easy calculation, the other cases are similar.

44

(ii) Suppose a "' o-3 and P = P( oo, a) corresponds to the oo-line {o-t,o-2,<13}. Assume 1r/3 = {o-t,o-2,<14}, then we have:

ul n Uoo ={<1t,<12,o-4,<1S,<16},

u2 n Uoo ={o-I,<12,<17,0"s,o-9},

U3 n Uoo ={o-3,o-1,o-9},

u4 nuoo ={o-s,o-6,o-s},

where U3 is of type A and U4 of type B. Again our claim follows with an easy calculation and the other cases can be treated similar again. (iii) Suppose 1ra = {o-1,o-2,o-4}, 1r/3 = {o-3,o-5 ,o-6}· Then we have

U1 n Uoo ={o-3,o-1,o-9},

U2 n Uoo ={ o-s, o-6, o-s},

U3 n Uoo ={o-4, o-1, o-9},

U4 n Uoo ={o-t, o-2, as},

where Ull U3 are of type A and U2, U4 are of type B. Our claim follows by an easy calculation, the other cases can be handled similar. D

Now let us fix a vertex a E r 2 ( oo) and fix a choice for U a.

LEMMA (3.28). We can define U f3 for all f3 E f2( oo) U f3( oo) such that: U f3 = U 'Y implies f3 = 'Y and U f3 "' U 'Y if and only if f3 ,....., 'Y for every pair f3, 'Y with d(f3, oo) ~ 3, d('Y, oo) ~ 3.

Proof. After defining U a, there is a unique way to define U f3 for all f3 E f 2( oo) n P( oo, a). When we have done this, there is a unique way to define U f3 for all f3 E r 2 ( oo), such that the distances between vertices of f 2( oo) is in accordance with the intersections of their images under U, as described in (*). This can be done by using the distances between vertices of the Petersen subgraphs containing oo. After this, we can define U 6 for every 6 E f 3 ( oo ), by using Lemma (3.27) and U'Y for some 'Y E f 2(oo) n r(6). It is straightforward to check that our claims hold, using Lemma (3.22). D

Finally, we have to define U'Y for 'Y E f4(oo). In order to do this, we once more have to study Petersen subgraphs on 'Y E f 4(oo).

45

LEMMA (3.29). Let fJ ,...., "'(, fJ,"'f E r4(co), fJ,"'f E P E P. Let {r} = r( co) n r 4("'1) and Pn r("Y) = {{J, 6, e }. Then 1re: U 1r6 is a pair of intersecting lines of U co with T as point of intersection. Furthermore, U g and U 6 are of the same type.

Fig.12 The vertices ofP in the proof of Lemma (9.29)

Proof. Let us take 1re = {ut,u2,u4}, T = u1. Lemmas (3.17) and (3.20) show 1r6 is in the same connected component of G as 1re. Also we have u 1 = r E 1r6 by Lemma (3.16)(i) and u5 ,u9 f/:. 1r6. Since 82,8a,84,86 E r2(co), we can consider P' = P(co,84) and P" = P(co,82 ). Then P' =f. P" (otherwise they would be both equal to P, implying co E P), hence P' and P" correspond to two different co-lines. Using the definition of 1re, 1r6 and the fact that 0'9 f/:. 1r6, these co-lines must be {ut,0'2,ua} and {ut,0'4,u7}. Now 1r6 = {ut,O'a,ur}, for 6""' 6a f ua (recall e,...., 62 ,...., ua) and 6 "'84 f 0'7 (e "'66 ""' 0'7 ), and 1re U 1r6 is the set of points on two lines intersecting in u1 = T.

c and 6 belong to different components of ra( co) by Lemma (3.20), in G we have d( 1r£, 1r6) = 3, so if U c is of type A, then U li', where li =f. 6' and 1r6 = 1rli', is of type B, by Lemma (3.27), hence U 6 is of type A. 0

46

LEMMA (3.30). Let 'Y"' ~' 'Y E f4(oo), ~ E fa(oo), ~,"( E PEP, p of type (3). If p n r('Y) = {~,81,82}, T E rz('Y) n rz(oo) n P, then exactly two elements of { U ~, U 81 , U 62 } are of the same type, viz. those corresponding to the vertices of {~,811 82 } \ f(r). Moreover, if {et,ez} = {~,8~,82} \ r(r), then 1rt1 and 1re2 are on the intersecting lines of Uoo as determined by e1 using Lemma (3.29). If e3 is the commpn neighbour of 'Y and r, then 7rea is on tbe other two lines of U oo through tbe intersection point.

fJ

Fig.19 The vertices ofP in the proof of Lemma (9.90}

Proof. Let us name the vertices of P as shown in Figure 13. Let us

take f(oo) n r 4('Y) = {0"1 } and 1r{3 = {O"ll0"2 ,0"4}. We have

T E rz(oo), ~,8t,8z,8a,84,81 E ra(oo) and ,,65,66 E r4(oo). Now Lemma (3.16)(i) gives O"t E 1r62 n 1r61 n 1r~. Since 1r82 n 1r67 = 0, we have 0"1 t/. 1r61, hence 0"1 t/. P' = P(oo,r), for 1r{3 n P' = 1r61 n P'. SoP' corresponds to the oo-line {0"2 ,0"4,0"9}. Let us denote by Ca(x) the connected component of G containing x. Then 1r61 t/. Co(1rP) by Lemma (3.17). So, 1r61 n {D"t,D"s,O"g} = 0, because {O"z,0"4} c 1r61. Take 1r67 = {O"z,0"4,0"6} (other cases can be treated similarly), then

1r6a = { O"a, D"s, 0"6} or {0"6, D"s, 0"9 }, for 1r6a n1r61 =/= 0 by Lemma (3.16)(i).

47

Furthermore we must have:

- 1r62 = {u11 as,a9} or {u17 ua,us}, for 62 "' 67 and hence 1r6z "' 1r67 in G;

- 1r64 = {u2,0'4 1 0's} or {u2,u4,u7}, for {u2,a4} C 1r64 and 1r64 ~ Ca( 1r/3), Ca( 1r61 );

- 1r6z n 1r63 n 1r64 ::j:. 0 by Lemma (3.16)(i).

Thus, 1r6z = {at,as,ag}, 1r63 = {u6,as,ag}, 1r64 = {uz,u4,as}, 1r61 = {u1 ,u6 ,ug}, the last one follows since u1 E 1r611 1r61 n 1r67 ::j:. 0. Now an easy calculation shows U /3, U 64 and U 61 are of the same type and hence U 61, U 62 and U t5a are of the other type. This proves the first claim of our lemma. Lemma (3.29) gives that the intersecting lines of U oo determined by 1r01 are the lines {ut, as, 0'9} and { a1, 0'6, as}. Thus, 1r /3 = { u1 , u2 , a 4 } is on the other two lines through the intersection point O'J. 0

LEMMA (3.31). Let 1 E f4(oo). Take /3 E f4(oo) n f(1),

/3,1 E P E P. Let P n f(1) n f3(oo) = {t5t,62}· If U61 and U6z are of type B, then let us define U 1 as the unique unital U such that U n U oo = 1r61 U 1r62 • If U 61 and U 62 are of type A, then U 1 is the unique unital U such that U n U oo = (U oo \ (1r61 U 1r62)) U (1r61 n 1r62).

Then, if 6 E f 3 (oo) U f 4(oo), 1 "'6 if and only ifU1"' U6 in f*.

Proo£ Straightforward calculations show that U 1 is well-defined and that the last statement holds. D

The lemmas in this subsection show that U : r -+ f* is an isomor­phism, hence we have shown

THEOREM (3.32). Up to isomorphism there is a unique distance regular graph with intersection array {9, 8, 6, 3; 1, 1, 3, 8}.

48

CHAPTER 4: THE HYPERPLANES OF THE REGULAR

NEAR HEXAGON ON 759 POINTS

4.1 Hyperplanes and embeddings.

Let (X, C) be a finite connected partial linear space such that all lines have exactly q + 1 points, where q is a prime power. A hyperplane of (X, C) is a subset H C X such that for each line L E C either

L is contained in H or L intersects H in exactly one point. (Such a set is called a projective hyperplane by Teirlinck [Te] and a geometric hyperplane by Ronan [Ro].) We will denote the collection of all hyper­planes of (X, C) by 1i. A (projective) embedding of (X, C) is a mapping cp : X --+ PW (not necessarily injective), where PW is the projective

space associated to the vector space W over GF(q), such that cp[L] is a line in PW for all lines L E C. If K is a hyperplane of the span of cp[X], then clearly H = {x E Xlcp(x) E K} is a hyperplane of (X,C). An embedding t/J : X --+ PW is called universal if for any embedding cp : X --+ PW there is a linear map a : V --+ V such that cp = at/J, where

V = span t/J[X] and V = span cp[X]. If a universal embedding t/J ex­ists, then the span of tfJ[X] is obviously unique up to isomorphism. We will give a universal embedding later for the case q = 2.

From now on we will take q = 2 (so we have AG(d, 2) = PG(d, 2)U{O} ).

DEFINITION. Let (X, C) be a connected partial linear space with exactly three points on each line and let 1i be the set of all hyperplanes of(X,C).

(1) Let F C 1i and let W be the binary vector space with basis {eHIH E F}. We define the mapping cp;:: X--+ PW by

{1ifxtf_H

(cp;:(x))H= OifxEH.

(2) For hyperplanes H, H' E 1i we define "their sum"

H ED H': = {x E XI('P1l(x))H + (cp1l(x))H' = 0}

= (H n H') U ((X\ H) n (X\ H')).

49

LEMMA (4.1). ('H,€9) is an elementary abelian 2-group with zero element X.

Proof. Straightforward. 0

THEOREM (4.2). Let V = span IPF[X], where :F is a subgroup of('H,€9). Then the correspondence HE :F +-+ {v E V!vn = 0} is a 1-1 correspondence between the elements of :F and the hyperplanes of v.

Proof. Everything is obvious after we have shown that each hy­perplane of V is of the form { v E V !v H = 0} for some hyperplane H E :F. So let us take a hyperplane K of V. K can be extended to a hyperplane of PW, where W is the binary vector space with basis

{eniH E :F}. Hence K is a set {v E Vlvn1 + · · · + vn .. = 0}, for some Ht, ... , Hn E :F. Now take H = Ht ffi · · · ffi Hn. Then H E :F and K = {v E V!vn = 0}. 0

THEOREM (4.3). (cf. Ronan [Ro]) Let us call a subset :F C 1t separating if for each line L E C there exist H, H' E :F that meet L in distinct points. Then:

(1) H :F is separating, then 'PF is an embedding of(X,C).

(2) If'H is separating, then 'P11 is a universal embedding of(X,C).

Proof. The first statement is straightforward to verify. In order to prove the second statement let us take an embedding 1.p and let V = span ~.p[X], with basis {e1, ... ,enl· Then Hl! . .. ,Hn E 1-£,, where Hi := {x E Xj(~.p(x))i = 0}. Now there is a natural isomorphism f between the span of 'PF[X] and V, where :F = {Hh ... ,Hn}· If we take for g : span 'P11[X] -+ span 'PF[X] the linear map which sends en to 0 whenever H ft. :F and to itself whenever H E :F, then fg is a linear map from span 'P11 [X] to V. 0

Observe that 'PF is injective if and only if for every two points x and y there is a hyperplane H E :F such that IH n { x, y} I = 1. For the remainder of this chapter we assume that (X, C) is a near polygon.

CONJECTURE. If (X, C) is a near polygon, then 1t is separating and 'P11 is injective.

The conjecture is obviously true if (X, C) is a regular near polygon.

50

By the near polygon axiom and Lemma ( 4.1) the subsets of X defined below are hyperplanes.

DEFINITION. Let (X, C) be a near polygon of diameter d. We define the following subsets of X:

(1) H(x) := {y E Xld(x,y) < d}, x EX;

(2) H(xt. ... , Xn) := H(xi) (f) .. • (f) H(xn), x1, ... , Xn EX.

Finally, let us define the subgroup :Fo.

DEFINITION. :F'o := {H(xJ, ... ,xn)lxt, ... ,xn EX}, i.e. :Fo is the subgroup of1-l generated by {H(x)lx EX}. The projective spaces V and V are defined by V = span lf'F0 [X] and V = span ~Pn[X].

In this chapter we will determine all hyperplanes of the (unique) regular near hexagon on 759 points. However, we will first consider the hyper­planes of a generalized quadrangle on 15 points, in which the situation is analogous to the situation in the hexagon on 759 points.

4.2 A small example.

Let (X,C) be the generalized quadrangle GQ(2,2) which can be con­structed as follows: take as point set X the set of all unordered pairs from a 6-set A, say A= {0, 1, 2,3,4, 5}, and take the triples of mutually disjoint pairs as lines. By Ronan [Ro] we have dim V = 4, dim V = 5. The embedding into V gives (X, C) as system of all points and all to­tally isotropic lines in Sp(4,2); the embedding into V gives (X,C) as system of all isotropic points and totally isotropic lines in 0(5, 2).

Let us denote the pair { i, j} by i j. The 32 hyperplanes of (X, C) are:

(a) X;

(b) H(ij) = {ij} U {mn E Xl{m,n,i,j} has cardinality 4} = {mn E Xld(mn,ij) < 2}, for all ij EX;

(c) [i] :={ikE Xli =/; k E A} for all i E A (the ovoids of (X, C));

(d) [i] (f) [j] ED [k] = {mn E Xlm E {i,j,k} and n E A\ {i,j,k} }, for all triples i,j, k of distinct elements of A.

We have one hyperplane of type (a), 15 of type (b), 6 of type (c) and 10 of type (d), for it is easy to see that:

- H(x) (f) H(y) = H(z) whenever {x, y, z} is a line;

- H(ij) (f) H(ik) = H(jk) whenever j =/; k;

51

- H(ij) = [i] ffi [j]; - [i] ffi [j] ffi [k] = [l] El1 [m] ffi [n], where {i,j,k} U {l,m,n} is a

partition of A into two triples.

Hence the hyperplanes of type (a) and (b) together form the collec­tion :Fo, and the remaining hyperplanes can be obtained by "adding" a fixed hyperplane not in :F0 to the ones in :F0 • The same situation will occur with respect to the hyperplanes of our near hexagon on 759 points, on which we will now focus our attention.

4.3 The near hexagon on 759 points.

Let r =(X, E) be the graph with as vertices the blocks of the Steiner system S(5, 8, 24), two blocks being adjacent whenever they are disjoint. r is distance regular with intersection array {30, 28, 24; 1, 3, 15} and its singular lines have size 3. Each singular line is a collection of three mutually disjoint blocks, i.e., each singular line is a trio of S(5, 8,24), cf. [CS]. Let C be the set of singular lines of r, then (X, C) is a regular near hexagon with parameters s = 2, tz = 2 and t3 = 14, cf. [SY]. (X,C) and rare uniquely determined by their parameters, cf. [Br1]. The automorphism group of S(5, 8, 24), the Mathieu group Mz4, is an automorphism group of (X, C). It acts distance transitively on (X, ..C), i.e., given ordered pairs of points ( x, y) and ( u, v) with d( x, y) =;= d( u, v ), there is an automorphism a E M24 such that ax u and ay = v. Our aim in this chapter is to find the hyperplanes of (X, C). In terms of r this means we look for subsets H of X such that each triangle of r either has exactly one point in H or is entirely contained in H.

We will have to perform some explicit computations in S(5, 8, 24) and M 24• An important tool in order to do so, is the Miracle Oc­tad Generator, or shortly, the MOG, which was invented by Curtis, see [Cu]. In Conway & Sloane [CS] one can find a lot of informa­tion on S(5, 8, 24), M24 , the MOG and computations using the MOG. Figure 14, taken from [CS], shows the MOG.

One can compute blocks of 8(5,8,24) using the MOG, due to the fact that for the intersection of a block and a trio there are only three possibilities:

- the block is one of the blocks of the trio;

- the block intersects two of the blocks of the trio ea.ch in four symbols;

52

~~ ~fJ ~~ ~m ~~ urn ~~ ~~ ~~ ~~ ~~ ~~ ~~ ~~ ~~ ~~ ~~ ~~ ~~ ~~ '~~ ~~ ~~ ~~ ~~ ~~ ~~ ~m ~B ~~• ~~0· ~~00· c~·· ~~0· ~~00· • • •0• • • •0• •00• ..

Fig.14 The Miracle Octad Generator

- the block intersects one of the blocks of the trio in four symbols and the other blocks of the trio each in two symbols.

The MOG gives a standard trio with the standard blocks

{0, oo, 19, 3, 15, 6, 5, 9} '

{1,11,20,4,14,16,21,13} and

{2,22,10,18,17,8, 7,12}.

Each of the 35 pictures shows some sextets: the left rectangle is a stan­dard block divided into two tetrads, the right quadrangle consists of the other two standard blocks and is divided into four tetrads. For example, the first picture on the top line gives the following three sextets:

{O,oo,6,9},{19,3,15,5},{1,4,17,12},

{2,18,14,13},{21,16,10,22},{7,8,20,11};

{1,11,16,13},{20,4,14,21},{0,3,17,12},

{19,oo, 7,8},{5,6,10,22},{15,9,2,18} and

53

{2,22,8,12},{10,18,17,7},{0,3,14,13},

{19,oo,21,16},{15,9,1,4},{5,6,20,11}.

Each sextet of 5(5,8,24) can be found in this way, so one can eas­ily perform computations on blocks of S( 5, 8, 24) using these pictures. For example, if we want to know the block containing the symbols 0, oo, 1, 22, 12, then we find that this block also contains the symbols 15, 3, 21 by the picture in the right bottom corner.

When we need explicit automorphisms of M24, then these usually will stabilize a block, have a given action on the block and will fix a symbol outside the block. Now the stabilizer of a block is a subgroup 24 : A8

of M 24 • The normal subgroup 24 consists of all elements that fix the block pointwise. A general element of 24

: As acts as an element of As on the symbols of the fixed block and is completely determined by this action as soon as we know its action on one symbol outside this block. For example, suppose we want u to act as follows: it fixes 0, oo, 19, 3, it interchanges 15 with 6 and 5 with 9, and it sends 1 to 4. Considering the action of u on the sextets

{O,oo,19,3},{15,6,5,9},{1,11,20,4},

{14,21,16,13},{2,22,10,18},{8,7,12,17}'

{O,oo,6,9},{19,3,15,5},{1,4,17,12},

{11,20,7,8},{21,16,10,22},{14,13,2,18}'

{0,19,15,5},{oo,3,6,9},{1,20,14,21},

{11,4,16,13},{2,10,17,7},{22,18,8,12}'

{0,3,5,9},{oo,19,15,6},{1,20,8,12},

{11,4,17,7},{2,10,16,13},{22,18,14,21} and

{0,3,6,9},{oo,19,15,5},{1,11,2,22},

{20,4,10,18},{14,16,17,8},{21,13,7,12}

54

we get

q = (0)( 00 )(19)(3)(15 6)(5 9)(1 4)(11 20)

(14 21 )(13 16)(2)(22)(10)(18)(17 8)(7 12).

In Conway & Sloane [CS] one can find more information about such computations.

The connection between the distance of two blocks in (X, C) and the size of their intersection is as follows:

- they have distance 1 if they are disjoint;

- they have distance 2 if their intersection has cardinality 4;

- they have distance 3 if their intersection has cardinality 2.

For a given block, there are 30 blocks at distance 1, 280 blocks at distance 2 and 448 blocks at distance 3.

A quad in (X, C) corresponds to a sextet as follows: each block of the quad is the union of two tetrads of the sextet, which is determined by the intersection of two blocks of the quad at distance two from each other. The distance of a block x to a quad Q can be given in terms of its relation to the corresponding sextet as follows:

- d( z, Q) = 0 if z intersects the sextet like 4204 , i.e., if z intersects two tetrads each in four symbols and is disjoint from the other four tetrads;

- d( z, Q) = 1 if z intersects the sextet like 2402 , in this case the block of Q disjoint from z is denoted by 1rz;

- d( z, Q) = 2 if z intersects the sextet like 1531 , in this case the tetrad of the sextet intersected three times by z is denoted by Oz.

In a fixed quad we can label the tetrads with the numbers 0, 1, 2, 3, 4 and 5 and identify any of its blocks by the unordered pair of labels of the tetrads in that block. This gives us the possibility to write [i] for the set of blocks in a quad containing the tetrad i. The meaning is the same as in section 4.2: a quad in (X, C) is just a generalized quadrangle GQ(2, 2). Thus, if z is a block at distance 2 from a quad Q, then [Oz] is the set of points of Q at distance 2 from z.

Finally, let us introduce one more useful notion, the line distribution of a hyperplane H.

55

DEFINITION. Let H be a hyperplane of (X,£) and let ai be the number of points of H which are on exactly i lines of H (0 ::; i ::; 15). Then the sequence oao ... 15415

, where i 4' is omitted if ai = 0, is called

the line distribution of H.

The line distribution turns out to be very useful to distinguish different types of hyperplanes. Given a hyperplane, its line distribution can be determined very easily with the help of a computer. Let us give the line distribution of the hyperplanes of (X,£) that we have seen sofar.

LEMMA ( 4.4). The hyperplanes of (X,£) generated by at most one point are:

(X) X, which has 759 points and line distribution 15759 ;

(H1) H(x), for all x EX, which has 311 points and line distribution 32801531.

Both of these types give one orbit under the action of M24 on the hyperplanes.

Proof. X is obviously a hyperplane, H ( x) is a hyperplane by the near polygon property. The line distribution of H(x) is as stated since x has 30 neighbours, 280 points at distance 2 and the remaining 448 points are at distance 3. The last statement is obvious. D

4.4 The hyperplanes generated by two points.

LEMMA ( 4.5). If {x, y, z} is a line of size 3 in a near polygon, then H(x) ffiH(y) = H(z).

Proof. This follows immediately from the near polygon axiom. D

LEMMA (4.6). The hyperplanes of(X,£) generated by two points (but not by fewer) are:

(H2) H(x, y) with 439 points and line distribution 38 t 24 9256 11144157 ,

where d(x, y) = 2;

(H3) H(x, y) with 367 points and line distribution 330572 7180940 1145 ,

where d(x, y) = 3.

Each of those types gives one orbit under the action of M24 on the hyperplanes.

56

Proof. By Lemma (4.5) the hyperplanes generated by two points can only be of type (H2) or (H3). The distance transitive action of M24

yields the last statement. 0

The last lemma of this section will be useful in the next sections.

LEMMA (4.7). Let x,y E X and suppose d(x,y) = 2. Then x = 01 and y = 02 in Q(x,y) and we have for i E {3,4,5}: H(01) E11 H(02) = H(i1) E11 H(i2).

Proof. If z E H(01) $ H(02), then:

- if z E Q(x, y), then z E H(mn) for all mn E Q(x, y);

- if z is at distance 1 from Q(x, y), then either z intersects all of the tetrads 0, 1 and 2 twice or 1r z = 12 in Q( x, y );

- if z is at distance 2 from Q(x, y), then either z intersects the tetrad 0 three times (i.e. Oz = 0) or z intersects all of the tetrads 0, 1 and 2 once, i.e. 0 z E { 3, 4, 5}.

In all these cases it is easy to verify that z E H( i1) $ H( i2). D

4.5 The hyperplanes generated by three points.

In order to find the hyperplanes generated by three points we add a point to the generating set of the hyperplanes of type (H2) and of type (H3). By the transitive action of M24 on both of those types it suffices to add a point to a fixed generating set. Let us first add a point to the generating set of a hyperplane of type {H2), in other words, we consider triples { x, y, z} for fixed x and y with d( x, y) = 2. If we are interested in hyperplanes generated by three points we have not met upto now, then by Lemma ( 4.5) we have to consider the following cases:

(i) d(x, z) = d(y, z) = 2 and z E [Oxny];

(ii) d(x, z) = d(y, z) = 2 and z E Q(x, y) \ [Oxny];

(iii) d(x, z) = d(y, z) = 2 and z has distance 1 to Q(x, y);

(iv) d(x, z) = d(y, z) = 2 and z has distance 2 to Q(x, y);

(v) d(x,z) = 2,d(y,z) = 3 and z has distance 1 to Q(x,y);

(vi) d(x,z) = 2,d(y,z) = 3 and z has distance 2 to Q(x,y);

(vii) d(x,z) = d(y,z) = 3 and z has distance 1 to Q(x,y);

(viii) d(x, z) = d(y, z) = 3 and z has distance 2 to Q(x, y).

Four of these cases, viz. (ii),(iii),(iv) and (vi), generate hyperplanes of a new type.

57

LEMMA ( 4.8). The hyperplanes generated by three points x, y and z (but not by fewer), where {x, y} generates a hyperplane of type (H2), are:

(H4) H(x, y, z ), where x, y, z satisfy (ii) above, which has 375 points and line distribution 7360 1515 ;

(H5) H(x, y, z), where x, y, z satisfy (iii) above, which has 375 points and line distribution 36596712691281118151;

(H6) H(x, y, z), where x, y, z satisfy (iv) above, which has 375 points and line distribution 318 5547189g781136;

(H7) H(x,y,z), where x,y,z satisfy (vi) above, which has 351 points and line distribution 3355126 7120g7o.

Each of those types gives one orbit under the action of Mz4 on the hyperplanes.

Proof. Computercalculations give the line distributions as stated, hence these hyperplanes must be of new types. The transitive action of Mz4 on each of those types follows from the following properties of M24, cf. Conway & Sloane [CS, Ch.lO]: M24 is transitive on sextets and the stabilizer of a sextet acts as the symmetric group 56 on the tetrads of this sextet. M 24 is also transitive on octads and the stabilizer of an octad is 6+ 1 transitive on the octad and its complement. The fact that no other new types arise in the cases (i ),( v ),(vii) and (viii) follows from the next lemma. 0

LEMMA (4.9). The set {x,y,z} generates a hyperplane of type (H2) if it satisfies (i), of type (H3) if it satisfies (v ), of type (H5) if it satisfies (vii) and of type (H6) if it satisfies (viii).

Proof. Suppose { x, y, z} satisfies (i ), then we may assume without loss of generality x = 01, y = 02, z = 03 in Q(x, y). Lemma ( 4.5) gives

H(01) $ H(02) $ H(03) = H(34) $ H(25) $ H(02) $ H(03)

= H(14) $ H(15).

Hence H(x,y,z) is of type (H2). If { x, y, z} satisfies ( v), then we may assume that in Q( x, y) we have x = 01, y = 02, 1rz = 23. Let v be the third point on the line containing z and 1rz, so d(v, 13) = 3. Then, using Lemma (4.5), we have

H(Ol) E& H(02) E& H(z) = H(23) E& H( 45) E& H(02) E& H(z)

= H(13) $ H(v).

58

Hence we have a hyperplane of type {H3). A similar argument can be used in case (vii) if 1r z E [ 0 zny]. Other­wise, using the 2+ 1 + 1 +0+0+0 transitivity of the subgroup 26 .3 of M 24

fixing the tetrads of a sextet individually ( cf. Conway & Sloane [CS, Ch.lO]), we may assume z n x = z n y = {0, oo}, 14,17 E z, x = {0, oo, 19,3, 15,6,5, 9} and y = {0, oo, 19, 3, 1, 11,20,4}. Then z = {O,oo,14,17,16,22,18,12} and using the computer it is easy to check that: H(x) EfJ H(y) EfJ H(z) = H(v1 ) EfJ H(v2) EfJ H(va), where

VI= {3,14,19,16,22,18,8,7};

v2 = {O,oo,22,18,8,7,13,21};

v3 = {O,oo,14,16,2,8,10,7}.

Hence we have a hyperplane of type (H5). In case (viii) we may assume x = 01, y = 02, [Oz] = [3] in Q(x, y), but, by Lemma (4.7), H(01) EfJ H(02) EfJ H(z) = H(31) EfJ H(32) EfJ H(z) and we are back in case (iv). 0

Now let us add a point to the generating set of a hyperplane of type (H3). Then we have a triple {x,y,z} and we may assume x, y and z have mutual distance 3, for otherwise we are back in a previous case. There are two distinct cases to consider, viz. the blocks x, y, z have either 0 or 1 symbol in common. (They cannot have two symbols in common, for this would give a word of weight 20 in the extended binary Golay code, which is impossible.)

LEMMA ( 4.10). The hyperplanes generated by at most three points are the hyperplanes of type (X), (Hl), ... ,(H7) or

(H8) H(x, y, z), where d(x, y) = d(x, z) = d(y, z) = 3 and the blocks have no symbol in common, which has 399 points and line dis­tribution 5247Isog120 1175.

I

(H9) H(x,y,z), where the distances are as in the previous type but the blocks have one symbol in common, which bas 383 points and line distribution 35 578 7120gtso11ao.

Each of those types gives one orbit under the action of M24 on the hyperplanes.

Proof. The previous Lemmas and computercalculations show ev­erything except the transitive action of M 24 on the types (H8) and (H9). Since M24 is transitive on octads and the stabilizer of an octad is

59

6+1 transitive, we may assume x = {O,oo,3,19,5,6,9,15}, O,oo E y, 3, 19 E z and 1 E z n y in the case of (H8). Now there are six octads satisfying the requirements on y and there are also six octads satisfying the requirements on z:

Yo= {O,oo,1,11,2,18,8,12}, Yl = {O,oo,1,11,22,10,17,7}, Y2 = {O,oo,1,2,22,4,16,13}, Ya = {O,oo,1,4,14,21,10,18}, Y4 = {O,oo,1,17,8,20,16,21}, Y5 = {O,oo,1,7,12,20,14,13}, z0 = {3,19,1,4,16,13,10,18}, Z1 = {3,19,1,11,2,18,17,7}, Z2 = {3,19,1,11,22,10,8,12}, za = {3,19,1,4,14,21,2,22}, Z4 = {3,19,1,20,16,21,7,12}, Z5 = {3,19,1,20,14,13,17,8}.

We have 24 pairs (y, z) with d(y, z) = 3 and the stabilizer of both x and the symbol! acts transitively on those pairs: take the automorphisms O'J, u2, 0'3 fixing 1 and x, acting on the symbols of x as (15 6 5), (15 59), (0 oo )(15 5). This shows the transitive action of M24 on the hyperplanes of type (H8). In case (H9) we may assume x = {0, oo, 19, 15, 5, 6, 9, 3}, 0 Ex n y n z, oo E x n y, 19 E x n z and 1 E y n z for the same reasons. We find the same six octads satisfying the requirements on y, and again six octads satisfying the requirements on z. For each of the first six octads there is exactly one of the other six octads such that they have distance 3. Using the same three automorphims as above we get the transitivity of the hyperplanes oftype (H9). 0

We will follow the same procedure in order to determine the hyper­planes generated by four points: add a point to the generating set of a hyperplane generated by three points and look if this set generates a hyperplane of a known type or of a new type. (Of course we will only add points which will give sets not studied before.)

60

4.6 The hyperplanes generated by four points.

We will start by adding a point to the generating set of a hyperplane of type (H4). So let {x, y, z} be a generating set of such a hyperplane, i.e. d(x,y) = d(x,z) = d(y,z) = 2 and z E Q(x,y) \ [Oxny], and let us add a point u. In order to find a new hyperplane, by Lemma ( 4.9) we have to consider the following cases (where Q = Q( x, y )):

(4.i) d(u,x) = d(u,y) = d(u,z) = 2 and d(u,Q) = 1;

(4.ii) d(u,x) = d(u,y) = 2, d(u,z) = 3 and d(u,Q) = 2;

(4.iii) d(u,x) = 2, d(u,y) = d(u,z) = 3 and d(u,Q) = 2;

(4.iv) d(u,x) = d(u,y) = d(u,z) = 3 and d(u,Q) = 1;

(4.v) d(u,x) = d(u,y) = d(u,z) = 3 and d(u,Q) = 2.

LEMMA ( 4.11). The set { x, y, z, u} generates a hyperplane of type (H5) if it satisfies (4.i) and of type (H9) if it satisfies (4.ii) or (4.v). The cases ( 4.iii) and ( 4.iv) do not occur.

Proof. Suppose we are in case ( 4.i). We may assume { u, v, 1ru} is a line and x = 01, y = 02, z = 12, 1ru = 34 in Q. Thus we have

H(u) Ef) H(01) Ef) H(02) Ef) H(12)

= H(v) Ef) H(1ru) Ef) H(23) Ef) H(45) Ef) H(02) Ef) H(12)

= H( v) Ef) H(05) Ef) H(23) Ef) H( 45) Ef) H(02)

= H(v) Ef) H(14) Ef) H(13)

by Lemma ( 4.5). Lemma ( 4.8) now implies that we have a hyperplane of type (H5). Suppose we are in case ( 4.ii). Since the subgroup 26 .3 of M 24 fixing the tetrads of a sextet individually is 2+ 1 + 1 +0+0+0 transitive on the tetrads in any order (cf. Conway & Sloane [CS, Ch.10]) we may assume

x = {O,oo,19,3,15,6,5,9},

y = {O,oo,19,3,1,11,20,4},

z = {15,6,5,9,1,11,20,4},

u = {0, oo, 19, 15, 1, 13, 17, 18} or u = {0, oo, 3, 15, 1, 22, 21, 12}.

However, we may assume 19 E u: take a in the stabilizer of x determined by the action (0 oo )(19 3) on x and fixing 1, then a fixes x, y, z and interchanges the two possibilities for u. Now we have

61

H(x) ED H(y) ED H(z) ED H(u) = H(vt) ED H(v2) ED H(v3), as can be checked with a computer, where

v1 = {O,oo,3,6,20,16,2,12},

V2 = {0, 3, 19, 5, 14, 4, 10, 8},

v3 = {oo,3,19,9,21,11,7,22}.

Hence case ( 4.ii) gives a hyperplane of type (H9). If we are in case (4.v), then we may assume x = 01, y = 02, z = 12 and Ou = 3 in Q. Now by Lemma (4.7) we are back in case (4.ii): H(01) ED H(02) ED H(12) ED H(u) = H(31) ED H(32) ED H(12) $ H(u). The cases (4.iii) and (4.iv) cannot occur: if d(u,Q) = 1, then u inter­sects four tetrads of Q twice and it is disjoint from the other two; if d(u, Q) = 2, then u intersects one tetrad of Q in three symbols and all the other tetrads once. But this means that the distances of u to x, y

and z cannot be as required. D

Now let {x, y, z} be a generating set of a hyperplane of type (H5), i.e., d(x,y) = d(x,z) = d(y,z) = 2, d(z,Q1) = d{y,Q2) = d(x,Q3) = 1, where Q1 = Q(x, y), Q2 = Q(x, z) and Q3 = Q(y, z). This means that x, y and z have exactly one common neighbour, which we call m, and the three quads and three lines through m form a Fano plane PG(2, 2), for the 15 lines and 35 quads on a given point of X form a PG:(3, 2), as was shown by Ronan & Smith in [RS]. We will add a point u to this set and in order to find new hyperplanes we only have to consider the cases where each triple of points gener­ate a hyperplane of type (H5), ... ,(H9), because we know already all "extensions" of hyperplanes of smaller type. This leaves us with the following cases:

(5.i) d(u,x) = d(u,y) = d(u,z) = 2,

d(u,QI) = 1;

(5.ii) d(u,x) = d(u,y) = d(u,z) = 2,

d(u, Qt) = d(u, Q2) = d(u, Q3) = 2;

(5.iii) d(u,x) = d(u,y) = 2, d(u,z) = 3, d(u,Q1) = 1, d(u,Q2) = d(u,Q3) = 2;

(5.iv) d(u,x) = d(u,y) = 2, d(u,z) = 3,

d(u,Qt) = d(u,Q2) = d(u,Q3) = 2;

(5.v) d(u,x) = 2, d(u,y) = d(u,z) = 3, d(u,Q1) = d(u,Q2) = 2, d(u,Q3) = 1;

62

(5.vi) d(u,x) = 2, d(u,y) = d(u,z) = 3,

d(u,Ql) = d(u,Qz) = d(u,Qa) = 2;

(5.vii) d(u,x) = d(u,y) = d(u,z) = 3, d(u,Qt) = d(u,Qz) = d(u,Q3) = 1;

(5.viii) d(u,x) = d(u,y) = d(u,z) = 3, d(u,Ql) = d(u,Qz) = 1, d(u,Q3) = 2;

(5.ix) d(u,x) = d(u,y) = d(u,z) = 3, d(u,QI) = 1, d(u,Qz) = d(u,Q3) = 2;

(5.x) d(u,x) = d(u,y) = d(u, z) = 3, d(u,Qt) = d(u,Q2) = d(u,Q3) = 2.

LEMMA ( 4.12). The hyperplanes generated by a set { x, y, z, u} satisfying one of the requirements mentioned above are already known or of one of the following three types:

(HlO) H(x,y,z,u), where {x,y,z,u} satisfies (o.i), m and u are on one line, which does not belong to the Fano plane formed by Q1 and Q2. H(x,y,z,u) has 407 points and line distribution 3147120g2241142157;

(Hll) H(x,y,z,u), where {x,y,z,u} satisfies (o.ii), which has 407 points and line distribution 5227165gnonno;

(H12) H(x,y,z,u), where {x,y,z,u} satisfies (o.ix) and 1ru fl. [Ozny], which has 367 points and line distribution 324564 71929641123 •

Each of those types gives one orbit under the action of M24 on the hyperplanes.

Proof. Computercalculations give the mentioned line distributions, so the three types are indeed new. M 24 acts transitive on the hyper­planes of type (H5), hence we may take fixed x, y and z. One can find all u satisfying the requirements for type (HlO), (Hll) or (H12) and as before find automorphisms fixing the set { x, y, z} and acting transi­tively on the collection of all possible choices of u in each of the three cases. The only thing left to prove is that in all the other cases of ( 5.i) upto (5.x) known hyperplanes are generated. In case (5.i) if 1ru :f. 1rz in Qt. then, using Lemma (4.7) on Q(u,x): H(u) El1 H(x) El1 H(y) El1 H(z) = H(v) El1 H(x') El1 H(y) El1 H(z), where { u, v, 1ru} and { x, x', 1ru} are lines of our near hexagon. But now H(x') El1 H(y) El1 H(z) satisfies case(v) of Lemma (4.9) and we get a known hyperplane. If 1ru = 1r z then either u E Qz, u E Q3 or u has

63

distance 1 to all of the three quads. H ( u) E9 H ( x) E9 H ( z) satisfies case(ii) of Lemma ( 4.8) or case (i) of Lemma ( 4.9) in the first case, hence H(u) E9 H(x) E9 H(y) E9 H(z) is a known one. The second case is analogous to the first one. In the last case let us fix x, y and z. There are two possible choices for u and both choices turn out to give known hyperplanes. In case (5.vi) let {x,a,m}, {y,b,m} and {z,c,m} be lines of the near hexagon. We may assume

x = 01, y = 02, a = 23, b = 13, m = 45 in Q1 ,

x = 01, z = 02, a = 23, c = 13, m = 45, in Q2 and

y = 01, z = 02, b = 23, c = 13, m = 45, in Q3.

Then we have Ou = 1 in Qt, Ou = 1 in Q2 and Ou = 3 in Q3. Now d(u,m) = 3 and, using Lemma (4.7) on Q3 , we get back in case (5.ii): H(x) E9 H(y) E9 H(z) E9 H(u) = H(x) E9 H(b) E9 H(c) E9 H(u). In all the other cases we fix x, y and z, determine all possible choices for u, for one of the choices we determine the hyperplane which turns out to be known. Also one can show that the setwise stabilizer of { x, y, z} is transitive on the collection of possible choices, therefore all hyperplanes in these cases are known ones. In case (5.viii) we have to distinguish between some possibilities: 1ru E Oxny in Q1 or not, and a similar one with respect to Q2. However, only the possibility 7rU E Oxny

for Q1 and the analogous possibility for the second quad turns out to be possible. 0

Using the same kind of arguments again and again one can handle the "extensions" of the hyperplanes of type (H6), ... ,(H9). Only two more types appear, we will give the results without the details.

LEMMA ( 4.13). The hyperplanes generated by at most four points are the hyperplanes of type (X), (Hl), ... , (H12) or

(H13) H(x,y,z,u), where d(x,y) = d(x,z) = d(y,u) = d(z,u) = 2, d(x,u) = d(y,z) = 3, both u and z have distance 2 to Q(x,y) and u has distance 2 to Q(x,z). H(x,y,z,u) has 407 points and line distribution 7330 1577 ;

(H14) H(x,y,z,u), where all distances between x, y, z and u are 3 and lx n y n z nul = 1, which has 343 points and line distribution 36051927301160151.

64

Each of those types gives one orbit under the action of M24 on the hyperplanes.

Proof. Similar to the previous proofs. 0

4. 7 The hyperplanes of :F0 •

In this section we will study the hyperplanes generated by 5 points of our near hexagon. In order to find new hyperplanes we only have to consider sets { x, y, z, u, v }, where each 4-subset generates a hyperplane

. of type (HlO), ... ,(H14). H we are looking for "extensions" of a hyper­plane of type (HlO), (HU) or (H12), then each triple of points from the generating set should generate a hyperplane of type (H5), ... ,(H9), for otherwise we get known hyperplanes. For similar reasons, if we are looking for "extensions" of a hyperplane of type (H13), then each triple should generate a hyperplane of type (H7),(H8) or (H9), and finally, if we are looking for "extensions" of a hyperplane of type (H14), then each triple should generate a hyper­plane of type (H9). Due to the transitive action of M24 on the collection of hyperplanes of a fixed type we may fix x, y, z and u in each of those cases. We can de­termine all v in each case and then proceed in the same way as before: find automorphisms of M24 fixing the set {x,y,z,u}, determine the or­bits on the collection of all possible v under the subgroup generated by these automorphisms and study the hyperplanes for a fixed v from each orbit. In some cases using Lemmas (4.5) and (4.7) we get back in previous cases finding known hyperplanes, in other cases with the help of a computer we find that the generated hyperplanes are known ones. No new hyperplanes are found, hence:

THEOREM (4.14). Let (X, C) be the regular near hexagon on 759 points. Then the subgroup :F0 of'H generated by {H(x)lx EX} con­sists of all hyperplanes of type (X),(Hl), ... ,(H14).

Proof. By the previous Lemmas we only have to prove that all hyperplanes generated by five points are also generated by at most four points. In the lines preceding this theorem it is said how this can be done. The details of these (somewhat) tedious calculations will be omitted. 0

65

4.8 The hyperplanes of 11..

By Ronan [Ro] we have dim V = 22, dim V = 23, so in order to find the hyperplanes of 1i it suffices to find one hyperplane of (X, .C) which is not in :Fo. There are some obvious candidates for such a hyperplane.

LEMMA (4.15). Let H[i] := {x E Xlx contains the symbol i} for each symbol i. Then H[i] is a hyperplane of (X, .C) not contained in :Fo, it has 253 points and line distribution 0253 .

Proof. Each line of the near hexagon is the collection of a triple pairwise disjoint octads, hence each line contains exactly one! octad containing the symbol i. Now everything is clear. 0

LEMMA (4.16). Let i and j be two symbols. Then H[i] ED H[i] is X if i = j and it is a hyperplane of type (H13) otherwise.

Proof. The case i = j is trivial, just as the fact that H[i] El1 H[i] belongs to :Fo (:Fo is a subgroup of 1{, of index 2 and both H[i] and H[i]

. do not belong to :Fo ). There are 77 blocks containing both i and j, and all the neighbours of these blocks do not contain i or j, hence there are at least 77 points x with 15 lines on x contained in H[i] El1 H[i]. On the other hand, there are blocks containing i but not j. Thus H[i] El1 H[j] :f. X. From previous results our claim is clear pow. 0

THEOREM ( 4.17). The hyperplanes of the unique regular near hexagon on 759 points are:

(X) X, which has 759 points and line distribution 15759 ;

(Hl) H(x), for all x, which has 311 points and line distribution 32801531;

(H2) H(x, y), for all x, y with d(x, y) = 2, which has 439 points and line distribution 387249256 11144 157

;

(H3) H(x, y), for all x, y with d(x, y) = 3, which has 367 points and line distribution 3305727180g401145;

(H4) H(x, y, z), for all x, y, z such that d(x, y) = d(x, z) = d(y, z) = 2 and z E Q(x, y) \ [Oxny], which has 375 points and line distri­bution 7360 1515 ;

(H5) H(x, y, z), for all x, y, z such that d(x, y) = d(x, z) = d(y, z) = 2 and z has distance 1 to Q(x, y), which has 375 points and line distribution 36596 712691281118 151;

66

(H6) H(x, y, z), for all x, y, z such that d(x, y) = d(x, z) = d(y,z) = 2 and z has distance 2 to Q(x, y), which bas 375 points and line distribution 3185547189978 1136 ;

(H7) H(x,y,z), for all x,y,z such that d(x,y) = d(x,z) = 2 d(y, z) = 3, and z bas distance 2 to Q(x, y), which has 351 points and line distribution 3355 1267120g7o;

(H8) H(x, y, z), for all x, y, z such that d(x, y) = d(x, z) = d(y, z) = 3 and the octads x, y and z have no symbol in common, which has 399 points and line distribution 5 24 7 180 9 1201175;

(H9) H(x, y, z), for all x, y, z such that d(x, y) = d(x, z) = d(y, z) = 3 and the octads x, y and z have one symbol in common, which has 383 points and line distribution 35 5 78 7 120g150113o;

(HlO) H(x,y,z,u), for all x,y,z,u such that all mutual distances are 2, z and u have distance 1 to Q( x, y) and the line through u and the common neighbour of x, y, z and u does not belong to the FB.no plane formed by the quads Q(x, y) and Q(x, z), which has 407 points and line distribution 314 7120 9224 1142157;

(Hll) H(x,y,z,u), for all x,y,z,u such that all mutual distances are 2, z has distance 1 to Q( x, y) and u has distance 2 to the quads determined by the three pairs from { x, y, z}, which bas 407 points and line distribution 522 7165 gno uno:

(H12) H(x,y,z,u), for all x,y,z,u such that x, y and z have mutual distance 2, d(u,x) = d(u,y) = d(u,z) = 3, z and u have dis­tance 1 to Q( x, y), u bas distance 2 to the quads Q( x, z) and Q(y, z) and 1ru ~ [Oxn11], which has 367 points and line distri­bution 32456471929641123;

(H13) H(x,y,z,u), for all x,y,z,u such that d(x,u) = d(y,z) = 3, d(x,y) = d(x,z) = d(y,u) = d(z,u) = 2, z and u have dis­tance 2 to Q( x, y) and u has distance 2 to Q( x, z), which has 407 points and line distribution 7330 1577 ;

(H14) H(x, y, z, u), for all x, y, z, u such that all mutual distances are 3 and the oct ads x, y, z and u have a symbol in common, which bas 343 points and line distribution 360 5192730 1160 151;

(C) H[i], for all symbols i, which bas 253 points but no lines;

(HC) H $ H[i], for all H of the form H( x1, ... , xn) and all symbols i. Such a hyperplane H' contains points x such that x is on an even number of lines contained in H'.

67

'~

Proof. Everything, except the last statement, follows immediately from Theorem (4.2), Theorem (4.14), Lemma (4.15) and the observa­tion dim V = dim V + 1. In order to show the last statement, let H E :F0 , H' = H[i] for some symbol i, ii = H EB H'. Take a line 1 in H, then 1 contains a point x of H'. Now a line m on x is contained in ii if and only if m n H =/= m, hence x is on an even number of lines contained in fl. 0

REMARKS.

1. In the previous sections we have seen that M24 acts transitively on the collection of hyperplanes of a fixed type from :Fo. It is possible to determine the orbits of the action of M24 on 1-l. First one has to determine the "sums" H ElJ H', where H runs through a system of representatives of the orbits from :Fo and H' runs through the collection of 276 hyperplanes of type (H13). Let us just observe that we know at least two orbits (an immediate consequence of the 5-transitive action of M24 on the 24 symbols):

- the hyperplanes of type (C);

- the hyperplanes H[i] EB H[j] EB H[k] with i =/= j =/= k =/= i.

2. In the small example of Section 4.2 1t is obviously generated by the hyperplanes [i]. This is not the case for the regular near hexagon on 759 points. H {it, ... , is} is an octad, then [i1] EB · · • EB [is] = X. It follows that the hyperplanes [i] generate a subgroup 1-lo of 1t that only contains the following hyperplanes.

- The hyperplane X of type (X);

- the hyperplanes [i] for a symbol i;

- the hyperplanes [i] EB [j] of type (H13);

- the hyperplanes [i] EB [j] EB [k] for any three distinct symbols

i,j, k; - the hyperplanes [i]EB[j]EB[k]EB[l] of type (H4) (such a hyperplane

is the set of points at distance ::::; 1 from the quad determined by the four symbols i,j, k, 1).

1-lo has eo4) + e14) + e24) + e:)+ te44) = 212 elements. 3. Using the results of this chapter we can give an alternative proof of Lemma (3.5). Indeed, suppose Y is a 253-coclique. Then Y is an hyperplaneof1t\F0 • Let H = YEBH[i] E Fo (and hence Y = HEBH[i]),

· where i is any symbol. If H = X, then Y = H[i]. Otherwise we have

68

to showY = H[j] for some symbol j f= i. Take a line l C H, then l contains an element x of H[i]. Now x E Y, all lines on x intersect Y only in x, i.e., all those lines are contained in H. This observation gives us the inequality 1h5 $; a 15 , where h is the number of lines contained in H. From this inequality it follows that H is either of type (Hl) or of type {H13). The first possibility is obviously ridiculous, hence H is of type {H13). We have equality in our inequality in this case, and this yields H = H[i] E9 H[.i] for some symbol j.

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CHAPTER 5: CONVEX SUBGRAPHS OF CLASSICAL GRAPHS

5.1 Introduction.

The objects of this chapter are the classical distance regular graphs, viz. the Hamming graphs, the Johnson graphs, the Grassmann graphs, the dual polar graphs, the bilinear forms graphs, the alternating forms graphs, the Hermitean forms graphs and the quadratic forms graphs. For these graphs we will show that 2-convexity usually implies con­vexity and we want to find all convex subgraphs of these graphs. It turns out that a convex subgraph C of a graph r is of the same type as r, when r is one of the graphs mentioned above. There is only one exception: the Hermitean forms graph sometimes has one other convex subgraph; see Section 5.10. This research was inspired by a question of A.A. Ivanov, who asked for the classification of convex subgraphs of the alternating forms graphs.

Let us fix some notation. If Cis a subgraph of the graph r, then de( x, y) and dr(x, y) will denote the distance between two vertices x andy in the graphs C and r, respectively. We will delete the index if it is dear from the text which distance is meant. Note that if C is a connected induced subgraph, then dc(x, y);::: dr(x, y) for any two vertices x and y of C. Sometimes we will identify a subset of the vertex set with its induced subgraph. Let us define subsets C( x, y ).

DEFINITION. If x and y are two vertices of the graph r, then we define the set C(x,y) by C(x,y) := {zld(x,z) +d(y,z) = d(x,y)}, in other words, C( x, y) is the union of all geodesics (all shortest paths) between x and y.

Finally, let us recall the definition of convexity. If C( x, y) C C for all vertices x andy of C with dr(x, y) ~ t, then we call C t-convex. Cis called convex if it is t-convex for all t.

5.2 When 2-convexity implies convexity.

In this section we will give conditions under which all connected 2-convex subgraphs of a graph are convex. Of course some restriction is necessary; for instance a path Pn+I of length n in the circuit C2n is ( n - 1 )-convex, but not convex.

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Let us give four properties a graph r =(X, E) can have. (C1) For all X E X' for all j > 1 and for all y, z E r;(x) the

following holds. If y ""' z or r(y) n r(z) n rj+l(x) =f:. 0, then r(y) n r(z) n r;-t(x) =f:. 0.

(C2) For all X E X, for all j > 1 and for all y,z E r;(x) the following holds. If y ""' z or r(y) n r( z) n r j+t (X) =f:. 0, then C(x,y)nC(x,z)\{x} :f:.0.

(C3) For all X E X, for all j > 1 and for all y,z E r;(x) the following holds. If y "' z or (y and z lie in different connected components of r;(x) and y ""' w ......, z for some w E r;+t(x)), then C(x, y) n C(x, z) \ {x} =f:. 0.

(C4) For all X E X, for all j > 1 and for all y,z E r;(x) the following holds. If y ......, z, then C(x,y) n C(x,z) \ {x} =/= 0. If y and z lie in different components of r ;(X) and y ""' w ,...., z for some w E r i+ 1 (X), then there are vertices Zo = y' Zl' ••• ' Zn-l'

Zn = z in r(w) n r;(x) such that C(x, Zi) n C(x, Zi+d \ {x} =F 0 for i = 0, 1, ... , n - 1.

THEOREM (5.1).

(i) (Cl) # (C2) => (C3) => (C4).

(ii) Suppose the graph r has property (C4). Then every connected 2-convex subgraph C of r is convex.

Proof. Obviously (i) holds, so let us show (ii). Suppose (C4) holds. Let C be a connected 2-convex subgraph of r, and let a, b E C. We show by induction on n = de (a, b) that C( a, b) C C. This is true by hypothesis when dr(a, b) :5 2, so let i := dr(a, b) :2:: 3. Let a "' c1 "" • • • ......, Cn-1 = c "" b be a shortest path in C. By induction on n we have C(a,c) C C. If dr(a,c) > i, then C(a,b) C C(a,c) C C, and we are done. If dr (a, c) = i, then by repeated application of as­sumption (C4) we find a point X E r(c) n r(b) n ri-t(a), and we have C(a,x) C C(a,c) C C, contradicting de( a, b)= n. Thus dr(a,c) = i 1 and n = i. Let y E C(a, b)\ C(a,c), y =f:. b, and let Uo E C(y, b) n r(b). Let F = ri-t(a). Let u E r(b) n F, where either (i) u "" c and dp(u,u0 ) = dp(c,u0 ) -1 or (ii) u = u0 and c and u lie in different connected components of F. In case (i), by repeated application of (C4), we can find a point v E r(c) n r(u) nr1-2(a), so that v E C(a,c) C C. By 2-convexity of C also u E C(v, b) C C, and by the induction on n also C(a, u) C C. Now

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by induction on dF(c,u0 ) it follows that y E C(a,u0 ) C C. In case (ii), by repeated application of assumption (C4), we can find points c = UJ, U2, ... 'Ut-I, Ut = u E r(b) n F and points v; E r(u;) n r(u;+t)n ri_2(a), so that v; E C(a,u;) C C, by 2-convexity of C also u i+t E C( v;, b) C C and by the induction on n also C(a,u;+t) C C for j = 1,2, ... ,l. Hence y E C(a,u) C C. 0

A connected graph r is bipartite if its vertex set can be partitioned into two cocliques. The halved graphs r+ and r- are the connected components of the graph with as vertex set the vertex set of r and joining two vertices when they have distance 2 in r. A distance regular graph r with intersection array { bo, ... , bd-1 j Ct, ••• , Cd} and diameter d E {2m, 2m+ 1} is bipartite if and only if ai = 0 for all i. In this case the halved graphs are distance regular of diameter m with intersection array

Some of the classical graphs we will meet are bipartite and have prop­erty (C3). The following proposition shows that their halved graphs also have property (C3).

PROPOSITION (5.2). Let r be a bipartite graph which has property (C3). Then the same holds for its halved graphs.

Proo£ Let~ be r+ orr-, X a vertex of~ and y,z E ~i(x), i.e. y,z E r 2i(x). If y and z are adjacent in~. then they have a common neighbour u in r and u is in r2i-I(x) u r2i+l(x). If u E r2i+I(x), then by repeated application of assumption (C3) we find a vertex v E r(y) n r(z) n r2i-t(x). Hence CA(x,y) n CA(x,z) \ {x} :/: 0 in all cases. Suppose y,z E ~(w) for some w E ~i+t(x) such that y and z are nonadjacent. Then dr(y,w) = 2, dr(z,w) = 2, dr(x,w) = 2i +2 and dr(y,z) = 4. Let U} E r(y) n r(w) and u2 E r(z) n r(w). Repeated application of assumption ( C3) gives points V} E r( u 1 ) n r( U2) n r 2i (X)' UJ E r(y) n r(vt) n r2i-l(x), u4 E r(z) n r(vt) n r2i-t(x) and v2 E r(u3) n r(u4) n r2i-2(x). Now V2 E ~(y) n ~(z) n ~i-I(x). 0

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5.3 Subspaces.

Let r =(X, E) be a graph. A subset C of X is called a subspace when {x,y}.l.l C C for any two~adjacent vertices x,y E C. A J..t-graph in r is a subgraph { u, v }.l, where d( u, v) = 2.

Convex subgraphs need not be subspaces. Indeed, any complete sub­graph is convex, and, e.g., ford :2: 2 the dual polar graphs 2 Dd+1(q) (with singular lines of size q2 + 1) contain as convex subgraphs dual polar graphs Bd(q) (with singular lines of size q + 1), and these in turn contain as convex subgraphs dual polar graphs Dd(q) (with sin­gular lines of size 2), see Section 5.7. In these graphs the J..t-graphs are cocliques pf size q + 1. When the J..t-graphs are fatter, then convex subgraphs must be subspaces.

THEOREM (5.3). Let r have locally noncomplete j..t-graphs. Then any noncomplete 2-convex subgraph of r is a subspace.

Proof. Let C be a noncomplete 2-convex subgraph of r, and let x,

y be two adjacent vertices of C. If there is a vertex u E C n r( x) \ r(y ), then by assumption x has two nonadjacent neighbours p, q in the J..t­graph {u,y}.l C C, and now {x,y}.l.l C {p,q}.l C C. Thus, we may assume that x.l n C = y.L n C, and that x.l n C is complete. If there is au E C with dr(u,x) = 2, then the subgraph {x,u}.l is complete and hence locally complete, a contradiction. Thus, we have C C x.L, and C = x.L n C is complete, a contradiction. 0

For example, in the Grassmann graphs [;] (see Section 5.6), the J..t­graphs are ( ( q + 1) x ( q + 1) )-grids, and the theorem applies.

We now consider each of the mentioned classical families of distance regular graphs. For more information on these families than given below, one is referred to [BCN]. Also the information contained in [BB] is useful.

5.4 Johnson graphs.

Let r = (X,E) be the Johnson graph J(n,e) with as vertex set the collection of e-subsets from a fixed n-set n, where two e-subsets are adjacent when they meet in a ( e - 1 )-subset, so that d( x, y) = i if and only if lx n Yl = e- i, or equivalently, d(x,y) = i if and only if lx U Yl = e + i. r is distance regular of diameter d = min( e, n- e) with

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intersection array given by

b i = ( e - j )( n - e - j)} Cj =j2

for 0$ j $d.

Finally, the map x 1-+ 11\x gives the isomorphism J(n, e)£:: J(n, n- e).

For A c B c n, let I( A, B):= {x EX lAc X c B}. Clearly we have I( A, B) £:: J( m, d), where m = !B \AI and d = e - IAJ.

LEMMA (5.4). Let x, y EX, d(x, y) = i > 0. Then ri-i(x) n r(y) is a (i X i)-grid, and in particular is connected.

Proof. Straightforward. 0

LEMMA (5.5). Forx,yEX wehaveC(x,y)=I(xny,xUy).

Proof. Straightforward. D

PROPOSITION (5.6). r has property (C3).

Proof. Ify,z E rj(x), then the set C(x,y)nC(x,z)nr(x) consists

of the vertices xU {o"} \ {r}, where u E y n z \ x andrE. x \ (y U z). Thus, this set is nonempty, unless y n z C x or x C y U z. But if this is the case, and x, y, z are as in the hypothesis of property ( C3), then 1 = d(y, z) ;:::: d(y, x) = j > 1, a contradiction. D

PROPOSITION (5.7). The noncomplete convex subgraphs ofr are precisely the subgraphs I( A, B).

Proof. Clearly, the I( A, B) are convex. Conversely, we show that if II( A, B)l > 1, then the smallest convex subgraph ~containing I( A, B) and x is I( A n x, B U x ). Indeed, suppose x rJ. I( A, B). By walking along a geodesic from x to some vertex of I( A, B), we see that we may assume that either A C x and lx \BI = 1, or x C Band lA \x! = 1. By the isomorphism mentioned above we may assume that we are in the former case. Now let y E I(Anx, BUx) \I( A, B). If B is not contained in xUy, then y E C(x,z) for some z E I(A,B), as desired. Otherwise, let p E x\y, u E y\x, and put w = yU {p} \ {u}, then wE C(x,z) for

·some z E I(A,B), andy E C(w,z') for some z' E I(A,B). 0

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5.5 Hamming graphs.

Let n be a set of cardinality q ~ 2 and let r = (X, E) be the Hamming graph H(d,q) = q11 with as vertex set 0 11 , the Cartesian product of d copies of n, two vertices being adjacent whenever they differ in exactly one coordinate, i.e. whenever they have Hamming distance 1. Then d(x, y) = i if and only if X andy have Hamming distance i. r is distance regular of diameter d with intersection array given by

bj = ~d- j)(q- 1)} Cj = J

for 0 '5 j '5 d.

r is a direct product of d cliques, the following results hold more gen­erally for any direct product IIXi of a number of cliques.

LEMMA (5.8). Let x,y EX, d(x,y) = i > 0. Then ri-t(x)nr(y) is an i-coclique.

Proof. Straightforward. 0

LEMMA (5.9). Forx,yEX wehaveC(x,y)~H(d(x,y),2).

Proof. Straightforward. IJ

PROPOSITION (5.10). r has property (C2).

Proof. lfy,z E rj(x), then the set C(x,y)nC(x,z)nr(x) consists of the vectors (xi)i that agree with x in all positions except one, and agree with both y and z in this one position. If this set is empty, then d(y, z) ~ d(y, x) and as before we find that ( C2) holds. 0

PROPOSITION (5.11 ). The convex subgraphs ofr are precisely the direct products Il}i, with Yi c xj.

Proof. Clearly, the subgraphs Illi are convex. Conversely, we show that if Illi =f: 0, then the smallest convex subgraph 6. containing Illi and X= (xi)i is nzi, where zi = Yi u {xi}· Indeed, suppose X rf. Il}i. By walking along a geodesic from x to some point z E Il}i, we see that we may assume that x; E Yi for all i =f: io, and x;0 rf. }i0 • If y E IIZi \Illi, then Yio = x;0 , andy E C(x, z) for some vector z E Illi agreeing with y in all positions except the i 0-coordinate. 0

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5.6 Grassmann graphs.

Let V be an n-dimensional vector space over the field G F( q) and let

r = (X, E) be the Grassmann graph [ ~] with as vertex set the collec­tion of e-subspaces, two vertices being adjacent whenever their inter­section has dimension e - 1. Then d( x, y) = i if and only if x n y has dimension e- i. r is distance regular of diameter d = min(e, n- e) with intersection array given by

for 0 ~ j ~d.

The Grassmann graphs [ ~] and [ n ~e J are isomorphic. Furthermore, we will write Z + Y for the subspace of V spanned by the subspaces Z and Y, 0 for the 0-dimensional subspace of V, Z ~ Y to denote Z is a subspace of Y and by abuse of notation sometimes we will write

r = [:]. For A ~ B ~ V, let I(A,B) = {x E XIA ~ x ~ B}. Clearly,

!(A, B)£:! [ B}A], where f = e- dim A. Ford 2::2 the singular lines in

rare the sets I(A,B) with dimA = e -1, dimB = e + 1, they have size q + 1.

LEMMA (5.12). Let x,y EX, d(x,y) = i > 0. Then r;-t(x)nr(y) is a ([;] x [;])-grid, and in particular is connected.

Proof. Straightforward. 0

PROPOSITION (5.13). r has property (C3).

Proof. Ify,z E r;(x), then the set C(x,y)nC(x,z)nr(x) consists of the e-spaces W + p with (X n y) + (X n Z) ~ W ~ X, dim W = e - 1, p ~ y n z, x n p = 0 and dimp = 1. Thus, this set is nonempty, unless (xny)+(xnz) = x or ynz :$ x. But if this is the case, and x,y,z are as in the hypothesis of property (C3), then 1 = d(y, z) 2:: d(y, x) = j > 1, a contradiction. 0

PROPOSITION (5.14). The noncomplete convex subgraphs ofr are precisely the subgraphs I( A, B).

Proof. Clearly, the I( A, B) are convex. Conversely we show that if 'Cis convex and I( A, B) U {x} C C, then I( An x, B + x) C C, provided

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that I( A, B) has at least two elements. (Note that since C contains more than one point, it contains a singular line.) We may assume that d( x, I( A, B)) = 1. We may assume that A ~ x or x ~ B, for other­wise x has a unique neighbour y = (A + x) n B in I( A, B), and, with y "' z E I( A, B), C( x, z) contains elements u which contain A, but are

not contained in B. Because [~] ~ [n~e], we may assume A~ x. Let y E I(A n x,B + x). We have to showy E C. Considering a path x "' Zt "' • • • "' Zt "' y, where Zi E I( A, B) (1 ~ i ~ t), we can find a common neighbour x' of x and z2 not contained in I( A, B). Now x' E C, so by induction on t we may assume that x andy have a com­mon neighbour z in I( A, B). If x nyC Band x,..., y, then the singular line I(x n y,x + y) contains x and meets I(A,B), hence is contained in c. Thus, y E I(x n y, X+ y) c c. If X"" y and X n y is not contained in B, then we take u E I(A,B) with u n x ~ y n z ~ u and we have y E C( x, u) C C. Finally, if x -f y, then we can find a vertex u E r( X) n r( y) n r( z) \ I( A, B). The preceding arguments show con­sequently u E Candy E C. 0

Note that here and in the following "thick" cases, C( x, y) is not convex as soon as d(x,y);:::: 2.

5. 7 Dual polar graphs.

Let V be one of the following spaces equipped with a specified form (d > 0):

- [Cd(q)] = GF(q)2d with a nondegenerate symplectic form;

- [Bd(q)J = GF(q)2d+ 1 with a nondegenerate quadratic form;

- [Dd(q)] = GF(q)2d with a nondegenerate quadratic form of {maximal) Witt index d;

- (2 D d+ 1 ( q)] = G F( q )2d+2 with a non degenerate quadratic form of (non-maximal) Witt index d;

- [2A2d(r)] = GF(q)2d+I with a nondegenerate Hermitean form, where q = r 2;

- [2 A2d-t(r)] = GF(q)2d with a nondegenerate Hermitean form, where q = r2.

A subspace of V is called isotropic whenever the form vanishes com­pletely on this subspace. Maximal isotropic subspaces have dimen­sion d. The dual polar graph on V has as vertices the maximal isotropic subspaces, two vertices x, y are adjacent whenever dim X n y = d 1,

77

, __

more general they have distance i whenever dimx n y = d- i. Dual polar spaces are near polygons, characterized by having classical point­quad relations (see [Ca] and [Sh]). More general, they have classical point-subspace relations.

Let r = (X, E) be a dual polar graph and let e be 1, 1, 0, 2, !, ! if r is Gd(q), Bd(q), Dd(q), 2 Dd+t(q), 2 A2d(r), 2 A2d-t(r) respectively. (By abuse of notation we will denote r = Gd(q) if V = [Gd(q)], etc.) r is distance regular of diameter d with intersection array given by

for 0:5 j :5 d

and has singular line size s + 1 = qe + 1.

LEMMA (5.15). Let x,y EX, d(x,y) = i > 0. Then ri-t(x)nr(y) is a U] -coclique.

Proo£ Straightforward. 0

PROPOSITION (5.16). r has property (G2).

Proo£ If y,z E ri(x), and there is a point p E y n z \ x, then p + (p.L n X) E r( X) n G( x, y) n G( X' z ). If there is no such p, but the hypothesis of property (C2) is satisfied, then y n z c y n x, so that d(y, z) 2: j > 1, i.e., j = 2 and y n z = y n x, and every common neighbour w of y and z contains y n z so cannot lie in rJ+t(x). 0

PROPOSITION (5.17). Let (X,£) beanearpolygon, andC a convex subset. Then ( C, .Cc) is a near polygon, where .Cc is the collection of lines from£ meetingC in at least two points. If(X,.C) has quads, then so has (C,.Cc). If in (X,£) all point-quad relations are classical, then the same holds in (C,.Cc). The convex subspace spanned by (C,.Cc) in (X, C) has the same diameter as (C, .Cc).

Proo£ Immediate from the definitions. 0

PROPOSITION (5.18). Let r be a dual polar graph. Then the non­complete subgraphs C of r are themselves dual polar graphs.

Proo£ r together with its singular lines is a near polygon with quads and classical point-quad relations, so by the previous proposition

78

a convex subset C induces a similar near polygon (not necessarily a subspace), hence is a classical near polygon in its own right. 0

Now, by Tits (Ti], all polar spaces of rank d;::: 3 are known, and one may check for inclusions among the corresponding dual polar spaces. Here we restrict ourselves to the finite case. The finite (dual) polar spaces of rank d ;::: 3 are the spaces [Cd(q)], [Bd(q)], [Dd(q)], [2 Dd+t(q)], [2 A2d(r)], [2 A2d-t(r)].

PROPOSITION (5.19). Let r be a dual polar graph Cd(q), Bd(q), · Dd(q), 2 Dd+t(q), 2 A2d(r) or 2 A2d-t(r) where d;::: 2. Let C be a non­complete convex subgraph of r. Then we have one of the following cases:

r subs!!aces C other convex sub!;!!Qhs C

Dd(q) n,(q)

Bd(q) Bt(q) n,(q)

Cd(q), q odd c,(q) 2 Dd+t(q) 2 Dt+t(q) B 1(q), n 1(q)

2 A2d-t(r) 2 A2t-t(r) 2 A2d(r) 2 A21(r) 2 A2t-t(r)

where 2 :5 f :5 d. (Note that when q is even, Cd(q) ~ Bd(q).)

Proof. It is clear that the cases listed do indeed occur. Since an inclusion of dual polar spaces implies an inclusion of their quads, it suffices to treat the case d = 2. Now a generalized quadrangle is a convex subgraph of another if and only if the dual of the first is a sub­quadrangle (with full lines) of the dual of the second. Thus we have to prove that the only (connected) subquadrangles of the generalized quadrangles Os(q), Sp4(q) (q odd), Ot(q), 06(q), U4(r), Us(r) are those given by Ot(q) C 05(q) C 06(q) and U4(r) C Us(r). Now if a GQ(s, t) has a proper subquadrangle GQ(s, t 0 ), then t ;::: st0 (since if t0 > 0 then any point outside the subquadrangle is collinear to st0 + 1 points inside the subquadrangle) and Sp4 (q) (of order (q,q)) does not have subquadrangles of order (q, 1) for odd q, so it has no (proper, con­nected) subquadrangles at all. Since any nonincident point, line pair in U5 (r) is contained in a U4(r) subquadrangle, and the intersection of two subquadrangles again is a subquadrangle, it follows that U4(r) (of

79

order ( r 2 , r)) has no subquadrangles, and U5( r) (of order ( r 2 , r 3 )) has no subquadrangles other than U4(r). Similarly, since any nonincident point, line pair in O;(q) is contained in a Ot(q) subquadrangle, and any three concurrent lines in O;(q) are contained in a 0 5 (q) subquad­rangle, it follows that ot(q) (of order (q, 1)) has no subquadrangles, that 0 5 (q) (of order (q, q)) has no subquadrangles other than Ot(q), and that O;(q) (of order (q,q2 )) has no subquadrangles other than Ot(q) and Os(q). 0

If r is D4(q), C4(q) or 2 A2d_1(r), then from the preceding proposition it is an immediate consequence that the only noncomplete subgraphs of rare the subgraphs I( A), where I( A) = {x E X lA ~ X} and A is some isotropic subspace. (We have I(A) ~ D1(q), C1(q), 2 A2,_1(r) in these cases respectively, where f = d - dim A.)

5.8 Bilinear forms graphs.

Set V = Fd and W =Fe, where F = GF(q). Let B be the vector space (of dimension de over F) of bilinear maps from V x W to F. The null space off E B in V is defined as { v E Vlf( v, W) = 0} and the rank rk(f) of f is the codimension of each of its null spaces (in V and W). If we define f ""g whenever rk(J- g) = 1, then we get the bilinear forms graph over F with dimensions d and e, more general we have d(f,g) = i if and only if rk(J- g)= i. We might also consider B as the set of d x e matrices over F, where the distance of two matrices is the rank of their difference.

A useful geometric description of the bilinear forms graphs is that of collinearity graphs of so-called attenuated spaces (or affine Grassmann spaces). Let V be a vector space of dimension d + e over F and W a subspace of V of dimension e. Then the corresponding attenuated space is the collection of subspaces U of V with U n W = 0, where subspaces U of dimension d are called "points" and those of dimension d - 1 "lines", and incidence is symmetrized inclusion. In this way a bilinear forms graph can be viewed as a subgraph of a Grassmann graph.

Let r =(X, E) be the bilinear forms graph over GF(q) with dimensions d and e (where d ~ e). Then r is distance regular of diameter d with intersection array given by

80

b; =q'i(q-l) [d~jl [·~jl}

Cj=qj-1[~] for 0 :5 j :5 d.

LEMMA (5.20). Let x,y EX, d(x,y) = j. Then C(x,y) is isomor­phic to the graph with as vertices the pairs (E, F) of subspaces of a j -space U such that E n F = 0, E + F = U, where ( E, F) "' ( E', F') if and only ifdim(E+E')/(EnE')+dim(F+F')/(FnF') = 2. The ver­tices ofr;-i(x)n ri(Y) correspond to the pairs (E,F) with dimE= i, dimF = j- i. In particular, r;-1(x) n r(y) is connected.

Proof. See [BB, §8). 0

PROPOSITION (5.21). r has property (C3).

Proof. We will use the description of r in terms of attenuated spaces. Let us take Y,Z,U E r withY"' Z and Y,Z E rj(U), i.e., dim U n Y = dim U n Z = d - j, dim Y n Z = d - 1 and U, Y, Z are disjoint from W. We want to show r(Y) n r(Z) n r;- 1(U) =/= 0. We may assume U n Y n Z = 0 and we have to consider two cases:

(i) u n Y = u n z = o, (ii) U n Y and U n Z are 1-spaces, say A and B.

Case (i) was settled in [BB, §8]. In the other case we have a 1-space (A+ B + (Y n Z)) n W = C and we -ean take a hyperplane H of Y n Z such that His disjoint from the projection of Con Ynz. The required vertex is A+ B +H. 0

Since r is a subgraph of the Grassmann graph [ d1e] , the t.t-graphs are

( q + 1) x ( q + 1 )-grids with an ovoid deleted, hence the {t-graphs are locally noncomplete.

Let us define the sets I( A, B) = {x E X lA :5 x :5 B} and I0(A, B) = {x :5 VI dimx = d and A :5 x :5 B}, where A and B are subspaces of V, A n W = 0. The singular lines of r are the subsets I(A,B), where dim A= d -1, dimE= d + 1, they have size q pro­vided d ~ 2. Observe that I( A, B) is again a bilinear forms graph over GF(q), this time with dimensions d- dim A and dimE- d.

PROPOSITION (5.22). The noncomplete convex subgraphs ofr are precisely the subgrapbs I( A, B).

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Proof. The proof of this proposition is exactly the same as the proof

of Proposition (5.141, except for the case x "'y; x, y "'z E I( A, B) and

X nyc B: in [ d1e the singular line Io(x n y,x + y) meets Io(A,B)

in one point, so it is possible that in r the singular line I( X n y' X + y ),

which misses exactly one vertex from I0 (x n y,x + y), does not meet I( A, B). In this case, however, we can find u E r(x) n r(y) \ I0 (A,B) and ver­tices v E r(u) n I(A,B) and wE r(y) n I(A,B) such that (x "'v and xnu ¢.B) or (d(x, v) = 2), and (u"' wand uny ¢.B) or (d(u, w) = 2). This gives u E C and y E C. 0

5.9 Alternating forms graphs.

Set V = GF(q)n and let A be the n(n -1)/2-dimensional vector space of (bilinear) alternating forms on V. Thus, f E A if and only if f is a bilinear form on V and f ( x, x) = 0 for all x E V. We define Radf = {x E Vlf(x, y) = 0 for ally E V} and rk(f) = dim(V/ Radf). Note that rk(f) takes even values only for f E A.

The alternating forms graph Alt(GF(q)n) = r =(X, E) on Vis de­fined to have A as vertex set, where two vertices x and y are adjacent whenever their difference has rank 2, more general d( x, y) = i if and only if rk( X - y) = 2i. r is distance regular of diameter d :::: [ ~] with intersection array given by

bj = l~(qn-2~ -1)(qn-2j-1 _ 1)/(q2 _ 1)}

Cj = q2J-2(q2J _ 1)j(q2 _ 1) for 0 :$ j :$ d.

We could also have defined A (and r) as the set of all skew symmetric n X n matrices over GF(q) with zero diagonal, where two matrices are adjacent whenever their difference has rank 2.

The alternating forms graph can be described inside the dual polar

graph Dn(q) = ~ = (Y,F). Let oo be a vertex of ~ and put D = ~n( oo ). Then D is a coclique in ~' but the graph included on D by joining two vertices whenever they have distance 2 in ~ is isomorphic to the alternating forms graph on GF(q)n. Let us define Io(A) = {y E YIA :$ y} and I(A) = Io(A) n ~n(oo), where A is an isotropic subspace of [Dn(q)] such that An oo = 0. This dual polar space has classical point-subspace relations only, hence

· I0 (A) ~ Dm(q) and I(A) ~ Alt(GF(q)m), where m = n- dim A. The

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singular lines of r are the sets I( A) with dim A = n - 2, they have stze q.

LEMMA (5.23). Let x,y EX, d(x,y) = j. Then C(x,y) is isomor­phic to the graph with as vertices the nondegenerate subspaces Y of a 2j-space V provided with a nondegenerate symplectic form, where Y1 , l2 are adjacent when dim(Y1 + Y2)/(Y1 n Y2) = 2. The vertices of rj-i(x) n ri(Y) correspond to the subspaces Y with dim Y = 2i.

Proof. See [BB, §9]. 0

COROLLARY (5.24).

(i) rj-t(X) n r(y) is connected whenever d(x, y) = j.

(ii) r has locally noncomplete p.-graphs.

Proof. Immediate consequence of Lemma (5.23). 0

PROPOSITION (5.25). r has property (C3).

Proof. Let y,z E rj(x), y "'z and j 2 2. Supposed 2 3, i.e., n 2 6. In dual polar terms this means we have dim y n z = n - 2, dimx n y = dimx n z = n- 2j. The only convex subgraphs of Dn(q) are the subgraphs D1(q) (Proposition (5.19)), hence we may assume xnynz = 0. Thisgivesn-2j:::; 2. Ifn-2j = 2, theny = (ynz)+(ynx), but this gives X n Z C ((y n z).L n (xn y).J..) = y.L = y, a contradiction. Thus we may assume j = d and by Van Bon & Brouwer [BB, §9] we have rd-t(Y) n rd-t(z) n r(x) =/::. 0. 0

PROPOSITION {5.26). The noncomplete convex subgraphs ofr are precisely the subgraphs I( A).

Proof. Clearly the I(A) are convex. Conversely we show that if C is a convex subgraph containing both I(A) and x, then it contains I(A n x), provided that I(A) has at least two elements. (Note that since C contains more than one point, it contains a singular line.) We may assume d(x, I( A))= 1, An x = 0 and C C I( An x). Now there are two possibilities (note that n 2 4):

(i) dimA = 1 and Io(A) has diameter n- 1 2 3;

(ii) dimA 2 and I0 (A) has diameter n- 2 2 2.

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Suppose we are in case (i). Let y E I(A n x) \ I(A). Then there is a unique a 11 E Io(A)n.6.(y), such that da(y, z) = da(y, a 31 )+da(a11 , z) for all z E Io(A). So we can find Zb ••• , Zt such that x""' Zt ""' • • • ""'Zt ""'y is a shortest path with Zi E I( A), i = 1, ... , t. We can find a common neighbour x1 of x and z2 not in I( A). Now x' E C and by induction on t we may assume x,...., z,...., y with z E I(A). If x,...., y, a:~: =f:. a 11 , then we can find u E (.6.(a11 ) \D.( ax)) n I(A), and y E C(u,x) C C. If x,.... y and ax = a 11 , then we can find {3 E .6.(z) n Io(A) \ {az} and u E .6.({3) n .6.2(x) n I(A n X)\ I( A). Now twice the previous argument gives us u E C and y E C. If X f y, then we can find u E r(x) n r(y) n r(z) \I( A), since we have locally noncomplete p-graphs. This gives u E Candy E C. If we are in case (ii), then da(x,I0 (A)) = 2, .6.2(x) n I0 (A) = {z }, z E I(A). Let a E .6.(x)n.6.(z), then a¢: I0 (A). Take {3 E 10 (A)n.6.(z), y E .6.({3) n .6.(a) n I(A n x) \ I(A) and v E .6.({3) n I(A) \ {z}. Now y E C(v,x) C C. By (i) we get I(A n y) C C and, again by (i), I(A n x) =I(( Any) n x) c C, unless I0 (A) has diameter 2. If I0 (A) has diameter 2 and dim A= 2, then we still have to show that I( Any) C C, after which again we can use (i). Note that u E C whenever . u E I(Any)nt:.(a) by the argument above. Also we have a E I0 (Any). lfu E I(Any) \I(A), u ¢: .6-(a), then take z' E I(A) \ {z}. We have da(u,z') = da(u,z) = 2. Let us take 6 E .6.(u) n .6.(z') \ I0 (A) and x' E .6.(6)n.6.2(x) ni(Anx) \I(Any). Then we have x' E C(x, z') c C and u E C(x', z) c C. Thus, I( Any) c C. 0

5.10 Hermitean forms graphs.

Set V = GF(q)d, where q = r2 with r a prime power, and letH stand for the ~-dimensional vectorspace over G F( r) of the Hermitean forms on V. Thus, f E H if and only iff( u, v) is linear in v, and f(v, u) = f(u, v) for all u, v E V.

The Hermitean forms graph r = Her( G F( q )d) = (X, E) on V is defined to have H as vertex set, where two vertices x and y are adjacent when­ever rk( x - y) = 1. Here, rk( x) and Rad x have the same meaning as in the previous section. More general we have d(x,y) = rk(x- y). Note that we could also have defined H as the set of Hermitean d x d matri­ces over GF(q). r is distance regular of diameter d with intersection

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array given by

b; = (q11- qi)j(r + 1) }

c; = ri- 1(ri- ( -l)i)j(r + 1) for 0 '5 j '5 d.

As before we can describe the Hermitean forms graph inside a dual polar graph: let~ = 2 A2d-t(r) and let oo be a vertex of~. then the subgraph ~~~(oo) of~ is isomorphic to Her(GF(q)11 ).

Define 10 (A) := {x E 2A2d-t(r)IA '5 x} and I(A) := 10(A) n ~a(oo), ·where A is a subspace of [2 A2d_1(r)] such that An oo = 0. Observe I( A)~ Her(GF(q)m), where m = d- dim A.

LEMMA (5.27). Let x, y EX, d(x, y) = j. Then C(x, y) is isomor­phic to the graph with as vertices the non degenerate subspaces Y of a j­space V provided with a nondegenerate Rermitean form, where Yt "' Y2 whenever dim(Yi + Y2)/(Y1 n Y2) = 1. The vertices ofr;_,(x) n ri(Y) correspond to the subspaces Y with dim Y = i.

Proof. See [BB, §10]. 0

PROPOSITION (5.28). Ifr > 2, then r has property (C4).

Proof. The previous lemma shows that C(x, y) is bipartite. For j 2: 3, the distance two graph of C(x,y) \ {x,y}, with d(x,y) = j, in­duced on r;-t(x)nr(y) is isomorphic to the graph on the nonisotropic points of a j-space V provided with a nondegenerate Hermitean form, adjacent when the line joining them is hyperbolic. This graph is con­nected precisely when r > 2, as was noticed by Van Bon & Brouwer [BB, §10). This yields the second requirement of (C4). If X E X, y,z E r;(x), y "" z, then, looking in 2A2d-t(r), as in the previous section we get y n x = z n x = 0, and, by [BB, §10], r(x) n r;-t(Y) n r;-t(z) =F 0. 0

PROPOSITION ( 5.29). If r = 2 and d 2 3, then r contains a con­nected 2-convex subgraph, which is not convex.

Proof. Let x,y E X, d(x,y) = 3. Then C(x,y) \ {x,y} is the disjoint union of 4 6-cycles, see the proof of Proposition (9.5.14) in [BCN]. Let ~be the graph induced by x, y and one of those 6-cycles. Then ~ is connected and 2-convex, but obviously not convex. 0

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PROPOSITION (5.30). The noncomplete convex subgraphs ofr are

precisely the subgraphs I(A), unless r = 2, in which case r also has convex subgraphs K2,2·

Proof. First of all, the mentioned subgraphs are obviously convex. Now let us show that a convex subgraph C containing both x and I( A) contains I(A n x), provided I(A) has at least diameter 2. As before we may assume d(x, I( A)) = 1 and An x = 0, i.e., I( A) has diameter d -1. We have to showy E C whenever y E I( An x ), but as before we may assume x andy have a common neighbour z in I(A). We have to deal with 3 cases (note that the p-graphs are cocliques):

(1) X"' y. We can find u E r2(x) n r2(Y) n r(z) \ I(A). Use one of the following two cases;

(2) x f y, but x andy have a common neighbour u at distance 1 from I(A). Then we have u "' v E I(A), v "' z, v f= z. This gives u E C, hence y E C;

(3) x f y and u has distance 2 to I(A) for all common neighbours u f= z of x and y. Take such a u, we can find v E Io(A) with da(u,v) = 1. For w E ~(v) n ~2 (z) n ~d(oo) n J0 (A) we have da(z,v) = 1, d(u,w) = 2, d(u,x) = 1, d(x,w) = 3, hence u E C(x,w) C C. Now we have d(u,z) = 2 and y E C(z,u) C C.

We still have to show that r has no proper noncomplete convex sub­graphs if d = 2 and r ;::: 3, r has only subgraphs K 2 ,2 as proper noncom­plete convex subgraphs if r = d = 2, and r has as proper noncomplete convex subgraphs the subgraphs K2,2 or I( A) if r = 2, d;::: 3. If r = d = 2, then it is straightforward to check that r has only sub­graphs K2,2 as proper noncomplete convex subgraphs. If r = 2 and d ;::: 3, then we only have to study the convex subgraphs C containing a K 2 ,2 and a vertex x outside this K 2 ,2 . We may assume x has a neighbour in this K 2 ,2 , which means that x and K 2,2 are in a convex subgraph I(A) of diameter :S 3, i.e., we may assume r = 2 and d = 3. In this case r has intersection array {21, 20, 16; 1, 2, 12} and was shown to be unique by Ivanon & Shpectorov ([ISl]), see also [BCN, Th. (11.3.6)]. In the proof of [BCN, Th. (11.3.6)] it was shown that r 2(x) is the disjoint union of 21 Petersen subgraphs. Each Pe­tersen subgraph P belongs to a I( Ap) of diameter 2. If d( x, u) :S 2 for all u E Kz,2 , then I(A) has diameter 2 and we are done. So we may

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assume d(x,u) = 3 for some u E K2,2. Now r(u)nr2(x) consists of 12 vertices at mutual distance 2. They appear in 4 groups of 3, in each group the common neighbours of two of them are u and a vertex con­tained in r 2( u) n r( X) (recall C( X' u) \ {X' u} ~ 4C6 ). Since JL-graphs were cocliques, we find (~) .32 other common neighbours of a pair of those 12 vertices, each such common neighbour is at distance 2 from u

and at distance ~ 2 from x. If they are all in r2(x), then we find at least two points in one Petersen subgraph P at distance 2 from x and C contains !(A,). Otherwise, there is avEC, d(u,v) = 2, d(x,v) = 3

. and jr( v)nr2(x) \r( u)l = 10 and again at least two points of cnr2(x) are in one Petersen subgraph. Remains the case d = 2, r ~ 3. In this case each coclique of size 3 in ~ is contained in a unique Kr+l,r+t· Our aim is to show IC n r2(x) n r2(u)l > tlr2(x)l whenever d(x,u) = 2 and x,u E c. This will implie r C C, since all intersections of convex subgraphs are again convex. Let us suppose x,u E C, d(x,u) = 2. Then v E r2(x) n r2(u) is inC whenever lr(v) n r(x) n r(u)l ~ 2. Thus, when v ftC we must have jr(v)nr(x)nr(u)l $1 and v in the Kr+t,r+t determined by x, u and oo (for: l~(v)n~(x)n~(u)n~(oo)l ~ r ~ 3). There are r-2 such ver­tices, hence ICnr2(x)nr2(u)l = 1r2(x)nr2(u)l-(r-2) > !lr2(x)l. 0

5.11 Quadratic forms graphs.

Set V = GF(qt. A quadratic form on V (over GF(q)) is a map 1: V-+ GF(q) such that

1(-\x) = -\21(x) for all,\ E GF(q) and x E V

and such that B.., : V x V -+ G F( q) defined by

B...,(x, y) = 1(x + y)- 1(x)- I(Y) (x, y E V)

is a symmetric bilinear form (the symmetric bilinear form associated

with 1). Let Q denote the n(n + 1)/2-dimensional vector space of all quadratic forms on V. The radical of 1, denoted by Rad 1, is defined by

Rad1 ={x E RadB"fh'(x) = 0}

={x E VII(Y) = 1(x + y) for ally E V},

where Rad B.., is defined as in the previous sections. We observe that Rad B.., = Rad 1 if q is odd and that dim(Rad B..,) $ dim(Rad 1) + 1

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in general. As before the rank of 'Y E Q, denoted rk('Y), is the number

rk('Y) = dim(V/Rad'Y)·

The quadratic forms graph r = (X, E) = Quad(GF(q)n) has as ver­tex set Q and /, 8 E Q are called adjacent if rk('Y - 8) E {1, 2}. More general, d( 'Y, 8) = i if and only if either rk( 'Y - 8) = 2i - 1 or rk('Y- 8) = 2i. r is distance regular of diameter d = [(n + 1)/2] with intersection array given by

bj = q4~(qn-2~+1 -1)(qn-2j -1)/(q2 -1)}

Cj = q2J-2(q2J -1)/(q2 _ 1) for 0 ~ j ~d.

We need an auxiliary graph :E, the symmetric bilinear forms graph. Let S be the n(n + 1)/2-dimensional vector space of symmetric bilinear forms on V, then :E has S as vertex set, and two vertices 'Y and 8 are adjacent whenever rk( 'Y - 8) = 1.

We could also have defined S (and :E) as the set of all symmetric n x n matrices over GF(q) (where two matrices are adjacent whenever their difference has rank 1 ).

The graph :E can be described inside the dual polar graph~= Cn(q): let oo be a vertex of~' then the subgraph ~n( oo) of ~ is isomorphic to the symmetric bilinear forms graph on V.

Let A be a subspace of [Cn(q)] such that An oo = 0. Then we de­fine Io(A) = {x E ~lAc x} and I(A) = Io(A) n ~n(oo). Clearly, I(A) ~ Quad(GF(q)m), where m = n- dim A. A quad in~ meeting ~n(oo) is just a subgraph Q(x,y) = I0 (x n y) where da(x,y) = 2. In a quad Q every 3-coclique {x, y, z} satisfies l~(x) n ~(y) n ~(z)l ~ 2, every two vertices at distance 2 in Q have exactly q + 1 common neigh­bours and if da(x,y) = 2, x E ~n(oo), y E ~n-t(oo), then x andy have one common neighbour in ~n -t ( oo) and q common neighbours in ~n(oo).

When q is odd, then we may canonically identify the vertex sets of r and :E, and the quadratic forms graph r is the distance-1-or-2-graph

of :E, i.e. r( 'Y) = :E('Y) u :E2 ( 'Y) for all 'Y.

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LEMMA (5.31). Let q be odd.

( 1) Let 1, 6 be two vertices at distance j in a symmetric bilinear forms graph E. Then ClJ(/,6) is isomorphic to the graph with as vertices the nondegenerate subspaces Y of a j-space V pro­vided with a nondegenerate quadratic form, where two sub­spaces Yt, l2 are adjacent when dim(Y1 + Y2 )/(Yi n Y2) = 1. The vertices ofrj-i('y) n ri(6) correspond to the subspaces Y with dim Y = i.

(2) If j =dE( 7, 6) is even, then

Cr(T,6) = {e E ClJ{T,6)IdE('y,c:) is even};

if j is odd, then

Cr('y,S) = CI:('Y,6) U

{c:ld!:{e,/) + dlJ(e,S) = j + 1 and dE(e,'Y) is even}.

(3) If dr(T,S) = j, then rj-t('Y) n r(6) is connected.

Proof. See Van Bon & Brouwer [BB, §11] (note that they omitted the requirement that the distances should be even in the second set of item (2)). 0

LEMMA (5.32). Let q beodd. r has property (C3), except for some edges yz in r 2(x). For those edges we have r(y) n r(z) n r 3 (x) = 0.

Proof. Let y,z E ri(x), dr(y,z) = 1 anq j > 1 and let us write d.= (d1J(x,y),d1J(x,z),dE(Y,Z)). We have to deal with the following 6 cases:

(1) d.= (2j,2j, 1);

{2) d.= {2j,2j -1, 1);

(3) d.= (2j -1,2j -1, 1);

(4) d.= (2j -1,2j -1,2);

(5) d.= (2j, 2j - 1, 2);

(6) d.= (2j, 2j, 2).

In dual polar terms we may assume x n y n z = 0 (otherwise we can divide it out). Furthermore, if {a,.B,'Y} = {x,y,z}, then we have {3 n 1 c ((an {3) +(an 'Y))J.. and (an {3) +(an 7) c a= al. with

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equality if and only if dim(an/1)+dim(an-y) = n. This leads ton= 2j in the cases (1), (2) and (5), n = 2j -1 in case (3), n E {2j- 1, 2j} in case (4) and n E {2j,2j + 1} in case (6). Suppose u E ~n-k( 00) n ~i( X) with i > k. Let A be a k-space in X \ u disjoint from ( u n oo )..L and let v = (A ..L n u) + A. Then we have

(*) v E ~n(oo) n ~i-k(x) n ~k(u).

Now let us try to show Cr(x,y) n Cr(x,z) \ {x} f:: 0 in each of the cases. (1),(2),(3) Take the line yz in ~. There is a unique point u on this line nearest to x (recall ~ is a near polygon with classical point-quad relations only). If u E ~n( oo ), then we are done, otherwise use (*). ( 4) Consider the quad Q = Q(y, z) in ~' then dt:.( x, Q) is either 2j- 3 or 2j-2. H dt:.(x, Q) = 2j-2, ~2i_2 (x )nQ = {'1rx }, then we are done if 1f'X E ~n(oo), otherwise use(*) (note that in this case 1f'X E ~n-l(oo)).

If da(x, Q) = 2j -3, {1rx} = ~2j-a(x)nQ, then again we have nothing to show if 1f'X E ~n( oo ). Otherwise, if j > 2 use (*), and if j = 2 take u E ~(z) n ~(1rx) n ~n(oo). (5) Consider the quad Q = Q(y,z) in~' then dt:,.(x,Q) = 2j- 2. Let {1fx} = ~2j-2(x) n Q. If 'TrX E ~n(oo), then we are done. Otherwise 1rx E ~n-1 (oo) andy has a neighbour u on the line z1rx in~. Now we have u E ~n(oo) n ~2j-t(x) and we can take v E I:(u) n I:2j-2(x). (6) Consider the quad Q = Q(y, z) in~' then dt:.(x, Q) is either 2j- 2 or 2j - 1. Let 1rx be the unique point in Q nearest to x. Suppose dt:.(x,Q) = 2j- 1. We are done if 1f'X E ~n(oo), otherwise use(*). If dt:.(x, Q) = 2j - 2, then again we are done if 1rx E ~n( oo) and otherwise again we can use(*), unless j = 2, i.e.,'n E {4, 5}. If y, z, 1rx have a common neighbour in~. then again we can use(*). If this is not the case then we must haven= 4 (for n = 5 would give y n z n 1rx has dimension 3, thus 1rx n x has dimension at most 2, ridiculous). This remaining case is just the exceptional case of our lemma. However, if r(y) n r(z) n ra(x) f:: 0, then the situation above cannot occur: if u is in this set, then it follows that Q contains a point v with dt:. ( x, v) = 5, hence da(x, Q) =J 2 = 2j- 2. In the exceptional case it really can happen that r(x)nr(y)nr(z) = 0, take for example x, y, z E ~4( oo ), y, z E ~4(x ), dt:.(Y, z) = 2, such that the quad Q = Q(y, z) meets ~2( x) n ~2( oo) in the point 1rx, where ~(y) n ~(z) n ~(1rx) = 0. If (C3) would hold in this case, then there would be a common neighbour of x, y and z in r, say u. We would

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have u E 62(x) n 62(y) n 62(z) n 64(oo), which means da(u, Q) = 1, say v E 6(u) n Q. But now v E 6(y) n 6(z) n Aa(x), i.e., v E A(1rx), a contradiction. D

PROPOSITION (5.33). Let q be odd. Every connected 2-convex subgraph of r is convex.

Proof. In the proof of Theorem (5.1) we used the fact that C(x,y) n C(x,z) \ {x} :/:: 0 for y"' z in rj(x), j > 1 in two ways. If r(y) n r(z) n rj+t(x) :/:: 0, then we can still use the old arguments. Otherwise, use the old arguments to find u E r2(y) n r2(z) n rj-2(x) and afterwards use the 2-convexity. With this adjustment the proof of Theorem ( 5.1) shows our claim. D

LEMMA (5.34). Let q be odd, dE(x,y) = 3 and u E E(x) n E2(y).

(1) r(x) n r(y) is locally noncomplete.

(2) If a convex subgraph C of r contains x and y, then it contains Q(u,y) n An(oo).

Proof. We may assume n = 3. (1) Let v E r(x) n r(y), then we may assume v E E(x) n E2(y) or v E E2(x) n E2(y). In the first case, let w1 E E( v) n E(y) and Q = Q( x, w 1 ). We can find w2 E E(v) n E(y) \ {wt} (hence w2 ~ Q) and wa E E(x) n E(w2) \ {v} with Wa ~ Q. We have dr( wl! wa) = 2 and wb Wa E r(x) n r(y) n r( v ).

If v E E2(x) n E2(y), then let us take Qt = Q(x,v), Q 2 = Q(y,v). Then Q1 n Q2 is either the point v or a line l on v. But Q1 n Q2 :/:: { v }: da(y, Qt) = 1 and 1ry E A(y )nQ1 satisfies da( 1ry, v) = 1, i.e., 1ry E Q2, a contradiction. Hence Q1 n Q2 is a line l, which contains points 1rx and 1ry, where 1rx and 1ry are A-neighbours of x and y respectively. Take w1 E E(x) n A(1ry) and w2 E E(y) n A(1rx) such that wbw2 ~ l. Then da(wt,w2) = 3 and Wt,W2 E E2(v) n r(x) n r(y). (2) Let v E E(u) n E(y) C Q(u,y) = Q. Then every point of the line uv, which is in An(oo), is in r(x) n r(y) c C. Any other point z of Q n An(oo) has a unique neighbour on uv. If z E A(u), then z E E2(x) n E9(y), thus z E r(x) n r(y) C C. There are vertices u' E E(x) n E(v) \ Q, v' E E(u') n E(y) \ Q and we have u', v' E r(x) n r(y) C C. The points in 6n(oo) on the lines vy and xu' are inC, for da(v',u) = 3. If z E A(v) n An(oo), then da(u',z) = 2, da(z, w) ~ 2 for all wE A(u) nAn(oo)n Qn A2(v) c C, da(w, u') = 3

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and z E r(u') n r(w) c c. If z E 6.(wJ), u =I= WJ =I= v, wl on the line uv, then z and u have a common neighbour w2 E 6.n( oo) \ { wt}. Now we have w2 E C, dA(u',w2) = 3, dA(x,w2) = 2, hence any vertex w3 of ~(x) n ~(w2 ) \ Q is in Cr(u',w2) C C. Thus, in ~~ wa ""w2 ""u"' v and dr;(wa,v) = 31 hence z E Cr(w3,v) C C. 0

LEMMA (5.35). Let q be odd, dr;(x, y) = 4.

( 1) r( X) n r( y) is locally noncomplete.

(2) In r(x) n r(y) = ~2(x) n ~2(Y) there are vertices u,v,w such that dr;(u,v) = 2 and u,v E ~4 (w).

(3) If Cis a convex subgraph of r containing both x andy, then it contains the set Q n 6.n ( oo) for some quad Q of 6. with dA(Q,oo) = n- 2.

Proof. We may assume n = 4. (1) and (2) follow immediately from the description of CE(x,y). (3) Suppose q ~ 5. We have· u, v, w E C and we can find 1ru E ~(u) n ~a(w) n ~a(v). Consider I= Io(?ru n w), then I has di­ameter 3 and dA(v,I) = 1, say 6.(v) n I= {1rv}. Then dA(?ru, 1rv) = 2 and dA(?rv,w) = 3. Let Q' = I0 (1ru n 1rv), then dA(w,Q') = 1, say 6.( w) n Q' = { ?rW}. Now { ?rtt, ?rV, 1rw} is a 3-codique in Q' and we can find a vertex z E 6.2( ?l'V) n 6.( ?l'U) n 6.( ?l'W) n 6.n( 00 ). Now dE( z, v) = 3, z E C and we can apply the previous lemma. If q = 3 then we use the matrix description of ~: the vertices are the symmetric 4 x 4 matrices over GF(3), two matrices being adjacent whenever they differ in a rank one matrix. By [BCN, §9.6] we may assume x is the all zero matrix and y is either

(~ 0 0

~) (~ 0 0

~} 1 0 1 0 0 1

or 0 1

0 0 0 0

In the first case take

U= 0 0 0

~) (~ 0 0

~) 1 0 0 0 0 0 ' v 0 1 0 0 0 0

92

(

1 1 2 1 1 2

w= 2 2 0

1 1 1

In the second case we take

0 1 2) 1 2 1 2 2 1 1 1 2

1 2 2 0 0 1 0 0

(0 0 0 0) 0 1 0 0

,v= 0 0 1 0 ' 0 0 0 0

(

0 0 0 0) 0 1 2 1 and z = 0 2 1 2 ·

0 1 2 0

Now in both cases we have u,v,w E r(x) n r(y) C C, di.:(u,w) = 4, z E r(u) n r(w) C C and di.:(v,z) = 3. We can apply the previous lemma. D

PROPOSITION (5.36). Let q be odd. The noncomplete convex sub­graphs ofr are precisely the subgraphs induced by I(A).

Proof. First of all, the mentioned subgraphs are obviously convex. We have to show that a convex subgraph C containing both x and I( A) contains I(A n x ), provided lo(A) has at least diameter 2. As before we may assume dr(x, I( A))= 1, say w E I(A) n r(x). We have to show y E C whenever y E I( A n x ). We have to deal with two cases:

(i) da(x,lo(A)) = 2. We can find xo E l:(x)nL:(w) and we can apply the next case twice, for I( An x) = I((A n x0 ) n x).

(ii) da(x, Io(A)) = 1.

In the last case as in the previous sections we may assume x and y have a common neighbour z in I( A). Moreover, we may assume x and y are adjacent in r' because r has locally non complete p-graphs. Let 1rx E fo(A) n .6-(x) and 1ry E fo(A) n .6-(y). We have to distinguish several cases:

(1) 1rx = z. If1ry E !(A)\ {z}, then take u E 2:(7ry)nL:2(z)ni(A). We have y E Cr(x, u) C C. If xyz is a line then we are done, for we have locally noncomplete p-graphs ..

93

If 1ry = z, then we can find u E .6.(x) n .6.(y) n .6.n(oo) n .6.2(z) and v E Io(A) n .6.(u) n .6-(z). Now use the previous arguments twice in case we have v E .6.n( oo ), otherwise use (3) and the next argument. If 1ry E .6.n-1(oo), then take u E .6.(1ry) n .6.2(z) n I(A). Now y E Cr(x,u) C C.

(2) 1rx E I(A) \ {z}. If 1ry E I(A) and 1ry is on the line z1rx, then actually we are in case (1 ). If 1ry E I( A), but 1ry is not on the line z1rx, then d~ ( 1rx, 1ry) = 2 andy E Cr(x, 1ry) C C. If 1ry E .6-n-1 ( oo ), then take u E l(A) n .6.( 1ry) n .6.2( 1rx ), which gives y E Cr(x,u) C C.

(3) 1rx E .6.n-1(oo). If 1ry E I(A) n .6.(1rx), then we can take u E I(A) n .6.(1ry) n .6.2(1rx) to get y E Cr(x,u) c C. If 1ry E I(A) \ .6.(1rx), then d~(1rx,1ry) = 2, which implies y E Cr(x, 1ry) C C. If 1ry E .6-n-1 ( oo) \ { 1rx }, then take u E .6.( 1ry) n .6.2( 1rx) n I( A), which yields y E Cr(x,u) C C. If 1rx = 1ry, but d~ ( x, y) = 2, then we can find vertices u,v E .6-n(oo)\I(A) with u E .6.(x)n.6.(z) and v E .6.(y)n.6.(z). Now use previous arguments to get u, v, y E C. Finally, if 1rx is on the line xy, then take a neighbour u E .6.(z)n.6.(x) n.6.n(oo). We have d~(y,u) = 2 and again we can use previous arguments twice to conclude u, y E C. 0

If q is even, then we have no description of r inside a dual polar graph, as was the case for the alternating forms graphs, the Hermitean forms graphs and the odd case of the quadratic forms graphs. Nevertheless, it is quite reasonable to expect that again there is an affirmative answer to the question whether a connected 2-convex subgraph is convex or not. Also one should expect that the convex subgraphs of the quadratic forms graphs in the even case are again quadratic forms graphs: obvi­ously Quad(GF(q)m) is a subgraph of Quad(GF(q)n) if m ~ n.

94

APPENDIX

A.l Graphs.

A graph is a pair r = (V, E) consisting of a set V, the vertex set of r, and ~ set E of 2-subsets of V, the edge set of r. This means that our graphs are undirected, without loops or multiple edges. Elements of V are called verticea or pointJ, elements of E are called edgea. If h, 8} is an edge, then we will write 'Y "" 8 and we will say that 'Y and 8 are adjacent, joined or neighbours. A path of length i from "( to 8 is a sequence 'Y = 'Yo ,...._ 'Yt ,..., • • • ,...._ 'Yi = 8. Being joined by a path is an equivalence relation on the vertices, its equivalence classes are the (connected) components ofr. If there is only one connected component, then the graph is called connected. The diJtance between the vertices 'Y

and 6 is the length of the shortest path from 'Y to 6. H there is no path from 'Y to 6, then we will take the distance to be oo. The diameter of a graph is the maximal distance occurring between pairs of vertices.

The subgraph of r induced on X, where X C V, is the graph with X as vertex set and whose edges are the edges of r contained in X. A matching (or 1-factor) is a partition of the vertex set into edges. Two graphs (V, E) and (W, F) are iaomorphic when there exists a bijection f: V-+ W such that {f("f),f(6)} E F if and only if {1,6} E E.

A graph is called complete when every pair of its vertices is an edge. The complete graph on n vertices is denoted by K n. A complete subgraph of a graph r is called a clique. A coclique in a graph r is a set of vertices such that no two of them are adjacent. A direct product of two graphs (V, E) and (W, F) is a graph with as vertex set V x W, where (a, (3) ,...._ ('Y, 6) if and only if (a "" "( and (3 = 6) or (a = 1 and (3 "" 6). Ann x m-grid is the direct product of ann-clique and an m-clique.

As mentioned in Chapter 5, a graph r is called bipartite if the vertex set can be partitioned into two cocliques M and N. If IMI = m, INI = n, and every vertex of M is adjacent to every vertex of N, then r is called complete bipartite and usually one writes r = Km,n· An antipodal graph is a connected graph r of diameter d > 1 for which the relation "being at maximal distance" is an equivalence relation on the vertices. In this case one can construct a new graph r, called the folded graph whosever­tices are the equivalence classes, where two classes are adjacent if their union contains an edge of r. A distance regular graph r of diameter dE {2m, 2m+ 1} with intersection array {bo, bt, ... , bd-1; c1, c2, •.. , cd}

95

is antipodal if and only if bi = Ca-i for i = 0, ... , d, i :f. m. In this case f is an antipodal r-cover of its folded graph 'f, where r = 1 +bm/Cd-m, and, if d > 2, then r is distance regular of diameter m with intersection array { bo, b11 .. • , bm-1; Ct, ~~ ... , Cm-b ")'Cm }, where ")' = r if d = 2m and 1' = 1 if d = 2m + 1.

A.2 Steiner systems.

A design is an ordered pair (X, 8) with point set X and set of blocks 8 such that 8 is a collection of subsets of X. A t-(v,k,A) design is a design (X, 8) with lXI = v, IBI = k for all B E 8 and. such that each t-subset of X is in exactly A blocks. A t-( v, k, A) design is an i-( v, k, Ai) design too for 0 ::; i ::; t, where Ai = A(~:;)/(~:;). H (X, 8) is a t-( v, k, A) design and x E X, then the derived design at x, i.e. (X \ { x}, { B \ { x} lx E B E 8} ), is a ( t - 1 )-( v - 1, k - 1, A) design.

A Steiner system S( t, k, v) is a t-( v, k, 1) design. A finite projective plane of order n is a Steiner system S{2, n + 1, n2 + n + 1) and a finite affine plane of order n is a Steiner system S(2, n, n2 ). Arbitrary projective and affine planes of order n are denoted by PG(2, n) and AG(2, n ), respectively. The planes PG{2, 4) and AG(2, 3) are unique. For d > 2, PG( d, q) will denote the d-dimensional projective spae,e over GF(q), i.e. the lattice of subspaces of a (d +!)-dimensional vector space over GF(q).

The Steiner system S(5, 8, 24) is unique; it has 759 blocks (also called octads) that correspond to the 759 words of weight 8 in the extended binary Golay code. The Steiner system has the Mathieu group M24 as automorphism group, which acts 5-transitively on the set of 24 sym­bols. The only possible intersection sizes for two octads are 0,2,4 and 8. Given a group of four symbols, the 24 symbols fall in six groups offour symbols (the given group and fl.ve more) such that every pair of groups forms an octad. Such a set of 6 groups of 4 symbols is called a sextet and the groups are called the tetrads of the sextet. An octad intersects the 6 tetrads of a sextet either 3115 , 4204 or 24 02 . (The first of these means that the octad intersects one tetrad in 3 symbols and the other five in one symbol.) A lot of information on 8(5,8,24) and its auto­morphism group M24 can be found in Chapters 10 and 11 of Conway and Sloane [CS].

96

A.3 Near polygons.

A partial linear space is a design (X, .C) in which the blocks -are called lines and such that the lines have size at least 2 and two distinct points are joined by at most one line. A near polygon is a connected partial linear space (X,£) such that for any point p E X and any line L E .C there is a unique point on L nearest to p, where the distance between two points is the distance they have in the collinearity graph, i.e. the graph with the points as vertices, two vertices being adjacent whenever they are on a common line.

A near polygon is called regular with parameters ( s, t2, ts, ... , td) if it is a finite near polygon of diameter d such that all lines have size s+ 1, each point is on t + 1 lines and any point at distance i from any given point x0 is adjacent to ti + 1 points at distance i ~ 1 from xo. Obviously one has to= -1, it= 0, td = t and t; 2:: ti-l (0 ~ i ~ d-1). A geodetically closed subset of X of diameter 2 which is nondegenerate, i.e. not all points are adjacent to one fixed point, is called a quad. A quad together with the lines contained in it turns out to be a generalized quadrangle

GQ( s, t ), i.e. a regular near polygon of diameter 2 with parameters (s, t). Shult & Yanushka {SY] showed the existence of quads: any two points x, y EX with at least two common neighbours determine a unique quad Q(x, y) containing them, provided that lines have at least three points.

Given a point x and a quad Q, there are two possibilities:

(1) there is exactly one point y in Q at distance i from x and all other points z of Q satisfy d(x, z) = d(x, y) + d(y, z);

(2) there is a number i such that each point of Q has distance either i or i + 1 to x.

In the first case x is said to be of classical type with respect to Q. In the second case x is said to be of ovoidal type with respect to Q. In this case the points of Q at distance i from x form an ovoid of Q, i.e. a set of points of Q that meets every line of Q exactly once. For more information on near polygons one is referred to the papers by Shult & Yanushka [SY] and by Brouwer & Wilbrink [BW].

97

REFERENCES

[BB] Bon, J.T.M. van, and A.E. Brouwer, The distance-regular an­tipodal covers of classical distance-regular graphs, pp.141-166 in: Colloq. Math. Soc. Janos Bolyai, Proc. Eger 1987, 1988.

[BBS] Biggs, N.L., A.G. Boshier and J. Shawe-Taylor, Cubic dis­tance-regular graphs, J. London Math. Soc. (2) 33 (1986) pp.385-394.

[BCN] Brouwer, A.E., A.M. Cohen and A. Neumaier, Distance-Regu­lar Graphs, Ergebnisse der Mathematik und ihrer Grenzgebiete 3.Folge Band 18, Springer-Verlag, Berlin Heidelberg, 1989.

[BG] Biggs, N .L. and A. Gardiner, The classification of distance tran­sitive graphs, unpublished manuscript (1974).

[BI] Bannai, E. and T. Ito, Algebraic Combinatorics I: Association Schemes, Benjamin-Cummings Lecture Note Ser. 58, The Ben­jamin/Cummings Publishing Company, Inc., London, 1984.

[Bi1] Biggs, N.L., Intersection matrices for linear graphs, pp.15-23 in: Combinatorial Mathematics and its applications (Proc. Oxford, 7-10 July 1969) (D.J.A. Welsh, ed.), Acad. Press, Lon­don, 1971.

(Bi2] Biggs, N.L., Algebraic Graph Theory, Cambridge Tracts in Math. 67, Cambridge University Press, Cambridge, 1974.

(BL] Brouwer, A.E. and E.W. Lambeck, An inequality on the param­eters of distance regular graphs and the uniqueness of a· graph related to M23 , Ann. Discr. Math. 34 (1987) (C.J. Colbourn & R.A. Mathon, eds.) pp.l01-106.

[BN] Boshier, A. and K. Nomura, A remark on the intersection arrays of distance-regular graphs, J. Combin. Th. (B) 44 (1988) pp.147-153.

[Brl] Brouwer, A.E., The uniqueness of the near hexagon on 759 points, pp.47-60 in: Finite Geometries, T.G. Ostrom Conf. Pullman 1981, Lecture Notes in Pure and Applied Math. 82 (N.L. Johnson, M.J. Kallaher & C.T. Long, eds.), Marcel Dekker, New York, 1982.

98

[Br2] Brouwer, A.E., On the uniqueness of a certain thin near oc­tagon (or partial 2-geometry, or parallelism) derived from the binary Golay code, IEEE Trans. Inform. Theory, IT-29 (1983) pp.370-371.

[Br3] Brouwer, A.E., Uniqueness and nonexistence of some graphs related to M22, Graphs Combin. 2 (1986) pp.21-29.

[BW] Brouwer, A.E. and B.A. Wilbrink, The structure of near poly­gons with quads, Geom. Dedicata 14 (1983) pp.145-176.

[Ca] Cameron, P.J., Dual polar spaces, Geom. Dedicata 12 (1982) pp.75-85.

[CS] Conway, J.H. and N.J.A. Sloane, Sphere Packings, Lattices and Groups, Grundlehren der mathematischen Wissenschaften 290, Springer-Verlag, New York, 1988.

[Cu] Curtis, R.T., A new combinatorial approach to M24, Math. Proc. Camb. PhiL Soc. 79 (1976) pp.25-42.

[De] Delsarte, Ph., An algebraic approach to the association schemes of coding theory, Philips Research Reports Suppl. 10, 1973.

[Eg] Egawa, Y., Characterization of H(n,q) by the parameters, J. Combin. Th. (A), 31 (1981) pp.108-125.

[EKR] Erdos, P., Chao Ko and R. Rado, Intersection theorems for systems of finite sets, Quart. J. Math. Oxford (2) 12 (1961) pp.313-320.

[FII] Faradjev, I.A., A.A. Ivanov and A.V. Ivanov, Distance-transi­tive graphs of valency 5, 6 and 7, Europ. J. Combinatorics 7 (1986) pp.303-319.

[Go] Godsil, C.D., Bounding the diameter of distance regular graphs, Combinatorica 8 (1988) pp.333-343.

[IS1] Ivanov, A.A. and S.V. Shpectorov, Characterization of the asso­ciation schemes of Hermitian forms over GF(22 ), Geom. Dedi­cata 30 (1989) pp.23-33.

[IS2] Ivanov, A.A. and S.V. Shpectorov, The 2-fundamental group of the M2a-geometry is trivial, preprint.

[Li] Lint, J.H.van, Introduction to coding theory, Graduate Texts in Math. 86, Springer-Verlag, New York Heidelberg Berlin, 1982.

[MS] MacWilliams, F.J. and N.J.A. Sloane, The Theory of Error­Correcting Codes, North Holland Publ. Co., Amsterdam, 1977.

99

[No1] Nomura, K., An inequality between intersection numbers of a distance-regular graph, J. Combin. Th. (B) 43 (1987) pp.358-359.

[No2] Nomura, K., On Local Structure of a Distance-Regular Graph of Hamming Type, J. Combin. Th. (B) 47 (1989) pp.120-123.

[Ro] Ronan, M.A., Embeddings and Hyperplanes of Discrete Geome­tries, Europ. J. Combinatorics 8 (1987) pp.179-185.

(RS] Ronan, M.A. and S.D. Smith, 2-Local Geometries for some Sporadic Groups, Proceedings of Symposia in Pure Mathe­matics 37 (Santa Cruz 1979) (B. Cooperstein and G. Mason, eds. ), pp. 283-289, Amer. Math. Soc., Providence, Rhode Island, 1980.

(Sh] Shult, E.E., Characterizations of the Lie incidence geometries, pp.157-186 in: Surveys in Combinatorics (Invited papers for the Ninth British Com bin. Conf. ), London Math. Soc. Lecture Note Ser. 82 (E. Keith Lloyd, ed.), Cambridge University Press, Cambridge, 1983.

[Su] Suzuki, H., On a Distance Regular Graph with be= 1, preprint.

(SY] Shult, E.E. and A. Yanushka, Near n-gons and line systems, Geom. Dedicata 9 (1980) pp.1-72.

[Tc] Tchuda, F.L., Const~ction of an automorphic graph on 280 vertices using finite geometries, pp.169-17 4 in: Investigations in the Algebraic Theory of Combinatorial Objects (M.H. Klin & I.A. Faradjev, eds.), Institute for System Studies, Moscow, 1985. (In Russian)

[Te] Teirlinck, L., On projective and afline hyperplanes, J. Combin. Th. (A) 28 (1980) pp.290-304.

[Ti] Tits, J., Buildings of spherical type and :6.nite BN-pairs, Lecture Notes in Math. 386, Springer-Verlag, Berlin, 1974.

100

SAMENVATTING

Een graaf r van diameter d heet afstandsregulier met parameters { bo, bt, ... , bd-1; Ct, ... , cd} als voor iedere i ~ d en voor ieder twee­tal punten X en y Van r op afstand i geldt: 1ri-1(x) n r(y)l = Ci

en lri+t(x) n r(y)l = bi. Hierbij is ri(x) de verzameling punten op af­stand j van X en r(y) = r1(y). We schrijven vaak bo = k, c2 = jt,

ai = k - bi - Ci en a1 = ..\. Voor een tweetal punten x en y op afstand i geldt dan ook nog lri(x)nr(y)l = ai. In dit proefschrift wordt aandacht besteed aan nonexistentie, uniciteit en deelstructuren van afstandsreguliere grafen.

In hoofdstuk 2 wordt een aantal restricties afgeleid voor de parameters, bijvoorbeeld:

- als voor een i > 0 geldt Ci = bt, dan geldt bi ~ 1;

- als d > 2i, i ~ 2, (ci-1,ai-bbi-1) = (c1,at,bt), ..\ = 0, Ci = 1, ai > 0, a2i < ai en (c2i-1, a2i-t, b2i-1) = (ci, ai, bi), dan geldt k ~ 1 + 2bi;

- als ai f=. 0, 1 ~ i ~ d, dan

waarbij gelijkheid equivalent is met: voor ieder viertal punten

a, /3, 1 en 8 van r, zodanig data"" /3, 1 ""8 en d('Y,a) = i, zijn de drie afstanden d(a,8), d(/3,/), en d(/3,8) niet allemaal gelijk.

In hoofdstuk 3 wordt de uniciteit van een tweetal grafen bewezen: op isomorfie na is er slechts een afstandsreguliere graaf met parameters {15, 14, 12; 1, 1, 9}, terwijl er ook precies een afstandsreguliere graafmet parameters {9, 8, 6, 3; 1, 1, 3, 8} is op isomorfie na. Hierbij wordt gebruik gemaakt van het feit, dat deze afstandsreguliere grafen Petersen deel­grafen hebben, hetgeen een gevolg is van gelijkheid in de laatste van bovengenoemde ongelijkheden. Voor een afstandsreguliere graaf met bi = 1, k > 2 geldt voor de diameter de volgende bovengrens: d ~ 3i - 1. In hoofdstuk 3 wor­den alle grafen gevonden die deze bovengrens halen. In het bijzonder geldt dat als i > 1, dan moet i = 2 en de graaf moet de dodecaeder zijn, een gevolg van de tweede van de bovengenoemde restricties.

101

In hoofdstuk 4 staat een afstandsreguliere gr.aaf op 759 punten cen­traal. Deze graaf heeft als punten de blokken van het Steinersysteem 8(5,8,24), waarbij twee blokken verbonden zijn als ze disjunct zijn. De graaf heeft de structutir van een bijna veelhoek, de lijnen zijn de trio's van het Steinersysteem. In dit hoofdstuk bepalen we de hypervlakken van deze bijna veelhoek, d. w .z. de deelverzamelingen H van de pun­tenverzameling zodanig dat iedere lijn of bevat is in H of precies een punt van H bevat.

Voor een aantal families afstandsreguliere grafen worden in hoofdstuk 5 de convexe deelgrafen bepaald. Als r = (X, E) een graaf is en C C X, dan heet C t-convex als voor ieder tweetal punten x en y van C met een afstand ~ t in r de punten van ieder kortste pad tussen X en y in rook tot C behoren. Als C t-convex is voor iedere t, dan is C convex. Het blijkt dat voor de bestudeerde families grafen de convexe deel­grafen tot dezelfde familie behoren, er is slechts een uitzondering: de grafen op de Hermitese vormen op een vectorruimte over G F( 4) hebben ook de vierhoek (K2,2) als convexe deelgraaf. Voor de bestudeerde families grafen is tevens de vraag beantwoord of een samenhangende 2-convexe deelgraaf convex is. Het antwoord hierop is ja, met dezelfde uitzondering: de grafen op de Hennitese vormen op een vectorruimte over GF(4) hebben samenhangende 2-convexe deelgrafen die niet con­vex zijn.

102

CURRICULUM VITAE

Ernst Willem Lambeck werd op 1 juni 1959 geboren te Siddeburen (gemeente Slochteren). Van 1971 tot 1977 bezocht hij de Rijksscho­lengemeenschap te Appingedam, waar hij in mei 1977 het V.W.O.­diploma behaalde. Gedurende deze zes jaren was drs. H.J.C. Vlogtman zijn wiskundeleraar. In september 1977 begon Ernst Lambeck wiskunde te studeren aan de Rijksuniversiteit van Groningen. Na het kandidaats­examen te hebben behaald in september 1981, legde hij het doctoraal

· examen (cum laude) af op 28 maart 1985. Zijn afstudeerhoogleraar was Prof. dr. M. van der Put. Van oktober 1985 tot oktober 1990 was de auteur van dit proefschrift als assistent-onderzoeker werkzaam in de vakgroep Discrete Wiskunde van de Technische Universiteit Eindhoven.

103

STELLING EN

behorende bij het proefschrift

CONTRIBUTIONS TO THE THEORY

OF DISTANCE REGULAR GRAPHS

door E.W. Lambeck

1. Het bewijs van Stelling 7 in het hieronder genoemde ar­tikel van Brouwer en Wilbrink is fout. In feite wordt het volgende bewezen. Zij (X, .C) een reguliere bijna acktkoek met parameters (s,t2 ,t3 ,t4 ). Dan geldt tenminste een van de volgende vijf beweringen:

(i) s = 1;

(ii) t2 = 0;

(iii) t2 = 1;

(iv) t3 = t2(t2 + 1) en t4 = t2(t3 + 1): (X, .C) is klassiek;

(v) t2 ~ 3, s = t2 2 en t24 +t23 +2t2 ~ t3 ~ t24 +t23 + t22 + 2t2 + 18.

Ref.: A.E. Brouwer & H.A. Wilbrink, Tke structure of near polygons with. quads, Geometriae Dedicata 14 (1983) pp. 145-176.

2. Vermoedelijk kunnen in stelling 1 de beweringen (iii) en ( v) worden weggelaten.

3. De Fostergraaf is de unieke 3-overdekking van de inciden­tiegraaf van het gegeneraliseerde vierkant GQ(2, 2) zonder achthoeken.

4. Laat m een geheel getal zijn, m ~ 1. De diophantische vergelijking

heeft niet-triviale gehele oplossingen dan en slechts dan als m oneven is.

Ref.: E.W. Lambeck & V.D. Tonchev, Probleem voor Nieuw Arckief voor Wiskunde (aangenomen).

2

5. Laat (X,C) een reguliere bijna veelhoek met parameters ( s, t2, ta, ... , td), waarbij t2 ::::: 1 en s ::::: 1. Analoog aan de definities van N;,o( Q) en N;,c( Q) in het hieronder ge­noemde artikel van Brouwer en Wilbrink defini(:;ren we voor een punt x E X en 0 $ i $ d - 1 de volgende verzamelingen:

MO;(x) := {QIQ is een quad van X met x E Ni,o(Q)} en

MCi(x) := {QIQ is een quad van X met x E Ni,c(Q)}. Zij verder

a = si _t-'-( t_+~l );..,.. t2(t2 + 1)

i t enb= IT -

i=2 1 + tj

Dan geldt voor x EX en 0$ i $ d -1:

ab I MOi(x)l =

1 (ti+l- t2(1 + t;)) ;

+stz IMC;(x)l=ab(t-ti+I) (irfO) en

IMCo(x)l = t(t+ 1) . t2(t2 + 1)

Ref.: A.E. Brouwer & H.A. Wilbrink, The &tructure of

near polygons with quads, Geometriae Dedicata 14 (1983) pp. 145-176.

6. Zij K = Q[9] een cyclisch cubisch getallenlichaam, d.w.z. een derdegraads Galoisuit brei ding van Q, waarin aileen het priemgetal p, p =F 3, vertakt. Laat x + y9 + z92 een element van A, de ring van gehelen van K, zijn met norm M. Dan zijn er c1. c2 , c3 , a:fhangend van M maar niet van x, y en z, met de eigenschap dat er een element x 0 + y0 9 + z092

in A is van norm M met lxol $ell IYol $ c2 en lzol $ca.

Ref.: E.W. Lambeck, Cyclisch cubische lichamen en hun klassegetal, afstudeerscriptie Rijksuniversiteit Groningen, februari 1985.

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7. Voor ied~r geheel getal i ~ 2 definieren we een permutatie '1ri op de natuurlijke getallen als volgt: '1ri(ki + j) = (k + 1)i + 1- j, k ~ 0, 1 ~ j ~ i. Laat, voor n ~ 2, B! = 7rk+l ... 7rn(1), 1 ~ k < n en B: = 1. Dan is voor vaste k de rij {B!}~=2 een strict toenemende rij.

Ref.: Problem E3163 van The American Mathematical Monthly, 93(7) 1986 en 96(1) 1989.

8. In tegenstelling tot wat de meeste mensen denken is sjoelen wei degelijk een sport.

9. Paul Erd8s zei eens: "Een wiskundige is een machine die koffie omzet in stellingen". Voor de werkruimte van een wiskundige zou derhalv~, naast de thans ingeburgerde per­sonal computer, een koffiezetapparaat tot de standaard­inventaris moeten behoren.

Ref.: Paul Hoffman, The man who loves only numbers, The Atlantic Monthly, november 1987.

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