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Distance-Transitive Graphs Submitted for the module MATH4081 Robert F. Bailey (4MH) Supervisor: Prof. H.D. Macpherson May 10, 2002

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Distance-Transitive Graphs

Submitted for the module MATH4081

Robert F. Bailey (4MH)

Supervisor: Prof. H.D. Macpherson

May 10, 2002

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2

Robert BaileyDepartment of Pure MathematicsUniversity of LeedsLeeds, LS2 9JTMay 10, 2002

The cover illustration is a diagram of the Biggs-Smith graph, a distance-transitivegraph described in section 11.2.

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Foreword

A graph is distance-transitive if, for any two arbitrarily-chosen pairs of vertices atthe same distance, there is some automorphism of the graph taking the first paironto the second.

This project studies some of the properties of these graphs, beginning withsome relatively simple combinatorial properties (chapter 2), and moving on to dis-cuss more advanced ones, such as the adjacency algebra (chapter 7), and Smith’sTheorem on primitive and imprimitive graphs (chapter 8).

We describe four infinite families of distance-transitive graphs, these being theJohnson graphs, odd graphs (chapter 3), Hamming graphs (chapter 5) and Grass-mann graphs (chapter 6). Some group theory used in describing the last two ofthese families is developed in chapter 4.

There is a chapter (chapter 9) on methods for constructing a new graph froman existing one; this concentrates mainly on line graphs and their properties.

Finally (chapter 10), we demonstrate some of the ideas used in proving thatfor a given integerk > 2, there are only finitely many distance-transitive graphsof valencyk, concentrating in particular on the casesk = 3 andk = 4. We also(chapter 11) present complete classifications of all distance-transitive graphs withthese specific valencies.

Acknowledgements

I would like to thank my supervisor, Prof. H.D. Macpherson, for his assistanceand encouragement throughout the duration of the project, and for suggesting thetopic in the first place. Thanks are also due to my tutor, Dr. R.B.J.T. Allenby, for(voluntarily!) reading a preliminary version of the project, to Profs. J.K. Truss andJ.C. McConnell for answering my “Do you know anything about....” questions,and to the assessor for his positive comments during the preliminary assessments.I should also thank Prof. E.R. Vrscay of the University of Waterloo for supplying[26].

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Contents

1 Introduction 61.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . 61.2 A Little Group Theory . . . . . . . . . . . . . . . . . . . . . . . 71.3 Automorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.4 Different Kinds of ‘Transitive’ . . . . . . . . . . . . . . . . . . . 10

2 Introducing the Distance-Transitive Graph 142.1 Basic Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.2 Distance Partitions . . . . . . . . . . . . . . . . . . . . . . . . . 172.3 Intersection Numbers and Intersection Arrays . . . . . . . . . . . 182.4 Distance-Regular Graphs . . . . . . . . . . . . . . . . . . . . . . 24

3 Uniform Subset Graphs 273.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.2 The Johnson GraphsJ(n,2,1) . . . . . . . . . . . . . . . . . . . 283.3 The Johnson GraphsJ(n,k,k−1) . . . . . . . . . . . . . . . . . 303.4 Pretty Pictures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333.5 The Odd Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

4 Some Permutation Group Theory 394.1 Primitive and Imprimitive Actions . . . . . . . . . . . . . . . . . 394.2 Direct and Semi-direct Products . . . . . . . . . . . . . . . . . . 404.3 Wreath Products . . . . . . . . . . . . . . . . . . . . . . . . . . . 414.4 Projective Groups . . . . . . . . . . . . . . . . . . . . . . . . . . 41

5 Hamming Graphs 435.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435.2 Distance-Transitivity . . . . . . . . . . . . . . . . . . . . . . . . 445.3 Thek-Cubes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

6 Grassmann Graphs 476.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476.2 Distance-Transitivity . . . . . . . . . . . . . . . . . . . . . . . . 496.3 Intersection Arrays . . . . . . . . . . . . . . . . . . . . . . . . . 52

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CONTENTS 5

6.4 Linking the Grassmann and Johnson Graphs . . . . . . . . . . . . 54

7 Linear Algebra and Distance-Transitive Graphs 567.1 The Spectrum and the Adjacency Algebra . . . . . . . . . . . . . 567.2 Distance Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . 607.3 The Intersection Matrix . . . . . . . . . . . . . . . . . . . . . . . 627.4 Algebraic Constraints on the Intersection Array . . . . . . . . . . 67

8 Primitive and Imprimitive Graphs 728.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 728.2 Antipodal Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . 758.3 Bipartite Distance-Transitive Graphs . . . . . . . . . . . . . . . . 778.4 Smith’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 80

9 New Graphs from Old 849.1 Line Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 849.2 Automorphisms of Line Graphs . . . . . . . . . . . . . . . . . . 869.3 Eigenvalues of Line Graphs . . . . . . . . . . . . . . . . . . . . . 889.4 Distance-Transitive Line Graphs . . . . . . . . . . . . . . . . . . 909.5 Bipartite Doubles . . . . . . . . . . . . . . . . . . . . . . . . . . 95

10 Bounding the Diameter 9910.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9910.2 Cubic Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9910.3 Tetravalent Graphs . . . . . . . . . . . . . . . . . . . . . . . . . 10410.4 Extending to Higher Valencies . . . . . . . . . . . . . . . . . . . 107

11 Graphs of Low Valency 11011.1 Smith’s Program . . . . . . . . . . . . . . . . . . . . . . . . . . 11011.2 Cubic Distance-Transitive Graphs . . . . . . . . . . . . . . . . . 11011.3 Tetravalent Distance-Transitive Graphs . . . . . . . . . . . . . . . 117

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Chapter 1

Introduction

1.1 Basic Definitions

There are many simple definitions from graph theory that the reader is probablyfamiliar with already (if not, consult an introductory text such as Wilson [40]).However, we will include these, if only to demonstrate our terminology and nota-tion, which varies considerably between texts.

Definitions 1.1.1– Graph Theory Definitions

• Let Γ be a graph, withVΓ denoting the set of vertices ofΓ andEΓ the set ofedges ofΓ.

• A graph issimpleif all edges join two distinct vertices, and between any pairof verticesu,v there is at most one edge. In this project, we will always beconsidering simple graphs.

• Two verticesu,v ∈ VΓ areadjacentif there is a single edge joining them.We write u∼ v to denote this. (Note that this relation∼ is symmetric, butnot reflexive or transitive.) Also, two edges are adjacent if they are incidentwith a common vertex.

• Thedegree, or valencyof a vertexv is the number of edges incident withv,denoted deg(v).

• A graphΓ is said to beregular if deg(u) = deg(v) for all u,v ∈ VΓ. It isk-regular if deg(v) = k for all v∈VΓ. In this case, we refer to thevalency ofΓ. A 3-regular graph is frequently described as acubicgraph, or sometimesas atrivalent graph.

• A pathπ in Γ is a finite sequence of edges from vertexu to vertexv whereall the intermediate vertices are distinct.

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1.2 A Little Group Theory 7

• Γ is said to beconnectedif for any u,v∈VΓ there exists a pathπ from u tov. Otherwise, we sayΓ is disconnected. A maximal connected subgraph ofa disconnected graphΓ is called acomponentof Γ.

• A geodesicin Γ is a path fromu to v containing the least number of edges.

• Thedistancefrom u∈VΓ to v∈VΓ is the least number of edges in a pathfrom u to v. This is denoted byd(u,v).

• The maximum distance in a graphΓ is called thediameterof Γ.

• A circuit in Γ is a path fromv to v.

• Thegirth of Γ is the length of the shortest circuit inΓ.

• A graph isbipartite if VΓ = V1∪V2 and each edge ofΓ has one end inV1 andthe other end inV2. It can be shown thatΓ is bipartite if and only if it has nocircuits of odd length.

• An isomorphismfrom a graphΓ to a graph∆ is a bijective functionϕ : VΓ→V∆ such that foru,v∈VΓ, ϕ(u)∼ ϕ(v) (in ∆) if and only if u∼ v (in Γ). Inthis case we sayΓ and∆ areisomorphic, denoted byΓ∼= ∆.

• An isomorphism from a graphΓ to itself is called anautomorphismof Γ.

1.2 A Little Group Theory

It is assumed that the reader is already familiar with basic group theory, for examplegroups, Abelian groups, subgroups, direct products, cosets, Lagrange’s Theorem,homomorphisms, isomorphisms and factor groups. (If not, see a book such as Al-lenby [2] or Gallian [18].) In this project, we are particularly concerned with theconcept of agroup actionon a set (specifically on the set of vertices of a graph).Formally, this is defined as follows:

Definition 1.2.1A group actionof a groupG on a setX is a functionρ : X×G→ X satisfying

• for all x∈ X and for allg,h∈G, x(gh) = (xg)h;

• for all x∈ X, xe= x (wheree is the identity element ofG).

An action is said to befaithful if the only element ofG fixing all elements ofX isthe identity.

(The basics of group actions are covered well in Slomson [31]). This definitionleads us straight away to a number of others:

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8 Introduction

Definition 1.2.2Theorbit of an elementx∈ X is the set OrbG(x) = xg

∣∣g∈ G (i.e. all elementsof X that are the image ofx under some permutationg∈ G). OrbG(x) is a subsetof X.

Definition 1.2.3SupposeG is a permutation group acting on a setX. We sayG is transitive on Xif, for all x,y∈ X, there existsg∈G such thatxg= y.

We can also characterise transitivity in this way:

Proposition 1.2.4G is transitive onX if and only if, for all x∈ X, OrbG(x) = X.

Proof:SupposeG is transitive onX. Then, by definition 1.2.3, for anyx∈ X there existssomeg∈G such thatxg= y for all y∈ X. Thusy∈OrbG(x) for all x∈ X.

Conversely, suppose OrbG(x) = X for all x∈ X. Then, by definition 1.2.2, foranyx,y∈ X, there existsg∈G such thatxg= y. SoG is transitive onX .

Definition 1.2.5Thestabiliserof an elementx∈ X is the set StabG(x) = g∈ G

∣∣xg= x (i.e. allpermutations ofx that mapx to itself).

It is easy to show the following:

Proposition 1.2.6StabG(x) is a subgroup ofG.

Proof:StabG(x) is closed, as for anyg,h ∈ StabG(x), we havex(gh) = (xg)h = xh = x.Associativity is automatic, as StabG(x) is a subset ofG. The identity is obviouslyin StabG(x), asxe= x is always true. Every element of StabG(x) clearly has aninverse in StabG(x), as ifxg= x, thenx = xg−1. Hence StabG(x)≤G.

We can now determine the orbit and stabiliser of an element under the actionof a small permutation group, and determine whether this action is transitive.

Example 1.2.7Consider the permutation groupS4 acting on the setX = 1,2,3,4. The orbitof 1 ∈ X, OrbS4(1) = 1,2,3,4, asS4 contains permutations that map any ele-ment ofX to any other element. Thus we can seeS4 is transitive onX (but noton 1,2,3,4,5, for example). The stabiliser of 1 is the group of permutationsStabG(x) = e,(2 3),(2 4),(3 4),(2 3 4),(2 4 3).

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1.3 Automorphisms 9

You might have noticed that, in the above example,|OrbS4(1)| = 4, and that|StabS4(1)| = 6, while |S4| = 24= 4×6. This is explained by the following theo-rem.

Theorem 1.2.8– The Orbit-Stabiliser TheoremLet G be a group acting on a setX. Then for anyg∈G and anyx∈ X,

|OrbG(x)|× |StabG(x)|= |G|.

In particular, the size of an orbit divides|G|.

Proof:Lagrange’s Theorem tells us that, for any subgroupH ≤G,

|G|= |H|× |G : H|

where|G : H| is the size of[G : H], the set of right cosets ofH in G. So to prove theOrbit-Stabiliser Theorem, we just have to establish a 1-1 correspondence betweenOrbG(x) and[G : StabG(x)]. So supposeg1,g2∈G, and letH = StabG(x) for clarity.Then

Hg1 = Hg2 ⇔ H = Hg2(g−11 ) (by standard group theory)

⇔ g2g−11 ∈ H

⇔ x(g2g−11 ) = x (sinceH is the stabiliser ofx)

⇔ (xg2)g−11 = x (by the definition of a group action)

⇔ xg2 = xg1.

So we have the 1-1 correspondence we require, so|OrbG(x)|= |G : StabG(x)|, andthe result follows.

In the following section, we bring groups and graphs together.

1.3 Automorphisms

We have already defined an automorphism ofΓ to be an isomorphism fromΓ to it-self. This can also be thought of as follows: an automorphism ofΓ is a permutationof the vertices ofΓ that maps edges to edges (and non-edges to non-edges). Thereis an identity automorphism (put simply: doing nothing) and each automorphismhas an inverse. We can therefore see quite easily that the set of all automorphismsof Γ forms a group.

Definition 1.3.1The group of all automorphisms of a graphΓ is called theautomorphism groupofΓ, denoted by Aut(Γ).

For a graph onn vertices, Aut(Γ) is a subgroup ofSn.

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10 Introduction

Examples 1.3.2– Some graphs with familiar groups as their automorphism groups

Figure 1.1: The complete graphK4

Aut(K4)∼= S4, as every permutation ofVK4 is an automorphism.

Figure 1.2: The square, or 4-circuitC4

Aut(C4)∼= D4 (the dihedral group on 4 vertices), as we can rotate or reflectC4.

Figure 1.3: A graph with Aut(Γ) = e

This graphΓ has only one automorphism, the identity, so Aut(Γ) = e.

1.4 Different Kinds of ‘Transitive’

In this project, we are going to consider a class of graphs that have special condi-tions on their automorphism groups. Such a condition is defined below:

Definition 1.4.1We say thatΓ is vertex-transitiveif Aut(Γ) acts transitively onVΓ.

A similar definition is as follows:

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1.4 Different Kinds of ‘Transitive’ 11

Definition 1.4.2Γ is said to beedge-transitiveif Aut(Γ) acts transitively onEΓ.

These two properties are not interchangable; there exist graphs that are vertex-transitive but not edge-transitive, and vice-versa, also graphs that are both vertex-and edge-transitive and graphs that satisfy neither property. We now give examplesof each.

Examples 1.4.3

Figure 1.4: A vertex- and edge-transitive graph

Figure 1.5: A graph that is neither vertex- nor edge-transitive

Examples 1.4.4– Graphs with only one of the two properties

Figure 1.6:∆ (left) andK1,4 (right)

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12 Introduction

∆ is clearly vertex-transitive, but is not edge-transitive. (This is an example of acirculant graph. Consult Godsil & Royle [22] for details of these.)K1,4 (a com-plete bipartitegraph) is clearly edge-transitive, but is not vertex-transitive as it isnot regular (it has one vertex of degree 4, and four of degree 1).

Examples 1.4.5– Two graphs with the same automorphism group

Figure 1.7:C5 (left) andW5 (right)

BothC5 andW5 (theW stands for “wheel”) haveD5 as their isomorphism group,but it does not act vertex-transitively onW5, as the vertexv is fixed by all elementsof D5. This is because it is the only vertex of degree 5, while all other vertices havedegree 3, so OrbD5(v) = v. It does not act edge-transitively onW5, as we canpartitionEW5 into two orbits underD5: the “rim” of the wheel and the “spokes”.No automorphism moves an edge between these two sets.

A stronger condition than either of the above is that ofs-arc-transitivity. Toexplain this, we first need to know what ans-arc is.

Definition 1.4.6An s-arc in a graphΓ is a sequence ofs+1 verticesv0, . . . ,vs, such that

• vi ∼ vi−1, and

• vi−1 6= vi+1

for 0< i < s.

Notice the subtle difference between the definitions ofs-arcandpath: ans-arcis not necessarily a path. Consider the following example:

Example 1.4.7In figure 1.8,abdehgi, abdgi andabd f i are all paths, of which the last two are

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1.4 Different Kinds of ‘Transitive’ 13

geodesics (sod(a, i) = 4). The first path is also a 6-arc, and the last two are 4-arcs.abdc f dgiis a 7-arc, but not a path, as the vertexd is repeated.

Figure 1.8: Examples of paths, geodesics and arcs

The next definition is a natural one, bearing in mind the previous one.

Definition 1.4.8A graphΓ is s-arc-transitiveif Aut(Γ) acts transitively on the set of alls-arcs ofΓ.

A 0-arc transitive graph is just another name for a vertex-transitive graph. A1-arc transitive graph is both vertex- and edge-transitive and is often known, ac-cording to Biggs [8], as asymmetricgraph. It is clear that ans-arc-transitive graphis also(s− 1)-arc-transitive, and thus is(s− 2)-arc-transitive, and so on induc-tively. Going the other way, however, we require the following definition:

Definition 1.4.9Γ is strictly s-arc-transitiveif it is s-arc-transitive but not(s+1)-arc-transitive.

We are going to consider in detail a class of graphs with a symmetry conditionthat is stronger than any of the above, namely distance-transitive graphs. They aredefined as follows:

Definition 1.4.10For any verticesu,v,u′,v′ ∈VΓ satisfyingd(u,v) = d(u′,v′), Γ isdistance-transitiveif there exists someg∈ Aut(Γ) satisfyingug= u′ andvg= v′.

Examples of distance-transitive graphs include the complete graphsKn, then-circuitsCn, the platonic graphs and many others, as we will discover.

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Chapter 2

Introducing theDistance-Transitive Graph

2.1 Basic Properties

In this chapter, we begin to determine the properties of distance-transitive graphsthat we are going to use. We start by finding relationships between the differentnotions of transitivity we saw in 1.4, specifically concerning distance-transitivegraphs. However, we also need some properties ofs-arc-transitive graphs, whichwe can then relate to the distance-transitive case. We start with a fairly obviousproperty.

Proposition 2.1.1Any distance-transitive graph is vertex-transitive.

Proof:Let Γ be a distance-transitive graph. Take verticesu = v andu′ = v′ ∈VΓ (so thatd(u,v) = d(u′,v′) = 0). We know there existsg ∈ Aut(Γ) such thatug = u′ andvg= v′ (asΓ is distance-transitive). HenceΓ is also vertex-transitive.

However, the converse is not true in general; there exist vertex-transitive graphsthat are not distance-transitive. Consider the following counterexample:

Example 2.1.2The cyclic n-ladderLn (L6 is shown in Figure 2.1) is clearly vertex-transitive –we can rotate and reflectLn (so thatDn ≤ Aut(Γ)); we can also move any of the‘inner ring’ to anywhere on the ‘outer ring’. In fact, Aut(Γ) = 〈r,v,s : rn = v2 =s2 = e, vrn−1 = rv, rs= sr, vs= sv〉. However,Γ is not distance-transitive: consideru,v,u′,v′ as shown. Clearlyd(u,v) = d(u′,v′) = 2. But there is no automorphismthat movesu,v to u′,v′, as there is only one geodesic path fromu to v, whilethere are two fromu′ to v′ .

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2.1 Basic Properties 15

Figure 2.1: A vertex-transitive graph that is not distance-transitive

We now take a detour into the theory ofs-arc-transitive graphs, with the aim offinding when such a graph is also distance-transitive. The following two lemmasare both due to W.T. Tutte [37], although Godsil & Royle [22] explain them usingmore modern terminology.

Lemma 2.1.3(Tutte 1947)SupposeΓ is s-arc-transitive, with valency≥ 3 and girthg. Theng≥ 2s−2.

Proof:By assumption,Γ contains a circuit of lengthg, say(v0,v1, . . . ,vg−1,v0). This is,in particular, ag-arc. Since each vertex ofΓ has degree≥ 3, we can change theterminal vertex to obtain a differentg-arc (v0,v1, . . . ,vg−1,vg). Clearly, no auto-morphism ofΓ cam map the firstg-arc onto the second, sos< g.

Any circuit in Γ contains ans-arc, so by thes-arc-transitivity ofΓ, anys-arclies in some circuit of lengthg. (?)

Now let α = (v0, . . . ,vs) be ans-arc inΓ. By (?) above,α lies in some circuitof lengthg, C1 say. Since all vertices ofΓ have degree≥ 3, vs−1 is adjacent tosome vertexw, which is not one ofvs−2 or vs. (In fact, w cannot be any of thevertices inα, as that would give a circuit of length< s, contradictings< g.) Soβ = (v0, . . . ,vs−1,w) is anothers-arc of Γ, with α∩ β as an(s− 1)-arc. By (?)above,β is in some other circuit of lengthg, C2 say.

So we have the following situation:

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16 Introducing the Distance-Transitive Graph

Figure 2.2: Schematic for 2.1.3

We can construct a new circuit,C = (C1 \α)∪(vsvs−1)∪(vs−1w)∪(C2 \β), whichhas length(g−s)+1+1+(g−s) = 2g−2s+2. Since the girth ofΓ is g, we have2g−2s+2≥ g, and sog≥ 2s−2.

An obvious follow-up question is: what happens when the girth of ans-arc-transitive graph is at its minimum value of 2s−2? The next lemma helps in an-swering this.

Lemma 2.1.4(Tutte)SupposeΓ is s-arc-transitive with girth 2s− 2. Then diam(Γ) = s− 1, andΓ isbipartite.

Proof:Since the girth ofΓ is 2s−2, anys-arc lies in at most one circuit of length 2s−2.SinceΓ is s-arc-transitive, everys-arc lies in such a circuit, also any such circuitmust contain ans-arc. (∗)

Now diam(Γ)≥ s−1, the distance of opposite vertices in a circuit. Then sup-pose we have verticesu,v with d(u,v) = s. By definition, there is ans-arc fromu tov, so these lie in a circuit of length 2s−2 (from (∗) above). Hence there is a path oflengths−2 fromu to v, contradictingd(u,v) = s. Hence we have diam(Γ)≤ s−1,and so diam(Γ) = s−1.

Now supposeΓ is not bipartite (so thereforeΓ contains an odd circuit) and thatdiam(Γ) = s−1. LetC be an odd circuit of minimal length. Since the girth ofΓ is2s−2, which is even,C must have length 2s−1.

Figure 2.3: Schematic for 2.1.4

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2.2 Distance Partitions 17

Now let u,v,v′ be as shown, so we have ans-arc α = (u, . . . ,v,v′). From (∗)above, this must lie in a circuitC′ of length 2s−2. So(C\α)∪(C′ \α) is a circuitof length(2s− 1− s) + (2s− 2− s) = 2s− 3, contradicting the girth ofΓ being2s−2. HenceΓ must be bipartite.

We now have the tools to show that certains-arc-transitive graphs are alsodistance-transitive.

Theorem 2.1.5A connecteds-arc-transitive graph with girth 2s−2 is distance-transitive.

Proof:Let Γ bes-arc-transitive with girth 2s−2, and choose(u,u′) and(v,v′) to be twopairs of vertices at distancei. By Lemma 2.1.4, diam(Γ) = s− 1, so we havei ≤ s−1. By assumption, there is a path of lengthi, and thus ani-arc, fromu tou′ and fromv to v′. SinceΓ is s-arc-transitive, it is alsoi-arc-transitive (fori ≤ s),so there is some automorphism ofΓ mapping the first arc to the second, and thusmapping(u,u′) to (v,v′). HenceΓ is distance-transitive.

2.2 Distance Partitions

Definition 2.2.1Let Γ be any connected graph. Then for some vertexv∈VΓ, we define

Γi(v) = u∈VΓ∣∣d(u,v) = i,

known ascellsof Γ. The set of all these cells form adistance partitionof Γ. (NotethatΓ0(v) = v for all v∈VΓ.)

If Γ has a fairly small number of vertices, we can easily drawΓ by putting thevertices ofΓ0(v), Γ1(v), etc. in columns from left to right. For example:

Example 2.2.2Consider the graphΓ shown in the figure below:

Figure 2.4: A distance partition ofΓ

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18 Introducing the Distance-Transitive Graph

Observation 2.2.3If Γ is vertex-transitive, theΓi(v) are independent ofv. So in this case we can justwrite Γi . For example, consider the cyclic 6-ladder we saw in example 2.1.2:

Figure 2.5: A distance partition ofL6

These distance partitions give us an alternative characterisation of distance-transitivity.

Lemma 2.2.4SupposeΓ is connected, has diam(Γ) = d and automorphism group Aut(Γ) = G.ThenΓ is distance-transitive if and only if it is vertex-transitive and StabG(v) istransitive onΓi(v) for i = 1, . . . ,d and for allv∈VΓ.

Proof:First, supposeΓ is distance-transitive. By 2.1.1,Γ is also vertex-transitive. Con-sideru,u′ ∈ Γi(v), i.e. with d(u,v) = d(u′,v) = i. Then there exists an automor-phismg∈G with vg= v andug= u′. Thusg∈StabG(v), and StabG(v) is transitiveon Γi(v).

Conversely, supposeΓ is vertex-transitive and that StabG(v) is transitive onΓi(v). Consideru,w,u′,w′ ∈ VΓ, such thatd(u,w) = d(u′,w′) = i. Let g ∈ G besuch thatwg= w′ and chooseh∈ StabG(w′) so that(ug)h = u′. Then for the com-positiongh, we getu(gh) = u′ andw(gh) = (w′)h = w′. Sogh is an automorphismtakingu,w to u′,w′. HenceΓ is distance-transitive.

2.3 Intersection Numbers and Intersection Arrays

Definition 2.3.1– Intersection NumbersFor any connected graphΓ, any verticesu,v,∈VΓ and forh, i ∈ N, define

Shi(u,v) = w∈VΓ |d(u,w) = h,d(v,w) = i

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2.3 Intersection Numbers and Intersection Arrays 19

That is,Shi is the set of all verticesw such that are distanceh from u and distanceifrom v. We denote the number of suchw by shi(u,v) = |Shi(u,v)|.

In a distance-transitive graph, theseshi are not dependent directly on the ver-ticesu,v but on the distanced(u,v) = j. This should be obvious from the definitionof distance-transitivity, as any pair of vertices distancej apart is equivalent to anyother such pair. So in this case, we just writeshi j . These are called theintersectionnumbersof a distance-transitive graph.

As h, i, j are all distances, we haveh, i, j ∈ 0,1, . . . ,d, whered = diam(Γ).From this observation, we can see that there are(d + 1)3 intersection numbers.However, some are more interesting than others.

Definition 2.3.2– The Intersection ArraySupposeΓ is distance-transitive. Fixh = 1 and choose two verticesu,v such thatd(u,v) = j. Now s1i j is the number of verticesw that are distance 1 fromu anddistancei from v, i.e. w∼ u andd(w,v) = i. So we have the following possiblevalues fori:

i = j−1i = j

or i = j +1

(This can easily be seen on a distance partition of a distance-transitive graph.) Thusall the intersection numberss1i j other thans1( j−1) j , s1 j j ands1( j+1) j must be zero.To simplify our notation, denotes1( j−1) j = c j , s1 j j = a j ands1( j+1) j = b j .

The intersection arrayof a distance-transitive graphΓ is as follows:

ι(Γ) =

∗ c1 · · · cd−1 cd

a0 a1 · · · ad−1 ad

b0 b1 · · · bd−1 ∗

Note thatc0 andbd are not defined (that’s what the∗ represent), as having pairs ofvertices distance−1 ord+1 apart is nonsense.

Alternatively, we can definec j ,a j ,b j as follows:Fix v∈VΓ andu∈ Γ j(v). Then

c j = |Cj |, where Cj = w∈VΓ |d(u,w) = 1,d(v,w) = j−1 a j = |A j |, where A j = w∈VΓ |d(u,w) = 1,d(v,w) = j b j = |B j |, where B j = w∈VΓ |d(u,w) = 1,d(v,w) = j +1

Observation 2.3.3ι(Γ) can be obtained from a distance partition ofΓ, as if we fixv to be the singlevertex ofΓ0, StabG(v) acts transtively on each ofΓ1, . . .Γd (see 2.2.4), thus givingthe samec j ,a j ,b j for any vertex inΓ j . (However, the converse of this is not true in

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20 Introducing the Distance-Transitive Graph

general: it is possible for two non-isomorphic graphs to have the same intersectionarray.)

Now we have an easy way to determine the intersection arrayι(Γ) for somefairly small graphs, by looking at their distance partitions.

Examples 2.3.4

ι(K4) =

∗ 10 23 ∗

ι(Oct) =

∗ 1 40 2 04 1 ∗

ι(O3) =

∗ 1 10 0 23 2 ∗

Figure 2.6: The complete graphK4, the octahedronOct and the Petersen graphO3

Observations 2.3.5In general, for any distance-transitive graph:

• The sum of the entries of each column ofι(Γ) is alwaysk, as any distance-transitive graph isk-regular for somek (as it is also vertex-transitive by

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2.3 Intersection Numbers and Intersection Arrays 21

1.4.5).

• a0 = 0, b0 = k (for k as above).

• c1 = 1 – this should be obvious.

This means that if we know the top and bottom rows, we can calculate the middlerow from these, by subtracting these fromk. So we can rewrite the intersectionarray as

k,b1, . . . ,bd−1; 1,c2, . . . ,cd .

We now identify some properties of the entries of the intersection array.

Theorem 2.3.6SupposeΓ is a distance-transitive graph, with diam(Γ) = d and deg(v) = k for allv ∈ VΓ. Let Γ have intersection arrayι(Γ) = k,b1, . . . ,bd−1;1,c2, . . . ,cd andwith ki vertices inΓi(v) for anyv∈VΓ. Then we have the following:

1. ki−1bi−1 = kici (for 1≤ i ≤ d);

2. 1≤ c2≤ ·· · ≤ cd;

3. k≥ b1≥ ·· · ≥ bd−1.

Proof:

1. Each vertex inΓi−1(v) is adjacent tobi−1 vertices inΓi(v). Also each vertexin Γi(v) is adjacent toci vertices inΓi−1(v). Thus the total number of edgesjoining Γi−1(v) andΓi(v) is ki−1bi−1 = kici .

2. Fix somev ∈ VΓ. Chooseu ∈ Γi+1(v), so thatd(u,v) = i + 1 (for i =1, . . . ,d−1). Choose some pathv,x, . . . ,u of lengthi +1, so thatd(x,u) = i.Then choose somew∈ Γi−1(x)∩Γ1(u). (See figure 2.7.)

Hencew∈ Γi(v)∩Γ1(u) also, so we have

(Γi−1(x)∩Γ1(u)) ⊆ (Γi(v)∩Γ1(u))and thus |Γi−1(x)∩Γ1(u)| ≤ |Γi(v)∩Γ1(u)| .

As Γ is distance-transitive, we have

|Γi−1(x)∩Γ1(u)| = ci−1

and |Γi(v)∩Γ1(u)| = ci .

Henceci−1≤ ci for all i.We already know thatc0 is undefined andc1 = 1, so we have 1≤ c2≤ ·· · ≤ cd.

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22 Introducing the Distance-Transitive Graph

Figure 2.7: Schematic for 2.3.6, part 2

3. Fix somev∈VΓ. Choosew∈Γi(v) (for i = 0, . . . ,d−2) and somezadjacentto v. Then for anyu∈ Γi+1(v)∩Γ1(w), by 2.3.2 we have either

d(z,u) = i +2d(z,u) = i +1

or d(z,u) = i

Supposed(z,u) = i + 2. Any z-u geodesic must take inv, otherwise a pathfrom z∈ Γ1(v) to u ∈ Γi+1(v) could only pass throughΓ1(v), . . . ,Γi+1(v),giving a shorter path fromz to u, which we assumed didn’t exist. So we have

(Γi+2(z)∩Γ1(w)) ⊆ (Γi+1(v)∩Γ1(w))and thus |Γi+2(z)∩Γ1(w)| ≤ |Γi+1(v)∩Γ1(w)|

As Γ is distance-transitive, we have

|Γi+2(z)∩Γ1(w)| = bi+1

and |Γi+1(v)∩Γ1(w)| = bi .

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2.3 Intersection Numbers and Intersection Arrays 23

Hencebi ≥ bi+1 for all i.We already know thatb0 = k andbd is undefined, so we have

k≥ b1≥ ·· · ≥ bd−1

This concludes the proof.

Figure 2.8: Schematic for 2.3.6, part 3

Corollary 2.3.7Assuming we knowι(Γ), a recursion formula forki (the number of vertices inΓi)is

ki =ki−1bi−1

ci

with initial conditionk0 = 1.

Proof:For the formula, just rearrange 2.3.6, part 1. For the initial condition, observe thatΓ0 has only one vertex by definition, sok0 = 1.

The two previous results give us some combinatorial constraints on whether anarbitrary array can be the intersection array of a distance-transitive graph. Further

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24 Introducing the Distance-Transitive Graph

such constraints are given below.

Proposition 2.3.8Suppose we have have an array of integers of the form

ι =

∗ 1 c2 · · · cd−1 cd

a0 a1 a2 · · · ad−1 ad

k b1 b2 · · · bd−1 ∗

.Then if ι is the intersection array of a distance-transitive graph, the following hold:

1. For 2≤ j ≤ d, the numbers

k j =kb1 · · ·b j−1

1c2 · · ·c j

are positive integers.

2. If part 1 holds andn = 1+k+k2 + · · ·+kd, thennk is even.

Proof:

1. In a distance-transitive graphΓ of valencyk with intersection arrayι(Γ), thenumberk j denotes the number of vertices inΓ j(v) (for somev ∈ VΓ and0≤ j ≤ d). By repeated application of the recursion formula in 2.3.7, weobtain

k j =kb1 · · ·b j−1

1c2 · · ·c j.

These numbers must be positive integers.

2. Clearly, sincek0=1 andk1 = k, |VΓ| = 1+ k+ k2 + · · ·+ d = n, say. SinceΓ is k-regular, the sum of the degrees of the vertices ofΓ is nk. By thehandshaking lemmafrom basic graph theory (Wilson [40], p.12), this mustbe an even integer.

Thus we now have some necessary conditions for an arbitrary array to be theintersection array of a distance-transitive graph. We will return to this problem insection 7.4.

2.4 Distance-Regular Graphs

Definition 2.4.1We can always find a distance partition for any graph (e.g. example 2.2.2). Sup-pose that it happens that for some graphΓ, say, theith cell Γi(v) has the same sizefor anyv∈VΓ and for anyi, and that each vertex inΓi(v) is adjacent to the samenumber of vertices inΓi−1(v), Γi(v) andΓi+1(v). Clearly we can write down an

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2.4 Distance-Regular Graphs 25

intersection array forΓ, as all the numerical condtions in the previous section aresatisfied. Under these cirumstances,Γ is said to be adistance-regulargraph.

Distance-regularity is a purely combinatorial condition: nowhere in the abovedefinition do we consider any automorphisms ofΓ. Clearly, any distance-transitivegraph is distance-regular, but the converse is certainly not true. Unlike exam-ple 2.1.2, however, there isn’t such a simple counterexample. This is taken fromAdel’son-Vel’skii et al [1].

Example 2.4.2– The Adel’son-Vel’skii Graph1

a_05

b_07

b_09

b_11

a_12

b_12

a_00

b_04

b_10

b_00

a_02

a_04

a_09

a_03

a_10

a_01

b_06

b_05

b_03

b_02

b_01

a_11

a_08

a_07

a_06

b_08

Figure 2.9: The Adel’son-Velskii Graph (drawn using MAPLE)

Let VΓ = a0, . . . ,a12,b0, . . . ,b12. Then let the following vertices be adjacent:

• ai ∼ a j ⇔ |i− j|= 1,3,4

• bi ∼ b j ⇔ |i− j|= 2,5,6

• ai ∼ b j ⇔ i− j = 0,1,3,9

(all taken modulo 13). Then the graph in figure 2.9 is obtained. It is distance-

1Not are all edges appear, due to limitations in the software – it draws edges in the same cell overthe top of each other!

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26 Introducing the Distance-Transitive Graph

regular, with intersection array

ι(Γ) =

∗ 1 40 3 6

10 6 ∗

.However, it is not distance-transitive or even vertex-transitive: there is no auto-mophism taking anyai to anyb j . For a proof, see [1]; it uses some quite advancedgroup theory.

We conclude this chapter with a diagram relating most of the different kinds ofgraph we have seen so far:

DISTANCE-TRANSITIVE =⇒ Distance-Regular

⇓Edge-Transitive ⇐= Symmetric (1-arc-transitive)

⇓Vertex-Transitive

Figure 2.10: A hierarchy of conditions

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Chapter 3

Uniform Subset Graphs

3.1 Introduction

These graphs form our first large class of examples. They are defined as follows:

Definition 3.1.1Consider the setΩ = 1, . . . ,n. Let Xn,k be the set of all subsets ofΩ of sizek(which we shall callk-subsets). We define theuniform subset graph J(n,k, i) hasvertex setXn,k, and two verticesu = i1, . . . , ik, v = j1, . . . , jk being adjacent ifand only if|u∩v|= i.

First, we make some observations about the uniform subset graphs that followfairly quickly from the definition.

Proposition 3.1.2

1. J(n,k, i) has

(nk

)vertices.

2. J(n,k, i) is regular, each vertex having valency

(ki

)(n−kk− i

).

Proof:

1. The number of vertices ofJ(n,k, i) is just the number of subsets of sizek of

Ω, which has sizen, which is given by

(nk

).

2. Supposeu∼ v. Then (regardingu,v ask-sets)v containsi elements fromu(of which there arek to choose from) and alsok− i elements that arenot inu (of which there aren−k).

This is clearly a very large family of graphs. However, it is not as large as itfirst appears, as we can assume thatn≥ 2k from the following result.

27

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28 Uniform Subset Graphs

Lemma 3.1.3For n≥ k≥ i, we haveJ(n,k, i)∼= J(n,n−k,n−2k+ i).

Proof:Supposeu = j1, . . . , jk ∈ Xn,k. Define a function

ψ : Xn,k→ Xn,n−k

where ψ(u) = (Ω\u) = uc.

Supposeu∼ v in J(n,k, i). Then, by definition,|u∩v|= i. So we have

|ψ(u)∩ψ(v)| = |uc∩vc|= |(u∪v)c| by De Morgan’s Laws= n−|u∪v|= n− (2k− i)= n−2k+ i.

Thereforeψ(u)∼ ψ(v) in J(n,n−k,n−2k+ i), soψ preserves adjacency. Also,ψis clearly a bijection. Soψ is an isomorphism fromJ(n,k, i) toJ(n,n−k,n−2k+ i),henceJ(n,k, i)∼= J(n,n−k,n−2k+ i).

For n≥ 2k, the graphsJ(n,k,k−1) are known asJohnson graphs, the graphsJ(n,k,0) are known asKneser graphsand the graphsJ(2k−1,k−1,0) are knownas theodd graphs. These families had been investigated before the general def-inition of a uniform subset graph was given by Chen and Lih [15] in 1987. TheJohnson and odd graphs are the ones which this chapter is devoted to, as we shallsee that they are distance-transitive.

3.2 The Johnson GraphsJ(n,2,1)

Let us first consider a simple example.

Example 3.2.1

Figure 3.1:J(4,2,1)

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3.2 The Johnson GraphsJ(n,2,1) 29

J(4,2,1) is the smallest Johnson graph. We have

X4,2 = 1,2,1,3,1,4,2,3,2,4,3,4 .

In fact,J(4,2,1) is isomorphic to the octahedron.

As a preview of the next section, we have this result:

Theorem 3.2.2The Johnson graphJ(n,2,1) is distance-transitive.

Proof:First, note that any permutation ofΩ = 1, . . . ,n immediately induces a permuta-tion onXn,2. For example, withn = 4, σ =

(1 2 3 4

)∈ S4, we get

(1,2)σ = ((1)σ,(2)σ) = 2,3(1,3)σ = 2,4(1,4)σ = 2,1= 1,2(2,3)σ = 3,4(2,4)σ = 3,1= 1,3(3,4)σ = 4,1= 1,4

Clearly, Sn ≤ Aut(J(n,2,1)) and acts transitively onXn,2. HenceJ(n,2,1) isvertex-transitive.

Secondly, notice that diam(J(n,2,1)) = 2: if i1, i2 6∼ j1, j2, we know thatbetween them lie the verticesi1, j1,i1, j2,i2, j1 and i2, j2, since are alladjacent to bothi1, i2 and j1, j2. Thus to determine distance-transitivity, weonly have two distances to consider: 1 and 2.

1. Supposed(i1, i2, j1, j2) = 1That is,i1, i2 ∼ j1, j2⇔ |i1, i2∩ j1, j2|= 1Assume WLOG1 that i1 = j1, i2 6= j2.Then, for anyσ ∈ Sn, (i1)σ = ( j1)σ,(i2)σ 6= ( j2)σ.So |(i1, i2)σ∩ ( j1, j2)σ|= 1⇔ (i1, i2)σ∼ ( j1, j2)σ⇔ d((i1, i2)σ,( j1, j2)σ) = 1.

2. Supposed(i1, i2, j1, j2) = 2.Theni1, i2 6∼ j1, j2, soi1, i2, i1 ,i2 are all distinct.So for anyσ ∈ Sn, (i1)σ, (i2)σ, ( j1)σ, ( j2)σ are also all distinct.Hence(i1, i2)σ 6∼ ( j1, j2)σ, sod((i1, i2)σ,( j1, j2)σ) = 2.

So for any two adjacent vertices (i.e. vertices of distance 1) we can find a per-mutationσ ∈ Sn which maps them onto any other two adjacent vertices. Likewise,

1Without Loss Of Generality

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30 Uniform Subset Graphs

any two non-adjacent vertices (i.e. vertices of distance 2) can be mapped onto anyother non-adjacent vertices. HenceJ(n,2,1) is distance-transitive.

Note that the graphsJ(n,2,1) are sometimes known as thetriangle graphs,denoted∆n.

3.3 The Johnson GraphsJ(n,k,k−1)

Now we want to generalise 2-subsets ofΩ to k-subsets, adjacent as vertices whenintersecting in a(k−1)-set; that is, the Johnson graphsJ(n,k,k−1) (for n≥ 2k,from 3.1.3). We start with a weaker result than the one we actually want.

Proposition 3.3.1J(n,k,k−1) is vertex-transitive.

Proof:As with 2-subsets,Sn has an induced action onk-subsets, so for allσ ∈ Sn,(i1, . . . , ik)σ = (i1)σ, . . . ,(ik)σ. All the (iα)σ are distinct. So for vertices

u = i1, . . . , ik and v = j1, . . . , jk, the permutationσ =(

i1 · · · ikj1 · · · jk

)takes

vertexu to vertexv. SoJ(n,k,k−1) is vertex-transitive.

We now outline our strategy for showing distance-transitivity:

• Show that permutations fix the size of the intersection of two vertices (whenviewed ask-sets);

• Show that there is a 1-1 correspondence between intersect size and distanceof two vertices;

• Conclude that permutations fix distance.

So we start with this lemma.

Lemma 3.3.2For any permutationσ ∈ Sn and any pairu,v∈ Xn,k, |u∩v|= |(u)σ∩ (v)σ|.

Proof:Suppose|u∩ v| = |i1, . . . , ik∩ j1, . . . , jk| = m say. Then WLOG assume thatiα = jα for 1≤ α≤m, iα 6= jα otherwise. So for anyσ ∈ Sn,

(iα)σ = ( jα)σ for 1≥ α≥m,(iα)σ 6= ( jα)σ otherwise.

Hence|(i1, . . . , ik)σ∩ ( j1, . . . , jk)σ| = m also. Soσ preserves the “intersectsize” for all σ ∈ Sn.

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3.3 The Johnson GraphsJ(n,k,k−1) 31

The key step is this next one:

Lemma 3.3.3For any two verticesu,v of J(n,k,k−1), d(u,v) = m if and only if |u∩v|= k−m(when regardingu,v ask-sets).

Proof – by induction onm:Basis:Supposed(u,v) = 1, i.e.u∼ v, so by definition we have|u∩v|= k−1.Induction Hypothesis (IH):Suppose the theorem holds form≤ r, i.e. for eachm≤ r, d(u,v) = m if and only if |u∩v|= k−m.

Choose verticesu,v,w such thatd(u,v) = r andd(u,w) = 1. By the IH, wehave|u∩v|= k− r, so

u = s1, . . . ,sk−r , i1, . . . , irv = s1, . . . ,sk−r , j1, . . . , jr

where thei’s and j ’s are all distinct. Also, by definition, we have|v∩w| = k−1.So we must have one of the following possibilities forw:

(i) w = s1, . . . ,sk−r , j1, . . . , jr−1, iα for someiα ∈ i1, . . . , ir;(ii) w = s1, . . . ,sk−r , j1, . . . , jr−1,a wherea∈Ω\ (u∪v);(iii) w = s1, . . . ,sk−r−1, iα, j1, . . . , jr .(iv) w = s1, . . . ,sk−r−1,a, j1, . . . , jr;

In case (i), we have|u∩w|= k− (r−1), so by the IH,d(u,w) = r−1. In cases (ii)and (iii), |u∩w| = k− r, so again by the IH we haved(u,w) = r. Thus it remainsto check case (iv), where|u∩w|= k− (r +1).

We know that there exists a path of lengthr + 1 from u to w, as there is apath of lengthr from u to v and becausev∼ w. Also, we know that there is nopath of length≤ r, as by the IH that would give|u∩w| ≥ k− r, contradicting|u∩w| = k− (r + 1). Hence the shortest path fromu to w is of lengthr + 1, sod(u,w) = r +1, and the result follows by induction.

Thus we can now prove our main result of the section:

Theorem 3.3.4The Johnson graphJ(n,k,k−1) is distance-transitive.

Proof:By Lemmas 3.3.2 and 3.3.3, any permutationσ of 1, . . . ,n takes two vertices atdistancem to some other pair of distancem. This is clearly bijective, so we canconclude that for any two pairs of verticesu,v andu′,v′ with d(u,v) = d(u′,v′),there is some permutationσ that takesu,v to u′,v′. HenceJ(n,k,k−1) is distance-transitive.

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32 Uniform Subset Graphs

As the Johnson graph is distance-transitive, it has an intersection array, whichwe shall now determine.

Theorem 3.3.5The intersection array ofJ(n,k,k−1) is given byι(J(n,k,k−1)) = ∗ 12 · · · i2 · · · (k−1)2 k2

0 · · · · · · · · · · · · · · · nk−2k2

k(n−k) (k−1)(n−k−1) · · · (k− i)(n−k− i) · · · n−2k+1 ∗

.

Proof:We know from 2.3.5 that it is sufficient to determinebi for i = 0,1, . . . ,d−1 andci for i = 1, . . . ,d.

First, we’ll find theci :Recall that for a fixedv∈VΓ, and someu∈VΓ such thatd(u,v) = i,

ci = |w∈VΓ : d(v,w) = i−1,d(u,w) = 1| .

Regarding vertices ask-sets,d(u,v) = i⇔ |u∩v|= k− i (by 3.3.3), so

v = α1, . . . ,αk−i ,β1, . . . ,βi, andu = α1, . . . ,αk−i ,γ1, . . . ,γi.

We wantw such thatd(v,w) = i−1 andd(u,w) = 1, i.e. |v∩w|= k− (i−1) and|u∩w|= k−1. Thenw must contain one ofβ1, . . . ,βi in place of one ofγ1, . . . ,γi .There arei choices ofβ j , and alsoi choices for whichγ j it replaces.Henceci = i2, for i = 1, . . . ,d.

Now we’ll find thebi :Recall that (foru,v as above),

bi = |w∈VΓ : d(v,w) = i +1,d(u,w) = 1| .

This time we wantw such thatd(v,w) = i + 1 andd(u,w) = 1, i.e. |v∩w| =k−(i +1) and|u∩w|= k−1. So we must havew = α1, . . . ,αk−i−1,δ,γ1, . . . ,γi,where we’ve replaced one of thek− i α j ’s with someδ chosen from outside ofuandv, of which there aren− (2k− (k− i)) = n−k− i.

Hencebi = (k− i)(n− k− i), for i = 0, . . . ,d−1. A quick check shows thatb0 = k(n−k), which is the valency ofJ(n,k,k−1) as required (see 2.3.5 and 3.1.2)and thatc1 = 1 as required (2.3.5 again). So we have the result we require.

So, for example, the intersection array ofJ(5,2,1) is

ι(J(5,2,1)) =

∗ 1 40 3 26 2 ∗

.

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3.4 Pretty Pictures 33

3.4 Pretty Pictures

The notationJ(n,k,k− 1) is a very succinct notation for a collection of objectsthat get very complicated very quickly. For example,J(6,3,2) is only the fourthJohnson graph (afterJ(4,2,1), J(5,2,1) andJ(6,2,1)), but it has 20 vertices and90 edges, as the diagram shows.

Figure 3.2:J(6,3,2)

Its diameter is 3, and it has intersection array

ι(J(6,3,2)) =

∗ 1 4 90 4 4 09 4 1 ∗

.3.5 The Odd Graphs

The odd graphsare another family of uniform subset graphs: using familiar no-tation, they are the graphsJ(2k− 1,k− 1,0). That is, their vertices are all the(k−1)-subsets of a setΩ of size 2k−1, adjacent if the two sets are disjoint. By

3.1.2, they have valency

(k−1

0

)(k

k−1

)= k. We shall use the less cumbersome

symbolOk to denoteJ(2k−1,k−1,0). O3 is the Petersen graph, shown below.

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34 Uniform Subset Graphs

Example 3.5.1

Figure 3.3: The Petersen graphO3

A (realistic?) construction ofO6 was given by Biggs [5]:

“In the little English hamlet of Croam the consuming passion of theinhabitants is Association Football. In fact, the members of the villagefootball team have become so ruthless in their will to win that no otherteam will play against them.

“Thus the eleven footballers of Croam (who are, incidentally, the onlyable-bodied men in the village) are forced to arrange their own math-ches between two teams of five, with the eleventh man as referee. Fur-ther, such is the bitterness of recrimination which follows even thesematches, that is has proved necessary to rule that only one match canbe played with the same teams and the same referee. This rule wasoriginally regarded with some misgiving, as it was felt that it mightseriously limit the number of matches which could be played. How-ever, a villager who has a head for figures worked out that there are1386 different ways of splitting the eleven men into two teams of fiveand a referee. This number is thought to be adequate but not generous,for the footballers of Croam are dedicated men.”

The odd graphs have several interesting properties, such as whetherOk ishamiltonian, or how many colours are needed for an edge-colouring (which wasthe motivation for Biggs’ description ofO6 above). See Holton & Sheehan [29]for more details on these problems. We shall prove, of course, thatOk is distance-transitive.

We have completed a great deal of the work for this already in this chapter.Just as with the Johnson graphs, the group Sym(Ω)∼= S2k−1 has an induced actionon the vertices ofOk when regarded as subsets ofΩ, soS2k−1 ≤ Aut(Ok). (Usingthe Erdos-Ko-Rado Theorem, it can be shown that Aut(Ok)∼= S2k−1: see [29].) By3.3.2, we have that all elements ofS2k−1 fix the size of|u∩v| for any two vertices

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3.5 The Odd Graphs 35

(i.e. (k−1)-sets)u andv. So it remains to show that there is a direct relationshipbetweend(u,v) and|u∩v|.

Analagous to 3.3.3, we have:

Lemma 3.5.2Let u,v be vertices ofOk. Then form≥ 0,

d(u,v) =

2m ⇔ |u∩v|= (k−1)−m2m+1 ⇔ |u∩v|= m.

(Note that this is only well-defined ifm 6= (k−1)−m for all m, i.e. if (k−1) 6= 2mfor all m. This is achieved when 2m< (k−1).)

Proof:We proceed by induction onm in the two cases separately.

1. First supposed(u,v) is even, sod(u,v) = 2m for somem.Basis:supposem= 0. Thend(u,v) = 0, i.e.u = v, so|u∩v|= |u|= k−1 =(k−1)−0.Induction Hypothesis (IH):suppose the theorem holds form≤ r. Sod(u,v) =2r if and only if |u∩v|= (k−1)− r.

Now choose a vertexw such thatd(u,w) = 2(r + 1) andd(v,w) = 2. (Ad-ditionally, in the caser = 1 we require explicitly thatu 6= w.) By the IH,|u∩v|= (k−1)−r. Also, there exists a vertexx such thatd(v,x) = d(x,w) =1. So by construction,|v∩ x| = |x∩w| = 0, so we havev ⊂ Ω \ x andw ⊂ Ω \ x. Since|v| = |w| = k− 1 and |Ω \ x| = (2k− 1)− (k− 1) = k,it must be that|v∩w|= (k−1)−1.

So what about|u∩w|? We know (by the IH) that|u∩ v| = (k− 1)− rand (by the above) that|v∩w| = (k− 1)− 1. So w has exactly one el-ement different from those inv, i.e. one element chosen fromΩ \ v re-placing an element chosen fromv. This can be done in four ways. Sup-pose (for clarity) that|u∩ v| = s, that u = x1, . . . ,xs,us+1, . . . ,uk−1 andv = x1, . . . ,xs,vs+1, . . . ,vk−1. Then we have the following possibilities:

(a) w = x1, . . . ,xs,vs+1, . . . ,vk−2,u j for somej.Then|u∩w|= |u∩v|+1.

(b) w = x1, . . . ,xs,vs+1, . . . ,vk−2,α for someα ∈Ω\ (u∪v).Then|u∩w|= |u∩v|.

(c) w = x1, . . . ,xs−1,u j ,vs+1, . . . ,vk−1 for somej.Then|u∩w|= |u∩v|−1+1 = |u∩v|.

(d) w = x1, . . . ,xs−1,α,vs+1, . . . ,vk−1 for someα ∈Ω\ (u∪v).Then|u∩w|= |u∩v|−1.

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36 Uniform Subset Graphs

In case (a), we have|u∩w| = (k−1)− r + 1 = (k−1)− (r−1), so by theIH we haved(u,w) = 2(r−1), contradicting our assumption thatd(u,w) =2(r + 1). Similarly, in cases (b) and (c) we obtain|u∩w| = (k− 1)− r,giving d(u,w) = 2r, another contradiction. So we are left with case (d),where|u∩w|= (k−1)− r−1 = (k−1)−(r +1). Henced(u,w) = 2(r +1)if and only if |u∩w|= (k−1)− (r +1), and the result follows by induction.

2. Now we have thatd(u,v) is odd, so equals 2m+1 for somem.Basis: supposem = 0. Thend(u,v) = 1, i.e. u ∼ v, so by construction|u∩v|= 0.Induction Hypothesis (IH):Suppose the theorem holds form≤ r. In partic-ular, this impliesd(u,v) = 2r +1 if and only if |u∩v|= r.

Now choosew such thatd(u,w) = 2(r +1)+1 andd(v,w) = 2. By the samearguments as in the first part of the proof, we know that|v∩w|= (k−1)−1,and by the IH we have|u∩v|= r. So we have to check the same four cases(a,b,c,d) as in part 1. (Only this time, we’ll do them backwards!)

In case (d), we have|u∩w| = |u∩ v|−1 = r−1. But by the IH, this givesd(u,w) = 2(r−1)+1, contradicting the assumption thatd(u,w) = 2(r +1)+1. In cases (c) and (b) we obtain|u∩w| = r, giving d(u,w) = 2r, anothercontradiction. Thus the only way out is case (a), where|u∩w| = r + 1.Therefored(u,w) = 2(r +1)+1 if and only if |u∩w|= r +1, and the resultfollows by induction.

A consequence of the requirement that 2m< (k−1) is that we obtain a boundfor the distance between two vertices. Because 2m< (k−1), we haved(u,v) <(k−1) if the distance is even, andd(u,v)−1< (k−1), sod(u,v)≤ (k−1) if thedistance is odd. Hence the diameter ofOk is (k−1).

We now have the facts we require to prove the next theorem.

Theorem 3.5.3The odd graphOk is distance-transitive.

Proof:We know from Lemma 3.5.2 that ifd(u,v) = d(w,x), then|u∩v|= |w∩x| (regard-ing u,v,w,x as(k−1)-sets). As with the Johnson graphs, from Lemma 3.3.2 weknow that, for any two such pairs, there existsg ∈ S2k−1 such that(u)g = w and(v)g = x. HenceOk is distance-transitive.

All that remains in this chapter is to calculate the intersection array ofOk.

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3.5 The Odd Graphs 37

Theorem 3.5.4The intersection array ofOk, is given by

ι(Ok) =

∗ 1 1 · · · m m+1 · · · 12(k−1) 1

2(k−1)0 0 0 · · · 0 0 · · · 0 1

2(k+1)k k−1 k−1 · · · k−m k−m−1 · · · 1

2(k+1) ∗

for k odd, and

ι(Ok) =

∗ 1 1 · · · m m+1 · · · 12k−1 1

2k−1 12k

0 0 0 · · · 0 0 · · · 0 0 12k

k k−1 k−1 · · · k−m−1 k−m · · · 12k+1 1

2k+1 ∗

for k even.

Proof:As before, we only calculateci andbi as these determine the intersection arraycompletely. However, we have to consider the cases wherei is even andi is oddseparately.

1. Supposei is even, soi = 2m for somem. Fix u,v∈VOk such thatd(u,v) =2m, so by Lemma 3.5.2,|u∩v|= k−m−1. Recall that

c2m = |w∈VOk : d(v,w) = 2m−1,d(u,w) = 1|= |w∈VOk : |v∩w|= m−1, |u∩w|= 0| .

Now w containsm−1 elements ofv and 0 elements ofu. So thesem−1elements ofv must be chosen from thosenot in u∩ v, of which there are

(k− 1)− (k− 1−m) = m. Hence there are

(m

m−1

)= m choices. The

remaining(k− 1)− (m− 1) = k−m elements ofw must be chosen fromthose outside ofu andv, of which there are(2k−1)− (k−1+ m) = k−m,so we must include all of them. Hencec2m = m.

Now recall that

b2m =∣∣w∈VOk : d(v,w) = 2m+1,d(u,w) = 1

∣∣=

∣∣w∈VOk : |v∩w|= m, |u∩w|= 0∣∣.

This time,w must includeall m elements ofv not in u, andk−m−1 of the

k−m elements outside. So we have

(k−m

k−m−1

)= k−m choices, and so

b2m = k−m.

2. Now supposei is odd, soi = 2m+ 1 for somem. Fix u,v∈VOk such thatd(u,v) = 2m+1, so by Lemma 3.5.2,|u∩v|= m. Hence we have

c2m+1 = |w∈VOk : |v∩w|= k−m−1, |u∩w|= 0| .

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38 Uniform Subset Graphs

So w containsk−m− 1 elements ofv and no elements ofu. But v onlycontainsk−m−1 elements not inu, so we must include all of these inw.The remainingm elements ofw can be chosen from them+ 1 outside ofu

andv in

(m+1

m

)= m+1 ways. Hencec2m+1 = m+1.

Finally, we have

b2m+1 = |w∈VOk : |v∩w|= (k−1)− (m+1), |u∩w|= 0| .

This time,w containsk−melements of thek−m−1 elements ofv that aren’t

in u, so we have

(k−m−1

k−m

)= k−m−1 choices. We must also include all

of them+1 elements from outsideu andv. Henceb2m+1 = k−m−1.

So if k is odd, diam(Ok) = k− 1 is even, so we havek− 1 = 2m and sock−1 =12(k−1). If k is even, diam(Ok) is odd, so then we havek−1 = 2m+ 1 and thenck−1 = 1

2(k−2)+1 = 12k. Hence we have proved the required result.

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Chapter 4

Some Permutation Group Theory

4.1 Primitive and Imprimitive Actions

In this section, we always supposeG is a permutation group acting transitively ona setX.

Definition 4.1.1A G-congruenceonX is aG-invariant equivalence relation onX.

That is, for allg ∈ G and for allx,y ∈ X, we have an equivalence relation≡such thatx≡ y if and only if (xg) ≡ (yg). The equivalence classes of≡ have aspecial name.

Definition 4.1.2The equivalence classes of≡ are calledblocks. The set of all blocks is referred toas ablock systemof X.

Informally, a block is a subset ofX whose elements are ‘equivalent’ in someway that is unaffected by the action ofG onX. Every set always has the followingtwo G-congruences:

• thetrivial G-congruence: where the blocks are all singletons;

• the universal G-congruence: where there is only one block, namely thewhole ofX, sox≡ y for all x,y∈ X.

In addition, the empty set Ø is always a block. These three types of block are calledthetrivial blocks.

Definition 4.1.3If the only blocks are the trivial blocks listed above, we sayG actsprimitively onX. If there is a non-trivial block system, we sayG actsimprimitivelyonX.

39

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40 Some Permutation Group Theory

We will encounter examples of primitive and imprimitive groups in chapter 8,where we meet the concept of primitive and imprimitive graphs, whereG is theautomorphism group of a distance-transitive graph.

4.2 Direct and Semi-direct Products

Recall thedirect productof two groupsG andH is

G×H =

(g,h)∣∣g∈G,h∈ H

with binary operation(g,h)(g′,h′) = (gg′,hh′).

The semi-direct product can be viewed as a generalisation of this.

Definition 4.2.1Let H,K be groups, withH acting onK in such a way that the group structure ofK is preserved, so for eachu∈ K andx∈ H the actionu 7→ ux is an automorphismof K. (Note that the action ofH onK is not specified directly.)

We define thesemi-direct product of K by Has

KoH =

(u,x)∣∣u∈ K,x∈ H

with binary operation(u,x)(v,y) = (uvx−1

,xy).

Check that this is a group:

• Associativity is the hardest axiom to check – for(u,x),(v,y),(w,z) ∈ KoH,we get (

(u,x)(v,y))

(w,z) =(

uvx−1, xy)

(w,z)

=(

uvx−1w(xy)−1

, xyz)

=(

u(vwy−1)x−1

, xyz)

= (u,x)(

vwy−1, yz)

= (u,x)(

(v,y)(w,z)).

• The identity is obviously(eK ,eH), as we have

(u,x)(eK ,eH) =(

uex−1

K , xeH

)= (u,x)

and(eK ,eH)(u,x) =(

eKue−1H , eHx

)= (u,x).

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4.3 Wreath Products 41

• The inverse of(u,x) ∈ KoH is(

(u−1)x,x−1)

, since

(u,x)((u−1)x,x−1)=

(u(u−1)xx−1

,xx−1)

= (uu−1,xx−1) = (eK ,eH)

and similarly for((u−1)x,x−1

)(u,x).

Observation 4.2.2The direct product is a special case of the semi-direct product, when eachx ∈ Hacts as the identity map onK, i.e. for allu∈ K, ux = u.

4.3 Wreath Products

The wreath product is an example of the semi-direct product.

Definition 4.3.1Suppose we have a groupG acting on a setX. Then consider the cartesian productY = X1×X2×·· ·×Xn where eachXi = X (the subscripts are just labels). Clearlythe direct productGn = G1×G2×·· ·×Gn (where eachGi = G, with G j being thecopy ofG acting onXj ) has an induced action onY.

Now letH be a group of permutations acting on the labels1, . . . ,n, soH hasan induced action on bothY (soH ≤ Sym(Y)) andGn. We can form a semi-directproduct ofGn by H, Gn

oH, where the action ofH on Gn is as above. This iscalled thewreath productof G by H, writtenGWr H. It has two actions onY: the“G” part acts on each of theXi in turn (which is an imprimitive action) and the “H”part moves around theXi , takingXi to Xj , where j = ih for someh∈H (a primitiveaction).

A classic example of a wreath product is the automorphism group of the Ham-ming graph, which we will meet in the next chapter. For more on semi-direct andwreath products in general, consult Cameron [13] or Dixon & Mortimer [17].

4.4 Projective Groups

When dealing with a vector spaceV, the first group one should think of is thegen-eral linear group, GL(V). In particular, ifV is ann-dimensional vector space overFq, the finite field withq elements, this would beGLn(Fq), which we shall denoteby GL(n,q). This groups acts onV; this action is not transitive on the whole space,but is transitive onV \0. (The zero vector is fixed by all linear transformationsof V.)

This group has an induced action on thek-dimensional subspaces ofV. Moreprecisely, it acts transitively on ordered bases of ak-space. However, this action isnot faithful: matrices of the formλIn, i.e. scalar multiples of the identity matrix,

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42 Some Permutation Group Theory

each have the effect of multiplying all basis vectors by the constantλ, thus keep-ing the space spanned by that basis fixed. These matrices form thecentre, Z, ofGL(n,q), which is a normal subgroup. So we have:

Definition 4.4.1The factor groupGL(n,q)/Z is known as theprojective general linear group,denoted byPGL(n,q).

Recall thatGL(V) has a normal subgroupSL(V), thespecial linear groupofV. In terms of matrices, this is the set of linear transformations ofV that havedeterminant 1. Again, ifV = F

nq, we denote this group bySL(n,q). As with

the GL(n,q), the centre ofSL(n,q) consists of diagonal matrices, but only thosewith determinant 1. Because the determinant of a diagonal matrix is the productof the diagonal elements, this is those matricesλIn, such thatλn = 1. That is,Z(SL(n,q)) = SL(n,q)∩Z = N (whereZ is the centre ofGL(n,q) as above), say.Hence the next definition:

Definition 4.4.2The factor groupSL(n,q)/N is called theprojective special linear group, denotedby PSL(n,q).

We will encounter these groups later, as the automorphism groups of the Grass-mann graphs (chapter 6) and also as the groups of some other ‘sporadic’ distance-transitive graphs in chapter 11.

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Chapter 5

Hamming Graphs

5.1 Introduction

The Hamming graphs are defined in a similar way to the Johnson graphs we saw inchapter 3. Again, we consider a setΩ of sizen (typically, we think ofZn, the set ofintegers modulon, or the finite field withq elementsFq), but this time we considerΩd, the cartesian product ofd copies ofΩ.

Definition 5.1.1The Hamming graph, denotedH(d,n), has vertex setΩd, with two vertices be-ing adjacent if, when regarded as orderedd-tuples, they differ in exactly one co-ordinate.

It is clear that two vertices are at distancek if and only if they differ in exactlykco-ordinates. (Recall the definition of theHamming distanceof two vectors beingthe number of co-ordinates in which they differ.) Let us consider the followingexample:

Figure 5.1: The Hamming graphH(2,3)

43

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44 Hamming Graphs

Two properties of the Hamming graphs are immeditaely apparent. First, thediameter ofH(d,n) is d, as the maximum distance occurs when the two vertices(regarded as orderedd-tuples) differ in alld co-ordinates. Secondly, the valency ofH(d,n) is d(n−1). So for our exampleH(2,3), the diameter is 2 and valency is2(3−1) = 4.

5.2 Distance-Transitivity

To show that the Hamming graphs are distance-transitive, we first need to knowtheir automorphism groups. As was mentioned in section 4.3, we have:

Lemma 5.2.1

SnWr Sd ≤ Aut(H(d,n)).

Proof:It is quite clear that the direct product ofd copies ofSn, i.e. S d

n = Sn×Sn×·· ·×Sn

acts onΩd in a way that preserves adjacency. Suppose we haveu,v ∈ Ωd withu∼ v. Regardingu andv asd-tuples, we haveu= (u1, . . . ,ud) andv= (v1, . . . ,vd),with

ui 6= vi for i = jui = vi for i 6= j

for 1≤ i ≤ d and some particularj. Take someσ = (σ1, . . . ,σd) ∈ S dn . Then we

get that

uσ = (u1,u2, . . . ,ud)(σ1,σ2, . . . ,σd)= (u1σ1,u2σ2, . . . ,udσd)

and likewise forvσ. As theσi are all permutations, we then have

uiσi 6= viσi for i = juiσi = viσi for i 6= j

and so adjacency is preserved.At first it may seem that these are the only automorphisms. (These are sufficient

to show transitivity.) However, consider a permutationρ on d symbols, permutingthe order of thed co-ordinates. So we have

uiρ 6= viρ for iρ = jρuiρ = viρ for iρ 6= jρ

for 1≤ i ≤ d. SoSd also acts onΩd, also preserving adjacency. Hence we get

S dn oSd

∼= SnWr Sd ≤ Aut(H(d,n)).

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5.2 Distance-Transitivity 45

It is clear from the above discussion thatSnWr Sd acts vertex-transitively onH(d,n). To show the action is distance-transitive is also fairly straightforward.Formally, we have:

Theorem 5.2.2The Hamming graphsH(d,n) are distance-transitive.

Proof:We just have to recall that two vertices are at distancek if and only if they differ inexactlyk co-ordinates. Then we can the proceed using exactly the same argumentas above, with (for someσ ∈ S d

n )

uiσi 6= viσi for i = j1, . . . , jkuiσi = viσi for i 6= j1, . . . , jk

for 1≤ i ≤ d, and somej1, . . . , jk ∈ 1, . . . ,d with 1≤ k≤ d. (We also get a sim-ilar result forρ ∈ St .) SoSnWr Sd acts transitively on pairs of vertices at distancek. HenceH(d,n) is distance-transitive.

As with the Johnson graphs and odd graphs, we know thatH(d,n) has an in-tersection array, which is determined below.

Theorem 5.2.3The intersection array ofH(d,n) is given byι(H(d,n)) =

∗ 1 · · · j · · · d−1 d0 · · · · · · · · · · · · · · · d(n−2)

d(n−1) (d−1)(n−1) · · · (d− j)(n−1) · · · 1(n−1) ∗

.Proof:Consider two verticesu,v of H(d,n), with d(u,v) = j, sou = (u1, . . . ,ud) andv =(v1, . . . ,vd) differ in exactly j co-ordinates. Suppose WLOG thatu1 6= v1, . . . ,u j 6=v j andu j+1 = v j+1, . . . ,ud = vd.

Then consider somew∈ Γ j−1(v)∩Γ1(u), i.e. w differs in one co-ordinate (msay) fromu, and in j−1 co-ordinates fromv. Hence thatmth co-ordinate must beone ofu1, . . . ,u j , so there arej choices for this. Once that co-ordinatem is chosen,what is put there is fixed; it must bevm, so that we haved(v,w) = j −1. Hencethere arej choices forw, soc j = |Γ j−1(v)∩Γ1(u)|= j.

Now consider somex∈ Γ j+1(v)∩Γ1(u), i.e. x differs in one co-ordinate (l say)from u, and in j +1 co-ordinates fromv. This time, thel th co-ordinate must be oneof u j+1 = v j+1, . . . ,ud = vd, so there ared− j choices for this. It can be changed toanything, so there aren−1 choices for this. Hence there are(d− j)(n−1) choicesfor x (d− j places,n−1 choices of what to put there). Sob j = |Γ j+1(v)∩Γ1(u)|=(d− j)(n−1).

Combining these two facts, we obtain the required result.

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46 Hamming Graphs

For example, we have

ι(H(2,3)) =

∗ 1 20 1 24 2 ∗

.5.3 Thek-Cubes

Thek-cubes, Qk, are a well-known and frequently studied family of graphs. Theyare in fact the Hamming graphsH(k,2), so their vertices can be regarded as binaryk-tuples, adjacent if their Hamming distance is 1. By the preceeding sections, wealready know that they are distance-transitive, have both diameter and valencyk,and their intersection arrays are given by

ι(Qk) =

∗ 1 · · · k−1 k0 0 · · · 0 0k k−1 · · · 1 ∗

.The more common alternative construction forQk is given recursively as fol-

lows: starting withQ1∼= K2, take two copies ofQk−1 (which we’ll label Qk−1

andQ′k−1). Then for each vertexv ∈ VQk−1, join it to the corresponding vertexv′ ∈VQ′k−1.

ThusQ2 is just the square, andQ3 is what would normally be referred to as thecube, andQ4 is the graph shown below.

ι(Q4) =

∗ 1 2 3 40 0 0 0 04 3 2 1 ∗

Figure 5.2: The 4-cubeQ4

We’ll meetQ3 andQ4 again in chapter 11.

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Chapter 6

Grassmann Graphs

6.1 Introduction

The Grassmann graphs are another family of graphs defined similarly to the John-son graphs. This time, however, we are concerned with subspaces of a vector space,rather than with subsets of a set.

Definition 6.1.1Let V be ann-dimensional vector space overFq, the finite field withq elements,whereq is a power of a prime. Then letVk be the set of allk-dimensional sub-spaces ofV. We define theGrassmann graph, denoted byG(q,n,k), as havingvertex setVk, and two vertices (i.e. subspaces)U,W being adjacent if and only ifdim(U ∩W) = k−1.

There are numerous similarities between the Grassmann and Johnson graphs(more so than with the Hamming graphs). For example, analogous to Lemma3.1.3, we have the following:

Lemma 6.1.2The graphsG(q,n,k) andG(q,n,n−k) are isomorphic.

Proof:Let 〈 , 〉 be some inner product onV. Then for any subspaceW ⊆V, defineW⊥ =v∈V | 〈v,w〉= 0 ∀w∈W, the orthogonal complement ofW in V. By standardlinear algebra, dim(W⊥) = dim(V)−dim(W) and dim(W⊥)−dim(W⊥ ∩U⊥) =dim(V)− dim(W∩U). Now suppose dim(V) = n, dim(U) = dim(W) = k anddim(U ∩W) = k− 1. Then we have dim(W⊥) = n− k and dim(W⊥ ∩U⊥) =(n−k)−k+(k−1) = n−k−1.

So we have shown that the function

ψ : Vk→Vn−k

where ψ(W) = W⊥

47

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48 Grassmann Graphs

is a bijection which preserves adjacency betweenG(q,n,k) andG(q,n,n−k).HenceG(q,n,k)∼= G(q,n,n−k).

As a consequence of the above lemma, we need only consider the cases wheren≥ 2k. Again, this is analogous to what we had for the Johnson graphs. As withthe Johnson graphs, we first want to determine properties such as the number ofvertices and the degree of each vertex for the Grassmann graphs. To do this, wemake use of the following definition:

Definition 6.1.3Theq-aryGaussian binomial coefficientis given by[

nk

]q

=(qn−1) · · · (qn−k+1−1)(qk−1) · · · (q−1)

.

For a deeper discussion, and further examples, of this, see Goulden & Jackson [23].

Theorem 6.1.4

1. The number of vertices ofG(q,n,k) is

[nk

]q

.

2. The degree of each vertex ofG(q,n,k) is q

[n−k

1

]q

[k

k−1

]q

.

Proof:

1. The number of vertices ofG(q,n,k) is, by definition, the number ofk-dimensional subspaces (to be succinct, we shall call thesek-spaces) of V(which has dimensionn). This is given by the number of orderedk-tuplesof linearly independent vectors inV (?), divided by the number of possibleordered bases for ak-space, which is the number of orderedk-tuples of lin-early independent vectors in(Fq)k (†).

(?) is given by(qn−1)(qn−q)(qn−q2) · · ·(qn−qk−1). (For our first vectorwe are allowed anything except the zero vector; for the second we are al-lowed anything except theq scalar multiples of the first; for the third we areallowed anything except theq2 linear combinations of the first two, and soon.)

(†) is given by(qk−1)(qk−q)(qk−q2) · · ·(qk−qk−1) by exactly the samearguments as with (?) above.

Dividing (?) by (†), we get

(qn−1) (qn−q) (qn−q2) · · · (qn−qk−1)(qk−1) (qk−q) (qk−q2) · · · (qk−qk−1)

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6.2 Distance-Transitivity 49

then by cancelling various powers ofq, we obtain

(qn−1) (qn−1−1) (qn−2−1) · · · (qn−k+1−1)(qk−1) (qk−1−1) (qk−2−1) · · · (q−1)

=[nk

]q

.

Hence there are

[nk

]q

k-spaces inV, soG(q,n,k) has

[nk

]q

vertices.

2. The degree of a vertexU of G(q,n,k) is the number ofk-spacesW (withU 6= W) such that dim(U ∩W) = k−1.

Let Z = U ∩W. By the above, there are

[k

k−1

]q

choices ofZ.

We can then generate ak-spaceW by adjoining any element ofV \U to abasis forZ. There areqn−qk such elements. However, several of these ele-ments will generate the samek-space–in fact, any of theqk−qk−1 elementsof W \Z will do this. So we divide through byqk−qk−1, giving us

qn−qk

qk−qk−1 =qn−k+1−q

q−1= q

qn−k−1q−1

= q

[n−k

1

]q

.

Hence there areq

[n−k

1

]q

[k

k−1

]q

k-spacesW such thatU ∩W is a(k−1)-

space, and so this number is the degree of the vertexU .

The aim of this chapter is to show that the Grassmann graphs are distance-transitive. We will do this asing the same broad strategy as in section 3.3, wherewe dealt with the Johnson graphs.

6.2 Distance-Transitivity

To begin with, we need to know some details about the automorphism group of theGrassmann graphG(q,n,k). From section 4.4, this should be quite straightforward.

Lemma 6.2.1

PGL(n,q)≤ Aut(G(q,n,k)).

Proof:In section 4.4, we saw thatPGL(n,q) acts faithfully on bases ofk-dimensional sub-spaces ofV. Hence it has an induced action on the vertices ofG(q,n,k).

Because the action ofPGL(n,q) is transitive onVk (the set of allk-spaces inV), we can see thatG(q,n,k) is vertex-transitive.

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50 Grassmann Graphs

As with the Johnson graphs, we prove the distance-transitivity of the Grass-mann graph by showing that there is a direct relationship between the distance oftwo vertices and the dimension of the intersection of two vertices when regardedask-spaces. The following lemma is the most crucial step.

Lemma 6.2.2LetU,W be two vertices ofG(q,n,k) such thatd(U,V) = m. Then, regardingU,Vask-spaces, dim(U ∩V) = k−m.

Proof – by induction onm:Basis: supposem = 0. That is,d(U,V) = 0, i.e. U = W, so dim(U ∩W) =dim(U) = k = k−0.Induction Hypothesis (IH):suppose the theorem holds form≤ r, that is for eachm≤ r, d(U,W) = m if and only if dim(U ∩W) = k−m.

ChooseU,W,X such thatd(U,W) = r, d(W,X) = 1 andd(U,X) = r + 1. Bythe IH, dim(U ∩W) = k− r and by definition, dim(W∩X) = k−1. We want tofind dim(U ∩X). Clearly, we have dim(U ∩X) < k− r, as otherwise that wouldcontradict the IH.

Recall that for any vector spacesA,B over the same field,

dim(A+B) = dim(A)+dim(B)−dim(A∩B). (?)

Consider the spaceU +W. By (?), we have

dim(U +W) = dim(U)+dim(W)−dim(U ∩W)= k+k− (k− r)= k+ r .

Also by (?), dim(W + X) = k+ k− (k−1) = k+ 1. Take a basisw1, . . . ,wk forW and extend it to bases forU +W andW +X:

U +W = spanw1, . . . ,wk,u1, . . . ,ur, W +X = spanw1, . . . ,wk,x.

Becausex is linearly independent fromu1, . . . ,ur , we have

U +W +X = spanw1, . . . ,wk,u1, . . . ,ur ,x,

so dim(U + W + X) = k + r + 1. Now (U + X) ⊆ (U + W + X), so thereforedim(U +X)≤ k+ r +1. Again using(?), we have

dim(U ∩X) = dim(U)+dim(X)−dim(U +X)≤ k+k− (k+ r +1)= k− (r +1).

But we know already that dim(U ∩X)< k− r. Hence it must be that dim(U ∩X) =k− (r +1), and the result follows by induction.

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6.2 Distance-Transitivity 51

A consequence of this result is that we now know that the diameter ofG(q,n,k):

Corollary 6.2.3The diameter ofG(q,n,k) is k.

Proof:From Lemma 6.2.2, the maximum distance between two verticesU,W occurs whenthe dimension of(U ∩W) is at its minimum, i.e. whenk−m= 0, som= k.

It is now a relatively straightforward matter to prove the following:

Theorem 6.2.4G(q,n,k) is distance-transitive.

Proof:From Lemma 6.2.2, it follows immediately that for any verticesU,W,X,Y ofG(q,n,k), d(U,W) = d(X,Y) if and only if dim(U ∩W) = dim(X∩Y).

Take a basisv1, . . . ,vk−m for U ∩W, and extend it to bases

v1, . . . ,vk−m,uk−m+1, . . . ,uk and v1, . . . ,vk−m,wk−m+1, . . . ,wk

for U andW respectively. Combine these to get a basis forU +W, and then extendthis to obtain as basis forV,

v1, . . . ,vk−m,uk−m+1, . . . ,uk,wk−m+1, . . . ,wk,an−(k−m)+1, . . . ,an.

Similarly, we take a basisz1, . . . ,zk−m for X∩Y, extend it to bases

z1, . . . ,zk−m,xk−m+1, . . . ,xk and z1, . . . ,zk−m,yk−m+1, . . . ,yk

for X andY and then obtaining another basis forV,

z1, . . . ,zk−m,xk−m+1, . . . ,xk,yk−m+1, . . . ,yk,bn−(k−m)+1, . . . ,bn.

Now, PGL(n,q) acts transitively on ordered bases ofV. Therefore there is anelementg∈ PGL(n,q) that takes the first set of basis vectors onto the second. Inparticular,

(v1, . . . ,vk−m,uk−m+1, . . . ,uk)g = z1, . . . ,zk−m,xk−m+1, . . . ,xkand (v1, . . . ,vk−m,wk−m+1, . . . ,wk)g = z1, . . . ,zk−m,yk−m+1, . . . ,yk.

That is,(U)g = X and(W)g = Y.Hence, for anyU,W,X,Y with d(U,W) = d(X,Y), there existsg∈PGL(n,q)≤

Aut(G(q,n,k)) such that(U)g = X and (W)g = Y. So G(q,n,k) is distance-transitive.

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52 Grassmann Graphs

6.3 Intersection Arrays

Of course, the next step is to calculate the intersection array ofG(q,n,k). As be-fore, we only need calulateb j andc j , as these determine the whole array.

Theorem 6.3.1For the Grassmann graphG(q,n,k), the intersection arrayι(G(q,n,k)) is given by:

1. c j =[

j1

]2

q

, and

2. b j = q2 j+1

[n−k− j

1

]q

[k− j

1

]q

.

Proof:

1. Recall that ther vertices ofG(q,n,k) are thek-dimemsional subspaces ofan n-dimensional vector spaceV overFq, two vertices being adjacent if assubspaces they intersect as a(k−1)-space. Fix a vertex (i.e.k-space)U , andsuppose we have a vertexW at distancej from U ; that is, dim(U ∩W) =k− j. Thenc j is precisely the number ofk-spacesX such that dim(U ∩X) =k− ( j−1) and dim(X∩W) = k−1.

Let Z = U ∩W, and take a basisz1, . . . ,zj for Z. Extend this to basesz1, . . . ,zk− j ,u1, . . . ,u j andz1, . . . ,zk− j ,w1, . . . ,w j for U andW respec-tively. We constructX from W by “throwing away” a 1-dimensional sub-space ofW\Z and replacing it with a 1-space fromU \Z. Now dim(W\Z) =

dim(U \Z) = j, so by 6.1.4, part 1, there are

[j1

]q

choices for the 1-space

we discard, and

[j1

]q

choices for its replacement.

Hencec j =[

j1

]2

q

.

2. Fix a k-spaceU as above. Letk j be the number of verticesW such thatd(U,W) = j (in the sense of 2.3.6). We shall calculateb j by calulatingk j

and using the formula (cf. 2.3.6)

b j =k j+1c j+1

k j. (‡)

So we want to find the number ofk-spacesW such that dim(U ∩W) = k− j.

Let U ∩W = Z0. By 6.1.4, part 1, there are

[k

k− j

]q

=[kj

]q

choices forZ0.

We constructW by adjoining toZ0 an orderedj-tuple of linearly independentvectors inV \U . There are(qn−qk) choices for the first vector, which we

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6.3 Intersection Arrays 53

adjoin toZ0 to obtainZ1 say. For the second vector, we choose one of the(qn−qk+1) vectors outsideU +Z1. We continue this until we have adjoinedj vectors. So altogether, there are

(qn−qk)(qn−qk+1) · · ·(qn−qk+ j−1)

possiblej-tuples of linearly independent vectors inV \U .

However, having generated a particulark-spaceW, we now have to countthe number of differentj-tuples which give rise to that particular space. Anyj-tuple of linearly independent vectors inW \Z0 will do this, and there are

(qk−qk− j)(qk−qk− j+1) · · ·(qk−qk−1)

of these. Hence there are

(qn−qk) (qn−qk+1) · · · (qn−qk+ j−1)(qk−qk− j) (qk−qk− j+1) · · · (qk−qk−1)

choices for extendingZ0 to ak-spaceW.

Rearranging this, we obtain

qk(qn−k−1) qk+1(qn−k−1−1) · · · qk+ j−1(qn−k− j+1−1)qk− j(q j −1) qk− j+1(q j−1−1) · · · qk(q−1)

=q j(qn−k−1) q j(qn−k−1−1) · · · q j(qn−k− j+1−1)

(q j −1) (q j−1−1) · · · (q−1)

= (q j) j[n−k

j

]q

and hencek j = q j2[n−k

j

]q

[kj

]q

.

All we have to do now is apply the formula(‡) to calculateb j . (This is easier

said than done!) We havek j+1 = q( j+1)2

[n−kj +1

]q

[k

j +1

]q

, c j+1 =[

j +11

]2

q

andk j = q j2[n−k

j

]q

[kj

]q

.

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54 Grassmann Graphs

So we have

b j =q( j+1)2

q j2

[n−kj +1

]q

[k

j +1

]q

[j +1

1

]2

q[n−k

j

]q

[kj

]q

= q2 j+1 ∏ ji=0(qn−k−i−1)

∏ j+1i=1 (qi−1)

∏ ji=0(qk−i−1)

∏ j+1i=1 (qi−1)

(q j+1−1)2

(q−1)2∏ j

i=1(qi−1)

∏ j−1i=0 (qn−k−i−1)

∏ ji=1(qi−1)

∏ j−1i=0 (qk−i−1)

= q2 j+1 ∏ ji=0(qn−k−i−1)

∏ j−1i=0 (qn−k−i−1)

∏ ji=0(qk−i−1)

∏ j−1i=0 (qk−i−1)

∏ j+1i=1 (qi−1)2

∏ j+1i=1 (qi−1)2

1(q−1)2

= q2 j+1 (qn−k− j −1)(q−1)

(qk− j −1)(q−1)

= q2 j+1[n−k− j

1

]q

[k− j

1

]q.

(I’ll bet you’re glad that’s over with!)

6.4 Linking the Grassmann and Johnson Graphs

Throughout this chapter, we have been reminded of the analogy between the Grass-mann graphs and the Johnson graphs we met in chapter 3. In fact, the Johnsongraphs can be thought of as a limiting case (or the ‘thin case’ – see [11] §9.1) ofthe Grassmann graphs. This is in part explained by the following proposition1:

Proposition 6.4.1

limq→1

[nk

]q

=(

nk

).

Proof:Recall that [

nk

]q

=k

∏i=1

qn−i+1−1qi−1

,

so therefore we have

limq→1

[nk

]q

= limq→1

k

∏i=1

qn−i+1−1qi−1

=k

∏i=1

limq→1

qn−i+1−1qi−1

1I had to put some Real Analysis insomewhere, didn’t I(!)?

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6.4 Linking the Grassmann and Johnson Graphs 55

=k

∏i=1

limq→1

ddq(qn−i+1−1)

ddq(qi−1)

using l’Hopital’s rule

=k

∏i=1

limq→1

(n− i +1)qn−i

iqi−1

=k

∏i=1

n− i +1i

=n(n−1) · · ·(n−k+1)

k!

=n!

k!(n−k)!

=(

nk

).

So by taking the limit asq tends to 1 in all our formulas involving

[nk

]q

(e.g. in

the intersection array) associated with the Grassmann graphsG(q,n,k), we obtainexactly those we had for the Johnson graphsJ(n,k,k−1). Although consideringa vector space over a field with one element is nonsense (as such a field cannotexist: a field must have at least two elements), we can (non-rigorously) think ofthis n-dimensional object as merely being a set withn elements, and its subspacesas subsets. In other words, we have exactly the construction of a Johnson graph.One could probably investigate a so-calledq-analogue of the odd graphs, too.

Brouwer, Cohen & Neumaier [11] refer to the Johnson, odd, Hamming andGrassmann graphs as examples offamilies of graphs with classical parameters.They give many more properties and related families than we have here; in effect,we have only covered the ‘bare essentials’.

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Chapter 7

Linear Algebra andDistance-Transitive Graphs

7.1 The Spectrum and the Adjacency Algebra

In the previous chapter, we defined a family of graphs using linear algebra. In thischapter, we go the other way round: we start with a graph, and obtain some linearalgebra from it. SupposeΓ is simple with vertex setVΓ = v1, . . . ,vn.

Definition 7.1.1Theadjacency matrixof Γ is then×n matrixA(Γ), with entries

A i j =

1 if vi ∼ v j

0 otherwise.

Clearly,A(Γ) is a symmetric matrix. We will regardA as a matrix overR, butin some contexts it also makes sense to regardA as a matrix overF2, as the entriesare always 0 or 1. (See Godsil & Royle [22] for more on this.)

Examples 7.1.2

A(K4) =

0 1 1 11 0 1 11 1 0 11 1 1 0

Figure 7.1: The complete graphK4

56

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7.1 The Spectrum and the Adjacency Algebra 57

A(Oct) =

0 1 1 1 1 01 0 1 0 1 11 1 0 1 0 11 0 1 0 1 11 1 0 1 0 10 1 1 1 1 0

Figure 7.2: The octahedronOct

The next definitions follow naturally from the above.

Definition 7.1.3The eigenvalues ofΓ are defined to be the eigenvalues ofA(Γ). Similarly, theeigenvectors ofΓ are the eigenvectors ofA(Γ).

SinceA(Γ) is symmetric,Γ hasn real eigenvalues, but these are not necessarilydistinct. Hence the following definition is non-trivial.

Definition 7.1.4Thespectrum ofΓ, denoted by Spec(Γ), is the set of distinct eigenvalues ofA(Γ),together with their (algebraic) multiplicities as roots of the characteristic polyno-mial χ(Γ) = det(λI −A(Γ)).

Examples 7.1.5We can now calculate the spectra of our two examples in 7.1.2:

• In the case ofK4, we find thatχ(K4) = (λ−3)(λ +1)3, so

Spec(K4) =(

3 −11 3

).

• In the case of the octahedron, we obtainχ(Oct) = (λ−4)(λ +2)2λ3, so

Spec(Oct) =(

4 −2 01 2 3

).

Note: to calculate the spectrum of a graph on more than three vertices, it is usefulto have access to a computer algebra package, such as MAPLE1. It’s not much funtrying to calculate eigenvalues of a matrix bigger than 3 by 3 by hand!

1MAPLE is a registered trade mark of Waterloo Maple Inc., 57 Erb St. West, Waterloo, Ontario,Canada N2L 6C2.

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58 Linear Algebra and Distance-Transitive Graphs

In the case whereΓ is connected andk-regular, we have the following spectralproperties.

Proposition 7.1.6Let Γ be a connected,k-regular graph onn vertices. Then the following hold:

1. k∈ Spec(Γ);

2. k has multiplicity 1;

3. for all λ ∈ Spec(Γ), |λ| ≤ k.

Proof:

1. Letv = (1,1, . . . ,1)T . SinceΓ is k-regular, the row sum of each row ofA(Γ)is k. By inspection, we see thatAv = (k,k, . . . ,k)T = kv. Hencek∈Spec(Γ).

2. SupposeAx = kx, and letx j be an entry ofx with largest absolute value.Then

(Ax) j = kxj

i.e.n

∑i=1

a ji xi = kxj .

Now letv1, . . . ,vk be thek vertices adjacent to the vertexj. By the definitionof A, we have

n

∑i=1

a ji xi = xv1 +xv2 + · · ·+xvk = kxj .

Sincex j is maximal, we havexvi ≤ x j for each 1≤ i ≤ k and it follows thatxv1 = xv2 = · · ·= xvk = x j .

We then apply this technique to all vertices adjacent to eachxvi , then repeat-ing until all vertices ofΓ are covered. (We can do this sinceΓ is connected.)In all cases, we can deduce thatxh = x j for 1≤ h≤ n. Thusx is a scalarmultiple ofv = (1,1, . . . ,1)T , so the multiplicity ofk is 1.

3. Letλ be an eigenvalue ofΓ andx the corresponding eigenvalue, soAx = λx.As before, letx j be an entry ofx with largest absolute value. Similarly toabove, we haveλx j = xv1 +xv2 + · · ·+xvk. Thus

|λ||x j | = |xv1 +xv2 + · · ·+xvk|≤ |xv1|+ |xv2|+ · · ·+ |xvk|≤ |x j |+ |x j |+ · · ·+ |x j |= k|x j |.

Hence|λ| ≤ k.

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7.1 The Spectrum and the Adjacency Algebra 59

Things become slightly more abstract with this next definition.

Definition 7.1.7Theadjacency algebra, A(Γ), is the algebra of polynomials inA(Γ).

In other words,A(Γ) is the vector space of such polynomials, but with an addi-tional operation of multiplication, so there is a ring structure as well as the vectorspace structure (this is what is meant by analgebra). As a vector space,A(Γ) isfinite-dimensional; by the Cayley-Hamilton Theorem,A satisfiesχ(A) = 0, so thisdimension is at mostn.

Every element ofA(Γ) is a linear combination of powers ofA. Because of this,the algebra is commutative under multiplication. Also, these powers ofA have agraph-theoretical interpretation, as the following theorem shows.

Theorem 7.1.8The number of paths fromvi to v j of lengthl in Γ is thei, jth entry inA l .

Proof – by induction onl :Basis: if l = 0, we haveA0 = I , so the result holds (as we define there to be a pathof length 0 fromvi to vi for all i, and noviv j path of length 0 fori 6= j).Induction hypothesis (IH):if l = 1, we haveA1 = A(Γ), the adjacency matrix ofΓ, so the result holds by the definition of this.Induction step: assume the result holds ofl = k; that is, the number of paths oflengthk from vi to v j is given by thei, jth entry ofAk.Now consider the set of paths of lengthk+ 1 from vi to v j . There is a 1-1 corre-spondence between this set and the set of all paths of lengthk from vi to vh, wherevh∼ v j . So the number of such paths is given by:

∑vh∼v j

(Ak)ih =n

∑h=1

(Ak)ihah j (sinceah j = 1 if vh∼ v j , ah j = 0 otherwise)

= (Ak+1)i j , thei, jth entry ofAk+1.

Hence, by induction, the number of paths of lengthl from vi to v j is given by thei, jth entry ofA l .

We now use this combinatorial condition to determine a lower bound for thedimension ofA(Γ).

Proposition 7.1.9SupposeΓ is connected, with diam(Γ) = d. Then the dimension ofA(Γ) is at leastd+1.

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60 Linear Algebra and Distance-Transitive Graphs

Proof:Choosex,y∈ VΓ such thatd(x,y) = d. Then there exists a geodesicw0w1 · · ·wd

(wherew0 = x andwd = y). (We also use these labels inA(Γ).) Then for eachi (1≤ i ≤ d) there is at least one path of lengthi from w0 to wi and no shorterpath. By 7.1.8 above, this implies there is a non-zero entry in the 0, i th entry ofA i ,while the 0, i th entries ofA0 = I , A, A2, . . . ,A i−1 are all zero. Consequently,A i isnot linearly dependent onA0, . . . ,A i−1 and henceI ,A,A2, . . . ,Ad is a linearlyindependent set of sized+1 in A(Γ).

Thus dim(A(Γ))≥ d+1.

Corollary 7.1.10A has at leastd+1 distinct eigenvalues.

Proof:This follows immediately from 7.1.9.

In the next section, we will see that ifΓ is distance-regular (and thus also ifΓis distance-transitive), then the dimension of the adjacency algebra isexactly d+1.

7.2 Distance Matrices

In this section, we will construct a basis for the adjacency algebra of a distance-regular graph in terms of thedistance matrices, as found by Damerell [16] in 1973.These are defined as follows:

Definition 7.2.1Let Γ be a connected graph with diameterd. Then, fori = 0, . . . ,d the i th distancematrix A i has entries

(A i)rs =

1 if d(vr ,vs) = i0 otherwise

for all v j ,vk ∈VΓ.

It is clear from the definition thatA0 = I andA1 = A(Γ), the adjacency matrix.Also, we haveA0 + A1 + · · ·+ Ad = J, whereJ is the all-ones matrix. It is clearthatA0,A1, . . . ,Ad is a linearly independent set, as exactly one of the matriceshas a non-zero entry in ther,sth position for allr,s.

We now determine a useful property of the distance matrices of a distance-regular graph.

Lemma 7.2.2(Damerell 1973)SupposeΓ is distance-regular, with adjacency matrixA(Γ). Then

AA i = bi−1A i−1 +aiA i +ci+1A i+1, for 1≤ i ≤ d−1;

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7.2 Distance Matrices 61

alsoAA0 = a0A0 +c1A1, AAd = bd−1Ad−1 +adAd.

Proof:Consider ther,sth entry of the matrices on each side.(AA i)rs is the number of verticesw such thatd(vr ,w) = 1 andd(vs,w) = i, i.e.|Γ1(vr)∩ Γi(vs)|. So there are three possibilities ford(vr ,vs): i − 1, i or i + 1.Hence

(AA i)rs =

bi−1 if d(vr ,vs) = i−1ai if d(vr ,vs) = i

ci+1 if d(vr ,vs) = i +10 otherwise

= bi−1(A i−1)rs +ai(A i)rs +ci+1(A i+1)rs

= (bi−1A i−1 +aiA i +ci+1A i+1)rs

and so the result follows. ForAA0 andAAd, the proof is almost identical.

(Note thatAA0 = a0A0 +c1A1 = 0A0 +1A1 = A = AI = AA0, as required.)

A consequence of the lemma above is the following:

Lemma 7.2.3(Damerell 1973)The distance matrixA i (for 0≤ i ≤ d) of a distance-regular graphΓ is a polynomialpi(A) of degreei in A, and so is an element ofA(Γ).

Proof – by induction oni:By definition (see 7.2.1),A0 = I = A0 andA1 = A1 are polynomials inA of degree0 and 1 respectively, and are elements ofA(Γ) (so p0(A) = I andp1(A) = A).

Now supposeA0 = p0(A), . . . ,A j = p j(A) ∈ A(Γ). Then, by 7.2.2 above,

A j+1 =1

c j+1

(AA j −a jA j −b j−1A j−1

)=

1c j+1

(Ap j(A)−a j p j(A)−b j−1p j−1(A)

),

which is a polynomial inA of degreej + 1. HenceA j+1 = p j+1(A) ∈ A(Γ) andthe result follows by induction.

We now have the tools we need to prove the main theorem of this section.

Theorem 7.2.4(Damerell 1973)SupposeΓ is a distance-regular graph of degreek and diameterd. ThenA(Γ) hasdimensiond+1 and has basisA0, . . . ,Ad.

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62 Linear Algebra and Distance-Transitive Graphs

Proof:First consider the matrixA−kI . Since the row sums ofA are allk and the diagonalelements are all 0, the row sums ofA−kI are all 0. Thus we have(

A−kI)(

A0 +A1 + · · ·+Ad)

=(A−kI

)J = 0 (†)

By the lemma above, eachA i is a polynomial inA of degreei, so the left-handside of(†) is a polynomial inA of degreed + 1. However, any polynomial inAof degree≤ d is non-zero, as theA i are all linearly independent (by 7.1.9), so thepolynomial min(A) =

(A− kI

)(A0 + A1 + · · ·+ Ad

)is the minimum polynomial

of A in A(Γ). Hence the dimension ofA(Γ) is d+1.SinceA0,A1, . . . ,Ad is a linearly independent set of sized + 1 in A(Γ), it

forms a basis.

However, this isn’t the only basis forA(Γ) that we know.

Corollary 7.2.5I ,A,A2, . . . ,Ad is a basis forA(Γ).

Proof:By 7.1.9,I ,A,A2, . . . ,Ad is a linearly independent of sized + 1 in A(Γ), andsince dim(A(Γ)) = d+1 by the theorem above, it must form a basis forA(Γ).

7.3 The Intersection Matrix

Let X be an element ofA(Γ). Because we know two bases,I ,A,A2, . . . ,Ad andA0,A1,A2, . . . ,Ad, for A(Γ), we can writeX in two ways,

X =d

∑i=0

r iA i =d

∑i=0

siA i ,

wherer i ,si are some constants. It is the second of these that interests us here.

Recall from Lemma 7.2.2 thatA iA = bi−1A i−1 + aiA i + ci+1A i+1 (†). (Usingthe first basis, it is clear thatA(Γ) is commutative.) Consider the linear transfor-mationτ of A(Γ) which sendsX to XA . By (†), we have

XA =( d

∑i=0

siA i)A

=d

∑i=0

si(A iA)

=d

∑i=0

si(bi−1A i−1 +aiA i +ci+1A i+1).

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7.3 The Intersection Matrix 63

Thus we have representedτ by the matrix

B(Γ) =

0 1k a1 c2

b1 a2 .b2 . .

. . cd−1

. ad−1 cd

bd−1 ad

with respect to the second basis,A0,A1, . . . ,Ad.

The matrixB(Γ) bears an uncanny resemblance to to the intersection arrayι(Γ)of Γ. The next definition is therefore unsurprising.

Definition 7.3.1The matrixB(Γ), as shown above, is called theintersection matrixof Γ.

SinceA,B both correspond to the same linear transformationτ of A(Γ), theymust therefore have the same eigenvalues. A consequence of Theorem 7.2.4 is thatA has exactlyd + 1 distinct eigenvalues, and asB has at mostd + 1, then eacheigenvalue ofB has multiplicity 1. However, as eigenvalues ofA, they have mul-tiplicities≥ 1, as the multiplicities must sum ton. We will show that it is possibleto calculate these multiplicites directly fromB.

Let λ be an unspecified eigenvalue ofB. Then there is aright eigenvectorv(λ) = (v0(λ),v1(λ), . . . ,vd(λ)) corresponding toλ. That is,v(λ) is a solution ofthe system

Bv(λ) = λv(λ).

Assume thatv(λ) is the uniquestandard2 eigenvector corresponding toλ, i.e. theunique solution withv0(λ) = 1. (We can do this becauseB is tridiagonal, wherev0 = 0 would implyv(λ) = 0). So we have

Bv(λ) =

0 1k a1 c2

b1 a2 c3

. . .. . .

bi−1 ai ci+1

. . .. . .

bd−2 ad−1 cd

bd−1 ad

1v1(λ)v2(λ)..

vi(λ)..

vd−1(λ)vd(λ)

2A vector isstandardif its first entry is 1.

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64 Linear Algebra and Distance-Transitive Graphs

=

v1(λ)k+a1v1(λ)+c2v2(λ)

b1v1(λ)+a2v2(λ)+c3v3(λ)..

bi−1vi−1(λ)+aivi(λ)+ci+1vi+1(λ)..

bd−2vd−2(λ)+ad−1vd−1(λ)+cdvd(λ)bd−1vd−1(λ)+advd(λ)

= λ

1v1(λ)v2(λ)..

vi(λ)..

vd−1(λ)vd(λ)

So we have a sequencevi(λ) as follows:

v0(λ) = 1 (by assumption,v(λ) is standard)

v1(λ) = λ...

λvi(λ) = bi−1vi−1(λ)+aivi(λ)+ci+1vi+1(λ)

and so on. Note that theith term of the sequencevi(λ) is a polynomial inλ ofdegreei. This gives us the following lemma:

Lemma 7.3.2

A i = vi(A).

Proof:SubstitutingA for λ in the definition ofvi(λ) above, we obtain

Avi(A) = bi−1vi−1(A)+aivi(A)+ci+1vi+1(A).

Comparing this with Lemma 7.2.2, we haveA i = vi(A).

There are alsod+1 left eigenvectors ofB; corresponding to each eigenvalueλof B we haveuT(λ) satisfyinguTB = λu(λ).

Let us now give names to the specific eigenvalues ofB. SinceΓ is k-regular,by Proposition 7.1.6k is the largest eigenvalue ofA and thus also ofB. Then labelthe remaining eigenvalues as follows:

k = λ0 > λ1 > λ2 > · · ·> λd.

Similarly, letv0, . . . ,vd andu0, . . . ,ud be the corresponding standard right and lefteigenvectors.

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7.3 The Intersection Matrix 65

It seems natural that theseui andvi are related in some way. In fact, we havethe following result.

Lemma 7.3.3For 0≤ j ≤ d, (vi) j = k j(ui) j , wherek j is as in 2.3.6.

Proof:DefineK to be the diagonal matrix

K =

k0 0k1

k2

..

0 kd

.

From Theorem 2.3.6, we haveb j−1k j−1 = c jk j , so the matrixBK is symmetric.Hence

BK = (BK)T = KTBT = KBT . (∗)

Supposevi anduTi are corresponding right and left eigenvectors ofB, that is

Bvi = λivi

and uTi B = λiuT

i (†).

Hence

(BK)ui = (KBT)ui (from (∗))= K(uT

i B)T

= K(λiuTi )T (from (†))

= λKu i

i.e. B(Ku i) = λi(Ku i). In other words,Ku i is a right eigenvector ofB correspond-ing to λ.

Sincek0 = 1, we get that(Ku i)0 = 1 also, soKu i is standard. Butvi is theunique standard right eigenvector corresponding toλi , so it follows thatvi = Ku i ,and hence(vi) j = (Ku i) j = k j(ui) j .

Now suppose〈 , 〉 is the usual inner product onRd+1. We now prove two usefulresults about the inner products of our eigenvectorsui ,vi .

Proposition 7.3.4

1. Fori 6= h, 〈ui ,vh〉= 0.

2. 〈u0,v0〉= n, the number of vertices ofΓ.

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66 Linear Algebra and Distance-Transitive Graphs

Proof:

1. We haveλi〈ui ,vh〉= λiuTi vh = uiBvh = uT

i (λhvh) = λh〈ui ,vh〉. But for i 6= h,λi 6= λh as the eigenvalues are all distinct, so the only way out is to have〈ui ,vh〉= 0.

2. First, we notice thatuT0 = (1,1,1, . . . ,1):

(1,1,1, . . . ,1)B = (k,1+a1 +b1,c2 +a2 +b2, . . . ,cd +ad)= (k,k,k, . . . ,k)= kuT

0 .

By Lemma 7.3.3 above,(v0) j = k j(u0) j , sov0 = (1,k,k2, . . . ,kd), thus〈u0,v0〉=1+k+k2 + · · ·+kd = n.

We can now move on to one of the most important results of the project. It gives usa formula for the multiplicities of the eigenvalues ofA, but in terms of informationprovided solely byB.

Theorem 7.3.5(Biggs 1970)Supposek = λ0 > λ1 > .. . > λd are the eigenvalues of bothA andB, andui ,vi

are the standard left and right eigenvectors ofB corresponding toλi . Then themultiplicity m(λi) of λi as an eigenvalue ofA is given by the formula

m(λi) =〈u0,v0〉〈ui ,vi〉

.

Proof:Define elementsL i ∈ A(Γ) as follows:

L i =d

∑j=0

(ui) jA j .

We will calculate the trace ofL i in two ways.

1. We have

tr(L i) = tr

(d

∑j=0

(ui) jA j

)=

d

∑j=0

(ui) j tr(A j).

Now, for j = 1, . . . ,d, tr(A j) = 0 as all diagonal entries are 0, and tr(A0) =tr(I) = n. Hence tr(L i) = (ui)0n.

But eachui is standard, so(ui)0 and thus tr(L i) = n.

2. Recall from Lemma 7.3.2 thatA j = v j(A). Thus the eigenvalues ofA j arethe eigenvalues ofv j(A), which arev j(λh) for 0≤ h≤ d. Clearly (and cru-cially!), the multiplicity of v j(λh) as an eigenvalue ofA j is equal to themultiplicity of λh as an eigenvalue ofA.

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7.4 Algebraic Constraints on the Intersection Array 67

Also, recall that the sum of the eigenvalues of any square matrix is preciselythe trace of that matrix. Hence we have

tr(A j) =d

∑h=0

m(λh)v j(λh)

=d

∑h=0

m(λh)(vh) j

and so tr(L j) =d

∑j=0

(ui) j

(d

∑h=0

m(λh)(vh) j

)

=d

∑h=0

m(λh)

(d

∑j=0

(ui) j(vh) j

)

=d

∑h=0

m(λh)〈ui ,vh〉.

But (by 7.3.4) fori 6= h, 〈ui ,vh〉= 0, so

tr(L i) = m(λi)〈ui ,vi〉.

By equating parts (1) and (2) and by using 7.3.4 we have

tr(L i) = m(λi)〈ui ,vi〉= n = 〈u0,v0〉,

so we obtain

m(λi) =〈u0,v0〉〈ui ,vi〉

,

as required.

The above result was obtained by Biggs around 1970 (as he puts it in [3]),“sneaking in at the back door of a theory developed by J.S. Frame, H. Wielandtand D.G. Higman, for which a basic reference is [27]”. The treatment given hereis an amalgam of those found in [3], [4], [8] and [30].

7.4 Algebraic Constraints on the Intersection Array

The result obtained in Theorem 7.3.5 has the following interpretation. Suppose wehave an arbitrary array of integers

ι =

∗ 1 c2 · · · cd−1 cd

a0 a1 a2 · · · ad−1 ad

k b1 b2 · · · bd−1 ∗

.

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68 Linear Algebra and Distance-Transitive Graphs

We can rewrite this array to form the matrix

B =

0 1k a1 c2

b1 a2 .b2 . .

. . cd−1

. ad−1 cd

bd−1 ad

.

If the ι is the intersection array of a distance-regular graphΓ, thenB is its intersec-tion matrix, and thus the numbers

〈u0,v0〉〈ui ,vi〉

are the multiplicities of the eigenvalues ofΓ, so must be positive integers. Thisgives an important necessary condition forι to be an intersection array. Combiningthis with the results of 2.3.6 and 2.3.8, we have the following definition:

Definition 7.4.1A (d+ 1)× (d+ 1) tridiagonal matrixB (with non-negative integer entries) of theform

B =

0 1k a1 c2

b1 a2 .b2 . .

. . cd−1

. ad−1 cd

bd−1 ad

is said to befeasibleif the following conditions are satisfied:

1. ci +ai +bi = k (for 1≤ i ≤ d−1), cd +ad = k;

2. 1≤ c2≤ ·· · ≤ cd andk≥ b1≥ bd−1;

3. For 2≤ j ≤ d, the numbers

k j =kb1 · · ·b j−1

1c2 · · ·c j

are positive integers;

4. Forn = 1+k+k2 + · · ·+kd, nk is even;

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7.4 Algebraic Constraints on the Intersection Array 69

5. Forui ,vi as given in 7.3.5, the numbers

〈u0,v0〉〈ui ,vi〉

are positive integers.

Similarly, an array of positive integers

ι =

∗ 1 c2 · · · cd−1 cd

a0 a1 a2 · · · ad−1 ad

k b1 b2 · · · bd−1 ∗

is said to befeasibleif it corresponds to a feasible matrix.

It should be emphasised that feasibility is not asufficientcondition forι to bethe intersection array of some distance-regular graph. Hence the following defini-tions are non-trivial.

Definitions 7.4.2

• A feasible arrayι is said to berealisableif it is the intersection array of somedistance-regular graph.

• An array is calledredundantif it is feasible but not realisible.

We now give some examples.

Example 7.4.3The array

ι =

∗ 1 1 30 0 1 03 2 1 ∗

is not feasible: we havek = 3, so clearly (1) and (2) are satisfied. We obtaink2 = 6, k3 = 2 andn = 1+ 3+ 6+ 2 = 12, so (3) and (4) are also satisfied. Itremains to check (5). The characteristic polynomial of the tridiagonal matrixBcorresponding toι is (λ−3)(λ + 1)(λ2 + λ−3), so the eigenvalues are 3, 1, and12(−1±

√13). Corresponding toλ0 = 3, we obtainuT

0 = (1,1,1,1) and vT1 =

(1,3,6,2), so〈u0,v0〉= 1+3+6+2 = 12.Now let λ1 = 1

2(−1+√

13). After several lines of calculations, we obtain

u1 =

1

16(−1+

√13)

−16(−1+

√13)

−1

and v1 =

1

12(−1+

√13)

−12(−1+

√13)

−1

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70 Linear Algebra and Distance-Transitive Graphs

so therefore

〈u1,v1〉 = 1+112

(−1+√

13)2 +112

(−1+√

13)2 +1

= 2+16

(1−2√

13+13)

=13

(13−√

13)

and so

〈u0,v0〉〈ui ,vi〉

=12

13(13−

√13)

which is clearly not an integer. Hence the arrayι is not feasible, so there can be nograph corresponding to it.

Virtually every source (e.g. [10]) mentions that there are very few redundantmatrices. However, without using a computer to enumerate large numbers of pos-sible examples, this is very difficult to verify. But there are still some examplesof feasible arrays (or matrices) that are not realisable. This example was found bySmith [36].

Example 7.4.4The array

ι =

∗ 1 2 2 40 0 0 0 04 3 2 2 ∗

is redundant: it is feasible, with eigenvalues 4,

√6,0,−

√6,−4, left eigenvectors

11111

,

1√6

416− 1√

6−2

3

,

10−1

3013

,

1

−√

64

161√6−2

3

,

1−1

1−1

1

,right eigenvectors

14663

,

1√6

1−√

6−2

,

10−2

01

,

1−√

61√6−2

,

1−4

6−6

3

,and the quotients

〈u0,v0〉〈ui ,vi〉

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7.4 Algebraic Constraints on the Intersection Array 71

are calculated to be 4, 10, 4 and 1.Any graphΓ corresponding toι would have girth 4, so any two verticesx,y

of Γ such thatd(x,y) = 2 must lie on exactly one circuit of length 4. Fixu∈VΓ.Now, becausec2 = 2, each vertex ofΓ2(u) is adjacent to two vertices ofΓ1(u).Thus we have the following situation:

Figure 7.3: Schematic for example 7.4.4

Becaused(v1,v2) = 2, v1 andv2 must lie on a circuit of length 4, so we joinv2

to w1 to create this. Likewise, we joinv3 to w2. Now, sinced(v2,v3) = 2, v2 andv3

must also lie in a 4-circuit, so we join them both tow3. We also haved(v1,v4) = 2,so these two vertices must also lie on a 4-circuit. However, we havec3 = 2, sowe cannot joinv4 to eitherw1 or w2, as both of these already are adjacent to twovertices ofΓ2(u). Hencev1 andv4 cannot lie on a 4-circuit.

So a graph corresponding toι cannot exist.

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Chapter 8

Primitive and ImprimitiveGraphs

8.1 Introduction

In chapter 4, we gave the definitions of primitive and imprimitive group actionsfor permutation groups in general. We now relate this to the study of distance-transitive graphs. Our first definitions are very natural.

Definition 8.1.1A block of a graphΓ is a block of the automorphism group Aut(Γ) acting on thevertex setVΓ.

As always, we have the trivial blocks Ø,v andVΓ.

Definition 8.1.2If a distance-transitive graphΓ has only trivial blocks, we sayΓ is a primitivegraph. Otherwise,Γ has a non-trivial block system and is called animprimitivegraph.

We now give examples of each of these.

Example 8.1.3The complete graphs,Kn, are all primitive distance-transitive graphs. (K5 is illus-trated overleaf.) This is because Aut(Kn) ∼= Sn, so every possible permutation ofthe vertices is an automorphism. Thus there is no non-trivial subset ofVKn that isfixed unde the action ofSn. HenceKn has no non-trivial blocks, so is a primitivegraph.

Some more sophisticated examples of primitive graphs will be shown later, insection 8.4.

72

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8.1 Introduction 73

Figure 8.1:K5, a primitive graph

Example 8.1.4The complete bipartite graphsKn,n are impimitive distance-transitive graphs. (K3,3

is illustrated).

Figure 8.2:K3,3, an imprimitive graph

The relation≡ whereu≡ v if and only if u,v ∈ V1 or u,v ∈ V2 is clearly anequivalence relation onVKn,n that is invariant under the action of Aut(Γ), with thebipartitionV1,V2 forming the block system. For example, as labelled in figure 8.2,we haveu≡ v but v 6≡ w. As Kn,n are clearly distance-transitive and have a non-trivial block system, they are imprimitive.

The complete characterisation of imprimitive graphs was one of the first majorresults obtained in the study of distance-transitive graphs. It is due to Smith [32] in1971. Proving this result, and some applications of it, forms the rest of this chapter.

We now prove some basic lemmas which start to give us an idea of what a non-trivial block system can look like.

Lemma 8.1.5(Smith 1971)SupposeB is a block of a distance-transitive graphΓ and letu∈ B. Then ifB con-tains a vertex ofΓ j(u), Γ j(u)⊆ B.

Proof:Let v∈B∩Γ j(u), and choose somew∈ Γ j(u) with v 6= w. By distance-transitivity,there exists someg∈ Aut(Γ) such that(u)g = u and(v)g = w. By the definition

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74 Primitive and Imprimitive Graphs

of a block, there is some equivalence relation≡ onVΓ such thatu≡ v. But≡ isinvariant under the action of Aut(Γ), so(u)g≡ (v)g, i.e. u≡ w, and thusw∈ B.HenceΓ j(u)⊆ B.

An alternative way of stating the result above is to say that “a blockB of adistance-transitive graphΓ is a disjoint union of cells of a distance partition ofΓ”.This is a very useful result, as it narrows down the possibilities for a non-trivialblock system. It also gives us an alternative way of showing thatKn is primitive.

Example 8.1.6Choose a vertexu of Γ = Kn. If u is in some non-trivial blockB, then there mustbe some other vertexv∈ B. Since diam(Γ) = 1, we havev∈ Γ1(u), so by Lemma8.1.5,Γ1(u)⊆ B. But u∪Γ1(u) = VΓ, soB = VΓ, which is also a trivial block.HenceΓ = Kn has no non-trivial blocks, so is primitive.

Another useful result is the following.

Lemma 8.1.7(Smith 1971)SupposeΓ is distance-transitive,B a block ofΓ, with u,v∈B andd(u,v) = 1. ThenB = VΓ.

Proof:By assumption,v ∈ B∪ Γ1(u), so by Lemma 8.1.5, we haveΓ1(u) ⊂ B. Also,u∈ B∪Γ1(v), soΓ1(v)⊂ B.

Now choosew∈ Γ2(u)∩Γ1(v) (this exists by distance-transitivity). Sincew∈Γ1(v), we havew∈ B, so thereforew∈ Γ2(u)∩B and by the lemma,Γ2(u) ⊂ B.Repeating this process, we see thatΓ3(u)⊂ B, . . . ,Γd(u)⊂ B.

Hence every vertex ofΓ is contained inB, so thereforeB = VΓ.

The next lemma is a similar result. First we let

η =

d if d is evend−1 if d is odd

(in other words,η is the longest possibleevenlength of a geodesic path inΓ).Similarly, we let

θ =

d−1 if d is evend if d is odd

(the longest possibleodd length of a geodesic path inΓ).

Lemma 8.1.8SupposeΓ is distance-transitive,B a block ofΓ, with u,v∈B andd(u,v) = 2. ThenB = Γ0(u)∪Γ2(u)∪·· ·∪Γη(u).

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8.2 Antipodal Graphs 75

Proof:By assumption,v∈B∪Γ2(u), so by Lemma 8.1.5, we haveΓ2(u)⊂B, and likewiseΓ2(v) ⊂ B. Now choose somew ∈ Γ4(u)∩Γ2(v) (again, this exists by distance-transitivity). Becausew∈ Γ2(v), w∈ B, sow∈ B∩Γ4(u) and by Lemma 8.1.5 wehaveΓ4(u)⊂ B.

Repeating this process, we obtain the required result.

8.2 Antipodal Graphs

Antipodal graphs are a class of imprimitive distance-transitive graphs. They aredefined as follows:

Definition 8.2.1A distance-transitive graphΓ with diameterd is said to beantipodalif, for all dis-tinct v,w∈ Γ0(u)∪Γd(u), we haved(v,w) = d.

We now clarify this definition with some examples.

Example 8.2.2

Figure 8.3:K3,3, an antipodal graph

Γ = K3,3 is antipodal: forv,w∈ Γ2(u), we haved(v,w) = 2. (Clearly, this extendsto Kn,n.)

Example 8.2.3TheHeawood graph, H, is not antipodal: the diameter ofH is 3, but the verticesv10,v12∈ H3(v1) are at distance 2.

Definition 8.2.4Notice that ifΓd(u) consists of a single vertexv, then the graph is automaticallyantipodal, because we haveΓ0(u)∪Γd(u) = u,v, whered(u,v) = d. In this case,we callΓ trivially antipodal.

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76 Primitive and Imprimitive Graphs

Figure 8.4: The Heawood graph, a non-antipodal graph

Examples familiar to us include thek-cubesQk (see section 5.4), the octahe-dron (figure 2.6) and our ‘pretty picture’,J(6,3,2) (figure 3.2).

Our next result is another step in the characterisation of imprimitive graphs weare aiming for.

Lemma 8.2.5(Smith 1971)A distance-transitive graphΓ is antipodal if and only ifB = Γ0(u)∪ Γd(u) is ablock of Γ.

Proof:SupposeB = Γ0(u)∪ Γd(u) is a block ofΓ, and choose verticesv,w ∈ B withd(v,w) = j. By distance-transitivity, there existsg∈Aut(Γ) such that(v)g = u and(w)g = x, wherex∈ Γ j(u). By the definition of a block,Bg= (x)g|x∈ B ⊆ B,sou,x∈ B and consequently (by Lemma 8.1.5)Γ j(u)⊂ B. Therefore, becauseu,xare distinct andd(v,w) = d(u,x), the only possibility for this distance isd. HenceΓ is antipodal.

Conversely, supposeΓ is antipodal. Consequently, for anyx,y∈ B = Γ0(u)∪Γd(u), d(x,y) = 0 or d, soB = Γ0(x)∪Γd(x). Therefore, for anyg ∈ Aut(Γ), if(v)g∈ B∩Bg, thenB = Γ0((v)g)∪Γd((v)g). But

Bg = (Γ0(v)∪Γd(v))g

= (Γ0((v)g)∪Γd((v)g))

becauseΓ is distance-transitive,g preserves distances. ThusB = Bg, soB is invari-ant under the action of Aut(Γ) and so is a block ofΓ.

The second part of the above proof tells us that any antipodal graph is imprim-itive. The converse is partially true, as we shall see later.

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8.3 Bipartite Distance-Transitive Graphs 77

8.3 Bipartite Distance-Transitive Graphs

It is assumed that the reader is familiar with the concept of a bipartite graph (as de-fined in 1.1.1). So we just state the following result (Theorem 5.1 in Wilson [40]):

Proposition 8.3.1A graph is bipartite if and only if it contains no odd circuits.

A consequence of this is the following:

Proposition 8.3.2SupposeΓ is distance-transitive, with intersection array

ι(Γ) =

∗ c1 · · · cd−1 cd

a0 a1 · · · ad−1 ad

b0 b1 · · · bd−1 ∗

.ThenΓ is bipartite if and only ifai = 0 for 1≤ i ≤ d.

Proof:We prove the contrapositive in both directions.

First, supposea j 6= 0 for some j, so there are adjacent verticesv,w ∈ Γ j(u).Also, there are pathsπ,ρ of length j from u to v and tow respectively. Letx be thelast vertex whereπ andρ meet and letm= d(x,v) = d(x,w). Then we can form acircuit of length 2m+1, an odd number. SoΓ is not bipartite.

Conversely, supposeΓ is not bipartite. ThenΓ contains some odd circuitσ,of length 2δ + 1 say. Choose some vertexu in σ. Then there exist two adjacentverticesv,w in σ such thatd(u,v) = d(u,w) = δ, sov,w∈ Γδ(u) as shown.

Figure 8.5: An odd circuit in a non-bipartite graphΓ

Therefore there is at least oneai that is non-zero.

The above proposition is important in the next result, which is another majorstep in our quest to classify all imprimitive distance-transitive graphs.

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78 Primitive and Imprimitive Graphs

Lemma 8.3.3(Smith 1971)SupposeΓ is distance-transitive with diameterd > 2 and valencyk > 2. ThenΓ0(u)∪Γ2(u)∪·· ·∪Γη(u) is a block ofΓ if and only if it is bipartite.

Proof:SupposeΓ is bipartite, and choosev,w∈ Γ0(u)∪Γ2(u)∪·· ·∪Γη(u). BecauseΓ isbipartite, it follows thatd(u,v) must be even:

Let π be a geodesic path fromu to v. By Lemma 8.3.2,a j = 0 for all j, so thereare no edges withinΓ j(u). Thereforeπ is composed of pairs of edges as follows:

Figure 8.6: Three types of path

(becausev,w ∈ Γ0(u)∪Γ2(u)∪ ·· · ∪Γη(u)). Thusπ contains an even number ofedges and sod(u,v) is even.

Now let g∈ Aut(Γ), and suppose towards a contradiction that(u)g∈ Γ0(u)∪Γ2(u)∪·· ·∪Γη(u) and that(v)g∈ Γ1(u)∪Γ3(u)∪·· ·∪Γθ(u). Let ρ be a geodesicpath from(u)g to (v)g. By Lemma 8.3.2,a j = 0 for all j, so since(v)g∈ Γ1(u)∪Γ3(u)∪ ·· ·∪Γθ(u), the penultimate vertexx of ρ must lie inΓ0(u)∪Γ2(u)∪ ·· ·∪Γη(u). By the above arguments,d((u)g,x) is even so therefored((u)g,(v)g) mustbe odd. However, by distance-transitivity,d((u)g,(v)g) = d(u,v). But a numbercan’t be both odd and even at the same time, so we have a contradiction.

ThusΓ0(u)∪Γ2(u)∪ ·· · ∪Γη(u) is a subset ofVΓ that is invariant under theaction of Aut(Γ), so it forms a block ofΓ.

Conversely, suppose thatΓ0(u)∪Γ2(u)∪ ·· · ∪Γη(u) is a block ofΓ, sayB,and thatΓ is not bipartite. Then by 8.3.1,Γ contains an odd circuit and by 8.3.2,there exists somej such thata j 6= 0. However, thisj must be odd, as ifj was eventhere would be two adjacent vertices inΓ j ⊂ B, implying (by 8.1.7) thatB = VΓ,contradicting our assumption thatB is non-trivial.

Let j = 2m+ 1 be the smallest integer such thata2m+1 6= 0. Then there existv,v′ ∈ Γ2m+1(u) such thatd(v,v′) = 1. This gives us two cases to consider:

1. The case where 2m+ 1 ≥ 3. Choosew,w′ ∈ Γ2m(u) satisfyingd(v,w) =

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8.3 Bipartite Distance-Transitive Graphs 79

d(v′w′) = 1, and also choosex,x′ ∈ Γ1(u) satisfyingd(w,x) = d(w′,x′) =2m−1. So we have the following situation:

Figure 8.7: Schematic for case 1

Sinced(u,w) = d(x,v) = 2m, by distance-transitivity there existsg∈Aut(Γ)such that(u)g = x,(w)g = v. Similarly, sinced(u,w′) = d(x′,v′) = 2m, thereexistsh∈ Aut(Γ) such that(u)h = x′,(w)h = v′. By the induced action ofgandh onB, we have thatx,v∈ Bgandx′,v′ ∈ Bh. Bg is the image ofB undersome automorphism, so for any vertex inBg (for example,x), it contains allvertices ofΓ at distance 2 from it. But becausea1 6= 0, we haved(x,x′) = 2,giving x′ ∈ Bg and consequentlyBh⊆ Bg. Similarly, x ∈ Bh so Bg⊆ Bh.Therefore we haveBg= Bh= b′, say. However,v,v′ ∈B′ andd(v,v′) = 1, soby Lemma 8.1.7,B′ = VΓ. But this is absurd, as it contradictsB′ being theimage of the non-trivial blockB under the automorphismsg andh. Henceour assumption thatΓ is not bipartite must be false.

2. The case where 2m+ 1 = 1. That is,a1 6= 0, so any vertexy ∈ Γ1(u) isadjacent to some other vertexz∈ Γ1(u), with the edgesuy,yz,zu forming atriangle. By the distance-transitivity ofΓ, any edge ofΓ must lie in a triangle.

Fix v∈ Γ1(u) and choosew∈ Γ2(u) such thatd(v,w) = 1. Thus the edgevwmust lie in a triangle. Sincea2 = 0, there existsv′ ∈ Γ1(u) with d(v,v′) =d(v′w) = 1 so that the edgesvw,wv′,v′v form the required triangle. Finally,choosex ∈ Γ3(u) (this is where we use the assumption thatd > 2) withd(w,x) = 1. The following diagram should clarify the above description.

Figure 8.8: Schematic for case 2

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80 Primitive and Imprimitive Graphs

Sinced(u,w) = d(x,v) = 2, then by distance-transitivity there existsg ∈Aut(Γ) satisfying(u)g = x,(w)g = v. Becausev ∈ Γ2(x)∩Bg, by Lemma8.1.5 we haveΓ2(x) ⊂ Bg. However, this implies that the adjacent verticesv,v′ are also inBg, so by Lemma 8.1.7,Bg = VΓ, which contradicts thenon-triviality of Bg.

In both cases we obtain a contradiction, so therefore it must be the case thatΓ isbipartite.

8.4 Smith’s Theorem

We now arrive at the primary goal of this chapter: the complete characterisation ofimprimitive distance-transitive graphs, which is known as Smith’s Theorem.

Theorem 8.4.1(Smith 1971)Let Γ be a distance-transitive graph of valencyk > 2 and diameterd > 2, withintersection array

ι(Γ) =

∗ c1 · · · cd−1 cd

a0 a1 · · · ad−1 ad

b0 b1 · · · bd−1 ∗

.ThenΓ is imprimitive if and only if it is bipartite or antipodal.

Proof:We’ll show that the only possible non-trivial blocks ofΓ are Γ0(u)∪ Γd(u) orΓ0(u)∪Γ2(u)∪·· ·∪Γη(u), for someu in VΓ.

Let B be a non-trivial block ofΓ, with u∈B. By Lemma 8.1.5,B is the disjointunion of some of theΓ j(u). Obviously,Γ0(u)⊂ B. Then either(i) B = Γ0(u)∪Γd(u), or(ii) B = Γ j(u)⊂ B for somej where 0< j < d.

Now consider the intersection arrayι(Γ).If a j 6= 0, then there existv,w∈B with d(v,w) = 1, so by Lemma 8.1.7B=VΓ,

contradictingB being non-trivial. Thereforea j = 0 if Γ j(u)⊂ B.If b j−1 ≥ 2 or c j+1 ≥ 2, then there existx,y∈ Γ j(u) ⊂ B with d(x,y) = 2, as

shown:

If b j−1≥ 2: If c j+1≥ 2:

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8.4 Smith’s Theorem 81

Therefore, by Lemma 8.1.8, we haveΓ0(u)∪ Γ2(u)∪ ·· · ∪ Γη(u) ⊆ B. But novertex inΓ1(u)∪Γ3(u)∪ ·· ·∪Γθ(u) is in B, as it would be adjacent to a vertex inΓ0(u)∪Γ2(u)∪ ·· · ∪Γη(u) ⊆ B which would implyB = VΓ, again contradictingthe non-triviality ofB. HenceB = Γ0(u)∪Γ2(u)∪·· ·∪Γη(u).

The remaining case to check is whenb j−1 = 1 andc j+1 = 1. However, byTheorem 2.3.6,c j ≤ c j+1 andb j−1 ≥ b j , so thereforec j = b j = 1. From above,a j = 0 so this givesk = c j + a j + b j = 2, contradicting our initial hypothesis thatk> 2.

Now, using lemmas 8.2.5 and 8.3.3, we know thatΓ0(u)∪Γd(u) is a block ifand only ifΓ is antipodal and thatΓ0(u)∪Γ2(u)∪·· ·∪Γη(u) is a block if and onlyif Γ is bipartite (ford > 2 andk> 2). By the argument above, these are the onlypossible non-trivial block systems that can occur in a distance-transitive graph.

Hence a distance-transitive graph (of diameter> 2, valency> 2) is imprimitiveif and only if it is bipartite or antipodal.

The most useful property of Smith’s Theorem is that it tells us a group-theoreticproperty ofΓ without making any reference to the automorphism group ofΓ what-soever. We finish this chapter by using it to ascertain whether certain graphs areprimitive or imprimitive.

Proposition 8.4.2The Hamming graphsH(d,n) are primitive forn> 2.

Proof:We check under what circumstancesH(d,n) is bipartite or antipodal. Recall fromsection 5.3 that the intersection array ofH(d,n) is

∗ 1 · · · j · · · d−1 d0 · · · · · · · · · · · · · · · d(n−2)

d(n−1) (d−1)(n−1) · · · (d− j)(n−1) · · · 1(n−1) ∗

.By 8.3.2,H(d,n) is bipartite if and only if the middle row ofι(H(d,n)) consists

only of zeroes. That is, for allj ∈ 0· · ·d, we have

d(n−1)− j− (d− j)(n−1) = 0⇔ nd−d− j−nd+d+n j− j = 0⇔ n j−2 j = 0⇔ n = 2.

Hence the only bipartite Hamming graphs areH(d,2), the d-dimensional cubesfrom section 5.4.

Checking antipodality is slightly more tricky. First, we notice thatH(d,2) is

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82 Primitive and Imprimitive Graphs

trivially antipodal (in the sense of 8.2.4) for everyd. By 2.3.8, we have

kd =b0b1 · · ·bd−1

c1c2 · · ·cd

=d(n−1)(d−1)(n−1) · · ·1(n−1)

1×2×·· ·×d

=d!(n−1)d

d!= (n−1)d.

So kd = 1 when(n− 1)d = 1, son− 1 = 1 and thusn = 2. Therefore thed-dimensional cubes are all antipodal. However no other Hamming graphs are an-tipodal: for n ≥ 3, we have verticesu,v,w with u = 00· · ·0, v = 11· · ·1 andw = 21· · ·1. Clearlyd(u,v) = d(u,w) = d, but d(v,w) = 1. Thus these graphsare not antipodal.

Consequently, by Smith’s Theorem the only imprimitive Hamming graphs areH(d,2) (they are both bipartite and antipodal). Therefore all other Hamminggraphs,H(d,n) wheren> 2, must be primitive.

Proposition 8.4.3The Johnson graphsJ(n,k,k−1) are primitive forn 6= 2k.

Proof:As with the Hamming graphs, we check ifJ(n,k,k−1) is bipartite and/or antipodal.Recall thatι(J(n,k,k−1)) is given by ∗ 12 · · · i2 · · · (k−1)2 k2

0 · · · · · · · · · · · · · · · nk−2k2

k(n−k) (k−1)(n−k−1) · · · (k− i)(n−k− i) · · · n−2k+1 ∗

.

For J(n,k,k−1) to be bipartite, we need to have the middle row of the inter-section array to be all zeroes. That is, for eachi ∈ 1, . . . ,k−1 we need

k(n−k)− i2− (k− i)(n−k− i) = 0

and (from the last column)nk− 2k2 = 0. This last condition impliesn = 2k, sosubstituting this gives us, for alli,

k(2k−k)− i2− (k− i)(2k−k− i) = 0⇔ k2− i2− (k− i)2 = 0⇔ 2ki−2i2 = 0⇔ i = k,

which is absurd. ThereforeJ(n,k,k−1) is never bipartite.In figure 3.2, we see thatJ(6,3,2) is trivially antipodal. The obvious question

is: does this apply to all Johnson graphs? The answer, in fact, is no. Consider

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8.4 Smith’s Theorem 83

two verticesu = u1, . . . ,uk andv = v1, . . . ,vk of J(n,k,k−1) and suppose thatthey are at the maximum possible distance, which isk. This occurs when the twosets are disjoint (soui 6= v j for all i, j). If n = 2k, u andv form a partition of thesetΩ, sov is the unique vertex at distancek from u and consequently the graphis trivially antipodal. However, ifn> 2k we have the vertexw = v1, . . . ,vk−1,x(wherex 6= ui and x 6= v j for all i, j). Clearly, d(u,v) = k and d(u,w) = k, butd(v,w) = 1 so the graph is not antipodal.

Using Smith’s Theorem, we can see that the only imprimitive Johnson graphsareJ(2k,k,k−1) (as they are antipodal), so all others must be primitive.

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Chapter 9

New Graphs from Old

In this chapter, we shall investigate some methods for constructing a new graphfrom a given one, and consider whether these new graphs are distance-transitive.

9.1 Line Graphs

The basic construction is simple enough:

Definition 9.1.1The line graph, L(Γ), of a connected graphΓ is the graph whose vertices corre-spond to the edges ofΓ, with two vertices ofL(Γ) being adjacent if the corre-sponding edges ofΓ are adjacent, i.e. if they are both incident with some vertex ofΓ. A graphΓ is a line graphif it is the line graph of some other graph.

This is best illustated with an example.

Example 9.1.2

Figure 9.1: Constructing the line graph ofL6

84

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9.1 Line Graphs 85

We start by deducing some basic properties of line graphs.

Proposition 9.1.3Let Γ be a graph andL(Γ) its line graph. Then:

1. |VL(Γ)|= |EΓ|;

2. |EL(Γ)|= ∑(

d(v)2

);

3. If Γ is k-regular, thenL(Γ) is (2k−2)-regular.

Proof:

1. This is immediate from the definition ofL(Γ).

2. The number of edges ofL(Γ) is the number of pairs of adjacent verticesof L(Γ), which is the number of pairs of adjacentedgesof Γ. Each vertex

v ∈ VΓ hasd(v) edges incident with it, giving

(d(v)

2

)pairs of mutually

adjacent edges passing throughv. Hence the result follows.

3. SupposeΓ is k-regular, and letuvbe an edge ofΓ. Then there arek−1 otheredges incident withu and also withv. Thus there are 2(k−1) edges ofΓadjacent touv, so the degree ofuv as a vertex ofL(Γ) is also 2(k− 1) =2k−2. HenceL(Γ) is (2k−2)-regular.

It should be noted thatΓ is cannot be uniquely determined byL(Γ). That is,there exist non-isomorphic graphs that have isomorphic line graphs. Consequently,the line graph is a “one-way” construction. An example of this is as follows:

Example 9.1.4The graphsK3 andK1,3 are clearly not isomorphic. However we haveL(K3) ∼=L(K1,3)∼= K3.

Figure 9.2: Two non-isomorphic graphs with the same line graph

However, all is not lost: H.R. Whitney showed that this is theonly such pair.We state this fomally below:

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86 New Graphs from Old

Theorem 9.1.5(Whitney 1932)Suppose we have non-isomorphic graphsΓ and∆ such thatL(Γ) ∼= L(∆). ThenΓ∼= K3 and∆∼= K1,3 (or vice-versa).

Proof:See Harary [24], Theorem 8.3.

9.2 Automorphisms of Line Graphs

In this project, an underlying theme has been investigating the automorphism groupsof graphs, so it is inevitable that we now consider the automorphism groups of linegraphs. It seems obvious that Aut(L(Γ)) should be related to Aut(Γ) in some way:as Aut(Γ) acts on the edges ofΓ, it has an induced action on the vertices ofL(Γ). Infact it turns out that, with four small exceptional cases, the two groups Aut(Γ) andAut(L(Γ)) are isomorphic. To prove this, we first need a definition and a lemma.

Definition 9.2.1A star in Γ is a subsetS⊆ EΓ such that each edge inS is incident with a commonvertex.

Lemma 9.2.2(Hemminger 1972)Supposeσ is a permutation of the edges ofΓ (i.e. the vertices ofL(Γ)). Thenσ isinduced by an automorphismσ∗ of Γ if and only if σ andσ−1 preserve stars.

Proof:See Hemminger [25] for the details.

We can now prove the desired result:

Theorem 9.2.3Let Γ be a connected, simple graph with automorphism group Aut(Γ), |VΓ| ≥ 3and with line graphL(Γ). Then

Aut(Γ)≤ Aut(L(Γ)).

Furthermore, ifΓ is notK4, K4 with an edge deleted orK4 with two adjacent edgesdeleted, then we have

Aut(Γ)∼= Aut(L(Γ)).

Proof:By construction, the vertices ofL(Γ) correspond to edges ofΓ, so can be labelled

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9.2 Automorphisms of Line Graphs 87

as unordered pairs of vertices ofΓ. For clarity, letG= Aut(Γ) andH = Aut(L(Γ)),and define a function

θ : G → Hg 7→ θ(g)

whereθ(g) has the following action on pairsu,v:

θ(g) : u,v 7→ (u)g,(v)g.

First, we will show that this is an injective homomorphism fromG to H.

• θ is well-defined: the pair(u)g,(v)g exists as an edge ofΓ (and thus as avertex ofL(Γ) sinceg is an automorphism ofΓ and thus preserves adjacency.

• θ preserves the group operation: chooseg,h∈G. Then

(u,v)θ(gh) = (u)gh,(v)gh= ((u)g)h,((v)g)h= ((u)g,(v)g)θ(h)= (u,v)θ(g)θ(h).

Henceθ(gh) = θ(g)θ(h).

• To showθ is injective is where we need that|VΓ| ≥ 3 (so that|VLΓ| =|EΓ| ≥ 2). We’ll show that Ker(θ) = eG.Supposeh ∈ Ker(θ). Then for allu,v ∈ VΓ, (u,v)θ(h) = (u,v)eH =u,v. (Note that if we have only two vertices, then the non-identity elementof G will fix the only possible pair: this is why we require|VΓ| ≥ 3.) Henceby the faithfulness ofH, we have thath= eG, so Ker(θ) = eG. By the FirstIsomorphism Theorem (Gallian [18] p.199),G∼= Im(θ)≤ H.

We now want to determine whenθ is also surjective, thus becoming an isomor-phism. Suppose that we have Im(θ) = φ ∈ H |φ = θ(g) ∀g ∈ G 6= H. That is,there exists someσ ∈ H that is not induced by an element ofG. By Lemma 9.2.2above,σ must not preserve stars inΓ. But becauseσ is an automorphism ofL(Γ),it preserves adjacency of vertices inL(Γ) (i.e. edges ofΓ), so this only happens ifσ maps the edges of a 3-starK1,3 to those of a triangleK3 (as subgraphs ofΓ).

Now, any edgeuv of Γ incident with the subgraphK3 must be adjacent toexactly two edges ofK3. So becauseσ is an automorphism,σ−1(uv) must beincident with exactly two edges ofK1,3, as shown below:

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88 New Graphs from Old

This gives the first of our three exceptional graphs.Furthermore, no edge can be incident with the other end ofuv if it is not inci-

dent withK3, as under the automorphismσ−1 it would have nowhere to go:

Hence we can only add another two more edges; adding one and then another givesthe second and third of the exceptional graphs.

Figure 9.3: The three exceptional graphs

So therefore, if Im(θ) 6= Aut(L(Γ)), it must be that (if|VΓ| ≥ 3), Γ is one of thethree graphs:K4, K4 with an edge deleted, orK4 with two adjacent edges deleted.So in all other cases, Im(θ)∼= Aut(Γ)∼= Aut(L(Γ)).

Theorem 9.2.3 was proved by G. Sabidussi in 1961; our proof follows thatgiven by Hemminger [26] in 1975.

9.3 Eigenvalues of Line Graphs

Line graphs also have interesting spectral properties. Recall (from 7.1.1) the defi-nition of the adjacency matrix of a graph. A related definition is this one.

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9.3 Eigenvalues of Line Graphs 89

Definition 9.3.1SupposeΓ is a graph with vertex setVΓ = v1, . . . ,vn and edge setEΓ = e1, . . . ,em.Then theincidence matrixM(Γ) is then×m matrix with entries

M i j =

1 if the edgeej is incident with the vertexvi

0 otherwise.

This is useful when working with the line graph ofΓ, as it involves edges aswell as vertices. In fact, we have the following result.

Lemma 9.3.2Let Γ be a simple graph withn vertices,medges and incidence matrixM(Γ) = M .Then the adjacency matrix ofL(Γ) is A(L(Γ)) = MTM −2I .

Proof:We have

(MTM)i j =n

∑a=1

(MT)ia(M)a j

=n

∑a=1

(M)ai(M)a j.

Now, for the summand we have

(M)ai(M)a j =

1 if the edgesei ,ej are both incident with the vertexva

0 otherwise.

If a given pair of edges ofΓ are adjacent they can meet at exactly one vertex;therefore ifi 6= j we have

(MTM)i j =

1 if ei is adjacent toej

0 otherwise.

However, if i = j, we are only dealing with a single edge, which is incident withtwo vertices. Hence(MTM)ii = 2.

Now let A = A(L(Γ)) be the adjacency matrix ofL(Γ). By definition, wehaveA i j = 1 if the verticeswi ,w j of L(Γ) are adjacent, that is if the correspondingedgesei ,ej of Γ are adjacent, i.e. if(MTM)i j = 1. Thus fori 6= j, we haveA i j =(MTM)i j . Also,A must have zeroes in all the diagonal entries, howeverMTM hasall diagonal entries equal to 2. So to allow for this, we just have to subtract 2I fromMTM to obtainA.

HenceA(L(Γ)) = MTM −2I .

A consequence of this is the following theorem about the spectrum of a linegraph.

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90 New Graphs from Old

Theorem 9.3.3If λ is an eigenvalue ofL(Γ), thenλ≥−2.

Proof:Let 〈 , 〉 denote the usual inner product inRm and letx be an eigenvector ofMTMwith corresponding eigenvalueµ. That is,

(MTM)x = µx.

So we have

xT(MTM)x = xT(µx)⇔ (Mx)TMx = µ(xTx)⇔ 〈Mx ,Mx〉 = µ〈x,x〉.

Since the inner product of a vector with itself is always non-negative, it fol-lows thatµ≥ 0. So we know that the eigenvalues ofMTM , i.e. the solutions ofdet(MTM −µI) = 0, are all≥ 0. (∗)

Now, if λ is an eigenvalue ofL(Γ), i.e. an eigenvalue ofMTM −2I , it is a so-lution of det(MTM −2I −λI) = det(MTM − (λ+2)I) = 0. So if we letµ= λ+2,then from(∗) we haveµ≥ 0; thereforeλ +2≥ 0 and henceλ≥−2.

The study of which graphs of least eigenvalue−2 are the line graphs of othergraphs is quite extensive: for example, Godsil & Royle [22] devote an entire chap-ter to the subject and its generalisations.

9.4 Distance-Transitive Line Graphs

The problem of “when is a line graph distance-transitive?” was addressed by Biggs[7] and is summarised by Holton & Sheehan [29]. Distance-transitive line graphsare very rare, as we shall see. First, we need some definitions.

Definition 9.4.1A (k,g)-cage is a simple graph of valencyk≥ 3 and girthg≥ 3 with the minimumpossible number of vertices.

We requireg≥ 3 because we want the graph to be simple. Also, we requirek≥ 3 because the casek = 2 is not very interesting and is usually ignored. Wecan always construct a graph of degree 2 and any girthg≥ 3 with minimal numberof vertices: these are just theg-circuitsCg. The notion of a cage was introducedby Tutte [37], when he investigated cubic cages. The table below gives a list of(3,g)-cages known to be unique. Many of these are graphs that are familiar to usalready.

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9.4 Distance-Transitive Line Graphs 91

Graph Common name/symbol |VΓ| Reference(3,3)-cage K4 4(3,4)-cage K3,3 6(3,5)-cage Petersen graphO3 10 3.5.1(3,6)-cage Heawood graph 14 11.2.1(3,7)-cage McGee graph 24 [38](3,8)-cage Tutte’s 8-cage 30 11.2.1(3,12)-cage Tutte’s 12-cage 126 [37]

Figure 9.4: Unique(3,g)-cages

The next result gives an idea of the size of a particular(k,g)-cage.

Proposition 9.4.2For anyk-regular graph of girthg (k,g≥ 3), definen0(k,g) to be

n0(k,g) =

1+ k((k−1)

12(g−1)−1)

k−2wheng is odd

2

((k−1)

12g−1

k−2

)wheng is even.

Thenn0(k,g) is a lower bound for|VΓ|.

Proof:First, supposeg is odd, so thereforeg = 2d + 1 for somed ≥ 1. Now, for somev∈VΓ, we can start to form a distance partition ofΓ. Since the girth ofΓ is 2d+1and the valency ofΓ is k, we have|Γ0(v)| = 1, |Γ1(v)| = k, |Γ2(v)| = k(k− 1)and so on until|Γd(v)|= k(k−1)d−1. (See the diagram, which illustrates the casek = 3, g = 7.)

Figure 9.5: Part of a (3,7)-cage

Hence we have|VΓ| ≥ 1+k+k(k−1)+ · · ·+k(k−1)d−1. On the right-hand side

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92 New Graphs from Old

is a finite geometric series, so we have

|VΓ| ≥ 1+k+k(k−1)+ · · ·+k(k−1)d−1

= 1+d−1

∑i=0

k(k−1)i

= 1+k((k−1)d−1)

k−2.

Substitutingd = 12(g−1), we obtain|VΓ| ≥ n0(k,g).

Secondly, supposeg is even, sog = 2d for somed ≥ 2. Choose some edgeuv∈ EΓ. Then at the verticesu andv, we can construct a tree in a similar fashionas above (k = 3, g = 8 is shown below) as far as the vertices distanced−1 from uandv.

Figure 9.6: Part of a (3,8)-cage

So we have|VΓ| ≥ 2+ 2(k− 1) + 2(k− 1)2 + · · ·+ 2(k− 1)d−1, another finitegeometric series. Thus

|VΓ| ≥ 2d−1

∑i=0

(k−1)i

= 2

((k−1)d−1

k−2

),

so substitutingd = 12g, we again obtain|VΓ| ≥ n0(k,g), as required.

Consequently, a(k,g)-cage must haveat least n0(k,g) vertices. In the casewhere this lower bound is actually achieved, we have the following:

Definition 9.4.3A (k,g)-cage with exactlyn0(k,g) vertices is called a(k,g)-Moore graph.1

1Some texts, e.g. [8], [22], only use the term Moore graph wheng is odd; if g is even, the termgeneralised polygonis used instead.

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9.4 Distance-Transitive Line Graphs 93

(For example, of the(3,g)-cages listed in figure 9.3, only the (3,7)-cage is not aMoore graph, sincen0(3,7) = 22< 24.)

There has been an extensive study of Moore graphs, especially concerning forwhich values ofk andg a Moore graph exists and whether such a graph is unique.We summarise various authors’ work in the next theorem.

Theorem 9.4.4

1. If a Moore graph with even girthg exists, theng∈ 4,6,8,12. Furthermore,such graphs exist when:

(a) g = 4, k ≥ 3: these are the complete bipartite graphsKk,k and areunique;

(b) g = 6,8,12,k = q+1 for some prime powerq.

2. If a Moore graph with odd girthg exists, theng∈ 3,4. Furthermore, suchgraphs exist when:

(a) g = 3, k≥ 3: these are the complete graphsKk+1 and are unique;

(b) g = 4, k = 3: the Petersen graphO3, which is unique;

(c) g = 5, k = 7: the Hoffman-Singleton graph, see [13] or [28], which isunique;

(d) g = 5, k = 57: ?

Proof:See Holton & Sheehan [29] for references.

The big question mark in 2(d) is not a typographical error! Whether a (57,5)-Moore graph exists is a long-standing and quite famous open problem in graphtheory. (Cases 2(b), (c), (d) are rather sporadic.) Note that the (3,6)-cage, (3-8)-cage and (3,12)-cage all fall into category 1(b), where the graphs are all related toprojective geometry; see [22] or [29] for an explanation.

This is all very interesting. But what does it have to do with line graphs ordistance-transitivity?

Theorem 9.4.5(Biggs 1974)Let Γ be a connected graph of valencyk≥ 3. Then if its line graphL(Γ) is distance-transitive,Γ is a(k,g)-Moore graph for somek≥ 3 andk≥ g.

Proof:See [7].

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94 New Graphs from Old

Consequently, all we have to do to identify a distance-transitive line graph isto consider each(k,g)-Moore graph, look at its line graph and check whether itis distance-transitive. The next two theorems are examples of when the answer ispositive.

Theorem 9.4.6The line graphs ofKn, for n≥ 3, are the Johnson graphsJ(n,2,1).

Proof:Label the vertices ofKn as 1,2, . . . ,n. The edges ofKn are all the possible un-ordered pairs of distinct integers chosen from 1,2, . . . ,n. Two edges are adjacentif and only if both are incident with a common vertex, i.e. if the pairs have anelement in common.

Therefore the vertices ofL(Kn) are all the possible 2-subsets chosen from1,2, . . . ,n, two subsetsx,y,z,w being adjacent if and only if|x,y∩z,w|=1.This is precisely the definition ofJ(n,2,1) (see chapter 3), which we know fromsection 3.2 to be distance-transitive.

As a special case, we see that the line graph of the tetrahedron (isomorphic toK4) is the octahedron (isomorphic toL(K4)∼= J(4,2,1)), a standard exercise aboutline graphs.

Theorem 9.4.7The line graphs ofKn,n are the Hamming graphsH(2,n).

WARNING: Don’t confuseH(2,n) (known as thelattice graphs) with H(k,2) (thek-cubes)!!

Proof:The vertex set ofKn,n has bipartitionV1,V2. Label the vertices ofKn,n in V1 as11,21, . . . ,n1 and those inV2 as 12,22, . . . ,n2. The edges ofKn,n are all the possiblepairs of the formi1 j2, that is all possibleorderedpairs of integers chosen from1,2, . . . ,n. Two edges ofKn,n are adjacent if and only if they are incident with acommon vertex, that is if the two pairs agree in the first co-ordinate or the secondco-ordinate.

Therefore the vertices ofL(Kn,n) are all the possible ordered 2-tuples of el-ements of the set1,2, . . . ,n, adjacent if and only if they agree in exactly oneco-ordinate (equivalently, if they differ in exactly one co-ordinate). This is pre-cisely the definition (see 5.1.1) of the Hamming graphH(d,n), with d = 2, whichwe know to be distance-transitive (by section 5.2).

For example,H(2,3) (see figure 5.1) is the line graph ofK3,3.

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9.5 Bipartite Doubles 95

9.5 Bipartite Doubles

This is another straightforward construction:

Definition 9.5.1Thebipartite double, D(Γ), of a graphΓ has vertex setV1∪V2, whereV1,V2 are twocopies ofVΓ, and a vertexv1 ∈V1 is adjacent to a vertexw2 ∈V2 if and only if thecorresponding verticesv,w∈VΓ are adjacent inΓ.

By construction, thisD(Γ) is bipartite and has double the number of verticesand edges asΓ (hence the name). Also, ifv1,v2 ∈VD(Γ) are the vertices ofD(Γ)corresponding tov∈VΓ, then deg(v1) = deg(v2) = deg(v). Thus ifΓ is k-regularthen so isDΓ. Enough theory-now for an example.

Examples 9.5.2(i) The bipartite double of the tetrahedron (K4) is the cube (Q3). (The labels areused to help us keep track of what’s happening.)

(a)K4 (b) D(K4)

(c) Q3

Figure 9.7:K4 and its bipartite double,Q3

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96 New Graphs from Old

(ii) The bipartite double ofK3,3 is two copies ofK3,3.

Figure 9.8:K3,3 (left) andD(K3,3) (right)

What we observed in the example ofK3,3 is not unusual, as we have this nextresult.

Proposition 9.5.3The bipartite double,D(Γ), of a connected graphΓ is connected if and only ifΓ isnot bipartite. Furthermore, ifΓ is bipartite, thenD(Γ) is isomorphic to two copiesof Γ.

Proof:SupposeΓ is bipartite. ThenVΓ = U∪W, soVD(Γ) = (U1∪W1)∪(U2∪W2), whereU1∼= U2

∼= U and likewise forW. The only edges ofΓ are betweenU andW, sotherefore the only edges ofD(Γ) are betweenU1 andW2 and betweenU2 andW1,as shown in the diagram.

Figure 9.9: Schematic for 9.5.3, part 1

ThereforeD(Γ) has two components, as there is no path from any vertex inU1 toW1, or fromU2 to W2, etc. BecauseU1

∼= U2 andW1∼= W2, it is clear that these two

components are each isomorphic toΓ.Now supposeΓ is not bipartite, and thatVD(Γ) = V1∪V2. ThenΓ contains an

odd circuit, sayC, of length 2m+ 1. HenceD(Γ) contains a circuitC of length4k+2, which contains vertices in bothV1 andV2. (The casek = 1 is shown below.)

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9.5 Bipartite Doubles 97

Figure 9.10: Schematic for 9.5.3, part 2

If Γ contains no other vertices than those inC, then we are done. So suppose not,i.e. that there existsx∈VΓ such thatx /∈C. BecauseΓ is connected, there is a pathπ = xab· · ·v that joinsx to a vertex ofC, say (WLOG)v. Then inD(Γ) we havethe path

π1 =

x1a2b1 · · ·v2 if π has odd lengthx1a2b1 · · ·v1 if π has even length.

Therefore any vertex inV1 is connected to the circuitC, and by symmetry everyvertex inV2 is also. Thus given any two vertices ofD(Γ), we can form a path fromone to the other (viaC, although there may well be other, shorter ones). HenceD(Γ) is connected.

Relating this to distance-transitive graphs, a slightly negative consequence ofthe above result is that the bipartite double of a distance-transitive graph need notitself be distance-transitive: it may not even be connected. Worse still, even somenon-bipartite distance-transitive graphs have non-distance-transitive bipartite dou-bles, as we see below.

Example 9.5.4The bipartite double of the octahedron is not distance-transitive, as it does not havea well-defined intersection array.

Figure 9.11: The octahedronOct and its bipartite doubleD(Oct)

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98 New Graphs from Old

If Γ = D(Oct), then we have|Γ1(1)∩Γ1(6)| = 4, but |Γ1(1)∩Γ1(2)| = 2, so theintersection numbers are dependent on the choice of vertices and we cannot writedown an intersection array. HenceD(Oct) cannot be distance-transitive, or evendistance-regular.

To end this chapter on a positive note, there is one family of graphs familiarto us whose bipartite doubles are all distance-transitive. There are the odd graphswe met in chapter 3. Brouwer, Cohen & Neumaier [11] claim that to show thesegraphs are distance-transitive is “trivial” and “straightforward”. I have to disagree.

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Chapter 10

Bounding the Diameter

10.1 Introduction

For any integerk≥ 2, there are only finitely many finite distance-transitive graphsof valencyk. This was shown by Cameron [12] and Weiss [39], by showing thatfor any suchk, the diameter of ak-valent distance-transitive graph is bounded bysome function ofk. However, for small values ofk (such ask = 3 or 4), the earliercase-specific work produced more manageable bounds. It is these that will weinvestigate first.

10.2 Cubic Graphs

In this section, we always assume thatΓ is cubic.

Now, any distance-transitive graph iss-arc-transitive for somes≥ 1. Tutte [37]proved a number of results regarding cubics-arc-transitive graphs, which can thenbe used to place a bound on the diameter of cubic distance-transitive graphs, as weshall see. We roughly follow the treatment given in Holton & Sheehan [29]. First,we need the following definition.

Definition 10.2.1Let π = (u0, . . . ,us) be ans-arc ofΓ, and choose any vertexw adjacent tous otherthanus−1. Then thes-arcπ∗ = (u1, . . . ,us,w) is called asuccessorof π.

In general, it is not particularly straightforward to tell if a given graph iss-arc-transitive or not. However, using successors we have the following result.

Lemma 10.2.2(Tutte 1947)Let π = (u0,u1, . . . ,us) be ans-arc of a connected graphΓ. Suppose that for everysuccessorπ∗ of π, there is someg∈ Aut(Γ) such that(π)g = π∗. ThenΓ is s-arc-transitive.

99

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100 Bounding the Diameter

Proof:Let Aπ be the set of alls-arcs ofΓ that are the image ofπ under an automorphismof Γ, i.e. Aπ = (π)g|g ∈ Aut(Γ). Choose someρ = (v0,v1, . . . ,vs) ∈ Aπ, soρ = (π)h for someh∈Aut(Γ). and letρ∗ = (v1,v2, . . . ,vs,vs+1) be some successorof ρ.

Then because(ρ)h−1 = π, we have(ρ∗)h−1 = (u1,u2, . . . ,us,(vs)h−1), a suc-cessor ofpi. So by assumption there exists someg ∈ Aut(Γ) satisfying(π)g =(ρ∗)h−1. Henceρ∗ = (π)hg, so thereforeπ∗ ∈ Aπ.

We can then repeat this argument to show that a successor ofπ∗ is in Aπ, anda successor of thats-arc is inAπ also, and so on. BecauseΓ is connected, this willeventually coverall s-arcs ofΓ, so they must all be elements ofAπ. Hence everys-arc ofΓ is the image ofπ under an automorphism ofΓ, soΓ is s-arc-transitive.

This result allows us to prove the next one, which applies specially to the cubiccase.

Lemma 10.2.3(Tutte 1947)Let Γ be a strictlys-arc-transitive, cubic graph (for somes), with π,ρ somes-arcsin Γ. Then there is aunique g∈ Aut(Γ) such that(π)g = ρ.

Proof:Suppose, for a contradiction, that we haveg1,g2 ∈ Aut(Γ), g1 6= g2, such that(π)g1 = ρ = (π)g2. Then (π)g1g−1

2 = π, and sinceg1g−12 6= e, we have a non-

identity element, sayh, of Aut(Γ) that fixesπ.Let π = (u0,u1, . . . ,us). SinceΓ is cubic,π has exactly two successors, say

(u1, . . . ,us,x) and(u1, . . . ,us,y). By symmetry,π is itself the successor of the twos-arcs(z,u0, . . . ,us−1) and(w,u0, . . . ,us−1). So we have the following situation:

Figure 10.1: Schematic for 10.2.3

We can assume thath does not fix the successors ofπ, so WLOG assume that(x)h = y. Now, becauseΓ is s-arc-transitive, there existsf ∈ Aut(Γ) such that(z,u0, . . . ,us−1) f = (u0, . . . ,us). Hence(us) f ∈ x,y. Again WLOG, let(us) f =x. So we now have(z,u0, . . . ,us) f = (u0, . . . ,us,x), and also(z,u0, . . . ,us) f h =(u0, . . . ,us,x)h = (u0, . . . ,us,y). That is, the group elementsf and f h map an(s+ 1)-arc to both of its successors. Hence by 10.2.2 above,Γ is (s+ 1)-arc-transitive, contradicting our supposition thatΓ wasstrictly s-arc-transitive.

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10.2 Cubic Graphs 101

Therefore our initial assumption must be false, so there must be a uniqueg∈ Aut(Γ) satisfying(π)g = ρ.

We can now us this to determine the size of the automorphism group of a cubicgraphΓ.

Lemma 10.2.4(Tutte 1947)SupposeΓ is connected, strictlys-arc-transitive cubic graph (fors≥ 1), with girthg≥ 3. ThenΓ contains exactly|VΓ|.3.2s−1 s-arcs and|Aut(Γ)|= |VΓ|.3.2s−1.

Proof:Sinceg≥ 2s−2 (by 2.1.3) andg≥ 3, we gets≤ 1

2(g+ 2) < g, so therefore anys-arc is ans-path (it can’t have any repeated vertices as that would give a circuit oflength< g).

Now let S be the set of alls-arcs inΓ. By 10.2.3 above,|S| = |Aut(Γ)| (i.e.there are as many automorphisms ofΓ as there ares-arcs inΓ). SinceΓ is cubic,from each vertex we can form 3.2s−1 s-arcs (which are alls-paths). Also,Γ isvertex-transitive, so we can move each of these paths to start from any of the|VΓ|vertices.

Thus|S|= |VΓ|.3.2s−1 = |Aut(Γ)|.

The next lemma is a direct consequence of the previous one.

Lemma 10.2.5(Tutte 1947)SupposeΓ is as in 10.2.4 and letG = Aut(Γ). Then for any vertexv∈VΓ,

|StabG(v)|= 3.2s−1.

Proof:SinceΓ is s-arc-transitive, it is also vertex-transitive, so for anyv∈VΓ, OrbG(v) =VΓ (from 1.2.4). By the Orbit-Stabiliser Theorem (1.2.8), we have

|StabG(v)|= |G||OrbG(v)|

=|VΓ|.3.2s−1

|VΓ|= 3.2s−1 .

The next step is the crucial one. It is another of Tutte’s results [37]; the proofis quite long. So we shall just state the result here.

Theorem 10.2.6(Tutte 1947)For anys-arc-transitive cubic graph,s≤ 5.

Proof:See [37], or a version using more modern terminology can be found in Biggs [8].

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102 Bounding the Diameter

We return to the domain of distance-transitive graphs with this corollary toTutte’s theorem, due to Biggs & Smith [10].

Corollary 10.2.7 (Biggs & Smith 1971)For any cubic distance-transitive graphΓ, eachki (the size ofΓi , a cell of a distancepartition) is a divisor of 48.

Proof:Fix somev∈VΓ, and letH = StabG(v). By 2.2.4,H acts transitively onΓi , so eachΓi is an orbit ofH. By the Orbit-Stabiliser Theorem (1.2.8), the size of an orbit ofH divides the order ofH, i.e. |Γi |

∣∣|H|, or ki∣∣3.2s−1, for s= 1,2,3,4,5.

Henceki∣∣48.

So we have a bound on each of theki . We now convert this into a bound onthe diameter ofΓ, which we shall denote byd. If Γ is distance-transitive, we canobviously find its intersection arrayι(Γ). BecauseΓ is cubic, by 2.3.5 the sum ofeach column ofι(Γ) is 3. Thus these columns (apart from the first and the last)must take one of these three forms:

Type A:102

Type B:111

Type C:201

We’ll denote theith column ofι(Γ) by pi . By 2.3.6, we know that the top row ofthe array must be increasing and the bottom row must be decreasing, so we canonly have so many columns of type A, then so many of type B and finally some oftype C.

Lemma 10.2.8(Biggs & Smith 1971)p1, . . . ,p6 cannot all be of type A. Also,p1, . . . ,p5 cannot all be of type A withp6

of type B.

Proof:By the recurrence relation in 2.3.7, we have

ki =ki−1bi−1

ci.

If p1, . . . ,p6 are all of type A, or ifp1, . . . ,p5 are of type A withp6 of type B (notethatk0 = 1 andb0 = 3), we havek1 = 3, k2 = 6, k3 = 12,k4 = 24 andk5 = 48 (alldivisors of 48), butk6 = 96 which does not divide 48. Hence the result follows.

Supposep j is the first column that is not of type A. Then we have two possiblecases to consider:

1. p j is of type C, with 1≤ j ≤ 6;

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10.2 Cubic Graphs 103

2. p j is of type B, with 1≤ j ≤ 5.

Lemma 10.2.9(Biggs & Smith 1971)In case 1,d< 2 j. In case 2,d< 3 j.

Proof:

1. Supposed≥ 2 j. Then choose verticesv∈ Γ j(u) andw∈ Γ2 j(u), so we haved(u,v) = d(v,w) = j.

Figure 10.2: Schematic for 10.2.9, part 1

Sincep j is of type C, we can form two pathsπ,ρ of length j from u to v, sowe can form a circuit of length 2j containingu andv.

SinceΓ is distance-transitive andd(u,v) = d(v,w), v,w must also lie in acircuit of length 2j. But this is impossible, as there is only one vertex inΓ j+1(u) adjacent tov. Sod< 2 j.

2. Supposed≥ 3 j. Then choose verticesv∈ Γ j(u), w∈ Γ2 j(u) andx∈ Γ3 j(u),so we haved(u,v) = d(v,w) = d(w,x) = j.

Figure 10.3: Schematic for 10.2.9, part 2

Now p j is of type B, sou,v lie in a circuitσ of length 2j + 1. By distance-transitivity, v,w must also lie in such a circuit, sayτ. Thusp2 j must be oftype C (asw is adjacent to two vertices inΓ2 j−1(u). Also, w,x should lie ina circuit of length 2j + 1, but by the same arguments as in part 1 above, thisis impossible. Henced< 3 j.

So we have laid all the groundwork for this next result, the target of this section.

Theorem 10.2.10(Biggs & Smith, 1971)In all cases,d< 15.

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104 Bounding the Diameter

Proof:In case 1,d< 2 j and 1≤ j ≤ 6. Sod< 12< 15.In case 2,d< 3 j and 1≤ j ≤ 5. Sod< 15.

We are now in a position to carry out what Cameron ([12], [13]) refers to asSmith’s programfor the case of cubic distance-transitive graphs. We will do this inthe next chapter.

10.3 Tetravalent Graphs

A tetravalentgraph is a graph of valency 4.

After classifying all distance-transitive graphs of valency 3 (cubic graphs), thisis the obvious next step. However, things aren’t quite so straightforward. First, thisis because Tutte’s results on cubics-arc-transitive graphs don’t generalise easilyto higher valencies (for example, 10.2.3 relies onΓ being cubic). A number ofresearchers in the late 1960s and early 1970s worked on this problem, culminatingin the following results of Gardiner [19], which are analagous to 10.2.6; see alsoSmith [34].

Theorem 10.3.1(Gardiner 1973)Let p be a prime, and letΓ be a connecteds-arc-transitive graph of valencyp+ 1.Thens∈ 1,2,3,4,5,7.

Also, this time analagous to 10.2.5, we have:

Theorem 10.3.2(Gardiner 1973)Let Γ be as above, and letG = Aut(Γ). Then for anyv∈VΓ,

|StabG(v)| ≤ (p+1)ps−1(p−1)2 for s∈ 4,5,7,|StabG(v)| ≤ (p+1)!p! for s∈ 1,2,3.

Consequently, in the special case we’re interested in (wherep + 1 = 4, sop = 3), we have:

Corollary 10.3.3For a tetravalent,s-arc-transitive graphΓ, with automorphism groupG = Aut(Γ)and anyv∈VΓ,

|StabG(v)| ≤ 24.36.

Proof: For the casep = 3, by 10.3.2 we have(p+ 1)!p! = 4!.3! = 24.32, and(p+1)ps−1(p−1)2 = 4.3s−1.22. By 10.3.1 we haves≤ 7, so combining these tworesults, we obtain

|StabG(v)| ≤ 24.3s−1≤ 24.36.

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10.3 Tetravalent Graphs 105

From now on in this section, we always assume thatΓ is tetravalent.

We now want to convert this bound on the order of the stabiliser of a vertexinto a bound on the diameter, as we did in the previous section in the cubic case.Analagous to 10.2.7, we again quote the Orbit-Stabiliser Theorem (1.2.8) to showthatki divides|StabG(v)|, so thereforeki ≤ 24.36.

BecauseΓ is tetravalent, the sum of each column ofι(Γ) is 4, so (apart fromthe first and last column) the columns ofι(Γ) are of the following types:

Type A:103

Type B:112

Type C:121

Type D:202

Type E:211

Type F:301

There are some immediate restrictions on the way the columns appear in an inter-section array. For example, the top row must be increasing and the bottom rowdecreasing (by Theorem 2.3.6), so the columns must appear in the order above.Also, not all of these column types can occur in the same graph: again because ofTheorem 2.3.6, we cannot have both of types C and D in the same array.

First, we’ll consider the case when the graph has girth 3 (as Smith did in [33]).

Theorem 10.3.4(Smith 1973)SupposeΓ is a tetravalent distance-transitive graph with girth 3 and diameterd.Thend≤ 8.

Proof:We know that the entrya1 is non-zero. Ifa1 = 3, we must have the complete graphK5, which has diameter 1. Ifa1 = 2, we can construct the following:

Figure 10.4: Schematic for 10.3.4

Because each edge must lie in exactly two triangles, it follows that we have to in-clude edgesv3v6, v4v6 andv5v6, in which case the graph obtained is the octahedron,which has diameter 2. The casea1 = 1 is more tricky, however Smith [33] showedthat in this case the diameter of the graph isd≤ 8.

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106 Bounding the Diameter

Now suppose the girth is greater than 3. Thena1 = 0, so the first column mustbe of type A. As in section 10.2, letpi denote theith column ofι(Γ). By theformulakibi = ki+1ci+1, we have the following:

1. If pi is of type A andpi+1 is of types A, B or C, thenki+1 = 3ki .

2. If pi is of type B andpi+1 is of types B or C, thenki+1 = 2ki .

3. If pi is of type A andpi+1 is of types D or E, thenki+1 = 32ki .

4. If pi is of type A andpi+1 is of type F, ifpi is of type B andpi+1 is of typesD or E, if pi andpi+1 are both of types C or D, or ifpi is of type D andpi+1

is of type E, thenki+1 = ki .

5. If pi is of type C or E andpi+1 is of type E, or if Ifpi is of types B, C, D, E,or F andpi+1 is of type F, thenki+1 < ki .

Loosely, this means that for column types A and B, the graph is ‘growing’ (i.e.the numberski are increasing), for column types C and D, the graph is ‘static’ (i.e.the numberski are constant) and for column types E and F, the graph is ‘contract-ing’ (i.e. the numberski are decreasing). Thus it is sufficient to find an upper boundfor the numberj, wherep j is the last column of type C or D, as this will enforcean upper bound on the diameter.

Now, the intersection array is one of the following four types:

1. ι(Γ) =

∗ 1 · · · 1 2 · · · 2 · · ·0 0 · · · 0 0 · · · 0 · · ·4 3 · · · 3 2 · · · 2 · · · ∗

2. ι(Γ) =

∗ 1 · · · 1 1 · · · 1 2 · · · 2 · · ·0 0 · · · 0 1 · · · 1 0 · · · 0 · · ·4 3 · · · 3 2 · · · 2 2 · · · 2 · · · ∗

3. ι(Γ) =

∗ 1 · · · 1 1 · · · 1 · · ·0 0 · · · 0 2 · · · 2 · · ·4 3 · · · 3 1 · · · 1 · · · ∗

4. ι(Γ) =

∗ 1 · · · 1 1 · · · 1 1 · · · 1 · · ·0 0 · · · 0 1 · · · 1 2 · · · 2 · · ·4 3 · · · 3 2 · · · 2 1 · · · 1 · · · ∗

Note that although there is necessarily a column of type A if the girth is greaterthan 3, the others types do not necessarily occur. So we now have:

Theorem 10.3.5(Smith 1974)SupposeΓ is a tetravalent distance-transitive graph with girth> 3 and diameterd.

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10.4 Extending to Higher Valencies 107

Thend≤ 29.

Proof:We consider the four cases listed above.

1. Let ps andps+t be the first and last columns of type D. By 10.3.3 above,we haveks≤ 24.36, so this impliess≤ 8. Smith [35] used quite a lengthygraph-theoretic argument to showt ≤ s, so therefores+ t ≤ 16, andks+t+1

must divide 2.37.

Now, for s+ t +1≤ i ≤ d−1, we haveki+1 = 12ki , ki+1 = 1

3ki or ki+1 = 14ki ,

sod− (s+ t +1)≤ 8 and therefored≤ 25.

2. Let pr be the first column of type B, andps, pt be as in case 1. Again by10.3.3, we havekr ≤ 24.36, so because there must be at least one column oftype A beforepr , we have 2≤ r ≤ 8. Using similar graph-theoretic argu-ments to those in case 1, Smith [35] showed thats+ t < 17.

Again, we have that fors+ t + 1 ≤ i ≤ d− 1, ki+1 = 12ki , ki+1 = 1

3ki orki+1 = 1

4ki . This then leads on to imply thatd< 29.

3. This time, letps andps+t be the first and last columns of type C. Similar tocase 1, due to Smith [35] we haves≤ 8 andt < s, sos+ t ≤ 16. This time,however, we haveks+t+1 divides 22.37, so it follows thatd≤ 26.

4. Similar to case 2, we letpr be the first column of type B, andps, pt be asin case 3. Again we have 2≤ r ≤ 8 andt < r. By further graph-theoreticarguments, we haves+ t < 16, and that fors+ t +1≤ i ≤ d−1, ki+1 = 1

2ki ,ki+1 = 1

3ki or ki+1 = 14ki . Consequently, this givesd< 27.

Combining all these cases together, we can conclude that in all cases,d≤ 29.

So we can now carry out Smith’s program for the case of tetravalent graphs,which we will do in the next chapter.

10.4 Extending to Higher Valencies

The major stumbling block in extending the arguments we have seen in the previ-ous two sections to graphs of arbitrary valencyk was finding a bound on the orderof the stabiliser of a vertex. However, in 1983 Cameronet al [14] published a proofof the following result.

Theorem 10.4.1– The Sims Conjecture1

SupposeG is a finite permutation group actingprimitivelyon a setΩ, and for somex ∈ Ω, StabG(x) has an orbit of sizek (other thanx). Then|StabG(x)| ≤ f (k),

1It’s still referred to as a conjecture, even though it has been proved for nearly 20 years!

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108 Bounding the Diameter

where f (k) is an integer.

Proof:See [14].

This proof is very deep, as it is dependent on the celebrated Classification ofFinite Simple Groups and uses the O’Nan-Scott Theorem (see [13]), which is an-other major achievement of late-20th century group theory. Translated into ourterms, we haveΩ = VΓ, G = Aut(Γ), x a vertex ofΓ and the orbit of sizek beingΓ1(x), whose size is, of course, the valency ofΓ. It is also important to notice thatthe theorem only refers toprimitivepermutation groups. As we know from chapter8, many distance-transitive graphs are imprimitive2, so the next theorem does notfollow immediately.

Theorem 10.4.2(Cameron 1982)There are only finitely many finite distance-transitive graphs of given valencyk> 2.

Proof:See [12].

Cameron’s argument roughly follows those in the previous two sections, ini-tially making use of the assumption thatΓ is primitive to show that because theorder of the stabiliser is bounded (by 10.4.1) then the diameter ofΓ is bounded.Consequently there can only be a finite number of such graphsΓ. However, ifΓ isimprimitive, then it is bipartite or antipodal (recall Theorem 8.4.1). Using a con-struction of Smith called thederived graph(see [32]) (for the antipodal case) andanother construction known as thedistance-two graph(for the bipartite case), itis possible to reduce an imprimitive graph to a primitive one. Cameron then goeson to agrue that if there were infinitely many finite imprimitive distance-transitivegraphs, then there would have to be infinitely many primitive ones, which con-cludes the proof.

The other important missing detail is what the value of the bound on the diameteractually is. Unforunately, it’s not a very practical number for performing calcula-tions with. This result can be found in Brouwer, Cohen & Neumaier [11]:

Theorem 10.4.3The diameter of a distance-transitive graph of valencyk≥ 3 is at most(k6)!22k.

Proof:See [11].

2In fact, in the casesk = 3 andk = 4, most are imprimitive: see chapter 11

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10.4 Extending to Higher Valencies 109

The order of magnitude of this bound is beyond comprehension. For example,in the casek = 3, we have(36)!26, which fills up twenty lines of MAPLE output!Obviously, this doesn’t compare very favourably to Biggs & Smith’s original bound(see 10.2.10) of 15.

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Chapter 11

Graphs of Low Valency

11.1 Smith’s Program

Smith’s program is the method originally used to determine all distance-transitivegraphs of a given valency, first used by Biggs and Smith [10] to find all cubicdistance-transitive graphs, and then by Smith ([33],[35],[36]) to find the tetravelantones.. It is something of a ‘brute force’ method; later methods, such as those ofGardiner [20], use more algebraic techniques.

Smith’s program works like this. First, use the required valencyk to find abound for the order of the stabiliser of a vertex. Second, use this bound to find anupper boundD for the diameter. (This, of course, is what we did in chapter 10.)Next, use a computer to find all the(D + 1)× (D + 1) tridiagonal matrices thatare feasible in the sense of 7.4.1. Finally, determine which of these matrices arerealisable (as defined in 7.4.2). The number of redundant matrices is, apparently,very small.

In this chapter, we give the results of applying Smith’s program to the cases ofcubic and tetravalent graphs.

11.2 Cubic Distance-Transitive Graphs

Using Theorem 10.2.10, we know that the diameter of a cubic distance-transitivegraph is less than 15. Biggs and Smith [10] then applied Smith’s program, obtain-ing the twelve graphs listed below. Gardiner [20] later obtained this classificationusing group-theoretic methods and without using a computer. (This presentationcontains material from a number of sources, such as Smith’s 1974 survey article[34], Biggs’ investigation of the primitive examples [6] and Gardiner’s classifica-tion [20]).

110

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11.2 Cubic Distance-Transitive Graphs 111

Theorem 11.2.1(Biggs & Smith 1971, Gardiner 1975)The only cubic distance-transitive graphs are the twelve listed below.

1. The complete graphK4

Number of vertices: 4Diameter: 1Girth: 3Automorphism group: S4

Primitive

Intersection array:

∗ 10 23 ∗

(See figure 1.1)

2. The complete bipartite graphK3,3

Number of vertices: 6Diameter: 2Girth: 4Automorphism group: S3Wr Z2

Bipartite and antipodal

Intersection array:

∗ 1 30 0 03 2 ∗

(See figure 8.3)

3. The Petersen graphO3 (also known as: (3,5)-cage)Number of vertices: 10Diameter: 2Girth: 5Automorphism group: S5

Primitive

Intersection array:

∗ 1 10 0 23 2 ∗

(See figure 3.3)

4. The cubeQ3Number of vertices: 8Diameter: 3Girth: 4Automorphism group: Z2Wr S3

Bipartite and antipodal

Intersection array:

∗ 1 2 30 0 0 03 2 1 ∗

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112 Graphs of Low Valency

Figure 11.1: The cubeQ3

5. The Heawood graphH (also known as: (3,6)-cage)Number of vertices: 14Diameter: 3Girth: 6Automorphism group: PGL(2,7), see [20]Bipartite, not antipodal (see 8.2.3)

Intersection array:

∗ 1 1 30 0 0 03 2 2 ∗

Figure 11.2: The Heawood graph

6. The Pappus graphNumber of vertices: 18Diameter: 4Girth: 6Bipartite and antipodal

Intersection array:

∗ 1 1 2 30 0 0 0 03 2 2 1 ∗

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11.2 Cubic Distance-Transitive Graphs 113

Figure 11.3: The Pappus graph

7. The Coxeter graphNumber of vertices: 28Diameter: 4Girth: 7Automorphism group: PGL(2,7), see [6]Primitive

Intersection array:

∗ 1 1 1 20 0 0 1 13 2 2 1 ∗

Figure 11.4: The Coxeter graph

8. Tutte’s 8-cageTNumber of vertices: 30Diameter: 4Girth: 8Automorphism group: PGL(2,9), see [20]Bipartite

Intersection array:

∗ 1 1 1 30 0 0 0 03 2 2 2 ∗

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114 Graphs of Low Valency

Figure 11.5: Tutte’s 8-cage

9. The dodecahedronNumber of vertices: 20Diameter: 5Girth: 5Automorphism group: S5

Antipodal

Intersection array:

∗ 1 1 1 2 30 0 1 1 0 03 2 1 1 1 ∗

Figure 11.6: The dodecahedron

10. The Desargues graphD(O3): the bipartite double of the Petersen graphNumber of vertices: 20Diameter: 5Girth: 6Automorphism group: S5×Z2

Bipartite and antipodal

Intersection array:

∗ 1 1 2 2 30 0 0 0 0 03 2 2 1 1 ∗

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11.2 Cubic Distance-Transitive Graphs 115

Figure 11.7: The Desargues graph

11. The Biggs-Smith graphNumber of vertices: 102Diameter: 7Girth: 9Automorphism group: PSL(2,17), see [6]Primitive

Intersection array:

∗ 1 1 1 1 1 1 30 0 0 0 1 1 1 03 2 2 2 1 1 1 ∗

(See below.)

12. The Foster graphNumber of vertices: 90Diameter: 8Girth: 10Automorphism group: see [32]Bipartite and antipodal

Intersection array:

∗ 1 1 1 1 2 2 2 30 0 0 0 0 0 0 0 03 2 2 2 2 1 1 1 ∗

This graph is described in Smith [32]: it can be constructed from Tutte’s8-cage.

ExcludingK4, the three primitive graphs (3, 7 and 11 above) were singled outby Biggs [6] as worthy of special atttention: he describes them as being “remark-able”. (Whether this is a formal definition rather than just an adjective is unclear.)They can each be constructed in a similar way.

We now give the construction for the Biggs-Smith graph, as originally given inBiggs & Smith [10]. Start with four 17-gons with verticesa0, . . . ,a16, b0, . . . ,b16,c0, . . . ,c16, d0, . . . ,d16 and edgesaiai+1, bibi+2, cici+4, didi+8 (all modulo 17).Then take seventeen copies of the “H”-configuration below:

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116 Graphs of Low Valency

Figure 11.8: “H”-configuration

The verticesai ,bi ,ci ,di are identified with those in the four 17-gons. Then thegraph below is obtained:

Figure 11.9: The Biggs-Smith graph

Similar constructions can be given for the Coxeter graph (using three heptagonsand seven copies of a “Y”-configuration), and for the Petersen graph (using twopentagons and five copies ofK2). These can be found in Biggs & Smith [10] orHolton & Sheehan [29]. Biggs [6] investigated several properties of these threeremarkable graphs, such as vertex- and edge-colourings, embeddings in orientablesurfaces and spectral properties.

In 1986, Biggs, Boshier & Shawe-Taylor [9] extended the classification to

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11.3 Tetravalent Distance-Transitive Graphs 117

determine all cubic distance-regular graphs. They showed that there is only onecubic graph that is distance-regular but not distance-transitive, this being Tutte’s(3,12)-cage (see [37]).

11.3 Tetravalent Distance-Transitive Graphs

Using the results of section 10.3, the maximum diameter of a tetravalent distance-transitive graph is 29. Applying Smith’s program to all tridiagonal matrices withthe appropriate column types, and analysing which feasible matrices are realisable,yields the fifteen graphs listed below. This was done by Smith: see [33], [35], [36].Later, an algebraic proof of this classification was obtained: see [11].

Theorem 11.3.1(Smith 1974)The only tetravalent distance-transitive graphs are the fifteen listed below.

1. The complete graphK5

Number of vertices: 5Diameter: 1Girth: 3Automorphism group: S5

Primitive

Intersection array:

∗ 10 34 ∗

(See figure 8.1)

2. The complete bipartite graphK4,4

Number of vertices: 8Diameter: 2Girth: 4Automorphism group: S4Wr Z2

Bipartite and antipodal

Intersection array:

∗ 1 40 0 04 3 ∗

Figure 11.10: The complete bipartite graphK4,4

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118 Graphs of Low Valency

3. The octahedron (also known as: Johnson graphJ(4,2,1), line graph ofK4)Number of vertices: 6Diameter: 2Girth: 3Automorphism group: S4×Z2

Antipodal

Intersection array:

∗ 1 40 2 04 1 ∗

(See figure 2.6)

4. L(K3,3): line graph ofK3,3 (also known as: Hamming graphH(2,3))Number of vertices: 9Diameter: 2Girth: 3Automorphism group: S3Wr Z2

Primitive

Intersection array:

∗ 1 20 1 24 2 ∗

(See figure 5.1)

5. The line graph of the Petersen graph,L(O3)Number of vertices: 15Diameter: 3Girth: 3Automorphism group: S5

Antipodal

Intersection array:

∗ 1 1 40 1 2 04 2 1 ∗

Figure 11.11: The line graph of the Petersen graph,L(O3)

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11.3 Tetravalent Distance-Transitive Graphs 119

6. The line graph of the Heawood graph,L(H)Number of vertices: 21Diameter: 3Girth: 3Automorphism group: PGL(2,7)Primitive

Intersection array:

∗ 1 1 20 1 1 24 2 2 ∗

Figure 11.12: The line graph of the Heawood graph,L(H)

7. K5,5 minus a perfect matching: see [36].Number of vertices: 10Diameter: 3Girth: 4Automorphism group: S5×Z2

Bipartite and antipodal

Intersection array:

∗ 1 3 40 0 0 04 3 1 ∗

Figure 11.13:K5,5 minus a perfect matching

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120 Graphs of Low Valency

8. Distance-three graph of the Heawood graphH: see [36].Number of vertices: 14Diameter: 3Girth: 4Bipartite

Intersection array:

∗ 1 2 40 0 0 04 3 2 ∗

This graph, denoted byH3, is constructed from the Heawood graph, by tak-ing the same 14 vertices asH, and joiningx,y in H3 if d(x,y) = 3 in H.

Figure 11.14:H3: the distance-three graph ofH

9. The (4,6)-cageNumber of vertices: 26Diameter: 3Girth: 6Bipartite

Intersection array:

∗ 1 1 40 0 0 04 3 3 ∗

Figure 11.15: The (4,6)-cage

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11.3 Tetravalent Distance-Transitive Graphs 121

10. The Gewirtz graphO4 (see [21])Number of vertices: 35Diameter: 3Girth: 6Automorphism group: S7

Primitive

Intersection array:

∗ 1 1 20 0 0 24 3 3 ∗

11. The line graph of Tutte’s 8-cage,L(T)

Number of vertices: 45Diameter: 4Girth: 3Automorphism group: PGL(2,9)Primitive

Intersection array:

∗ 1 1 1 20 1 1 1 24 2 2 2 ∗

Figure 11.16: The line graph of Tutte’s 8-cage,L(T)

12. The 4-cubeQ4Number of vertices: 16Diameter: 4Girth: 4Automorphism group: Z2Wr S4

Bipartite and antipodal

Intersection array:

∗ 1 2 3 40 0 0 0 04 3 2 1 ∗

(See figure 5.2)

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122 Graphs of Low Valency

13. 4-fold covering ofK4,4 (see [36])Number of vertices: 32Diameter: 4Girth: 6Bipartite and antipodal

Intersection array:

∗ 1 1 3 40 0 0 0 04 3 3 1 ∗

14. (4,12)-cage

Number of vertices: 728Diameter: 6Girth: 12Bipartite

Intersection array:

∗ 1 1 1 1 1 40 0 0 0 0 0 04 3 3 3 3 3 ∗

15. D(O4): the bipartite double ofO4

Number of vertices: 70Diameter: 7Girth: 6Automorphism group: S7×Z2

Bipartite and antipodal

Intersection array:

∗ 1 1 2 2 3 3 40 0 0 0 0 0 0 04 3 3 2 2 1 1 ∗

Note that to calculate the automorphism groups of the graphsL(O3), L(H)

andL(T), we can use Theorem 9.2.3, as we know (from the previous section) theautomorphism groups ofO3, H andT.

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