Graphs and Semibiplanes by Jennifer A. Muskovin

On (0,2)-graphs and Semibiplanes

Jennifer A. Muskovin

(Under the direction of Lenny Chastkofsky)

Abstract

A semibiplane is a connected point-block incidence structure such that any two points are in either 0 or 2 common blocks and any two blocks have either 0 or 2 common neighbors. A (0,2)-graph is a connected graph such that any pair of vertices has either 0 or 2 common neighbors. The incidence graph of a semibiplane is a bipartite (0,2)-graph. In this paper we construct (0,2)-graphs from known graphs by taking Cartesian products, quotients and adding matchings and construct semibiplanes by taking quotients of ﬁnite projective planes.

Index words: (0,2)-graph, Biplane, Hypercube, Projective Plane, Semibiplane On (0,2)-graphs and Semibiplanes

Jennifer A. Muskovin

B.S., Benedictine University, 2007

A Thesis Submitted to the Graduate Faculty of The University of Georgia in Partial Fulﬁllment of the Requirements for the Degree

Master of Arts

Athens, Georgia

Jennifer A. Muskovin

Approved:

Major Professor: Lenny Chastkofsky

Committee: Dino Lorenzini Robert Varley

Electronic Version Approved:

Maureen Grasso Dean of the Graduate School The University of Georgia May 2009 Dedication

To Mom and Papa

iv Acknowledgments

I would like to thank the faculty, staff and students within the Mathematics Department at UGA for encouraging me to finish writing. Specifically, thank you Lenny for giving me so much of your time this past year and for staying patient with me during my many struggles. I would also like to thank Joe for putting so much work into writing his paper. I admit that at times my motivation was knowing that you were consistently working until 2 a.m. Finally, thank you to Matt and Whitney for making me smile so many times this year.

v Table of Contents

Page

Acknowledgments ...... v

List of Figures ...... viii

List of Tables ...... ix

Chapter

1 Introduction and Background ...... 1 1.1 Graph Definitions and Examples ...... 1 1.2 Design Definitions and Examples ...... 4

2 Graph Constructions ...... 7 2.1 General Results of (0,2)-graphs ...... 7 2.2 Semibiplanes and (0,2)-graphs ...... 8 2.3 Cartesian Product of Graphs ...... 11 2.4 Quotient Graphs ...... 13 2.5 Bipartite Double ...... 16 2.6 Matchings ...... 18

3 Biplanes ...... 20 3.1 Existence of Symmetric Block Designs ...... 20 3.2 Existence of Biplanes ...... 22

4 Constructing Semibiplanes from Finite Projective Planes . . . . 26 4.1 Collineations of a Finite Projective Plane ...... 28 4.2 Semibiplane Construction ...... 29

vi vii

4.3 Using GAP to Investigate PG(n,q) ...... 31

5 Hypercubes ...... 37

5.1 The Automorphism Group of Qk ...... 37 5.2 Quotients of Hypercubes Resulting in Nonbipartite Graphs 38

5.3 Quotients of Folded Qk ...... 40

5.4 Quotients of coHeawood ×Qk ...... 43

Appendix

A Code Used to Find “Good Conjugacy” Class Relations ...... 47

B Code to Construct Graphs of CoHeawood × Qk ...... 50

C Code to Construct SemiBiplane and Incidence Graph from Finite Projective Plane ...... 53

D Sample List of Bipartite Graphs from [2] ...... 55

Bibliography ...... 56 List of Figures

1.1 The octahedral graph with colored antipodal vertices ...... 2 1.2 Petersen Graph ...... 3 1.3 The Heawood Graph and coHeawood Graph ...... 4 1.4 The coHeawood Graph ...... 5 2.1 Given u, v, w in a (0,2)-graph we can ﬁnd a unique y completing the quadrangle. 8 ∼ 2.2 K2 × Q2 = Q3 ...... 12

2.3 Q3 and K4 ...... 14

2.4 Bipartite Double of Q2 ...... 17

2.5 Q3 and K4 × K2 ...... 19 4.1 Projective Plane of order n = 2: Fano Plane ...... 27 4.2 An Elation of the Fano Plane ...... 28 4.3 A Homology of a Projective Plane of Order 3 ...... 29 4.4 Constructing the incidence graph of the square from the Fano Plane . . . . . 31

5.1 Quotients of Q6 ...... 41 5.2 Shrikhande graph with code labels ...... 42

viii List of Tables

3.1 Using Theorem 3.1.1 to ﬁnd v given k, supposing such a design existed . . . 22 3.2 Known biplanes from [6]; * indicates the dual of a biplane ...... 24 4.1 Points and Lines of the Fano Plane ...... 27 4.2 Points and Lines of PG(2,4) ...... 32 4.3 Points and Blocks of the semibiplane ...... 33 4.4 Semibiplanes Constructed from Finite Projective Planes ...... 36

5.1 An Automorphism of Q3 ...... 37

5.2 Graphs Isomorphic to Quotients of Λ × Qk, k = 2, 3, 4...... 46

D.1 ∆0.1 − ∆6.4 from [2] ...... 55

ix Chapter 1

Introduction and Background

1.1 Graph Definitions and Examples

In this paper, we will assume that all graphs are ﬁnite, connected, simple undirected graphs.

Unless otherwise speciﬁed we will use ∼ to denote adjacency. We will use Qk to denote the graph of the k-hypercube with valency k, V (Γ) and E(Γ) to denote the vertex set and edge set of a graph Γ and ∂(u, v) to denote the distance between vertices u and v. The remainder of this section contains the deﬁnitions we will need in the following chapters. A (0,2)-graph Γ is a connected graph with the property that any pair of vertices has 0 or 2 common neighbors. A (0,2)-graph without any triangles is called a rectagraph. In other words a rectagraph is a connected triangle-free graph with the property that every path of length two is in a unique quadrangle. Notice that a bipartite (0,2)-graph is a rectagraph. If Γ is a bipartite (0,2)-graph then it cannot contain any triangles. If it did, then it would force an adjacency between 2 vertices of the same partition of the vertex set. In particular, Qk is a rectagraph. An isomorphism between the graphs Γ and Γ0 is a bijection f : V (Γ) → V (Γ0) such that any two vertices u, v ∈ Γ are adjacent in Γ if and only if f(u) and f(v) are adjacent in Γ0. An isomorphism from Γ to itself is an automorphism of Γ. In other words, an automorphism of a simple graph Γ is just a permutation of its vertices with the same adjacency property. The set of all automorphisms of Γ along with the operation of composition forms a group called the automorphism group of Γ, denoted Aut(Γ). For some graphs, especially those of the hypercubes, it will be useful to label the vertices as binary code words or vectors. Let S=F2 and k ∈ N.A binary code of length k over S is a

1 2

Figure 1.1: The octahedral graph with colored antipodal vertices

subset of Sk. The elements of the code are called code words and the support of a code word

u is {i : ui 6= 0}. The distance between two code words is given by the number of positions in which the two diﬀer. For instance, the distance between (0,1,0) and (1,1,1) is 2 while the distance between (0,1,0) and (1,0,1) is 3. We can alternatively deﬁne Qk as the graph with vertices all binary codes words of length 2k with an edge connecting two vertices when the distance between their respective code words is exactly 1. When the distance between two length k code words is k we say the code words are antipodal. The antipodal graph of a graph Γ has the same vertex set as Γ with an edge connecting u, v ∈ V (Γ) if the distance between u and v in Γ is equal to the diameter of Γ. We say a graph Γ with diameter d is antipodal if for any vertices u, v, w such that ∂(u, v) = ∂(u, w) = d either ∂(v, w) = d or v = w. The graphs of the hypercubes are antipodal graphs because every vertex has a unique vertex at a maximum distance from it. Example: The octahedral graph as pictured in Figure 1.1 is an example of an antipodal graph. The red vertices are antipodal as are the blue and green vertices.

Let Γ be a graph with u ∈ V (Γ) we will use Γi(u) to denote the set of vertices v such that ∂(v, u) = i. A connected graph Γ is called distance regular if there are integers bi, ci 3

Figure 1.2: Petersen Graph

(i ≥ 0) such that for any two vertices u, v of Γ at distance i from each other, there are

precisely ci neighbors of v in Γi−1(u) and bi neighbors of v in Γi+1(u). The intersection array of a distance regular graph Γ is the sequence ι(Γ) := {b0, b1, ..., bd−1; c1, c2, ..., cd}. Note that the numbers in the intersection array are independent of u, v. Further, Γ is k − regular with k = b0 as b0 is by deﬁnition the number of vertices adjacent to a given vertex. It will also always be the case that bd = 0 because Γd+1(u) will be empty. Similarly, c0 is undeﬁned as

Γ−1(u) is empty. Clearly, Γ0(u) = 1 as u can only be of distance 0 away from itself so if

∂(u, v) = 1 then the number of neighbors of v in Γ0(u) is also 1 so c1 = 1. Example : Let Γ be the Petersen graph as labeled in Figure 1.2. The diameter of Γ is 3.

Let u = 1. Then Γ1(u) = {2, 6, 5} and Γ2(u) = {3, 4, 7, 8, 9, 10}. Let v = 2 then ∂(v, u) = 1

0 0 so b1 = 2 as 2 ∼ 3, 7. Let v = 3 then d(v , u) = 2 so c1 = 1 as 3 ∼ 2. Thus the intersection array of Γ is {3, 2; 1, 1}. If Γ is a connected graph of diameter d then the distance-k graph for k = 1, ..., d is the graph with the same vertex set of Γ in which two vertices are connected if they are distance k apart in Γ. Example: The Heawood graph is a distance-regular graph on 14 vertices with intersection array {3, 2, 2; 1, 1, 3}. The distance-3 graph of the Heawood graph, called the coHeawood, 4

Figure 1.3: The Heawood Graph and coHeawood Graph

is distance regular with intersection array {4, 3, 2; 1, 2, 4}. The coHeawood is a (0,2)-graph that will be studied further in Chapter 3; note that the Heawood is not a (0,2)-graph.

1.2 Design Definitions and Examples

A t − (v, k, λ) design is a point-block incidence structure where v is the number of points, k is the number of points contained in each block and every subset of t points is incident with λ blocks. The incidence graph of a design with point set X and block set B is the bipartite graph with vertex set X ∪ B with a point x ∈ X adjacent to a block B if x ∈ B. Example: The coHeawood Graph is the incidence graph of a 2-(7,4,2) biplane. It is con- sidered to be the non-incidence graph of the Fano Plane. The labels in Figure 1.4 correspond to the adjacency list found at Andries Brouwer’s website1. A 2-design is called a balanced incomplete block design (BIBD). A design with the same number of blocks as points is called a symmetric design. When λ = 2 a symmetric design is called a biplane.

1http://www.win.tue.nl/ aeb/graphs/recta/02graphs.html. 5

Figure 1.4: The coHeawood Graph

When λ = 1 a design is called a Steiner system. A finite projective plane of order n is a Steiner system. It is a symmetric 2 − (n2 + n + 1, n + 1, 1) design. A design is called semisymmetric when any two distinct blocks meet in exactly 0 or λ points, and any two distinct points are in 0 or λ blocks. When λ = 2 a semisymmetric design is called a semibiplane. It is clear that the incidence graph of a semibiplane is a bipartite (0,2)-graph and conversely a bipartite (0,2)-graph is the incidence graph of a semibiplane. Formal proofs can be found in Chapter 2. In [2] Andries Brouwer found and classified all (0,2)-graphs with valency at most 7 and in [3] he and Patric Osterg˙ardextended¨ the classification of (0,2)-graphs up to valency 8. In [2] and [3] the (0,2)-graphs are listed as bipartite or nonbipartite, with the bipartite graphs representing the incidence graph of a semibiplane. We will use ∆k.i when referring to a bipartite graph with valency k and serial number i in the tables of [2] or [3] and Γk.i as the nonbipartite graph with the same parameters. For the graphs in each list the authors provide the valency, diameter, automorphism group size, orbit description and for some, a 6 construction or comment about the graph. For each graph the number of points at each distance from a given point is also given under the ’Distribution’ column. A sample of the bipartite list from [2] is in Appendix D. In Chapter 2 we will illustrate several graph constructions and describe when the use of these constructions will yield a (0,2)-graph. In Chapter 3 we will give the conditions for the existence of a biplane. In Chapter 4 we will construct semibiplanes from finite projective planes and find the semibiplanes from this construction up to block size 11. In Chapter 5 we will discuss the automorphism group of Qk and construct (0,2)-graphs by starting with a (0,2)-graph Γ and taking a quotient of Γ × Qk. Chapter 2

Graph Constructions

This chapter contains examples of constructing a graph from another graph. Since the graphs of the hypercubes are easier to understand than most other graphs, they are used in most of the examples, however unless otherwise speciﬁed the construction can be used by starting with any ﬁnite simple graph.

2.1 General Results of (0,2)-graphs

We will begin with an alternative deﬁnition of a (0,2)-graph that will be used in several of the proofs in this chapter and follow it with a general result about (0,2)-graphs.

Lemma 2.1.1. A connected graph Γ is a (0,2)-graph if and only if for all u, v, w ∈ V (Γ) such that u ∼ w ∼ v, there exists a unique y ∈ V (Γ), y 6= w with u ∼ y ∼ v.

Proof. First suppose Γ is a (0,2)-graph. Let u, v, w ∈ V (Γ) with u ∼ w ∼ v. Then since u and v have at least one common neighbor, it must have 2 common neighbors. Thus there exists a unique y 6= w such that u ∼ y and v ∼ y. Conversely, suppose that for a graph Γ with u, v ∈ V (Γ), {x ∈ V (Γ)|x ∼ u} ∩ {x ∈ V (Γ)|x ∼ v}= 6 ∅. Let w ∈ {x ∈ V (Γ)|x ∼ u} ∩ {x ∈ V (Γ)|x ∼ v} then by our assumption there exists a unique y 6= w such that y is also an element of {x ∈ V (Γ)|x ∼ u} ∩ {x ∈ V (Γ)|x ∼ v}. Thus |{x ∈ V (Γ)|x ∼ u} ∩ {x ∈ V (Γ)|x ∼ v}| = 2 so Γ is a (0,2)-graph.

Theorem 2.1.2. A (0,2)-graph is regular and consequently so is a rectagraph.

Proof. Let Γ be a (0,2)-graph with u, v ∈ V (Γ) with u ∼ v. If for some x ∈ V (Γ), x 6= v, x ∼ u then from Lemma 2.1.1 we know there exists a unique vertex x0 ∈ V (Γ) with x0 ∼ v

7 8

Figure 2.1: Given u, v, w in a (0,2)-graph we can ﬁnd a unique y completing the quadrangle.

thus there is an injective map from {x ∈ V (Γ)|x ∼ u} to {x ∈ V (Γ)|x ∼ v}. Similarly, if for some y ∈ V (Γ), y 6= u, y ∼ v then we know there exists a unique vertex y0 ∈ V (Γ) with y0 ∼ u so there is an injective map from {x ∈ V (Γ)|x ∼ v} to {x ∈ V (Γ)|x ∼ u}. Thus u and v have the same number of neighbors. Further, since Γ is a (0,2)-graph, it is connected so we can ﬁnd a path from u to any other vertex u0 so u and u0 have the same number of neighbors. Hence every vertex has the same degree.

2.2 Semibiplanes and (0,2)-graphs

Theorem 2.2.1. The incidence graph of a semibiplane is a bipartite (0,2)-graph.

Proof. Let Γ be the incidence graph of a semibiplane B with point set X and block set B. Let X˜ and B˜ denote the set of vertices from the point set and block set of B respectively. Then V (Γ) = X˜ ∪ B˜ with x ∈ X˜ adjacent to a block b if x ∈ b. By deﬁnition, X˜ ∩ B˜ = ∅. Further, by construction, no two points are adjacent and no two blocks are adjacent. Thus Γ is bipartite. 9

Let x, y ∈ X˜ then by construction of B, x and y are in either 0 or 2 common blocks. Thus in Γ, x and y are adjacent to either 0 or 2 vertices in B˜. Similarly, for any two vertices in B˜ their corresponding blocks have either 0 or 2 common points in B so in Γ they have 0 or 2 common neighbors. Thus Γ is a (0,2)-graph.

Theorem 2.2.2. A bipartite (0,2)-graph is the incidence graph of a semibiplane.

Proof. Let Γ be a bipartite (0,2)-graph. Then V (Γ) = A∪˙ B. Let A be the set of points and B

the set of blocks. Then since Γ is a (0,2)-graph, we know that for any two points a1, a2 ∈ A,

a1 and a2 are incident with either 0 or 2 common blocks in B. Similarly, any two blocks in B are adjacent to either 0 or 2 common points in A. Thus we can construct a semibiplane from Γ.

Example: Using GAP and the package GRAPE [10] we can construct a semibiplane from

∆5.1. First we start with the adjacency list of ∆5.1 found at Brouwer’s website. Using the list along with the Graph function from GRAPE, we ﬁrst construct the graph. The graph has 22 vertices, labeled 1 through 22, each of valency 5. We know this graph is bipartite so using the Bicomponents function from GRAPE returns a list of length 2 containing the vertices in each class of the bipartition. The output from this function is [ [ 1, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 ], [ 2, 3, 4, 5, 6, 17, 18, 19, 20, 21, 22 ] ].

Let A = [1, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16] and B = [2, 3, 4, 5, 6, 17, 18, 19, 20, 21, 22] where A represents 11 points and B represents 11 blocks. If we look at the list of adjacencies of the graph, we should ﬁnd that each point in A is adjacent to 5 blocks in B and each block in B is adjacent to 5 points in A. AdjList:=[ [ 2, 3, 4, 5, 6 ], [ 1, 7, 8, 9, 10 ], [ 1, 7, 11, 12, 13 ], [ 1, 8, 11, 14, 15 ], [ 1, 9, 12, 14, 16 ], [ 1, 10, 13, 15, 16 ], [ 2, 3, 17, 18, 19 ], [ 2, 4, 17, 20, 21 ], [ 2, 5, 18, 20, 22 ], [ 2, 6, 19, 21, 22 ], [ 3, 4, 18, 21, 22 ], [ 3, 5, 19, 20, 21 ], [ 3, 6, 17, 20, 22 ], [ 4, 5, 17, 19, 22 ], [ 4, 6, 18, 19, 20 ], [ 5, 6, 17, 18, 21 ], [ 7, 8, 13, 14, 16 ], [ 7, 9, 11, 15, 16 ], [ 7, 10, 12, 14, 15 ], [ 8, 9, 12, 13, 15 ], [ 8, 10, 11, 12, 16 ], [ 9, 10, 11, 13, 14 ] ] 10

We read the list as the point 1 is adjacent to the blocks [ 2, 3, 4, 5, 6 ]. We know ∆5.1 is a (0,2)-graph, but by inspection of the list we can also verify that each pair of points of A are on 0 or 2 common blocks and vice versa. Thus ∆5.1 is the incidence graph of a semibiplane with 11 points and blocks such that each point is on 5 blocks and each block contains 5 points. Usually a semibiplane is represented by a list of blocks so if we isolate the sublists corresponding to the blocks in B then reindex the points by labeling them 1 through 11 we get Blocks:=[[ 1, 2, 3, 4, 5 ], [ 1, 2, 6, 7, 8 ], [ 1, 3, 6, 9, 10 ], [ 1, 4, 7, 9, 11 ], [ 1, 5, 8, 10, 11 ], [ 2, 3, 8, 9, 11 ], [ 2, 4, 6, 10, 11 ], [ 2, 5, 7, 9, 10], [ 3, 4, 7, 8, 10 ], [ 3, 5, 6, 7, 11 ], [ 4, 5, 6, 8, 9 ] ].

In the previous example, it holds that each pair of points are actually on 2 common blocks and each pair of blocks share 2 common points so the semibiplane we constructed is a biplane.

Hence by the following result, a stronger statement is that ∆5.1 is the incidence graph of a biplane.

Theorem 2.2.3. A bipartite graph is the incidence graph of a symmetric 2-design if and only if it is distance regular with diameter 3.

Proof. Let D be a symmetric 2-(v, k, λ) design with incidence graph X. Any two points lie at distance two in X, and similarly for blocks since D is symmetric. By construction, X is bipartite. Therefore, a block lies at distance three from a point not on the block, thus the diameter of X is three. Since two points lie in λ blocks, we know c2 = λ. Further, we know that every point is incident to k blocks and every block is incident to k points thus b0 = k.

Similary, if pi is a point then X2(pi) is the set of points of distance 2 away from it. To get to any of these points from pi we must use a block that pi is adjacent to. Thus if li is a block adjacent to pi, X2(pi) will contain all of the points adjacent to li except pi so b1 = k −1. Now suppose pj is of distance two away from pi. We know that pi is adjacent to k blocks but there are only k − λ that are of distance 3 away from pi because λ of the blocks are of distance 1 from pi so b2 = k − λ. If pj is a point then X1(pj) is all blocks adjacent to pj since any other 11

point is of distance two away from pj and every subset of 2 points is incident with λ blocks

we know c2 = λ. Since any two points are of distance 2 from each other, all k neighbors of

any block of distance 3 from p1 will be in X2(p1) so c3 = k. Thus the intersection array of X is {k, k − 1, k − λ; 1, λ, k}. Since D is symmetric, we will get the same intersection array if we started with a line thus X is distance regular. Now suppose that X is a bipartite distance regular graph with diameter three and inter-

section array {b0, b1, b2; c1, c2, c3}. Declare one part of the bipartition to be points and the

other to be blocks. Then by deﬁnition we see that every point lies in b0 blocks, and every

two points lie in c2 blocks, hence we can construct a 2-design with b0 being the number of

blocks containing each point and λ = c2. Further, every block contains b0 points and every

two blocks meet in c2 points. Thus, we can construct a 2-design with the number of blocks equal to the number points. [5]

From the previous theorem we know ∆5.1 is also a distance regular graph with diameter 3.

2.3 Cartesian Product of Graphs

Suppose G and H are simple graphs. The Cartesian product G×H is the simple graph on the vertices V (G) × V (H) and edges E(G × H) = {{(u, v), (u0, v0)} : u = u0 and {v, v0} ∈ E(H), or v = v0 and {u, u0} ∈ E(G)}.

Example: Let G = Q1 = K2 and H = Q2 with vertices of each labeled as in ﬁgure 2.2. Then the eight vertices of G × H are (a,1), (a,2), (a,3), (a,4), (b,1), (b,2), (b,3), (b,4). Since a ∼ b ∈ G and 1 ∼ 2, 4 ∈ H we know (a,1)∼(b,1),(a,2) and (a,4) in G × H. Similarly we can ∼ ﬁnd the remaining edges of G × H to see that G × H = Γ3.

In some graph theory texts, Qk is deﬁned as the Cartesian product of k copies of K2.

k From this deﬁnition it not hard to see that the number of vertices of Qk is 2 . Thus from

1 k−1 the Handshaking Lemma we know the number of edges of Qk is 2 · k.

1 P Handshaking Lemma: For a graph G we have u∈V (G) dG(u) = 2|E(G)|. 12

∼ Figure 2.2: K2 × Q2 = Q3

Theorem 2.3.1. If Γ and ∆ are (0,2)-graphs, then Γ × ∆ is a (0,2)-graph. Further if both graphs are bipartite, then the resulting graph will also be bipartite.

Proof. Let Γ × ∆ = Λ then V (Λ) = {(i, j)|i ∈ V (Γ), j ∈ V (∆)}. Suppose (a, b), (c, d) are two distinct vertices of Λ, then they are connected by an edge if a ∼ c in Γ and b = d or b ∼ d in ∆ and a = c. Since (a, b) and (c, d) are distinct we cannot have both a = c and b = d. If a ∼ c in Γ and b = d then Λ is a (0,2)-graph because Γ is a (0,2)-graph. Otherwise, b ∼ d in ∆ and a = c and Λ is a (0,2)-graph because ∆ is a (0,2)-graph. Now suppose Γ and ∆ were bipartite graphs with V (Γ) = A∪˙ B and V (∆) = C∪˙ D. Then V (Γ×∆) = (A×C)∪(A×D)∪(B ×C)∪(B ×D) = (A×C ∪B ×D)∪˙ (A×D∪B ×C). Since A and B are disjoint and C and D are disjoint, the graph with vertex set (A × C ∪ B × D) is empty. Similarly, the graph with vertex set (A × D ∪ B × C) is empty. Thus (A × C ∪ B × D)∪˙ (A × D ∪ B × C) forms a bipartition of the vertices.

From Theorem 2.1.2 we know that the graph Λ will be a regular graph. The valency of Λ will be the sum of the valencies of Γ and ∆ and the diameter will be the sum of the diameters of Γ and ∆. 13

A large portion of the graphs in Brouwer’s list can be constructed by a Cartesian product of graphs. The next result allows for some understanding of the automorphism group of the newly constructed graphs.

Theorem 2.3.2. For graphs Γ1, Γ2, Aut(Γ1)×Aut(Γ2) ≤Aut(Γ1 × Γ2).

Proof. Aut(Γ1)×Aut(Γ2) contains the identity so it is nonempty. Suppose (σ, τ) ∈Aut(Γ1)×Aut(Γ2)

and (a, b), (c, d) ∈ (Γ1 × Γ2) with (a, b) ∼ (c, d). We need to check that (σ, τ)(a, b) ∼ (σ, τ)(c, d). If (a, b) ∼ (c, d) then either a = c and b ∼ d or b = d and a ∼ c. If a = c and b ∼ d then

σa = σc and since τ ∈Aut(Γ2) and b, d ∈ Γ2 we know τb ∼ τd. Similarly if b = d and a ∼ c we know τb = τd and σa ∼ σc. Hence (σ, τ)(a, b) ∼ (σ, τ)(c, d).

2.4 Quotient Graphs

Let Γ be a ﬁnite graph and G be a group of automorphisms of Γ. The quotient graph Γ/G is constructed as follows. The vertices of Γ/G are the orbits of G and two vertices are adjacent if the orbits contain elements adjacent in Γ.

Example: Let Q3 be the graph of the cube labeled as in Figure 2.3 and σ be the

permuation that interchanges vertices in Q3 of distance 3. The orbits of < σ > are [1,8],

[2,7], [3,6], [4,5] so Q3/ < σ > will have four vertices each corresponding to an orbit. Since the ﬁrst orbit contains 1 and the second orbit contains 2, which are adjacent vertices in

Γ3, we know [1,8]∼[2,7] in Q3/ < σ >. If we continue, we will see that all the vertices of ∼ Q3/ < σ > are adjacent to each other so Q3/ < σ >= K4, the complete graph on four vertices. When a graph Γ with diameter d is antipodal we can construct the folded graph of Γ as the quotient graph of Γ with equivalence relation u ∼ v if u and v are antipodal vertices. The resulting graph Γ0 will have half as many vertices and its diameter will be bd/2c. From

the previous example, we saw that the folded graph of Q3 is isomorphic to K4. By similar 14

Figure 2.3: Q3 and K4

constructions, we see that the folded graph of Q2 is isomorphic to the the path graph on 2 vertices; the folded graph of Q4 is isomorphic to the complete bipartite graph K4,4; and the folded Q5 is isomorphic to Γ5.2 which is also known as the Clebsch graph. Note that all but the folded Q4 resulted in a (0,2)-graph.

Theorem 2.4.1. If Γ is a (0,2)-graph and G a group of automorphisms of Γ and no two vertices of any G-orbit are joined by a path of length 1,2 or 4, then Γ/G is also a (0,2)-graph.

Proof. Let a,b and c be distinct orbits of G with a ∼ b and b ∼ c in Γ/G. Then for some x ∈ a and y ∈ b x ∼ y in Γ and some z ∈ c with y ∼ z. From Lemma 2.1.1 since Γ is a (0,2)-graph we know there must be a unique vertex w ∈ V (Γ), w 6= y with w ∼ x and w ∼ z. Let d be the orbit in G that contains w. Then a ∼ d ∼ c in Γ/G. Suppose d = b then y and w would be in the same orbit, but the path between y and w is of length 2 so d 6= b. Similarly, if d = a or d = c then the orbit would contain a path of length 1 so d 6= a and d 6= c. Thus d is distinct from a, b and c. Now suppose there is another orbit e with a ∼ e and e ∼ c in Γ/G. Then there exists vertices r, s ∈ V (Γ) with r ∼ x and s ∼ z, but r ∼ x ∼ y ∼ z ∼ s. Since no orbit of G contains vertices with a path of length 4 between them, it must be that r = s. Thus r ∼ x 15

and r ∼ z. By the uniqueness of w it must be that z = w thus e = d so d is unique. Thus by Lemma 2.1.1, Γ/G is a (0,2)-graph.

Corollary 2.4.2. Let C be a code in 2k viewed as a binary vector space. If the minimum

distance between two code words in C is at least 5, then Qk/C is a rectagraph.

If the minimum distance between two code words in C is 5 then we are just adding the case that no orbits contain paths of length 3 to Theorem 2.4.1 so we are omitting any triangles that could possibly occur in the quotient graph. The following theorem will allow us to take a representative from the conjugacy class of the automorphism group of a graph when we construct quotient graphs using GAP.

Theorem 2.4.3. Let Γ be a graph with G a group of automorphisms of Γ. If τ, σ ∈ G are

conjugate then Γ/ < σ >∼= Γ/ < τ >.

Proof. Suppose the order of τ, σ is k. Since τ, σ are conjugate, we know there exists some π ∈ G such that π−1τπ = σ. Supposev ¯ ∈ Γ/ < σ > deﬁne φ :Γ/ < σ >→ Γ/ < τ >, φ(¯v) = πv¯ .

i φ is well-deﬁned: Supposev ¯1 =v ¯2 then for some i v¯2 = σ (v1) thenπv ¯ 2 =πv ¯ 1 so

i i πv2 = πσ v1 = τ (πv1) which meansπv ¯ 2 =πv ¯ 1. φ is surjective: Since φ(π−1w) = π(π−1w) =w ¯, φ is surjective.

i φ is injective: Suppose φ(¯v1) = φ(¯v2) thenπv ¯ 1 =πv ¯ 2 so for some i we have πv2 = τ πv1

−1 i i so v2 = π τ πv1 = σ v1 thusv ¯2 =v ¯1.

i i i −1 j φ is a homomorphism: Suppose v1 ∼ σ v2 then πv1 ∼ πσ v2 = πσ v2π πv2 = τ πv2 thus

πv1 ∼ πv2 so φ(v ¯1) ∼ φ(v ¯2).

After constructing a quotient graph, we can use the following result to gain some understanding of the automorphism group of the new graph.

Theorem 2.4.4. If H ≤Aut(Γ) = G and σ ∈ CG(H) then σ induces an automorphism of Γ/H. 16

Proof. Deﬁneσ ¯ :Γ/H → Γ/H byσ ¯(¯a) = σ(¯a). To see thatσ ¯ is well-deﬁned, supposea ¯ = ¯b. Then for h ∈ H, σb = h−1σha = σa since σ commutes with h. Supposeσ ¯(¯a) =σ ¯(¯b) then for some h ∈ H σa = hσb = σhb. By applying σ−1 to both sides we see that a = hb soa ¯ = ¯b which means thatσ ¯ is injective. Clearly,σ ¯ is surjective so to check that it is a graph homomorphism supposea ¯ ∼ ¯b. Then for some h ∈ H, a ∼ hb soσa ¯ ∼ σhb¯ = hσb¯ so σ(¯a) ∼ σ(¯b).

Example: Let Γ = Q6/(000111), this is Γ6.10 in Brouwer. The automorphism group of Γ

are the elements in Aut(Q6) that commute with Γ. There are 36 translations in Aut(Q6) of the form (abc)(def) that commute with Γ. We can permute the three coordinates of (abc)

and the three coordinates of (def) so we get 36 more elements from Aut(Q6) that commute with Γ. Thus |Aut(Γ)| = 36 · 36 = 1152.

2.5 Bipartite Double

The bipartite double of a graph is the graph constructed by connecting two copies of the original vertex set, removing the original edges and constructing edges u+v− and u−v+ for every uv edge of the original graph. By construction, the bipartite double of a graph will have the same degree as the original graph and twice the number of vertices, hence the name. Note that a bipartite graph Γ0 can be the bipartite double to multiple graphs.

Example: The Shrikhande graph (Γ6.2) and K4 × K4 (Γ6.1) each have 16 vertices and valency 6. They are not isomorphic graphs but both have the same bipartite double, the

folded Q6.

Example: To construct the bipartite double of Q2 we will use the vertex labels in Figure 2.4. For each vertex on the graph of the square on the left, we get two new vertices, distin- guished by a + or -. Since A ∼ 1 in the original graph, the bipartite double will have edges A+1− and A−1+. Similarly we ﬁnd the remaining edges to see that the resulting graph is two copies of the original graph so in this case the bipartite double is not a connected graph. 17

Figure 2.4: Bipartite Double of Q2

Lemma 2.5.1. The bipartite double of Γ is connected if and only if Γ is a connected nonbipartite graph.

Proof. Suppose Γ is nonbipartite with V (Γ) = A. Then the bipartite double Γ0 has vertex set A+∪˙ A−. Since Γ is connected we can ﬁnd a path in Γ from u to v. Thus in Γ0 we can ﬁnd a path from u+ to v− and a path from u− to v+ so Γ0 is connected. Conversely, suppose Γ is a bipartite graph with V (Γ) = A∪˙ B. Then V (Γ0) = (A+∪˙ B+) ∪ (A−∪˙ B−) where vertices in A+ are connected to vertices in B− and vertices in A− are connected to vertices in B+. Hence Γ0 has two connected components, each the graph of Γ, so Γ0 is not connected.

Theorem 2.5.2. If Γ is a nonbipartite (0,2)-graph then the bipartite double of Γ denote Γ0 is a (0,2)-graph.

Proof. Let Γ be a nonbipartite (0,2)-graph with bipartite double Γ0 with V (Γ0) = Γ+ ∪ Γ− where Γ+ = {u+|u ∈ V (Γ)} and Γ− = {u−|u ∈ V (Γ)}. Suppose u+, v− ∈ V (Γ0) with u+ ∼ v−. If for some x− 6= v−, x− ∼ u+ we want to show there exist a unique y+ 6= u+ with v− ∼ y+ and x− ∼ y+. 18

In Γ, u ∼ v and u ∼ x so since Γ is (0,2) from Lemma 2.1.1 we know there exists a unique y 6= u with y ∼ v and y ∼ x. In Γ0 there are two vertices y+, y− corresponding to y. Since no two vertices in Γ− are connected, it must be that y+ ∼ v− and y+ ∼ x−. Suppose that z+ 6= u+ is another vertex such that z+ ∼ v− and z+ ∼ x− then in Γ there must be a vertex z 6= u with z ∼ v and z ∼ x. By uniqueness of y in Γ, z = y thus z+ = y+.

The following is a special case of Theorem 2.4.1.

Corollary 2.5.3. Given a bipartite (0,2)-graph Λ with an automorphism, σ, of order two without ﬁxed edges that interchanges the two classes of the bipartition, the graph Λ/σ with σ−orbits as vertices, where two σ−orbits are adjacent when there are edges between them, is a (0,2)-graph.

Since σ interchanges classes of the bipartition, the path lengths in the orbit must all be odd. Since there are no ﬁxed edges, there are no orbits containing paths of length one.

2.6 Matchings

Suppose Γ is a graph of valency k. Let σ be an involution of Γ such that no two vertices of any σ-orbit are joined by a path of length 0,1 or 3. Then we can construct a new graph Λ by adding edges (i, σ(i)) to Γ for all i ∈ V (Γ).

Theorem 2.6.1. If Λ is the graph constructed by adding a matching to Γ then for σ ∈Aut(Γ), ∼ (Γ × K2)/(σ, 1) = Λ.

Proof. Denote the vertices of K2 as 0 or 1. Deﬁne a map from (Γ × K2)/(σ, 1) → Λ with {(v, 0), (σv, 0)} = v and {(v, 1), (σv, 1)} = σv. It can be easily checked that this map is a graph isomorphism.

Since a path of length 0,1 or 3 between two vertices in Γ becomes a path of length 1,2

or 4 in (Γ × K2), by Theorem 2.4.1 we know that if Γ is a bipartite (0,2)-graph then Λ will also be a (0,2)-graph. 19

Figure 2.5: Q3 and K4 × K2

Clearly, Λ will have the same number of vertices as Γ and since the matching adds an edge to each vertex, Λ will have valency k + 1.

As an illustration of such a matching we will start with Q3 with vertices labeled by binary vectors of length 3. Then for x ∈ V (Q3) we have three options for the map σ; σ: x → x + 101, x → x + 011, x → x + 110. Since the weights of 011, 101 and 110 are the same they are conjugate in Aut(Q3) and thus give rise to isomorphic graphs. Without loss of generality, suppose σ: x → x + 101. Then we must add the following edges to Q3: (000, 101), (010, 111), (100, 001), (110, 011). After adding these edges we get a new graph that is isomorphic to K4 × K2, which is a (0,2)-graph. We can also go the other way by starting with a (0,2)-graph Λ of valency k + 1 and an automorphism σ with the property that all orbits are edges, then removing these edges will produce a graph in which its connected components are (0,2)-graphs of valency k. For

instance, if we started with the graph of K4 and let σ be the antipodal map then the orbits

are the diagonal edges. By removing those edges we would obtain Q2 which is the graph of the square. Chapter 3

Biplanes

It has been conjectured that only a finite number of symmetric designs with λ ≥ 2 exist and hence that only a finite number of biplanes exists. Biplanes are known to exist for k = 2, 3, 4, 5, 6, 9, 11 and 13. In fact, exactly one biplane exists for k = 2, 3, 4 and 5, exactly 3 for k = 6, exactly four for k = 9, at least three for k = 11 and at least two for k = 13. It is not the case, however that there are a finite number of semibiplanes. As we will see in Chapter 4 we can construct semibiplanes from finite projective planes.

3.1 Existence of Symmetric Block Designs

We begin this section by determining the number of points in a design given the number of points in a block. As we will see, the existence of a design is determined by the number of points in a block.

Lemma 3.1.1. In a (v, k, 2) design where k is the number of points in a block and v is the k(k − 1) + 2 number of points, k(k − 1) = 2(v − 1) so v = . 2

Proof. Fix a point p. We will count the number of pairs of points (p, p0) such that p and p0 are in a block together. We know p appears in k blocks and can be paired with k − 1 other points in that block so there are k(k − 1) of these pairs. There are v − 1 choices for a point to pair with p and each pair will appear in 2 blocks so there are 2(v − 1) of these pairs.

20 21

The incidence matrix of a block design is A = (aij), i = 1, ..., v, j = 1, ..., b where if   1 if pi ∈ Bj p1, ..., pv are the points and B1, ..., Bb are the blocks we have aij = . If the the  0 if pi ∈ Bj design is symmetric with v points and v blocks, then the incidence matrix has the property that the diagonal entries of   k λ . . . λ      λ k . . . λ       ...  T   AA =   .    ...       ..      λ . . . k

If A is the incidence matrix of a design D then AT is the incidence matrix of the

0 0 dual design D . The points of D are b1, ..., bv and the blocks are A1, ..., Av. A biplane is self-dual if D0 is D. Of the known biplanes only two are not self-dual.

T th th Let AA = B with entries bij. Then bij is the dot product of the i row of A with the j

T th column of A . Then bii counts the number of 1’s in the i row of A, where each 1 signiﬁes pi was in a block. Since the design is symmetric we know the number of blocks pi is contained in is k. The oﬀ-diagonal entries count the number of blocks a pair of points pi and pj are contained in, which is λ. After performing some row and column operations, we see that   k + (v − 1)λ 0 ... 0      0 k − λ . . . 0       ...  T   v−1 det(AA ) =   = (k − λ) (k + vλ − λ).    ...       ..      0 . . . k − λ

The following theorem is called the Bruck-Ryser-Chowla Theorem. A proof of the ﬁrst condition is provided. 22

Table 3.1: Using Theorem 3.1.1 to ﬁnd v given k, supposing such a design existed

k 2 3 4 5 6 7 8 9 10 11 12 13 v 2 4 7 11 16 22 29 37 46 56 67 79

Theorem 3.1.2. If a symmetric block design exists with parameters v, k, λ then

1. If v is even, k − λ is a square.

2. If v is odd, z2 = (k − λ)x2 + (−1)(v−1)/2λy2 has a nonzero integral solution.

Proof. Since the design is symmetric, the number of blocks is equal to the number of points so the incidence matrix A of the design will be a square matrix. Thus det(A) = det(AT ) so det(AAT ) = det(A) det(A) = (det(A))2 = (k − λ)v−1(k + vλ − λ). From Theorem 3.1.1 we know v(λ − 1) = k(k − 1) so k + vλ − λ = k + k2 − k = k2. This implies that (k − λ)v−1 must also be a square. Since v is even, it must be that k − λ is a square [6].

3.2 Existence of Biplanes

The following is a consequence of the Bruck-Ryser-Chowla Theorem and will be used to determine for k ≤ 19 when a biplane can exist.

Theorem 3.2.1. For a (v, k, 2) biplane (i) if k ≡ 2, 3 (mod 4), then v is even and k − 2 is a square.

2 2 v−1 2 (ii) if k ≡ 0, 1 (mod 4), then z = (k − 2)x + 2(−1) 2 y has a nonzero integral solution.

If k = 7 then k ≡ 3 (mod 4) but k − 2 = 5 is not a square so by Theorem 3.2.1.(i) there is not a (22,7,2) biplane. Similarly, when k = 10, k ≡ 2 (mod 4) but k − 2 = 8 is not a square so there is not a (46,10,2) biplane. 23

Suppose k = 8 then k ≡ 0 (mod 4) so if a (29,8,3) biplane existed then by 3.2.1.(ii)

z2 = 6x2 + 2y2 (3.1) has a nonzero integral solution. Suppose that (3.1) has a solution. Since any multiple of that solution is also a solution we know that there is a solution with x,y and z relatively prime. Since 6x2 + 2y2 is even and square roots of even squares are even, we know z = 2c for some integer c. Now we have 2c2 = 3x2 + y2 then 2c2 ≡ y2 (mod 3). Since 2c2 6≡ 1 (mod 3) and 2c2 6≡ 2 (mod 3) it must be that 2c2 ≡ 0 (mod 3) so c is a multiple of 3. Thus y must be a multiple of 3. Then from 3.1 we have 9|6x2 so 3|x contradicting x,y and z relatively prime. Thus there cannot be a nonzero integral solution to 3.1. Suppose k = 12 then k ≡ 0 (mod 4) to if a (67,12,2) biplane existed then similarly,

z2 = 10x2 − 2y2 (3.2) has a nonzero integral solution. Again, if a solution exists then there is a solution with x,y and z relatively prime. Note that if (3.2) does not have a solution modulo 5 then it will not have a solution at all. Consider (3.2) modulo 5, that is z2 = 10x2 − 2y2 (mod 5) so z2 ≡ 3y2 (mod 5). Since squares modulo 5 are equivalent to 0, 1 or 4 we know z2 ≡ 3y2 ≡ 0 (mod 5) so y and z are multiples of 5. Then we can write 3.2 as 25i2 + 50j2 = 10x2 for some i, j so 5(i2 + 2j2) = 2x2 which means 5|x2 so 5|x. From Theorem 3.2.1.i we know there are no biplanes when k = 14, 15. We cannot rule out the existence of a (121,16,2) biplane because z2 = 14x2 + 2y2 has solution (1,1,4) or the existence of a (154,18,2) biplane because 16 is a square. We can, however rule out the existence of biplanes when k = 17 and 19. In [6], Marshall Hall constructs the known biplanes for k ≤ 13 and their respective automorphism groups. From Theorem 2.2.1 it follows that the incidence graph of a biplane is a bipartite (0,2)-graph so for each of the biplanes in Table 3.2 there is a bipartite (0,2)- graph on 2v vertices of valency k. 24

Table 3.2: Known biplanes from [6]; * indicates the dual of a biplane

Biplane Order of Automorphism Group (2,2,2) 2 (4,3,2) 24 (7,4,2) 168 (11,5,2) 660 (16,6,2) 11,520 (16,6,2) 768 (16,6,2) 384 (37,9,2) 54 (37,9,2) 1512 (37,9,2)* 1512 (37,9,2) 333 (56,11,2) 80,640 (56,11,2) 288 (56,11,2) 144 (56,11,2) 64 (56,11,2) 24 (78,13,2) 110 (78,13,2)* 110 25

For all of the known biplanes in Table 3.2, the size of the automorphism group of the incidence graph is twice the size of the automorphism group of the biplane. It seems likely that this should hold for all biplanes, if more exist, however at this time we have yet to prove the claim. Chapter 4

Constructing Semibiplanes from Finite Projective Planes

Formally, a ﬁnite projective plane of order n is a set of n2 + n + 1 points with the properties:

(i) Any two points determine a line, any two lines determine a point.

(ii) Every point has n + 1 lines on it.

(iii) Every line contains n + 1 points.

Note that for the purpose of this paper we will only be considering desarguesian1 planes. Projective planes are only known when n = pk, where p is prime and k ≥ 1. Let a V be a vector space of dimension 3 over a field F such that for x, y ∈ V , x + y = (x1 + y1, x2 + y2, x3 + y3) and for λ ∈ F, λx = (λx1, λx2, λx3). To construct a finite projective plane we define an equivalence relation on V − 0 by two non-zero vectors x,y ∈ V being equivalent if there exists a non-zero scalar λ ∈ F such that x = λy. The set of equivalence classes is called the projective space associated to the vector space V and is denoted P(V). The dimension of P (V ) is defined to be one less than the dimension of V .A subspace of P (V ) is a projective space associated to some subspace of V . Further, the 0-dimensional subspaces of P (V ) are projectivizations of the 1-dimensional subspaces of V and are called the points of projective space. The 1-dimensional subspaces of P (V ) are the projectivizations of the 2-dimensional subspaces of V and are called the straight lines of projective space. The smallest projective plane, known as the Fano Plane, is of order n = 2 and can be constructed as followed. Let V be a vector space of dimension 3 with V = F2 ⊕ F2 ⊕ F2 = 1See [4]

26 27

Figure 4.1: Projective Plane of order n = 2: Fano Plane

3 {(a1, a2, a3)|ai ∈ F2}. Since there are 2 choices for each ai there are 2 − 1 nonzero 3-tuples so P(V) has 7 points. Since the plane is a symmetric design we know there are also 7 lines which can be found in Table 4.1. The incidence graph of the Fano Plane is the Heawood graph in Figure 1.3 and can be constructed as followed. The graph is bipartite with vertices {1, 2, 3, 4, 5, 6, 7} ∪

{`1, `2, `3, `4, `5, `6, `7}. If i ∈ {1, 2, 3, 4, 5, 6, 7} and j ∈ {`1, `2, `3, `4, `5, `6, `7} then i is adja-

cent to j if i ∈ j so 1 ∼ `1, `2, `4; 2 ∼ `1, `3, `5; 3 ∼ `2, `3, `6; 4 ∼ `1, `6, `7; 5 ∼ `2, `5, `7; 6 ∼

`3, `4, `7; 7 ∼ `4, `5, `6.

Table 4.1: Points and Lines of the Fano Plane

Points Lines 1 (1,0,0) `1 (1,0,0), (0,1,0), (1,1,0) 2 (0,1,0) `2 (1,0,0), (0,0,0), (1,0,1) 3 (0,0,1) `3 (0,1,0), (0,0,1), (1,0,1) 4 (1,1,0) `4 (1,0,0), (0,1,1), (1,1,1) 5 (1,0,1) `5 (0,1,0), (1,0,1), (1,1,1) 6 (0,1,1) `6 (0,0,1), (1,1,0), (1,1,1) 7 (1,1,1) `7 (1,1,0), (1,0,1), (0,1,1) 28

Figure 4.2: An Elation of the Fano Plane

4.1 Collineations of a Finite Projective Plane

A collineation of a projective geometry is a permutation of its points which maps lines onto lines. A point p is a center of a collineation σ fixes all all subspaces containing p. A hyperplane H is an axis if σ fixes every point in H. We call σ a Baer map if the points and lines that are fixed by σ form a Baer subplane2. A Baer collineation of order two is called a Baer involution. A collineation is central if has a center and is called an elation if the center and axis are incident and a homology otherwise. An involutorial collineation of a projective plane P of order n is either an elation (n even), a homology (n odd) or a Baer involution (n square) [4]. Example: Let P be the Fano Plane. Let σ be the elation that fixes the red line and the red point as seen in Figure 4.2. Then clearly all other points on the red line are fixed by σ. Further, the dashed lines are also fixed but not point-wise. Thus the green point must map to the other green point and the blue point must map to the other blue point. Since P has seven lines and three points on each, there are 21 elations of P all of which are conjugate in the group of collineations.

2A subplane C of P is a Baer subplane if every point of P is incident with a line of C and every line of P is incident with a point of C 29

Figure 4.3: A Homology of a Projective Plane of Order 3

Example: Let P be the projective plane of order 3. We know P has 13 lines with four points on each line and each point incident with four lines. In Figure 4.3 we can see all 13 points of the plane but not all the lines. Let φ be the homology that fixes a the red line and the red point. Then φ fixes the four dashed lines. Again the lines are not fixed point-wise, thus φ maps the remaining two points on each of the four lines onto each other. Since there are 13 lines and 9 points to choose not on a given line, there are 117 homologies of P.

4.2 Semibiplane Construction

Let P be a ﬁnite projective plane of order n, and let α be an involutary collineation of P. To construct a semibiplane B = B(P, α) let the points be the unordered pairs (p, pα) for all p ∈ P such that p 6= pα and let the blocks be the unordered pairs (`, `α) where similarly ` is a line of P such that ` 6= `α. The point (p, pα) is on the block (`, `α) if and only if p is on ` or p is on `α. 30

Theorem 4.2.1. B(P, α) is a semibiplane and has the following parameters where v is the number of points (and blocks) and k is the number of points on every block: (a) if α is an elation then v = n2/2, k = n. (b) if α is a homology then v = (n2 − 1)/2, k = n. √ (c) if α is a Baer involution3 then v = (n2 − n)/2, k = n. Further, in each case, the diameter of the incidence graph of B is at most 4.

α α Proof. Suppose (p1, p1 ) and (p2, p2 ) are two points in B. Then p1 and p2 are in a unique line

α α α α `1 of P and similarly p1 and p2 are in another unique line `2 with p2 and p1 on `2 . If `1 = `1

α α then the points on `1 are ﬁxed so p1 and p2 must also be on `1 so `2 = `1 which means in

α α α α α B,(p1, p1 ) and (p2, p2 ) are not any common blocks. If `1 6= `1 then `2 6= `2 so (`1, `1 ) and

α α α (`2, `2 ) are both blocks of B and are the only two that are common to (p1, p1 ) and (p2, p2 ).

By construction of P we know there is a unique line ` ∈ P that contains p1 and p2. If

α α α α α α α ` 6= ` then (`, ` ) ∈ B and (p1, p1 ) and (p2, p2 ) are on (`, ` ). If ` = ` then (`, ` ) ∈/ P.

However, in P we can ﬁnd another point p3 with unique lines `1 and `2 that contains p1,p3 and

α α p2,p3 respectively such that `1 6= `1 and `2 6= `2 . Thus, B is connected so it is a semibiplane

α α and in the incidence graph of B the path between (p1, p1 ) and (p2, p2 ) can be at most 4. By construction a point (p, pα) is on n lines (`, `α) because in P p is on n + 1 lines and we are just taking away the line at which ` = `α. Thus in each case k = n. If α is an elation, we are ﬁxing a line in P and the n + 1 points on the line. Since P has n2 + n + 1 points that leaves us with n2 points which leaves us with n2/2 pairs (p, pα). Thus in this case v = n2/2. Similarly for the cases in which α is a homology or a Baer involution we can see √ that v = (n2 − 1)/2 or v = (n2 − n)/2 respectively.

Example: Let P be the Fano plane and α be the elation as illustrated in Figure 4.2. Under α the pink lines in Figure 4.2 map to each other as do the orange lines. The semibiplane

3If P is a projective plane of order n and C a proper subplane of order m then m2 = n if and only if C is a Baer subplane [4] 31

Figure 4.4: Constructing the incidence graph of the square from the Fano Plane

will have two points corresponding to the green and blues pairs of points in P and two blocks corresponding to the orange and pink pairs of lines in P. The incidence graph of the semibiplane created from the Fano plane and α will have four vertices, each corresponding to a point or a line of the semibiplane. Since at least one of the green points in P is on at least one of the pink lines then we know there will be an edge connecting the green vertex and pink vertex. We can similarly ﬁnd the remaining edges of the incidence graph to see that it is the graph of the square as seen in Figure 4.2.

4.3 Using GAP to Investigate PG(n,q)

As motivation we will first construct PG(2,4) by hand then find an elation to use in the construction of a semibiplane. To construct PG(2,4) we must first construct a field of order

2 4 denoted F4. Let f(x) = x + x + 1 then since f(x) does not have a root in F2[x], f(x) is irreducible in F2[x]. Thus F2[x]/< f(x) >= {0, 1, x, 1 + x} is a ﬁnite ﬁeld of order 4.

2 Let V = F4. Then a point in PG(2,4) is a non-zero equivalence class of a 3-tuple of the scalars {0, 1, x, 1 + x}. Since there are four choices for each coordinate in the 3-tuple there are 43 − 1 = 63 points. Since there are 3 non-zero scalars each tuple is represented by 3 32 non-zero tuples. Thus the number of points is 63/3 = 21. As a design, we know that the projective plane is symmetric thus the number of points is equal to the number of lines so there are also 21 lines in PG(2,4). Further, we know that each line must contain 5 points. Using Mathematica we were able to construct the 21 points and lines as seen in Table 4.2.

Suppose α is an elation of PG(2,4) with `1 as its axis and the point a its center. Then all

Table 4.2: Points and Lines of PG(2,4)

a (0,0,1) `1 (0, 0, 1), (0, 1, 0), (0, 1, 1), (0, 1, x), (0, 1, 1 + x) b (0,1,0) `2 (0, 0, 1), (1, 0, 0), (1, 0, 1), (1, 0, x), (1, 0, 1 + x) c (0,1,1) `3 (0, 0, 1), (1, 1, 0), (1, 1, 1), (1, 1, x), (1, 1, 1 + x) d (0,1,x) `4 (0, 0, 1), (1, x, 0), (1, x, 1), (1, x, x), (1, x, 1 + x) e (0,1,1+x) `5 (0, 0, 1), (1, 1 + x, 0), (1, 1 + x, 1), (1, 1 + x, x), (1, 1 + x, 1 + x) f (1,0,0) `6 (0, 1, 0), (1, 0, 0), (1, 1, 0), (1, x, 0), (1, 1 + x, 0) g (1,0,1) `7 (0, 1, 0), (1, 0, 1), (1, 1, 1), (1, x, 1), (1, 1 + x, 1) h (1,0,x) `8 (0, 1, 0), (1, 0, x), (1, 1, x), (1, x, x), (1, 1 + x, x) i (1,0,1+x) `9 (0, 1, 0), (1, 0, 1 + x), (1, 1, 1 + x), (1, x, 1 + x), (1, 1 + x, 1 + x) j (1,1,0) `10 (0, 1, 1), (1, 0, 0), (1, 1, 1), (1, x, x), (1, 1 + x, 1 + x) k (1,1,1) `11 (0, 1, 1), (1, 0, 1), (1, 1, 0), (1, x, 1 + x), (1, 1 + x, x) l (1,1,x) `12 (0, 1, 1), (1, 0, x), (1, 1, 1 + x), (1, x, 0), (1, 1 + x, 1) m (1,1,1+x) `13 (0, 1, 1), (1, 0, 1 + x), (1, 1, x), (1, x,1), (1, 1 + x, 0) n (1,x,0) `14 (0, 1, x), (1, 0, 0), (1, 1, x), (1, x, 1 + x), (1, 1 + x, 1) o (1,x,1) `15 (0, 1, x), (1, 0, 1), (1, 1, 1 + x), (1, x, x), (1, 1 + x, 0) p (1,x,x) `16 (0, 1, x), (1, 0, x), (1, 1, 0), (1, x, 1), (1, 1 + x, 1 + x) q (1,x,1+x) `17 (0, 1, x), (1, 0, 1 + x), (1, 1, 1), (1, x, 0), (1, 1 + x, x) r (1,1+x,0) `18 (0, 1, 1 + x), (1, 0, 0), (1, 1, 1 + x), (1, x, 1), (1, 1 + x, x) s (1,1+x,1) `19 (0, 1, 1 + x), (1, 0, 1), (1, 1, x), (1, x, 0), (1, 1 + x, 1 + x) t (1,1+x,x) `20 (0, 1, 1 + x), (1, 0, x), (1, 1, 1), (1, x, 1 + x), (1, 1 + x, 0) u (1,1+x,1+x) `21 (0, 1, 1 + x), (1, 0, 1 + x), (1, 1, 0), (1, x, x), (1, 1 + x, 1)

lines through a are fixed (non-pointwise). Let α be the natural map that sends a point x to x+(0,0,1) (in characteristic 2). Then we can find 8 points and 8 blocks as in figure 4.3 to use in the construction of a semibiplane B.

To construct the semibiplane we must remove the 5 points on `1 and the 5 blocks that contained the point a. Recall that a point (p, pα) in the semibiplane is on a block (`, `α) if in P (V ) either ` or `α contained p. Without loss of generality take the ﬁrst entry in the points in Table 4.3 to be p. Then by looking at Table 4.2 we see that, for example, (f,g) is on blocks

(`6,`7), (`10,`11), (`14,`15) and (`18,`19). Similarly we can see that each point is on 4 blocks and each block contains 4 points. Let G be the incidence graph of B. Then the vertices of 33

Table 4.3: Points and Blocks of the semibiplane

Points Blocks (f,g) (`6,`7) (h,i) (`8,`9) (j,k) (`10,`11) (l,m) (`12,`13) (n,o) (`14,`15) (p,q) (`16,`17) (r,s) (`18,`19) (t,u) (`20,`21)

G are {points of B} ∪ {blocks of B}. Thus G is a bipartite graph with 16 vertices each of ∼ degree 4 so G = Q4. Now we will construct PG(2,4) using GAP then find an involution to use in the construction of a semibiplane. By using the DESIGN package in GAP we are able to construct and study block designs. In the package, the built-in function PGPointFlatBlockDesign(n,q,d) constructs the block design whose points are the projective points of PG(n,q) and whose blocks are the d-flats of PG(n,q) where a d-flat is a set of projective points [9]. In this example PG(2,4) consists of the subspaces of the vector space V(3,4) with the projective points being the 1-dimensional subspaces and the 1-flats being the 2- dimensional subspaces. After loading the design package we can construct PG(2,4) by PGPointFlatBlockDesign(2,4,1). GAP labels the points as 1 through 21 and the blocks are given as lists of 5 points, as seen below.

blocks := [ [ 1, 2, 3, 4, 5 ], [ 1, 6, 7, 8, 9 ], [ 1, 10, 11, 12, 13 ], [ 1, 14, 15, 16, 17 ], [ 1, 18, 19, 20, 21 ], [ 2, 6, 10, 14, 18 ], [ 2, 7, 11, 15, 19 ], [ 2, 8, 12, 16, 20 ], [ 2, 9, 13, 17, 21 ], [ 3, 6, 11, 16, 21 ], [ 3, 7, 10, 17, 20 ], [ 3, 8, 13, 14, 19 ], [ 3, 9, 12, 15, 18 ], [ 4, 6, 12, 17, 19 ], [ 4, 7, 13, 16, 18 ], [ 4, 8, 10, 15, 21 ], [ 4, 9, 11, 14, 20 ], [ 5, 6, 13, 15, 20 ], [ 5, 7, 12, 14, 21 ], [ 5, 8, 11, 17, 18 ], [ 5, 9, 10, 16, 19 ] ] 34

After constructing PG(2,4) we can use another built-in function AutGroupBlockDesign that returns the automorphism group of a given block design. In this case we ﬁnd that the automorphism group is a permutation group with 8 generators. The size of the automorphism group is 120,960. Using the built-in function StructureDescription we see that the automor-

phism group of PG(2,4) is isomorphic to (PSL(3, 2) o C3) o C2. To ﬁnd the involutions in the automorphism group we can use

Filtered(List(ConjugacyClasses(autpg24),Representative),x → Order(x) = 2)

where autpg24 is the name we have given to the automorphism group of PG(2,4). In the case of PG(2,4) there are two types of involutions, elations and Baer involutions. By examining the number of ﬁxed points, we can determine which are the elations and which are Baer involutions. The command above ﬁrst takes a representative of each conjugacy class in the automorphism group then returns those that have order 2. For this example suppose GAP returned

[ (1,20)(2,10)(4,7)(5,17)(6,14)(8,11)(13,16)(19,21), (1,9)(2,20)(3,11)(5,14)(10,15)(13,18)(17,19) ].

Note that a representative of a conjugacy class in GAP is chosen at random. Since the first entry in the list moves 16 of the 21 points, or in other words fixes 5 points, we know that it corresponds to an elation. Similarly, since the second entry fixes 7 points we know it must correspond to a Baer involution. We will refer to the Baer involution as bInv. The fixed points in bInv are the points that do not appear in the permutation so the points 4,6,7,8,12,16,21 are fixed. To construct a semibiplane using bInv we must first remove the subplane in PG(2,4) corresponding to these points. We know which points to remove now we must figure out the blocks to remove. We could write a function in GAP that finds the fixed blocks in PG(2,4) under bInv or in this case since there are only 21 blocks we can find the fixed blocks by inspection. When we generalize this example, we will do the former. Since three points determine a unique line in PG(2,4), the fixed blocks must contain three of the fixed points in bInv. Thus the fixed blocks in this case are 35

[ [ 1, 6, 7, 8, 9 ], [ 2, 8, 12, 16, 20 ], [ 3, 6, 11, 16, 21 ], [ 4, 6, 12, 17, 19 ], [ 4, 7, 13, 16, 18 ], [ 4, 8, 10, 15, 21 ], [ 5, 7, 12, 14, 21 ] ].

Now that we know which points and blocks of PG(2,4) to remove we can use bInv to construct the points and blocks of the semibiplane B. The points of B are clearly (1,9),(2,20),(3,11),(5,14),(10,15),(13,18),(17,19). To construct the points in GAP we will take a representative of each point. A natural choice for the representative is the smaller number in each pair so now the points of B are [1,2,3,5,10,13,17]. Note that after replacing the points in each block by its representative the blocks b, bα will contain the same points so after this step we must take out the extra copy of each block within our list. To construct the blocks of B in GAP we must first remove the fixed points from the remaining 14 blocks in PG(2,4). At this stage we have the points and blocks we need to construct the semi-biplane, however, after removing the fixed points the numbering of the points in B is not ideal. We can rename the points as [1,2,3,4,5,6,7] by replacing the existing name by its position in the (ordered) list of points. After doing so we see that the blocks of B are

[ [ 1, 2, 3, 4 ], [ 1, 3, 5, 6 ], [ 1, 4, 5, 7 ], [ 1, 2, 6, 7 ], [ 2, 4, 5, 6 ], [ 2, 3, 5, 7 ], [ 3, 4, 6, 7 ] ].

Now we see that B is a 2-(7,4,2) design. Further, we can construct the incidence graph of B to see that it is isomorphic to the coHeawood graph. In order to construct more semibiplanes similar to B we can use the code in Appendix D

along with the DESIGN package in GAP. In [3] only two graphs Γ7.10 and Γ8.2 are described as the incidence graphs of semibiplanes constructed from a projective plane, as indicated by Π\σ. A complete list of the graphs in [3] along with some of valency greater than 8 can be found in table 4.4. 36

Table 4.4: Semibiplanes Constructed from Finite Projective Planes

Plane k v |G| Distribution Structure Description of G Structure Description of B PG(2,4) 4 14 336 1 + 4 + 6 + 3 PSL(3, 2) o C2 PSL(3, 2) PG(2,4) 4 16 384 1 + 4 + 6 + 4 + 1 (((C2 × D8) o C2) o C3) o C2) o C2 (((C2 × D8) o C2) o C3) o C2) PG(2,5) 5 24 480 1 + 5 + 10 + 7 + 1 (C2 × C2 × A5) o C2) A5 × C4 PG(2,7) 7 48 2016 1 + 7 + 21 + 17 + 2 (PSL(3, 2) o C2) × S3 C3 × (PSL(3, 2) × C2) PG(2,8) 8 64 10752 1 + 8 + 28 + 24 + 3 - - PG(2,9) 9 78 11232 1 + 9 + 36 + 30 + 2 PSL(3, 3) o C2 PSL(3, 3) PG(2,9) 9 80 11520 1 + 9 + 36 + 31 + 3 (S6 × D8) o C2 (A6 o C8) o C2 PG(2,11) 11 120 13200 1 + 11 + 55 + 49 + 4 (PSL(2, 11) o C2) × D10 C5 × (PSL(2, 11) o C2) Chapter 5

Hypercubes

5.1 The Automorphism Group of Qk

We will begin this section with an example to motivate our claim about the automorphism group of Qk. Example: Let π = (1, 2, 3) ∈ Sym(3) and u = 111. Then if σ = (π, u), Table 5.1 illustrates how σ acts on the vertices of Q3.

Table 5.1: An Automorphism of Q3

x 000 001 010 011 100 101 110 111 σ(x) 111 011 110 010 101 001 100 000

Clearly, as expected, the structure of Q3 remains unchanged under σ so it is easy to see that maps like σ are in the automorphism group of Q3.

∼ k Theorem 5.1.1. Aut(Qk) = Sym(k) o (Z2) .

First notice that a permutation of the coordinates of all the vertices, as binary vectors, preserves the number of diﬀerences in position. Thus any permutation of this type is in the

automorphism group of Qk. Since there are k positions in each vector to permute, we know there are exactly k! automorphisms of this type.

Other elements in the automorphism group of Qk are obtained by adding a constant

k nonzero vector u ∈ (Z2) to a vertex. The eﬀect of adding u to a vertex interchanges some 0s and 1s in the original vertex. Clearly there are 2k of these elements in the automorphism group.

37 38

k So far, we can see that the automorphism group of Qk has order at least 2 · k!, because to obtain an automorphism we can combine a permutation of the coordinates of a binary vector with interchanging some 0s and 1s in some coordinate positions.

We can also put an upper bound on the amount of automorphisms of Qk. Choose a vertex

k u ∈ V (Qk). There are 2 choices for u to be sent. After the position of u is decided, we must choose where to send the k vertices adjacent to u. Since an automorphism must preserve adjacencies, our only choice is to permute the k vertices. After the positions of the vertices adjacent to u are decided, the rest of the vertices are decided through the preservation of

k adjacencies. Thus the automorphism group of Qk cannot be greater than 2 · k!. Now we can

k conclude that Aut(Qk) = {(π, u)|π ∈ Sym(k), u ∈ (Z2) }.

k k Let G =Aut(Qk). To show that G = Sym(k)o(Z2) it remains to show that {1}×(Z2) E k −1 −1 −1 G and {1}×(Z2) ∩Sym(k)×{1} = {1}. Suppose (π, u) ∈ G then (π, u) = (π , −π (u)).

−1 −1 −1 −1 Let x ∈ V (Qk) then (π, u) (1, x)(π, u) = (π , −π (u))(π, u + x) = (1, π (u + x) −

−1 −1 k k π (u)) = (1, π (x)) ∈ {1} × (Z2) . Thus {1} × (Z2) E G. k k Clearly {1} × (Z2) ∩ Sym(k) × {1} = (1, 1). Thus G = Sym(k) o (Z2) .

5.2 Quotients of Hypercubes Resulting in Nonbipartite Graphs

We begin this section with an intuitive result about the vertices of bipartite graphs.

Lemma 5.2.1. The partition of vertices of Qk into vectors of odd weight and vectors of even

weight makes Qk into a bipartite graph.

Proof. To connect two binary vectors by an edge, they must diﬀer in exactly one coordinate. Suppose u, v are two binary vectors of even weight. If u 6= v then u and v must diﬀer in at least 2 coordinates. The same will hold if u and v were both odd. Thus no two binary vectors with weight having the same parity are connected so we can partition the vertices in this way. 39

Theorem 5.2.2. Let σ = (π, u) ∈Aut(Qk), π ∈ Sym(k) and u ∈ Qk, be an involution. Then

σ interchanges the two classes of the bipartion of Qk without ﬁxing any edges if and only if π is a permutation of order 1 or 2 and u has odd weight with π(u) = u and π ﬁxes at least three coordinates in the support of u.

Proof. First suppose π is a permutation of order 1 or 2 and u has odd weight with π(u) = u and π ﬁxes at least three coordinates in the support of u. If u is a translation, x → x + u preserves the classes if u has an even weight and interchanges the classes if u has odd weight.

If π has order 2 then for x ∈ V (Qk) the weight of x and the weight of σ(x) diﬀer by an

even number so σ preserves the two classes of the bipartition of Qk then x → σ(x) + u interchanges the two classes of the bipartition. Suppose an edge (x, π(x) + u) is fixed. Then the distance between x and π(x) + u is 1 which means they differ in exactly one coordinate so π(x) + u − x is a vector containing exactly one nonzero entry and thus has weight one so π fixes a unique coordinate in the support of u. Now suppose the converse. If σ interchanges the two classes of the bipartion then |σ| = 2 then x = σ(σ(x)) = σ(π(x) + u) = π(π(x) + u) + u = π(π(x)) + π(u) + u for all x but since

k 2 (Z2) has characteristic 2 π(u) = u = −u so π = 1 or π = 1. Thus π maps the vertices to the same class of the bipartition so u must change an odd number of coordinates in order to interchange the two classes. Since there are no fixed edges we know π fixes more than one coordinate in the support of u so it must fix at least 3.

From Corollary 2.5.3 if σ is an automorphism as in the previous result, Qk/σ is a (0,2)- graph.

Example: Let σ = (π, u) ∈Aut(Q5). The conjugacy classes of π ∈ Sym(5) are of the form: (1),(ab),(abc),(abcd),(abcde),(ab)(cd),(ab)(cde). In order for σ to be an involution we

2 2 5 need π = 1 and u = 1 which is all u because (Z2) has characteristic two. Thus we need to ﬁnd combinations of (1),(ab) or (ab)(cd) and u such that at u has odd weight and at least three coordinates of u stay ﬁxed. Clearly, ((1),11111) is such a combination and the other 40

two are ((1),00111) and ((ab),(00111)). As veriﬁed in [2] the resulting quotient graphs are

Γ5.2,Γ5.3 and Γ5.4 respectively.

5.3 Quotients of Folded Qk

Theorem 5.3.1. For even k, the folded Qk is bipartite. It follows that for odd k, the folded

Qk is nonbipartite.

Proof. If k is even, then antipodal points in Qk are an even distance apart. Thus the orbits are pairs of vertices from the same class of the bipartition so the vertices of the quotient graph with maintain the same bipartition of vertices. If k is odd, then antipodal points are an odd distance apart so the orbits are pairs of vertices with one vertex in the class of the bipartion with even weight and the other vertex from the class of vertices with odd weight.

The ﬁrst interesting quotients of a folded Qk occur when k = 6. Note that by interesting we mean the resulting graph will be a (0,2)-graph.

∼ Theorem 5.3.2. If u is of the form (0...01...1) with n 0’s then Qk/u = Qn × Qk−n/1 where

Qk−n/1 is the folded (k − n)-hypercube.

Deﬁne φ : Qk/u → Qn × Qk−n/1, φ(u1, ..., un, un+1, ..., uk) = [(u1, ..., un), (un+1, ..., uk)]. It can be easily checked that φ is an isomorphism. ∼ ∼ Example: If u = (0111) then Q4/u = Q1 × Q3/1 = K2 × K4.

Theorem 5.3.3. Suppose k is even. Let σ = (π, u) ∈Aut(Qk/1), π ∈ Sym(k) and u ∈ Qk,

be an involution. Then σ interchanges the two classes of the bipartion of Qk without ﬁxing any edges π is a permutation of order 1 or 2 and u has odd weight, and either π(u) = u+1, or π(u) = u where π ﬁxes at least three coordinates inside the support of u and three coordinates outside the support of u. 41

Figure 5.1: Quotients of Q6

If k is even then by Theorem 5.3.1 we know Qk/1 is bipartite. The proof is similar to that of Theorem 5.2.2.

The folded graph of Q6 is a bipartite graph with 32 vertices of degree 6 and diameter 3. Recall that if we take a quotient by σ ∈ Aut(Γ) such that no two elements in a σ-orbit are joined by a path of 1, 2 or 4 then the resulting graph is a (0,2)-graph so in this case we should look at automorphisms with distance set 3. It turns out that there are only two such

automorphisms and the resulting quotients are K4 × K4 and the Shrikhande graph.

The Shrikhande Graph K4 × K4 with each copy of K4 colored

We can also obtain both graphs by taking diﬀerent quotients of Q6. Let σ = (π, u) where

π = (12)(34)(45) and u = (101010) and x = (abcdef) ∈ V(Q6). Then σ(x) = badcfe + 42

Figure 5.2: Shrikhande graph with code labels

(101010) = b + 1 a d + 1 c f + 1 e so σ(σ(x)) = a b + 1 c d + 1 e f + 1 +(101010) = a + 1

2 b + 1 c + 1 d + 1 e + 1 f + 1. Thus σ = (id, 111111) which means |σ| = 4. Since Q6 has

64 vertices and valency 6 we know Q6/σ will have 16 vertices and valency 6 and in fact ∼ ∼ Q6/σ =Shrikhande. If we start with Q6 and let u = (000111) then Q6/u = Q3 × K4. If we take take another quotient that folds Q3 and keeps the vertices in K4 ﬁxed then we will again obtain K4 × K4. The operation of switching a graph Γ with respect to a set of vertices S replaces the edges between S and its complement Γ S with non-edges and all the non-edges between S and its complement with edges, leaving the edges inside and outside S unaltered. The Shrikhande graph can be obtained from the lattice graph, L2(4) by switching with respect to the set of vertices on the diagonal. If we label the vertices of the resulting version of the Shrikhande graph with the code words 000000, 110000, 010111, 011011 and those obtained by cyclic permutation of each, connecting two vertices if they in two entries as in Figure 5.2, it is easier to see the connection between the Shrikande graph and its bipartite double, the folded

Q6. Note that the folded Q6 is also the bipartite double of L2(4). Other constructions and general properties of the Shrikhande graph are outlined in [8]. 43

5.4 Quotients of coHeawood ×Qk

From [2] the results are known of taking quotients of hypercubes so a natural question to ask ourselves is if similar results hold when combining the hypercube with another graph.

To explore this question we start by looking at quotients of the coHeawood graph ×Qk. With the aid of the program GAP and the package GRAPE we were able to easily inspect the resulting quotient graphs. The code used is in the Appendix A. After reading in the ﬁle containing the code into GAP and another ﬁle containing the code to create the graphs as seen in Appendix B we were able to quickly test graph isomorphisms, compute the sizes of the automorphism groups and use other functions built into GRAPE to explore these graphs. Note that for when k = 4 the graph has 224 vertices so for k > 4 the size of the graph became too large to work with. The general outline of the code in Appendix A is:

1. Find the conjugacy classes of the automorphism group G of the graph Γ

2. Take a representative from each class (by Theorem 2.4.3 this is valid)

3. For each class compute the orbits

4. Find the distance between each pair of vertices in the orbit

5. Create a list of the orbits not containing paths of length 1, 2 or 4

6. Create the quotient graphs of Γ mod the list of orbits from the previous step

Example: Let Λ denote the coHeawood graph. In this example we will show how to construct quotient graphs of Λ × Q3. We can think of this graph as the graph of the cube with a copy of Λ attached to each of its 8 vertices. Since Λ has 14 vertices, a natural labeling of the vertices is 1-14 on vertex 1 of the cube, 15-28 on vertex 2, 29-42 on vertex 3,... and 99-112 on vertex 8.

To ﬁnd the automorphisms of Λ × Q3 that will allow for the quotient to be a (0,2)-graph we use the function relsForQuot. Using this function, GAP returns a list of lists with each 44

sublist containing the orbits of an automorphism acting on the vertices of Λ × Q3. In the case of Λ × Q3 GAP returns a list of length 3 so there are 3 automorphisms σ1, σ2, σ3 such that (Λ × Q3)/σi is a (0,2)-graph.

The ﬁrst sublist, corresponding to σ1 is

R1:=[ [ 1, 99 ], [ 2, 100 ], [ 3, 101 ], [ 4, 102 ], [ 5, 103 ], [ 6, 104 ], [ 7, 105 ], [ 8, 106 ], [ 9, 107 ], [ 10, 108 ], [ 11, 109 ], [ 12, 110 ], [ 13, 111 ], [ 14, 112 ], [ 15, 85 ], [ 16, 86 ], [ 17, 87 ], [ 18, 88 ], [ 19, 89 ], [ 20, 90 ], [ 21, 91 ], [ 22, 92 ], [ 23, 93 ], [ 24, 94 ], [ 25, 95 ], [ 26, 96 ], [ 27, 97 ], [ 28, 98 ], [ 29, 71 ], [ 30, 72 ], [ 31, 73 ], [ 32, 74 ], [ 33, 75 ], [ 34, 76 ], [ 35, 77 ], [ 36, 78 ], [ 37, 79 ], [ 38, 80 ], [ 39, 81 ], [ 40, 82 ], [ 41, 83 ], [ 42, 84 ], [ 43, 57 ], [ 44, 58 ], [ 45, 59 ], [ 46, 60 ], [ 47, 61 ], [ 48, 62 ], [ 49, 63 ], [ 50, 64 ], [ 51, 65 ], [ 52, 66 ], [ 53, 67 ], [ 54, 68 ], [ 55, 69 ], [ 56, 70 ] ].

In R1 vertices 1-14 are mapping to 99-112 (and vice versa); vertices 15-28 to 85-98; vertices 29-42 to 71-84; and vertices 43-56 to 57-70. Further vertex i is mapping to vertex j ≡ i (mod 14). Thus σ1 ﬁxes vertices in the coHeawood graph and maps the vertices in the graph of the cube to their antipodal points. Using the IsIsomorphicGraph function in

GRAPE we ﬁnd that (Λ × Q3)/σ1 is isomorphic to Γ7.49 from [2].

The second sublist, corresponding to σ2 is

R2:=[ [ 1, 99 ], [ 2, 100 ], [ 3, 101 ], [ 4, 103 ], [ 5, 102 ], [ 6, 104 ], [ 7, 106 ], [ 8, 105 ], [ 9, 108 ], [ 10, 107 ], [ 11, 109 ], [ 12, 111 ], [ 13, 110 ], [ 14, 112 ], [ 15, 85 ], [ 16, 86 ], [ 17, 87 ], [ 18, 89 ], [ 19, 88 ], [ 20, 90 ], [ 21, 92 ], [ 22, 91 ], [ 23, 94 ], [ 24, 93 ], [ 25, 95 ], [ 26, 97 ], [ 27, 96 ], [ 28, 98 ], [ 29, 71 ], [ 30, 72 ], [ 31, 73 ], [ 32, 75 ], [ 33, 74 ], [ 34, 76 ], [ 35, 78 ], [ 36, 77 ], [ 37, 80 ], [ 38, 79 ], [ 39, 81 ], [ 40, 83 ], [ 41, 82 ], [ 42, 84 ], [ 43, 57 ], [ 44, 58 ], [ 45, 59 ], [ 46, 61 ], [ 47, 60 ], [ 48, 62 ], [ 49, 64 ], [ 50, 63 ], [ 51, 66 ], [ 52, 65 ], [ 53, 67 ], [ 54, 69 ], [ 55, 68 ], [ 56, 70 ] ].

In R2 the same is happening with respect to the graph of the cube, but notice that vertex 4 is mapping to 103 ≡ 5 (mod 4) and 5 is mapping to 102 ≡ 4 (mod 14) which are of distance 2 from each other in the coHeawood graph, see Figure 1.4. There are more 45 orbits of exhibiting the same behavior as well as those that illustrate that some vertices are staying ﬁxed in the coHeawood graph for instance 1 is mapping to 99 ≡ 1 (mod 14).

Thus σ2 either ﬁxes vertices in the coHeawood graph or maps them to vertices of distance 3 away and maps vertices in the cube to their antipodal points. Using the IsIsomorphicGraph function in GRAPE we ﬁnd that (Λ × Q3)/σ2 is isomorphic to Γ7.50 from [2].

The last sublist, corresponding to σ3 is

R3:=[ [ 1, 44 ], [ 2, 43 ], [ 3, 48 ], [ 4, 49 ], [ 5, 50 ], [ 6, 45 ], [ 7, 46 ], [ 8, 47 ], [ 9, 54 ], [ 10, 55 ], [ 11, 56 ], [ 12, 51 ], [ 13, 52 ], [ 14, 53 ], [ 15, 30 ], [ 16, 29 ], [ 17, 34 ], [ 18, 35 ], [ 19, 36 ], [ 20, 31 ], [ 21, 32 ], [ 22, 33 ], [ 23, 40 ], [ 24, 41 ], [ 25, 42 ], [ 26, 37 ], [ 27, 38 ], [ 28, 39 ], [ 57, 100 ], [ 58, 99 ], [ 59, 104 ], [ 60, 105 ], [ 61, 106 ], [ 62, 101 ], [ 63, 102 ], [ 64, 103 ], [ 65, 110 ], [ 66, 111 ], [ 67, 112 ], [ 68, 107 ], [ 69, 108 ], [ 70, 109 ], [ 71, 86 ], [ 72, 85 ], [ 73, 90 ], [ 74, 91 ], [ 75, 92 ], [ 76, 87 ], [ 77, 88 ], [ 78, 89 ], [ 79, 96 ], [ 80, 97 ], [ 81, 98 ], [ 82, 93 ], [ 83, 94 ], [ 84, 95 ] ].

In R3 we see vertices 1-14 are mapping to 43-56; vertices 15-28 to 29-42; vertices 57-

70 to 99-112; and vertices 71-84 to 85-98. Thus in the graph of the cube σ3 is mapping vertices to those of distance two away. In the coHeawood graph, we see again that some vertices are being sent to vertices of distance 3 away as illustrated by vertex 14 mapping to 53 ≡ 11 (mod 14). However, we also see, as is the case in vertex 3 mapping to 48 ≡ 6 (mod 14), that vertices of the coHeawood are mapping to vertices of distance 1 away. Using the IsIsomorphicGraph function in GRAPE we ﬁnd that (Λ × Q3)/σ3 is isomorphic to Γ7.48 from [2].

Through a similar process, we found that the automorphism group of Λ × Q1 does not contain any interesting elements. In the case of Λ × Q2 the automorphism group contains one good element that maps vertices in ∆ to vertices of distance 1 or 3 away and vertices in Q2 to those of distance 2 away. Lastly, in the case of Λ × Q4, we found 5 elements of the automorphism group ﬁxing vertices in Λ with moving vertices distance 3 in Q4, 0 or 2 in Λ 46

Table 5.2: Graphs Isomorphic to Quotients of Λ × Qk, k = 2, 3, 4

Original Graph Isomorphic Graph in [2] or [3] Quotient Graph Isomorphism Λ × Q2 ∆6.12 Γ6.8 Λ × Q3 ∆7.39 Γ7.48 Λ × Q3 ∆7.39 Γ7.49 Λ × Q3 ∆7.39 Γ7.50 Λ × Q4 ∆8.103 Γ8.188 Λ × Q4 ∆8.103 Γ8.189 Λ × Q4 ∆8.103 Γ8.190 Λ × Q4 ∆8.103 Γ8.191 Λ × Q4 ∆8.103 Γ8.182

with 3 in Q4, 1 or 3 in Λ with 4 in Q4, 1 or 3 in Λ with 2 in Q4 and 1 or 3 in Λ with 2 or 4 in Q4. Appendix A

Code Used to Find “Good Conjugacy” Class Relations

#grp is automorphism group of graph. This function returns #a list of conjugacy classes of the group. ccOrbits:=function(grp) local r,i;r:=[]; for i in [1..NrConjugacyClasses(grp)] do r[i]:= OrbitsPerms([Representative(ConjugacyClasses(grp)[i])] ,[1..LargestMovedPoint(grp)]);od;return r;end;

#The following two functions are used to return #a list whose elements are of distance 1 apart isLengthgt1:=function(l) return Length(l)>1;end; orbitsLengthgt1:=function(orbs) return Filtered(orbs,isLengthgt1); end;

#This function takes in a list and filters out elements that are not pairs 2combs:=function(l) return Combinations(l,2);end;

#This function returns a list whose elements are of distance 1 apart #and that are pairs orbitRel:=function(orbs) return Concatenation(List(orbitsLengthgt1(orbs),2combs));end;

#This function returns a list of orbits created in the first function #with the previous function applied to them orbitRelsGrp:=function(grp) return List(ccOrbits(grp),orbitRel);end;

47 48

# l is a list of vertex adjacencies of a graph. This #function returns the set of vertices adjacent to any vertex in the list l1 adjacentVertsList:=function(l,l1) local l2,i,k; l2:=[];k:=0; for i in l1 do k:=k+1; l2[k]:=l[i];od; return Unique(Concatenation(l2)); end;

#g is a graph given as an adjacency list graphDist:=function(g,v1,v2) local d,l;d:=0;l:=[v1];while not v2 in l do d:=d+1;l:=adjacentVertsList(g,l);od;return d;end;

#If l is a list of pairs of vertices this function returns a #list of the distances between each element of the pair #g is a graph distSet:=function(g,l) local r,i;r:=[]; for i in [1..Length(l)] do r[i]:=graphDist(g,l[i][1],l[i][2]);od; return r;end; distrels:=function(grph,grp) local r,i;r:=[]; for i in [1..NrConjugacyClasses(grp)] do r[i]:=distSet(grph,orbitRelsGrp(grp)[i]); od; return r;end;

#The following functions are used to filter out orbits #containing paths of 1,2 or 4 eq2:=function(n) return n=2;end; eq1:=function(n) return n=1;end; eq4:=function(n) return n=4;end; isnot124:=function(n) return not eq1(n) and not eq2(n) and not eq4(n);end; 49 listnot124:=function(l) return ForAll(l,isnot124);end; posnsnot124:=function(l) local r,t,j;r:=0;t:=[]; for j in [1..Length(l)] do if listnot124(l[j]) then r:=r+1;t[r]:=j;fi;od; return t;end; posnsForQuot:=function(grph,grp) return posnsnot124(distrels(grph,grp));end;

#This function returns the relations needed for GRAPE #to create a quotient graph relsForQuot:=function(grph,grp) local r,t,j;r:=0;t:=[]; for j in [2..Length(orbitRelsGrp(grp))] do if listnot124(distrels(grph,grp)[j]) then r:=r+1;t[r]:=orbitRelsGrp(grp)[j];fi;od; return t;end;

#This function returns a list of the distances #between the vertices in each orbit distsForQuot:=function(grph,lst) local t,j;t:=[]; for j in [1..Length(lst)] do t[j]:=distSet(grph,lst[j]);od; return t;end; Appendix B

Code to Construct Graphs of CoHeawood × Qk

#below is the adjacency matrix for the coHeawood graph #it was copied from the site http://www.win.tue.nl/~aeb/graphs/recta/02graphs.html b41:=[[1,2,3,4],[ 0,5,6,7],[ 0,5,8,9],[ 0,6,8,10],[ 0,7,9,10],[ 1,2,11,12],[ 1,3,11,13],[ 1,4,12,13],[ 2,3,12,13],[ 2,4,11,13],[ 3,4,11,12],[ 5,6,9,10],[ 5,7,8,10],[ 6,7,8,9]]+1;

LoadPackage("grape");

G:=Group(());

#Below are function built into GRAPE or GAP used to #construct the graph and automorphism group of the graph graphb41:=Graph(G,[1..14],OnPoints,function(x,y) return y in b41[x];end,true); autb41:=AutGroupGraph(graphb41);

#Given the adjacency matrix of a graph the function below returns #the adjacency matrix of the Cartesian product of that graph with K2 cartProdWithK2:=function(m)

50 51 local a,i,j; a:=IdentityMat(2*Length(m)); for i in [1..Length(m)] do for j in [1..Length(m)] do a[i][j]:=m[i][j]; od;od; for i in [1..Length(m)] do for j in [1..Length(m)] do a[i+Length(m)][j+Length(m)]:=m[i][j]; od;od; for i in [1..Length(m)] do for j in [1..Length(m)] do if i=j then a[i+Length(m)][j]:=1;else a[i+Length(m)][j]:=0; fi; od;od; for i in [1..Length(m)] do for j in [1..Length(m)] do if i=j then a[i][j+Length(m)]:=1;else a[i][j+Length(m)]:=0; fi; od;od; return a; end;

#Given the adjacencies of a graph as a list, as in b41 above, #this function returns the binary adjacency matrix adjListToMat:=function(l) local m,i,j; m:=IdentityMat(Length(l)); for i in [1..Length(l)] do for j in [1..Length(l)] do if j in l[i] then m[i][j]:=1; else m[i][j]:=0; fi; od;od;return m; end; adjb41:=adjListToMat(b41);

#The adjacency matrices for the graphs of the #coHeawood with the hypercubes are created below k2:=[[0,1],[1,0]]; cart41K2:=cartProdWithK2(adjb41); cart41cube2:=cartProdWithK2(cart41K2); cart41cube3:=cartProdWithK2(cart41cube2); cart41cube4:=cartProdWithK2(cart41cube3); cart41cube5:=cartProdWithK2(cart41cube4);

#The graphs and automorphism groups are created below graphCart41K2:=Graph(G,[1..28],OnPoints,function(x,y) return cart41K2[x][y]=1; end,true); 52 graphCart41cube2:=Graph(G,[1..56],OnPoints,function(x,y) return cart41cube2[x][y]=1; end,true); graphCart41cube3:=Graph(G,[1..112],OnPoints,function(x,y) return cart41cube3[x][y]=1; end,true); graphCart41cube4:=Graph(G,[1..224],OnPoints,function(x,y) return cart41cube4[x][y]=1; end,true); autGraphCart41K2:=AutGroupGraph(graphCart41K2); autGraphCart41cube2:=AutGroupGraph(graphCart41cube2); autGraphCart41cube3:=AutGroupGraph(graphCart41cube3); autGraphCart41cube4:=AutGroupGraph(graphCart41cube4);

#The functions below are used to create the adjacency #list of a graph given a binary adjacency matrix adjacentVertsFromAdjMat:=function(m,i) local l,j,k; l:=[];k:=0; for j in [1..Length(m[i])] do if m[i][j]=1 then k:=k+1;l[k]:=j;fi;od; return l; end; adjMatToList:=function(m) local l,i; l:=[]; for i in [1..Length(m)] do l[i]:=adjacentVertsFromAdjMat(m,i);od; return l;end;

#The adjacency lists for the new graphs #are created below listCart41K2:=adjMatToList(cart41K2); listCart41cube2:=adjMatToList(cart41cube2); listCart41cube3:=adjMatToList(cart41cube3); listCart41cube4:=adjMatToList(cart41cube4); Appendix C

Code to Construct SemiBiplane and Incidence Graph from Finite Projective Plane

mPoints:=function(inv) return MovedPoints(inv);end;

#n is number of points fixedPoints:=function(inv,n) return Difference([1..n],mPoints(inv));end;

#blk is a list of blocks(ordered lists). inv is an involution (or any automorphism), acting on points isfixedblockQ:=function(blk,inv) return OnSets(blk,inv)=blk;end;

#returns fixed blocks fxdblcks:=function(blcks,inv) return Filtered(blcks,x->isfixedblockQ(x,inv));end;

#removes fixed blocks rmvdblcks:=function(blcks,inv) return Difference(blcks,fxdblcks(blcks,inv));end;

#removes fixed points remfxdpts:=function(blcks,pnts) return List(blcks,x->Difference(x,pnts));end; pairrep:=function(inv,k) local r; if k^inv

#blcks should be, fp is fixed points preblocks:=function(blcks,fp,inv) return Unique(List(remfxdpts(blcks,fp),

53 54 x->SortedList(List(x,y->pairrep(inv,y)))));end;

#following two functions used to renumber the points in the blocks ren:=function(l,i) return Position(l,i);end; psn:=function(blcks) return Unique(Flat(blcks));end;

#returns blocks to use when constructing design desblocks:=function(blcks) return Unique(List(blcks, x->SortedList(List(x,y->ren(psn,y)))));end;

#use B=desblocks in BlockDesign(v,B) to construct the design

#below is to construct the incidence graph list psns:=function(l,i) local r,j; r:=[]; for j in [1..Length(l)] do if i in l[j] then Add(r,j); fi; od; return r;end;

#b is blocks of design and n is the number of vertices on incidence graph graphList:=function(b,n) local l; l:=[]; for i in [n/2+1..n] do l[i]:=b[i-n/2];od; for i in [1..n/2] do l[i]:=psns(b,i)+n/2;od;return l;end;

#to construct graph; n is the number of vertices incGraph:=function(blcks,n) return Graph(G,[1..n],OnPoints,function(x,y) return y in graphList(blcks,n)[x];end,true);end; Appendix D

Sample List of Bipartite Graphs from [2]

Table D.1: ∆0.1 − ∆6.4 from [2]

# k v d distribution gsz orbits graph 1 0 1 0 1 1 tra 20 1 1 2 1 1+1 2 tra 21 1 2 4 2 1+2+1 8 tra 22 1 3 8 3 1+3+3+1 48 tra 23 1 4 14 3 1+4+6+3 336 tra 2-(7,4,2) 2 4 16 4 1+4+6+4+1 384 tra 24 1 5 22 3 1+5+10+6 1320 tra 2-(11,5,2) 2 5 24 4 1+5+10+7+1 480 tra bipd(ico) 3 5 28 4 1+5+10+9+3 672 tra 2 × ∆4.1 4 5 32 5 1+5+10+10+5+1 3840 tra 25 1 6 32 3 1+6+15+10 1536 tra 2-(16,6,2) 2 6 32 3 1+6+15+10 768 tra 2-(16,6,2) 3 6 32 3 1+6+15+10 23040 tra 2-(16,6,2), 26/1 4 6 36 4 1+6+15+12+2 96 12+24 -

In Table D.1 # is the serial number given to the graph by Brouwer. The columns k, v, and d are the valency, number of vertices and diameter of the graph. The distribution column gives the distance distribution of a vertex of the graph while the gsz column is the size of the automorphism group of the graph. The graph column gives a description or a construction of the graph. In the table, 2k denotes the graph of the k-hypercube, tra means the graph is transitive and bipd(ico) stands for the bipartite double of the icosahedron.

The coHeawood graph is ∆4.1 and is the incidence graph of a a (7,4,2) biplane. Notice that

∆6.1, ∆6.2 and ∆6.3 are all incidence graphs of (16,6,2) biplanes where ∆6.3 is also the graph of the folded Q6.

55 Bibliography

[1] Brouwer, Andries E., A.M. Cohen, A. Neumaier, Distance-Regular Graphs, Springer- Verlag, Berlin, 1989.

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[3] Brouwer, Andries E., P.R.J. Osterg˙ard,¨ Classiﬁcation of small (0,2)-graphs of valency 8 (2007) http://www.win.tue.nl/ aeb/preprints.html.

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[8] Pedersen,Ryan M., The Shrikhande Graph (2007) http://www- math.cudenver.edu/ wcherowi/courses/m6023/ryan.pdf.

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56 57

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