# STRONGLY REGULAR GRAPHS

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1 STRONGLY REGULAR GRAPHS By: BAHMAN AHMADI A Project Report for The Course ALGEBRAIC COMBINATORICS Instructor: Dr. Karen Meagher Department of Mathematics and Statistics University of Regina April, 2009.

2 In The Name of GOD, The Most Beneficent, The Most Merciful.

3 1 Introduction One of the most important family of regular graphs is the family of Strongly Regular Graphs (abbreviated SRG) which has so many beautiful properties. One of these well-known properties is the fact that the eigenvalues of all SRGs depend only the parameters of these graphs. There are many SRGs arising from combinatorial concepts such as orthogonal arrays, Latin squares, conference matrices, designs and geometric graphs. In this report, we try to familiarize the reader with the concept of strongly regular graphs and their nice properties and connections with these combinatorial objects and to describe an interesting open problem related to the cores of SGRs. The main reference for this report is [1].

4 2 1. Definition and basic properties Definition 1.1. Let X be a graph which is neither empty nor complete. Then X is said to be a strongly regular graph (SRG) with parameters (n, k, a, c) if X is a k regular graph on n vertices in which every pair of adjacent vertices has a common neighbors and every pair of non-adjacent vertices has c common neighbors. Throughout the text, when X is supposed to be a SRG, unless otherwise stated, we assume that the parameters of X are (n, k, a, c). We will denote the adjacency matrix of any graph X with A(X). It easily follows from the definition that if X is a bipartite SRG, then a = 0; but as in the following example, the converse is not true. Example 1.2. i) The cycle C 4 is a (bipartite) SRG with parameters (4, 2, 0, 2). The cycle C 5 is a (non-bipartite) SRG with parameters (5, 2, 1, 0). It is easily seen that for any n 6 the cycle C n is not a SRG. ii) The Petersen graph P 10 is a SRG with parameters (10, 3, 0, 1); indeed since the diameter is 2, every nonadjacent pair of vertices has at least a common neighbor, and since the girth is 5, this common neighbor is unique. Also a = 0 follows from the fact that the girth is 5 and so P 10 has no triangle. iii) A connected graph X is said to be a Moore graph if diam(x) = d and girth(x) = 2d + 1. It can be seen ([1], p.90) that X is regular. Now if X is a Moore graph of diameter 2 on n vertices with valency k, then by similar argument as in part (ii), we can see that X is a SRG with parameters (n, k, 0, 1). iv) Let X be L(K n ), the line graph of K n. Then the number of vertices of X, is the number of edges of K n, namely n(n 1) 2. Let x, y V (K n ). Then the edge e = xy is adjacent to n 2 edges through x and to n 2 other edges through y. Therefor X is (2n 4) regular. Let e = xy and e = xz be two adjacent edges in

6 4 i) X is not connected; ii) c = 0; iii) a = k 1; iv) X is isomorphic to mk k+1, for some m > 1. Proof. Suppose X is not connected and let x, y V (X) be such that there is no path joining them to each other. So x is not adjacent to y, and they cannot have a common neighbor; therefore c = 0. Let c = 0, and let x, y V (X) be adjacent. If z V (X) and z y, then x must be adjacent to z; indeed, if x is not adjacent to z, since they have y as a common neighbor, then c 0. Thus x is adjacent to all the neighbors of y and so a = k 1. Obviously a = k 1 implies that each component of X has to be a complete graph, and since the degree is k, each component has to be isomorphic to K k+1, and the result follows. The part (iv) (i) is trivial. Corollary 1.6. Suppose X is an imprimitive SRG. Then either X = mk k+1 or X = mk k+1, for some m > 1. It is reasonable to ask: if n, k, a, c are some positive integers such that k < n and a, c k, whether they are parameters of some SRG or not. The following lemma provides us with a simple feasibility condition. Lemma 1.7. If X is a SRG with parameters (n, k, a, c), then k(n a 1) = (n k 1)c. Proof. Fix a vertex u V (X), and let A be the set of all neighbors of u, and B = V (X) \ A. Note that A = k and B = n k 1. We count t, the number of all edges between A and B, in two ways.

7 5 First, each vertex in A is adjacent to a vertices in A and so is adjacent to k 1 a vertices in B. Thus t = A (k a 1) = k(k a 1). Second, any vertex x B, is adjacent to c vertices in A, since x is not adjacent to u. Therefor t = B c = (n k 1)c, and the result follows. 2. Eigenvalues In this section we try to determine all the eigenvalues of the strongly regular graphs using their parameters. We shall show that all SRGs have at most three distinct eigenvalues and evaluate their multiplicities. Let X be a SRG and let A be the adjacency matrix of X. Note that the ij th entry of the matrix A 2 is the number of walks of length 2 joining i to j. Consider the following cases: If i = j, then (A 2 ) ij is the degree of i. So all the diagonal entries of A 2 are equal to k. If i j and they are adjacent, then (A 2 ) ij is the number of common neighbors of i and j. Thus (A 2 ) ij = a. If i j and they are nonadjacent, then (A 2 ) ij is the number of common neighbors of i and j. Thus (A 2 ) ij = c. Consequently we have the following identity: A 2 = ki + aa + c(j I A), where J is the n n all 1 s matrix. We may rewrite it as A 2 (a c)a (k c)i = cj.

8 6 Since X is k regular, k is an eigenvalue of A with the eigenvector 1, which is the all 1 s vector of length n. Furthermore, if z is another eigenvector of A with eigenvalue λ, then z has to be orthogonal to 1. Therefore A 2 z (a c)az (k c)iz = cjz = 0, so λ 2 (a c)λ (k c) = 0. This means that the other eigenvalues of A are the roots of the monic quadratic equation (2.1) x 2 (a c)x (k c) = 0, Set = (a c) 2 + 4(k c). Then the roots of (2.1), i.e. the eigenvalues of A other than k, are (2.2) θ = (a c) + 2, τ = (a c). 2 Now θτ = (c k) and therefore, for the cases in which c < k, θ and τ have opposite signs. We always assume that τ < 0. Note that c = k if and only if X = mk k+1, for some m > 1. Thus for a strongly regular graph X, which is not isomorphic to mk k+1, (m > 1), we have τ < 0 and θ > 0. Next we try to find the multiplicities of the eigenvalues. We do the work for connected SRGs and the result for all SRGs will follow easily. Let X be a connected SRG with eigenvalues k, θ and τ. Since X is connected, the multiplicity of k is 1. Denote by m θ and m τ the multiplicities of θ and τ, respectively. Because tr(a) = 0, we have that m θ +m τ = n 1 and m θ θ+m τ τ = k. Thus following equalities hold: m θ = (n 1)τ + k, m τ = θ τ (n 1)θ + k. θ τ

9 7 On the other hand, (θ τ) 2 = (θ + τ) 2 4θτ = (a c) 2 + 4(k c) =. We can, therefore, write m θ and m τ as m θ = 1 2k + (n 1)(a c) [(n 1) ], 2 (2.3) m τ = 1 2k + (n 1)(a c) [(n 1) + ]. 2 Note that in equations (2.3), right hand sides of the equations have to be integers. Therefore, (2.3) also can be regarded as other feasibility conditions on a parameter set (to be parameters of some SRGs). Following proposition shows that having three distinct eigenvalues, is indeed a sufficient condition for connected regular graphs to be strongly regular. Proposition 2.1. If X is a connected k regular graph with exactly three distinct eigenvalues k, θ and τ, then X is strongly regular. Proof. Consider the matrix (2.4) M = 1 (A θi)(a τi), (k θ)(k τ) where A is the adjacency matrix of X and n is the number of vertices of X. Since M is a polynomial in A, we can see that all the eigenvalues of M are 0 or 1. Let z be an eigenvector of A with eigenvalue θ. Then 1 Mz = (k θ)(k τ)(a θi)(a τi)z 1 = (k θ)(k τ)(a θi)(az τz) 1 = (k θ)(k τ)(a θi)(θz τz) = 1 (k θ)(k τ) (θaz τaz θ2 z + θτz) = 1 (k θ)(k τ) (θ2 z τθz θ 2 z + θτz) = 0. Therefore all eigenvectors of A corrosponding to the eigenvalue θ are in the kernel of M. The same argument shows that this is also the case for all the eigenvectors of A corresponding to the eigenvalue τ. Since the set of all eigenvectors of A is

10 8 a basis for R n, this implies that the rank of M is equal to the multiplicity of the eigenvalue k which is 1 (since X is connected). That is, all the rows of M are linear combinations of one row. On the other hand, 1 is an eigenvalue of M with eigenvector 1, and so M1 = 1. Therefor M = 1 nj. Substituting this in (2.4) yields A 2 = θτi (θ + τ)a (k θ)(k τ) J. n Since we could write A 2 as a linear combination of I, A and J, this shows that A is a strongly regular graph (i.e. the parameters are well-defined). Definition 2.2. Let X be a SRG with m θ = m τ. Then X is said to be a conference graph. This name arises from the fact that these graphs are associated to so-called symmetric conference matrices. A conference matrix C of order n is an n n, (0, 1, 1) matrix whose diagonal entries are all 0 and CC T = (n 1)I. If C is a symmetric conference matrix, then by permuting rows and columns (if necessary) we may assume that the first row and the first column of C are all 1 s except the element C 11 = 0. Now delete the first row and the first column and switch all +1 s to 0 and all 1 s to 1 to obtain the (n 1) (n 1) matrix A C. Let X C be a graph whose adjacency matrix is A C. It is not hard to show that X C is a conference graph. It has been shown that if n 2 (mod2), then a conference matrix of order n exists if and only if n 1 is the sum of two squares. (For a proof of this, you may refer to [2] or [3]). Therefore we have the following result. Corollary 2.3. A conference graph with n vertices exists if and only if n is the sum of two squares. Example 2.4. Dose a SRG with parameters (21, 10, 4, 5) exists?

11 9 Answer: If X is a SRG with parameters (21, 10, 4, 5), then using (2.3), we see that m θ = m τ = 10, and so X would be a conference graph; but this is contradiction, since 21 is not sum of two squares. Recall that a graph X is said to be vertex-transitive if for any pair {x, y} V (X), there is an automorphism g on X such that g(x) = y. One of the obvious properties of vertex-transitive graphs is that they are regular. Example 2.5. In this example we introduce an important (infinite) family of conference graphs, namely the Paley graphs. Let q be a prime power such that q 1 (mod4), and let GF (q) be a filed with q elements. Then the Paley graph P (q) is a graph with vertex set GF (q), in which two vertices are adjacent if and only if their difference is a square of an element of GF (q). To show that P (q) is not directed, we should show that 1 is a square. Since GF (q) = GF (q) \ {0} is a cyclic (multiplicative) group, there is an α GF (q) such that GF (q) =< α >= {α, α 2,..., α q 1 = 1}. Let C = {(α 1 ) 2, (α 2 ) 2, (α 3 ) 2,..., (α (q 1)/2 ) 2 } be the set of all squares in GF (q). Obviously C is a subgroup of GF (q) of order (q 1)/2. We claim that 1 C. Note that (α (q 1 )/4) GF (q), (since q 1 (mod4)), and ((α (q 1)/4 ) 2 ) 2 = α q 1 = 1 (α (q 1)/4 ) 2 = ±1; and since GF (q) =< α >, we must have 1 = (α (q 1)/4 ) 2 C (since otherwise we would have < α > < GF (q) ). Now for any β GF (q) consider the homomorphism x x + β. It is easily seen that this homomorphism is an automorphism, and therefore P (q) is vertextransitive. Consequently P (q) is regular. On the other hand the vertex 0 is adjacent to all the elements of C. Thus P (q) is ( q 1 2 ) regular.

12 10 We claim that P (q) is self-complementary; i.e. P (q) P (q). Let 0 β / C and define φ : P (q) P (q) by x βx. This map is clearly well-defined and bijection. We claim that φ is an isomorphism. To show that, first note that for some odd number s, β = α s. If x, y V (P (q)) are adjacent in P (q), then x y = (α i ) 2, for some 1 i (q 1)/2. Therefore φ(x) φ(y) = βx βy = βα 2i = α 2i+s / C; that is, φ(x) and φ(y) are not adjacent in P (q), and so are adjacent in P (q). Also if x and y are not adjacent in P (q), then x y = α t for some odd number t. So φ(x) φ(y) = βx βy = βα t = α s+t C; that is, φ(x) and φ(y) are adjacent in P (q), and so are not adjacent in P (q). Consequently φ is an isomorphism and P (q) is self-complementary. Now let γ C, and consider the map x γx. It is easily seen that this is an automorphism of P (q), which fixes 0 and maps neighbor 1 of 0 to γ. We can deduce that the number of common vertices of 0 and γ is independent from the choice of γ C. Thus the number of triangles on an edge containing 0 is independent of the edge. Since P (q) is vertex-transitive, it follows that any pair of adjacent vertices in P (q) has the same number of common vertices. On the other hand since P (q) is selfcomplementary, every pair of nonadjacent vertices of P (q) has the same common neighbors as well. Therefor P (q) is a strongly regular graph with parameters (n = q, k = q 1 2, a, c). Finally we determine a and c. Since a = ā, from Proposition 1.3, we have a c = n 2k 2 = q 2 q = 1, and using the fact that k(k a 1) = kc, we have a + c = k 1. Together these imply that c = k/2 and a = c 1; thus a = (q 5)/4 and c = (q 1)/4, and so P (q) is a SRG with parameters (q, q 1 2, q 5 4, q 1 4 ).

13 11 Using these parameters we see that θ = ( 1 + q)/2 and τ = ( 1 q)/2. Furthermore the multiplicities are m θ = m τ = (q 1)/2. Consequently P (q) is a conference graph. As we saw in the example above, all the parameters of the Paley graph can be written in terms of q, the number of vertices. We show that this is the case for all conference graphs. We need the following lemma. Lemma 2.6. Let X be a SRG with distinct eigenvalues k, θ and τ. Then m θ m τ = nk k (θ τ) 2. Proof. Let 1 k k P = 1 θ 1 θ 1 τ 1 τ in which first, second and third column are the eigenvalues of I, X and X. We show that (2.5) P T 0 m θ 0 P = n 0 k m τ 0 0 k Denote the left hand side of (2.5) by M. Then we have M 12 = k + m θ θ + m τ τ = 0, M 13 = k + m θ (1 θ) + m τ ( 1 τ) = tr(a( X)) = 0, M 23 = k k + m θ θ( 1 θ) + m τ τ( 1 τ) = tr(a(x)a( X)) = 0, where A(X) is the adjacency matrix of X. To see that tr(a(x)a( X)) = 0, it may help to note that tr(bc) = ij B ij C ji. Given that M is symmetric, these equations show that all the off-diagonal entries of M are zero. Similar calculations show that the diagonal entries of M are n, tr(a(x) 2 ) and tr(a( X) 2 ). But tr(a(x) 2 ) is the number of closed walks of length

14 12 two in X, namely twice the number of edges. Therefore tr(a(x) 2 ) = nk. Similarly tr(a( X) 2 ) = n k and the claim is proved. Now if we take determinant of both sides of (2.5), we will find (det(p )) 2 m θ m τ = n 3 k k (n(τ θ)) 2 m θ m τ = n 3 k k; and the lemma follows. Proposition 2.7. Let X be a conference graph with parameters (n, k, a, c). Then k = (n 1)/2, a = (n 5)/4 and c = (n 1)/4. Proof. Since m θ + m τ = n 1 and m θ = m τ, we have m := m θ = m τ = (n 1)/2. Obviously gcd(m, n) = 1 (i.e. m and n are relatively prime). By Lemma 2.6, we have m 2 nk k m 2 k k. On the other hand k + k = n 1, and so k k (n 1) 2 /4 = m 2 and equality holds if and only if k = k. Therefore k = k = m. Since m(θ + τ) = mθ + mτ = k, we see that a c = θ+τ = k/m = 1 and so a = c 1. Finally because k(k a 1) = kc, we find c = k a 1, and hence c = k/2. Putting these all together shows that X has the stated parameters. If a SRG, X is not conference graph, then m τ m θ will be a non-zero integer. But using (2.3), we have m τ m θ = 2k + (n 1)(a c). Therefore must be a perfect square, and so θ and τ are rational numbers. On the other hand θ and τ are the roots of the monic quadratic equation (2.1). Consequently θ and τ are integers. Therefore we have the following interesting result. Corollary 2.8. Let X be a SRG with eigenvalues k, θ and τ. Then either

15 13 i) X is a conference graph, or ii) (θ τ) 2 is a perfect square and θ and τ are integers. When a SRG has p or 2p vertices, where p is a prime number, using the facts above, we would have very interesting information about the parameter set; in what follows we prove these results. Proposition 2.9. If X is a SRG on p vertices, where p is a prime number, then X is a conference graph. Proof. By Lemma 2.6, we have (2.6) (θ τ) 2 = pk k m θ m τ. By Corollary 2.8, if X is not a conference graph, then (θ τ) 2 will be a perfect square. Thus pk k m θ m τ is a perfect square having p as a factor. So p 2 must divide pk k m θ m τ ; that is, p must divide k k m θ m τ. But this cannot be the case, since all of the numbers k, k, m θ and m τ are non-zero numbers less than p. This contradiction shows that X has to be a conference graph. Lemma Let X be a primitive SRG with an eigenvalue λ of multiplicity n/2. If k < n/2, then the parameters of X are ((2λ + 1) 2 + 1, λ(2λ + 1), λ 2 1, λ 2 ). Proof. Since X is connected, λ k, and so λ is either θ or τ. Since m θ = n 1 m τ, we see that m θ m τ and hence θ and τ are integers. We will show by contradiction that λ = θ. Suppose λ = τ and so m τ = n/2. Note that m τ = (n 1)θ + k θ τ = nθ + k θ θ τ m τ (nθ + k θ).

16 14 On the other hand m τ n. Therefore m τ must divide k θ. Now we claim that 0 < θ < k. Indeed if θ = 0, then using the fact that θτ = (c k), we will have c = k. Therefore by Proposition 1.3, ā = n k 2 = k 1, and so by Lemma 1.5, X would be disconnected which is a contradiction. Consequently θ > 0. Also if θ = k, then using the fact that m θ = n 1 m τ, we will have that the multiplicity of k as an eigenvalue of X is n 2 1 which contradicts with the connectivity of X. So the claim is proved; that is 0 < k θ < n/2 and so m τ (= n/2) cannot divide k θ. This contradiction shows that we must have λ = θ and so m θ = n/2. Next, note that m θ = (n 1)τ + k θ τ = nτ + k τ θ τ m θ (nτ + k τ), and since m θ n, m θ must divide k τ. But since k τ, k τ 2k < n, and so we must have m θ = k τ. Set m = m θ = n/2, so m τ = m 1. Since tr(a(x)) = 0, we have k + mθ + (m 1)τ = 0 k + mθ + mτ τ = 0 k + mθ + mτ k + m = θ + τ = 0, and since tr(a(x) 2 ) = nk, we will have k 2 + mθ 2 + (m 1)τ 2 = 2mk k 2 + mθ 2 + mτ 2 τ 2 = 2mk k 2 + mθ 2 + mτ 2 k 2 + 2mk m 2 = 2mk θ 2 + τ 2 = m. Combining these we will get m = θ 2 + (θ + 1) 2 and hence k = θ(2θ + 1). Also since c k = θτ = (θ 2 + θ), we have c = θ 2. Finally using a c = θ + τ = 1, we get a = θ 2 1 and we are done.

17 15 Lemma If X is a SRG with k and k coprime, then X is imprimitive. Proof. Note that k(k a 1) = kc. Thus k divides kc. Since gcd(k, k) = 1, k must divide c. Since c k, either c = 0 or k = c. If c = 0, then Lemma 1.5 implies that X is not connected, and if k = c, using Proposition 1.3, we will find ā = k 1, and again by Lemma 1.5, we deduce that X would not be connected. Therefore, in any case X is imprimitive. It can be inferred from the last lemma that if p is a prime number, then all SRGs on p + 1 vertices are imprimitive. Proposition Let X be a primitive SRG on 2p vertices. Then the parameters of X or its complement are ((2θ + 1) 2 + 1, θ(2θ + 1), θ 2 1, θ 2 ). Proof. By taking complement if necessary, we may assume that k (2p 1)/2. Note that X cannot be a conference graph, since otherwise we would have 2m θ + 1 = 2p which is contradiction. Therefor θ and τ are integers. By Lemma 2.6 we have (θ τ) 2 m θ m τ = 2pk k, thus p divides (θ τ) 2 m θ m τ. Hence p must divide either (θ τ) 2 or m θ m τ. In the former case, (θ τ) 2 must be divisible by p 2 as well; that is p 2 must divide 2pk k. So p must divide k k. By assumption, k (2p 1)/2, thus p would divide k, and hence k = p and k = p 1; but by Lemma 2.11, this is impossible, since X is primitive. Consequently p must divide m θ m τ and so either m θ = p or m τ = p and the result follows form Lemma Latin Square Graphs In this section we will discuss about the graphs arising from Latin squares, orthogonal arrays, generalized quadrangles and etc. and will try to investigate their properties. We will, also, discuss the problem of cores of SRGs.

18 16 Definition 3.1. A Latin square L = [l ij ] of order n is an n n matrix with entries from a set of size n (called symbols) such that each row and each column contains each symbol exactly once. Example 3.2. The matrix L = is a Latin square of size 4. If G is a group of order n, the the multiplication table of G is a Latin square of size n. Given a Latin square L = [l ij ] of size n, define the graph X to be a graph whose vertex set is the set of all ordered triples (i, j, l ij ) and two vertices are adjacent if and only if they agree exactly in one position. The graphs associated to Latin squares are special cases of the graphs associated to orthogonal arrays. Definition 3.3. An orthogonal array with parameters k and n, OA(k, n), is a k n 2 matrix with entries from a set of size n, in which n 2 ordered pairs defined by any pair of rows of the array are all distinct. Example 3.4. The matrix is an OA(3, 4) Given a Latin square L of order n, we can build an OA(3, n) by taking the three rows of the array to be row numbers, column numbers and symbols. By reversing this process, we can build a Latin square of order n from an OA(3, n).

19 17 Given an orthogonal array OA(k, n), define the graph X to be a graph whose vertices are the column vectors of OA(k, n) and two vertices are adjacent if and only if their corresponding vectors have the same entry exactly in one position. Theorem 3.5. The graph X defined by an OA(k, n) is a SRG with parameters (n 2, (n 1)k, n 2 + (k 1)(k 2), k(k 1)). Proof. Let x = (x 1,..., x k ) V (X). Since for each 0 i k, the element x i happens n times in the i th row, x is adjacent to n 1 other vertices (i.e. vectors) which have the same entry in their i th position. Therefore x is adjacent to k(n 1) vertices and so X is regular of valency k(n 1). Let x = (x 1,..., x k ), y = (y 1,..., y k ) V (X) be adjacent and assume they agree in i th position. For an arbitrary vertex z = (z 1,..., z k ) V (X) to be adjacent to both x and y, we must have either z i = x i = y i or z s = x s and z t = y t, for some s t {1,..., k} \ {i}. For the first case, there are n 2 possibilities, since the entry x i = y i occurs n 2 times in the i th row and columns other than x and y. Also in the second case, there are (k 1)(k 2) possible choices for z, since (s, t) can be chosen in (k 1)(k 2) ways and because of the property of OA(k, n), z is identified by choosing (s, t). Therefor every two adjacent vertices have a = n 2 + (k 1)(k 2) common neighbors. Let x = (x 1,..., x k ), y = (y 1,..., y k ) V (X) are not adjacent. For an arbitrary vertex z = (z 1,..., z k ) V (X) to be adjacent to both x and y, we must have z i = x i and z j = y j, for some i j {1,..., k}. We can choose such i and j in k(k 1) ways, and because of the property of OA(k, n), z is identified by choosing (i, j). Therefore every pair of nonadjacent vertices has c = k(k 1) common neighbors and the theorem is proved.

20 18 Note that the graph defined by an OA(2, n) is isomorphic to L(K n,n ). Now we study the chromatic numbers of the graphs associated to Latin squares and orthogonal arrays. Proposition 3.6. Let L be a Latin square of order n, X be the graph corresponding OA(3, n), the orthogonal array defined by L. Then α(x) n χ(x). Proof. Identify the n 2 vertices of X with the cells of L. Since all the entries in a row (or in a column) are mutually adjacent, any independent set of vertices of X, can contain at most one element form each row. Therefore α(x) n. On the other hand, if there is a coloring of X with fewer than n classes, then at least one of color classes will contain more than n vertices, and since a color class is an independent set, this is a contradiction and so χ(x) n. An OA(k, n) is said to be extendible if we can extend it to an OA(k + 1, n) by adding a new row to it. The following result states when the graph corresponding to an OA(k, n) is n colorable. Theorem 3.7. Let X be the graph corresponding to an OA(k, n). Then χ(x) = n if and only if OA(k, n) is extendible. Proof. Assume first that χ(x) = n and that V (X) = V 1 V n is the partition of V (X) to the color classes of a proper n coloring. Obviously V i = n, for each 1 i n. Define the (k + 1)st row R k+1, such that its entry in j th column is i, where j V i. To show that R k+1 extends OA(k, n) to OA(k + 1, n), let R l be ( ) r the l th row of the OA(k, n), and suppose the ordered pair appears in two s columns j 1 and j 2 by two rows R l and R k+1. Then two vertices of X corresponding to the j 1 th and j 2 th columns cannot be in the same color class, since they are adjacent. Also since R k+1 has element s in both j 1 th and j 2 th columns, by

21 19 definition, these two columns are in the same color class. This contradiction shows that R k+1 extends OA(k, n) to a OA(k + 1, n). Now assume that we can extend OA(k, n) ro an OA(k + 1, n) and let R k+1 be the new added row. Then color each vertex j in X by the color i {1,..., n} where the entry of column j in the row R k+1 is i. It is easily seen that this coloring is proper and each color class contains n vertices. In the rest of this section we study SRGs arising from other combinatorial objects, namely generalized quadrangles. Definition 3.8. An incidence structure I = (P, L, I) is a set P of points, along with a set L of lines (disjoint from P), and a relation I P L called incidence. If (p, L) I, then we say that the point p and the line L are incident. A partial linear space is an incidence structure in which any two points are incident with at most one line. The definition of partial linear space implies that also any two lines are incident with at most one point. From these we can see that the girth of any partial linear space is at least six. Given an incidence structure I, we define the incidence graph, X(I), of I, to be a graph with vertex set P L, where two vertices are adjacent if and only if they are incident. From the definition it can be easily inferred that X(I) is bipartite with bipartition (P, L). Conversely, we can build an incidence structure from a given bipartite graph: define P to be one part, and L to be the other part. Terminologies.

22 20 In an incidence structure, when a point p and a line L are incident, we usually say that L contains p and p lies on L. Also when two points are incident to a common line, we say that these two points are joined by a line or are collinear; similarly, when two lines are incident to a common point, we say that these two lines meet at a point or are concurrent. Definition 3.9. A generalized quadrangle of order (s, t), GQ(s, t), is a partial linear space in which every line contains s + 1 points and every point lies on t + 1 lines; in addition if p is a point not in a line L, then there is a unique point on L collinear with p. Example Let P be the set of all edges of K 6 and L be the set of all 1-factors of K 6. We define the incidence of an e P and an F L to be simply e F. Since any 1-factor of K 6 is identified by two of its elements, we conclude that every two points can be joined at most by one line. Therefore this incidence structure is a partial linear space. Furthermore, if an edge e is not in a 1-factor F, then e has no common vertices with exactly one element e F ; thus e and e belong to one other 1-factor. The translation of this fact to the language of incident structures is that, any point e not in a line L is collinear with exactly one point on L. Finally, we see that every line contains 3 points and every points lies on 3 lines. The conclusion is that this incident structure is a GQ(2, 2). The generalized quadrangles are special cases of some other incidence structures called partial geometries in which given a line L and a point p not in L, there are exactly α points on L which are collinear with p. Definition The point graph of a generalized quadrangle GQ(s, t) is a graph whose vertex set is P, the set of points of the GQ(s, t), and two vertices are adjacent if and only if they are collinear.

23 21 The number of vertices of the point graph of a GQ(s, t) is (s + 1)(st + 1). To see that, let L be a fixed line in the GQ(s, t). It has s + 1 points. Each of these points lies on (t+1) 1 lines other than L, each of which has (s+1) 1 points other than the point which lies on L. Therefore we have counted (s+1)+(s+1)(ts) = (s+1)(st+1) points. Now note that since each line meets L, we have considered all lines and that no two of the mentioned lines meet out of L, since otherwise there would be a point not in L, which is collinear with two distinct points of L, which is a contradiction, and so the claim is proven. The point graph of the GQ(2, 2) defined in Example 3.10 is L(K 6 ) which is a SRG. We show that this is the case for the point graphs of all nontrivial generalized quadrangles. Proposition Let X be a point graph of a GQ(s, t), with s, t > 1. Then X is a SRG with parameters ((s + 1)(st + 1), s(t + 1), s 1, t + 1). Proof. Let p be an arbitrary vertex of X, i.e. a point in GQ(s, t). Then p lies on t + 1 lines, any two of which have exactly the point p in common. So p is collinear with s(t + 1) points; thus X is regular of valency s(t + 1). Now let p and p be two collinear points, and let both lie on the line L. there are s 1 other points on L which are so collinear with both p and p. It is clear that no point out of L can be collinear with both of p and p. Therefore a = s 1. Finally let p and p be two non-collinear points. There are t + 1 lines through p, and since p is not on any of them, p is collinear with exactly one point on each of these lines. In other words, the vertices p and p have exactly t + 1 common neighbors, and so c = t + 1, and the proposition is proved.

25 23 References [1] C. Godsil, G. Royle, Algebraic Graph Theory, Graduate Texts in Math. 207 Springer-Verlag (2001). [2] Belevitch, V. (1950), Theorem of 2n-terminal networks with application to conference telephony. Electr. Commun., vol. 26, pp [3] van Lint, J.H., and Seidel, J.J. (1966), Equilateral point sets in elliptic geometry. Indagationes Mathematicae, vol. 28, pp [4] P.J. Cameron, P.A. Kazanidis, Cores of symmetric graphs. Preprint, [5] C. Godsil, G.F. Royle, Cores of geometric graphs. Preprint, 2008.

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