Conditions on the parameters

Let G be a strongly regular graph with parameters (n,k,,) and adjacency matrix A. Then the (x,y) entry of A2 is the number of vertices adjacent to x and y. This number is k,, according as x and y are equal, adjacent or non-adjacent. Thus:

(2.13) A2 = kI +A + (J - I - A).

Also, since G is regular:

(2.14) AJ = JA = kJ.

Conversely, a strongly regular graph can be defined as a graph (not complete or null) whose adjacency matrix satisfies (2.13) and (2.14).

The all 1 vector j is an eigenvector of both A and J with eigenvalues k and n respectively. Applying (2.13) to this vector, we obtain

jA2 = (kj)A = k2j
j( kI + A + (J - I - A)) = kj + kj + (nj -j -kj)
k2 = k + k + (n -1 -k)
k(k - -1) = (n - k - 1).
(See (2.6))

As A and J are commuting real symmetric matrices, they can be simultaneously diagonalized by an orthogonal matrix. Thus any further eigenvectors of A are orthogonal to j, and so, are eigenvectors of J with eigenvalue 0. Let m be such an eigenvector, with eigenvalue þ (wrt A). Then

þ2m = km +þm + (-m - þ m)
þ2 = (k - ) + ( - )þ.
Now let the roots of this quadratic equation be r and s and we have that
r,s = ½( - ± [(- )2 + 4(k - )]1/2 ).
If r and s have multiplicities f and g, then we have
n = f + g + 1,
0 = Trace(A) = k + fr + gs.
We can solve these equations for f and g since k = = is not possible, to obtain the integrality conditions.

(2.16) Theorem. The numbers

are non-negative integers.

(2.17)Example: The integrality conditions applied to a (6u-3, 2u, 1,u) srg yields that

f,g = 3u - 2 ± (3u-1)(u-2)/(u+1),
thus to get integral values we must have u+1 dividing (3u-1)(u-2). As (3u-1)(u-2) = (u+1)(3u-10) + 12, we get u = 2,3,5 or 11 (u = 1 is excluded since a K3 is not a srg).

Def: (n,k,,) is a Type I parameter set if (n-1)( -) = 2k. Then since n > 1 + k and , we have that n = 4 + 1, k = 2 and = - 1.

(2.18) Theorem. If a type I srg on n vertices exists, then n is the sum of two integer squares.

Note: The Paley graphs are of type I.

Def: Type II parameters are those satisfying the integrality condition "normally". A Type I srg is of Type II iff n is a square.

An example of this last possibility is given by the graph L2(3) which is isomorphic to P(9).

Note that a graph Γ is regular if and only if j is an eigenvector of A. Furthermore, for regular graphs will be A-invariant, that is, vectors orthogonal to j are mapped by A to vectors orthogonal to j. When we restrict A to the subspace , if Γ is strongly regular then there are just two eigenvalues (r and s). If an adjacency matrix A has just two eigenvalues on , then (A - rI)(A - sI) is a multiple of J. This leads to:

(2.19) Proposition. The regular graph Γ with adjacency matrix A is strongly regular if and only if A restricted to has exactly two eigenvalues.

Proof: By the above remarks, (A - rI)(A - sI) = mJ, and so we have A2 = -rsI + (r+s)A + mJ = (k - m)I + (λ - m)A + mJ = kI + λA + m(J - I - A), thus satisfying (2.13).

We now consider some further necessary conditions on the parameters of an srg.

Definition: A square matrix A is positive semidefinite if for all x ε V, xtAx 0. All the eigenvalues of a positive semidefinite matrix are non-negative. Also note that QA(x) = xtAx is a quadratic form.

(2.22) Lemma. Let A = (aij) be a positive semidefinite symmetric real n × n matrix of rank d. Then there are vectors v1,v2,...,vn in d such that aij = <vi,vj> for i,j = 1,...,n.

Proof: There is a basis with respect to which the quadratic form QA(x) is the sum of squares. Letting P be the change of basis matrix (real and invertible) gives,

Putting S = P-1, we have
where S1 is the n × d matrix consisting of the first d columns of S. Now take v1,v2, ..., vn to be the rows of S1.

A is called the Gram matrix of the set v1,...,vn of vectors. A set of vectors is uniquely determined, up to isometry of d, by its Gram matrix.

(2.23) Theorem. Let Γ be a strongly regular graph on n vertices, having the properties that Γ and its complement are both connected and that the adjacency matrix of Γ has an eigenvalue of multiplicity f (greater than 1). Then n ½ f(f+3).

Proof: The adjacency matrix A has 3 distinct eigenspaces, and any matrix having these eigenspaces is a linear combination of I, A and J-I-A. In particular, there is such a linear combination E having eigenvalue 1 on the given f-dimensional eigenspace and 0 on its complement. Then E is positive semidefinite, and so is the Gram matrix of a set S of vectors of f. Since E = aI + bA + c(J-I-A), any vector in S has length a½, and two vectors in S make an angle cos-1(b/a) or cos-1(c/a). The vectors are all distinct since neither Γ nor its complement is a disjoint union of complete graphs. We may normalize to assume that a = 1, that is, S is a subset of the unit sphere Ω.

For v ε S, let fv : Ω be the function defined by

Now fv is a polynomial function of degree 2; and the functions fv, for v ε S are linearly independent, since
But these functions live in the space spanned by the f linear and ½f(f+1) homogeneous quadratic functions on Ω. Thus,
n = |S| f + ½f(f+1) = ½f(f+3),
as required.

This bound is called the absolute bound.

Example: Continuing with example (2.17). If u = 11 we obtain f,g = (33 - 2) ± (33 - 10 + 1) = 31 ± 24. Taking the smaller value for f, that is f = 7, the absolute bound gives n 35, but n = 55 + 7 + 1 = 63, so this value of u is excluded.

Definition: The Hadamard product of two n × n matrices A and B, denoted AºB, is the entrywise product, that is, the (i,j)th entry is aijbij.

(2.25) Lemma. Let A and B be positive semi-definite real symmetric matrices. Then AºB is positive semi-definite.

Proof: The Kronecker product of A and B is positive semidefinite since its eigenvalues are all products of an eigenvalue of A with an eigenvalue of B. The Hadamard product is a principal submatrix of the Kronecker product, and so, represents the Kronecker product restricted to a subspace. Thus, AºB is positive semidefinite.

(2.26) Theorem. Let Γ be a strongly regular graph which is connected and whose complement is connected. If Γ has eigenvalues k,r and s, then
(a) (r+1)(k+r+2rs) (k+r)(s+1)2;
(b) (s+1)(k+s+2rs) (k+s)(r+1)2.

Proof: The idempotent matrix E of the proof of (2.23) is positive semidefinite, so EºE is positive semidefinite. But

EºE = a2I + b2A + c2(J - I - A),
so we can find its eigenvalues. The result follows from this calculation.

These bounds are known as the Krein conditions.

Polarity Graphs

Let D = (X,B) be a symmetric 2-(v,k,λ) design, with 1 < k < v-1 (to avoid trivial cases) and a polarity σ. A point x is absolute if x ε xσ. We form a graph Γ associated with σ on the vertex set X by joining x to y whenever x and y are distinct and y ε xσ. Note that since a polarity is an involution, this joining rule is symmetric. The graph Γ has the property that Γ(x) = xσ\{x} for all x. In general, Γ is not regular since absolute points have degree k-1 while non-absolute points have degree k. However, if one or the other type of point is missing, then Γ is strongly regular.

(2.27) Proposition. (a) The graph Γ is associated with a polarity of a symmetric 2-design with no absolute points if and only if it is strongly regular with μ = λ.
(b) Γ is associated with a polarity of a symmetric 2-design with every point absolute if and only if it is strongly regular with μ = λ + 2.

Proof: If no point is absolute, then xσ yσ = Γ(x) Γ(y) has cardinality λ, whether or not x and y are adjacent. Conversely, if Γ is strongly regular with μ = λ, then (X, {Γ(x): x ε X}) is a symmetric 2-design and σ : x Γ(x) a polarity without absolute points. (b) is obtained by applying (a) to the complementary design and graph.

Definition: A strongly regular graph with parameters (v,k,λ,λ) is called a (v,k,λ) graph.

(2.28) Proposition. For fixed λ, there are only finitely many (v,k,λ) graphs.

Proof: Let Γ be such a graph. Since μ = λ, Γ must be of type II with respect to the integrality condition. This implies that 4(k- λ) = s2 for some integer s, and s divides -2k. Letting t = s/2, we have k = λ + t2 with t dividing k. But this implies that t divides λ, so we have that t λ. Thus, k λ + λ2 and therefore it follows from (2.6) that v λ2(λ + 2).

Note: The extremal case v = λ2(λ+2), k = λ(λ+1) occurs for all prime power values of λ and we will see examples later. Also note that L2(4) is a (16,6,2) graph. We shall see later that there is just one other (16,6,2) graph, the Shrikhande graph. It turns out that these two graphs are associated with different polarities of the same 2-(16,6,2) design.

For the strongly regular graphs associated with polarities in case (b) of (2.27), no such finiteness condition is known. In the smallest case, λ = 0 and μ = 2, there are only three known graphs:
(a) CP(2), having parameters (4,2,0,2);
(b) the Clebsch graph, with parameters (16,5,0,2); and
(c) the Gewirtz graph, with parameters (56,10,0,2).

All these graphs are uniquely determined by their parameters. They are associated with polarities of the biplanes 2-(4,3,2), 2-(16,6,2) and 2-(56,11,2). It is known that there are exactly 3 non-isomorphic biplanes with k = 6 and at least five with k = 11, but the other designs do not admit polarities with every point absolute.

Quadratic Form Graphs

Let V be a vector space over the binary field F = GF(2), and let Q be a quadratic form on V (a function from V to F which is homogeneous of degree 2 in the coordinates). We obtain a bilinear form f(x,y) on V called the polarization of Q by:
f(x,y) = Q(x+y) - Q(x) - Q(y).
Q is called non-singular if the only vector x satisfying Q(x) = 0 and f(x,y) = 0 for all y in V, is x = 0.

Example: Let V be a 3-dimensional vector space and define Q(x1,x2,x3) = x1x2 + x32. The polarization of Q is

f(x,y) = (x1+y1)(x2+y2) + (x3 + y3)2 + x1x2 + x32 + y1y2 + y32 = x1y2 + x2y1.

This Q is non-singular, since f(x,y) = 0 for y = (0,1,0) and (1,0,0) shows that x1 = x2 = 0 and Q(0,0,x3) = 0 shows that x3 = 0.

For a non-singular quadratic form Q, we obtain a graph Γ with vertex set {x ε V : Q(x) = 0, x not 0}, in which x and y are adjacent iff f(x,y) = 0.

Example (cont.): For our example, the vertex set is a = (0,1,0), b = (1,0,0) and c = (1,1,1). Since f(a,b) = f(a,c) = f(b,c) = 1, the graph obtained is a null graph on 3 vertices.

It can be shown that, except for some small trivial examples (such as ours), Γ is strongly regular. Furthermore, Γ has the triangle property: any edge {x,y} is contained in a triangle {x,y,z} having the property that any further vertex is joined to one or all of x,y,z. (Since Q(x) = Q(y) = f(x,y) = 0, Q(x+y) = 0 and x+y is the required third vertex of the triangle. The triangle property is a consequence of the equation f(x,w) + f(y,w) + f(x+y, w) = 0.)

(2.30) Theorem (Shult, Seidel): A non-null graph with the triangle property, in which no vertex is joined to all others, is obtained as above from a non-singular quadratic form over GF(2).