When does Delsarte’s bound hold?

Philippe Delsarte (An algebraic approach to the association schemes of coding theory. Philips Res. Rep. Suppl., 10:vi+97, 1973) generalised the “MacWilliams transform” of coding theory to the theory of association schemes, and there have been numerous powerful applications of this object, particulary in finding the maximum size of a clique in some particular kinds of graphs. These graphs are regular and they are in fact a union of graphs from some association scheme. Let \omega(\Gamma) be the maximum size of a clique of a graph \Gamma.

Theorem (Delsarte): If \Gamma is the graph of an association scheme with valency k, and if \lambda is a negative eigenvalue of \Gamma, then

\omega(\Gamma)\le 1-k/\lambda.

I will give a proof I’ve seen in one of Chris Godsil’s notes as it bypasses the usual theory on inner distribution vectors and the MacWilliams transform.

Proof: Suppose we have an association scheme \mathcal{A} with primitive idempotents J=E_0,E_1,\ldots,E_d and suppose \Gamma is a graph for one of the relations of \mathcal{A}, of order n. Let C be a clique of \Gamma and let \theta be an eigenvalue of \Gamma with multiplicity m_\theta. Let us denote by E_\theta the primitive idempotent whose rowspace is the eigenspace corresponding to \theta. Then the submatrix of E_\theta with rows and columns indexed by the vertices of C has the form

\alpha I + \beta J.

Since E_\theta is positive semidefinite, the eigenvalues of \alpha I + \beta J are non-negative. The constant row-sum value of E_\theta is an eigenvalue, and so \alpha + (|C| - 1)\beta is an eigenvalue and thus is non-negative. This implies that

|C|\le 1-\alpha/\beta.

Since \alpha I is the Hadamard product I\circ E_\theta we have n\alpha = \mathrm{Trace}(E_\theta) = m_\theta. Let A be the adjacency matrix for \Gamma. Since AE_\theta = \theta E_\theta and A\circ E_\theta = \beta A, we have

\beta n k = \sum(\beta A) = \sum(A\circ E_\theta) = \mathrm{Trace}(AE_\theta) = \theta m_\theta. Therefore \beta=\theta \alpha/k. Hence, if we consider all cliques and the least eigenvalue \lambda, we have \omega(\Gamma)\le 1-k/\lambda.

\square

This bound doesn’t hold for all regular graphs, as we shall see.

Generalised quadrangles

Consider a generalised quadrangle of order (s,t) and its collinearity graph \Gamma. Then \Gamma is regular of valency s(t+1) and a maximal clique of \Gamma is simply the set of points incident with some line. So \omega(\Gamma)=s+1. The eigenvalues of \Gamma are s(t+1), s-1 and -t-1. Now \Gamma is strongy regular and so is the graph of an association scheme, and Delsarte’s bound applies and is tight:

\omega(\Gamma)\le 1-s(t+1)/(-t-1)=s+1.

Now consider what we get when we remove a line \ell of \Gamma. If s\ge 2, then the biggest clique of \Gamma-\ell is still the set of points of a line of the original generalised quadrangle. The valency of \Gamma-\ell is s(t+1)-1. So if Delsarte’s bound applies, then the least negative eigenvalue \tau of \Gamma-\ell satisfies s+1\le 1-(s(t+1)-1)/\tau, and hence,

\tau\ge 1/s-(t+1).

However, by the theory of interlacing eigenvalues, we must have \tau\le -(t+1); a contradiction.

Question

For what families of regular graphs does this bound hold?

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5 thoughts on “When does Delsarte’s bound hold?

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  1. I don’t understand the argument in the last sentence, as the minimal eigenvalue of an induced subgraph is always bigger than or equal to the minimal eigenvalue of the graph (by interlacing!).

    By the way, is there a description of eigenvectors of collinearity graphs of GQs which correspond to eigenvalue s-1 ?

    1. By “generalised interlacing”, deleting m vertices from a graph alters the minimum eigenvalue from \lambda_n to somewhere between \lambda_{n-m} and \lambda_n. As the original collinearity graph has got lots of eigenvalues equal to -t-1 (that is, more than s+1) the minimum eigenvalue of the graph after deleting the vertices of a line is still equal to -t-1, and the bound fails.

      As for the eigenvectors of the collinearity graph, consider the following set: take two intersecting lines and put a weight of +1 on each point of the first line, -1 on each point of the second line, 0 on the point of intersection and 0 elsewhere. Unless I screwed up, this is an eigenvector with eigenvalue s-1. I don’t know if this is all of them though.

      1. Yes, in a way there is. Let S be a set of points of a GQ of order (s,t). Now consider the average valency of S, that is, the average number of points a of S collinear with a given point of S. Then it can be shown that a\le s-1+|S|/(s+1), and if equality is attained, then S is what is called a tight set of the GQ. If we let A be the collinearity matrix of the GQ, we have an equivalent definition of a tight set in terms of the characteristic vector of S:
        A\chi_S=(s-1)\chi_S+ \frac{|S|}{s+1}\mathbf{j} where {\mathbf{j}} is the all-ones vector. So -(st+1)\chi_S-\frac{|S|}{s+1}\mathbf{j} is an eigenvector of A with eigenvalue s-1.

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