### The Erdos-Ko-Rado Theorem for Graphs

Fred Holroyd
*The Open University*

I shall report on work in progress, conducted jointly with John Talbot and
Antony Johnson, on the question of extending the Erdos-Ko-Rado Theorem to
independent sets of vertices of graphs.
We shall say that a graph G is r-EKR if the size of a family of pairwise
intersecting independent vertex r-sets is maximised by choosing all those
containing some particular vertex; it is strictly r-EKR if all the extremal
families are of this type. If G = N_n (the null graph of order n), then
the original EKR Theorem states that G is r-EKR if and only if n is at
least 2r (and shortly thereafter it was shown that G is strictly r-EKR if n >
2r). Bollobas and Leader (1997) have extended this result to the case where G
is the lexicographic product N_n[K_l], and Talbot (2002, unpublished) has shown
that if G is a power of a cycle graph, then G is r-EKR for each r such that
independent r-sets exist (and strictly so except when G = C_{2r+2}), thus
settling a conjecture by Johnson and myself (1997).
After powers of cycles, life gets tough! One might expect vertex-transitivity
to ensure the EKR property for all r for which independent sets exist (except
for the null graphs), but this is false, and I shall present some quite
pleasant counterexamples. A more convincing possibility is that if m(G) is the
minimum size of a maximal independent set, then G should be EKR if m(G) is at
least 2r. I shall describe reasons why it is plausible that this may be true.
If it is, then it is in a sense 'best possible', as for any r > 1, the union of
two copies of the complete bipartite graph K_{r,r} fails to be (r+1)-EKR (and, if
r is odd, is r-EKR but not strictly so).
The Hilton-Milner Theorem does, however, provide an upper bound on the vertex
order of a graph of maximum degree d that can fail to be r-EKR. This bound can
be expressed as a polynomial of degree r - 1 in d.
The 'EKR analysis' of graphs that are unions of copies of a complete
multipartite graph is already quite a non-trivial problem, and I shall discuss
progress on this.