Propagation in Graphs Modulo a Prime Jacob Steinhardt

Background Given a graph X, adjacency matrix A(X), its spectrum is defined to be all of the eigenvalues of the adjacency matrix, that is all k for which Av=kv has a solution for non-zero vectors v. The spectrum of a graph over the complex numbers gives us bounds regarding many of the properties of a graph.

My Idea What information can the spectrum of a graph over finite fields yield? One answer: it can tell us about the behavior of the graph modulo p a prime (and, perhaps for composite n as well). Theorem If A is a linear operator on a finite- dimensional vector space V (over an arbitrary field F), then A is nilpotent iff all of its eigenvalues are zero.

One Case: A Grid Consider a graph consisting of all points (x1,...,xk ) with 0≤xi≤di, where two points are connected if all but one of the coordinates are the same, and the remaining coordinate differs by 1. We call this a d1×d2×···×dk grid. Theorem The d1×···×dk grid is nilpotent mod p if and only if one of the following hold: p=2,and each di is either one less than a power of 2 or 2. Furthermore, the number of indices i with di=2 is even. di=1 for all i.

Nilpotency Note: nilpotency mod p is equivalent to saying that, for any starting assignment of numbers to each vertex, if we repeatedly replace the number at each vertex with the sum of the numbers on the neighboring vertices, all numbers will eventually be divisible by p. Other possible questions: When can we say that the number at a single vertex will always be eventually divisible by p? Can we tie the spectrum mod p to some property that exists outside of mod p?

Other Questions Eigenvalue techniques over the complex numbers order the eigenvalues by magnitude. Can we define some sort of norm on the eigenvalues over finite fields? Idea: Let dF(x,y) = [F(x)F(y):F]-[F(x)∩F(y):F]. Is this a meaningful distance? Would have to prove that [F(x)F(y):F]-[F(x)∩F(y):F]=[F(x-y):F] or possibly [F(x-y):F]-1. Possibly |x| = [F(x):F] could define a sort of distance.