A Recursive Proof of Aldous’ Spectral Gap Conjecture

Aldous’ spectral gap conjecture asserts that on any graph the random walk process and the random transposition (or interchange) process have the same spectral gap. We prove the conjecture using a recursive strategy. The approach is a natural extension of the method already used to prove the validity of the conjecture on trees. The novelty is an idea based on electric network reduction which reduces the problem to the proof of an explicit inequality for a random transposition operator involving both positive and negative rates. The proof of the latter inequality uses suitable coset decompositions of the associated matrices on permutations. 1. Aldous’ conjecture Aldous’ conjecture concerns the spectral gap, a quantity that plays an important role in the analysis of the convergence to equilibrium of reversible Markov chains. We begin by reviewing some well known facts about Markov chains and their spectral gaps. For details we refer to [2]. 1.1. Finite state, continuous time Markov chains. Let us consider a continuous time Markov chain Z = (Zt)t > 0 with finite state space S and transition rates (qi,j : i 6= j ∈ S) such that qi,j > 0. We will always assume that the Markov chain is irreducible and satisfies qi,j = qj,i for all i 6= j. Such a Markov chain is reversible with respect to the uniform distribution ν on S, which is the unique stationary distribution of the chain. The infinitesimal generator L of the Markov chain is defined by Lg(i) = ∑ j∈S qi,j(g(j) − g(i)) , where g : S → R and i ∈ S. The matrix corresponding to the linear operator L is the transition matrix Q = (qi,j)i,j, where qi,i := − ∑ j 6=i qi,j, and the Date: June 26, 2009. 2000 Mathematics Subject Classification. 60K35; 60J27; 05C50.