Spectral gap for the interchange process in a box

We show that the spectral gap for the interchange process (and the symmetric exclusion process) in a $d$-dimensional box of side length $L$ is asymptotic to $\pi^2/L^2$. This gives more evidence in favor of Aldous's conjecture that in any graph the spectral gap for the interchange process is the same as the spectral gap for a corresponding continuous-time random walk. Our proof uses a technique that is similar to that used by Handjani and Jungreis, who proved that Aldous's conjecture holds when the graph is a tree.


Aldous's conjecture
This subsection is taken (with minor alterations) from Davis Aldous's web page. Consider an nvertex graph G which is connected and undirected. Take n particles labeled 1, 2, ..., n. In a configuration, there is one particle at each vertex. The interchange process is the following continuous-time Markov chain on configurations. For each edge (i, j), at rate 1 the particles at vertex i and vertex j are interchanged.
The interchange process is reversible, and its stationary distribution is uniform on all n! configurations. There is a spectral gap λ IP (G) > 0, which is the absolute value of the largest non-zero eigenvalue of the transition rate matrix. If instead we just watch a single particle, it performs a continuous-time random walk on G (hereafter referred to simply as "the continuous-time random walk on G"), which is also reversible and hence has a spectral gap λ RW (G) > 0. Simple arguments (the contraction principle) show λ IP (G) ≤ λ RW (G).
Discussion. Fix m and color particles 1, 2, ...., m red. Then the red particles in the interchange process behave as the usual exclusion process (i.e., m particles performing the continuous-time random walk on G, but with moves that take two particles to the same vertex suppressed). But in the finite setting, the interchange process seems more natural.

Results
Aldous's conjecture has been proved in the case where G is a tree [7] and in the case where G is the complete graph [5]; see also [12]. In this note we prove an asymptotic version of Aldous's conjecture for G a box in Z d . We show that if B L denotes a box of side length L in Z d then as L → ∞.
Remark: After completing a draft of this paper, I learned that Starr and Conomos had recently obtained the same result (see [14]). Their proof uses a similar approach, although the present paper is somewhat shorter.
Connection to simple exclusion. Our result gives a bound on the spectral gap for the exclusion process. The exclusion process is a widely studied Markov chain, with connections to card shuffling [16,1], statistical mechanics [8,13,2,15], and a variety of other processes (see e.g., [10,6]); it has been one of the major examples behind the study of convergence rates for Markov chains (see, e.g., [6,3,16,1]). Our result implies that the spectral gap for the symmetric exclusion process in B L is asymptotic to π 2 /L 2 . The problem of bounding the spectral gap for simple exclusion was studied in Quastel [13] and a subsequent independent paper of Diaconis and Saloff-Coste [3]. Both of these papers used a comparison to Bernoulli-Laplace diffusion (i.e., the exclusion process in the complete graph) to obtain a bound of order 1/dL 2 . Diaconis and Saloff-Coste explicitly wondered whether the factor d in the denominator is necessary; in the present paper we show that it is not.

Background
Consider a continuous-time Markov chain on a finite state space W with a symmetric transition rate matrix Q(x, y). The spectral gap is the minimum value of α > 0 such that for some f : W → R. The spectral gap governs the asymptotic rate of convergence to the stationary distribution. Define where If f is a function that satisfies Qf = −λf for some λ > 0, then Furthermore, if α is the spectral gap then for any non-constant f : W → R we have Thus the spectral gap can be obtained by minimizing the left hand side of (3) over all non-constant functions f : W → R.

Main result
Before specializing to the interchange process, we first prove a general proposition relating the eigenvalues of a certain function of a Markov chain to the eigenvalues of the Markov chain itself. Let X t be a continuous-time Markov chain on a finite state space W with a symmetric transition rate matrix Q(x, y). Let T be another space and let g : We shall need the following proposition, which generalizes Lemma 2 of [7].
Proposition 1 Let X t , g and Q ′ be as defined above. Suppose f : W → R is an eigenvector of Q with corresponding eigenvalue −λ and define h : That is, either h is an eigenvector of Q ′ with corresponding eigenvalue −λ, or h is identically zero.
The following Lemma is a weaker version of Aldous's conjecture. The proof is similar to the proof of Theorem 1 in [7].
Lemma 2 Let G be a connected, undirected graph with vertices labeled 1, . . . , n. For 2 ≤ k ≤ n let G k be the subgraph of G induced by the vertices 1, 2, . . . , k. Let λ RW (G k ) be the spectral gap for the continuous-time random walk on G k , and define α k = min 2≤j≤k λ RW (G j ). Then Proof: Our prooof will be by induction on the number of vertices n. The base case n = 2 is trivial, so assume n > 2. Let W and Q be the state space and transition rate matrix, respectively, for the interchange process on G. Let f : W → R be a function that satisfies Qf = −λf . We shall show that λ ≥ α n . Note that a configuration of the interchange process can be identified with a permutation π in S n , where if particle i is in vertex j, then π(i) = j. For positive integers m and k with m, k ≤ n, we write f (π(m) = k) for π:π(m)=k f (π).
We consider two cases. Then h is not identically zero, and using Proposition 1 with g defined by g(π) = π(m) gives that if Q ′ is the transition rate matrix for continuous time random walk on G, then Q ′ h = −λh. It follows that λ is an eigenvalue of Q ′ and hence λ ≥ λ RW (G) = α n .
Case 2: For all m and k we have f (π(m) = k) = 0. Define the suppressed process as the interchange process with moves involving vertex n suppressed. That is, the Markov chain with the following transition rule: For every edge e not incident to n, at rate 1 switch the particles at the endpoints of e.
(Note that for every k we have π∈W k f (x) = 0.) By the induction hypothesis, the spectral gap for the interchange process on G n−1 is at least α n−1 . Hence for every k with 1 ≤ k ≤ n we have
Remark: Theorem 2 is optimal if the vertices are labeled in such a way that λ RW (G k ) is nonincreasing in k, in which case it gives λ IP (G) = λ RW (G). Since any tree can be built up from smaller trees (with larger spectral gaps), we recover the result proved in [7] that λ IP (T ) = λ RW (T ) if T is a tree.
Our main application of Lemma 2 is the following asymptotic version of Aldous's conjecture in the special case where G is a box in Z d .
. . , L} d be a box of side length L in Z d . Then the spectral gap for the interchange process on B L is asymptotic to π 2 /L 2 .
Proof: In order to use Lemma 2 we need to label the vertices of B L in some way. Our goal is to label in such a way that for every k the quantity λ RW (G k ) (i.e., the spectral gap corresponding to the subgraph of B L induced by the vertices 1, . . . , k) is not too much smaller than λ RW (B L ). So our task is to build B L , one vertex at a time, in such a way that the spectral gaps of the intermediate graphs don't get too small.
We shall build B L by inductively building B L−1 and then building B L from B L−1 . Since where β L is the minimum spectral gap for any intermediate graph between B L−1 and B L . For a graph H, let V (H) denote the set of vertices in H. For j ≥ 0, let L j = {0, . . . , j} be the line graph with j + 1 vertices. Define γ L = λ RW (L L ). It is well known that γ L is decreasing in L and asymptotic to π 2 /L 2 as L → ∞. It is also well known that if H and H ′ are graphs and × denotes Cartesian product, then λ RW (H × H ′ ) = min(λ RW (H), λ RW (H ′ )). Since B L = L d L , it follows that λ RW (B L ) = γ L .
We construct B L from B L−1 using intermediate graphs We obtain H k from H k−1 by adding vertices to lengthen H k−1 by one unit in direction k. The order in which the vertices in V (H k ) − V (H k−1 ) are added is arbitrary.
Fix k with 1 ≤ k ≤ d, and define G ′ = G ′ (L, k) as follows. Let and let G ′ = (V ′ , E ′ ). It is well known and easily shown that if H is a graph, then adding edges to H cannot decrease λ RW (H), nor can removing pendant edges. Since each intermediate graph G between H k−1 and H k can be obtained from G ′ by adding edges and removing pendant edges, it follows that for any such graphG we have λ RW (G) ≥ λ RW (G ′ ). Thus, it is enough to bound λ RW (G ′ ) from below. We shall show that for any ǫ > 0 we have λ RW (G ′ (L, k)) ≥ (1 − ǫ)γ L if L is sufficiently large. Let e k be the unit vector in direction k. Let Let X t be the continuous-time random walk on G ′ , with transition rate matrix Q. Fix f : V ′ → R with Qf = −λf for some λ > 0. For x ∈ Z d , let g k (x) denote the component of x in the kth coordinate. Note that g k (X t ) is the continuous-time random walk on L L . Let Q ′ be the transition rate matrix for g k (X t ). Proposition 1 implies that if h : {0, . . . , L} → R is defined by Thus if λ < γ L , then g is identically zero and hence and let E S (f, f ) be defined analogously, but with only vertices in S included in the sum. Note that where the second inequality follows from the fact that H k−1 is a Cartesian product of d graphs, each of which is either L L−1 or L L . Fix ǫ > 0 and let M be a positive integer large enough so that say that x is good if there is a y ∈ S such that x = y + ie k for some i ≤ M and |f (y)| ≤ |f (x)|/2. Otherwise say that x is bad. Let G and B denote the set of good and bad vertices, respectively, in ∂S. Note that if x is bad and M ≤ L then f (x) 2 ≤ 4 M M j=1 f (x − je k ) 2 . Summing this over bad x gives Note that if x is good, then there must be an x ′ ∈ S of the form x−ie k such that |f (x ′ )−f (x ′ +e k )| > f (x)/2M . It follows that Since V ′ = S ∪ B ∪ G, combining equations (11), (9) and (10) gives and hence Recall that (1−4M −1 ) −1 ≤ (1−ǫ) − 1 2 , and note that since γ L → 0 as L → ∞, we have γ −1 L +4M 2 ≤ (1 − ǫ) − 1 2 γ −1 L for sufficiently large L. Combining this with equation (12) gives for sufficiently large L. It follows that λ RW (G ′ ) ≥ (1 − ǫ)γ L for sufficiently large L and so the proof is complete.