Return probabilities for random walks on infinite Schreier graphs

Question

Question: Is there a sequence $(\delta_n)_n$ of real numbers with $\delta_n \to 0$ as $n \to \infty$, such that the following holds:

Let $F$ be a free group on two generators, let $F \curvearrowright X$ be a transitive action on an infinite set, and let $x \in X$. Then, the probability that a random word of length $n$ in $F$ fixes $x \in X$ is smaller than $\delta_n$.

I would like to consider random unreduced words (so that there are $4^n$ such words of length $n$ and each is equally likely), but probably this does not matter much. It corresponds to the nearest neighbor random walk on the Schreier graph corresponding to the action of $F$ on $X$.

It is clear that for each individual action $x \in X$, the return probability decays; and it seems plausible that this happens uniformly over all actions.

Answer 1

So, if I understand things right, we have a random walk on an undirected connected graph with possibly multiple edges and degree 4 at each vertex that starts at some vertex $x$.

The usual way to proceed is to consider the vector $P=P_n$ whose components $P_n(y)$ are the probabilities to end at $y$ after $n$ steps. The dynamics is $P_{n+1}=AP_n$ where $A$ is one quarter of the incidence matrix. The key is that the $\ell^2$ norm of $P_n$ drops at each step by some constant multiple of the "gradient energy" $E=\sum_y \sum_{z,w\in N(y)}|P_n(z)-P_n(w)|^2$ where $N(y)$ is the set of neighbors of $y$. This allows to control the $L^\infty$ norm of $P_n$ (that also can go only down) very easily. If we made $n$ steps, the energy at some step was at most $C/n$. Now, take any vertex $v$ and join it with a faraway vertex by a shortest path $v=v_0--v_1--v_2--v_3--\dots$. Then the energy is at least $\sum_{k=0}^\infty |P(v_{2k})-P(v_{2k+2})|^2$. Now it remains to note that the $\ell^1$ norm of $P$ is always $1$, so we can estimate $P(v)$ by $1/M+\sqrt{M/n}$ with any integer $M$ we want (among first $M$ vertices with even indices, we have at least one with the probability $1/M$ or less and the sum of differences on the return path to $v$ from there is bounded by $\sqrt{M/n}$ by Cauchy-Schwarz. Taking $M=n^{1/3}$ gives $\delta_n=O(n^{-1/3})$. This is by no means optimal but it gives you the general idea of how such proofs go.

I leave it to the professional probabilists to refer you to the relevant literature.

Answer 2

While Fedja has given a concrete argument, it occured to me that it also follows from (more or less) abstract nonsense, by compactness-and-contradiction.

Let me explain in more detail. Let $\delta_n$ be the supremum over all actions $G \curvearrowright X$ and $x \in X$ of the return probability after $n$ steps.

We need to show that $\delta_n \to 0$ as $n \to \infty$. It is clear that $\delta_{2n}$ is decreasing. This follows since the return probabilities after 2n steps for each individual action is decreasing, since they are given by the moments of positive contraction on a Hilbert space. Now, if $\inf_n \delta_{2n} = \delta >0$, then we find a sequence of actions $G \curvearrowright X_n$ and $x_n \in X_n$, such that the return probability after $2n$ (and hence $2k$ for $k<n$) steps to $x_n$ is greater or equal $ \delta/2$.

Now, the space of transitive actions with chosen point is compact! Indeed, it is just the Chabauty space of subgroups of the free group, which is obviously compact. Moreover, the return probabilities are continuous functions on the space of transitive actions with chosen point. Hence, we find a limit point in this space and hence a limit action, for which the return probability after $2n$ steps is greater or equal $\delta/2$ for all $n$. Now, the only thing we can quickly say about the limit is that it is not finite. Indeed, the finite transitive actions are all isolated in the space of actions. This is a contradiction since we know already that for each individual action the return probabilities tend to zero.