A Class of Random Walks in Reversible Dynamic Environments: Antisymmetry and Applications to the East Model

Abstract

We introduce via perturbation a class of random walks in reversible dynamic environments having a spectral gap. In this setting one can apply the mathematical results derived in Avena et al. ($L^2$-Perturbed Markov processes and applications to random walks in dynamic random environments, Preprint, 2016). As first results, we show that the asymptotic velocity is antisymmetric in the perturbative parameter and, for a subclass of random walks, we characterize the velocity and a stationary distribution of the environment seen from the walker as suitable series in the perturbative parameter. We then consider as a special case a random walk on the East model that tends to follow dynamical interfaces between empty and occupied regions. We study the asymptotic velocity and density profile for the environment seen from the walker. In particular, we determine the sign of the velocity when the density of the underlying East process is not 1 / 2, and we discuss the appearance of a drift in the balanced setting given by density 1 / 2.

Appendix: Useful Facts on the East model

Definition 3

(Graphical representation of the East model) Starting from a configuration $\eta \in \Omega$, the East dynamics $(\eta _t)_{t\ge 0}$ can be constructed as follows. With every $x\in {\mathbb Z}$ independently we associate a Poisson process with parameter 1 that will be called the (Poisson) clock at x. The process can then be constructed in the following way:

• Check the constraint: if the clock at site x rings at time t, look at the constraint at x in $\eta _t$, the configuration at time t.

• If $c^{\text {east}}_x(\eta _t)=1$, the constraint is satisfied and the occupation variable at site x is replaced by a Bernoulli variable of parameter $\rho$ independent of all the rest. The ring at time t is said to be a legal ring.

• If $c^{\text {east}}_x(\eta _t)=0$, the constraint is not satisfied and the system is left unchanged.

The following lemma is a consequence of reversibility and the orientation property of the East model which we use to prove Proposition 4.2. We write ${\mathbb E} ^\mathrm{east}_{\eta }$ for the expectation of the East dynamics starting at $\eta$, and we define ${\mathbb E} ^\mathrm{east}_{\nu _\rho }$ similarly.

Lemma 5.3

The following holds:

• 1. Let $0\le t_1\le t_2\cdots \le t_k$ and let $f_1,...,f_k$ be functions on $\{0,1\}^{\mathbb Z}$ such that the convex envelopes of their supports $\mathrm{Conv}( Supp(f_{1}) ), \dots , \mathrm{Conv} ( Supp(f_k) )$ are disjoint. Then

  $$\begin{aligned} {\mathbb E} ^\mathrm{east}_{\nu _\rho }\left[ f_1\left( \eta _{t_1}\right) \ldots f_k\left( \eta _{t_k}\right) \right] =\prod _{i=1}^k\nu _\rho \left( f_i\right) . \end{aligned}$$

• 2. Let $0\le t_1\le t_2 \le \cdots \le t_k$, $f_1,...,f_k$ functions on $\{0,1\}^{\mathbb Z}$ and $i_0\in \{1,...,k\}$ such that $\nu _\rho (f_{i_0})=0$ and $x <y$ for all $x\in Supp(f_{i_0})$ and $y\in \cup _{i\ne i_0}Supp(f_i)$ (i.e. the support of $f_{i_0}$ is to the left of all the other supports). Then

  $$\begin{aligned} {\mathbb E} ^\mathrm{east}_{\nu _\rho }\left[ f_1\left( \eta _{t_1}\right) \ldots f_k\left( \eta _{t_k}\right) \right] =0. \end{aligned}$$

  (63)

Proof

The first statement is a consequence of the second one by iteration (let $i_0$ be the index of the function with left-most support and apply the second statement replacing $f_{i_0}$ by $f_{i_0}-\nu _\rho (f_{i_0})$).

Notice that by reversibility we can construct the process at equilibrium also for negative times by mirroring the graphical construction. The process obtained is invariant by time translation. In particular, we have

$$\begin{aligned} {\mathbb E} ^\mathrm{east}_{\nu _\rho }\left[ f_1\left( \eta _{t_1}\right) \ldots f_k\left( \eta _{t_k}\right) \right]= & {} {\mathbb E} ^\mathrm{east}_{\nu _\rho }\left[ f_1\left( \eta _{t_1-t_{i_0}}\right) \ldots f_k\left( \eta _{t_k-t_{i_0}}\right) \right] \\= & {} \nu _\rho \Big (f_{i_0}(\eta ){\mathbb E} ^\mathrm{east}_{\eta }\Big [\prod _{i\ne i_0}f_i\Big (\eta _{t_i-t_{i_0}}\Big )\Big ]\Big ). \end{aligned}$$

Now notice that ${\mathbb E} ^\mathrm{east}_{\eta }\left[ \prod _{i\ne i_0}f_i\left( \eta _{t_i-t_{i_0}}\right) \right]$ has disjoint support from $f_{i_0}$ thanks to the orientation property of the East model. The two terms in the $\nu _\rho$-mean are therefore decorrelated. Hence the result. $\square$