Random walk and the liouville property Hello,
How can one prove that the Lamplighter group on $G=\mathbb{Z}$ is Liouville. I have seen a stronger claim which states that the Lamplighter group over all recurrent graphs is Liouville. How can either of these claims be proved?
Thank you very much!
 A: Indeed, as Adam Smith pointed out, you can do this using coupling. Very briefly, since I don't have the time to elaborate, for any two vertices $x,y\in G$ and any two finite configurations of lamps $u,w$, you can couple the two random walks in $LL(G)$, started at $(x,u)$ and $(y,w)$ in the following manner: First you let the first random walk $X_i$ walk until it hits $y$, say at time $i_0$ (which happen almost surely, since $G$ is recurrent). Then the two random walks walk together with a time shift of $i_0$, that is $Y_{i}=X_{i+i_0}$. As they walk together, they also change the states of the lamps together. Again, since $G$ is recurrent we see that after a finite time, the two lamp configurations will be the same and will stay the same forever after (this is called shift coupling). From here it is easy to see, using a martingale argument, that the value of the harmonic function at  $(x,u)$ and $(y,w)$ had to be the same.
In fact, the existence of shift coupling from any two starting points is equivalent to the Liouville property, but we don't need this direction here.
A: It follows from the following fact: for any Markov chain its trace on a recurrent subset has the same space of bounded harmonic functions (equivalently, the same Poisson boundary). In particular, the Liouville property for bounded harmonic functions is preserved when passing to a recurrent subset. In the context of random walks this observation was first made by Furstenberg, in particular, if $G'$ is a recurrent subgroup for a random walk $(G,\mu)$, then the induced random walk $(G',\mu')$ is Liouville iff the original one is. For a lamplighter group recurrence of the quotient random walk on the base means precisely that the group of configuartions is recurrent. Now the rest of the argument depends on what assumptions you impose on the "lamp group" (the group of values of a single lamp). For instance, if it is abelian, then the group of configurations is also abelian, so that it is Liouville by Blackwell's theorem (which is more often attributed to Choquet and Deny although they did it several years later). This proof is contained in the 1983 Kaimanovich and Vershik paper. 
PS If you want to do it by coupling, then take your favorite coupling proof of the Liouville property for general abelian groups and recast it in terms of the ambient chain on the lamplighter group. However, it does not seem to be a natural thing in this situation. The recurrent subset argument I am referring to is totally elementary.
A: I have been trying to prove this using simpler tools. Namely if f is a bounded harmonic function then for (x,w)\in LL(Z) it is suffice to show that f(x,w) does not depend on w. I think this can be done by comparing f(x,w) with f(x,w^y) (where w^y(z)=w(z) \forall z\neq y and w^y(z)=(w(z)+1) mod 2 for z=y) for any y. I have been trying to prove this using a coupling of random walks started at (x,w) and (x,w^y) but haven't manage to find the right coupling. Any ideas of how can I complete such a proof will be very welcome. Thanks again!
