How does a quasi-isometry affect Poisson or Martin boundaries?

Question

Take two graphs (of bounded valency) or manifolds (f bounded geometry) $G$ and $G'$. Assume there is a quasi-isometry $f:G \to G'$, and assume the Poisson or Martin boundary of $G$ is known, what may one say about the Poisson or Martin boundary of $G'$ (or how does $f$ transforms Poisson or Martin boundaries)?

EDIT: a map $f:X \to Y$ between two metric spaces $(X,d_X)$ and $(Y,d_Y)$ is a quasi-isometry if $\exists C \geq 0$ and $D \geq 1$ such that

• $\forall x,x' \in X$, $D^{-1} d_Y \big( f(x), f(x') \big) - C \leq d_X(x,x') \leq D d_Y \big( f(x), f(x') \big) + C $

• $\forall y \in Y, d_Y \big(f(X),y \big) < C$

Answer 1

1. Terry Lyons in "Instability of the Liouville property for quasi-isometric Riemannian manifolds and reversible Markov chains", Journal of Diffe Geometry, 1987, constructs examples of graphs (and Riemannian manifolds) $G, G'$ which are quasi-isometric, but one admits no (nonconstant) positive harmonic functions while the other admits bounded (nonconstant) harmonic functions. Thus, quasi-isometries cannot respect Martin (and Poisson, as Ori noted) boundaries in any meaningful way.

2. It is well-known that, in general, even for CAT(0) spaces, quasi-isometries do not induce maps of geometric boundaries. The easiest example to consider is the case of quasi-isometries of the Euclidean plane ${\mathbb R}^2$. Let $f(x)$ be any $L$-Lipschitz function ($L\ge 1$) of one variable (you can take, for instance, $f(x)=|x|$). Define a self-map $F(x,y)= (x, f(x)+y)$ of ${\mathbb R}^2$. Then $F$ is an $2L$-bilipschitz homeomorphism of ${\mathbb R}^2$. Considering images of the $x$-axis under $F$, we see that they are (in general) not close to geodesics (say, the graph of $y=|x|$). Moreover, Croke and Kleiner constructed example ("Spaces with nonpositive curvature and their ideal boundaries", Topology, 2000) of a CAT(0) group $G$ so that $G$ acts geometrically (discretely, isometrically, cocompactly) on two CAT(0) spaces $X, X'$ so that the geometric boundaries of $X, X'$ are not homeomorphic. Existence of geometric action of $G$ on $X, X'$ implies that these spaces are quasi-isometric to each other.

3. Still, for some interesting classes of spaces, quasi-isometries act naturally on all three boundaries (geometric, Poisson, Martin); this happens for instance, in the case of Cayley graphs of hyperbolic groups.

Answer 2

$\newcommand{\Z}{\mathbb Z}$ $\newcommand{\T}{\mathbb T}$

It is well know that the Poisson boundary of simple random walk on graph is not invariant under quasi isometries. Here's a construction:

Take $\Z^4$ and notice that (1) Its Poisson boundary is trivial (i.e. it is Liouville) and (2) a random walk starting anywhere on the line $L=\{(n,0,0,0)\mid n\in\Z \}$ has positive probability to never hit this line.

Let $\T=\{0,1\}^*$ be the infinite binary tree and let $S$ be the set of balanced vertices, that is, vertices with the same number of 0s and 1s in their name. Connect the vertices of $S$ with those of $L$ in an arbitrary 1-1 manner. Call the resulting graph $G$.

Now, a random walk on $G$, started anywhere in the tree, will hit $S$ almost surely. When it does, there is a positive probability for it to be absorbed in the $\Z^4$ part (i.e. to stay there forever). If this doesn't happen, then the walk will just hit $S$ again and again until it is absorbed. Since $\Z^4$ was Liouville, we get that $G$ is also Liouville.

However, consider the graph $G'$ which is identical to $G$ except that we replace each right going edge of the tree $\T$ with a path of length 2. This new graph is clearly quasi-isometric to $G$, but now there is a positive probability that the random walk will never hit $S$ and stay in the tree part of the graph forever. This give rises to a non-constant, bounded harmonic function $f(v)$ which is simply the probability of a random walk started at $v$ to be absorbed in $\Z^4$.