Is there an infinite (finite degree) transitive amenable hyperbolic graph ?
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1$\begingroup$ The Cayley graph of $\mathbb Z$ with the standard generating set. $\endgroup$– user6976Commented Nov 19, 2010 at 13:38
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1$\begingroup$ it's a pity the question hasn't been rephrased to take into account the previous comment and eliminate the trivial cases. For instance, it could have been added "with no proper cocompact action of $\mathbf{Z}$", or the weaker "with at least 3 boundary points". The answer is no in any case, ans follows (for instance?) from arxiv.org/abs/1202.3585 $\endgroup$– YCorCommented Jan 16, 2016 at 2:15
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2 Answers
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I think the answer is that every such graph should be quasi-isometric to the infinite line. Look at this paper. Theorem 9 there states that even a quasi-transivite amenable graph should have trivial Poisson boundary (no non-constant harmonic functions). On the other hand, a hyperbolic transitive graph, if it is not quasi-isometric to the Cayley graph $\mathbb Z$, always has non-trivial Poisson boundary. Here it is proved for groups, but I think it works for transitive hyperbolic graphs as well.
Edit. As R W says, the first reference in my answer is not relevant. It is nice that the statement is still true, and that it can be proved using Poisson boundaries.
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1$\begingroup$ A small correction: should be (no non-constant bounded harmonic functions). $\endgroup$ Commented Nov 27, 2010 at 0:48
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The paper by Elek and Tardos quoted by Mark is not relevant here as it is about bounded harmonic functions with finite Dirichlet integral only. Otherwise it is not true as numerous examples of amenable groups with non-trivial Poisson boundary (e.g., the lamplighter group in higher dimensions) show.
However, one can still answer the question by using the Poisson boundary in the following way. Amenability of a transitive graph is equivalent to amenability of its group of isometries (that's a theorem of Soardi and Woess). Now, on one hand, any amenable group carries a random walk with trivial Poisson boundary (Kaimanovich-Vershik-Rosenblatt), and on the other hand the Poisson boundary of any non-degenerate random walk on a non-elementary group of isometries of a proper hyperbolic space is non-trivial (this is the second reference given by Mark).
Here "non-elementary" means that the group does not fix any finite subset of the hyperbolic boundary (actually such a fixed subset can not consist of more than 2 points). In fact, this notion can be used to give a much more direct answer to the original question, as a group of isometries of a proper hyperbolic space is amenable if and only if it is elementary.
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$\begingroup$ Should be read: "any amenable locally compact group carries a random walk with trivial Poisson boundary", and "a closed group of isometries of a proper hyperbolic space is amenable iff it's elementary". $\endgroup$– YCorCommented Jan 16, 2016 at 2:02
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$\begingroup$ By the way only one half (the half you use) of result attributed to Soardi and Woess is correct: while the isometry group of a transitive locally finite connected amenable group is amenable (as a locally compact group), the converse does not hold. $\endgroup$– YCorCommented Jan 16, 2016 at 8:04
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2$\begingroup$ ... so finally what you prove is that if $X$ is an infinite transitive hyperbolic amenable graph, then its isometry group $G$ (which is a locally compact group) is amenable and acts on the boundary with a finite orbit. How does this answer the question? $\endgroup$– YCorCommented Jan 16, 2016 at 11:05