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Let $P=(p(x,y))_{x, y\in N}$ be the transition matrix over countable states $N$.

Consider the generating Green function $G(x, y|t)=\sum_{0}^{\infty} p^n(x, y) t^n$, where $p^n(x,y)$ is the $(x,y)$-entry of the matrix $P^n$.

The spectral radius is given by $\rho(P)=\limsup_{n\to\infty} p^n(x,y)^{1/n}$. It is clear that $G(x,y|t)$ has convergence radius $1/\rho(P)$.

I'm now interested in when $G(x,y|t)$ diverges at $t=1/\rho(P)$. In particular, I'm considering that $P$ induces a random walk on a non-amenable group $N$. Does there any results related to this divergence in some interesting class of groups?

I'm newbie to random walk on graphs. Any comments would be extremely welcome!

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Concerning random walk on non-amenable groups, the Green function converges at the spectral radius: this is a result of Guivarc'h, quoted in Wolgang Woess's book (Random walks on infinite graphs and groups), chapter IIB (in particular Theorem 7.8).

For a bit more explicit statement, one can laso look at the introduction of this paper of Gouezel and Lalley: https://arxiv.org/pdf/1107.5591.pdf

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  • $\begingroup$ Sorry, small question. I've read many times that this is due to Guivarc'h, as you say, but I do not know a precise reference. It seems that what Woess states in his book (in particular Theorem 7.8) is more general and combines the results of Varapoulos. Do you know where to find the original result of Guivarc'h ? $\endgroup$ – M. Dus Aug 17 '18 at 6:17
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    $\begingroup$ The reference is Astérisque 74 - MR0588157, page 85, remark b). It is written as a remark and the proof is quite elliptical (at least for me). $\endgroup$ – user120527 Aug 27 '18 at 13:53
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I would advise you to first look on random walk on trees. I suspect non-amenable group has been treated a lot as well but I can't find a good reference.

For example the random walk on the 4 regular tree (which is the free group $F_2$) can be also as a random walk on $\mathbb{N}$ with drift (the probability to go away from 0 is $3/4$ and to go toward $0$ is $1/4$). At $t=1/\rho(P)$, the drift disappears and we should obtain $p^{2n}(0,0)t^{2n}\sim \frac{1}{2^n}\frac{1}{n+1}\begin{pmatrix}2n \\n\end{pmatrix}\sim n^{-3/2}$

My intuition is then the following. We note $S_n= \{ y:d(y,x_0)=n \}$ with $d(x_0,y)$ the graph distance from a root point $x_0$. Because for any non-amenable graph, there should exist $r>1$ such that $|S_n|\approx r^n$. If one focus then only on the distance between the walker $X_n$ and $x_0$, one should recover somehow the previous picture of a random walk on $\mathbb{N}$ with a drift away from zero. (but this is only an educated guess).

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    $\begingroup$ Almost every sentence in what you have written is wrong. 4-regular tree is not the group $Z_2\times Z_2$, the polynomial term in the asymptotic formula is $n^{-3/2}$, and exponential growth does not imply positivity of the drift. $\endgroup$ – R W Feb 7 '18 at 18:41

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