0
$\begingroup$

Let $\Gamma$ be a finitely generated group and let $\mu$ be a probability measure on $\Gamma$. Consider the Green function $G(x,y)=\sum_{n\geq 0}\mu^{*n}(x^{-1}y)$, where $\mu^{*n}$ is the $n$th convolution power of $\mu$ and $x,y\in \Gamma$. Denote by $e$ the neutral element of $\Gamma$. The Martin kernel is then defined as $K(x,y)=\frac{G(x,y)}{G(e,y)}$. As a function of the variable $x$, $K(x,y)$ is $\mu$-harmonic everywhere except at $y$.

Now, the Martin compactification is defined as follows: it is the smallest compact metrizable space $M$, such that $\Gamma\subset M$ is open and dense and such that the Martin kernel $K(\cdot, \cdot)$ continuously extend to $\Gamma \times M$. In other words, a sequence $y_n$ converges to a point $\tilde{y}$ in the Martin compactification if and only if the function $K(\cdot,y_n)$ converges pointwise. In this case, denote by $K(\cdot,\tilde{y})$ the limit function. The Martin boundary is then defined as $\partial \Gamma=M\setminus \Gamma$.

Consider a point $\tilde{y}$ in the Martin boundary. Then, $K(\cdot,\tilde{y})$ is a limit a super-harmonic function and it is thus super-harmonic. Moreover, if $z\in \Gamma$ is fixed and if $y_n$ converges to $\tilde{y}$, then for large enough $n$, $y_n\neq z$, so $K(\cdot,\tilde{y})$ is a limit a harmonic functions at $z$. This does not ensure however that $K(\cdot, \tilde{y})$ is harmonic at $z$. It does when the measure $\mu$ is finitely supported, or when it has super-exponential moments (see for example the preprint Martin boundary covers Floyd boundary, Gekhtman, Gerasimov, Potyagailo, Yang, https://arxiv.org/abs/1708.02133).

In the paper harmonic measures versus quasiconformal measures for hyperbolic groups (https://arxiv.org/abs/0806.3915), Blachère, Haïssinsky and Mathieu claim that for every point $\tilde{y}\in \partial \Gamma$, the Martin kernel $K(\cdot,\tilde{y})$ is harmonic. They do not really use this property, so it's not a real issue, but I wonder if their claim is true.

So my question is the following: Is it true that Martin kernels $K(\cdot, \tilde{y})$ is always harmonic, when $\tilde{y}$ is in the Martin boundary ? If not, can we improve the assumption of super-exponential moments ?

Another question, more vague though: if $\tilde{y}$ is in the minimal-Martin boundary, then the function $K(\cdot,\tilde{y})$ is indeed harmonic. Is there a link between non-minimality of the Martin boundary and non-harmonicity of Martin kernels ? For example, is $K(\cdot,\tilde{y})$ harmonic if and only if $\tilde{y}$ is minimal ?

$\endgroup$
1
$\begingroup$

There are explicit examples of radially symmetric random walks on free groups for which the Martin boundary contains potentials, see Theorems 2 and 4 in Cartwright and Sawyer.

Concerning your second question, there are finitely generated groups $G$ for which the set of minimal harmonic functions is not closed in the Martin boundary for any random walk on $G$, see Remark 6.10 in Kaimanovich (the reason is that if the set of minimal harmonic functions is closed, then the group action on it is topologically amenable, whereas it is known that there are groups which do not have such actions). Thus, if you take a finitely supported random walk on a group like this, then the closure of minimal harmonic functions necessarily contains non-minimal harmonic functions as well.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.