3
$\begingroup$

Suppose $\Gamma$ is a discrete, finitely generated, non-amenable group, and consider a random walk given by a measure $\mu$. Assume the measure is symmetric, finitely generated, and the support of $\mu$ generates $\Gamma$.

Let $H<\Gamma$ be an infinite index subgroup.

Consider what I call the "induced random walk" on $H$ defined as follows (note $H$ is transient for the original random walk).

Let $G$ be the Green's function for $\Gamma$.

Define a Markov operator $P_{H}$ on $H$ by

$P_{H}(x,y)=G(x,y,H^c)/G(x,H)$,

in other words for $x,y \in H$, $P_{H}(x,y)$ is the probability that the $\mu$ random walk starting at walk makes its first return to $H$ at $y$ given that it ever returns to $H$. This Markov chain is $H$ invariant and so defines a random walk given by a measure $\mu_{H}$ on $H$.

(Here $G(x,y,H^c)$ denotes the total weight of all the paths starting at $x$ ending at $y$ and not going through $H$ in between, and $G(x,H)$ the weight of all paths starting at $x$ and ending at some point of $H$).

Have these induced random walks been studied before? (I could not find references in the literature).

In case they have, I have two specific questions:

  1. Is it true that $\mu_{H}$ approximates the asymptotic behavior of the $\nu$ random walk restricted to $H$: in other words for $A\subset H$, is it true that $\mu^{*n}_{H}(A)-\mu^{*n}(A\cap H)/\mu^{*n}(H)\to 0$ as $n\to \infty$?

  2. Is there a relation between the Green's function for $\mu_{H}$ and the restriction to $H$ of the Green's function for $\Gamma$?

  3. (Under suitable conditions?)Does the Martin boundary for $(H,\mu_{H})$ embed into the Martin boundary for $(G,\mu)$? What about for the Poisson boundary?

(Also, if this definition not the right one, is there a more suitable definition for an induced random walk on a transient subgroup of a group?)

Any illumination is greatly appreciated.

$\endgroup$
  • $\begingroup$ When defining $p_H$, the problem is normalization. When computing the convolution powers, you have powers of $G(x,H)$ and it is hard to compare the convolution powers of $p_H$ and those of $p$. Another way to do things is not to normalize $p_H$, so that the Green function of $p_H$ is the same as the Green function of $p$, but now you have to deal with a strictly sub-Markov chain and new issues appear. $\endgroup$ – M. Dus Jul 14 '17 at 9:24

Your Answer

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

Browse other questions tagged or ask your own question.