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Let $\mathbb{F}_2$ be the free group on two generators. By a result of Kaimanovich and Vershik, for each measure $\mu$ on $\mathbb{F}_2$ such that the support of $\mu$ generates $\mathbb{F}_2$, we have that the random walk is not $\mu$-Liouville, i.e. there is a bounded $\mu$-harmonic function on $\mathbb{F}_2$ which is non-constant. Can one construct this function geometrically without involving KV result? I see this for finitely supported measures. Likely this has been clarified somewhere, I would like to have a citation in the latter case.

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    $\begingroup$ What do you mean by "construct"? What do you mean by "this function"? One can always "find" a stationary measure on the geometric boundary, fix a non-constant $L^\infty$ function on this boundary (say the characteristic function of the boundary of a half tree) and take its Poisson transform. This will give a non-constant bounded harmonic function. Is this a satisfying "construction"? $\endgroup$ – Uri Bader Apr 2 '16 at 17:08
  • $\begingroup$ Let me add that one can "construct" the stationary measure as limit of averages of convolution operators applied to a given fixed measure (say the delta measure at the identity). Combining the two "constructions" above, one can find an actual limiting process that will converge to a bounded harmonic function. $\endgroup$ – Uri Bader Apr 2 '16 at 17:29
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If I understand correctly the question is a request for a formula expressing a specific non-constant bounded $\mu$-harmonic function on $\mathbb{F}_2$, where $\mu$ is a fixed generating probability measure.

I assume below that $\mathbb{F}_2$ is freely generated by $\{a,b\}$. Let $A$ be the set consisting of all words starting with $a$ in $\mathbb{F}_2$. I claim that $$ h(x)=\lim_{n\to\infty}\frac{1}{n+1} \sum_{k=0}^n\mu^k(xA), \quad x\in \mathbb{F}_2 $$ is such a $\mu$-harmonic, where $\mu^k$ stands for the $k$-th convolution power of $\mu$ ($\mu^0=\delta_e$).

In fact, it is not hard to see that the formula for $h$ above converges pointwise to a $[0,1]$-valued $\mu$-harmonic. With some work you can see that this function is non-constant.

A slightly more sophisticated way to see this, which is in line with my comments above, is as follows. Consider the compact space $\bar{\mathbb{F}}_2=\mathbb{F}_2\cup \partial\mathbb{F}_2$. Then $\nu=\lim_{n\to\infty}\frac{1}{n+1} \sum_{k=0}^n\mu^n*\delta_e$ is a (in fact, the unique) stationary measure on $\bar{\mathbb{F}}_2$. It is supported on $\partial\mathbb{F}_2$, because $\mathbb{F}_2$ supports no stationary measure. It is fully supported there, by minimality. It is not hard to see that $(\partial \mathbb{F}_2,\nu)$ is a $\mu$-boundary in the sense of Furstenberg, thus the Poisson transform of any non constant $L^\infty$ function is non-constant. The expression for $h$ above is the Poisson transform for $\chi_\bar{A}$ wrt $\nu$ on $\bar{\mathbb{F}}_2$, which is the same as the Poisson transform for $\chi_{\partial A}$ wrt $\nu$ on $\partial \mathbb{F}_2$. The latter is non-constant by the fact that $\nu$ is fully supported.

Of course, the choice of $A$ in the construction above was quite arbitrary.

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  • $\begingroup$ Actually, there is no need to take the Cesaro averages as the convolution powers themselves already converge to the stationary measure. $\endgroup$ – R W Apr 2 '16 at 20:27
  • $\begingroup$ Thanks for this. In fact I had in mind is something which does not really involve convolutions. For instance, if the measure is fin supported, then $f(x)=\mathbb{P}_x(\exists K \forall n\geq K\text{ } X_n \text{is in the left tree})$ is harmonic, non-constant. $\endgroup$ – Kate Juschenko Apr 3 '16 at 15:05
  • $\begingroup$ @Kate Juschenko This is precisely the same harmonic function as the one user89334 is talking about, and there is no need to require finiteness of support of the step distribution $\mu$ for its construction. No matter what the step distribution is, $X_n$ almost surely converges to the boundary of the free group, and the arising hitting distribution is precisely the unique stationary measure evoked by user89334. $\endgroup$ – R W Apr 3 '16 at 18:22
  • $\begingroup$ I agree with both comments of R W. Kate, what exactly is it that you're looking for? the simplest expression? an easy to prove example? An exact citation? $\endgroup$ – Uri Bader Apr 3 '16 at 19:43
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    $\begingroup$ @Kate Juschenko Honestly, don't see much difference. The averages are a vestige from what is used in the proof of the Bogolyubov-Krylov theorem on existence of invariant (in this context stationary) measures. What does make difference with the Schreier graph you have in mind is the very notion of a stationary measure - it does not really make sense for the graph itself, as a result of which the usual Furstenberg's technique for proving boundary convergence (hinged on using the martingale theorem) does not work. $\endgroup$ – R W Apr 4 '16 at 8:04

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