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Let $\Gamma$ be a non-elementary Gromov's $\delta$-hyperbolic group. Let $B(1,n)$ be the set of elements at distance at most $n$ from the identity and let $\partial B(1,n)$ be the elements at distance $n$ (with the word metric). Consider the probability measures $\mu_n$ and $\nu_n$ defined as $$ \mu_n := \frac{1}{|B(1,n)|}\sum_{\gamma\\, \in B(1,n)}{\delta_{\gamma}} $$ and $$ \nu_n := \frac{1}{|\partial B(1,n)|}\sum_{\gamma\\,\in\partial B(1,n)}{\delta_{\gamma}}. $$ It is clear that these measures converge weakly (there is a convergent sub-sequence) on the compact space $\Gamma\cup\partial\Gamma$. Moreover, the limit measure is supported on $\partial\Gamma$.

My question are:

  • Does anybody have study the limit measures?
  • Are these related with the Hausdorff measure on the boundary? to the Patterson-Sullivan measure? to the harmonic measure?
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up vote 6 down vote accepted

I don't know about the specific sums you suggest, but here are some well established alternatives.

Try Kaimanovich's paper "The Poisson boundary of hyperbolic groups", which is about boundaries arising from random walks, which are shown to coincide with the Gromov boundary.

Another somewhat different approach is found in Coornaert's paper Patterson-Sullivan measures on the boundary of a hyperbolic space in the sense of Gromov. The idea, which goes back to the original Patterson-Sullivan measures on limit sets of Kleinian groups, is to construct measures on the boundary using sums over the whole group, where group elements are weighted by a function that decays exponentially in the distance; the base of the exponential is chosen delicately, based on geometric properties of the group. The resulting measures on the boundary are ``quasiconformal'' measures.

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Thanks. I think that my question is more related with the Sullivan's quasi-conformal measures but it isn't clear to me what exactly is the relationship. On the other hand it seems strange nobody has looked at this question before, it seems very natural. –  ght Mar 8 '12 at 12:23
It's not so much that people have not looked at this exact question, but more that by thinking about a broader class of questions, they have discovered nearby, different, but very fruitful questions. Still, though, maybe there's some interesting undiscovered fruit regarding your exact question. –  Lee Mosher Mar 8 '12 at 12:41
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