Consider a Gromov-hyperbolic group $\Gamma$ and let $\mu$ be a finitely supported probability measure on $\Gamma$. Assume that the support of $\mu$ generates $\Gamma$ as a semi-group, in other words, the random walk $X_n$ driven by $\mu$ can visit the whole group $\Gamma$.

Fact : the random walk $X_n$ almost surely converges to a point in the Gromov boundary $\partial X$ of $X$. Let $\nu$ be the exit measure on $\partial X$. Then, $(\partial X,\nu)$ is a model for the so-called Poisson boundary. The measure $\nu$ is called the harmonic measure (with respect to $\mu$).

Now consider the reverse measure $\check{\mu}$ defined by $\check{\mu}(g)=\mu(g^{-1})$ and let $\check{\nu}$ be the corresponding harmonic measure.

Question : Are $\nu$ and $\check{\nu}$ equivalent ?


No - you are completely off. There is no reason whatsoever for the harmonic measures of the original and of the reflected random walks to be equivalent (unless these random walks coincide, i.e., unless the step distribution of the random walk is symmetric). The fact that these harmonic measures have the same Hausdorff dimension implies their equivalence only in the case when this dimension is maximal (i.e., coincides with the Hausdorff dimension of the hyperbolic boundary). For the simplest explicit counterexamples just look at the nearest neighbour random walks on the free group.

  • $\begingroup$ Okay thanks, I removed the fake startegy then. $\endgroup$ – M. Dus Nov 6 '19 at 5:53

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