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Let $\Gamma$ be a Gromov hyperbolic group coming endowed with a word metric coming from some finite generating set. Let $\nu$ be an associated Patterson-Sullivan measure (quasi-conformal density).

I am interested, for each $g\in \Gamma$, the continuity properties of the Radon-Nikodym derivative $\frac{d\nu\circ g}{d\nu}$ as a function on the Gromov boundary $\partial \Gamma$ (the set of equivalence classes of geodesic rays, where two are equivalent of they stay a bounded distance apart).

Certainly, if $\Gamma$ is free, and a free generating set is chosen, then the word metric is $CAT(-1)$ and so the Busemann functions are continuous which implies that $\nu$ is actually conformal and continuous.

What about when $\Gamma$ is not free? The Busemann functions for a word metric on $\Gamma$ are then discontinuous.

Does it necessarily follow that for each $g\in \Gamma$, the function $\zeta\to \frac{d\nu\circ g}{d\nu}(\zeta)$ must be discontinuous on $\partial \Gamma$?.

Moreover, in Remark 2.15 of the paper "ENTROPY AND DRIFT IN WORD HYPERBOLIC GROUPS" by Gouezel, Matheus, and Maucourant the authors claim that "for nice groups, such as surface groups this function $\frac{d\nu\circ g}{d\nu}$ is continuous at all but finitely many points" of $\partial \Gamma$.

http://web.univ-ubs.fr/lmam/matheus/articles/hlv.pdf

Can anyone provide a proof or reference for this fact, or some clarification of what "nice" groups are?

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  • $\begingroup$ I'm confused where and how you define Busemann functions. If you define them on $\Gamma$, then saying that they are continuous is void. If you define them on the 1-skeleton ok it makes sense, but if you define Busemann functions as pointwise limit of functions $x\mapsto d(a_n,x)-d(a_n,1)$, when $a_n$ tends to infinity, they are obviously 1-Lipschitz hence continuous on the 1-skeleton. Possibly you have another definition of Busemann function in mind but on the first hand you should say where you define them. $\endgroup$
    – YCor
    Mar 8, 2015 at 14:47
  • $\begingroup$ For each $x,y \in \Gamma$ and $\zeta \in \partial \Gamma$ (the Gromov boundary, the set of equivalence classes of geodesic rays) I define $\beta_{zeta}(x,y)=\lim \inf_{z\to \zeta} d(z,y)-d(z,x)$. The fact that this is discontinuous as a function of $\zeta \in \partial\Gamma$ follows from the fact that there can be geodesic rays going to the same boundary point that stay a nonzero distance apart. In other words the Gromov boundary differs from the horofunction boundary. $\endgroup$
    – Yellow Pig
    Mar 8, 2015 at 15:08
  • $\begingroup$ I am fixing the pair of points $x,y$ on the 1-skeleton, and considering it as a function of $\zeta \in \partial \Gamma$, $\zeta \to \lim \inf_{z\to \zeta}d(z,y)-d(z,x)$. Then I believe it is discontinuous for reasons I said in my last comment. $\endgroup$
    – Yellow Pig
    Mar 8, 2015 at 15:19
  • $\begingroup$ OK, Here's an example on the free group on 2 generators $x,y$ (but with a non-free generating subset) where the Busemann kernel at 2 given points is not continuous as a function of the boundary point. The generating set is $\{x,x^2,y\}$. Take the infinite words $u=x^\infty$ and $u_n=x^{2n+1}y^\infty$. Then $u_n$ tends to $u$ in the boundary. But $\beta_u(x,1)=0$ (use that $d(x^{2k},x)-d(x^{2k},1)=k-k=0$ and $x^{2n}\to u$), while $\beta_{u_n}(x,1)=1$ (because the geodesic from any point close enough to $u_n$ to $1$ or $x$ goes through $x^{2n+1}$, and $d(x^{2n+1},1)-d(x^{2n+1},x)=(n+1)-n=1$). $\endgroup$
    – YCor
    Mar 8, 2015 at 23:18
  • $\begingroup$ Thanks for the correction! Of course, the Cayley group of a free group is only CAT(-1) (a tree) if you choose a free generating set. I edited the post accordingly. $\endgroup$
    – Yellow Pig
    Mar 9, 2015 at 1:07

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