Let $(X,d)$ be Gromov-hyperbolic space and let $\Gamma$ be a finitely generated group acting on $\Gamma$ by isometries. Recall the following two definitions.

  • Say that the action is acylindrical if for every $\epsilon$, there exist $R,N$ such that for every two points $x,y\in X$ with $d(x,y)\geq R$, there are at most $N$ elements $g\in \Gamma$ such that both $d(x,g\cdot x)\leq \epsilon$ and $d(y,g\cdot y)\leq \epsilon$.
  • Let $h\in \Gamma$ be a loxodromic element with respect to the action. Say that $h$ is WPD if for every $\epsilon$ and every $x\in X$, there exists $m\in \mathbb{N}$ such that the set of elements $g\in \Gamma$ satisfying both $d(x,g\cdot x)\leq \epsilon$ and $d(h^m\cdot x,gh^m\cdot x)\leq \epsilon$ is finite.

Of course if the action is acylindrical, every loxodromic element is WPD. One of the main result of Osin's paper Acylindrically hyperbolic groups is that if $\Gamma$ is not virtually cyclic and acts on a hyperbolic space $X$ with a WPD element, then it acts acylindrically on a hyperbolic space $Y$.

We thus have two boundaries for $\Gamma$, namely its limit set $\Lambda_X\Gamma$ in the Gromov boundary of $X$ and its limit set $\Lambda_Y\Gamma$ in the Gromov boundary of $Y$. My question is as follows.

Question. Can we construct the space $Y$ such that $\Lambda_X\Gamma$ and $\Lambda_Y\Gamma$ equivariantly homeomorphic ? At least can we construct the space $Y$ such that there is an equivariant embedding $\Lambda_Y\Gamma\hookrightarrow \Lambda_X\Gamma$ ? For example, does Osin's construction of $Y$ yield an embedding $\Lambda_Y\Gamma\hookrightarrow \Lambda_X\Gamma$ ?

Some motivation. On the one hand, usually, when $\Gamma$ acts on a space $X$ with WPD elements, one has a geometric interpretation of $\Lambda_X\Gamma$. This is for example the case for the group $Out(F_n)$ acting on the free factor complex or on the sphere complex. On the other hand, when one has an acylindrical action, one can derive a lot of properties from this action.

One thing I'm interested in is the following. Given a random walk on $\Gamma$, the set $\Lambda_Y\Gamma$ endowed with the harmonic measure is a model for the Poisson boundary.

Thus, if $\Gamma$ acts on $X$ with a WPD element and if $\Lambda_Y\Gamma\hookrightarrow \Lambda_X\Gamma$ (or even better $\Lambda_Y\Gamma\simeq \Lambda_X\Gamma$) then one has a nice geometric interpretation of the Poisson boundary.

Basically, the proof of Osin's theorem goes as follows. Recall that a subgroup $H$ of $\Gamma$ is hyperbolically embedded in $\Gamma$ (with respect to a subset $S$ of $\Gamma$) if

  • the group $\Gamma$ is generated by $S$ and $H$ and the Cayley graph $\mathrm{Cay}(\Gamma, S\cup H)$ is hyperbolic,
  • the subgroup $H$, endowed with the induced metric $d_H$ is a proper metric space. This induced metric $d_H(h_1,h_2)$ is basically given by the smallest possible length of a path from $h_1$ to $h_2$ staying outside of $H$.

Now, assume that $\Gamma$ acts on $X$ with a WPD element. First, if $h$ is WPD, then it is contained in a maximal virtually cyclic subgroup $E(h)$ and $E(h)$ hyperbolically embeds into $\Gamma$. Second, if $H$ is hyperbolically embedded into $\Gamma$ with respect to a set $S$, define a set $S'$ consisting of all elements $g$ such that a geodesic from 1 to $g$ does not stay longer than $D$ inside $H$, for some fixed constant $D$. Then, the Cayley graph $\mathrm{Cay}(\Gamma,S'\cup H)$ is hyperbolic and the action of $\Gamma$ on this Cayley graph is acylindrical.

Let us look at an easy example now. Consider the free group $\Gamma=F_2$ and let $X$ be its Cayley graph with respect to the standard system of generators. Denote by $a$ and $b$ these two generators. Then, $a$ is WPD and one can take $E(a)=\langle a\rangle$. Clearly, $\langle a \rangle$ hyperbolically embeds into $\Gamma$ with respect to the set $S=\{b\}$.

Now, Osin's construction yields $S'=\langle b\rangle$. Thus, the space $Y$ on which $\Gamma$ acylindrically acts is $\mathrm{Cay}(\Gamma,\langle a\rangle \cup \langle b\rangle)$. This Cayley graph $\mathrm{Cay}(\Gamma,\langle a\rangle \cup \langle b\rangle)$ is quasi-isometric to the coned-off graph of $\Gamma$, considered as hyperbolic relative to its free factors. Its Gromov boundary is the set of conical limit points and it embeds into the Gromov boundary of $\Gamma$. Precisely, the boundary of $\mathrm{Cay}(\Gamma,\langle a\rangle \cup \langle b\rangle)$ is the set of infinite words alternating elements of $\langle a \rangle$ and elements of $\langle b \rangle$.

In particular, in this example, we indeed have $\Lambda_Y\Gamma\hookrightarrow\Lambda_X\Gamma$. Note however that one could take $Y$ to be the standard Cayley graph of $\Gamma$ and then one would get $\Lambda_Y\Gamma\simeq\Lambda_X\Gamma$.

  • $\begingroup$ What you say about the boundaries of $X$ and $\mathrm{Cay}(\Gamma,\langle a\rangle \cup \langle b\rangle)$ is not true. For one, the infinite word $aaaaa...$ represents an element of $\partial X$, but not an element of $\partial \mathrm{Cay}(\Gamma,\langle a\rangle \cup \langle b\rangle)$. Also $\partial X$ is compact but $\partial \mathrm{Cay}(\Gamma,\langle a\rangle \cup \langle b\rangle)$ is not: in the latter, the sequence $ababab...$, $a^2babab...$, $a^3babab..., $a^4babab...$ has no convergent subsequence. $\endgroup$
    – Lee Mosher
    Sep 16, 2019 at 23:34
  • $\begingroup$ @LeeMosher Of course you're right, thank you very much. I edited the question accordingly. $\endgroup$
    – M. Dus
    Sep 17, 2019 at 7:14
  • $\begingroup$ I haven't thought too much about it but there is a theorem that a group is acylindrically hyperbolic iff it acts acylindrically on a quasi-tree, and the quasi-tree is essentially Osin's construction you give above. The flexibility(totally disconnected) of such a boundary might be helpful to test equivariant embeddings. $\endgroup$
    – user35370
    Sep 20, 2019 at 20:26

1 Answer 1


I think the equivariant embedding you ask for is given in Theorem 3.2 of this paper: https://arxiv.org/pdf/1601.00101.pdf.

Actually, there is such an embedding any time you cone off uniformly quasiconvex subspaces of a hyperbolic space.


Theorem 3.2 states that if $X$ and $Y$ are hyperbolic and $f \colon X \to Y$ is coarsely Lipschitz, coarsely surjective, and alignment preserving, then there is a subspace $\partial_Y X$ of $\partial X$ that is homeomorphic to $\partial Y$. The inverse of the homeomorphism is the embedding $\partial Y \to \partial X$ that you are looking for.

One source of examples of such a map $f \colon X \to Y$ comes from work of Kapovich--Rafi. See Corollary 2.4 of this paper https://arxiv.org/pdf/1206.3626.pdf, where the alignment preserving property is exactly their `Moreover' statement. In fact, Corollary 2.4 is the result that Osin uses to show that his space (i.e. $\mathrm{Cay}(\Gamma,S'\cup H)$ in your notation) is hyperbolic.

As for a precise statement about coning off quasiconvex subspaces, see Proposition 2.6 in the Kapovich--Rafi paper above. In that setting, the resulting map $X \to Y$ is alignment preserving and so gives an embedding $\partial Y \to \partial X$ by Theorem 3.2 of my paper with Spencer.

  • $\begingroup$ Thank you. I'm going to take a look at the proof, but I can't see right now how this answers the question. I don't really see how the coning-off of $S'$ relates to the first space $X$ and how you get an alignement-preserving map from $Y$ to $X$ (using the terminology of the theorem you state). Also, could you be more precise about your second statement ? $\endgroup$
    – M. Dus
    Sep 23, 2019 at 7:48
  • $\begingroup$ I think this is just a matter of unpacking what's in Osin's paper, but I'll add more detail. $\endgroup$
    – staylor
    Sep 23, 2019 at 12:27
  • $\begingroup$ Thank you for the details. So you made it very clear why $\partial \mathrm{Cay}(\Gamma,S'\cup E(h))$ embeds into $\partial \mathrm{Cay}(\Gamma,S\cup E(h))$, but I'm not sure why $\partial \mathrm{Cay}(\Gamma,S\cup E(h))$ should be homeomorphic to $\partial X$, the initial space on which $\Gamma$ acts with a WPD element. Maybe this is obvious from Dahmani-Guirardel-Osin's paper, but I'm not too familiar with it. $\endgroup$
    – M. Dus
    Sep 23, 2019 at 17:50
  • $\begingroup$ @M.Dus If you act coboundedly you can get a generating set which is quasi-isometric(standard argument). If it isn't cobounded things are not clear, as you can have surface group acting on hyperbolic three space where the boundary is a space filling curve(Cannon-Thurston maps). $\endgroup$
    – user35370
    Sep 23, 2019 at 21:05
  • $\begingroup$ @PaulPlummer Right, so this basically answers the question. You need to add cobounded to the assumptions, and then styalor answered the question. Thanks $\endgroup$
    – M. Dus
    Sep 24, 2019 at 6:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.