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

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