# Relations between boundaries of groups acting on hyperbolic spaces with WPD elements

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 $$\Lambda_Y\Gamma\simeq\Lambda_X\Gamma$$.

• 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. Sep 16, 2019 at 23:34 • @LeeMosher Of course you're right, thank you very much. I edited the question accordingly. Sep 17, 2019 at 7:14 • 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. – user35370 Sep 20, 2019 at 20:26 ## 1 Answer 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. Added: 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. • 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 ofS'$relates to the first space$Xand how you get an alignement-preserving map fromY$to$X$(using the terminology of the theorem you state). Also, could you be more precise about your second statement ? Sep 23, 2019 at 7:48 • I think this is just a matter of unpacking what's in Osin's paper, but I'll add more detail. Sep 23, 2019 at 12:27 • Thank you for the details. So you made it very clear why$\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. Sep 23, 2019 at 17:50
• @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).
– user35370
Sep 23, 2019 at 21:05
• @PaulPlummer Right, so this basically answers the question. You need to add cobounded to the assumptions, and then styalor answered the question. Thanks Sep 24, 2019 at 6:01