# On what proper Gromov hyperbolic space does a free product act?

Per Bowditch, a group is relatively hyperbolic if it acts geometrically finitely on a proper geodesic Gromov hyperbolic space. A free product of two (or finitely many) finitely generated groups is well known to be relatively hyperbolic. It also acts on an associated Bass-Serre tree which is however locally infinite. My question: Is there a nice explicit description of a proper geodesic Gromov hyperbolic space on which the free product acts geometrically finitely?

• So, how does the proof of relative hyperbolicity for free products work, if not by exhibiting an action? – ThiKu Mar 3 at 9:01
• In his paper "Relatively Hyperbolic Groups" Bowditch proved that relative hyperbolicity (with the above definition) is equivalent to admitting an action on a (locally infinite) hyperbolic graph K such that the following condition hold. 1) All edge stabilizers are finite. 2) The number of orbits of edges is finite. 3) The graph K is fine, that is, for every n ∈ N, any edge of K is contained in finitely many circuits of length n. (Here circuit means a cycle without self– intersections). – Yellow Pig Mar 3 at 11:29
• But I wanted a very explicit description of the action on a proper hyperbolic metric space. It feels like there should be some action on a finite valence tree lurking... – Yellow Pig Mar 3 at 11:37
• @ThiKu probably because in the definition of relative hyperbolicity, no properness (of the space) is required. – YCor Mar 3 at 12:18
• @YellowPig, I think your intuition isn't right here. A proper action on a finite-valence tree would imply that your group is virtually free. – HJRW Mar 3 at 18:13

I think that the characterisation of relatively hyperbolic groups given in Groves and Manning's article Dehn filling in relatively hyperbolic groups answers your question. I give a few details:

Let $$G:=\underset{1 \leq i \leq n}{\ast} A_i$$ be a free product of $$n$$ finitely generated groups. For every $$1 \leq i \leq n$$, fix a finite generating set $$S_i$$ of $$A_i$$. Clearly, $$S:= \bigcup\limits_{i=1}^n S_i$$ is a finite generating set of $$G$$. The Cayley graph $$\mathrm{Cayl}(G,S)$$ is naturally a tree of spaces, the vertex-spaces being Cayley graphs of $$A_i$$'s. Now, the idea is to glue "horoballs" on the vertex-spaces. More precisely, consider $$X:= \left( \mathrm{Cayl}(G,S) \cup \bigcup_\limits{g \in G} \bigcup\limits_{i=1}^n \mathcal{H}(gA_i) \right) / \sim,$$ where the combinatorial horoball $$\mathcal{H}(gA_i)$$ over $$\mathrm{Cayl}(A_i,S_i)$$ is glued on $$gA_i$$.

The combinatorial horoball $$\mathcal{H}(Y)$$ over a graph $$Y$$ is defined as follows. The vertex-set of $$\mathcal{H}(Y)$$ is $$Y \times \mathbb{N}$$. If $$u$$ and $$v$$ are two adjacent vertices of $$Y$$, connect $$(u,0)$$ and $$(v,0)$$ with an edge. Also, for every $$k \geq 0$$ and for every vertex $$u \in Y$$, connect $$(u,k)$$ and $$(u,k+1)$$ with an edge. Finally, for every $$k \geq 0$$, if $$u,v \in Y$$ are two vertices satisfying $$d_Y(u,v) \leq 2^k$$, connect $$(u,k)$$ and $$(v,k)$$ with an edge.

It turns out that $$X$$ is a proper hyperbolic space, and $$G$$ naturally acts on it.

• This is a nice answer, but I think the OP was hoping for another hyperbolic space whose description is well-known. Note that you can also glue ideal triangles rather than combinatorial horoballs, using Bowditch's construction. – M. Dus Mar 3 at 16:57
• Gluing horoballs is a pretty well known operation, going back pretty far into the history of Gromov hyperbolic spaces. – Lee Mosher May 19 at 2:12

Too long for a comment. Here is another approach when the free factors are torsion-free abelian groups. Let $$G=\mathbb{Z}^{d_1}*\mathbb{Z}^{d_2}...*\mathbb{Z}^{d_n}$$. You can realize $$G$$ as a generalized Schottky group, acting on the real hyperbolic space $$H^n$$ for some $$n$$ via a geometrically finite action. You first take a geometrically finite kleinian group $$G_0$$ whose parabolic subgroups are exactly $$\mathbb{Z}^{d_1}$$,...,$$\mathbb{Z}^{d_n}$$. Now for each subgroup $$\mathbb{Z}^{d_i}$$, you choose some large $$k_i$$ and consider powers $$e_1^{k_i},...,e_{d_i}^{k_i}$$ of the standard generators $$e_1$$,...,$$e_{d_i}$$. If those $$k_i$$ are chosen sufficiently large, then an easy application of the ping-pong lemma shows that the subgroup $$G_1$$ generated by these elements is a free product. Since the subgroup of $$\mathbb{Z}^{d_i}$$ generated by $$e_1^{k_i}$$,...,$$e_{d_i}^{k_i}$$ is still isomorphic to $$\mathbb{Z}^{d_i}$$, $$G_1$$ is isomorphic to $$G$$.

Note that you can play the same game with torsion-free nilpotent groups, replacing the real hyperbolic plane by some simply connected Riemannian manifold of pinch negative curvature, but there are some differences. To simplify, assume that $$G$$ is of the form $$G=G_1*\mathbb{Z}$$, where $$G_1$$ is torsion-free nilpotent. Then, $$G_1$$ can be realized as the stabilizer of a cusp in a finite volume manifold of pinched negative curvature. This is a consequence of several results, including those of P. Ontaneda's spectacular paper Pinched smooth hyperbolization, see this question on mathoverflow for more details. You then choose a loxodromic element $$g$$ and applies the ping-pong lemma to $$g$$ and a subgroup of $$G_1$$ generated by elements that are far from the identity. You thus get a free product of the form $$\mathbb{Z}*G_1'$$. However, the main difference is that it is not clear to me if you can always have $$G_1'$$ isomorphic to $$G_1$$. To my knowledge, different lattices in the same nilpotent Lie group can be non-isomorphic, see for instance there.