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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?

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  • $\begingroup$ So, how does the proof of relative hyperbolicity for free products work, if not by exhibiting an action? $\endgroup$ – ThiKu Mar 3 at 9:01
  • $\begingroup$ 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). $\endgroup$ – Yellow Pig Mar 3 at 11:29
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    $\begingroup$ 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... $\endgroup$ – Yellow Pig Mar 3 at 11:37
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    $\begingroup$ @ThiKu probably because in the definition of relative hyperbolicity, no properness (of the space) is required. $\endgroup$ – YCor Mar 3 at 12:18
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    $\begingroup$ @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. $\endgroup$ – HJRW Mar 3 at 18:13
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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.

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  • $\begingroup$ 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. $\endgroup$ – M. Dus Mar 3 at 16:57
  • $\begingroup$ Gluing horoballs is a pretty well known operation, going back pretty far into the history of Gromov hyperbolic spaces. $\endgroup$ – Lee Mosher May 19 at 2:12
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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.

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