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It is an important and deep fact of geometric group theory that if the Gromov boundary of a hyperbolic group $G$ is a circle, then $G$ is virtually Fuchsian [Tukia, Gabai, Casson-Jungreis...]. I am interested in the following slightly more general question:

Let $X$ be a connected, locally finite, hyperbolic graph such that $\partial X$ is a circle. Suppose that $G$ is a group acting cocompactly (but not properly) on $X$.

Question. Does there necessarily exist Fuchsian $H \leq G$ acting properly and cocompactly on $X$?

Perhaps this can be seen by piecing together known results, but I have not had much luck.

As an aside, some evidence this should be true is the following theorem [1]:

Theorem. If $X$ is a connected, locally finite graph with quadratic growth and $Aut(G)$ acts on $X$ cocompactly, then there exists $\mathbb Z^2 \leq Aut(G)$ acting properly and cocompactly on $X$.

Any references are appreciated!

[1] Seifter, N., & Trofimov, V. I. (1997). Automorphism groups of graphs with quadratic growth. Journal of combinatorial theory, Series B, 71(2), 205-210.

[Edit: I have corrected the statement of Seifter-Trofimov's theorem]

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    $\begingroup$ I think $G$ is an extension of a Fuchsian group by a profinite group. Such extension need not split. Now consider Cayley graphs of nontrivial central extensions of Fuchsian groups by finite groups. $\endgroup$
    – Moishe Kohan
    Commented Jun 29 at 15:30
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    $\begingroup$ The question doesn’t say so, but presumably you want to assume that $G$ acts faithfully. $\endgroup$
    – HJRW
    Commented Jun 29 at 16:20
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    $\begingroup$ @MoisheKohan: every extension of a Fuchsian group by a finite group is residually finite (because Fuchsian groups are good in the sense of Serre) and hence virtually torsion-free. So I believe these kinds of examples do contain a surface subgroup acting cocompactly. $\endgroup$
    – HJRW
    Commented Jun 29 at 16:26
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    $\begingroup$ @HJRW thanks for pointing out the necessity of assuming faithfulness. I’m secretly interested in the automorphism group of $X$, so faithful but not necessarily finitely generated. I agree the f.g. case is interesting in its own right, however. $\endgroup$
    – jpmacmanus
    Commented Jun 30 at 17:03
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    $\begingroup$ @HJRW Well, it's not "just" this. The compact kernel preserves a partition of the graph with bounded finite components. So it embeds into some power of a finite group, and in particular is locally finite of bounded torsion. But conversely there is no reason to believe that any extension of this form possesses a faithful action on a graph preserving such a partition. $\endgroup$
    – YCor
    Commented Jun 30 at 17:10

1 Answer 1

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What is true is that $G$ has a compact normal subgroup with Fuchsian quotient.

First, let $G$ be the isometry group and $W$ the kernel of the action on the Gromov boundary. Then $W$ is compact (as true for every nonelementary hyperbolic group). So $G/W$ acts faithfully continuously on the circle, and the 1-dimensional case of the Smith conjecture, which is an easy theorem, implies that $G/W$ is Lie. But here $G$ is also totally disconnected, so $G/W$ is discrete. Hence $G/W$ is a discrete hyperbolic group whose boundary is a topological circle, and by the work of the authors you're mentioning (Tukia, Gabai, Casson-Jungreis), $G/W$ is isomorphic to a cocompact lattice in $\mathrm{Isom}(\mathbf{H}^2)=\mathrm{PGL}_2(\mathbf{R})$.

If $G$ is an arbitrary totally disconnected locally compact Gromov-hyperbolic group with circle Gromov boundary, however, not true that there exists a cocompact lattice in $G$. For instance, let $\Gamma$ be the inverse image of a genus $g\ge 2$ surface group in $\widetilde{\mathrm{SL}_2}$. This is the group with presentation with $2g$ generators $a_1,\dots,b_g$ and the relators expressing that $z=\prod [a_i,b_i]$ is central. Let $G$ be the quotient of $G\times\mathbf{Z}_p$ ($p$-adics) identifying $z$ with the generator of $\mathbf{Z}_p$. Then $G$ has no cocompact lattice.

I have to think twice to get such $G$ to be the cocompact automorphism group of a graph.

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    $\begingroup$ But the question asks for a subgroup, not a quotient… $\endgroup$
    – HJRW
    Commented Jun 29 at 16:18
  • $\begingroup$ That said, if the conclusion is that $G$ is Fuchsian-by-finite, then one can conclude that $G$ contains a cocompact surface subgroup of finite index, as I explained in a comment above. $\endgroup$
    – HJRW
    Commented Jun 29 at 16:27
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    $\begingroup$ @HJRW the conclusion is that $G$ is compact-by-(discrete Fuchsian). This can fail to have a cocompact lattice (I provided an example). $\endgroup$
    – YCor
    Commented Jun 29 at 16:29
  • $\begingroup$ Thank you - does this Fuchsian quotient also act on $X$? $\endgroup$
    – jpmacmanus
    Commented Jun 29 at 17:17
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    $\begingroup$ @jpmacmanus Not necessarily. It acts on a quotient of $X$, namely by some equivalence relation with uniformly bounded equivalence classes (so that the quotient map is a quasi-isometry). $\endgroup$
    – YCor
    Commented Jun 29 at 17:20

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