3
$\begingroup$

I'am wondering whether there exists a non-discrete hyperbolic totally disconnected locally compact group such that the boundary is a finite-dimensional sphere. If the answer is positive, could you please provide few examples?

$\endgroup$
0

1 Answer 1

6
$\begingroup$

Non-discrete is not a reasonable assumption: for any hyperbolic group take the direct product with a compact group. Every hyperbolic group locally compact group $G$ has a unique maximal compact normal subgroup $W(G)$, which for $G$ non-elementary is the kernel of the $G$-action on the boundary $\partial G$. A reasonable assumption is to assume that $W(G)=1$, and in this case "non-discrete" is a reasonable assumption. I'll assume so, to make the question non-trivial. I will also assume that by "is a sphere" you mean "is homeomorphic to a sphere" (this is not an anecdotical point: for instance I don't know if there's a single (discrete) hyperbolic group whose boundary is homeomorphic, but not Hölder homeomorphic to a round sphere).

The Hilbert-Smith conjecture asserts that a non-discrete totally disconnected locally compact group cannot act continuously and faithfully on a connected topological manifold.

If one believes this conjecture, then the answer to your question is no. Indeed assume that $W(G)=1$, and that $\partial G$ is homeomorphic to a sphere (which is equivalent to $\partial G$ being a topological manifold):

  • if $G$ is non-elementary, then it acts faithfully and continuously on its boundary $\partial G$, by the Hilbert-Smith conjecture we deduce that $G$ is discrete;
  • if $G$ is any elementary hyperbolic group with $W(G)=1$, then it is known that $G$ has an open subgroup of index $\le 2$ that is isomorphic to one of the following three groups: $\{1\}$, $\mathbf{Z}$ and $\mathbf{R}$. Hence for $G$ totally disconnected, this again implies discrete.

For small-dimensional spheres the Hilbert-Smith conjecture is known to hold:

  • for the 1-sphere it's an exercise;
  • for the 2-sphere it's done in Montgomery-Zippin (although I think the proof has a gap, but anyway using facts from a 2006 survey of Kolev (MR link) in l'Ens. Math., it's fine)
  • for the 3-sphere it follows from the more solution in dimension 3 by J. Pardon.

So the answer (of the corrected question) is no for spheres of dimension $\le 3$, and no in general assuming the Hilbert-Smith conjecture.

It is maybe possible to prove the Hilbert-Smith conjecture in the very specific setting of such actions: they are bilipschitz actions for some (non-Riemannian) metric, they are minimal, etc. Probably the natural setting would be convergence groups, so the question becomes whether if $G$ is a totally disconnected locally compact group acts continuously and faithfully on a sphere so that the action on the set of distinct triples is proper and cocompact, is $G$ discrete?

$\endgroup$
4
  • $\begingroup$ PS the Hilbert-Smith conjecture is known for Lipschitz actions on Riemannian manifolds (Repovš-Ščepin 1997 mathscinet.ams.org/mathscinet-getitem?mr=1464908). The action of a hyperbolic group on its boundary is Lipschitz for a given choice visual distance. This is not enough to conclude, because even if it's homeomorphic to a sphere, the distance is not in general not Lipschitz equivalent to a Riemannian distance (they can have distinct Hausdorff dimension). $\endgroup$
    – YCor
    Feb 2, 2018 at 21:35
  • $\begingroup$ PPS beware that the above MR review by S. Antonyan of the Repovš-Ščepin paper claims that it's been generalized to the Hölder case (for actions on Riemannian manifolds) by Malešič (mathscinet.ams.org/mathscinet-getitem?mr=1480156) but the review of that paper says something weaker, namely bounds the Hölder exponent in terms of the dimension of the manifold. $\endgroup$
    – YCor
    Feb 2, 2018 at 21:41
  • 1
    $\begingroup$ A relevant partial result regarding the Hilbert-Smith conjecture is given in M. Mj, Pattern rigidity and the Hilbert-Smith conjecture, Geom. Topol. 16 (2012), no. 2, 1205–1246. $\endgroup$
    – Uri Bader
    Feb 3, 2018 at 19:33
  • $\begingroup$ @UriBader thanks! If I understand correctly, Theorem 2.24 of Mj's paper (msp.org/gt/2012/16-2/gt-v16-n2-p13-s.pdf, unrestricted access) implies the following result: For $G$ (totally disconnected locally compact, hyperbolic with $W(G)=1$), if the boundary $\partial G$ is homeomorphic to an $n$-sphere and admits a visual metric of Hausdorff dimension $<n+2$ [or equivalently has Ahlfors regular conformal dimension $<n+2$], then $G$ is discrete. $\endgroup$
    – YCor
    Feb 4, 2018 at 2:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.