Welcome to MathOverflow

Question

4

Let G be a word-hyperbolic group acting on its boundary, which is homeomorphic to $S^n$ (n-sphere), effectively. Does this imply that G acts on the boundary as a convergence group of $S^n$?

If this is true in general, is it easy to see, at least for n = 1 or 2?

flag	asked Oct 26 at 18:10 Harry Baik 93●4

Andy Putman · Accepted Answer · 2012-10-26 18:24:15Z UTC

Bowditch proved much more. Namely, if a group $\Gamma$ acts properly discontinuously on a $\delta$-hyperbolic space $X$, then $\Gamma$ acts as a convergence group on $\partial X$. See Lemma 1.11 of his paper

B.H. Bowditch, Convergence groups and configuration spaces, in ``Geometric Group Theory Down Under, proceedings of a Special Year in Geometric Group Theory, Canberra, Australia'' (ed. J.Cossey, C.F.Miller III, W.D.Neumann, M.Shapiro), de Gruyter (1999), 23-54.

which is available here. To get the result you want, consider the action of $\Gamma$ on its Cayley graph.

Igor Rivin · Answer 2 · 2012-10-26 18:16:06Z UTC

3

This appears to be a theorem of Brian Bowditch.

answered Oct 26 at 18:16

Igor Rivin
31.2k●3●37●71

2

I think it was first written by P.Tukia in "Convergence groups and Gromov's metric hyperbolic spaces", New Zealand J. Math. 23 (1994), no. 2, 157–187. – Misha Oct 26 at 19:57

1

I guess Brian's result is really the converse... – Igor Rivin Oct 26 at 20:11

convergence action on the boundary of hyperbolic groups

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

2 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.