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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?

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2 Answers 2

up vote 4 down vote accepted

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.

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This appears to be a theorem of Brian Bowditch.

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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 '12 at 19:57
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I guess Brian's result is really the converse... –  Igor Rivin Oct 26 '12 at 20:11

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