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It is known that a group $G$ acts geometrically on $\mathbb{H}^2$ if and only if $G$ is word-hyperbolic and its boundary $\partial G$ is homeomorphic to $S^1$.

The analogous statement for $\mathbb{H}^3$ and $S^2$ is open and is a conjecture of Cannon.

I read somewhere that this fails in higher dimensions, but I can't find an explicit counterexample. Could somebody provide one (in dimension 4)?

Thanks.

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it's "iff $G$ is word-hyperbolic and its boundary $\partial G$ is homeomorphic to $S^1$" (the boundary doesn't make sense for an arbitrary group) –  YCor Aug 11 '13 at 16:33

1 Answer 1

up vote 10 down vote accepted

There are various compact manifolds of negative curvature which are not homnotopy-equivalent to closed hyperbolic manifolds: Locally symmetric ones (complex hyperbolic, etc) as well as Gromov-Thurston and Mostow-Siu examples. Their $\pi_1$'s are Gromov-hyperbolic, boundary is a topological sphere. The examples exist in all dimensions $\ge 4$ (Gromov-Thurston).

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