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When the interior of an n-manifold $M$ has a pinched negative curvature metric of finite volume, then its fundamental group $\Gamma=\pi_1M$ is relatively hyperbolic relative to the parabolic groups $\pi_1\partial M$, so one can consider $\partial\Gamma$ which is defined as the ideal boundary of the coned off Cayley-Graph (i.e. the Cayley-Graph of $\pi_1M$ coned off over the right cosets of $\pi_1\partial M$).

Question: Is $\partial\Gamma$ homeomorphic to the (n-1)-sphere and, if yes, where has this been proved?

In the closed case this is immediate from the quasiisometry between $\Gamma$ and the universal covering $\widetilde{M}$. In the cusped case of course $\Gamma$ is quasiisometric to the complement of the horoballs in $\widetilde{M}$, but the cones over the parabolic subgroups seem not to be quasiisometric to the horoballs, so it is not clear to me how to adapt the argument.

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    $\begingroup$ You are using a nonstandard definition of a relatively hyperbolic boundary, and I think for this definition the boundary is usually not a sphere. The standard definition is in Bowditch'es paper "Relatively hyperbolic groups" and for that definition the homeomorphism to a sphere is obvious. $\endgroup$
    – Igor Belegradek
    Commented Feb 22, 2014 at 13:47
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    $\begingroup$ Here is a sketch: suppose that we have real hyperbolic finite volume manifold whose cusps are sufficiently thick. Let's chopp them off. Then Mosher-Sageev (in a paper available at Mosher's homepage) showed that coning off the boundary results in a locally CAT(-1) pseudomanifold. Its universal cover is qi to the coned off Cayley graph of the group, so it is enough to understand its ideal boundary. It won't be a sphere. Essentially, the ideal boundary is the inverse limit of metric r spheres as $r\to\infty$ and the metric spheres are connected sums of links at cone points. $\endgroup$
    – Igor Belegradek
    Commented Feb 22, 2014 at 15:30
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    $\begingroup$ (cont.) So the metric spheres are not topological spheres, and their inverse limit isn't a sphere either. If memory serves, a recent JDG paper by Fujiwara and Manning worked out this (and much more) in detail. $\endgroup$
    – Igor Belegradek
    Commented Feb 22, 2014 at 15:32
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    $\begingroup$ ...or use Floyd boundary to get the sphere as the boundary of such group. $\endgroup$
    – Misha
    Commented Feb 22, 2014 at 16:44
  • $\begingroup$ As Igor says, this is not the correct definition of the boundary. To get the right definition, you don't cone off the parabolic subgroups, but glue combinatorial horoballs to them. As these are qi to genuine horoballs, it's obvious that you recover a space qi to the universal cover $M$ (equipped with a complete metric with cusps) and hence the boundary is a sphere. $\endgroup$
    – HJRW
    Commented Feb 22, 2014 at 19:58

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As pointed out in the comments, the boundary of the coned-off graph won't be a sphere in this case. In fact, it won't even be compact.

Bowditch's paper on relatively hyperbolic groups (recently published in IJAC) gives some details about the relationship between the two boundaries in the last few sections. In particular see Propositions 8.5 and 9.1.

To summarize Bowditch's results: If $\hat{\Gamma}$ is the coned-off Cayley graph for a relatively hyperbolic pair, and $B$ is the standard boundary for the relatively hyperbolic pair, then $\partial\hat{\Gamma}$ sits inside $B$ as a dense subset. The complement is a countable (also dense) set which can be identified with the set of maximal parabolic subgroups.

In the case you are interested in $B$ is an $(n-1)$-sphere. Thus $\partial\hat{\Gamma}$ is an $(n-1)$-sphere with a countable dense subset removed.

(EDIT: I was going to add this as a comment, but I don't have enough reputation. Let $X$ be the space obtained by coning off the boundary of $M$, and let $Y$ be the universal cover of $X$. The space $Y$ isn't quasi-isometric to $\hat{\Gamma}$. Rather the normal closure $N$ of the parabolic subgroups acts on $\hat{\Gamma}$, with a quotient quasi-isometric to $Y$.)

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