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I'm trying to convince myself that if $M\cong\mathbb{H}^3/G$ is a closed hyperbolic 3-manifold then the limit set $\Lambda(G)$ equals the whole Riemann sphere $S_\infty^2$. My idea of the proof goes as follows: since $M$ is compact, then it has finite volume, so that $G$ must be finitely generated. This would imply, by Ahlfors' finiteness theorem, that the quotient of the regular set $\Omega(G)=S^2_\infty\setminus \Lambda(G)$ must be a Riemann surface with finitely many components and finitely many punctures. On the other hand, we can identify the (conformal) boundary of $M$ with $\Omega(G)/G$ thus, by closedness of $M$, $\Omega(G)$ must be empty and $\Lambda(G)$ is the entire sphere.

The only part that makes me doubtful is the one about the conformal boundary: could I treat the conformal boundary exactly as if it were the boundary of the manifold $M\cong \mathbb{H}^3/G$?

Thank you.

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1 Answer 1

In general, there is a homeomorphism between the conformal boundary and the boundary of the convex core of $M$, see Epstein-Marden "Convex hulls in hyperbolic space, a theorem of Sullivan, and measured pleated surfaces". I didn't check whether their proof goes still through when the conformal boundary is empty, but in any case what you are asking for is a well-known theorem. Morally it is true because ${\mathbb H}^3$ can be tesselated by copies of a compact fundamental domain and their compactness implies that you have infinitely many copies of a fundamental domain (and hence infinitely many points of any orbit) in any given neighborhood of any given point $x\in S^2_\infty$, so $x$ must belong to the limit set.

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