Is the conformal compactification of a convex-cocompact hyperbolic manifold $M$ conformally diffeomorphic to the convex core of $M$?

Question

I am not a hyperbolic geometer, so I apologize if I get anything wrong here, and please correct me.

The conformal compactification $\overline {\mathbb{H}^n}$ of hyperbolic $n$-space $\mathbb{H}^n$ can be obtained by viewing hyperbolic space as a subspace of projective space, since the boundary, which is the projectivized Minkowski null cone, is the conformal sphere $\partial \mathbb{H}^n \cong \mathbb{S}^{n-1}$. From this perspective, it is tautological that isometries of $\mathbb{H}^n$ extend to the boundary. As I understand it, the conformal compactification can alternatively be obtained by attaching the visual boundary to hyperbolic space.

Now, for a complete hyperbolic manifold, we know that $M \cong \Gamma \backslash \mathbb{H}^n$ where $\Gamma := \pi_1(M)$. The hyperbolic manifold $M$ is convex-cocompact if and only if it is conformally compact. In this case, the conformal compactification $\overline{M}$ can be obtained by $\Gamma \backslash (\overline {\mathbb{H}^n} - \Lambda(\Gamma))$ where $\Lambda(\Gamma)$ is the subset of the boundary which is limit set of $\Gamma$.

We also know that $M$ is convex-cocompact if and only if it contains a convex compact subset $K$ such that the interior $\operatorname{int}K$ is homeomorphic to $M$. In this case, we obtain $K$ by taking the convex core of $M$.

I am not sure what the boundary $\partial K$ of the convex core $K$ of a convex-cocompact manifold $M$ looks like in general, but at least in the case where $M$ has Fuchsian ends and is not Fuchsian, then I know from this paper of Kerckhoff and Storm that the convex core is a manifold with totally geodesic boundary. Thus, we can restrict to this special case if necessary.

I know that the conformal compactification $\overline{M}$ is homotopy equivalent to its convex core $K$, and moreover the boundary of the convex core $\partial K$ is homotopy equivalent to the boundary of $\partial \overline{M}$ thanks to the answers on this question here.

However, I would like to know if in fact we can say more: do we have a conformal diffeomorphism from $\overline{M}$ to the convex core $K$ of $M$?

Answer 1

No, this follows from Liouville's theorem. A conformal diffeomorphism $\phi: \overline{M}\to K$ would have to be a restriction of an $n$-dimensional Möbius transformation in any chart. Lifting to the universal cover, we would get a conformal map $\mathbb{H}^n \to \mathbb{H}^n$, with image in the convex hull of $\Lambda(\Gamma)$, a proper subset of $\mathbb{H}^n$ whose closure contains $\Lambda(\Gamma)$. By Liouville's theorem, the image of this map must be a ball (since Möbius transformations take balls to balls). Moreover, it must be equivariant under the action of $\Gamma$, and the closure of the image must contain $\Lambda(\Gamma)$. Since the convex hull is $n$-dimensional, the image must be all of $\mathbb{H}^n$, a contradiction.