2
$\begingroup$

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 $M$?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks for taking the time to answer! For my purposes I could settle for the weaker statement that the conformal boundary $\partial \overline{M}$ is conformally diffeomorphic to the boundary $\partial K$ of the convex core of $M$ (if not for general convex cocompact hyperbolic manifolds, then at least for the ones with Fuchsian ends as considered in the Kerckhoff and Storm paper). Would you happen to know if this weaker statement holds? $\endgroup$ – ಠ_ಠ Aug 31 '18 at 0:16
  • 1
    $\begingroup$ Yes, the boundary is fuchsian iff the convex core boundary is conformally equivalent to the boundary at infinity. $\endgroup$ – Ian Agol Aug 31 '18 at 2:56
  • $\begingroup$ Awesome! That's perfect. Would you happen to know of a reference for this? $\endgroup$ – ಠ_ಠ Aug 31 '18 at 3:02
  • 1
    $\begingroup$ Maybe in the 3D case there’s a reference, but the proof is not hard if you are familiar with Möbius transformations. The boundary of $H^n$ is an $n-1$ sphere, and a Fuchsian group preserves a geodesic $H^{n-1}$ in $H^n$. The boundary is an $n-2$ sphere, and it bounds two hemispheres in the $n-1$ sphere. There is a Möbius transformation fixing the $n-2$ sphere and taking the $H^{n-1}$ to one of the hemispheres. This transformation commutes with the Fuchsian group preserving the $n-2$ sphere, so it gives a conformal equivalence between the Fuchsian group and its conformal boundary. $\endgroup$ – Ian Agol Aug 31 '18 at 3:22
  • $\begingroup$ Great! Thanks so much for your help. $\endgroup$ – ಠ_ಠ Aug 31 '18 at 4:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.