

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer

I have an hyperbolic manifold with boundary a conformal sphere. Can I extend any conformal transformation of boundary to the interior of the ball? I know how to do with Moebius but I wonder if there is a general prescription for generic holomorphic transformations on the boundary

share|cite|improve this question
Like Igor, I'm not sure I understand the question. If the boundary of a hyperbolic manifold is a "conformal sphere", then isn't the hyperbolic manifold just the hyperbolic ball itself? If so, then any conformal transformation of the boundary (sphere) is a Mobius transformation. And, as you apparently already know, this can be extended uniquely to an isometry of the interior. Did you mean to ask something else? – Deane Yang Jul 13 '11 at 22:35
If your hyperbolic space has dimension $>3$ (so the sphere has dim $>2$), then it follows from Liouville's theorem that every conformal transformation is the restriction of a Mobius transformation, and therefore of a hyperbolic isometry. However, since you use the term "holomorphic", I suspect you are referring to hyperbolic 3-space, with boundary $S^2$. – Ian Agol Aug 11 '11 at 0:29

I assume you mean that you want to extend a holomorphic map $\phi:S^2\to S^2$ to a map $\Phi: \overline{\mathbb{H}^3}\to \overline{\mathbb{H}^3}$. Since a holomorphic map $\phi:S^2\to S^2$ is a rational map, and in particular a branched cover, you may extend it over the 3-ball (e.g. just by coning off). I assume, however, that you want to do it in some canonical way which is natural under conjugation by the Mobius group. For example, map $z\mapsto z^n$ gives a map of $S^2=\hat{\mathbb{C}}$ which extends in a natural way to $\mathbb{H}^3$ as a branched cover over a geodesic connecting $0$ and $\infty$ in $\mathbb{H}^3$. I think one may also be able to use the Douady-Earle extension as Igor suggests. The higher-dimensional version of this is given by Besson-Courtois-Gallot, and is called the "natural map". The rough idea is to associate to each point in $\mathbb{H}^3$ the visual measure on $\partial \mathbb{H}^3=S^2$, then push forward this measure by the map at infinity, and then take the barycenter of this measure to determine where the point goes. I don't see why this couldn't work for a rational map, although it may be tricky to compute. You might also have a look at a paper of Lyubich and Minsky who show how to associate a hyperbolic lamination to a rational map.

share|cite|improve this answer

I dont really understand the question, but I suspect the keywords are "Douady-Earle extension"'; see

Quasiconformal harmonic extension of a quasi-symmetric map on $S^1$

or the original paper:

share|cite|improve this answer

Your Answer


