2
$\begingroup$

Let $M$ be a hyperbolic $3$-manifold and $X$ a negatively curved, simply connected space with geometric boundary $\partial X$.

If $\rho:\pi_1M\rightarrow Isom(X)$ does not fix a point in $\partial X$, then by Korevaar-Schoen (or Corlette-Donaldson, Labourie) there exists a $\rho$-equivariant harmonic map $$H^3=\widetilde{M}\rightarrow X$$

What can be said about the induced boundary map between the geometric boundaries $$S^2=\partial H^3\rightarrow \partial X?$$ Is it quasiconformal? injective? an immersion?

| cite | improve this question | | | | |
$\endgroup$
  • $\begingroup$ Measurable and nothing else in general. $\endgroup$ – Misha Oct 24 '13 at 21:18
2
$\begingroup$

I will assume, in addition, that $M$ is compact and that the target space is again ${\mathbb H}^n$, $n\ge 1$. Now, here are some examples to ponder:

  1. Suppose that the image of $\rho$ is a Schottky group (the limit set is a Cantor set $C$), then the boundary map $f$ sends $S^2$ to $C$. Can such a map be continuous?

  2. The image of $\rho$ is a dense subgroup of $Isom({\mathbb H}^n)$. Can the map $f$ be continuous in this case?

  3. Theorems you are quoting are "nonabelian generalizations" of the existence theorem for harmonic functions on hyperbolic plane. Such functions, in general, have no continuous extension to the boundary circle, only a measurable extension (in the sense of convergence a.e. along rays).

Thinking about such examples will help you to appreciate how complicated the boundary map $f$ is. In general, this map is only measurable. Sometimes, you can get a better conclusion. For instance, if $X$ is Gromov-hyperbolic and $\rho$ is an isomorphism to a quasiconvex isometry group, then the boundary map will be quasisymmetric. However, this has nothing to do with $h: {\mathbb H}^3\to X$ being harmonic, all you need is equivariance.

| cite | improve this answer | | | | |
$\endgroup$

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.