Are harmonic maps quasiconformal at the boundary of hyperbolic spaces? 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?
 A: 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:


*

*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?

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

*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. 
