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Let $\Sigma$ be a closed surface of genus at least $2$.

Put a quasi-Fuchsian $\mathbb{CP}^1$-structure (i.e. complex projective structure) on $\Sigma$. Thus the universal cover $\tilde{\Sigma}$ is identified with a domain $\Omega\subset\partial\mathbb{H}^3$, and the fundamental group $\Gamma$ identified with a subgroup of $\mathrm{PSL}(2,\mathbb{C})$.

By Donaldson's theorem on twisted harmonic maps, there is a unique $\Gamma$-equivariant harmonic map $\varphi:\Omega\rightarrow\mathbb{H}^3$. $\varphi(\Omega)$ is a surface in $\mathbb{H}^3$ with asymptotic boundary $\partial\Omega$. Assume that $\varphi$ is an immersion. Then there is a hyperbolic Gauss map $g$ sending $\varphi(\Omega)$ back to $\partial\mathbb{H}^3$. (To be precise, for each point $m$ on the surface $\varphi(\Omega)$, we let $v(m)\in T_m\mathbb{H}^3$ be the unit normal vector of the surface pointing towards $\Omega$. Then we follow the geodesic with initial velocity $v(m)$ to the infinity, the limit is defined to be $g(m)$.)

The map $g\circ \varphi: \Omega\rightarrow\partial\mathbb{H}^3$ is of course $\Gamma$-equivariant and is just the identity if the quasi-Fuchsian structure is actually Fuchsian.

Question: What can we say about $g\circ \varphi$ in general? Is it conformal? Is its image $\Omega$ itself? Is it always the identity?

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  • $\begingroup$ this does not answer the question, but may be of interest. By results of F. Labourie and G. Smith, (under some weak hypotheses on S) for every map $f:\tilde S\to\partial \mathbb H^3$ there is a unique map $\tilde S\to\mathbb H^3$ whose image is a surface of constant curvature and whose Gauss map is $f$. What can be said about the curvature of the image of an harmonic map? $\endgroup$
    – user126154
    Jun 7, 2014 at 11:00
  • $\begingroup$ You might find the papers of Epstein and Marden interesting. Rather than consider $\phi(\Omega)$ and projecting that back to $\partial\mathbb{H}^3$, they consider the boundary of the $\mathbb{H}^3$ convex hull of $\Omega$, and they project that back to $\Omega$. They construct these maps to be uniformly quasiconformal. That might be a better, or at least easier, guess for $g \circ \phi$ than that it be conformal. $\endgroup$
    – Lee Mosher
    Jun 7, 2014 at 12:07
  • $\begingroup$ Thanks, these are important informations. Now it seems to me that this map is not as good as I thought. B.T.W., the principal curvatures of the image of an harmonic map can be quite arbitrary. $\endgroup$
    – Xin Nie
    Jun 7, 2014 at 17:23
  • $\begingroup$ The paper to read is D.Dumas "Holonomy limits of complex projective structures". $\endgroup$
    – Misha
    Jun 7, 2014 at 23:42

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