Let $S$ be a closed surface of genus $g\geq 2$ and $h$ a hyperbolic Riemannian metric (constant curvature $-1$) on $S$. A conformal diffeomorphism $f \colon (S,h) \to (S,h)$ is a diffeomorphism that preserves angles, in technical terms: for all $p\in S, v,w \in T_pS$ we have
$$
h_p\left(\frac{v}{\sqrt{h_p(v,v)}},\frac{w}{\sqrt{h_p(w,w)}}\right) =
h_{f(p)}\left(\frac{D_pf v}{\sqrt{h_{f(p)}(D_p v,D_pf v)}},\frac{D_pf w}{\sqrt{h_{f(p)}(D_pfw,D_pfw)}}\right).
$$
We note that every isometry $g$, i.e. $h_p(v,w)=h_p(D_pgv, D_pgw)$, is a conformal map.
The question is: Is every conformal diffeomorphism an isometry?
I know that every conformal map has to satisfy $h_p(v,w) = \alpha(p) \cdot h_{f(p)}\left( D_pfv , D_pf w \right)$ for a positive function $\alpha \colon S\to \mathbb{R}_{>0}$. For an isometry, $\alpha = 1$. Perhaps Gauss-Bonnet could be used to solve this question?
Best Answer
A simple corollary of the Riemann mapping theorem states that a conformal bijection $g$ from the unit disk $D \subset \mathbb C$ to itself is uniquely determined by $g(0)$, the complex argument of $g'(0)$, and whether $g$ is orientation preserving. As these values also uniquely determine Möbius transformations preserving $D$, and Möbius transformations are conformal, we see that all conformal self-maps of $D$ are Möbius transformations.
Now in the (conformal) Poincaré disk model of the hyperbolic plane $H^2$, hyperbolic isometries are exactly those Möbius transformations preserving $D$, hence conformal self-diffeomorphisms of $H^2$ are isometries.
Finally, a conformal diffeomorphism $f: S \to S$ lifts to a conformal diffeomorphism $\tilde f: \tilde S \to \tilde S$, where $\tilde S = H^2$ is the universal cover of $S$. By the above argument, $\tilde f$ is an isometry, therefore so is $f$.