# Are conformal maps on a hyperbolic surface isometries?

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?

• Yes, of course. Oct 20, 2021 at 13:30
• Thanks. But why? Oct 20, 2021 at 13:37
• math.stackexchange.com/questions/886542/… Oct 20, 2021 at 14:32

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 $$f$$.