Let $X$ be a compact oriented surface of genus at least two, equipped with a Riemannian metric $g$. By the uniformization theorem for Riemann surfaces, there is a conformal universal covering map $p:D\to X$, where $D$ is the unit disc. The standard hyperbolic metric on $D$ is thus $e^{2f\_0}p^\*(g)$ for some function $f\_0$ on $D$. This is invariant under the group of deck transformations, and so descends to a function $f$ on $X$. The hyperbolic metric has Gaussian curvature $-1$, and we deduce that $e^{2f}g$ also has curvature $-1$.


I think it is true that $f$ is the **unique** function such that $e^{2f}g$ has curvature $-1$, but this seems to be a bit harder to prove than I thought. Does anyone know a proof or reference? 



Uniqueness of hyperbolic rescaling
Here is an answer which I learned from a set of notes titled "Conformal Geometry Seminar - The Poincare Uniformization Theorem" by Gilbert Weinstein (DOI: 10.13140/RG.2.1.4644.9767). Suppose we have found one metric $g$ in the original conformal class with curvature $K(g)=-1$. Suppose also that $K(e^{2f}g)=-1$; we need to show that $f=0$. A fairly standard formula tells us that $K(e^{2f}g)=(K(g)-\Delta(f))/e^{2f}$, so when $K(g)=K(e^{2f}g)=-1$ we obtain $\Delta(f)=e^{2f}-1$, so $f\Delta(f)=f(e^{2f}-1)$. We now integrate over $X$. It is another standard fact that $\int\_Xa\Delta(b)=-\int\_X\langle\nabla(a),\nabla(b)\rangle$, so we get 
$$ \int\_X \|\nabla(f)\|^2 = - \int\_X f(e^{2f}-1). $$
Note that the integrand $f(e^{2f}-1)$ is nonnegative, and can only be zero where $f=0$. This is only consistent if $f$ is identically zero, as required.


