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?

  • $\begingroup$ If you precompose $p$ with a hyperbolic isometry, the function $f$ will have to change. Are you asking whether $f$ is uniquely determined by $p$? $\endgroup$ – Igor Belegradek Jan 7 '16 at 18:35
  • $\begingroup$ I don't think $f$ depends on $p$. If you change $p$ to $pq$ then $f_0$ becomes $f_0q$, but $f$ is characterised by $f_0=fp$ or $f_0q=fpq$ and so $f$ does not change. $\endgroup$ – Neil Strickland Jan 7 '16 at 19:16
  • $\begingroup$ I do not see how you concluded that $f_0$ becomes $f_0q$. You are not claiming $e^{2f_0}q^*p^*g=e^{2f_0q}p^* g$, are you? $\endgroup$ – Igor Belegradek Jan 7 '16 at 19:29
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    $\begingroup$ Let $h$ be the hyperbolic metric. We can apply $q^*$ to the identity $e^{2f_0}p^*(g)=h$ to get $e^{2f_0q}(pq)^*(g)=q^*(h)=h$, so $f_0q$ is related to $pq$ in the same way that $f_0$ is related to $p$. $\endgroup$ – Neil Strickland Jan 7 '16 at 20:20
  • $\begingroup$ I think what you want is the standard fact that on a closed surface of negative Euler characteristic every conformal class contains a unique hyperbolic metric (for if $e^{2f_1}g$ and $e^{2f_2}g$ are hyperbolic, then these two hyperbolic metrics are pointwise conformal to $g$ and hence to each other).. The fact follows from the uniformization because the hyperbolic metric becomes standard in the universal cover. $\endgroup$ – Igor Belegradek Jan 8 '16 at 0:09

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.


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