We approach the problem of finding a metric of constant curvature on a surface (i.e. a $C^\infty$ 2-manifold). Specifically, what we want to do is, given a surface $M$ and a metric $g_0$, show that there exists a new metric $g$ of the form $g=e^{2u}g_0$ for some $u\in C^\infty (M)$ such that $g$ has constant curvature. It is well known that if we let $K_0$ and $K$ be the curvatures associated to $g_0$ and $g$ respectively, then they must satisfy the curvature equation
\begin{equation}
K=(K_0-\triangle u)e^{-2u}
\end{equation}
or equivalently,
\begin{equation}
\triangle u-K_0+Ke^{2u}=0
\end{equation}
where $\triangle$ stands for the Laplace-Beltrami operator associated to the original metric $g_0$. Thus, solving our problem amounts to solving the curvature equation (i.e. finding $u$) for different values of $K$.
We have essentialy three cases (thanks to the Gauss-Bonnet theorem), which are $K=0$, $K=-1$, and $K=1$. The first two cases are fairly straightforward (simple solutions can be found in an article by Melvin S. Berger 1971, or the book on PDE's by Taylor), but the case when $K=1$ is more challenging. I have read solutions which use the Riemann-Roch theorem on one hand or Ricci flow on the other.
The question is: is there any other way to find a solution to the curvature equation (in the case $K=1$)? The reason I want to know this is because I am writing my thesis and I would like to give the simplest proof of this result to make it more accessible. Thanks!
Best Answer
If you like the case $K=-1$ better, one way to do this is to choose 3 points $\{x,y,z\} \subset S^2$, and use the uniformization theorem to find a complete conformally equivalent metric on $P=S^2-\{x,y,z\}$ with constant curvature $K=-1$. There is a unique such metric on $P$, which is conformally equivalent to $\mathbb{CP}^1-\{0,1,\infty\}$. Then fill in the punctures to get a conformally equivalent metric on $S^2$. One way to understand why uniformization of $S^2$ is a bit harder than the other cases is that there are Mobius transformations of $S^2$ which are conformal transformations but not isometries, so there is not a unique metric with $K=1$. By choosing three points, though, one gets rid of these conformal symmetries.
Addendum: Incidentally, the first proof that the Ricci flow on $S^2$ converges to the round metric was due to Bennett Chow, using a type of entropy defined specially on $S^2$ (Chow finished off a case not resolved by work of Hamilton). I think Perelman's work gives a new proof in the case of $S^2$, which I'm sure he realized, but I'm not sure has been properly disseminated. The idea is that if a singularity forms in finite time for Ricci flow on $S^2$, then one may take a rescaled limit to get a $\kappa$-non-collapsed positive curvature ancient solution (Perelman proof of this works in arbitrary dimensions, so is not special to $S^2$). In two dimensions, the only such solutions are solitons, which are either Hamilton's cigar or $S^2$ with the round metric (plus some non-orientable examples). But the cigar is collapsed, so the only possibility is $S^2$, which implies that the metric converges at the singular time to the round metric on $S^2$.