Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 to 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!

share|improve this question
    
A variant of this question was asked in February: mathoverflow.net/questions/14548/…. In an answer there, Dmitri mentions notes of Donaldson www2.imperial.ac.uk/~skdona/GEOMETRICANALYSIS.PDF which sketch (p48) a non-complex-analysis, non-Ricci proof. The argument is apparently: show that a Riemannian manifold diffeo to $S^2$ must admit a "harmonic spinor with one pole," then take the norm of this harmonic spinor as a conformal rescaling and get a flat metric on the complement of the pole. –  macbeth Nov 17 '10 at 3:05

4 Answers 4

up vote 8 down vote accepted

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 $S^2$.

share|improve this answer
    
This idea of dealing with $S^2-\{x,y,z\}$ instead of $S^2$ is interesting. But can you clarify how to prove "directly" (i.e., without complex analysis or Ricci flow) that any metric on $S^2-\{x,y,z\}$ is conformal to a hyperbolic one? As far as I can tell, it doesn't immediately follow from the references (Berger, Taylor) cited by Betan -- they solve the constant-curvature equation for compact hyperbolic-type surfaces only. –  macbeth Nov 17 '10 at 2:23
    
@macbeth: you're right, I hadn't looked at the references closely before answering the question. So this suggestion probably needs some work to complete, given your constraints. I think one could approach this by first making a conformal change of the metric around $x,y, z$ to look like a hyperbolic cusp (which is possible, e.g. using isothermal coordinates), getting a new metric $g'$ on $S^2-\\{x,y,z\\}$. Then one could search for solutions $u$ to the scalar curvature equation which are asymptotically constant near $x,y,z$. Such boundary conditions may enable the techniques to go through. –  Ian Agol Nov 18 '10 at 7:11

If you relax the fact the metrics $g$ and $g_0$ have to be point-wise conformal to be globally conformal, i.e there exists $\phi$ a diffeomorphism of $S^2$ such that $g=e^u \phi^*(g_0)$. Then the existence of a metric conform to $g_0$ with constant curvature is equivalente to the fact that there is only one conformal class on $S^2$. Which can be proved using the fact every surface is locally conformaly flat and the fact that $S^2$ is simply connected.

share|improve this answer

Try looking in Alice Chang's ETH notes. Chapters 1 and 2 deals with the prescribed Gaussian curvature equation in 0 and positive Euler characteristics. Presumeably there's some mention of where you can find a reference for a proof of the uniformization theorem.

share|improve this answer
    
These notes are available at math.princeton.edu/~chang/zur.pdf. Incidentally, I don't think they do solve this problem. There are some partial results (see eg. Theorem 2.3) on the problem of characterizing the functions which can be realised as the scalar curvature of a metric conformal to the round metric. But this isn't quite the same thing! –  macbeth Nov 17 '10 at 2:35

One does not need the full Riemann-Roch theorem: It is enough to know the existence of a holomorphic map of degree $1$ to $CP^1$ for every compact Riemann surface $M$ of genus $0.$ There is the concept of log singularities for real functions: the local behaviour of a function f with log-sing. of weight a is $f(z)=a \ln\parallel z\parallel+g(z,\bar z)$ which is independent of the choosen holomorphic coordinated centered at the singularity. Globally $\Delta f$ is well-defined (as a $2$-form) and $\int\Delta f=2\pi\sum weights.$ With Hodge therory one can easily show that there is a harmonic function on $M$ with log-singularities of weight $\pm 1$ exactly at two points $N,S\in M.$ Because $M$ has genus $0,$ $\partial f\in H^0(M,K)$ cannot have other zeros or poles than simple poles in $N,S.$ Then $*df$ defines a map $h$ to $S^1$ away from $N,S$ by $\exp{i\int*df},$ and $\Phi=[e^f h,1]\colon M\to CP^1$ is well-defined, holomorphic of degree $1.$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.