One of the fundamental theorems of
(I rewrote this question, hopefully it's more clear now. It's still the theory of surfaces is thatsame question, for any surface but I reordered its parts.)
Let S, given be a surface (possibly non-compact, but no boundary). It seems that there are three different theorems that people call "Uniformization Theorem":
There is some constant-curvature metric on S.
Given any conformal structure (or a complex structure)on S, there is a constant-curvature complete metric on S that represents the same conformal structure (and itis unique up to scaling). For the purposes of this questionMore precisely,here's the exact statement I have in mind:
Given any metric g on S, there is a unique (unique) complete metric g0 of constant curvature that represents the same conformal structure, i.e. there is a strictly positive function $f\in C^\infty (S)$ such that g'=fg.
The
Up to diffeomorphism, there is only one conformal structure on the sphere, and two on a topological disk (namely, the disk with the standard hyperbolic metric or the plane with the Euclidean metric).
The first of these is easy, see proof below. It's also easy to see that I'm aware of uses the Uniformization theorem3 implies 2, see below. However
Henceforth, the "Uniformization theorem Theorem" will mean number 3. It is quite very powerful, but also rather mysteriousto me. At the same time, this feels excessive - Its proofs are rather complicated; here is a discussion about it.
I think feel that , philosophically2, on the theorem I'm talking about other hand, should be simpler. Moreover, I feel that the fact that the conformal structure on a disk/sphere/plane is unique and not require the fact that any conformal structure full power of uniformization to prove it. I also think it is represented by quite distinct in spirit from 3, and so having a constant-curvature metric are two unrelated factsdifferent proof for it could be illuminating. So, I wonder, is my basic questions are:
Are there a any nice proof proofs of 2. that does do not use Uniformization3.?Is there any way to prove 3. from 2.? (I'm willing this would imply that the answer to compromise on the above question is no. This is certainly possible unless we remove the work "unique" and "complete" parts, although from the statement of course it would be best to have them2., especially completeness. Uniqueness would probably be equivalent to the actual Uniformization... so we'll do that)(It would also be great nice if this the proof worked for non-compact surfaces, i.e. with cusps or funnels)funnels.
Clarification: apparentlyThen, the title of this post is an oxymoron to somehowever, since they call it could be tricky to make the theorem I'm talking about Uniformization. In my mind, there are three different theorems. The statement that there is some constant curvature metric is easier than the statement that complete - or maybe there is one conformally equivalent some trick to a given one, which in turn (I hopemake any constant-curvature metric complete?)should be much easier than the statement that there are only two possible conformal structures on a topological open disk (and only one on a sphere). When I say "Uniformization", I mean the last one; the theorem I'm talking about is the middle one.
The fact that people call all of these "Uniformization" suggests they might be equivalent, but this completely contradicts my intuition, and I certainly can't imagine a proof of how the first would imply the last. Is there one?
What follows are the promised proofs together with some vague thought thoughts on various approaches that came how to my mindprove number 2 using them or otherwise. Ideally, it would be great to have two proofs, along one for Ricci flow, and another one with something like the lines proof of numbers theorem 1and 2 below. Also, perhaps there are other/better approaches?
Proof #0: Uniformization
Consider the universal cover $U=\tilde S$ of S with the metric g pulled back from S. By uniformization, there is a conformal map $f: (U,g) \to (X,g_{standard})$ where $(X,g_{standard})$ is one of three: the Poincare disk with the standard hyperbolic metric, the plane with the Euclidean metric, or the sphere with the standard spherical metric. In any case, it's a standard space with a constant-curvature metric. So, we can pull $g_{standard}$ back to $U$, and then down to $S$ to get $g_0$.
This argument has a small problem with it: it's not clear that the metric we get on U is invariant under the covering transformations. One way to avoid this problem is to include this fact in the mystery that is Uniformization (in other words, the Uniformization theorem can be done in an equivariant way); I don't know if there are others.
Proof idea #1: Ricci-like flow
It seems that we could try to apply something like the Ricci flow to the metric g on S to make its curvature uniform. I don't know enough analysis to fill in the details; I'm wondering whether it's possible to create such a flow so that it does not change the conformal structure. Also, getting a complete metric could be tricky1.
Proof #2 (of a simpler fact): Construct it by handabove
Note that we can get many other conformal structures on our surface by replacing our regular *4g*-gon with an irregular one (we can pick any length for each side, I think, as long as the sides we glue have the same length). However, it's not clear that we could get any conformal structure this way. So, this is also a vague approach to proving number 2.
Proof idea for number 2: Ricci-like flow (used to be Proof idea #1, sorry.)
It seems that we could try to apply something like the Ricci flow to the metric g on S to make its curvature uniform. I don't know enough analysis to fill in the details; I'm wondering whether it's possible to create such a flow so that it does not change the conformal structure. Also, getting a complete metric could be tricky.
(Many people answered that this is possible. In particular, Dmitri's answer pretty much settled for me this part of the question. Only one small question remains: will these flows work for non-compact surfaces and result in a complete metric?)
Appendix: Proof of 2. from Uniformization (#3.)
Consider the universal cover $U=\tilde S$ of S with the metric g pulled back from S. By uniformization, there is a conformal map $f: (U,g) \to (X,g_{standard})$ where $(X,g_{standard})$ is one of three: the Poincare disk with the standard hyperbolic metric, the plane with the Euclidean metric, or the sphere with the standard spherical metric. In any case, it's a standard space with a constant-curvature metric.
As Tom's comment pointed out, at least in the hyperbolic case, all conformal maps on the disk preserve the constant-curvature metric (we can list what they all are). Since the covering transformations become conformal maps on X, they preserve the metric. So, we can pull $g_{standard}$ back to $U$, and then down to $S$ to get $g_0$. The non-hyperbolic cases can also be dealt with.
