(I rewrote this question, hopefully it's more clear now. It's still the same question, but I reordered its parts.)

Let S be a surface (possibly non-compact, but no boundary). It seems that there are three different theorems that people sometimes call "the Uniformization Theorem":

  1. There is some constant-curvature metric on S.

  2. Given any conformal structure on S, there is a constant-curvature complete metric on S that represents it. More precisely,

    Given any metric g on S, there is a (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.

  3. 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 3 implies 2, see below.

Henceforth, "Uniformization Theorem" will mean number 3. It is very powerful, but also rather mysterious. Its proofs are rather complicated and usually involve solving a somewhat complicated differential equation, here is a very interesting discussion about that.

I feel that 2, on the other hand, should be simpler, and not require the full power of uniformization to prove it. I also think it is quite distinct in spirit from 3, and so having a different proof for it could be illuminating. So, my basic questions are:

  • Are there any nice proofs of 2. that do not use 3.?
  • Is there any way to prove 3. from 2.? (this would imply that the answer to the above question is no. This is certainly possible unless we remove the work "unique" from the statement of 2., so we'll do that)

(It would be nice if the proof worked for non-compact surfaces, i.e. with cusps or funnels. Then, however, it could be tricky to make the metric complete - or maybe there is some trick to make any constant-curvature metric complete?)

What follows are the promised proofs together with some vague thoughts on how to prove number 2 using them or otherwise. Ideally, it would be great to have two proofs, one for Ricci flow, and another one with something like the proof of theorem 1. Also, perhaps there are other/better approaches?

Proof of theorem 1. above (used to be Proof #2, sorry)

Here's a proof that just constructs some constant-curvature metric on any compact surface. This basically comes from Thurston's book, and could probably be adapted to surfaces with boundary, although it wouldn't be trivial.

Namely, a genus-g surface can be thought of as a regular 4g-gon with sides glued in the appropriate fashion. (If g=0, it's already a sphere, so we ignore this case). For example, a torus is a square with opposite sides glued. So, we can tile the plane with squares, by taking one square and translating it around so that new squares are adjacent to old ones. We notice that these translations do not change the standard metric, and conclude that the plane modulo a $\mathbb Z^2$ action by translations is a torus with a constant-curvature metric.

Similarly, if $g>1$, we can present a regular 4g-gon in the hyperbolic plane. By varying its size, we can make the angles at vertices as small as we want (if it's tiny, the angles will be almost as large as they would be for a Euclidean regular 4g-gon. If it's so huge that the vertices are almost at the line at infinity, the angles will be almost zero. So, we can get any intermediate value). If we make these angles precisely $\frac{2\pi}{4g}$, then translating the 4g-gons around just like we translated squares before will give us a tiling of the hyperbolic plane. Just like before, quotienting out the hyperbolic plane by the appropriate group will give us our surface with a constant-curvature metric.

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)

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? Also, it's not quite clear to me whether the proofs with flows are powerful enough to show #3. My guess would be that they aren't, but Hamilton's paper seems to claim that the Ricci flow proof is.)

Appendix: Proof of 2. from Uniformization (#3.) (Used to be Proof #0)

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.

Usual disclaimer: there might be mistakes, point them out. Thank you!

  • $\begingroup$ Proof idea #1 has been carried through, by C. Chen, P. Lu and G. Tian, improving on earlier work by R. Hamilton. $\endgroup$
    – engelbrekt
    Feb 7, 2010 at 21:06
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    $\begingroup$ Regarding your concern in Proof 0: for D the unit disk, the conformal automorphisms Aut(D) are the Mobius transformations preserving D (conjugate to PSL(2,R)), and by Schwartz-Pick these preserve the Poincare (hyperbolic) metric. It's fun to prove this yourself. For Aut(C) this is not true, since Aut(C) is the affine transformations {z -> az+b}. But deck transformations act freely, and only transformations of the form z -> z+b have no fixed point (otherwise -b/(a-1) is fixed), and translations preserve the Euclidean metric. The sphere has no orientable quotients so this case is automatic. $\endgroup$
    – Tom Church
    Feb 7, 2010 at 21:48

4 Answers 4


I don't know that you can make a good distinction between the Riemann uniformization theorem, and the theorem that a surface has a constant-curvature metric in the same conformal class. Each of these results is a relatively easy corollary of the other one, and you can view them as the same theorem.

The old methods of Riemann and Weierstrass give you one difficult proof of the uniformization theorem. Ricci flow is another, beautiful proof of the uniformization theorem. I find it conceptually easier, but it's still a formidable proof. The Ricci flow proof requires various non-trivial results in differential geometry and parabolic PDEs.

Your proof #2 is indeed a proof of a much simpler fact. I think that the main lesson of this proof is the great distance between the existence of a constant-curvature metric on a surface, and the bijection between all such metrics and conformal structures.

  • $\begingroup$ How can you prove that having a constant-curvature metric in a conformal class implies the full Uniformization? $\endgroup$ Feb 7, 2010 at 20:58
  • $\begingroup$ By the developing map, any complete simply connected surface with constant curvature (-1 / 0 / 1) is isometrically covered by the (hyperbolic plane / Euclidean plane / round sphere). If your surface is simply connected, this must be a homeomorphism, i.e. an isometry, and so your Riemann surface is isomorphic to the (disk / plane / Riemann sphere). $\endgroup$
    – Tom Church
    Feb 7, 2010 at 21:39
  • $\begingroup$ @Tom: I don't think the developing map will give us the standard conformal structure on the hyperbolic plane, will it? The way I see it, it'll give us some constant curvature metric on the plane, but to see that this is in fact the hyperbolic plane you need actual uniformization. $\endgroup$ Feb 7, 2010 at 21:51
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    $\begingroup$ There is only one complete constant curvature -1 metric on the plane. Proof: fix a basepoint, and parametrize your space with "geodesic polar coordinates": for each other point you specify a) a direction, and b) the distance to travel in that direction. Now doing the same thing in the hyperbolic plane gives you an identification of your space with the hyperbolic plane which is an isometry on geodesics through the basepoint. (So far this works for any nonpositively curved metric.) Now analyzing the curvature tensor and using the constant curvature -1 you can check this really is an isometry. $\endgroup$
    – Tom Church
    Feb 8, 2010 at 1:23
  • $\begingroup$ @Tom: thank you, this seems to solve it. Why didn't you write this as an answer, so that I & other people could give you credit? $\endgroup$ Feb 8, 2010 at 2:10

As for the Ricci flow bit, Ricci flow does it. I sketch the proof in high genus here, and here's the paper that that post is based on. So, this may not be quite what you want (and it needs patching in genus 0, and is much harder in the non-hyperbolic cases) but it proves uniformization through Ricci flow, which seems to be what you speculated might be sufficient.

  • $\begingroup$ Is the resulting metric conformally equivalent to the original one? It seems to me that the statement that there is some constant curvature metric is easier than the statement that there is one conformally equivalent to a given one, which in turn (I hope) 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). Is that wrong? $\endgroup$ Feb 7, 2010 at 20:48
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    $\begingroup$ Miraculously, Ricci flow in 2D evolves a metric without changing its conformal class. $\endgroup$ Feb 7, 2010 at 20:50
  • $\begingroup$ Yeah, what Greg said. You can actually check this pretty easily, as I recall, though I've lost most of my Ricci flow knowledge since moving in a more algebraic direction. $\endgroup$ Feb 7, 2010 at 20:51
  • $\begingroup$ I'll put my previous comment in the question proper, since apparently I'm not using the standard terminology for Uniformization. $\endgroup$ Feb 7, 2010 at 20:56

There is a different PDE proof that does not use uniformisation, it is sketched on one page (34) of Donaldson's notes. http://www2.imperial.ac.uk/~skdona/GEOMETRICANALYSIS.PDF This proof goes via continuity method.

It is intriguing (as is remarked on page 35 of the notes), that in the case of $CP^1$ the proof does not work (i.e. becomes much harder), due to non-compactness of the group of conformal symmetries of $CP^1$. For higher-dimensional Kahler manifolds in general it is harder to construct Kahler-Einstein metrics of positive curvature comparing to KE metrics of negative curvature.

  • $\begingroup$ Thank you, this settles a large part of my question. I wonder, however, can that proof be adapted to non-compact surfaces? (Of course, this will probably be clear once I read it). $\endgroup$ Feb 7, 2010 at 23:39
  • $\begingroup$ Ilya, unfortunately I can not tell this straight away, but I'll try to find out $\endgroup$ Feb 8, 2010 at 0:15
  1. In your statement you said "let S be a surface". First of all one has to be careful about what is a "surface". It must be a) orientable, and b) separable (have a countable base of topology). (All Reimann surfaces automatically satisfy this but b) is non-trivial).

  2. So suppose that S is a Riemann surface. All of the statements 1, 2, 3 contain a "soft part" and a "hard part".

The hard part is about simply connected Riemann surface. (Sometimes THIS is called the uniformization theorem).

The reduction of the general case to the simply connected case is by the universal covering. (You can pull back all metrics, conformal structures etc. to the universal cover). There are several ways to prove the theorem for a simply connected surface.

One of the simplest ways is Lavrentjev's proof which can be found in Goluzin's book Geometric theory of functions of a complex variable. This proof assumes that we already know that the surface is triangulable. (This is a purely topological consequence of a) and b) above, but again, not completely trivial.

Of course, if you are interested only in COMPACT case, then it is not necessary to pass to the universal covering, many topological difficulties disappear, and more methods are available. Like the Ricci flow, for example.


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