## Ricci flow with surgery in dimension 2

Is it possible to define the Ricci flow with surgery in dimension 2 and use it to classify the surfaces?

I know this is overkill, there are simpler ways to classify surfaces, but I would like to understand the Ricci flow with surgery in dimension 3 and perhaps that this is simpler in dimension 2.

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I'm definitely not an expert, but as far as I know you never need surgery in dimension 2. If you start with a compact (probably oriented, as well) 2-manifold, the Ricci flow always has long-time existence and convergence to a constant curvature metric. This does indeed give (I believe) the classification of compact oriented surfaces. – Spiro Karigiannis Dec 2 2010 at 2:36
You can find this in the book of Ben Chow and Dan Knopf, by the way. – Spiro Karigiannis Dec 2 2010 at 2:36
To add to what Spiro said, the long-time existence result is for the volume-renormalized Ricci flow. Otherwise the sphere is a shrinking soliton and there is no convergence of the flow. See also arxiv.org/abs/math/0505163 – Willie Wong Dec 2 2010 at 3:28
While one doesn't need surgery to prove the uniformization theorem, there is some issue classifying the solitons in dimension 2. My understanding (also not an expert) is that for a long time the proofs that the only soliton metric on $S^2$ was the round sphere used the uniformization theorem. Recently, Chen, Lu and Tian proved this, though I don't know about other topologies. – Rbega Dec 2 2010 at 3:32
@RBega: that's the arxiv paper I linked to above :-p. – Willie Wong Dec 2 2010 at 3:43

(1) The normalized Ricci flow (NRF) on compact surfaces always exists for all time and does not have singularities. Moreover, NRF fixes the conformal class of the metric.

(2) Let $r$ be the integral of the curvature, which is constant under any flow. Hamilton and Osgood-Phillips-Sarnak (independently, I think) showed that if $r\leq 0$, then the NRF converges to a metric of constant curvature. Hamilton also proved that if the curvature is positive everywhere, then NRF converges to a metric of constant curvature. This part of the argument unfortunately assumes the uniformization theorem.

(3) Chow showed that if $r>0$, then eventually the curvature will become positive everywhere and then Hamilton's argument applies.

(4) Much later, Chen, Lu, and Tian wrote a 2 page paper explaining how to remove the uniformization theorem from Hamilton's argument. http://arxiv.org/abs/math/0505163

(5) Putting it all together, Ricci flow gives an new proof of uniformization, i.e. every metric on a compact surface is conformal to a metric of constant curvature. Since the only constant curvature metrics are quotients of the standard sphere, Euclidean space, and hyperbolic space, one can then deduce the classification of surfaces.

As mentioned by Spiro, this whole story, except for (4), is told in the book by Chow and Knopf.

Note: A lot of this information is already in comments by others, but not in answer form. I'm writing this mainly because the accepted answer does not seem complete to me.

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 Thank you, this is rather this kind of answer I wanted. – Guillaume Brunerie Dec 9 2010 at 19:06

The Ricci flow in dimension two is (in essence) the gradient flow of the "Polyakov action" (renormalized $\log \det \Delta$). B. Osgood, R. Phillips, and P. Sarnak proved in the late eighties (using Polyakov's trace formula) that $\log \det \Delta$ is convex on conformal classes, the critical points are metrics of constant curvature, the function is proper (which also shows that isospectral sets of metrics are compact -- the most celebrated corollary of their result at the time), and hence the uniformization theorem follows. As pointed out in the previous comments, no surgery is necessary in dimension two.

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Where is this documented? Is it in one of Ben Chow's books? I confess I didn't know this. – Deane Yang Dec 2 2010 at 4:40
I assume that the only part you don't know is that Ricci flow is the gradient flow of log det. For this, you can look at a (one page) paper by Kokotov and Korotkin Letters in Mathematical Physics (2005) 71:241–242 though the observation was known to Sarnak (and Hamilton, I would imagine) long before. I haven't read Chow's books, but I don't believe it is there (do correct me if I am wrong...) – Igor Rivin Dec 2 2010 at 6:42
@Deane: you may also be interested in the recent work of Albin, Aldana, and Rochon extending the analysis to the non-compact case. That the Laplacian has continuous spectrum and the volume can be infinite gives rise to some interesting technical restrictions. See math.jussieu.fr/~albin (I'd link to arXiv, but I think arXiv is broken at the moment: I cannot get any abstract to open up.) At the very least, their paper also contains references to the facts Igor Rivin mentioned. – Willie Wong Dec 2 2010 at 12:00