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Hi, I have a question about the positive Euler characteristic case. My question is: why is it so difficult as compared to the zero and negative cases? I am more interested in a pictorial/intuitive answer as compared to a very rigorous analytic answer. In other words, I want to get an intuitive "feel" of what goes wrong..... Thanks in advance.

flag	asked Dec 1 2011 at 3:51 anonymous 63●2

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You're talking about 2-dimensional manifolds... so perhaps you could quantify what you mean by difficult? – Ryan Budney Dec 1 2011 at 6:54

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"Difficult" in this case means the opposite of "easy" -- it took Ben Chow many years to fill in this case (Hamilton did the negative curvature case). – Igor Rivin Dec 1 2011 at 12:19

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I think it's possible now using Perelman's work to prove Ricci flow on the 2-sphere converges to the round metric replacing the entropy of Hamilton/Chow with Perelman's entropy. The idea is that if there's a finite-time singularity, then it should be kappa-non-collapsed from Perelman's entropy monotonicity, and a blow-up limit should be a Ricci soliton with positive curvature. But the only solitons in dimension 2 with positive curvature are the cigar (which is collapsed) and the round 2-sphere. What I don't remember is whether the proof of uniqueness of solitons depends on uniformization. – Agol Dec 1 2011 at 17:25

Vitali Kapovitch · Accepted Answer · 2011-12-01 16:54:35Z UTC

I'm not an expert on Ricci flow but I believe the rough general reason for this is as follows. In dimension 2 the normalized Ricci flow gives the following evolution equation for scalar curvature $$ \frac{\partial R}{\partial t}=\Delta_t R+R(R-r) $$ Where $r=\frac{\int_MR}{vol M}=2\pi \chi(M)$. The analysis of this equation is closely linked (via maximum principle) to that of the ODE $\frac{\partial R}{\partial t}=R(R-r) $. The nonzero stationary point of the ODE is $R=r=const$ (which is also the limit of $R$ under the Ricci flow as $t\to\infty$); it is stable when $r<0$ and unstable when $r>0$. Thus for $r<0$ the behaviors of the ODE and the PDE agree as both the diffusion laplacian term and the ODE term work in the "same direction" toward the stationary solution. This makes the convergence estimates in this case quite easy. In contrast for $r>0$ the laplacian term and the ODE term work in "opposite directions" (with the laplacian term ultimately winning) which makes the analysis in this case more delicate.

Igor Rivin · Answer 2 · 2011-12-01 14:18:05Z UTC

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There is a very nice exposition of the whole argument here. (or here, the spaces freak out some browsers)As that paper points out, Ricci flow did not produce a proof of Uniformization until 2009, although this is not entirely true: Ricci flow is the gradient flow of $\log \det \Delta$ in two dimensions, and Osgood-Phillips-Sarnak proved uniformization by optimizing $\log \det \Delta$ in a conformal class in the late '80s.

edited Dec 1 2011 at 14:18

answered Dec 1 2011 at 12:25

Igor Rivin
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	Link doesn't work for me – Deane Yang Dec 1 2011 at 14:14
	OK, try the second link (I don't know why space confuse things...) – Igor Rivin Dec 1 2011 at 14:18
	Another (possibly) useful link: dx.doi.org/10.1090/S0002-9939-06-08360-2 – YangMills Dec 2 2011 at 4:11
	Did the Osgood-Phillips-Sarnak paper tackle the positive Euler characteristic case? – anonymous Dec 2 2011 at 22:57
	Yes, it did, as I recall (and there, too it was the hardest case...) I – Igor Rivin Dec 3 2011 at 10:19

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uniformization theorem via ricci flows

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