uniformization theorem via ricci flows

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

Given a closed surface $M$ with Euler characteristic $\chi$ and a fixed pointwise conformal class $\mathcal{C}$, consider the problem of finding a metric $g\in\mathcal{C}$ with $R(g)\equiv r\doteqdot\operatorname{sign}(\chi )$. If $r=-1,$ then there is a unique such metric. If $r=0,$ then such a $g$ is unique up to scaling; one can see this from the fact that if $g=e^{u}h$, then $R_{g}=e^{-u}\left( R_{h}-\Delta_{h}u\right) $. On the other hand, the group of conformal diffeomorphisms of $S^{2}$ is a noncompact group of linear fractional transformations. So pulling back such a $g$ on $S^{2}$ yields a noncompact set of such metrics. So, on the torus, the noncompactness is due only to scaling, whereas on the sphere, the noncompactness is insidious.

For $r=-1$, Hamilton used the potential $f$ defined by $\Delta f=r-R$ and proved that $H=R-r+\left\vert \nabla f\right\vert ^{2}$ satisfies $(\frac{\partial}{\partial t}-\Delta)H=-2|\operatorname{Ric}+\nabla^{2}% f-\frac{r}{2}g|^{2}+rH$. Since $r<0$, with consequent higher derivative estimates, this implies that the modified flow $\frac{\partial}{\partial t}% g=-2(\operatorname{Ric}+\nabla^{2}f-\frac{r}{2}g)$ converges exponentially fast in each $C^{k}$-norm to an expanding gradient Ricci soliton. This expander, since it is on a closed manifold, must have constant curvature, which in turn, implies that the original normalized Ricci flow converges to a metric $g_{\infty}$ with $R(g_{\infty})\equiv r,$ where $g_{\infty}$ is in the same conformal class as the initial metric $g\left( 0\right) $.

See also the paper of X. Chen, P. Lu and G. Tian.