TITLE: uniformization theorem via ricci flows
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: 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.