$C_0$ Convergence of Metrics under Ricci Flow when $\chi(M) = 0$.

Question

I am a beginning graduate student, and I am trying to learn basic aspects of Ricci Flow via the uniformization theorem for compact surfaces. I am reading Chow and Knopf's introductory book as well as Hamilton's original paper on the subject. I have a question regarding the following theorem:

Theorem: If $(M^2, g_0)$ is a compact surface with metric $g_0$, then a 1-parameter family of metrics $g(t)$ with $g(0) = g_0$ under the Normalized Ricci-Flow equation: $$ \frac{\partial g}{\partial t} = - (R_g - r) g $$ exists for all time and converges in every $C^k$-norm to a metric of constant scalar curvature. Here $R$ is the scalar curvature and $r = \int_M R_g d\mu_{g}/\int_{M}d \mu_{g} $ is the average scalar curvature. $r$ is constant by the Gauss-Bonnet Theorem.

The proof is divided into three cases, each of increasing difficulty: $r < 0$, $r = 0$, and $r>0$. I have a question only about the $r = 0$ case. The idea for both $r = 0$ and $r<0$ is to apply the maximum principle to quantities related to gradient-ricci solitons.

In Knopf and Chow's book, they prove that the metrics are equivalent for all time when $r \leq 0$ (Proposition 5.15). Then they say that it is enough to that $R$ and all of it's derivatives go to zero as $t \to \infty$. However, they show that $|R| \leq C/(1+t)$, where $C>0$. This function is not in $L^1([0, \infty))$ and the hypothesis: $$ \int_0^\infty \left|\frac{\partial g}{\partial t}\right|_{g(t)} dt < \infty $$ does not apply (See Lemma 6.49 in Chow and Knopf). This hypothesis is used to show the metrics converge uniformly.

On the other hand Hamilton's paper, he shows that integral bounds of $R$ and its first two derivatives decay exponentially with time. Then he uses a Sobolev Inequality to get that $|R_{\max}|$ decays expontially, too. Then the hypothesis above is satisfied by the Normalized-Ricci Flow equation.

Question: Is there a basic fact that Chow and Knopf are exploiting that implies that $g$ converges in $C_0$? They are excellent at including relevant details and have a good view of this subject, so I am eager to know how they are thinking about this.

Answer 1

You might be able to get by using Poincaré and Sobolev inequalities, given the better rate of decay of $|\nabla R|$: Since $\int_M R = 0$ by Gauss-Bonnet, you can apply the Poincaré inequality and get that $$||R_{g(t)}||_{L^p(M)} \leq C||\nabla R_{g(t)}||_{L^p(M)}$$ for any $p$. But for the gradient of $R$, we have the improved decay $|\nabla R_{g(t)}| \leq C/(1+t)^{3/2}$ which is integrable in time. But then we also have the Sobolev inequality (Morrey's inequality according to Wikipedia) which says that $$ ||R||_{C^{0,\alpha}(M)} \leq C||R||_{W^{1,p}(M)} $$ for $p$ large enough. Putting this all together we get a bound $$||R_{g(t)}||_{L^{\infty}(M)} \leq C ||R_{g(t)}||_{W^{1,p}(M)} \leq C||\nabla R_{g(t)}||_{L^p(M)} \leq C||\nabla R||_{L^{\infty}(M)} \leq \frac{C}{(1+t)^{3/2}}.$$ Now that you know $R$ is integrable, you can integrate $\frac{\partial g}{\partial t}$ to get the limit metric, since $$\frac{\partial g(t)}{\partial t} = -R_{g(t)}g(t)$$ and the $g(t)$'s are uniformly equivalent.