On page 105 of Chow--Knopfs "Ricci Flow: An Introduction", it reads: "$r = \int_M R d\mu / \int_M d\mu$ ... is determined by the Euler characteristic $\chi(M^2)$ of the surface, hence is independent of the metric $g$."

Now this is wrong on the face of it. For instance, $r=1$ for the round metric with constant curvature $1$ and $r < 1$ for the round metric of constant curvature $< 1$. I think what they mean to say is that that the integral of the scalar curvature is determined by the Euler characteristic and, since the normalized Ricci flow leaves volume invariant, $r$ is constant in time.

So far, no question. But, I see this mistake being made ALL THE TIME. People keep writing "let $r = 2\pi\chi(M)$ be the average scalar curvature" and similar things, which (as someone who is far from being or wanting to be a professional when it comes to Ricci flow) got me wondering: Do people really mean by $r$ the average scalar curvature, or do they mean $2\pi\chi$?

This came to me when I saw the formula for the evolution of scalar curvature on compact surfaces: $$ \partial_t R = \Delta R + R(R-r).$$ If $r$ is the average scalar curvature, then this means that if the metric has $-1 \leq R \leq 1$, then it might happen that $R > 1$ for small $t>0$. Is this really the case?

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    $\begingroup$ Of course, $r$ depends on the topological type of the underlying surface as well as the volume of $(M,g)$ for a fixed metric. I don't see what the question is, people mean $r$ by the average of scalar curvature. If you assume that the metric has volume $1$, then you may evaluate $r=4\pi\chi(\Sigma)$. Your final paragraph does not make sense, can you please clarify it? If what metric has $-1\leq R \leq 1$? It might happen that $R > 1$ at some point? Everywhere? What is the underlying surface? $\endgroup$ Sep 16 '13 at 14:34
  • $\begingroup$ Thank you for the answer. Sorry the second question wasn't clear. The topological type of the surface is not important, execpt that it is supposed to be compact, and the metric is supposed to have scalar curvature $-1 \leq R(x) \leq 1$ everywhere. The question is: If you apply the normalized Ricci flow to such a surface, then can it happen that for $t>0$ the curvature is $> 1$ at some point? $\endgroup$
    – user35946
    Sep 16 '13 at 14:43
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    $\begingroup$ I don't see why this can't happen. Herustically speaking, you could have a large region of $R \equiv 1$ on a torus (i.e. $r=0$), and then there the PDE looks like it will increase $R$ to something $>1$ there for small $t$, because it looks like $\partial_t R \approx R^2$. This is, of course, not a proof. But, I don't see any reason to expect that e.g. the maximum of $R$ is non-increasing in time with some positive curvature. The usual ODE trick yields $\frac{d}{dt} R_{max} \leq R_{max}(R_{max} - r)$, which doesn't say much in this direction unless $R_{max} <0$. $\endgroup$ Sep 16 '13 at 15:04
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    $\begingroup$ In the case that $R_{max} <0$, then the equivalent formulation to what you want is true, i.e. if $R \leq -c$ at $t=0$, then it remains so for all time under the normalized flow. $\endgroup$ Sep 16 '13 at 15:05

Answering the initial question, you are right; what was written on p. 105 of my book with Dan Knopf is erroneous (sorry for this error). By $r$, we meant the average scalar curvature. To confirm what Otis Chodosh said, what should have been written is the following. Under the normalized Ricci flow in any dimension on a closed manifold, the volume is preserved. Since, by the Gauss--Bonnet formula, $\int_M Rd\mu$ on a closed surface is the topological invariant $4\pi \chi (M)$ (the scalar curvature $R$ is twice the Gauss curvature), $r$ is constant under the normalized Ricci flow on a closed surface (this is true also on a closed 2-orbifold since the Gauss--Bonnet formula still holds with a fractional Euler characteristic, etc.). What is most relevant to the nature of the analysis of the Ricci flow on closed surfaces, however you wish to proceed, is the sign of $r$, which is equivalent to the sign of $\chi (M)$. The same can be said in higher dimensions for the Kaehler--Ricci flow, with $\chi (M)$ being replaced by $c_1(M)$.

In addition, as Otis Chodosh said, if you take a torus with an initial metric with $R=1$ in some open set $U$, then at $t=0$ we have under the normalized Ricci flow that $\frac{\partial R}{\partial t} = \Delta R +R(R-r) = 1>0$ at points in $U$ since $r=0$ and there we have $\Delta R = 0$. So, for sufficiently small $t>0$ we have $R(x,t)>1$ for $x\in U$.

Heuristically, eventually the diffusion (i.e., the $\Delta R$ term) dominates the reaction (i.e., the $R(R-r)$ term). One thing that actually helps you here is that the Laplacian is with respect to the evolving metric. To wit, the Ricci flow is more linear than it appears (this is borne out by the plethora of apriori estimates proven by the maximum principle/sharp Bochner formulas for the Ricci flow in all dimensions). To show that the geometry of the metric improves under the Ricci flow relies on monotonicity formulas, many motivated by the analogy of the Ricci flow with the heat equation. This is a recurring theme in the works of Hamilton, Perelman, Boehm, Wilking, Brendle, Schoen, Nyugen, Andrews, etal. in Riemannian Ricci flow. The same is true for the Kaehler--Ricci flow. In addition, important coarser estimates in Riemannian Ricci flow are due to Bando, Shi, etal.


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