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?

whatmetric has $-1\leq R \leq 1$? It might happen that $R > 1$ at some point? Everywhere? What is the underlying surface? $\endgroup$ – Otis Chodosh Sep 16 '13 at 14:34