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Consider the Ricci flow on a compact surface of negative curvature. It has been proved by Hamilton that the flow in this case exists for all time. My question is, is the curvature uniformly bounded in space-time? I am guessing the answer is negative in the positive curvature case, for if one starts with the usual sphere $S^2$, the Ricci flow will scale it into a smaller sphere with higher curvature. With negative curvature, I get the intuitive feeling that the opposite will happen, but cannot prove it. I have just started learning about Ricci flow, so any help is appreciated, thanks!

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Your question is basically answered here (or in Hamilton's paper). The normalized Ricci flow on a compact surface with negative curvature, $$\frac{\partial}{\partial s} \hat g_s = -2 Ric_{\hat g}+ \hat S _s\hat g_s$$

(here $\hat S_s$ is the average of scaler curvature) not only exist for all time, but also converges to a metric of constant negative curvarture. Thus for $s$ large enough, the curvature of $\hat g_s$ is clsoed to a negative constant. Now to relate this to the Ricci flow, note that the solution to Ricci flow $g_t$ is related to $\hat g_s$ by

$$(*)\ \ \ \ \hat g(s) = \frac{g(t(s))}{\text{Vol}(t(s))}$$

(Please see the formula relating $t(s)$ and $s$ in the linked answer. Note that $t(s) \to \infty$ when $s\to \infty$ and the volume grows linearly in $t$). Thus using $(*)$,

$$Ric_{g(t)} = \frac{1}{\text{Vol}(t(s))} Ric_{\hat g_s} \to 0$$

(as $Ric_{\hat g_s} \sim K$ for large $s$).

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