Why is it important to stay positive Ricci curvature during Ricci flow?

Question

Why is it important to stay positive Ricci curvature during Ricci flow? Isn't it possible to start from mixed Ricci curvature and arrive to a positive Ricci curvature and/or a Riemannian metric with perfect symmetry?

Answer 1

Yes, it is entirely possible that $(M,g_t)$ is a Ricci flow such that $g_0$ has mixed Ricci curvature but $g_\lambda$ has positive Ricci curvature for some $\lambda>0$ (and hence, if $M$ is closed and three-dimensional, that $g_t$ has positive Ricci curvature for all $t>\lambda$).

The importance of Hamilton's 1982 theorem is simply that it identifies a whole range of Riemannian metrics for which convergence can be concluded, and for 40 years nobody has managed to do better (in the three-dimensional closed context). For instance, it just could not be true that if $M$ is closed and three-dimensional, and $\operatorname{Ric}(g_0)>-\frac{1}{10}g_0$, then the Ricci flow smoothly converges to a constant-curvature metric. And it could not be true that positive scalar curvature implies convergence. The problem is that there are manifolds supporting such metrics which in principle cannot support metrics of constant curvature.

And the importance of positive Ricci curvature is simply that it's the assumptions of that theorem.

So it's not just that nobody's done better - nobody knows what a better formulation of the same theorem might look like, i.e. what systematic conditions a Riemannian metric on a closed 3-manifold would have to satisfy in order for the normalized Ricci flow to smoothly converge to a constant curvature metric. (edit- see comment thread below)