Hamilton's program for proving the Poincaré conjecture involves first putting a Riemannian metric on the unknown simply connected closed 3-manifold. The basic idea is to try to "improve" this metric; for example, if the metric can be improved enough so that it has constant positive curvature, then according to classical results in Riemannian geometry, it must be the 3-sphere. Hamilton prescribed the "Ricci flow equations" for improving the metric;

$$\partial {t}g_{ij}=-2R_{ij} $$

where g is the metric and R its Ricci curvature, and one hopes that as the time t increases the manifold becomes easier to understand.

Then I read from Wikipedia said:

Ricci flow expands the negative curvature part of the manifold and contracts the positive curvature part.

Why contracts the positive curvature part? If we want to flow to 3-sphere, should not we consider to use the Ricci flow that contracts the negative curvature part and expands the positive curvature part?

here is the link of the quoted text https://en.wikipedia.org/wiki/Poincaré_conjecture#Ricci_flow_with_surgery

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    $\begingroup$ Not necessarily. Consider what happens to a manifold of constant negative sectional curvature: It expands, but the sectional curvature approaches zero. $\endgroup$
    – Kajelad
    Sep 4, 2021 at 2:44
  • $\begingroup$ I am not sure I understand. You mean the manifold expands under Ricci flow, but the sectional Gaussian curvature goes to 0? that means one of the $k_1 k_2$ goes to zero? How is that approaching to a sphere? How is that relevant to my question? Can you write an answer? $\endgroup$ Sep 4, 2021 at 3:24
  • $\begingroup$ I merely mean that regions of negative curvature may increase in volume, but they become less negatively curved, so in a sense they do "approach" a spherical metric. The dynamics in the general case are much more complicated, of course. $\endgroup$
    – Kajelad
    Sep 4, 2021 at 3:30
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    $\begingroup$ in my opinion that Wiki paragraph has been said very roughly just for understanding the RF general behavior. I think there are some situations (like neck pinch) that this doesn't happen. $\endgroup$
    – C.F.G
    Sep 4, 2021 at 6:34
  • $\begingroup$ Could you provide a link to the text you quoted? I can't find it $\endgroup$
    – Deane
    Sep 4, 2021 at 15:57

1 Answer 1


Looking at the quote in context, it appears to be saying only what is obvious from the equation. In a region where the Ricci tensor is negative definite, then, accounting for the $-2$ factor, the Ricci flow will expand the metric. And in a region where the Ricci tensor is positive, the metric will contract. This, however, says little about what's happening globally, since those regions themselves change. It is also difficult to see what is happening where the Ricci tensor is indefinite.

Also, note that the Ricci curvature is an average of sectional curvatures and therefore is not easily interpreted geometrically. The sectional curvature is more easily to understand geometrically, but how it evolves under the Ricci flow is quite difficult to see. One has to study the evolution equataion satisfied by the Riemann curvature tensor.

Even in the original result of Hamilton on the Ricci flow of a metric with positive Ricci curvature on the $3$-sphere, it is nontrivial to show that positive curvature is preserved and that eventually the metric has positive sectional curvature. In higher dimensions, stronger assumptions are needed to ensure this.


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