Consider the Ricci curvature satisfying $Ric(v,v)>0$ for $v\neq 0$. We know that Ricci curvature can be computed as the sum of sectional curvatures of the planes containing $v$ so that if the sectional curvature $K$ satisfies $K>0$ then necessarily $Ric>0$. But, is the converse true?

The reason I ask is we can have manifolds of positive scalar curvature $R>0$ which do not have $Ric>0$. This can be done by taking a product manifold of say a manifold of positive sectional and negative sectional curvatures (so take product of $S_{r}^{n}$ the n-sphere of radius $r$ and $H^n$, the hyperbolic plane with the standard metrics). Then the Riemann tensor of the product, decomposes as the sum of Riemann tensors of the factors, and taking the trace, so does the Ricci tensor. So taking vectors tangent to $S_r^n$ we get a positive value for $Ric$ and taking vectors tangent to $H^n$ we get negative $Ric$. On the other hand, if we take another trace, we see that the scalar curvature is the sum of the scalar curvatures of the factors, so depending on how we choose $r$ we can have $R>0$, $R<0$, or $R=0$.

However, this trick of using product manifolds doesn't work when comparing Ricci and sectional curvatures.

So, my question is, does $Ric>0$ imply that $K>0$ ($K\geq 0$)? If so how would I prove it? If not, is there some simple counterexample?


1 Answer 1


No, $Ric > 0$ does not imply $K >0$ or even $K\geq 0$, except in dimensions at most $3$.

For a counterxample to $Ric > 0\implies K > 0$, one can look at products. For example, $S^2\times S^2$ equipped with the product of round metrics has positive Ricci curvature, but each point has $0$ curvature planes.

Unfortunately, I don't know of an explicit counterexample to $Ric > 0\implies K\geq 0$, but here are two reasons that such examples must exist.

First, note that $Ric > 0$ is an open condition - small perturbations of the metric preserve this property. On the other hand, $K\geq 0$ is a closed condition, so one should expect that small perturbations of $K\geq 0$ ruin this property. So, it should be possible to slightly deform the product metric on $S^2\times S^2$ to get a counterexample. (I admit this is not a proof, just an idea).

Now, here is a proof that such examples must exist. We begin with a theorem of Sha and Yang (see here) that any number of connect sums of $S^3\times S^4$ with itself admit a metric of positive Ricci curvature.

On the other hand, a theorem of Gromov (see here) asserts a Betti number bound (depending only on the dimension) in the presence of non-negative sectional curvature.

Now, if we connect sum enough $S^3\times S^4$s together, we can achieve arbitrarily large Betti numbers. By Gromov's result, such a manifold does not admit a metric with $K\geq 0$, but by Sha and Yang's result, such a manifold does admit a metric with $Ric > 0$.

  • $\begingroup$ I see, so it's not something that is trivial to prove. Thanks! $\endgroup$ Feb 15, 2017 at 1:50
  • $\begingroup$ Well, it may be trivial to cook up counterexamples, I just don't know any off the top of my head. $\endgroup$ Feb 15, 2017 at 2:33
  • $\begingroup$ @JasonDeVito: Am I right that the Gromov's Betti number bound is trivially hold for closed manifolds? $\endgroup$
    – C.F.G
    Nov 7, 2020 at 14:36
  • $\begingroup$ @C.F.G.: No. The example in my last paragraph is a counterexample $\endgroup$ Nov 7, 2020 at 14:43

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