Given a closed Riemannian manifold $(M,g)$ with non-negative Ricci curvature and $dim\geq 3$, when can we deform the metric to a positive Ricci curved one?

I know it's impossible in general due to the flat factor in the universal covering. But what about we add some topological restrictions on $M$ like simply connectedness? Are there any positive or negative results on this problem?

( Besides, are there now any examples of simply connected closed manifold with positive scalar curved metric by do not admit a positive Ricci curved metric? )

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Thanks to the answer by Robert, I may simplify the problem in the following sense,

Given a simply connected flat Einstein manifold $(M,g)$ with $\hat{A}$-genus non-vanishing, and set $(\mathbb{S}^2,h)$ be the standard unit sphere, can $(M\times \mathbb{S}^2, g+h)$ be perturbed to a Ricci-positive manifold? $\ \ $In general, what about changing $(\mathbb{S}^2,h)$ to an arbitrary closed Ricci-positive manifold?

  • $\begingroup$ It is better to ask the follow up question separately. $\endgroup$ Nov 29, 2018 at 22:36
  • $\begingroup$ On the bottom of p.3 of arxiv.org/pdf/1607.00657.pdf D.Wraith discusses a related question. He considers the product of a K3 surface with a homotopy $(4n-1)$-sphere $\Sigma$ that bounds a parallelizable manifold. (Wraith proved elsewhere that any such sphere admits a metric of $Ric>0$ so $K3\times\Sigma$ has a metric of $Ric\ge 0$). Wraith then remarks: "There are no known obstructions to positive Ricci curvature for these manifolds... Nevertheless, the author is tempted to conjecture that no Ricci positive metrics exist". $\endgroup$ Nov 30, 2018 at 1:35

2 Answers 2


There are obstructions. Perhaps the most famous comes from the theorem that, if a compact spin manifold has a metric of positive scalar curvature, then its $\hat A$-genus must vanish.

If you take a compact Riemannian spin manifold $(M,g)$ with special holonomy $\mathrm{G}_2$ (in dimension $7$), $\mathrm{Spin}(7)$ (in dimension $8$), or holonomy in $\mathrm{SU}(n)$ (in dimension $2n$) whose $\hat A$-genus is nonzero (and there are lots of these, even simply-connected ones), then $g$ will be Ricci-flat and hence will have non-negative Ricci curvature. However, by the above theorem, it cannot carry any metric with positive scalar curvature, let alone a metric with positive Ricci curvature.

  • $\begingroup$ This is why I asked the second question. So what about supposing Ricci non-negative and scalar positive on the metric, can it be deformed to a Ricci positive one? In general, what Riemannian manifold could be on the boundary of the space of non-collapsing positive Ricci curved manifolds? $\endgroup$
    – H_Wang
    Nov 27, 2018 at 5:03

This is not a complete answer but would be helpful. Here are a few facts:

Theorem (T. Aubin 1970 and P. Ehrlich 1976). If the Ricci curvature of a compact Riemannian manifold is non-negative and positive somewhere, then the manifold carries a metric with positive Ricci curvature.

Also vanishing the first Betti number is a necessary condition in compact case for admitting strictly positive Ricci curvature.

Relation with scalar curvature:

  1. There are still no known examples of simply connected manifolds that admit positive scalar curvature but not positive Ricci curvature

  2. If a manifold $M$ cannot have a metric with positive (or zero) Scalar curvature, then it certainly does not admit a metric with positive (zero res.) Ricci curvature.

This paper of G. Perelman is also useful: "Construction of manifolds of positive Ricci curvature with big volume and large Betti numbers". Comparison Geometry. 30: 157–163 Click here for pdf


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