# Deforming metrics from non-negative to positive Ricci curvature

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? )

------------------------------------------------------------update 1------------------------------------------------------------

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?

• It is better to ask the follow up question separately. – Igor Belegradek Nov 29 '18 at 22:36
• 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". – Igor Belegradek Nov 30 '18 at 1:35

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.

• 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? – ZHans Wang Nov 27 '18 at 5:03

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

Theorem (T. Aubin 1979). 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 (see on Google books: A Course in Differential Geometry, By Thierry Aubin).

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