Vrftjkry

Given a closed Riemannian manifold $(M,g)$ with non-negative Ricci curvature and $dimgeq 3$, 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 $(Mtimes mathbb{S}^2, g+h)$ be perturbed to a Ricci-positive manifold? $ $In general, what about changing $(mathbb{S}^2,h)$ to an arbitrary closed Ricci-positive manifold?

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 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