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?



Deforming metrics from non-negative to positive Ricci curvature
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.


