It is discussed in this [question](https://mathoverflow.net/questions/316209/deforming-metrics-from-non-negative-to-positive-ricci-curvature) whether a simply-connected closed Riemannian manifold with non-negative Ricci curvature admits positive Ricci curvature, and the answer appears to be "no, there are counter-examples and known obstructions".


My question is this: are there any known examples of closed, simply-connected manifolds of non-negative **sectional** curvature that do not admit positive Ricci curvature? Are there known obstructions?



Is it known whether a closed simply-connected manifold of non-negative curvature admits positive Ricci?
No there are no such examples known. Most known examples of manifolds of nonnegative sectional curvature come from biquotients or cohomogeneity one manifolds. If these are simply connected they are known to admit metrics of positive Ricci curvature by a [result of Schwachhoefer and Tuschmann](https://link.springer.com/article/10.1007/s00208-004-0538-x). Also by a result of Aubin mentined in the thread you referenced a manifold of nonnegative Ricci curvature with Ricci positive somewhere admit positive Ricci everywhere. In the simply connected case the only known obstruction to positive Ricci is the scalar curvature one (also mentioned in that thread). If a closed spin manifold has nonvanishing $\hat A$ genus then it does not admit a metric of positive scalar curvature and it particular also doesn't admit a metric of positive Ricci curvature. But simply connected manifolds of nonnegetaive sectional curvature have scalar curvature positive somewhere (else they'd be flat and not simply connected). Their scalar curvature is also nonnegative. This implies that the $\hat A$ is zero in the spin case.


