Assume that $(M^n,g)$ is an $n$ dimensional ($n \geq 3$) closed Riemannian manifold with constant scalar curvature and $Ric\_g$ nonnegative. Then is $g$ Einstein?



when constant scalar curvature implies Einstein?
There is no reason for this, and the answer is indeed **no**.


The simplest example I can think of is the product of two $\mathbb{S}^2$, each endowed with round metrics *of different radius* (added in edit). This manifold is homogeneous and thus has constant scalar curvature, its sectional curvature is non-negative so its Ricci tensor also is (and is in fact even positive), but the Ricci curvature in a direction $u$ is not constant (directions tangent to the largest radius sphere have smallest Ricci curvature).


