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?
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 the round metric. This manifold is homogeneous and thus has constant scalar curvature, its sectional curvature is nonnegative so its Ricci tensor also is (and is in fact even positive), but the Ricci curvature in a direction $u$ depends on the angle between $u$ and the tangent spaces to the fibers of the projection on each factor (i.e., on whether $u$ is close to be horizontal or vertical or not).

$\begingroup$ Yes, I think your example has scalar curvature equal to 4 and Ricci curvature nonnegative. But some sectional curvature is zero. Thank you very much. $\endgroup$ – Mathboy Apr 17 '13 at 12:46

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2$\begingroup$ Just a caveat, for this metric on $\mathbb{S}^2\times\mathbb{S}^2$ not to be Einstein, the two $\mathbb{S}^2$ need to have different radius. And a remark about Agol's comment : what I like in it is that it in fact doesn't admit any Einstein metric, just because in dimension 3 Einstein is equivalent to constant sectional curvature. I wonder wether $\mathbb{S}^1\times\mathbb{S}^3$ enjoys the same property or not... $\endgroup$ – Thomas Richard Apr 18 '13 at 9:36

1$\begingroup$ This comment comes very late but it is maybe worth to post it: $S^1\times S^3$ does not admit an Einstein metric because the Euler Characteristic is zero. In this case (this is special in dimension $4$), the manifold must be flat, and since it is simplyconnected, it must be $\mathbb{R}^4$. $\endgroup$ – Klaus Kröncke Dec 12 '14 at 16:56

1$\begingroup$ @Malkoun: ChernGaussBonnet gives $$\chi(M) = \frac{1}{32\pi^2}\int_M(\operatorname{Riem}^2  4\operatorname{Ric}^2 + R^2)d\mu = \int_M(\operatorname{Riem}^2  4\operatorname{Ric}\limits^{\circ}^2)d\mu$$ where $\operatorname{Ric}\limits^{\circ}$ denotes the tracefree Ricci curvature. So on an Einstein fourmanifold, $\chi(M) = \frac{1}{32\pi^2}\int_M \operatorname{Riem}^2d\mu$. In particular, $\chi(M) = 0$ if and only if $M$ is flat. $\endgroup$ – Michael Albanese Jun 4 '19 at 15:26
As an example where this does hold, for $\omega$ a Kähler metric of constant scalar curvature with $\pi c_1(M) = \lambda [\omega]$, then $\omega$ is KählerEinstein. This is Proposition 2.12 in Tian's "Canonical metrics in Kähler Geometry".

$\begingroup$ P.S. If anyone would like to remove the e from "Kaehler" and insert an umlaut, they're more than welcome to, as I was unable to. $\endgroup$ – Ruadhaí Dervan Apr 17 '13 at 14:52

$\begingroup$ Thank you for telling me this interesting reference. $\endgroup$ – Mathboy Apr 18 '13 at 9:14
There's a conclusion but I don't know the proof.
For $\omega$ a Kähler metric of constant scalar curvature with positive bisectional curvature, $\omega$ is KählerEinstein.