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?
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3 Answers
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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).
answered Apr 17, 2013 at 11:56
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4$\begingroup$ 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... $\endgroup$ Commented Apr 18, 2013 at 9:36
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4$\begingroup$ 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 simply-connected, it must be $\mathbb{R}^4$. $\endgroup$ Commented Dec 12, 2014 at 16:56
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4$\begingroup$ @Malkoun: Chern-Gauss-Bonnet 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 trace-free Ricci curvature. So on an Einstein four-manifold, $\chi(M) = \frac{1}{32\pi^2}\int_M |\operatorname{Riem}|^2d\mu$. In particular, $\chi(M) = 0$ if and only if $M$ is flat. $\endgroup$ Commented Jun 4, 2019 at 15:26
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2$\begingroup$ @R.Rankin: It means that it is flat. Note that $|\operatorname{Riem}|^2 \geq 0$ so $\int_M|\operatorname{Riem}|^2d\mu = 0$ if and only if $|\operatorname{Riem}|^2 \equiv 0$ and hence $\operatorname{Riem} \equiv 0$. $\endgroup$ Commented Nov 29, 2020 at 12:08
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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 Kähler-Einstein. This is Proposition 2.12 in Tian's "Canonical metrics in Kähler Geometry".
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$\begingroup$ Thank you for telling me this interesting reference. $\endgroup$– MathboyCommented Apr 18, 2013 at 9:14
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To have a wide familly of counter example lets consider the Yamabe problem which says every compact manifold admite a metric of constant scalar curvature. So every manifold which does not admite any Einstein structure is a counter example to your question. For example every 4 manifold which violates the Hitchin Thorp in equality
