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Assume that $(M^n,g)$ is an $n$ dimensional ($n>=3$)closed Riemannian manifold with constant scalar curvature and $Ric_g$ nonnegative. Then is $g$ Einstein?

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2 Answers 2

up vote 6 down vote accepted

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 non-negative 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).

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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. –  Mathboy Apr 17 '13 at 12:46
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$S^1\times S^2$ works too. –  Ian Agol Apr 17 '13 at 21:04
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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... –  Thomas Richard Apr 18 '13 at 9:36
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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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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. –  Ruadhaí Dervan Apr 17 '13 at 14:52
    
Thank you for telling me this interesting reference. –  Mathboy Apr 18 '13 at 9:14
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