# when constant scalar curvature implies Einstein?

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 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).

• 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
• $S^1\times S^2$ works too. – Ian Agol Apr 17 '13 at 21:04
• 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
• 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 $\mathbb{R}^4$. – Klaus Kröncke Dec 12 '14 at 16:56
• @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. – 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 Kähler-Einstein. This is Proposition 2.12 in Tian's "Canonical metrics in Kähler Geometry".

• 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

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 Kähler-Einstein.