6
$\begingroup$

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?

$\endgroup$
0

2 Answers 2

11
$\begingroup$

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).

$\endgroup$
10
  • 2
    $\begingroup$ $S^1\times S^2$ works too. $\endgroup$
    – Ian Agol
    Apr 17, 2013 at 21:04
  • 3
    $\begingroup$ 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... $\endgroup$ Apr 18, 2013 at 9:36
  • 3
    $\begingroup$ 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 $\mathbb{R}^4$. $\endgroup$ Dec 12, 2014 at 16:56
  • 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. $\endgroup$ Jun 4, 2019 at 15:26
  • 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 $\operatorname{Riem} \equiv 0$. $\endgroup$ Nov 29, 2020 at 12:08
5
$\begingroup$

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".

$\endgroup$
1
  • $\begingroup$ Thank you for telling me this interesting reference. $\endgroup$
    – Mathboy
    Apr 18, 2013 at 9:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.