Topological consequences of negative and zero Einstein condition

Question

Let $(M,g)$ be a complete Riemannian manifold which is Einstein, i.e. $\mathrm{Ric}=kg$ for some constant $k\in \mathbb{R}$.

1) If $k<0$, is $M$ then necessarily noncompact? If so, does the condition $k<0$ give any other topological restraints?

All examples of negative Einstein manifolds that I know are noncompact (e.g. hyperbolic space). Note that when $k>0$, $M$ must be compact by Myers' theorem. I know that any Riemannian manifold admits a complete metric with $\mathrm{Ric}<0$, but I suspect that the Einstein condition is much more restrictive.

2) Does the condition $k=0$ give any topological information? For example, can $S^n$ admit a metric with $k=0$?

Answer 1

Everything here is from Besse's book "Einstein manifolds". If you are interested in the subject, you owe it to yourself to have a copy nearby.

(1) is false for a number of reasons.

In dimension 2, negative Ricci curvature is the same thing as negative scalar curvature (and Einstein is the same as constant scalar curvature), so every closed surface of genus $g > 1$ admits a negative Einstein metric by the uniformization theorem. Besse give a more elementary argument in terms of pants decompositions.

In dimension 3, Einstein is the same as constant sectional curvature; see eg here; the Riemann curvature tensor is determined by (and determines) the Ricci tensor. many hyperbolic manifolds in dimension 3; these are in fact the majority of 3-manifolds, in some sense. It is not true that negative Ricci curvature is the same as negative sectional curvature (that is, if we do not assume constancy). A counterexample is given by $S^2 \times S^1$.

In dimensions larger than 3 constant sectional curvature manifolds are still Einstein manifolds (with the same sign). So an example of a closed hyperbolic manifold still gives you an example of a closed Einstein manifold with negative sign. These are less universal than in dimension 3, but many still exist. So there are counterexamples to your question.

There are even simply-connected closed Einstein manifolds; Aubin proved, before Yau's theorem on the existence of Kahler-Einstein metrics when $c_1 = 0$, that if $(M,J)$ is a Kahler manifold with $c_1^n < 0$, then $M$ admits a Kahler-Einstein metric of negative sign. Such manifolds exist in great supply in all even dimensions. (See Besse, chapter 11.) The existence of a Ricci-flat metric on the K3 surface is the simplest version of Yau's result on the $c_1 = 0$ case, and is very impressive.

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For (2), I will simply quote to you what Besse have to say on the subject.

In dimensions greater than $4$, we do not know of any topological restriction for a manifold to be Einstein. It may very well be that any manifold with dimension greater than $4$ admits a negative Einstein metric - or, that most manifolds do.

I do not know what has changed on the subject, but I don't think much has. It is still open whether or not $S^n$ has an Einstein metric of negative Einstein constant for $n > 3$, I think. (By the discussion above it cannot for $n \leq 3$.)