# Converse to Milnor's theorem on manifolds with nonnegative Ricci curvature

Disclaimer : I suspect the question I am about to ask is really hard, but I just want to know the status of such questions.

Thanks to Milnor, we know that the $\pi_1$ any compact manifold with nonnegative Ricci curvature has polynomial growth.

I want to know if anything is known about the opposite direction : which manifolds whose $\pi_1$ has polynomial growth admit metric with nonnegative Ricci curvature ?

At least if you allow non compact manifolds, the answer cannot be 'all of them' (because of Whitehead's three manifold), and I feel like there are compact counter examples but I am not able to cook one.

• "cook one UP"... – YangMills May 8 '13 at 15:24
• The introduction of arxiv.org/abs/math/0109167 lists a number of examples (maybe all known ones?). I have not been following the subject for the last several years but I do not recall any new examples since that paper was written. – Igor Belegradek May 9 '13 at 0:32

For a compact counterexample, take any nilmanifold $N/H$ modulo the action of a freely acting cocompact lattice $\Lambda$, assuming $N/H$ is not just Euclidean space and $\Lambda$ is not just virtually abelian. For instance, $N=N/H$ is the $3 \times 3$ real Heisenberg group and $\Lambda$ is the $3 \times 3$ integer Heisenberg group. The proof that this is a counterexample is to apply Wilking's theorem in Anton's answer, and the theorem that a finitely generated nilpotent group has polynomial growth.
• I don't know what you mean by "$N/H$ is not just Euclidean space". – YCor May 7 '19 at 9:51
• Hah, hard to tell what I might have meant 6 years ago. But, I think I simply intended to rule out the case that $N = \text{Isom}_+(\mathbb E^n) \approx \mathbb R^n \rtimes SO(n)$ and $H = SO(n)$. – Lee Mosher May 7 '19 at 14:17
Given a group $$\Pi$$, is there a compact manifold $$M$$ with non-negative Ricci curvature such that $$\pi_1(M)=\Pi$$?