2

Let $(M,g)$ be a smooth compact Riemannian manifold of dimension at least 3. Does the Hopf-Rinow theorem imply that $(M,g)$ is a complete Riemannian manifold? I.e. is a compact Riemannian manifold always complete?

If so, then can we conclude from the Cartan-Hadamard theorem that if $(M,g)$ is a compact and simply connected Riemannian manifold, then it must have positive sectional curvature?

If so, this seems to contradict the answer to this question, which says that any Riemannian manifold of dimension at least 3 admits a complete Ricci negative metric.

Where is the error in my reasoning?

What is an example of a sphere with negative sectional or Ricci curvature?

Michael Albanese
  • 99,526
rpf
  • 2,385
  • 12
  • 21
  • 1
    Any compact metric space is complete. Nothing special about Riemannian manifold. Second paragraph is wrong. And for the last paragraph, Cartan Hadamard will give you a contradiction, i.e, if that is true then exponential map gives you a diffeomorphism with Euclidean space. – Anubhav Mukherjee Mar 23 '18 at 19:26

1 Answers1

6

The Hopf-Rinow Theorem states that if $(M, g)$ is a Riemannian manifold such that the closed and bounded subsets are compact, then $(M, g)$ is complete. If $M$ is compact, then every closed subset is compact, so $(M, g)$ is complete.

The Cartan-Hadamard Theorem states that if $(M, g)$ is a Riemannian manifold with non-positive sectional curvature, then its universal cover is diffeomorphic to $\mathbb{R}^n$. If $M$ is compact and simply connected, then it is its own universal cover which is not diffeomorphic to $\mathbb{R}^n$, so $M$ cannot admit a Riemannian metric with non-positive sectional curvature. However, that does not imply that $M$ must admit a metric of positive sectional curvature. For example, the $K3$ surface is a compact, simply connected four-manifold which cannot admit a metric with positive sectional curvature (it can't even admit a metric with positive scalar curvature). All we can deduce from the Cartan-Hadamard Theorem is that for any metric on a compact, simply connected manifold $M$, there is a point $p \in M$ and a two-plane $\Pi \in \operatorname{Gr}_2(T_pM)$ such that $K(\Pi) > 0$.

A sphere can't have negative sectional curvature, otherwise it would have universal cover $\mathbb{R}^n$ by Cartan-Hadamard.

Michael Albanese
  • 99,526