# Existence of incomplete Riemannian metrics

During this question a manifold $M$ is meant to be a smooth connected (second countable, finite dimensional) manifold.

A Riemannian metric on a manifold $M$ is called complete if every geodesic is defined for all times. It is well-known that this is equivalent to completeness of the metric space $(M,d)$ where $d$ is the geodesic distance.

I want to show the following equivalence: A manifold $M$ is compact if and only if every Riemannian metric on $M$ is complete. (keep in mind that $M$ is connected)

It is easy to show the implication from left to right, but I have no clue for the reverse implication.

I tried the following argument: Every smooth manifold $M$ can be embedded as a bounded submanifold of $\mathbf R^n$. Assume that $M$ is not compact, then $M$ is not closed in $\mathbf R^n$. So, as a metric space $M$ is not complete when given the metric of the surrounding space $\mathbf R^n$. Then I realized that this does not imply that $M$ with the induced Riemannian metric is not complete since the geodesic distance on $M$ may be totally different from the induced distance function.

So my question is: If $M$ is non-compact, does there exist an incomplete Riemannian metric on $M$?

• As you said, in the Riemannian case it can be proved that a compact manifold is complete, invoking the fact that a vector field on a compact manifold is complete. However, one needs to mention that this argument brakes down in the (generic) pseudo-Riemannian case. In the same case, compactness does not guarantee completeness either. For example, the so-called Clifton-Pohl torus is a compact, geodesically incomplete Lorentz manifold. – 314159. Nov 1 '16 at 16:02
• If $M$ is non-compact, then there exists a sequence of points with no convergent subsequence. Fix a curve passing through all of the points. It suffices to construct a Riemannian metric such that this curve has finite length. – Deane Yang Nov 1 '16 at 17:29
• @DeaneYang You should wrtie it as an answer. – Anton Petrunin Nov 1 '16 at 18:52
• Anton, with your endorsement, I will. I don't trust any of my answers these days, so I always post them as comments. Thanks. – Deane Yang Nov 1 '16 at 19:04

If $M$ is non-compact, then there exists a sequence of points with no convergent subsequence. Fix a curve passing through all of the points. It suffices to construct a Riemannian metric such that this curve has finite length.