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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 $M$?

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    $\begingroup$ 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. $\endgroup$
    – 314159.
    Nov 1, 2016 at 16:02
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    $\begingroup$ 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. $\endgroup$
    – Deane Yang
    Nov 1, 2016 at 17:29
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    $\begingroup$ @DeaneYang You should wrtie it as an answer. $\endgroup$
    – Anton Petrunin
    Nov 1, 2016 at 18:52
  • $\begingroup$ Anton, with your endorsement, I will. I don't trust any of my answers these days, so I always post them as comments. Thanks. $\endgroup$
    – Deane Yang
    Nov 1, 2016 at 19:04

1 Answer 1

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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.

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  • $\begingroup$ Thanks for your answer, but unfortunately, I dont quite understand that approach. Do you mean that the curve should be a geodesic or just a curve with finite length? And how do I construct the Riemannian metric such that this curve has finite length? If the curve is a closed 1-dimensional submanifold, then I just pick a metric on it and extend it using a partition of unity argument, but what if this curve looks somehow ugly? $\endgroup$
    – Tom
    Nov 2, 2016 at 7:39
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    $\begingroup$ It is a standard exercise to show the existence of a "ray" e.g. Exercise 3 mathematik.uni-muenchen.de/~rschmidt/riemgeometry/…. So you can take the points of Dean Yang answer on the ray. Then just re-scale the metric by a suitable conformal multiple to make the ray to have finite length as suggested by Dean Yang. $\endgroup$
    – Holonomia
    Nov 2, 2016 at 11:45
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    $\begingroup$ First, note that the curve does not have to be a geodesic. A length-minimizing geodesic joining two points in the sequence will only be shorter. To get a curve of finite length, rescale the metric inside a tubular neighborhood of the curve such that the scale factor is 1 on the boundary and a function decreasing to 0 fast enough along the curve. This is probably most easily done, with a minimum of machinery and theorems, using a partition of unity subordinate to a covering of the curve by open balls. $\endgroup$
    – Deane Yang
    Nov 2, 2016 at 18:15
  • $\begingroup$ Thanks for the answer! I think I understand the construction now! $\endgroup$
    – Tom
    Nov 7, 2016 at 10:16

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