After proving the existence of Riemannian metrics on manifolds, one of the students asked if the "paracompactness" is necessary. Of course the standard proof with the partition of unity uses this assumption, and for the only non-paracompact manifold I know, namely the long line, a similar argument seems to show the existence of a metric. So, just out of curiosity, does there exist a "manifold" (of course, non-paracomapact) which does not admit any Riemannian metric?


On the contrary, the long line does not have a Riemannian metric. Every countable subset of the long line has a least upper bound, so if it were Riemannian then a geodesic ray in the long direction would have to stop short of the end. But then the Riemannian metric would break down at the "endpoint" of this geodesic.

More generally, it is a theorem of Stone that every metric space is paracompact, hence every Riemannian metric space is too.

  • $\begingroup$ @Greg Kuperberg: Thank you for clarification. I had carelessly assumed that it is possible to cover the long line with an uncountable number of open intervals in a locally finite way, similar to ${\mathbb R}$. But this cannot be done all the way to the end. $\endgroup$ – Keivan Karai Feb 8 '11 at 14:18
  • $\begingroup$ Indeed, that is exactly the paracompactness condition, that every open cover has a locally finite refinement. $\endgroup$ – Greg Kuperberg Feb 8 '11 at 14:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.