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I've seen lots of casual claims that Riemannian manifolds (even without assuming second-countability) are metrizable. In the path-connected case, we can use arc-length to create a distance function. But how can one prove this without assuming path-connectedness (and second-countability)?

(Motivated by Problem 2-D of Milnor-Stasheff.)

Kyle
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  • I don't see how to define a metric for non-connected manifolds in general. – user39082 Jul 27 '14 at 06:19
  • @user39082: I think you could handle each connected component separately, then use something like $d(x,y)=1$ for all pairs $x,y$ lying in distinct connected components. – Kyle Jul 27 '14 at 06:25
  • No, this usually will not satisfy the triangle inequality. – user39082 Jul 28 '14 at 07:18
  • Anyway, the Smirnov metrization theorem mentioned in the answer below, of course does the job. – user39082 Jul 28 '14 at 14:17
  • @user39082: Absolutely! I was being silly before. – Kyle Jul 28 '14 at 15:53
  • @user39082 Of course this does not satisfies the triangle inequality, but you can first replace the distance by another one which is bounded from above by 1. – ACL May 31 '16 at 16:18

2 Answers2

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By the Smirnov metrization theorem, a space that is locally metrizable and paracompact is metrizable. So as each path component of a Riemannian manifold is metrizable as you explain, local metrizability clearly follows. Paracompactness is often required in the definition of a manifold (or that it is second countable, which implies paracompactness), which is why this claim is often made without additional hypotheses.

But even in the non-second countable case, as each path component of a Riemannian manifold is metrizable, it must be paracompact, again by Smirnov. An arbitrary disjoint union of paracompact spaces is paracompact, a quick proof is here, so a Riemannian manifold must be paracompact.

See also this related question.

Community
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cws
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Assume the manifold $M$ is connected. If there exists a piecewise-smooth path from $p$ to $q$ in $M$, then there exists such a path between $p$ and $q'$ for any $q'$ in a small closed neighborhood of $q$, since $M$ is locally path-connected. The set of such $q$ is thus open and closed in $M$, and so must be $M$ itself. It follows that $M$ is path-connected, and you can define the metric on $M$ by arc-length.

anomaly
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  • Wonderfully simple argument. – Kyle Jul 27 '14 at 06:22
  • I guess I forgot that a connected and locally path-connected space is path-connected! – Kyle Jul 27 '14 at 06:45
  • @squirrel: This is not yet a complete argument. "Arc-length" between two points is not well-defined, it depends on the choice of the piecewise smooth path; take, for example, the complement of a point in the Euclidean plane. Typically what one does it to take the infimum of arc length over all paths between $p$ and $q$. – Lee Mosher Jul 28 '14 at 20:47
  • @LeeMosher: Sure, but it sounds from the initial question that the poster can handle that part given path-connectedness. – anomaly Jul 28 '14 at 21:10