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Is it possible to find a smooth manifold on which it is impossible to define a metric function?

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  • $\begingroup$ It depends. How do you define “smooth manifold”? $\endgroup$ Commented Nov 7, 2018 at 13:31
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    $\begingroup$ A smooth manifold can be given a metric using a partition of unity. $\endgroup$
    – deb
    Commented Nov 7, 2018 at 13:37
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    $\begingroup$ Do you want to talk about metric tensor or distance function? $\endgroup$
    – edm
    Commented Nov 7, 2018 at 13:38
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    $\begingroup$ For example the long line? $\endgroup$
    – freakish
    Commented Nov 7, 2018 at 14:24
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    $\begingroup$ By the Whitney embedding theorem, any smooth manifold can be thought of as being inside some Euclidean space $R^N$ (you can take $N$ to be twice the dimension of the manifold). You can now define your distance function on the manifold as the restriction of the usual distance function in Euclidean space. $\endgroup$ Commented Nov 9, 2018 at 0:01

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It depends on the definition of a smooth manifold $M$. Usually one requires that $M$

1) is Hausdorff,

2) is second countable,

3) has a smooth atlas.

The "minimal" requirement for a smooth manifold would be 3), but obviously 1) is a necessary condition for the existence of a metric. See https://en.wikipedia.org/wiki/Non-Hausdorff_manifold for examples of non-Hausdorff manifolds.

That 2) is necessary for the for the existence of a metric is less obvious. As a counterexample take the long line https://en.wikipedia.org/wiki/Long_line_(topology).

If 1) - 3) are satisfied, then deb's and Aleksandar Milivojevic's comments show that $M$ is metrizable.

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