# Non-metrizable smooth manifold? [closed]

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

• It depends. How do you define “smooth manifold”? Commented Nov 7, 2018 at 13:31
• A smooth manifold can be given a metric using a partition of unity.
– deb
Commented Nov 7, 2018 at 13:37
• Do you want to talk about metric tensor or distance function?
– edm
Commented Nov 7, 2018 at 13:38
• For example the long line? Commented Nov 7, 2018 at 14:24
• 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. Commented Nov 9, 2018 at 0:01

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.