# Why is important for a manifold to have countable basis?

I've seen there are few questions similar, but I haven't seen anyone so precise or with a good answer.

I'd like to understand the reason why we ask in the definition of a manifold the existence of a countable basis. Does anybody has an example of what can go wrong with an uncountable basis? When does the problem arise? Does it arise when we want to differentiate something or does it arise before? Thank You

• Do you like to have partition of unity? Do you like your manifolds to embed in some $R^N$? Do you like to have the invariance of domain theorem? Do you like being able to classify surfaces? All these require 2nd countability. – Moishe Kohan Feb 6 '17 at 9:02
• Thank you Moishe, if you expand a little bit the comment in an answer I will certainly upvote it and accept it – Dac0 Feb 6 '17 at 10:54

## 2 Answers

Do you like to have a partition of unity? (For instance, in order to integrate differential forms.) Do you like your manifolds to embed in some ${\mathbb R}^N$? Admit a Riemannian metric? Do you like your orientable surfaces to admit a complex structure? (If you are a geometer or an analyst, you surely do.) Do you like to have the invariance of domain theorem? Do you like being able to classify noncompact surfaces? (A classification of connected 1-dimensional manifolds not satisfying the 2nd countability axiom is possible, see here.) All these require 2nd countability, typically in the form of paracompactness.

Edit:

1. Suppose that $(M,g)$ is a connected Hausdorff Riemannian manifold (no 2nd countability assumption). Then $M$ is 2nd countable.

Proof. It suffices to show that $M$ is metrizable, see e.g. Spivak's reference in this question. To prove that $M$ is metrizable we define the Riemannian distance function on $M$ as usual: $$d(p,q)=\inf_c L(c),$$ where $L(c)$ is the length of the path $c$ and the infimum is taken over all piecewise-smooth paths $c$ connecting $p$ to $q$. (To see that such a path exists repeat the proof of the fact that a connected manifold is necessarily path connected.)

1. Connected Riemann surfaces are necessarily 2nd countable, as proven by Rado. An aside: There are connected complex manifolds which are not 2nd countable, see here. You can find a proof of Rado's theorem in section 2.11 of the book

T. Napier, M.Ramachandran, "An Introduction to Riemann Surfaces", Birkhauser, 2012.

• I understand the need of countable basis for partition of unity. I kind of remember this was also needed in Whitney theorem... But why do I need it for admitting a Riemannian Metric? – Dac0 Feb 6 '17 at 14:56
• @Dac0: See an edit. – Moishe Kohan Feb 6 '17 at 17:31
• Why does invariance of domain need paracompactness? Or, which formulation of invariance of domain needs paracompactness? That a continuous injective map between two manifolds of the same dimension is open holds without paracompactness, since it's a local property, unless I'm overlooking something. I think that two homeomorphic connected manifolds have the same dimension doesn't need it either. – Daniel Fischer Feb 6 '17 at 20:10
• @DanielFischer: "A bijective continuous map of manifolds is a homeomorphism" fails if you allow a manifold to have continuum of connected components. – Moishe Kohan Feb 6 '17 at 20:39
• Ah, right. That requires second countability directly, paracompactness isn't sufficient for that. Well, I guess if $2^{\aleph_0} > \aleph_1$, something strictly between second countability and paracompactness suffices. – Daniel Fischer Feb 6 '17 at 20:47

There is one point that is mentioned in passing in Moishe Cohen's nice answer that deserves a bit of elaboration, which is that a lot of the time it is not important for a manifold to have a countable basis. Rather, what is important in most applications is for a manifold to be paracompact: this is what gives you partitions of unity, which are essential to an enormous amount of the theory of manifolds (for instance, as the other answer mentioned, proving that any manifold admits a Riemannian metric).

Paracompactness follows from second-countability, which is the main reason why second-countability is useful. Paracompactness is weaker than second-countability (for instance, an uncountable discrete space is paracompact), but it turns out that it isn't weaker by much: a (Hausdorff) manifold is paracompact iff each of its connected components is second-countable. To put it another way, a general paracompact manifold is just a disjoint union of (possibly uncountably many) second-countable manifolds. So if you care mainly about connected manifolds (or even just manifolds with only countably many connected components), you lose no important generality by assuming second-countability rather than paracompactness.

There are also a few situations where it really is convenient to assume second-countability and not just paracompactness. For instance, in the theory of Lie groups, it is convenient to be able to define a (not necessarily closed) Lie subgroup of a Lie group $G$ as a Lie group $H$ together with a smooth injective homomorphism $H\to G$. If you allowed your Lie groups to not be second-countable, you would have the awkward and unwanted example that $\mathbb{R}$ as a discrete space is a Lie subgroup of $\mathbb{R}$ with the usual $1$-dimensional smooth structure (via the identity map). For instance, this example violates the theorem (true if you require second-countability) that a subgroup whose image is closed is actually an embedded submanifold.

• Thank you for your precisation on paracompactness. Since my seminar it will be Lie Groups related, I think I will keep 2nd countability, but good to know! Thank you – Dac0 Feb 7 '17 at 6:12