Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've been studying Differential Geometry on Spivak's Differential Geometry book. Since Spivak just works with notions of metric spaces and analysis, I'm doing fine. The point is that Spivak presents the following definition of a manifold:

A manifold $M$ is a metric space such that for every $p \in M$ there's some neighbourhood $V$ of $p$ and some integer $n \geq 0$ such that $V$ is homeomorphic to $\Bbb R^n$

Now, there's another definition, usually given in texts that assume the reader knows general topology, and the definition is:

A manifold $M$ is a topological space such that:

  1. $M$ is Hausdorff;
  2. $M$ has a countable basis for its topology;
  3. $M$ is locally Euclidean.

For now I'm happy with Spivak's definition because I've not seen general topology yet, but I'm curious with one thing: these two definitions are equivalent? In other words, every topological space with those three properties is metrizable, so that it can be put in terms of the first definition? Is there any other way in which these definitions can be said to be equivalent?

Thanks very much in advance for the help.

share|improve this question
    
Spivak's requirement is a little bizarre [and it's something of a pain, though an interesting one, to check that say $\mathbb{P}^n$ has a metric]. I think he envisions the reader having gone through his Calculus and Calculus on Manifolds, which don't talk about general topological spaces. [Actually, I can't remember if the second talks about metric spaces other than $\mathbb R^n$]. In his defense, it's an old book. –  TTS Jul 2 '13 at 21:02

1 Answer 1

up vote 7 down vote accepted

In other words, every topological space with those three properties is metrizable, so that it can be put in terms of the first definition?

Yes, that's right. Properties 1. and 3. give that $M$ is Hausdorff and locally compact, hence regular by a standard exercise in general topology. Together with Property 2. this gives that $M$ is regular and second countable, hence metrizable by Urysohn's Theorem.

Thus your second definition implies Spivak's definition with metrizable in place of metric. (It is slightly strange to define a manifold as a metric space in the absence of a Riemannian metric. It is unlikely that the specific metric is ever used.)

Conversely, suppose $M$ is a metrizable locally Euclidean space. Then of course $M$ is Hausdorff. However, it need not be second countable. A cheap way for it not to be is just to take a direct sum of uncountably many connected components. However it is also possible for a connected Hausdorff locally Euclidean space not to be second countable. Here the condition is equivalent to something called paracompactness and the standard counterexample of a connected Hausdorff locally Euclidean space which is not paracompact is the long line. In fact I am pretty sure I learned about this from an Appendix to Volume I of Spivak's Comprehensive Introduction to Differential Geometry! The equivalence of paracompactness, second countability and metrizability for connected, Hausdorff locally Euclidean spaces should also be found there, if memory serves.

Let me also say that absolutely everyone agrees that a "topological manifold" should be a locally Euclidean topological space. Most people agree that it should also be Hausdorff, but the minority who does not has its reasons: as soon as you start taking quotient of manifolds by group actions you will start to meet non-Hausdorff guys. Whether one should impose paracompactness or (stronger) second-countability is really not standard. I would say that neither should probably figure in the definition of a topological manifold but one should expect that these hypotheses will often be imposed in practice.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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