1
$\begingroup$

I have come across two different defintions of a topological manifold -

Def 1: A topological manifold of dimension n is a second-countable Hausdorff space M such that for all $p\in $M, there exists open neighborhood $U$ at $p$ and a homeomorphism $x:U\to x(U)\subseteq \mathbb{R}^{n}$

Def 2: A topological manifold M of dim. n is a Hausdorff topological space with an open cover $C$ with countable elements $U_i\in C$ and a collection of homeomorphism $\phi_i:U_i\to \phi_i(U)\subseteq\mathbb{R}^{n}$ where $\phi_i(U)$ is an open subset in $\mathbb{R}^{n}$.

  1. Are these two equivalent? If not, which one of them is the correct one (if any of them is)?

  2. Is second-countable same as to have an open cover $C$ with countable elements?

  3. Does the target of chart map ($x/\phi$) need to be an open subset in $\mathbb{R}^{n}$?

$\endgroup$
2
  • $\begingroup$ All the neighborhoods in definition 1 are the open cover from definition 2, and an open cover like in definition 2 contains neighborhoods of all points, so it also fulfills definition 1. $\endgroup$ – Vercassivelaunos Oct 15 '20 at 16:40
  • 1
    $\begingroup$ If we do not require $x(U)$ to be open in $\mathbb R^n$, then all subsets $M \subset \mathbb R^n$ would be manifolds (take the identity on $M$ as a universal chart around all $p$). This doesn't make much sense. $\endgroup$ – Paul Frost Oct 15 '20 at 23:39
2
$\begingroup$
  1. Definition 1 is missing (or assuming) the requirement that $x(U)$ be open. With that addition, both definitions are equivalent.

  2. Yes. This is because $\mathbb{R}^n$ is itself second countable. To show a countable cover implies a second countable manifold, choose a countable basis $\mathcal{B}$ for $\mathbb{R}^n$ (e.g. balls of rational center/radii), and let $\mathcal{B}'=\{\varphi_i^{-1}(B):B\in\mathcal{B},i\in\mathbb{N}\}$. I claim this is a countable basis for $M$.

  3. Yes, removing that requirement allows for objects which are not conventionally thought of as manifolds, such as graphs or $\mathbb{Q}\subset\mathbb{R}$.

$\endgroup$
2
  • $\begingroup$ Does paracompact + Hausdorff imply second countability? $\endgroup$ – aneet kumar Oct 16 '20 at 11:44
  • $\begingroup$ @aneetkumar Not quite. Paracompact + Hausdorff + locally Euclidean + countably many connected components $\implies$ second countable, but all four are necessary if I'm not mistaken. Also see here. $\endgroup$ – Kajelad Oct 16 '20 at 13:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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