2
$\begingroup$

I’ve been trying to teach myself differential geometry using Will Merry’s notes (found here) and am struggling to prove Proposition 1.17 on page 7. Here’s the statement…

Let $M$ be a set. Suppose we are given a collection $\{U_a\ | \ a\in A\}$ of subsets of $M$ together with bijections $x_a: U_a \to \mathcal{O}_a$ where $\mathcal{O}_a$ is an open subset of $\mathbb{R}^m$. Assume in addition that:

(I) For any $a,b \in A$, $x_a(U_a \cap U_b)$ is open in $\mathbb{R}^m$

(II) If $U_a \cap U_b \neq \varnothing$, the map $x_b \circ x^{-1}_a: x_a(U_a \cap U_b) \to x_b(U_a \cap U_b)$ is a diffeomorphism

(III) Countably many of the $U_a$ cover $M$

(IV) If $p\neq q$ are points in $m$ then either $\exists a \in A$ so that $p,q \in U_a$ or $\exists a,b \in A$ so that $p \in U_a$, $q \in U_b$, and $U_a \cap U_b=\varnothing$.

Then $M$ has a smooth manifold structure for which the collection $\{x_a : U_a \to \mathcal{O}_a\}$ is an atlas.

For context, Merry defines (1) a topological manifold of dimension $m$ to be a separable and metrizable topological space which is locally homeomorphic to $\mathbb{R^m}$ and (2) a smooth manifold of dimension $m$ to be a topological manifold of dimension $m$ equipped with an atlas which satisfies (II).

Thus I really only need to show (1), i.e. that there is a well-defined separable and metrizable topology on $M$ which makes the $x_a$’s homeomorphisms.

My immediate intuition is to consider the topology generated by the sets: $$\bigcup_{a \in A} \{x_a^{-1}(O) \ | \ O \text{ is an open set of } \mathcal{O}_a\}$$

While this topology clearly makes each $x_a$ continuous, is it possible to further show that the $x_a$’s are actually homeomorphisms? Furthermore, how do I demonstrate the topology is separable and metrizable?

Any help is appreciated, thanks in advance :)

$\endgroup$
5
  • 1
    $\begingroup$ That is indeed the right topology. You can show each $x_a$ is an open map, hence a homeomorphism. This makes use of (I) and (II). Then, each $U_a$ is separable, so (III) tells you $M$ is the countable union of separable spaces, hence separable. (IV) implies that $M$ is Hausdorff. You can argue by hand that $M$ is a regular space, hence metrizable by Urysohn metrization. (Quite frankly, I'm not a big fan of requiring manifolds to be metrizable by definition as it needlessly complicates things.) $\endgroup$
    – Thorgott
    Mar 31, 2023 at 16:33
  • $\begingroup$ Thanks @Thorgott, that’s really helpful. Do you mind explaining a bit further how I would show that M is a regular space? $\endgroup$ Mar 31, 2023 at 18:08
  • 1
    $\begingroup$ If $A$ is closed in $M$ and $x\not\in A$, then first show you can find a small open subset $x\in U$ that is disjoint from $A$ and homeomorphic to an open ball in $\mathbb{R}^n$. Then, use the existence of "bump function" in $\mathbb{R}^n$, take such a function, pull it back to the open subset of $M$ and extend it by zero. This function separates $A$ and $x$. $\endgroup$
    – Thorgott
    Mar 31, 2023 at 18:51
  • 1
    $\begingroup$ Sorry, I slightly got ahead of myself in that comment. It's easier than that. Of course you don't need the existence of smooth bump functions, since we only need a continuous function. So it suffices to take some makeshift construction (e.g. the function on $\mathbb{R}^n$ that has value $1$ at $0$, then has decreases linearly as the radial component increases at a steep enough rate and stays $0$ once it hits $0$). $\endgroup$
    – Thorgott
    Mar 31, 2023 at 19:35
  • $\begingroup$ @PaulFrost my mistake $\endgroup$
    – pancini
    Mar 31, 2023 at 20:07

1 Answer 1

1
$\begingroup$

You correctly state that you only need to show that there is a well-defined separable and metrizable topology $\mathscr T$ on $M$ which makes the $x_a$ homeomorphisms.

The idea is that each $x_a$ should be a homeomorphism between an open subset of $M$ and $\mathcal{O}_a$. Thus you correctly argue that as a subbasis of this topology $\mathscr T$ we can take $$\mathscr S =\bigcup_{a \in A} \{x_a^{-1}(O) \ | \ O \text{ is an open set of } \mathcal{O}_a \} = \bigcup_{a \in A} \mathscr S_a$$ with $$\mathscr S_a = \{x_a^{-1}(O) \ | \ O \text{ is an open set of } \mathcal{O}_a \} .$$ Clearly $\mathscr S_a$ is the unique topology on the set $U_a$ making the $x_a : (U_a, \mathscr S_a) \to \mathcal O_a$ homeomorphisms. Moreover, the topology $\mathscr T$ generated by $\mathscr S$ is the coarsest topology making the $x_a : (U_a,\mathscr T_a) \to \mathcal O_a$ continuous, where $\mathscr T_a$ is the subspace induced by $\mathscr T$ on $U_a$ (which is finer than $\mathscr S_a$). It remains to show that $\mathscr T_a = \mathscr S_a$.

  1. If $V_a \in \mathscr S_a, V_b \in \mathscr S_b$, then $V_a \cap V_b \in \mathscr S_a \subset \mathscr S$. In particular, $\mathscr S$ is a basis for $\mathscr T$.

Write $V_a = x_a^{-1}(O_a), V_b = x_b^{-1}(O_b) $ with an open $O_a \subset \mathcal O_a, O_b \subset \mathcal O_b$. The bijection $x_b \circ x_a^{-1} : x_a(U_a \cap U_b) \to x_b(U_a \cap U_b)$ is a homeomorphism between open subsets of $\mathbb R^n$. The set $O_b \cap x_b(U_a \cap U_b)$ is open in $x_b(U_a \cap U_b)$, thus $(x_a \circ x_b^{-1})(O_b \cap x_b(U_a \cap U_b)) $ is open in $x_a(U_a \cap U_b)$ and therefore open in $\mathcal O_a$. We conclude that $O_a^* = O_a \cap (x_a \circ x_b^{-1})(O_b \cap x_b(U_a \cap U_b))$ is open in $\mathcal O_a$. But we have $$x_a^{-1}(O^*_a) = x_a^{-1}(O_a) \cap x_b^{-1}(O_b \cap x_b(U_a \cap U_b)) = x_a^{-1}(O_a) \cap x_b^{-1}(O_b) \cap (U_a \cap U_b) \\= x_a^{-1}(O_a) \cap U_a \cap x_b^{-1}(O_b) \cap U_b = x_a^{-1}(O_a) \cap x_b^{-1}(O_b) = V_a \cap V_b .$$ Thus $V_a \cap V_b \in \mathscr S_a$.

  1. $\mathscr T_a = \mathscr S_a$.

Trivially $\mathscr S_a \subset \mathscr T_a$. Now consider $V \in \mathscr T_a$. By 1. we can write $V = \bigcup_{j \in J} V_j$ with $V_j \in \mathscr S_{b_j}$. Since $V \subset U_a \in \mathscr S_a$ and $\mathscr S_a$ is a topology, 1. implies $V = \bigcup_{j \in J} V_j \cap U_a \in \mathscr S_a$.

  1. $M$ is second countable (i.e. has a countable base).

This follows from the fact that countably many $U_a$ cover $M$ and each $U_a$ has a countable base (because it is homeomorphic to $\mathcal O_a$).

  1. $M$ is separable.

It is a well-known theorem of general topology that second countable spaces are separable.

  1. $M$ is Hausdorff.

This follows immediately from (IV).

Now there is a problem: The author claims that the proof of Proposition 1.17 is essentially trivial. As Thorgott writes in a comment, the standard definition of a topological manifold requires that is a locally Euclidean Hausdorff second countable space. This was proved above, and this was easy.

The author decided to require that a topological manifold is a locally Euclidean separable metrizable space, and here we have a gap: We must prove that $M$ is metrizable. This can be done by Urysohn's metrization theorem. It says that a regular second countable space is metrizable. This is a non-trival result.

Therefore it remains to prove that $M$ is regular. This is easy. We have to show that for each $ \in M$ and each open neigborhood $U$ of $p$ in $M$ there exists a closed neigborhood $V$ of $p$ in $M$ such that $V \subset U$. W.lo.g. we may assume that $U$ is contained in some $U_a$. The set $U_a$ is locally compact because it is homeomorphic to $\mathcal O_a$. Thus there exist a compact neigborhood $V$ of $p$ in $U_a$ such that $V \subset U$. Since $M$ is Hausdorff, $V$ is a closed neigborhood $V$ of $p$ in $M$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .