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This might be a silly question, but I just wanted to confirm my understanding. Does every smooth manifold have a countable smooth atlas (it obviously has a countable atlas since every manifold is second-countable)? My initial reaction is no, but my reasoning is a bit shaky. I am thinking that this would imply a countable basis of smooth coordinate balls, which doesn't seem to be possible (the best I've seen is a countable basis of regular coordinate balls - which I'm pretty sure doesn't require a countable smooth atlas). However, I'm not quite able to get the argument straight in my head.

MandelBroccoli
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  • I'm not sure I understand your reasoning. Manifold is second countable, in particular Lindelof. And so any open cover has countable subcover. – freakish Jun 13 '23 at 06:19
  • Oh right. Silly question indeed. – MandelBroccoli Jun 13 '23 at 06:22
  • See also https://math.stackexchange.com/questions/2812455/manifolds-equivalence-of-second-countability-vs-countable-atlas – subrosar Jun 13 '23 at 06:23
  • @subrosar what additional structure would i need to get a basis of smooth coordinate balls? – MandelBroccoli Jun 13 '23 at 06:27
  • @MandelBroccoli That comes for free. For every $x\in M$ and neighborhood $U$ of $x$ in $M$, choose a coordinate ball $B$ with $x\in B\subseteq U$. Then these balls $B$ form a basis of coordinate balls. – subrosar Jun 13 '23 at 07:11

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Let $\{A_i\}$ be a smooth atlas for a manifold $M$. Using the countable basis, there is a cover of $M$ by countably many basis open sets $B_j$ such that for each $j$ there is some $i$ with $B_j\subseteq A_i$. Then restricting charts, $\{B_j\}$ is a countable smooth atlas for $M$.

subrosar
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