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If a smooth manifold admits a finite atlas, then for many technical purposes it is as good as compact. I was surprised to learn that every connected smooth manifold actually has a finite atlas. This result can be found in

A. Solecki. "Finite atlases on manifolds", Annales Societatis Mathematicae Polonae. Series I: Commentationes Mathematicae XVII (1974)

A Google search reveals only a few references to this paper in the whole net.

Question: Why is this result/paper so unpopular? Are there alternative proofs that are more popular/cited?

Please note that this question is very different from the existence of a finite covering atlas for a topological manifold. The atlas here has to be in the given smooth structure.


Theorem: For every smooth connected manifold $M^m$ of dimension m>1 there exists a finite atlas consisting of at most $2\cdot 3^{2m}$ full charts. The number of charts can be restricted to $3^{2m}$ if $\partial M=\emptyset$, to $2\cdot 3^m$ if $M$ is compact, and to $3^m$ is $M$ is compact and unbounded.

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    $\begingroup$ Wow, mathscinet doesn't list a single citation to this paper. $\endgroup$ Commented Sep 8, 2022 at 17:41
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    $\begingroup$ I don't know what you mean by "for many technical purposes, it is as good as compact". I don't know any applications of having a finite atlas. It's also not a particularly hard result to prove; in fact, I've assigned it as a challenge problem when teaching smooth manifolds. Of course, I don't have access to the paper in question, so I don't know what other properties the author establishes. $\endgroup$ Commented Sep 8, 2022 at 18:07
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    $\begingroup$ @AndyPutman why of course? You can Google the title to get access to the paper right away. That's how I got it in the first place. Also, having a finite atlas allows one to easily obtain many factorization properties of partial differential equations. What formulation exactly did you give as a problem that your topology students managed to prove? $\endgroup$
    – Bedovlat
    Commented Sep 8, 2022 at 18:27
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    $\begingroup$ @AndyPutman Every smooth scheme over the field of complex numbers has an associated smooth complex analytic space. This is what most people call the "associated complex manifold." It need not be second countable and paracompact. It need not be Hausdorff. When I teach about complex manifolds, I do not assume that they are Hausdorff, second countable, paracompact. I state the hypotheses that I need when I prove a theorem. $\endgroup$ Commented Sep 8, 2022 at 20:21
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    $\begingroup$ @JasonStarr: You're made of sterner stuff than me! I'm more pro-algebraic geometry than most geometric topologists, but in my heart a smooth scheme over $\mathbb{C}$ is always separated and of finite type. In any case, I personally always include Hausdorff and second countable whenever I teach manifolds, and I think that most books do too. $\endgroup$ Commented Sep 8, 2022 at 21:01

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