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So, generally, a manifold is defined using only finitely many charts. A sphere is the graph of 2n different continuous functions, or just 2 stereographic projections. Obviously, there's no compact manifold that requires infinitely many charts, and there can't be more than countably many charts since manifolds are 2nd countable, but is there any manifold nasty enough to require countably many charts?

Furthermore, if the answer is no, what if we narrow down the class of allowable charts, and demand that our manifold be smooth, analytic, Riemannian, or otherwise. Will any of these change the answer?

edit: Because of Antonios-Alexandros Robotis counter-example, I'm going to add the stipulation that a chart is defined as "a homeomorphism between an open subset of $M$ and an open subset of $\mathbb{R}^n$".

Duncan Ramage
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  • Consider a surface of infinite genus. I am not sure, but it looks like requiring countably many charts. – lisyarus Jun 14 '17 at 17:05
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    I don't think it does. This was one of the first counter examples offered when I discussed it with friends. You can, speaking very loosely, cover it with five different charts. Two which just project the top and bottom "halves" down to the plane, one which snakes along the outside and picks up the outer seam, and two which snake along the inside, one collecting the left half of the inner seams, and the other the right halves. – Duncan Ramage Jun 14 '17 at 17:09
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    Here's an obstacle to a positive answer: given a $d$-manifold $M$, if possible picture a "checkerboard tiling" of $M$, where the "squares" are small enough for this to look appropriate. Thicken each "square" a tiny bit; in the result, there may be a bound on the number of squares which any one square can touch. (E.g. if $M$ is the plane, this number is $8$ if we use the usual tiling.) If so we can cover it with finitely many charts (with disconnected domain of course, so if we demand connected charts this doesn't work). So a counterexample is "hard to tile reasonably." Not sure how helpful ... – Noah Schweber Jun 14 '17 at 17:27
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    https://math.stackexchange.com/questions/75594/surface-where-number-of-coordinate-charts-in-atlas-has-to-be-infinite – Pete Caradonna Jun 14 '17 at 17:39
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    Mariano's answer (which for some strange reason was not accepted) to the duplicate question settles it in all dimensions: $n+1$ charts always suffice where $n$ is the dimension of your manifold. This works in any category you want (TOP, PL, DIFF,...). – Moishe Kohan Jun 14 '17 at 20:03

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