Timeline for Least number of charts to describe a given manifold
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| Oct 27, 2009 at 5:03 | comment | added | Aaron Mazel-Gee | Right. (I also just realized that my last statement is clearly false; e.g., the unit disk has trivial cohomology.) But maybe there's a weaker Cech-ish cohomology that can address this anyways. What happens if we work with a complex where only the individual open sets must be contractible? It seems possible that there's a statement along the lines of "the cohomology of this complex injects into the cohomology of a good cover", and maybe we could get a lower bound on the difference in rank (based on higher-dimensional cohomology of M and/or the cohomology of the nonempty intersections...?). | |
| Oct 26, 2009 at 19:48 | comment | added | Eric Wofsey | Cech cohomology is too weak to give a full answer, though, because it only gives you a bound on how to cover a space such that all nonempty intersections are contractible, rather than just the individual open sets themselves. For example, a sphere S^n can be covered by 2 contractible sets, even though it takes n+1 to give a good cover. Also, you could have an acyclic space where you see no cohomological obstructions to the space itself being contractible, but \pi_1 is nonzero so you certainly need at least two contractible sets to cover. | |
| Oct 26, 2009 at 19:18 | history | answered | Aaron Mazel-Gee | CC BY-SA 2.5 |