Timeline for Can the nth projective space be covered by n charts?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Nov 27, 2021 at 14:32 | answer | added | Mark Grant | timeline score: 16 | |
| Nov 27, 2021 at 12:18 | comment | added | Z. M | To supplement the comment of @R.vanDobbendeBruyn: there is also a difference between the "cover by affines" in AG and the question here: the affine cover versions have few to do with the multiplicative structure, if I understand correct — given the coherent acyclicity of affines, the coherent cohomology could be computed via Čech cohomology. On the other hand, an étale analogue might be much closer to the question here. It is still unclear to me whether the coincidence of local cohomology and the cohomology of the cone is essential in the proof of the question. | |
| Nov 26, 2021 at 22:46 | comment | added | R. van Dobben de Bruyn | To put @DavidESpeyer's comment in perspective: the analogous statement in algebraic geometry is true: a proper variety of dimension $n$ (e.g. $\mathbf P^n$) cannot be covered by $n$ affine open subvarieties. One argument is to construct a sheaf $\mathscr F$ with $H^n(X,\mathscr F) \neq 0$. The important difference here is that intersections of affine opens are affine, whereas the intersection of contractible open submanifolds need not be contractible. | |
| Nov 26, 2021 at 22:16 | review | Close votes | |||
| Nov 30, 2021 at 14:11 | |||||
| Nov 26, 2021 at 16:30 | history | edited | Manfred Weis | CC BY-SA 4.0 |
fixed a glitch
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| Nov 26, 2021 at 12:20 | comment | added | Pedro | See the theorem on page 9 here for example. | |
| Nov 26, 2021 at 10:01 | history | became hot network question | |||
| Nov 26, 2021 at 6:30 | comment | added | Ryan Budney | @DavidESpeyer: This is a standard homework problem, and it's clearly false for $n=1$. | |
| Nov 26, 2021 at 3:10 | vote | accept | Saúl RM | ||
| Nov 26, 2021 at 2:59 | comment | added | Saúl RM | Thanks! Recently in a course I´ve been learning about the cup product and the cohomology rings of projective spaces, and I suspected they could somehow be used for the problem but I hadn´t thought of cup length. | |
| Nov 26, 2021 at 2:59 | answer | added | Steven Landsburg | timeline score: 37 | |
| Nov 26, 2021 at 2:41 | comment | added | user127776 | I think the answer is negative. If a manifold can be covered by $n$ contractible charts then its cup length will be less than $n$. The cup-length for the projective space $\mathbb{RP}^n$ is $n$. | |
| Nov 26, 2021 at 2:34 | comment | added | David Roberts♦ | It's not true for $n=1$, at least... | |
| Nov 26, 2021 at 2:29 | comment | added | Aleksandar Milivojević | A relevant concept here that you can look up goes by “Lusternik-Schnirelmann category”. | |
| Nov 26, 2021 at 2:09 | comment | added | David E Speyer | I'm baffled as to why this got downvoted. I imagine it is straightforward with some alg. top. tool I don't know, but it can't be that easy: $S^n$ is coverable with two copies of $\mathbb{R}^n$, so it isn't like compact $n$-folds always need $n+1$ charts. | |
| Nov 26, 2021 at 1:57 | history | asked | Saúl RM | CC BY-SA 4.0 |