Timeline for Union of Schubert cells being affine
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| May 27, 2020 at 23:33 | comment | added | Igor Makhlin | Oh, okay, I see now. Yes, if we leave out the cases where $S$ is not convex in the Bruhat order as ambiguous, the rest should be clear. | |
| May 27, 2020 at 23:21 | comment | added | dhy | If it is open in its closure and is not an antichain. The point being that then it contains cells $w, w'$ with $l(w)=l(w')+1$, and the union of two such cells is a product of an affine space and a $\mathbb{P}^1.$ | |
| May 27, 2020 at 23:14 | comment | added | Igor Makhlin | @dhy Not sure if I understand the "if Z is open in its closure, then it will contain a P1" part. It would seem that a $Z$ consisting of a single cell is open in its closure but contains no $\mathbb P^1$? | |
| May 27, 2020 at 16:59 | comment | added | dhy | (To answer your parenthetical question at the end: It is indeed possible, already for surfaces. I don't remember the construction though - maybe it's something like remove a divisor from a Hirzebruch surface? In any case, these examples shouldn't work here, because if $Z$ is open in its closure, then it will contain a $\mathbb{P}^1$ and so can't be affine. On the other hand, if $Z$ is not open in its closure, then the question is a bit ambiguous because there is no natural scheme structure on $Z$.) | |
| May 27, 2020 at 15:18 | history | edited | Igor Makhlin | CC BY-SA 4.0 |
rephrased the third sentence
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| May 27, 2020 at 14:15 | history | answered | Igor Makhlin | CC BY-SA 4.0 |