Timeline for when is a section of vector bundle determined by its zero locus?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jun 3, 2011 at 1:55 | history | edited | Dmitry Kerner | CC BY-SA 3.0 |
added 145 characters in body
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| Jun 3, 2011 at 1:50 | vote | accept | Dmitry Kerner | ||
| Jun 3, 2011 at 0:54 | comment | added | Theo Johnson-Freyd | I suppose you're looking for generalizations of the fact that a polynomial is determined up to scaling by its zeros? Note that this is very special to the algebraic situation, and also to the situation where the base field is algebraically closed, and also to the situation where you know the zero locus as a subscheme and not just a subset (so that you can distinguish $x$ from $x^2$ as sections of the trivial rank-1 bundle over the affine line). | |
| Jun 2, 2011 at 22:59 | comment | added | J.C. Ottem | Which resolution are you talking about qui-vadis? The structure sheaf of $Z$? | |
| Jun 2, 2011 at 22:29 | answer | added | Jorge Vitório Pereira | timeline score: 4 | |
| Jun 2, 2011 at 21:59 | comment | added | Dmitry Kerner | Yes, certainly in general we do not have such a property. But maybe with some restrictions on the resolution of this bundle? | |
| Jun 2, 2011 at 21:56 | comment | added | Mohan | The question is too general. For example, take $V=\mathcal{O}(1)^2$ on $\mathbb{P}^2$. Then $H^0(V)$ is six dimensional and an open set of a four dimensional subspace of this vector space has precisely just one fixed point as its zeroes. So, the zero does not determine the section. | |
| Jun 2, 2011 at 21:23 | history | asked | Dmitry Kerner | CC BY-SA 3.0 |