Timeline for Blow-up along singularity of the degeneracy locus
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Apr 25, 2017 at 12:53 | comment | added | user43198 | @JasonStarr Thanks a lot. That clears my doubt. | |
| Apr 25, 2017 at 12:42 | comment | added | Jason Starr | My copy of '3264 and all that' does not say that for Proposition 7.4, but I have an early draft, and things may have changed in the published version. Certainly a "sufficiently generic" section of a vector bundle has smooth zero locus. But $Z$ is not the zero locus of a sufficiently generic section of a vector bundle. It is the zero locus of the section $s_1\wedge \dots \wedge s_r$ of the invertible sheaf $\bigwedge^r (E(d))$. That is not a general section. For yourself, compute the singular locus of the determinant $s_{1,1}s_{2,2}-s_{1,2}s_{2,1}$ of a general $2\times 2$ matrix. | |
| Apr 25, 2017 at 10:25 | comment | added | user43198 | @JasonStarr I found a Bertini type theorem (Proposition 7.4 of Eisenbud and Harris's intersection theory) which states that if the determinant $L$ of $E$ is very ample then the zero locus $Z$ mentioned above is smooth. Do I understand this correctly or I am missing something? | |
| Apr 24, 2017 at 12:16 | comment | added | Jason Starr | If $\text{dim}(X)\geq 9$, and if the sections are sufficiently generic, then $F\to D$ is smooth away from a closed subset that has codimension $\geq 9$ in $X$. However, the singular locus of $F\to D$ will be nonempty. | |
| Apr 22, 2017 at 18:35 | history | asked | user43198 | CC BY-SA 3.0 |