Timeline for Explicit locally free resolution of a perfect complex $E\oplus F\to (E\oplus F)\otimes \mathcal{O}_X(D)\to (E\otimes \mathcal{O}_X(D))|_D$
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 12, 2020 at 16:06 | vote | accept | Arkadij | ||
| Apr 12, 2020 at 16:01 | comment | added | Sasha | Because the projective dimension of a locally free sheaf on a Cartier divisor is 1. | |
| Apr 12, 2020 at 15:59 | comment | added | Arkadij | Thank you for the edit Sasha. Could you please add some argument explaining why $K$ is locally free? I am struggling to see why that is true. | |
| Apr 12, 2020 at 15:51 | vote | accept | Arkadij | ||
| Apr 12, 2020 at 15:51 | |||||
| Apr 12, 2020 at 14:43 | history | edited | Sasha | CC BY-SA 4.0 |
added 406 characters in body
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| Apr 12, 2020 at 12:36 | comment | added | Arkadij | I have changed the question to be more explicit now. It is important to me that it goes the way that I have mentioned, otherwise it is trivial as your answer shows. | |
| Apr 12, 2020 at 12:33 | comment | added | Sasha | For me a resolution is a quasiisomorphism, no matter which direction it goes. | |
| Apr 12, 2020 at 11:46 | comment | added | Arkadij | For me a resolutions is a complex of vector bundle $L^\bullet$ with a map $L^\bullet \to \mathcal{E}^\bullet$ which is a quasi-isomorphism. The map goes the wrong way in your case. | |
| Apr 12, 2020 at 11:40 | history | answered | Sasha | CC BY-SA 4.0 |