Timeline for Local Ext for reflexive sheaves on surfaces
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 12, 2023 at 19:09 | answer | added | Karl Schwede | timeline score: 2 | |
| Feb 27, 2023 at 18:40 | history | became hot network question | |||
| Feb 27, 2023 at 13:15 | answer | added | R. van Dobben de Bruyn | timeline score: 8 | |
| Feb 27, 2023 at 12:53 | comment | added | R. van Dobben de Bruyn | That's right, I was just trying to say that your comment about the smooth case is not very relevant because the smooth case is trivial. I'm actually working out the very same example you're giving, because it also gives a counterexample to the question. | |
| Feb 27, 2023 at 12:51 | comment | added | Francesco Polizzi | @R.vanDobbendeBruyn: but the OP is supposing $X$ normal Gorenstein, that is a bit less than regular (=smooth). For instance, it seems to me that the Weil divisor of a line in a quadric cone in $\mathbb{P}^3$ defines a rank 1 reflexive sheaf that is not locally free. Or am I missing something? | |
| Feb 27, 2023 at 12:45 | comment | added | R. van Dobben de Bruyn | @FrancescoPolizzi if $X$ is regular, then any reflexive sheaf of rank $1$ is locally free (so then the answer to the question is positive). | |
| Feb 27, 2023 at 12:41 | comment | added | Francesco Polizzi | Perhaps you already know this: if $X$ is smooth, a coherent sheaf $F$is locally free if and only if $$\mathcal{Ext}^i(F, \, G)=0$$ for all $\mathcal{O}_X$-module $G$ and for all $i \geq 1$. | |
| Feb 27, 2023 at 11:18 | history | edited | Jooh | CC BY-SA 4.0 |
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| Feb 27, 2023 at 10:55 | history | edited | Jooh | CC BY-SA 4.0 |
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| Feb 27, 2023 at 10:34 | history | asked | Jooh | CC BY-SA 4.0 |