Timeline for Tangent space to Hilbert schemes of points
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 11, 2019 at 18:40 | vote | accept | Ron | ||
| Oct 11, 2019 at 6:45 | answer | added | Sasha | timeline score: 4 | |
| Oct 9, 2019 at 21:00 | comment | added | Yosemite Stan | Hmmm, I guess I applied the dimension vanishing on the wrong side. | |
| S Oct 9, 2019 at 19:36 | history | suggested | Emily | CC BY-SA 4.0 |
typo: hilbert ---> Hilbert
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| Oct 9, 2019 at 19:31 | comment | added | R. van Dobben de Bruyn | @YosemiteStan I just deleted a wrong answer along those lines (because I had made a mistake). Using Serre duality, one can show that $\operatorname{Ext}^1_X(\mathcal I_Z,\mathcal O_X) \neq 0$ (in fact it has dimension equal to the length of $Z$). But it can still be the case that the map to $\operatorname{Ext}^1_X(\mathcal I_Z,\mathcal O_Z)$ is injective, which would also be enough. | |
| Oct 9, 2019 at 19:27 | review | Suggested edits | |||
| S Oct 9, 2019 at 19:36 | |||||
| Oct 9, 2019 at 19:04 | comment | added | Yosemite Stan | Morally Ext1(I,I) is the tangent space to the module of sheaves at I, so it should be true because Pic0(X)={0}. If you take the long exact sequence coming from deriving Hom(I,-) applies to the ideal sequence of Z you see that Hom(I,Oz)=Ext1(I,I) exactly when Ext1(I,Ox)=0. Taking the long exact sequence coming from deriving Hom(-,Ox) applied to the ideal sequence shows Ext1(Ox,Ox)=Ext1(I,Ox). Ext1(Ox,Ox)=H1(Ox)=0 because Pic0(X)={0} and H1(Ox) parametrizes infinitesimal deformations of Ox. | |
| Oct 9, 2019 at 17:09 | history | asked | Ron | CC BY-SA 4.0 |