Timeline for Cohomology of tangent bundles
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 8, 2018 at 6:05 | comment | added | Sasha | @TaisongJing: Of course. Sorry for this typo. | |
| Jan 7, 2018 at 2:26 | comment | added | Taisong Jing | I suppose you meant $\mathcal{E}xt^1(i_*F, O_M) = i_*F^\vee(E)$ @Sasha | |
| Oct 25, 2013 at 19:43 | vote | accept | Puzzled | ||
| Oct 25, 2013 at 19:28 | comment | added | Sasha | This is what derived functors do. One can check that for a divisorial embedding $i:E \to M$ and a locally free sheaf $F$ on $E$ one has $\mathcal{H}om(i_*F,O_M) = 0$, $\mathcal{E}xt^1(i_*F,O_M) = i_*F(E)$. This is an instance of the Grothendieck duality. | |
| Oct 25, 2013 at 18:55 | comment | added | Puzzled | I am sorry but I did not understand how I get the second exact sequence from the first. It seems to me that dualizing the first exact sequence you wrote one has $$0\mapsto T_{E/Z}\rightarrow T_{\widetilde{X}}\rightarrow \pi^{*}T_{X}\mapsto 0$$ instead of $(\ast)$. | |
| Oct 25, 2013 at 17:47 | comment | added | Puzzled | Yes, I know. For that equality one needs to have $H^{i}(j_{*}N) = 0$. | |
| Oct 25, 2013 at 17:39 | comment | added | Sasha | I never wrote that $H^i(\pi_*T_{\tilde X}) = H^i(T_X)$. | |
| Oct 25, 2013 at 17:25 | comment | added | Puzzled | Thank you. The isomorphism $H^{i}(T_{\widetilde{X}})=H^{i}(\pi_{*}T_{\widetilde{X}})$ can be also seen as a consequence of the fact that $R^{i}\pi_{*}T_{\widetilde{X}} = 0$. From the last exact sequence you wrote to prove that $H^{i}(\pi_{*}T_{\widetilde{X}}) = H^{i}(T_{X})$ we need to have $H^{i}(j_{*}N) = 0$. | |
| Oct 25, 2013 at 6:25 | history | answered | Sasha | CC BY-SA 3.0 |