Timeline for Non-trivial extension and tangent bundle isotropic Grassmannian
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 3, 2022 at 8:39 | comment | added | Sasha | Do you know how the equivalence between equivariant bundles on a homogeneous variety and representations of the stabilizer of a point works? | |
| Aug 3, 2022 at 8:01 | comment | added | Bobech | Because I don't know who are the representations associated with the vector bundles appearing in the sequence | |
| Aug 2, 2022 at 18:23 | comment | added | Sasha | Why don't you apply Borel--Bott--Weil instead? | |
| Aug 2, 2022 at 14:46 | comment | added | Bobech | I tensor the short exact sequence by $\operatorname{Hom}(S^2 \mathcal S^\vee,\cdot)$ and then induce the long exact sequence in cohomology. Since $S^2 \mathcal S^\vee$ is globally generated, its dual is not. Hence $H^0(X, \operatorname{Hom}(S^2\mathcal S^\vee,S^2 \mathcal S^\vee))$ is zero. So I can conclude that $$ H^1(X,\operatorname{Hom}(S^2\mathcal S^\vee,\mathcal S^\vee \otimes \mathcal K)) \hookrightarrow H^1(X,\operatorname{Hom}(S^2\mathcal S^\vee,T_X)). $$ But I cannot say something more for the moment being | |
| Aug 2, 2022 at 14:39 | comment | added | Sasha | How did you try to compute $H^1$? | |
| Aug 2, 2022 at 14:25 | history | asked | Bobech | CC BY-SA 4.0 |