Timeline for Sheaf cohomology of the universal sub and quotient bundles of the Grassmannian
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 6, 2017 at 22:33 | vote | accept | DCT | ||
| Mar 4, 2017 at 23:07 | answer | added | Allen Knutson | timeline score: 5 | |
| Mar 4, 2017 at 22:59 | comment | added | Jason Starr | This does indeed follow from Borel-Weil-Bott. However, it can also be proved directly with Cech covers. It can also be proved by Leray spectral sequences: for the projective subbundle $\mathbb{P}_{G(k,V)}(S) \subset G(k,V)\times_k \mathbb{P}(V)$, the projection of $\mathbb{P}_{G(k,V)}(S)$ to $\mathbb{P}(V)$ is the bundle $G(k-1,Q')$, so you can use induction on $k$ to compute cohomology. | |
| Mar 4, 2017 at 22:56 | comment | added | Jason Starr | @LiviuNicolaescu. If $G(k,V)$ denotes the Grassmannian as a $k$-scheme with the Zariski topology, and if the subbundle, resp. quotient bundle, denotes the usual coherent sheaf on this scheme, then the dimensions of the respective cohomology groups are independent of the field $k$. For the universal short exact sequence $0\to S \to V\otimes_k \mathcal{O}_{G(k,V)} \to Q \to 0$, all of the cohomology groups of $S$ are zero, and all higher cohomology groups of $Q$ are zero. The natural map $V\to H^0(G(k,V),Q)$ is an isomorphism. | |
| Mar 4, 2017 at 22:47 | comment | added | Liviu Nicolaescu | Your question does specify not over what field you're working. | |
| Mar 4, 2017 at 22:33 | comment | added | DCT | I mean we regard the sections of the two vector bundles as a locally free sheaves, and take sheaf cohomology in the Zariski topology. From more googling, I'm guessing this question is answered by the Borel-Weil-Bott theorem, but I still need to figure what that says, or if there is an easier way in this special case. | |
| Mar 4, 2017 at 22:24 | comment | added | Liviu Nicolaescu | The cohomology of what sheaf? | |
| Mar 4, 2017 at 22:14 | history | asked | DCT | CC BY-SA 3.0 |