How do we compute the sheaf cohomology of the universal subbundle and universal quotient bundle of the Grassmannian $G(k,V)$?

$\begingroup$ The cohomology of what sheaf? $\endgroup$– Liviu NicolaescuMar 4, 2017 at 22:24

$\begingroup$ 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 BorelWeilBott theorem, but I still need to figure what that says, or if there is an easier way in this special case. $\endgroup$– DCTMar 4, 2017 at 22:33

$\begingroup$ Your question does specify not over what field you're working. $\endgroup$– Liviu NicolaescuMar 4, 2017 at 22:47

3$\begingroup$ @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. $\endgroup$– Jason StarrMar 4, 2017 at 22:56

3$\begingroup$ This does indeed follow from BorelWeilBott. 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 $G(k1,Q')$, so you can use induction on $k$ to compute cohomology. $\endgroup$– Jason StarrMar 4, 2017 at 22:59
1 Answer
See them as pushforwards of line bundles from the flag manifold by using BorelWeil fiberwise. Then use BorelWeil up on the flag manifold. Jerzy Weyman's book is the source I know of for these techniques.