The Borel--Bott--Weil Theorem is usually stated for the complete flag manifold of $SU(N)$. Does an analogue hold for the other flags, for example the Grassmannians?

More precisely, suppose $G(\mathbf C)$ is a complex reductive group, and $P(\mathbf C)$ is a parabolic subgroup. Characters $\lambda$ of $P(\mathbf C)$ give rise to line bundles $\mathcal{L}(\lambda)$ on $G(\mathbf C)/P(\mathbf C)$. When is $H^i(G(\mathbf C)/P(\mathbf C),\mathcal{L}(\lambda))$ nonzero, and, in terms of the parabolic $P(\mathbf C)$, which irreducible representations of $G(\mathbf C)$ arise from this construction?

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    $\begingroup$ mathoverflow.net/questions/178783/… $\endgroup$ – Carlo Beenakker Mar 16 '16 at 20:04
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    $\begingroup$ Yes. The theorem is exactly the same. The highest weights which appear are the ones that extend to characters of $P$. The proof is just pushing forward under the obvious map, and noting that those line bundles are trivial on the fibers. $\endgroup$ – Ben Webster Mar 16 '16 at 20:55
  • $\begingroup$ Also, there is a useful extension for vector bundles corresponding to the irreducible representations of the parabolic subgroup. $\endgroup$ – Sasha Mar 17 '16 at 13:55

Kostant, Lie algebra cohomology and the generalized Borel-Weil theorem, Ann. Math. 74 (1961), 329-387.

W. Schmid, Homogeneous complex manifolds and representations of semisimple Lie groups, Proceedings of the International Congress of Mtahematicians: Helsinki 1978 (ed. O. Lehto) 195-208.

You could also look at R. Baston, M. Eastwood, The Penrose Transform: Its Interaction with Representation Theory, Oxford, 1989.


Lars, search for a paper of Kostant ora paper of Griffiths-Schmid, you will find a complete answer to your question, even when \lambda is just an irreducible representation. best regards


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