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Nov 20, 2020 at 16:56 vote accept Avi Steiner
Nov 20, 2020 at 14:13 answer added jorge vargas timeline score: 1
Nov 10, 2020 at 12:14 comment added Will Sawin @AviSteiner Because the fibers are connected (and have no coherent cohomology, if you're interested in the derived pushforward). This is a local question, so we can work locally on $G/P$, and in particular assume the line bundle is trivial - it's a question about $f_* \mathcal O_{G/B}$. We know this is trivial for a proper separable map with reduced connected fibers.
Nov 10, 2020 at 3:31 comment added Avi Steiner @WillSawin Ok, I see why that answers my question. However, it’s not immediately clear to me why the natural map from a line bundle on $G/P$ to its pull-then-push is an isomorphism rather than just injective. (Though injectivity is enough to give me what I want)
Nov 10, 2020 at 3:27 comment added LSpice @Jef's reference: Gruson, Sam, and Weyman - Moduli of Abelian varieties, Vinberg theta-groups, and free resolutions.
Nov 9, 2020 at 22:48 comment added Jef Section 2.3 of arxiv.org/abs/1203.2575 and references therein should be useful.
Nov 9, 2020 at 22:10 comment added Will Sawin The cohomology of any equivariant line bundle on $G/P$ is equal to the cohomology of its pullback to $G/B$, which can be calculated using the usual Borel-Weil-Bott, by the Leray spectral sequence.
Nov 9, 2020 at 22:01 comment added Avi Steiner @WillSawin How does that help?
Nov 9, 2020 at 21:54 comment added Will Sawin Doesn't this follow from the usual Borel-Weil-Bott by noting that the (derived) pushforward from $G/B$ to $G/P$ of a line bundle pulled back from $G/P$ is again that line bundle?
Nov 9, 2020 at 21:46 history asked Avi Steiner CC BY-SA 4.0