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Let $G$ be a semi-simple algebraic group over $\mathbb C$, $B$ be a Borel and P is be parabolic containing $B$ and let $\mathcal L$ be an ample line bundle on $G/P$. Is it true that $H^0(G/B, L^{\otimes d})=H^0(G/P, L^{\otimes d})$ for all d > 0?

From the projection map from $G/B \rightarrow G/P$ we get an injection from $H^0(G/P, L^{\otimes d})$ to $H^0(G/B, L^{\otimes d})$. But why is it a surjection ?

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  • $\begingroup$ The fiber is $P/B$. $\endgroup$
    – Wilberd van der Kallen
    Commented Aug 12, 2016 at 7:05
  • $\begingroup$ Use the projection formula. $\endgroup$
    – Sasha
    Commented Aug 12, 2016 at 7:08
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    $\begingroup$ The morphism from $G/B$ to $G/P$ is rational. The derived pushforward of the structure sheaf of $G/B$ is naturally equivalent to the (complex supported in degree $0$ of the) structure sheaf of $G/P$. Combine this with the projection formula. $\endgroup$
    – Jason Starr
    Commented Aug 12, 2016 at 11:19
  • $\begingroup$ @Pitor: Do you allow $P$ to equal $G$? That is technically possible for a parabolic subgroup. $\endgroup$
    – Jim Humphreys
    Commented Aug 12, 2016 at 22:17
  • $\begingroup$ No, I will assume P to be a proper parabolic subgroup of G. $\endgroup$
    – Pitor
    Commented Aug 13, 2016 at 10:31

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Q. Why is it a surjection? A. It's a nonzero $G$-equivariant map to (by Borel-Weil) a $G$-irrep.

(Of course this has punted the difficulty, solved geometrically by the commenters, into the Borel-Weil theorem.)

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