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For a projective space one has Bott formula to compute $h^q(ℙ^n,Ω^p(k))$, where $Ω^p(k)$ is the k-twisted sheaf of sections in the p-th power of the cotangent bundle of $ℙ^n$. I am wondering if there is a similar formula to compute these dimensions when you consider a projective bundle instead of a projective space.

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    $\begingroup$ The analogue of Bott's formula computes the (derived) direct images of these sheaves to the base of the projective bundle. $\endgroup$
    – Sasha
    Apr 17, 2014 at 16:29

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As Sasha says, the best analogy is probably the following. Let $B$ be a complex variety (or analytic space), $E$ a vector bundle on $B$ of rank $r+1$, $\pi:P=\mathbb{P}(E)\rightarrow B$ the corresponding projective bundle. Then $R^p\pi_*(\Omega ^q_P)\cong \Omega _B^{q-p}$ for $p\leq r$, $=0$ for $p>r$, and there are vanishing statements for $R^q\pi_*(\Omega ^q_P(n))$. For these statements and a simple proof (if you can read french) I recommend Verdier's lecture "Théorème de Le Potier" in Astérisque no. 17 (1974).

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  • $\begingroup$ I think a more close analogue is given by the formulas for relative differentials --- $R^p\pi_*(\Omega^q_{P/B}) = \mathcal{O}_B$ for $p = q$ and $0$ otherwise. Then the formulas for the pushforwards of $\Omega^q_P$ can be obtained from (exterior powers) of the standard sequence $0 \to \pi^*\Omega_B \to \Omega_P \to \Omega_{P/B} \to 0$. $\endgroup$
    – Sasha
    Apr 18, 2014 at 7:15
  • $\begingroup$ Yes indeed, this how Verdier does it. Which one is closer is probably a matter of taste... $\endgroup$
    – abx
    Apr 18, 2014 at 8:11
  • $\begingroup$ @abx, what do you mean with "... there are vanishing statements for $R^qπ_∗(Ω^qP(n))$"? Unfortunately I have not access to Verdier's reference. $\endgroup$
    – boristenes
    Apr 20, 2014 at 17:49
  • $\begingroup$ With the above notations, the precise statements are: $R^p\pi _*\Omega ^q_P(n)=0$ for $p\neq 0$, $n\neq 0$, $n\geq q-r$, and for $p\neq r$, $n\neq 0$, $n\leq q$. $\endgroup$
    – abx
    Apr 20, 2014 at 18:54
  • $\begingroup$ Many thanks! that is what I was looking for $\endgroup$
    – boristenes
    Apr 22, 2014 at 14:25

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