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Let $L$ be a linear subspace of $P^n$ defined by say $x_1=x_2=\cdots=x_m=0$. Consider the projection of $P^n$ to $P^m$ from $L$. This is a rational map $f:P^n\rightarrow P^m$ defined on the complement of $L$ by $(x_0:\cdots:x_n)\mapsto (x_0:\cdots:x_m)$.

Let $P'$ be the blow up of $P^n$ along L and consider $f':P'\rightarrow P^m$. I am trying to understand this morphism.

Edit: Let $p:P'\rightarrow P^n$ the blow up morphism. Then what is the direct image of the line bundles $p^*O(i)$ under $f'$. I would in particular like to know what is $f'_*O_{P'}$.

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  • $\begingroup$ What do you mean by $\mathcal{O}_{P'}(i)$? $\endgroup$
    – Sasha
    Aug 3 at 12:56
  • $\begingroup$ For $\mathcal{O}_{P'}(i)$ cf. Hartshorne, "Algebraic geometry", pp. 117, 161, 163. $\endgroup$ Aug 3 at 13:23
  • $\begingroup$ @ThomasPreu: it is better to give an actual explanation instead of a reference to Hartshorne. $\endgroup$
    – Sasha
    Aug 3 at 13:59
  • $\begingroup$ The Picard group of the variety $P'$ has rank 2, so there are more than one choice of an ample line bundle, and therefore the notation $\mathcal{O}_{P'}(1)$ is ambiguous. Consequently, the answer to your question depends on the choice of this line bundle. $\endgroup$
    – Sasha
    Aug 3 at 15:56
  • $\begingroup$ @Sasha. I am sorry. If $p:P'\rightarrow P^n$ is the blow up morphism, then I denoted $p^*O(i)$ by $O_{P'}(i)$. I would in particular like to know what is $f'_*O_{P'}$. $\endgroup$
    – user52991
    Aug 3 at 16:10

1 Answer 1

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The morphism $f' \colon P' \to \mathbb{P}^m$ is a projective bundle; more precisely $$ P' \cong \mathbb{P}_{\mathbb{P}^m}(\mathcal{O}^{\oplus (n-m)} \oplus \mathcal{O}(-1)).,$$ hence $f'_*\mathcal{O}_{P'} \cong \mathcal{O}_{\mathbb{P}^m}$. Moreover, $\mathcal{O}_{P'}(1)$ is the relative hyperplane bundle for this projective bundle, therefore if $i > 0$ then $$ f'_*\mathcal{O}_{P'}(i) \cong \mathrm{Sym}^i(\mathcal{O}^{\oplus (n-m)} \oplus \mathcal{O}(1)). $$

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  • $\begingroup$ thank you! Here are you still using the notation $O_{P'}(1) =p^*O_{P^n}(1)$ ? $\endgroup$
    – user52991
    Aug 3 at 17:30
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    $\begingroup$ Yes, I am using your notation. $\endgroup$
    – Sasha
    Aug 3 at 17:35

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