# Morphism from blow up to a linear projection

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 $$f'_*O_{P'}$$.

• What do you mean by $\mathcal{O}_{P'}(i)$? Aug 3 at 12:56
• For $\mathcal{O}_{P'}(i)$ cf. Hartshorne, "Algebraic geometry", pp. 117, 161, 163. Aug 3 at 13:23
• @ThomasPreu: it is better to give an actual explanation instead of a reference to Hartshorne. Aug 3 at 13:59
• 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. Aug 3 at 15:56
• @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 $f'_*O_{P'}$. Aug 3 at 16:10

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 $$f'_*\mathcal{O}_{P'}(i) \cong \mathrm{Sym}^i(\mathcal{O}^{\oplus (n-m)} \oplus \mathcal{O}(1)).$$
• thank you! Here are you still using the notation $O_{P'}(1) =p^*O_{P^n}(1)$ ? Aug 3 at 17:30