# Ample line bundle of Blow up

Let $P$ be a point in $\mathbb{P}^n$. Then we have a regular map $\mathbb{P}^n\setminus P\rightarrow \mathbb{P}^{n-1}$, which is not defined only at $P$. Then We know that the graph closure of this map is the Blow Up of $\mathbb{P}^n$ at $P$. $Blow_P \mathbb{P}^n\hookrightarrow \mathbb{P}^n\times \mathbb{P}^{n-1}$. Now Let $p_2$ be the projection to the second factor. $p_2:Blow_P \mathbb{P}^n\rightarrow \mathbb{P}^{n-1}$. My question is what is $p_2^*\mathcal{O}(1)$? Is it $\mathcal{O}(-E)$ or $p_1^*\mathcal{O}(1)\otimes \mathcal{O}(-E)$?

Here $E$ is the exceptional divisor and $p_1$ is the projection onto the first factor.

• It is the latter. To see, just interpret the sections of $p_1^*O(1)\otimes O(-E)$ as the sections of $p_1^*O(1)$ vanishing at $E$; these correspond to linear polynomials vanishing at $P$. – byu Jul 2 '17 at 22:04
• It is the latter. You can convince yourself by calculating the degree of this bundle on proper transforms of lines on $P^n$ that do or don't pass through the blown-up point. – Bertie Jul 2 '17 at 22:05
• @byu the explanation is convincing....@bartie I was doing like that, but wasn't confident. – user100841 Jul 2 '17 at 22:20
• @byu So the line bundle $p_1^∗\mathcal{O}(1)⊗\mathcal{O}(−E)$ is not very ample on $Blow_P \mathbb{P}^n$, even though $\mathcal{O}(1)$ is very ample on $\mathbb{P}^n$ and $\mathcal{O}(−E)|_E$ is very ample line bundle on E. – user100841 Jul 2 '17 at 23:51
• I din't say that that line bundle was very ample - in fact I claim that the global sections of this line bundle gives the projection map. – byu Jul 3 '17 at 13:30

You can undertstand this by thinking about the simplest case $n=2$, the case $n \geq 3$ being analogous.
The blow-up of $\mathbb{P}^2$ at a point $P$ is the Hirzebruch surface $\mathbb{F}_1$, and the two projections $$p_1 \colon \mathbb{F}_1 \longrightarrow \mathbb{P}^2, \quad p_2 \colon \mathbb{F}_1 \longrightarrow \mathbb{P}^1$$ are the blow-up morphism and the $\mathbb{P}^1$-fibration induced by the pencil of lines through $P$, respectively. So, denoting by $f$ the fibre of $p_2$, we have $$p_1^*\, \mathcal{O}(1) = E+f, \quad p_2^* \,\mathcal{O}(1) = f,$$ that is $p_2^*\,\mathcal{O}(1)=p_1^*\,\mathcal{O}(1)\otimes \mathcal{O}(-E).$