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Let $X$ be a compact complex manifold. According to Fulton and Lazarsfeld, a vector bundle $E$ on $X$ is called ample if the Serre line bundle $\mathcal{O}_{\mathbb{P}(E)}(1)$ on the projectivized bundle $\mathbb{P}(E)$ is ample.

This notion should be a generalization of ampleness of line bundles, but I don't quite understand. Assume that $E$ is a line bundle, then its projectivization is isomorphic to $X$. In this case how can we conclude that $\mathcal{O}_{\mathbb{P}(E)}(1)\cong E$?

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Isn't this just the relative form of the tautological isomorphism $\mathscr{O}_{\operatorname{Proj} S}(1) \cong S_1$? – Zhen Lin Sep 9 '12 at 8:30
@ZhenLin, you have to take the global section on the LHS (if $S_1$ denotes elements of degree $1$ of $S$), and this holds for certain $S$ (e.g. polynomial rings). – user18119 Sep 9 '12 at 14:41

This isomorphism comes from the canonical isomorphism $$ E\simeq \pi_*(\mathcal O_{\mathbb P(E)}(1))$$ (in any rank), where $\pi : \mathbb P(E)\to X$ is the canonical morphism.

By definition $\mathbb P(E)$ is the Proj of the sheaf of symmetric algebras $$\mathcal{Sym}(E)=\mathcal O_X\oplus E\oplus ...$$ In particular, for any affine open subset $U$ of $X$, $$\pi^{-1}(U)=\mathrm{Proj}\ \mathrm{Sym}(E(U))$$ and the sheaf $\mathcal O_{\mathbb P(E)}(1)|_{\pi^{-1}(U)}$ is the $\mathcal O(1)$ on the projective scheme.

We thus have a canonical homomorphism $$ E\to \pi_* (\mathcal O_{\mathbb P(E)}(1)).$$ It is an isomorphsim because its restriction to any $U$ small enough (so that $E|_U$ is free) is an isomorphism by usual computations on a projective space over a ring.

When $E$ is rank one, as $\pi$ is an isomorphism, we get $\mathcal O_{\mathbb P(E)}(1)\simeq \pi^*E$.

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