I encounter a problem when I read compact complex surfaces.Let $X$ be a projective surface,$E$ is a vector bundle of rank 2 over $X$, $S \subset \mathbb{P}(E)$ is an irreducible subvariety such that $\pi |_S :S\rightarrow X$ is birational. It say that $\mathcal{O}_{\mathbb{P}(E)}(1) \otimes \mathcal{O}_{\mathbb{P}(E)}(-S)$ is trivial on all fibres of $\mathbb{P}(E)$, why?
1 Answer
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Since $\pi|_S$ is birational, it follows that $S$ is a section of $\pi$. Hence we can write $$\mathcal{O}_{\mathbb{P}(E)}(S) \cong \mathcal{O}_{\mathbb{P}(E)}(1) \otimes \pi^* \mathcal{L},$$ where $\mathcal{L}$ is a line bundle on $X$. This implies $$\mathcal{O}_{\mathbb{P}(E)}(1) \otimes \mathcal{O}_{\mathbb{P}(E)}(-S) \cong \pi^* \mathcal{L}^{-1}.$$ Since $\pi^*{\mathcal{L}}^{-1}$ comes from the base $X$ of the projective bundle $\mathbb{P}(E) \to X$, it is trivial when restricted to any fibre.
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$\begingroup$ Thank you very much, can you tell me where to find: $\mathcal{O}_{\mathbb{P}(E)}(S) = \mathcal{O}_{\mathbb{P}(E)}(1) \bigotimes \pi ^*\mathcal{L} $ $\endgroup$– swalkerCommented Sep 24, 2013 at 14:12
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$\begingroup$ This is rather standard stuff on projective bundles. One has $$\textrm{Pic}\, \mathbb{P}(E)= \pi^*\textrm{Pic}\, X \times \mathbb{Z},$$ where the $\mathbb{Z}$ is generated by $\mathcal{O}_{\mathbb{P}(E)}(1)$. Have a look at Hartshorne's Algebraic Geometry, Chapter II, Section 7. $\endgroup$ Commented Sep 24, 2013 at 14:18