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Supose $L$ and $L'$ are holomorphic line bundles over $\mathbb{CP}^n$ such that $L|_{H} \simeq L'|_{H}$ for every hyperplane $H \subset \mathbb{CP}^n$. Does it follow that $L \simeq L'$?

Using the fact that every $x \in \mathbb{CP}^n$ is contained in a hyperplane one gets that $L_x \simeq L'_x$ for every $x$ but I don't know how to prove that the isomorphisms glue to a global one.

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up vote 1 down vote accepted

Yes. Actually if two holomorphic line bundles $L,L'$ on $\mathbb C \mathbb P^n \quad (n\geq 2)$ coincide on just one hyperplane, then they are already isomorphic.
This is because a holomorphic line bundle on $\mathbb C \mathbb P^n$ is of the form $\mathcal O_{\mathbb C \mathbb P^n}(r) \; (r\in \mathbb Z)$ and its restriction to $H$ is $\mathcal O_H(r)\simeq \mathcal O_{\mathbb C \mathbb P^{n-1}}(r) .\;$
So that $r$, an integer which suffices to characterize the line bundle, is determined by its restriction to $H$.

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Do you know any kind of generaliztion of this fact, e.g, if the restriction of a line bundle to hyperplane secions of a projective manifold determines it? – Lucas Kaufmann Dec 7 '11 at 7:02
Another question Georges. Do you know any other proof that doesn't use the fact that any line bundle over $\mathbb{P}^n$ is isomorphic to some $\mathcal{O}(r)$? – Lucas Kaufmann Dec 8 '11 at 13:43

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