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Is it possible to know whether the given line bundle over compact complex manifold or projective variety is trivial or not from its sheaf cohomology data?

I found this question when I trying to solve the exercise problem in voisin's book Hodge theory and complex algebraic geometry 1

(I haven't solve that problem yet)

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up vote 2 down vote accepted

There is one useful fact that might by relevant to you. A line bundle $L$ and a complex projective variety $X$ is trivial if and only if $H^0(X,L) \neq 0$ and $H^0(X,L^{-1})\neq0$.

Indeed one implication is clear, and for the other choose two non-zero elements $s \in H^0(X,L)$ and $s' \in H^0(X,L^{-1})$. Then $ss' \in H^0(X,\mathcal{O})$ is non-zero. Since $H^0(X,\mathcal{O}) = \mathbb{C}$, we see that in fact $ss'$ is nowhere vanishing. Thus $L$ admits a nowhere vanishing section, and so is trivial.

Naturally a similar result holds for more general fields.

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Clear! Thank you so much! – Choa May 8 '11 at 9:41


Let $X$ be the blow up of a smooth variety at a closed point and $E\simeq \mathbb P^r$ the exceptional divisor. Let $\mathscr L=\mathscr O_X(E)$. Then the section $E$ corresponds to a morphism $\mathscr O_X\to\mathscr L$ that sits in a short exact sequence:

$$ 0\to \mathscr O_X \to \mathscr L\to \mathscr O_{\mathbb P^r}(-1)\to 0. $$

All the cohomology groups of $\mathscr O_{\mathbb P^r}(-1)$ are zero, so all the cohomology groups of $\mathscr O_X$ and $\mathscr L$ are isomorphic, but $\mathscr L$ is not trivial.

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Thank you so much! Sorry for add one more question; What kind of (minimal) condition do we need to know that it is trivial? For example, in voisin's book,the followings are given 1. H0(X,L^n) is not zero 2. Hn(X,L*Kx) is not zero I still don't know how the triviality comes from these but anyway it is sufficient, she says. – Choa May 8 '11 at 9:34
@Choa: those conditions combined with Serre duality show that $L^n$ is trivial by Daniel's answer. That implies that $L^{-1}$ has a section whose $n^\text{th}$ power is nowhere vanishing, but then it has to be nowhere vanishing and hence $L^{-1}$ and therefore $L$ are trivial. – Sándor Kovács May 8 '11 at 17:46

Given any holomorphic line bundle $L$ on a complex manifold $X$ , then we can get a element in $H^1(X,\mathcal{O}^\ast)$ . Then if it is zero in $H^1(X,\mathcal{O}^\ast)$ , it is trivial .

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Thank you for your answer. Is there any effective way to determine whether it is zero or not in H1(X,O∗)? Some kind of exact sequence will be helpful, I think. – Choa May 8 '11 at 9:01

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