# Triviality of line bundle over complex manifold

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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## 3 Answers

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

No.

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