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Let $X$ be a proper curve, not necessarily smooth nor reduced, and $E$ a vector bundle on $X$ of rank $r$. Assume we know that $H^0(X,E)\geq r$ and $H^0(X,E^{\vee})\geq r$, can we conclude that $E$ is trivial? Is there a cohomological criterion to decide whenever a vector bundle is trivial?

vonbrand
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jikki
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  • In general a vector bundle $\rm E$ of rank $r$ is trivial iff $\rm h^0(X, E) = r$. –  Mar 15 '13 at 15:20
  • Dear @Adeel, what you write is completely false. – Georges Elencwajg Mar 15 '13 at 15:46
  • @Adeel the generic line bundle of degree equal to the genus has trivial $H^1$, so that the $H^0$ has dimension one but it is not trivial. – jikki Mar 15 '13 at 15:50
  • On the other hand for a line bundle $L$ it is true that $h^0(X,L)>0$ and $h^0(X,L^{\vee})>0$ implies $L=\mathcal{O}$. – jikki Mar 15 '13 at 15:52
  • This is a well-known fact though, see page 8 of Hatcher's book on vector bundles for example. –  Mar 15 '13 at 16:02
  • Also http://mathoverflow.net/questions/124624/cohomological-criterion-of-triviality – Ehsan M. Kermani Mar 15 '13 at 16:47
  • @Adeel: I think you misunderstand what you read. Linear independence as elements of H^0(X,E) doe not imply that the sections are linearly independent at every point (which is what Hatcher is talking about). –  Mar 15 '13 at 18:10
  • @user64687 Ah, thanks for this clarification. –  Mar 15 '13 at 18:15

1 Answers1

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No, you cannot conclude that $E$ is trivial: take $X=\mathbb P^1_k$ and $E=\mathcal O_{\mathbb P^1 _k}(-1) \oplus \mathcal O_{\mathbb P^1 _k} (1)$.
The rank $2$ self-dual bundle $E$ is not trivial, even though $h^0(\mathbb P^1_k, E)=h^0(\mathbb P^1_k, \check {E})=2$.

EDIT
1) This example is probably the simplest one, but it can be varied in innumerable ways by taking sums $\mathcal O_X(-D) \oplus \mathcal O_X(D)$ of line bundles associated to a sufficiently ample effective divisors $D$ on a projective variety $X$ .

2) There is however no counterexample of rank one on a projective variety $X$:
If a line bundle $L$ and its dual bundle $\check L$ have each a non-trivial section $s\in \Gamma(X,L),\sigma \in \Gamma(X,\check L) $ , then the section $s\otimes \sigma \in \Gamma(X,L\otimes\check L )=\Gamma(X,\mathcal O)=k$ is a non-zero constant.
This forces $s$ to also have no zero and thus shows that $L$ is trivial.

Georges Elencwajg
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