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Question. Is it known/easy to see that every smooth projective variety $X$ (over an algebraically closed field), except for the point and $\mathbb{P}^1$, has a vector bundle which is not a direct sum of line bundles?

I have a result which trivially shows the above fact in positive characteristic, but I don't know whether I should state it as a corollary as it might be well-known or obvious for reasons I don't see.

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Of arbitrary rank $>1$? For $\mathbb{P}^n$ itself, $n>1$, the tangent bundle is indecomposable. –  Mahdi Majidi-Zolbanin Mar 24 '13 at 3:31
    
@Mahdi: yes. One idea of construction would be to embed $X$ into $\mathbb{P}^N$ and restrict $\Omega^1_{\mathbb{P}^N}$ to $X$ (or pull it back along a finite flat map $f:X\to \mathbb{P}^n$), but I don't know if it can work... –  Piotr Achinger Mar 24 '13 at 3:57
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Restriction fails for curves in $\mathbb P^2$. Of course curves of positive genus have a two-dimensional vector bundle which is a nontrivial extension of the trivial line bundle with itself. –  Will Sawin Mar 24 '13 at 4:02
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Even if it is well-known, isn't it still fairly standard to include these types of things as corollaries? It shows that your work has applications to things that people care about and gives alternate proofs of known results. Both these things seem worth it. –  Matt Mar 25 '13 at 2:21
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1 Answer 1

up vote 16 down vote accepted

Yes, it is true that (over an algebraically closed field) the only positive-dimensional smooth projective variety on which every algebraic vector bundle splits as a sum of line bundles is $\mathbb P^1$.
More generally Ballico has proved that if every algebraic vector bundle splits as a sum of line bundles on a reduced but maybe reducible positive-dimensional projective variety , then that variety is a chain of $\mathbb P^1$'s.
For holomorphically inclined readers, here is a related result in complex geometry.

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Perfect, thanks!! –  Piotr Achinger Mar 24 '13 at 20:33
    
You are welcome, Piotr. –  Georges Elencwajg Mar 24 '13 at 20:49
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