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Serre-Swan's theorem (see the MO discussion) says that any locally free sheaf over an affine variety is a direct summand of a free sheaf. However, this is not true on projective varieties. It is not hard to check that a non-trivial line bundle with non-zero global sections can not be a direct summand of a free sheaf. The reason is that, being a direct summand of a free sheaf implies that the dual line bundle also has non-zero global sections. But that implies the line bundle is trivial.

I am wondering if the same holds for vector bundles, i.e. if a vector bundle and its dual over a projective variety both have non-zero global sections, then the vector bundle is trivial.

Another I think related question is the following:

Is a direct summand of a direct sum of line bundles on a projective variety also a direct sum of line bundles?

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1 Answer

up vote 4 down vote accepted

1) The vector bundle $E=\mathcal O(1) \oplus \mathcal O(-1)$ over $\mathbb P^1$ has non trivial sections and so has its dual (which happens to be the same bundle $E$ ).
However $E$ is non trivial because each of its global sections has a zero.

2) Yes, a direct summand of a direct sum of line bundles on a complete variety is also a direct sum of line bundles.
This follows from Atiyah's general version of the Krull-Schmidt theorem .

Edit
Since Fei asks, let me remark that the trick in 1) works also for bundles which are not direct sums of line bundles:
If $T$ is the tangent bundle to $\mathbb P^2$, then $E=T \oplus T^* $ has non-trivial global sections , and so has $E^*=E$
However, since $T$ is indecomposable $E$ is not a sum of line bundles by Atiyah's result and is thus a fortiori not trivial.

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Thank you so much for your answer. For the first question, do you have a counterexample that $E$ is not a direct sum of line bundles? –  Fei YE Mar 25 '12 at 11:07
    
Dear @Fei,yes: I have added such a counterexample in an Edit. –  Georges Elencwajg Mar 25 '12 at 11:26
    
Dear George, Thank you so much for the example. I saw you used a fact $E^\ast=E$ for $E=T\oplus T^\ast$. Is there are general relation between $E$ and $E^\ast$. I saw somewhere that $E^\ast\otiems \det(E)=E$. Is this really true in general? Thanks again for you help! –  Fei YE Mar 25 '12 at 12:14
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Dear @Fei, I have just used that the dual of a direct sum is the direct sum of the duals. The formula you quote is already false for every non trivial line bundle $L$ since $L^* \otimes det(L)=L^* \otimes L$ is trivial. –  Georges Elencwajg Mar 25 '12 at 12:24
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Dear @Fei, yes, for rank $2$ it is correct: Hartshorne, Ex.II 5.16 –  Georges Elencwajg Mar 25 '12 at 14:06
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