# Does the dual of a vector bundle with ample determinant have global sections

Let $E$ be a locally free sheaf on a smooth projective variety $X$ over $\mathbb C$. Suppose that $\det E$ is an ample line bundle on $X$.

Is $$H^0(X,E^\vee) =0?$$

In fact, if $E$ of rank $1$, it is easy to see that $h^0(X,E^\vee) =h^0(X,\omega_X\otimes E)=0$ by Serre duality and Kodaira's vanishing theorem. (I think there should be an easier argument, though.)

If $E$ is of rank at least two, I thought about either

1) an induction argument, or

2) a strengthening of Kodaira's vanishing theorem for vector bundles (is there such a theorem?), or

3) an easy elementary argument that I'm simply not seeing...

Finally, if it helps, please do assume the canonical sheaf $\omega_X$ is ample.

-
The example given by @Georges Elencwajg in your other question shows that this cannot be true: if $X=\mathbf{P}^1$ and $E=O(-1) \oplus O(2)$, then $\mathrm{det} \, E$ is ample, but $E^{\vee} = O(1) \oplus O(-2)$ does have global sections. – user64687 Apr 14 '13 at 21:57
By the way, there is a much simpler proof that the dual of an ample line bundle cannot have global sections. Suppose E is ample and E* has a nonzero global section s. Choose a curve C intersecting the vanishing set of s properly. Then $E^*.C \geq 0$. But $E.C >0$ too, since E is ample. This contradicts additivity of degree. – user64687 Apr 15 '13 at 15:45