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Let $C$ be a smooth projective curve and $E$ a vector bundle of rank $r$ on $C$. We say that $E$ is nef/ample if $\mathcal{O}_{\mathbb{P}(E)}(1)$ is so. Equivalently (see Hartshorne's papers on 'Ample vector bundles' and 'Ample vector bundles on curves'), $E$ is ample if and only if for any coherent $F$, $S^m(E)\otimes F$ is globally generated for all $m\geq n_0$.

The statement I'm slightly stuck on is the following comment in a paper of Fujita 'On Kahler fibre spaces over curves': If $C$ has genus $g\geq2$ and $H^1(C, E)=0$ for some vector bundle $E$, then $E$ is ample.

This follows easily in the rank $1$ case from Riemann-Roch. I suspect the general case will also be easy but I have been through Lazarsfeld's book and the standard references with no luck so far. Any help appreciated!

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  • $\begingroup$ Riemann-Roch tells you that $\mu(E) \geq g - 1$. Does it help? $\endgroup$
    – Piotr Achinger
    Commented Dec 16, 2010 at 13:30
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    $\begingroup$ Have you forgotten to mention the condition on $H^1(C,E)$? $\endgroup$
    – Rex
    Commented Dec 16, 2010 at 13:46

3 Answers 3

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Here's a partial answer. Suppose we're in characteristic 0 (Fujita would be assuming this), and that $rank(E)=2$. By cor 7.6 of Hartshorne's ample vector bundles paper, it suffices to check that $deg(E)>0$ and $deg(L)>0$ for ay quotient line bundle. From Riemann-Roch as in Piotr's comment, we get $$deg(E) + rank(E)(1-g) = h^0(E)\ge 0$$ which implies positivity of $deg(E)$. On a curve $H^1(E)=0$ implies the vanishing for any quotient bundle, and so in particular for $L$. Combing this with the above argument, gives $deg(L)>0$.

I think this can be pushed, but I'd better back to the less fun things that I'm supposed to be doing now.

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  • $\begingroup$ Thank you! Your comment about $H^1$ of a quotient bundle also being trivial is spot on, as from Theorem 2.4 from Hartshorne's 'Ample vector bundles on curves', (over $\mathbb{C}$) $E$ is ample if and only if every quotient bundle has positive degree, and by Riemann-Roch (or at least the vector bundle equivalent $\chi(E) = \deg(E) + rk(E)\chi(\mathcal{O}_X)$) it follows that if $E\to E'\to0$, then $\deg(E')>0$. If you'd like to incorporate this in your answer for completeness I'd be glad to accept it. $\endgroup$
    – Frank
    Commented Dec 16, 2010 at 16:23
  • $\begingroup$ I didn't see this comment before I posted my answer, but it is essentially the same argument. $\endgroup$
    – J.C. Ottem
    Commented Dec 16, 2010 at 16:45
  • $\begingroup$ Frank & J.C.: Good work! $\endgroup$
    – Donu Arapura
    Commented Dec 17, 2010 at 14:10
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This builds on the idea of Donu Arapura: We use the following criterion of Hartshorne (Thm. 2.4 in this article): $E$ is ample iff $Q$ has positive degree on $C$ for any quotient bundle $Q$ of $E$. Let $r=rank(E)$ and $s=rank(Q)$. Here

$$deg(Q)=s(g-1)+\chi(Q)> \chi(Q)$$

so it suffices to show that $\chi(Q)\ge 0$. But this is elementary, since $H^1(E)$ implies $H^1(Q)=0$ by the exact sequence $$ 0\to L \to E\to Q \to 0. $$

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  • $\begingroup$ Thank you, just posted this as a comment to Donu's answer. $\endgroup$
    – Frank
    Commented Dec 16, 2010 at 16:44
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It follows from Gieseker's Criterion- we have to show that E can't have t any trivial quotient. Use the Cohomology seq for 0 -> K -> E -> O_C -> 0

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