# A funny condition for ampleness on a curve

Let $E$ be a locally free sheaf on a complete nonsingular curve $C$ over $\mathbb C$.

Suppose that, for all points $P$ in $C$, the locally free sheaf $$E\otimes \mathcal{O}_C(P)$$ is an ample locally free sheaf on $C$. Also, suppose that $E$ is not isomorphic to $\mathcal{O}_C^{r}$, where $r$ is the rank of $E$.

Does it follow that $E$ is ample?

It's true when $r=1$.

I think it should be true, but I don't see how to do it. Intuitively, I keep thinking about the $r=1$ case. In that case you could just compute the degree, but you could also use that for all rational sections of $E$ there exists a point $P$ not in the support of the divisor of that section. This helps you conclude that the divisor of your section is effective, and thus ample. Does this line of thought help when $r>1$?

Final remark. By induction, we may and do assume $E$ is indecomposable. In particular, it's true for all $r\geq 1$ when $C =\mathbf P^1$.

Edit: Asal shows that it is not true under the above conditions when $g=1$. Let me show that it is also false when $g\geq 2$.

Let $$0 \to \omega_C\to E\to \mathcal O_C\to 0$$ be a non-trivial extension $\mathcal O_C$ by $\omega_C$. This exists because $H^1(\omega_C) \cong \mathbb C \neq 0$. Then $E$ is not ample, because $E$ surjects onto $\mathcal O_C$. But clearly, $E\otimes L$ is an extension of $L$ by $L\otimes \omega_C$. If $L$ is ample, then $L\otimes \omega_C$ is ample. In particular, $E\otimes L$ is an extension of an ample line bundle by an ample line bundle. Thus, $E\otimes L$ is ample. (Now apply this to $L=\mathcal {O}_C(P)$...)

• It's not true for $r=1$: say $C$ is an elliptic curve, and $E$ is a line bundle of the form $O_C(p) \otimes O_C(q)^*$ where $p$ and $q$ are distinct points on $C$. Then $E$ is not isomorphic to $O_C$, but for any $p$, $E \otimes O_C(p)$ has degree 1, hence is ample. Is there an extra hypothesis you missed? – user64687 Apr 17 '13 at 18:29
• I guess I should have said in the previous comment: $E$ is certainly not ample either, since it has degree zero. – user64687 Apr 17 '13 at 18:37
• Ow no, I completely overlooked that. Anyway, I just came up with another example when $C$ is a curve of genus at least two; see the edited question. I'm going to think about what extra condition we need. Maybe we should impose that the $H^0(E)$ is of dimension at least two? – Tom Apr 17 '13 at 18:43
• Well, if $H^0(E)$ has dimension at least 2, then $E$ is a nontrivial effective line bundle, but any such bundle on a curve is ample (because it has positive degree). I'm not sure that any such condition will work on a curve other than $\mathbf{P}^1$. But I might be wrong... – user64687 Apr 17 '13 at 18:50
• When you say "any such condition", you mean the condition of the question, right? Not the condition of $H^0(E)$ being of dimension at least two, no? Also, why is $E$ a line bundle if $h^0(E) \geq 2$? Vector bundles of rank at least two can't have more than one global section? – Tom Apr 17 '13 at 19:09