Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

M. Atiyah in "VECTOR BUNDLES OVER AN ELLIPTIC CURVE" defined ample line bundle $E$ on $X$ as satisfying the following conditions:

  1. Canonical map $H^0(X, E)\to E_x$ is surjective for any $x\in X$.

  2. $H^q(X,E)=0$ for $q>0$.

But in standard textbooks like Hartshorne there is another definition:

$E^{\otimes n}$ is very ample for some $n\in \mathbb N$.

Are these definitions really equivalent? Where can I find proof of this?

share|improve this question

2 Answers 2

up vote 2 down vote accepted

Even a very ample line bundle does not need to be ample in the sense of the first definition (consider $\mathcal O(1)$ on a smooth plane curve of degree $>4$ or, more generally, on any smooth hypersurface of high enough degree). I do not know whether ampleness in the sense of the 1st definition implies ampleness in the sense of teh second definition.

Nowadays the standard definition of ampleness is as given in Hartshorne's book.

share|improve this answer
    
Wouldn't $E=\mathcal O_X$ satisfy the first condition on $X=\mathbb P^1$, without being ample? –  J.C. Ottem Apr 19 '13 at 10:00
    
Yes, I already realised that. Very ample line bundles usually don't have vanishing non-zero-th cohomologies. Thank you for your answer. –  zroslav Apr 19 '13 at 10:12
    
@J.C.Ottem: $\mathcal O_X$ is not globally generated, so it does not meet the requirements of the 1st definition. –  Serge Lvovski Apr 19 '13 at 10:14
2  
@Serge: $\mathcal{O}_X$ is always globally generated! –  Laurent Moret-Bailly Apr 19 '13 at 12:32
    
@Laurent: What exactly do we mean by "globally generated"? OP gives a condition "$H^0(X,E)$ surjects onto $\mathcal O_x$ for each $x\in X$"). If $X$ is proper and positive-dimensional (and, say, connected), this condition does not hold: a local ring of $X$ cannot be generated by constants only! –  Serge Lvovski Apr 19 '13 at 13:05

Let $X$ be an elliptic curve.

Then a divisor $E$ which is ample in the sense of the first definition is necessarily ample in the usual sense. Conversely, if it is ample in the usual sense and $\deg E \geq 2$ then it is ample in the sense of the first definition.

In fact, assume that $E$ ample in the sense of the first definition. Then $(2)$ implies that $\mathcal{O}_X(E) \neq \mathcal{O}_X$ and $(1)$ implies that $\dim H^0(X, E) \geq 2$, because a non-trivial divisor with $\dim H^0(X, E) =1 $ on an elliptic curve cannot be globally generated. Then by $(2)$ and Riemann-Roch it follows $$\deg E = \dim H^0(X, E) \geq 2,$$ hence $E$ is ample in the usual sense by [Hartshorne, Algebraic Geometry, Corollary 3.3 page 308].

Conversely, assume that $E$ is ample in the usual sense and $\deg E \geq 2$. This last condition implies that $|E|$ is base-point free on $X$, hence $E$ is globally generated and $(1)$ holds. Moreover, since $\omega_X = \mathcal{O}_X$, by Kodaira Vanishing Theorem and Serre duality one deduces $(2)$. So $E$ is ample in the sense of the first definition.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.