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

Let $X$ be a non-singular complex variety with a big line and base point free bundle $M$ on it. My question is can we say that for any locally free sheaf $F$ on $X$, $F \otimes M^n$ is globally generated for $n \gg 0$.

Motivation: If $M$ were an ample line bundle then all we need is that $F$ is coherent sheaf. But since we are given a much stronger condition on $F$ (which is local freeness) can we say the same thing with $M$ just being big and base point free.

I tried to use the fact that any big line bundle is tensor product of an ample line bundle and an effective line bundle.

I am not even sure that this has to be true but am unable to find a counterexample.

share|improve this question

3 Answers 3

up vote 5 down vote accepted

In general the answer is no, even $F$ is a line bundle itself. It is easy to see that a globally generated line bundle is nef, and if $F$ is not nef, and the segment between $F$ and $M$ does not intersect with the ample cone in $N^{1}(X)$, then $F \otimes M^{n}$ is numerically propotional to a divisor lies in the interior of the segment, thus is not nef.

share|improve this answer
1  
Don't you need to choose F anti-ample? (if you want him to intersect negatively curves C satisfying $(M\cdot C)=0$, which exist since $M$ is semi-ample but non ample?) –  Henri Dec 2 '11 at 13:57
2  
In fact, many line bundles satisfy the property that the segment does not intersect with the ample cone. Anti-ampleness is not required. –  Zhengyu Hu Dec 2 '11 at 14:01
    
Ok, thank you. (I think you edited your answer while I wrote my comment) –  Henri Dec 2 '11 at 15:01
    
Oh... I think your comment is a good counter example. It seems that works for any $M$. –  Zhengyu Hu Dec 2 '11 at 15:10

Here is a simple counterexample (of the form Zhengyu Hu suggested). Take $X$ to be the blowup of a point in $\mathbb P^2$, $M$ the pullback of $\mathcal O(1)$ under the blowup map, and $F$ the line bundle associated to the exceptional divisor $E$. Sections of $F\otimes M^n$ are rational functions which may have a pole of order 1 along $E$ and a pole of order up to $n$ along a line not meeting $E$. However, any rational function $f$ actually having a pole along $E$ (i.e. generating $F\otimes M^n$ at the points of $E$) must have a pole along some other divisor passing through $E$ (since $X$ has the same rational functions as $\mathbb P^2$).

share|improve this answer
    
Thanks for this concrete counterexample. –  Amit Dec 2 '11 at 15:17

In my view being big corresponds to being "generically ample", that is being ample outside a closed subset. This closed subset is called the augmented base locus of $M$ and it is often denoted by $\mathbb{B}_+(M)$ ( see http://de.arxiv.org/PS_cache/math/pdf/0308/0308116v2.pdf). So, if for example you take $M$ big and $F$ to be a line bundle you can prove that $M^m\otimes F$ is always globally generated (i.e. base point free) outside the augmented base locus of $M$ (note that you don't need global generation of $M$ for this). In the example of Anton Geraschenko in fact $\mathbb{B}_+(M)$ is exactly the exceptional divisor. Something similar might hold if you consider locally free sheaves of higher rank.

share|improve this answer
    
Thanks for this reference –  Amit Dec 3 '11 at 10:51

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.