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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.

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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.

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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
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 $\mathbb P^2$).

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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.

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Thanks for this reference –  Amit Dec 3 '11 at 10:51

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