# Is the pullback of an ample bundle minus the exceptional divisor ample?

If X is a normal projective variety with an ample line bundle $$L$$, and $$\pi:Y\to X$$ a resolution of $$X$$ and $$E$$ be the exceptional divisor, then is it true that $$A\pi^{\star}L-[E]$$ is always ample for $$A$$ sufficiently large?

## 1 Answer

You should probably state what you want a little more precisely. As it is currently stated, it allows for the possibility that $$\pi$$ is not projective in which case there is no chance. The statement also allows a small resolution in which case $$E$$ is empty and $$a\pi^*L$$ is not ample for any $$a$$.

Unfortunately, even if you assume that $$\pi$$ is projective and $$E$$ is the entire exceptional set(!), it is possible that this fails. The condition you need is that $$-E$$ has to be relatively ample for $$\pi$$. This can fail already for surfaces. For instance, assume that $$E$$ has two components, both with negative self-intersection, say $$-n$$ and $$-m$$, and let's say that the intersection number of the two components is $$n+e$$ for some positive number $$e$$. The intersection matrix has to be negative definite, which means that we need $$nm> (n+e)^2$$, but this is easy to satisfy by making $$m$$ really big. In this case, $$E$$ restricted to the component with self-intersection $$-n$$ has positive degree, so $$-E$$ cannot be ample.

This suggests that you cannot allow $$E$$ to have more than one component. On the other hand, in that case you are OK. Alternatively, if you allow different coefficients for the components of $$E$$ then a similar statement holds.

Addendum: In response to @freidtchy's comment-question below here is an explanation of the last sentence above. Yes, I meant exactly that there exist $$a_i>0$$ such that $$A\pi^*L-\sum a_iE_i$$ is ample. Since $$\pi$$ is ample, there exists a $$\pi$$-ample Cartier divisor on $$Y$$. With a little bit of work one may assume that there is one which is entirely supported on $$E$$ (the possible components that are a priori not can be exchanged to something pulled back from $$X$$ plus something supported on $$E$$ and the pull-back of anything is $$\pi$$-trivial). In other words, one has a $$\pi$$-ample Cartier divisor $$\sum b_iE_i$$. Then the Negativity Lemma [Kollár-Mori-98, 3.39] tells us that all the $$b_i$$ are negative and then choosing a large enough $$A$$ gives the claimed statement.

• Can you clarify what you mean by if you allow different coefficient for the components of E then a similar statement holds? Do you mean something like if $E_i$ are the irreducible components of $E$, then there always exists $a_i>0$ such that $A\pi^{\star}L-\sum a_iE_i$ is ample? Can you point me to some reference? – freidtchy Nov 13 '19 at 1:49
• Yes. I added some explanation and a reference in the body of my answer. – Sándor Kovács Nov 13 '19 at 18:50