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?



Is the pullback of an ample bundle minus the exceptional divisor ample?
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 that if $X$ is $\mathbb Q$-factorial, then 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 $Y$. One may assume that there is one which is entirely supported on $E$; indeed suppose that $H+\sum\_ia\_iE\_i$ is $\pi$-ample. Consider $\pi\_\*H$ and observe that if $X$ is $\mathbb Q$-factorial, then some multiple of this is Cartier. So, replace $H+\sum\_ia\_iE\_i$ with that multiple and write $H=\pi^\*\pi\_\*H+\sum a'\_iE\_i$ as the pull-back of anything is $\pi$-trivial, this shows that $\sum (a\_i+a'\_i)E\_i$ is $\pi$-ample. 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.


