0
$\begingroup$

Let $X, Y$ be $\mathbb{Q}$-factorial, projective, normal varieties. Let $f: X --> Y$ be a small birational map. I have two related questions about pushforward of an ample divisor:

(1) Let $H_X$ be an arbitrary ample $\mathbb{Q}$-divisor on $X$, and $H_Y:= f_*(H_X)$ be its pushforward, then is $H_Y$ nef on $Y$?

(2) If $H_X$ is a general ample $\mathbb{Q}$-divisor, is $H_Y$ nef (or even ample)?

I want to prove (1) as follows:

Let $p: W \to X, q: W \to Y$ be a resolution of $f$, and $H = p^*H_X$ be the pull back of $H_X$. Because $X,Y$ are $\mathbb{Q}$-factorial, the exceptional locus are divisors; and since $f$ is small, $p$-exceptional divisor is the same as $q$-exceptional divisor. Then, by the negativity lemma, and the fact that if $E$ is a exceptional divisor there must be a curve $C$, such that $E \cdot C < 0$, we can show $H = p^*H_X = q^* H_Y$.

Let $i: C \to Y$ be a curve on $Y$.

(i) If $C \not\subset q(Exc(q))$ (that is $C$ is not contained in the image of exceptional locus of $q$), we take the strict transform $C'$, and we have $$0 \leq C' \cdot H = C' \cdot q^*H_Y= q_* C' \cdot H_Y = C \cdot H_Y .$$

(ii) If $C \subset q(Exc(q))$, there should exist a curve $C' \subset Exc(q)$, such that the $p_* C' =C$, then again, we have $$0 \leq C' \cdot H = C' \cdot q^*H_Y= q_* C' \cdot H_Y = C \cdot H_Y .$$

I am not very confident about the case(ii) (i.e. the existence of $C'$).

$\endgroup$
3
$\begingroup$

Both statements are essentially never true. For example, the strict transform of an ample divisor under a simple flop is not ample anymore; this is partly worked out here: projection formula for birational map . The problem is that the divisor $D$ is going to be negative on the indeterminate curve of the inverse map.

Indeed, if the strict transform of an ample under a small birational map is ample, the map must be an isomorphism! I vaguely remember this being explained in "Cones of divisors on Calabi-Yau fiber spaces" by Kawamata.

In your proof, $p^* H_X = q^* H_Y$ will never hold if $H_X$ is an ample divisor, and holds for a nef one only if $H_X$ is $0$ on the flopping curve. See the comments in the other thread for an example.

It's also worth pointing out that taking strict transform under a small birational map gives a well-defined map even on numerical classes. So for your question (2), it doesn't make any difference whether you use a general representative or not.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Dear Mark: Thank you for your answer, I understand it now. My I ask one more question: could you explain more about "$p^*H_X = q^*H_Y $ holds for a nef one only if $H_X$ is 0 on the flopping curve"? $\endgroup$ – Li Yutong Oct 13 '14 at 2:08
  • $\begingroup$ Well, let me just explain this for the standard flop (there is probably a more general statement, but I'm not sure what it is). We can write $p^*H_X +aE =q^* H_Y$, where $E$ is the exceptional divisor. Taking the intersection with the ruling of $E$ contracted by $q$, we get that the value $a = H_X \cdot C$. Hence $p^*H_X = q^*H_Y$ if and only if $H_X \cdot C = 0$ (which can happen if $H_X$ is nef, but not if it's ample). $\endgroup$ – user47305 Oct 13 '14 at 12:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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