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 $C'$).


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.

  • $\begingroup$ 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

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