Take the 2-minute tour ×

Let $X$ be a smooth complex projective variety of dimension $n$.

Under the duality between $N_1(X)$ and $N^1(X)$ we know that closure of cone of effective curves $\overline{NE}(X)$ is dual to closure of ample cone $\overline{Amp}(X)$.

It was proved in 2004 that the closure of cone of effective divisors $\overline{Eff}(X)$ is dual to the closure of cone of movable curves $\overline{Mov}(X)$. A movable curve by definition is a curve class $C \in N_1(X)$ such that $C=\pi_*(H_1 H_2 \cdots H_{n-1})$, where $\pi: X' \rightarrow X$ is a birational morphism and $H_i$'s are ample classes on $X'$.

My question: Let $Q(X)$ be the cone obtained by curve classes $H_1 H_2 \cdots H_{n-1}$ where $H_i$ are ample divisors on $X$ itself. Is it true/false that $\overline{Q}(X)=\overline{Mov}(X)$? i.e. as long as I am interested only in the closure of these cones; do I really miss some curve class if I only restrict my self to intersection of ample classes on $X$ itself.

Can any body give an example where $\overline{Q}(X) \neq \overline{Mov}(X)$?

Meanwhile, I am only interested in $n=3$ case.

share|improve this question
add comment

3 Answers

up vote 3 down vote accepted

A more direct approach is the following:

Let $X$ be the projective bundle $\pi:\mathbb{P}(\mathcal{E})\to \mathbb{P}^1$ where $\mathcal{E}=\mathcal{O}\oplus \mathcal{O}(-1) \oplus \mathcal{O}(-2)$. Let $M$ be the tautological bundle of $X$. It is easily checked that the ample line divisors $H_i$ on $X$ correspond to line bundles of the form $M^a\otimes \pi^*\mathcal{O}(b)$ with $b>2a$. We show that $Mov(X)$ is not spanned by products of the form $H_1\cdot H_2$.

Consider the line bundle $L=M\otimes \pi^*\mathcal{O}(-1)$. Using the Leray spectral sequence for the morphism $\pi$ we easily see that $L$ is not pseudoeffective. However, it is also straightforward to check that $$L\cdot H_1\cdot H_2=b_1+b_2-4>0$$ for $H_i=M\otimes \pi^*\mathcal{O}(b_i)$. Hence $L$ lies in the dual cone of $\overline{Q}(X)$ (using your notation). Now, if $\overline{Mov}(X)$ was generated by the $H_1\cdot H_2$'s this would imply that $L$ is pseudoeffective (by BDPP), a contradiction.

share|improve this answer
add comment

I think following post in the mathoverflow gives an answer:

Effective versus movable cones of curves

There, people mention that there is an example where the Ample cone is rational polyhedral but movable cone is not. But if Ample cone is polyhedral then $\overline{Q}(X)$ would be polyhedral too and so they can not be equal.

share|improve this answer
add comment

The definition you give for movable is actually the definition for strongly movable. A movable curve on X is one that lives in a family that covers a dense open subset of X, or the closure of the convex span of this concept. That the definitions are equivalent is a strong result.

I am mentioning this because I needed the other definition, and Google brought me here. Cheers.

share|improve this answer
add comment

Your Answer


