Q:

Cone of movable curves

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.

A:

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. 

A:

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.

