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Let $\overline{NE}(X)$ be the closure of the cone generated by the numerical classes of effective curves and $\overline{\mathrm{Mov}}(X)$ the closure of the cone of moving curves.

(Q1) Q) Is there an example of a smooth projective variety $X$ such that

• $\overline{NE}(X)$ is (finite) polyhedral, but
• $\overline{\mathrm{Mov}}(X)$ is not?

Here are some trivial observation:

## 1

If $X$ is a surface then the two cones are dual to each other and hence they are either both polyhedral or not.

## 2

If $X$ is a Fano variety then

• $\overline{NE}(X)$ is polyhedral by the Cone Theorem, and
• $\overline{\mathrm{Mov}}(X)$ is polyhedral if $\dim \leq4$ by a result of Barkowski if $\dim \leq4$ (see here) .

So, perhaps I should really ask:

(Q2) Is it true that $\overline{NE}(X)$ is polyhedral if and only if $\overline{\mathrm{Mov}}(X)$ is polyhedral?

If true, this would (for example) provide a proof of Barkowski's result by [BCHM] in all dimensionsgeneral.

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(Q1) Is there an example of a smooth projective variety $X$ such that

• $\overline{NE}(X)$ is (finite) polyhedral, but
• $\overline{\mathrm{Mov}}(X)$ is not?

Here

Let $\overline{NE}(X)$ is be the closure of the cone generated by the numerical classes of effective curves and $\overline{\mathrm{Mov}}(X)$ is the closure of the cone of moving curves.

(See the mentioned MO question for more details)

The way this relates to the question asked in that MO question is that if Q1) Is there is such an $X$, then the first condition implies that in the definition example of a smooth projective variety $Q(X)$ (see the definition of X$such that •$Q(X)$here) it \overline{NE}(X)$ is enough to take finitely many $H_i$'s and hence $\overline Q(X)$ will also be (finite) polyhedralso it could not equal $\overline{\mathrm{Mov}}(X)$.

On the other hand, regardless of that question it seemed to me that this but

• $\overline{\mathrm{Mov}}(X)$ is interesting on its own.not?

Here is how much I see right awayare some trivial observation:

## 1

If $X$ is a surface then the two cones are dual to each other and hence they are either both polyhedral or not.

## 2

If $X$ is a Fano variety then

• $\overline{NE}(X)$ is polyhedral by the Cone Theorem, and
• $\overline{\mathrm{Mov}}(X)$ is polyhedral if $\dim \leq4$ by a result of Barkowski (see here).

So, perhaps I should really ask:

(Q2) Is it true that $\overline{NE}(X)$ is polyhedral if and only if $\overline{\mathrm{Mov}}(X)$ is polyhedral?

If true, this would (for example) provide a proof of Barkowski's result in all dimensions.

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