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) 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 (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.

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.

(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 by a result of Barkowski if $\dim \leq4$ (see here) and by [BCHM] in general.

This question occurred to me while thinking about this one.

(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 $\overline{NE}(X)$ is 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 there is such an $X$, then the first condition implies that in the definition of $Q(X)$ (see the definition of $Q(X)$ here) it is enough to take finitely many $H_i$'s and hence $\overline Q(X)$ will also be polyhedral so it could not equal $\overline{\mathrm{Mov}}(X)$.

On the other hand, regardless of that question it seemed to me that this is interesting on its own.

Here is how much I see right away:

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.

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) 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 (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.