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This is an answer to the two more questions asked in the comments. I started it out as a comment, but got tired of the space restriction....

1. I am not certain, but you are right, $NE(X)$ for this $X$ has rank at least $2$. I think that depending on the actual cover you choose the Picard number of $X$ may or may not be larger than $2$. Here is how you can study this: I think the key is to understand the Picard group of the fibers. It seems that in the general case all of these Picard groups should have rank $1$. If that's so, then using cohomology and base change (Hartshorne, Chapter III, Section 12) you can prove that then the Picard number of $X$ is $2$. In that case $NE(X)$ can be of only one type.

If the Picard number of $X$ is larger than $2$, then it gets more complicated. For a description of what can happen for a K3 you could look at this and this papers. I think that if you identify the relative cone of the fibration, you have a good chance at figuring out $NE(X)$ or at least enough about it so you can do whatever you need this for. On the cone of CY manifolds there are some results by Wilson, Morrison, Kawamata, Totaro.

2. Yes, the Kähler cone is the dual of $NE(X)$ in the sense you wrote it. This follows from Kleiman's criterion for ampleness: If $D\subset X$ is a divisor on a smooth projective variety, then it is ample if and only if $D\cdot\sigma>0$ for any $\sigma\in\overline{NE}(X)\setminus\{0\}$. This can be found in pretty much any book dealing with higher dimensional (birational) geometry, for example here.

This is an answer to the two more questions asked in the comments. I started it out as a comment, but got tired of the space restriction....

1. I am not certain, but you are right, $NE(X)$ for this $X$ has rank at least $2$. I think that depending on the actual cover you choose the Picard number of $X$ may or may not be larger than $2$. Here is how you can study this: I think the key is to understand the Picard group of the fibers. It seems that in the general case all of these Picard groups should have rank $1$. If that's so, then using cohomology and base change (Hartshorne, Chapter III, Section 12) you can prove that then the Picard number of $X$ is $2$. In that case $NE(X)$ can be of only one type.

If the Picard number of $X$ is larger than $2$, then it gets more complicated. For a description of what can happen for a K3 you could look at The cone of curves of a K3 surface and The Cone of Curves of K3 Surfaces Revisited. I think that if you identify the relative cone of the fibration, you have a good chance at figuring out $NE(X)$ or at least enough about it so you can do whatever you need this for. On the cone of CY manifolds there are some results by Wilson, Morrison, Kawamata, Totaro.

2. Yes, the Kähler cone is the dual of $NE(X)$ in the sense you wrote it. This follows from Kleiman's criterion for ampleness: If $D\subset X$ is a divisor on a smooth projective variety, then it is ample if and only if $D\cdot\sigma>0$ for any $\sigma\in\overline{NE}(X)\setminus\{0\}$. This can be found in pretty much any book dealing with higher dimensional (birational) geometry, for example Lazarsfeld's Positivity in Algebraic Geometry I.

This is an answer to the two more questions asked in the comments. I started it out as a comment, but got tired of the space restriction....

1. I am not certain, but you are right, $NE(X)$ for this $X$ has rank at least $2$. I think that depending on the actual cover you choose the Picard number of $X$ may or may not be larger than $2$. Here is how you can study this: I think the key is to understand the Picard group of the fibers. It seems that in the general case all of these Picard groups should have rank $1$. If that's so, then using cohomology and base change (Hartshorne, Chapter III, Section 12) you can prove that then the Picard number of $X$ is $2$. In that case $NE(X)$ can be of only one type.

If the Picard number of $X$ is larger than $2$, then it gets more complicated. For a description of what can happen for a K3 you could look at this paper. I think that if you identify the relative cone of the fibration, you have a good chance at figuring out $NE(X)$ or at least enough about it so you can do whatever you need this for. On the cone of CY manifolds there are some results by Wilson, Morrison, Kawamata, Totaro.

2. Yes, the Kähler cone is the dual of $NE(X)$ in the sense you wrote it. This follows from Kleiman's criterion for ampleness: If $D\subset X$ is a divisor on a smooth projective variety, then it is ample if and only if $D\cdot\sigma>0$ for any $\sigma\in\overline{NE}(X)\setminus\{0\}$. This can be found in pretty much any book dealing with higher dimensional (birational) geometry, for example here.

This is an answer to the two more questions asked in the comments. I started it out as a comment, but got tired of the space restriction....

1. I am not certain, but you are right, $NE(X)$ for this $X$ has rank at least $2$. I think that depending on the actual cover you choose the Picard number of $X$ may or may not be larger than $2$. Here is how you can study this: I think the key is to understand the Picard group of the fibers. It seems that in the general case all of these Picard groups should have rank $1$. If that's so, then using cohomology and base change (Hartshorne, Chapter III, Section 12) you can prove that then the Picard number of $X$ is $2$. In that case $NE(X)$ can be of only one type.

If the Picard number of $X$ is larger than $2$, then it gets more complicated. For a description of what can happen for a K3 you could look at this and this papers. I think that if you identify the relative cone of the fibration, you have a good chance at figuring out $NE(X)$ or at least enough about it so you can do whatever you need this for. On the cone of CY manifolds there are some results by Wilson, Morrison, Kawamata, Totaro.

2. Yes, the Kähler cone is the dual of $NE(X)$ in the sense you wrote it. This follows from Kleiman's criterion for ampleness: If $D\subset X$ is a divisor on a smooth projective variety, then it is ample if and only if $D\cdot\sigma>0$ for any $\sigma\in\overline{NE}(X)\setminus\{0\}$. This can be found in pretty much any book dealing with higher dimensional (birational) geometry, for example here.

This is an answer to the two more questions asked in the comments. I started it out as a comment, but got tired of the space restriction....

1. I am not certain, but you are right, $NE(X)$ for this $X$ has rank at least $2$. I think that depending on the actual cover you choose the Picard number of $X$ may or may not be larger than $2$. Here is how you can study this: I think the key is to understand the Picard group of the fibers. It seems that in the general case all of these Picard groups should have rank $1$. If that's so, then using cohomology and base change (Hartshorne, Chapter III, Section 12) you can prove that then the Picard number of $X$ is $2$. In that case $NE(X)$ can be of only one type.

If the Picard number of $X$ is larger than $2$, then it gets more complicated. For a description of what can happen for a K3 you could look at this paper. I think that if you identify the relative cone of the fibration, you have a good chance at figuring out $NE(X)$ or at least enough about it so you can do whatever you need this for. On the cone of CY manifolds there are some results by Wilson, Morrison, Kawamata, Totaro.

2. Yes, the Kähler cone is the dual of $NE(X)$ in the sense you wrote it. This follows from Kleiman's criterion for ampleness: If $D\subset X$ is a divisor on a smooth projective variety, then it is ample if and only if $D\cdot\sigma>0$ for any $\sigma\in\overline{NE}(X)$. This can be found in pretty much any book dealing with higher dimensional (birational) geometry, for example here.

This is an answer to the two more questions asked in the comments. I started it out as a comment, but got tired of the space restriction....

1. I am not certain, but you are right, $NE(X)$ for this $X$ has rank at least $2$. I think that depending on the actual cover you choose the Picard number of $X$ may or may not be larger than $2$. Here is how you can study this: I think the key is to understand the Picard group of the fibers. It seems that in the general case all of these Picard groups should have rank $1$. If that's so, then using cohomology and base change (Hartshorne, Chapter III, Section 12) you can prove that then the Picard number of $X$ is $2$. In that case $NE(X)$ can be of only one type.

If the Picard number of $X$ is larger than $2$, then it gets more complicated. For a description of what can happen for a K3 you could look at this paper. I think that if you identify the relative cone of the fibration, you have a good chance at figuring out $NE(X)$ or at least enough about it so you can do whatever you need this for. On the cone of CY manifolds there are some results by Wilson, Morrison, Kawamata, Totaro.

2. Yes, the Kähler cone is the dual of $NE(X)$ in the sense you wrote it. This follows from Kleiman's criterion for ampleness: If $D\subset X$ is a divisor on a smooth projective variety, then it is ample if and only if $D\cdot\sigma>0$ for any $\sigma\in\overline{NE}(X)\setminus\{0\}$. This can be found in pretty much any book dealing with higher dimensional (birational) geometry, for example here.