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maybe it's true.
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user47305
user47305

No. Let $X = E \times E$ with $E$ an elliptic curve and let $G = \mathbb Z_2 \oplus \mathbb Z_2$, with each factor acting on one of the $E$'s by the involution and fixing the other. The quotient is $\mathbb P^1 \times \mathbb P^1$. The effective cone of $E \times E$ is round, while the effective cone of $\mathbb P^1 \times \mathbb P^1$ is polyhedral.

Offhand I don't see an example where $Y$ has infinitely many rays and $X$ doesn't, but it wouldn't surprise me. It'sThe converse seems true, though: if $X$ is a Mori dream space (which implieshas polyhedral effective cone), then so isdoes $Y$, spanned by the pushforwards of the generators of the cone for $X$.

No. Let $X = E \times E$ with $E$ an elliptic curve and let $G = \mathbb Z_2 \oplus \mathbb Z_2$, with each factor acting on one of the $E$'s by the involution and fixing the other. The quotient is $\mathbb P^1 \times \mathbb P^1$. The effective cone of $E \times E$ is round, while the effective cone of $\mathbb P^1 \times \mathbb P^1$ is polyhedral.

Offhand I don't see an example where $Y$ has infinitely many rays and $X$ doesn't, but it wouldn't surprise me. It's true $X$ is a Mori dream space (which implies polyhedral effective cone), then so is $Y$.

No. Let $X = E \times E$ with $E$ an elliptic curve and let $G = \mathbb Z_2 \oplus \mathbb Z_2$, with each factor acting on one of the $E$'s by the involution and fixing the other. The quotient is $\mathbb P^1 \times \mathbb P^1$. The effective cone of $E \times E$ is round, while the effective cone of $\mathbb P^1 \times \mathbb P^1$ is polyhedral.

The converse seems true, though: if $X$ has polyhedral effective cone, then so does $Y$, spanned by the pushforwards of the generators of the cone for $X$.

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user47305
user47305

No. Let $X = E \times E$ with $E$ an elliptic curve and let $G = \mathbb Z_2 Z \oplus \mathbb Z_2$$G = \mathbb Z_2 \oplus \mathbb Z_2$, with each factor acting on one of the $E$'s by the involution and fixing the other. The quotient is $\mathbb P^1 \times \mathbb P^1$. The effective cone of $E \times E$ is round, while the effective cone of $\mathbb P^1 \times \mathbb P^1$ is polyhedral.

Offhand I doubt there aredon't see an example where $Y$ has infinitely many positive results of this sort in either directionrays and $X$ doesn't, but it wouldn't surprise me. It'sIt's true $X$ is a Mori dream space (which implies polyhedral effective cone), then so is $Y$.

No. Let $X = E \times E$ with $E$ an elliptic curve and let $G = \mathbb Z_2 Z \oplus \mathbb Z_2$, with each factor acting on one of the $E$'s by the involution and fixing the other. The quotient is $\mathbb P^1 \times \mathbb P^1$. The effective cone of $E \times E$ is round, while the effective cone of $\mathbb P^1 \times \mathbb P^1$ is polyhedral.

I doubt there are many positive results of this sort in either direction. It's true $X$ is a Mori dream space (which implies polyhedral effective cone), then so is $Y$.

No. Let $X = E \times E$ with $E$ an elliptic curve and let $G = \mathbb Z_2 \oplus \mathbb Z_2$, with each factor acting on one of the $E$'s by the involution and fixing the other. The quotient is $\mathbb P^1 \times \mathbb P^1$. The effective cone of $E \times E$ is round, while the effective cone of $\mathbb P^1 \times \mathbb P^1$ is polyhedral.

Offhand I don't see an example where $Y$ has infinitely many rays and $X$ doesn't, but it wouldn't surprise me. It's true $X$ is a Mori dream space (which implies polyhedral effective cone), then so is $Y$.

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user47305
user47305

No. Let $X = E \times E$ with $E$ an elliptic curve and let $G = \mathbb Z_2 Z \oplus \mathbb Z_2$, with each factor acting on one of the $E$'s by the involution and fixing the other. The quotient is $\mathbb P^1 \times \mathbb P^1$. The effective cone of $E \times E$ is round, while the effective cone of $\mathbb P^1 \times \mathbb P^1$ is polyhedral.

I doubt there are many positive results of this sort in either direction. It's true $X$ is a Mori dream space (which implies polyhedral effective cone), then so is $Y$.