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Let $X$ be a projective variety and $G$ a finite group acting on $X$. We consider the quotient $\pi:X\rightarrow Y :=X/G$.

I'm interested in the relation between $Eff(X)$ and $Eff(Y)$. In particular, is it true that if $Eff(X)$ has infinitely many extremal rays then $Eff(Y)$ has infinitely many extremal rays as well?

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Here is an example with $Pic(X)$ finitely generated. Take $X = \overline{M}_{1,n}$ the moduli space of genus one curves with $n$ marked points. Then $Pic(\overline{M}_{1,n})$ is finitely generated.

Now, by Theorem $1.1$ of this paper:

https://www2.bc.edu/dawei-chen/Extremal.pdf

we have that $Eff(\overline{M}_{1,n})$ is not finitely generated. Now, consider the action of the symmetric group $S_n$ on $\overline{M}_{1,n}$ permuting the markings, and the quotient $\widetilde{M}_{1,n}:=\overline{M}_{1,n}/S_n$. Then, by Theorem $5.1$ of the same paper we get that $Eff(\widetilde{M}_{1,n})$ is the closed, simplicial cone generated by the boundary divisors.

Therefore, $Eff(\overline{M}_{1,n}/S_n)$ is finitely generated while $Eff(\overline{M}_{1,n})$ is not.

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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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  • $\begingroup$ Thanks! Do you have an example with $Pic(X)$ finitely generated? $\endgroup$
    – user68440
    Commented Feb 24, 2015 at 16:09
  • $\begingroup$ You could try $Y$ the blow-up of $\mathbb P^2$ at $4 \leq k \leq 8$ points, and $X$ the double-cover branched over a conic missing all the points. Certainly $Y$ has polyhedral cone, but $X$ is $\mathbb P^1 \times \mathbb P^1$ blown up at $2k$ points, which is isomorphic to $\mathbb P^2$ blown up at $2k+1$ points. My guess is that $X$ can have infinitely many rays, but there's something to check since the resulting $2k+1$ points aren't general. For $k=8$ I think it's easy to see this works, though I'm not sure about $k=4$. $\endgroup$
    – user47305
    Commented Feb 24, 2015 at 17:05

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