I'm reading on $\overline M_{0, n}$ and $\widetilde M_{0, n}$. I know that $\overline M_{0, n}$ is a Mori Dream space for $n \leq 6$ and not a Mori Dream space for $n \geq 13$. Is there a similar result about $\widetilde M_{0, n}$ ? I thought there should be some obvious statement of the type "GIT quotients of Mori Dream spaces are themselves Mori Dream spaces" but somehow I'm missing something. I would also like to understand a counterexample to the previous statement.

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    $\begingroup$ What is $\tilde M_{0,n}$? $\endgroup$ Mar 28, 2017 at 10:09
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    $\begingroup$ Very briefly: there is a $S_n$ action on $\overline M_{0, n}$ (permuting marked points). $\widetilde M_{0, n}$ is then just a GIT quotient with respect to this action. $\endgroup$
    – Alex
    Mar 28, 2017 at 11:06
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    $\begingroup$ Then the answer to the implied question in this case is easy: since the group is finite and the variety is quasiprojective, the GIT quotient is just the usual geometric quotient, so there is a finite surjective map $\bar M_{0,n}\to \tilde M_{0,n}$. Now (normal, $\mathbb{Q}$-factorial) images of MDS are MDS, so $\tilde M_{0, n}$ is a MDS whenever $\bar M_{0, n}$ is. $\endgroup$ Mar 28, 2017 at 12:37
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    $\begingroup$ @PiotrAchinger. That is not the implication that Alex is asking about. The moduli space $\overline{M}_{0,n}$ is not a Mori dream space for $n\geq 13$. However, it remains possible that the quotient $\widetilde{M}_{0,n}$ is a Mori dream space. $\endgroup$ Mar 28, 2017 at 13:16
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    $\begingroup$ @PiotrAchinger. The divisor theory of $\widetilde{M}_{0,n}$ is considerably simpler than the divisor theory of $\overline{M}_{0,n}$. For instance, the pseudoeffective cone of $\widetilde{M}_{0,n}$ equals the effective cone, and this is simplicial with extremal rays spanned by the $\lfloor (n-2)/2 \rfloor$ irreducible components of the boundary. $\endgroup$ Mar 28, 2017 at 13:39


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