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Mar 29, 2017 at 0:28 comment added user47305 According to Maksym Fedorchuk in arxiv.org/abs/1407.7839 (pg 2), this is an open question in general (or was, as of the last update in June 2015). Obviously this is a busy area lately, so maybe things have changed.
Mar 28, 2017 at 13:39 comment added Jason Starr @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.
Mar 28, 2017 at 13:24 comment added Piotr Achinger That comment was referring to the part "GIT quotients of Mori Dream spaces are themselves Mori Dream spaces" but maybe I misunderstood. I should have written "one of the implied questions"; the other direction seems more difficult. Maybe one could follow the original strategy of Castravet–Tevelev and reduce to the (quotient by $S_{n-3}$ of the) Losev–Manin toric variety blown up at a point in the open torus orbit?
Mar 28, 2017 at 13:16 comment added Jason Starr @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.
Mar 28, 2017 at 12:42 comment added Piotr Achinger (For the statement about images of MDS cf. S. Okawa "On images of Mori dream spaces" Math. Ann. (2016) 364:1315.)
Mar 28, 2017 at 12:37 comment added Piotr Achinger 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.
Mar 28, 2017 at 11:06 comment added Alex 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.
S Mar 28, 2017 at 11:03 history suggested Lazzaro Campeotti CC BY-SA 3.0
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Mar 28, 2017 at 11:01 review Suggested edits
S Mar 28, 2017 at 11:03
Mar 28, 2017 at 10:09 comment added Piotr Achinger What is $\tilde M_{0,n}$?
Mar 28, 2017 at 9:57 history asked Alex CC BY-SA 3.0