In Kapranov's marvelous paper Chow quotients of Grassmannian I, he proves that $\overline{M}_{0,n}$ is isomorphic to both the Hilbert quotient and Chow quotient $(\mathbb{P}^1)^n//\text{SL}_2$. These quotients are by definition closed subvarieties of the Hilbert scheme and Chow variety (respectively) of $(\mathbb{P}^1)^n$. Every Chow variety has a natural polarization coming from Chow forms, and Hilbert schemes have a family of polarizations. Is it known what ample divisors on $\overline{M}_{0,n}$ we get by restricting these polarizations via Kapranov's construction?
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asked Apr 5, 2011 at 15:06
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$\begingroup$ There are some papers on this question. For starters, have you seen Alexeev and Swinarski's paper math.uga.edu/~davids/0812.0778.pdf? $\endgroup$– J.C. OttemCommented Apr 5, 2011 at 20:55
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$\begingroup$ Indeed, it is a nice paper. The Chow quotient maps to all the GIT quotients, so the polarizations on the latter pull back to the former, namely M_{0,n}, but the Chow-GIT morphisms are contractions (for n > 6), so come from nef but not ample divisors. So the Alexeev-Swinarski divisors cannot be the Hilbert or Chow polarizations (though perhaps their convex hull contains these ample divisors?...) $\endgroup$– Noah GiansiracusaCommented Apr 5, 2011 at 22:37
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