Let XX be a Deligne-Mumford stack and let XX \to X be a coarse moduli space. Suppose that X is smooth. Is XX smooth? If not, what is an example? What if XX is of finite type over C (the complex numbers)? What are conditions we can put on XX to make this true?
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asked Oct 21, 2009 at 3:05
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$\begingroup$ What does it mean for XX \to X to be a coarse moduli space? $\endgroup$– Kevin H. LinCommented Oct 21, 2009 at 14:21
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$\begingroup$ The map XX --> X is a coarse space if: 1) It is universal for maps to algebraic spaces. 2) It induces a bijection on geometric points (so kbar points for algebraically closed fields kbar). Examples: the coarse space of M_1,1 is A^1, given by the j-invariant map (this is actually a little hard to prove). Easier: the coarse space of BG is a point (or for BG over S, S). This one follows directly from the definitions. Quotients by finite groups are also easy to work out. $\endgroup$– David Zureick-BrownCommented Nov 2, 2009 at 18:03
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$\begingroup$ Links: Anton's notes (section 38 on Keel-Mori): math.berkeley.edu/~anton/written/Stacks/Stacks.pdf. Conrad: notes on Keel-Mori: math.stanford.edu/~conrad/papers/coarsespace.pdf. Alper: math.columbia.edu/~jarod/stacks_guide.pdf has a lot good pointers. $\endgroup$– David Zureick-BrownCommented Nov 2, 2009 at 18:05
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$\begingroup$ One more link: the appendix of Kai-Wei's thesis has some good stuff in it too: math.princeton.edu/~klan/academic.html $\endgroup$– David Zureick-BrownCommented Nov 2, 2009 at 19:21
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2 Answers
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The answer is yes, a singular DM stack can have a smooth coarse space. Let U=Spec(k[x,y]/(xy)) be the union of the axes in A2, and consider the action of G=Z/2 given by switching the axes: x→y and y→x. Then take XX to be the stack quotient [U/G]. This is a singular Deligne-Mumford stack (since it has an etale cover by something singular), but its coarse space is A1, which is smooth.
answered Oct 21, 2009 at 4:43
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$\begingroup$ what's the difference between this stack quotient and the quotient of the affine line (char is not 2) by x --> -x? $\endgroup$– shenghaoCommented Jan 3, 2010 at 21:16
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1$\begingroup$ The quotient of the line by that action is smooth (since it has an etale cover by something smooth, namely the affine line), but this stack is not smooth (since it has an etale cover by something non-smooth, namely the axes). $\endgroup$ Commented Jan 3, 2010 at 22:25
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I think if the coarse moduli space is smooth, so is the DM stack, because XX --> X is a gerbe, which is always smooth (since smoothness can be checked fppf locally on X, and B(G/X) is smooth over X). A stack (or a morphism of stacks, not necessarily representable) is defined to be smooth if one can find a presentation which is smooth over the base. And if it is smooth, then any presentation is smooth. That's why I got confused on Anton's example. Maybe someone can explain this to me. Thanks in advance.
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$\begingroup$ Why is XX --> X a gerbe? $\endgroup$ Commented Jan 3, 2010 at 23:07
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$\begingroup$ I seem to remember this from the book <champs algebrique> by LMB. 1. In the book, a coarse moduli space of a stack XX is defined to be the functor X sending U to the set of isom classes in XX(U). And XX->X is a gerbe. When XX is DM over k, I guess (without checking details) these two notions agree. 2. At least for any field-valued pt x:Spec k -> X, the fiber f^{-1}(x) is a fppf-gerbe over x (fppf-locally there exists an obj over x, and any two objects are isomorphic over \bar{k}, hence isom over some finite ext'n of k), and therefore smooth. Does this imply that f is smooth? $\endgroup$– shenghaoCommented Jan 4, 2010 at 0:43
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$\begingroup$ Ah, so in that case the fibers are gerbes (and I think the total space is a gerbe). In general, the coarse space does not represent the functor you wrote down; it is an intermediate step in proving that a coarse space exists, but you may still need too modify that functor more. Even for M_1,1 I believe the coarse space differs from this functor. For instance, two families of curves with the same `j-invariant' give the same map to M_1,1, but there isn't necessarily a flat cover of the base where they become isomorphic (though there is on some open subset of the base). $\endgroup$ Commented Jan 4, 2010 at 23:53
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$\begingroup$ (No guarantee on that last sentence). $\endgroup$ Commented Jan 4, 2010 at 23:54