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This question was prompted by my almost-answer to the question Does smooth and proper over $\mathbb Z$ imply rational? , but I never got around to asking this until now.

It is well known that $M_g$, the moduli space of curves, is a smooth stack. In fact it is globally a quotient of a smooth variety by the action of a finite group: if we take curves with level $n \geq 3$ structure, then we get a fine moduli space, and the quotient by $\mathrm{Sp}(2g,\mathbf{Z}/n)$ recovers $M_g$. However, we know that $M_g$ is in fact smooth over $\mathrm{Spec}(\mathbf Z)$, but this construction does not work over the integers: we need $n$ not divisible by the characteristic. This leads to the first question:

Question (a): Could it be true that $M_g = [X/G]$ where $X$ is a scheme, smooth over the integers, and $G$ is a finite group?

It is very possible that this question is too naive. Maybe there is a standard argument for why this can not be true, involving structure of group schemes and wild inertia and whatnot. Nevertheless one can ask for something weaker:

Question (b): Is there a scheme $X$, smooth over the integers, with a finite map $X \to M_g$? What if we only ask it to be proper and generically finite?

A (presumably harder) question is what happens over the boundary, i.e. if one replaces $M_g$ with $\overline M_g$ or $\overline M_{g,n}$. As I mentioned in my answer to the question linked above, it is known that over certain $\mathrm{Spec}(\mathbf Z[\frac 1 d])$ one can write $\overline M_g$ and $\overline{M}_{g,n}$ as global quotients by actions of finite groups, now using non-abelian level structures.

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  • $\begingroup$ In the second part of Question (b), I recommend you change "generically finite" to "proper and generically finite". $\endgroup$
    – Jason Starr
    Commented Jun 12, 2012 at 12:56
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    $\begingroup$ For g=1 the answer is negative. The moduli stack of elliptic curves over $\mathbb Z$ is simply connected: no etale covers. It becomes non-simply connected if you invert 2 ($\pi_1^{et}$ has order 24), or if you invert 3 ($\pi_1^{et}$ has order 12). $\endgroup$
    – André Henriques
    Commented Jun 12, 2012 at 15:16
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    $\begingroup$ Erratum: the groups whose order I listed above are the kernels of $\pi_1^{et}(M_1[1/2])\to \pi_1^{et}(Spec(\mathbb Z[1/2]))$ and $\pi_1^{et}(M_1[1/3])\to \pi_1^{et}(Spec(\mathbb Z[1/3]))$. $\endgroup$
    – André Henriques
    Commented Jun 12, 2012 at 15:34
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    $\begingroup$ Dear André, the question does not require $X$ to be étale over the stack. $\endgroup$
    – Angelo
    Commented Jun 12, 2012 at 16:08
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    $\begingroup$ @Angelo: you are right. Computing $\pi_1^{et}$ isn't enough. Is there some version of $\pi_1$ than controls this problem? $\endgroup$
    – André Henriques
    Commented Jun 12, 2012 at 18:38

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Here's a suggestion in the genus $g=2$ case. If one considers a genus 2 Riemann surface, then it is well-known that it is hyperelliptic, and the fixed points of the hyperelliptic involution are 6 Weierstrauss points. If one quotients by the hyperelliptic involution, the quotient is a Riemann sphere with 6 distinguished points which are the images of the Weierstrauss points. So moduli space is isomorphic to the space of 6 points on a sphere, up to conformal equivalence. There is a finite cover of this, which is the space of 6 marked points, up to conformal equivalence. Thus, moduli space is $$ S_6 \backslash ((\mathbb{CP}^1)^6- \Delta)\ /\ PSL(2,\mathbb{C}),$$ where $\Delta$ denotes the large diagonal, and is cut out by $\Delta = \{ (z_1,\ldots, z_6) | z_i \neq z_j\ for\ i\neq j\}$, and $S_6$ acts by permuting coordinates and $PSL(2,\mathbb{C})$ acts coordinatewise . Since $PSL(2,\mathbb{C})$ acts faithfully on ordered triples, we may normalize the last three coordinates to be $0, 1, \infty$, so that $$ ((\mathbb{CP}^1)^6- \Delta) / PSL(2,\mathbb{C}) \cong \{ (z_1,z_2,z_3)\in \mathbb{C}-\{0,1\} | z_i\neq z_j, i\neq j\}.$$ This is isomorphic to an affine variety defined over $\mathbb{Z}$ by the usual trick of introducing three new coordinates and equations $(z_i-z_j)t_{ij}=1$, $i\neq j \in \{1,2,3\}$ and has a quotient by $S_6$ which is isomorphic to $M_2$. The group elements of $S_6$ act as integral fractional transformations.

So the suggestion is to take the affine scheme defined by the spectrum of the coordinate ring of this affine variety defined over $\mathbb{Z}$. But I don't know enough about schemes or stacks to know if this works for what you want to do.

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  • $\begingroup$ You might as well start with $M_{0,4}\sim M_{1,1}$. $\endgroup$
    – Ben Wieland
    Commented Jun 13, 2012 at 2:39
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    $\begingroup$ This unfortunately doesn't work. There are two problems here, one minor and one larger. The minor problem is that $M_{0,6}/S_6 \cong M_2$ is not actually an isomorphism of stacks, only of coarse moduli spaces. (Every point on the right hand side has twice as big isotropy group because of the hyperelliptic involution, which is not visible on the left hand side). The more serious problem is that these assertions are false in characteristic two. To prove that the canonical map $C \to \mathbf P^1$ has exactly six branch points you need Riemann--Hurwitz, and if the characteristic two there might... $\endgroup$
    – Dan Petersen
    Commented Jun 13, 2012 at 7:20
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    $\begingroup$ ...be wild ramification which throws things off. The same thing happens for elliptic curves in characteristic two. Incidentally, this is a special case of the level structures I mentioned in the beginning of my question, since for a level two curve an ordering of the six Weierstrass points is exactly the same thing as a full level two structure. Proof: the divisors $\{ \pm p \mp q\}$, where $p$ and $q$ varies over the Weierstrass points of the curve, give rise to distinct 2-torsion points in the Jacobian. There are $\binom 6 2$ such divisors, which is exactly $2^4-1$, so this exhibits all... $\endgroup$
    – Dan Petersen
    Commented Jun 13, 2012 at 7:23
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    $\begingroup$ ...of them. So the "minor problem" is really the fact that you need level at least three to get a fine moduli space, and the "major problem" is that level structures won't cut it unless the level and characteristic are coprime. $\endgroup$
    – Dan Petersen
    Commented Jun 13, 2012 at 7:25
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    $\begingroup$ I don't think you should delete it. It's a nice example of the issues one runs into even if it doesn't work. (Also, "level two curve" in my second comment above should of course be "genus two curve".) $\endgroup$
    – Dan Petersen
    Commented Jun 13, 2012 at 15:12

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