I am trying to understand the statement that a Deligne-Mumford stack is locally a quotient $[U/G]$, where $G$ is a finite group. I don't understand why you can make $G$ a finite group, instead of a general group scheme. For example, if $G$ is a nonconstant étale group scheme over $S$, and $\mathcal{X}=[S/G]$ a quotient stack over $S$, how to make $\mathcal{X}$ a group quotient?

Intuitively, I feel the difference between a finite group and a finite étale group scheme is like the difference between a constant sheaf and a local system. But I am not sure whether it is true. Hope someone can help me to understand these group scheme things. It would be helpful if you can provide some reference.

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    $\begingroup$ A Deligne-Mumford stack is only a finite quotient locally in the etale topology of the coarse moduli space (cf. Abramovich-Vistoli). Thus, since you are working locally in the etale topology anyway, you may as well assume that your etale-locally-constant group scheme is a constant group scheme. $\endgroup$ – Jason Starr Feb 18 '15 at 22:50
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    $\begingroup$ Yes, my intuition is this. However, I am not very familiar with group schemes, and I lack the reference. I want to check with the exports here. $\endgroup$ – user38276 Feb 19 '15 at 0:52

The way I think about finite etale group schemes over a connected scheme S, is that it's a group object in the category of schemes finite etale over S. This latter category is well-known to be equivalent to the category of finite sets X together with a action of $\pi_1(S,s)$ (pick some geometric point $s\in S$). This is described in many sources, though the canonical reference would probably be SGA 1, Expose V. Thus, a finite etale group scheme is just a finite (abstract) group $G$, together with an action of $\pi_1(S,s)$ acting on $G$ via group automorphisms. Expose XI of SGA 1 also talks a little about group schemes.

A finite group is just the group $G$ without the action of $\pi_1(S,s)$. Of course, you can always have $\pi_1(S,s)$ act on any group trivially, giving $G$ the structure of a constant group scheme over $S$ (ie, just $|G|$ disjoint copies of $S$, corresponding to a trivial degree $|G|$ cover of $S$), so in that sense any abstract group can be thought of as a constant group scheme.

As an example, you can consider the difference between $\mathbb{Z}/N\mathbb{Z}$ (integers mod $N$) and $\mu_N$ ($N$th roots of unity). They can both be thought of as group schemes over $\mathbb{Q}$, but whereas the former doesn't have a natural action by $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$, the latter does. Indeed, usually people think of $\mu_N$ over $\mathbb{Q}$ as $\text{Spec }\mathbb{Q}[x]/(x^N-1)$, and $x^N-1$ will only factor into degree 1 polynomials over $\mathbb{Q}$ if $N \le 2$. When $N\ge 3$, the irreducible factors of degree $> 1$ correspond to connected components of the cover $\mu_N\rightarrow \text{Spec }\mathbb{Q}$ which themselves are nontrivial covers of $\mathbb{Q}$, also showing that $\mu_N$ is nonconstant. The action of $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ can be seen as acting on the geometric fiber of $\mu_N$ over $\overline{\mathbb{Q}}$, where you can see explicitly that the action is just the typical action sending primitive $N$th roots of unity to other primitive $N$th roots of unity. Contrast this with the situation over $\mathbb{C}$, where for any $N$, $x^N-1$ will factor into linear polynomials (corresponding to the fact that $\pi_1(\text{Spec }\mathbb{C})$ is trivial), making $\mu_N/\mathbb{C}$ the trivial degree $N$ cover, hence constant.

A DM stack can be thought of as locally being $[U/G]$ ($G$ a finite group) because etale locally, any finite etale group scheme is constant (and quotienting out by a constant group scheme is the same as quotienting by a group - think about what the quotient means on functors of points). To see that every finite etale group scheme is locally constant, note that an action of $\pi_1(S)$ on $G$ corresponding to some group scheme $\mathcal{G}$ is the same as a homomorphism $\pi_1(S)\rightarrow S_{|G|}$. The kernel of this homomorphism is a finite index subgroup of $\pi_1(S)$, corresponding to a scheme $T$ finite etale over $S$. The pullback of this homomorphism gives an exact sequence $$\pi_1(T)\rightarrow\pi_1(S)\rightarrow S_{|G|}$$ since $T$ was defined by the property that $\pi_1(T)$ is the kernel, so the composition is 0. Thus the homomorphism $\pi_1(T)\rightarrow S_{|G|}$ thought of as the pullback of the homomorphism $\pi_1(S)\rightarrow S_{|G|}$ to $T$ corresponds to the pullback of $\mathcal{G}$ to $T$, which is constant because the associated $\pi_1(T)$-action is trivial.

  • $\begingroup$ Thank you very much. This is exactly the explanation I am looking for. $\endgroup$ – user38276 Feb 19 '15 at 0:52

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