I've been reading this really nice paper by Alper http://math.columbia.edu/~jarod/good_moduli_spaces.pdf, and there's a question that doesn't seem to be answered (perhaps it's not relevant).

Any stack F has a corresponding `sheaf of connected components' (or sheaf of isomorphism classes), by taking $\pi_0^{pr}(F)(S) = \pi_0 (F(S))$ and then sheafifying. (where $\pi_0$ of a groupoid, or more generally a category, is the set of isomorphism classes)

If $X$ is an Artin stack (although I'm currently more interested in DM stacks) and $X$ admits a good moduli space, then is $X \to \pi_0(X)$ a good moduli space?

Also, when is the good moduli space a scheme (and not just an algebraic space)?

And finally, if $X$ = $Spec R$ is affine and $G$ acts on it (I'm mainly interested in the $G$ finite case), is $\pi_0([X/G]) = Spec R^G$?

  • $\begingroup$ Why do you expect there to be any kind of relation between moduli spaces and connected components ? If $X$ is a connected scheme over a field $k$, then I guess that $X$ is a good moduli space of itself while $\pi_0(X)$ is something like $Spec(k)$. $\endgroup$ Mar 14, 2012 at 20:18
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    $\begingroup$ apologies for the confusing notation: if X is a sheaf on th, say, etale site of affine schemes. Then X(R) is a set for all rings R. Understand this set as a discrete topological space, therefore it makes sense to take the connected components of it. This gives me back the same set!(as you wrote: a scheme is a good moduli space of itself). If X is a stack of groupoids, then X(R) is a groupoid which (by taking nerves) can be interpreted as a (1-connected) topological space: it makes sense to consider $\pi_1$ (automorphisms), $\pi_0$ (connected components), so $\pi_0(X)$ is the shf closest to X. $\endgroup$ Mar 14, 2012 at 20:42
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    $\begingroup$ I'm guessing the OP means pi_0 in the groupoid sense, so that pi_0 of a scheme is itself. $\endgroup$
    – David Roberts
    Mar 14, 2012 at 20:43
  • $\begingroup$ Beat me to it... $\endgroup$
    – David Roberts
    Mar 14, 2012 at 20:44
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    $\begingroup$ I think the question is yes, but I'll check properly before I answer. $\endgroup$
    – David Roberts
    Mar 16, 2012 at 0:06

1 Answer 1


You have probably already come up with the answer yourself, but I just thought the question shouldn't hang around unanswered in the forum.

What you call "the sheaf of connected components", I would call the coarse sheaf of the stack or the sheaf associated to the stack. It is usually not representable by an algebraic space. When it is, the stack is called a gerbe and the map to the coarse sheaf is called the structure morphism of the gerbe.

A gerbe with structure morphism $X \to Y$ is fppf locally on $Y$ of the form $B_YG := [Y/G]$, where $G$ is a group-algebraic space which is fppf over $Y$, and the action on $Y$ is trivial. (In fact, it is even étale locally on this form, since the structure morphism $X \to Y$ of a gerbe is smooth.)

Although gerbes certainly have at lot of good properties, the structure morphisms of a gerbe need not be a good moduli space in the sense of Alper. It is, exactly when $G$ above is linearly reductive.

The property of being a gerbe as a very strong property. Thus good (or a coarse) moduli spaces are seldom coarse sheaves. In particular $Spec\ R^G$ will usually not be the coarse sheaf of $[Spec\ R / G]$. Taking the stack quotient and then taking the associated sheaf, is the same as taking the sheaf quotient directly. The result is seldom (never?) representable unless the action of $G/N$ is free, where $N$ is the kernel of the action.


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