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 $\pi\_0([X/G]) = Spec R^G$? 



coarse moduli space and $\pi_0$
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.


