Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $\mathcal{M}$ be an algebraic stack, and let $X$ be its coarse moduli space (assume it exists as a scheme).

We know that $h_X(Spec(k))=\mathcal{M}(Spec(k))$ if $k$ is algebraically closed. Is there anything intelligent we can say about $h_X(U)$ for a general scheme $U$?

For example, can you come up with an algorithm for knowing what $h_X(U)$ is that would be considerably easier than constructing $X$?

share|improve this question
    
You didn't need to make that question Community Wiki. It's a perfectly valid question. –  André Henriques Jul 16 '11 at 21:54
    
I'm no expert, but maybe is it something like $h_X(U)=\pi_0(\mathcal{M}(U))$ i.e. the set of isomorphism classes of objects of the groupoid $\mathcal{M}(U)$? Or is my guess totaly mistaken? –  Qfwfq Jul 16 '11 at 22:08
    
@Andre: I think I can't change it now. The reason I made it community wiki is because the question is pretty vague. There is no precise definition of an algorithm that is "considerably easier" than some other algorithm. –  James D. Taylor Jul 16 '11 at 22:17
    
@unknowngoogle: that sounds interesting! Can anyone confirm? –  James D. Taylor Jul 16 '11 at 22:18
add comment

2 Answers 2

up vote 6 down vote accepted

This is more an answer to the comment of unknowngoogle.
• The object U ↦ π0(ℳ(U)) that you described is the initial presheaf to which the stack ℳ maps.
• One can also consider the sheafification of U ↦ π0(ℳ(U)), which is the initial sheaf to which ℳ maps.
• Finally, there is the (possibly non-existent) initial representable sheaf (=scheme) to which ℳ maps.

In general, all those are different.
They are already different in the case of the affine line modded out by the involution x ↦ -x.

share|improve this answer
4  
Dear Andre -- The last statement you make here is sensitive to the topology. What you say is correct for the etale topology, but your second and third objects are the same for the fppf topology. This is because the quotient map (to the scheme quotient) is a flat cover (though not an etale cover) and the corresponding equivalence relation is the one induced by your involution. Therefore this scheme quotient is the sheaf quotient. –  JBorger Jul 17 '11 at 12:28
    
Thank you James. Very relevant comment. I was indeed thinking of the étale topology. –  André Henriques Jul 17 '11 at 20:51
3  
If you take the plane instead of the line, with the involution $(x,y) \mapsto (-x,-y)$, the quotient map is not flat anymore. –  Angelo Jul 17 '11 at 21:03
add comment

What André says is absolutely correct. The functor represented by the moduli space does not have any reasonable description, except in very particular cases. Given a map $U \to X$, it can be hard work to decide whether it comes from an object of $\mathcal M(U)$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.