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Well, the title almost says it all. I would like to list as many examples as possible of moduli functors, for which a coarse moduli space does not exist (and maybe explain why). So, examples such as $[\mathbb{A}^1/G_m]$ are not what I mean. I would really like to see moduli functors that come from some classifying problem.

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3 Answers 3

up vote 7 down vote accepted

$\overline{\mathfrak{M}}_{0,0}$, the stack of pre-stable genus zero curves. This parameterizes genus zero curves having at worst nodal singularities. There are an infinite number of isomorphism classes of such curves, but they all occur as a specialization of a trivial family (just do repeated blowups of the central fiber of the family $\mathbb{P}^1\times \mathbb{A}^1 \to \mathbb{A}^1$ ).

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There is an example in Harris-Morrison Moduli Of Curves (Exercise 1.7): the moduli functor $F:\mathfrak{Sch}/\mathbb C\to\textrm{Sets}$ that sends a scheme $B$ to the set of isomorphism classes of (flat $B$-families of) reduced plane curves of degree $2$. The point is to show that there exists a "universal" natural tranformation $\eta:F\to \hom_{\mathbb C}(-,\textrm{Spec }\mathbb C)$. So by the uniqueness of the moduli solution, $\textrm{Spec }\mathbb C$ is the unique reasonable candidate as a moduli space for this problem. But unfortunately $F(\textrm{Spec }\mathbb C)$ has two $\mathbb C$-points (the class of $\mathbb P^1$ and the class of a double line), while $\textrm{Spec }\mathbb C$ has only one.

It's likely that one can also find an example where a classifying space exists, but is not universal.

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Classifying space: you mean a scheme that parametrizes the whole thing? Like $\mathbb{P}^5 - Ven_2(\mathbb{P}^2)$ in your case. –  IMeasy Feb 10 '13 at 17:49
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I mean a scheme $M$ such that $M(\mathbb C)$ is in bijection with $F(\textrm{Spec }\mathbb C)$. I'm sorry if it is not the right word. –  Brenin Feb 10 '13 at 19:06
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I think Brenin's example is valid: isomorphism class probably means "up to projective equivalence". In this case, this stack is an open substack of Jim Bryan's stack $\mathfrak{M}_{0,0}$ above, and fails to have a coarse moduli space for the same reason. –  Jason Starr Feb 11 '13 at 0:57
    
@Brenin: no problem, it is a cute example! –  IMeasy Feb 11 '13 at 12:41

Here is a classifying problem without a coarse moduli space: Let $F$ be the stack of line bundles with section, say over an algebraically closed field k. That is $F(X)$ is the category of pairs $(L,s)$ where $L$ is a line bundle on $X$ and $s \in \Gamma(X,L)$ is a section of $L$. Then $F$ has no coarse moduli space: if it did, there would be two points, corresponding to $(k,1)$ and $(k,0)$ with the former specializing to the latter. This would be an affine scheme with a closed point and a generic point, both with residue field $k$. That's impossible.

Of course, $F$ is just a geometric description of $[\mathbf{A}^1 / \mathbf{G}_m]$, so we already knew it didn't have a coarse moduli space.

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