Given a moduli problem, it appears that nonexistence of automorphisms is a necessary condition for existence of a fine moduli space(is this strictly true?).

In any case, assuming the above, what additional condition on a moduli problem in algebraic geometry will make sure that a coarse moduli space is in fact a fine moduli space?

In the n-lab page on Deligne-Mumford, the following appears.

Deligne-Mumford stacks correspond to moduli problems in which the objects being parametrized have finite automorphism groups.

Also, a few problematic examples I have heard of, had infinite automorphism groups.

Therefore, is it true that for a moduli problem in which the stack is Deligne-Mumford, and where there are no automorphisms, existence of a coarse moduli space would imply the existence of a fine moduli space?

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    $\begingroup$ Anweshi's usage of moduli problem is standard. $\endgroup$ Jan 3, 2010 at 2:51
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    $\begingroup$ Fine, fine, I guess I'm just getting cranky in my old age. I think I just found the question generally a bit hard to read, and was trying to put a finger on why. $\endgroup$
    – Ben Webster
    Jan 3, 2010 at 3:31
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    $\begingroup$ Well, for me the issue is that I can't quite sort out what it is you really want to know, and perhaps more importantly, which facts you take for granted. For example, reading the question, I have no idea whether you would view "a coarse moduli space is fine if and only if a fine moduli space exists" as annoyingly obvious, or genuinely helpful. $\endgroup$
    – Ben Webster
    Jan 3, 2010 at 5:42
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    $\begingroup$ @Ben: Be nice; that's not genuinely helpful to anyone, and these are totally legit questions. $\endgroup$ Jan 3, 2010 at 5:54
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    $\begingroup$ @Anweshi: One thing that helped me learn stacks was doing very basic exercises about fibered categories; once you're comfortable with the basic definitions many things (like parts of your question, or that say BG is the moduli space of G-torsors) become tautological. One source of these is the course notes from the Deformation Theory Workshop at MSRI (only some of them are note relevant): msri.org/people/members/defthy07/exercises.html $\endgroup$ Jan 3, 2010 at 19:27

4 Answers 4


I think this is an instructive question. Here are some partial answers.

A category fibered in groupoids whose fibers are sets (e.g. no automorphisms) is a presheaf. Strictly speaking, I mean equivalent to the fibered category associated to a presheaf. This truly follows from the definitions, and is a good exercise to do when one is learning the basic machinery behind stacks. Similarly, a stack which is equivalent to a presheaf is a sheaf (i.e. the descent condition collapses to the sheaf condition; this is a little harder but still tautological). And a pre-stack with no automorphisms is a separated pre-sheaf.

The other claim is more interesting and not tautological, and reflects the fact that the diagonal of a morphism of stacks is way more interesting than in the case of schemes. For instance, an algebraic stack is DM iff its diagonal is unramified. (I believe this is in Champs Algebriques, but there's a nice discussion in Anton's notes from Martin Olsson's stacks course.

The point is that the diagonal of a stack carries information about automorphisms of the objects it parameterizes. So for instance the condition (in the definition of a stack) that the diagonal is representable is equivalent to the statement that Isom(X,Y) (and thus Aut(X)) is representable by an algebraic space. Also tautological is the statement that the diagonal being unramified is equivalent to the statement that there are no infintesmal automorphisms (e.g. non-trivial automorphisms of an object over $k[\epsilon]/\epsilon^2$ which reduce to the identity map over $k$). So here it is now clear where one uses finiteness: the $k[\epsilon]/\epsilon^2$ points of a finite groups scheme are the same as the $k$ points; on the other hand this fails if the automorphism scheme is, say, $\mathbf{G}_m$.

Finally, while the tautological answer above answers your question, it is instructive to see how automorphisms cause $M_g$ (moduli of genus g curves) to not be representable. Let H be a hyperelliptic curve given by $y^2 = f(x)$ defined over a field $k$. Then the curve $H_d$ given by $dy^2 = f(x)$ is not isomorphic to $H$ over $k$ if d is not a square in k. Call this a `twist' of H; in general twists of a variety X are given by the Galois cohomology group $H^1(G_k,Aut X)$ which is non-zero in non-trivial situations (alternatively you can use torsors and etale cohomlogy), and a generic hyperelliptic curve has automorphism group $\{\pm 1\}$. Thus H and $H_d$ give two different $k$ points of $M_g$ which become the same point over a finite extension; thus $M_g$ fails the sheaf condition in the etale topology, (and so in general existence of automorphisms causes, for cohomological reasons, failure of your moduli problem to even be a sheaf).

Last comment (to clarify others' comments): fine moduli space should certainly allow algebraic spaces for a correct answer to your question.

  • $\begingroup$ To be pedantic: maybe one should say that for etale group schemes, the $k[\epsilon]/\epsilon^2]$ points are the same as the $k$-points. One could imagine having a non-DM stack with stabilizer groups that are finite non-etale group schemes. (Of course, this distinction only occurs in positive characteristic, so may be esoteric for non-number theorists.) $\endgroup$
    – Emerton
    Jan 3, 2010 at 5:22
  • $\begingroup$ Ah, you are right, and not just being pedantic. I was thinking of the special case of G a discrete group (so actually a disjoint union of |G| many copies of the base). $\endgroup$ Jan 3, 2010 at 5:57
  • $\begingroup$ Those are called "constant group schemes". $\endgroup$
    – S. Carnahan
    Jan 3, 2010 at 18:27
  • $\begingroup$ @David. Thanks for your answer. I accepted it because it is the best of so far, but I hope if someone answers it more fully then I can change the accepted answer. $\endgroup$
    – Anweshi
    Feb 5, 2010 at 12:58
  • $\begingroup$ @Also thanks a lot for your helpful comments in getting this question(and me) accepted in MO. $\endgroup$
    – Anweshi
    Feb 5, 2010 at 12:58

Since nobody gave a reference yet, in my paper "Artithmetic moduli of generalized elliptic curves" I included a proof that an Artin stack whose geometric points have trivial automorphism schemes is necessarily an algebraic space. See Theorem 2.2.5(1) there; I am sure this is a folklore fact (which I inserted there because I didn't know a reference, and to my surprise seems to not be stated in the L-MB book). So that answers the original question: if the moduli problem is an Artin stack and a coarse moduli space exists then it is a fine moduli space (meaning that the moduli problem is an algebraic space) if and only if objects over algebraically closed fields have trivial automorphism schemes (stronger than just trivial automorphism groups, except in the DM case when equivalent since then such groups are etale).

I wrote that paper in the days before I realized that non-qs algebraic spaces made sense, so I had the convention throughout (following the L-MB book) that diagonals are separated and especially quasi-compact. I have not revisited the proof to see the effect of weakening these assumptions (especially the q-c assumption) on the diagonal. I should do that some day.

  • $\begingroup$ @BCnrd Sorry for reviving an old thread, and I'm still somewhat new to the notation, but when you say "if the moduli problem is an Artin stack and a coarse moduli space exists then it is a fine moduli space (meaning that the moduli problem is an algebraic space) if and only if objects over algebraically closed fields have trivial automorphism schemes " Are your moduli "space"s assumed to be schemes, or algebraic spaces? $\endgroup$
    – Will Chen
    Aug 14, 2012 at 12:54

IIRC there's an example due to Gabber in the book by Katz and Mazur of a representable moduli problem where objects have automorphisms (I forget the trick---perhaps he rigged it so that every object had precisely 2 automorphisms). But on the other hand there are theorems of the form "a 'reasonable' moduli problem which is representable has the property that objects have no automorphisms" on the same page. If memory serves "reasonable" in this case is "relatively representable over the stack of elliptic curves". I'm not at work so can't follow this up and say anything more precise.

My understanding is that if you have a D-M stack and objects have no automorphisms then you have an algebraic space. But there are people here who know a lot more about this sort of stuff than I do.


If your question is "if I have a Deligne-Mumford stack where all points have trivial automorphism group, is it isomorphic to any coarse moduli space for that stack?" then the answer is yes. As Kevin says, a D-M stack with no automorphisms is just an algebraic space, and the map to a coarse moduli space, by definition, is universal amongst maps to algebraic spaces.

I'll note, if this seems too easy, this is because assuming that a particular moduli problem has a solution given by a D-M stack is asking a lot. It's not necessarily an easy thing to prove.

  • $\begingroup$ But the coarse moduli space isn't necessarily a scheme, right? $\endgroup$ Jan 3, 2010 at 2:04
  • $\begingroup$ That is true. I think Anweshi is assuming that there is a coarse moduli space which is a scheme, which we can thus conclude is a fine moduli space. If you're happy with a coarse moduli space with is an algebraic space, you should be just as happy with a fine moduli space which is an algebraic space. $\endgroup$
    – Ben Webster
    Jan 3, 2010 at 2:49
  • $\begingroup$ Yes, Webster is right. Anweshi gave the citation of GIT, which uses schemes in its definition. $\endgroup$
    – Anweshi
    Jan 3, 2010 at 3:03
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    $\begingroup$ The reason it seems too easy doesn't seem to be the hypothesis that the modulo problem is a DM stack, but rather the fact that DM stacks with trivial geometric automorphism groups are algebraic spaces is not a definition but requires a proof; that is where the work lies. $\endgroup$
    – BCnrd
    Feb 13, 2010 at 5:54

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