When is a coarse moduli space also a fine moduli space? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T09:13:28Z http://mathoverflow.net/feeds/question/10532 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/10532/when-is-a-coarse-moduli-space-also-a-fine-moduli-space When is a coarse moduli space also a fine moduli space? Anweshi 2010-01-02T21:48:45Z 2010-02-25T05:36:37Z <p>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?). </p> <p>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?</p> <p>In the n-lab page on Deligne-Mumford, the following appears.</p> <p>Deligne-Mumford stacks correspond to moduli problems in which the objects being parametrized have finite automorphism groups.</p> <p>Also, a few problematic examples I have heard of, had infinite automorphism groups. </p> <p>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?</p> http://mathoverflow.net/questions/10532/when-is-a-coarse-moduli-space-also-a-fine-moduli-space/10544#10544 Answer by Kevin Buzzard for When is a coarse moduli space also a fine moduli space? Kevin Buzzard 2010-01-02T22:51:36Z 2010-01-02T22:51:36Z <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/10532/when-is-a-coarse-moduli-space-also-a-fine-moduli-space/10554#10554 Answer by Ben Webster for When is a coarse moduli space also a fine moduli space? Ben Webster 2010-01-03T01:28:37Z 2010-01-03T02:50:30Z <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/10532/when-is-a-coarse-moduli-space-also-a-fine-moduli-space/10562#10562 Answer by David Zureick-Brown for When is a coarse moduli space also a fine moduli space? David Zureick-Brown 2010-01-03T03:19:59Z 2010-01-03T03:31:56Z <p>I think this is an instructive question. Here are some partial answers.</p> <p>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.</p> <p>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 <a href="http://math.berkeley.edu/~anton/index.php?m1=writings" rel="nofollow">notes</a> from Martin Olsson's stacks course.</p> <p>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 <em>infintesmal</em> 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, $\mathbf{G}_m$.</p> <p>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 <strong>H</strong> 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 <strong>H</strong>; 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).</p> <p>Last comment (to clarify others' comments): fine moduli space should certainly allow algebraic spaces for a correct answer to your question.</p> http://mathoverflow.net/questions/10532/when-is-a-coarse-moduli-space-also-a-fine-moduli-space/16367#16367 Answer by BCnrd for When is a coarse moduli space also a fine moduli space? BCnrd 2010-02-25T05:01:54Z 2010-02-25T05:36:37Z <p>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).</p> <p>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. </p>