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Notation

The term "field of moduli" comes in up in different scenarios, but let's consider the following: Let X->ℙ1 be a G-Galois cover, where everything is over the algebraic closure of some field L. Assume that X->ℙ1 descends (without group action -- as a cover) to XL->ℙL1. Then I define the field of moduli to be the intersection of all finite extensions of L for which base change of XL->ℙL1 becomes G-Galois.

Question

There is the saying that the field of moduli is the function field of the (coarse?) moduli space of when you let the branch points vary. What is the precise statement of that? (and why is it true?)

Thoughts

It would seem that we should fix a dedekind ring whose quotient field is L (ℤ if L is ℚ), and call it D. Then descend to a D-model of ℙ1 (for a D-model of X take the integral closure of ℙ1 in the function field of X). Then do something like look at the moduli space of all covers of ℙ1 with that number of (distinct) branch points, and in it look at the subscheme of all covers that can be achieved by deforming any of the fibers of our XD->ℙD1 (pick a fiber such that there's no coalescence of branch points) by a family. But there's a lot missing here, even in terms of making this precise. For example: IS there a coarse moduli space of all covers with n branch points over ℙD1 (where by n branch points, I mean n branch point on each geometric fiber)? What does it look like? Why should the function field of said subscheme be the field of moduli?

Thanks in advance.

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    $\begingroup$ Since you are in Philadelphia, why not ask David Harbater? $\endgroup$ Jan 24, 2010 at 14:29
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    $\begingroup$ Harbater is actually my advisor. Our meetings usually fill up pretty tightly with both of our ideas, so things that don't require his feedback I prefer to do separately. I figured this question would be an easy one for mathoverflow, but I guess the field of moduli is not in the common vernacular of most people? Anyway, if this won't resolve itself soon (either by mathoverflow or me), I'll bring it up with him. $\endgroup$
    – H. Hasson
    Jan 24, 2010 at 18:09
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    $\begingroup$ It may also help the rest of the community if there is some answer posted (if someone who can explain this well can post). Not everyone has access to David Harbater or other experts. $\endgroup$
    – Ben Weiss
    Jan 24, 2010 at 19:03

2 Answers 2

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A few responses to different parts of your question.

  1. In my experience, the phrase "field of moduli" doesn't usually refer to the function field of a coarse moduli space. Rather: the base change of your cover to Lbar corresponds to a point of M(Lbar), where M/L is the coarse moduli space. This point has a well-defined field of definition, which is by definition the field of moduli of your cover. The phrase "field of moduli" is usually used in distinction with "field of definition" -- if your cover is actually defined over L', then the field of moduli is certainly contained in L', but it may not be equal to L'. This phenomenon isn't restricted to Hurwitz spaces; there are abelian varieties over Qbar whose field of moduli is Q (that is, they correspond to points of A_g(Q)) but which don't descend to Q. This can only happen when g is even. Off the top of my head I don't remember a reference for an example, nor for the assertion of the previous sentence; maybe somebody can help me out in comments. Certainly when g=1 you don't have this problem; given a rational number j, there is an elliptic curve E/Q with j(E) = j. But you prove this by writing it down -- it's not completely obvious "by pure thought" that it should be so.

  2. The most complete description of the Hurwitz stack (the moduli stack of finite covers with fixed combinatorial invariants) its associated coarse moduli space, etc., is in the Ph.D. thesis of Stefan Wewers, which is unfortunately not available online. However, the survey paper of Romagny and Wewers should give you most of what you need.

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  • $\begingroup$ Interesting. I will take a look. $\endgroup$
    – H. Hasson
    Jan 24, 2010 at 22:40
  • $\begingroup$ From a cursory look it seems that he's again looking only at simple covers. Is this what he's doing, or am I missing something? $\endgroup$
    – H. Hasson
    Jan 24, 2010 at 22:44
  • $\begingroup$ It depends; what do you mean by "simple covers"? If you mean "simply branched," no -- for instance, Wewers allows an arbitrary finite group as Galois group, not just S_n. $\endgroup$
    – JSE
    Jan 24, 2010 at 23:05
  • $\begingroup$ By simple cover I mean that over every branched point there's only one ramification point, and it ramifies with multiplicity 2 (and some additional technical requirements, if we're not over C). $\endgroup$
    – H. Hasson
    Jan 24, 2010 at 23:11
  • $\begingroup$ No, Romagny and Wewers certainly don't restrict themselves to this case. $\endgroup$
    – JSE
    Jan 24, 2010 at 23:14
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What I usually see called field of moduli is the following. Suppose $C$ is some member of a collection which admits a coarse moduli space $M$ say defined over $Q$. E.g. $C$ could be a curve or a $G$ cover, as in your example, and so on. So $C$ corresponds to a point $[C]$ of $M$. The field generated by the coordinates of $[C]$ is the field of moduli of $C$. But in your post, it seems you might be taking $C$ to be a generic element of your collection and what you get as field of moduli is $Q(M)$, the function field of $M$, but that sounds tautological to me.

If $C$, as above, is defined over a field $K$, then $K$ contains the field of moduli of $C$ but it's not always true that $C$ can be defined over the field of moduli.

Moduli spaces of branched covers are usually called Hurwitz schemes. I don't know the precise conditions for their existence offhand.

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  • $\begingroup$ My familiarity with Hurwitz schemes is mostly from Fulton's "Hurwitz Schemes and Irreducibility of Moduli of Algebraic Curves" - but there he shows that if you fix the genus of the coverings, and the degree of the covers, AND you only look at simple covers (heuristically meaning that over each branch point there's only one ramified point, and it has ramification degree=2) then you have a FINE Hurwitz scheme. But I'm not sure that if you look at all covers you get a coarse moduli space. $\endgroup$
    – H. Hasson
    Jan 24, 2010 at 19:16
  • $\begingroup$ I see. So your definition of "moduli space" would be closely related to the second (function field) definition. Can you see why it would be equal to the intersection of all finite extension of L where it's still Galois? $\endgroup$
    – H. Hasson
    Jan 24, 2010 at 19:21
  • $\begingroup$ Sorry, I meant "field of moduli", not "moduli space". $\endgroup$
    – H. Hasson
    Jan 24, 2010 at 19:29
  • $\begingroup$ Sorry, I don't think I can answer your more precise question about the intersection of fields of definition. If I really needed the answers, I'd do as Pete suggested and ask Harbater :-) $\endgroup$ Jan 24, 2010 at 20:08

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