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  1. For the moduli problem of a curve of genus $g$ with $n$ marked points, how large an $n$ is needed to ensure the existence of a fine moduli space? For this question, terminology is that of Mumford's GIT.

  2. For the following three moduli problems, how big an $N$ is required for existence of a fine moduli space? The terminology is from the exposes of Deligne-Rapoport and Katz-Mazur, or Shimura. The first is in French, the second is too big, and the third is using old language and never mentions the modern terminology of universal elliptic curve, etc.. Therefore it is not possible for me to dig up the information myself.

i) Elliptic curves equipped with a cyclic subgroup of order $N$ -- this moduli problem corresponds to the modular group $\Gamma_0(N)$.

ii) Elliptic curves equipped with a point of order $N$ -- this moduli problem corresponds to the modular group $\Gamma_1(N)$.

ii) Elliptic curves equipped with a symplectic pairing on $N$-torsion points -- this moduli problem corresponds to the modular group $\Gamma (N)$.

References other than the above, will be appreciated.

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To get a $Gamma_0(N)$-structure, the subgroup of order N needs to be cyclic (in a sense that is precisely explained in Katz-Mazur, chapter 3). –  S. Carnahan Jan 10 '10 at 2:28
    
Yes, of course. Thanks for pointing out. I have added. –  Anweshi Jan 10 '10 at 2:29
    
If your two questions had mated, you would have asked what value of N guarantees the existence of a fine moduli space of genus-g curves endowed with various N-level structures on the Jacobian... –  JSE Jan 10 '10 at 2:43
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3 Answers

If you want to work over a base ring such as $\mathbf{Z}[1/n]$ rather than over $\mathbf{Q}$ or $\mathbf{C}$ then the relevant numerical condition is that the part of $N$ coprime to $n$ not be "too small" in the $\Gamma_1$ and full level cases. For an extreme example, if $N$ is a $p$-power and you work over $\mathbf{Z}_{(p)}$ then you'll always have problems in characteristic $p$ at the supersingular points.

On the other hand, if you're willing to go beyond schemes and work with algebraic spaces or Deligne-Mumford or Artin stacks then these issues go away (at the expense of more technical background) in the sense that one has a reasonable "moduli space" over $\mathbf{Z}$ with nice regularity properties for all $N$ (even incorporating degenerations in the sense of generalized elliptic curves with level structure). It has better properties than a coarse moduli space (aside from perhaps not being a scheme...).

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The first is unrepresentable for arbitrary large $N$ (it depends on the residue class of $N$ mod 12), the second is representatble for $N \geq 4$ (if you are considering $Y_1(N)$) or $N \geq 5$ (if you are considering $X_1(N)$, i.e. including the cusps), the third is representable for $N \geq 3$.

The references you mentioned are the standard ones. Probably Silverman discusses these in his books somewhere too (maybe the 2nd). If you look in Gross's Duke paper on companion forms (A tameness criterion ... ) you will find a summary of the story for $X_1(N)$. In the $\Gamma_0(N)$ case, Mazur has a careful discussion in the beginning of section 2 of his Eisenstein ideal paper. Both Gross and Mazur refer back to Deligne--Rapoport for proofs.

It is also just a matter of computing the torsion in each of the $\Gamma$'s (plus epsilon more if you want to understand representability at the cusps), which is an exercise. (Although you have to do a little work to see why this is the necessary computation.)

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Regarding the answer to 2.i), I think the answer should be no for all N because a pair (E,C) with E an elliptic curve and C a cyclic subgroup always has at least the automorphism -1, no matter what N is. Maybe we are not all thinking about the same moduli problem? –  Bjorn Poonen Jan 10 '10 at 7:06
    
You're right; I was rather just thinking about when $\Gamma_0(N)/\langle \pm 1 \rangle$ is torsion free. –  Emerton Jan 10 '10 at 10:29
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So this is another example of a moduli problem having no fine solution because of existence of automorphims. .. –  Anweshi Jan 10 '10 at 14:17
    
@Emerton. I am unable to accept this answer because the question was in two parts, and the other part was answered by Jordan Ellenberg. See meta here .. tea.mathoverflow.net/discussion/178 –  Anweshi Jan 22 '10 at 19:12
    
Dear Anweshi, I think I can speak for Jordan as well as myself in telling you not to worry about it. –  Emerton Jan 22 '10 at 19:37
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Here is a thought on the first question. What you need to know (at least to get an algebraic space; I'll let others be more careful than I if you want a scheme) is how large n must be to ensure that an automorphism of a smooth genus g curve X which fixes n points must be the identity. Let G be the cyclic group generated by this automorphism: then the map X -> X/G is totally ramified at your n fixed points. So by Riemann-Hurwitz, g(X) [NO, 2g(X)-2, THANKS, BJORN) is at least -2|G| + n(|G|-1). If G is nontrivial, in other words, g is at least n-4 [NO, 2g+2, THANKS, BJORN]. So I think g+5 [NO, 2g+3, THANKS, BJORN] marked points should be enough. That this is necessary can be seen by taking g=2; on M_{2,6} you'll have a bunch of loci with an extra involution, parametrizing curves whose marked points are precisely the Weierstrass points.

[NO MORE LATE-NIGHT RIEMANN-HURWITZ: THANKS TO BJORN FOR CORRECTING THE ERRORS]

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Thanks, this was useful. However a solution to the scheme situation also will be appreciated(perhaps from the next person). –  Anweshi Jan 10 '10 at 2:39
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@JSE: I think you meant to have 2g(X)-2 on the LHS of your Riemann-Hurwitz, in which case you need n greater than 2g+2 (the number of fixed points of the hyperelliptic involution on a hyperelliptic curve of genus g). As for algebraic space vs. scheme, it's going to be a scheme since if you include level structure what you have is quasi-projective. –  Bjorn Poonen Jan 10 '10 at 6:59
    
The second correction could use a little more correcting: maybe (n-2)/2. Also, the html "strike" tag can be useful for demarcating deprecated parts in a readable way. –  S. Carnahan Jan 10 '10 at 19:26
    
@JSE. I am unable to accept this answer because the question was in two parts, and the other part was answered by Emerton. See meta here .. tea.mathoverflow.net/discussion/178 –  Anweshi Jan 22 '10 at 19:12
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@JSE: There should be a law against texting while applying Riemann-Hurwitz. –  Bjorn Poonen Feb 13 '10 at 6:18
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