So, when people say, "the moduli problem of classifying elliptic curves over $\mathbb{C}$ with level $N$ structure", there are usually two associated functors I've seen:

  1. $P_N : \textbf{Ell}\rightarrow\textbf{Sets}$, where $\textbf{Ell}$ is the category of elliptic curves $E\rightarrow S$ over $S$ and morphisms are cartesian squares, and

    $P_N(E/S) = \text{set of isomorphisms } \alpha : E[N]\rightarrow (\mathbb{Z}/N\mathbb{Z})^2 \text{ of determinant 1}$

  2. $F_N : \textbf{Sch}\rightarrow\textbf{Sets}$, where

    $F_N(S) = \text{set of isomorphism classes of pairs } (E/S,\alpha) \text{ with } \alpha\in P_N(E/S) $

I apologize for the length of this post, but this has been terribly confusing for me.

Alright, so I know that for $N\ge 3$, both functors are representable by the modular scheme $Y(N) := \Gamma(N)\backslash\mathcal{H}$ which are fine moduli schemes, and that's got something to do with the fact that there are no automorphisms of elliptic curves $E/S$ fixing any $\alpha\in P_N(E/S)$.

However, in the case $N = 1,2$, the modular curve $Y(2) := \Gamma(2)\backslash\mathcal{H}$ only gives you a coarse moduli scheme.

How should I think of the relation between the two above functors? In a way, a representing object $E/S$ for $P_N$ gives you both the universal elliptic curve $E$ and the base moduli scheme $S$ in one fell swoop. However, the functor $P_N$ doesn't seem like a naturally phrased moduli problem, since being able to represent $P_N$ just says:

"there is an elliptic curve $E/S$ such that for any other elliptic curve $E'/S'$, a level structure on $E'/S'$ is equivalent to a morphism $S'\rightarrow S$ such that $E'\cong E\times_S S'$."

(after thinking about it for a bit, it seems you can show that $P_N$ representable $\Longrightarrow$ $F_N$ representable)

On the other hand, the functor $F_N$ is much more natural, in that a representing object for $F_N$ much more clearly parametrizes elliptic curves with level structure. However, Peter Bruin's article (http://user.math.uzh.ch/bruin/moduli.pdf) and Katz/Mazur's book (specifically thm's 3.6 and 4.7) both seem to imply that if $P_N$ is not rigid (eg, $N = 1,2$), then even if $F_N$ is representable by an object $M$, the object $M$ might not carry a universal family. I am further confused by the fact that wikipedia (in the section on Fine Moduli Spaces) says that the universal family exists and must correspond to $\text{id}_M\in\text{Hom}(M,M)$.

I'm assuming wikipedia is wrong.

If wikipedia is wrong, then in the case $N = 2$, is the functor $F_2$ representable? If it is, is $\Gamma(2)\backslash\mathcal{H}$ the representing object?

When people talk of the moduli problem of classifying elliptic curves with full level $N$ structure, which functor are they referring to?

...onto stacks...

For $N = 1,2$, there is no fine moduli scheme, and hence at least $P_N$ is not representable. However, is there a fine moduli stack? (does that mean anything?)

In general, would I be correct in saying that a stack for a moduli problem is basically just the moduli functor itself? Can you make this more precise? (though I guess you'd have to replace "isomorphism classes of..." with the objects themselves)

Are there meaningful moduli problems that aren't stacks?

Thanks for bearing with me

  • will
  • $\begingroup$ There's to say here, but let's start with a couple questions. Something's clearly wrong with your $P_N$ - is $\mathbb{Z}/N\mathbb{Z}$ supposed to be $(\mathbb{Z}/N\mathbb{Z})^2$ or something? Are you ultimately working over $\mathbb{C}$-schemes here? This is the only way $\Gamma(N)\backslash\scr{H}$ will be a "moduli scheme" after all. $\endgroup$ – Ramsey Aug 14 '12 at 18:47
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    $\begingroup$ Wow, that $H$ came out fancier than I intended. Let's try $\Gamma(N)\backslash \mathcal{H}$. $\endgroup$ – Ramsey Aug 14 '12 at 18:48
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    $\begingroup$ On second thought, I think I misspoke. I didn't quite understand $P_N$ correctly (in particular I missed the bit about the morphisms in $\mathbf{Ell}$ being Cartesian squares). It seems to me that $P_N$ should be representable iff $F_N$ is. You've stated one direction. For the other: suppose that $F_N$ is representable by the $\mathbb{C}$-scheme $X$. The identity map $X\to X$ gives rise to a universal elliptic curve $E\to X$ which ought to represent $P_N$. $\endgroup$ – Ramsey Aug 14 '12 at 20:13
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    $\begingroup$ cont'd: The K-M proof that representability implies rigidity uses constructions specific to elliptic curves. For example, in char. 0 choose $(E,\alpha) \in F_2(K)$ and let $E'$ be the twist of $E$ by a quadratic extension $L/K$. Quadratic twisting is invisible on 2-torsion (!), so $E'$ has an associated full level-2 structure $\alpha'$. Now $(E,\alpha), (E',\alpha')$ are distinct in $F_2(K)$ (check!) but the same in $F_2(L)$, so $F_2(K) \rightarrow F_2(L)$ is not injective. But $X(k) \rightarrow X(k')$ is injective for any scheme $X$ and field extension $k'/k$, so $F_2$ isn't representable. $\endgroup$ – user22479 Aug 15 '12 at 8:05
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    $\begingroup$ @oxeimon: One should not think about elliptic curves in terms of those affine coordinate rings; best to always think in terms of the proper curve. Also, when you say that one "usually" views elliptic curves in terms of $\overline{K}$-points...once you learn about schemes, you can and should think entirely in terms of $K$-schemes. It is much clearer. Mumford's book on abelian varieties, Chapters 2 and 3, gives an excellent presentation of how to use scheme-theoretic methods in both a classical "variety" setting as well as a scheme-theoretic setting (to deal with inseparable isogenies, etc.). $\endgroup$ – user22479 Aug 16 '12 at 1:04

Your $F_N$ is the functor people would usually mean when they talk about the functor classifying elliptic curves with full level N structure (though it's a bit nicer if you replace $(Z/nZ )^2$ with $\mu_n \times Z/nZ$, so that the determinant takes values in $\mu_n$ on both sides.)

Wikipedia is not wrong. (Wikipedia is surprisingly seldom wrong!) Your F_2 is indeed not representable by a scheme, which is to say that the scheme known as Y(2) is not a fine moduli space. To say it was a fine moduli space would be precisely to say it represents the functor in question. That's why Y(2) doesn't have to have a universal family over it.

Yes -- the stack is the same thing as the functor; but we're keeping track of certain facts about that functor when we say it's a stack. For that matter, in the case where the functor is representable by a scheme, the scheme is also the same thing as a functor; but in that case we don't think of it as or refer to it as a functor, because that would make us seem very pretentious.

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    $\begingroup$ I don't think it is quite correct to say "the stack is the same thing as the functor". Unless I am misreading, the OP is talking about a set-valued functor. Although often a smooth stack such as this one can be uniquely recovered from a closely related set-valued functor, I think you have to be quite careful how you make that precise. $\endgroup$ – Jason Starr Aug 15 '12 at 1:44
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    $\begingroup$ Jason is right, of course! Better to say the stack is the functor to Cat which sends S to the category of pairs (E,alpha)/S, but even then you had better be a bit more careful about what you mean by "functor to Cat." Jason is wrong even more seldom than Wikipedia is wrong. $\endgroup$ – JSE Aug 15 '12 at 1:51
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    $\begingroup$ +1 for resisting the pretensions of publicly thinking of schemes of functors (it's okay to do it privately). $\endgroup$ – Jim Bryan Aug 15 '12 at 19:05

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