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, 2012 at 18:47
  • 4
    $\begingroup$ Wow, that $H$ came out fancier than I intended. Let's try $\Gamma(N)\backslash \mathcal{H}$. $\endgroup$
    – Ramsey
    Aug 14, 2012 at 18:48
  • 1
    $\begingroup$ 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, 2012 at 20:13
  • 2
    $\begingroup$ 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, 2012 at 8:05
  • 2
    $\begingroup$ @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, 2012 at 1:04

1 Answer 1


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.

  • 1
    $\begingroup$ 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. $\endgroup$ Aug 15, 2012 at 1:44
  • 1
    $\begingroup$ 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, 2012 at 1:51
  • 1
    $\begingroup$ +1 for resisting the pretensions of publicly thinking of schemes of functors (it's okay to do it privately). $\endgroup$
    – Jim Bryan
    Aug 15, 2012 at 19:05
  • $\begingroup$ For the benefit of someone trying to improve their understanding of stacks, would you mind saying a tiny bit more about what facts you're tracking when you regard the functor as a stack? $\endgroup$ May 8, 2022 at 19:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.