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Let $Y_1(N)=\Gamma_1(N)/H$, where $H$ is the upper half plane.

In these lecture notes http://math.uga.edu/~pete/modularandshimura.pdf , the author makes the following statement:

"$Y_1(N)$ parameterizes isomorphisms $(E,P) \mapsto (\Lambda_\tau,1/N \cdot 1)$, or, informally, elliptic curves together with a distinguished point of exact order $N$."

I would like to know:

i) What precisely does this mean? What does 'parameterize' mean?

ii) Can anyone sketch the main ideas in proving this result? (A reference would also be appreciated. I have found the paper by Deligne-Rapoport mentioned in the link below, but it looks a little intimidating. I will tackle it if need be, but would like a more informal explanation first!)

This is a related question, but I don't think it's quite the same as what I am asking: The modular curve X(N)

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  • $\begingroup$ I edited the question because the OP used quotation marks but didn't quote accurately from the source. If it had just been a matter of making the spelling British and replacing "infomally" with "that is" I would have let it go, but there was also a confusion between $Y_1(N)$ and $Y(N)$. In general though I would advise that when you quote someone directly, really do quote them -- i.e., copy and paste and then reformat if necessary. $\endgroup$ – Pete L. Clark Apr 11 '15 at 8:24
  • $\begingroup$ Thanks Pete; please accept my apologies for the inaccuracy. $\endgroup$ – user3131035 Apr 11 '15 at 16:53
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So, as you might have guessed, this is a complicated question. I don't know what your background is, so feel free to ask me to clarify some things.

So, suppose that you wanted to study elliptic curves over a field--maybe over $\mathbb{C}$ or over $\mathbb{Q}_p$. A sort of natural question you might begin to ask is whether there is a geometric way to study all elliptic curves at once. Namely, is there a geometric space $X$ such that the following dictionary holds:

$$\left\{\begin{array}{c}\text{Geometric information}\\ \text{on }X\end{array}\right\}\longleftrightarrow\left\{\begin{array}{c}\text{Information about}\\\text{all elliptic curves}\end{array}\right\}$$

This would be great since we could then study the whole kit and caboodle by just studying this one object $X$.

A common way to this is to try and understand the moduli space of elliptic curves. Here's where things get heavy (hopefully it's useful to you). To make this space right, we need to not only focus our attention on some field like $\mathbb{C}$ or $\mathbb{Q}_p$, but on ALL schemes (if you aren't familiar with schemes, replace them with rings).

In particular, for any scheme $X$ there is a notion of an elliptic scheme $E/X$ (roughly it's a family of elliptic curves indexed by the points of $X$). We could then consider the functor

$$\mathcal{Ell}:\mathbf{Sch}\to\text{Set}$$

given by

$$\mathcal{Ell}\left(X\right)=\left\{\text{Elliptic schemes over }X\right\}/\sim$$

where $\sim$ means 'up to isomorphism'. So, for example, $\mathcal{Ell}(\text{Spec}(\mathbb{C}))$ would just be the isomorphism classes of elliptic curves over $\mathbb{C}$.

We might then ask whether this functor $\mathcal{Ell}$ is representable, meaning that, as a functor, it's isomorphic to $\text{Hom}(-,X_0)$ for some scheme $X_0$. This would mean that to give an elliptic scheme $E/X$ would be the same thing as giving a morphism $X\to X_0$. Algebraic geometry in the raw tells us that $X_0$ would be the right candidate for a space whose geometry tells us about elliptic curves.

Alas, the functor $\mathcal{Ell}$ is not representable! The issue, in short, is the existence of 'automorphisms'. The non-rigidity of elliptic curves makes this impossible. It's best to understand this by asking yourself the following question: "is it possible that I could have a bijection between isomorphism classes $E/X$ and morphisms $X\to X_0$ if $E$ had non-trivial automorphisms?" (By and by, this is the impetus for the so-called theory of stacks, $\mathcal{Ell}$ is not represented by a scheme, but is represented by [more correctly is] a stack, called $\mathcal{M}_{1,1}$).

One way to try and make $\mathcal{Ell}$ more likely to be representable is to impose extra structure on the objects we're trying to study. This extra structure will make it more difficult to have automorphisms, and so make it more likely that such a representing object (moduli space), whose geometry tells us things about the objects, likely to exist.

One way of adding this extra structure for elliptic curves is by adding in the data of a point of order $N$. So, instead of elliptic curves, we try and study pairs $(E,P)$ of elliptic curves $E$, together with a point of exact order $N$. Once again, to understand the geometry, to try and define such a moduli space, we need to move to all schemes. Namely, for a scheme $X$, there is an appropriate notion of an elliptic scheme $E/X$, and a point $P$ of order $N$ 'on' $E$. We can then, roughly, consider the functor

$$Y_1(N):\mathbf{Sch}\to\text{Set}:X\mapsto \left\{(E,P)\text{ over }X\right\}/\sim$$

It turns out that we've sufficiently rigidified the situation, and $Y_1(N)$ is, in fact, representable! We call the representing scheme also $Y_1(N)/\text{Spec}(\mathbb{Z})$ (the Yoneda philosophy tells us to not distinguish between an object and its associated Hom functor).

So, this is great, the scheme $Y_1(N)$ has geometry which tells us about pairs $(E,P)$. But, what does this have to do with $\mathbb{H}/\Gamma_1(N)$?

Well, suppose we wanted to study not all pairs $(E,P)$ over all schemes, but just the ones over $\mathbb{C}$. Well, think about precisely how we made $Y_1(N)$. We made it such that

$$Y_1(N)(\text{Spec}(\mathbb{C}))=\left\{(E,P)\text{ over }\mathbb{C}\right\}/\sim$$

So, the $\mathbb{C}$-points of $Y_1(N)$ literally do parameterize (bijectively correspond with) isomorphism classes of pairs $(E,P)$.

That said, by the general theory of 'analytification' the set $Y_1(N)(\text{Spec}(\mathbb{C})$ naturally has the structure of a complex manifold, whose points are literally the points in $Y_1(N)(\text{Spec}(\mathbb{C}))$. It turns out that this manifold is biholomorphic to $\mathbb{H}/\Gamma_1(N)$!

So, how does $\mathbb{H}/\Gamma_1(N)$ 'index' or 'parameterize' pairs $(E,P)$? Well, as a complex manifold, it can be naturally identified with the 'analytification' of the moduli space $Y_1(N)$ which, by the very definition of a moduli space, has points which parameterize pairs $(E,P)$.

I hope this was useful. I fear that I may have taken a big highfalutin approach (although I see that you have at least studied some algebraic geometry). If so, please let me know, and I'l try and make it simpler.

EDIT: For fear of possibly alienating someone reading this, let me give the much more down-to-earth (but honestly less fulfilling) answer:

i) There is a 'natural' bijection between $\mathbb{H}/\Gamma_1(N)$ and the set $S$ of isomorphism of elliptic curves over $\mathbb{C}$ with a specified point of order exactly $N$. Of course, in this more naive approach (not using the machinery above), $S$ has no inherent structure other than being a set, so it doesn't make sense to say they're, perhaps, bioholomorphic. The bijection is just

$$\mathbb{H}/\Gamma_1(N)\xrightarrow{\approx}S:\tau\mapsto (E_\tau,\frac{1}{N})$$

where $E_\tau=\mathbb{C}/\Lambda_\tau$, and $\Lambda_\tau=\mathbb{Z}+\mathbb{Z}\tau$. Note that, indeed, $\displaystyle \frac{1}{N}$ is a point of order $N$ in the group $E_\tau$.

The seemingly unnecessary notation $\Lambda_\tau$ is actually quite natural. In some philosophical sense, $\mathbb{H}/\text{SL}_2(\mathbb{Z})$, the 'moduli space' of all elliptic curves over $\mathbb{C}$ (it's the analytification of a coarse moduli space for elliptic curves, if that means anything to you), actually more naturally indexes lattice up to homothety. The bijection being $\tau\mapsto\Lambda_\tau$. The fact that lattices up to homothety uniformize elliptic curves over $\mathbb{C}$ is then just a happy coincidence.

2) If you mean the actual explanation (above the edit), then no. This would require a prodigious amount of space. If you just want the simpler explanation as in 1), you can probably prove it yourself, or, instead, you could look at the relevant parts of Diamond and Shurman's Elliptic Curves and Modular Forms.

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  • $\begingroup$ Just discovered this post now while trying to read through Katz's article on p-adic modular forms and this is great! It was nice to be able to hear things in a way that is so down to earth (despite you thinking that this was already too "high falutin" :)) $\endgroup$ – MCT Feb 28 '18 at 18:52

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