Lecture 12: Modular curves over C

$$ \newcommand{\SL}{\mathrm{SL}} \def\mat#1#2#3#4{\left( \begin{array}{cc} #1 & #2 \\ #3 & #4 \end{array} \right)} \newcommand{\bZ}{\mathbf{Z}} \newcommand{\bC}{\mathbf{C}} \newcommand{\fh}{\mathfrak{h}} \newcommand{\bP}{\mathbf{P}} \newcommand{\bR}{\mathbf{R}} \newcommand{\bQ}{\mathbf{Q}} \let\ol\overline \DeclareMathOperator{\Aut}{Aut} \DeclareMathOperator{\End}{End} $$

In this lecture, we begin the study of modular curves. I introduce the basic modular curves $Y(N)$, $Y_0(N)$, and $Y_1(N)$ and describe them as quotients of the upper half-plane. I then discuss more general upper half-plane quotients, their complex structures, and their compactifications. Finally, I prove the genus formula and give some examples of it.

Sets of elliptic curves

Cast of characters

Let $Y(1)$ be the set of isomorphism classes of elliptic curves over the complex numbers. It turns out to be useful to consider some generalizations of this set as well; we mention here the three most common. Let $N$ be a positive integer. Define $Y_1(N)$ to be the set of isomorphism classes of pairs $(E, P)$ where $E$ is an elliptic curve and $P$ is a point of exact order $N$; an isomorphism $(E, P) \to (E', P')$ is an isomorphism $f \colon E \to E'$ such that $f(P)=P'$. Define $Y_0(N)$ to be the set of isomorphism classes of pairs $(E, G)$ where $E$ is an elliptic curve and $G \subset E$ is a cyclic subgroup of order $N$. And define $Y(N)$ to be the set of isomorphism classes of pairs $(E, i)$ where $E$ is an elliptic curve and $i$ is an isomorphism $(\bZ/N\bZ)^2 \to E[N]$.

Description of Y(1)

The set $Y(1)$ is not hard to describe directly: the $j$-invariant gives a bijection $Y(1) \to \bC$. However, this description does not easily generalize to the related sets, and so we seek an alternate way to get at $Y(1)$.

Every elliptic curve is of the form $\bC/\Lambda$ for some lattice $\Lambda$. Scaling $\Lambda$ by an element of $\bC^{\times}$ leads to an isomorphic curve, so we may as well assume that 1 is a generator of $\Lambda$. We can choose another generator $\tau$ that has positive imaginary part, i.e., $\tau$ belongs to $\fh$, the upper half-plane.

For $\tau \in \fh$, let $\Lambda_{\tau}$ be the lattice spanned by 1 and $\tau$, and let $E_{\tau}=\bC/\Lambda_{\tau}$. We have thus shown that the map $\fh \to Y(1)$ given by $\tau \mapsto E_{\tau}$ is surjective. However, it is not injective. Indeed, suppose

$$ \gamma=\mat{a}{b}{c}{d} $$

is an element of $\Gamma(1)=\SL_2(\bZ)$. Then $az+b$ and $cz+d$ also form a basis for $\Lambda_{\tau}$. Define

$$ \gamma(z)=\frac{az+b}{cz+d}. $$

Then $\Lambda_{\tau}$ is just $\Lambda_{\gamma(\tau)}$ scaled by $cz+d$, and so $E_{\tau}$ and $E_{\gamma(\tau)}$ are isomorphic. It is not difficult to see that $(\gamma, z) \mapsto \gamma z$ defines an action of $\Gamma$ on $\fh$. We have just shown that our map $\fh \to S$ descends to a map $\fh/\Gamma(1) \to Y(1)$. In fact:

Theorem. The map $\fh/\Gamma(1) \to Y(1)$ is a bijection.

Proof. Suppose that $\Lambda_{\tau}$ and $\Lambda_{\tau'}$ define isomorphic elliptic curves, so $\Lambda_{\tau}=\alpha \Lambda_{\tau'}$ for some scalar $\alpha$. Then $1=\alpha (c\tau'+d)$ for some $c,d \in \bZ$. In fact, $c$ and $d$ are coprime: if $e \gt 1$ divides both then we would have $1/e= \alpha ( (c/e) \tau + (d/e)) \in \alpha \Lambda_{\tau'}=\Lambda_{\tau}$, which is not the case. We can therefore find integers $a$ and $b$ such that $ad-bc=1$. Let $\gamma$ be the corresponding element of $\Gamma(1)$. Then $\alpha \Lambda_{\tau'}$ is exactly $\Lambda_{\gamma(tau')}$, and so $\Lambda_{\tau}=\Lambda_{\gamma(\tau')}$. But this implies $\tau=\tau'+n$ for some integer $n$, and so $\tau=\gamma'(\gamma(\tau'))$ with

$$ \gamma'=\mat{1}{n}{}{1}. $$

Description of Y_1(N)

This description does generalize well. For example, we can define a map $\fh \to Y_1(N)$ by $\tau \mapsto  (E_{\tau}, 1/N)$, where by $1/N$ we really mean its image in $E_{\tau}=\bC/\Lambda_{\tau}$. Then it is not difficult to see that this map descends to a bijection $\fh/\Gamma_1(N) \to Y_1(N)$, where $\Gamma_1(N)$ is the subgroup of $\Gamma(1)$ consisting matrices $\gamma$ for which

$$ \gamma = \mat{1}{\ast}{}{1} \pmod{N}. $$

Similarly, we have a bijection $\fh/\Gamma_0(N) \to Y_0(N)$ and $\fh/\Gamma(N) \to Y(N)$, where $\gamma$ belongs to $\Gamma_0(N)$ if

$$ \gamma = \mat{\ast}{\ast}{}{\ast} \pmod{N} $$

and $\gamma$ belongs to $\Gamma(N)$ if $\gamma=1 \pmod{N}$.

Modular curves

Motivated by the above examples, we consider an arbitrary finite-index subgroup $\Gamma$ of $\Gamma(1)$ and define $Y_{\Gamma}$ to be the quotient $\fh/\Gamma$. We now study these spaces.

Complex structure

Write $\pi$ for the quotient map $\fh \to Y_{\Gamma}$. We give $Y_{\Gamma}$ a complex structure by declaring a function $f \colon U \to \bC$ to be holomorphic if $\pi^*(f) \colon \pi^{-1}(U) \to \bC$ is, where $U$ is an open subset of $Y_{\Gamma}$. One can show that this gives $Y_{\Gamma}$ the structure of a Riemann surface.


The space $Y_{\Gamma}$ is never compact: for example, $Y(1)$ is missing the point at infinity. There is an elegant way to fill in the missing points, as follows.

Think of $\fh$ as the open half of the Riemann sphere $\bP^1(\bC)$ above the equator $\bP^1(\bR)$. Define $\fh^*$ to be the union of $\fh$ and $\bP^1(\bQ)$, the set of cusps. Then $\Gamma(1)$ naturally acts on $\fh^*$. We give $\fh^*$ a topology by declaring the circles tangent to $\bP^1$ at cusps to be open neighborhoods of cusps. More precisely, a neighborhood basis of $\infty$ consists of the subsets of $\fh^*$ of the form $U_K = \{\infty\} \cup \{z \in \fh \mid \Im(z) \gt K \}$, as $K$ varies. A neighborhood basis of $\gamma(\infty)$ is given by $\gamma(U_K)$. As $\Gamma(1)$ acts transitively on the set of cusps, this defines the topology everywhere. We give $\fh^*$ a complex structure by defining a function to be holomorphic at $\infty$ if it is holomorphic and bounded on $U_K$, for $K$ sufficiently large, and then moving this condition to other cusps as before.

We now define $X_{\Gamma}$ to be $\fh^*/\Gamma$. This is again a reasonable topological space, and can be given a complex structure in a manner similar to $Y_{\Gamma}$. It is a compact Riemann surface. We call a point of $X_{\Gamma}$ a cusp if it is the image of a cusp under the quotient map $\pi$. We note that the set of cusps is $\bP^1(\bQ)/\Gamma$, which is finite since $\Gamma(1)$ acts transitively on $\bP^1(\bQ)$ and $\Gamma$ has finite index in $\Gamma(1)$.


To understand $X_{\Gamma}$, it will be important to understand the stabilizers $\Gamma_z$ of points $z \in \fh^*$, as this is where $\pi$ ramifies. We being by considering the case $\Gamma=\Gamma(1)$. As $-1$ stabilizes everything, we let $\ol{\Gamma(1)}$ be the quotient $\Gamma(1)/\{-1\}$, which is sometimes easier to deal with.

Proposition. Suppose $z \in \fh$. There is a natural identification of $\Gamma(1)_z$ with $\Aut(E_z)$.

Proof. Suppose $\gamma \in \Gamma(1)$ stabilizes $z$. Via the basis $1, z$ of $\Lambda_z$, we can regard $\gamma$ as an automorphism $g$ of $\Lambda_z$. Precisely, if

$$ \gamma = \mat{a}{b}{c}{d} $$

then $g$ is defined by $g(z)=az+b$ and $g(1)=cz+d$. The condition $\gamma(z)=z$ exactly says $g(z)=(cz+d) z$, and so $g$ is $\bC$-linear. Thus $g$ induces an automorphism of $E_z$. The converse reasoning is similar.

We thus see that $\Gamma(1)_z=\End(E_z)^{\times}$. We know that $\End(E_z)$ is either $\bZ$ or an order in a quadratic imaginary field; we thus get non-trivial units only if $\End(E_z)$ is $\bZ[i]$ or $\bZ[\rho]$, with $\rho=e^{2\pi i/3}$.

Proposition. Suppose $E$ is an elliptic curve with $\End(E)=\bZ[i]$. Then $E$ is isomorphic to $E_i$.

Proof. Write $E=\bC/\Lambda$ where $\Lambda$ is a lattice stable under $\bZ[i]$. Since $\Lambda$ is a torsion-free $\bZ[i]$-module of $\bZ$-rank 2, it is a projective $\bZ[i]$-module of rank 1, and therefore free since the class group of $\bQ(i)$ is trivial. We thus see that $\Lambda=\bZ[i] \alpha$ for some $\alpha \in \bC^{\times}$, and so $E$ is isomorphic to $E_i$.

Proposition. Suppose $E$ is an elliptic curve with $\End(E)=\bZ[\rho]$. Then $E$ is isomorphic to $E_{\rho}$.

Proof. Exactly the same.

Proposition. Let $z \in \fh$. We have one of three cases:

  • $z$ belongs to $\Gamma(1)i$. In this case, $\ol{\Gamma(1)}_z \cong \bZ/2\bZ$.

  • $z$ belongs to $\Gamma(1) \rho$. In this case $\ol{\Gamma(1)}_z \cong \bZ/3\bZ$.

  • $z$ does not belong to $\Gamma(1) i$ or $\ol{\Gamma(1)} \rho$. In this case, $\Gamma(1)_z \cong 1$.

Proof. This follows from what we have shown above.

Determining the stabilizers of cusps is easy:

Proposition. Let $z$ be a cusp. Then $\ol{\Gamma(1)}_z/\{\pm 1\} \cong \bZ$.

Proof. It suffices to handle the case $z=\infty$. If

$$ \gamma=\mat{a}{b}{c}{d} $$

then $\gamma(z)=a/b$. Thus $\gamma(z)=z$ if and only if $b=0$, which implies $a=d=1$ or $a=d=-1$. Thus $\ol{\Gamma(1)}$ is the set of matrices of the form

$$ \mat{1}{n}{}{1}. $$

Now suppose $\Gamma \subset \Gamma(1)$ has finite index. Let $\ol{\Gamma}$ be the image of $\Gamma$ in $\ol{\Gamma(1)}$. We say that $z \in \fh$ is an <em>elliptic point</em> of order 2 (resp. 3) for $\Gamma$ if $\ol{\Gamma}_z$ has order 2 (resp. 3). Since $\ol{\Gamma}_z$ is a subgroup of $\ol{\Gamma(1)}_z$, this implies $z$ belongs to $\Gamma(1) i$ (resp. $\Gamma(1) \rho$). We thus see that $\ol{\Gamma}_z$ is non-trivial only when $z$ is a cusp or elliptic point, and these form finitely many orbits of $\Gamma$.


Proposition. Let $\Gamma \subset \Gamma'$ be finite index subgroups of $\Gamma(1)$. Let $z \in \fh^*$ and let $p$ be its image in $X_{\Gamma}$. Then the ramification index of $f \colon X_{\Gamma} \to X_{\Gamma'}$ at $p$ is $[\ol{\Gamma}'_z : \ol{\Gamma}_z]$.

Proof. First suppose $z \in \fh$. The map $\fh \to Y_{\Gamma}$ has ramification index $e=\# \ol{\Gamma}_z$ at $z$. Indeed, in a small neighborhood about $z$, the action of $\ol{\Gamma}_z$ looks like multiplication by $e$th roots of unity, and so $\pi$ looks like $z \mapsto z^e$. Consider the diagram $\fh \to Y_{\Gamma} \to Y_{\Gamma'}$. At $z$ the first map has ramification index $\# \ol{\Gamma}_z$, while the composite has ramification index $\# \ol{\Gamma}'_z$; thus the second has ramification index $[\ol{\Gamma}'_z : \ol{\Gamma}_z]$.

Now suppose $z$ is a cusp; we may as well assume $z=\infty$. The $\ol{\Gamma}_z$ is the intersection of $\ol{\Gamma(1)}_z$ and $\ol{\Gamma}$, and therefore consists of all matrices of the form

$$ \mat{1}{an}{}{1} $$

for some fixed $a$ and $n$ variable. Similarly, $\ol{\Gamma}'_z$ is of the form

$$ \mat{1}{bn}{}{1} $$

with $b$ a divisor of $a$. Then $e^{2 \pi i z/a}$ descends to a function on $X_{\Gamma}$ with a zero of order 1 at $p$ (i.e., a uniformizer), while $e^{2 \pi i z/b}$ descends is a uniformizer at $f(p)$. Since $e^{2\pi i z/b}=(e^{2 \pi iz/a})^{a/b}$, we find that the ramification index is $a/b=[\ol{\Gamma}_z \colon \ol{\Gamma'}_z]$.

The genus formula

Let $\Gamma \subset \Gamma(1)$ have finite index $d$. Let $\nu_2$ (resp. $\nu_3$) be the number of $\Gamma$-orbits of elliptic points of order 2 (resp. 3), and let $\nu_{\infty}$ be the number of cusps. Let $g$ be the genus of $X_{\Gamma}$. Then

Proposition. Let $g$ be the genus of $X_{\Gamma}$. Then

$$ g=1+\frac{d}{12}-\frac{\nu_2}{4}-\frac{\nu_3}{3}-\frac{\nu_{\infty}}{2}. $$

Proof. We apply the Riemann--Hurwitz formula to the map $f \colon X_{\Gamma} \to X(1)$. Since $X(1)$ has genus 0, this states

$$ (2-2g) = 2d - \sum (e_p-1) $$

where the sum is over all points $p$ of $X_{\Gamma}$, and $e_p$ denotes the ramification index of $p$ along the map $f$. Let $q_2$, $q_3$, and $q_{\infty}$ be the images of $i$, $\rho$, and $\infty$ in $X(1)$. Then $e_p=1$ unless $p$ lies above $q_2$, $q_3$ or $q_{\infty}$.

The points of $f^{-1}(q_2)$ which are elliptic are unramified, while those which are not have ramification index 2. Since the total number of points, with multiplicity, is $d$, we see that the number of ramified points is $(d-\nu_2)/2$. We thus find

$$ \sum_{f(p)=q_2} (e_p-1) = \frac{d-\nu_2}{2}. $$

Similarly, we find

$$ \sum_{f(p)=q_3} (e_p-1) = \frac{2(d-\nu_3)}{3}. $$

It is clear that

$$ \sum_{f(p)=q_{\infty}} (e_p-1) = d - \# f^{-1}(q_{\infty}) = d-\nu_{\infty}. $$

We thus have

$$ 2-2g=2d-\frac{d-\nu_2}{2} - \frac{2(d-\nu_3)}{3}-(d-\nu_{\infty}), $$

from which the result follows.

Example. Suppose $N$ is prime, and consider the case $\Gamma=\Gamma_0(N)$. One can show that $d=N+1$, $\nu_2$ is 1 if $N=2$, 2 if $N=1 \pmod{4}$, and 0 otherwise, $\nu_3$ is 1 if $N=3$, 2 if $N=1 \pmod{3}$, and 0 otherwise, and $\nu_{\infty}=2$. This amounts to

$$ g = \left\lfloor \frac{N}{12} \right\rfloor + \begin{cases} -1 & N = 1 \pmod{12} \\ 1 & N = 11 \pmod{12} \end{cases} $$

For example, if $N \le 13$ then $g=0$, except if $N=11$ when $g=1$.

Example. The genus of $X_1(N)$ is 0 if and only if $N \le 12$ and $N \ne 11$. These are exactly the values of $N$ for which an elliptic curve over $\bQ$ can have an $N$-torsion point. This is not a coincidence!