Deriving a formula for the genus of the modular curve X(N)

Hi all! In this post, we’ll show how we can use the Riemann-Hurwitz formula to derive the genus of the modular curve X(N). But what is X(N)?

The group \mathrm{SL}_2(\mathbf{Z}) acts on the complex upper half-plane, denoted \mathfrak{H}. Let \Gamma be a congruence subgroup of this group, i.e., a group containing the principal congruence subgroup of level N, \Gamma(N), defined as \Gamma(N)=\left\{ \begin{pmatrix} a & b\\ c & d\\ \end{pmatrix} : \ a \equiv d \equiv \pm 1 \mod N \text{ and } b, c \equiv0 \mod N \right\}. The quotient \Gamma\backslash \mathfrak{H} is a noncompact Riemann surface, denoted Y(\Gamma). Its compactification, denoted X(\Gamma), is the quotient \Gamma\backslash (\mathfrak{H}\cup\mathbf{Q}\cup\{\infty\}). We remark that \mathfrak{H}\cup\mathbf{Q}\cup\{\infty\} is called the extended complex upper half-plane, and is given a topology by choosing a particular basis. The cusps of \Gamma are elements of the orbit of the action of \Gamma on \mathbf{Q}\cup\{\infty\}.

If \Gamma=\Gamma(N), the principal congruence subgroup of level N, then X(\Gamma) is denoted X(N). If \Gamma=\mathrm{SL}_2(\mathbf{Z}), X(\Gamma) is isomorphic as a Riemann surface to \mathbf{C}\cup\{\infty\}\cong\mathbf{P}^1\mathbf{C}\cong S^2. Now, \Gamma(1)\simeq \mathrm{SL}_2(\mathbf{Z}), so X(1)\cong S^2. It therefore has genus 0. Now, we have a map X(\Gamma)\to X(\mathrm{SL}_2(\mathbf{Z})) coming from the inclusion \mathrm{P}\Gamma\subset\mathrm{PSL}_2(\mathbf{Z}). We will prove, using the Riemann-Hurwitz formula, the following equation for N>2:

g(X(N))=g(X(N))=1+\dfrac{N^2(N-6)}{12}\prod_{p|N}\left(1-\dfrac{1}{p^2}\right)

Where e_i is the number of elliptic points of order i, and c is the number of cusps.

How do we go about proving this? First recall the Riemann-Hurwitz formula: if X\to Y is a surjective morphism of Riemann surfaces of degree d, let e_x denote the ramification degree at x. Then

2g(X)-2 = d(2g(Y) - 2) + \sum_{x\in X}(e_x - 1).

There’s a statement in the following proof which I don’t understand, and would like to.  If X\to Y is Galois, the e_x in the fibers of each y\in Y is equal, this becomes

2g(X)-2 = d\left(2g(Y) - 2 + \sum_{y\in Y}\left(1-\dfrac{1}{e_y}\right)\right).

The map X(N)\to X(1) is Galois, so the indices e_x are the same for x over a y\in X(1). The degree of this map is [\mathrm{PSL}_2(\mathbf{Z}):\mathrm{P}\Gamma] (the number of preimages of generic points of X(\mathrm{SL}_2(\mathbf{Z}))). If N>2, this is explicitly \dfrac{1}{2}N^2\prod_{p|N}\left(1-\dfrac{1}{p^2}\right).

We need to compute the ramification degree e_\infty. The fiber of X(N)\to X(1) over \infty is the \Gamma(N)-equivalence class of cusps, which, by the orbit stabilizer theorem, number \#(\mathrm{SL}_2(\mathbf{Z})/\Gamma(N))/\# R where R is the stabilizer of \Gamma(N)\infty (a coset) in \mathrm{SL}_2(\mathbf{Z})/\Gamma(N) (which is isomorphic to \mathrm{SL}_2(\mathbf{Z})/N\mathbf{Z})). Thus R is the subgroup consisting of matrices \begin{smallmatrix} 1&k\\ 0&1 \end{smallmatrix} such that 0\leq k\leq N-2, and \# R=2N. Consequently, e_\infty=N.

Cool. Now, the ramfications of X(N)\to X(1) are over \infty, 0, and 12^3=1728. For the map in consideration, e_0=3 and e_{1728}=3 (I don’t think I understand why, perhaps it’s a simple error on my part). Thus, because X(N)\to X(1) is Galois, using the Riemann-Hurwitz formula, we get:

2g(X(N)) - 2 = \dfrac{1}{2}N^2\prod_{p|N}\left(1-\dfrac{1}{p^2}\right)\left(0-2+\left(1-\dfrac{1}{e_0}+1-\dfrac{1}{e_{1728}}+1-\dfrac{1}{N}\right)\right)=\dfrac{1}{2}N^2\prod_{p|N}\left(1-\dfrac{1}{p^2}\right)\left(\dfrac{1}{N}-\dfrac{1}{6}\right)

Solving for g(X(N)) gives us the required formula. QED.

Here are the first few values of g(X(N)): g(X(1))=g(X(2))=g(X(3))=g(X(4))=g(X(5))=0, g(X(6))=1, g(X(7))=3, g(X(8))=5. It’s some fun stuff, and I hope to continue to learn more of it! If anyone can tell me why e_0=3 and e_{1728}=3, it would be greatly appreciated!

Have fun!
SKD

 

 

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