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We write $SL_2(\mathbb{Z})$ as $\coprod\alpha_i\Gamma_0(4)$, where $\coprod$ means the disjoint union, and the $\alpha_i$ are the coset representatives of $SL_2(\mathbb{Z})\diagup\Gamma_0(4)$.

We were told that the cusps are then the points $\alpha_i\infty$, as $\alpha_i$ varies.

I understand this, but then I am having trouble working out what the coset reps are for this particular example.

Since $$\Gamma_0(4)=\left\lbrace \begin{pmatrix} a&b\\4c&d\\ \end{pmatrix}:a,b,c,d\in \mathbb{Z}\right\rbrace $$

I thought the representatives could be something like $\left\lbrace\begin{pmatrix}a&b\\c'&d\\ \end{pmatrix}\right\rbrace$, where $0\leq c'\leq 3$ and $a,b,d$ are arbitrary.

But then I know that $SL_2(\mathbb{Z})\diagup\Gamma_0(4)$ has order $6$, due to the formula $\left(SL_2(\mathbb{Z}):\Gamma_0(N)\right)=N\prod_{p|N} (1+\frac 1p)$.

This may be more of an algebraic problem than a modular forms one specifically, but I am struggling here!

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Let me begin with some generalities:

Let $R$ be a ring with identity and $\mathbf{P}^1(R)$ the projective line over $R$. To define this gadget, consider the equivalence relation $\sim$ on $R \times R$ defined by $(a, b) \sim (c, d)$ if there is a unit $u$ such that $ua = c$ and $ub = d$. Note now that the ideal $aR + bR$ generated by $(a, b)$ depends only on the equivalence class of $(a, b)$. Put $\mathbf{P}^1(R)$ for the set of all equivalence classes for which this ideal generated is the full ring $R$.

In the case of the ring $\mathbf{Z}/N\mathbf{Z}$, there is the following fact:

Fact. If $c, d$ are positive integers with $(c, d, N) = 1$, then, there is $c' \equiv c \bmod{N}$ and $d' \equiv d \bmod{N}$ so that $(c' , d') = 1$.

This fact allows us to choose representatives for our class $(c: d)$ so that $(c, d) = 1$.

Now, we have the following general proposition:

Proposition. The map $$(c: d) \mapsto \begin{pmatrix} \ast & \ast \\ c & d\end{pmatrix}: \mathbf{P}^1(\mathbf{Z}/N\mathbf{Z}) \to [SL_2(\mathbf{Z}):\Gamma_0(N)]$$ between $\mathbf{P}^1(\mathbf{Z}/N\mathbf{Z})$ and coset representatives $[SL_2(\mathbf{Z}):\Gamma_0(N)]$ for $\Gamma_0(N)$ in $SL_2(\mathbf{Z})$ is a bijection.

Proof. The map is firstly well-defined: as remarked, we work with "points" $(c :d)$ where $(c, d) = 1$; thus, there are integers $a, b$ so that $ad - bc = 1$ and these integers are what the $\ast$ represents. Check that the class that $(c :d )$ maps to is independent of the choice of $(a, b)$ chosen above!

The map is clearly surjective. Showing injectivity is a small computation which also I am happy to leave for you.

In this case, we have

$$\mathbf{P}^1(\mathbf{Z}/4\mathbf{Z}) = \{(0:1), (1:0), (1:1), (1:2), (1:3), (2:1)\} $$ and a set of coset representatives for $\Gamma_0(4)$ in $SL_2(\mathbf{Z})$ is: $$\left\{ \begin{pmatrix}1 &0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 & -1 \\ 1& 0\end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 1 & 1\end{pmatrix}, \begin{pmatrix} 0 & -1 \\ 1 & 2\end{pmatrix}, \begin{pmatrix} 0 & -1 \\ 1 & 3\end{pmatrix}, \begin{pmatrix}1 & 0 \\ 2 & 1 \end{pmatrix}\right\}.$$

Edit. Indeed, there are theorems that give coset representatives for other important congruence subgroups (viz. $\Gamma_1(N)$ and $\Gamma(N)$). See L. Kilford's book starting from Proposition 2.11. He discusses this at length. (Let me add that you'll find the following group theory fact useful: if $K \subseteq H \subseteq G$ are groups, then, there is an explicit bijection between $[G: K]$ and $[G: H] \times [H: K]$.)

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  • $\begingroup$ Accepted since the set of matrices is exactly what I was looking for, albeit using an interesting proposition. Do you know of anywhere I can find this proposition? Also, will I be able to generalise this for $\Gamma_1(N)$ and $\Gamma(N)$? It looks like it has the potential to. I'm not quite sure the image of the map is necessarily the set of coset reps, but I think I could figure it out with a bit of thought.. $\endgroup$ Commented May 18, 2015 at 20:52
  • $\begingroup$ I will try to add a proof of this but this is to be found in L. Kilford's book "Modular forms: A classical and computational introduction" published by the Imperial College Press (cf. with proof of Proposition 2. 12 there.) $\endgroup$
    – knsam
    Commented May 18, 2015 at 20:53
  • $\begingroup$ @Ramified_Minds: I gave some details (little short of a complete proof mainly because Kilford does this in detail and it is not difficult to do this anyway). Moreover, I added a small remark in the recent edit. Hope that is helpful. :-) $\endgroup$
    – knsam
    Commented May 18, 2015 at 23:20
  • $\begingroup$ Ah yes the fact that $c$ and $d$ are coprime so we can choose $a,b$ s.t. ... Got it now. Thanks a lot for your help! On a side note, I wasn't aware Imperial College press existed, which is funny as I'm studying there.. $\endgroup$ Commented May 18, 2015 at 23:36

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