Showing the cusps in $\Gamma_1(4)$: [$\infty$], [0], [1/2] equal $\mathbb{P_1(Q)}$

Question

I am trying to show that $\Gamma_1(4)$ only has three cusps: $0, 1/2$ and $\infty$. To do so, I need to show $\Gamma_{0},\Gamma_{\infty}, \Gamma_{1/2}$ equal $\mathbb{P_1(Q)}$.

Here is what I have so far:

$\Gamma_{\infty}= \{\frac{4a+1}{4c} | a,c \in \mathbb{Z}, (4a+1,4c)=1 \}$

$\Gamma_{0}= \{\frac{b}{4d+1} | b,d \in \mathbb{Z}, (4d+1,b)=1 \}$

$\Gamma_{1/2}= \{\frac{4a+1+2b}{4c+8d+2} | a,b,c,d \in \mathbb{Z}, (4a+1)(4d+1)-4bc=1 \}= \{\frac{4a+1}{4c} | a,b,c,d \in \mathbb{Z}, (4a+1)(4d+1)-4bc=1 \}$

Now how do I show their union equals $\mathbb{P_1(Q)}$?

Answer 1

Write a cusp as $r/s$ in lowest terms. If $r$ is odd and $4\mid s$ then $r/s=(-r)/(-s)$ is in your $\Gamma_\infty$. Likewise if $s$ is odd, then $r/s\in\Gamma_0$. What's left? $r$ odd and $s=2t$ with $t$ odd. This had better be in $\Gamma_{1/2}$.

Acting on $r/(2t)$ by powers of $\pmatrix{1&1\\0&1}$ takes it to $(r+2kt)/(2t)$ for $k\in\Bbb Z$. We can choose $k$ to make $|r+2kt|\le|t|$. Acting on $r/(2t)$ by powers of $\pmatrix{1&0\\4&1}$ takes it to $r/(2(t+2kr))$ for $k\in\Bbb Z$. We can choose $k$ to make $|t+2kr|\le|r|$. Eventually we reduce $r/s$ to $ \pm1/2$ and if we get $-1/2$ we add one to get $1/2$. So $r/s$ is an equivalent cusp to $1/2$.

Answer 2

Let $\frac{r}{s}$ be an element of $\mathbb{Q}$ such that $\gcd(r,s) = 1$.

We shall distinguish the following cases:

1. $r$ odd, $4\mid s$
2. $r$ any number, $s$ odd
3. $r$ odd, $s\equiv 2 \mod 4$

Note that if $A \equiv 3 \mod 4$, then $A^2 \equiv 1\mod 4$.

For the first case: If $r\equiv 1 \mod 4$, choose $a = r, b = $anything, $ c = s, d = 1$. Then $$\begin{pmatrix} a&b\\c&d \end{pmatrix}.\infty = \frac{a}{c} = \frac{r}{s}.$$ Similarly, if $r\equiv 3\mod 4$, then $a = r^2, b = $anything, $c = rs, d = 1$ does the job: $$\begin{pmatrix} a&b\\c&d \end{pmatrix}.\infty = \frac{a}{c} = \frac{r^2}{sr} = \frac{r}{s}.$$

Now consider the second case: Let $r$ be any integer, $s\equiv 1\mod 4$. Then $$\begin{pmatrix} a&b\\c&d \end{pmatrix}.0 = \frac{b}{d} = \frac{r}{s}$$ if $a=1,b=r, c=0,d=s$. If $s\equiv 3\mod 4$, choose $a=1,b=rs,c=0,d=s^2$ instead and proceed as above: $$\begin{pmatrix} a&b\\c&d \end{pmatrix}.0 = \frac{b}{c} = \frac{rs}{s^2} = \frac{r}{s}.$$

Finally, if $s\equiv 2\mod 4$, and $r\equiv 1 \mod 4$, then we can write $s = 4l+2, r = 4k+1$. Hence, $$\begin{pmatrix} a&b\\c&d \end{pmatrix}.\frac{1}{2} = \frac{a/2+b}{c/2+d} = \frac{1/2+2k}{4l/2+1} = \frac{1+4k}{4l+2} = \frac{r}{s}$$ for $a = 1, b=2k, c=4l, d=1$. If $s\equiv 2\mod 4$ and $r\equiv 3\mod 4$, then we can write $s=4l+2, r = 4k+3$. Therefore, $$\begin{pmatrix} a&b\\c&d \end{pmatrix}.\frac{1}{2} = \frac{a/2+b}{c/2+d} = \frac{1/2+2k+1}{4l/2+1} = \frac{1+4k+2}{4l+2} = \frac{r}{s}$$ for $a=1,b=2k+1,c=4l,d=1$.

We have accounted for all possible cases for $r$ and $s$ and thus we are done showing that the cusps are $[0],[\frac{1}{2}],$ and $[\infty]$.