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Let $\Gamma_0(4)$ be a congruence subgroup of $SL(2,\mathbb{Z})$ defined as $$\Gamma_0(4)=\Big\{M=\begin{pmatrix} a &b\\ c& d \end{pmatrix}\in SL(2,\mathbb{Z}) | c \equiv 0\bmod 4\Big\}.$$ How to show that cusp of $\Gamma_0(4)$ is $0,\frac{1}{2},\infty$?I have looked up the answer: Inequivalent cusps of $\Gamma_0(4)$, but I do not understand. We define the group action: if $\forall \gamma \in SL(2,\mathbb{Z})$, $$ \gamma \infty= \begin{cases} \frac{a}{c}, c\neq 0\\ \infty, c=0 \end{cases}$$

$$ \gamma z= \begin{cases} \frac{az+b}{cz+d}, cz+d \neq 0\\ \infty, cz+d=0 \end{cases}$$

I try to use the same method of proof $\Gamma_0(2)$ cusp $0,\infty$.

For $\forall \frac{p}{q}$, if $q$ is even, there exists $r,s\in \mathbb{Z}$, s.t. $rp-sq=1$, we have

$$\begin{pmatrix} p &s\\ q& r \end{pmatrix} \infty = \frac{p}{q},$$ where $\begin{pmatrix} p &s\\ q& r \end{pmatrix} \in \Gamma_0(2)$.

if $q$ is odd, there exists $r,s\in \mathbb{Z}$, sinece $\gcd(2p,q)=1$, s.t. $-2rp+sq=1$, we have

$$\begin{pmatrix} s &p\\ 2r& q \end{pmatrix} 0 = \frac{p}{q}, $$ where $\begin{pmatrix} s &p\\ 2r& q \end{pmatrix} \in \Gamma_0(2)$.

Similarly, for $\Gamma_0(4)$, I have known $0, \infty$ and $0,\frac{1}{2}$ are not equivalent. $\textbf{But for $\forall \frac{p}{q}\in \mathbb{Q}$, how to find the matrix $\gamma \in \Gamma_0(4)$}$, s.t. $$ \gamma \frac{1}{2}=\frac{p}{q}.$$

If $q \equiv 0 \mod4$, $\gcd(p,q)=1$, it is easy to seek the $$\begin{pmatrix} p &s\\ q& r \end{pmatrix} \infty = \frac{p}{q},$$ where $\begin{pmatrix} p &s\\ q& r \end{pmatrix} \in \Gamma_0(4)$.

How about the $q \equiv 1,2,3 \mod4$?

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This is addressed in the third erratum of the book at the page 103 note http://people.reed.edu/~jerry/MF/errata3.pdf

You should check all four errata if you have not done so yet, as there are quite a few mistakes that can make things confusing.

You might also want to look at this Modular Forms: Find a set of representatives for the cusps of $\Gamma_0(4)$

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  • $\begingroup$ Sorry, I do not understand what is your meaning? $\endgroup$ – user469065 Jun 3 '18 at 6:33
  • $\begingroup$ @Tinzoe-Yui The pdf I linked has a paragraph starting with "A procedure to list the cusps of $\Gamma_0(N)$ is as follows..." and ending with the cusps of $\Gamma_0(4)$. $\endgroup$ – häxq Jun 3 '18 at 7:37
  • $\begingroup$ I see, thanks a lot! $\endgroup$ – user469065 Jun 4 '18 at 8:38
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I am using definitions from Milne's notes on modular forms. The relevant part is on "classification of linear fractional transformations".

A point $s \in \mathbb{R} \cup \{\infty\}$ is called a cusp of $\Gamma_0(4)$ if it is fixed by a parabolic element $\gamma \in \Gamma_0(4)$ with $s$ fixed by $\gamma$.

I found $\alpha = \begin{pmatrix}-1&1\\-4&3\end{pmatrix}$ is in $\Gamma_0(4)$ because $c = -4 \equiv 0 \mod 4$. It is also parabolic since $\text{Tr}(\alpha) = -1 + 3 = 2$. Moreover, $\alpha\frac{1}{2}$ is $\frac{1}{2}$.

I found this $\alpha$ by first finding a matrix $\beta$ which sends $\frac{1}{2}$ to $\infty$. And then I took $\alpha = \beta^{-1} T \beta$ where $T = \begin{pmatrix}1&1\\0&1\end{pmatrix}$.

The $\beta$ which I found is $\begin{pmatrix}1&0\\-2&1\end{pmatrix}$. An easy calculation shows that the negative of the square of the lower-left entry gives you the lower-left entry of the $\beta^{-1}T\beta$. Thus, you are assured here that $c \equiv 0 \mod 4$.

I'm not sure if you're using an equivalent definition, but that's how I would show that $\frac{1}{2}$ is a cusp.

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  • $\begingroup$ why is the $s\in \mathbb{R}\cup \{\infty\}$ but not the $\mathbb{Q}\cup \{\infty\}$? $\endgroup$ – user469065 Jun 3 '18 at 6:31
  • $\begingroup$ The definition says it's in the former a priori. A remark below shows it's in the latter for $\Gamma$. $\endgroup$ – JGA Jun 3 '18 at 9:19

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