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Recently, I saw such a result in Kohnen's paper "New forms of half-integral weight" at page 26:

Since $N$ is odd and squarefree, the cusps of $\Gamma_{0}(4 N)$ are represented by the numbers $\frac{1}{t}$, where $t$ runs over all positive divisors of $4 N$. For such a $t$ put $A_{t}=\left(\begin{array}{cc}1+t & -1 \\ -t & 1\end{array}\right)$ and $\delta_{t}=\left(A_{t},(-t z+1)^{k+1 / 2}\right) .$

I know that the representatives of $SL_2(\mathbb{Z})/\Gamma_0(N)$ is bijective to $\mathbb{P}^1(\mathbb{Z}/N\mathbb{Z})$. But how can we make the cusps to be of the form $\frac{1}{t}$ as in Kohnen's paper? If there are some suitable reference or proof? Another thing I want to consider is about the elements $A$ in $SL_2(\mathbb{Z})/\Gamma_0(N)$ such that $A\infty$ equivalent to $1/t$ for a given $t$. Can we make explicitly the forms of $A$?

Thank you very much!

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For your second question: what is the image of $\infty$ under $\begin{bmatrix}1 &0\\t &1\end{bmatrix}$?

For your first question: as can be seen in the book A first course in modular forms by Diamond and Shurman, Chapter 3.8, two rational-or-infinity numbers $a/c$ and $a’/c’$ written in irreducible form represent the same cusp of $\Gamma_0(N)$ iff for some $y$ coprime to $N$ and some integer $j$, $(a’,c’)$ and $((a+jc)/y,yc’)$ are congruent mod $N$.

Now, given a certain cusp $u/v$ we can change it with a certain $y$ and $j=0$ so that $v|N$. We want to show that it is then represented by $1/v$. To do that, we look for some $y$ coprime to $N$ and some $j$ such that $N/v | y-1$ and $(u+jv)/y$ is congruent to $1$. Now, the gcd of $u,v,N$ is $1$, and $jv/y$ is congruent to $jv$ mod $N$, so all we need to do is find some $y$ coprime to $N$ such that $N/v|y-1$ and $v|u-y$.

We can do it with the CRT (because the gcd of $v$ and $N/v$ is $1$ or $2$ and, in the second case, $u$ has to be odd), thus we’re done.

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  • $\begingroup$ I know that $\left(\begin{array}{cc}1 & 0\\ t & 1\end{array}\right)\infty = 1/t$, but how about the other representatives in $SL_2(\mathbb{Z})/\Gamma_0(4N)$ ? Because $Ax=Bx$ for a point x doesn't mean $A=B$, so I have no idea how to find all different elements A of $SL_2(\mathbb{Z})/\Gamma_0(4N)$ such that $A\infty$ equivalent to $1/t$ under $\Gamma_0(4N)$. Would you mind say more about it? $\endgroup$
    – Strange
    Commented Nov 5, 2021 at 1:13
  • $\begingroup$ A matrix $A \in SL_2(\mathbb{Z})$ maps $\infty$ to $1/t$ (as cusps for $\Gamma_0(4N)$) iff it is in the double coset $\Gamma_0(4N)\begin{bmatrix}1&0\\t&1\end{bmatrix}\begin{bmatrix}1&1\\0&1\end{bmatrix}^{\mathbb{Z}}$. $\endgroup$
    – Aphelli
    Commented Nov 5, 2021 at 7:15

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