# On the cusps of $\Gamma_0(4N)$

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!

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.
• 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? Commented Nov 5, 2021 at 1:13
• 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 $\Gamma_0(4N)\begin{bmatrix}1&0\\t&1\end{bmatrix}\begin{bmatrix}1&1\\0&1\end{bmatrix}^{\mathbb{Z}}$. Commented Nov 5, 2021 at 7:15