Modular interpretation of the modular curve associated to $\Gamma_0(N)\cap\Gamma_1(p^s)$

Question

Let $p$ prime, $N\ge 1$ with $p\nmid N$, $s\ge 1$. Let $\Phi_s=\Gamma_0(N)\cap\Gamma_1(p^s)$. Let $Y_s$ be the open curve $\mathcal{H}/\Phi_s$, where $\mathcal{H}$ is the complex upper half plane. Why is it true that $Y_s$ is the space of triples $(E,C,P)$ with $E/\mathbb{C}$ elliptic curve, $C$ subgroup of $E$ of order $N$ and $P$ point of $E$ of order $p^s$ (modulo isomorphism)?

Moreover, why the natural degeneracy map $Y_{s+1}\to Y_s$ given by the inclusion $\Phi_{s+1}\subseteq \Phi_s$ corresponds to the map $(E,C,P)\mapsto (E,C,pP)$?

I am reading this in an article, and I'm familiar with the modular interpretations of the classical modular curves $Y_0(N)$, $Y_1(N)$, $Y(N)$ (from Diamond-Shurman). Of course the fact that $Y_s$ must have this modular interpretation is believable, but I can't get a precise proof of this.

Answer 1

Consider $P: \tau \longmapsto (E_{\tau}=\mathbb{C}/(\mathbb{Z}{\tau}\oplus \mathbb{Z}),\langle 1/N\rangle,1/p^s)$.

Let $g=\begin{bmatrix}a&b\\c&d\end{bmatrix} \in SL_2(\mathbb{Z})$. The isomorphism $E_{\tau} \rightarrow E_{g\tau}$ is given (lifted to $\mathbb{C}$ by dividing by $c\tau+d$. Then $P(g\tau)$ and $P(\tau)$ are isomorphic iff under this isomorphism,

1. $\frac{\mathbb{Z}}{N}+\tau\mathbb{Z}$ is mapped to $\mathbb{Z}{N}+(g\tau)\mathbb{Z}$,

2. $\frac{1}{p^s}+\mathbb{Z}+\tau\mathbb{Z}$ is mapped to $\frac{1}{p^s}+\mathbb{Z}+(g\tau)\mathbb{Z}$.

3. is equivalent to $\frac{\mathbb{Z}}{N}+\tau\mathbb{Z}=\frac{\mathbb{Z}(c\tau+d)}{N}+(a\tau+b)\mathbb{Z}$, which is equivalent (these lattices have the same index in $\mathbb{Z}+\tau\mathbb{Z}$) to $\frac{c\tau+d}{N} \in \frac{\mathbb{Z}}{N}+\tau\mathbb{Z}$. This, in turn, is equivalent to $N|c$ ie $g \in \Gamma_0(N)$.

4. is equivalent to $\frac{c\tau+d}{p^s}+(c\tau+d)\mathbb{Z}+(a\tau+b)\mathbb{Z}=\frac{1}{p^s}+\mathbb{Z}+\tau\mathbb{Z}$. This is equivalent to $\frac{c\tau+d}{p^s}+\Lambda=\frac{1}{p^s}+\Lambda$ ($\Lambda= \mathbb{Z}+\tau\mathbb{Z}$) ie to $\frac{c\tau+d}{p^s} \in \frac{1}{p^s}+\Lambda$. This is equivalent to $p^s|c$ and $p^s|d-1$, ie $g \in \Gamma_1(p^s)$.

Thus $P(\tau)$ and $P(g\tau)$ are isomorphic iff 1) and 2) hold ie iff $g \in \Gamma_0(N) \cap \Gamma_1(p^s)$.

The compatibility with the degeneracy map is more or less straightforward.

Now all we need to show is that $P$ is “onto”, that is, given a lattice $L \subset \mathbb{C}$, and two points $A,B \in \mathbb{C}$ with orders $N$ and $p^s$ respectively in $\mathbb{C}/L$, there exists a point $D \in \mathbb{C}$ such that, in $\mathbb{C}/L$, $p^sD=A$ and $ND=B$ (so that $D$ has order exactly $Np^s$ and $(Np^sD, E)$ for some point $E$ is a basis of $L$).

Now consider two integers $u,v$ such that $uN+vp^s=1$, then take $D=vA+uB$.