Why $\Delta(N\tau)\in\mathcal{S}_{12}(\Gamma_0(N))$ There is an argument saying that because $\Delta(\tau)\in\mathcal{S}_{12}(\text{SL}_2(\mathbb{Z}))$, then we have $\Delta(N\tau)\in\mathcal{S}_{12}(\Gamma_0(N))$.
I don't understand the logic here.
If I let $\gamma\in\Gamma_0(N)$, then I can say nothing about $\Delta(N\gamma(\tau))$ because $\tau\mapsto N\gamma(\tau)$ is not in $\text{SL}_2(\mathbb{Z})$.
Can anyone help?
 A: In your other questions I see that you know about the operator $[\gamma]$ acting on functions on upper half plane for $\gamma \in SL_2(\mathbb{Z})$. This operator can be generalized to $GL_2(\mathbb{Z})$ with $(det \gamma)^{k/2}$ in front, with $k$ being the weight. (The exponent is either $k/2$ or $-k/2$ - I don't remember it right now)
One can show that this still defines a right action. Thus up to a scalar factor, your $\Delta (N\tau)$ is really $\Delta [\gamma]$, with $\gamma = \begin{bmatrix} N & \\ & 1 \end{bmatrix}$. Your question then follows from the identity
$$\begin{bmatrix}N & \\ & 1 \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} N^{-1} \\ & 1 \end{bmatrix} = \begin{bmatrix} a & Nb \\ c/N & d\end{bmatrix}$$
which shows that if $\sigma = \begin{bmatrix}a & b \\ c& d\end{bmatrix} \in \Gamma_0(N)$, then $\sigma' = \begin{bmatrix} a & Nb \\ c/N & d\end{bmatrix} \in SL_2(\mathbb{Z})$ and
$$\Delta [\gamma][\sigma] = \Delta [\sigma'][\gamma] = \Delta [\gamma]$$
since $\Delta$ is modular with respect to $SL_2(\mathbb{Z})$.
