(Sorry for my poor english...)
Let $\triangle(z)=q\prod_{n=1}^{\infty}(1-q^n)^{24}$ be a cusp form of weight $12$. Let $N$ be a positive integer, $\delta \mid N$ and $\triangle_{\delta}(z)=\triangle(\delta z)$. Let $s=\frac{b}{d}$ be a cusp on $\Gamma_0(N)$ with $(b,d)=1$ and $\rho\in SL_2(\mathbb{Z})$ with $\rho(s)=i\infty$.
Lemma 3.2.5 in Ligozat's paper ("Courbes modulaires de genre 1) says that \begin{equation} \triangle_{\delta}(\rho^{-1}(z))=\left(\frac{d}{\delta}(d,\delta)z+\beta\right)^{12}\triangle\left(\frac{(d,\delta)^2}{\delta}z+\gamma \right) \end{equation} with some integers $\beta$ and $\gamma$.
In the proof of this lemma, let $\rho=\begin{pmatrix} c & -a\\ d & -b \end{pmatrix}. $ Then, \begin{equation} \triangle_{\delta}(\rho^{-1}(z))=\triangle\left(\begin{pmatrix} -b\delta & a\\ -d & c \end{pmatrix}z\right). \end{equation} I think that \begin{equation} \triangle_{\delta}(\rho^{-1}(z))=\triangle\left(\begin{pmatrix} -b\delta & a\delta\\ -d & c \end{pmatrix}z\right). \end{equation} What's wrong in my idea..?
Furthermore, I want to show Lemma 3.2.5 myself as follow.. \begin{equation} \begin{aligned} \triangle_{\delta}(\rho^{-1}(z))&=\triangle_{\delta}\left(\frac{-cz+a}{-dz+b} \right)\\ &=\triangle\left(\frac{-c\delta'(d,\delta)z+a\delta}{-d'(d,\delta)z+b} \right)\\ &=\triangle\left(\frac{-c\delta'u+a\delta}{-d'u+b} \right) \end{aligned} \end{equation} with $\frac{d}{d'}=\frac{\delta}{\delta'}=(d,\delta)$ and $u=(d,\delta)z$. Then, \begin{equation} \begin{pmatrix} -c\delta'& a\delta\\ -d' & b \end{pmatrix}=\begin{pmatrix} -c\delta' & \alpha\\ d' & \beta \end{pmatrix} \begin{pmatrix} 1 & \gamma \\ 0 & \delta' \end{pmatrix} \end{equation} with $\begin{pmatrix} -c\delta' & \alpha\\ d' & \beta \end{pmatrix} \in SL_2(\mathbb{Z})$ and $\alpha,\beta$ and $\gamma$ are integers. Since $\triangle(z)$ is a cusp form, \begin{equation} \begin{aligned} \triangle\left(\frac{-c\delta'u+a\delta}{-d'u+b} \right)&=\triangle\left(\frac{-c\delta'(\frac{u+\gamma}{\delta'})+\alpha}{-d'(\frac{u+\gamma}{\delta'})+\beta} \right)\\ &=\left(-d'\left(\frac{u+\gamma}{\delta'}\right)+\beta \right)^{12}\triangle\left(\frac{u+\gamma}{\delta'} \right)\\ &=\left(-\frac{d}{\delta}(d,\delta)z+\frac{b}{\delta'} \right)^{12}\triangle\left(\frac{(d,\delta)^2}{\delta}z+\frac{\gamma}{\delta'} \right). \end{aligned} \end{equation} However, $\frac{b}{\delta'}$ and $\frac{\gamma}{\delta'}$ are not integers..
Where is my mistake...?