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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?

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up vote 2 down vote accepted

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 $SL_2(\mathbb{Z})$.

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You mean we can define an weight-$k$ operator $[\gamma]$ on functions on upper half plane for $\gamma\in GL_2(\mathbb{Z})$ such that $f[\gamma](\tau)=f(\gamma(\tau))\cdot j(\gamma,\tau)^{-k}\cdot |\gamma|^{k/2}$(or $-k/2$)? – hxhxhx88 Jan 10 '13 at 9:36
I don't know why do we need to add a factor $\det{\gamma}^{k/2}$ in front? I can verify that without such factor, we still have $[\gamma_1][\gamma_2]=[\gamma_1\gamma_2]$ for any $\gamma_1,\gamma_2\in\text{GL}_2(\mathbb{Z})$, so isn't it enough for well-definition? – hxhxhx88 Jan 10 '13 at 9:44
One can define the action of $SL_2(\mathbb{R})$ similar to $SL_2(\mathbb{Z})$. The reason for adding determinant factor is for consistency. For example, $\begin{bmatrix}N & \\ & 1\end{bmatrix}$ should have the same action as $\begin{bmatrix} \sqrt{N} & \\ & 1/\sqrt{N}\end{bmatrix}$. – user27126 Jan 10 '13 at 16:45
OK, thank you!. – hxhxhx88 Jan 11 '13 at 1:12

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