In many literatures, I can find the definition of the congruence subgrorup $\Gamma_{0}(2)$. It is defined by

$\Gamma_{0}(2) = \left \{ \left(\begin{array}{cc} a & b \\ c & d \end{array} \right) \in \text{SL}(2,\mathbb{Z}), \ c\equiv0 \ (\text{mod} \ 2) \right \}$

And its cusp form can be found by SAGE.

On the other hand, now I want to consider the different subgroup $\Gamma^{0}(2)$ which is defined by

$\Gamma^{0}(2) = \left \{ \left(\begin{array}{cc} a & b \\ c & d \end{array} \right) \in \text{SL}(2,\mathbb{Z}), \ b\equiv0 \ (\text{mod} \ 2) \right \}$

Is there any known cusp forms for this subgp $\Gamma^{0}(2)$? Is it possible to generate them by SAGE?

Thank you.

  • 12
    $\begingroup$ Note $\Gamma^0(2)=\alpha^{-1}\Gamma_0(2)\alpha$ for $\alpha=[[1,0],[0,2]]$. $\endgroup$
    – Wojowu
    Sep 25, 2017 at 9:19

1 Answer 1


If $F(z)$ is a modular (cusp) form for $\Gamma_0(N)$, then $F(\frac zN)$ is a modular (cusp) form for $\Gamma^0(N)$. The converse is also true.

This follows directly from the observation that $\Gamma^0(N)=\alpha_N^{-1}\Gamma_0(N)\alpha_N$, where $\alpha_N=\begin{pmatrix}1&0\\0&N\end{pmatrix}.$ In particular, if $\gamma\in \Gamma^0(N)$, then let $\gamma'=\alpha_N\gamma\alpha_N^{-1}\in \Gamma_0(N).$ If $F$ is modular for $\Gamma_0(N)$ with weight $k$, let $G(z)=F|_k\alpha_N(z)=F(z/N).$ Then we have $$F|_k(\alpha_N\gamma\alpha_N^{-1})=F|_k\gamma'=f$$ By acting on th right by $\alpha_N$, this becomes $$F|_k\alpha_N\gamma=F|_k\alpha_N,$$ or $$G|_k\gamma=G.$$ The other direction follows similarly.

SAGE only computes modular forms with integral powers of $q$, so no it won't compute these directly, but as shown, all you need to do is compute the expansions for modular forms for $\Gamma_0(N)$ and replace $q$ with $q^{\frac 1N}.$


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