Show that $S_6(\Gamma_0(4))=\eta^{12}(2z)\mathbb{C}$
where $\eta(z)=e^{2\pi i z/24}\prod_{n=1}^{\infty}(1-e^{2\pi i nz})$ is the eta-function .
$S_6(\Gamma_0(4))$ is the space of cusp forms of weight 6 on $$ \Gamma_0(4))=\Big\lbrace\begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}\in Sl_2(\mathbb{Z}):c\equiv0 \ mod \ 4 \Big\rbrace $$
Proof. If $ f\in S_6(\Gamma_0(4))$ , we must have a zero of order at least $1$ at each of $i\infty$ , $1/2$ and $0$ . Then for some $\lambda$, $f(z)-\lambda\eta^{12}(2z)$ has a zero of order $2$ at $i\infty$ . By Sturm’s bound we deduce $f(z)=\lambda\eta^{12}(2z) .$
Is it from defintion of cusp forms that $f $ must have a zero of order at least $1$ at $i\infty,0,1/2$ ?
Why does $f(z)-\lambda\eta^{12}(2z)$ have a zero of order 2 at $i\infty$ ?
Thanks for the help .
The Fourier coefficients of $\, \eta^{12}(2z) = q - 12q^3 + 45q^5 + \dots\,$ where $\, q:=\exp(2\pi i z)\,$ is the OEIS sequence A000735. The three inequivalent cusps are $0, 1/2, i\infty$. The Fourier coefficients in $A000735$ give the behavior at $\,z = i\infty\,$ since $\,q=0\,$ there. By definition of Cusp form the Fourier series must have zero constant term and this refers to $\,q=0\,$ or equivalently to $\,z=i\infty.\,$ Use $\, f(-1/(4z)) = 64 (z/i)^6 f(z)\,$ to get the behavior at $\,z=0.\,$ Use $\,f(1/2+z) = -f(z)\,$ to get the behavior at $\,z=1/2.\,$