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Let $\Gamma_0(4)$ be a congruence subgroup of $SL(2,\mathbb{Z})$ defined as

$$\Gamma_0(4)=\{M=\begin{pmatrix} a &b\\ c& d \end{pmatrix}\in SL(2,\mathbb{Z})\mid c=0\bmod 4\}.$$ Dedekind eta-function is defined as $$\eta(z)=q^{1/24}\prod_{n\geq1}(1-q^n).$$

How to prove that the ideal of cusp forms for $\Gamma_0(4)$ is principal and generated by

$$f(z)=\eta(2z)^{12}?$$

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The form $f(z)$ generates the $\Gamma_0(4)$ cuspforms (as an ideal over the graded ring of all modular forms for $\Gamma_0(4)$) for much the same reason that $\Delta(z) = \eta(z)^{24}$ generates the cuspforms for the full modular group $\Gamma(1) = {\rm SL}_2({\bf Z})$: this $f$ is a cuspform with no zeros in the finite plane, and it vanishes at each cusp only to the multiplicity needed to make it a cuspform. Hence if $\varphi$ is any other cuspform, say of weight $w$, then $\varphi_1 := \varphi/f$ is a modular form of weight $w-6$, and we have written $\varphi$ as $f \varphi_1$ for some modular form $\varphi_1$ as desired.

The fact that there are no zeros in the finite plane is clear from the product formula for $\eta(z)$. The vanishing order at the cusps is easier to see if we change variables to $w := 2z$, because then $f(z) = \eta(w)^{12}$ is a cuspform for the action of $\Gamma(2)$ on $w$, and the three cusps of $\Gamma(2)$ all have the same width (namely $2$) because they're transitively permuted by $\Gamma(1)$, so $\eta(w)^{12}$ vanishes to order $1$ at each cusp.

Similar examples for other groups are $\eta(3z)^8$ for $\Gamma_0(9)$ (equivalently $\eta(z)^8$ for $\Gamma(3)$), and $(\eta(z)\eta(pz))^w$ for $\Gamma_0(p)$ with $p=2,3,5,11$ (i.e. the primes with $p+1\mid12$) and $(p+1)w=24$. The last of these also works for $p=7$ and $p=23$ if we allow modular forms with quadratic character.

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  • $\begingroup$ Noam, thank you! There is a little misprint in your answer: it seems that it should be $\phi_1=\phi/f$. $\endgroup$ – vitaliy Jun 17 '15 at 21:39
  • $\begingroup$ You're welcome, and thank you for the correction; I'll fix it next (though retaining $\varphi$ in place of $\phi$ [\varphi vs. \phi in TeX). $\endgroup$ – Noam D. Elkies Jun 17 '15 at 21:43
  • $\begingroup$ Dear Noam, can you explain please why the coefficients of the expansion of $f(z)$ are multiplicative? I know that this fact is related to the notion of Hecke operators for the group $\Gamma_0(4)$ but I haven't found the suitable reference. Actually as I understand I need to prove that $f(z)$ (which is normalized) is the eigenform of Hecke operators for $\Gamma_0(4)$. Thank you. $\endgroup$ – vitaliy Jun 18 '15 at 20:54
  • $\begingroup$ In Zagier's "1-2-3" the case of congruence subgroups of $SL_2(\mathbb{Z})$ is not explained. $\endgroup$ – vitaliy Jun 18 '15 at 20:56
  • $\begingroup$ The Hecke operators act on the space of cusp forms, and since that space is one-dimensional $f$ is automatically an eigenform. Sorry I don't have a reference to hand (maybe somebody else here can suggest one) but you might be able to figure out for yourself how to generalize the definition of $T_m$ that you learned from Zagier to congruence groups $\Gamma_0(n)$ with $\gcd(m,n)=1$ (and how this yields multiplicativity at least for coefficients of $q^k$ with $\gcd(k,n)=1$). $\endgroup$ – Noam D. Elkies Jun 18 '15 at 22:02

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