# product of modular forms of half integral weight on different congruence subgroups.

my question is about the theorem 5.11 in the book "complex analysis" from Freitag. It claims that if $SL_2(\mathbb{Z}) = \bigcup_{i=1}^n \Gamma M_i$ for a given congruence subgroup $\Gamma$. And we look after $F:= \prod_{i=1}^n f \vert M_i$ for $f \in [\Gamma, r/2, v]$, then $F$ is a modular form of weight $kr/2$ for $SL_2(\mathbb{Z})$. I don't understand how $F$ can transform respecting all matrices in $SL_2(\mathbb{Z})$. As I see it just transform respecting the intersection of all congruence groups for which one of the forms $f \vert M_i$ transforms. Thanks for your help.

Hari.

• Your $n$ should be a $k$ (or the other way around).
– user301452
Jul 23 '17 at 13:43
• yes, thank you. Jul 23 '17 at 16:06
• That's the same idea as Hecke operators. Jul 26 '17 at 5:47

Let $A\in\text{SL}_2(\Bbb Z)$. Then the cosets $\Gamma M_i A$ are the $\Gamma M_i$ in some order: $\Gamma M_i A=\Gamma M_{\tau(i)}$ for some permutation $\tau$. Thus $M_iA=C_i M_{\tau(i)}$ where $C_i\in\Gamma$. Then $$F|A=\prod_i f|M_iA=\prod_i f|C_iM_{\tau(i)}=\prod_i f|M_{\tau(i)}=F.$$
• Thanks for the answer, but why are $\Gamma M_i A$ for different indices disjoint, so that there needs to exist such a permutation? And your proof only hold for integral weight, but it should work similarly for halfintegral weight! Jul 23 '17 at 16:02