0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Your $n$ should be a $k$ (or the other way around). $\endgroup$
    – user301452
    Jul 23 '17 at 13:43
  • $\begingroup$ yes, thank you. $\endgroup$
    – Hari.M.S.
    Jul 23 '17 at 16:06
  • $\begingroup$ That's the same idea as Hecke operators. $\endgroup$
    – reuns
    Jul 26 '17 at 5:47
0
$\begingroup$

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 $$F|A=\prod_i f|M_iA=\prod_i f|C_iM_{\tau(i)}=\prod_i f|M_{\tau(i)}=F.$$

$\endgroup$
1
  • $\begingroup$ 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! $\endgroup$
    – Hari.M.S.
    Jul 23 '17 at 16:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.