Show that $(f|M)(z) := (cz+d)^{-r/2}f(Mz)$ has a weight $r/2$.

98 Views Asked by At

The following text is from Complex Analysis by Freitag : enter image description here For $r ∈ \mathbb{Z}$ the modified Petersson notation is defined : $$(f|M)(z) := \sqrt{cz+d}^{-r}f(Mz)$$ for $M ∈ SL(2, \mathbb{Z})$.

In the theorem above (Proposition VI.5.11), f is a modular form of weight $r/2$ which means $f(Mz)=(cz+d)^{r/2}f(z)$ so, all $(f|M)(z)$'s must have weight $0$ because $(f|M)(z) = (cz+d)^{-r/2}f(Mz)=(cz+d)^{-r/2}(cz+d)^{r/2}f(z) =f(z).$ So how $(f|M)(z)$ has weight $r/2$ or/and consequently $F = \Pi (f|M)$ weight $kr/2$? Simple clear explanation would be much appreciated.