I am reading the book representation theory of semisimple groups. On page 33, I tried to verify that $\mathcal{P}^{k,iv}$ is unitary. We need to verify that $$ \left|\left| \mathcal{P}^{k,iv}\left(\begin{matrix} a & b \\ c & d \end{matrix} \right)f \right|\right| = ||f||, \quad f \in L^2(\mathbb{C}). $$ We have \begin{align} & \left|\left| \mathcal{P}^{k,iv}\left(\begin{matrix} a & b \\ c & d \end{matrix} \right)f \right|\right| \\ & = \int_{\mathbb{C}} |-bz+d|^{-2-iv} (\frac{-bz+d}{|-bz+d|})^{-k} f(\frac{az-c}{-bz+d}) \overline{|-bz+d|^{-2-iv} (\frac{-bz+d}{|-bz+d|})^{-k} f(\frac{az-c}{-bz+d}) } dz \\ & = \int_{\mathbb{C}} |-bz+d|^{-4} f(\frac{az-c}{-bz+d}) \overline{ f(\frac{az-c}{-bz+d}) } dz. \end{align} How to show that \begin{align} \int_{\mathbb{C}} |-bz+d|^{-4} f(\frac{az-c}{-bz+d}) \overline{ f(\frac{az-c}{-bz+d}) } dz = \int_{\mathbb{C}} f(z) \overline{ f(z) } dz? \end{align}
Let $$ z_1 = \frac{az-c}{-bz+d}. $$ Then $$ z = \frac{c+z_1d}{a+z_1b}. $$ We have $$ dz = \frac{1}{(a+z_1b)^2}dz_1, \\ |-bz+d| = |a+z_1b|^{-1}. $$ Therefore \begin{align} & \int_{\mathbb{C}} |-bz+d|^{-4} f(\frac{az-c}{-bz+d}) \overline{ f(\frac{az-c}{-bz+d}) } dz = \int_{\mathbb{C}} f(z) \overline{ f(z) } dz \\ & = \int_{\mathbb{C}} |-bz+d|^{-4} f(\frac{az-c}{-bz+d}) \overline{ f(\frac{az-c}{-bz+d}) } dz = \int_{\mathbb{C}} \frac{|a+z_1b|^4}{(a+z_1b)^2} f(z_1) \overline{ f(z_1) } dz_1. \end{align} But I didn't get \begin{align} \int_{\mathbb{C}} |-bz+d|^{-4} f(\frac{az-c}{-bz+d}) \overline{ f(\frac{az-c}{-bz+d}) } dz = \int_{\mathbb{C}} f(z) \overline{ f(z) } dz. \end{align}
Any help will be greatly appreciated!
