Why $\int_{U} d(m n m^{-1}) = \int_{mUm^{-1}} dn = \omega^2_E(t_1) \int_U dn$?

92 Views Asked by At

I am reading the lecture notes.

On page 5, formula (1.24) is
$$ \int_{U} d(m n m^{-1}) = \int_{mUm^{-1}} dn = \omega^2_E(t_1) \int_U dn, $$ where $dn$ is the Haar measure on $N$, $U \subset N$ is open and compact, $$ m=m(t_1, s_1), \\ m(t,s)=\left( \begin{matrix} t & 0 & 0 \\ 0 & s & 0 \\ 0 & 0 & \bar{t}^{-1} \end{matrix} \right), $$ $t\in E^{\times}$, $s \in E^{1}$, $E^1=\ker N_{E/F}$, $E/F$ is a quadratic extension.

Why $\int_{U} d(m n m^{-1}) = \int_{mUm^{-1}} dn = \omega^2_E(t_1) \int_U dn$?

Thank you very much.