Question: I am working through the proof of the following theorem 3.1.19 (Page 86) from Deitmar's Automorphic Forms,
(Iwasawa Integral formula) Let $G = SL_2(\mathbb{R})$ and $A,N,K$ be subgroups of $G$ defined as $$A= \left\{ a_t:= \begin{pmatrix} e^{t} & 0\\ 0 & e^{-t} \end{pmatrix} : t\in \mathbb{R} \right\}$$ $$N \left\{n_x := \begin{pmatrix} 1 & x\\ 0 & 1 \end{pmatrix} : x \in \mathbb{R} \right\}$$ and $$K = SO(2) = \left\{ k_\theta := \begin{pmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{pmatrix} : \theta \in \mathbb{R} \right\}$$ For any Haar measures on three of the groups $G,A,N,K$ there is a unique Haar measure on the fourth, so that for every $f \in L^1(G)$ we have $$\int_G f(x) \, dx = \int_A \int_N \int_K f(ank) \, dk \, dn \, da.$$ The author chosen fixed Haar measures defined as follows
- On the compact group $K$, its Haar measure $\mu_K$ satisfy $\mu_K(K)= 1$.
- On $A$, it's measure is $2 \, dt$ where $t = \underline{t}(a)$ so that $$a \in A \implies a = \begin{pmatrix} e^{ \underline{t}(a)} & 0\\ 0 & e^{-\underline{t}(a)}\end{pmatrix}$$
- On $N$, we define its Haar measure as $\int_{\mathbb{R}} f(n_s) \, ds.$ Next, he consider the Borel subgroup $ B:= AN$ consisting of all upper triangular matrices with positive diagonal entries and $d b:= da \, dn$ is a Haar measure on $B$ and $B$ is not unimodular. As $\Delta_B(a_t) = e^{-2t}$ since $a_t n_x a_s n_y = a_{t+s} n_{y + e^{-2s} x}$
My question is how did he obtain the express for the modular function $\Delta_B$, as I am aware for any measurable $E\subset B$, $E$ can be written as $E = A_E N_E$ where $A_E$ is measurable in $A$ and $N_E$ is measurable in $N$. So, given $a_t \in A$ $$\mu_B(E) = \mu_A(A_E) \mu_N(N_E)$$ and $$\mu_B(E a_t) = \Delta_B(a_t) \mu_B(E).$$ How did the author use the identity $a_t n_x a_s n_y = a_{t+s} n_{y + e^{-2s} x}$ to obtain $\Delta_B(a_t) = e^{-2t}$?