Iwasawa decomposition and Haar integration formula for SL2(R)

Write $g=QR$ by [F3]. Since $\det g=1$ and $R$ has positive diagonal, $\det Q=1$, so $Q\in K$. Writing $a=R_{11}>0$ and $R_{12}=ax$, the determinant condition gives $R_{22}=a^{-1}$ and $R=\mathrm{diag}(a,a^{-1})n_x$. The first column gives $a=\sqrt{p^2+q^2}$ and $Q=a^{-1}\left(\begin{array}{cc}p& -q\\ q& p\end{array}\right)$; the second gives $x=(pr+qs)\mathrm{/}{a}^2$. QR uniqueness proves the factorization unique.

Multiplication is smooth. Its inverse is given by the displayed formulas, with $a>0$ because the first column of a determinant-one matrix is nonzero; $a$, $a^{-1}$, $k$, and $x$ therefore depend smoothly on $g$. The coordinate $t=2\log a$ is smooth as well, so the multiplication map is a diffeomorphism.

For fixed $g_0\in G$, write uniquely $g_0{k}_{\theta }=k_{{\theta }_1}{a}_{u(\theta )}{n}_{v(\theta )}$. With $e_1=(1,0{)}^T$, put $w(\theta )=g_0{k}_{\theta }{e}_1=r(\theta )k_{{\theta }_1}{e}_1$, where $r=e^{u\mathrm{/}2}>0$. Since $\det (k_{\theta }{e}_1,(k_{\theta }{e}_1{)}^{\mathrm{\prime }})=-1$ and $\det g_0=1$, differentiating this identity gives $-1=-r^2{\theta }_1^{\mathrm{\prime }}$, hence ${\theta }_1^{\mathrm{\prime }}=e^{-u}>0$. Left translation sends $(k_{\theta },t,x)$ to $(k_{{\theta }_1},t+u,x+e^{-t}v)$; its $(t,x)$-Jacobian is $1$, so its full orientation-preserving Jacobian is $e^{-u}$. The target density is $e^{t+u}dkdtdx$, and its pullback is $e^{t+u}{e}^{-u}dkdtdx=e^{t}dkdtdx$. Thus the smooth density is left invariant; the change-of-variables theorem applies to compactly supported smooth test densities.

Let $\eta =e^{t}dkdtdx$ be the positive smooth density in the global $K\times {\mathbb{R}}^2$ chart. By [F8] it defines a Radon Borel measure ${\mu }_{\eta }$ finite on compact sets, and [F9] identifies its Borel integral with the displayed coordinate integral. Step 3.1 makes ${\mu }_{\eta }$ left invariant. A nonnegative smooth bump supported in a nonempty coordinate box has positive integral because $\eta$ is positive, so $I(f):=\int fd{\mu }_{\eta }$ is nonzero. It is positive and real-linear by the Lebesgue integral properties, and [F10] gives left invariance of $I$; hence $I$ is a left Haar integral and ${\mu }_{\eta }$ is a left Haar measure by [F6].

Put $H=\mathrm{diag}(1,-1)$, $e=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, $f=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$, $J=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, and $S=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Direct conjugation gives $\mathrm{Ad}(k_{\theta })J=J$, $\mathrm{Ad}(k_{\theta })H=\cos (2\theta )H-\sin (2\theta )S$, and $\mathrm{Ad}(k_{\theta })S=\sin (2\theta )H+\cos (2\theta )S$, so its determinant is $1$. Also $\mathrm{Ad}(a_t)$ has eigenvalues $1,e^t,e^{-t}$ on $(H,e,f)$. On $(e,H,f)$, $\mathrm{Ad}(n_x)e=e$, $\mathrm{Ad}(n_x)H=H-2xe$, and $\mathrm{Ad}(n_x)f=f+xH-x^{2}e$, so its determinant is $1$. Since $\mathrm{Ad}$ is multiplicative and $G=KAN$, $\det \mathrm{Ad}(g)=1$ for every $g\in G$. For a left-invariant density $\eta$, comparing $d{R}_g$ with $d{L}_{g^{-1}}$ at the identity gives $d{L}_{g^{-1}}d{R}_g=\mathrm{Ad}(g^{-1})$, hence $R_g^{\ast }\eta =\mathrm{\mid }\det \mathrm{Ad}(g^{-1})\mathrm{\mid }\eta =\eta$. Thus ${\mu }_{\eta }$ is right invariant and $G$ is unimodular by [F7].

Inversion pulls the right-invariant density $\eta$ back to a left-invariant density; its differential at the identity is $-I$ on the three-dimensional tangent space, whose absolute determinant is $1$. Since a left-invariant density is determined by its value at the identity, inversion preserves ${\mu }_{\eta }$. Applying inversion invariance to the KAN integral and using $(k{a}_t{n}_x{)}^{-1}=n_{-x}{a}_{-t}{k}^{-1}$, then substituting $(k,t,x)\mapsto (k^{-1},-t,-x)$, gives the same measure in NAK coordinates with density $e^{-t}dkdtdx$; here $dk=d\theta \mathrm{/}(2\pi )$ is inversion invariant by [F2].