Haar measure on $PSL_2(\mathbb{R})$ via Iwasawa decomposition

Question

I am missing something basic about the relation between the Haar measure on the group $G = KAN$ and the haar measures on the subgroups $K$, $A$, and $N$. Specifically, let $G=PSL_2(\mathbb{R})$ then we can write in explicit coordinates the subgroups $$ K = \left\{\left(\begin{array}{cc} \cos(\theta) & \sin(\theta)\\ -\sin(\theta)&\cos(\theta) \end{array}\right):\theta\in[0,\pi)\right\} $$ $$ A = \left\{\left(\begin{array}{cc} y^{1/2}& 0\\ 0&y^{-1/2} \end{array}\right):y\in(0,\infty)\right\} $$ $$ N = \left\{\left(\begin{array}{cc} 1& x\\ 0&1 \end{array}\right):x\in\mathbb{R}\right\} $$ In these coordinates, it would seem to me the Haar measures of these subgroups is $d\theta$, $y^{-1}dy$ and $dx$ respectively. However I know the Haar measure on $G$ is $y^{-2}\, d\theta \, dy \, dx$. I am confused because it is often stated that $dg = dk \, da \, dn$ where it seems those measures are the Haar measures on the subgroups, but somehow in explicit coordinates a new factor of $y^{-1}$ appears. Thank you.

Answer 1

Your problem is that $\mathrm{PSL}(2,\mathbb{R})$ is not a reductive Lie group (at least by the definitions I know). You are correct that the measures on each subgroup are as you say, but note that $AN$ is the subgroup isomorphic to the group of matrices $$ \begin{pmatrix} a& b \\ 0&1 \end{pmatrix} $$ (for some reason the TeX will not work) with $a>0$.

The left and right Haar measures on this group are different! Your measure on $G$ corresponds to taking the right Haar measure on $AN$, the one you think you have found corresponds to taking the left one.

Do you know a reference where the Iwasawa decomposition is stated for non-redcutive groups?

Edit: The set $K$ in your question is not closed under multiplication.