TITLE: Haar measure on $PSL_2(\mathbb{R})$ via Iwasawa decomposition
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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.