Does anyone know how the Iwasawa decomposition of $SU(1,1)=KAN$ and the centralizer $M$ of $A$ in $K$ looks like? Thanks in advance!
Edit: $SU(1,1)\cong SL(2,\mathbb{R})$ so one can optain the Iwasawa decomposition by using the Iwasawa decomposition of $SL(2,\mathbb{R})$, right?
Edit: My definition of $SU(n,1)$ is $$SU(n,1):=\{f:\mathbb{C}^{n+1} \rightarrow \mathbb{C}^{n+1}\text{ linear}\mid Q(f(x))=Q(x)\ \forall x\in\mathbb{C}^{n+1}\text{ and }\det(f)=1\}$$ where $Q$ is the quadratic form defined by $Q(x):=|x_1|^2+\ldots+|x_n|^2-|x_{n+1}|^2$.
The maximal parabolic subgroup is of the form $$ M(a,b)=A N =\left(\begin{array}{cc} a&b\\ 0&a^{-1}\end{array}\right) $$ with $$ A=\left(\begin{array}{cc} a & 0 \\ 0&a^{-1}\end{array}\right)\, ,\qquad N= \left(\begin{array}{cc} 1 & b \\ 0&1\end{array}\right)\, . $$
$SL(2,\mathbb{R})$ elements are then of the form of $SO(2)M(a, b)$ where $SO(2)$ contains elements of the form $$ K=\left(\begin{array}{cc} \cos\theta&\sin\theta\\ -\sin\theta&\cos\theta\end{array} \right) $$ There are some details in this student project by S. Jana:
As to your comment regarding a look up table, see table 1 of this paper: