General form of Matrix in SU(1,1),PSU(1,1) and Unit Disk Automorphisms I have the definition of $SU(1,1)$ as
$SU(1,1)=\{M\in M_{2}(\mathbb{C}) ; M^*JM=J, detM=1\}$ where
$J=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$,
and $PSU(1,1)$ as the image of $SU(1,1)$ in $PGL_2(\mathbb{C})$ (Though I don't quite understand what that means).
I'm a little confused as to how to get the general form of a matrix in $SU(1,1)$ -
I've read it's supposed to be something like $\begin{pmatrix} a & b \\ b^* & a^* \end{pmatrix}$ but I can't seem to be able to show it.
The purpose of the question is to eventually show:
$PSU(1,1)\cong \operatorname{Aut}(D(0,1))=\{e^{i\theta}\frac{z-\alpha}{1-\bar\alpha z};|\alpha|<1,\theta\in\mathbb{R}/2\pi\mathbb{Z}\} $ by
$\begin{pmatrix} a & b \\ c & d \end{pmatrix}\mapsto \frac{az+b}{cz+d}$.
And I struggle to show any matrix in PSU(1,1) is mapped to a function of the form $e^{i\theta}\frac{z-\alpha}{1-\bar\alpha z}$.
 A: Suppose $\big(\begin{smallmatrix} a & b \\ c & d\end{smallmatrix}\big)\in\mathrm{SU}(1,1)$. Writing out the condition $M^\ast J M=J$ explicitly gives
$$ \begin{cases} 
|a|^2-|b|^2=1 \\
\overline{a}c-\overline{b}d=0 \\
|c|^2-|d|^2=-1
 \end{cases} $$
Suppose $(a,b)$ is given and we want to know $(c,d)$. The second equation above and be combined with the determinant condition $ad-bc=1$. This is a $2\times 2$ linear system, so the solution $(c,d)$ is unique, whatever it is. Then we can check $(c,d)=(\overline{b},\overline{a})$ is the solution. Therefore we can say
$$ M=\begin{pmatrix} a & b \\ \overline{b} & \overline{a}\end{pmatrix}, \qquad |a|^2-|b|^2=1. $$
We can check all matrices of the above form are indeed elements of $\mathrm{SU}(1,1)$.
There is a projection map $\mathrm{GL}_2\mathbb{C}\to\mathrm{PGL}_2\mathbb{C}$. The word "image" means "range," so $\mathrm{PSU}(1,1)$ is the range of this projection map restricted to the subset $\mathrm{SU}(1,1)\subset \mathrm{GL}_2\mathbb{C}$. Equivalently, $\mathrm{PSU}(1,1)\cong\mathrm{SU}(1,1)/\{\pm I_2\}$.
Given $M$ as above, we may rewrite its Mobius transformation as
$$ \frac{az+b}{\overline{b}z+\overline{a}} = \frac{a}{\overline{a}} \frac{z+(b/a)}{1+(\overline{b/a})z} $$
and then solve $e^{i\theta}=a/\overline{a}$ and $\alpha=-b/a$ to get the form you want.
