$\mathcal{Remember:}$
$i)$ $\mathbb{D}={\{z\in\mathbb{C}: |z|<1}\}$
$ii)$ $\text{SU}(1,1)=\left\{\left( \begin{array}{ccc}
\alpha & \beta \\\overline\beta & \overline\alpha \end{array} \right)\mid \alpha,\beta\in \mathbb C,|\alpha|^2-|\beta|^2=1\right\}$
$={\{\left( \begin{array}{cc}
e^{i\theta} & 0\\
0 & e^{-i\theta}
\end{array} \right)\left( \begin{array}{cc}
\text{cosh}(t) & \text{sinh}(t)\\
\text{sinh}(t) & \text{cosh}(t)
\end{array} \right)\left( \begin{array}{cc}
e^{i\phi} & 0\\
0 & e^{-i\phi}
\end{array} \right):\phi,\theta\in\mathbb{R}, t\geq 0}\}$
$iii)$ $\rho: \text{SU}(1,1)\to \text{SU}(1,1)/U(1)$, $\rho(g)=gU(1)$ (canonical projection)
$\mathcal{Problem:}$
I need to find a section of the form:
$G:\mathbb{D}\longrightarrow \text{SU}(1,1)$, $z\ \ \mapsto \left( \begin{array}{ccc}
e^{i\theta(z)} & 0 \\\ 0 & e^{-i\theta(z)} \end{array} \right)\left( \begin{array}{ccc}
C(z) & S(z) \\\ S(z) & C(z) \end{array} \right)$ Such that $C^2(z)-S^2(z)=1$, $S(z)\geq 0$, $C(z)\geq 1$
$\mathcal{Solution:}$
Let $z\in\mathbb{D}$ be then $z=re^{i\psi}$, where $0\leq r<1$ and $\psi= \text{arg}(z)$.
In this way I can define,
$G:\mathbb{D}\longrightarrow \text{SU}(1,1)$, $G(z)=G(re^{i\psi})=\left( \begin{array}{ccc}
e^{i\psi} & 0 \\\ 0 & e^{-i\psi} \end{array} \right)\left( \begin{array}{ccc}
\sqrt{1+r^2} & r\\\ r & \sqrt{1+r^2} \end{array} \right)$. Note that $\sqrt{1+r^2}\geq 1$, $r\geq 0$ and also $(\sqrt{1+r^2})^2-r^2=1$
${Claim:}$ $G:\mathbb{D}\longrightarrow \text{SU}(1,1)$ define a section.
We must show that $\rho \circ G = I_d $, i.e $[G(z)]=[z]$, $\forall z\in\mathbb{D}$
In effect
$[G(z)]=[G(re^{i\psi})]=\left( \begin{array}{ccc}
\sqrt{1+r^2}e^{i\psi} & re^{i\psi}\\\ re^{-i\psi} & \sqrt{1+r^2}e^{-i\psi} \end{array} \right)U(1)$
$=\left( \begin{array}{ccc}
\sqrt{1+r^2}e^{i\psi} & re^{i\psi}\\\ re^{-i\psi} & \sqrt{1+r^2}e^{-i\psi} \end{array} \right)\left( \begin{array}{ccc}
e^{-i\psi} & 0 \\\ 0 & e^{i\psi} \end{array} \right)U(1)$
$=\left( \begin{array}{ccc}
\sqrt{1+r^2} & re^{2i\psi}\\\ re^{-2i\psi} & \sqrt{1+r^2} \end{array} \right)U(1)$
$=re^{i\psi}U(1)$?
My question is: can I identify the elements of $\mathbb{D}$ with those of $ \text{SU}(1,1)$ in such a way that what I intend to prove is fulfilled?
Best Answer
Identify $SU(1,1)/U(1)$ with $\Bbb D$ via $g\mapsto g(0)$, where we interpret $g$ as a Mobius transformation.
The section condition now reads $G(g)(0)= e^{i\theta(z)}S(z)/C(z)=z$. Clearly $\theta(z)=\psi$. Taking absolute values, we may substitute $|S(z)|=|z||C(z)|$ into $C(z)^2-S(z)^2=1$ to solve for $C(z)$ and $S(z)$.