Let $\text{SU}(1,1)=\left\{\left[ \begin{array}{ccc} \alpha & \beta \\\overline\beta & \overline\alpha \end{array} \right]\mid \alpha,\beta\in \mathbb C,|\alpha|^2-|\beta|^2=1\right\}$ and $\mathbb S^1=\{z\in\mathbb C\mid |z|=1\}$.

I have already proved that the mapping $f:\text{SU}(1,1)\times \mathbb S^1\to \mathbb S^1$ given by $\biggr(\left[ \begin{array}{ccc}\alpha & \beta \\\overline\beta & \overline\alpha \end{array} \right], z\biggr)\mapsto\dfrac{\alpha z+\beta}{\overline\beta z+\overline\alpha}$ is a group action. Is $f$ transitive?


I can connect $i$ and $-i$ using the matrix $A=\left[ \begin{array}{ccc}i & 0 \\0 & -i \end{array} \right].$ Using the same matrix, $1$ can be sent to $-1$. As a next step, I was trying to find a matrix $B=\left[ \begin{array}{ccc}\alpha & \beta \\\overline\beta & \overline\alpha \end{array} \right]\in\text{SU}(1,1)$ which takes $i$ to $1$, that is $$\dfrac{\alpha i+\beta}{\overline\beta i+\overline\alpha}=1\tag1.$$ Writing $\alpha=x_1+iy_1$ and $\beta=x_2+iy_2$ and substituting in $(1)$ gives me $$x_1+y_2=x_2-y_1.$$ Now since $|\alpha|^2-|\beta|^2=x_1^2+y_1^2-(x_2^2+y_2^2)=1$, I have \begin{align*}x_1^2+y_1^2-(x_2^2+y_2^2)&=x_1^2-y_2^2-(x_2^2-y_1^2)\\&=(x_1+y_2)(x_1-y_2)-(x_2-y_1)(x_2+y_1)\\&=(x_1+y_2)(x_1-y_2-x_2-y_1)\\&=1,\end{align*}which seems to be of no use. So is $f$ transitive?

Hints are welcome.


For $e^{i\theta_1}$, $e^{i\theta_2} \in \mathbb{S}^1$, let $\alpha = e^{i\frac{\theta_2 - \theta_1}{2}}$ and $\beta = 0$, then $|\alpha|^2 - |\beta|^2 = 1$ and \begin{align} \frac{\alpha e^{i\theta_1} + \beta}{ \overline{\beta}e^{i\theta_1} + \overline{\alpha}} = \frac{e^{i\frac{\theta_2 - \theta_1}{2}}e^{i\theta_1}}{e^{-i\frac{\theta_2 - \theta_1}{2}}} = e^{i\theta_2} \end{align} which proves transitivity.

Following your line of thought, you can also try to find a matrix $A\in SU(1,1)$ that sends $1$ to a unit complex number with irrational angle, say $e^{i}$, then prove that the set $$ \{ A^n(1) \mid n\in \mathbb{N} \} $$ is dense in $\mathbb{S}^1$. A bit overkill but very fun.

It might also be interesting for you to know that, when acting on $\mathbb{S}^2 = \mathbb{P}^1(\mathbb{C})$, $SU(1,1)$ has exactly $3$ orbits, one of which is $\mathbb{S}^1$.


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