# Does $\text{SU}(1,1)$ act transitively on $\mathbb S^1=\{z\in\mathbb C\mid |z|=1\}$?

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?

Observations:

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 $$\mathbb{S}^1$$.