The Limit Set of a Cyclic Group Generated by an Irrational Rotation and Acting on the Complex Projective Line

Question

Let $\gamma$ be an irrational rotation (elliptic element) of ${\rm PSL} (2,\mathbb{C})$, $\Gamma$ the group generated by $\gamma$ $\left({\rm i.e.,} \Gamma= \langle \gamma \rangle \right)$ and $\mathbb{C}\mathbb{P}^1$ the complex projective line.

Suppose the group $\Gamma$ acting on $\mathbb{C}\mathbb{P}^1$.

First of all, it is well-known that the elliptic element $\gamma$ possesses two fixed points $x$ and $y$. Meanwhile, the Limit Set, by definition, $\Lambda(\Gamma)$ of the group $\Gamma$ is the minimal closed subset of $\mathbb{C}\mathbb{P}^1$ that is invariant under the action of the group $\Gamma$.

Under the assumption above:

1. what is the limit set of the group $\Gamma$?
2. Do the two fixed points $x$ and $y$ of $\gamma$ belong to the limit set $\Lambda(\Gamma)$ or its complement?

Answer 1

A more terrestial definition of $\Lambda(\Gamma)$:

Let $l\in\mathbb{CP}^1$. If there is a sequence $\{T_n\}\subset\Gamma$ of distinct elements of $\Gamma$ and $z\in\mathbb{CP}^1$ such that $$T_n(z)\rightarrow l$$ then $l$ is a limit point of $\Gamma$. The set of all such $l$'s is called a limit set $\Lambda(\Gamma)$.

Note for the remainder of this comment we refer to $\mathbb{CP}^1$ as the Riemann sphere $\Sigma$.

Let $\pi:\mathbb{S}^2/\{N\}\rightarrow \mathbb{C}$ be the stereographic projection of the $2$-sphere less the North pole onto the complex plane. We can extend $\pi$ to a bijection $\pi:\mathbb{S}^2\rightarrow\Sigma$ by defining $\pi(N)=\infty$. Let $J:\Sigma\rightarrow\Sigma$ be the inverse map $J(w)=w^{-1}$, with the convention $J$ sends $\infty$ to $0$ and vice versa. The following diagram

$$\require{AMScd}\begin{CD}\mathbb{S}^2 @>{\pi^{-1}J\pi}>> \mathbb{S}^2\\ @V{\pi}VV @VV{\pi}V \\\Sigma @>>{J}> \Sigma\end{CD}$$

shows that we can work more simply in $\mathbb{S}^2$.

Lemma: For any $z\in\mathbb{S}^1$, the orbits of $z$ under the irrational rotation $r_\alpha$ are dense in $\mathbb{S}^1$.

Proof: Euler's formula $$e^{i\theta}=\cos\theta+i\sin\theta$$ maps the line $i\mathbb{R}$ isometrically onto the unit circle $\mathbb{S}^1$ modulo $2\pi$. Hence it suffices to show the additive group $(G,+)$ generated by $\alpha$ and $2\pi$ is dense in $\mathbb{R}$.

Assume, for $n\neq m$, $n\alpha$ and $m\alpha$ are congruent mod $2\pi$, then $$\alpha=\frac{2k\pi}{n-m},$$ which implies $e^{i\alpha}$ is a root of unity, a contradiction. Hence the sequence $$\alpha,2\alpha,3\alpha,\cdots\pmod{2\pi}$$ is unique. Let $$A=\{(n\alpha-2k\pi)\in(0,2\pi):n=1,2,\cdots\}\subset G,$$ then $A$ is a bounded infinite sequence and so, for any $\epsilon\gt 0$ there are $\beta,\phi\in A$ such that $$\lvert\beta-\phi\rvert\lt\epsilon.$$ Since $A\subset G$, $\beta-\phi\in G$, thus, for any interval $I\subset\mathbb{R}$ of width $\epsilon$, we can find $\psi\in G$ such that $n\psi\in I$. $\qquad\square$

A rotation of $\alpha$ around antipodal points $x,y\in\mathbb{S}^2$ leaves invariant any $\mathbb{S}^2$-circle $\mathscr{C}_x$ centred on $x$ (equivalently, $y$). Hence, take any point $z\in\mathbb{S}^2$, then $z\in\mathscr{C}_x$. Take any other $u\in\mathscr{C}_x$, then, by the Lemma above, orbits of $u$ under $\gamma$ can be found arbitrarily close to $z$ and we can thus build a sequence $\{T_n\}\subset\Gamma$ such that $$T_n(u)\rightarrow z.$$ Since $x$ and $y$ are their own limit points for any sequence $\{T_n\}\subset\Gamma$, we conclude $$\Lambda(\Gamma)=\Sigma.$$