Let $\mathbb{Z}$ acts on $\mathbb{S}^1$ with the action $\phi (n)(z)=e^{i \alpha n}z$ defined for $\alpha\in\mathbb{R}$ and $z\in\mathbb{S}^1$.
First we easily see that $\phi$ is stable on $\mathbb{S}^1$.
What are the orbits of this action ? Can we describe all of them ? Are they related to $O(2)$ group ?
I already did some special cases for any point $P$ in $\mathbb{S}^1$:
- For $\alpha=0$, $\phi(n)(z)=Id_{\mathbb{S}^1}$
- For $\beta\in\mathbb{Z}^*$ such that $2\pi/\alpha=\beta$ we have $\lvert\mathbb{Z}(P)\rvert=\beta$
- For $p/q\in\mathbb{Q}^*$ such that $2\pi / \alpha=p/q$ we have $\lvert\mathbb{Z}(P)\rvert=p$
- I think for $2\pi / \alpha$ as an irrational, the orbit may be $\mathbb{S}^1$ but can't prove it.
Best Answer
There are two cases. If $\frac{\alpha}{2\pi} = \frac{p}{q}$ is rational and in lowest terms then every orbit has size $q$ (as you say). Otherwise, if $\frac{\alpha}{2\pi}$ is irrational then every orbit has trivial stabilizer so is a copy of $\mathbb{Z}$. (The orbit can't be all of $S^1$ because it has to be countable. What's true in the irrational case is that none of the orbits are closed, and any of their closures is all of $S^1$.)