if possible, analyze my proof. (I don't know if it's correct).

Exercise is here

An action of $\mathbb{Z}_n$ on complex $S^1$ and its geometric meaning.

proof: Type $\xi = e^{\frac{2 \pi i}{n}}$. Note that for all $k \in \mathbb Z_n $ \begin{align*} \varphi_k(z) = \left[e^{\frac{2 \pi i}{n}} \right]^k z = e^{\frac{2k \pi i}{n}} z, \end{align*} furthermore, it is easily verified that $\varphi_k$ is an action as follows

i) $\varphi_0(z) = e^0z = z$

ii) $\varphi_{k + l}(z) = e^{\frac{2\pi i (k + l)}{n}} z = e^{\frac{2\pi ik}{n}} e^{\frac{2\pi il}{n}} z =\varphi_k(\varphi_l(z))$

Therefore $\varphi_k$ is a action of $\mathbb Z_n$ in $\mathbb S^1$. Furthermore, the equivalence relation induced by the action is given by \begin{align*} x\mathcal{R}y & \Leftrightarrow \exists k \in \mathbb Z_n: \xi^k x = y \\ &\Leftrightarrow e^{\frac{2k\pi i}{n}} x = y \\ &\Leftrightarrow e^{\frac{2k\pi i}{n}} e^{i \theta} = e^{i \lambda} \end{align*} Let $\pi: \mathbb S^1 \rightarrow \mathbb S^1 / \mathbb Z_n$ be the projection, and $p: \mathbb S^1 \rightarrow \mathbb S^1$, $$z \mapsto z^n.$$ Note that $p$ is continuous, and in particular satisfies \begin{align*} p(t) = p(t') \Leftrightarrow t^n = t'^n &\Leftrightarrow e^{i \theta} = e^{i \lambda} \end{align*} but then there is $k \in \mathbb Z_n$ such that \begin{align*} e^{\frac{2k\ pi i}{n}} e^{i \theta} = e^{i \lambda} \Leftrightarrow \xi^k t = t ' \end{align*} therefore there is a continuous bijection $h: \mathbb S^1 / \mathbb Z_n \rightarrow \mathbb S^1$. But being $\mathbb S^1 / \mathbb Z_n$ compact (compact image by continuous application), and $\mathbb S^1 $ Hausdorff, $h$ is a homeomorphism.

  • 1
    $\begingroup$ the proof is right $\endgroup$
    – ali
    Oct 24, 2020 at 21:02
  • $\begingroup$ Thank you very much @ali. $\endgroup$
    – PaulichenT
    Oct 25, 2020 at 4:17


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