Fundamental group of the quotient space of the $\Bbb Z$-action on $X=\Bbb C-\{0\}$ given by $n\cdot z=\lambda^nz$

Question

Let $\lambda$ be a fixed nonzero complex number whose modulus is not $1$. Consider the $\Bbb Z$-action on $X=\Bbb C-\{0\}$ given by $n\cdot z=\lambda^nz$. I want to compute the fundamental group of the quotient space $X/\Bbb Z$. I think it would be easier if I could show this action is a covering action (i.e., each point $z\in X$ has a neighborhood $U$ such that $n\cdot U\cap U$ is empty for all $n\in \Bbb Z-\{0\}$), but I can't see how to show this action is a covering space action. Or there is another way to compute $\pi_1(X/\Bbb Z)$?

Answer 1

Considering $\lambda$ or $\lambda^{-1}$ without loss of generality, we may assume that $\lvert \lambda \rvert > 1$. Let $A$ be the annulus $$ A = \{z \in \mathbb{C} \mid 1 <\lvert z \rvert < \lvert \lambda \rvert\} $$ We claim that $A$ is a fundamental region for the action of $\mathbb{Z}$. That is, no two distinct points of $A$ are equivalent under $\mathbb{Z}$, and for every $w \in \mathbb{C}-\{0\}$, there is a point $z$ in the closure of $A$ that is equivalent to $w$.

Firstly, suppose that $z, z' \in A$ are equivalent under $\mathbb{Z}$. We will show that $z = z'$. Then there exists $n \in \mathbb{Z}$ such that $z' = \lambda^nz$, and without loss of generality $n \geq 0$. Taking absolute values, we have $\lvert z' \rvert = \lvert\lambda\rvert^n \lvert z\rvert \geq \lvert\lambda\rvert^n$, where the inequality comes from $\lvert z \rvert \geq 1$ since $z \in A$. Clearly the $n = 0$ since $z' \in A$, which means that $z = z'$.

For the second condition, let $w \in \mathbb{C} - \{0\}$. By basic real analysis, there is some $n \in \mathbb{Z}$ such that $\lvert \lambda \rvert ^n \leq \lvert w\rvert < \lvert \lambda\rvert ^{n+1}$. Therefore $1\leq \lvert \lambda^{-n}w\rvert < \lambda$ so $n\cdot w = \lambda^{-n}w \in \bar A$.

Therefore, the topological space $X /\mathbb{Z}$ is homeomorphic to $\bar A / \mathbb{Z}$. The good news is that we can now forget about the group action, and just think about this in terms of elementary quotient spaces.

Now, $\bar A$ is a closed annulus, and the action of $\mathbb{Z}$ identifies each point on its inner boundary circle with a single point on its outer boundary circle. In particular, if $\lambda = re^{i\theta}$, then the point $e^{it}$ is identified with $re^{i(\theta + t)}$. It is now visually clear that this quotient space is homeomorphic to a torus.