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 $\pi_1(X/\Bbb Z)$?
Fundamental group of the quotient space of the $\Bbb Z$-action on $X=\Bbb C-\{0\}$ given by $n\cdot z=\lambda^nz$
algebraic-topologycovering-spacesfundamental-groupsgroup-actionshomotopy-theory
Related Solutions
Well, since $X$ is simply connected, then by construction $q : X \to X/G$ is the universal cover of $X / G$.
Now, if we have a basepoint $[x_0] \in X / G$, then the fibre $q^{-1}([x_0])$ is $\{ x \mid qx = [x_0] \}$, and (again basically by construction) we see that this is $\{ g x_0 \mid g \in G \}$. That is, we can identify the fibre over $[x_0]$ with $G$. (As an aside, this identification is noncanonical in the sense that it requires us to choose a representative of $[x_0] \in X/G$ in order to make the identification. Properly, then, we should think of the fibre as a $G$ torsor, but this is a fairly subtle distinction that we can gloss over for now).
But now, what is the fundamental group of $X/G$? Well, it's all the paths in $X/G$ which start and end at $[x_0]$. But we can lift one of these paths to $X$, and when we do this we get a path from $x_0$ to some other point in the fibre of $[x_0]$ -- this is because if $\gamma$ is a loop in $X/G$ and $\tilde{\gamma}$ is its lift, then $q\tilde{\gamma} = \gamma$ by the definition of a lift, and so $q \tilde{\gamma}(1) = \gamma(1) = [x_0]$, so that $\tilde{\gamma}(1) \in q^{-1}([x_0])$.
At this point, you should draw pictures of:
- $S^1$, with its universal cover $\mathbb{R}$
- $S^1 \times S^1$, with its universal cover $\mathbb{R}^2$ (you'll probably want to draw this as a square with opposite sides identified)
- $S^1 \vee S^1$, with its universal cover "the cayley graph of $F_2$"
and convince yourself that loops in the base space lift to paths in the universal cover whose endpoints lie in the fibre of the basepoint.
But we know that points in the fibre $q^{-1}([x_0])$ are in bijection with $g x_0$ for group elements $g$! If we're feeling optimistic (and if we know a handful of examples) we might guess that two loops in $X/G$ are homotopic if and only if they lift to the same point in the fibre $q^{-1}([x_0])$. If this is true, then it means we get a bijection $\pi_1(X/G) \to q^{-1}([x_0]) \cong G$, which (with some work) we can show to be an isomorphism of groups).
Now you should look at the pictures you drew before and convince yourself that two loops are homotopic in the base space if and only if they lift to the same endpoints in the universal cover. This will be easiest to see for $S^1 \vee S^1$, since two loops basically never lift to the same endpoint. It will be very instructive to see this for $S^1 \times S^1$, since now we might get two a priori different loops that lift to the same endpoint, and you'll have to convince yourself that they're actually the same. This will be related to the abelian-ness of $\pi_1(S^1 \times S^1) = \mathbb{Z}^2$.
And importantly, we can show this! If $\gamma_1$ and $\gamma_2$ are loops in $X/G$ whose lifts $\tilde{\gamma_1}$ and $\tilde{\gamma_2}$ both end at the same $gx_0$, then $\tilde{\gamma_1} \tilde{\gamma_2}^{-1}$ is a loop in $X$, and thus there's a homotopy $H$ contracting this loop to the identity (since $X$ is simply connected). Then $qH$ gives the desired homotopy between $q\tilde{\gamma_1} = \gamma_1$ and $q \tilde{\gamma_2} = \gamma_2$.
Now, it takes a little bit of work to show that composing loops downstairs is compatible with the $G$ structure on the fibre. The idea, though, is that even if the identification of the fibre with $G$ is noncanonical, the change $gh^{-1}$ is well defined. But this is all we need, since we're really identifying the loop in $X/G$ with the group element that changes the fibre by $g$. So if we do a loop that moves $x_0$ to $gx_0$, that loop also moves $hx_0$ to $ghx_0$ -- all the changes are the same! Then if we compose two loops, one changing $x_0 \to hx_0$ and one changing $x_0 \to g x_0$, then the composite makes the change $x_0 \to hx_0 \to ghx_0$, so that composing loops is the same thing as composing the $G$ action on the fibre.
At this point you should again consult your pictures. Take two loops downstairs and concatenate the paths. Where do the paths lift to? Convince yourself by trying a couple examples that if you lift $\gamma_1 \cdot \gamma_2$ then the endpoint of the lift is the group product in the fibre of where we would get by lifting $\gamma_1$ and lifting $\gamma_2$.
To summarize, then:
The fibre of a basepoint $[x_0] \in X/G$ can be identified $G$ after we pick a basepoint $x_0 \in X$ with $q x_0 = [x_0]$. Now given a loop $\gamma : [x_0] \to [x_0]$ in $X/G$, this lifts to a path $\tilde{\gamma} : x_0 \to x_1$ in $X$, where $qx_1 = [x_0]$ so that $x_1 = g x_0$ for some (unique!) $g \in G$.
Now it turns out that the map $\pi_1(X/G) \to G$ sending $\gamma$ to this $g$ is an isomorphism of groups. We sketched an argument above, and you can find a more precise treatment in any algebraic topology book of your choosing. I'm fond of Hatcher's or Rotman's books, personally, but you might also like the treatment in Pierre Albin's recorded lectures (available on youtube). The video relevant to this particular question is probably this one, but you may have to go $\pm 1$ lecture (I didn't rewatch them, I'm going purely off the video title).
I hope this helps ^_^
Best Answer
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.