# Compute the fundamental group of $\mathbb{C}^*/\Gamma$, where $\Gamma=\{\varphi^n:\varphi(z)=4^nz,n\in\mathbb{Z}\}$

Problem Find the fundamental group of the orbit space $$\mathbb{C}^*/\Gamma$$, where $$\mathbb{C}^*=\mathbb{C}\backslash\{0\}$$, and $$\Gamma=\{\varphi^n:\varphi(z)=4^nz, n\in\mathbb{Z}\}$$ acts on $$\mathbb{C}^*$$ in the natural way.

Idea: We claim that $$\pi_1(\mathbb{C}^*/\Gamma)=\mathbb{Z}\times\mathbb{Z}.$$ We will find a space $$X$$, a group $$G$$ and a normal subgroup $$H\unlhd G$$ such that $$X$$ is a simply-connected $$G$$-space, $$X/H\cong \mathbb{C}^*$$, and $$G/H=\Gamma$$. We will assume the following theorems:

Theorem $$1$$: Suppose that $$X$$ is a $$G$$-space and $$H$$ is a normal subgroup of $$G$$, then $$X/H$$ is a $$(G/H)$$-space and $$(X/H)/(G/H)\cong X/G.$$

Definition: If $$G$$ acts on $$X$$, then the action is a covering space action if every point $$x$$ in $$X$$ has a neighbourhood U such that $$\{g\in G:g\cdot U\cap U\neq \emptyset \}=e$$.

Theorem $$2$$: Suppose that $$X$$ is path-connected and a group $$G$$ acts on $$X$$ as a covering space action, if $$X$$ is simply connected, then $$\pi_1(X/G)\cong G.$$

From Theorem $$1$$, we can deduce that $$X/G\cong(X/H)/(G/H)\cong \mathbb{C}^*/\Gamma.$$ Finally, we observe that the action of $$G$$ is a covering space action and so by Theorem $$2$$, $$G\cong\pi_1(X/G)\cong\pi_1(\mathbb{C}^*/\Gamma)$$.

Proof $$1$$: Let $$X=\mathbb{R}\times \mathbb{R}_{>0}$$ be the upper half-plane, let $$G=\mathbb{Z}\oplus\Gamma$$, and let $$H=\mathbb{Z}\unlhd G$$ be the first factor of $$G$$. Then the action of $$G$$ on $$X$$ is given by $$(k,\varphi^n)\cdot(a,b)=(a+k,\varphi^n(b))=(a+k,4^nb)$$. This is clearly a covering space action. Also, $$X/H=(\mathbb{R}\times \mathbb{R}_{>0})\big/(\mathbb{Z}\times\{1\})\cong S^1\times \mathbb{R}_{>0}\cong \mathbb{C}^*$$ and $$G/H=\Gamma$$. By the argument explained in Idea, we conclude that $$\pi_1(\mathbb{C}^*/\Gamma)\cong G=\mathbb{Z}\oplus\Gamma\cong\mathbb{Z}\times \mathbb{Z}$$.

Proof $$2$$: Actually, we have a simpler geometric proof: the space $$\mathbb{C}^*/\Gamma$$ is just $$\{z\in\mathbb{C}: 4^{-1}\leq|z|\leq 1\}$$ with some extra gluing: $$z_1\sim z_2 \iff$$ Arg$$(z_1)$$=Arg$$(z_2)$$ and $$\{|z_1|,|z_2|\}=\{4^{-1},1\}$$, while the latter space is clearly a $$2$$-torus.

Questions: Could anybody please verify my proofs? Does my second proof make sense (is it rigorous enough)?

• Your set-builder notation for $\Gamma$ doesn't make sense. I assume you mean to say $\phi(z):=4z$ (outside of the set notation) and $\Gamma=\langle\phi\rangle=\{\phi^n\mid n\in\mathbb{Z}\}$. Jul 28, 2020 at 5:34

Another way to put it: there is an equivalence $$S^1\times\mathbb{R}\simeq \mathbb{C}^{\ast}$$, defined by $$(z,x)\mapsto z\cdot4^x$$. It is $$\Gamma$$-equivariant, where $$\phi(z,x):=(z,x+1)$$ defines an action of $$\Gamma=\langle\phi\rangle$$ on $$S^1\times\mathbb{R}$$. So you pretty much want the quotient $$(S^1\times\mathbb{R})/\Gamma=S^1\times(\mathbb{R}/\mathbb{Z})$$, which is a torus.