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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)?

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  • $\begingroup$ 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 $\Gamma=\langle\phi\rangle=\{\phi^n\mid n\in\mathbb{Z}\}$. $\endgroup$
    – runway44
    Jul 28, 2020 at 5:34

1 Answer 1

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Both arguments look correct to me.

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.

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