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I have the following problem (in some slightly different context, but I think it works here too):

Suppose we have $G$ and $H$ two cyclic infinite subgroups of the homeomorphisms of $\mathbb{R}^n$, say $G=\{\alpha^k\}_{k\in\mathbb{Z}}, H=\{\beta^\ell\}_{\ell\in\mathbb{Z}}$, which act freely and properly discontinously on $\mathbb{R}^n$ such that both projections $\pi:\mathbb{R}^n\to\mathbb{R}^n/G$ and $\pi':\mathbb{R}^n\to\mathbb{R}^n/G$ are covering maps.

Moreover, suppose we have an homeomorphism $f:\mathbb{R}^n/G\to\mathbb{R}^n/H$. Then, I know that, as $\mathbb{R}^n$ is simply-connected, I can lift $f\circ\pi:\mathbb{R}^n\to\mathbb{R}^n/H$ to a function $\tilde{f}:\mathbb{R}^n\to\mathbb{R}^n$ such that $\pi'\circ \tilde{f}=f\circ\pi$. Moreover, using $f^{-1}$ we can see that $\tilde{f}$ is an homeomorphism

Problem: I have to prove that the generator $\alpha$ goes to $s_{\alpha}$ where $s_{\alpha}$ is a generator of $H$, i.e. $s_{\alpha}=\beta$ or $\beta^{-1}$. More explicitly, $\pi'\circ \tilde{f}(\alpha x)=f\circ \pi(\alpha x)=f\circ \pi(x)=\pi'\circ \tilde{f}(x)$ so $\tilde{f}(\alpha x)=\beta^k \tilde{f}(x)$. I have to prove that $k=\pm 1$.

I tried to prove that there is an isomorphism between $G$ and $H$ given by $\varphi(\alpha^k)=\beta^\ell$ where $\ell$ is the only that satisfies $\tilde{f}(\alpha^k x)=\beta^\ell \tilde{f}(x)$ but I couldn't succeed.

Any help is welcome.

Alejandro Tolcachier
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1 Answers1

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The following feels overcomplicated, but it should do what you want as I understand your question.

For $y \in \mathbb{R}^n$, applying $\pi' \circ \tilde{f}$ to $y$ and $\alpha(y)$ gives $$\{\beta^{\ell_1}(\tilde{f}(y)) : \ell_1 \in \mathbb{Z}\} = \{\beta^{\ell_2}(\tilde{f} \circ \alpha(y)) : \ell_2 \in \mathbb{Z}\}.$$ Thus for all $x \in \mathbb{R}^n$, there exists some $\ell \in \mathbb{Z}$ such that $\beta^\ell(x) = \tilde{f} \circ \alpha \circ \tilde{f}^{-1}(x)$. Set $$E_\ell := \{x \in \mathbb{R}^n : \beta^\ell(x) = \tilde{f} \circ \alpha \circ \tilde{f}^{-1}(x)\}.$$ Clearly $E_\ell$ is closed. We've just shown $\cup_\ell E_\ell = \mathbb{R}^n$. Since $\beta$ acts freely, the $E_\ell$ are also disjoint. It is a fact that the only way to write $\mathbb{R}^n$ as a countable union of disjoint closed sets is the trivial way, so some $E_\ell = \mathbb{R}^n$. Hence $\beta^\ell = \tilde{f} \circ \alpha \circ \tilde{f}^{-1}$ for some $\ell \in \mathbb{Z}$.

Symmetrically, $\alpha^k = \tilde{f}^{-1} \circ \beta \circ \tilde{f}$ for some $k \in \mathbb{Z}$. Hence $\beta^{\ell k} = (\beta^\ell)^k = \beta$. Again since $\beta$ is free, this forces $\ell k = 1$, so $\ell = k \in \{\pm 1\}$, i.e. $\beta^{\pm 1} = \tilde{f} \circ \alpha \circ \tilde{f}^{-1}$.

Joshua P. Swanson
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