Prove that the group is properly discontinuous

Question

Consider $G= \langle f,g \rangle$, such that, $f,g: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ given by $f(x,y)=(x,y+1)$ and $g(x,y)=(x+1,1−y)$.

$h \in G$ if and only if there are $m,n \in \mathbb{Z}$ such that $h = f^mg^n$ on what $gf =fg^{-1}$.

Can anyone help me prove that the $G$ group is properly discontinuous?

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Remark:

Definition 1: $G$, a group of diffeomorphims, act freely in M if for all $g \in G −{e}$ , g has not fixed points, where e is the identity.

Definition 2: $G$ is properly discontinuous in $M$ if:

$(i)$ $G$ act freely in M.

$(ii)$ for all $x,y \in M$, such that $Gx \neq Gy$, there are open subsets of $M$, $x \in U$, $y \in V$ such that $U \cap g(V) = \emptyset$, for all $g \in G$.

$(iii)$ for each $x \in M$, there is an open subset of $M$, $x \in V$, such that $g(V) \cap V= \emptyset$ for all $g \in G−{e}$.

Answer 1

First note that because $f$ and $g$ are both isometries, every element of $G$ is an isometry (for the norm $\Vert \cdot\Vert=\Vert \cdot\Vert_\infty$).

• Let $h$ be an element of $G$ which has a fixed point $(x,y)$. Write $h=f^mg^n$. We have $$(x,y)=h(x,y)=(x+n,\text{something})$$ so $n=0$ and $(x,y)=h(x,y)=f^m(x,y)=(x,y+m)$ so $m=0$. Therefore $h=id=e$ and the action is free.

• Let $a,b\in\Bbb R^2$ such that $G\cdot a\neq G\cdot b$. You can prove, for example by contradiction, that the set $\{h\in G~\vert~\Vert a-h(b)\Vert \leq 1\}$ is finite. Therefore if $\delta:=\inf_{h\in G}\Vert a-h(b)\Vert$, then $\delta>0$. Let $U=B(a,\delta/2)$ and $V=B(b,\delta/2)$.

Suppose by contradiction that there is $h\in G$ such that $U\cap h(V)\neq \emptyset$. Take $z\in U\cap h(V)$. Then from the triangle inequality $$\Vert a-h(b)\Vert\leq \Vert a-z\Vert+\Vert z-h(b)\Vert$$ But $\Vert z-h(b)\Vert=\Vert h(h^{-1}(z))-h(b)\Vert=\Vert h^{-1}(z)-b\Vert$ because $h$ is an isometry. Also by assumption $h^{-1}(z)\in V$ so we get $$\Vert a-h(b)\Vert< \delta/2+\delta/2=\delta$$ which is in contradiction with the definition of $\delta$.

• Let $x\in \Bbb R^2$ and let $V:=B(x,1/3)$. Let $h=f^mg^n\in G$ such that $h(V)\cap V\neq \emptyset$. Take $z\in V$ such that $f(z)\in V$ ($z$ is just $f^{-1}(a)$ for $a\in h(V)\cap V$). By triangle inequality we have $$\Vert z-h(z)\Vert\leq\Vert z-x\Vert+\Vert x-h(z)\Vert<1$$ Because $z-h(z)=(-n,\text{something})$ we have $\vert n\vert \leq \Vert z-h(z)\Vert=\Vert z-h(z)\Vert_\infty <1$ so $n=0$. Now we have $z-h(z)=(m,0)$ so $m=0$. Finally $h=id=e$.

I hope this helps.