Let \(\varphi: \mathbb{R}^2 \to \mathbb{R}^2\) be the linear transformation \(\varphi(x,y) = (2x, y/2)\). This generates an action of \(\mathbb{Z}\) on \(X = \mathbb{R}^2 - \{0\}\). Show this action is a covering space action and compute \(\pi_1(X/\mathbb{Z})\). Show the orbit space \(X/\mathbb{Z}\) is non-Hausdorff, and describe how it is a union of four subspaces homeomorphic to \(S^1 \times \mathbb{R}\), coming from the complementary components of the \(x\)-axis and the \(y\)-axis.
Solution
Write \[ \varphi^n(x,y)=(2^n x,2^{-n}y). \]
1. The \(\mathbb Z\)-action generated by \(\varphi\) is free.
If \(\varphi^n(x,y)=(x,y)\) for \((x,y)\ne(0,0)\), then \[ (2^n-1)x=0,\qquad (2^{-n}-1)y=0. \] At least one of \(x,y\) is nonzero, so \(2^n=1\), hence \(n=0\).
2. The action is a covering space action.
Fix \(z=(x,y)\in X\). If \(x\ne0\), then the first coordinates of \(\varphi^n(z)\) are \(2^nx\), so only finitely many orbit points lie in any compact horizontal annulus \[ c\le |u|\le C. \] Choose a small rectangle \(U\) about \(z\) contained in such an annulus and disjoint from the finitely many distinct points \(\varphi^n(z)\) that remain in a slightly larger rectangle. Shrinking \(U\) if necessary gives \[ \varphi^n(U)\cap U=\varnothing \qquad(n\ne0). \] The case \(y\ne0\) is identical using the second coordinate \(2^{-n}y\). Thus every point has a neighborhood disjoint from all its nontrivial translates.
3. The quotient map\[ p:X\to X/\mathbb Z \]is a normal covering with deck group \(\mathbb Z\).Moreover\[ \pi_1(X)\cong\mathbb Z. \]
2.Since\[ X=\mathbb R^2\setminus\{0\}\simeq S^1, \]its fundamental group is infinite cyclic.
4. One has\[ \boxed{\pi_1(X/\mathbb Z)\cong\mathbb Z^2.} \]
For a normal covering there is an exact sequence \[ 1\to\pi_1(X)\xrightarrow{p_*}\pi_1(X/\mathbb Z)\to\mathbb Z\to1. \] Let \(a\) be a positively oriented meridian around the origin. The linear map \(\varphi\) has determinant \(1\), so it is orientation preserving and induces the identity on \[ \pi_1(X)\cong\mathbb Z. \] Hence conjugation by a lift \(t\) of the generator of the deck group fixes \(a\): \[ tat^{-1}=a. \] Since the quotient group is infinite cyclic, choose \(t\) mapping to its generator. The extension then has presentation \[ \langle a,t\mid [a,t]=1\rangle\cong\mathbb Z^2. \]
5. The orbit space is not Hausdorff.
Let \[ p=(1,0),\qquad q=(0,1). \] Their orbits are distinct because every iterate of \(p\) lies on the \(x\)-axis and every iterate of \(q\) lies on the \(y\)-axis. Suppose their orbit classes had disjoint neighborhoods in the quotient. Their inverse images would be disjoint saturated open neighborhoods \(U\) of the orbit of \(p\) and \(V\) of the orbit of \(q\). Choose \(\epsilon>0\) such that \[ (1,s)\in U\quad\text{for }|s|<\epsilon, \] and choose \(\delta>0\) such that \[ (r,1)\in V\quad\text{for }|r|<\delta. \] For \(n\) large enough, set \[ z_n=(1,2^{-n}). \] Then \(z_n\in U\), while \[ \varphi^{-n}(z_n)=(2^{-n},1)\in V. \] Since \(U\) is saturated, \(\varphi^{-n}(z_n)\in U\) as well, contradicting \(U\cap V=\varnothing\). Thus the two orbit classes cannot be separated.
6. Each open quadrant is invariant under the action, and its quotient is homeomorphic to \(S^1\times\mathbb R\).
On any quadrant introduce logarithmic coordinates \[ u=\log|x|,\qquad v=\log|y|. \] Then \(\varphi\) becomes translation by \[ (u,v)\longmapsto(u+\log2,v-\log2). \] Use coordinates \[ s=u+v,\qquad r=u-v. \] The action fixes \(s\) and translates \(r\) by \(2\log2\). Therefore the quotient is \[ \mathbb R_s\times(\mathbb R_r/2\log2\,\mathbb Z) \cong\mathbb R\times S^1. \] The four quadrants are precisely the complementary components of the union of the two coordinate axes, so their images give the four stated \(S^1\times\mathbb R\) subspaces of the orbit space.