Linear transformation and Covering space action

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Exercise 1.3.25. Exercise 1.3.25 Consider the action on $X=\mathbb{R}^2 \backslash\{0\}$ sending $(x, y) \mapsto(2 x, y / 2)$. Show that this generates a covering space action of $\mathbb{Z}$ on $X$, and that $X / \mathbb{Z}$ is not Hausdorff. Show that $X / \mathbb{Z}$ contains four subspaces homeomorphic to $S^1 \times \mathbb{R}$ coming from the complement of the $x$-axis and $y$-axis. Compute $\pi_1(X / \mathbb{Z})$.

Proof. Consider $(x, y) \in X$, and assume that $x \neq 0$. Then the neighborhood $U=(x-$ $|x| / 3, x+|x| / 3) \times \mathbb{R}$ around $(x, y)$ is sent to disjoint open neighborhoods under the ac-
The action is a covering space action because it is free and properly discontinuous: for any point \((x, y) \in X\) with \(x \neq 0\), a neighborhood \(U = (x - |x|/3, x + |x|/3) \times \mathbb{R}\) is mapped to disjoint neighborhoods under the action, and similarly for \(y \neq 0\) using \(\mathbb{R} \times (y - |y|/3, y + |y|/3)\).

\(\pi_1(X/\mathbb{Z}) \cong \mathbb{Z} \oplus \mathbb{Z}\).

To arrive at \(\pi_1(X/\mathbb{Z})\), note that the quotient map \(p: X \to X/\mathbb{Z}\) is a covering map with deck transformation group \(\mathbb{Z}\). This yields the exact sequence \(\pi_1(X) \to \pi_1(X/\mathbb{Z}) \to \mathbb{Z} \to 1\). Since \(\pi_1(X) \cong \mathbb{Z}\) and \(p_*(\pi_1(X)) \cong \mathbb{Z}\), it follows that \(\pi_1(X/\mathbb{Z}) \cong \mathbb{Z} \rtimes \mathbb{Z}\). The semidirect product is direct because the generators—a loop around the origin and a path along the hyperbola \(xy = 1\)—commute.

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Why the semidirect product \(\mathbb{Z} \rtimes \mathbb{Z}\) in the short exact sequence \(1 \to \pi_1(X) \to \pi_1(X/\mathbb{Z}) \to \mathbb{Z} \to 1\) is actually a direct product \(\mathbb{Z} \oplus \mathbb{Z}\)? This happens precisely when the action of the deck transformation group \(\mathbb{Z}\) on \(\pi_1(X) \cong \mathbb{Z}\) is trivial, which implies that the two generators of \(\pi_1(X/\mathbb{Z})\) commute.

To see this explicitly, consider the following representatives for the generators:

One generator, call it \(\alpha\), is the image under the quotient map \(p: X \to X/\mathbb{Z}\) of a loop around the origin in \(X\), such as the unit circle \(\gamma(t) = (\cos(2\pi t), \sin(2\pi t))\) for \(t \in [0,1]\). This generates the subgroup \(p_*(\pi_1(X)) \cong \mathbb{Z}\) in the exact sequence.

The other generator, call it \(\beta\), comes from a choice of section to the exact sequence (lifting the generator of the deck group \(\mathbb{Z}\)). Fix a basepoint in \(X\), say \((1,1)\). The deck transformation \(\phi\) sends this to \((2,1/2)\). Connect these by a path in \(X\) along the hyperbola \(xy=1\), such as \(\sigma(t) = (e^{t \log 2}, e^{-t \log 2})\) for \(t \in [0,1]\) (which starts at \((1,1)\) and ends at \((2,1/2)\)). In the quotient \(X/\mathbb{Z}\), the endpoints are identified, so \(p(\sigma)\) is a loop \(\beta\).

In \(\pi_1(X/\mathbb{Z})\), the relation is \(\beta \alpha \beta^{-1} = \phi_*(\alpha)\), where \(\phi_*\) is the induced map on \(\pi_1(X)\). Here, \(\phi\) is linear with positive determinant 1, so it is orientation-preserving and \(\phi_*\) is the identity on \(\pi_1(X) \cong \mathbb{Z}\). Thus, \(\beta \alpha \beta^{-1} = \alpha\), or equivalently, \(\alpha\) and \(\beta\) commute: \(\alpha \beta = \beta \alpha\).