4
$\begingroup$

I'm working on the following exercise from Hatcher's algebraic topology:

Let $\phi:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ be the linear transformation $\phi(x,y) = (2x,y/2)$. This generates an action of $\mathbb{Z}$ on $X = \mathbb{R}^2\setminus \big\{0\big\}$. Show that this action is a covering space action and compute $\pi_1(X/\mathbb{Z})$. Show that 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.

I have been able to prove everything but the fundamental group. Here is what I have so far on this:

I let $p$ be the covering map $p:X \rightarrow X/\mathbb{Z}$. $X$ is path connected and locally path connected. Given this, we know from Proposition 1.40 in Hatcher that $\mathbb{Z}\cong \pi_1(X/\mathbb{Z})/p_*(\pi_1(X))$. Since $p$ is a covering map, $p_*$ is injective and, therefore, $p_*(\pi_1(X)) \simeq \pi_1(X) \simeq \mathbb{Z}$. Here we used the fact that $X \cong S^1$. Hence, we conclude that $\pi_1(X/\mathbb{Z}) \simeq \mathbb{Z} \rtimes \mathbb{Z}$. This semi-direct product is isomorphic to $\mathbb{Z} \times \mathbb{Z}$ or a non-abelian group. We claim that $\pi_1(X/\mathbb{Z}) \cong \mathbb{Z} \times \mathbb{Z}$. It is sufficient to show that $\pi_1(X/\mathbb{Z})$ is abelian, which can be shown via its generators. The generators of $\pi_1(X/\mathbb{Z})$ are given by paths in $X/\mathbb{Z}$ as follows: one is the image of the loop $\gamma$ in $X$ which generates $\pi_1(X,x_0)$, and the other is the image of a path $\alpha$ in $X$ joining the basepoint $x_0$ to $\phi(x_0)$.

From this point it seems like we need to show that $\phi(\gamma) \alpha \overline{\gamma}\overline{\alpha}$ is homotopic to the constant map. This, along with the injectivity of $p_*$, would give us the abelian property we're after. However, I'm not sure how to do this. Would it be sufficient to say that $\phi(\gamma) \alpha \overline{\gamma}\overline{\alpha}$ is not homotopic to $\gamma$ since $\gamma$ is the only non-trivial path up to homotopy?

I have one more idea. Is it true that $X/\mathbb{Z}$ is a topological group? This would tell us that $\pi_1(X/\mathbb{Z})$ is abelian.

$\endgroup$
2
  • 1
    $\begingroup$ The covering group acts by trivial automorphisms on $\pi_1(X)$, hence, the fundamental group of the quotient is the direct product. $\endgroup$ Commented Feb 17, 2022 at 19:08
  • $\begingroup$ @MoisheKohan I'm not familiar with this type of argument. Could you please expand? $\endgroup$
    – slowspider
    Commented Feb 18, 2022 at 0:01

1 Answer 1

1
$\begingroup$

Consider a generator $\gamma\in \pi_1(X)$ as shown in blue:

enter image description here

Here $\gamma$ is represented by the unit circle, taking $(1,0)$ as base point.

We will next pick $\alpha\in \pi_1(X/\mathbb{Z})$ which maps to $\phi$ the generator of $\mathbb{Z}$, in the short exact sequence:$$1\to \pi_1(X)\to\pi_1(X/\mathbb{Z})\to \mathbb{Z}(=\langle\phi\rangle)\to 1$$

The lift of $\alpha$ to $X$ will be a path from the basepoint $(1,0)$ to $\phi(1,0)=(2,0)$. We may thus represent some such $\alpha$ by the following path in $X$, indicated in red:

enter image description here

Now in $\pi_1(X/\mathbb{Z})$ we have the composition: $\alpha\gamma\alpha^{-1}$. The lift of this to $X$ will be the path $\alpha$ followed by $\phi$ applied to the loop $\gamma$ followed by the path $\alpha$ in reverse:

enter image description here

By considering the winding number about $(0,0)$ we see that: $$\alpha\gamma\alpha^{-1}=\gamma,$$ as required.

$\endgroup$
3
  • $\begingroup$ This is a nice explanation. Instead of appealing to the winding number, could we use the fact that the loop you have pictured in your third diagram is homotopic to $\gamma$? $\endgroup$
    – slowspider
    Commented Feb 18, 2022 at 0:55
  • $\begingroup$ Yes absolutely. $\endgroup$
    – tkf
    Commented Feb 18, 2022 at 0:56
  • 1
    $\begingroup$ Oh nice! The result would follow from this fact as well. Thank you! $\endgroup$
    – slowspider
    Commented Feb 18, 2022 at 0:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .