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if the group $G=\mathbb{Z}$ acts on $Y=\mathbb{R}^2\setminus \{0\}$ using by the action $m\cdot (x,y)=(2^m x, 2^m y)$ then I was able to show that

$Y \to Y/ \mathbb{Z}$ is a covering map.

Can I conclude that $\pi_1(Y/\mathbb{Z})=G=\mathbb{Z}?$

I was able to show that the map is covering map using the fact that a properly discontinuous map gives us a covering map.

Sine of the Time
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permutation_matrix
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  • To prove that $\pi_{1}(Y/\Bbb{Z})\cong\Bbb{Z}$ you need to produce an explicit isomorphism . What theorem/result are you using that is letting you "conclude"? – Mr.Gandalf Sauron Dec 27 '22 at 08:26
  • @Mr.GandalfSauron, we have the result that $\pi_(Y/G)\equiv G$ given it is a covering map. – permutation_matrix Dec 27 '22 at 08:29
  • That is not true. You have $\frac{\pi_{1}(Y/G,y)}{p_{*}(\pi_{1}(Y,y))}\cong G$. $p$ denotes the covering map .Ofcourse you are dealing with an infinite group. So isomorphism is not an equality. What you only have is that the quotient map is a covering map from $Y$ to the orbit space. – Mr.Gandalf Sauron Dec 27 '22 at 08:34
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    An aside (I don't know if it is pertinent, it depends upon your context): usually the group action of $\mathbb{Z}$ on $\mathbb{R}^2-0$ is expected to do a loop around the origin when incrementing $m$ in $\mathbb{Z}$. Here with $m\cdot (x,y)=(2^m x, 2^m y)$ this is not the case. Your reasoning probably applies to $\mathbb{R}^d-0$ for any $d$, which means you should not expect to find anything special to $d=2$, notably the fact that its fundamental group is non-trivial, while fundamental groups of $\mathbb{R}^d-0$ for $d>2$ are trivial. – Jean-Armand Moroni Dec 27 '22 at 08:53
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    you missed one condition for $\pi_1(Y/G)=G$: $Y$ needs to be 1-connected, and your $Y$ is not (and in fact the quotient is a torus, with the fundamental group $\Bbb Z^2$) – user8268 Dec 27 '22 at 09:34

1 Answers1

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You need to state what results are letting you conclude that $\pi_{1}(Y/\Bbb{Z})\cong \Bbb{Z}$ . Ofcourse it is not at all true that $\pi_{1}(Y/G)\cong \Bbb{Z}$. Technically you need to produce an explicit isomorphism between the two groups. Ofcourse if you already know results then fine but still try to state them clearly. In doing so many wrong things will get cleared out.

For example the fact you are using is incorrect. You have $\frac{\pi_{1}(Y/G,y)}{p_{*}(\pi_{1}(Y,y))}\cong G$ and not what you state.See Hatcher Proposition $1.40$ pg $72$ for a proof .What you only have is that the quotient map is a covering map from Y to the orbit space. Also try not to write $\pi_{1}(X,x)$ is equal to $\Bbb{Z}$ . Write it as $\pi_{1}(X,x)\cong \Bbb{Z}$. That is because you have an isomorphism and not necessarily an equality of fundamental groups. Sometimes we just abuse notation and write equal to.

Solution:

We will solve the problem by proving that $Y/G$ is homeomorphic to the Torus $S^{1}\times S^{1}$.

$\Bbb{R}^{2}\setminus 0\cong (0,\infty)\times [0,2\pi]/0\sim 2\pi = (0,\infty)\times S^{1}$ by using a variable change to polar coordinates.

Then the action of $\Bbb{Z}$ becomes $m\cdot(r,\theta)=(2^{m}r,\theta)$ .

Now $\log_{2}(x):(0,\infty)\to\Bbb{R}$ is a homeomorphism.

Under this homeomorphism the action becomes $m\cdot(r,\theta)=(\log_{2}(r\cdot 2^{m}),\theta)=(m+\log_{2}(r),\theta)$.

That is the action of $\Bbb{Z}$ on $\Bbb{R}$ just becomes $m\cdot x = x+m$ .

Using this, the action of $\Bbb{Z}$ just factors to $\Bbb{R}^{2}\setminus\{0\}/\Bbb{Z}\cong (0,\infty)/\Bbb{Z}\times S^{1}\cong \Bbb{R}/\Bbb{Z}\times S^{1}$ .

Now it's an easy excercise to show that $\Bbb{R}/\Bbb{Z}$ such that $x\sim y\iff x-y\in\Bbb{Z}$ is just homeomorphic to $S^{1}$.

Define $e:\Bbb{R}\to S^{1}$ by $e(x)=e^{i2\pi x}$ . Then $x\sim y\iff x-y\in\Bbb{Z}\iff e(x)=e(y)$ . Then this map descends to a unique homeomorphism from the quotient to prove that $\Bbb{R}/\Bbb{Z}\cong S^{1}$

So $Y/G\cong S^{1}\times S^{1}$ which is a torus.

So the fundamental group is isomorphic to $\Bbb{Z}\times \Bbb{Z}$.

Mr.Gandalf Sauron
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