# Can the torus and the Klein bottle be thought as $\mathbb{R}^2/G$ with $G$ a finite group acting freely

I have recently proved the following exercise,

Let $$G$$ be a finite group acting freely on a (compact) topological manifold of dimension $$n$$. Then, $$X/G$$ is a (compact) manifold of dimension $$n$$.

I have done this in the context of an introductory course on topology: thus, we have not talked about anything much more profound than the definition of manifold itself. More so, the aforementioned exercise was guided.

In the same spirit, there is an excersice which asks to prove that the torus and the Klein bottle are $$2$$-dimensional manifolds. I presume this can be proved 'manually' i.e. by looking at an explicit construction of the quotient of $$[0,1]^2$$. However, I want to know whether there is a proof involving an free action from a finite group acting on $$\mathbb{R}^2$$. Any hints would be much appreciated.

• $[0, 1]^2$ is not a manifold. – Qiaochu Yuan Oct 11 '18 at 6:51
• @QiaochuYuan my bad! I now realize how that fails at the edges. Thanks. Is there something that we can say about quotients of the plane? – Guido A. Oct 11 '18 at 6:58
• The torus is a quotient $\mathbb{R}^2/\mathbb{Z}^2$. The Klein bottle is also a quotient of $\mathbb{R}^2$ but in a slightly more complicated way. – Qiaochu Yuan Oct 11 '18 at 6:59
• I (somewhat) remember the usual construction of the torus, but can any of these be obtained via quotients of free actions via finite groups? That is, can the quoted result be of any use in this case? – Guido A. Oct 11 '18 at 7:02
• The torus is the orientation double cover of the Klein bottle, so you can use this result to prove that the Klein bottle is a manifold once you know the torus is a manifold. But every finite cover of the torus is itself a torus. – Qiaochu Yuan Oct 11 '18 at 7:10

1. There is some lovely topological machinery you can learn about called covering space theory, and it tells you a lot about quotients like this. Among other things, it tells you that if $$X$$ is simply connected and $$G$$ is a finite group acting freely on it, then $$X$$ is a finite covering space of $$X/G$$, and $$X/G$$ has fundamental group $$G$$.

2. The torus and the Klein bottle both have infinite fundamental groups. Also, any quotient $$\mathbb{R}^2/G$$ would be noncompact, and the torus and the Klein bottle are compact. So these are two ways of seeing that neither can occur as a quotient $$\mathbb{R}^2/G$$ for $$G$$ finite; they both occur as such a quotient, but for $$G$$ infinite.

3. Every finite covering space of the torus is another torus, meaning that if $$X$$ is any manifold and $$G$$ a finite group acting freely on it such that $$X/G \cong T^2$$, then $$X$$ itself must already be a torus. So it's impossible to use this technique to prove non-circularly that the torus is a manifold. On the other hand, the torus is the orientation double cover of the Klein bottle, meaning there's an action of the cyclic group $$C_2$$ of order $$2$$ on the torus whose quotient is the Klein bottle.

4. Fortunately it is easier than this to prove that the torus and the Klein bottle are manifolds.

What the covering space arguments further reveal is the following perhaps surprising fact: no nontrivial finite group acts freely on $$\mathbb{R}^2$$, because the only connected surface with nontrivial finite fundamental group is $$\mathbb{RP}^2$$, whose universal cover is $$S^2$$, not $$\mathbb{R}^2$$.

• Wow, thanks! This was a super interesting and understandable read. We're supposed to eventually cover covering spaces (no pun intended) in my course, so this will be of great use. Thanks for dedicating the time for such a thorough answer! – Guido A. Oct 11 '18 at 17:11

While not being a finite group, $$\Bbb{Z}^2$$ acts on $$\Bbb{R}^2$$ by translation (concretely $$(n,m) \cdot (a,b) :=(a+n,b+m)$$) and the torus arises as the quotient space for this action.

Similarly, let $$f,g:\Bbb{R}^2\to \Bbb{R}^2$$ be the bijections given by $$(a,b)\mapsto (a,b+1)$$ and $$(a,b)\mapsto (a+1, 1-b)$$ respectively. If $$G$$ denotes the (infinite!) group generated by $$f$$ and $$g$$, the quotient $$\Bbb{R}^2/G$$ is the Klein bottle. You may check this by inspecting how the points in $$[0,1]^2$$ are identified by the action.

In both cases, the action is properly discontinuous: this ensures that each point in the quotient has a locally euclidean neighborhood (since the quotient map is a local homeomorphism). The fact that $$\Bbb{R}^2/G$$ is Hausdorff is not automatic, but is implied by the hypothesis

for each $$x,y\in\Bbb{R}^2$$ with disjoint orbits, there exist neighborhoods $$U\ni x$$, $$V\ni y$$ such that $$gU\cap V$$ is empty for each $$g\in G$$,

which is easily verified in both examples (and is in fact automatic provided $$G$$ is finite).

• While not a direct application of the original theorem, this goes along the lines of my intuition when trying to solve this problem. After all, the finiteness of the group is used to prove that $X/G$ id Hausdorff but unnecessary otherwise. Thanks for the insight! I wish I could accept more than one answer. – Guido A. Oct 12 '18 at 18:06