# Different constructions of Klein's Bottle

Consider the following constructions for the Klein's bottle:

1) $$\mathbb{R}^{2}/G$$ : where $$G=\langle f_1, f_2\rangle$$, such that, $$f_1, f_2:\mathbb{R}^{2}\rightarrow \mathbb{R}^{2}$$ given by $$f_1(x, y)=(x, y+1)$$ and $$f_2(x, y)=(x+1, 1-y)$$.

2) $$(S^{1}\times \mathbb{R})/H$$ : where $$H$$ is the ciclic group spanned by $$h:S^{1}\times \mathbb{R} \rightarrow S^{1}\times \mathbb{R}$$ given by $$h((x, y), z)=h(x, y, z)=(x, -y, z+1)$$.

3) $$(S^{1}\times S^{1})/K$$ : where $$K=\{ I_{S^{1}\times S^{1}}, k\}$$, such that $$I_{S^{1}\times S^{1}}$$ is the identity and $$k:S^{1}\times S^{1}\rightarrow S^{1}\times S^{1}$$ is given by $$k(x, y)=(\overline{x}, -y)$$.

I want to prove the following :

The 3 smooth quotient manifolds $$\mathbb{R}^{2}/G$$, $$(S^{1}\times \mathbb{R})/H$$ and $$(S^{1}\times S^{1})/K$$ are diffeomorphic.

Edit: This is what I have done so far : I verified that the groups $$G$$, $$H$$ and $$K$$ act properly discontinuous over $$\mathbb{R}^{2}$$, $$S^{1}\times \mathbb{R}$$ and $$S^{1}\times S^{1}$$, respectively. Then I applied a theorem to conclude that there exist a unique smooth structure such that those quotients are smooth manifolds and their respectives quotient maps are, in fact, covering maps.

Remark:

Definition 1: $$G$$, a group of diffeomorphims, act freely in $$M$$ if for all $$g\in G\ -\{e\}$$ , $$g$$ has not fixed points, where $$e$$ is the identity.

Definition 2: $$G$$ is properly discontinuous in $$M$$ if:

(i) $$G$$ act freely in $$M$$.

(ii) for all $$x, y\in M$$, such that $$Gx\neq Gy$$, there are open subsets of $$M$$, $$x\in U$$, $$y\in V$$ such that $$U \cap g(V)= \emptyset$$, for all $$g\in G$$.

(iii) for each $$x\in M$$, there is an open subset of $$M$$, $$x\in V$$, such that $$g(V)\cap V=\emptyset$$ for all $$g\in G-\{ e \}$$.

Can anyone give me some help? I still need to find the diffeomorphisms.

Thanks.

• Your comment that you have "concluded that they have such structure" is intriguing. But that comment is too vague, so the best I can suggest is that whatever you have done to show they "have such structure" should be useful for producing a diffeomorphism. If you can explain your thoughts regarding that comment in more detail, perhaps it might be easier to assist you. – Lee Mosher Jun 12 at 1:18
• I see. I'll specify what I've done so far. – ArkPDEnational Jun 12 at 1:52

Let me denote $$p_1 : \mathbb R^2 \mapsto \mathbb R^2 / G$$ as the quotient map given in 1), and $$p_2 : (S^1 \times \mathbb R) / H$$ as the quotient map given in 2). Both of $$p_1,p_2$$ are covering maps, and both are smooth. Another way to express this is that the smooth structures on the quotients $$\mathbb R^2 / G$$ and $$(S^1 \times \mathbb R) / H$$ are induced, or inherited, from the smooth structures on $$\mathbb R^2$$ and $$S^1 \times \mathbb R$$, respectively.
There is a universal covering map $$q : \mathbb R^2 \mapsto S^1 \times \mathbb R$$ which is defined by $$q(x,y) = ((\cos 2 \pi y,\sin 2\pi y),x)$$. The deck transformation group of $$q$$ is generated by your map $$f_1$$. There is also a surjective homomorphism $$\alpha : G \mapsto H$$ defined by $$f_1 \mapsto \text{Id}$$ and $$f_2 \mapsto h$$. The map $$q$$ has the property that it is "equivariant with respect to $$\alpha$$", meaning that for each $$g \in G$$ and each $$p \in \mathbb R^2$$ we have $$q(g \cdot p) = \alpha(g) \cdot q(p)$$. It follows that $$q$$ induces a well-defined function $$Q : \mathbb R^2 / G \mapsto (S^1 \times \mathbb R) / H$$ In words, $$Q$$ takes the $$G$$-orbit of a point $$p \in \mathbb R^2$$ to the $$H$$ orbit of the point $$q(p) \in S^1 \times \mathbb R$$. Now you just have to chase through the diagrams and verify for yourself that $$Q$$ is a diffeomorphism: it's one-to-one; it's onto; it's smooth; and its inverse is smooth.
• Sorry fr bothering, but do you know any reference that I could found something about the deck transformation group of $q$. I read something about deck transformations on Lee's - Smooth Manifolds but I didn't get what is the idea here. Specially because there he works define a deck transformation from $X$ to $X$ only. – ArkPDEnational Jun 13 at 0:03
• You can read about the theory of deck transformations in many topology books, for example Munkres' "Topology". But I'll say that if $q : X \to Y$ is a covering map then a deck transformation of $q$ is a deck transformation from $X$ to $X$. – Lee Mosher Jun 13 at 18:19