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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.

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  • $\begingroup$ 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. $\endgroup$ – Lee Mosher Jun 12 at 1:18
  • $\begingroup$ I see. I'll specify what I've done so far. $\endgroup$ – ArkPDEnational Jun 12 at 1:52
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Here's how to do it with what you've accomplished so far. Once you see the method for 1) and 2), you can probably extend it to the other cases.

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.

The idea is to express the diffeomorphism of the quotients first as a smooth map between the covers, and then to check whatever needs to be checked in order to verify that there is an induced diffeomorphism on the quotients.

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.

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  • $\begingroup$ 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. $\endgroup$ – ArkPDEnational Jun 13 at 0:03
  • $\begingroup$ 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 $X$. $\endgroup$ – Lee Mosher Jun 13 at 18:19
  • $\begingroup$ Even though I am not being able to complete the proof, I appreciate your help. Thank you. $\endgroup$ – ArkPDEnational Jun 13 at 23:04

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