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Disclaimer: This question is somewhat of a duplicate. However, I do not feel it was answered adequately in previous questions.

We wish to show that the mapping $$ H(x,y) = ((r \cos{y} + a) \cos{x}, (r \cos{y} + a)\sin{x}, r \sin{y} \cos{\frac{x}{2}},r \sin{y} \sin{\frac{x}{2}}) $$ induces an embedding of the Klein bottle into $\mathbb{R}^4$. Here the Klein bottle is defined as the quotient space $T^2 /G$ where $G$ is the group of diffeomorphisms on $T^2$ (torus of revolution) generated by the identity map and the antipodal map $A(p) = -p$.

  1. Is this a valid definition of the Klein bottle? I've seen someone say this is not an accurate description.
  2. Is it enough to 'restrict' $G$ to the equivalence classes $$ [(x,y)]:= \{ \pm (x+2\pi z_1, y+2 \pi z_2) \} $$ and show that this mapping is continuous, an immersion, and injective, if we assume compactness of the Klein bottle?
Andrew Whelan
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  • I don't know what the "antipodal map" on $T^2$ is. You should give an explicit formula for the intended map. – Lee Mosher Oct 17 '16 at 16:00
  • For that matter, you should say what definition of $T^2$ you are using. – Lee Mosher Oct 17 '16 at 16:01
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    You have $G$ wrong. It is generated by a translation in one direction and a glide reflection in the other. – Ted Shifrin Oct 17 '16 at 16:24
  • Thanks for confirming, Ted Shifrin. What manifold, up to homeomorphism, do we get by quotient-ing the torus by the group I mentioned? – Andrew Whelan Oct 17 '16 at 19:26
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    Oh sorry, I was letting the group act on $\Bbb R^2$ (which is actually easier). Your $G$ acts by orientation-preserving diffeomorphisms, so the quotient cannot be the Klein bottle. It will again be a torus. If you want to do it with an action on $S^1\times S^1$, you need a diffeomorphism like $f(z,w)=(z,w^{-1})$, I believe. – Ted Shifrin Oct 17 '16 at 20:55
  • Which duplicate are you referring to? –  Oct 17 '16 at 21:09
  • http://math.stackexchange.com/questions/759264/klein-bottle-embedding-on-mathbbr4, and http://math.stackexchange.com/questions/330856/how-to-embed-klein-bottle-into-r4 – Andrew Whelan Oct 18 '16 at 08:21

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We have known that the $2$-torus $T^2\simeq S^1\times S^1\simeq\mathbb{R}^2/\mathbb{Z}^2$. To see that $T^2$ is diffeomorphic to the torus of revolution in $\mathbb{R}^3$ obtained as the inverse image of zero of the function $f:\mathbb{R}^3\to\mathbb{R}$ $$ f(x,y,z)=z^2+(\sqrt{x^2+y^2}-a)^2-r^2, $$ we construct parametrization of $S:=f^{-1}(0)$ as follows. Define $\Psi:[0,1]\times[0,1]\to\mathbb{R}^3$ by: $$ (x,y)\mapsto\left( (a+r\cos 2\pi y)\cos2\pi x,(a+r\cos 2\pi y)\sin2\pi x, r\sin 2\pi y \right). $$ Then the antipodal mapping $A:S\to S$ can be expressed as $(x, y)\mapsto(x+1/2,-y)$. We slightly twist $G:\mathbb{R}^2\to\mathbb{R}^4$ by multiplying $4\pi$ for $x, y$ such that it can be well-defined on $T^2$ and we still denote it by $G$: $$ \begin{aligned} G(x,y)=(&(r\cos 4\pi y+a)\cos4\pi x,(r\cos4\pi y+a)\sin4\pi x,\\ &r\sin4\pi y\cos(2\pi x),r\sin4\pi y\sin(2\pi x)). \end{aligned} $$ Then for any $p\in S$, we have $$ G(p)=G(\Psi(x,y))=G(\Psi(x+1/2,-y))=G(-p). $$ Thus $G$ induces an mapping $\tilde{G}:K\to\mathbb{R}^4$, where $K=S/\{A,\operatorname{Id}_{\mathbb{R}^3}\}$ is the Klein bottle, by defining $\tilde{G}([p])=G(p)$ for each $p\in S$. It is easily testified that $\tilde{G}$ is an injective immersion. By compactness of $K$, we conclude that $\tilde{G}$ is an embedding of $K$ in $\mathbb{R}^4$.

User1999
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