Differential Geometry – Klein Bottle Embedding in $\mathbb{R}^4$

Question

First of all, I am aware of the question in How to embed Klein Bottle into $R^4$ , which was inconclusive. Anyway, I've made some progress, but I still have a question.

I am using Do Carmo's Riemannian Geometry, and struggling to solve a problem.

The problem is:

Show that the mapping $G:\mathbb{R}^2\to\mathbb{R}^4$ given by

$$G(x,y)=((r\cos (4\pi y)+a)\cos (4\pi x),(r\cos (4\pi y)+a)\sin (4\pi x),r\sin (4\pi y)\cos (2\pi x),r\sin (4\pi y)\sin (2\pi x)))$$

induces an embedding of the Klein bottle into $\mathbb{R}^4$ (It is a slightly different function from the one in the book, but works in the same way).

First of all, it's not hard to see that
$$G(x+n,y+m)=G(x,y)\text{ whenever }m,n\in\mathbb{Z}.$$
Therefore, this mapping is well-defined over the torus $\mathbb{T}^2$. What I need now is to show that $G(-x,-y)=G(A(x,y))=G(x,y)$, where $A$ is the antipode mapping. If this were true, then the mapping G would be well-defined over the Klein Bottle, but it's obvious that this is false.

Am I working wrong here somewhere?

Answer 1

Consider the embedding $\psi:\Bbb R\times\Bbb R\big/{\Bbb Z\times\Bbb Z}\hookrightarrow\Bbb R^3$ $$\psi([\theta,\tau])= \begin{pmatrix} \cos(2\pi\theta) &-\sin(2\pi\theta) & 0\\ \sin(2\pi\theta) &-\sin(2\pi\theta) & 0\\ 0&0&1 \end{pmatrix} \begin{pmatrix} 2+\cos(2\pi\tau)\\ \sin(2\pi\tau)\\ 0 \end{pmatrix}$$ Then one verifies (on the picture and then by doing the math) that $-\psi([\theta,\tau])=\psi([\theta+\frac12,-\tau])$

Thus the Klein bottle is the quotient of $\Bbb R\times \Bbb R$ by the (non-commutative) group $G=\langle v,t\rangle$ of homeomorphisms generated by vertical displacement by one $$v(\theta,\tau)=(\theta,\tau+1)$$ and a twist $$t(\theta,\tau)=(\theta+\frac12,-\tau)$$ (Notice that $t^2=h$ the horizontal displacement by one $h(\theta,\tau)=(\theta+1,\tau)$) You'll verify that $G(v(\theta,\tau))=G(\theta,\tau)=G(t(\theta,\tau))$ so $G$ descends to a map $\tilde{G}:K\to \Bbb R^4$. $\tilde{G}$ is an injective (easy verification) immersion ($G$ already was). By compactness of $K$, it is an embedding.