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A question about topology from an ignorant logician, so please be kind if this is obvious!

We all know that the Klein bottle, unlike the torus, cannot be embedded in 3-space. And we all know (because we were told) that it can be embedded in 4-space. I can see how to embed the torus in 3-space, using only quadratic stuff (I can do that with stuff I learnt at school) but how does one embed the Klein bottle in 4-space? I am asking because I think if I had an equation to pour over I might start to get a feel for how the injection works and what the bottle looks like. In particular, I would like to know if the isometry group of the embedded bottle looks like the isometry group of the embedded torus in 3-space. I'm hoping it does, because that would put flesh on the assertion that there isn't really any self-intersection.

I asked this of a visiting topologist but she didn't know off the top of her head so I'm trusting that this isn't a stupid question!

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    $\begingroup$ You know how to embed it in three space with only a bit of self-intersection. Push the intersecting piece one centimeter in the blue direction. $\endgroup$
    – Ville Salo
    Mar 18, 2020 at 11:55
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    $\begingroup$ A nicely parametrized embedding into $4$-space, very similar to the standard parametrization of the torus in $3$-space, can be found here: mathcurve.com/surfaces.gb/klein/klein.shtml. $\endgroup$
    – Will Brian
    Mar 18, 2020 at 12:26
  • $\begingroup$ Excellent: lots to think about there - thank you very much! What about the group of isometries..? $\endgroup$ Mar 18, 2020 at 14:21
  • $\begingroup$ I was hoping to be shown a set of quadruples in 4-space satisfying some quadratic condition. The first thing on that mathcurve page looks hopeful, but it's parametrized, so it's the set of all quadruples <x,y,z,t> such that there exist u such that etc etc. Presumably there is a way of getting rid of the parameter..? So that it's the set of all quadruples <x,y,z,t> satisfying a quadratic condition..? Or is that too much to expect? This is making me think about whether we have quantifier elimination for the elementary theory of trig functions and i've never heard of such a thing.. $\endgroup$ Mar 18, 2020 at 14:33
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    $\begingroup$ There is no "the Klein bottle" if you talk of it as a metric space. You have plenty of Riemannian metrics on it, most of which have a trivial isometry group. Maybe you mean with a flat metric, and still there is a whole moduli space of such structures. $\endgroup$
    – YCor
    Mar 18, 2020 at 16:18

4 Answers 4

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The parametrization for the Klein bottle provided by Will Brian is \begin{align} x &= (a+b\cos v)\cos u\\ y &= (a+b\cos v)\sin u\\ z &= (b\sin v)\cos(u/2)\\ t &= (b\sin v)\sin(u/2)\\ \end{align} where $a>b>0$.

This leads to the defining conditions: \begin{align} 4a^2(x^2+y^2)&=(a^2-b^2+t^2+x^2+y^2+z^2)^2\\ y(z^2-t^2)&=2txz\\ tyz&>0 \end{align}

The first equation comes from expressing $b \cos v$ both in $x,y$-terms and in $z,t$-terms.

The second equation comes from expressing $\tan u$ both in $x,y$-terms and in $z,t$-terms.

The inequality comes from $(a+b \cos v)(b \sin u \sin v)^2 > 0$.

There may be one other independent inequality also.

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If embedding means smooth embedding, the non-orientability of the Klein bottle $K$ implies that its stable normal bundle will also not be trivial. This implies that $K$ cannot be smoothly described as the solutions to an equation $f(\bf x)=\bf b$, with $\bf b \in \mathbb R^2$ a regular value of a smooth function $f: \mathbb R^4 \rightarrow \mathbb R^2$, i.e. solutions to two equations in four unknowns, which seems to be your goal.

(Guilliman and Pollack's differential topology book is a friendly place to read about the mathematics here.)

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Step by step. First, we have the unit circle in the complex plane: $$ S\ :=\ \{s\in\Bbb C:\ |s|=1 \} $$ We may even have a larger circle:

$$ C\ :=\ \{5\!\cdot\! s:\ s\in S\} $$

Then, a torus $\ T\ $ is a surface around circle $\ C\times\{0\}\subseteq \Bbb C\times\Bbb R:$

$$ T\ :=\ \{ ((5+\Re s)\cdot c,\, \Im s)\,:\,\ (c\ s)\in S\times S\} \quad\subseteq\quad\Bbb C\times\Bbb R $$

Of course:

$$ T\ =\ \{ ((5+\Re s)\cdot c^2,\, \Im s)\,:\,\ (c\ s)\in S\times S\} \quad\subseteq\quad\Bbb C\times\Bbb R $$

Finally, let's obtain Klein bottle $\ K\subseteq\Bbb C\times\Bbb C\ $ by twisting $\ T\ $ half as fast as points of $\ T\ $ rotate around $\ (\mathbf 0\ 0)\in\Bbb C\times\Bbb R:$

$$ K\ :=\ \{ ((5+\Re s)\cdot c^2,\, \Im s\cdot c)\,: \,\ (c\ s)\in S\times S\} \quad\subseteq\quad\Bbb C\times\Bbb C $$

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    $\begingroup$ Thank you. That looks what i am after. I shall try to get my head round it. Something to d while i self-isolate out in the NZ countryside. Thank you!! $\endgroup$ Mar 22, 2020 at 0:55
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$\bigg\{ (z_1,z_2)\in \mathbb{C}^2 | |z_1|^2+|z_2|^2=1\bigg\}$ is 3-dimensional sphere $S$. When $ \Sigma =\bigg\{ (z_1,z_2)\in \mathbb{C}^2 | |z_1| = |z_2| = \frac{1}{\sqrt{2}} \bigg\}$ is a torus in $S$, then we know that $S$ is a union of two solid torus.

Here $((R+a\cos\ t)\cos\ \theta,(R+a\cos\ t)\sin\ \theta,a\sin\ t)$ is a parametrization for torus $\Sigma$ in $\mathbb{R}^3$. Further, a circle $$\bigg((R+a\cos\ \frac{\theta }{2} )\cos\ \theta, (R+a\cos\ \frac{\theta }{2} )\sin\ \theta,a\sin\ \frac{\theta }{2} \bigg) $$ in $\Sigma$ can be a boundary of Mobius band, which is in solid torus. Hence since Klein bottle $K$ is a union of two Mobius bands, then $K$ is in a $3$-dimensional sphere.

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