Can the Klein bottle $K$ be embedded into $S^{2} \times S^{1}$?
If so, how does it work? If not, what is the obstruction?
Thanks in advance.
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is this different from is $S^3 = S^2 \times S^1$? – JMP Mar 23 '15 at 09:51
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Not! Fundamental groups are different. – user223515 Mar 23 '15 at 09:52
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The best I can do is to embed it into $RP^{2} \times S^{1}$. – user223515 Mar 23 '15 at 09:54
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Or an 3-manifold with a free $S^{1}$-action. However there are two double orbits. The quotient space is $S^{2}$ in the orbifold sense. – user223515 Mar 23 '15 at 09:56
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can you embed a 3-manifold into a Klein bottle? – JMP Mar 23 '15 at 10:04
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@JonMarkPerry, the Klein bottle is a 2-manifold. You certainly can't embed a 3-manifold in a 2-manifold. – Christoph Mar 23 '15 at 10:07
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No. Klein bottle is 2-dimesion. – user223515 Mar 23 '15 at 10:07
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Thanks, John. That's great! – user223515 Mar 23 '15 at 10:25
2 Answers
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The Klein bottle can be given by $S^1 \times [0,2\pi] $ with the identificition
$$(\theta, 0) \sim (-\theta, 2\pi)$$
(where I used $\theta$ to parametrizes the circle). Now define
$$S^1 \times [0,2\pi] \to S^2 \times S^1, \ f(\theta, t) =\big( a(\theta, t), e^{it} \big),$$
where
$$a(\theta, t ) = \sin\theta (\cos\frac{t}{2} , \sin \frac{t}{2}, 0)+ \cos \theta (0,0,1).$$
Note that
$$f(-\theta, 2\pi) = (a(-\theta, 2\pi ), e^{i2\pi}) = (a(-\theta, 2\pi ), e^{i0}) $$
and
\begin{equation} \begin{split} a(-\theta, 2\pi ) &=\sin(-\theta) (\cos \pi , \sin \pi, 0)+ \cos (-\theta) (0,0,1)\\ &= \sin \theta (\cos 0 , \sin 0, 0)+ \cos \theta (0,0,1) \\ &= a(\theta, 0) \end{split} \end{equation}
Thus $f(\theta, 0) = f(-\theta, 2\pi)$ and so $f$ descends to a map
$$\tilde f: K \to S^2 \times S^1. $$
Note that $\tilde f$ is injective.
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(I haven't read your answer, so can't say anything about whether or not I agree with your construction.) I don't see how this answer is contradicted by Joel's; $H_1(S^1 \times S^2; \Bbb Z/2\Bbb Z) = \Bbb Z/2\Bbb Z$. – Mar 23 '15 at 10:43
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Ar ha!! I dunno why I was thinking about $S^3$ all the time (probably by some weird comment by someone below the question) Thanks. @MikeMiller – Mar 23 '15 at 10:45
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The Klein bottle is the space $S^1 \times I/\sim$, where the equivalence relation identifies $(z,0) \sim (\bar z, 1)$, where $\bar z$ denotes complex conjugation (aka reflection across the x-axis).
Denote by $\text{rot}_\theta$ the rotation of 3-space which moves the z-axis down theta degrees toward the y-axis (this is a rotation whose fixed set is the x-axis). Write $g: S^1 \to S^2$ for the standard inclusion of the equator.
$f: S^1 \times I \to S^1 \times S^2$, given by $f(z,t) =(2\pi t, \text{rot}_{\pi t} g(z))$ descends to an embedding of the Klein bottle.
Jay Qish
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