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There is a question on this site about the distinctions between the full-twisted Mobius band and the cylinder, but I would like to ask something different, so I start a new question.

Let us call $C$ the standard cylinder embedded in $\mathbb{R}^3$, and call $F$ the full-twisted Mobius band (aka. the Mobius band with two "half-twists", or the Mobius band with a 360 degree twist), also embedded in $\mathbb{R}^3$. $C$ and $F$ are topologically homeomorphic to each other, but they are not isotopic within $\mathbb{R}^3$. Now think of $\mathbb{R}^3$ as a subspace of some higher dimensional Euclidean space $\mathbb{R}^n$ where $n\geq 3$, and ask if $C$ and $F$ are isotopic in $\mathbb{R}^n$. Obviously if $n \geq 6$ then $C$ and $F$ are isotopic. But what about $n = 4$ or $5$? Is there any chance that they are still isotopic, or can one prove that $n=6$ is the smallest number of dimensions in which $C$ and $F$ are isotopic? Any help or idea is appreciated. Thanks!

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  • $\begingroup$ I can prove they're isotopic in $\Bbb R^5$. I don't know about $\Bbb R^4$ yet. $\endgroup$ Commented Mar 12, 2016 at 5:12
  • $\begingroup$ One can show that the $\pm 1$-twisted bands are isotopic. I feel like these should both be isotopic to the $0$-twisted one since I've never heard of a well-defined $\mathbb Z_2$ framing number for curves in $4$-space, but I'm not seeing the right moves right now. $\endgroup$ Commented Mar 12, 2016 at 5:23
  • $\begingroup$ @GrumpyParsnip: Well, what's worse, I have no earthly idea how I would prove they're not isotopic. $\endgroup$ Commented Mar 12, 2016 at 5:46
  • $\begingroup$ @MikeMiller: I believe I found a way to trivialize any band. What do you think? $\endgroup$ Commented Mar 12, 2016 at 15:52

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I think they are isotopic in $\mathbb R^4$. Consider a circle embedded smoothly in $\mathbb R^4$. It has a normal bundle with fiber $D^3$ and boundary $S^2\times S^1$. Given a band, such as the twisted or untwisted ones in the question, you can think of one boundary component of the band as an embedded circle in $\mathbb R^4$, while the other boundary component lies in the boundary of the normal bundle. This gives a loop in $\pi_1(S^2)$. Such a loop can be homotoped to a point, since $S^2$ is simply connected, and this homotopy will give a fiber-preserving isotopy untwisting any band.

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  • $\begingroup$ We can null-homotope one of the boundary components, but can we do both at once? Because they're two-sided bands, perhaps we should think of the loops as loops in $\pi_1(\Bbb{RP}^2)$... EDIT: Ah, but we're twisting an even number of times, so these two-sided bands represent the trivial element. So they're trivializable. I agree! $\endgroup$ Commented Mar 12, 2016 at 16:04
  • $\begingroup$ Oh, now I understand your argument, no need to talk about $\Bbb{RP}^2$ whatsoever. I doubly agree. $\endgroup$ Commented Mar 12, 2016 at 16:22
  • $\begingroup$ @MikeMiller: thanks. $\endgroup$ Commented Mar 12, 2016 at 16:40
  • $\begingroup$ @Grumpy Parsnip: Thanks for your great answer! I think that I can get the main idea. But when saying "this gives a loop in $\pi_1(S^2)$", is it implied that we somehow identify all the fibers of the bundle? $\endgroup$ Commented Mar 13, 2016 at 4:41
  • $\begingroup$ Well Okay the bundle itself is the direct product $D^3\times S^1$, so it is a trivial matter to identify all the fibers. $\endgroup$ Commented Mar 13, 2016 at 4:45

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