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!
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.