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Abstractly, on the topological circle $S^1$ there are only two real line bundles, up to isomorphism: the trivial one $\mathcal{O}$ and the Moebius strip $\mathcal{O}(1)$ (thinking of $S^1$ as $\mathbb{RP}^1$). So we have $\mathcal{O}(2k)\cong\mathcal{O}$ and $\mathcal{O}(2k+1)\cong\mathcal{O}(1)$ for any $k$.

Let's embed $S^1\hookrightarrow\mathbb{R}^3$ as the set $x^2+y^2=1$ in the $z=0$ plane. We can embed the total spaces $\iota_n:\mathrm{tot}(\mathcal{O}(n))\hookrightarrow\mathbb{R}^3$, respecting the zero section, for example as the union, for $t\in [0,2\pi]$, of the (open) segments joining the two points

$((1\pm\frac{1}{2}\cos(nt))\cos(t), (1\pm\frac{1}{2}\sin(nt))\cos(t), \pm\frac{1}{2}\sin(nt))\in\mathbb{R}^3$.

Call $\mu(n)\subset\mathbb{R}^3$ the so obtained $n$-twisted Moebius strips.

Is $\mu(2k)$ (ambient) isotopic to $\mu(0)$, and $\mu(2k+1)$ to $\mu(1)$?

(of course, properly, we're talking about the embeddings $\iota_n$)

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$x^2+y^2=1$, not $=0$. –  Ben McKay Dec 22 '11 at 12:02
Yep. Fixed, thanks. –  Qfwfq Dec 22 '11 at 12:07

2 Answers 2

up vote 18 down vote accepted

No. The unit normal bundle to the circle is a torus. Each fiber of the Mobius strip contains a unit vector going in some choice of direction, lying on that torus. That vector travels around the torus representing a homology class, which is clearly not trivial. If we can isotope to the trivial strip, we get a homotopy to the trivial homology class.

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Very nice! thanks –  Qfwfq Dec 22 '11 at 12:04

You could look at the knot type of the boundary of these twisted bands. (Substitute the unit sphere bundle if you want the noncompact version.) For each of your $\mu(k)$ you get a $T(2,k)$ torus link as the boundary. Each of these is non-isotopic. For example, $k=2$ gives the Hopf link and $k=3$ the trefoil. This implies that no two would be ambiently isotopic.

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