Removing zero sections of n-twisted strips

Question

First of all, a note: This question is motivated by my friend, who is a physics PhD student (I'm a math PhD student) and arose when I was trying to explain to him what line bundles are. (I wanted to tell him that a wavefunction is really a global section of a complex line bundle.) I was using a toy example of a Mobius strip. These questions arose from experiments that we did using paper strips.

Let us consider a strip of paper and we glue the two ends of the strip after doing $ n $ half twists. So for example, $ n=0 $ gives a cylinder, $ n=1$ gives an ordinary Mobius strip, and so on. We start cutting the paper at the center of the strip, or in mathematical terms, we delete the zero section of the corresponding line bundle, call the resulting manifold $ Z_n $. Hence,

For $ n= 0 $ we get two (unlinked) cylinders, i.e. the trivial line bundles over two unlinked circles. So two connected components, each of which is orientable. For $ n=1 $ we get a single $ 2 $-half twisted strip. So one connected component and orientable again. For $ n=2 $, we get two $ 2 $-half twisted strips, linked together. So two connected components and orientable. For $ n=3 $, it starts getting weird: we get a trefoil knotted strip whose twists we can't really keep track of, but tracing through one side shows that it's orientable again and one connected component. For $ n=4 $, two connected orientable components knotted together again. For $ n=5 $, one very complicated knotted strip, orientable.

Here's my incomplete math explanation: I know that the Picard group of (real) $ C^{\infty} $-line bundles on the circle is $ \mathbb{Z}/2 $, with generator given by an ordinary Mobius strip (the $ n=1 $ case). So the diffeomorphism class of the $ n $-half twisted strip is [cylinder] for $ n $ even and [Mobius strip] for $ n $ odd, explaining why the number of connected components after deleting the zero section alternates between $ 2 $ and $ 1 $. This also explains why any component is orientable by only verifying the $ n = 0 $ and $ n=1 $ cases. I hope this explanation is correct.

Our questions begin here and I can't explain why these happen:

(1) Is there a reason why the trefoil knot shows up for $ n=3 $? In general what kinds of knots and what kinds of links show up as $ n \ge 4 $?

(2) We can consider negative integers $ n $ as well, where the interpretation is that an $ n $-half twist is a $ (-n) $-half twist but with ends glued oppositely. What is the relationship between $ Z_n $ and $ Z_{-n} $? Our conjecture is that the knot(s) appearing in $ Z_n $ and $ Z_{-n} $ are mirror knots. (linked identically when $ n $ is even.) Is this correct and can someone prove this?

Differential topology is not my field of study as I mostly deal with complex algebraic geometry. So feel free to correct me and I'd be grateful to have a rigorous explanation.

Answer 1

One particularly concrete way to get at this issue is to fix real numbers $0 < r < R$, and to parametrize the solid torus of major radius $R$ and minor radius $r$ by $$ X(\rho, \theta, \phi) = ((R + \rho\cos\theta)\cos\phi, (R + \rho\cos\theta)\sin\phi, \rho\sin\theta) $$ for $0 \leq \rho \leq r$ and $0 \leq \theta, \phi \leq 2\pi$. (Geometrically, use polar coordinates $(R + r\cos\theta, r\sin\theta)$ to parametrize the disk of radius $r$ centered at $(R, 0)$, then revolve about the $z$-axis. Naturally, this map is defined on all of three-space.)

If $n$ is an integer, the image of $$ x(\rho, \phi) = X(\rho, \tfrac{1}{2}n\phi, \phi),\qquad -r \leq \rho \leq r,\ 0 \leq \phi \leq 2\pi, $$ is a strip with $n$ half-twists. (Geometrically, we're taking the diameter from $(R - r, 0, 0)$ to $(R + r, 0, 0)$ in the longitudinal half-plane where $\phi = 0$ and revolving it $\frac{n}{2}$ times about the center of the disk (in the revolving longitudinal plane) as the disk revolves once about the $z$-axis.)

The upward-pointing unit vector $(0, 1, 0)$ at the center of the disk extends locally to a unit normal field to the strip.

If $n$ is even, the diameter rotates through a whole number of turns; the unit normal vector returns to its initial direction $(0, 1, 0)$, and the strip is orientable. The are two components to the complement of the zero section because each radius "joins back to itself". Each endpoint $(R \pm r, 0, 0)$ traces an $(n, 1)$ torus knot.

If $n$ is odd, the diameter rotates through a half-integer number of turns; the unit normal vector returns to the opposite direction $(0, -1, 0)$, and (mod details) the strip is consequently non-orientable. There is one component to the complement of the zero section because each radius "joins back to the other". The endpoints $(R \pm r, 0, 0)$ together trace a single $(n, 2)$ torus knot. (Equivalently, consider one endpoint and run $\phi$ two full turns, from $0$ to $4\pi$.)

In summary:

1. The trefoil is a $(3, 2)$ torus knot; generally, we get torus knots from the boundary curves of the strips, though there is "extra" twisting of the half-strips, loosely because we're seeing twisting around the central curves of the cut strip(s).

2. If we replace $n$ by $-n$, we're equivalently post-composing $x$ with the reflection $(x, y, z) \mapsto (x, y, -z)$, so your mirror-image conjecture is correct.