general-topologyknot-theory
A Möbius strip with one half twist has the unknot as its boundary. One with two half twists has a link of two unknots. One with three half twists has the trefoil knot as its boundary.
Years ago, I remember hearing more about this. Does anyone know of any resources or theorems or anything pertaining to this idea? (Specially which knots/links you obtain from the boundary of a Möbius strip with k half twists?)
Suppose that the strip, on both of its sides, has two stripes on either side of the centre line. After folding up and cutting along the centre line, treat the stripe(s) as a knotted rope (or ropes).
For the regular Möbius strip (left), the resulting rope can be deformed to a simple ring. For the three-twist Möbius strip (right), a trefoil (overhand) knot results. For a five-twist Möbius strip, the cinquefoil (double overhand) knot is obtained, and so on.
In general, a strip with $n$ twists when cut along the centre line yields the $(n,2)$ torus knot/link. The cut gives one strip if $n$ is odd and two if $n$ is even; as examples of the latter, the two-twist strip (the result of cutting the one-twist strip) when cut yields the Hopf link (two linked rings), while the four-twist strip when cut yields Solomon's link, distinct from the Hopf link.
Here are the definitions I'm using:
(Since we are working in dimension at most $3$, I'm not being careful about technicalities of smooth/piecewise-linear embeddings.)
Take a look at Rolfsen's book "Knots and Links" for the classification of knots in the torus $T^2$. (Summary: for every pair of coprime numbers $p,q\in\mathbb{Z}$, there is a unique nontrivial knot up to equivalence in the homology class $(p,q)\in\mathbb{Z}^2\approx H_1(T^2)$. There are no nontrivial knots in any other homology classes.)
For a surface $\Sigma$, then each oriented knot $K$ in $\Sigma$ gives a conjugacy class of $\pi_1(\Sigma)$. This is because the choice of basepoint matters, but it only changes the element of $\pi_1(\Sigma)$ by conjugation. There is a theorem that if two knots are freely homotopic (that is, if they correspond to the same conjugacy class in $\pi_1(\Sigma)$), then they are equivalent. See Proposition 1.10 of "A primer on mapping class groups."
Let's consider $\mathbb{R}\mathrm{P}^2$. We know $\pi_1(\mathbb{R}\mathrm{P}^2)\approx \mathbb{Z}/2\mathbb{Z}$, which has exactly two conjugacy classes. By the previous discussion, nontrivial knots, if they exist, must be freely homotopic to the generator. Using a Hatcher-style identification diagram, we can draw a loop $\alpha$ representing the generator like so:
Using covering spaces, since $\mathbb{R}\mathrm{P}^2$ is double covered by $S^2$, and since this $\alpha$ is the image of a path between antipodal points of $S^2$, it must be a generator.
Because $\alpha$ is a simple closed curve, it is a knot, and since it is not nullhomotopic, it is not equivalent to a trivial knot.
Compare this to $S^2$: since $\pi_1(S^2)=1$, every knot is trivial. (This is the 2D Schoenflies theorem.)
Removing a disk from $\mathbb{R}\mathrm{P}^2$ gives a Möbius strip, which has a fundamental group of $\mathbb{Z}$. The $\alpha$ loop corresponds to the core circle in the strip, which is a nontrivial knot. The only other nontrivial knot is from manipulating $\alpha\cdot\alpha$ a bit; this is two parallel curves in the strip, but they join up to form a single loop, or, more simply stated it is the boundary of the Möbius strip.
There is no way to intrinsically count the number of twists in a Möbius strip, and the best you can say is "an odd number." However, if you actually embed the Möbius strip in $\mathbb{R}^3$, then you can count the number of twists, but it is more complicated than you might initially think. Given an embedding, consider the core circle of the Möbius strip, which is a knot in $\mathbb{R}^3$. Every knot has a canonical framing, which is a continuous choice of normal vectors at each point of the knot. The canonical framing (also known as the $0$-framing) has the property that if you take the knot $K$ and take a copy $K'$ that is from moving $K$ in the direction of each of its normal vectors (called a push-off), then the linking number between $K$ and $K'$ is $0$. With this framing, you can count how many times the Möbius strip twists as you go around $K$.
If the core circle of the Möbius strip is a trivial knot, then this definition matches the intuitive notion of the number of twists.
You bring up how if you take an unknotted embedding of a Möbius strip with three twists, then its boundary is a trefoil knot, which is nontrivial. This is a specific example of a more general construction called a satellite operation that is known as a cable, where you take a knot and place along it a Möbius strip with some number of twists. If you start with the unknot, then the cables are all nontrivial knots unless the Möbius strip is given $\pm 1$ twists; collectively they are all known as $(2,t)$ torus knots (not to be confused with knots in a torus like at the beginning!), where $t$ is the odd number of twists. If you start with any nontrivial knot, then every cabling of it will also be a nontrivial knot.
Maybe this will clear something up: if $K$ is a knot in a surface $\Sigma$ and $\Sigma$ is embedded in $\mathbb{R}^3$, then if $K$ is trivial in $\Sigma$, it is trivial in $\mathbb{R}^3$, but just because $K$ might be trivial in $\mathbb{R}^3$, it is not necessarily trivial in $\Sigma$. That is, just because there might be an disk in $\mathbb{R}^3$ whose boundary is $K$ doesn't mean there's also an embedded disk in the surface!
I don't really know about what happens to knots from immersed surfaces. You ought to take some care because knot equivalence in the surface might lead to the knot passing through itself from the point of view of $\mathbb{R}^3$. All I do know is that Boy's surface, which is an immersion of $\mathbb{R}\mathrm{P}^2$ in $\mathbb{R}^3$, gives an embedded trefoil coming from the $\alpha\cdot\alpha$ knot in the Möbius strip: https://commons.wikimedia.org/wiki/Boy%27s_surface
I wonder if there's anything to be said about taking a knot $K$ in a surface $\Sigma$ that is immersed in $\mathbb{R}^3$, then looking at the set of all knot types in $\mathbb{R}^3$ that come from knots equivalent to $K$ in $\Sigma$? The interesting part here is that the singularities are places where you can pass the knot through itself. Is there an immersion of $\mathbb{R}\mathrm{P}^3$ where its nontrivial knot is always nontrivial in $\mathbb{R}^3$?
Best Answer
This answer is going to be less useful than I would wish, because I don't have a reference. But I can at least tell you the answer.
These knots are clearly all elements of the braid group on two strands, called $B_2$. Elements of $B_n$ are generated by taking $n$ separate strands, switching the places of one end of each a pair of strands without switching the other ends,thus half-twisting the two strands together. The group $B_2$ on two strands is particularly simple: it is isomorphic to the integers. We can identify its elements just by saying which integer it corresponds to. This is simply the number $k$ of half-twists of the two strands, which is the same as the number of half-twists of your strip. (Or in the case of negative $k$, half-twists in the other direction.)
If you take a strip and give it $k$ half-twists before gluing the edges, the result is a knot with one component if $k$ is odd, two components if $k$ is even.
When $k$ is odd, the knot is an unknot when $k=1$, a trefoil when $k=3$, a cinquefoil when $k=5$, and so forth. Knot notations usually express the common structure of this family of knots. For example, in Conway notation these are $[1], [3], [5], [7],\ldots$. They are all torus knots (meaning they can be embedded in a torus) and in the special torus knot notation they are written $(2, k)$.
When $k$ is even the edge forms two linked circles, linked $\frac k2$ times. (Or in the trivial $k=0$ case two unlinked circles.) For $k=2$ this is the Hopf link, two circles linked in the simplest possible way, and for $k=4$ it is sometimes called Solomon's knot, just like the Hopf link except linked $\left(\frac k2=2\right)$ twice instead of once.
These are also torus knots, again written $(2,k)$ in the special torus knot notation. There is a theorem of torus knots that $(a,b)$ has a single component exactly when $a$ and $b$ are relatively prime, so for your family of knots, there is a single component exactly when $k$ is relatively prime to $2$; that is when $k$ is odd.
It might also be worth pointing out that the knotting of the edges is exactly what determines the behavior when you cut the strip down the middle. Cutting a (single-half-twist) Möbius strip down the middle famously produces a single strip, because the edge is a single unknot. Cutting a double-half-twist strip down the middle produces two linked strips, because the boundary is the Hopf link. Cutting a three-half-twist strip produces a single strip tied in a trefoil knot.
I hope this collection of miscellanea contains something helpful. I suggest you look into the braid groups, because the braid group concept corresponds exactly to what you want to look at: what happens if you take $n=2$ separate strands and cross them exactly $k$ times before joining the ends together.