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Start with a closed, self-intersecting curve, where every crossing is transverse. Now form something like the opposite of an alternating knot diagram as follows. Starting anywhere, traverse the curve, and at each previously unvisited crossing, go over/above. If the crossing has been previously visited, leave the assigned crossing designation.

Two examples are shown below. (a) is clearly the unknot. (b) is also the unknot, perhaps not as obviously.


Knots

Red circle indicates starting point, arrow the traversal direction.


I expected these diagrams to obviously represent the unknot, but I am not seeing a clear proof. So:

Q. Prove (or disprove) that such a knot diagram always represents the unknot.

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A diagram such as you describe is called a descending diagram, and indeed always results in the trivial knot. For a proof, see Lemma 3.2.10 of http://www.math.ucsd.edu/~justin/Roberts-Knotes-Jan2015.pdf. The previous answer has the right idea.

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  • $\begingroup$ Nice: "The resulting diagram is called the standard descending projection. (I call it the “lazy man’s diagram”, because when you draw it, you only take your pencil off the page when you have to!)" $\endgroup$ Commented Mar 12, 2021 at 21:17
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This is the always the unknot. I was introduced to this by my advisor but I don't think it is originally his argument either, so I don't know who did this first.

To see this, we will use the fact that the bridge number of a knot is one iff the knot is the unknot.

Draw your projection of the knot and choose your starting point. We will make this projection into a diagram by only making over crossings as we traverse the projection. If the projection is drawn in the $x,y$ plane where $z=0$, we can create a knot in $\mathbb{R}^3$ by making every $i$-th new crossing we come to at the level $z=i$. Thus, when we have met every crossing in the projection and are about to come back to the first crossing crossing, our knot in 3-space must drop back from some high $z$ value back to $z=0$.

What we have is a height function where the the knot is strictly increasing everwhere except the small segment between the last crossing and the first crossing. Thus, there is one maximum and one minimum and therefore a bridge number 1 knot, the unknot.

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  • $\begingroup$ Thanks for this. I did not know that bridge #1 = unknot. One question. If you look at (b), one returns to a previously lifted crossing before returning to the start. So it is not up-up-up. $\endgroup$ Commented Aug 16, 2020 at 16:30
  • $\begingroup$ @JosephO'Rourke Whoops. Yes, I see that and I missed this in your description of crossing assignment. I am now not sure I believe it will always be the unknot. I will see if I can cook up a counter-example. $\endgroup$
    – N. Owad
    Commented Aug 16, 2020 at 22:07
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Not sure how helpful, as I am not an expert, but here is an idea that might be right.

First, introduce the third dimension, perpendicular to your drawing, and make sure the "initial" point is a projection of a segment going straight "up". Then, it should be possible to place the rest of the knot so that, while going alongside the line, you only go down. Imagine a helter skelter (with an almost vertical staircase going up), and you will have a good idea what I mean. Now this is a bit handwavy, but I believe you can just assign fixed heights to each of the intersections, as you go through them on the way "down", and then extend to all the other points on the knot. (E.g. if the "staircase" part rises from height $0$ to $1$, for $n$ intersections, as you go through each one twice, you can reserve the heights $\frac{k}{2n+1}, k=1,2,\ldots,2n$ for the "intersecting" points on the knot.)

The rest should be the simple calculation to show that this knot can be deformed into unknot. If the equation of the original knot (the "slide" part) is parameterised as $(\rho(t)\cos\phi(t),\rho(t)\sin\phi(t),1-t), t\in[0,1]$, with $\rho(0)=\rho(1)=0$, then deform it, for $\lambda\in[0,1]$ into $(\rho(t)\cos\lambda\phi(t),\rho(t)\sin\lambda\phi(t),1-t)$. $\lambda=1$ gives the original knot, while $\lambda=0$ gives an obvious unknot in $x-z$ plane.

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