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(I asked this on MSE a few days ago without a clear resolution.)

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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    $\begingroup$ This construction can be found in some textbooks on knot theory in the context of the unknotting number of a knot. For example it is described on pp. 58-59 of The Knot Book by Colin Adams and on p. 133 of Knot Theory by Charles Livingston. I have a dim memory of seeing it somewhere else as well. In Adams' book he gives an argument which is essentially the one in the answer by Wojowu below, except that Adams makes the unstated hypothesis that the starting point lies on the boundary of the convex hull of the knot projection, which simplifies his argument significantly. $\endgroup$
    – Allen Hatcher
    Aug 19, 2020 at 15:33
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    $\begingroup$ I believe this fact was also used in the original proof that the skein-theoretic definition of the Homfly polynomial is well-defined. $\endgroup$
    – Peter Samuelson
    Aug 19, 2020 at 17:49

1 Answer 1

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Let us parametrize the plane curve by $\gamma:[0,1]\to\mathbb R^2$ and assume $\gamma(0)=\gamma(1)=(0,0)$. Then your curve is the knot diagram of the knot which is parametrized by $K:[0,2]\to\mathbb R^3$ given by $$K(t)=\begin{cases}(\gamma(t),1-t)&\text{if }t\leq 1,\\(0,0,t-1)&\text{if }t>1.\end{cases}$$ (essentially, imagine suspending your knot on a stick, such that the rope goes down at a uniform speed.) Then we can "unwind" this knot. Namely, since $\gamma$ only goes through $(0,0)$ at the endpoints, we can write $\gamma(t)$ in polar coordinates by $(r(t),\phi(t))$ with $r,\phi$ continuous on $(0,1)$. We can then unknot $K$ by the following sequence of knots $K_s$, which starts with an unknot and ends with $K$, written in cylindrical coordinates: $$K_s(t)=\begin{cases}(r(t),s\phi(t),1-t)&\text{if }t\leq 1,\\(0,0,t-1)&\text{if }t>1.\end{cases}$$

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    $\begingroup$ Likely this is my understanding. I am a bit concerned that your $\gamma(t)$ rises consistently, $t \le 1$, but in fact in (b) of the figure the curve encounters a crossing previously assigned, and so then it must go under; and then later above. $\endgroup$
    – Joseph O'Rourke
    Aug 18, 2020 at 23:50
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    $\begingroup$ I think the parametrization here is in the opposite direction than in the figure. $\endgroup$
    – Jan Kyncl
    Aug 19, 2020 at 0:25
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    $\begingroup$ I think this is essentially correct. We can construct $K$ as the union of a finite number of arcs $\{K_i\}_{i=1}^n$ such that the $z$-coordinate of $K_i$ is monotonically increasing for each $i$, and then $K_{i+1}$ is strictly below $K_i$ for every $i$, as well as arcs connecting the "top" point of $K_i$ with the "bottom" point of $K_{i+1}$ for every $1 \leq i \leq n$, with $K_{n+1} = K_1$. Each of the $K_i$ can be isotoped rel endpoints to a straight line, so the result should be unknotted. $\endgroup$
    – Rohil Prasad
    Aug 19, 2020 at 1:05
  • $\begingroup$ @JanKyncl Oh yes, good point. I meant the curve to go down, but wrote it as going up. $\endgroup$
    – Wojowu
    Aug 19, 2020 at 9:02
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    $\begingroup$ @RohilPrasad If we start drawing our knot in $\mathbb R^3$ at $z=1$, and then continually go down, then for any intersection point on the diagram, we will go over every intersection point of the original diagram twice, and the first visit will always be at a higher $z$ coordinate. If you then project the resulting knot onto the $z=0$ plane, the resulting diagram will be the non-alternating diagram as defined by the OP. $\endgroup$
    – Wojowu
    Aug 19, 2020 at 15:31

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