TITLE: Non-alternating knot diagrams
QUESTION [5 upvotes]: (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.


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.

REPLY [6 votes]: 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:
$$K_s(t)=\begin{cases}(r(t),s\phi(t),1-t)&\text{if }t\leq 1,\\(0,0,t-1)&\text{if }t>1.\end{cases}$$