# Knots in three space: non-ambient isotopy equivalence classes

Here, our objects will be knots. Now I don’t want to be too pedantic but we do need some precision here. I will NOT worry about orientation as that will not be important for this type of knot theory.

We have two knots, $K_1 = g(K)$ where $K$ is, say, the standard circle in the plane and $g$ is a embedding (one to one continuous map), and $K_2 = h(K)$ is possibly some other knot.

We say that $K_1$ and $K_2$ are isotopic if there exists an isotopy $F: S^1 \times [0,1] \rightarrow S^3$ where $F(K,0) = g(K) = K_1$ and $F(K,1) = h(K) = K_2$. As usual, we require $F(_,t)$ to be an embedding for all $t \in [0,1]$ (if we allow for non embeddings, then we have a homotopy, not an isotopy).

NOTE it is very common in the literature to denote “ambient isotopy” by “isotopy”; usually it is clear from context what is meant. But beware of this practice.

Now in this setting (isotopy, possibly non-ambient), it IS important to note if the isotopy $F$ is smooth, piecewise linear, or merely topological.
1. If we insist that $F$ is smooth, then it is a theorem of differential topology that these smooth isotopies of compact sets “can be covered by an ambient isotopy”; that is, given a smooth $F$ we can find an ambient isotopy $G$ that accomplishes the same deformation. Note: this is only true for compact sets (in our case, knots). This is NOT true for lines in space.

2. If we insist that $F$ be piecewise linear (roughly speaking: take polygonal objects to polygonal objects), then the classification of knots is rather boring. Here is why: the following process is the result of a piecewise linear isotopy:

Think of shrinking the local knot with time until it disappears. Now remember that we are working in $S^3$; the complement of a piecewise linear ball is another piecewise linear ball on “the other side”.

Now one can find an interesting theory for links (knots with more than one component) using the piecewise linear isotopies; basically it is the usual ambient isotopy classes for links with the “local knots” removed.

3. If we allow for the isotopy to be topological, we CAN get interesting things happening, provided our knots are wild….”very wild”.
Now if the knots in question “pierce a disk” at a point (that is, there is a disk whose boundary links the knot and whose interior touches the knot “honestly”; that is, the knot enters the product neighborhood of the disk in one side and exits it out the other side; (to you experts: the Bing approximation theorem says we can use a disk that is p. l. everywhere but the point of entry).

Now if the knot pierces a disk: we can thicken that disk to “stretch out” the point into a nice, smooth arc, and then we can employ what we did earlier:

So if there is a knot that is NOT isotopic to the unknot (a simple, smooth simple closed curve that projects to a curve with no crossings), that knot has to be wild enough to fail to pierce a disk at any of its points.

Do such beasts exist? Yes; we’ve seen one in the previous post:

One obtains a knot by iterating this knotted solid tori construction over and over again, and taking the intersection. Yes, one does get a knot out this, and yes, this requires proof. R. H. Bing came up with this example: A Simple Closed Curve Which Pierces no Disk, J. Math. Pures and Appl. (9) 35, (1956), 339.

You can find that paper reproduced here in Bing’s “complete works.

Open Problem
It is unknown if this example is isotopic to the standard smooth unknot and it is unknown if there is ANY knot that is isotopic to the standard smooth unknot!

Is is known that there ARE simple knots which fail to pierce a disk at any of their points but ARE isotopic to the unknot:

see M. Brinn: Curves Isotopic to Plane Curves in Continua, Decompositions and Manifolds, Ed. Bing, Eaton, Starbird, University of Texas Press, 1980, pp. 163-166. Note: this article is mostly diagrams.