Can distinct open knots correspond to the same closed knot?

Question

A topological ("closed") knot is an embedding of a circle in $\mathbb{R}^3$. It's possible for a knot to be distinct from the unknot because there are no free ends to move around and untie the knot. An alternative notion of a knot is an embedding of a closed line segment in $\mathbb{R}^2 \times [0,1]$ such that the two boundary points of the line segment get mapped to the two boundary planes of $\mathbb{R}^2 \times [0,1]$ respectively (one end on each plane), and all other points get mapped to the interior of $\mathbb{R}^2 \times [0,1]$. I'll call this an "open knot". It's possible for an open knot to be distinct from the unknot because, although it has two ends that can move around, they are stuck on the boundary planes, and no part of the knot can get around them, so it can't get untied.

My question is whether these two kinds of knot are fundamentally equivalent. Clearly there are some differences related to chirality/reversibility issues, for example left- and right-handed "overhand" open knots are distinct even though the closed trefoil knot is equivalent to its mirror image (just turn it over). I want to know if that's the full extent of the difference. Whoops, don't know what I was thinking here. If you turn a right-handed trefoil over it remains right-handed.

There's an obvious function mapping from open knots to closed knots - just connect the two ends together. But if you try to invert this function you have to make a choice of where to cut the knot, and you might get different results depending on where you cut it.

Do all choices of where to cut a closed knot result in the same open knot? If not, what's an example?

Answer 1

I will ignore many important questions about wildness and the like, and instead suppose that you have enough regularity to make work the following argument (which I learned from John H. Conway).

Take a closed knot. Place a small bead somewhere along it, and make the bead out of a very shiny material: silver, say. Clearly, moving the bead along the knot doesn't make a difference up to isotopy of knots. Now, the point of making the bead very shiny is that if you look at it, what you see is the mirror image of the knot, now drawn inside the sphere of the bead. (Implicitly I have compactified $\mathbb R^3$ to $S^3$ by adding a point at infinity, which the mirror maps to the center of the bead. Which is to say, what I'm really doing is geometric inversion through the surface of the bead.)

Then inside the bead you see what you've called an "open" knot, and is also called a "long" knot. Any isotopy of the original knot can be performed so that the strings never have to pass through the very small bead, and so these reflect through to isotopies of the open knot. It follows that the isotopy class of the open knot is the same as the isotopy class of its closure.

As an aside: it is a theorem of Dehn's that the left and right trefoils are not equivalent.

Answer 2

Here is a slightly different point of view.

Consider two "open" knots in $\mathbb{R}^2\times [0,1]$ with isotopic closures. By assumption there is a positive diffeomorphism $h$ of $\mathbb{S}^3$ taking the closure of one knot to the other. It can clearly be assumed that $h$ coincides with the identity on a close contractible neighborhood of the infinity. Now for any close contractible subset $X\subset\mathbb{S}^3$ the space $\operatorname{Diff}^+(\mathbb{S}^3,X)$ (of orientation-preserving diffeomorphisms that induce the identity on $X$) is path-connected. Hence there is an isotopy from $h$ to the identity, that induces a long knot isotopy.