If two long knots give the same knot, are the long knots isotopic?

Question

A long knot is a smooth embedding $f\colon \mathbb{R} \to \mathbb{R}^3$ such that $f(t)=(t, 0, 0)$ for all $|t| > 1$. Given a smooth oriented knot $g\colon S^1 \to \mathbb{R}^3$, we can cut it at a point and extend it to a long knot, and given any long knot, we can add a closure strand to recover the knot. Connected sums for knots correspond to a kind of concatenation for long knots.

My question is this: If we cut a knot $g$ at one position to get a long knot $f_1$ and cut $g$ at a different position to get a long knot $f_2$, can $f_1$ be continuously deformed into $f_2$? Formally, must there be an isotopy from $f_1$ to $f_2$ that is constant over time outside some compact set?

As an example, the leftmost and rightmost figures in the image below are two long knots arising from the figure-eight knot $4_1$. In this case, they are indeed isotopic (you can reach one from the other using Reidemeister moves), but I don't know how to prove this in general (or if it's true). It's unclear to me if/why such an isotopy can be kept "local" and not require some loop to go "around the outside of the long strand". Does this have something to do with why knot theorists are always using $S^3$ rather than $\mathbb{R}^3$?

Answer 1

I think I figured it out: the answer is yes. Suppose $f_0$ and $f_1$ are long knots with isotopic closures (call them $\overline{f_0}$ and $\overline{f_1}$, respectively). The idea is this: if we can make another isotopy from $\overline{f_1}$ to $\overline{f_2}$ that fixes some neighborhood, then we can transform $S^3$ so that that region is at infinity.

Suppose $G\colon S^1\times [0,1] \to \mathbb{R}^3$ is a smooth isotopy: with $G(\cdot, 0) = \overline{f_0}$, with $G(\cdot, 1) = \overline{f_1}$ and with $G(\cdot, t)$ an embedding for each $t$. For $i\in\{0, 1\}$ let $s_i$ be such that $\overline{f_i}(s_i)$ is the middle of the closure strand:

Now define a new isotopy $G'$ by $$ G'(s, t) := G(s + t(s_1 - s_0), t). $$

This is still a smooth isotopy, but $G'$ gradually reparameterizes the second knot to look like this:

We can make another adjustment to ensure the image of $s_0$ is constant over time, not just for $t\in\{0, 1\}$: $$ G''(s, t) := G'(s, t) - G'(s_0, t) + G'(s_0, 0). $$

Now we can smoothly adjust each timestep of $G''$ is constant over time on some neighborhood of $s_0$, and in particular, $G''$ constantly a straight line in that neighborhood.

Finally, as alluded to before, there is some smooth orientation-preserving homeomorphism $H$ of $S^3$ that sends $G(s_0, 0)$ to $\infty$ and transforms our constant line segment to the appropriate long-knot-line $(t, 0, 0)$ in a neighborhood of $\infty$. Applying $G''$ as conjugated by $H$ gives an isotopy from $f_1$ to $f_2$ that is constant in a neighborhood of infinity.

I got less precise toward the end of that, but hopefully the idea is clear.

Answer 2

This isn't materially different from user513093's answer, but it goes into a bit more detail with a diagrammatic point of view.

Due to the boundary conditions for a long knot, one observation is that a long knot is a stereographic projection of a knot $K\subset S^3$ that passes through the projection point $p\in S^3$ along an open arc of the geodesic $\gamma$ that maps to the $x$-axis through the projection.

In other words, after choosing a point $p\in S^3$ and a geodesic $\gamma\subset S^3$ containing $p$, a long knot $K\subset S^3$ is a knot containing $p$ such that there is an open neighborhood $U\subseteq S^3$ of $p$ where $K\cap U=\gamma\cap U$. The closure of a long knot in $S^3$ is just $K$ from this point of view, and the cut operation of a knot $K\subset S^3$ at a point $p\in K$ corresponds to making $K$ be locally a geodesic through $p$.

One can go back and forth between this and the usual notion of a long knot by stereographically projecting from $p$ so that $\gamma$ maps to the $x$-axis. Equivalence of long knots in $S^3$ is given by isotopies of $S^3$ that fix some open neighborhood of $p$. These are the ones that after stereographically projecting will be isotopies that are compactly supported.

Let's define an unmoored long knot to be an oriented knot $K\subset S^3$ with a choice of point $p\in S^3$ such that in a neighborhood of $p$, $K$ is geodesic. Equivalence of unmoored long knots $(K,p)$ and $(K',p')$ is given by isotopies $f_t: S^3\to S^3$ carrying $K$ to $K'$ (preserving orientation) such that for each $t$, $f_t$ is locally an isometry at $p$. After composing with a family of isometries of $S^3$, one gets an isotopy of long knots. In particular, an isotopy of unmoored long knots between two long knots is, after such a composition, an isotopy of long knots, since the family of rotations will form a loop of isometries of $S^3$ (note: this hints at a subtle difference between the geometry of unmoored long knots and long knots, which is that the space of unmoored long knots has this additional $\pi_1(SO(4))$ involved).

We could work out a theory of diagrams for unmoored long knots from this point of view. They are simply the usual knot diagrams on $S^2$ but with a choice of point along the knot, with the point denoting where the knot is locally geodesic. There are two additional types of "Reidemeister" moves, which are that the point can isotope through a crossing. Here is the move involving the overstrand, with the other being for the understrand:

Note that the strands must also be oriented, so there are four moves in total.

If long knots $K,K'\subset S^3$ have equivalent closures (i.e, there is an isotopy of $S^3$ carrying $K$ to $K'$), then we can come up with a sequence of Reidemeister moves representing this isotopy, given diagrams of the long knots. We can then consider the knots as unmoored long knots, updating the diagrams to include the point, and including the necessary additional Reidemeister moves wherever the point passes through crossings. Finally, we can append a sequence of these additional Reidemeister moves to "push" the point around the knot until it coincides with where it is supposed to be in the second long knot's diagram. This establishes an isotopy of unmoored long knots, and thus an isotopy of long knots.

This can be done without diagrams, but I think this illuminates some of the structure of the problem.