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A few weeks ago, my friend radii was puzzling in his server hall. He asked if it was possible to prove that what he wanted to do was impossible, or if he had to remain with his gut feeling. I asked him, and got the following explanation:

He had two strands of something ropelike, both fixed at large furnishings at one end, and fixed in a fixed sized loop at the other. He wanted to take these, and link them fast to each other in this fashion:

I started thinking about the problem, and am now convinced I can prove the impossibility he asked for by basic techniques of knot theory. The argument is what I’ll fill this blog post about.

First observation is that due to the size of the fixtures, we can essentially consider the endpoints fixed. Not only that, but since we cannot thread the loops over the fixtures, there’s no way to just stick that loose end through any of the loops. So we can basically extract a cube of space and require that all of our modifications be contained completely within this cube, and then stick the endpoints just outside that cube.

This is something called a *framed knot*, and a popular object of study. My argument is going to manage to steer basically entirely clear of this, though, and I’ll just mention it as an extra property to remember.

The second observation is that the size of the loops is basically irrelevant. So we can make the length of the single strand part as small as we like, and as close to the fixed end as we like. Hence, for all purposes, we essentially want to produce a link like this:

And this is essentially the problem I will approach. Now, I’m going to prove the impossibility by a method similar to how we prove that the trefoil and unknot are different; using three-colorability.

Equality of knots is basically equivalent to being able to go from the drawing of one knot to the drawing of another knot by the Reidemeister moves. I warmly recommend the wikipedia page for a first glance – these turn out to be absolutely central for constructing knot invariants.

For the trefoil argument, we start with discussing coloration of knots. Given three colors: red, green and blue, we may color each strand in a knot diagram with one of the colors. We require that at every crossing, each strand is either the same color, or different color. Obviously, every knot, including the unknot, can be colored in all the same color. And it turns out that the Reidemeister moves are compatible with knot coloration. The proof of this is entirely pictorial; up to a permutation of the colors, the following are all the options we have:

Reidemeister I:

Reidemeister II:

Reidemeister III:

The take home message is that for every crossing configuration, with fixed frame outside, all Reidemeister moves respect the colors at the frame.

So, how may we apply this to our problem? We still have the characterization of knot equality by finding sequences of Reidemeister moves. So we’ll start with the unknotted version of radii’s problem, and a chosen coloration:

Notice that in the diagrams above, the only way to introduce a new color is to already have at least two colors present. Hence, any Reidemeister moves made on this untied configuration will stay entirely green.

However, the target configuration does have this coloration:

Notice that the fixture coloration is identical to the one in the unknotted version, but that this version has all colors represented. However, since the Reidemeister moves cannot possibly introduce a new color starting in the untied version, they can never arrive at this particular configuration.

Of course, I have to jump in and point out that you’re using colored

tangles, not just knots.