Un-knotting an Un-knot

In a recent project, I was thinking about curves in 3D space. For example, think of the slow-motion replays of the trajectory of a tennis ball that they show at Wimbledon.

The idea is to take a fixed curve in space. Then, to stand at a particular location in space and photograph the curve. This gives a 2D image of it. We are interested in if this 2D image curve intersects itself.

One example is this 3D curve:

We want to map out all places in 3D space where there is a crossing. This question is tangentially related to real tensor decomposition (reading the paper will reveal that this last sentence is true and also a pun!).

To separate the viewpoints for which there is a crossing, from those for which there isn’t, we need to find the boundary cases between the two. There turn out to be only two ways, locally speaking, that a curve can transition from having a crossing to not having one, as we change the viewpoint slightly.

The first move, the T-move, gradually untwists a single loop, which we can see happening for the curve above. The second move, the E-move, starts with two arcs of the curve that are overlapping, and our position changes so that they cease to overlap. This second case is more elusive than the first case.

For some curves, both moves occur:

Looking carefully at this example, we see that a crossing appears for an E-move reason. But that the curve becomes un-twisted again for a T-move reason.

So far, we’ve seen a curve which transitions from crossed to uncrossed, or vice-versa, only via a T-move. We also saw a curve that crosses/un-crosses itself both via a T-move and via an E-move. What about the other case? Does there exist a curve that can only cross and un-cross itself via E-moves?

If so, what would this curve look like?

• T-moves could still exist: we can have loops that appear and then untwist themselves. The crucial thing is that such an untwisting cannot cause there to be no crossings. It can only happen if there is another crossing elsewhere on the curve that stops this from being a true transition point.
• The curve has to have some viewpoints from which it looks completely un-tangled (no crossings). If a curve crosses over itself, as seen from every possible angle, then we wouldn’t have an E-move boundary point between the self-intersecting and non-self-intersecting parts. One example of a 3D curve that has crossings, regardless of which way you look, is a knot such as your shoelaces.

I thought about the question of making an “E-move-only” curve for a day or two. One morning I sat in a cafe with a friend and constructed possibilities using plastic straws. So, if you see someone playing with straws at a cafe: just think! they could be a maths phd student. Or, you know, a child.

And here it is!

This example is different than the ones above – it’s not anything like the trajectory of a tennis ball (unless the tennis ball is navigating a complex architectural construction in zero-gravity). The curve is made of a collection of straight lines. If we wanted a smooth curve, we could smooth the corners slightly without changing the E-move property. But it remains to be seen if we can find a low-degree algebraic example like the ones above.

And here’s some code I used to generate it.

``````[sourcecode language="mathematica"]
szl = -1; sz = 4;
a = ParametricPlot3D[{{v, 0, 0}, {v + 2, 0, 0}, {1, 1.5*v, 0}, {v + 1,
1.5, 0}, {2, 1.5*v, 0}, {3, v, 0}, {3, v + 2, 0}, {3, 1,
1.5*v}, {3, v + 1, 1.5}, {3, 2, 1.5*v}, {3, 3, v}, {3, 3,
v + 2}, {1.5*(v + 1), 3, 1}, {1.5, 3, v + 1}, {1.5*(v + 1), 3,
2}, {v, 3, 3}, {v + 2, 3, 3}, {1, 1.5*(v + 1), 3}, {v + 1, 1.5,
3}, {2, 1.5*(v + 1), 3}, {0, v, 3}, {0, v + 2, 3}, {0, 1,
1.5*(v + 1)}, {0, v + 1, 1.5}, {0, 2, 1.5*(v + 1)}, {0, 0,
v}, {0, 0, v + 2}, {1.5*v, 0, 1}, {1.5, 0, v + 1}, {1.5*v, 0,
2}}, {v, 0, 1}, PlotRange -> {{szl, sz}, {szl, sz}, {szl, sz}},
ViewPoint -> {1.3, -2.4, 2}];
[/sourcecode]``````

P.S. I know, I know, there are 101 ways to export an animation/gif/table/3dplot… from mathematica and embed it into a WordPress post. All of them are better than an embedded YouTube video, but none of them work. If you know one that works, get in touch!