We finally reconvened our maths club after a mid-term break (children not available) combined with the university exams period (me not available). I correctly assumed nobody remembered anything we talked about before. At least not in detail. Also, we had a significant number of new children joining our group, who had no idea about what we have been up to.
In the previous sessions we discovered and invented our notation, which consisted of three combinators (X, C, I) and three operations: composition (“gluing”), tensoring (“stacking”) and dual (“flipping”).
This time we just took this as a given, the starting point of our development. If the previous sessions were analytic, trying to decompose knots into basic constants and operations, this session was synthetic, starting with formulas and drawing them up and seeing what we get.
The first formula we contemplated was C;C*, which everyone immediately identified as a circle. The knot-theoretic name for this is the unknot, word which caused a certain amount of amusement. One student cleverly observed, unprompted, that (C;C*)* would have been the same thing, also an unknot, and we had a short discussion about how symmetry of a knot K boils down to K=K*. That was quite rewarding and fun and prepared the ground, somewhat serendipitously, for the next formula C;X;C*:
The same student rashly pointed out that this shape is also symmetric, i.e. C;X;C*=(C;X;C*)*. Is this true or false, qua knots?
The ensuing discussion was the most interesting and important part of the session. Initially the students were in majority supportive of the validity of the equation, then someone pointed out that X≠X* so, they reckoned, it must be that
Indeed, the shapes are not identical. The opinion of the group swung so they all agreed that the equation is not valid and the two shapes (original and flipped version) are not equal.
But aren’t they? What do you think? We are talking knots here, and two knots should be equal when they are topologically equal, i.e. they can be deformed into each other without any tearing and gluing. So in fact they are equal because:
So here we have a genuinely interesting equation, where two formulae are equal not because the shapes they correspond to are geometrically equal (i.e. “identical”) but because they are topologically equal, i.e. there is a smooth deformation between them. Also note that the deformation must be in 3D by “twisting” the left (or right) side of the unknot — a 2D transformation would make the wire “pinch” which may or may not be allowable, depending on how we precisely set things up (if we were very serious and careful about what we are doing).
The point was quickly grasped and we moved on. It is a subtle point which I think can be damaged by over-explaining if the initial intuitions are in the right place, which they seemed to be.
Next we looked at a much more complicated formula that took a long time to draw correctly: C;(I⨂C⨂I);(X⨂X*);(I⨂C*⨂I);C*.
As you may see, if we are drawing this as a knot and tidy things up a bit we actually have this:
Which is really just two unknots — they don’t interlock so they can be pulled apart with no tearing and gluing, i.e. topologically. The formula for this is the much simpler C;C*;C;C*!
This point was not easy to make, and it seemed difficult for the students to appreciate it. By this time they were quite tired and the drawing of the more complex formulation of the two unknots diagram drained them of energy and patience — they made quite a few errors and had to restart several times. Showing them a much easier way to achieve the same knot diagram almost made them angry!
I hope to use this as a motivation for the future: how to simplify a formula before we draw it. Until now we have looked at equality semantically, using the underlying model. In due course, as students become more used with manipulating knot diagrams using formulas, I will start introducing some general equational properties.