# The Unapologetic Mathematician

## More knot coloring

Two weeks ago I went over how to color knots with three colors. Of course, being mathematicians, we want to generalize the hell out of this.

Really, as is often the case with this sort of thing, “color” is just a metaphor. We labeled each arc of a diagram with one element of a set with three elements and put a condition on the labels attached to the arcs at each crossing. So, what sets do we know of with three elements. I’ll give you some time to think.

Back? Did you come up with “the abelian group $\mathbb{Z}_3$“? If so, great. If not, this is one of the first examples of groups I mentioned waaaaaay back in January. Basically, it’s made of the numbers in the set $\{0,1,2\}$. We add and subtract as usual, but we loop around from $2$ to ${}0$. For example, $2+2=1$ and $0-1=2$.

So let’s imagine we’re “coloring” a knot with $\mathbb{Z}_3$. What’s the condition at a crossing? Imagine we’ve got an overcrossing arc colored $a$, and we approach it along an undercrossing arc colored $b$. If $a=b$ then we have two arcs of the same color, and we have to leave along an arc with the same color. On the other hand, if $a\neq b$ we have to use the third color. So how do we recognize the third color? It turns out that there’s an easy way to do this: the third color is $2a-b$. In fact, this also works if $a=b$. Try writing down all nine combinations of $a$ and $b$ and seeing that $2a-b$ is always a valid coloring for the third arc.

Now there’s nothing in our notion of coloring that relies on the number $3$. All we’re really using in this new condition is the fact that we have an abelian group. So take your favorite abelian group $G$ and try coloring knots with it, requiring that if an arc colored $b$ undercrosses an arc colored $a$, it comes out the other side colored $2a-b$.

Does this new form of $G$-coloring depend on the knot diagram we use, or only on the knot. Well, it turns out just to depend on the knot. To show this, we again use Reidemeister moves.   In each of these, I’ve colored some of the end and worked across the diagram seeing what I’m required to color the other edges. In the first move, for example, if I color the top of the left $a$, the bottom will also get $a$. If I color the top of the right $a$, the arc crosses under itself so the bottom must be colored $2a-a$. But we see that $2a-a=a$, so this is the same end coloring. The same sort of argument works for the second and third moves.

So for every abelian group $G$, we have an invariant: the number of ways of $G$-coloring the arcs of any diagram of the knot. The colorings we did the last time were $\mathbb{Z}_3$-colorings.

But why does this rule work so well? It turns out that we’re not interested in $G$ as an abelian group. We take the underlying set of $G$ and equip it with the operation $a\triangleright b=2a-b$. This makes $G$ into an involutory quandle, called the “dihedral quandle” of $G$. The name comes from the fact that $b\mapsto2a-b$ is like “reflection through $a$“, and such reflections generate all symmetries of regular polygons in the plane. A polygon in the plane, of course, is a sort of degenerate polyhedron in space with only two sides: “dihedron”.

Anyhow, recall the axioms for an involutory quandle:

• $a\triangleright a=a$
• $a\triangleright(a\triangleright b)=b$
• $a\triangleright(b\triangleright c)=(a\triangleright b)\triangleright(a\triangleright c)$

First check that these axioms really do hold for the operation we defined on $G$. Then imagine using any other involutory quandle $Q$ to color knots. Go back to the Reidemeister diagrams and do just what we did for $G$-coloring, but use the quandle operation instead: if an arc colored $b$ undercrosses one colored $a$, it leaves colored $a\triangleright b$. Show that the number of $Q$-colorings is an invariant of the knot, not just of its diagrams.

May 2, 2007 Posted by | Knot theory | 10 Comments

## A thought

I think I need to go lie down. I’ve been feeling a bit wobbly all day.