Notes for AKT-140106/0:43:23

Claim: The number of legal 3-colorings of a knot diagram is always a power of 3.

This is an expansion on the proof given by Przytycki (https://arxiv.org/abs/math/0608172).

We'll show that the set of legal 3-colorings ${\mathcal {S}}$ forms a subgroup of $Z_{3}^{r}$, for some r, which suffices to prove the claim. First, label each of the segments of the given diagram 1 through n, and denote a 3-coloring of this diagram by $x=(x_{1},x_{2},...,x_{n})$, where each $x_{n}$ is an element of the cyclic group of order 3 $Z_{3}=<a|a^{3}=1>$ (each element representing a different colour). It is clear that ${\mathcal {S}}$ is a subset of $Z_{3}^{n}$. To show it is a subgroup, we'll take $x=(x_{1},x_{2},...,x_{n}),y=(y_{1},y_{2},...,y_{n})\in {\mathcal {S}}$, and show that $xy^{-1}=(x_{1}y_{1}^{-1},x_{2}y_{2}^{-1},...,x_{n}y_{n}^{-1})\in {\mathcal {S}}$. It suffices to restrict our attention to one crossing in the given diagram, so we can without loss of generality let n = 3.

First, we (sub)claim that a crossing (involving colours $x_{1},x_{2},x_{3}$ is legal if and only if $x_{1}x_{2}x_{3}=1$ in $Z_{3}$. Indeed, if the crossing is legal, either it is the trivial crossing in which case their product is clearly 1, or each $x_{i}$ is distinct, in which case $x_{1}x_{2}x_{3}=1aa^{2}=a^{3}=1$. Conversely, suppose $x_{1}x_{2}x_{3}=1$, and suppose $x_{1}=x_{2}$. It suffices to show that $x_{3}=x_{1}$. This follows by case checking: if $x_{1}=1$, then $1=x_{1}x_{2}x_{3}=x_{3}$; if $x_{1}=a$, then $1=a^{2}x_{3}$, implying that $x_{3}=a^{-2}=a$; and if $x_{1}=a^{2}$, then $1=a^{4}x_{3}=ax_{3}$, implying that $x_{3}=a^{-1}=a^{2}$. Thus, the subclaim is proven.

As a result, $xy^{-1}=(x_{1}y_{1}^{-1},x_{2}y_{2}^{-1},x_{3}y_{3}^{-1})$ satisfies $x_{1}y_{1}^{-1}x_{2}y_{2}^{-1}x_{3}y_{3}^{-1}=(x_{1}x_{2}x_{3})(y_{3}y_{2}y_{1})^{-1}=1$ since both $x,y\in {\mathcal {S}}$. This implies that $xy^{-1}\in {\mathcal {S}}$, and hence shows that ${\mathcal {S}}$ is a subgroup of $Z_{3}^{n}$ for n = the number of line segments in the diagram. By Lagrange's theorem, the number of legal 3-colorings (the order of ${\mathcal {S}}$) is a power of 3.

Using linear Algebra: Idea from class on Wednesday 23 May, 2018

Let D be a knot diagram for the knot K with n crossings. There are n arcs. Let a_1, a_1, \ldots, a_n \in {\mathbb Z}_3 represent the arcs. Now let a,b,c \in {\mathbb Z}_3. Define \wedge : {\mathbb Z}_3 \times {\mathbb Z}_3 \rightarrow {\mathbb Z}_3 by

a\wedge b = \left\{ \begin{array}{cc} a, & a = b\\ c, & a\not= b \end{array} \right., so that a\wedge b + a + b \equiv 0\mod 3.

Then, with the above definition, we get a linear equation a_{i_1} + a_{i_2} + a_{i_3} \equiv 0\mod 3 for each each of the n crossings, where i_1, i_2, i_3 \in \{1, 2, \ldots, n\}. Thus we get a system of n linear equation, from which we get a matrix M. The nullspace \mathrm{Null}(M) of M is the solution to this system of equation and this is exactly the set of all 3-colourings of D. This is a vector space of size \lambda(K) =|\mathrm{Null}(M)| = 3^{\dim(\mathrm{Null}(M))}