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 $\lambda (K)=|\mathrm {Null} (M)|=3^{\dim(\mathrm {Null} (M))}$