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 [math]\displaystyle{ \mathcal{S} }[/math] forms a subgroup of [math]\displaystyle{ Z_3^r }[/math], 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 [math]\displaystyle{ x = (x_1, x_2, ..., x_n) }[/math], where each [math]\displaystyle{ x_n }[/math] is an element of the cyclic group of order 3 [math]\displaystyle{ Z_3 = \lt a|a^3=1\gt }[/math] (each element representing a different colour). It is clear that [math]\displaystyle{ \mathcal{S} }[/math] is a subset of [math]\displaystyle{ Z_3^n }[/math]. To show it is a subgroup, we'll take [math]\displaystyle{ x = (x_1, x_2, ..., x_n), y = (y_1, y_2, ..., y_n) \in \mathcal{S} }[/math], and show that [math]\displaystyle{ xy^{-1} = (x_1y_1^{-1}, x_2y_2^{-1}, ..., x_ny_n^{-1}) \in \mathcal{S} }[/math]. 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 [math]\displaystyle{ x_1, x_2, x_3 }[/math] is legal if and only if [math]\displaystyle{ x_1x_2x_3 = 1 }[/math] in [math]\displaystyle{ Z_3 }[/math]. Indeed, if the crossing is legal, either it is the trivial crossing in which case their product is clearly 1, or each [math]\displaystyle{ x_i }[/math] is distinct, in which case [math]\displaystyle{ x_1x_2x_3 = 1aa^2 = a^3 = 1 }[/math]. Conversely, suppose [math]\displaystyle{ x_1x_2x_3 = 1 }[/math], and suppose [math]\displaystyle{ x_1 = x_2 }[/math]. It suffices to show that [math]\displaystyle{ x_3 = x_1 }[/math]. This follows by case checking: if [math]\displaystyle{ x_1 = 1 }[/math], then [math]\displaystyle{ 1 = x_1x_2x_3 = x_3 }[/math]; if [math]\displaystyle{ x_1 = a }[/math], then [math]\displaystyle{ 1=a^2x_3 }[/math], implying that [math]\displaystyle{ x_3 = a^{-2} = a }[/math]; and if [math]\displaystyle{ x_1 = a^2 }[/math], then [math]\displaystyle{ 1 = a^4x_3 = ax_3 }[/math], implying that [math]\displaystyle{ x_3 = a^{-1} = a^2 }[/math]. Thus, the subclaim is proven.

As a result, [math]\displaystyle{ xy^{-1} = (x_1y_1^{-1}, x_2y_2^{-1}, x_3y_3^{-1}) }[/math] satisfies [math]\displaystyle{ x_1y_1^{-1}x_2y_2^{-1}x_3y_3^{-1} = (x_1x_2x_3)(y_3y_2y_1)^{-1} = 1 }[/math] since both [math]\displaystyle{ x, y \in \mathcal{S} }[/math]. This implies that [math]\displaystyle{ xy^{-1} \in \mathcal{S} }[/math], and hence shows that [math]\displaystyle{ \mathcal{S} }[/math] is a subgroup of [math]\displaystyle{ Z_3^n }[/math] for n = the number of line segments in the diagram. By Lagrange's theorem, the number of legal 3-colorings (the order of [math]\displaystyle{ \mathcal{S} }[/math]) 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))}