Notes for AKT-140106/0:43:23

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(Note by User:Cameron.martin):

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_1y_1^{-1}, x_2y_2^{-1}, ..., x_ny_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_1x_2x_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_1x_2x_3 = 1aa^2 = a^3 = 1. Conversely, suppose x_1x_2x_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_1x_2x_3 = x_3; if x_1 = a, then 1=a^2x_3, implying that x_3 = a^{-2} = a; and if x_1 = a^2, then 1 = a^4x_3 = ax_3, implying that x_3 = a^{-1} = a^2. Thus, the subclaim is proven.


As a result, xy^{-1} = (x_1y_1^{-1}, x_2y_2^{-1}, x_3y_3^{-1}) satisfies x_1y_1^{-1}x_2y_2^{-1}x_3y_3^{-1} = (x_1x_2x_3)(y_3y_2y_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.


(Note by User:Leo algknt):

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))}