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 forms a subgroup of , 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 , where each is an element of the cyclic group of order 3 (each element representing a different colour). It is clear that is a subset of . To show it is a subgroup, we'll take , and show that . 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 is legal if and only if in . Indeed, if the crossing is legal, either it is the trivial crossing in which case their product is clearly 1, or each is distinct, in which case . Conversely, suppose , and suppose . It suffices to show that . This follows by case checking: if , then ; if , then , implying that ; and if , then , implying that . Thus, the subclaim is proven.


As a result, satisfies since both . This implies that , and hence shows that is a subgroup of for n = the number of line segments in the diagram. By Lagrange's theorem, the number of legal 3-colorings (the order of ) is a power of 3.


(Note by User:Leo algknt):

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

Let be a knot diagram for the knot with crossings. There are arcs. Let represent the arcs. Now let . Define by


, so that .

Then, with the above definition, we get a linear equation for each each of the crossings, where . Thus we get a system of linear equation, from which we get a matrix . The nullspace of is the solution to this system of equation and this is exactly the set of all 3-colourings of . This is a vector space of size