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(Note by [[User:Cameron.martin]]): |
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'''Claim:''' The number of legal 3-colorings of a knot diagram is always a power of 3. |
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'''Claim:''' The number of legal 3-colorings of a knot diagram is always a power of 3. |
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(Note by [[User:Leo algknt]]): |
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'''Using linear Algebra: Idea from class on Wednesday 23 May, 2018''' |
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'''Using linear Algebra: Idea from class on Wednesday 23 May, 2018''' |
(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