Notes for AKT-140106/0:43:23 - Revision history

Drorbn at 18:36, 25 August 2018

2018-08-25T18:36:27Z

← Older revision		Revision as of 14:36, 25 August 2018
Line 1:		Line 1:
			(Note by [[User:Cameron.martin]]):

	'''Claim:''' The number of legal 3-colorings of a knot diagram is always a power of 3.		'''Claim:''' The number of legal 3-colorings of a knot diagram is always a power of 3.

Line 14:		Line 16:


			(Note by [[User:Leo algknt]]):

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

Leo algknt at 06:06, 24 May 2018

2018-05-24T06:06:33Z

← Older revision		Revision as of 02:06, 24 May 2018
Line 17:		Line 17:
	'''Using linear Algebra: Idea from class on Wednesday 23 May, 2018'''		'''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		Let <math>D</math> be a knot diagram for the knot <math>K</math> with <math>n</math> crossings. There are <math>n</math> arcs. Let <math>a_1, a_1, \ldots, a_n \in {\mathbb Z}_3</math> represent the arcs. Now let <math>a,b,c \in {\mathbb Z}_3</math>. Define <math>\wedge : {\mathbb Z}_3 \times {\mathbb Z}_3 \rightarrow {\mathbb Z}_3</math> by


			<math>a\wedge b =
	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.

			\left\{
⚫	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))}
			\begin{array}{cc}
			a, & a = b\\
			c, & a\not= b
			\end{array}
			\right.</math>,
			so that <math>a\wedge b + a + b \equiv 0\mod 3</math>.

		⚫	Then, with the above definition, we get a linear equation <math>a_{i_1} + a_{i_2} + a_{i_3} \equiv 0\mod 3</math> for each each of the <math>n</math> crossings, where <math>i_1, i_2, i_3 \in \{1, 2, \ldots, n\}</math>. Thus we get a system of <math>n</math> linear equation, from which we get a matrix <math>M</math>. The nullspace <math>\mathrm{Null}(M)</math> of <math>M</math> is the solution to this system of equation and this is exactly the set of all 3-colourings of <math>D</math>. This is a vector space of size <math>\lambda(K) =\|\mathrm{Null}(M)\| = 3^{\dim(\mathrm{Null}(M))}</math>

Leo algknt at 06:05, 24 May 2018

2018-05-24T06:05:20Z

← Older revision		Revision as of 02:05, 24 May 2018
Line 12:		Line 12:

	As a result, <math>xy^{-1} = (x_1y_1^{-1}, x_2y_2^{-1}, x_3y_3^{-1})</math> satisfies <math>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>x, y \in \mathcal{S}</math>. This implies that <math>xy^{-1} \in \mathcal{S}</math>, and hence shows that <math>\mathcal{S}</math> is a subgroup of <math>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>\mathcal{S}</math>) is a power of 3.		As a result, <math>xy^{-1} = (x_1y_1^{-1}, x_2y_2^{-1}, x_3y_3^{-1})</math> satisfies <math>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>x, y \in \mathcal{S}</math>. This implies that <math>xy^{-1} \in \mathcal{S}</math>, and hence shows that <math>\mathcal{S}</math> is a subgroup of <math>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>\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))}

Cameron.martin: Created page with "'''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/06081..."

2018-05-15T15:53:53Z

Created page with "'''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/06081..."

New page

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

As a result, <math>xy^{-1} = (x_1y_1^{-1}, x_2y_2^{-1}, x_3y_3^{-1})</math> satisfies <math>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>x, y \in \mathcal{S}</math>. This implies that <math>xy^{-1} \in \mathcal{S}</math>, and hence shows that <math>\mathcal{S}</math> is a subgroup of <math>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>\mathcal{S}</math>) is a power of 3.