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AKT-140106 Video

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Videography by Iva Halacheva troubleshooting

Notes on AKT-140106: [edit, refresh]

Course introduction, knots and Reidemeister moves, knot colourings.

#	Week of...	Notes and Links
1	Jan 6	About This Class (PDF). Monday: Course introduction, knots and Reidemeister moves, knot colourings. Tricolourability without Diagrams Wednesday: The Gauss linking number combinatorially and as an integral. Friday: The Schroedinger equation and path integrals. Friday Introduction (the quantum pendulum)
2	Jan 13	Homework Assignment 1. Monday: The Kauffman bracket and the Jones polynomial. Wednesday: Self-linking using swaddling. Friday: Euler-Lagrange problems, Gaussian integration, volumes of spheres.
3	Jan 20	Homework Assignment 2. Monday: The definition of finite-type and some examples. Wednesday: The self-linking number and framings. Friday: Integrating a polynomial times a Gaussian. Class Photo.
4	Jan 27	Homework Assignment 3. Monday: Chord diagrams and weight systems. Wednesday: Swaddling maps and framings, general configuration space integrals. Friday: Some analysis of $d^{-1}$ .
5	Feb 3	Homework Assignment 4. Monday: 4T, the Fundamental Theorem and universal finite type invariants. The Fulton-MacPherson Compactification (PDF). Wednesday: The Fulton-MacPherson Compactification, Part I. Friday: More on pushforwards, $d^{-1}$ , and $d^\ast$ .
6	Feb 10	Homework Assignment 5. Monday: The bracket-rise theorem and the invariance principle. Wednesday: The Fulton-MacPherson Compactification, Part II. Friday: Gauge fixing, the beginning of Feynman diagrams.
R	Feb 17	Reading Week.
7	Feb 24	Monday: A review of Lie algebras. Wednesday: Graph cohomology and $\Omega_{dR}^\ast(\Gamma)$ . Friday: More on Feynman diagrams, beginning of gauge theory. From Gaussian Integration to Feynman Diagrams (PDF).
8	Mar 3	Homework Assignment 6 (PDF) Monday: Lie algebraic weight systems. Wednesday: Graph cohomology and the construction of $Z_0$ . Graph Cohomology and Configuration Space Integrals (PDF) Friday: Gauge invariance, Chern-Simons, holonomies. Mar 9 is the last day to drop this class.
9	Mar 10	Homework Assignment 7 (PDF) Monday: The $gl(N)$ weight system. Wednesday: The universal property, hidden faces. Friday: Insolubility of the quintic, naive expectations for CS perturbation theory.
10	Mar 17	Homework Assignment 8 (PDF) Monday: $W_{\mathfrak g}\colon{\mathcal A}(\uparrow)\to{\mathcal U}({\mathfrak g})$ and PBW. Wednesday: The anomaly. Friday: Faddeev-Popov, part I. Gaussian Integration, Determinants, Feynman Diagrams (PDF).
11	Mar 24	Homework Assignment 9 (PDF) Monday: ${\mathcal A}$ is a bi-algebra. Wednesday: Understanding and fixing the anomaly. Friday: class cancelled.
12	Mar 31	Monday, Wednesday: class cancelled. Friday: A Monday class: back to expansions.
E	Apr 7	Monday: A Friday class on what we mostly didn't have time to do.
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Dror's Notebook

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0:27:16 [edit] A green knot.

0:43:23 [edit] (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 $\lambda(K) =|\mathrm{Null}(M)| = 3^{\dim(\mathrm{Null}(M))}$

0:46:29 [edit] (Note by User:Cameron.martin). The unknot $0_1$ and the figure-eight knot $4_1$ both have 3 legal 3-colorings, i.e. $\lambda(0_1) = \lambda(4_1) = 3$. 3-coloring fails to distinguish the unknot from the figure-eight. See http://katlas.math.toronto.edu/wiki/The_Rolfsen_Knot_Table for more information on specific knots.

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Videography by Iva Halacheva	troubleshooting