\documentclass[12pt]{article}
\usepackage{amsmath,amssymb,graphicx,txfonts,import}
\usepackage[a5paper,margin=3mm]{geometry}
\usepackage[setpagesize=false]{hyperref}
\usepackage[usenames,dvipsnames]{xcolor}
% Following http://tex.stackexchange.com/a/847/22475:
\usepackage[setpagesize=false]{hyperref}\hypersetup{colorlinks,
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}

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\def\chunk#1#2{{\begin{minipage}{#1\textwidth}#2\end{minipage}}}

\def\blue{\color{blue}}
\def\green{\color{green}}
\def\red{\color{red}}

\def\bbC{{\mathbb C}}
\def\bbR{{\mathbb R}}
\def\calD{{\mathcal D}}
\def\calE{{\mathcal E}}
\def\frakg{{\mathfrak g}}

\def\hol{{\operatorname{hol}}}
\def\tr{{\operatorname{tr}}}

\begin{document}
\href{http://www.math.toronto.edu/~drorbn/}{Dror Bar-Natan}:
\href{http://www.math.toronto.edu/~drorbn/classes/}{Classes}:
\href{http://www.math.toronto.edu/~drorbn/classes/20-1350-KnotTheory/}{20-1350}:
\hfill\href{http://drorbn.net/20-1350}{drorbn.net/20-1350}

\vskip 3mm
{\Large Class of November 2: A Quick Introduction to Feynman Diagrams}

\vskip 1mm \rule{\textwidth}{0.5mm} \vskip 1mm

We wish to understand

\vskip 2mm
$\displaystyle \int\limits_{A\in\Omega^1(\bbR^3,\frakg)}\hspace{-3mm}\calD A\,\hol_\gamma(A)
  \exp\left[
    \frac{ik}{4\pi}\int_{\bbR^3}\tr\left(
      A\wedge dA+\frac23A\wedge A\wedge A
    \right)
  \right]
$

\newpage

\chunk{0.6}{As a warm up, suppose $(\lambda_{ij})$ is a symmetric positive definite
matrix and $(\lambda^{ij})$ is its inverse, and $(\lambda_{ijk})$ are the
coefficients of some cubic form. Denote by $(x^i)_{i=1}^n$ the
coordinates of $\bbR^n$, let $(t_i)_{i=1}^n$ be a set of ``dual''
variables, and let $\partial^i$ denote $\frac{\partial}{\partial t_i}$.
Also let $C\coloneqq\frac{(2\pi)^{n/2}}{\det(\lambda_{ij})}$. Then
}

\chunk{0.5}{
\[
  \int\limits_{\bbR^n}
    \exp\left({-\frac12\lambda_{ij}x^ix^j+\frac16\lambda_{ijk}x^ix^jx^k}\right)
\]
\[
  = \int\limits_{\bbR^n}
    \exp\left(\frac16\lambda_{ijk}x^ix^jx^k\right)
    \exp\left(-\frac12\lambda_{ij}x^ix^j\right)
\]
}

\vskip 1mm \rule{\textwidth}{0.5mm} \vskip 1mm

\chunk{0.5}{
{\red The Fourier Transform.}
\[ (f\colon V\to\bbC)\Rightarrow(\tilde{f}\colon V^\ast\to\bbC) \]
via $\tilde{f}(\varphi)\coloneqq\int_Vf(v)e^{-i\langle\varphi,v\rangle}dv$.
Some facts:
\begin{itemize}
\item $\tilde{f}(0)=\int_Vf(v)dv$.
\item $\frac{\partial}{\partial\varphi_i}\tilde{f}\sim\widetilde{v^if}$.
\item $\widetilde{(e^{Q/2})}\sim e^{Q^{-1}/2}$, where $Q$ is quadratic,
$Q(v)=\langle Lv,v\rangle$ for $L\colon V\to V^\ast$, and
$Q^{-1}(\varphi)\coloneqq\langle\varphi,L^{-1}\varphi\rangle$. (This is the
key point in one of the proofs of the Fourier inversion formula!)
\end{itemize}
}

\vskip 1mm \rule{\textwidth}{0.5mm} \vskip 1mm

\chunk{0.5}{
\[
  = \left. C
      \exp\left(\frac16\lambda_{ijk}\partial^i\partial^j\partial^k\right)
      \exp\left(\frac12\lambda^{\alpha\beta}t_\alpha t_\beta\right)
    \right|_{t_\alpha=0}
\]
\[
  = \sum_{\substack{m,l\geq 0 \\ 3m=2l }} \frac{C}{6^mm!2^ll!}
      \left(\lambda_{ijk}\partial^i\partial^j\partial^k\right)^m
      \left(\lambda^{\alpha\beta}t_\alpha t_\beta\right)^l
\]
}

$\displaystyle
  = \sum_{\substack{m,l\geq 0 \\ 3m=2l }} \frac{C}{6^mm!2^ll!}
    \left[\begin{array}{c}\import{./}{Pitchforks.pdf_t}\end{array}\right]
$

\vskip 1mm \rule{\textwidth}{0.5mm} \vskip 1mm

Examples.

$
\def\Da{$\lambda_{i_1j_1k_1}\lambda_{i_2j_2k_2}
  \lambda^{i_1i_2}\lambda^{j_1j_2}\lambda^{k_1k_2}$}
\def\Db{$\lambda_{i_1j_1k_1}\lambda_{i_2j_2k_2}
  \lambda^{i_1j_1}\lambda^{k_1k_2}\lambda^{i_2j_2}$}
\displaystyle\begin{array}{c}\import{./}{MarkedDiagrams.pdf_t}\end{array}
$

\vskip 1mm \rule{\textwidth}{0.5mm} \vskip 1mm

$\displaystyle
  = \sum_{\substack{m,l\geq 0 \\ 3m=2l }} \frac{C}{6^mm!2^ll!}
    \sum_{\substack{m\text{-vertex fully marked} \\ \text{Feynman diagrams }D}}
      \hspace{-18pt}\calE(D)
$

\end{document}