\documentclass[10pt,notitlepage]{article}
\usepackage{amsmath,graphicx,amssymb,color,datetime,stmaryrd}
\usepackage[setpagesize=false]{hyperref}
\usepackage[all]{xy}
\usepackage{txfonts}	% for the likes of \coloneqq.

\paperwidth 8.5in
\paperheight 11in
\textwidth 8.5in
\textheight 11in
\oddsidemargin -1in
\evensidemargin \oddsidemargin
\topmargin -1in
\headheight 0in
\headsep 0in
\footskip 0in
\parindent 0in
\setlength{\topsep}{0pt}
\pagestyle{empty}

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

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

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

\def\qed{{\hfill\text{$\Box$}}}

\def\navigator{{Dror Bar-Natan: Classes: 1314: AKT-14:}}
\def\webdef{{\url{http://drorbn.net/index.php?title=AKT-14}}}

\def\wish{{\raisebox{-3mm}{\parbox[t]{5in}{
We wish to understand
$\displaystyle \int\limits_{A\in\Omega^1(\bbR^3,\frakg)}\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].
$
}}}}

\def\main{{\raisebox{0mm}{\parbox[t]{5.4in}{
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
\begin{multline*}
  \int\limits_{\bbR^n}
    e^{-\frac12\lambda_{ij}x^ix^j+\frac16\lambda_{ijk}x^ix^jx^k} 
  = \int\limits_{\bbR^n}
    e^{\frac16\lambda_{ijk}x^ix^jx^k} 
    e^{-\frac12\lambda_{ij}x^ix^j} \\
  = \left. C
      e^{\frac16\lambda_{ijk}\partial^i\partial^j\partial^k}
      e^{\frac12\lambda^{\alpha\beta}t_\alpha t_\beta}
    \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 \\
  = \sum_{\substack{m,l\geq 0 \\ 3m=2l }} \frac{C}{6^mm!2^ll!}
    \left[\begin{array}{c}\input{Pitchforks.pstex_t}\end{array}\right] \\
  = \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)
      \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{$\lambda_{i_1j_1k_1}\lambda_{i_2j_2k_2}
        \lambda^{i_1j_1}\lambda^{k_1k_2}\lambda^{i_2j_2}$}
      \qquad\begin{array}{c}\input{MarkedDiagrams.pstex_t}\end{array} \\
  = C\sum_{\substack{\text{unmarked Feynman} \\ \text{diagrams }D}}
      \frac{\calE(D)}{|\Aut(D)|}.
    \qquad\parbox[t]{3in}{
      {\bf Claim.} The number of pairings that produce a given unmarked
      Feynman diagram $D$ is $\frac{6^mm!2^ll!}{|\Aut(D)|}$.
    }
\end{multline*}
{\bf Proof of the Claim.} The group $G_{m,l}\coloneqq[(S_3)^m\rtimes
S_m]\times[(S_2)^l\rtimes S_l]$ acts on the set of pairings, the action
is transitive on the set of pairings $P$ that produce a given $D$,
and the stabilizer of any given $P$ is $\Aut(D)$. \qed
}}}}

\begin{document}
\setlength{\jot}{3pt}
\setlength{\abovedisplayskip}{0ex}
\setlength{\belowdisplayskip}{0ex}
\setlength{\abovedisplayshortskip}{0ex}
\setlength{\belowdisplayshortskip}{0ex}

\begin{center}
\null\vfill
\input{FD.pstex_t}
\vfill\null
\end{center}

\end{document}

\endinput

