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% Following http://tex.stackexchange.com/a/847/22475:
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\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\Aut{{\operatorname{Aut}}}
\def\hol{{\operatorname{hol}}}
\def\tr{{\operatorname{tr}}}

\def\act{\!\sslash\!}
\def\qed{{\hfill\text{$\Box$}}}

\def\navigator{{Dror Bar-Natan: Talks: Louvain-1506:}}

\def\w#1{{\href{http://www.math.toronto.edu/drorbn/Talks/Louvain-1506/#1}{$\omega$/#1}}}
\def\webdef{{$\omega:=$\url{http:drorbn.net/Louvain-1506}}}
\def\webnote{{Handout, video, and links at \w{}}}

\def\Fourier{{\raisebox{2mm}{\parbox[t]{2.5in}{
{\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:

\vskip 1mm
$\bullet$ $\tilde{f}(0)=\int_Vf(v)dv$.

\vskip 1mm
$\bullet$ $\frac{\partial}{\partial\varphi_i}\tilde{f}\sim\widetilde{v^if}$.

\vskip 1mm
$\bullet$ $\widetilde{(e^{Q/2})}\sim e^{Q^{-1}/2}$, 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 the proof of the Fourier inversion formula!)
}}}}

\def\main{{\raisebox{9pt}{\parbox[t]{5.4in}{
{\red Gaussian Integration.} $(\lambda_{ij})$ is a symmetic 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+\frac{\epsilon}{6}\lambda_{ijk}x^ix^jx^k} 
  = \sum_{m\geq 0}\frac{\epsilon^m}{6^mm!}
    \int\limits_{\bbR^n}
    (\lambda_{ijk}x^ix^jx^k)^m e^{-\frac12\lambda_{ij}x^ix^j} \\
  = \left. \sum_{m\geq 0}\frac{C\epsilon^m}{6^mm!}
      (\lambda_{ijk}\partial^i\partial^j\partial^k)^m
      e^{\frac12\lambda^{\alpha\beta}t_\alpha t_\beta}
    \right|_{t_\alpha=0}
  = \sum_{\substack{m,l\geq 0 \\ 3m=2l }} \frac{C\epsilon^m}{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\epsilon^m}{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\epsilon^m}{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{\epsilon^{m(D)}\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
}}}}

\def\perturbeddets{{\raisebox{7pt}{\parbox[t]{2.5in}{
{\red Perturbing Determinants.} If $Q$ and $P$ are matrices and $Q$ is
invertible,
\begin{multline*}
  |Q|^{-1}|Q+\epsilon P| = |I+\epsilon Q^{-1}P| \\
  = \sum_{k\geq 0}\epsilon^k\tr\left(
      \bigwedge\nolimits^kQ^{-1}P
    \right) \\
  = \hspace{-6pt}\sum_{k\geq 0,\,\sigma\in S_k}\hspace{-6pt}
    \frac{\epsilon^k(-)^\sigma}{k!}
    \tr\left(
      \sigma(Q^{-1}P)^{\otimes k}
    \right) \\
  = \hspace{-6pt}\sum_{k\geq 0,\,\sigma\in S_k}\hspace{-6pt}
    \frac{(-\epsilon)^k(-)^{\#\text{cycles}}}{k!}
    \begin{array}{c}\input{MultiTrace.pstex_t}\end{array}
\end{multline*}
}}}}

\def\dets{{\raisebox{7pt}{\parbox[t]{3.9in}{
{\red Determinants.} Now suppose $Q$ and $P_i$ ($1\leq i\leq n$) are
$d\times d$ matrices and $Q$ is invertible. Then
\begin{multline*}
  |Q|^{-1}I_{\epsilon,\lambda_{ij},\lambda_{ijk},Q,P_i}
  = |Q|^{-1}\int\limits_{\bbR^n}
    e^{-\frac12\lambda_{ij}x^ix^j+\frac{\epsilon}{6}\lambda_{ijk}x^ix^jx^k}
    \det(Q+\epsilon x^iP_i) \\
  = \hspace{-6pt}\sum_{m,k\geq 0,\,\sigma\in S_k}\hspace{-6pt}
    \frac{C\epsilon^{m+k}(-)^\sigma}{6^mm!k!}
    \int\limits_{\bbR^n}
    (\lambda_{ijk}x^ix^jx^k)^m
    \tr\left(\sigma(x^iQ^{-1}P_i)^{\otimes k}\right)
    e^{-\frac12\lambda_{ij}x^ix^j} \\
  = \hspace{-6pt}
      \sum_{\substack{\text{fully marked} \\ \text{Feynman diagrams}}}
    \hspace{-6pt}
    \frac{C\epsilon^{m+k}(-)^\sigma}{6^mm!k!}
    \calE\left(
      \begin{array}{c}\input{DetDiagram1.pstex_t}\end{array}
    \right) \\
  = \hspace{-6pt}
      \sum_{\text{Feynman diagrams}}
    \hspace{-6pt}
    C\epsilon^{m+k}(-)^k(-)^l
    \calE\left(
      \begin{array}{c}\input{DetDiagram2.pstex_t}\end{array}
    \right),
\end{multline*}
where $l$ is the number of purple (``Fermion'') loops.
}}}}

\def\Berezin{{\raisebox{7pt}{\parbox[t]{4in}{
\parshape 6 0in 3.25in 0in 3.25in 0in 3.25in 0in 3.25in 0in 3.25in 0in 4in
{\red The Berezin Integral} ({\blue physics} / math language, formulas from
\href{http://en.wikipedia.org/wiki/Grassmann_integral}{\tt Wikipedia:Grassmann
integral}).
\newline{\blue The {\em Berezin Integral} is linear on functions of
anti-commuting variables, and satisfies $\int\theta d\theta=1$,
and $\int 1d\theta=0$, so that
$\int\frac{\partial f(\theta)}{\partial\theta}d\theta=0$.}

Let $V$ be a vector space, $\theta\in V$, $d\theta\in V^\ast$
s.t.{} $\langle d\theta,\theta\rangle=1$. Then $f\mapsto\int fd\theta$
is the interior multiplication map $\bigwedge\!V\to\bigwedge\!V$:
$\int fd\theta \coloneqq i_{d\theta}(f)$ $\left(=\frac{\partial
f}{\partial\theta}\right)$.

{\blue Multiple integration via ``Fubini'': $\int f_1(\theta_1)\cdots
f_n(\theta_n)d\theta_1\ldots d\theta_n \coloneqq \left(\int
f_1d\theta_1\right)\cdots\left(\int f_nd\theta_n\right)$.} $\int
fd\theta_1\ldots d\theta_n \coloneqq f \act i_{d\theta_1} \act\cdots\act
i_{d\theta_n}$.

{\blue Change of variables. If $\theta_i=\theta_i(\xi_j)$, both $\theta_i$
and $\xi_j$ are odd, and $J_{ij}\coloneqq\partial\theta_i/\partial\xi_j$,
then
\[ \int f(\theta_i)d\theta
  = \int f(\theta_i(\xi_j))\det(J_{ij})^{-1}d\xi.
\]
}

Given vector spaces $V_{\theta_i}$ and $W_{\xi_j}$,
$d\theta=\bigwedge\!d\theta_i\in\bigwedge^{\text{top}}(V^\ast)$, 
$d\xi=\bigwedge\!d\xi_i\in\bigwedge^{\text{top}}(W^\ast)$, and $T\colon
V\to\bigwedge^{\text{odd}}(W)$. Then $T$ induces a map
$T_\ast\colon\bigwedge\!V\to\bigwedge\!W$ and then
\[ \int fd\theta
  = \int(T_\ast f)
    \det\left(\frac{\partial(T\theta_i)}{\partial\xi_j}\right)^{-1}
    d\xi.
\]

{\blue Gaussian integration. For an even matrix $A$ and odd vectors
$\theta$, $\eta$,
\[
  \int e^{\theta^TA\eta}d\theta d\eta=\det(A),
  \quad
  \int e^{\theta^TA\eta+\theta^TJ+K^T\eta}d\theta d\eta
  = \det(A)e^{-K^TA^{-1}J}.
\]
}

}}}}

\def\ghosts{{\raisebox{7pt}{\parbox[t]{3.9in}{
\parshape 4 0in 3.4in 0in 3.4in 0in 3.4in 0in 3.9in
{\red Ghosts.} Or else, introduce ``ghosts'' $\bar{c}_a$ and $c^b$, write
\[ I_{\epsilon,\lambda_{ij},\lambda_{ijk},Q,P_i}
  = \hspace{-3pt}\int\limits_{\bbR^n}\hspace{-3pt}dx
    \hspace{-6pt}\int\limits_{\bar{\bbR}^d\times\bar{\bbR}^d}\hspace{-6pt}
    e^{
      -\frac12\lambda_{ij}x^ix^j+\frac{\epsilon}{6}\lambda_{ijk}x^ix^jx^k
      +\bar{c}_a(Q^a_b+\epsilon x^iP^a_{ib})c^b
    }
\]
and use ``ordinary'' perturbation theory.
}}}}

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