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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{{
  \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/\#2122}{2021-22}:
  \href{http://drorbn.net/21-1350}{MAT 1350 Knot Theory}:
}}
\def\webdef{{\url{http://drorbn.net/21-1350}}}

\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 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+\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{figs/Pitchforks.pdf_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{figs/MarkedDiagrams.pdf_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
}}}}

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