\documentclass[11pt,notitlepage]{article}
\usepackage{amsmath,graphicx,amssymb,xcolor,stmaryrd,datetime,amscd,txfonts,picins}
% Following http://tex.stackexchange.com/a/847/22475:
\usepackage[setpagesize=false]{hyperref}\hypersetup{colorlinks,
  linkcolor={green!50!black},
  citecolor={green!50!black},
  urlcolor=blue
}

\def\myurl{http://www.math.toronto.edu/~drorbn/}

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

\def\act{{\hspace{-1pt}\sslash\hspace{-0.75pt}}}
\def\diag{{\operatorname{diag}}}
\def\remove{\!\setminus\!}
\def\tr{\operatorname{tr}}
\def\yellow#1{{\setlength{\fboxsep}{0pt}\colorbox{yellow}{#1}}}

\paperwidth 8in
\paperheight 10.5in
\textwidth 8in
\textheight 10.5in
\oddsidemargin -0.75in
\evensidemargin \oddsidemargin
\topmargin -0.75in
\headheight 0in
\headsep 0in
\footskip 0in
\parindent 0in
\setlength{\topsep}{0pt}
\pagestyle{empty}
\dmyydate

% Following http://tex.stackexchange.com/questions/23521/tabular-vertical-alignment-to-top:
\def\imagetop#1{\vtop{\null\hbox{#1}}}

\begin{document}
\setlength{\jot}{0ex}
\setlength{\abovedisplayskip}{0.5ex}
\setlength{\belowdisplayskip}{0.5ex}
\setlength{\abovedisplayshortskip}{0ex}
\setlength{\belowdisplayshortskip}{0ex}

\noindent\parbox[b]{4.6in}{
  %\url{http://drorbn.net/AKT-17}\newline
  \bf\href{\myurl}{Dror Bar-Natan}:
  \href{\myurl/classes/}{Classes}:
  \href{\myurl/classes/\#1617}{2017}:
  \href{http://drorbn.net/?title=AKT-17}{MAT 1350 AKT}:
  \newline Handout for January 31, 2017:
}
\hfill\text{\huge\bf $\Gamma$-Calculus}

\vskip 3mm

{\footnotesize
  Derived from Cheat Sheet Meta-Calculi, in \url{http://drorbn.net/AcademicPensieve/Projects/MetaCalculi/}.
}

%\vskip -5pt
\rule{\textwidth}{1pt}
\textbf{$\sigma$-calculus.}
\hfill $\sigma_1\ast\sigma_2=\sigma_1\cup\sigma_2$,
\hfill $m^{ab}_c(\sigma)=(\sigma\remove\{a,b\})\cup(c\to\sigma_a\sigma_b)\,/\,({T_a,T_b\to T_c})$,
\hfill $\tr_c(\sigma)=\sigma\remove c$,
\hfill $R^\pm_{ab}\mapsto (a\to 1,\,b\to T_a^{\pm 1})$

\rule{\textwidth}{1pt}

\textbf{Gassner calculus / $\Gamma$-calculus.}
\hfill Preserves $C_1\coloneqq[\text{col sum}=1]$ ($\Leftrightarrow$OC) and $\green\checkmark$ $C_2\coloneqq[\forall a,b,\, (T_a-1)\mid(A_{ab}-\delta_{ab}\sigma_b)]$
\newline\null\hfill$\bullet$ Except under $\tr_c$, at $T_\ast=1$, $\omega=1$ and $A=I$.

\vskip -3mm
$\displaystyle
  \begin{CD}
    \begin{array}{c|ccc}
      \omega & a & b & S \\
      \hline
      a & \alpha & \beta & \theta \\
      b & \gamma & \delta & \epsilon \\
      S & \phi & \psi & \Xi
    \end{array}
    @>{m^{ab}_{c}}>{\mu\coloneqq 1-\beta \atop T_a,T_b\to T_c}>
    \begin{array}{c|cc}
      \mu\omega & c & S \\
      \hline
      c & \gamma+\alpha\delta/\mu & \epsilon+\delta\theta/\mu \\
      S & \phi+\alpha\psi/\mu & \Xi+\psi\theta/\mu
    \end{array}
  \end{CD}
$
\hfill$\displaystyle
\begin{CD}
  \begin{array}{c|ccc}
    \omega & c & S \\
    \hline
    c & \alpha & \theta \\
    S & \psi & \Xi
  \end{array}
  @>\tr_c>\mu\coloneqq 1-\alpha>
  \begin{array}{c|cc}
    \mu\omega & S \\
    \hline
    S & \Xi+\psi\theta/\mu
  \end{array}
\end{CD}$
\hfill$R^{\pm}_{ab} \underset{\Gamma}{=} \begin{array}{c|cc}
  1 & a & b \\
  \hline
  a & 1 & 1-T_a^{\pm 1} \\
  b & 0 & T_a^{\pm 1}
\end{array}$

$\displaystyle
  \begin{CD}
    \begin{array}{c|cc}
      \omega & a & S \\
      \hline
      a & \alpha & \theta \\
      S & \phi & \Xi
    \end{array}
    @>\Delta^a_{bc}>{\mu\coloneqq T_a-1\atop\nu\coloneqq \alpha-\sigma_a}>
    \left(\begin{array}{c|ccc}
      \omega & b & c & S \\
      \hline
      b &
        (\sigma_a-\alpha T_a-\nu T_c)/\mu &
        (T_b-1)T_c\nu/\mu &
        (T_b-1)T_c\theta/\mu \\
      c &
        (T_c-1)\nu/\mu &
        (\alpha-\sigma_a T_a-\nu T_c)/\mu &
        (T_c-1)\theta/\mu \\
      S & \phi & \phi & \Xi
    \end{array}\right)_{T_a\mapsto T_bT_c}
  \end{CD}
$
\hfill\parbox{2.1in}{Satisfies:
$\green\checkmark$ $R^+_{13}\act \Delta^1_{12}=R^+_{23}\#R^+_{13}$.
\newline$\green\checkmark$ $R^-_{13}\act\Delta^1_{12}=R^-_{13}\#R^-_{23}$.
\newline$\green\checkmark$ $\Delta^a_{a_1a_2}\act\Delta^b_{b_1b_2}\act m^{a_1b_1}_{c_1}\act m^{a_2b_2}_{c_2}$ \newline\null\quad$=m^{ab}_c\act\Delta^c_{c_1c_2}$.
}

$\displaystyle
  \begin{CD}
    \begin{array}{c|cc}
      \omega & a & S \\
      \hline
      a & \alpha & \theta \\
      S & \phi & \Xi
    \end{array}
    @>S^a>>
    \left(\begin{array}{c|cc}
      \alpha\omega/\sigma_a & a & S \\
      \hline
      a & 1/\alpha & \theta/\alpha \\
      S & -\phi/\alpha & (\alpha\Xi-\phi\theta)/\alpha
    \end{array}\right)_{T_a\to T_a^{-1}}
  \end{CD}
$\hfill\parbox{4in}{Satisfies:
$\green\checkmark$ $R^\pm_{12}\act S^{1\text{ or }2}=R^\mp_{12}$.
\hfill$\green\checkmark$ $m^{ab}_c\act S^c=S^a\act S^b\act m^{ba}_c$.
\newline$\green\checkmark$ $S^a\act S^a=I$.
\hfill$\green\checkmark$ $\Delta^a_{bc}\act S^b\act S^c=S^a\act\Delta^a_{cb}$.
\newline$\green\checkmark$ Assuming $C_2$, $\eta^a\act \epsilon_a=\Delta^a_{bc}\act S^c\act m^{bc}_a$ (also 3 variants).
}

The map (tangle $T$ $\mapsto$ matrix $A$) is anti-multiplicative.
\hfill The MVA mod units:
$L\mapsto(\omega,A) \mapsto \omega\det'(A-I)/(1-T')$\ ${\green\checkmark}$

\vskip -2mm
\rule{\textwidth}{0.5pt}
\vskip -1mm

\end{document}

\endinput