\documentclass[11pt,notitlepage]{article}
\usepackage{amsmath,graphicx,amssymb,color,stmaryrd,datetime,amscd}
\usepackage[setpagesize=false]{hyperref}

\def\sheeturl{{\url{http://drorbn.net/AcademicPensieve/2013-05/}}}

\def\tbd{\text{\color{red} ?}}

\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
\newcounter{linecounter}
\newcommand{\cheatline}{\vskip 1mm\refstepcounter{linecounter}\thelinecounter. }
% 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}

{\LARGE\bf Cheat Sheet $\beta$}\hfill
\parbox[b]{5.2in}{\small
  \null\hfill\sheeturl
  \newline\null\hfill initiated 24/3/13;
  continues \href{http://drorbn.net/AcademicPensieve/2013-04/}{2013-04};
  continued \href{http://drorbn.net/AcademicPensieve/2013-06/}{2013-06}; modified \today, \ampmtime
}

\rule{\textwidth}{1pt}

The original $\beta$-calculus: With $\epsilon:=1+\alpha$, $\langle\alpha\rangle:=\sum_v\alpha_v$,
and $\langle\gamma\rangle:=\sum_{v\neq u}\gamma_v$,

\vskip 1mm  $\displaystyle
  \begin{array}{c|c}\omega_1&H_1\\\hline T_1&A_1\end{array}
  \ast
  \begin{array}{c|c}\omega_2&H_2\\\hline T_2&A_2\end{array}
  \underset{\beta}{=}
  \begin{array}{c|cc}
    \omega_1\omega_2 & H_1 & H_2 \\
    \hline
    T_1 & A_1 & 0 \\
    T_2 & 0 & A_2
  \end{array}
$
\hfill $\displaystyle
  \begin{CD}
    \begin{array}{c|c}
      \omega & H \\
      \hline
      u & \alpha \\
      v & \beta \\
      T & \gamma
    \end{array}
    @>tm^{uv}_w>\beta>
    \begin{array}{c|c}
      \omega & H \\
      \hline
      w & (\alpha+\beta)\sslash({u,v\atop\to w}) \\
      T & \gamma
    \end{array}
  \end{CD}
$
\hfill $\displaystyle
  \rho^\pm_{ux} \underset{\beta}{=}
  \begin{array}{c|cc}
    1 & x \\
    \hline
    u & t_u^{\pm 1}-1
  \end{array}
$

\vskip 1mm $\displaystyle
  \begin{CD}
    \begin{array}{c|ccc}
      \omega & x & y & H \\
      \hline
      T & \alpha & \beta & \gamma
    \end{array}
    @>{hm^{xy}_z}>\beta>
    \begin{array}{c|cc}
      \omega & z & H \\
      \hline
      T & \alpha+\beta+\langle\alpha\rangle\beta & \gamma
    \end{array}
  \end{CD}
$
\hfill $\displaystyle
  \begin{CD}
    \begin{array}{c|cc}
      \omega & x & H \\
      \hline
      u & \alpha & \beta \\
      T & \gamma & \delta
    \end{array}
    @>{sw^{ux}_{th}}>\beta>
    \begin{array}{c|cc}
      \omega\epsilon & x & H \\
      \hline
      u & \alpha(1+\langle\gamma\rangle/\epsilon)
        & \beta(1+\langle\gamma\rangle/\epsilon) \\
      T & \gamma/\epsilon & \delta-\gamma\beta/\epsilon
    \end{array}
  \end{CD}
$

\vskip 1mm Constraints. $\bullet$ Column sums are monomials minus 1.

\rule{\textwidth}{1pt}

$\beta$-better calculus: \hfill
Constraints. $\bullet$ Sum of column $x$ is $(\sigma_x-1)w$.
\quad $\bullet$ $\omega^{k-1}\mid\Lambda^kA$.
\quad $\bullet$ At $t_\ast=1$, $\omega=1$ and $A=0$.

\vskip 1mm  $\displaystyle
  \begin{array}{c|c}\omega_1&H_1\\ - & \sigma_1\\\hline T_1&A_1\end{array}
  \ast
  \begin{array}{c|c}\omega_2&H_2\\ - & \sigma_2\\\hline T_2&A_2\end{array}
  \underset{\beta_b}{=}
  \begin{array}{c|cc}
    \omega_1\omega_2 & H_1 & H_2 \\
    - & \sigma_1 & \sigma_2 \\
    \hline
    T_1 & \omega_2A_1 & 0 \\
    T_2 & 0 & \omega_1A_2
  \end{array}
$
\hfill $\displaystyle
  \begin{CD}
    \begin{array}{c|c}
      \omega & H \\
      - & \sigma \\
      \hline
      u & \alpha \\
      v & \beta \\
      T & \gamma
    \end{array}
    @>tm^{uv}_w>\beta_b>
    \begin{array}{c|c}
      \omega & H \\
      - & \sigma \\
      \hline
      w & (\alpha+\beta)\sslash({u,v\atop\to w}) \\
      T & \gamma
    \end{array}
  \end{CD}
$
\hfill $\displaystyle \rho^\pm_{ux}
  \underset{\beta_b}{=}
  \begin{array}{c|cc}
    1 & x \\
    - & t_u^{\pm 1} \\
    \hline
    u & t_u^{\pm 1}-1
  \end{array}
$

\vskip 1mm $\displaystyle
  \begin{CD}
    \begin{array}{c|ccc}
      \omega & x & y & H \\
      - & \sigma_x & \sigma_y & \sigma \\
      \hline
      T & \alpha & \beta & \gamma
    \end{array}
    @>{hm^{xy}_z}>\beta_b>
    \begin{array}{c|cc}
      \omega & z & H \\
      - & \sigma_x\sigma_y & \sigma \\
      \hline
      T & \alpha+\sigma_x\beta & \gamma
    \end{array}
  \end{CD}
$
\hfill $\displaystyle
  \begin{CD}
    \begin{array}{c|cc}
      \omega & x & H \\
      - & \sigma_x & \sigma \\
      \hline
      u & \alpha & \beta \\
      T & \gamma & \delta
    \end{array}
    @>{sw^{ux}_{th}}>\beta_b>
    \begin{array}{c|cc}
      \omega+\alpha & x & H \\
      - & \sigma_x & \sigma \\
      \hline
      u & \sigma_x\alpha & \sigma_x\beta \\
      T & \gamma & \delta+\frac{\alpha\delta-\gamma\beta}{\omega}
    \end{array}
    =: \begin{array}{c|c}
      \cdot & - \\
      \cdot & - \\
      \hline
      \mid & \begin{pmatrix}\sigma_x\\1\end{pmatrix}\cdot A^{ux}
    \end{array}
  \end{CD}
$

\vskip 1mm
Note. $A^{ux}=\begin{pmatrix}
  \alpha & \beta \\
  \gamma & \delta+\frac{\alpha\delta-\gamma\beta}{\omega}
\end{pmatrix}
= \begin{pmatrix}
  \alpha & \beta \\
  \gamma & \frac{(\omega+\alpha)\delta-\gamma\beta}{\omega}
\end{pmatrix}
= \frac{1}{\omega}\left[
  (\omega+\alpha)\begin{pmatrix}
    \alpha & \beta \\ \gamma & \delta
  \end{pmatrix}
  - \begin{pmatrix}\alpha\\\gamma\end{pmatrix}
    \begin{pmatrix}\alpha&\beta\end{pmatrix}
  \right]
  =\frac{1}{\omega}\left[(\omega+a_{ux})A-a_{\ast x}a_{u\ast}\right].
$

\vskip 1mm
{\bf Claim.} $\omega^{k-1}\mid \Lambda^kA$ and $\omega^k\mid \Lambda^{k+1}A$ implies $(\omega+\alpha)^{k-1}\mid \Lambda^kA^{ux}$, with $\alpha=a_{ux}$.

{\bf Proof.} With $\bar{u}\in T^k$ and $\bar{x}\in H^k$, $\omega^k$ divides
$\left|\begin{array}{cc} \omega & 0 \\ 0 & a_{\bar{u}\bar{x}}\end{array}\right|$
and
$\left|\begin{array}{cc} a_{ux} & a_{u\bar{x}} \\ a_{\bar{u}x} & a_{\bar{u}\bar{x}} \end{array}\right|$
and hence their sum,
$\left|\begin{array}{cc} \omega+\alpha & a_{u\bar{x}} \\ a_{\bar{u}\bar{x}} & a_{\bar{u}\bar{x}} \end{array}\right|$
$= (\omega+\alpha)\left|\begin{array}{cc} 1 & 0 \\ 0 & a_{\bar{u}\bar{x}}-\frac{1}{\omega+\alpha}a_{\bar{u}x}a_{u\bar{x}} \end{array}\right|$
$=\frac{1}{(\omega+\alpha)^{k-1}}\left|(\omega+\alpha)a_{\bar{u}\bar{x}}-a_{\bar{u}x}a_{u\bar{x}}\right|$.
So
$\frac{1}{(\omega+\alpha)^{k-1}} \left|\frac{1}{\omega}\left[(\omega+\alpha)a_{\bar{u}\bar{x}}-a_{\bar{u}x}a_{u\bar{x}}\right]\right|$
is integral. \quad$\Box$

That is, with $A_{\bar{u};\bar{x}}$ denoting minors, if $\omega^{k-1}\mu_{\bar{u};\bar{x}}=A_{\bar{u};\bar{x}}$ and $\omega^k\mu_{u\bar{u};x\bar{x}}=A_{u\bar{u};x\bar{x}}$, then $(\omega+\alpha)^{k-1} (\mu_{\bar{u};\bar{x}}+\mu_{u\bar{u};x\bar{x}}) = A^{ux}_{\bar{u};\bar{x}}$.

\rule{\textwidth}{1pt}

{\bf Relations.} $\bullet$ $\rho^+_{ux}\rho^-_{vy}\sslash tm^{uv}_w\sslash
hm^{xy}_z=t\epsilon_w h\epsilon_z$.
\quad$\bullet$
$
  \rho^{s_1}_{ux}\rho^{s_2}_{vy}\rho^{s_2}_{wz}
    \sslash tm^{vw}_v\sslash hm^{xy}_x\sslash tha^{uz}
  = \rho^{s_2}_{vx}\rho^{s_2}_{wz}\rho^{s_1}_{uy}
    \sslash tm^{vw}_v\sslash hm^{xy}_x.
$

\rule{\textwidth}{1pt}

$\Lambda$-calculus: $\Lambda(T;H)=R(T)\otimes\left(\Lambda(T)\otimes\Lambda(H)\right)_= \times \Sigma(T)^H$, with $R(T)$ rational functions in $\{t_u\}_{u\in T}$ and $\Sigma(T)$ its units. Generic element is $L=(\lambda, (x\to\sigma_x))$.

$tm^{uv}_w:\ u,v\to w,\,t_u,t_v\to t_w$
\hfill$hm^{xy}_z:\ x\to z,\, y\to\sigma_xz,\, \sigma_z:=\sigma_x\sigma_y$
\hfill$tha^{ux}:L\mapsto(((1-i_u\otimes i_x)\lambda\sslash(u\to\sigma_xu),\sigma)$

\hfill$L_1\cdot L_1 = (\lambda_1(\wedge\otimes\wedge)\lambda_2, \sigma_1\cup\sigma_2)$
\hfill$\rho^\pm_{ux}=((t_u^{\pm 1}-1)ux, (x\to t_u^{\pm 1}))$
\hfill\null

\rule{\textwidth}{1pt}

\vfill
\rule{\textwidth}{1pt}

{\bf To do.}
$\bullet$ Consider a verification program.
\quad$\bullet$ Add $dm$ formulas.
\quad\colorbox{yellow}{$\bullet$ Add Burau calculus.}
\quad$\bullet$ Add the MVA formula

\end{document}

\endinput