# Binomial tree using matrices?

I have the following code for a particular group, now I wish to show the matrix representations for each permutation in the same manner as:

\documentclass[11pt]{article}
\usepackage{forest}
\begin{document}

\textit{Generator $(12)$}

$\begin{forest} for tree={grow'=east,l sep=8em,s sep=3em,circle,inner sep=2pt,fill} [,label=left:{(12)} [,label=above:{()},edge label={node[midway,sloped,above]{(12)}} [,label=right:(12),edge label={node[midway,sloped,above]{(12)}}] [,label=right:(13),edge label={node[midway,sloped,below]{(13)}}] ] [,label=below:{(123)},edge label={node[midway,sloped,below]{(13)}} [,label=right:(23),edge label={node[midway,sloped,above]{(12)}}] [,label=right:(12),edge label={node[midway,sloped,below]{(13)}}] ] ] \end{forest}$
\end{document}


further question, how do I align this whereby the starting points for both trees are aligned. and i wish to draw a right arrow centred between them.

\textit{Generator $(132)$}
$\begin{forest} for tree={grow'=east,l sep=5em,s sep=2em,circle,inner sep=2pt,fill} [,label=left:{(132)} [,label=above:{(12)},edge label={node[midway,sloped,above]{(23)}} [,label=right:(132),edge label={node[midway,sloped,above]{(23)}}] [,label=right:(23),edge label={node[midway,sloped,below]{(132)}}] ] [,label=below:{(123)},edge label={node[midway,sloped,below]{(132)}} [,label=right:(13),edge label={node[midway,sloped,above]{(23)}}] [,label=right:(),edge label={node[midway,sloped,below]{(132)}}] ] ] \end{forest} \rightarrow \begin{forest} for tree={grow'=east,l sep=7em,s sep=4em,circle,inner sep=2pt,fill} [,label=left:{\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}} [,label=above:{\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}},edge label={node[midway,sloped,above]{(23)}} [,label=right:\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix},edge label={node[midway,sloped,above]{(23)}}] [,label=right:\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix},edge label={node[midway,sloped,below]{(132)}}] ] [,label=below:{\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}},edge label={node[midway,sloped,below]{(132)}} [,label=right:\begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix},edge label={node[midway,sloped,above]{(23)}}] [,label=right:\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix},edge label={node[midway,sloped,below]{(132)}}] ] ] \end{forest}$

• The representation matrices of cycles depend, unsurprisingly, on the representation, I feel you should explain which matrices you want, and also what prevents you from adding them to the labels. – Schrödinger's cat Oct 15 at 11:40
• for example the permutation (12) has a representation as [010: 100: 001] – Math Oct 15 at 11:42
• What prevents you from adding that matrix to the tree? – Schrödinger's cat Oct 15 at 11:48
• I tried but I get errors, for example at the 'start' of the tree, I used \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{bmatrix} – Math Oct 15 at 11:50
• but it gave errors – Math Oct 15 at 11:50

I do not get errors (of course after loading amsmath for bmatrix).

\documentclass[11pt]{article}
\usepackage{forest}
\usepackage{amsmath}
\begin{document}

\textit{Generator $(12)$}

$\begin{forest} for tree={grow'=east,l sep=8em,s sep=3em,circle,inner sep=2pt,fill} [,label=left:{(12)=\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{bmatrix}} [,label=above:{()},edge label={node[midway,sloped,above]{(12)=\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{bmatrix}}} [,label=right:(12),edge label={node[midway,sloped,above]{(12)}}] [,label=right:(13),edge label={node[midway,sloped,below]{(13)}}] ] [,label=below:{(123)},edge label={node[midway,sloped,below]{(13)}} [,label=right:(23),edge label={node[midway,sloped,above]{(12)}}] [,label=right:(12),edge label={node[midway,sloped,below]{(13)}}] ] ] \end{forest}$
\end{document}


(I may see an off-topic error: these matrices are not any permutation matrices I know of, they are not group elements since not invertible, but it may just be that I do not know the notation.)

As for the additional request: use \vcenter{\hbox{...}}. (I think you could ask a follow-up question, this is done very quickly and I am sure one can avoid the \hspace* somehow but now I don't know how.)

\documentclass[11pt]{article}
\usepackage{amsmath}
\usepackage{forest}
\usepackage{pdflscape}
\begin{document}
\begin{landscape}
\textit{Generator $(12)$}
$\hspace*{-10em}\vcenter{\hbox{ \begin{forest} for tree={grow'=east,l sep=5em,s sep=2em,circle,inner sep=2pt,fill} [,label=left:{(132)} [,label=above:{(12)},edge label={node[midway,sloped,above]{(23)}} [,label=right:(132),edge label={node[midway,sloped,above]{(23)}}] [,label=right:(23),edge label={node[midway,sloped,below]{(132)}}] ] [,label=below:{(123)},edge label={node[midway,sloped,below]{(132)}} [,label=right:(13),edge label={node[midway,sloped,above]{(23)}}] [,label=right:(),edge label={node[midway,sloped,below]{(132)}}] ] ] \end{forest}}} \rightarrow\hspace*{-10em} \vcenter{\hbox{\begin{forest} for tree={grow'=east,l sep=7em,s sep=4em,circle,inner sep=2pt,fill} [,label=left:{\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}} [,label=above:{\begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}},edge label={node[midway,sloped,above]{(23)}} [,label=right:\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix},edge label={node[midway,sloped,above]{(23)}}] [,label=right:\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix},edge label={node[midway,sloped,below]{(132)}}] ] [,label=below:{\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}},edge label={node[midway,sloped,below]{(132)}} [,label=right:\begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix},edge label={node[midway,sloped,above]{(23)}}] [,label=right:\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix},edge label={node[midway,sloped,below]{(132)}}] ] ] \end{forest}}}$
\end{landscape}
\end{document}


• well they are matrix representations of group elements :) – Math Oct 15 at 12:04
• But thank you for your solution, I was being rather silly for not loading in the package aha! – Math Oct 15 at 12:05
• @Math You're welcome! Just out of curiosity: what kind of representations are these? Usually they should at least be invertible to form a group, shouldn't they? – Schrödinger's cat Oct 15 at 12:07
• well I used the wrong terminology aha, they are just descriptions of the elements in matrices. groupprops.subwiki.org/wiki/… – Math Oct 15 at 12:11
• a little side question, suppose I want the matrix description in the same equation, can I do so? I mean, I have the permutations from my code in the question on the left and the matrix description to lhe right – Math Oct 15 at 12:14