Binomial tree using matrices?

Question

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. — , Oct 15 '19 at 11:40
for example the permutation (12) has a representation as [010: 100: 001] — Math, Oct 15 '19 at 11:42
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 '19 at 11:50

score 3 · Accepted Answer · 2019-10-15T15:51:24.170

3

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}

edited Oct 15 '19 at 15:51

answered Oct 15 '19 at 12:00

well they are matrix representations of group elements :) – Math Oct 15 '19 at 12:04
But thank you for your solution, I was being rather silly for not loading in the package aha! – Math Oct 15 '19 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? – Oct 15 '19 at 12:07
well I used the wrong terminology aha, they are just descriptions of the elements in matrices. https://groupprops.subwiki.org/wiki/Element_structure_of_symmetric_group:S3 – Math Oct 15 '19 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 '19 at 12:14
I think I've got it, I'll attach it in the question – Math Oct 15 '19 at 12:19
never mind, that was disastrous aha – Math Oct 15 '19 at 12:22
@Math Sorry, what is a "matrix description"? – Oct 15 '19 at 12:35
what I defined [010:100:001], this is a matrix description of the permutation (12) – Math Oct 15 '19 at 13:07
@Math For that you need double braces, e.g. ,label=below:{{$(123)=[010:100:001]$}}. Why? Because you need to hide the [ and ] from the forest parser, they have special meanings there. – Oct 15 '19 at 13:11
@Math or at least braces braces, e.g. ,label=below:{$(123)=[010:100:001]$}. – Oct 15 '19 at 13:17
sorry i was in a meeting, i did something but its a mess, have a look at the edited question – Math Oct 15 '19 at 14:41
@Math I added something but recommend also asking a follow-up question. – Oct 15 '19 at 15:51
Thank you for the help :) – Math Oct 16 '19 at 11:33
is this page landscape?? I need it to be vertical as I'll need to explain what's going here – Math Oct 16 '19 at 11:35
i'll ask a separate question as advised :) – Math Oct 16 '19 at 11:42

Binomial tree using matrices?

1 Answers1