2

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}
\]
5
  • 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.
    – user194703
    Oct 15, 2019 at 11:40
  • for example the permutation (12) has a representation as [010: 100: 001]
    – Math
    Oct 15, 2019 at 11:42
  • What prevents you from adding that matrix to the tree?
    – user194703
    Oct 15, 2019 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, 2019 at 11:50
  • but it gave errors
    – Math
    Oct 15, 2019 at 11:50

1 Answer 1

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}

enter image description here

(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}

enter image description here

16
  • well they are matrix representations of group elements :)
    – Math
    Oct 15, 2019 at 12:04
  • But thank you for your solution, I was being rather silly for not loading in the package aha!
    – Math
    Oct 15, 2019 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?
    – user194703
    Oct 15, 2019 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, 2019 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, 2019 at 12:14

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