---
title: "The permutations package and representation theory"
author: "Robin K. S. Hankin"
date: "`r Sys.Date()`"
vignette: >
  %\VignetteIndexEntry{representation theory}
  %\VignetteEngine{knitr::rmarkdown}
  %\VignetteEncoding{UTF-8}
---


<p style="text-align: right;">
![](`r system.file("help/figures/permutations.png", package = "permutations")`){width=10%}
</p>



```{r set-options, echo = FALSE}
knitr::opts_chunk$set(collapse = TRUE, comment = "#>", dev = "png", fig.width = 7, fig.height = 3.5, message = FALSE, warning = FALSE)
options(width = 80, tibble.width = Inf)
```

# Introduction

We will consider $S_9$, the symmetric group on 9 elements, and
representations of its elements.  First we will load the package and
change the default print method:

```{r}
library("permutations")
options("print_word_as_cycle" = FALSE)
```


Now we will generate a couple of permutations, `a` and `b`:


```{r}
a <- as.word(char2cycle("(175296)(348)"))
b <- as.word(char2cycle("(27)(45)(89)"))
a
b
```

Now we will show a representation of $S_9$, using permutation matrices:

```{r}
M <- diag(9)
rownames(M) <- 1:9
colnames(M) <- 1:9
M
```

We will use the map $\phi\colon S_9\longrightarrow\operatorname{Aut}\left(\mathbb{R}^9\right)$ given by $\phi(a)=$ `M[a,]`.  Then


```{r}
M[a,]
```

permutes the rows of $M$ with permutation `a`.  Note how the row
names are permuted as well as the elements of $M$.  Verifying that
$\phi$ is indeed a homomorphism---that is, $\phi(\iota)=I_9$ and
$\phi(a)\phi(b)=\phi(a\circ b)$---is straightforward:

```{r}
all(M[as.word(id,9),] == diag(9))
all(M[a*b,] == M[a,] %*% M[b,])
```

In the second line above, note that the left hand side of the equality
is group composition, while the right hand side is matrix
multiplication.  We can further verify that
$\phi\left(a^{-1}\right)=\phi(a)^{-1}$:

```{r}
all(M[inverse(a),] == solve(M[a,]))
```

again with group inversion on the left and matrix inversion on the
right.

The map $\psi(a)=$ `M[,a]` is not a homomorphism:


```{r}
all(M[,a*b] == M[,a] %*% M[,b])
```

but we can "rescue" it by considering a group operation $\star$
defined by $a\star b=b\circ a$:

```{r}
all(M[,b*a] == M[,a] %*% M[,b])
```
