---
title: "The permutation group: active and passive permutations, and the order of operations"
author: "Robin K. S. Hankin"
date: "`r Sys.Date()`"
vignette: >
  %\VignetteIndexEntry{order of operations}
  %\VignetteEngine{knitr::rmarkdown}
  %\VignetteEncoding{UTF-8}
---

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


Order of operations can be confusing when considering permutation
groups.  Here I discuss active and passive transforms, order of
operations, prefix and postfix notation, and associativity from the
perspective of the `permutations` R package.

Consider the following:

```{r,label=setup,include=FALSE}
library("permutations")
library("magrittr")
```

```{r,label=definea}
a <- as.cycle("(145)(26)")
a
```

Thus we can see that $a$ has a three-cycle $(145)$ and a two-cycle
$(26)$.  We can express $a$ in word form (note that we have to stop
the print method from printing in cycle form by changing the default):


```{r}
options("print_word_as_cycle" = FALSE)
(a <- as.word(a))
```

showing that 3 is a fixed point (indicated with a dot).  In matrix
form we would have:

\[
\left(
\begin{array}{ccccccccc}
1&2&3&4&5&6\\
4&6&3&5&1&2
\end{array}
\right)
\]

Expressed in functional notation, we have a function $a\colon
[6]\rightarrow [6]$ (here $[n]=\left\{1,2,\ldots,n\right\}$); and we
have $a(1)=4$, $a(2)=5$, $a(3)=3$, and so on.  If these were objects,
or people, we might want to keep track of where they are.  We would
say: "at the start, object $i$ sits in place $i$, $i\in[6]$.  Then,
after the move, object 1 is in place 4, object 2 in place 5, object 3
in place 3, and so on".  This information is encapsulated by
`as.word(a)`.  In R matrix form we would have

```{r}
a_active <- rbind(1:6,as.word(a))
rownames(a_active) <- c("place before move","place after move")
a_active
```

(in the above R chunk, note how the top row is `1:6`.  We give the
objects persistent labels: each object is named according to the place
it sits in, before any moving).  On the other hand, we might be more
interested in the *places*.  We might want to know which object is
sitting in place 4.  We would say: "at the start, object $i$ sits in
place $i$, $i\in[6]$.  Then place 1 is occupied by object 4, place 2
occupied by object 5, and so on".  This information is technically
represented by permutation `a` but in an obscure form.  To answer
the question "which object is in place $i$?" in a convenient way, we
need to rearrange the permutation:

\[
\left(
\begin{array}{ccccccccc}
1&2&3&4&5&6\\
4&6&3&5&1&2
\end{array}
\right)
{\mbox{swap rows}\atop\longrightarrow}
\left(
\begin{array}{ccccccccc}
4&6&3&5&1&2\\
1&2&3&4&5&6
\end{array}
\right){\mbox{shuffle columns}\atop\longrightarrow}
\left(
\begin{array}{ccccccccc}
1&2&3&4&5&6\\
5&6&3&1&4&2
\end{array}
\right)
\]

In the above, it is easy to see that the rearrangement of permutation
is equivalent to taking a group-theoretic inverse:

```{r}
inverse(a)
```

(even now I find the R idiom for inversion to be unreasonably elegant:
`a[a] <- seq_along(a)`).  In R idiom, the two-row matrix form
would use `inverse()`:

```{r}
a_passive <- rbind(1:6,as.word(inverse(a)))
rownames(a_passive) <- c("place after move","place before move")
a_passive
```

(in the above R chunk, note how the top row---the *place* an object
sits in after the move, is `1:6`).  Thus from the first column we see
that the object currently in place 1 was originally in place 5.  If
the people subsequently move again, the mathematics and the R idiom
depend on whether we are interested in people, or places.  We need to
specify use of *active* or *passive* transformations, much as in the
Lorentz package.

### Active permutations

An active permutation $\pi$ moves an object from place $i$ to place
$\pi(i)$.  Textbooks and undergraduate courses usually use this
system, and is used above.

### Passive permutations

A passive permutation $\pi$ replaces an object in position $i$ by that
in position $\pi(i)$.

## Composition of active permutations


Suppose we have (active) permutation `a` as above, and another active
permutation `b`:

```{r}
b <- as.word(c(5, 2, 3, 4, 6, 1))
b
```

(note the three dots representing three fixed points of `b`).
Note carefully that the operations $a$ and $b$ do not commute and we
will discuss this in the context of active and passive transforms.
What is the result of executing $a$, followed by $b$?  Symbolically we
have:

\[
\overbrace{
\left(
\begin{array}{ccccccccc}
1&2&3&4&5&6\\
4&6&3&5&1&2
\end{array}
\right)
}^{a}
\circ
\overbrace{
\left(
\begin{array}{ccccccccc}
1&2&3&4&5&6\\
5&2&3&4&6&1\\
\end{array}
\right)
}^{b}=
\overbrace{
\left(
\begin{array}{ccccccccc}
1&2&3&4&5&6\\
4&1&3&6&5&2\\
\end{array}
\right)
}^{a\circ b}
\]

Thus, for example, $4\longrightarrow 5\longrightarrow 6$.  Considering
the operation $a\circ b$, this means that we perform permutation `a`
first, and then perform permutation `b`.  Taking this one step at a
time we would have, for example: "the person in place 4 (this is
object 4) moves to place 5 (but is still object 4) $\ldots$ and then
the object in place 5 (this is still object 4) moves to place 6".  See
how we track the object that started in place 4 (that is, object 4)
over two permutations, and so overall object 4 ends up in place 6.  We
see this on the right hand side: the fourth column of $a\circ b$ is
$\left(\begin{array}{c}4\\6\end{array}\right)$.  If we execute `a` and
then `b` using active language [explicitly: express `a` and `b` as
active permutations, and express the result of performing `a` then `b`
in active language], we can use standard permutation composition, in R
idiom the `*` operator:

```{r}
a*b
```

The `*` operator in R idiom is essentially carries out `b[a]`
to evaluate `a*b` (which is why indexing starts at 1, not 0).
Indeed we may verify that package idiom operates as expected:

```{r}
as.vector(b)[as.vector(a)]
```

showing agreement.  With functional notation (also known as prefix
notation) we can ask what happens to the object originally in place 1
(that would be object 1)

```{r}
fa <- as.function(a)
fb <- as.function(b)
fb(fa(1))
as.function(a*b)(1)  # should match fb(fa(1))
```

Note, however, the confusing order of operations: in functional
notation, if we want to operate on an element $x$ by function $f$ and
then by function $g$ we write $g(f(x))$ for the two successive
mappings $x\longrightarrow f(x)\longrightarrow g(f(x))$.  Postfix
notation would denote the same process as $xfg$, as shorthand for
$(xf)g$.  In R idiom, this is implemented by the excellent
`magrittr` package:

```{r}
1 %>% fa %>% fb  # idiom for fb(fa(1)), should match result above
```


# Composition of passive permutations

Now we consider the same operations `a` and `b` as discussed
above, and perform `a` and then `b`.  But this time we express
the permutations, and their composition, in passive form, and this
requires some modification.  First we will express `a` and `b`
in passive matrix form which we will call `a_passive` and
`b_passive`:


```{r}
a   # word form 
a_active  # matrix form (active)
a_passive  # matrix form (passive)
```

and then `b`, but we need to create equivalents `b_active` and
`b_passive` which we do as before:

```{r,echo=FALSE}
b_active <- rbind(1:6,as.word(b))
rownames(b_active) <- c("place before move","place after move")
b_passive <- rbind(1:6,as.word(inverse(b)))
rownames(b_passive) <- c("place after move","place before move")
```


```{r}
b
b_active
b_passive
```

(note again the relationship between `b_active` and
`b_passive`).  We want to perform `a` and then `b` as
before, but this time we want to use matrices `a_passive` and
`b_passive`:

```{r}
a_passive
b_passive
```

To work out the composition of `a_passive` and `b_passive`
[explicitly: give the passive transform corresponding to performing
`a_passive` first, and then `b_passive` second] we want a
passive transformation, that is, a matrix with the same row labels as,
say `a_passive` and first row `1:6`.  Let us call the result
of these two permutations `ab_passive`.  Given that
`ab_passive[1,1]=1`, we ask "what is `ab_passive[2,1]`?
This is equivalent to asking, "we have just
performed permutation `a` followed by `b`.  The object
currently in place 1: where was it before this composition of
permutations?"

To figure out which object is in place 1, we would look at
`b_passive`, being the most recent operation.  We would then look
at the first column of `b_passive` and say that the object that
was in place 6 was moved by `b_passive` to place 1.  And then you
would have to figure out which object was in place 6 before
`b_passive` was executed.  To answer *that*, you would look at
`a_passive` and see, from column 6 of `a_passive` that the
object in place 6 was moved from place 2 by `a`.  Thus the passive
transform `ab_passive` indicates that the object in place 1 after
the move was in place 2 before the move.  We would have
$1\longrightarrow 6\longrightarrow 2$ where in this case
"$\longrightarrow$" means "comes from".

We can thus represent the passive transformation by
`b_passive*a_passive`: see how the R idiom for permutation composition
"`*`" is used in exactly the same way for active and passive
permutations, but with a different meaning which requires us to
reverse the order of permutations.  To express the result,
`ab_passive` in active language we need to take the group-theoretic
inverse of the composition.  Recalling that passive transforms are the
group-theoretic inverses of the same active transformation, in
algebraic notation we would have

\[
\left(a^{-1}b^{-1}\right)^{-1}=ab;\qquad
a^{-1}b^{-1}=\left(ab\right)^{-1}
\]

and in R idiom we would have

```{r}
inverse(inverse(b) * inverse(a)) == a*b # both should be TRUE
inverse(b) * inverse(a) == inverse(a*b) # note b precedes a on LHS
```


# Permutation matrices

Now we will show how permutation matrices work and how they deal with
active and passive language.


```{r}
g <- as.cycle(c(1,2,6))
g
```

Then we can express `g` in terms of permutation matrices:


```{r}
pg <- perm_matrix(g)
pg
```

But it is convenient to relabel the rows and the columns:

```{r}
dimnames(pg) <- list(place_before_move=1:6,place_after_move=1:6)
pg
```

Row `n` of matrix `pg` shows where the object that was in
place `n` before the move ends up.  Thus, looking at the top row
(row 1), we see that the object that was in place 1 is now in place 2
[because the second column of row 1 is nonzero].  The second row (row
2) shows that the object that was in place 2 is now in place 6, the
object that was in place 3 is now in place 6, and so on.  This is
active language.

We can see that taking the transpose is equivalent to inverting the
matrix: a permutation matrix is orthogonal.  Now we can consider a
second permutation `h` and convert it to matrix form:

```{r}
h <- as.word(c(1,3,4,5,2,6))
h
ph <- perm_matrix(h)
dimnames(ph) <- list(place_before_move=1:6,place_after_move=1:6)
ph
```

We now consider what happens with successive permutations, as above,
but this time using permutation matrices.  We will permute first with
`g` and then with `h`, using matrix multiplication.

```{r}
pg %*% ph
```

```
 pg                                  ph                                  pg %*% ph
                  place_after_move                    place_after_move                    place_after_move
 place_before_move 1 2 3 4 5 6       place_before_move 1 2 3 4 5 6       place_before_move 1 2 3 4 5 6
                 1 0 1 0 0 0 0                       1 1 0 0 0 0 0                       1 0 0 1 0 0 0
                 2 0 0 0 0 0 1                       2 0 0 1 0 0 0                       2 0 0 0 0 0 1
                 3 0 0 1 0 0 0                       3 0 0 0 1 0 0                       3 0 0 0 1 0 0
                 4 0 0 0 1 0 0                       4 0 0 0 0 1 0                       4 0 0 0 0 1 0
                 5 0 0 0 0 1 0                       5 0 1 0 0 0 0                       5 0 1 0 0 0 0
                 6 1 0 0 0 0 0                       6 0 0 0 0 0 1                       6 1 0 0 0 0 0
```

(the above is hand-edited to put the matrices side-by-side).  Let us
consider the top row of `pg`.  This multiplies by each column of
`ph` but the only nonzero term is with column 3 of `ph` which
has row 2 nonzero.  Thus `(pg%*%ph)[1,3]==1`.  The process is,
symbolically, $1\longrightarrow 2\longrightarrow 3$.

### Passive language for permutation matrices

Alternatively, we could look at matrix `pg` in terms of columns.
Column `n` of this matrix shows where the object that ended up in
place `n` came from.  Thus, looking at column 1, we see that the
object that ended up in column 1 came from place 6, and the object
that ended up in place 2 came from place 1, and so on.  This is
passive language.

Thus the passive matrix is the *transpose* of the active matrix (we
could see this by swapping the dimension names).  Now we use the matrix rule

\[
AB=(B'A')'
\]

to show that permutation matrix multiplication has to be in the
opposite order for passive matrices.  Of course, we could observe that
permutation matrices are orthogonal and use

\[
AB=\left(B^{-1}A^{-1}\right)^{-1}
\]

instead.


[reference](https://en.wikiversity.org/wiki/Permutation_notation)
