# The Unapologetic Mathematician

## Characters of Permutation Representations

Let’s take $(\mathbb{C}S,\rho)$ to be a permutation representation coming from a group action on a finite set $S$ that we’ll also call $\rho$. It’s straightforward to calculate the character of this representation.

Indeed, the standard basis that comes from the elements of $S$ gives us a nice matrix representation:

$\displaystyle\rho(g)_s^t=\delta_{\rho(g)s,t}$

On the left $\rho(g)$ is the matrix of the action on $\mathbb{C}S$, while on the right it’s the group action on the set $S$. Hopefully this won’t be too confusing. The matrix entry in row $s$ and column $t$ is $1$ if $\rho(g)$ sends $s$ to $t$, and it’s $0$ otherwise.

So what’s the character $\chi_\rho(g)$? It’s the trace of the matrix $\rho(g)$, which is the sum of all the diagonal elements:

$\displaystyle\mathrm{Tr}\left(\rho(g)\right)=\sum\limits_{s\in S}\rho(g)_s^s=\sum\limits_{s\in S}\rho(g)_s^s=\sum\limits_{s\in S}\delta_{\rho(g)s,s}$

This sum counts up $1$ for each point $s$ that $\rho(g)$ sends back to itself, and $0$ otherwise. That is, it counts the number of fixed points of the permutation $\rho(g)$.

As a special case, we can consider the defining representation $V^\mathrm{def}$ of the symmetric group $S_n$. The character $\chi^\mathrm{def}$ counts the number of fixed points of any given permutation. For instance, in the case $n=3$ we calculate:

$\displaystyle\begin{aligned}\chi^\mathrm{def}\left((1)(2)(3)\right)&=3\\\chi^\mathrm{def}\left((1\,2)(3)\right)&=1\\\chi^\mathrm{def}\left((1\,3)(2)\right)&=1\\\chi^\mathrm{def}\left((2\,3)(1)\right)&=1\\\chi^\mathrm{def}\left((1\,2\,3)\right)&=0\\\chi^\mathrm{def}\left((1\,3\,2)\right)&=0\end{aligned}$

In particular, the character takes the value $3$ on the identity element $e\in G$, and the degree of the representation is $3$ as well. This is no coincidence; $\chi(e)$ will always be the degree of the representation in question, since any matrix representation of degree $n$ must send $e$ to the $n\times n$ identity matrix, whose trace is $n$. This holds both for permutation representations and for any other representation.

## 3 Comments	»

1. [...] pass from representations to their characters. Of course, this isn’t much of a stretch, since we saw that the character of a representation includes information about the dimension: [...]

Pingback by | October 25, 2010 | Reply

2. [...] The nice thing here is that it’s a permutation representation, and that means we have a shortcut to calculating its character: is the number of fixed point of the action of on the standard basis [...]

Pingback by | November 17, 2010 | Reply

3. [...] we’ve constructed. Since these come from actions of on various sets, we have our usual shortcut to calculate their characters: count fixed [...]

Pingback by | December 15, 2010 | Reply

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This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

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