2.7. Lecture 7

2.7.1. Characters

Throughout this section, $G$ is a finite group and $V$ is a finite-dimensional complex vector space.

Definition 2.7.1.

Let $(\pi,V)$ be a finite-dimensional complex representation of $G$ . The character of $(\pi,V)$ is the function $\chi_{\pi}:G\rightarrow{\mathbb{C}}$ defined by

\chi_{\pi}(g):=\operatorname{tr}\big{(}\pi(g)\big{)}.

Remark 2.7.2.

Here we define $\operatorname{tr}A$ ( $A:V\rightarrow V)$ as the sum of the eigenvalues of the matrix $A$ (with multiplicities). It is a theorem from linear algebra that this is equal to the sum of the diagonal elements of the matrix of $A$ with respect to any choice of basis of $V$

Example 2.7.3.

Let $({\,\mathrm{Id}},V)$ be the trivial representation of $G$ on $V$ . Then $\chi_{{\,\mathrm{Id}}}(g)=\operatorname{dim}V$ for all $g\in G$ .

Example 2.7.4.

Let $(\rho,{\mathbb{C}}^{2})$ be the defining representation of $D_{n}$ . Then

\rho(e)=\left(\begin{matrix}1&0\\ 0&1\end{matrix}\right),\quad\rho(s)=\left(\begin{matrix}1&0\\ 0&-1\end{matrix}\right),\quad\rho(r)=\left(\begin{matrix}\cos(2\pi/n)&-\sin(2% \pi/n)\\ \sin(2\pi/n)&\cos(2\pi/n)\end{matrix}\right),

hence $\chi_{\rho}(e)=2$ , $\chi_{\rho}(s)=0$ , $\chi_{\rho}(r)=2\cos(2\pi/n)$ .

Important: $\chi_{\pi}(gh)\neq\chi_{\pi}(g)\chi_{\pi}(h)!$ This is easily seen from the previous example, as $s^{2}=e$ .

The following lemma is quite important:

Lemma 2.7.5.

Isomorphic representations have the same character.

Proof.

By Problem 1.3.2, if $\pi$ and $\rho$ are isomorphic, then $\pi(g)$ and $\rho(g)$ have the same eigenvalues, hence also the same traces. ∎

It seems that we throw away a lot of information when we go from studying a representation to studying its character; the remarkable thing is that, in fact, the character completely determines the representation. Moreover, there is a lot of structure to the characters and it is often possible to find all of the characters of a group, even when it is not clear how to construct the representations!

Example 2.7.6.

If $(\psi,{\mathbb{C}})$ is a 1-dimensional representation, then $\psi$ is its own character (the trace of a scalar is just itself).

Example 2.7.7.

Let $(\pi,{\mathbb{C}}^{n})$ be the permutation representation of $S_{n}$ . Then

\chi_{\pi}(\sigma)=\#\big{\{}j\in\{1,2,\ldots,n\}\,|\,\sigma(j)=j\big{\}}.

Proof.

Given $\sigma\in S_{n}$ , consider the matrix of $\pi(\sigma)$ with respect to the standard basis ${\bf e}_{1},\ldots,{\bf e}_{n}$ . The $j$ -th column of this matrix is $\pi(\sigma){\bf e}_{j}={\bf e}_{\sigma(j)}$ . Thus, this column will contribute $+1$ to the trace if it has a $1$ in the $j$ -th row (i.e. if $\sigma(j)=j$ ), and zero otherwise. ∎

By a similar argument:

Example 2.7.8.

Let $(\lambda,{\mathbb{C}}G)$ be the regular representation of $G$ . Then

\chi_{\lambda}(g)=\begin{cases}|G|\quad&\mathrm{if}\;g=e\\ 0\quad&\mathrm{otherwise.}\end{cases}

Proof.

Let $h_{1},h_{2},\ldots,h_{|G|}$ be an enumeration of the elements of $G$ ; this forms a basis of ${\mathbb{C}}G$ . Then for any $j$ , we have

\lambda(g)h_{j}=gh_{j}=h_{k}

for some $k\in\{1,\ldots,|G|\}$ . The $j$ -th column of the matrix of $\lambda(g)$ is the coordinate vector of $h_{k}$ with respect to the choice of basis, i.e. it has a $1$ in row $k$ , and zeroes elsewhere. This gives a non-zero contribution to the trace (in in that case +1) if and only if $h_{k}$ is on the diagonal, i.e. $h_{k}=h_{j}$ . But $h_{k}=gh_{j}$ , so this happens if and only if $g=e$ . ∎

We collect several useful properties of characters in the following proposition:

Proposition 2.7.9.

Let $(\pi,V)$ and $(\rho,W)$ be finite-dimensional representations of a finite group $G$ . Then

(i)

$\chi_{\pi}(e)=\operatorname{dim}V$ .
(ii)

$\chi_{\pi}(gh)=\chi_{\pi}(hg)$ and $\chi_{\pi}(hgh^{-1})=\chi_{\pi}(g)$ for all $g,h\in G$ .
(iii)

$\chi_{\pi\oplus\rho}=\chi_{\pi}+\chi_{\rho}$
(iv)

$\chi_{\pi}(g^{-1})=\overline{\chi_{\pi}(g)}$ for all $g\in G$ .

Proof.

(i)

$\chi_{\pi}(e)=\operatorname{tr}({\,\mathrm{Id}})=\operatorname{dim}V$ .
(ii)

This follows from the linear algebra identity $\operatorname{tr}(AB)=\operatorname{tr}(BA)$ .
(iii)

Let $\{{\bf v}_{i}\}_{i=1,\ldots,\operatorname{dim}V}$ be an an eigenbasis of $V$ for $\pi(g)$ with eigenvalues $\lambda_{i}$ , and $\{{\bf w}_{j}\}_{j=1,\ldots,\operatorname{dim}W}$ an eigenbasis of $W$ for $\rho(g)$ with eigenvalues $\mu_{j}$ . Then $\{({\bf v}_{i},{\bf 0})\}_{i}\cup\{({\bf 0},{\bf w}_{j})\}_{j}$ is a basis of $V\oplus W$ . This is in fact an eigenbasis for $(\pi\oplus\rho)(g)$ , since

$(\pi\oplus\rho)(g)({\bf v}_{i},{\bf 0})=(\pi(g){\bf v}_{i},\rho(g){\bf 0})=(% \lambda_{i}{\bf v}_{i},{\bf 0})=\lambda_{i}({\bf v}_{i},{\bf 0}),$

and similarly $(\pi\oplus\rho)(g)({\bf 0},{\bf w}_{j})=\mu_{j}({\bf 0},{\bf w}_{j})$ . All the eigenvalues of $(\pi\oplus\rho)(g)$ are thus given by $\{\lambda_{i}\}\cup\{\mu_{j}\}$ , hence

$\chi_{\pi\oplus\rho}(g)=\sum_{i}\lambda_{i}+\sum_{j}\mu_{j}=\chi_{\pi}(g)+\chi% _{\rho}(g).$
(iv)

This follows from the fact that for any eigenvector ${\bf v}_{i}$ of $\pi(g)$ with eigenvalue $\lambda_{i}$ , we have

${\bf v}_{i}=\pi(e){\bf v}_{i}=\pi(g^{-1})\pi(g){\bf v}_{i}=\lambda_{i}\pi(g^{-% 1}){\bf v}_{i},$

so $\pi(g^{-1}){\bf v}_{i}=\lambda_{i}^{-1}{\bf v}_{i}$ . Since $\pi(g)$ has finite order, $\lambda_{i}^{|G|}=1$ , giving $\lambda_{i}^{-1}=\overline{\lambda_{i}}$ . This then yields

$\chi_{\pi}(g^{-1})=\sum_{i}\lambda_{i}^{-1}=\sum_{i}\overline{\lambda_{i}}=% \overline{\sum_{i}\lambda_{i}}=\overline{\chi_{\pi}(g)}.$

∎

2.7.2. Exercises

Problem 43. Let $(\pi,W_{0})$ denote the usual $(n-1)$ -dimensional irreducible subrepresentation of the permutation representation of $S_{n}$ on ${\mathbb{C}}^{n}$ . Compute the character $\chi_{(\pi,W_{0})}$ of this subrepresentation.

Hint: combine Example 2.7.7 with Proposition 2.7.9 (iii).

Problem 44. Let $g\in G$ be such that $g$ and $g^{-1}$ are both in the same conjugacy class. Show that $\chi_{\pi}(g)$ is real-valued for any representation $\pi$ of $G$ .

Problem 45. Let $(\pi,V)$ be a representation of a finite group $G$ . Show that

(a)

$|\chi_{\pi}(g)|\leq\operatorname{dim}(\pi)\quad\forall g\in G$ .
(b)

$\ker\pi=\chi_{\pi}^{-1}\big{(}\{\operatorname{dim}(\pi)\}\big{)}.$

Problem 46. Let the finite group $G$ act on the finite set $X$ , and consider the representation $\big{(}\pi,{\mathbb{C}}(X)\big{)}$ as in Definition 1.1.9. Show that

\chi_{\pi}(g)=\#\{x\in X\,|\,g\cdot x=x\}.