Characters

Section 2.7 Characters

Objectives

You should be able to:

Compute the character of a representation.
Recognize that the character is a function of class.
Calculate the character of direct sum and tensor product representations.

In this section we introduce the notion of the character of a representation. The idea is simple: we take the traces of the matrices in a given representation. Characters will give a clean way of studying and classifying irreducible representations, and determining whether a given representation is reducible or not.

Subsection 2.7.1 Definition

Definition 2.7.1. The character of a representation.

Let \(T: G \to GL(V)\) be a representation of \(G\text{.}\) The character of this representation is given by the map \(\chi: G \to \mathbb{C}\text{,}\) which assigns to every group element the trace of the corresponding matrix in the representation \(T\text{:}\)

\begin{equation*} \chi^{(T)}(g) = \Tr T(g). \end{equation*}

If \(T\) is irreducible, the character is called simple (or irreducible), while it is called compound otherwise. The degree of a character is the dimension of the corresponding representation.

Subsection 2.7.2 A few simple properties

The first thing to note about characters is that they are the same for all elements that are in the same conjugacy class. This is usually encapsulated in the statement that “character is a function of class”:

Lemma 2.7.2. Character is a function of class.

The character of a representation is a class function, meaning that all elements of a group belonging to the same conjugacy class have the same character.

Proof.

Let \(x,y \in G\) be conjugate. That is, there exists a \(g \in G\) such that \(x = g y g^{-1}\text{.}\) Since a representation is a group homomorphism, this means that \(T(x) = T(g) T(y) T(g^{-1})\text{.}\) Taking the trace, we get

\begin{equation*} \Tr T(x) = \Tr \left(T(g) T(y) T(g^{-1}) \right) = \Tr\left( T(y) T(g^{-1}) T(g) \right) = \Tr T(y), \end{equation*}

where we used the cyclic property of the trace of a product of matrices.

It is also important to note that equivalent representations have the same character:

Lemma 2.7.3. Equivalent representations have the same character.

If \(T:G \to GL(V)\) and \(S: G \to GL(V)\) are equivalent representations, then \(\chi^{(T)}(g) = \chi^{(S)}(g)\) for all \(g \in G\text{.}\)

Proof.

Two representation \(T\) and \(S\) are equivalent if their matrices are related by a similarity transformation \(T(g) = A S(g) A^{-1}\text{.}\) Taking the trace, and using cyclicity again, we get:

\begin{equation*} \Tr T(g) = \Tr \left( A S(g) A^{-1} \right) = \Tr \left( S(g) A^{-1} A \right) = \Tr S(g). \end{equation*}

What this tells us is that if two representations have different characters, then they must be inequivalent. We will see that the converse statement (that two inequivalent representations must have different characters) follows from the orthogonality theorem.

Another simple property of characters is how they behave for direct sums and tensor products of representations. We will leave the proof of the following lemma as an exercise:

Lemma 2.7.4. Character of a direct sum and a tensor product.

Let \(T: G \to GL(V)\) and \(S: G \to GL(V)\) be two representations of \(G\text{.}\) Then

\begin{equation*} \chi^{(T \oplus S)}(g) = \chi^{(T)}(g) + \chi^{(S)}(g), \qquad \chi^{(T \otimes S)}(g) = \chi^{(T)}(g) \chi^{(S)}(g). \end{equation*}

In particular, the character of a semisimple representation \(T\) can be written as

\begin{equation*} \chi^{(T)}(g) = \sum_{i=1}^n a_i \chi^{(T_i)}(g), \end{equation*}

where the \(T_i\) are the irreducible representations appearing in the direct sum with multiplicity \(a_i\text{.}\)