Characters

This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.

In this post we define character functions and examine their properties. They will be of great help in our goal of classifying representations.

Characters of Group Representations

Definition 1 For a representation \(\rho:G\rightarrow\Aut(V)\) of \(G\), the corresponding character \(\rchi_\rho:G\rightarrow\mathbb{C}\) is defined by

\[\rchi_\rho(g)=\tr(\rho(g))\]

.

In other words, this function takes each \(g\in G\) and returns the trace of the linear map \(\rho(g):V\rightarrow V\) it defines. As we shall see, this function plays a major role in describing representations of \(G\). For instance, we can immediately see that this function encodes the dimension of \(V\).

\[\rchi_\rho(e)=\tr(\rho(e))=\tr(\id_V)=\dim V.\]

Similarly, when two linear maps

\[L_V:V\rightarrow V,\qquad L_W:W\rightarrow W\]

are given, we know how their direct sum \(L_V\oplus L_W: V\oplus W\rightarrow V\oplus W\), their tensor product \(L_V\otimes L_W: V\otimes W \rightarrow V\otimes W\), and so on are defined, and we also know what their traces are (for instance, by computing with matrices). From this we obtain the following proposition.

Proposition 2 For representations \(V, W\), the following hold.

1. \[\rchi_{V\oplus W}=\rchi_V\oplus \rchi_W\]
2. \[\rchi_{V\otimes W}=\rchi_V\rchi_W\]
3. \[\rchi_{V^\ast}=\overline{\rchi}_V\]

In particular, by the first formula, since any representation admits an irreducible decomposition

\[V\cong V_1^{\oplus a_1}\oplus\cdots\oplus V_r^{\oplus a_r}\]

we know that the character of any representation can be expressed as

\[\rchi_V=a_1\rchi_{V_1}+\cdots+a_r\rchi_{V_r}\]

.

On the other hand, by definition,

\[\rchi_\rho(hgh^{-1})=\tr(\rho(h)\rho(g)\rho(h)^{-1})=\tr(\rho(g))=\rchi_\rho(g)\]

holds, so ([Linear Algebra] §Characteristic Polynomial, ⁋Theorem 5), we see that \(\rchi_\rho\) is constant on the conjugacy classes of \(G\). Such functions also have a name.

Definition 3 A function \(\rchi:G\rightarrow\mathbb{C}\) is called a class function if \(\rchi(hgh^{-1})=f(g)\) holds for all \(g,h\in G\). We denote the collection of all class functions defined on \(G\) by \(\mathbb{C}_\class(G)\).

By definition, class functions are determined by their values on each conjugacy class, and thus, as a vector space, \(\mathbb{C}_\class(G)\) has dimension equal to the number of conjugacy classes of \(G\). Meanwhile, the idea we considered important in the previous post was that given a value, we could average it over all of \(G\) to obtain a \(G\)-invariant value. Using this, we can give the following definition on \(\mathbb{C}_\class(G)\).

Definition 4 For any class functions \(\rchi_1,\rchi_2: G\rightarrow \mathbb{C}\), we define

\[\langle \rchi_1,\rchi_2\rangle=\frac{1}{\lvert G\rvert}\sum_{g\in G} \rchi_1(g)\overline{\rchi_2(g)}\]

.

This is merely the standard Hermitian product on the target space \(\mathbb{C}\) transferred to \(\mathbb{C}_\class(G)\). On the other hand, for the character \(\rchi_\rho\) of any representation \(\rho\), for any \(g\in G\),

\[\rchi_\rho(g^{-1})=\tr(\rho(g^{-1}))=\tr(\rho(g)^{-1})=\tr(\rho(g)^\dagger)=\overline{\tr(\rho(g))}=\overline{\rchi_\rho(g)}\]

holds, so for two characters \(\rchi_1,\rchi_2\) we know that the following formula

\[\langle \rchi_1,\rchi_2\rangle=\frac{1}{\lvert G\rvert}\sum_{g\in G}\rchi_1(g)\overline{\rchi_2(g)}=\frac{1}{\lvert G\rvert}\sum_{g\in G}\rchi_1(g^{-1})\overline{\rchi_2(g^{-1})}=\frac{1}{\lvert G\rvert}\sum_{g\in G}\overline{\rchi_1(g)}\rchi_2(g)=\langle \rchi_2,\rchi_1\rangle\]

holds. That is, when restricted to characters, this inner product takes real values.

Orthogonality of Characters

As we saw in the previous post, for any representation \(U\), the fixed-point subspace

\[U^G=\{u\in U\mid g\cdot u=u\text{ for all $g\in G$}\}\]

exists, and in this case

\[p:U\rightarrow U^G;\qquad u\mapsto \frac{1}{\lvert G\rvert}\sum_{g\in G}g\cdot u\]

defines a \(G\)-invariant projection from \(U\) to \(U\), and its image is \(U^G\). By its definition, the subrepresentation defined on \(U^G\) is precisely the trivial representation

\[G\rightarrow \Aut(U^G);\quad g\mapsto \id_{U_G}\]

so from this we can decompose the representation \(U\) into the trivial representation \(U^G\) and the remaining part \(W\), obtaining

\[U=U^G\oplus W\]

.

Moreover, we can also compute the dimension of \(U^G\). Using suitable bases of \(U^G\) and \(W\) in the above decomposition, we can represent this in the block matrix form

\[\begin{pmatrix}\id_{U^G}&0\\0&0\end{pmatrix}\]

so the trace of \(\varphi: U\rightarrow U\) equals \(\dim U^G\). Now by definition,

\[\dim U^G=\tr(\varphi)=\tr\left(\frac{1}{\lvert G\rvert}\sum_{g\in G}\rho(g)\right)=\frac{1}{\lvert G\rvert}\sum_{g\in G}\tr(\rho(g))=\frac{1}{\lvert G\rvert}\sum_{g\in G}\rchi(g)\tag{1}\]

.

More generally, in §Representations of Finite Groups, ⁋Definition 3 we defined, for any \(G\)-representations \(V,W\), a \(G\)-action on their \(\Hom\)-set \(\Hom_\mathbb{C}(V,W)\) (as underlying \(\mathbb{C}\)-vector spaces) by

\[(g\cdot f)(v)=g\cdot f(g^{-1}\cdot v)\qquad\text{for all $v\in V$}\]

.

Then the formula

\[\Hom_\mathbb{C}(V,W)^G=\Hom_G(V,W)\]

holds, and thus applying (1) to \(U=\Hom(V,W)\) and the corresponding trace map \(\varphi\), we know that

\[\dim \Hom_G(V,W)=\tr(\varphi)=\frac{1}{\lvert G\rvert}\sum_{g\in G}\rchi_{\Hom(V,W)}(g)\]

.

On the other hand, using \(\Hom_G(V,W)=V^\ast\otimes W\), the character on the right-hand side is obtained by the formula

\[\rchi_{\Hom_G(V,W)}(g)=\overline{\rchi_V(g)}\rchi_W(g)\]

so the above equation can be rewritten as

\[\dim\Hom_G(V,W)=\frac{1}{\lvert G\rvert}\sum_{g\in G}\overline{\rchi_V(g)}\rchi_W(g)=\langle \rchi_W, \rchi_V\rangle\]

.

Finally, assuming \(V,W\) are irreducible representations, by §Representations of Finite Groups, ⁋Lemma 8, \(\Hom_G(V,W)\) is \(1\)-dimensional if \(V\cong W\) and \(0\)-dimensional otherwise, so

\[\dim \Hom_\mathbb{C}(V,W)^G=\dim \Hom_G(V,W)=\begin{cases}1&\text{if $V\cong W$,}\\0&\text{otherwise}\end{cases}\]

and from this we obtain the formula

\[\langle \rchi_W,\rchi_V\rangle=\delta_{VW}\]

.

That is, with respect to the inner product of Definition 4, the irreducible characters form an orthonormal set. Since we know that \(\mathbb{C}_\class(G)\) has dimension equal to the number of conjugacy classes of \(G\), it follows that there cannot be more irreducible representations than the number of conjugacy classes of \(G\). Moreover, using this inner product we can compute the multiplicity of \(V_i\) inside \(V\) by taking the inner product of the character \(\rchi_V\) of any representation \(V\) with the character \(\rchi_{V_i}\) of a fixed irreducible representation \(V_i\).

Regular Representation

In this section we obtain the Artin–Wedderburn decomposition considered in the previous post (§Representations of Finite Groups, Equation (1)) using characters. To this end, first observe that \(\mathbb{C}[G]\) is a left \(\mathbb{C}[G]\)-module over itself, and hence by the categorical equivalence

\[\Rep_\mathbb{C}(G)\cong \lMod{\mathbb{C}[G]}\]

it is also a representation of \(G\). This is obtained simply by restricting the module structure on \(\mathbb{C}[G]\), that is, its multiplication structure as a ring, to \(G\). Explicitly, using the image \(\delta_g=\sum_{x\in X}\delta_g(x)x\) of any \(g\in G\) in \(\mathbb{C}[G]\), we can write

\[g\cdot \left(\sum_{y\in G} \phi(y)y\right)=\left(\sum_{x\in X}\delta_g(x)x\right)\left(\sum_{y\in G}\phi(y)y\right)=\sum_{z\in G}\left(\sum_{x\in G}\delta_g(x)\phi(x^{-1}z)\right)z=\sum_{z\in G}\phi(g^{-1}z)z=\sum_{z\in G}\phi(z)(gz)\]

and we call such a representation the regular representation.

Now, to decompose \(\mathbb{C}[G]\), we consider the character theory of the regular representation. Regarding \(\mathbb{C}[G]\) as the vector space having the elements \(g\in G\) (more precisely, the \(\delta_g\)) as a basis, and representing each linear operator \(\rho_\reg(g)\) given by the regular representation as a matrix via this basis, we find that its trace is

\[\rchi_{\mathbb{C}[G]}(g)=\begin{cases}\lvert G\rvert&\text{if $g=e$}\\0&\text{otherwise}\end{cases}\tag{2}\]

.

Now if \(V_i\) is an irreducible subrepresentation of \(\mathbb{C}[G]\), then

\[\langle\rchi_{\mathbb{C}[G]}, \rchi_{V_i}\rangle=\frac{1}{\lvert G\rvert}\sum_{g\in G}\rchi_{\mathbb{C}[G]}(g)\rchi_{V_i}(g)=\frac{1}{\lvert G\rvert} \rchi_{\mathbb{C}[G]}(e)\rchi_{V_i}(e)=\dim V_i\]

holds. Thus we obtain the decomposition

\[\mathbb{C}[G]\cong \bigoplus_{i=1}^r V_i^{\dim V_i}\]

.

Moreover, \(\mathbb{C}[G]\) acts on itself by multiplication, and under this action, thinking that by §Representations of Finite Groups, ⁋Lemma 8 each \(V_i\) maps only into \(V_i\), we know that each \(V_i^{\dim V_i}\) is exactly the matrix algebra \(\Mat_{d_i}(\mathbb{C})\), and by the uniqueness of the Artin–Wedderburn theorem we can verify that this is precisely

\[\mathbb{C}[G]\cong \bigoplus_{i=1}^r\Mat_{d_i}(\mathbb{C})\]

.

Projection Formula

Earlier we showed that the characters of irreducible representations form an orthonormal set inside \(\mathbb{C}_\class(G)\). Now we show that they form an orthonormal basis of \(\mathbb{C}_\class(G)\).

Lemma 5 Let any function \(\phi:G\rightarrow \mathbb{C}\) and any representation \(\rho:G\rightarrow\Aut(V)\) be given. If we define

\[\rho_\phi=\sum_{g\in G} \phi(g)\rho(g): V\rightarrow V\]

then \(\rho_\phi\) is a \(G\)-map if and only if \(\phi\) is a class function.

Proof

For \(\rho_\phi\) to be a \(G\)-map, the formula

\[\rho_\phi(h\cdot v)=h\cdot\rho_\phi(v)\]

must hold for any \(h\in G\) and any \(v\in V\). Computing the left-hand side directly,

\[\rho_\phi(h\cdot v)=\sum_{g\in G}\phi(g)\rho(g)(h\cdot v)\]

and since taking this sum over \(hgh^{-1}\) yields the same sum,

\[\rho_\phi(hv)=\sum_{g\in G}\phi(hgh^{-1})\rho(hgh^{-1})(h\cdot v)=\sum_{g\in G}\phi(hgh^{-1})\rho(h)\rho(g)(v)=\rho(h)\left(\sum_{g\in G}\phi(hgh^{-1})\rho(g)v\right)\]

we can write this. Now for this to equal

\[h\cdot\rho_\phi(v)=\rho(h)\rho_\phi(v)=\rho(h)\left(\sum_{g\in G}\phi(g)\rho(g)(v))\]

we must have exactly \(\phi(g)=\phi(hgh^{-1})\), that is, \(\phi\) must be a class function.

Now we use this to show that every class function can be expressed as a linear combination of irreducible characters. That is, we must show that if for a class function \(\phi\), \(\langle \rchi_V,\phi\rangle=0\) holds for all irreducible characters \(\rchi_V\), then \(\phi=0\).

To this end, apply the above lemma to a class function \(\phi\) and an irreducible representation \(\rho:G\rightarrow\Aut(V)\). Since \(\phi\) is a class function, so is \(\overline{\phi}\), and thus \(\rho_{\overline{\phi}}\) is a \(G\)-map; by §Representations of Finite Groups, ⁋Lemma 8, \(\rho_{\overline{\phi}}\) is of the form \(\lambda\id_V\). Now taking the trace here, we know that

\[(\dim V)\lambda=\tr(\rho_{\overline{\phi}})=\tr\left(\sum_{g\in G}\overline{\phi(g)}\rho(g)\right)=\sum_{g\in G}\overline{\phi(g)}\rchi_V(g)=\lvert G\rvert\langle \rchi_V,\phi\rangle=0\]

.

Now since any representation has an irreducible decomposition, we know that \(\sum \overline{\phi(g)}g\) must act as \(0\) on any representation, and in particular also on the regular representation \(\mathbb{C}[G]\). However, letting this element act on \(\delta_e\) in the regular representation gives

\[\left(\sum\overline{\phi(g)}g\right)\cdot \delta_e=\sum_{g\in G}\overline{\phi(g)}g\]

and therefore \(\overline{\phi(g)}=0\) must hold for all \(g\).

Example: \(S_3\)

We close this post with an example illustrating the theory we have built since the previous posts. First, for any abelian group \(G\) it is obvious that irreducible representations are only \(1\)-dimensional representations, so to test our theory we need a non-abelian group. For convenience of computation, let us consider \(S_3\), the smallest non-abelian group. Explicitly,

\[S_3=\{(\;),\,(1\;2),\,(1\;3),\,(2\;3),\,(1\;2\;3),\,(1\;3\;2)\}\]

.

First, it is obvious that the following two representations

\[\rho_0: S_3 \rightarrow \Aut(\mathbb{C})\qquad \sigma\mapsto \id_\mathbb{C}\]

and

\[\rho_\sgn: S_3 \rightarrow \Aut(\mathbb{C})\qquad \sigma\mapsto \sgn(\sigma)\id_\mathbb{C}\]

are two irreducible representations of \(S_3\). Meanwhile, \(S_3\) acts on \(\mathbb{C}^3\) by permutation via

\[\sigma\cdot(x_1,x_2,x_3)=(x_{\sigma(1)},x_{\sigma(2)},x_{\sigma(3)})\]

.

However, this action is trivial along the line spanned by \((1,1,1)\), and we can view all the action as taking place on the subspace orthogonal to this line,

\[V_\std=\{(x_1,x_2,x_3)\mid x_1+x_2+x_3=0\}\]

and we can verify that this subrepresentation is irreducible. We call this the standard representation of \(S_3\).

These three representations have dimensions \(1,1,2\) respectively, and since

\[\lvert S_3\rvert=6=1^2+1^2+2^2\]

we can verify that the dimensions in the irreducible decomposition match. To proceed further, let us compute the characters of these three irreducible representations. For this, consider the conjugacy classes of \(S_3\):

\[A_1=\{(\;)\},\qquad A_2=\{(1\;2),\,(1\;3),\,(2\;3)\},\qquad A_3=\{(1\;2\;3),\,(1\;3\;2)\}\]

.

For notational convenience, if a character \(\rchi\) takes values \(a_1,a_2,a_3\) on \(A_1,A_2,A_3\), we shall write it in the vector form \((a_1,a_2,a_3)\).

• \(\rho_0\) sends every element of \(S_3\) to \(\id_\mathbb{C}\in \Aut(\mathbb{C})\), and the trace of this \(1\times 1\) matrix is \(1\), so \(\rchi_0\) is \((1,1,1)\).
• \(\rho_\sgn\) sends only the odd permutations of \(S_3\) (that is, the elements of \(A_2\)) to \(-\id_\mathbb{C}\in\Aut(\mathbb{C})\), and the remaining elements to \(\id_\mathbb{C}\in \Aut(\mathbb{C})\), so \(\rchi_\sgn\) is \((1,-1,1)\).

There are two methods for computing the character \(\rchi_\std\) of the standard representation \(\rho_\std\), and we shall introduce both.

First, to compute it directly, choose the basis of \(V_\std\)

\[\{e_1=(1,0,-1), e_2=(0,1,-1)\}\]

.

Then \((\;)\) does not touch this basis, so of course \(\rchi_\std\) takes the value \(2\) on \(A_1\). On \(A_2\), for instance, acting by \((1\;2)\) swaps the basis elements, so it corresponds to the matrix

\[\rho_\std((1\;2))\begin{pmatrix}0&1\\1&0\end{pmatrix}\]

and its trace is \(0\). For reference, if \((1\;3)\) acts on this basis, \(e_1\) is sent to \(-e_1\) and \(e_2\) to \((-1,1,0)=-e_1+e_2\), so it corresponds to the matrix

\[\rho_\std((1\;3))=\begin{pmatrix}-1&-1\\0&1\end{pmatrix}\]

and since the trace of this matrix is also \(0\), we can verify by computation that a character function is a class function. In the case of \(A_3\), since it sends \(e_1\) to \((-1,1,0)=-e_1+e_2\) and \(e_2\) to \((-1,0,1)=-e_1\),

\[\rho_\std((1\;2\;3))=\begin{pmatrix}-1&-1\\1&0\end{pmatrix}\]

so we know that \(\rchi_\std\) is \((2,0,-1)\).

A more convenient computational method uses the decomposition

\[V_\perm=V_0\oplus V_\std\]

.

We already know that the character of \(V_0\) is \((1,1,1)\). Now considering the action of \(V_\perm\),

\[\rho_\perm((\;))=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix},\quad \rho_\perm((1\;2))=\begin{pmatrix}0&1&0\\1&0&0\\0&0&1\end{pmatrix},\quad \rho_\perm((1\;2\;3))=\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}\]

we know that \(\rchi_\perm\) is \((3,1,0)\), and now by Proposition 2 we have \(\rchi_\perm=\rchi_0+\rchi_\std\), so we can see that \(\rchi_\std\) is \((2,0,-1)\). These three characters obtained in this way,

\[\rchi_0=(1,1,1),\qquad \rchi_\sgn=(1,-1,1),\qquad \rchi_\std=(2,0,-1)\]

are (as expected) orthonormal.

From the computation of the character of the regular representation we know that

\[\rho_{\mathbb{C}[S_3]}(g)=\begin{cases}6&\text{if $g=e$}\\0&\text{otherwise}\end{cases}\]

(Equation (2)). This must be a \(\mathbb{Z}^{\geq 0}\)-linear combination of the above three characters, and indeed we can verify that

\[\rchi_{\mathbb{C}[S_3]}=\rchi_0+\rchi_\sgn+2\rchi_\std\]

.

Furthermore, this is a result consistent with the above discussion that in the regular representation, the multiplicity of each irreducible factor must equal its own dimension.