We continue our adventure in the land of representation theory of finite groups. Our goal in this post is to introduce characters and prove (one of) Schur’s orthogonality relations. In the previous post we never really used that we can interpret \( GL(V) \) as \( GL_n(\mathbb{C}) \) – this will figure more prominently in this text. We start of by concretizing this.

Let \( \rho: G \to GL(V) \) be a representation, where \( \dim(V)=n \). Choose a basis \( \mathcal{B} = \{\beta_1,\dots,\beta_n \} \) of \( V \). Let \( e_i \) denote the standard \(i\)-th unit vector of \( \mathbb{C}^n \). Then the linear map \( T: V \to \mathbb{C}^n \) defined by \( \beta_1 \mapsto e_i \) is an isomorphism, and so the representation \( \varphi: G \to GL_n(\mathbb{C}) \) defined by \( \varphi_g = T\rho_g T^{-1} \) for all \( g \in G \) is equivalent to \( \rho \). A direct sum of two representations \( \rho^{(1)}: G \to GL(V), \rho^{(2)}: G \to GL(W) \) where \(\dim(V)=n,\dim(W)=m \) can hence be interpreted as \( \rho^{(1)} \oplus \rho^{(2)} : G \to GL_{m+n}(\mathbb{C}) \) so that for all \(g \in G \) we have $$ (\rho^{(1)} \oplus \rho^{(2)})_g = \begin{pmatrix} \rho^{(1)}_g & 0 \\ 0 & \rho^{(2)}_g \end{pmatrix} $$

To study characters we introduce what is called the **group algebra** of \( G \). This is denoted as \( L(G) \) or \(\mathbb{C}^G \) and defined as \( \{ f | f: G \to \mathbb{C} \} \), that is *all* functions from \( G \) to \( \mathbb{C} \). Characters live inside this group algebra as we will see towards the end of this post. Defining addition by \( (f_1 + f_2)(g) = f_1(g) + f_2(g) \) and scalar multiplication by \( (cf)(g) = cf(g) \) we get a vector space structure on \( \mathbb{C}^{G} \). Actually we can turn \( \mathbb{C}^G \) into an inner product space, where the inner product is $$ \langle f_1,f_2 \rangle = \frac{1}{|G|}\sum_{g \in G } f_1(g) \overline{f_2(g)} $$ That this is indeed an inner product is a routine calculation and will be left to the reader.

In light of lemma 1.2 from the previous article, we focus on unitary representations. By \( U_n(\mathbb{C}) \) we will mean the group of unitary \( n \times n \)-matrices. Let \( \rho : G \to GL(V) \) and \( \varphi : G \to GL(W) \) be representations of \( G \) and \( T : V \to W \) be a linear transformation. Then we define \(T^{\sharp}: V \to W \) by $$T^{\sharp} = \frac{1}{|G|} \sum_{g \in G } \varphi_{g^{-1}}T\rho_g$$ It isn’t hard to see that \( T^{\sharp} \in \text{Hom}_G(\rho,\varphi) \) and this is a good exercise in the “change of variables”-trick from the previous article. The following is also true.

**Lemma 2.1:** Using the notation from the paragraph over:

- If \( T \in \text{Hom}_G(\rho,\varphi) \), then \( T^{\sharp} = T \).
- As a consequence, \(\mathcal{P}: \text{Hom}(V,W) \to \text{Hom}_G(\rho,\varphi) \) defined by \( T \mapsto T^{\sharp} \) is a linear map which is onto.

*Proof.* If \( T \in \text{Hom}_G(\rho,\varphi) \), then \( \varphi_g T = T \rho_g \) for all \( g \in G \). Using this we get $$ T^{\sharp} = \frac{1}{|G|} \sum_{g \in G} \varphi_{g^{-1}}T\rho_g = \frac{1}{|G|} \sum_{g \in G} \varphi_{g^{-1}}\varphi_g T = \frac{1}{|G|} \sum_{g \in G} T = T$$ The linearity of \( \mathcal{P} \) is clear and so is the surjectiveness by 1). \( \blacksquare \).

In the case where \( \rho \) and \( \varphi \) are irreducible representations, we can describe \( T^{\sharp} \) very specifically:

**Lemma 2.2:** Let \(\rho \) and \( \varphi \) be as before and also irreducible this time. Furthemore let \( T: V \to W \) be a linear transformation. If \(\rho \not \sim \varphi \) then \( T^{\sharp} = 0 \). If \( \rho = \varphi \) then \( T^{\sharp} = \frac{\text{Tr}(T)}{\deg(\rho)} I \)

*Proof.* By Schur’s lemma \( T^{\sharp} = 0 \) if \( \rho \not \sim \varphi \). If \( \rho = \varphi \) then Schur’s lemma tell us that \( T^{\sharp} = \lambda I \) for some \( \lambda \in \mathbb{C} \). Our goal is hence to find \( \lambda \). We use the well-known concept of trace from linear algebra to do this: $$\lambda\deg(\rho)=\lambda\dim(V)=\lambda \text{Tr}(I)=\text{Tr}(\lambda I)=\text{Tr}(T^{\sharp})$$ What is left to show is that \(\text{Tr}(T)=\text{Tr}(T^{\sharp}) \). By using that \(\text{Tr}(AB)=\text{Tr}(BA)\) we get that $$\begin{alignat*}{2} \text{Tr}(T^{\sharp}) &= \frac{1}{|G|} \sum_{g \in G} \text{Tr}(\rho_{g^{-1}}T\rho_g) \\ &= \frac{1}{|G|} \sum_{g \in G } \text{Tr}(T\rho_g\rho_{g^{-1}}) \\ &= \frac{1}{|G|} \sum_{g \in G } \text{Tr}(T) \\ &= \text{Tr}(T) \end{alignat*}$$ This finishes the proof. \( \blacksquare \)

We are closing in on the orthogonality relations. What is left is just one lemma. Before stating and proving it we write down a formula related to matrix multiplication. We will need this in the following proof of the lemma. The formula follows straight away from the definition of matrix multiplication, and here it is in all its glory: Let \(A \) be a \(r \times m \)-matrix, \(B \) a \(n \times s \)-matrix and \( E_{ki} \) a \( m \times n \)-matrix, where \( E_{ki} \) is the matrix with zeros everywhere except a \( 1 \) at \((k,i) \). Then if \( A = (a_{ij}), B=(b_{ij} \), the following formula holds $$ (AE_{ki}B)_{\ell j} = a_{\ell k}b_{ij} $$ With this scary-looking formula stated, we prove the last ingredient we need for Schur’s orthogonality relations.

**Lemma 2.3:** Let \( \rho:G \to U_n(\mathbb{C}) \) and \(\varphi: G \to U_m(\mathbb{C}) \) be unitary represenations of \( G \). Then \( (E_{ki}^{\sharp})_{\ell j} = \langle \rho_{ij}, \varphi_{k\ell} \rangle \)

*Proof.* Recall from linear algebra that for a unitary operator \( T \), we have \(TT^*=I=T^*T \). Here \( T^* \) denotes the adjoint of \( T \). Since \( \varphi_g \) is unitary we have \( I=\varphi_g\varphi_{g^{-1}} \) and so \( \varphi_{g^{-1}}=\varphi_g^{-1} = \varphi_g^{*} \). This implies that \(\varphi_{g^{-1}})_{\ell k} = \overline{(\varphi_{g})_{k\ell}} \). Using the formula from the paragraph over we get $$ \begin{alignat*}{2} (E_{ki}^{\sharp})_{\ell j} &= \frac{1}{|G|} \sum_{g \in G} (\varphi_{g^{-1}} T \rho_{g})_{\ell j} \\ &= \frac{1}{|G|} \sum_{g \in G} (\varphi_{g^{-1}})_{\ell k}(\rho_g)_{ij} \\ &= \frac{1}{|G|} \sum_{g \in G} \overline{(\varphi_{g})_{k\ell}}(\rho_g)_{ij} \\ &= \langle \rho_{ij}, \varphi_{k\ell} \rangle \end{alignat*}$$ We are done. \( \blacksquare \)

**Theorem 2.4: **(Schur’s orthogonality relations)

Let \( \rho: G \to U_n(\mathbb{C}), \varphi:G \to U_m(\mathbb{C}) \) be inequivalent unitary representations of \( G \). Then \( \langle \rho_{ij},\varphi_{k \ell} \rangle = 0 \). Furthermore $$\langle \rho_{ij},\rho_{k\ell}\rangle = \left\{\begin{matrix}

1/n & \text{ if }i=k,\ell=j \\

0 & \text{otherwise}

\end{matrix}\right.$$

*Proof.* By lemma 2.3, when \(E_{ki} \) is a \( m \times n\)-matrix, \( (E_{ki}^{\sharp})_{\ell j} = \langle \rho_{ij}, \varphi_{k\ell} \rangle \), and by lemma 2.2 \( E_{ki}^{\sharp}=0\), and so \( \langle \rho_{ij},\varphi_{k \ell} \rangle = 0 \). Now, suppose \( E_{ki} \) is a \( n \times n\)-matrix, so a matrix representing a linear transformation \( V \to V \). Then by lemma 2.2 we have $$ E_{ki}^{\sharp} = \frac{\text{Tr}(E_{ki})}{\deg(\rho)}I = \frac{\text{Tr}(E_{ki})}{n}I $$ By lemma 2.3 we then get $$\langle \rho_{ij}, \varphi_{k\ell} \rangle = (E_{ki}^{\sharp})_{\ell j} = \left(\frac{\text{Tr}(E_{ki})}{n}I \right)_{\ell j} = \left\{\begin{matrix}

1/n & \text{ if } i=k, \ell = j \\

0 & \text{otherwise}

\end{matrix}\right.$$ which was what we wanted to prove. \( \blacksquare \)

With this out of the way we finally introduce characters. To a representation \( \rho: G \to GL(V) \) we define the **character** \( \chi_\rho: G \to \mathbb{C} \) by \( g \mapsto \text{Tr}(\rho_g) \). This is well-defined as the trace is independent of basis. From a character we can get a lot of information about the representation. For example evaluating the character in the identity element we see $$\chi_\rho (e) = \text{Tr}(\rho_e) = \text{Tr}(I)=\dim(V)=\deg(\rho)$$ Thanks to \( \text{Tr}(AB)=\text{Tr}(BA) \), characters are preserved by equivalence of representations as the following calculation shows $$ \chi_{\rho}(g) = \text{Tr}(\rho_g) = \text{Tr}(T\varphi_gT^{-1})=\text{Tr}(\varphi_gT^{-1}T) = \chi_{\varphi}(g)$$ Another crucial property of the character is that it is constant on conjugacy classes, and this again follows by our good friend \( \text{Tr}(AB)=\text{Tr}(BA) \): $$\chi_{\rho}(hgh^{-1}) = \text{Tr}(\rho_{hgh^{-1}}) = \text{Tr}(\rho_h\rho_g\rho_{h^{-1}})=\text{Tr}(\rho_{h^{-1}}\rho_h\rho_g) = \text{Tr}(\rho_g)$$

From Schur’s orthogonality relations we get *the first orthogonality relations*. This is a central theorem in the theory of representations of finite groups as it shows that the (pairwise inequivalent) irreducible characters (a character is irreducible when its associated representation is irreducible) form an orthonormal set. It also gives a way to check whether a representation is irreducible or not. We will use the theorem to prove the uniqueness of the decomposition provided in Mashcke’s theorem from the previous article.

**Theorem 2.5: **(First orthogonality relations)

Let \( \rho: G \to GL(V) \), \( \varphi: G \to GL(W) \) be irreducible representations of \( G \). Then $$\langle \chi_\rho, \chi_\varphi \rangle = \left\{\begin{matrix}

1 & \text{ if } \rho \sim \varphi \\

0 & \text{ if } \rho \not\sim \varphi

\end{matrix}\right.$$

*Proof.* Without loss of generalization we can assume, by lemma 1.2 from the previous article, that \( \rho, \varphi\) are unitary representation, say \(\rho:G \to U_n(\mathbb{C}), \varphi:G \to U_m(\mathbb{C}) \). If \( \rho \not \sim \varphi \) then Schur’s orthogonality relations yield that \( \langle \rho_{ii},\varphi_{jj} \rangle = 0 \) for all \(i,j\). If \( \rho \sim \varphi \), then since \( \chi_{\rho} = \chi_{\varphi} \), there is no harm done is assuming that \( \rho = \varphi \). In this case Schur’s orthogonality relations tells us that \( \langle \rho_{ii} , \rho_{jj}\rangle = \frac{1}{n} \) if \(i = j \) and \(0 \) if \( i \neq j \). Now we compute $$\begin{alignat*}{2} \langle \chi_{\rho},\chi_{\varphi} \rangle &= \frac{1}{|G|} \sum_{g \in G} \chi_{\rho}(g)\overline{\chi_{\varphi}(g)} \\ &= \frac{1}{|G|}\sum_{g \in G} \sum_{i=1}^n \left(\rho_{g}\right)_{ii} \sum_{j=1}^m \overline{\left( \varphi_g \right)_{jj}} \\ &= \sum_{i=1}^n \sum_{j=1}^m \langle \rho_{ii},\varphi_{jj} \rangle \end{alignat*} $$ If \( \rho \not \sim \varphi \) then this is \(0 \) by the paragraph over. If \( \rho \sim \varphi \), then we get that the inner product is \( 1\), as desired. \( \blacksquare \)

We end this post by using our new theory to show the uniqueness of the decomposition in Maschke’s theorem. For this we introduce new notation: \(m V \) and \(m\rho \) is defined to be the direct sum of \(m \) copies of \(V \) and \( \rho \) respectively. We only need one lemma before attacking the problem at hand.

**Lemma 2.6:** Let \(\rho: G \to GL_n(\mathbb{C})\) and \(\varphi: G \to GL_m(\mathbb{C})\) be representations of \(G\). Then \(\chi_{\rho \oplus \varphi} = \chi_{\rho} + \chi_{\varphi} \)

*Proof.* We have $$ (\rho \oplus \varphi)_g = \begin{pmatrix} \rho_g & 0 \\ 0 & \varphi_g \end{pmatrix} $$ and so \( \chi_{\rho \oplus \varphi}(g) = \text{Tr}(\rho_g) + \text{Tr}(\varphi_g) = \chi_{\rho}(g)+\chi_{\varphi}(g) \). \(\blacksquare \)

Hence if \( \rho \sim m_1\varphi^{(1)} \oplus \cdots \oplus m_r\varphi^{(r)} \) then by induction on lemma 2.6 and evaluating in the identity element of \(G\) we see that \(\deg(\rho)=m_1\deg(\varphi^{(1)})+\dots+m_r\deg(\varphi^{(r)}) \).

**Theorem 2.7**: Let \(\varphi^{(1)},\dots,\varphi^{(r)} \) be a complete list of representatives of the equivalence classes of irreducible representations of \( G \). Then if \(\rho \) is a representation of \( G\), we have $$ \rho \sim m_1\varphi^{(1)} \oplus \cdots \oplus m_r\varphi^{(r)} $$ where \( m_i = \langle \chi_{\rho}, \chi_{\varphi^{(i)}} \rangle \).

*Proof.* By Mashcke’s theorem we can decompose \( \rho \) like \( \rho \sim m_1\varphi^{(1)} \oplus \cdots \oplus m_r\varphi^{(r)} \) By lemma 2.6 we have \( \chi_{\rho} = m_1\chi_{\varphi^{(1)}} + \dots + m_r\chi_{\varphi^{(r)}} \). By the first orthogonality relations \( \langle \chi_{\varphi^{(i)}}, \chi_{\varphi^{(j)}} \rangle = 1 \) if \(i = j \) and \( 0 \) otherwise. Now $$ \langle \chi_{\rho}, \chi_{\varphi^{(i)}} \rangle = m_1\langle \chi_{\varphi^{(1)}}, \chi_{\varphi^{(i)}} \rangle + \dots + m_r \langle \chi_{\varphi^{(r)}}, \chi_{\varphi^{(i)}} \rangle = m_i$$ which is exactly what we wanted. \( \blacksquare \)

This theorem gives us a very nice way to check whether a representation is irreducible or not.

**Corollary 2.8: **A representation \( \rho: G \to GL(V) \) is irreducible if and only if \( \langle \chi_{\rho},\chi_{\rho} \rangle = 1 \).

*Proof.* By theorem 2.7 we have \( \rho = m_1\varphi^{(1)} \oplus \cdots \oplus m_r\varphi^{(r)} \) for irreducible representations \(\varphi^{(i)} \) of \( G \). By the first orthogonality relations \( \langle \varphi^{(i)},\varphi^{(j)} \rangle = 1 \) if and only if \( i = j \) and \( 0 \) otherwise. Using this we compute the inner product: \(\langle \chi_{\rho}, \chi_{\rho} \rangle = m_1^2+\dots+m_r^2 \). Clearly this is \(1\) if and only if \( m_i = 1 \) for exactly one \( i \), and \( 0 \) for the rest, and then \( \rho \sim \varphi^{(i)} \) – that is \( \rho \) is irreducible. \( \blacksquare \)