Posts Tagged ‘class function’

Throughout G is a finite group. Recall that we defined the degree of a character \chi to be the degree of the representation which affords \chi.

Remark 1. Let \rho : G \longrightarrow \text{GL}(V) be a representation of G and let \chi : G \longrightarrow \mathbb{C} be its character. Then

i) \chi(1)=\deg \chi.

ii) \chi(g_1g_2)=\chi(g_2g_1), for all g_1,g_2 \in G.

iii) \chi(x)=\chi(gxg^{-1}), for all g,x \in G.

Proof. i) Let \deg \chi = \deg \rho=\dim V = n. By definition, \chi(1) = \text{Tr}(\rho(1))=\text{Tr}(\text{id}_V). But the matrix of \text{id}_V is the  n \times n identity matrix and so its trace is n. Thus \chi(1)=n=\deg \chi.

ii) Recall from linear algebra that for any n \times n matrices A and B we have \text{Tr}(AB)=\text{Tr}(BA). Thus

\text{Tr}(\rho(g_1g_2))=\text{Tr}(\rho(g_1) \rho(g_2))=\text{Tr}(\rho(g_2) \rho(g_1))=\text{Tr}(\rho(g_2g_1)).

Hence \chi(g_1g_2)=\chi(g_2g_1).

iii) By ii), \chi(gxg^{-1})=\chi(g^{-1}gx)=\chi(x). \ \Box

Remark 2. Let \rho_i: G \longrightarrow \text{GL}(V_i), \ i=1,2, be two representations of G. If \rho_1 and \rho_2 are equivalent, then \chi_{\rho_1}=\chi_{\rho_2}.

Proof. By (*) in here, there exists a \mathbb{C}[G]-module isomorphism \varphi : V_1 \longrightarrow V_2 such that \rho_2(g)=\varphi \rho_1(g) \varphi^{-1}, for all g \in G. Choose some \mathbb{C}-basis for V_1 and V_2 and look at \rho_1(g), \rho_2(g) and \varphi as matrices. Then, since \text{Tr}(AB)=\text{Tr}(BA) for any n \times n matrices A and B, we have

\text{Tr}(\rho_2(g))=\text{Tr}(\varphi \rho_1(g) \varphi^{-1})=\text{Tr}(\varphi^{-1} \varphi \rho_1(g))=\text{Tr}(\rho_1(g)).

Therefore \chi_{\rho_2}(g)=\chi_{\rho_1}(g), for all g \in G and so \chi_{\rho_2}=\chi_{\rho_1}. \ \Box

The converse of the statement in Remark 2 is also true, i.e. if \chi_{\rho_1}=\chi_{\rho_2}, then \rho_1 and \rho_2 are equivalent. We will prove this later.