Character of a representation; two remarks

Posted: March 24, 2011 in Characters of Finite Groups, Representations of Finite Groups
Tags: ,

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.

1. drexel28 says:

Yaghoub,

This is Alex Youcis, just out of curiosity what book are you using for rep. theory?

• Yaghoub says:

Hi Alex
Nothing yet.

• drexel28 says:

Well, if you happen to find any hidden gems, I would appreciate if you’d let me know 🙂

Good luck with your bloggings, I enjoy them and will enjoy seeings your take on rep. theory.

• Yaghoub says:

Well, I have just started writing these stuff up and, yes, there are very nice stuff in character theory of finite groups that I will definitely get to them in details. And, I’m glad you like my blog.