## Character of a representation; basic definitions and remarks

Posted: February 23, 2011 in Characters of Finite Groups, Representations of Finite Groups
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Throughout $G$ is a finite group.

Definition. Let $\rho : G \longrightarrow \text{GL}(V)$ be a representation of $G.$ The map $\chi : G \longrightarrow \mathbb{C}$ defined by $\chi(g) = \text{Tr}(\rho(g)), \ g \in G,$ is called the character of $\rho$ and is denoted by $\chi_{\rho}.$ We say that $\chi_{\rho}$ is an irreducible character if $\rho$ is irreducible. We also define the degree of $\chi_{\rho}$ to be $\deg \rho.$

Note 1. If $\rho$ is understood, we will just write $\chi$ instead of $\chi_{\rho}.$

Note 2. The term $"\rho$ (or $V$) affords $\chi"$ is also used.

Remark 1. In the above definition, $\text{Tr}(\rho(g))$ menas the trace of the linear transformation $\rho(g): V \longrightarrow V$ with respect to a basis of $V.$ To be precise, $\chi(g)$ is the trace of the matrix of $\rho(g)$ with respect to a basis of $V.$ Note that, since the trace of similar matrices are equal, $\chi(g)$ does not depend on the basis we choose for $V$ and so our definition makes sense.

Remark 2. If $\deg \rho = 1,$ then clearly $\rho(g)=\chi_{\rho}(g)$ and so $\chi_{\rho}(g) \neq 0,$ for all $g \in G.$

Remark 3. Recall from linear algebra that the trace of a linear transformation is the sum of its eigenvalues. So $\chi_{\rho}(g)$ is the sum of eigenvalues of $\rho(g).$

Theorem. If $|G|=n, \ \deg \rho = m$ and $g \in G,$ then $\chi_{\rho}(g)=\sum_{i=1}^m \lambda_i,$ where $\lambda_i$ are (not necessarily distinct) $n$-th roots of unity.

Proof. By Remark 3, we only need to show that every eigenvalue of $\rho(g)$ is an $n$-th root of unity. Since $\deg \rho = m,$ the degree of the characteristic polynomial of $\rho(g)$ is $m$ and thus this polynomial has $m$ roots, which are not necessarily distinct. Now if $\lambda$ is an eigenvalue of $\rho(g),$ then $\rho(g)(v)=\lambda v,$ for some $0 \neq v \in V.$ Since $|G|=n,$ we have $g^n=1$ and hence $v = \rho(g^n)(v)=(\rho(g))^n(v)=\lambda^n v.$  Thus $(\lambda^n - 1)v=0$ and hence $\lambda^n=1$ because $v \neq 0.$ So each eigenvalue of $\rho(g)$ is an $n$-th root of unity. $\Box$

Corollary 1. $\chi(g)$ is an algebraic integer for all $g \in G.$

Proof. By the theorem, $\chi(g)$ is a sum of some roots of unity. Obviously any root of unity is an algebraic integer and we know that the set of algebraic integers is a ring and so it is closed under addition. $\Box$

Corollary 2. $\chi(g^{-1})=\overline{\chi(g)},$ where $\overline{\chi(g)}$ is the complex conjugate of $\chi(g).$

Proof.  Let $\rho$ be a representation which affords $\chi.$ Suppose that $\deg \rho = n.$ By the theorem, $\chi(g)=\sum_{i=1}^n \lambda_i,$ where $\lambda_i$ are the eigenvalues of $\rho(g)$ and they are all roots of unity. Since $\rho(g^{-1})=(\rho(g))^{-1},$ the eigenvalues of $\rho(g^{-1})$ are $\lambda_i^{-1}.$ Also, $\lambda_i^{-1}=\overline{\lambda_i}$ because $\lambda_i$ are roots of unity. Thus

$\chi(g^{-1})=\sum_{i=1}^n \lambda_i^{-1}=\sum_{i=1}^n \overline{\lambda_i}=\overline{\sum_{i=1}^n \lambda_i}=\overline{\chi(g)}.$  $\Box$