Posts Tagged ‘algebraic integer’

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