Coro. to the Orthogonality Thm. for Chars. of Irreducible Group Reps.

Corollaries to the Orthogonality Theorem for Characters of Irreducible Group Representations

Recall from The Orthogonality Theorem for Characters of Irreducible Group Representations page that if $G$ is a finite group and $V$, $W$ are representations of $G$ then:

• 1. If $V$ and $W$ are irreducible representations that are not isomorphic then:

(1)

\begin{align} \quad \langle \chi_V, \chi_W \rangle = 0 \end{align}

• 2. If $V = W$ is an irreducible representation then:

(2)

\begin{align} \quad \langle \chi_V, \chi_V \rangle = 1 \end{align}

We will now look at some corollaries to the above theorem.

Corollary 1: Let $G$ be a finite group and let $W$ be a group representation of $G$. If $W$ has a decomposition $W = \bigoplus_{i=1}^{m} V_k^{\oplus n_k}$ into irreducibles $V_1$, $V_2$, …, $V_m$, and where $n_1$, $n_2$, …, $n_m$ denote the multiplicities of each irreducible, then $n_k = \langle \chi_{W}, \chi_{V_k} \rangle$ for each $1 \leq k \leq m$.

• Proof: Let $W = \bigoplus_{i=1}^{m} V_k^{\oplus n_k}$. Then the character of $W$ is given by:

(3)

\begin{align} \quad \chi_W = \sum_{i=1}^{m} n_i \chi_{V_i} \end{align}

• Thus for each $1 \leq k \leq n$ we have that:

(4)

\begin{align} \quad \langle \chi_W, \chi_{V_k} \rangle &= \left \langle \sum_{i=1}^{m} n_i \chi_{V_i}, \chi_{V_k} \right \rangle \\ &= \sum_{i=1}^{m} n_i \langle \chi_{V_i}, \chi_{V_k} \rangle \\ &= n_k \langle \chi_{V_k} ,\chi_{V_k} \rangle \\ &= n_k \end{align}

Corollary 2: Let $G$ be a finite group and let $V$ and $W$ be representations of $G$. Then $V$ and $W$ are isomorphic representations if and only if the characters $\chi_V$ and $\chi_W$ are the same.

Corollary 3: Let $G$ be a finite group and let $V$ be a representation of $G$. Then $V$ is irreducible if and only if $\langle \chi_V, \chi_V \rangle = 1$.

• Proof: $\Rightarrow$ Suppose that $V$ is irreducible. Then by the above theorem $\langle \chi_V, \chi_V \rangle = 1$.

• $\Leftarrow$ Suppose that $\langle \chi_V, \chi_V \rangle = 1$. Then write:

(5)

\begin{align} \quad \chi_V = \sum_{i=1}^{m} n_i \chi_{V_i} \end{align}

• So:

(6)

\begin{align} \quad 1 &= \langle \chi_V, \chi_V \rangle \\ &= \left \langle \sum_{i=1}^{m} n_i \chi_{V_i}, \sum_{i=1}^{m} n_i \chi_{V_i} \right \rangle \\ &= \sum_{i=1}^{m} n_i^2 \langle \chi_{V_i}, \chi_{V_i} \rangle \\ &= \sum_{i=1}^{m} n_i^2 \end{align}

• But each of the $n_i$s are nonnegative integers and thus there exists a $1 \leq k \leq m$ such that $n_k = 1$. So $V = V_k$, that is, $V$ is irreducible. $\blacksquare$