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:

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

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

\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:

\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:

\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:

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

- So:

\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. $\blacksquare$