The Orthogonality Theorem for Characters of Irreducible Group Reps.

The Orthogonality Theorem for Characters of Irreducible Group Representations

 Definition: Let $G$ be a group and let $\varphi, \psi : G \to \mathbb{C}$. The Inner Product of $\varphi$ and $\psi$ will be defined as $\displaystyle{\langle \varphi, \psi \rangle = \frac{1}{|G|} \sum_{g \in G} \overline{\varphi(g)} \psi(g)}$.

If $G$ is a group and $\varphi : G \to \mathbb{C}$ has the property that for all $g \in G$ and for all $h \in H$ that:

(1)
\begin{align} \quad \varphi(g) = \varphi(hgh^{-1}) \end{align}

Then $\varphi$ is said to be invariant under conjugation. If $\varphi$ and $\psi$ are both invariant under conjugation, then the inner product of $\varphi$ and $\psi$ is given by:

(2)
\begin{align} \quad \langle \varphi, \psi \rangle = \frac{1}{|G|} \sum_{C \subseteq G} |C| \overline{\varphi(C)} \psi(C) \end{align}

Where the sum above is over all conjugacy classes $C \subseteq G$ and where $\overline{\varphi(C)} := \varphi(c)$ and $\psi(C) = \psi(c)$ where $c \in C$.

 Lemma 1:Let $G$ be a group and let $(V, \rho_V)$ and $(W, \rho_W)$ be representations of $G$. Let $\phi : V \to w$ be a linear map and let $\phi_0 : V \to W$ be defined for all $v \in V$ by $\displaystyle{\phi_0(v) = \frac{1}{|G|} \sum_{g \in G} \rho_W(g) \circ \phi \circ \rho_V(g)(v)}$. Then: a) $\phi_0$ is a $G$-equivariant map. b) If $V$ and $W$ are irreducible representations of $G$ that are not isomorphic, then $\phi_0 = 0$. c) If $(V, \rho_V) = (W, \rho_W)$ are irreducible representations of $G$ then $\phi_0 = \frac{\mathrm{trace}(\phi)}{\mathrm{dim}(V))} \cdot \mathbf{id}_V$.

The lemma above will be used in order to prove the following major theorem:

 Theorem 2 (The Orthogonality Theorem): Let $G$ be a group. a) If $V$ and $W$ are irreducible group representations of $G$ that are non-isomorphic, then $\langle \chi_V, \chi_W \rangle = 0$. b) If $V$ is an irreducible group representation of $G$ then $\langle \chi_V, \chi_V \rangle = 1$.
• Proof of a): Equip $V$ and $W$ with inner products and let $\{ v_1, v_2, ..., v_m \}$ be an orthonormal basis for $V$ and let $\{ w_1, w_2, ..., w_n \}$ be an orthonormal basis for $W$. Let $\varphi : V \to V$ be a linear operator. The trace of $\varphi$ is then:
(3)
\begin{align} \quad \mathrm{trace} (\varphi) = \sum_{i=1}^{m} \langle v_i, \varphi(v_i) \rangle \end{align}
• Look at the inner product of the characters $\chi_W$ and $\chi_V$:
(4)
\begin{align} \quad \langle \chi_W, \chi_V \rangle &= \frac{1}{|G|} \sum_{g \in G} \overline{\chi_W(g)} \chi_V(g) \\ &= \frac{1}{|G|} \sum_{g \in G} \chi_W(g^{-1}) \chi_W(g) \\ &= \frac{1}{|G|} \sum_{g \in G} \mathrm{trace}_W(g^{-1}) \mathrm{trace}_V(g) \\ &= \frac{1}{|G|} \sum_{g \in G} \sum_{i=1}^{m} \sum_{j=1}^{n} \langle w_j, g^{-1} \cdot w_j \rangle \langle v_i, g \cdot v_i \rangle \\ &= \frac{1}{|G|} \sum_{g \in G} \sum_{i=1}^{m} \sum_{j=1}^{n} \langle w_j, \overline{\langle v_i, g \cdot v_i \rangle} g^{-1} \cdot w_j \rangle \\ &= \frac{1}{|G|} \sum_{g \in G} \sum_{i=1}^{m} \sum_{j=1}^{n} \langle w_j, \langle g \cdot v_i, v_i \rangle g^{-1} \cdot w_j \rangle \quad (*) \end{align}
• Now, since $W \otimes V^* \cong \mathrm{Hom}(V, W)$, we see that if $w \in W$ and $f \in V^*$, $f : V \to \mathbb{C}$ then $w \otimes f$ can be viewed as a linear operator from $V$ to $W$ defined for all $v \in V$ by:
(5)
\begin{align} \quad (w \otimes f)(v) = f(v) w \end{align}
• For each $v_0 \in V$, we can let $v_0^* : V \to \mathbb{C}$ be defined for all $v \in V$ by:
(6)
\begin{align} \quad v_0^*(v) = \langle v, v_0 \rangle \end{align}
• So for each $w \in W$ and each $v_0 \in V$ we have that $w \otimes v_0^*$ can be viewed as a linear operator from $V$ to $W$ defined for all $v \in V$ by:
(7)
\begin{align} \quad (w \otimes v_0^*)(v) = v_0^*(v)w = \langle v, v_0 \rangle w \end{align}
• For each $1 \leq i \leq m$ and for each $1 \leq j \leq n$, consider the linear operator $w_j \otimes v_i^* : V \to W$. For each $g \in G$ and for each $v \in V$ we see that:
(8)
\begin{align} \quad g^{-1} \circ (w_j \otimes v_i^*) \circ g(v) &= g^{-1} [\circ \langle g \cdot v, v_i \rangle w_j] \\ &= \langle g \cdot v, v_i \rangle g^{-1} \cdot w_j \end{align}
• So in the sum at $(*)$ we have that:
(9)
\begin{align} \quad \langle \chi_W, \chi_V \rangle &= \frac{1}{|G|} \sum_{g \in G} \sum_{i=1}^{m} \sum_{j=1}^{n} \langle w_j, \langle g \cdot v_i, v_i \rangle g^{-1} \cdot w_j \rangle \\ &= \frac{1}{|G|} \sum_{g \in G} \sum_{i=1}^{m} \sum_{j=1}^{n} \langle w_j, g^{-1} \circ (w_j \otimes v_i^*) \circ g(v_i) \rangle \end{align}
• If $V \not \cong W$ then by the above lemma the above sum is zero. $\blacksquare$
• Proof of b) If $V = W$ then we have by the above lemma that:
(10)
\begin{align} \quad \langle \chi_V, \chi_V \rangle &= \frac{1}{|G|} \sum_{g \in G} \sum_{i=1}^{m} \sum_{j=1}^{m} \langle v_j, g^{-1} \circ (v_j \otimes v_i^*) \circ g(v_i) \rangle \\ &= \sum_{i=1}^{m} \sum_{j=1}^{m} \left \langle v_j, \frac{1}{|G|} \sum_{g \in G} g^{-1} \circ (v_j \otimes v_i^*) \circ g(v_i) \right \rangle \\ &= \sum_{i=1}^{m} \sum_{j=1}^{m} \left \langle v_j, \frac{\mathrm{trace}(v_j \otimes v_i^*)}{\mathrm{dim}(V)} \mathrm{id}_V(v_i) \right \rangle \\ &= \sum_{i=1}^{m} \sum_{j=1}^{m} \left \langle v_j, \left\{\begin{matrix} \frac{1}{\dim (V)} & \mathrm{if} \: i = j \\ 0 & \mathrm{if} \: i \neq j \end{matrix}\right. v_j \right \rangle \\ &= \sum_{i=1}^{m} \left \langle v_i, \frac{1}{\dim (V)} v_i \right \rangle \\ &= \frac{1}{\dim (V)} \sum_{i=1}^{m} \langle v_i, v_i \rangle \\ &= \frac{1}{\dim (V)}\cdot \mathrm{dim}(V) \\ &= 1 \quad \blacksquare \end{align}