Character orthogonality

On the space $\mathrm{Cl}(G)$ of complex class functions, define

This is a Hermitian inner product (compare inner product and orthogonality).

Let $\chi_1,\dots,\chi_r$ be the distinct irreducible characters of $G$ over $\mathbb C$. Then:

1. Row orthogonality (orthonormality):
   $$\langle \chi_i, \chi_j\rangle = \delta_{ij}.$$
   In particular, for any character $\chi$,
   $$\langle \chi,\chi\rangle \in \mathbb Z_{\ge 0},$$
   and $\chi$ is irreducible iff $\langle \chi,\chi\rangle=1$ (see also character orthonormality).

1. Completeness: The set $\{\chi_1,\dots,\chi_r\}$ is an orthonormal basis of $\mathrm{Cl}(G)$. Hence $r=\dim_\mathbb C \mathrm{Cl}(G)$, which equals the number of conjugacy classes (cf. number of irreducibles equals number of conjugacy classes).

1. Column orthogonality (one common form): for $g,h\in G$,
   $$\sum_{i=1}^r \chi_i(g)\,\overline{\chi_i(h)} = \begin{cases} 0, & \text{if } g \text{ and } h \text{ are not conjugate},\\[4pt] |C_G(g)|, & \text{if } g \text{ and } h \text{ are conjugate}, \end{cases}$$
   where $C_G(g)$ is the centralizer of $g$. In particular,
   $$\sum_{i=1}^r |\chi_i(g)|^2 = |C_G(g)|.$$

These identities are proved using complete reducibility (via Maschke's theorem), the decomposition of tensor products, and Schur's lemma.

1. Cyclic group $C_n$: discrete Fourier orthogonality. Let $C_n=\langle t\rangle$. Its irreducible characters are $\chi_j(t^m)=\zeta_n^{jm}$ for $j=0,\dots,n-1$. Then
   $$\langle \chi_j,\chi_\ell\rangle = \frac{1}{n}\sum_{m=0}^{n-1} \zeta_n^{jm}\overline{\zeta_n^{\ell m}} = \frac{1}{n}\sum_{m=0}^{n-1} \zeta_n^{(j-\ell)m} = \delta_{j\ell}.$$

1. $S_3$: checking row orthogonality from the character table. $S_3$ has three conjugacy classes: $()$, transpositions, and 3-cycles. Let $\chi_{\mathrm{triv}},\chi_{\mathrm{sgn}},\chi_{\mathrm{std}}$ be the irreducible characters (degrees $1,1,2$). Using the class sizes $1,3,2$ and values
   $$\chi_{\mathrm{triv}}=(1,1,1),\quad \chi_{\mathrm{sgn}}=(1,-1,1),\quad \chi_{\mathrm{std}}=(2,0,-1),$$
   one computes for example
   $$\langle \chi_{\mathrm{std}},\chi_{\mathrm{sgn}}\rangle =\frac{1}{6}\Big(1\cdot 2\cdot 1 + 3\cdot 0\cdot(-1) + 2\cdot (-1)\cdot 1\Big)=0,$$
   and
   $$\langle \chi_{\mathrm{std}},\chi_{\mathrm{std}}\rangle =\frac{1}{6}\Big(1\cdot |2|^2 + 3\cdot |0|^2 + 2\cdot |-1|^2\Big)=1,$$
   confirming orthogonality and irreducibility.

1. Column orthogonality in $S_3$: centralizer sizes. In $S_3$, a transposition $\tau$ has centralizer size $|C_{S_3}(\tau)|=2$. Column orthogonality predicts
   $$|\chi_{\mathrm{triv}}(\tau)|^2 + |\chi_{\mathrm{sgn}}(\tau)|^2 + |\chi_{\mathrm{std}}(\tau)|^2 = |1|^2 + |-1|^2 + |0|^2 = 2,$$
   matching $|C_{S_3}(\tau)|$.