Orthonormality of irreducible characters

Define an inner product on $\mathrm{Cl}(G;\mathbb C)$ by

Equivalently, if the sum is taken over conjugacy classes $C\subseteq G$,

since $f,g$ are constant on each class.

For a (finite-dimensional complex) representation $V$ with character $\chi_V(g)=\mathrm{tr}(V(g))$ (using trace), the inner product $\langle \chi_V,\chi_W\rangle$ measures overlap between $V$ and $W$.

Let $\chi,\psi$ be irreducible characters of $G$ (over $\mathbb C$). Then

(i.e. $1$ if $\chi=\psi$ and $0$ otherwise).

This is often presented as the “character orthogonality relations”; see character orthogonality.

1. Multiplicity formula. If $V$ is a complex representation with character $\chi_V$ and $\chi_i$ is irreducible, then the multiplicity $m_i$ of the corresponding irreducible representation in $V$ is
   $$m_i \;=\; \langle \chi_V,\chi_i\rangle \in \mathbb Z_{\ge 0}.$$
   This uses Maschke's theorem / complete reducibility over $\mathbb C$.

1. Orthonormal basis of class functions (over $\mathbb C$). The irreducible characters form an orthonormal basis of $\mathrm{Cl}(G;\mathbb C)$. In particular, every class function $f$ has a unique expansion
   $$f \;=\; \sum_i \langle f,\chi_i\rangle\, \chi_i.$$
   The spanning/basis part is tied to the number of irreducibles equals the number of conjugacy classes.

1. Character tables as unitary matrices (after normalization). Writing the character table with rows $\chi_i$ and columns indexed by conjugacy classes, orthonormality implies the rows are orthonormal with respect to the weights $|C|/|G|$. (There is also a “column orthogonality” relation, equivalent to the same set of facts.)

Example 1: Cyclic group $C_n$

Let $G=C_n=\langle a\rangle$ with $|G|=n$, and fix $\zeta=e^{2\pi i/n}$. The irreducible characters are 1-dimensional:

Then

since the sum is a geometric series.

Example 2: $S_3$

The group $S_3$ has three conjugacy classes: $1$, transpositions, and 3-cycles, with sizes $1,3,2$. Its irreducible characters are:

| class | size | representative | $\chi_{\mathrm{triv}}$ | $\chi_{\mathrm{sgn}}$ | $\chi_{\mathrm{std}}$ | |---:|---:|:---:|---:|---:|---:| | $C_1$ | 1 | $e$ | 1 | 1 | 2 | | $C_2$ | 3 | $(12)$ | 1 | $-1$ | 0 | | $C_3$ | 2 | $(123)$ | 1 | 1 | $-1$ |

Check orthonormality using $\langle \chi,\psi\rangle=\sum_C \frac{|C|}{6}\chi(C)\overline{\psi(C)}$:

• \(\langle \chi_{\mathrm{triv}},\chi_{\mathrm{sgn}}\rangle

=\frac16(1\cdot 1\cdot 1 + 3\cdot 1\cdot (-1) + 2\cdot 1\cdot 1)=0.\)

• \(\langle \chi_{\mathrm{std}},\chi_{\mathrm{std}}\rangle

=\frac16(1\cdot 2^2 + 3\cdot 0^2 + 2\cdot (-1)^2)=\frac16(4+0+2)=1.\)

• \(\langle \chi_{\mathrm{std}},\chi_{\mathrm{triv}}\rangle

=\frac16(1\cdot 2\cdot 1 + 3\cdot 0\cdot 1 + 2\cdot (-1)\cdot 1)=0.\)

Thus the three irreducible characters are orthonormal.

Example 3: Dihedral group $D_4$ of order $8$

Let $D_4=\langle r,s \mid r^4=s^2=1,\; srs=r^{-1}\rangle$. Its conjugacy classes can be taken as $\{1\}, \{r^2\}, \{r,r^3\}, \{s,r^2s\}, \{rs,r^3s\}$ with sizes $1,1,2,2,2$.

The unique 2-dimensional irreducible character $\chi_{2}$ has values

(and hence $0$ on both reflection classes).

Then

Also, against the trivial character $\mathbf{1}$,

so $\chi_2$ is orthogonal to the 1-dimensional characters, as orthonormality predicts.