Orthonormality of irreducible characters

With respect to the standard inner product on class functions, irreducible characters are orthonormal (and over ℂ they form an orthonormal basis).
Orthonormality of irreducible characters

Let GG be a finite group, and let Cl(G;C)\mathrm{Cl}(G;\mathbb C) denote the C\mathbb C-vector space of complex-valued on GG.

The standard inner product on class functions

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

f,g  :=  1GxGf(x)g(x). \langle f,g\rangle \;:=\; \frac{1}{|G|}\sum_{x\in G} f(x)\,\overline{g(x)}.

Equivalently, if the sum is taken over CGC\subseteq G,

f,g  =  CCGf(C)g(C), \langle f,g\rangle \;=\; \sum_{C} \frac{|C|}{|G|}\, f(C)\,\overline{g(C)},

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

For a (finite-dimensional complex) VV with χV(g)=tr(V(g))\chi_V(g)=\mathrm{tr}(V(g)) (using ), the inner product χV,χW\langle \chi_V,\chi_W\rangle measures overlap between VV and WW.

Orthonormality statement

Let χ,ψ\chi,\psi be of GG (over C\mathbb C). Then

χ,ψ  =  δχ,ψ \langle \chi,\psi\rangle \;=\; \delta_{\chi,\psi}

(i.e. 11 if χ=ψ\chi=\psi and 00 otherwise).

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

Consequences

  1. Multiplicity formula.
    If VV is a complex representation with character χV\chi_V and χi\chi_i is irreducible, then the multiplicity mim_i of the corresponding irreducible representation in VV is

    mi  =  χV,χiZ0. m_i \;=\; \langle \chi_V,\chi_i\rangle \in \mathbb Z_{\ge 0}.

    This uses / over C\mathbb C.

  2. Orthonormal basis of class functions (over C\mathbb C).
    The irreducible characters form an orthonormal basis of Cl(G;C)\mathrm{Cl}(G;\mathbb C). In particular, every class function ff has a unique expansion

    f  =  if,χiχi. f \;=\; \sum_i \langle f,\chi_i\rangle\, \chi_i.

    The spanning/basis part is tied to .

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

Examples

Example 1: Cyclic group CnC_n

Let G=Cn=aG=C_n=\langle a\rangle with G=n|G|=n, and fix ζ=e2πi/n\zeta=e^{2\pi i/n}. The irreducible characters are 1-dimensional:

χk(am)=ζkm(k=0,1,,n1). \chi_k(a^m)=\zeta^{km}\qquad (k=0,1,\dots,n-1).

Then

χk,χ=1nm=0n1ζkmζm=1nm=0n1ζ(k)m={1,k=,0,k, \langle \chi_k,\chi_\ell\rangle =\frac1n\sum_{m=0}^{n-1}\zeta^{km}\overline{\zeta^{\ell m}} =\frac1n\sum_{m=0}^{n-1}\zeta^{(k-\ell)m} =\begin{cases} 1,&k=\ell,\\ 0,&k\ne \ell, \end{cases}

since the sum is a geometric series.

Example 2: S3S_3

The group S3S_3 has three conjugacy classes: 11, transpositions, and 3-cycles, with sizes 1,3,21,3,2. Its irreducible characters are:

classsizerepresentativeχtriv\chi_{\mathrm{triv}}χsgn\chi_{\mathrm{sgn}}χstd\chi_{\mathrm{std}}
C1C_11ee112
C2C_23(12)(12)11-10
C3C_32(123)(123)111-1

Check orthonormality using χ,ψ=CC6χ(C)ψ(C)\langle \chi,\psi\rangle=\sum_C \frac{|C|}{6}\chi(C)\overline{\psi(C)}:

  • χtriv,χsgn=16(111+31(1)+211)=0.\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.

  • χstd,χstd=16(122+302+2(1)2)=16(4+0+2)=1.\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.

  • χstd,χtriv=16(121+301+2(1)1)=0.\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 D4D_4 of order 88

Let D4=r,sr4=s2=1,  srs=r1D_4=\langle r,s \mid r^4=s^2=1,\; srs=r^{-1}\rangle. Its conjugacy classes can be taken as {1},{r2},{r,r3},{s,r2s},{rs,r3s}\{1\}, \{r^2\}, \{r,r^3\}, \{s,r^2s\}, \{rs,r^3s\} with sizes 1,1,2,2,21,1,2,2,2.

The unique 2-dimensional irreducible character χ2\chi_{2} has values

χ2(1)=2,χ2(r2)=2,χ2(r)=χ2(r3)=0,χ2(s)=χ2(rs)=0 \chi_{2}(1)=2,\quad \chi_{2}(r^2)=-2,\quad \chi_{2}(r)=\chi_{2}(r^3)=0,\quad \chi_{2}(s)=\chi_{2}(rs)=0

(and hence 00 on both reflection classes).

Then

χ2,χ2=18(122+1(2)2+202+202+202)=18(4+4)=1. \langle \chi_{2},\chi_{2}\rangle =\frac18\Big(1\cdot 2^2 + 1\cdot (-2)^2 + 2\cdot 0^2 + 2\cdot 0^2 + 2\cdot 0^2\Big) =\frac18(4+4)=1.

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

χ2,1=18(121+1(2)1+0)=0, \langle \chi_{2},\mathbf{1}\rangle =\frac18\Big(1\cdot 2\cdot 1 + 1\cdot (-2)\cdot 1 + 0\Big)=0,

so χ2\chi_2 is orthogonal to the 1-dimensional characters, as orthonormality predicts.