Let G be a finite group, and let Cl(G;C) denote the C-vector space of complex-valued class functions
on G.
The standard inner product on class functions
Define an inner product
on Cl(G;C) by
⟨f,g⟩:=∣G∣1x∈G∑f(x)g(x).Equivalently, if the sum is taken over conjugacy classes
C⊆G,
⟨f,g⟩=C∑∣G∣∣C∣f(C)g(C),since f,g are constant on each class.
For a (finite-dimensional complex) representation
V with character
χV(g)=tr(V(g)) (using trace
), the inner product ⟨χV,χW⟩ measures overlap between V and W.
Orthonormality statement
Let χ,ψ be irreducible characters
of G (over C). Then
⟨χ,ψ⟩=δχ,ψ(i.e. 1 if χ=ψ and 0 otherwise).
This is often presented as the “character orthogonality relations”; see character orthogonality
.
Consequences
Multiplicity formula.
If V is a complex representation with character χV and χi is irreducible, then the multiplicity mi of the corresponding irreducible representation in V is
mi=⟨χV,χi⟩∈Z≥0.This uses Maschke's theorem
/ complete reducibility
over C.
Orthonormal basis of class functions (over C).
The irreducible characters form an orthonormal basis of Cl(G;C). In particular, every class function f has a unique expansion
f=i∑⟨f,χi⟩χi.The spanning/basis part is tied to the number of irreducibles equals the number of conjugacy classes
.
Character tables as unitary matrices (after normalization).
Writing the character table with rows χi 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.)
Examples
Example 1: Cyclic group Cn
Let G=Cn=⟨a⟩ with ∣G∣=n, and fix ζ=e2πi/n. The irreducible characters are 1-dimensional:
χk(am)=ζkm(k=0,1,…,n−1).Then
⟨χk,χℓ⟩=n1m=0∑n−1ζkmζℓm=n1m=0∑n−1ζ(k−ℓ)m={1,0,k=ℓ,k=ℓ,since the sum is a geometric series.
Example 2: S3
The group S3 has three conjugacy classes: 1, transpositions, and 3-cycles, with sizes 1,3,2. Its irreducible characters are:
| class | size | representative | χtriv | χsgn | χstd |
|---|
| C1 | 1 | e | 1 | 1 | 2 |
| C2 | 3 | (12) | 1 | −1 | 0 |
| C3 | 2 | (123) | 1 | 1 | −1 |
Check orthonormality using
⟨χ,ψ⟩=∑C6∣C∣χ(C)ψ(C):
⟨χtriv,χsgn⟩=61(1⋅1⋅1+3⋅1⋅(−1)+2⋅1⋅1)=0.
⟨χstd,χstd⟩=61(1⋅22+3⋅02+2⋅(−1)2)=61(4+0+2)=1.
⟨χstd,χtriv⟩=61(1⋅2⋅1+3⋅0⋅1+2⋅(−1)⋅1)=0.
Thus the three irreducible characters are orthonormal.
Example 3: Dihedral group D4 of order 8
Let D4=⟨r,s∣r4=s2=1,srs=r−1⟩. Its conjugacy classes can be taken as
{1},{r2},{r,r3},{s,r2s},{rs,r3s}
with sizes 1,1,2,2,2.
The unique 2-dimensional irreducible character χ2 has values
χ2(1)=2,χ2(r2)=−2,χ2(r)=χ2(r3)=0,χ2(s)=χ2(rs)=0(and hence 0 on both reflection classes).
Then
⟨χ2,χ2⟩=81(1⋅22+1⋅(−2)2+2⋅02+2⋅02+2⋅02)=81(4+4)=1.Also, against the trivial character 1,
⟨χ2,1⟩=81(1⋅2⋅1+1⋅(−2)⋅1+0)=0,so χ2 is orthogonal to the 1-dimensional characters, as orthonormality predicts.