Questions: Orthogonality Relations

Question 1

Using the orthogonality relations, how do you compute the multiplicity of an irreducible representation Vᵢ in a representation V?

Model answer: The multiplicity is nᵢ = ⟨χ_V, χᵢ⟩ = (1/|G|) Σ_{g∈G} χ_V(g) conjugate(χᵢ(g)). Since χ_V = Σⱼ nⱼχⱼ and the χⱼ are orthonormal, the inner product extracts the coefficient nᵢ.

This is the Fourier-analytic viewpoint: just as a function's Fourier coefficient is computed by an inner product with a basis function, the multiplicity of an irreducible in a representation is computed by inner product with the corresponding character. This reduces decomposition from an algebraic problem to a numerical one.

Question 2

The column orthogonality relations state that Σᵢ χᵢ(g) conjugate(χᵢ(h)) = |C_G(g)| if g and h are conjugate, and 0 otherwise. What is C_G(g)?

• A. The center of G
• B. The centralizer of g — the set of elements commuting with g
• C. The conjugacy class of g
• D. The commutator subgroup of G

Answer: (B)

C_G(g) = {h ∈ G : hg = gh} is the centralizer of g. Its order relates to the conjugacy class size by |C_G(g)| = |G|/|Cl(g)|. The column orthogonality relations are dual to the row relations and express orthogonality across different conjugacy classes rather than across different representations. Both sets of relations follow from Schur's lemma applied to specific intertwining operators.

Question 3

If a character χ satisfies ⟨χ, χ⟩ = 1, then the corresponding representation is irreducible.

• T. True
• F. False

Answer: True

Write χ = Σ nᵢχᵢ where the χᵢ are irreducible characters. Then ⟨χ, χ⟩ = Σ nᵢ² by orthonormality. The equation Σ nᵢ² = 1 with non-negative integers nᵢ forces exactly one nᵢ = 1 and all others zero. So χ = χₖ for some k, meaning the representation is irreducible. This gives a quick test for irreducibility.

Question 4

The inner product ⟨χ, ψ⟩ = (1/|G|) Σ_{g∈G} χ(g)ψ(g) (without conjugation) works for characters over ℂ.

• T. True
• F. False

Answer: False

Complex conjugation is essential. The correct formula is ⟨χ, ψ⟩ = (1/|G|) Σ χ(g) conjugate(ψ(g)). For representations of finite groups over ℂ, characters satisfy χ(g⁻¹) = conjugate(χ(g)) (since eigenvalues of ρ(g) are roots of unity), so the inner product can also be written (1/|G|) Σ χ(g)ψ(g⁻¹). Without conjugation, the inner product would not be positive definite and the orthonormality statement would fail.