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The irreducible complex characters form an orthonormal basis of $c f (G)$

Statement

Let $G$ be a finite group. The irreducible complex characters of $G$ , $χ_{1}, \dots, χ_{r}$ , form an orthonormal basis of the complex vector space $c f (G)$ of class functions on $G$ , with respect to the standard inner product.

Facts & Assumptions

Given: A finite group $G$ and its irreducible complex characters $χ_{1}, \dots, χ_{r}$ .

[F1]

The space $c f (G)$ consists of the functions $G \to C$ constant on conjugacy classes, with pointwise operations (Class functions and the complex vector space $c f (G)$ ).

[F2]

The standard inner product on $c f (G)$ is the Hermitian form $⟨ φ, ψ ⟩ = \frac{1}{∣ G ∣} \sum_{g} φ (g) \overline{ψ (g)}$ (The standard inner product on $c f (G)$ ).

[F3]

The irreducible characters are orthonormal: $⟨ χ_{i}, χ_{j} ⟩ = δ_{i j}$ (The first orthogonality relation for irreducible complex characters).

[F4]

Over an algebraically closed field of characteristic not dividing $∣ G ∣$ , the number of irreducible representations up to equivalence equals the number of conjugacy classes (If $k$ is algebraically closed and $char k ∤ ∣ G ∣$ , the number of irreducible representations of $G$ equals the number of conjugacy classes).

[A1]

A class function is determined by its values on one representative of each conjugacy class, and the indicator functions of the distinct conjugacy classes form a basis of $c f (G)$ ; in particular $\dim_{C} c f (G)$ equals the number of conjugacy classes of $G$ .

Proof

technique · direct

1.1

By [F3] the family $χ_{1}, \dots, χ_{r}$ is orthonormal, hence linearly independent: any relation $\sum_{i} c_{i} χ_{i} = 0$ has inner product with $χ_{j}$ equal to $c_{j}$ , because $⟨ χ_{i}, χ_{j} ⟩ = δ_{i j}$ and the inner product is linear in its first argument by [F2].

F2F3given

1.2

The group $G$ is finite, $C$ is algebraically closed, and $char C = 0$ does not divide $∣ G ∣$ , so [F4] applies and $r$ equals the number of conjugacy classes of $G$ .

F4given

2.1

By [A1], $\dim_{C} c f (G)$ equals the number of conjugacy classes, which step 1.2 identified with $r$ . Hence the linearly independent family of $r$ elements from step 1.1 has as many elements as the dimension of the space.

A1step 1.1step 1.2

3.1

A linearly independent family whose size equals the dimension spans, so the orthonormal family of step 1.1 is a basis of $c f (G)$ .

F1step 1.1step 2.1algebra∎

Depends on

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Sources

Peter Webb, A Course in Finite Group Representation Theory, Section 3.3 (standard reference, not scraped)
Pavel Etingof et al., Introduction to Representation Theory, Section 3.5 (standard reference, not scraped)

The irreducible complex characters form an orthonormal basis of cf(G)cf(G)