Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-30
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Class sums act on an irreducible representation by central-character scalars

Statement

Let G be a finite group, let C be a conjugacy class of G, let C^C[G] be its class sum, and let V be an irreducible complex representation of G with character χ. Then C^ acts on V as the scalar ωχ(C^):

ρV(C^)=ωχ(C^)idV.

Facts & Assumptions

Given: A finite group G, a conjugacy class C of G, its class sum C^, and an irreducible complex representation V of G with character χ.

[F1]

The class sums form a basis of the center of C[G], so each class sum lies in the center (For a finite group, the class sums form a basis of Z(k[G])).

[F2]

The central character is ωχ(C^)=Cχ(g)/χ(1) for gC (The central character of an irreducible complex character).

[F3]

On an irreducible complex representation, every G-endomorphism is scalar (Over an algebraically closed field, every endomorphism of an irreducible representation is scalar).

Proof

technique · direct
1.1

Because C^ is central by [F1], the operator ρV(C^) commutes with ρV(g) for every gG, so it is a G-endomorphism of the irreducible representation V. By [F3], there is a scalar λC with ρV(C^)=λCidV.

F1F3given
2.1

Taking traces gives λCχ(1)=trρV(C^)=xCtrρV(x)=xCχ(x)=Cχ(g) for any gC, because χ is constant on C. Hence λC=Cχ(g)/χ(1)=ωχ(C^) by [F2].

F2step 1.1algebra
3.1

Substituting the scalar from step 2.1 into step 1.1 yields ρV(C^)=ωχ(C^)idV.

F2step 1.1step 2.1

Depends on

Used by

Dependency tree · two levels

11 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources