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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generated
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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.
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These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Distinct fusion-ring characters are orthogonal under duality

Statement

Let A be a fusion ring over Z with finite distinguished basis I, unit 0, and duality i↦i∗. If χ,ψ:A→C are distinct unital ring homomorphisms, then ∑i∈Iχ(i)ψ(i∗)=0.

Facts & Assumptions

Given: A finite fusion ring A with its distinguished basis, duality, and two distinct unital ring characters χ,ψ.

[F1]

Multiplication has structure constants Nabc in the distinguished basis, and Frobenius reciprocity gives Ny,x∗z=Ny∗,zx∗ (Fusion rules, Fusion and multifusion categories).

[F2]

A unital ring character is additive, multiplicative, and sends the unit basis element to 1.

Proof

Given: The data in the Statement.

1.1givenF1F2algebra

For a basis element y, form Pχ=∑x∈Iχ(x)x∗ in A⊗ZC. The coefficient of z in yPχ is ∑xχ(x)Ny,x∗z=∑uχ(u∗)Ny∗,zu by [F1], after setting u=x∗. By multiplicativity and the anti-involution identity (y∗z)∗=z∗y, this equals χ(z∗)χ(y), which is the coefficient of z in χ(y)Pχ. Hence yPχ=χ(y)Pχ.

2.1givenstep 1.1F2algebra∎

Applying ψ to the equality in step 1.1 gives ψ(y)ψ(Pχ)=χ(y)ψ(Pχ), where ψ(Pχ)=∑xχ(x)ψ(x∗). Since χ≠ψ, some y∈A has χ(y)≠ψ(y); subtracting the two sides forces ψ(Pχ)=0. This is the asserted sum.

Depends on

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Sources