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PropositionStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)
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Wilson traces are invariant under vertex gauge transformations

Statement

For every finite SU(2) graph configuration, continuous finite-dimensional unitary representation ρ, closed graph path ℓ, and vertex gauge assignment h, the Wilson trace of ℓ is unchanged by ge↦ht(e)gehs(e)−1.

Facts & Assumptions

Given: The edge orientation, path ordering, representation, and vertex action of Definition Finite SU(2) graph holonomy model.

[F1]

Finite SU(2) graph holonomy model: each edge variable transforms as ge′=ht(e)gehs(e)−1, and a path product is ordered with later edges on the left.

Proof

technique · direct cancellation
1.1givenF1algebra

The vertex action preserves the reverse-edge condition because geˉ′=hs(e)ge−1ht(e)−1=(ht(e)gehs(e)−1)−1=(ge′)−1. Thus the transformed variables are again a graph configuration.

1.2givenF1algebra

If the path is empty, its product is I2 before and after the gauge action, so its trace is unchanged. For the rest of the proof assume the closed path has n≥1 edges and write it as e1:v0→v1,…,en:vn−1→vn=v0.

2.1givenF1algebrastep 1.2

The transformed product is gen′⋯ge1′=hv0genhvn−1−1hvn−1gen−1⋯hv1−1hv1ge1hv0−1=hv0(gen⋯ge1)hv0−1; all adjacent intermediate factors cancel because hv−1hv=I.

3.1givenF1step 2.1algebra∎

Applying the homomorphism ρ carries this conjugation to conjugation by ρ(hv0). For finite matrices, Tr⁡(AB)=∑i,jAijBji=Tr⁡(BA); take A=ρ(hv0) and B=ρ(gen⋯ge1)ρ(hv0)−1 to obtain the same trace as before. Thus the Wilson trace is invariant.

Depends on

Dependency tree · 2 levels

Sources