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Wilson loop is invariant under vertex gauge transformations

Statement

Assume ACω. Let G be a group and let (W,q) be a finite-dimensional complex Hermitian inner-product space. Define U(W,q):={T∈GL⁡C(W):q(Tu,Tv)=q(u,v) for all u,v∈W}, and let ρ:G→U(W,q) be a group homomorphism, with d=dim⁡CW≥1. Let a finite oriented graph have vertices V and one chosen orientation E+ for each edge; for e∈E+ write s(e),t(e) for its source and target. A link configuration assigns Ue∈G to each e∈E+ and sets Ueˉ=Ue−1. A based closed edge word C=(e1,…,eℓ) has length ℓ≥0, consecutive endpoints match, and when ℓ>0 it starts and ends at a chosen vertex v0; for ℓ=0 it consists only of v0. Put UC:=Ue1⋯Ueℓ in traversal order, with the empty product equal to the identity, and put Wρ(C;U):=d−1Tr⁡W(ρ(UC)). For every vertex tuple k∈GV acting by (k⋅U)e=ks(e)Uekt(e)−1, one has UC(k⋅U)=kv0UC(U)kv0−1. Thus the normalized trace is gauge invariant and unchanged when the based word is cyclically rotated; for ℓ>0 reversal is C−1=(eˉℓ,…,eˉ1), while the empty word reverses to itself, and Wρ(C−1;U)=Wρ(C;U)‾.

Facts & Assumptions

Given: ACω, the finite link gauge action, a based closed edge word C=(e1,…,eℓ), and a finite-dimensional unitary representation ρ.

[A1]

ACω is inherited through the finite non-Abelian lattice datum. The Axiom of Countable Choice (ACω)

[F1]

Finite non-Abelian lattice gauge data: (h⋅U)e=hs(e)Ueht(e)−1 and a face or loop holonomy is the ordered product of its edge variables.

[F2]

Wilson loop on a finite lattice: Wρ(C;U)=d−1Tr⁡(ρ(Ue1⋯Ueℓ)) for d≥1.

Proof

technique · direct
1.1givenA1F1algebra

If ℓ=0, both the loop holonomy and its conjugate are the identity. For ℓ≥1, write the successive vertices as v0,v1,…,vℓ=v0. Substituting [F1] into the ordered product gives UC(h⋅U)=(hv0Ue1hv1−1)(hv1Ue2hv2−1)⋯(hvℓ−1Ueℓhv0−1)=hv0UC(U)hv0−1, because each adjacent factor hvi−1hvi is the identity. The set-theoretic input is the inherited ACω [A1].

2.1F1F2step 1.1algebra

Applying the homomorphism ρ preserves conjugation, and trace is cyclic, so Tr⁡(ρ(hv0)ρ(UC)ρ(hv0)−1)=Tr⁡(ρ(UC)). Thus [F2] is gauge invariant; a cyclic rebase is the same trace after a cyclic permutation of factors.

3.1F2step 2.1algebra∎

Reversal replaces each edge matrix by its inverse in reverse order, hence replaces ρ(UC) by its inverse. Since ρ(UC) is unitary, Tr⁡(ρ(UC)−1)=Tr⁡(ρ(UC))‾; dividing by the real integer d gives the stated complex-conjugation identity, including the empty-word case.

Depends on

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