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A non-Abelian zero-form symmetry and its representation labels

Physical scope

A finite three-state quantum-mechanical system with a unitary S3 zero-form symmetry; the representation decomposition is exact for this finite model and makes no field-theory or empirical claim.

Example

Let M=Sβ1 be Euclidean time of circumference β>0, let H=C3 with orthonormal basis e1,e2,e3, and let the Hamiltonian be H0=0. Use natural units ℏ=1. The finite model's normalized correlation functional for time-ordered operator insertions is C(O1,…,On)=13Tr⁡(O1⋯On). Let S3 be the group of permutations of {1,2,3}, with composition (στ)(j)=σ(τ(j)), and define the unitary action ρ(σ)ej=eσ(j). A symmetry defect at time t is Uσ(t)=ρ(σ). The complete admissible inputs are arbitrary finite lists of distinct points in a fixed cyclic time order, each carrying either an ordinary matrix O∈End⁡(C3) or a defect matrix Uσ, with the same ordered product and normalized trace rule for every such mixed list. The circle has no boundary and this model has no varying background, so its combined background/boundary-data set is a singleton. The value space is C with its usual topology. A symmetry g conjugates every insertion matrix by ρ(g), sends a defect label σ to gσg−1, and fixes the singleton background datum. These are bijections of the complete input domain.

The state space decomposes as H=C(e1+e2+e3)⊕V2, where V2={(z1,z2,z3)∈C3:z1+z2+z3=0}. The first summand is the trivial representation and V2 is the two-dimensional standard irreducible representation. The representation label of a state sector is therefore either the trivial representation or the standard two-dimensional representation; a permutation σ∈S3 is instead a group element implementing a symmetry transformation. This is an ordinary p=0 example and does not conflict with the exchange theorem for p≥1.

Facts & Assumptions

Given: The finite Hilbert space, zero Hamiltonian, normalized trace, and permutation action in the Example.

[A1]

H0=0, so time evolution is the identity and the stated normalized trace defines the model's time-ordered correlation functional.

[F1]

Group-like p-form global symmetry and backgrounds: at p=0, group elements label ordinary symmetry transformations and local operators may transform in representations.

[F2]

Exchangeability makes p-form symmetry groups Abelian and simple charges characters: the Abelianity conclusion requires an effective group-like symmetry with total fusion, 1≤p<d, and the stated pairwise exchange isotopies; the separate continuous-character conclusion additionally requires a unitary Hilbert-space crossing action and a nonzero bounded simple operator with continuous scalar action. Neither conclusion applies at p=0.

Verification

1.1A1F1algebra

For every σ,τ∈S3 and basis vector ej, ρ(σ)ρ(τ)ej=eσ(τ(j))=ρ(στ)ej, so ρ is a representation by unitary permutation matrices. With s=(12) and t=(23), ρ(s)ρ(t)e1=e2 while ρ(t)ρ(s)e1=e3; hence S3 is non-Abelian.

1.2givenalgebra

Every permutation fixes e1+e2+e3, and permutes coordinates, so it preserves the sum-zero subspace V2. These subspaces intersect only at zero and have dimensions one and two, so H=C(e1+e2+e3)⊕V2.

1.3givenalgebra

Let r=(123) and ζ=e2πi/3. The vectors v+=(1,ζ,ζ2) and v−=(1,ζ2,ζ) form eigenlines for ρ(r) with distinct eigenvalues ζ2 and ζ. The transposition s=(12) sends v+ to ζv− and v− to ζ2v+. Any invariant line for S3 would have to be one of the two eigenlines of ρ(r), but neither is preserved by ρ(s); thus V2 has no nonzero proper invariant subspace and is irreducible.

2.1A1F2step 1.1algebra

For insertions whose cyclic time order is fixed, moving a symmetry point without crossing another insertion leaves the ordered matrix product unchanged; the trace correlation is therefore unchanged. This applies to arbitrary mixed lists of matrices and other defects. Simultaneous conjugation of all factors telescopes to conjugation of their product, so cyclicity gives 13Tr⁡(ρ(g)O1⋯Onρ(g)−1)=13Tr⁡(O1⋯On), including all defect factors. The singleton boundary datum is fixed. Adjacent defect factors multiply as UσUτ=Uστ, independently of separation, proving fusion in every complete correlator. The operator assignment Uσ(t)=ρ(σ) is therefore topological and has the full symmetry covariance in this finite model. Since p=0, the hypothesis p≥1 of [F2] fails, so no Abelianity conclusion applies to S3.

3.1F1step 1.1step 1.2∎

The two summands in step 1.2 are representation spaces, while the elements s,t∈S3 in step 1.1 are transformations; their distinct roles are explicit in the action ρ(σ).

Depends on

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