CounterexampleConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)

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Permutation invariance alone does not select a statistics sector

Statement refuted

The false claim is: if a linear operator on $H_{2} = (C^{2})^{\otimes 2}$ commutes with the transposition that swaps the two tensor factors, then the whole space $H_{2}$ must lie in one exchange eigenspace, either the symmetric $+ 1$ eigenspace or the antisymmetric $- 1$ eigenspace. More generally, the claim that permutation invariance by itself chooses one of those two sectors is false.

Witness

Take the identity operator $A = I_{H_{2}}$ and the swap $s (v \otimes w) = w \otimes v$ . Then $A s = s A$ . But $s ∣ 00 ⟩ = ∣ 00 ⟩$ , while for the normalized singlet $∣ ω ⟩ = (∣ 01 ⟩ - ∣ 10 ⟩) / \sqrt{2}$ one has $s ∣ ω ⟩ = - ∣ ω ⟩$ . Both vectors are nonzero, so $H_{2}$ contains nonzero vectors in both eigenspaces and cannot be contained in either one. Thus invariance under permutations preserves the sector decomposition but does not select a sector.

Facts & Assumptions

Given: $H_{2} = (C^{2})^{\otimes 2}$ , the factor swap $s$ , and the identity operator $I_{H_{2}}$ .

[F1]

Finite spin-half tensor chains: the two-factor tensor space has the ordered basis $∣ 00 ⟩, ∣ 01 ⟩, ∣ 10 ⟩, ∣ 11 ⟩$ , and the swap interchanges the two factors. The product inner product is linear in its first argument and satisfies $⟨ v_{1} \otimes v_{2}, w_{1} \otimes w_{2} ⟩ = ⟨ v_{1}, w_{1} ⟩_{C^{2}} ⟨ v_{2}, w_{2} ⟩_{C^{2}}$ , where $⟨ (a, b), (c, d) ⟩_{C^{2}} = a \overset{ˉ}{c} + b \overset{ˉ}{d}$ ; thus the four displayed basis words are orthonormal.

[F2]

Commuting symmetric-group and linear actions on a tensor power: permutations act by left place permutations on tensor factors.

[F3]

Unitary exchange blocks for collective spins: the tensor space decomposes into Specht exchange factors and their nonzero multiplicity spaces; the transposition acts on the Specht factor and as identity on multiplicity.

Counterexample

1.1F1F2algebra

The identity commutes with every endomorphism, so in particular $I_{H_{2}} s = s I_{H_{2}}$ . By [F1], $s ∣ 00 ⟩ = ∣ 00 ⟩$ . Also $s ∣ ω ⟩ = (∣ 10 ⟩ - ∣ 01 ⟩) / \sqrt{2} = - ∣ ω ⟩$ ; the two vectors are nonzero and orthogonal because $∣ 00 ⟩$ is orthogonal to both one-zero/one-one basis vectors. By [F2], this swap is exactly the left place action of the transposition. Thus both exchange eigenspaces in $H_{2}$ are nonzero.

2.1F3step 1.1algebra∎

The operator $I_{H_{2}}$ satisfies the claimed permutation invariance, but step 1.1 shows that its underlying space is contained in neither single exchange eigenspace. This is a counterexample to the universal claim. The Schur–Weyl decomposition [F3] records both exchange factors and their multiplicity spaces; it does not add a rule that removes one of them.

Remarks

The physical assignment of a symmetric or antisymmetric sector remains the separate adopted physical postulate P6 in the identical-particle framework. This mathematical counterexample neither derives nor modifies that postulate; permutation invariance is an algebraic symmetry statement.

Depends on

Dependency tree · 2 levels

Sources

Chetan Nayak, Steven H. Simon, Ady Stern, Michael Freedman, and Sankar Das Sarma, Non-Abelian Anyons and Topological Quantum Computation, §II.A.1