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A topological interface between different models is not an internal symmetry

Physical scope

An exact finite-dimensional 0+1-dimensional transfer model with hbar=1 and zero Hamiltonians; it refutes only the stated universal identification of interfaces with internal symmetries.

Statement refuted

Every topological codimension-one interface between two selected quantum-mechanical models is an invertible internal symmetry of its source model.

Statement references

Topological interface between specified models

Carqueville, §2.1, PDF pp. 5–8

Facts & Assumptions

Given: The finite-dimensional models and interface constructed in the Counterexample.

[A1]

The source model has state space $C$ and zero Hamiltonian; the target has state space $C^{2}$ and zero Hamiltonian. Propagation over every interval is the identity, and the interface map is $J (z) = (z, 0)$ . The allowed boundary vectors are unit vectors $b \in C$ and $a \in C^{2}$ with $Z = ⟨ a, J b ⟩ \neq 0$ . For any finite ordered source-side list $0 < s_{1} < \dots < s_{k} < τ$ of matrices $A_{i} \in End (C)$ and target-side list $τ < t_{1} < \dots < t_{ℓ} < T$ of matrices $B_{j} \in End (C^{2})$ , the complete normalized correlator is $C_{τ} = ⟨ a, B_{ℓ} \dots B_{1} J A_{k} \dots A_{1} b ⟩ / Z$ , with an empty product interpreted as the identity.

[F1]

Topological interface between specified models: a topological interface preserves every declared complete correlator under allowed decorated isotopies, while an invertible symmetry must be an endodefect of the same complete model data, have two-sided identity fusion, and induce observable and combined background/boundary bijections preserving all correlators.

Counterexample

Use the interval $[0, T]$ , $T > 0$ , with $ℏ = 1$ . Let the source model have Hilbert space $H_{-} = C$ and zero Hamiltonian operator $H_{-} = 0 \in End (H_{-})$ , and let the target model have Hilbert space $H_{+} = C^{2}$ and zero Hamiltonian operator $H_{+} = 0 \in End (H_{+})$ . All matrices on each state space are allowed local observables; propagation over every interval is the identity. The interface is the point insertion $J : C \to C^{2}$ , $J (z) = (z, 0)$ . Boundary data are unit vectors $b \in C$ and $a \in C^{2}$ satisfying $Z = ⟨ a, J b ⟩ \neq 0$ . The complete admissible other insertions are arbitrary finite ordered source-side lists $(s_{i}, A_{i})$ , $0 < s_{1} < \dots < s_{k} < τ$ , $A_{i} \in End (C)$ , and target-side lists $(t_{j}, B_{j})$ , $τ < t_{1} < \dots < t_{ℓ} < T$ , $B_{j} \in End (C^{2})$ . The time parameters $T$ , $τ$ , $s_{i}$ , and $t_{j}$ have units of inverse energy, both Hamiltonians have units of energy, and the boundary vectors, insertion matrices, and interface map are dimensionless by convention. The normalized correlator for every such input is

$C_{τ} ((s_{i}, A_{i})_{i = 1}^{k}; (t_{j}, B_{j})_{j = 1}^{ℓ}) = \frac{⟨ a, B_{ℓ} \dots B_{1} J A_{k} \dots A_{1} b ⟩}{Z},$

where an empty list contributes the identity matrix. This defines all declared correlators for arbitrary finite ordered lists on both sides and all admissible boundary vectors.

Proof technique: direct.

Given: [A1], [F1], and the statement in Statement refuted.

1.1A1givenalgebra

Regard both models on $[0, T]$ , $T > 0$ , with a point interface at $τ \in (0, T)$ . For the complete source- and target-side lists in [A1], the correlator is $C_{τ} = ⟨ a, B_{ℓ} \dots B_{1} J A_{k} \dots A_{1} b ⟩ / Z$ , because both Hamiltonians vanish. Moving $τ$ within the gap between the fixed source-side and target-side lists crosses no insertion, leaves their order unchanged, and introduces no nonidentity propagation factor. The numerator and denominator are therefore independent of $τ$ for every allowed list and boundary pair. This is a topological interface under the full declared input set.

1.2A1algebra

The interface is not an endodefect: its source space is $C$ and its target space is $C^{2}$ , so the two complete model data are distinct. It also has no two-sided inverse. For any linear map $S : C^{2} \to C$ , the composite $J S$ has rank at most one, whereas $I_{C^{2}}$ has rank two; hence $J S \neq I_{C^{2}}$ .

2.1F1step 1.1step 1.2∎

By [F1], topologicality alone does not make this interface an internal symmetry; the explicit topological interface from step 1.1 fails both the same-model and two-sided-inverse conditions in step 1.2. It refutes the universal statement.

A topological interface between different models is not an internal symmetry

Statement refuted

Statement references

Facts & Assumptions

Counterexample

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