TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)

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Sector restriction as observable algebra

Statement

Assume Countable Choice. Let $H = ⨁_{j = 1}^{r} H_{j}$ be a finite orthogonal direct sum of nonzero complex Hilbert spaces, with $r \geq 1$ , and let $P_{j}$ be the orthogonal projection onto $H_{j}$ . Write $B (H)$ for the algebra of bounded linear operators on $H$ . Let $A$ be the block-diagonal subalgebra

$A = {A \in B (H) : A = \sum_{j = 1}^{r} P_{j} A P_{j}} .$

For every trace-class operator $ρ$ on $H$ , define $E (ρ) = \sum_{j = 1}^{r} P_{j} ρ P_{j}$ . Then, for every $A \in A$ ,

$tr (ρ A) = tr (E (ρ) A) .$

Thus $ρ$ and $E (ρ)$ define the same trace functional on the specified block-diagonal algebra. This is relative to the declared algebra; it implies no universal superselection law for a larger observable algebra.

Statement references

Nuclear bounded cyclicity for trace-class operators

The Axiom of Countable Choice ( ${A C}_{ω}$ )

Facts & Assumptions

Given: Countable Choice; the finite orthogonal decomposition $H = ⨁_{j = 1}^{r} H_{j}$ ; its orthogonal projections $P_{j}$ ; a trace-class operator $ρ$ ; and a bounded operator $A = \sum_{j} P_{j} A P_{j}$ .

[A1]

Countable Choice (AC $_{ω}$ ) is the choice-function principle for every countable family of nonempty sets. The Axiom of Countable Choice ( ${A C}_{ω}$ )

[F1]

A trace-class operator has a nuclear representation $T = \sum_{j} T_{u_{j}, v_{j}}$ with $\sum_{j} ∥ u_{j} ∥ ∥ v_{j} ∥ < \infty$ ; its trace is the absolutely convergent, basis-independent trace sum. For every bounded $B$ , both $B T$ and $T B$ are trace class and $tr (B T) = tr (T B)$ ; the trace is linear on finite sums. Nuclear bounded cyclicity for trace-class operators

Proof

1.1givenA1F1algebra

Since $I = \sum_{i} P_{i}$ and $A = \sum_{j} P_{j} A P_{j}$ , expansion gives $ρ A = \sum_{i, j} P_{i} ρ P_{j} A P_{j}$ . This is a finite sum, and every summand is trace class by [F1], so taking its trace term by term is legitimate.

2.1A1F1step 1.1algebra∎

If $i \neq j$ , cyclicity in [F1] gives $tr (P_{i} ρ P_{j} A P_{j}) = tr (P_{j} P_{i} ρ P_{j} A) = 0$ , since orthogonality gives $P_{j} P_{i} = 0$ . The remaining terms are $tr (P_{i} ρ P_{i} A P_{i})$ . Because $A$ is block diagonal, $P_{i} ρ P_{i} A = P_{i} ρ P_{i} A P_{i}$ , so their sum is exactly $tr ((\sum_{i} P_{i} ρ P_{i}) A)$ . This proves the statement.

Remarks

The paper cited above discusses additional physical assumptions behind a mass superselection argument. This theorem proves only equality of trace functionals for the explicitly chosen block-diagonal algebra.

Sector restriction as observable algebra

Statement

Statement references

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources