PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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A nontrivial weight anomaly obstructs forgetting the weights

Physical scope

Pure mathematics; no physical premises or physical claim.

Statement

Assume full AC and its inherited countable-choice consequences, and fix a weighted RT bordism $M$ with all its boundary data, ribbon graph, object Lagrangians, and coupons held fixed. If its integer weight is $n$ , then $Z (M, n + 1) = κ Z (M, n)$ . If $κ \neq 1$ and $Z (M, n)$ is a nonzero linear map, the weighted RT assignment cannot factor through the operation that forgets integer weights while keeping the same maps. This obstruction concerns ordinary forgetting of the weight; an anomaly-canceling enlargement or a different normalization requires separate data and is not asserted here.

Statement references

Weighted modular TQFT model and physical interpretation boundary

The weighted Reshetikhin–Turaev TQFT and modular-functor factorization

Modular ribbon fusion category and Hopf-link data

The Axiom of Choice

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

Facts & Assumptions

Given: A complex modular ribbon fusion category, its weighted RT model, a fixed underlying colored bordism with weight $n \in Z$ , and the conditional assumption $Z (M, n) \neq 0$ .

[A1]

Full AC is adopted and countable choice is retained as an inherited consequence of the weighted RT construction (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

Weighted modular TQFT model and physical interpretation boundary fixes $κ = p_{+} / D$ and the exact weighted RT map $Z (M, n)$ for the same underlying bordism data.

[F2]

The weighted Reshetikhin–Turaev TQFT and modular-functor factorization states that a unit increase of integer weight multiplies the map by $κ$ ; all other bordism, Lagrangian, boundary, and coupon data remain fixed in this comparison.

Proof

1.1A1F1F2givenalgebra

In the formula of [F2], increasing $n$ to $n + 1$ leaves the Maslov correction, surgery signature, colored graph operator, boundary data, and cap normalization unchanged, and replaces the factor $κ^{n}$ by $κ^{n + 1}$ . Hence $Z (M, n + 1) = κ Z (M, n)$ .

2.1F1F2step 1.1algebra∎

Suppose the assignment factored as $Z = \overline{Z} \circ U$ through the forgetful operation $U$ on weighted bordisms. Since $U (M, n + 1) = U (M, n)$ , factorization would give $Z (M, n + 1) = Z (M, n)$ . With step 1.1 this yields $(κ - 1) Z (M, n) = 0$ . A nonzero vector in the complex vector space of linear maps cannot be annihilated by a nonzero scalar, so $κ = 1$ , contradicting $κ \neq 1$ . Therefore no such factorization exists under the stated nonzero-map condition.

A nontrivial weight anomaly obstructs forgetting the weights

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources