3043 — Hadamard Admissibility Is Fiberwise and Does Not Select a Finite Scheme
Question
Can the Hadamard short-distance condition select one point in Ledger 3037's full finite-counterterm orbit?
Construction
Let denote the admitted renormalized field operators and the Hadamard states for . Hadamard admissibility defines a family
It tests membership in a fiber; it does not construct a section of the family.
For fixed , Hadamard two-point functions share the universal singular part and differ by smooth terms. There are infinitely many such states and no canonical smooth part. Separately, locality, covariance, and scaling leave a finite renormalization ambiguity in local fields and time-ordered products.
Result
A finite counterterm changes the renormalized operator or observable prescription, so its appropriate parametrix changes as well. Testing each scheme against its own parametrix can reject inadmissible singularities but cannot select a scheme point. Testing all schemes against one fixed parametrix imports that fixed operator as independent normalization authority.
Even after the operator is fixed, Hadamard regularity leaves smooth state freedom. Therefore it does not select the missing finite cosmological readout.
Combined with Ledgers 3037 and 3040, the surviving selector must contain operational normalization data: renormalized couplings or amplitudes and a state preparation/readout condition.
Scope
This excludes Hadamard regularity alone as the selector. A source that fixes the renormalized operator and sufficient observables may still select a unique state; those observables would be additional physical input.
Durable verification
research/benincasa/hadamard-renormalization-section-audit.md- Fewster, arXiv:1803.06836, Hadamard discussion around Eq. (12)
- Hollands--Wald, arXiv:gr-qc/0103074
- Collins--Holman--Vardanyan, arXiv:1408.4801
- ledger sequence claim:
seqclaim-16bdab402d0ec6d66e8e3e79 - epistemic graph event:
ev-000000006034-3c67ddb5-7981-476d-a4ce-df7cca7b7c10