[unpaidmoon]

Change and mathematical scope

Proof, under the paired-star hypotheses. No new κ. The selected record stays κ = 781451/250000000. This package does not edit that record.

For every integer q ≥ 1 and every field of characteristic 2, a g = 3 paired-star word with exactly one constant thin map misses the endpoint. The two full calls of a stage may be interleaved, and the off-diagonal block B may be nonzero. Every successful word at this leaf count therefore uses at least two constant thin maps. The adjacent word in research/coupled31-chronology/COMMUTING.md meets that count for every q.

The argument evaluates every stage variable at 1. Each full factor collapses to I + e R, so the single constant thin map is forced to be

K = I + T_0 + T_1 + T_2.

The three leaf columns of K are identical, and rank K = 4q. The one-sided determinant comparison from PR #5673 then requires (1 + x)^{3d} = (1 + x)^d with d = 2q. Those polynomials differ in characteristic 2. The same chain, restricted to any three leaves, shows that one constant thin map also fails for every g ≥ 3 on an arbitrary orthogonal resolution. At g = 2 there are not three leaves, and one thin map meets the endpoint.

Evidence class: proof. This is not an exhaustive census and not a search over q or over call orders. The F_2 census in PR #6922 is not a premise; a finite statement at block size 2 does not lift, and this package does not rerun it. A failed search is not used as nonexistence. The g ≥ 3 corollary stays inside the star contract. It is not unrestricted nonexistence, and it says nothing about five thin maps.

Task and overlap

Base commit 17469fff78c22941280f04d39dc5a1faee09c840 (canonical CrocSwap main, re-fetched at session start and before submission; both reads were this SHA). This fork draft uses cursor/crocswap-main-17469fff-1751 as its base because that branch is the same commit. The fork's own main is older. The upstream target is CrocSwap main.

Task card joint-query-controller, scope refined. The assigned question is open-next item 3 of PR #7010: whether one thin map fails at g = 3 for every q, which would say whether the smallest interleaved escape exists. The F_2 census does not lift.

Nearest overlap:

• PR Exclude the 31-leaf target with a chronological connector rank bound #5747 is the contract, the rank floor, the one-sided column theorem, and the two-map word. The new step is that one constant thin map must equal K, so the one-sided hypothesis applies for every q.
• Merged PR Record that one F2 thin map misses the three-leaf paired star. #6922 is the F_2 one-map census at block size 2. It is not rerun. It remains a finite census, not this proof.
• Merged PR Record that two F2 thin maps on the three-leaf paired star need total rank 8 #6940 is the F_2 two-map rank bound. Two or more maps are not re-proved here. The matching upper bound at g = 3 is the existing adjacent word.
• Open PR Prove that adjacent paired-star halves need ceil(g/2) thin maps. #7010 proved the adjacent-halves count ceil(g/2) and left this uniform question open. This package drops the adjacent-halves hypothesis at g = 3.
• Open PR Record that three F2 thin maps meet total rank 6 on the three-leaf paired star #7011, PR Record a partial F4 census of two thin maps on the three-leaf paired star #7006, and PR Prove one F8 nullity-7 paired-star class has rank sum at least 8. #7028 are the three-map F_2 word and the F_4 / F_8 two-map records. They are not taken.
• The five-map profiles at g = 13 and g = 31, and the internal-gap and nested schedules, are different counts. They are not investigated.

GitHub search on 2026-11-10 found no other pull request stating this uniform one-map proof.

Validation

Python 3.12, no third-party packages. From the repository root:

python3 -B scripts/research_package.py --package research/paired-star-one-thin-every-q

Registered result: PASS_REGISTERED_RESEARCH_PACKAGE with outcome negative-result. Both the python3 -B replay and the python3 -O -B replay print PASS_PAIRED_STAR_ONE_THIN_EVERY_Q. Checks raise SystemExit, so -O does not strip them.

Directly:

python3 -B research/paired-star-one-thin-every-q/verify.py

Observed counts: k_shapes=9, det_pairs=2, commutator_blocks=20, one_map_controls=18.

The checker locks the shape and rank of K, the cancellation of B in the summed commutator, the determinant gap, the telescope identity, and the adjacent two-map polynomial at (g, q) in {(2, 1), (3, 1), (3, 2)}, with B = 0 and with one nonzero B. It also checks that the forced matrix K misses representative adjacent and interleaved words. Those words are controls. The proof for every q is PROOF.md, not the list of controls.

Not run: a proof assistant, a search over call orders or over q, a native or supplier replay, and scripts/research_selected_baseline.py. The selected record is read only to pin κ.

Credit

The paired-star contract, the rank floor, the one-sided column theorem, and the two-map word are from PR #5747 (Ali Romman), including the predecessor statements in PR #5673, PR #5681, PR #5631, and PR #5617. The adjacent-halves count that left this question open is PR #7010. The F_2 one-map census that does not lift is PR #6922.

Prepared by unpaidmoon with substantial Cursor Grok assistance. Independent human review and a proof-assistant proof are unclaimed.

Open next questions

• Does any paired-star word at g = 5 meet the endpoint with exactly two constant thin maps, for some q? One constant thin map is impossible for every g ≥ 3, and at g = 4 the only positive count below ceil(g/2) is one map, so two maps at g = 5 are the smallest count this argument does not settle. A yes answer would be an interleaved escape below ceil(g/2). The first step is q = 1, one fixed legal pair, and one order that separates a stage: solve the two constant gaps by the specialization formula already used for the three-leaf census. Nothing in the star contract blocks that calculation. A failed search is not a proof. This is not the five-map question at g = 13 or g = 31.

Workflow feedback (optional)

The useful split was the evaluation at every D_j = 1. It names the only possible thin matrix before any search, so the F_2 census did not need to be repeated in a larger field.

Feedback discussion: #1200