Scoped obstruction + search: balanced ten-call four-SWAP residual, J^2 = J^3 kills the layer route (no new kappa)

[navinor]

Balanced ten-call four-SWAP residual: the J^2 layer cannot decide it (no new kappa)

Prepared by Aaron Wang, with help from opencode. Apache-2.0.

Scope

One focused package, research/four-swap-balanced-j3-layer/ (8 files; no other path touched, upstream/ and frozen pins unchanged). Outcome: partial — a scoped obstruction for one route plus a scoped search failure; the balanced question itself stays open.

Question. In the formal F2[GL_2(F2)] rank-one-idempotent register model (paid calls I + ([g]+1)uv^T, g in {a,b}, v^T u = 1, free scalar GL_4(F2) mixing normalized by augmentation, all four registers arbitrary dirty restored), can T = diag([S],[S],[S],[S]), S = aba = bab, at exactly r = 4 be realized by a ten-call word with exactly five a-labels and five b-labels?

This is the sole surviving sub-12 four-SWAP candidate after the concurrent label-saturation result in #1715 ("four swaps need ten or twelve calls; a ten-call word must use five of each label"), which states it claims "neither existence nor exclusion of that balanced case". It is also the second branch of the handoff #1166 (comment): "derive the J^2-coefficient-layer bound in F_2[G]/J^3 (sum E_i + sum_{i<j} E_i E_j = T - I) to exclude q = 10 in writing."

Results

1. Exact obstruction: the proposed layer bound is vacuous in this algebra. Augmentation powers of F2[G] have dimensions (5, 4, 4), so J^2 = J^3 and R/J^3 is just the sign quotient F2[z]/(z^2): the pair terms ([a]+1)([b]+1) and ([b]+1)([a]+1) vanish. The expansion identity prod(1+eps_i) = 1 + sum eps_i + sum_{i<j} eps_i eps_j (mod J^3 M_r) is derived and checked on seeded random words, and the whole layer equation is machine-checked equivalent to sum N_i = I_r (200/200 random instances) — the first-order sign invariant already known from Scoped proof: two coordinate swaps need six paid coordinate-CNOT calls (no new kappa) #1414. There is no second augmentation layer to read a coefficient bound off, so no written exclusion of q = 10 can come from F_2[G]/J^3; word order or a different quotient (natural M_2(F2) image, the commutator [ab]+[ba] in J^2) is required. This closes that route with the exact obstruction recorded rather than silently lowering the target.

2. Bounded exact search of the balanced profile. With sum N_i = I_r a justified lossless reduction, an ordered-tuple sampler over the 120 binary rank-one idempotents (seed 20261014) checks all 252 labelings per valid tuple, at J^3 layer and then at full formal level in M_4(F2[G]) (96-bit exact states). Recorded run (--full): 6,000,000 samples, 22,903 valid tuples, 5,771,556 labelings, 0 full-formal hits. This is a scoped search failure over the sampled domain, not an impossibility theorem.

Controls (all exact): four direct words (3/6/9/12 calls, incl. the balanced 6a6b twelve-call word) hit their targets; four mutation controls fail; the search finds known-solvable instances at r = 1 and r = 2 (balanced q = 6, 3,254 witnesses); a strict-relaxation honesty control exhibits a layer-feasible word failing full equality; conjugation closure of the idempotent family; two hand-derived algebra pins bind the group implementation.

Reproduce

From the repository root, Python 3.12 stdlib only (verify.py --full is about 6 minutes):

python3 -I -B research/four-swap-balanced-j3-layer/verify.py --full
python3 -I -B research/four-swap-balanced-j3-layer/test_verify.py
python3 scripts/research_package.py --package research/four-swap-balanced-j3-layer --check-only

Expected: PASS_FOUR_SWAP_BALANCED_J3_LAYER balanced_5a5b_unrealized_in_search layer_hits=5771556 no_new_kappa, 9 tests OK, PASS_PACKAGE_METADATA_AND_PINS. EXPECTED.json pins the deterministic counts for both budgets and any mismatch fails the replay.

Task card and overlap

Base 0b1cccf86e2f5132ed286a26ef0a681a495740ea (selected-record SHA-256 10b9e340fd8e354a37a0c4813c90e5de1dd42044b5f6aca4a3fc5a42b685fa32). Card bit-new-register-model / scope key 7084092dd726e7ba1f2f, priority revision 3, chooser nonce cd67986ec4d736b6ed683332001aecde, track bit-mechanisms; scope refined to the balanced ten-call residual (card allows "a fixed larger register count"). Full pre-computation card: research/four-swap-balanced-j3-layer/TASK-CARD.md; structured metadata in TASK.json/SOURCES.json/VERIFY.json (registered, --check-only passes).

Claim: #1317 (comment) (08:01Z). Refined-object overlap searches (find_overlap.py with 'balanced ten-call', 'five of each label', 'ten-call balanced', 'augmentation layer', 'F2[G]/J3'; snapshots 07:53–07:54Z at 410 open PRs/358 discussions and 08:45–08:46Z at 430 open PRs/386 discussions, complete pagination) matched only #1715, which excludes this case, and my own claim/package. No match is not a novelty certificate.

Limitations and remaining obligations

Not checked: Lean or independent human proof review; selected-supplier cold replay; physical/compiler/hosting/precision/tape admission. Not covered: q = 8 profiles, r = 5 with a restored dirty extra, other label profiles, and unsampled valid tuples of the balanced domain. No claim that a balanced ten-call word does or does not exist. No new supplier, recursive profile or kappa; the selected frontier is unchanged.

Open next questions

1. Order-sensitive exclusion (or construction) for the balanced (5,5) profile: natural M_2(F2)-image 5+5 meet (the image PR1705's census used), or a commutator-layer bound via [ab]+[ba] in J^2, which R/J^3 cannot see. Blocked by the J^2 = J^3 collapse recorded here.
2. The same balanced question at r = 5 with one restored dirty extra (smallest falsifying instance of "workspace cheapens the tariff"); the sign-quotient reduction carries over verbatim.

Credit

Predecessors: commdata2338 (#1414, #1491 — model, sign parity, three-SWAP finite result), Gareth Davidson (#1333 — free-scalar normalization), Cinna Davis (#1541 — arbitrary-workspace three-SWAP theorem), Aaron Wang (#1705 — four-SWAP r = 4 bounds and 4+4 obstruction measurement), Ross Richardson (#1715 — concurrent saturation result, complementary; it makes this package's case the sole remaining four-SWAP candidate). Substantial AI assistance disclosed as required ("with help from opencode"). This contribution does not repeat any predecessor census and does not promote anything into the selected record.

[Mister-Meeseeks]

The complete --full replay passed here (6,000,000 samples, 5,771,556 balanced labelings), as did the tests. The scoped J² = J³ conclusion is useful and remains distinct from a nonexistence theorem.

One small but consequential correction before incorporation: the final r=5 follow-up proposes diag([S],[S],[S],[S],1) with sum N_i = I_5. The sign quotient instead requires sum N_i = diag(I_4,0); the restored fifth role has zero target augmentation. Conditioning on I_5 would exclude every solution to that stated endpoint. Please correct that follow-up (and any sampler extension built from it). This does not affect the retained r=4 computation.

[navinor]

Correction applied on top of the reviewed head bf73c084 (commit f41de012, same branch), in response to the review comment #1755 (comment). The finding is verified correct in exact arithmetic.

Verified derivation. The sign quotient is the algebra map $\rho: F_2[G] \to F_2[z]/(z^2)$, $\rho([g]) = 1 + \mathrm{sgn}(g) z$, where $\mathrm{sgn}(g)$ is the parity of the permutation of $g \in GL_2(F_2)$ on the three nonzero vectors of $F_2^2$; it is an algebra map because $\mathrm{sgn}$ is a group homomorphism, and it agrees entrywise with the package's own J^2 = J^3 reduction on all six group elements. Both paid labels a and b are odd, so every paid factor 1 + ([g]+1) N reduces to $1 + z N$; since $z^2 = 0$, the whole word reduces to $1 + z \sum_i N_i$. For the restored-role endpoint T = diag([S],[S],[S],[S],1) at r = 5, S = aba is a transposition (odd) and the fifth role carries the identity boundary, so $\rho(T) = 1 + z \cdot \mathrm{diag}(I_4, 0)$: the fifth role's target augmentation is zero. Comparing z-coefficients gives sum N_i = diag(I_4, 0). The previously proposed sum N_i = I_5 differs from diag(I_4, 0) exactly in the (5,5) entry, so it contradicts the forced value and excludes every solution to that stated endpoint. The exact check exhibits the exclusion directly: the four disjoint aba/bab control words with the fifth role restored (12 calls, 6a6b) realize diag([S],[S],[S],[S],1) exactly in M_5(F2[G]) and have sum N_i = diag(I_4, 0) != I_5, and forcing the total to I_5 makes the endpoint layer equation fail.

Old vs new wording of the r = 5 follow-up. Old, in the README open question 2: "rerun the conditioned sampler at r = 5, target diag([S],[S],[S],[S],1), sum N_i = I_5 (the sign-quotient lemma is register-count independent)"; and in TASK.json: "the identical sum N_i = I_5 reduction, which the sign-quotient lemma still justifies". New: conditioning on sum N_i = diag(I_4, 0) — the restored fifth role carries the identity boundary, so its target augmentation is zero and the sign quotient forces the fifth diagonal entry of sum N_i to be 0; the earlier sum N_i = I_5 wording was wrong and would exclude every solution to that endpoint. Corrected in README.md (model parenthetical, controls paragraph, open question 2), TASK.json, and VERIFY.json; the package has no CERTIFICATE.json, so nothing needed correcting there. No sampler extension had been built on the I_5 condition — the r = 5 run was only a proposed first step — so it is withdrawn as stated and the exact open obligation is recorded instead: rerun the conditioned sampler at r = 5 with sum N_i = diag(I_4, 0). A regression control now pins the corrected condition in exact bit arithmetic: the permutation-sign quotient matching the J^2 = J^3 reduction; zero fifth-role augmentation while the retained r = 4 statement is still I_4; the witness above; the I_5 total failing the endpoint layer equation; and the r = 5 layer equivalence with diag(I_4, 0) on 200/200 random instances.

Untouched. The retained r = 4 computation and the J^2 = J^3 equivalence are unchanged and reproduce exactly: augmentation dims (5,4,4) with J^2 = J^3; layer-check equivalence with sum N_i = I_r (200/200); and the recorded search (6,000,000 samples, 22,903 valid tuples, 5,771,556 labelings, 5,771,556 layer hits, 0 full-formal hits) with expected line PASS_FOUR_SWAP_BALANCED_J3_LAYER balanced_5a5b_unrealized_in_search layer_hits=5771556 no_new_kappa. The reproduction commands are unchanged:

python3 -I -B research/four-swap-balanced-j3-layer/verify.py --full
python3 -I -B research/four-swap-balanced-j3-layer/test_verify.py
python3 scripts/research_package.py --package research/four-swap-balanced-j3-layer --check-only

Remaining obligations. Unchanged apart from the corrected condition: no Lean or independent human proof review; the q = 8 profiles; label profiles other than (5,5); unsampled valid tuples of the balanced domain; and the r = 5 balanced question at diag([S],[S],[S],[S],1), now conditioned on sum N_i = diag(I_4, 0) (open compute). No claim is made that a balanced ten-call word does or does not exist; no new supplier, recursive profile or kappa follows; the selected frontier is unchanged.

Aaron Wang, with help from opencode.