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… two For m = 2^k, the [R] -> 1+s character at order m and the first-order B character force N_B + E_1...E_m = D_w, a rank contradiction. This settles the power-of-two part of the open profile question in PR6089.
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Result
Claim type: scoped written proof with exact finite corroboration. No κ change.
This answers part of Open next question 2 of apoursh's #6089 (
research/bit-positive-frame-use): is the third necessary multiplicity profile(0,1,3,3,m)realizable for anym >= 2?Theorem. If
m = 2^kfor somek >= 0, no exact word in #6089's formal PR2085/2112 four-target model has the profile(0,1,3,3,m), at any finite widthw >= 4. The model has rank-one binary idempotent calls overF2[Λ], freeGL_w(F2)mixing and exactly restored dirty workspace.m = 1is #6089's exclusion;m = 2, 4, 8, ...are new. For thesem, the attained count profiles are exactly #6089's two subdivision profiles.Proof idea. After #6089's normal form, send
[R] -> 1+sand[B],[P] -> 1. Then[A],[Q] -> (1+s)^m = 1+s^mover F2. Thes^mcoefficient of the word is the single ordered productE_1...E_mof the R projectors plus the three Q projectors, and it must equaldiag(1,0,0,1). #6089's first-order B character givesN_B + ΣN_Q = diag(0,1,1,0). Adding the two equations givesN_B + E_1...E_m = D_w, which has rank at most 2 on the left and rank 4 on the right.The full argument is in
research/bit-positive-frame-third-profile/PROOF.md. It also explains why the argument stops whenmhas an odd factor.Evidence
verify.pyuses exact arithmetic and needs only the standard library. It checks:m = 2^k,k <= 10;w = 4, 5andm = 1, 2, 4, 8;w = 4, 5.The random words corroborate the lemma; the proof does not depend on them.
Locally (CPython 3.12), all four pass:
PASS_POSITIVE_FRAME_THIRD_PROFILE_POWERS_OF_TWOtwice,PASS_REGISTERED_RESEARCH_PACKAGE ... outcome=partialin about 1 s, andPASS_RESEARCH_COMMIT_BYTES.Limitations
m = 3, 5, 6, 7, ...the profile is neither constructed nor excluded.m+7, its length spectrum, the selected frontier and κ are unchanged.Open next questions
(0,1,3,3,m)realizable whenmhas an odd factoro >= 3? Atm = 3the two characters only giveE_1+E_2+E_3+N_B = D_w, which coordinate projectors satisfy, so a second-order character or an exact 10-call search atw = 4is the next step.m >= 2? The blocker is the rank of the elementary sume_{2^k}(E)onceo >= 3.Related open work
No open PR, including this account's later PRs #6664 to #6675, studies this profile or the positive added-frame classification.
CI notes
All checks passed at head
7b6bf6ff83e905f27c063edb07300daaa4ccc899:planandreplay (research/bit-positive-frame-third-profile), run https://github.com/CrocSwap/integer-mult-bounds/actions/runs/41670611518.Credit
Produced by Nick Petalas. apoursh's #6089 supplies the model reduction, symmetries, necessary profile list, B character and subdivision words, and poses the question. Cinna Davis's PR2085/2112 supply the model and endpoint, and PR2910/PR6032 the determinant and first-order methods.