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Price integer transvections on the shear target B^k. - #4146

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unpaidmoon:cursor/transvection-alphabet-moment-ae7c
Oct 20, 2026
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Mister-Meeseeks merged 1 commit into
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unpaidmoon:cursor/transvection-alphabet-moment-ae7c

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@unpaidmoon unpaidmoon commented Oct 20, 2026 •

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Change and mathematical scope

Scoped moment calculation for one larger shear alphabet. No new κ, no supplier admission, and no selected-frontier change. The selected conditional exponent remains 0.001672660197076918.

PR #4126 shows that the unit alphabet {X, X^{-1}, Y, Y^{-1}} makes every faithful word for B^k fail the strict supplier test. This package prices the explicitly larger alphabet of arbitrary integer transvections

X^t = [[1, t], [0, 1]],    Y^t = [[1, 0], [t, 1]],    t in Z, t != 0,

each letter one rank-1 child, with the same address width m = 2 and the same moment M(a) = (λ/W)·2^{a-1}. The three-bit quotient is not used as a bill.

For B = X^2 Y^2 and every integer k ≥ 1:

  • λ(B) = 2 via X^2 Y^2.
  • λ(B^k) ≥ 4 for every k ≥ 2, because a length-2 word needs a diagonal entry equal to 1 and a length-3 word needs v_k to divide u_{k-1}-1, while 0 < u_{k-1}-1 < v_k. The word (X^2 Y^2)^k gives λ(B^k) ≤ 2k.
  • At W = 2, the single matrix B satisfies M(a) = 2^{a-1} < 1 for every a ∈ [0, 1), and M(1) = 1. Every higher power fails the strict test.
  • At W = 1 the strict test already fails for B.
  • Re-expanding each coefficient into |t| unit shears restores length 4 for B and removes the W = 2 pass.

Status: partial. The length classification is a proof for every k; the verifier checks the identities it uses through B^{24}. This is not a register identity, not a native XOR bill, and not a multiplication theorem.

Task and overlap

Base commit bb4bccbf42844d2801487450727c2f63e10d1d0f. Catalogue card bit-new-register-model, priority revision 6. Scope key 7084092dd726e7ba1f2f for the card method “Derive a target-specific relaxation bound for that same explicit alphabet.”

Scope: changed. The card asks for a new register identity escaping the three-register obstruction. This package does not build that identity. It answers open question 1 of the unit-shear supplier inequality, a different object on the same bit-mechanism track. An optimistic pass for one matrix at one role count is not catalogue acceptance.

Nearest work:

Session selection: eligible open handoffs on Discussion #1166, after excluding work already delivered this round, numbered from 1 so sibling sessions keep items 1–10. Fewer than 27 eligible items remained. Item index (27-1) mod 12 = 2 is the shear supplier/recurrence inequality (comments 19031016 and 19032401). #4126 already performed that item, so this PR is the distinct larger-alphabet remainder.

Validation

Environment: Python 3.12.3, standard library only. No Lean, Docker, or selected cold replay.

python3.12 -B scripts/research_package.py --package research/sl2z-transvection-moment

Also run under python3.12 -I -O -Werror -B. Observed:

PASS_SL2Z_TRANSVECTION_MOMENT powers=24 lambda(B)=2 lambda(B^k)>=4_for_k>=2 W2_passes_on_[0,1) W1_fails powers_fail no_new_kappa
PASS_PACKAGE_METADATA_AND_PINS
PASS_REGISTERED_RESEARCH_PACKAGE research/sl2z-transvection-moment outcome=partial

The finite script checks the recurrence, the word (X^2 Y^2)^k, the determinant and companion identities, both closed forms on a sample of exponents, the unit breadth-first control (length 4), the strict moment on every rational in [0, 1] with denominator at most 8, rejection of the boundary M = 1, and a damaged coefficient. The claim for every k is the divisibility argument in research/sl2z-transvection-moment/README.md. Exact λ(B^k) = 2k is not claimed.

Credit

Prepared by unpaidmoon with substantial Cursor Grok assistance. Apache-2.0.

Open next questions

Package question 1: if a coefficient t = 2 is one shift plus a unit shear, rather than one child or two unit children, does M(a) < 1 still hold for B at W = 2? That price decides which of the two extremes this alphabet lands on. First step: name one native operation that implements the shift, charge its rank and any dirty auxiliary, and recompute the two-role moment. Blocker: the rank-only model has no such native shift. Inventing one would be a new alphabet.

Workflow feedback (optional)

Handoff-first selection on Discussion #1166, then a one-alphabet stop after both role readings, matched the package guide. Feedback thread: #1200

The unit alphabet already fails the supplier test. Allowing every nonzero
integer coefficient makes B itself pass at two roles for a in [0, 1), and
makes every higher power fail even before the idle role is dropped.

Co-authored-by: unpaidmoon <unpaidmoon@users.noreply.github.com>
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