The E47 moment selector cuts two whole Casimir sectors on the cube, 125 → 47. Every tensor order where it can even try sits on a Pell ladder, and on the next three rungs, n = 25, 243 and 2401, its two roots never meet.
Above them every rung allows at most finitely many spins, and through the ninth rung there are none below 101000. That answers the question the selector theorem left open on 15 October: proved through n = 2401, finite on every rung, and searched far beyond.
“For n = 25, 243, ... sigma is rational and the question is open.”
On three factors, Vj⊗Vj⊗Vj, the selector’s roots are j(j+1) and 5j(j+1), and both land exactly on the Fibonacci–Lucas spins. The other tensor orders were left open. Here are all of them.
Each vertex is a root coefficient s; the root itself is s · j(j+1). Each edge is a rung: a tensor order n where σ is rational, joining its lower root to its upper root. The upper root of every rung is the lower root of the next, so the rungs chain into one zigzag. Tap a rung.
On the carrier H = Vj⊗n with total Casimir C, write c = j(j+1) and let μ and σ² be the mean and variance of spec C under the normalized trace. The moment selector is Kmom = (C − (μ−σ)I)(C − (μ+σ)I), and the corpus lemma gives μ = nc, σ² = (2n(n−1)/3) c². A root lands when it equals a Casimir value J(J+1) of a sector present in H.
σ is rational exactly when n = (x+1)/2 with x² − 6y² = 1. The rungs are n = 1, 3, 25, 243, 2401, 23763, … and there σ = y c. Off the ladder no root is rational, so Kmom is invertible on every carrier.
On rung k the roots are sk−1 c and sk c, where sk = nk + yk = (1 + uk)/2 and v√2 + u√3 = (√2 + √3)2k+1. The upper root of each rung is the lower root of the next.
Both roots land, giving a kernel of two whole sectors, on rung 1 (n = 3) exactly when 2j+1 = Fk with k odd. On rungs 2, 3 and 4 (n = 25, 243, 2401) it never happens for j > 0. On every rung k ≥ 2 it happens for at most finitely many spins, and on rungs 5 to 9 for none with 2j+1 < 101000.
On the Fibonacci–Lucas family the upper sector has multiplicity Fk−2, so the kernel has dimension Fk² + Fk−2Lk. On the cube that is 47 = 5² + 2·11, and along the family Fk · Ω(k) → 3/φ.
Pick a spin and a number of factors. The page builds the Casimir census of Vj⊗n from the character (1 + q + ⋯ + q2j)n in exact integers, reads μ and σ² off it, checks them against the lemma, and tests whether each root lands on a sector.
47 = 5² + 2·11
On the cube the lower root keeps the spin-2 sector, five copies of V2, and the upper root keeps the spin-5 sector, two copies of V5. Across the family both counts are Fibonacci numbers: Fk copies of Vj below and Fk−2 copies of VJ above.
The fraction kept falls like 3/(φ Fk). The cube keeps the most of any proper member.
| k | 2j+1 = Fk | 2J+1 = Lk | copies of VJ | dim ker | carrier | Ω | Fk·Ω |
|---|---|---|---|---|---|---|---|
| 3 | 2 | 4 | 1 | 8 | 8 | 1 | 2.0000 |
| 5 | 5 | 11 | 2 | 47 | 125 | 0.376 | 1.8800 |
| 7 | 13 | 29 | 5 | 314 | 2197 | 0.14292 | 1.8580 |
3/φ = 1.8541019662… · multiplicities replayed from the exact census for k ≤ 13.
σ/c = √(2n(n−1)/3). If this is rational, then 6n(n−1) = 9(σ/c)² is an integer that is a rational square, hence an integer square r². Since 3 divides r², write r = 3y, so 2n(n−1) = 3y². Put x = 2n − 1; then x² − 1 = 4n(n−1) = 6y².
Off the ladder μ ± σ is irrational while every Casimir value J(J+1) is rational, so no root lands.∎
Consecutive solutions satisfy (xk, yk) = (5xk−1 + 12yk−1, 2xk−1 + 5yk−1). Then
so the lower root μ − σ on rung k is the upper root μ + σ of rung k − 1. Since (√2 + √3)² = 5 + 2√6, the coefficients are sk = (1 + uk)/2 with u = 1, 9, 89, 881, 8721, …∎
The roots are 5c and 45c, so a double landing needs J₂(J₂+1) = 9 J₁(J₁+1). With B = 2J+1 this is
Both factors are positive with equal parity, and the larger is at least 4. The only split is 2 · 4, which gives B₁ = B₂ = 1, so J₁ = J₂ = 0 and c = 0.∎
The coefficient 441 = 21² is the upper root of rung 3 and the lower root of rung 4. With a = 2j+1, the equation J(J+1) = 441 j(j+1) reads
For a ≥ 2 the factor 21a + B is at least 43. The even splits of 440 above that, 2·220, 4·110 and 10·44, give 42a = 222, 114 and 54, none divisible by 42. So this root never lands for j > 0, and neither rung can land twice.∎
A double landing on rung k is an integer point on two quadrics, with s = sk−1 and s′ = sk:
Their pencil degenerates at 0, ∞, −s′/s and −(s′−1)/(s−1). These are distinct whenever s, s′ ∉ {0, 1} and s ≠ s′, which holds on every rung k ≥ 2. The curve is then smooth of genus one, and Siegel’s theorem leaves finitely many integer points. On the cube s = 1: the first quadric splits into b = ±a (the lower root always lands, at J = j), and what remains is the conic b² − 5a² = −4 of genus zero, whose integer points are the Fibonacci–Lucas pairs.∎
The obstruction above the cube is integrality. The curves carry rational points of infinite order (rank 1 on rungs 2 and 3, rank 2 on rungs 4 and 5, by PARI 2-descent, E1), and on rung 2 the rational spin j = 98/1501 lands both roots, at 2J+1 = 2321/1501 and 5519/1501.
In Vj⊗3 = Vj ⊗ (V0 ⊕ ⋯ ⊕ V2j), a spin J with j ≤ J ≤ 3j occurs once for each L with J − j ≤ L ≤ 2j, so 3j − J + 1 times, and spin j occurs 2j+1 times. With 2j+1 = Fk, 2J+1 = Lk and Lk = Fk + 2Fk−1:
For odd k, Fk−2Lk = F2k−2 + 1, so FkΩ = 1 + (F2k−2 + 1)/Fk² → 1 + √5/φ² = 3/φ.∎
Claude reads the theorem and proofs above and can run this page’s exact arithmetic, weighing a carrier or re-searching a rung, before it answers. Questions run on your own Claude account.
Connecting to Claude…
pip install sympy python3 e47-moment-ladder/research/validation/e47_moment_ladder_certificate.py # rungs ≤ 6, about 10 s python3 e47-moment-ladder/research/validation/e47_moment_ladder_certificate.py --rungs 9 # rungs ≤ 9, about 2½ min