Phase-2 P2 (Route D v2: the float dress rehearsal, and why the naive Newton–Kantorovich does not close

Status: Level-1 tooling + a structural negative with a constructive repair) NOT a certificate. This is the brick TECHNICAL_P2_ROUTED.md §8 named as next: compute the radii-polynomial bounds Y₀, Z₀, Z₁, Z₂ for the gCLM two-scale traveling wave at the exact a=0 anchor, in plain floating point, across a truncation ladder, and see whether the certification ball closes before spending effort on interval hardening.

It does not close: and the measurements say it cannot, at any truncation, under any gauge, and with no positive weight able to repair it. The obstruction is located precisely and confirmed causally: the transport term c (1 + cos θ) ∂_θ degenerates at θ = ±π, i.e. at X = ∞. The same measurements then specify the fix: the linearized inverse loses exactly one power of spatial decay, and becomes uniformly bounded the moment the codomain is graded by that one power. Clay odds unchanged (~0.05%).

Rebuild the figure from committed data (no re-derivation): python writeup/4_p2_lottery/p2_route_d_dress_evidence.py → writeup/figures/fig20_p2_route_d_dress.png (reads writeup/data/p2_route_d_dress.json; regenerate the data (deterministic, a few seconds) with python experiments/p2_route_d_dress.py).

Code: solver/nk_fourier.py (the exact Fourier-basis operator) + test_nk_fourier.py (6/6). Full suite now 9 files green.

0. What was being tested, and why it is worth reporting

Route-D v1 established the framing: gauge the two scaling symmetries, work in the compactified Fourier basis where the line Hilbert transform becomes the circular conjugate, and exploit that the linearization DF is tridiagonal. It ended with one gating question, do the radii-polynomial bounds close?, and the honest warning that they might not.

A negative here is not a wasted brick. "The certification ball does not close, and here is exactly which structural feature blocks it, confirmed by a controlled ablation, together with the space in which it would close" is a more useful statement than a certificate of a solution that was already known in closed form. The a=0 anchor Ω₂ = −1/(1+X²) is an exact traveling wave; certifying it proves nothing new. Its value was always as a rehearsal: the machinery you build there is the machinery you would carry to a ≠ 0, where the answer is unknown. This leg reports what the rehearsal found.

1. The exact Fourier operator (solver/nk_fourier.py)

Route-D v1 verified the change of variable numerically (~10⁻⁷ on endpoint-vanishing combinations). It is in fact exact and unconditional. With X = tan(θ/2),

e^{iθ} = (1 + iX)/(1 − iX)

is holomorphic in the upper half X-plane (its only pole is X = −i) and tends to −1 at infinity, so G_k := e^{ikθ} − (−1)^k lies in the Hardy space and decays; H(Re G) = Im G then gives, for every k ≥ 0, with no decay assumption,

H(cos kθ) = sin kθ,        H(1) = 0.                                   (H)
f_X = (1 + cos θ) f_θ.                                                 (D)

So for Ω = Σ_{k≥0} a_k cos kθ the residual F(Ω,c) = Ω H(Ω) − c Ω_X is odd and its sine coefficients are closed-form polynomials in a:

Ω H(Ω)   = ½ Σ_{k≥0} Σ_{j≥1} a_k a_j [ sin((j+k)θ) + sin((j−k)θ) ]
−c Ω_X   = c Σ_{k≥1} k a_k [ sin kθ + ½ sin((k+1)θ) + ½ sin((k−1)θ) ]

with sin(−m) = −sin m folded and sin 0 = 0 dropped. No grid, no quadrature. The anchor is a = (−½, −½, 0, …), c = ½, and the two terms cancel exactly (sinθ/4 + sin2θ/8 each): residual returns identically zero at every truncation, a much sharper gate than the grid's 10⁻⁹.

Validation (test_nk_fourier.py, 6/6) uses three independent oracles: the exact anchor; central finite differences of the residual (Jacobian agrees to 1.4×10⁻¹⁰); and the banked grid residual on the sinh grid with the dense line-Hilbert matrix (agrees to 1.6×10⁻⁴ relative, discretization-limited). The two exact symmetry tangents are annihilated to 0.0.

A bug this caught. The v1 probe's closed-form band had an unfolded sin(−θ) in its k = 0 column (−1/4 instead of −1/2); its grid cross-check only ran k ≥ 1, so it never exercised the folding case. Fixed in experiments/p2_route_d_probe.py; the two independently-written closed forms now agree to 0.0, and the reported grid cross-check (3.9×10⁻²) is unchanged, since that residual is the θ = ±π endpoint correction, not this entry.

2. The gauged square system

Following v1 §Q2: fix the speed c = ½ and impose one scalar normalization. Unknowns a_0 … a_N (N+1); equations = the normalization row plus sine modes 1 … N. The two exact kernel directions of the un-gauged [DF | ∂F/∂c] are the scaling-valley tangents, now available in closed form by differentiating A/(1+BX²) (speed −A/(2√B)) at (A,B) = (−1,1):

With c fixed the surviving kernel is the combination with zero δc, which is w = −¼ cos θ (1 + cos θ) = (−⅛, −¼, −⅛, 0, …). All three normalizations tested (Σa_k = −1, a_0 = −½, a_1 = −½) are non-degenerate on w, so each isolates the zero. This is the analytic version of v1's singular-value count.

3. D1, the ladder: the inverse is unbounded (fig20-A)

Fitted over the upper half of the ladder:

‖A_N‖_{ℓ¹} ~ N^0.97,   σ_min ~ N^−0.98,   cond ~ N^2.03.

The norm of the finite-section inverse grows linearly with the truncation. In the unweighted Fourier (Wiener ℓ¹) space the linearized operator is therefore not boundedly invertible, and no truncation can certify: Z₂ = 2‖A‖ diverges while the contraction budget 1 − Z₀ − Z₁ is already negative.

Y₀ itself is 0.0 at every N: the anchor is an exact zero and a degree-1 trig polynomial, so both the finite defect and the convolution tail vanish identically. The informative quantity is therefore the certification budget

Y₀^max = (1 − Z₀ − Z₁)² / (4 Z₂),

the largest defect these bounds could tolerate. It is identically zero at every N. For comparison, the residual floor of the best GA profile at a ≈ 0.5 (the boundary this programme actually cares about) is ~10⁻² (TECHNICAL_P2_KLADDER.md). There is no gap to close; there is no budget at all.

4. D3, the gauge is exonerated

Sub-task G (v1 §7) was flagged as "the crux": get the gauge/Fredholm bookkeeping wrong and Z₀ is meaningless. All three normalizations give ‖A_N‖ ~ N^0.97: the three ladders lie on top of each other (fig20-A, open markers). G is not what blocks closure. It still has to be got right for a working certificate, but it is no longer a suspect for this failure.

5. D4, causal isolation: it is the far-field degeneracy (fig20-B)

The suspect was sub-task R, the θ = ±π (Cayley) endpoint. Test it by ablation: rebuild the identical ladder with the transport factor (1 + cos θ) replaced by 1 (so the transport term is −c h_θ instead of −c (1+cos θ) h_θ) and change nothing else (both Hilbert/product terms untouched). This surrogate is not a physical operator; it removes exactly one feature.

true operator:       ‖A_N‖_{ℓ¹} ~ N^0.97,  reaching 198.7 at N = 256
surrogate:           ‖A_N‖_{ℓ¹} ~ N^0.00,  FLAT at exactly 4.0

The growth vanishes completely. The vanishing of the transport symbol at θ = ±π is not a suspect but the cause. Sub-task R is promoted from footnote to blocker.

6. D2/D5: why the far field is exactly marginal, and why no weight fixes it (fig20-C)

Two independent mechanisms both fail, and both fail at exactly the boundary.

(a) Truncation coupling. Column N+1 of DF feeds row N with the sub-diagonal weight c(N+1)/2. Propagating it through A_N gives, exactly,

Z₁ ≥ ‖A_N (A† − DF)‖ ≥ N + 1                (measured: 5, 17, 65, 257 at N = 4, 16, 64, 256), the finite section is coupled to the modes it discards *more* strongly the

larger it gets. Bigger N is strictly worse. Even the fiction in which the far field is ignored entirely does not close at any N.

(b) Far-field column weight. The far-field columns of DF are exactly tridiagonal, (sub, diag, sup) = (ck/2, ck − ½, ck/2 − ½). With the standard diagonal tail model A_tail = diag(1/Λ_m), the discarded off-diagonal contributes a Z₁ column weight z(k). Two natural choices of Λ bracket it:

Both tend to 1, from opposite sides, at rate O(1/k), independent of c. So sup_{k>N} z(k) = 1 for every N: the marginality is not an artifact of the tail model. The reason is structural: the transport term is multiplication by the symbol 1 + cos φ, whose ℓ¹ norm is 2, exactly twice its own mean, because it vanishes at φ = π.

No weight repairs it. In a weighted ℓ¹ with domain weight w, the column weight becomes z_w(k) ≈ (u_{k−1} + u_{k+1}) / (2 u_k) with u_k := w_k / k. Demanding z_w ≤ 1 − δ for all large k forces u_{k+1} ≤ 2(1−δ) u_k − u_{k−1}, whose characteristic roots (1−δ) ± i√(1−(1−δ)²) lie on the unit circle with argument arccos(1−δ) > 0; every solution oscillates, so any positive sequence obeying it changes sign in finitely many steps. No positive weight achieves a uniform Z₁ < 1. The best possible is u affine (w_k = k(α + βk)), which gives z_w ≡ 1 exactly. The numerics agree: over k ∈ [5, 2000], w = k^0.5 → 1.0155, w = 1.05^k → 1.0327, w = k → 1.000000, w = k(1+k) → 1.000000, w = k^1.5 → 0.99999997, the last one dips below only because the window is finite; the theory says its supremum over all k is exactly 1, so no fixed margin δ > 0 exists.

Sharpening the tail model beyond a diagonal does not escape this: taking A_tail to be the exact far-field inverse zeroes that Z₁ block but moves the divergence into ‖A‖, which §3 already measures as unbounded. It is one fact wearing two hats.

7. D6, the repair: the inverse loses exactly one power of decay (fig20-D)

The far field of the linearization is, at large X (H(Ω₂) ~ −1/X, Ω₂ ~ −1/X²):

DF[h] ≈ −c h_X − h/X = g.

With c = ½ the integrating factor is X², so (X² h)' = −2X² g and

h(X) = 2 X^{−2} ∫_X^∞ s² g(s) ds.

For oscillatory g of unit amplitude this gives h ~ X: the inverse amplifies by one power of X, i.e. it loses one power of decay. Under X = tan(θ/2) a Fourier mode m resolves the scale X ~ m, so the prediction is ‖A e_m‖_{ℓ¹} ∝ m. Measured at N = 256:

||A e_m||_1  =  1.97 m       (fit on the interior window m <= N/8), slope `2` to within 2%. (The profile rolls over as `m → N`; that is the

finite-section edge, where ‖A e_N‖_{ℓ¹} = 4 exactly at every N, not the asymptotics.) The N^0.97 growth of ‖A_N‖ in §3 is the same fact: the worst column is the largest available mode.

So the operator is invertible, just not from a space to itself. Grade the codomain by one mode power (v_m = m on sine mode m, 1 on the gauge row) and re-measure:

‖A‖ = 3.000, constant from N = 8 to N = 384. That is the functional setting a working certificate must use: an asymmetric pair of spaces separated by exactly one power of spatial decay.

8. What this means for Route D, said plainly

• The naive uniform-Fourier radii-polynomial NK does not close at the a = 0 anchor, and the failure is structural rather than numerical: it survives every truncation, every gauge, and every weight.
• The blocker is R (the θ = ±π far field), demonstrated by ablation, not G (the gauge), which is exonerated. That reorders v1's open-risk list.
• The certification budget Y₀^max is identically zero, so the pessimistic forecast in v1 §7 (that an a ≈ 0.5 certificate would fail because Y₀ jumps to ~10⁻²) was, if anything, too optimistic. The programme does not fail at a ≠ 0 for lack of accuracy; it fails at a = 0 for lack of a space.
• The repair is identified and numerically validated: an asymmetric, decay-graded pair of spaces, in which ‖A‖ = 3.000 uniformly.

Next brick (specification, not a promise). Rebuild the bounds in the graded pair: domain ℓ¹ cosine coefficients, codomain graded by one mode power, with the far-field block handled by the exact ODE inverse above rather than a diagonal model. Two things must then be re-derived, and neither is free: the quadratic term D²F[h,h] = 2 h H(h) must be shown to land in the graded codomain (the Wiener algebra bound gives ℓ¹, not the graded space, so the domain norm likely has to move too), and the far-field inverse must be enclosed rigorously. This is the standard "compact core + explicit far field" two-region structure of the Chen–Hou / Gómez-Serrano genre; the honest reading of this leg is that the project has arrived at the point where that structure becomes necessary, and now knows exactly why.

Ceiling. Everything here is plain float64. Nothing in this note is interval-enclosed and nothing is claimed as rigorous: that was the whole point of running the rehearsal first, and it saved the interval hardening of a set of bounds that could never have closed. This leg does not climb the rigor ladder. Even the success it is scouting would be a computer-assisted toy-model certification, not a Clay solve. Overall Clay odds unchanged (~0.05%).