TECHNICAL (Route-M2CI v1 (leg 187): Object A against every certificate hypothesis

Gate (pre-committed, verbatim). "Does a full radii-polynomial certificate close on Chen's γ = 2 inviscid profile (Object A), using leg 125's own transcribed constants and recovered shape as the starting construction, with every hypothesis of the certificate framework satisfied (not just the budget comparison leg 125 already made)?"

Answer: NO. Three clauses fail) H3 (isolation), H5 (contraction, $Z_0+Z_1<1$), H7 ($Z_2$ bounded), while the clause leg 125 measured, $Y_0$, is not merely favourable but exactly zero.

0. Scope, stated before the numbers

• Object A is INVISCID, ν = 0, Chen arXiv:1908.09385 sec 1.5. This leg bears on the viscous "missing rung" question not at all. That question stands exactly where leg 125 left it. The γ = 2 in the route name is the dissipation exponent of the dynamics Chen's Theorem 1.1 concerns; the profile his theorem converges to is inviscid.
• Object A is a published closed form, Chen eq (2.2) p.4; the entire branch $a \le 1$ in HQWW arXiv:2305.05895. Recorded as novelty finding N2 PRE-EMPTED before construction. Consequence: $Y_0 \equiv 0$, and the YES branch would have carried nil existence content. See §6.
• The symmetry zero mode is published, Xu arXiv:2607.19762 gives point spectrum ${0,1}$ (scaling and time-shift modes) for the $a = 0$ sibling. Novelty finding N3 PRE-EMPTED in general form. This leg cites it; the contribution is the certificate-clause audit at $a = 1/2$ with magnitudes.

1. The object, and what was reused

$$\Omega(X) = \frac{-2bX}{(X^2+b^2)^2},\quad b^2=\tfrac38,\quad a=\tfrac12,\quad c_l=\tfrac13,\quad c_\omega=-1,\quad \nu=0.$$

Per the standing ban, capabilities.py was grepped before anything was written. Reused, not rebuilt: solver/dissipative_profile.py (leg 125's territory, read-only) for the transcribed constants, chen_profile, the grid, the whole-line Hilbert matrix, the 4th-order derivative/velocity operators and the analytic Jacobian at ν = 0; solver/nk_bounds.budget for the one budget; solver/certificate_guards.hypothesis_violations for the one shared hypothesis guard.

The new module adds only what did not exist: exact rational arithmetic on the defect, the exact dilation orbit and its generator, the far-field symbol clause, and the battery.

Convention check (load-bearing). The module's exact-arithmetic $H\Psi = (g-X^2)/D^2$ and $U = X/D$ are verified equal to leg 125's returned U_x and U to < 1e-13 (test_orbit_matches_leg125_transcription_at_chens_gamma). Without this the exact defect would be solving a different equation. chen_profile returns the 3-tuple (Omega, U_x, U): a first draft of the test compared against the tuple and produced a spurious 3.394 discrepancy.

2. H1: space membership

Weighted sup norm $|h|_s = \sup |(1+X^2)^{s/2}h(X)|$. Finite for $s \le 3$, divergent above. On a truncated grid this shows as growth with the truncation radius:

The $s = 3.5$ row is the control: the norm is being carried by the outermost nodes.

3. H2, the defect, in exact arithmetic

The profile family, with $g$ the squared length scale and $\kappa$ the amplitude:

$$\Psi = \kappa\frac{-2\sqrt{g}X}{D^2},\quad H\Psi = \kappa\frac{g-X^2}{D^2},\quad U = \kappa\frac{X}{D},\quad D = X^2+g.$$

Every term of $R(\Psi) = (c_\omega + H\Psi)\Psi - c_l X\Psi_X - aU\Psi_X$ carries exactly one factor $\sqrt g$, so $R/\sqrt g$ has rational coefficients and the whole computation stays in Fraction. The distinguished amplitude is $\kappa = 8g/3$.

Result: the numerator of $R/\sqrt g$ is the ZERO POLYNOMIAL at $g \in {3/8, 1, 2, 1/7, 9/4}$. $g = 3/8$ with $\kappa = 1$ is Chen eq (2.2) verbatim. Hence $Y_0 = 0$ exactly.

Controls, all of which must and do come out nonzero:

Leg 125's lead, for the record: $Y_0/\text{budget} \in [1.325\mathrm{e}{-09},\, 5.800\mathrm{e}{-05}]$ over 9 rows. Correct, and, as §4 shows, never the binding clause.

4. H3 / H4 / H5, the dilation orbit, and why no $A$ repairs it

The orbit. $\Psi_g(X) = -\frac{16}{3}g^{3/2}X/(X^2+g)^2$ solves the steady equation for every $g > 0$ at fixed $c_l, c_\omega$: the Hilbert transform commutes with dilation and the velocity's factor of $\mu$ cancels against $\Omega_X$. §3 proves it exactly.

H3 fails. Every ball around Object A contains other exact zeros. By bisection along the true (nonlinear) orbit at $s = 2$, $n = 401$:

The competitors' residuals sit at the centre's own discretisation floor, because they are exact solutions too. The theorem's conclusion (uniqueness in the ball) is false, so by contraposition its hypotheses cannot all hold, whatever $A$ or space is chosen.

H4 fails. $\varphi = d\Psi_g/dg = -\frac{16}{3}\sqrt g\,X(\frac32 X^2 - \frac g2)/D^3$ is an exact kernel element (checked against a central difference in $g$ to 1e-7 relative).

Kernel defect falls with the grid (tracking the 4th-order residual floor 1.889e-05 → 1.187e-06); the control is flat. This is an operator fact.

H5 fails, exactly and unrepairably. $DF\varphi = 0 \Rightarrow (I - A\,DF)\varphi = \varphi$ for every $A$, so

$$|I - A\,DF| \ \ge\ 1 \quad\Longrightarrow\quad Z_0 + Z_1 \ \ge\ 1 \quad\Longrightarrow\quad 1 - Z_0 - Z_1 \ \le\ 0,$$

in every norm, for every $A$, in every space containing $\varphi$, i.e. every $s \le 3$, which by §2 is every space where the centre itself lives. With $Y_0 = 0$: $p(r) = Z_2r^2 - (1-Z_0-Z_1)r \ge 0$ for all $r > 0$. No admissible radius exists.

4.1 The pinv trap: reported in two regimes on purpose

The default-rcond column appears to refute §4's argument. It does not: NumPy's cutoff is a relative $\sim n\varepsilon \approx 10^{-13}$, below $\sigma_{\min}/\sigma_{\max}$, so pinv retains the near-null direction and numerically inverts a kernel the continuum does not permit. Truncating it, which any rigorous $A$ is forced to do, returns the shadow to 1 from below, converging as the grid refines. A single-resolution number here would have been a statement about rcond (lesson 86). The clause is carried by the exact argument; the float columns are reported because the gap between them is the measurement.

5. H6 / H7

H6a: the standard repair is available. Bordering with a phase functional transversal to $\varphi$ (unknowns $(h, \delta c_l)$) lifts $\sigma_{\min}$:

Symmetry reduction / phase conditions are standard validated-numerics practice (novelty Q2); this leg claims none of it. Reported so the NO is not mistaken for "the repair was unavailable."

H6b: the far-field symbol, the one genuinely open sub-question (novelty N4). Beyond the profile's support, $L_\infty h = c_\omega h - c_l X h_X$, acting on $h \sim X^{-s}$ as

$$\sigma(s) = c_\omega + s\,c_l = \tfrac{s}{3} - 1, \qquad \sigma(3) = 0.$$

The profile's measured decay exponent is 2.999955 (log-log slope over the outer decade, n = 801); the $X^{-2}$ control measures −1.999911. The symbol vanishes exactly at the centre's own decay rate. This is the kernel obstruction seen from the far field: $X^{-3}$ is the homogeneous solution of the tail operator. The tail inverse norm $3/|3-s|$ diverges as $s$ approaches the only grading at which the centre is comfortably in the space.

H7 fails: $Z_2$ is not an operator constant here. $F$ is exactly quadratic, so $B(u,v) = (Hu)v + (Hv)u - a[(V!Hu)v_X + (V!Hv)u_X]$ is constant and $Z_2 = |A|\,|B|$. $B$ contains $v_X$, which the weighted sup norm does not control:

Least-squares slopes in $\log n$: $Z_2 \sim n^{2.36}$ (66.9× over the ladder, monotone at every step), $|A| \sim n^{1.36}$, $|B| \sim n^{1.00}$. The negative control that could have come out flat: $\sigma_{\min}/\sigma_{\max} \sim n^{-6.68}$, collapsing by 1.52e05×; it does not settle, confirming the kernel is the operator, not the grid.

6. Why the YES branch would have been hollow anyway

With $Y_0 = 0$ exactly, $p(r) = Z_2r^2 - (1-Z_0-Z_1)r$, so any bordered $Z_0+Z_1 < 1$ closes it, and would assert a zero at a point already published in closed form (Chen eq (2.2); HQWW for all $a \le 1$). The existence content is nil. Recorded in writeup/novelty/leg_187.md before any number here was computed.

This leg deliberately does not supply a bordered $Z_0 + Z_1$: bounding it rigorously needs an interval enclosure of the tail block, which this repository's banked ceiling (solver/interval_certificate.py) says it does not have. Reporting a float there as if it were a bound would be lesson 73. The JSON records it as NOT MEASURED with the reason.

7. What this banks

1. A characterized negative. Budget-under is not certificate-ready: now demonstrated with the budget pinned at exactly zero, the strongest form available. Leg 125's 1.325e-09 to 5.800e-05 was correct and load-free.
2. The failing clause, named with its constant. $Z_0 + Z_1 \ge 1$ for every $A$, from an exact symmetry, verified in exact rational arithmetic with controls that can fail.
3. The N4 sub-question, answered. The far-field symbol's zero coincides exactly with the profile's decay exponent ($\sigma(3) = 0$; measured 2.999955). Kernel and tail are one mechanism.
4. A reusable trap. pinv with default rcond silently inverts a near-kernel whenever $\sigma_{\min}/\sigma_{\max}$ exceeds $n\varepsilon$, which is the normal case for a discretised singular operator, not an edge case.

Not banked, and explicitly not claimed: anything about the viscous problem; any novel statement about the scaling zero mode (Xu's, cited); any credit for symmetry reduction (standard, cited); and any existence result about Object A (Chen's, analytic, and already in closed form).