Gap A is not a missing calculation, it is a forbidden constant

Hunt r_b9552d, run 2, run id 113786a8-1f1c-4220-b772-15160a0274fa, 2026-08-20. Instrument: probe.py (--quick in seconds, full in a few minutes). Data: results.json (and results_quick.json).

Run 1 of this hunt (37fb06a9-..., 2026-08-18) is preserved unchanged as RESULTS-37fb06a9.md, probe_37fb06a9.py, results_37fb06a9.json, HANDBACK-37fb06a9.json. Nothing it settled is re-derived here, and nothing it said is contradicted. Two of its four threads have since been answered on main by other sessions and this run starts from that:

• its thread 1 (the G4 irregular-occupancy counterexample did not reproduce) was withdrawn on main in a735965: the recorded 0.1200 was a two-sided number compared against a one-sided one. There is no recorded counterexample to lattice extremality.
• its thread 2 (the Cohn–Elkies certificate route, left open by a failing LP) was answered analytically rather than by LP, in 7efd506: gap B of hunts/frontier_math/LATTICE-EXTREMALITY-ROUTE.md closes with the explicit majorant v = K_1(0)·(sin(x/2)/(x/2))².

That leaves gap A, which the route document names as "where the work is": the argument bounds the centre-gas row J(T) by LP_v(rho) only for densities rho >= 1/(2*pi), and says nothing below. This run asks what a certificate would have to be to close the sparse side, and whether the family that closed gap B can be it.

Grade: measured (double precision, no enclosures, no Lean). Section 1 is an argument, not a formalised proof. Nothing here is evidence for or against RH (docs/08); no claim here uses the reserved word.

0. Notation, in the route document's conventions

f(s) = Dam(1,s) - Kpair(s) = -kappa(s) + K_1(s)^+ is the centre-gas summand; J(T) = (2/m) Σ' f(a_p - a_q + nP) is the per-centre row; L := 2*kappa(0) - 4*c2(0) = 0.11433003938654052... is its value on the uniform 2*pi lattice and the conjectured ceiling. A certificate is a g with g >= f pointwise and ghat <= 0 everywhere; it yields, for every P-periodic configuration of density rho,

J(T) <= B_g(rho) := 2*rho*ghat(0) - 2*g(0).

1. What a certificate that closes gap A must be, exactly

Suppose one g (independent of rho) proves J(T) <= L at every density. Then:

• B_g is affine in rho with slope 2*ghat(0) <= 0, so its supremum over rho > 0 is the limit at rho -> 0, namely -2*g(0). Hence -2*g(0) <= L.
• The 2*pi lattice attains J = L at rho = 1/(2*pi), so B_g(1/(2*pi)) >= L. But B_g(1/(2*pi)) <= -2*g(0) <= L.

Both inequalities therefore bind, and that pins four things at once:

(P) ghat(0) = 0; g(0) = -L/2; sup_x |g| <= (1/2*pi)∫|ghat| = -g(0) = L/2; and, from equality in the Poisson identity at the lattice, ghat(j) = 0 for every integer j and g(2*pi n) = f(2*pi n) for every n != 0.

ghat(0) = 0 is the whole of it: it makes B_g constant in rho, which is exactly what "closes both sides at once" means. The dense-side argument of the route document has ghat(0) = -4.943, and that negative slope is precisely why it must give up as rho falls.

A cheap necessary test falls out. g >= f and sup|g| <= L/2 force sup f <= L/2. Measured:

The test does not fire. A universal certificate is not excluded by it, with a factor of 3.05 to spare. That is a positive finding about the route: the obvious first obstruction is absent.

2. The family that closed gap B cannot close gap A, and by an exact amount

Take g = -kappa + c·s with s(x) = (sin(x/2)/(x/2))², shat = 2*pi*(1-|xi|)^+, the gap-B majorant. Then ghat(0) = -khat(0) + 2*pi*c = -4*pi*c2(0) + 2*pi*c, so ghat(0) = 0 if and only if

c = 2*c2(0) = 1.6984559986366083...

and at that c the bound is not merely constant in rho, it equals the lattice value: B_g(rho) = 2*kappa(0) - 4*c2(0) = L for every rho. Verified across rho·2*pi from 0.0063 to 6.28: spread 6.66e-16.

So gap A would close, for every density at once, at exactly one value of c. Admissibility forbids that value:

shortfall factor = 2*c2(0) / (cos²(√2/2)(1+cosh 1)) = 1.15554665... shortfall absolute = 0.22862699...

The witness is explicit: at c = 2*c2(0), ghat(xi) = -khat(xi) + c·shat(xi) is positive on a whole interval below xi = 1, measured maximum +0.08390546 at xi = 0.87493: so the certificate condition fails there. The route document already records the inequality c < 2*c2(0) as the reason LP_v "stays decreasing". The reading it does not draw is the one that matters here: decreasing is the failure mode. c = 2*c2(0) is the exact boundary between a bound that decays with density and a bound that is uniform in it, and the ghat <= 0 constraint sits 15.55% below that boundary.

3. Why this is a statement about a family and not about one ansatz

The Fejér kernel is not one guess among many. Suppose v = g + kappa is any majorant with vhat supported in [-1,1] (the band of khat, and the band that makes the frequency sum finite). Then (P) forces:

• v >= K_1^+ >= 0, so v >= 0;
• v(2*pi n) = K_1(2*pi n)^+ = 0 for n != 0 (K_1 < 0 at every non-zero lattice point, the route document's section 5 asymptotic lim K_1(2*pi n)(2*pi n)² = -0.6277706 and clip_is_idle_on_lattice);
• v is of exponential type 1 and integrable.

A non-negative L¹ function of type 1 factors as v = |h|² with h in the Paley–Wiener space of type 1/2 (Fejér–Riesz / Krein), and h must vanish at 2*pi n for n != 0. But {2*pi n} is the critical sampling set for that space, so h(x) = Σ_n h(2*pi n)·sinc((x - 2*pi n)/2) = h(0)·sin(x/2)/(x/2), whence v = c·s. Up to the constant, the Fejér kernel is the only band-limited candidate, and section 2 shows the constant it needs is forbidden.

Grade this argument honestly: the factorisation and sampling steps are standard but are quoted here, not proved, and probe.py checks none of them, it checks only the two constants that make them bite. What it establishes, subject to that, is that gap A is closed to the entire band-limited rectification family, not just to the ansatz that closed gap B. Any certificate that closes gap A must put Fourier mass outside [-1,1], where (P) requires ghat <= 0 and ghat(j) = 0 at every integer, while v stays non-negative with double zeros on 2*pi Z\{0}. That is a sharply specified target and it is the concrete shape of the remaining work.

4. What the family does give: a uniform bound at every density

Nothing above stops one from spending the largest admissible c instead of the smallest. At c = cos²(√2/2)(1+cosh 1) = 1.4698290, which is admissible (it clears the floor 0.9116 and meets the cap with equality only in the limit xi -> 1), B_g(rho) = LP(rho) + 2c(2*pi*rho - 1) is still decreasing, so its value at rho -> 0 bounds every density:

For every periodic configuration, at every density, J(T) <= 2*kappa(0) - 2*cos²(√2/2)(1+cosh 1) = **0.5715840115651507**, which is 4.99942 × L.

That is a factor 5.0 away from the conjectured ceiling and so proves nothing about extremality, but it is the first bound in this route that holds on the sparse side at all. The route document's own sparse-side illustration is +2.15 at rho·2*pi = 0.4, computed at c = K_1(0); the same formula at the largest admissible c gives 0.4573 there. Gap A's numerical distance shrinks by 4.7×, and its structural distance does not shrink at all: the slope is still negative, so no choice of c survives rho -> 0 with the truth.

Configurations measured against all three (full run, N = 1500 periods; --quick at N = 400 agrees to six decimals on every shared row):

Zero violations of L, zero violations of the uniform bound. The sparse configurations are not close to threatening: clustering pays f(0) = -0.844 per ordered coincident-ish pair and dilution pays almost nothing, so J falls away from L in both directions off the critical lattice.

5. Controls

The inflation ladder is the honest statement of this run's power: the necessary condition of section 1 has a 3.05× margin, so it would catch a kernel three times more positive than the real one and nothing weaker. It is a real test that happens to pass, not a test that cannot fail.

6. What this settles and what it does not

• Does not close k >= 3, T1, or lattice extremality. Gap A stands.
• Does not produce a certificate. It produces the specification of one, and a proof-shaped reason the obvious family cannot supply it.
• Does pin every constant a universal certificate must hit: ghat(0) = 0, g(0) = -L/2, sup|g| <= L/2, ghat vanishing on Z.
• Does show the first necessary condition passes with margin 3.05, so the route is not dead.
• Does show the gap-B family misses the required constant by exactly 1.15554665×, with an explicit frequency-domain witness at xi = 0.875, and argues the miss is the whole band-limited family's, not one ansatz's.
• Does give the first sparse-side bound in this route: J <= 0.5715840 at every density, 4.99942 × L.

On the harness loop. This row's brief carries the standing instruction to record an outcome in a ledger under harness/departments/. Checked, and reported rather than performed: harness.review.standing_reasons is empty for both ledger claims (blockpos-0.672529, urms2-0.51), graveyard.unguarded is empty, and guards.undemonstrated names one item, tests/test_doors.py, which is not this row. This task came from the operator roster (operator:2026-08-17:k-ge-3-gas-split), not from a ledger generator, and the review ledger's own docstring says a frontier-math thread is entered "when its subject lands, not while it moves", lattice extremality moved four times in the last two days. So no ledger entry was made, deliberately, and the generator output above is the record of the check. scripts/70_lab_state.py runs clean.

Loose threads

1. The out-of-band certificate. What: section 3 says any g closing gap A must have ghat supported beyond [-1,1], non-positive there, vanishing at every integer, with v = g + kappa >= 0 double-zero on 2*pi Z\{0}. Nothing here says such a g exists or does not. Why it might matter: it is the entire remaining content of lattice extremality, which is the gas half of T1, which is blocker 2 for k >= 3. First step: solve the LP min -g(0) over g = -kappa + c·s + w with what <= 0 supported on [1, 3] and w free-sign in space, on a grid sampling f at 2*pi Z exactly, and see whether -g(0) can reach L/2. Run 1's thread 2 conditioning advice applies: concentrate the space grid on [5.8, 60] where f > 0, and rescale the measure by cell width.

1. The 4.99942 coincidence. What: the uniform bound of section 4 is 4.99942 × L, which is 5 to four figures and is not 5. Why it might matter: if it were exactly 5, 2*kappa(0) - 2c_cap = 5(2*kappa(0) - 4*c2(0)) would be a closed-form identity among c2(0), kappa(0) and cos²(√2/2)(1+cosh 1), and identities among those three are what the whole route is made of. First step: evaluate both sides at 50 digits with mpmath; the difference is 6.6e-05 at double precision, so 50 digits settles it in one line. A near-miss is the likely answer and is worth recording as one so nobody chases it twice.

1. Uniqueness under a uniform certificate. What: the route's uniqueness argument (section 4 of the route document) needs ghat < 0 strictly on (0,1). A gap-A certificate has ghat(j) = 0 at every integer and ghat(0) = 0, so the Newton step must be re-derived, the strictness it uses is gone at xi -> 0. Why it might matter: extremality without uniqueness is still enough for T1, so this is a completeness question, not a blocker; but a proof written without noticing would be wrong. First step: redo section 4 assuming only ghat < 0 on (0,1) open and ghat(0) = 0, and check whether A_j = 0 for 1 <= j <= m-1 still follows.

1. Depth. Run 1's thread 4 is untouched and still open: everything here is y = 1/2. LATTICE-EXTREMALITY-ROUTE.md is stated at y = 1/2 too, so the whole certificate analysis inherits it. Why it might matter: T1 is an obligation for all y in (0, 1/2]. First step: recompute c2, khat and c_needed = 2*c2(0) at 2y = 0.5 and 2y = 0.2: the kappa in play is y-dependent, so both the required constant and the cap move, and whether they move together decides if this section is one calculation or a family of them.