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Library · hunts/outband_certificate/RESULTS.md

outband_certificate: a ceiling. The out-of-band positivity is worth zero to any certificate whose positivity input is the Hermitian form's, and what hunt #110 priced was the RH-conditional class

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Verdict, 2026-09-06: kill condition 2 fires, at prose grade. Section 8 has the argument; sections 1 to 7 are the measurements and the lemma it rests on. The question the hunt asked, whether an inertia or isolation argument can let a finite certificate spend BGSTB's positivity, is answered no for every argument whose positivity comes from Weil's Hermitian form, which is every unconditional argument in the field. What hunt #110 measured as "+0.005 to +0.009" is the worth of the information to the pointwise-positivity class, which is the class RH grants. The two inputs that would reopen the question are named at the end of section 8. Grade: proved for the edge lemma and the dichotomy, read for the structure of the paper's compression, measured for everything numerical. Nothing kernel-checked. Nothing here is evidence for or against RH (docs/08).

First session, 2026-09-05 to 06. Everything below reproduces hunt #110's numbers before it adds to them: the in-band control at X = 40, 80, 240 and 320 and the out-of-band values at X = 40 (A = 3) and X = 80 (A = 1.5) match ../outband_intake/RESULTS.md to the digit.

1. The certificate the information supports, read off the dual

dual.py rebuilds the configuration LP of ../frontier_math/configuration_lp.py and returns what that module does not: the dual solution. In the LP's own model the dual is the certificate: the multiplier profile on the data rows is a kernel Khat(alpha), signed in band and nonpositive on the strip, and the multipliers on tau >= -1 are its x-space partner. Strong duality is checked first on every case (gap 1e-13 or better); whatever argument eventually spends the information has to produce a kernel of this shape.

gridcasevalueg(x) on the tau gridKhat in bandKhat on the strip
X=40, J=200control0.6793882min 0 (to 5e-16), max 0.125min −0.020 at α=0.985, 5 of 201 nodes negative
X=40strip to 1.50.6911038min 0 (7e-16)min −0.032 at 0.995active on [1.06, 1.50], peak at 1.205, mass 0.0126
X=40strip to 3.00.6918387min 0 (6e-10)min −0.030 at 0.995active on [1.06, 3.0], peak at 1.23, mass 0.0164
X=80, J=320control0.6775676min 0 (1e-15)min −0.022 at 0.991, 8 of 321
X=80strip to 1.50.6855095min 0 (3e-9)min −0.022 at 0.991, 10 of 321active on [1.031, 1.50], peak at 1.119, mass 0.0112
X=80strip to 1.050.6792132min 0 (1e-15)min −0.025 at 0.991, 7 of 321active on [1.025, 1.05], peak at 1.028, mass 8.4e-5

Three things in that table.

The kernel is pointwise nonnegative in x-space. On every case g on the tau grid is nonnegative to numerical precision. That is the positivity a certificate needs for pairs of on-line zeros, whose differences are real. It is not the square structure.

The strip multiplier moves toward the band edge under refinement: active support begins at 1.06 at X = 40 and at 1.031 at X = 80, peak at 1.205 then 1.119. For the width-0.05 strip the whole multiplier mass is 8.4e-5, and the value still moves by +0.0016. That is #110's indirect channel made visible: a sliver of negative kernel just past the edge relaxes the adversary far more than its own weight.

The in-band kernel is slightly negative right at the band edge, at 5 to 10 nodes with α ≈ 0.99 and magnitude about 0.02, in the control as well as with the strip, at both grids. Whether that is a boundary layer of the discretised dual or a feature of the continuous one is not resolved here; it does not change the sign of g.

The adversary, in every case, is simple and double zeros only: p_3 .. p_6 = 0.

2. The LP cannot tell a double zero from an off-line pair, and that locates the missing argument

In configuration_lp.py, the off-line pair density q enters D with weight 4 and the density constraint with weight 2. So does p_2. The two columns are identical, the solver puts the mass wherever it likes (here in p_2, so q prints as zero), and q = 0 is not a finding. It is the model's statement, taken from the paper's section 7.5(b), that a depth-zero off-line pair costs the certificate exactly what a double zero costs.

Read against section 1 that says where the obstruction lives. A pointwise-nonnegative kernel handles every pair of on-line zeros on its own: for real differences the off-diagonal pair sum is a sum of nonnegative terms. The square structure g = |k|^2, spectral density v >= 0, is what gives positivity at complex points, and complex differences arise only from off-line zeros. So the value #110 measured is "what the information supports when off-line pairs cost what double zeros cost", and the only reason the unconditional proof cannot claim it is that the proof has to cover off-line zeros, for which pointwise positivity says nothing and the square structure forces Khat >= 0 everywhere.

The missing argument, stated as sharply as this session can state it: a bound for the off-line-zero contribution to the pair sum of a kernel that is nonnegative on the real line but not a square. The unconditional inputs of the right kind are zero-density estimates; Guth and Maynard's 2024 improvement (arXiv:2405.20552, fetched and confirmed) is the current one. This is a reading of the classical mechanism, not a measurement: it rests on the two facts that nonnegative kernels give nonnegative off-diagonal sums over real differences and that positive-definite kernels give positivity at complex points, and on the LP's degeneracy above. Whether the off-line term can actually be bounded to within 0.0006 with a density estimate is the research question, and it is a different question from #110's "find an inertia count for a non-Gram kernel". It is also, presumably, what Chirre, Gonçalves and de Laat's isolation input does with RH, where there are no off-line zeros.

3. How narrow a strip suffices

The record is 0.6734164909714992949 (AMTOPA), +0.00057 over this lab's four-point 0.6728470198 and +0.00092 over Theorem D. If a small strip past the band were worth that, the argument of section 2 would only have to control the off-line term on a small strip.

One grid says yes. At X = 80, J = 320, sweeping the strip's outer edge:

strip tovaluegain over control
1.020.67756760
1.050.6792132+0.00165
1.100.6801949+0.00263
1.150.6810415+0.00347
1.200.6819554+0.00439
1.300.6835537+0.00599
1.400.6847045+0.00714
1.500.6855095+0.00794

The ladder says not yet. #110's matched-ladder method, six rungs, ladder.py and pinned.py, exponent pinned rather than fitted because four to six points do not fix three parameters (the free fit diverged on the width-0.10 column):

widthgain per rung, X = 40, 80, 120, 160, 240, 320pinned-p limit, p = 0.5 … 2control's own fitted excess (true 0)limit / error
0.05+0.00137, +0.00165, +0.00147, +0.00137, +0.00113, +0.00099+0.00098 … +0.00129+0.00107 … +0.003650.4 to 0.9×
0.10+0.00314, +0.00263, +0.00231, +0.00203, +0.00174, +0.00153+0.00077 … +0.00190same0.5 to 0.7×
0.20 (four rungs)+0.00572, +0.00439, +0.00379, +0.00355+0.00130 … +0.00359same1.0 to 1.2×
#110, strip to 3.0+0.0065+0.00183.5×

So: a strip of width 0.05 is worth about the record gap, and by #110's own criterion that is inside the method error and not distinguishable from zero. The wide strip is established; the narrow one is not. The share of the full gain that sits inside (1, 1.05] also falls under refinement, from 20% at X = 80 to 16% at X = 240, so the one-grid front-loading partly washes out. The extrapolation is the only instrument short of the mathematics: a finite-X dual is not a continuous certificate, because its cosine sum is only known nonnegative on the grid up to X, and the adversary's room beyond X is exactly what the ladder is chasing.

What would settle it without the ladder is an explicit continuous kernel, nonnegative on the whole line with a transform nonpositive on the strip, whose bound in the LP model exceeds the in-band optimum by more than 0.00057; that is a lower bound, not an extrapolation, and it is the same object the argument of section 2 would consume. Nobody has written one.

4. What this changes in the mission

Nothing in the HuntSpec is withdrawn. The target stays the record. Two sharpenings:

5. Second pass, same session: a candidate, its death, and what killed it

The candidate. stable_rank_trace (lean/bridge/Zeta23Ext/StableRankTrace.lean) takes any Hermitian Q with a bounded positive index. For a signed spectral profile u = v_plus - v_minus the compression is P_plus - P_minus + Q; apply the theorem to V_plus and Q' = Q - P_minus (subtracting a positive semidefinite matrix cannot raise the positive index, Weyl). The Frobenius side becomes the pair sum with weight |FT u|^2, whose transform u * u is signed and could be nonpositive outside the band, where BGSTB's positivity drops it. The price sits on the trace side, 4 tr P_minus, four times the negative mass per zero. In the normalisation of paper_pin.py the candidate's value is 2 - min_u [(u*u)(0) + int |alpha| (u*u) + 4 int u_minus] over even u with int u_plus = 1 and u*u <= 0 on the strip. signed_window.py is the attempt to price it.

Its death, with proof. No grid u satisfied the sign condition on the whole strip, and that is a theorem, not an optimiser failure.

Edge lemma (real even factors). Let u be real and even in L^2, supported in [-s, s] with s the true edge, and let Khat = u * u. Then Khat is not nonpositive on any interval (2s - eta, 2s). Hence Khat cannot be nonpositive on the out-of-band part of its support, so a kernel of this class either has supp Khat inside [-1, 1] or fails the strip condition.

Proof. Near the edge only the two ends of u meet: writing W(v) = u(s - v) for the edge profile, Khat(2s - c) = (W * W)(c) for small c > 0, where * is convolution on [0, infinity). Suppose W * W <= 0 on (0, c_0). Since s is the true edge, 0 is in the support of W, so by Titchmarsh's convolution theorem 0 is in the support of W * W, and W * W is not zero on (0, c_1) for any c_1 < c_0. Take Laplace transforms at large real lambda: (L W)(lambda)^2 = L(W * W)(lambda). The left side is a square of a real number. The right side is int_0^{c_0} e^{-lambda c} (W * W)(c) dc + O(e^{-lambda c_0}), and the first term is at most -e^{-lambda c_1} int_0^{c_1} |W * W|, which dominates the error for large lambda. So the right side is negative for large lambda, a contradiction.

So the difference-of-squares candidate is dead for every real even u, however signed. The 4 int u_minus price was never reached: the sign condition fails first.

And the lemma is false for the class one needs, which is the finding. Krein's factorisation of a nonnegative kernel with compactly supported transform, K = |k|^2 with k of half the type, does not make k real and even. Take k(x) = sqrt2 sinc(x) sin(2 pi x), real and odd. Then K = sinc^2(x) (1 - cos 4 pi x) >= 0, and by the convolution theorem Khat = tri(alpha) - tri(alpha - 2)/2 - tri(alpha + 2)/2, which is compactly supported in [-3, 3] and nonpositive on |alpha| > 1. Verified on a grid of step 1e-3: min K = 0, max Khat on 1 < |alpha| < 3.05 equals 1.5e-6 (FFT noise), min Khat = -0.5 there, nothing beyond 3.05. With 1 - c cos in place of 1 - cos the same holds with K(0) = 1 - c > 0. For an odd factor the Laplace argument above flips sign, and the transform is nonpositive at its support edge automatically.

The dichotomy, which is #110's obstruction stated with both halves proved. A pointwise nonnegative kernel whose spectral factor is real even can never spend the strip (the lemma). A pointwise nonnegative kernel that does spend the strip has an odd or complex factor (the counterexample and its family). An odd factor is antisymmetric under the difference of two ordinates, so it is never the kernel of a Gram matrix, and a Gram matrix is what stable_rank_trace needs for the on-line block and what handles off-line zeros through Sylvester. Even factor: Gram-able, strip-blind. Odd factor: strip-capable, not Gram-able. The paper's v >= 0 is the even class, and that is why "the framework never has a free ghat".

Consequence for the LP's number, stated as a hypothesis. The LP's dual kernels are pointwise nonnegative and signed, which is the class the RH-conditional Montgomery argument uses (RH is what makes the zero side "a positive sum over real ordinates", the paper's abstract). So the class value [0.679, 0.682] is plausibly exactly what pointwise-positivity certificates reach under RH from bandwidth-one data plus BGSTB, and Chirre, Goncalves and de Laat's 0.6792 sits inside it. Hunt #110 withdrew that coincidence because two fits landed on two constants; this reading gives it a mechanism. Not measured, not claimed.

6. The lattice and the truncation, measured

Two checks on whether the LP's gain is an artifact of its own discretisation.

Lattice spacing. #110 imposed pair-measure positivity at spacing h = 1/16 and never varied h. At X = 80, J = 320, strip to 1.5:

hin-bandstrip 1.5gain
1/160.67756760.6855095+0.00794
1/320.67622840.6858172+0.00959
1/640.67614470.6855025+0.00936

The gain does not shrink as the lattice refines. It is not a spacing artifact at this X. At X = 160 the same holds and the direction is the same: gain +0.00693 at h = 1/16 and +0.00808 at h = 1/32; the finer lattice lowers the in-band control more than it lowers the strip value. The X = 320 rows at finer spacing were not obtained on this machine (RUNS.md).

Truncation. The dual kernels at X = 80 evaluated as continuous functions: on the LP's own lattice the minimum of g is -3e-9; off-lattice within (0, 80) it is -9e-5; beyond X = 80 both kernels go negative immediately (first negative point 80.07), the strip kernel to -0.077 against a maximum of 0.125, negative on 25% of [80, 400]; the in-band control kernel to -0.051, negative on 24%. So every finite-X value, control and strip alike, rests on the adversary being cut off at X, and the strip's gain is the differential use of that room. The large-X rungs at finer h are in RUNS.md when they land; the question the truncation leaves open is the X -> infinity limit at fixed h, which #110's extrapolation put at +0.0065 with the local decay exponent falling with X.

7. Dead routes added by this session

8. The ceiling, and why the LP priced the wrong class

What the compression is. From the paper's abstract and the Lean statement of stable_rank_trace: Weil's Hermitian form is compressed to a finite basis of test functions, and the explicit formula writes the compressed matrix as a sum over zeros. An on-line zero contributes a rank-one positive semidefinite piece v v^H (with multiplicity m, the same vector m times); an off-line pair contributes a rank-two indefinite piece with one positive eigenvalue, which is Sylvester's law and the source of posIndex Q <= b. So G = V V^H + Q exactly as the theorem takes it. The pair weight in the Frobenius norm is |<v_rho, v_rho'>|^2, and <v_rho, v_rho'> is a Gram pairing in a definite inner product (the standard one on the compression space). A Gram pairing of shifted windows is a positive-definite kernel by construction: in the paper it is FT(phi^2), whose transform phi^2 is nonnegative, and that is #110's "the framework has a free v >= 0, never a free ghat".

Why a positive-definite kernel cannot use the strip, in two lines. Its transform is nonnegative everywhere, so the out-of-band term int_{|alpha|>1} Khat F is nonnegative and, since F has no unconditional upper bound outside the band, it is unbounded above. The only way to keep the bound is Khat = 0 outside [-1, 1]. Bandwidth one is forced, and BGSTB's positivity has nothing to act on.

Why no other kernel can enter the argument. A kernel with a signed transform is, by the dichotomy of section 5, one whose spectral factor is odd or complex; equivalently it is a Gram pairing in an indefinite inner product, V S V^H with S indefinite. Hunt #110 tested exactly that weakening of the inertia inequality and refuted it with a two-by-two witness (rank half survives, inertia half fails). The two results are the same fact from two sides.

So the ceiling. Every unconditional certificate in this field takes its positivity from the Hermitian form: that is what replaces RH. Within that input the on-line block is a definite Gram matrix, its kernel is positive-definite, its transform is nonnegative, and the strip is worth zero. The unconditional room that remains is inside bandwidth one, between the current 0.6734 and the bandwidth-one configuration ceiling 0.6818, and it is reached by using (tr G, tr G^2) better, which is the n-point family this lab already runs.

What hunt #110 priced. Its LP is a depth-zero model of the zeros with pointwise positivity of the ordinate pair measure (tau >= -1) and Montgomery's in-band data. That is the model RH grants: no depth, and pointwise positivity over real ordinates, which is the very thing the paper's abstract says RH was "classically needed" for. The LP's optimal kernels are pointwise nonnegative with signed transforms (section 1), which is the pointwise-positivity certificate class, which is conditional. So "+0.005 to +0.009" is what the out-of-band fact is worth to a certificate that may assume RH, and Chirre, Goncalves and de Laat's conditional 0.6792 sits inside the measured range because it is a member of that class. #110's withdrawn coincidence was real. In band, the conditional and unconditional values coincide at Theorem D, which is the paper's achievement and why the LP's control converges to the right number.

The two inputs that would reopen this. Either would be a result far larger than the record: an unconditional evaluation of the ordinate pair correlation in band (which would make pointwise positivity usable without RH), or an unconditional upper bound on the form factor beyond the band (which would let a positive-definite kernel carry out-of-band support). Neither is a certificate trick, and the hunt's spec forbids presenting either as available.

What is proved and what is read, and then checked against the paper. The edge lemma and the odd counterexample are proved and verified. That a positive-definite kernel cannot use the strip is two lines. The structure of the compression was first read from the abstract, the theorem's statement and paper_pin.py; it was then checked against the paper's full text (the public PDF hunts/wide_search/HANDOFF.md points at, 2260 lines through pdftotext), which says it in its own words:

So the only sentence in this section that is not mathematics is that every unconditional argument in the field takes its positivity from the Hermitian form. The paper says as much of its own method, and names the CGdL route as the conditional other regime.

9. The neighbouring threads, placed by this verdict

Everything unconditional lives inside bandwidth one, between the record 0.6734165 and the configuration ceiling 0.68185 of Remark 1.1, and is reached by using (tr G~, tr G~^2) better. Read against that, the threads of 2026-08-24 to 09-05:

Not done, and why

Artifacts

artifacts/dual.json (X = 40, three cases), artifacts/dual-x80.json, artifacts/dual-x80-a105.json, artifacts/ladder-narrow-strip.json (18 solves, seconds per solve recorded). RUNS.md has the timings.