LATTICE-EXTREMALITY-ROUTE

A route to proving lattice extremality, worked out far enough to say exactly what is proved, under which hypotheses, and what is missing. It is not a completed proof, and section 6 is the honest part of this document.

Companion to lattice_extremality.py (the computations) and K2-TWO-SPECIES.md T1 (the obligation this would discharge). Written 2026-08-20, after the search in that module failed to find a counterexample. A search that finds nothing bounds the search. This is the attempt to bound the problem instead.

0. The statement

For a P-periodic centre configuration with offsets a_1..a_m and density rho = m/P, the per-centre cost at y = 1/2, in the one-sided convention two_species.centre_gas_row uses, is

J(T) = (2/m) sum'_{p,q,n} f(a_p - a_q + nP), f(s) = Dam(1,s) - Kpair(s),

the prime excluding only (p = q, n = 0). The claim to prove:

Lattice extremality. J(T) <= -4 c2(0) + 2 kappa(0) = 0.11433003938654052... for every configuration, with equality if and only if T is a translate of the 2*pi lattice.

The right-hand side is two_species.centre_gas_row_closed(), landed in 084f326. What follows shows it is not merely the lattice's value but a ceiling, subject to two hypotheses.

1. The identity

Everything rests on one exact rewriting. Poisson summation on the inner sum gives, for any h for which it is valid,

sum_{p,q} sum_n h(a_p - a_q + nP) = (1/P) sum_j hhat(2*pi*j/P) |A_j|^2, A_j = sum_p exp(2*pi*i*j*a_p/P),

so that

J_h(T) = 2*rho*hhat(0) - 2*h(0)

• (2/(mP)) sum_{j != 0} hhat(2*pi*j/P) |A_j|^2. (*)

Two things make (*) worth having. The left side is an infinite sum over space; the right side is a finite sum whenever hhat has compact support. And |A_j|^2 >= 0 always, so the sign of each term is decided entirely by the sign of hhat.

structure_factor_defect measures (*) against direct summation rather than trusting it. Residual below 4e-9 on five configurations, three of them not lattices, at h = -kappa.

2. Three facts about kappa_hat

mean_damage records -D(y,s) = int_{-1}^{1} c2(w) cosh(yw) cos(sw) dw. Summing at y = 0 and y = 1 gives counting_lemma.kappa, hence

kappa_hat(xi) = 2*pi * c2(|xi|) * (1 + cosh|xi|) on [-1,1], 0 outside.

(a) Compact support, on [-1,1]. This is what collapses (*) to finitely many terms.

(b) Non-negativity, and this one is proved rather than measured. c2 = g star g with g(u) = cos(sqrt2 u) on |u| <= 1/2. Since sqrt2/2 = 0.7071... < pi/2, g is strictly positive on the whole of its support, so c2(w) = int g(u) g(u+w) du is an integral of a product of positive functions and is strictly positive wherever the two supports overlap, that is on (-1,1). 1 + cosh is positive. So kappa_hat > 0 on (-1,1).

(c) Vanishing at the endpoints. c2(+-1) = 0, because the supports of g(u) and g(u+1) meet in a single point. This is the fact the whole argument turns on, and it is also the fact counting_lemma already used to get its Poisson limit.

3. The bound

Take h = -kappa in (*). Every j != 0 term is (-kappa_hat) * |A_j|^2 <= 0 by (b), so dropping them can only raise the value:

J_kappa(T) <= 2*rho*(-kappa_hat(0)) + 2*kappa(0) = -8*pi*rho*c2(0) + 2*kappa(0) =: LP(rho).

LP is strictly decreasing in rho because c2(0) > 0, and

LP(1/(2*pi)) = -4*c2(0) + 2*kappa(0) = 0.11433003938654052...,

exactly the lattice value. The lattice attains it because at spacing 2*pi its only non-zero-frequency mass sits at xi = +-1, where kappa_hat vanishes by (c). It is the one configuration that pays nothing.

Since LP decreases, LP(rho) <= LP(1/(2*pi)) for every rho >= 1/(2*pi). The dense side is therefore closed by the bound alone.

4. Uniqueness, and why the threshold is sharp

Equality in section 3 forces rho = 1/(2*pi) (else LP(rho) < LP(1/(2*pi)) strictly) and kappa_hat(2*pi*j/P) |A_j|^2 = 0 for every j != 0. At rho = 1/(2*pi) we have P = 2*pi*m, so kappa_hat(2*pi*j/P) > 0 exactly for 0 < |j| < m, and equality forces

A_j = 0 for j = 1, ..., m-1.

Write z_p = exp(2*pi*i*a_p/P), so A_j is the power sum p_j = sum_p z_p^j. Newton's identities turn p_1 = ... = p_{m-1} = 0 into e_1 = ... = e_{m-1} = 0, so the monic polynomial with roots z_p is z^m - c. Its roots are the m-th roots of c, which are equally spaced on the circle. So T is a translate of the 2*pi lattice.

The threshold matches exactly: the band [-1,1] and the density 1/(2*pi) conspire so that the constrained modes are precisely j = 1..m-1, which is precisely the Newton condition. One fewer and uniqueness would fail; one more and the lattice would not attain the bound.

5. What this proves, and under what hypotheses

Under H1 alone. Let T be a P-periodic configuration with rho = m/P >= 1/(2*pi). Then J(T) <= -4 c2(0) + 2 kappa(0), with equality if and only if rho = 1/(2*pi) and T is a translate of the 2*pi lattice.

The first version of this section also assumed H2, that K_1(d) <= 0 at every non-zero difference, which is what makes f = -kappa where it matters. Section 5a removes it. H2 is nonetheless true of the extremiser, and for a reason worth recording: clip_is_idle_on_lattice(5000) is negative throughout, and

lim_n K_1(2*pi*n)*(2*pi*n)^2 = 2*cos^2(sqrt2/2)*(1 - cosh 1) = -0.6277706...

K_1 has a 1/x^2 asymptotic with two sources. The endpoints of c2 at w = +-1 contribute 2*c2'(1-)*cosh(1)*cos(x), and the kink at w = 0 contributes -2*c2'(0+), which does not oscillate. Both slopes equal -cos^2(sqrt2/2), so at x = 2*pi*n, where the cosine is +1, they combine to the constant above. It is negative precisely because cosh 1 > 1, and that single inequality is why the rectification never fires at a lattice difference. Predicted -0.6277706 against a measured -0.6277713 at n = 5000.

5a. Gap B, closed: an explicit majorant

Gap B asked for v >= K_1^+ vanishing on 2*pi*Z minus the origin with vhat <= kappa_hat. There is a fourth constraint the first draft of this document missed. The bound from g = -kappa + v is LP(rho) + 2*rho*vhat(0) - 2*v(0), so tightness at rho = 1/(2*pi) needs

int v = 2*pi*v(0).

That kills the obvious ansatz. v = (1 - cos x)*psi(x) has the right zeros for free, but it also vanishes at the origin, forcing int v = 0 against v >= K_1^+ >= 0. What works instead is

v(x) = c * s(x), s(x) = (sin(x/2)/(x/2))^2,

because s >= 0, s(2*pi*n) = 0 for n != 0, shat(xi) = 2*pi*(1-|xi|)^+ is supported exactly on [-1,1], and int s = 2*pi = 2*pi*s(0), so the tightness condition holds for every multiple at once.

Feasibility is then two one-dimensional inequalities:

c >= sup_x K_1(x)^+ / s(x) = K_1(0) = 0.9115647... c <= inf_{|xi|<1} khat(xi)/shat(xi) = cos^2(sqrt2/2)*(1 + cosh 1) = 1.4698290...

Both hold at once, so the problem is feasible with margin 0.558. The supremum is attained at x = 0; the tail of K_1^+/s settles at 0.734916 by x = 3000, comfortably under. The infimum is the limit at xi -> 1, and it is in closed form because c2 vanishes linearly there with slope -cos^2(sqrt2/2).

Take c = K_1(0). Then g = -kappa + c*s satisfies g >= f everywhere, ghat = -kappa_hat + c*shat <= 0 everywhere, and g = f at every lattice difference. The bound becomes

LP_v(rho) = LP(rho) + 2c(2*pi*rho - 1),

equal to the lattice value at rho = 1/(2*pi) for any admissible c, and still strictly decreasing in rho because that needs only c < 2*c2(0) = 1.6985..., which the whole admissible interval satisfies. Uniqueness survives too: ghat < 0 strictly on (0,1) since c is strictly below the infimum, so the Newton argument of section 4 runs unchanged.

Measured end to end: 300 random configurations at rho >= 1/(2*pi), zero violations of J <= LP_v(rho).

6. What is missing

Four gaps as first written. Gap B is now closed; A remains the real one, and C and D are routine but unwritten. None should be described as detail.

Gap A, the sparse side. NARROWED, not closed. See sections 6a and 6b. For rho < 1/(2*pi), LP(rho) > LP(1/(2*pi)) and the bound says nothing. It is not close: at rho*2*pi = 0.4 the bound is +2.15 against a true value of -0.043, a margin of 2.2. This is where the work is.

Gap B, the rectification. CLOSED, see section 5a. The concrete majorant v = K_1(0)*(sin(x/2)/(x/2))^2 satisfies every constraint with margin 0.558, so hypothesis H2 is gone and section 5 holds for every configuration at rho >= 1/(2*pi).

What is not closed about it: the two inequalities in 5a are verified numerically on a finite range with a settled tail, not enclosed. rigor.py is the natural next step and would turn this into an enclosure-carrying statement. Both are one-dimensional and should be well within reach.

Gap C, periodic to general. (*) is stated for periodic configurations. General locally finite sets need the autocorrelation-measure form. Standard, and still a step.

Gap D, Poisson. (*) needs justification for the h used. With kappa continuous and L^1, kappa_hat continuous and compactly supported, the sum on the right is finite and the formula is a sampling theorem. Routine, and not written out here.

6a. Gap A: the only possible route, and half of it is already closed off

The bound 2*rho*ghat(0) - 2*g(0) is linear in rho. To hold at every density its slope must be non-negative, but ghat <= 0 everywhere forces ghat(0) <= 0. So there is exactly one way through:

ghat(0) = 0 and g(0) = -LAT/2 = -0.0571650197...,

and then the bound is density-independent and equals the lattice value everywhere at once, closing gap A and the dense side together. Writing g = -kappa + c*s + u with c = K_1(0), this is a problem about u alone:

u >= 0, u(2*pi*n) = 0 for n != 0, uhat <= M := khat - c*shat, uhat(0) = M(0) = 4.9441838, int uhat = uhat(0).

The headroom is exactly saturated at the origin: M(0) and the required uhat(0) are the same expression, so the constraint is tight there with no slack at all.

At bandwidth 1 this is infeasible, and not narrowly. Boas-Kac writes a non-negative band-limited u as |h|^2 with hhat supported on [-1/2,1/2]. That is exactly Nyquist for spacing 2*pi, so Shannon pins h to its samples at 2*pi*k, and u(2*pi*k) = 0 for k != 0 kills every sample but one. What survives is u = |h(0)|^2 * s, a multiple of the very function gap B already used, and uhat <= M caps it at

uhat(0) <= (1.4698290 - 0.9115647) * 2*pi = 3.5076780,

against the 4.9441838 needed. Short by 29%.

What remains open is whether bandwidth above 1 helps, since uhat may go negative outside [-1,1] where M vanishes. Poisson already constrains that hard: uhat(j) = 0 is forced at every non-zero integer, which makes the 2*pi-periodisation of u exactly constant. That is a great deal of rigidity for a function that must also be non-negative and vanish on the whole punctured lattice, but it is not a proof either way.

A method that does not work, recorded with its witness. The obvious attack is a linear program over uhat with u >= 0 imposed at sampled x. It is unsound here, and that is demonstrated rather than suspected, on the case whose answer is known: at bandwidth 1 the LP returns 4.908534, exceeding the proved ceiling by 40%. Its solution dips to -7.6e-4 between its own sample points, reaches -4.2e-2 beyond them, and violates the lattice zeros by up to 0.781 past the last n constrained. Sampling a continuum constraint at finitely many points relaxes it, and the optimiser finds every gap. The right tool is an SDP enforcing u >= 0 exactly through the Fejer-Riesz representation u = |h|^2, with the lattice zeros handled structurally by factoring u = s*q rather than by listing them.

6b. Gap A above bandwidth 1: the SDP built, and what it says

Section 6a leaves one question open, and names the tool: an SDP enforcing u >= 0 through Fejer-Riesz rather than by sampling, with the lattice zeros carried by factoring u = s*q. That is sparse_sdp.py, pinned by test_sparse_sdp.py. The question is still open. What follows is what the tool established and what it did not.

The reduction in 6a is now independently derived. This session reached ghat(0) = 0, g(0) = -LAT/2, and the forced vanishing of uhat at every non-zero integer from the Poisson side, before reading 6a, and landed on the same constants. Two derivations by different routes agree. Hunt #64 (r_b9552d run 2, 2026-08-20) reaches the bandwidth-1 exclusion a third way, through a quoted Paley-Wiener factorisation with critical sampling, and with the same constants in other units: the multiplier it needs, 1.6984560, against the admissible cap 1.4698290, is 6a's 4.9441838 against 3.5076780 once K_1(0) is subtracted and the difference multiplied by 2*pi. It also concludes that any g closing gap A must put Fourier mass outside [-1,1], which is the region this SDP searches.

Calibration first, because 6a records an LP that failed one. At bandwidth 1 the SDP returns u(0) = 0.558236 against the proved ceiling 0.5582643, at scale 1.0000 with residual 0 and no positive-semidefinite repair needed. It reproduces the case whose answer is known, which the sampled LP did not.

Above bandwidth 1, solutions are returned and none of them verify. Solves at B up to 4 return points, and every one collapses to 0 under verification. The reason is specific: after projecting the Gram matrix back onto the positive-semidefinite cone, uhat is strictly positive somewhere in |xi| > 1. There M vanishes, so no positive scaling recovers feasibility, unlike a violation inside [-1,1]. The solver output was not a witness, and reporting it as one would have repeated 6a's own lesson in a new place.

How close the raw solutions came is now on record. The first scan stored only verified values, so the figure this section first gave, raw solutions "reaching as high as 100.00% of the value needed", rested on printed output that was not kept. The scan was re-run at landing (2026-10-21, cvxpy 1.9.3 with Clarabel 0.11.1), and results_sparse_sdp.json now also stores each case's raw objective (raw_u0) and the largest out-of-band uhat of its repaired matrix (max_uhat_out_of_band). The raw u(0) comes to 82.25% of the value needed at B = 1, m = 4, 85.52% at B = 1, m = 6, 94.11% at B = 2, m = 6 and 100.00%, reached from below, at B = 4, m = 8, so the first figure is reproduced, and every one of these points still verifies to 0. The first scan's statuses and verified values reproduced exactly; only the psd_repair column moved, as it can for solves the solver itself reports as inaccurate. The stored figures are pinned by test_sparse_sdp.py, by a test that reads the JSON and needs no solver.

Why a generic solver will not settle this. The feasible set has empty interior. 6a's knife-edge is forced: the lattice attains the bound at rho = 1/(2*pi), so B_g(1) >= LAT, and with the zero-density requirement B_g(0) <= LAT this pins uhat(0) = M(0) and u(0) = M(0)/(2*pi) exactly. An interior-point method has no interior to find, which is what every optimal_inaccurate status is reporting, together with Gram eigenvalues as low as -1.2e-6 (the psd_repair column of results_sparse_sdp.json runs from 1.2e-7 to 1.2e-6 across the four wider solves). The scale-0 verdicts sit at that tolerance: the out-of-band uhat that forces them runs from 2.1e-8 to 4.8e-8 (max_uhat_out_of_band). So they show that the solver cannot decide the question, not that bandwidth above 1 fails. Settling it needs exact arithmetic or enclosures, or a formulation that carries the band-edge constraint structurally rather than as one more sampled inequality.

Three defects found on the way, recorded because each cost real time. The x-space route is unusable here: with the constant in the basis the integrand decays like 1/x^2, and truncation ripple of order 1/x_max lands exactly in |xi| > 1, driving every answer to zero. The constant must be in the basis at all: u = a*s has q constant, which no family of decaying sincs contains, so leaving it out excludes the one solution known to exist. And a scan that read problem.value without problem.status reported 2.4e167 as a number.

7. Grade and scope

The numerics are measured: double precision, no enclosure anywhere. The argument in sections 1 to 4 is a proof sketch and is not formalised; section 5 is what it yields, and section 6 is what it does not. Lattice extremality is not established, T1 is not discharged, and k >= 3 is untouched. Nothing here is evidence about RH.