# Approaches 11–15: constructive, game and phase tests

## Scope and notation

The target throughout is the **general** Chowla cosine conjecture: for every finite nonempty set of distinct positive integers \(A\), with \(N=|A|\),

\[
f_A(\theta)=\sum_{a\in A}\cos(a\theta),\qquad
K_A=-\min_{\theta\in\mathbb T}f_A(\theta),
\]

does \(K_A\ge c\sqrt N\) hold with an absolute positive constant? Haar integrals on the circle are normalized to have total mass 1.

The tests below investigate five particular routes to that universal assertion or to its disproof. A small counterexample can disprove an alleged universal intermediate step, but does not solve the original conjecture. A targeted large-size sample of a proposed counterexample family does not determine its asymptotics.

| ID | Route | Precise outcome |
|---:|---|---|
| 11 | Prefix-floor, noise-shaped rounding of Fejér weights | Exact bounded-prefix and grid identities proved; specified construction has substantial negative witnesses in the tested range, including one rationally certified witness; asymptotic route remains unproved and undisproved. |
| 12 | Nonlinear measure-preserving time reversal | Exact universal reversal is false; uniform approximate reversal with sublinear error is impossible for an infinite admissible family. |
| 13 | Randomize whole sets and interchange minimax | Convexified game has exact value 1/2 for every cardinality N on frequencies 1,…,2N; this relaxation loses the property needed for a growing universal lower bound. |
| 14 | Spin frustration, MaxCut, magnetic gauge | Independent phase and all-negative-gauge shortcuts fail; full distance-graph cuts transfer correctly, but the required universal cut surplus is equivalent to Chowla up to constants. |
| 15 | Clip negative values and restore the spectrum | Exact spectral projection fails to preserve positivity; the natural clip-and-restore iteration converges to the original unknown K_A and gives no new bound in terms of N. |

Executable checks and complete rational certificate: `constructive_checks.py` and `constructive_check_results.json` in this directory.

## 11. One-sided sigma-delta rounding of a positive taper

### Exact candidate

For \(M\ge3\), set \(p_a=1-a/M\), \(1\le a<M\). Their cosine polynomial is

\[
g_M(\theta)=\sum_{a=1}^{M-1}p_a\cos(a\theta)
=\frac{F_M(\theta)-1}{2}\ge-\frac12,
\]

where \(F_M\) is the nonnegative Fejér kernel of integral 1. Let

\[
S_a=\sum_{j=1}^ap_j=a-\frac{a(a+1)}{2M},\quad S_0=0,
\qquad b_a=\lfloor S_a\rfloor-\lfloor S_{a-1}\rfloor\in\{0,1\}.
\]

The proposed set is \(A_M=\{a:b_a=1\}\), with

\[
N_M=\left\lfloor\frac{M-1}{2}\right\rfloor.
\]

The hoped-for counterexample mechanism is that the rounding error could be concentrated inside the positive spikes of \(g_M\), leaving \(K_{A_M}=o(\sqrt{N_M})\). The ordinary L² size of the rounding error does not by itself rule this out: it does not determine which side of zero carries its energy.

### Bounded-prefix identity and its actual scope

Put \(r_a=\{S_a\}\), so \(0\le r_a<1\), \(r_0=0\). The coefficient error is

\[
e_a=b_a-p_a=r_{a-1}-r_a,\qquad
\sum_{a=1}^je_a=-r_j\in(-1,0].
\]

With \(L=M-1\), its complex polynomial satisfies exactly

\[
E_M(z)=\sum_{a=1}^Le_az^a
=(z-1)\sum_{a=1}^{L-1}r_az^a-r_Lz^L.
\]

Hence

\[
|E_M(e^{i\theta})|\le2(M-2)|\sin(\theta/2)|+1.
\]

This controls the error by O(1) on angles of size O(1/M). It does not give a useful small bound on the entire circle, where the Fejér background can be close to −1/2. A bounded internal rounding state therefore does not automatically preserve a one-sided trigonometric bound. This is the exact point at which the intended noise-shaping argument needs an additional estimate.

For comparison with the technique, Daubechies and DeVore's stable sigma-delta work concerns coarse quantization and reconstruction of bandlimited functions. The present full-circle, one-sided coefficient-rounding requirement is an additional assertion, not a conclusion taken from that theory.

### A further exact identity, valid for every odd M

For odd \(M\), write \(N=(M-1)/2\). Direct calculation gives

\[
S_{M-a}=N-\frac{a(a-1)}{2M},\qquad
b_{M-a}=1-b_a\quad(1\le a<M).
\]

Thus the set selects one representative from every pair \(\{a,M-a\}\). At every nontrivial M-th root,

\[
\tag{11.1}
f_{A_M}(2\pi j/M)=-\frac12\qquad(1\le j<M).
\]

Proof: cosines of the paired frequencies agree on that grid, and the sum of all nonconstant M-th root characters is −1. This exact discrete flatness does not bound the values between grid points.

### Targeted check of the specified construction

No frequency-set enumeration or optimization is used here. The bits are computed from the exact integer numerator \(2Ma-a(a+1)\), then the single resulting polynomial is sampled on a uniform grid of 16M points.

| M | N_M | Most negative sampled value | Sampled −f/√N_M |
|---:|---:|---:|---:|
| 255 | 127 | −22.00691086 | 1.95279756 |
| 256 | 127 | −13.57664904 | 1.20473279 |
| 257 | 128 | −19.33590623 | 1.70906880 |
| 1024 | 511 | −34.17097612 | 1.51163499 |
| 4096 | 2047 | −64.27028114 | 1.42053284 |
| 16384 | 8191 | −147.70099541 | 1.63198048 |
| 65536 | 32767 | −320.97726526 | 1.77319269 |
| 262144 | 131071 | −698.23475275 | 1.92862661 |

The floating point values are not certified continuous minima. A negative value at one angle is already a lower-bound witness for K, so a separate exact evaluation was made at

\[
M=262144,\qquad \theta=2\pi\frac{2023809}{4194304}.
\]

Machin's formula encloses π rationally; Taylor remainders enclose the sine and cosine of this angle. A fixed-point complex recurrence with denominator \(Q=2^{100}\) then sums all selected terms. If the approximated unit complex number has error at most \(2/Q\), rounding each multiplication gives an error at exponent a at most \(6a/Q\), because

\[
R_{a+1}\le(1+2/Q)R_a+3/Q,
\qquad (1+2/Q)^a\le\frac1{1-2a/Q}\le2.
\]

Adding these rational error bounds and performing a squared rational comparison proves

\[
\tag{11.2}
\frac{K_{A_{262144}}}{\sqrt{131071}}>\frac{241}{125}=1.928.
\]

The certificate does not rely on trusting the FFT output; the FFT only chose the angle.

**Status.** The simple construction fails to hide negative excursions in the tested range, and its bounded-prefix property does not prove the desired global bound. Neither this finite range nor (11.2) disproves the asymptotic statement \(K_{A_M}=o(\sqrt{N_M})\), and no such asymptotic disproof is claimed. Other noise-shaped rounding rules remain outside this check.

## 12. Nonlinear reversal of time

### Candidate and exact obstruction

The candidate is a Haar-measure-preserving map \(T:\mathbb T\to\mathbb T\) such that \(f_A\circ T=-f_A\). It would make the whole value distribution symmetric and force its maximum and negative minimum to agree.

Every odd moment would then vanish. But for \(A=\{1,2\}\), exact multiplication of the Laurent polynomial with coefficients 1/2 at ±1,±2 gives

\[
\int f_A^3=\frac34\ne0.
\]

Thus no such measure-preserving T exists for this valid frequency set. Also

\[
f_A=2\left(\cos\theta+\frac14\right)^2-\frac98,
\]

so its range is \([-9/8,2]\). Any transformation, even one not preserving measure, obeys

\[
\|f_A\circ T+f_A\|_\infty\ge2-\frac98=\frac78.
\]

The same statement holds for the essential supremum, since f is arbitrarily close to its maximum on sets of positive measure.

### Why approximate reversal cannot universally have a small uniform error

For every A and every T,

\[
\tag{12.1}
\|f_A\circ T+f_A\|_\infty\ge N-K_A.
\]

There is an infinite admissible family for which the right side is linear in N. Take \(B=\{b^j:0\le j<m\}\), \(b\ge4\), and the set A of all positive differences of distinct elements of B. The differences are distinct: their b-adic valuation identifies the smaller exponent, and then their value identifies the larger one. Hence

\[
N=\frac{m(m-1)}2,\qquad
f_A(\theta)=\frac{\left|\sum_{j=0}^{m-1}e^{ib^j\theta}\right|^2-m}{2}\ge-\frac m2.
\]

Equation (12.1) gives uniform reversal error at least \(N-m/2\). A universal reversal with error o(N), and therefore one with error O(√N), is impossible.

For an L¹ version on A={1,2}, measure preservation and the third moment imply

\[
\frac32=\left|\int\big((f\circ T)^3+f^3\big)\right|
\le12\int|f\circ T+f|,
\]

so even the mean absolute error is at least 1/8. A weak mean-error estimate would in any event require a separate argument to control an extreme value.

**Status.** The proposed exact reversal and the uniform sublinear-error variant are disproved. Replacing one common time transformation by independently choosing each frequency's phase changes the feasible set and is not a reversal of this signal.

## 13. Minimax with randomized whole frequency sets

### Candidate

Allow the frequency player to randomize over all N-element subsets of \(\{1,\ldots,2N\}\), and interchange the order of the two players' optimizations. Expected payoffs depend only on the inclusion probabilities \(p_a\in[0,1]\), \(\sum p_a=N\). Convex minimax theorems apply to this mixed game, but this changes the deterministic optimization.

### Exact value of the relaxed game, for every N

Set \(L=2N+1\) and \(p_a=1-a/L\), \(1\le a<L\). These probabilities have sum N, and

\[
\sum_{a=1}^{L-1}p_a\cos(a\theta)=\frac{F_L(\theta)-1}{2}\ge-\frac12.
\]

This proves that the randomized frequency player can hold the worst expected negative value to at most 1/2.

It is achievable by randomizing **whole N-element sets**, not merely allowing a random cardinality. Put \(S_a=\sum_{j\le a}p_j\), choose \(U\) uniformly in [0,1), and define

\[
b_a(U)=\lfloor S_a+U\rfloor-\lfloor S_{a-1}+U\rfloor.
\]

Every bit is 0 or 1, their sum is exactly N, and \(\mathbb E b_a=p_a\).

For the opposite bound, let the time player use the uniform distribution on the L−1 nontrivial L-th roots. Each frequency \(1\le a<L\) has average cosine \(-1/(L-1)\). Thus for **every** N-element set A the time player's expected payoff is

\[
\frac{N}{L-1}=\frac12.
\]

We have proved exactly

\[
\tag{13.1}
\min_{\operatorname{law}(A),\ |A|=N}
\max_\theta\mathbb E[-f_A(\theta)]
=\frac12
\qquad\text{on the universe }\{1,\ldots,2N\}.
\]

By contrast, already for N=1 every deterministic singleton has negative minimum 1. More generally, the full conjecture concerns choosing the time after seeing the realized set, which involves \(\mathbb E[\max_\theta(-f_A)]\), not \(\max_\theta\mathbb E[-f_A]\).

**Status.** This exact relaxation cannot prove any growing lower bound in N. The finite-universe calculation is a universal-in-N obstruction to the proposed relaxation, not a replacement target. An additional rounding theorem that preserves a suitable negative witness would be necessary; equation (13.1) supplies no such theorem.

## 14. Spin frustration, MaxCut and magnetic gauge

### The invalid independent-phase and gauge shortcuts

A time orbit has unit spins

\[
v_n(\theta)=(\cos n\theta,\sin n\theta),\qquad
v_j\cdot v_k=\cos((j-k)\theta).
\]

Its complex phases satisfy all integer relations, in particular \(z_{j+k}=z_jz_k\). Arbitrarily optimized unit vectors or separate frequency phases need not satisfy these constraints. For A={1,2}, independent phases π,π give a relaxed sum −2, but \(z_2=z_1^2\) forbids this assignment and the true minimum is −9/8.

A sign gauge making every edge coupling negative would require \(\sigma_j\sigma_k=-1\) on every edge. Such signs exist exactly when the graph is bipartite: signs provide the two classes, and conversely any bipartition supplies signs. The distance graph for A={1,2} contains the triangle 0,1,2. Multiplying its three required equalities yields the contradiction 1=−1. Thus this gauge cannot be imposed for arbitrary A.

### Whole-graph cuts nevertheless give valid witnesses

It would be incorrect to dismiss every MaxCut approach because a cut is not a single time orbit. For a window of length L and signs \(s_0,\ldots,s_{L-1}\), define

\[
q(\theta)=\frac1{\sqrt L}\sum_{n=0}^{L-1}s_ne^{in\theta}.
\]

The density \(|q|^2\) is nonnegative and has integral 1. Orthogonality gives exactly

\[
\tag{14.1}
\int f_A|q|^2
=\frac1L\sum_{\substack{a\in A\\a<L}}\sum_{n=0}^{L-1-a}s_ns_{n+a}.
\]

Consequently a cut with a sufficiently negative right side really does imply an at least equally negative value of f at some time. This transfer is by a nonnegative average, not by rounding every vertex phase to one common time.

Let \(E_{A,L}\) be the number of edges of the distance graph on the window. Its sign energy is \(E_{A,L}-2\operatorname{Cut}(s)\). A universal surplus

\[
\operatorname{Cut}(s)\ge E_{A,L}/2+cL\sqrt N
\]

for an appropriate window and cut would prove Chowla via (14.1).

### The required surplus is itself equivalent to the target up to constants

Suppose θ* minimizes f. Choose a uniform phase Φ, then conditionally independent signs with means

\[
\mathbb E[s_n\mid\Phi]=\cos(n\theta_*+\Phi).
\]

For distinct indices,

\[
\mathbb E[s_ns_{n+a}]=\frac12\cos(a\theta_*).
\]

It follows that the expected sign energy per vertex is

\[
\frac12\sum_{\substack{a\in A\\a<L}}(1-a/L)\cos(a\theta_*)
\longrightarrow -K_A/2
\quad(L\to\infty).
\]

For every fixed A and sufficiently large L there is therefore a cut with sign energy per vertex at most \(-K_A/4\). Its surplus is at least \(LK_A/8\). Thus a root-scale Chowla bound implies the proposed root-scale distance-graph surplus, while (14.1) proves the converse, with changed constants.

The Goemans–Williamson semidefinite relaxation and rounding theorem concerns approximation to the optimum cut. It does not, by itself, establish this stronger absolute surplus over half of all edges for every integer distance graph.

**Status.** Two shortcuts are rigorously blocked, while the complete MaxCut route remains valid but has an unresolved universal estimate equivalent in scale to the original problem. No new surplus theorem is claimed.

## 15. Rectify negative values and restore the Fourier support

### Spectral projection does not preserve positivity

Let \(\mathcal P_A\) retain Fourier frequencies \(\{0\}\cup A\cup(-A)\). It is convolution by

\[
D_A=1+2f_A.
\]

For every nonempty A this kernel has a negative value. Indeed, if \(D_A\ge0\), then for any \(a\in A\),

\[
\int D_A(\theta)(1-\cos a\theta)=1-1=0.
\]

The integrand would be nonnegative, and \(1-\cos a\theta\) is positive except on a finite set, forcing \(D_A=0\) almost everywhere, contrary to its integral 1.

Positive approximate identities witness failure of positivity preservation. In fact,

\[
\mathcal P_A F_M
=1+2\sum_{\substack{a\in A\\a<M}}(1-a/M)\cos(a\theta)
\longrightarrow D_A
\]

uniformly. For M larger than all a, the error is at most \(2\sum_{a\in A}a/M\). Therefore a negative value of D survives the projection of a sufficiently narrow nonnegative Fejér kernel.

An explicit exact example is A={1,2} and

\[
F_4=1+\frac32\cos\theta+\cos2\theta+\frac12\cos3\theta\ge0.
\]

After projection,

\[
\tag{15.1}
(\mathcal P_{\{1,2\}}F_4)(2\pi/3)
=1-\frac34-\frac12=-\frac14.
\]

This failure also occurs directly after clipping a cosine sum: for A={1}, \(g=(\cos\theta)_+\) has mean 1/π and first cosine coefficient 1/2, so

\[
\mathcal P_{\{1\}}g=1/\pi+(1/2)\cos\theta
\]

is negative at π. Resetting the coefficient to 1 makes that negative value still smaller.

### Exact behavior of a natural alternating clip-and-restore rule

There is a general analysis of the iteration that clips negative values, then restores all the prescribed cosine coefficients to 1 while retaining the new mean. Starting with \(c_0\le K_A\), its successive functions are \(c_j+f_A\), where

\[
\tag{15.2}
c_{j+1}=\int(c_j+f_A)_+
=c_j+\int(-c_j-f_A)_+.
\]

For every A,

\[
c_j\uparrow K_A.
\]

Proof: the last integral makes the sequence nondecreasing. If \(c_j\le K_A\), then pointwise \((c_j+f_A)_+\le K_A+f_A\), so \(c_{j+1}\le K_A\). Let its limit be c*. The map on the right of (15.2) is continuous. If \(c_*<K_A\), continuity of f gives a set of positive measure where \(-c_*-f_A>0\), so the next increment is strictly positive, contradicting the fixed-point equation at the limit. Hence \(c_*=K_A\).

This produces a nonnegative function with unit coefficients only by allowing the constant term to rise to exactly the originally unknown threshold. If instead a constant c<K_A is fixed in advance, positivity and the prescribed spectrum are incompatible. Alternating projection cannot be assumed to converge to a feasible point when the intersection is empty. If the support itself is allowed to change among binary choices, that additional constraint is nonconvex and the usual convex projection conclusions do not apply.

**Status.** Positivity preservation is disproved exactly, and the natural mean-adaptive iteration is understood for every A. It recovers K_A but supplies no estimate of K_A as a function of N; such an estimate would require a new bound on the accumulated mean increments.

## Primary references inspected

- I. Daubechies and R. DeVore, *Approximating a bandlimited function using very coarsely quantized data: A family of stable sigma-delta modulators of arbitrary order*, Annals of Mathematics 158 (2003), 679–710: <https://annals.math.princeton.edu/2003/158-2/p09>.
- M. Sion, *On general minimax theorems*, Pacific Journal of Mathematics 8 (1958), 171–176: <https://msp.org/pjm/1958/8-1/pjm-v8-n1-p14-p.pdf>, especially its explicit convexity hypotheses.
- M. X. Goemans and D. P. Williamson, *Improved approximation algorithms for maximum cut and satisfiability problems using semidefinite programming*, JACM 42 (1995), 1115–1145: <https://math.mit.edu/~goemans/PAPERS/maxcut-jacm.pdf>.

The algebraic derivations and the certificates in this note are supplied in full and do not depend on treating these references as statements of Chowla's conjecture or of the missing intermediate estimates.

## Independent cross-audit of physics approaches 1–5

The author of this note independently read the completed `physics.md` and checked the decisive calculations. No mathematical error was found within the stated scope.

- **Lee–Yang step:** the partition function is entire of exponential type, has real coefficients and value 1 at zero. If every zero were imaginary, conjugate pairs cancel the genus-one exponential factors; the remaining factor is exp(aβ). Its derivative at zero forces a=0. The resulting evenness contradicts the exact third moment 3/4. The argument also covers a finite or empty zero set.
- **Determinant/Mahler step:** for A={1,2,3}, the factorization with α=(√7−1)/6 and c=(1+2√7)/6 gives h=8(cos θ−α)²(cos θ+c). The logarithmic averages give G(h)=c+√(c²−1). The exact value L²=(2528+952√7)/729 is less than 7, while G(h)²+6 exceeds 7. The factor 2 in F=2f and the squared geometric mean are consistent.
- **Other steps:** the free-tree counterexample, positive definiteness of every finite Toeplitz matrix, the heat-kernel tail and N^(3/2) ratio divergence, and the Brownian variance/quadratic-variation and zero-hitting assertions were checked algebraically. The stochastic countermodel is explicitly outside the binary polynomial class, so its scope is correctly limited to the proposed scalar inference.

This was an independent mathematical audit of the text, not a new formal Lean verification or a claim that the original conjecture has been resolved.