# Unconventional approaches 1–5: exact checks of proposed mechanisms

Research checkpoint, 18 October 2026. These five approaches address the **general** Chowla cosine problem. A small or structured example below tests a proposed universal inference; no frequency subclass is substituted for the original problem.

Throughout,

\[
f_A(x)=\sum_{a\in A}\cos(ax),\quad N=|A|,\quad
K=-\min_x f_A(x),\quad F=2f_A,\quad L=2K,
\]

with normalized circle measure \(d\mu=dx/(2\pi)\). The target is \(N\le C K^2\) with one absolute constant for every finite nonempty \(A\subset\mathbb Z_{>0}\).

**Outcome:** none of these five attempts proves or disproves the general conjecture. Each has been checked beyond an analogy. Several specific transfers are rigorously false. Their failures do not prove that every possible refinement of the corresponding field is useless.

| ID | Proposed mechanism actually tested | Result |
|---|---|---|
| 1 | Transfer the free-group negative spectral edge to the commuting frequency operator | Exact transfer is false already for \(A=\{1,2\}\) |
| 2 | Obtain a particle/rank cutoff or a variance subtraction bound from full Toeplitz determinants | Cutoff and the tested determinant free-energy inequality are false |
| 3 | Apply a Lee–Yang imaginary-axis zero theorem to the one-angle partition function | Universal imaginary-axis property is false already for \(A=\{1,2\}\) |
| 4 | Control energy/minimum ratio by running the whole wave through heat flow | Natural flow-stable ratio fails by an unbounded factor |
| 5 | Force a large negative barrier using Brownian quadratic variation and optional stopping | Scalar martingale inference is false; the exact minimum barrier is almost surely not hit |

## 1. Free/noncommuting rotations and the Kesten spectral edge

### Exact candidate

On \(\ell^2(\mathbb Z)\), let \(U\) be the bilateral shift. The original self-adjoint operator is

\[
H_A=\sum_{a\in A}(U^a+U^{-a}),
\qquad
\inf\operatorname{spec}(H_A)=\min_x F(x)=-2K.
\]

Replace these commuting powers by the canonical left-regular generators \(V_1,\ldots,V_N\) of the free group:

\[
H_{\rm free}=\sum_{j=1}^{N}(V_j+V_j^*).
\]

Kesten's theorem gives

\[
\operatorname{spec}(H_{\rm free})
=[-2\sqrt{2N-1},\,2\sqrt{2N-1}].
\]

A proposed covering/quotient transfer would be

\[
\inf\operatorname{spec}(H_A)
\le \inf\operatorname{spec}(H_{\rm free}).
\tag{1.1}
\]

If true, this would give \(K\ge\sqrt{2N-1}\), much more than the desired order.

### Exact check and counterexample

For \(A=\{1,2\}\), write \(y=\cos x\). Then

\[
f_A(x)=2y^2+y-1
=2(y+1/4)^2-\frac98.
\]

Hence \(K=9/8\) and

\[
\inf\operatorname{spec}(H_A)=-9/4
>-2\sqrt3=\inf\operatorname{spec}(H_{\rm free}).
\]

Thus (1.1) is false **within the original torsion-free integer setting**. A counterexample in a finite group is unnecessary.

### Why the analogy fails

Kesten's quotient comparison concerns the **upper spectral edge/spectral radius**. Additional relations increase return probabilities and can increase that edge. They do not preserve spectral symmetry: the free tree is bipartite, while the relation \(1+1=2\) creates triangles in the quotient. The upper edge of \(H_A\) is always \(2N\); its growth tells us nothing sufficient about the negative edge.

The first quotient-comparison lemma and Theorem 3 were checked directly in Kesten's paper. The distinction between the two edges is present in the primary source, not an interpretation added to rescue the argument.

### Remaining mathematical candidate

A weaker transfer with an unknown absolute loss,

\[
-\inf\operatorname{spec}(H_A)
\ge c\,[-\inf\operatorname{spec}(H_{\rm free})],
\]

would suffice. No such transfer was proved. Establishing it would require controlling precisely the negative-edge effect of all additive relations; the free model alone has removed this information.

Primary source: [Kesten, Symmetric random walks on groups (1959), Lemma 3.1 and Theorem 3](https://sites.math.northwestern.edu/~auffing/papers/KRW.pdf).

## 2. Fermionic exterior powers and full Toeplitz determinants

### Exact construction

Put \(h=L+F\ge0\), and let

\[
T_m(h)=(\widehat h(j-i))_{0\le i,j<m}.
\]

These are the Gram matrices of the functions \(e^{ijx}\) in \(L^2(h\,d\mu)\). The fermionic squared volume is

\[
D_m=\det T_m(h)
=\frac1{m!}
\int_{\mathbb T^m}
\prod_{\ell=1}^m h(x_\ell)
\prod_{i<j}|e^{ix_i}-e^{ix_j}|^2\,d\mu^m.
\tag{2.1}
\]

Identity (2.1) follows directly by expanding both Vandermonde determinants and integrating the products: after the \(m!\) identical permutation terms are collected, the remaining sum is the determinant of the one-particle Gram matrix. Its Vandermonde factor is exactly the algebraic exclusion effect.

Two concrete proposed consequences were checked:

1. Reaching \(h=0\) somewhere should eventually collapse these volumes/ranks, giving an effective finite occupancy bound in terms of \(L\).
2. A natural variance subtraction for the geometric symbol average,
   \[
   G(h)^2+2N\le L^2,\qquad G(h)=\exp\!\int\log h\,d\mu,
   \tag{2.2}
   \]
   could turn exclusion plus the \(2N\) unit off-diagonal Fourier entries into a quadratic bound.

### Rank and integrality check: the proposed collapse is false

For \(A=\{1\}\), \(K=1\), \(L=2\), and

\[
h(x)=2+2\cos x
\]

vanishes at \(x=\pi\). Nevertheless, \(T_m(h)\) is the tridiagonal matrix with diagonal 2 and adjacent entries 1. Its determinants satisfy

\[
D_0=1,\quad D_1=2,\quad D_m=2D_{m-1}-D_{m-2},
\]

so

\[
D_m=m+1>0
\]

for **every** \(m\). The rank is \(m\), with no finite cutoff. This is an exact infinite sequence of full Toeplitz determinants.

More generally, a nonzero nonnegative trigonometric polynomial is positive except at finitely many points. For every nonzero polynomial \(p\),

\[
\int h|p|^2\,d\mu>0.
\]

Hence all the finite Toeplitz matrices remain positive definite even at the true minimum shift. The one-particle Hilbert space is still infinite-dimensional.

Integrality does not create the missing cutoff. At an integer shift \(\ell\ge L\), one only gets

\[
1\le\det T_m(\ell+F)\le\ell^m.
\]

At the actual critical shift, integrality need not hold at all. For example, the critical shift for \(A=\{1,2,3\}\), calculated below, is irrational.

### A new explicit determinant/variance candidate also fails

For \(A=\{1,2,3\}\),

\[
f_A(x)=4y^3+2y^2-2y-1,\qquad y=\cos x.
\]

Its global minimum on \([-1,1]\) occurs at

\[
\alpha=\frac{\sqrt7-1}{6},\qquad
K=\frac{17+7\sqrt7}{27}.
\]

This follows by checking the two derivative roots and the endpoints. With

\[
c=\frac{1+2\sqrt7}{6}>1
\]

the critical symbol factors exactly as

\[
h(x)=2K+2f_A(x)=8(\cos x-\alpha)^2(\cos x+c).
\]

The logarithmic averages can be evaluated without a numerical integral:

\[
\int\log|\cos x-\alpha|\,d\mu=-\log2,\qquad
\int\log(c+\cos x)\,d\mu
=\log\frac{c+\sqrt{c^2-1}}2.
\]

For the first identity, factor \(2(\cos x-\alpha)\) into two unit-circle linear factors. For the second, factor \(c+\cos x\) as a positive constant times \(|1+r e^{ix}|^2\), \(0<r<1\), and integrate its convergent logarithmic power series. Therefore

\[
G(h)=c+\sqrt{c^2-1}>1.
\]

But

\[
L^2
=\frac{2528+952\sqrt7}{729}
<7=1+2N<G(h)^2+2N.
\]

For an exact elementary check of \(L^2<7\), use
\(\sqrt7<8/3\), giving
\(2528+952\sqrt7<15200/3<5103=7\cdot729\).
Thus (2.2) is false. Numerically the two sides are approximately \(7.86067\) and \(6.92285\), but the displayed algebra, not that decimal comparison, proves failure.

### What exterior powers actually supply

The second elementary symmetric polynomial in the eigenvalues gives the valid inequality

\[
\operatorname{tr}(\wedge^2 T_m)
=\frac{m(m-1)}2L^2-
\sum_{\substack{a\in A\\a<m}}(m-a)\ge0.
\tag{2.3}
\]

The finite-index weights and the factor \(m(m-1)\) remain. As \(m\to\infty\) for a fixed \(A\), (2.3) does not become \(N\le CL^2\); its resulting lower bound for \(L^2\) tends to zero.

All exterior powers preserve genuine information, but no degree-free inequality extracting \(N\le CL^2\) from them was obtained. A weaker version of (2.2) with an unknown constant multiplying \(N\) remains an unproved candidate, not a consequence of exclusion. The exact counterexample only rejects the coefficient 2 and the proposed rank/integrality shortcuts.

## 3. Complex temperature, partition functions, and Lee–Yang zeros

### Exact candidate

Define the single-angle partition function

\[
Z_A(\beta)=\int e^{\beta f_A(x)}\,d\mu(x),\qquad \beta\in\mathbb C.
\]

It is entire of exponential type because \(|f_A|\le N\), and

\[
Z_A(0)=1,\quad Z'_A(0)=0,\quad Z_A''(0)=N/2.
\]

One candidate was to import the Lee–Yang imaginary-axis zero property, and use the resulting product formula to constrain the zero-temperature limit and hence \(K\).

The relevant limit is indeed

\[
K=\lim_{\beta\to+\infty}\frac{\log Z_A(-\beta)}{\beta}.
\tag{3.1}
\]

For completeness: the upper bound follows from \(f_A\ge-K\). For every \(\varepsilon>0\), continuity supplies a positive-measure neighborhood with \(f_A\le-K+\varepsilon\); integrating over it gives the matching lower exponential rate. Equation (3.1) by itself is just a reformulation.

### Exact analytic test: the universal zero-location premise is false

For \(A=\{1,2\}\), Fourier orthogonality gives

\[
Z_A'(0)=0,\qquad Z_A'''(0)=\int(\cos x+\cos2x)^3\,d\mu=\frac34.
\tag{3.2}
\]

Suppose all zeros of \(Z_A\) were purely imaginary. Real Taylor coefficients imply that the nonzero zeros occur in pairs \(it_j,-it_j\), with matching multiplicities. Hadamard factorization for an entire function of exponential type gives, after pairing,

\[
Z_A(\beta)=e^{a\beta}
\prod_j\left(1+\frac{\beta^2}{t_j^2}\right).
\tag{3.3}
\]

The product is convergent because an exponential-type entire function has
\(\sum_j |t_j|^{-2}<\infty\). The usual genus-one exponential factors cancel within each pair. The possible remaining polynomial in the exponential has degree at most one, and the constant is fixed by \(Z_A(0)=1\).

Differentiating (3.3) at zero and using (3.2) gives \(a=0\). Thus \(Z_A\) would be even, so \(Z_A'''(0)=0\), contradicting (3.2). This proves that \(Z_{\{1,2\}}\) has zeros **off** the imaginary axis. No numerical zero search is needed.

### Why the physical theorem does not transfer

The harmonics are all functions of the same angle. They are not independent spins with the required ferromagnetic pair interactions. In addition, \(\beta\) here couples to the full nonlinear observable \(f_A\), rather than to the standard magnetization variable in the usual circle theorem.

The Lieb–Sokal extension starts with a measure already possessing the Lee–Yang property and shows preservation under suitable ferromagnetic pair interactions. The pushforward of uniform angle under \(\cos x+\cos2x\) already fails that starting property, as the exact calculation proves.

### Remaining candidate

A genuinely new theorem controlling the off-axis zero configuration of this particular \(Z_A\), or a uniform bound such as

\[
Z_A(-c_1/\sqrt N)\ge e^{c_2}
\]

with fixed positive constants, would yield useful information. No such bound was established. The latter would imply \(K\ge(c_2/c_1)\sqrt N\) via \(Z_A(-\beta)\le e^{\beta K}\), so it must not be mistaken for an already-proved thermal identity.

Primary sources checked:

- [Lee–Yang, Statistical theory of equations of state and phase transitions II (1952)](https://journals.aps.org/pr/abstract/10.1103/PhysRev.87.410).
- [Lieb–Sokal, A general Lee–Yang theorem for one-component and multicomponent ferromagnets (1981)](https://link.springer.com/article/10.1007/BF01213009).
- [Hadamard, Étude sur les propriétés des fonctions entières (1893), scanned original](https://www.numdam.org/item/JMPA_1893_4_9__171_0.pdf).

## 4. Whole-wave heat flow, entropy, and the loss of binary data

### Exact evolution and identities

Evolve every frequency simultaneously:

\[
f_t(x)=\sum_{a\in A}e^{-a^2t}\cos(ax).
\]

Let \(K_t=-\min f_t\) and \(E(t)=\int f_t^2\,d\mu\). The heat kernel is positive, hence \(K_t\le K_0\) and \(K_0+f_t\ge0\). Parseval and differentiation give

\[
E(t)=\frac12\sum_{a\in A}e^{-2a^2t},
\quad
E'(t)=-2\int |f_t'|^2\,d\mu,
\quad
2\int_0^\infty\!\int|f_t'|^2\,d\mu\,dt=N/2.
\tag{4.1}
\]

For \(u_t=K_0+\varepsilon+f_t>0\), the entropy identity is

\[
\frac{d}{dt}\int u_t\log\frac{u_t}{K_0+\varepsilon}\,d\mu
=-\int\frac{|u_t'|^2}{u_t}\,d\mu.
\tag{4.2}
\]

It follows simply by integration by parts using \(\partial_t u_t=u_t''\), periodicity, and conservation of mass. Thus no uncertain entropy theorem is being assumed.

### Concrete candidate actually checked

A possible backward strategy needs a quantity retaining the desired variance/minimum relationship under smoothing. Test the natural claim that

\[
R(t)=\frac{E(t)}{K_t^2}
\]

is controlled by an absolute constant, or even by a constant multiple of \(R(0)\), along heat flows that begin with binary Fourier coefficients.

### Exact unbounded counterexample to that control

Take the actual unit-coefficient initial set \(A=\{1,\ldots,N\}\) and \(t=1/N\). The full circle heat kernel

\[
\Theta_t(x)=\sum_{n\in\mathbb Z}e^{-n^2t}e^{inx}
=\sqrt{\frac{\pi}{t}}\sum_{k\in\mathbb Z}
e^{-(x-2\pi k)^2/(4t)}
\]

is nonnegative. This identity can be verified by taking the Fourier coefficients of the periodic Gaussian. Hence

\[
\sum_{a=1}^{\infty}e^{-a^2/N}\cos(ax)\ge-\frac12.
\]

The omitted tail obeys

\[
\sum_{a=N+1}^{\infty}e^{-a^2/N}
\le e^{-N}\sum_{j=1}^\infty e^{-2j}
=\frac{e^{-N}}{e^2-1}.
\]

It follows that the finite, evolved wave satisfies

\[
K_{1/N}\le \frac12+\frac{e^{-N}}{e^2-1}.
\tag{4.3}
\]

At the same time,

\[
E(1/N)
=\frac12\sum_{a=1}^{N}e^{-2a^2/N}
\ge\frac{e^{-2}}2\lfloor\sqrt N\rfloor.
\tag{4.4}
\]

Thus \(R(1/N)\) is unbounded.

The failure relative to the initial value is even stronger. The Dirichlet identity at \(x=3\pi/(2N+1)\) gives

\[
K_0\ge
\frac{1}{2\sin(3\pi/(4N+2))}+\frac12
\ge\frac{2N+1}{3\pi}+\frac12.
\]

Since \(E(0)=N/2\), this yields \(R(0)=O(1/N)\), whereas
\(R(1/N)\ge c\sqrt N\). Consequently even a fixed-factor comparison of those ratios is false by a factor of order at least \(N^{3/2}\).

### What remains and what this does not show

The initial wave has a large trough and satisfies the conjectured lower bound comfortably. It is the proposed invariant through the **whole-wave transformation** that fails. The evolved coefficients \(e^{-a^2t}\) are no longer binary.

Keeping \(K_0\) fixed preserves positivity, but then (4.1)–(4.2) are valid for general nonnegative initial densities with arbitrarily concentrated peaks; they contain no established binary-specific reverse bound. A successful backward-flow proof must retain additional arithmetic information about the initial coefficients and quantify how it controls the complete trajectory. No such degree-free estimate was derived in this pass.

## 5. Brownian optional stopping for the analytic polynomial

### Exact martingale representation

Let

\[
P_A(z)=\sum_{a\in A}z^a,\qquad u(z)=\Re P_A(z).
\]

Let \(B_t\) be standard planar Brownian motion starting at zero, with generator \(\tfrac12\Delta\), and let \(\tau\) be its first exit from the unit disk. Then

\[
M_t=u(B_{t\wedge\tau})
\]

is a bounded stopped martingale. The exit angle is uniform, so

\[
\mathbb E M_\tau=0,\quad
\mathbb E M_\tau^2=N/2,\quad
-K\le M_t\le N.
\tag{5.1}
\]

The Cauchy–Riemann equations give \(|\nabla u|^2=|P_A'|^2\). Itô's formula/isometry therefore yields

\[
\mathbb E\langle M\rangle_\tau
=\mathbb E\int_0^\tau |P_A'(B_s)|^2\,ds
=N/2.
\tag{5.2}
\]

The proposed inference was that this much accumulated fluctuation forces a negative barrier of order \(\sqrt N\).

### Exact stopping countermodel with the same scalar data

Let \(W_t\) be ordinary one-dimensional Brownian motion starting at zero and stop it upon exiting \((-1/2,N)\), at time \(\sigma\). The terminal probabilities are

\[
\mathbb P(W_\sigma=N)=\frac1{2N+1},\qquad
\mathbb P(W_\sigma=-1/2)=\frac{2N}{2N+1}.
\]

Therefore, exactly,

\[
\mathbb E W_\sigma=0,\quad
\mathbb E W_\sigma^2=N/2,\quad
\mathbb E\langle W\rangle_\sigma=\mathbb E\sigma=N/2.
\]

These statements follow either from stopping \(W_t\) and \(W_t^2-t\), or by solving the elementary interval exit equations. This continuous martingale has the same mean, variance, upper bound and expected quadratic variation as (5.1)–(5.2), but its negative barrier is \(1/2\) for arbitrarily large \(N\).

Thus the proposed inference from those data is rigorously false. The countermodel is **not** asserted to arise from a binary-coefficient cosine polynomial.

### Analyticity alone does not fix the failure

For \(0<r<1\), consider the analytic function

\[
P_r(z)=\frac{rz}{1-rz}=\sum_{a\ge1}r^a z^a.
\]

Its boundary real part is half the Poisson kernel minus \(1/2\), and

\[
K_r=\frac{r}{1+r}<\frac12,\qquad
\int(\Re P_r)^2\,d\mu=\frac{r^2}{2(1-r^2)}\longrightarrow\infty.
\]

Hence even bounded analytic functions with positive Fourier coefficients at most one can exhibit precisely the rare-positive-exit mechanism. Finite truncations preserve it up to an arbitrarily small uniform error. Exact coefficient membership in \(\{0,1\}\), rather than positivity or an upper coefficient bound, must do essential work.

### A further exact barrier issue

Stopping \(u(B_t)\) at its **actual** global minimum \(-K\) does not produce a useful negative hitting event. The strong minimum principle gives \(u(z)>-K\) in the open disk. On the boundary, \(f_A(x)+K\) is a nonzero trigonometric polynomial and has finitely many zeros. A uniform exit angle hits that finite set with probability zero.

One can stop at a slightly higher negative level instead, but then the rare-positive-exit calculation still has to be controlled; optional stopping does not remove it.

### Precise remaining requirement

A useful new result would have to bound the Brownian occupation or positive-exit contribution by exploiting the exact binary Fourier coefficients, for example enough to force

\[
\mathbb E\int_0^\tau |P_A'(B_s)|^2\,ds\le C K^2.
\]

In light of (5.2), this displayed estimate is itself of the strength required to solve Chowla. It was not proved here, and labeling it a martingale inequality does not make it an available theorem.

Primary source for the stochastic integral/isometry used above:
[Itô, Stochastic Integral (1944), Theorem 2.1 and the final change-of-variables formula](https://www.jstage.jst.go.jp/article/pjab1912/20/8/20_8_519/_pdf).

## Verification record and scope

The file physics_check.py independently computes the small polynomial minima, exact signed-frequency moments, integer Toeplitz determinants through size 12, heat-flow energy bounds, and interval-exit moments using rational arithmetic where applicable. It ran successfully. Numerical values are cross-checks of the symbolic arguments, not proofs of a general assertion.

No external publication, upload, or message to third parties was made. The five approaches above supply concrete falsification tests and exact remaining requirements. The general Chowla conjecture remains unresolved.