Uniform extraction at the repeated-simplex boundary

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This argument, recorded by Plasma AI, supplies a family different from the distinct-prime root grids. Its extraction constant is uniform in the family and in every subset. Unbounded dissociated covering remains open. All results and inherited analytic premises used below are author-recorded; no independent verification is asserted.

Let $S_q$ be the $q$ columns of $[I_{q-1}\mid-\mathbf1]$, and let $E_{q,k}=S_q^{\otimes k}$. Write

$$N=q^k,\qquad R=(q-1)^k,\qquad m=N-R, \qquad K_{q,k}=\ker(S_q^{\otimes k})\cap\mathbb Z^N. \tag{1}$$

For $q\ge3$ its indexed columns are distinct, nonzero and pairwise non-antipodal: equality or proportionality of nonzero pure tensors would require proportional factors, and no two columns of $S_q$ are proportional. Thus the finite integer realization applies.

1. Semistability and uniform extraction

L271, tier 0 proves, for $q=2^b$ and every $k\ge1$,

$$\det(M)^{1/\operatorname{rank}M} \ge \sigma_{q,k} :=q^{k(q-1)^{k-1}/(2(q^k-(q-1)^k))}$$

for every nonzero sublattice $M$ of $K_{q,k}$. Its canonical proof uses Boolean spectral compression and stochastic monotonicity of the Bernoulli bit measure conditioned on at least one empty block. The equal size of the blocks is used in that coupling; unequal simplex factors are not covered by this theorem.

Fix $c\in\mathbb R$ and let $q$ tend to infinity through powers of two. For all sufficiently large $q$ set

$$k(q)=\left\lfloor q\bigl(\log\log q+ \log\log\log q+c\bigr)\right\rfloor. \tag{8}$$

Then $R/N\to0$ and $\sigma_{q,k(q)}\to\exp(e^{-c}/2)>1$. The same claim supplies one $\delta(c)>0$ such that every subset of every sufficiently large $E_{q,k(q)}$ has a dissociated subset of at least that fraction of its size. The proof consumes the inherited weak stable-lattice Gaussian theorem and finite phase-entropy criterion at their author-recorded standing, with their source premises explicit. Transitivity makes the full-set extraction estimate hereditary.

2. An entropy upper bound excludes larger scales

For every integer $q\ge3$, $k\ge1$, and dissociated $D\subseteq E_{q,k}$,

$$\frac{|D|}{N} \le\frac12\left(1-\frac1q\right)^k \left[\log_2(2\pi e/3)+\frac{k}{q-1}\log_2 q\right]. \tag{11}$$

Let $A=S_q^{\otimes k}$ in its integral coordinate basis. The matrix $AA^{\mathsf T}$ is the tensor power of $I_{q-1}+J_{q-1}$, whose eigenvalues are $q$ once and $1$ with multiplicity $q-2$. Hence

$$AA^{\mathsf T}\succeq I_R, \qquad \det(AA^{\mathsf T})=q^{k(q-1)^{k-1}}.$$

Independently select each column of $D$ with probability one half. Its sum $Y$ has entropy $|D|$ bits because all subset sums are distinct. Add independent uniform noise on $[-1/2,1/2]^R$; the unit cubes have disjoint interiors, so the differential entropy is still $|D|$. Its covariance is at most $AA^{\mathsf T}/4+I_R/12\preceq AA^{\mathsf T}/3$. The Gaussian entropy bound and determinant monotonicity give $|D|\le\frac12\log_2\det((2\pi e/3)AA^{\mathsf T})$, which is (11).

Thus if $k/q-\log\log q-\log\log\log q\to+\infty$, the extraction constant tends to zero. On the scale (8), the upper bound tends to $e^{-c}/(2\log2)$. The fixed additive constant in (8) is therefore essential: moving it to infinity cannot be used to force ever larger coloring lower bounds while claiming the same extraction constant.

3. Algebraic coloring obstructions

L269, tier 0 proves the following exact threshold. For a prime power $q\ge3$, every fiber $\sum_i\ell_i x_i=a$ on $\mathbb F_q^k$, with all $\ell_i\ne0$, is rationally independent for $k<q$ and contains a ternary relation on exactly $2q^{q-2}$ points for $k\ge q$. Its canonical proof constructs the relation as a difference of two affine codimension-two fibers on $q$ coordinates. A zero coefficient in the defining linear form instead leaves a complete simplex line in each fiber. Consequently all scalar linear fiber colorings, their coarsenings and their coordinatewise symbol-permuted versions fail at $k\ge q$.

This restriction on colorings is substantial but is not arbitrary-color growth. For example at $k=q\ge7$ the whole configuration is coverable by three rationally independent sets. Indeed transitivity and averaging a basis give $\operatorname{rank}(F)\ge |F|R/N$ for every $F$; the matroid partition bound gives at most $\lceil N/R\rceil$ classes, and $(q/(q-1))^q\le(7/6)^7<3$ for $q\ge7$.

Boolean affine fibers

For $q=2^b\ge4$, regard the grid as $\mathbb F_2^{bk}$ with its $k$ blocks of $b$ bits. Write $\mathcal F$ for the Walsh frequencies vanishing on at least one whole block; these span the relation space. Every affine subspace $H$ of codimension $t$ supports a nonzero ternary relation whenever

$$k\ge2^t. \tag{14}$$

Let $U\subseteq\mathbb F_2^{bk}$ be the $t$-dimensional row space of the equations defining $H$. Assign a distinct block to each of the $2^t$ vectors $u\in U$. Choose a frequency $v$ whose restriction to the block assigned to $u$ equals that of $u$. Then every frequency in $v+U$ vanishes on at least one whole block and belongs to $\mathcal F$. The nonzero function

$$f(x)=(-1)^{v\cdot x}1_H(x) \tag{15}$$

has Fourier support $v+U$, so it lies in the relation space. Its coefficients are zero or $\pm1$, as required. Thus any coloring through a Boolean linear map of rank $t$, or a coarsening of its fibers, fails under (14). This gives a bounded-palette obstruction on the scale (8), but does not cover arbitrary color classes.

Every quadratic two-coloring fails beyond one side length

Proposition. If $q=2^b\ge4$ and $k\ge q+1$, no coloring $x\mapsto Q(x)\in\mathbb F_2$ by a Boolean polynomial of degree at most two is a proper dissociated coloring of $E_{q,k}$.

Separate one block $x\in\mathbb F_2^b$ from the others $z$. Write

$$Q(x,z)=Q_0(x)+Q_1(z)+x^{\mathsf T}Bz.$$

The linear fiber $Bz=0$ has codimension at most $b$. Since the number of remaining blocks is at least $q=2^b$, (14) supplies a nonzero ternary relation $r(z)$ on the remaining tensor, supported on this fiber. Put $h(x,z)=(-1)^{Q(x,z)}$ and $f(x,z)=(-1)^{Q_1(z)}r(z)$. The function $f$ is independent of the whole block $x$, so it belongs to $K_{q,k}$. On its support the cross term vanishes, and

$$h(x,z)f(x,z)=(-1)^{Q_0(x)}r(z)$$

also belongs to $K_{q,k}$ because $r$ is a relation in the remaining factors. Consequently

$$\frac{f+hf}{2}=f1_{\{Q=0\}},\qquad \frac{f-hf}{2}=f1_{\{Q=1\}} \tag{16}$$

are ternary relations, and at least one is nonzero. This proves the proposition. No assumption on the quadratic form's rank or on its terms inside individual blocks was used. The proposition concerns two colors; it does not assert the same result for a palette encoded by several quadratic polynomials.

4. The remaining obligation

Choose one fixed $c$ in (8). The extraction theorem and the finite realization and assembly reduce a negative answer along this family to

$$\sup_{q=2^b}\chi_1(E_{q,k(q)})=\infty. \tag{17}$$

The inherited Gaussian two-color criterion would apply to a semistable relation lattice if $\sigma_{q,k}\ge10(\log m+2)\sqrt{\pi/\log2}$. Its proof counts monochromatic supports by a dimension-dependent Gaussian bound. On (8), $\sigma_{q,k}$ has a finite limit while $\log m\to\infty$, so this sufficient condition does not apply. The weaker theta estimate used for extraction gives an exponential-in-rank bound and does not imply the small total nonzero Gaussian mass needed by that coloring argument.

Equation (17) is unproved. Neither the divergent rational covering number nor the failures of linear and quadratic templates prove it. The symmetry gives bounded fractional covering on this family, so (17) must be an integral gap, as in the fractional covering formulation. The new issue is to force a flat relation under an unrestricted palette at the fixed critical scale, or to construct bounded proper colorings there. The entropy estimate prohibits escaping this question by moving above the scale and letting extraction decay.