Actual two-colorings from stable moment lattices

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Boundary extraction is proved by a different, dimension-uniform exponential theta bound. This does not change the actual two-color regime proved below or supply a boundary coloring theorem.

The finite theorem

For 1<=k<d, write

$$n=2^d,\quad r=\sum_{j=0}^k\binom dj,\quad m=n-r, \quad D=(d-k)\binom dk,\quad s=2^{D/(2m)}.$$

Then the following explicit condition implies a partition of the entire moment configuration into two fully dissociated classes:

$$\boxed{s\geq10(\log m+2)\sqrt{\pi/\log2} \quad\Longrightarrow\quad \chi_1(E_{d,k})=2.} \tag{1}$$

Unsubscripted logarithms are natural. This is a statement about all ternary relations and arbitrary support sizes, not just cube templates. Consequently these parameter pairs also have hereditary extraction constant at least 1/2. They cannot form the desired counterexample.

Checked external input

Regev--Stephens-Davidowitz, A Reverse Minkowski Theorem, full arXiv version 4, Theorem 1.2 and its definition of rho_s, states: for a rank-m Euclidean lattice K whose every sublattice has determinant at least one,

$$\sum_{z\in K}\exp(-\pi t^2\|z\|^2)\leq3/2, \qquad t=10(\log m+2). \tag{2}$$

The primary abstract was read first, followed by the full theorem and the Gaussian normalization. This page uses (2), not an unproved constant-parameter strengthening or a volume-only bound. The paper is also published in Annals of Mathematics 199 (2024), 1--49.

Proof of (1)

Let L be the integer relation lattice of E_(d,k). Multiplication by full parity identifies it isometrically with L_(d,d-k-1). Its rank is m and determinant is 2^(D/2), by the exact phase-lattice calculation. The semistability theorem proves that K=s^(-1)L satisfies every hypothesis of (2). Hence

$$\sum_{0\ne z\in L} \exp\left(-\pi(t/s)^2\|z\|^2\right)\leq1/2.$$

Under (1), pi(t/s)^2<=log 2, so

$$\sum_{0\ne z\in L}2^{-\|z\|^2}\leq1/2. \tag{3}$$

For every distinct inclusion-minimal support F of a ternary relation, choose one relation f_F, up to its overall sign. Its squared Euclidean norm is |F|. A uniform independent two-coloring makes F monochromatic with probability 2^(1-|F|), precisely the sum of the two weights in (3) contributed by f_F and -f_F. Distinct supports give distinct pairs. The expected number of monochromatic minimal supports is therefore at most 1/2. Some coloring has none, and then has no monochromatic ternary relation at all. This proves chi_1<=2. Alternation on any coordinate (k+1)-cube is a nonzero relation in the full configuration, so chi_1>=2.

This proof needs no estimate on the minimum relation support, and no independence between different monochromatic-support events.

A substantial excluded parameter range

For every fixed 0<a<1/2, along even d and

$$k=\left\lfloor d/2-a\sqrt{d\log d}\right\rfloor,$$

condition (1) holds for all sufficiently large d. In particular

$$\boxed{\chi_1(E_{d,k})=2\quad\hbox{eventually}.} \tag{4}$$

Here is the asymptotic check. Put u=d/2-k=a sqrt(d log d)+O(1). The central-binomial estimate and the product of successive binomial ratios give

$$2^{-d}\binom d{k} =\sqrt{\frac2{\pi d}} \exp(-2u^2/d+o(1)) =\sqrt{\frac2\pi}\,d^{-1/2-2a^2}(1+o(1)). \tag{5}$$

For precision, the logarithm of that ratio product is -2u^2/d+O(u/d+u^3/d^2), whose error tends to zero in the displayed range. The central coefficient is asymptotic to sqrt(2/(pi d)); equivalently apply the elementary Stirling estimate to the two factorials. The geometric bound on the lower binomial tail from the moment entropy page gives r/n=o(1). Therefore

$$\frac Dm=\frac{1+o(1)}{\sqrt{2\pi}} d^{1/2-2a^2}. \tag{6}$$

The logarithm of s is a positive constant times the positive power d^(1/2-2a^2), whereas log(10(log m+2)sqrt(pi/log2))=O(log d). This proves (1) eventually.

The earlier entropy theorem excludes uniform extraction for every fixed a>1/2. Thus neither side with a fixed strict separation from 1/2 can supply the counterexample: one side loses extraction, and the other has actual two-colorings.

Exact surviving gap

The boundary a=1/2 and narrower varying-a windows are not settled. At the exact boundary D/n tends to 1/sqrt(2 pi), so s tends to the finite number 2^(1/(2 sqrt(2 pi))); condition (1) fails. The existing entropy bound gives only eventual chi_1>=6 there, because alpha/n<=1/(2 sqrt(2 pi))+o(1)<1/5. This is a fixed lower bound, not unbounded coloring. Uniform extraction at that boundary is proved separately on the boundary extraction page.

More generally, (1) excludes every sequence satisfying its exact finite inequality; an intermediate range between positive constants and the displayed logarithmic threshold is not decided by these estimates. In particular, this does not include constants times log d already large enough to meet (1). No claim is made that failure of (1) forces chromatic growth. The original uniformly extracting line-cover candidate and its arbitrary-color gap remain intact.

Review and computation scope

The semistability proof and the Gaussian-to-two-color step each received independent internal review. The external theorem was checked in its full primary source. No numerical experiment enters (1)--(6). These results retire a region of the new route; they do not resolve E0774.