The remaining coloring problem for linear-side line families

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Let $L_t$ be the full axis-line indicators on $X_t=\prod_{i\le t}[i+2]$, let $N_t=|X_t|$, and put $s_t=\sum_{i\le t}1/(i+2)$. The direct extraction theorem proves one uniform full-dissociation extraction constant. The new sufficient target is

$$\boxed{\sup_t\chi_1(L_t)=\infty.} \tag{1}$$

It is unproved. This page states the actual-color consequences obtained so far and the precise limits of transferring earlier mechanisms.

1. Every three-direction slice has a connectivity criterion

Fix three directions and all the other coordinates. On $[a]\times[b]\times[c]$, identify a first-direction line with the edge between its second and third labels, and similarly cyclically. Thus lines correspond to the edges of $K_{a,b,c}$.

Proposition. A line subset $C$ is dissociated if and only if its complementary edge graph is connected. Equivalently, dissociation and rational independence coincide on this slice.

Indeed a line relation is exactly

$$A(y,z)+B(x,z)+C(x,y)=0.$$

Its solutions are exactly

$$A(y,z)=u_y-v_z,\qquad B(x,z)=v_z-w_x, \qquad C(x,y)=w_x-u_y.$$

To verify completeness, fix a reference $x_0$ in the first part; the equation at $x_0$ splits $A$ into a function of $y$ plus one of $z$, and the remaining equations determine $w_x$. These are graph gradients with the displayed cyclic edge orientations. A nonzero gradient supported on a given edge subset exists precisely when its complement is disconnected. A component indicator potential gives a supported gradient with coefficients $0,\pm1$, proving the claim.

Therefore any proper finite coloring must leave a connected complement after deleting each individual color in each three-direction slice. In a proper two-coloring both color graphs must be connected and spanning. This is a simultaneous actual-color constraint, but no global contradiction has been obtained from it.

Elementary planes alone cannot be forced for arbitrary colorings. The two-coloring $c_i(x_{-i})=\sum_{j\ne i}x_j\pmod2$ uses both colors among the direction-$i$ lines of every $(i,j)$-plane, since the fixed $j$-label varies over both parities. Hence no whole elementary plane trade is monochromatic. This is not a proper coloring of the full relation system; the three-direction criterion supplies larger relations that it fails to avoid.

The collapse to rational independence fails in four directions. The fixed exact certificate, checked by a finite verification script, gives 27 lines in $[3]^4$, rank 26, with one primitive relation having one coefficient two and all other coefficients $\pm1$. Thus that support is a dissociated rational circuit. The finite discovery and checking bounds are recorded on the certificate page.

2. A stronger fixed-height lower bound, not unbounded coloring

For general sides $d_i$, write $N=\prod_i d_i$ and $s=\sum_i1/d_i$. If all $q$ classes are height-$h$ independent, then

$$s\log(h+1)\leq \frac q2\log\left(2\pi e\left[ \frac{h(h+2)t}{12q}+\frac1{12}\right]\right). \tag{2}$$

Here and below height bounds coefficient magnitude with unrestricted support.

For each color class independently, choose independent uniform coefficients $U_\ell\in\{0,\ldots,h\}$. Its line sums are injective in these coefficients, so their joint discrete entropy is $|C|\log(h+1)$. At a point $x$, the sum has variance $\sigma_h^2 n_C(x)$, where $\sigma_h^2=h(h+2)/12$ and $n_C(x)$ counts its incident lines of that color.

For an integer-valued variable $Z$, add an independent uniform variable on $[-1/2,1/2]$. Its differential entropy equals $H(Z)$ and its variance is $\operatorname{Var}(Z)+1/12$. Comparing its density to the Gaussian of the same mean and variance, nonnegativity of relative entropy gives

$$H(Z)\le\tfrac12\log\bigl(2\pi e(\operatorname{Var}(Z)+1/12)\bigr).$$

For completeness, that nonnegativity follows by applying $\log u\le u-1$ to the ratio of the comparison density to the given density and integrating. Expanding the Gaussian logarithm gives the displayed variance bound.

Entropy subadditivity over points, followed by concavity of the logarithm over the colors, proves (2): every point lies on exactly $t$ lines, so $\sum_C n_C(x)=t$, and there are $N$ points.

For $d_i=i+2$, divide (2) by $\log t$ and pass to the limit. Any fixed $q,h$ valid for all stages must satisfy

$$q\ge2\log(h+1). \tag{3}$$

In particular, sufficiently large stages have $\chi_2(L_t)\ge3$. This is not infinite $\chi_2$, and (3) does not establish (1).

Keeping the full averaged covariance has the same leading bound

Let $A$ be the full line matrix. Applying the Gaussian bound to the whole jittered sum vector in each color, then using concavity of $\log\det$, gives the stronger finite necessary inequality

$$s_t\log(h+1)\le\frac q2\, \mathbb E\log\left(2\pi e\left[ \frac{\sigma_h^2}{q}\Lambda_t+\frac1{12}\right]\right), \qquad \Lambda_t=\sum_{i\le t}d_i B_i, \tag{3a}$$

where the $B_i$ are independent Bernoulli variables of probabilities $1/d_i$. Indeed $AA^{\mathsf T}=\sum_i I\otimes J_{d_i}\otimes I$; its eigenvalue $\sum_{i\in S}d_i$ has multiplicity $\prod_{j\notin S}(d_j-1)$. Dividing the log determinant by the row count gives precisely the expectation in (3a). The covariance of a color's jittered sum is $\sigma_h^2 A_cA_c^{\mathsf T}+I/12$, and averaging these positive definite matrices justifies the stated use of concavity.

For linear sides and fixed $a,b>0$, $\mathbb E\log(a\Lambda_t+b)/\log t\to1$. Jensen's inequality and $\mathbb E\Lambda_t=t$ give the upper bound. For the lower bound, some $B_i=1$ with $i\ge\varepsilon t$ occurs with probability tending to $1-\varepsilon$: the complementary product $\prod(1-1/(i+2))$ telescopes. On that event $\Lambda_t\ge\varepsilon t$; elsewhere the logarithm is bounded below by $\log b$. Let $\varepsilon\downarrow0$. Thus this averaged full-covariance estimate still gives only (3), not unbounded fixed-height coloring. No limitation is asserted for arguments retaining additional color-specific information before averaging.

3. Why the earlier focusing theorem cannot be imported automatically

The root-family fixed-height focusing theorem uses multiplicative closure and conductor-disjoint products. The linearly growing line family has no proved replacement for those properties. Its full tensor products actually lose uniform extraction.

For two stages $L_s,L_t$, the tensor family has $N_sN_t$ rows, $N_sN_t s_s s_t$ columns, and exactly $st$ ones in each full row. All subset sums lie in $[0,st]^{N_sN_t}$, so any dissociated subfamily $D$ satisfies

$$2^{|D|}\le(1+st)^{N_sN_t},\qquad \frac{|D|}{|L_s\otimes L_t|} \le\frac{\log_2(1+st)}{s_s s_t}. \tag{4}$$

For the linear schedule the right side tends to zero whenever both stages tend to infinity, even at different rates. This excludes full-product flattening as a uniformly extracting family, not a thinner selected support. Thus unbounded coloring at a fixed height greater than one for this new family would itself still require a coefficient-safe transfer to (1).

Ordinary product-Ramsey trimming also needs an additional idea. If every alphabet is replaced by a proper subset $A_i\subsetneq[d_i]$, retain the full lines whose fixed coordinates lie in the $A_j$. Each retained direction-$i$ line has a private point with its varying coordinate outside $A_i$. That point belongs to no other retained line. The retained family is therefore rationally independent. More generally, only untrimmed directions can participate in a relation. A monochromatic trimmed subbox alone cannot supply a line trade.

4. Exact actual-color extension test

Let $X$ be a fixed old box and add one coordinate $[d]$. For a given coloring, write $C_{s,c}$ for the old direction lines of color $c$ in layer $s$, and $P_c\subset X$ for points whose new-axis line has color $c$. Let $\Sigma_h(C)$ denote the sums on $C$ with integer coefficients in $[-h,h]$. Then the coloring is height-$h$ proper if and only if every $C_{s,c}$ is height-$h$ independent and

$$\left(\bigcap_{s=1}^d\Sigma_h(C_{s,c})\right) \cap\{-h,\ldots,h\}^{P_c}=\{0\} \quad\text{for every }c. \tag{5}$$

The last cube denotes vectors on $X$ vanishing outside $P_c$.

Indeed the new-axis coefficients give one common vector $b$ on all layers. A monochromatic relation requires each old layer to represent $-b$. When $b\ne0$, these representations and the new-axis coefficients yield precisely the forbidden common vector in (5). When $b=0$, a nontrivial relation is exactly an internal old-layer relation. This proof assumes no uniqueness of signed representations.

The test makes the missing coupling explicit: common old-layer signed sums must also have bounded coordinates and be supported on points carrying the actual new-axis color. No lower bound forces such a common vector.

There is an exact nested-alphabet model for the linear schedule: take three fixed roots numbered $0,1,2$; vertex $i+2$ chooses any smaller number as its parent. Complete choices are recursive forests, and a line leaves one parent choice unspecified. This is a bijective relabeling only. No Ramsey theorem for monochromatic trades follows just from that model.

The connectivity criterion, entropy calculation, and tensor-product bound received independent internal checks. None supplies the missing actual-color ternary relation for every finite palette.