A uniformly extracting axis-line subfamily

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Standing

This is a smaller sufficient target for the missing coloring theorem, not a proof of that theorem. Root slices provide one realization; the direct extraction theorem now also permits linearly growing sides without a root realization. It does not replace dissociation by rational independence.

For integers $d_i\ge2$, put $X_t=\prod_{i=1}^t[d_i]$ and $N_t=|X_t|$. Let $\mathcal L_t\subset\{0,1\}^{X_t}$ be the indicator vectors of all full axis-parallel lines: one coordinate varies over its entire alphabet, and all other coordinates are fixed.

1. Two ways to obtain uniform extraction

There is a direct connection with the original prime grid. After changing its last column's sign, the simplex factor is

$$S_p=\{e_1,\ldots,e_{p-1},-\mathbf1\} \quad\longmapsto\quad \{e_1,\ldots,e_{p-1},\mathbf1\}.$$

In a tensor of these factors, select exactly the columns with one all-ones factor. They are precisely $\mathcal L_t$ with $d_i=p_i-1$. Independent column sign changes preserve every bounded-height relation and coloring number. Thus the original prime-grid extraction constant works for every subset of every line family along this prime-minus-one schedule. This simpler realization needs no coefficient-two circuit.

The initially found realization remains valid: the linear-size coefficient-two root circuits, on consecutive disjoint prime triples, have signed rational-coordinate models

$$C_{d_i}=\{e_{i,1},\ldots,e_{i,d_i},w_i\}, \qquad w_i=\tfrac12\sum_a e_{i,a}, \qquad d_i=3p_i+2q_i+2r_i-14.$$

Their one-pivot slice becomes $\mathcal L_t$ after the common scale factor two. Coprime conductors give a faithful tensor map, so this larger-side schedule inherits uniform extraction too.

More substantially, the new direct line-extraction theorem proves uniform extraction for the linear schedule $d_i=i+2$, without realizing it in prime grids. Its positive-real Sidon lower bound is $e^{-\pi^2/2}$, followed by one application of Pisier's Theorem 2 to the nested union. This is the simpler current line-family candidate.

For any of these uniformly extracting schedules, a proof that

$$\sup_t\chi_1(\mathcal L_t)=\infty \tag{2}$$

would complete the finite counterexample obligation using the existing integer realization and separated assembly. Equation (2) is unproved. The statements do not assert uniform extraction for arbitrary side-length sequences; for example, constantly binary sides give hypercube edges and fail uniform extraction as the dimension grows.

2. Exact rank and an integral presentation

For arbitrary $d_i\ge2$, write

$$s_t=\sum_i\frac1{d_i},\qquad R_t=\prod_i\left(1-\frac1{d_i}\right).$$

The lines in different directions are distinct, and

$$|\mathcal L_t|=N_ts_t,\qquad \operatorname{rank}_{\mathbb Q}\mathcal L_t=N_t(1-R_t). \tag{3}$$

The column image is saturated in $\mathbb Z^{X_t}$. Its integer kernel is generated by the elementary two-direction plane trades: in a fixed coordinate plane, the sum of all lines in either one of its two directions equals the sum of all lines in the other direction. Each displayed trade has coefficients in $\{0,\pm1\}$.

Here is an integral proof of all but the immediate count. Put $V_i=\mathbb Z^{d_i}$, $v_i=(1,\ldots,1)$, and $U_i=\operatorname{span}_{\mathbb Z}\{e_{i,2},\ldots,e_{i,d_i}\}$. Then $V_i=\mathbb Zv_i\oplus U_i$ is a direct sum of integer lattices. The image of direction $i$ is

$$W_i=\mathbb Zv_i\otimes\bigotimes_{j\ne i}V_j.$$

Expanding the direct-sum decompositions of the $V_i$, the sum of the $W_i$ consists exactly of all tensor summands with at least one $\mathbb Zv_i$ factor. Its complementary summand is $\bigotimes_iU_i$, of rank $\prod_i(d_i-1)$. This proves the rank and saturation statements.

For the kernel consider the map

$$\bigoplus_i\bigotimes_{j\ne i}V_j \longrightarrow \bigotimes_iV_i$$

that inserts $v_i$ in the $i$-th summand. After the same integral decomposition, each nonempty set $S$ of constant-factor positions gives $|S|$ copies of one free lattice, mapping by addition to that lattice. Its kernel is generated by differences of two copies. These differences are images of a common tensor in $\bigotimes_{k\ne i,j}V_k$, with $v_j$ inserted in copy $i$ and $-v_i$ inserted in copy $j$. Expanding that tensor in the standard bases gives exactly integer combinations of elementary plane trades. Thus the claimed generation is over $\mathbb Z$, not only over $\mathbb Q$.

For each of the three extraction schedules in Section 1, $s_t\to\infty$: use prime-reciprocal divergence for the two root schedules, and the harmonic series for the linear one. Therefore

$$\chi_{\mathbb Q}(\mathcal L_t) \ge\frac{s_t}{1-R_t}\longrightarrow\infty. \tag{4}$$

This rational lower bound is not (2). The uniform extraction also implies that these line families have dissociated subfamilies whose cardinality-to-rational-rank ratios tend to infinity. In particular no universal rational-rank bound for dissociated line subfamilies can prove the desired coloring theorem.

3. Actual-color obligation

The integral presentation gives a concrete way to seek the missing relation. Given a finite coloring of the lines, one needs a nonzero integer combination of plane trades whose collected coefficients lie in $\{0,\pm1\}$ and whose nonzero lines all have one actual color. Generation of the full integer kernel by ternary trades does not imply that a monochromatic rational dependence has such a representative.

This page records a proved smaller uniformly extracting family and its exact algebra. It asserts neither a coloring lower bound beyond (4) nor that rational dependence and dissociation coincide for line families. The one-pivot reduction, rank, saturation, and integral presentation received independent internal review. The direct extraction theorem has a separate proof and review record. No coloring-growth claim is established.