Unbounded quotient coloring and the colored lifting obligation

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Standing

Let $X_t=\prod_{i\le t}[d_i]$, $V_t=\mathbb Z^{X_t}$, and let $W_t$ be the integer span of the full axis-line indicators. The line family itself is uniformly extracting when $d_i=i+2$. Its dissociated coloring growth remains unproved. This page gives a positive, quantitatively unbounded coloring statement in a quotient, and states exactly what must be lifted to use it.

1. The integral quotient is a full simplex tensor

Put $V_i=\mathbb Z^{d_i}$ and $v_i=(1,\ldots,1)$. There is a natural integral isomorphism

$$V_t/W_t\ \cong\ \bigotimes_{i\le t}(V_i/\mathbb Zv_i). \tag{1}$$

Indeed $v_i$ is primitive and $V_i=\mathbb Zv_i\oplus U_i$, where $U_i$ is generated by all but one standard basis vector. Tensor expansion splits off every summand with a $v_i$ factor; their sum is exactly $W_t$. The complementary summand $\bigotimes_iU_i$ proves (1) over $\mathbb Z$, as well as saturation of $W_t$.

The projected point vectors $e_x+W_t$, indexed by $x\in X_t$, are exactly the columns of

$$Q_t=S_{d_1}\otimes\cdots\otimes S_{d_t}. \tag{2}$$

Each factor consists of the $d_i$ standard basis images with their sole relation that their sum is zero. For $d_i\ge3$ these indexed columns are nonzero and distinct, with no antipodal pairs.

2. Linear sides force unbounded coloring in the quotient

Take $d_i=i+2$. Write

$$N=\prod_i d_i=(t+2)!/2,\qquad D=\prod_i(d_i-1)=(t+1)!,\qquad L=\sum_{i=1}^t\frac{\log(d_i-1)}{d_i}.$$

The small-column counting lemma from the tensor schedule analysis gives

$$\chi_1(Q_t)> \frac{N}{4D(1+2L/\log2)} \ge \frac{t+2}{8\bigl(1+(2/\log2)\log^2(t+2)\bigr)}. \tag{3}$$

In particular $\chi_1(Q_t)\to\infty$.

Here is the finite argument, to make the coefficient requirement explicit. In standard simplex coordinates the average logarithmic column $\ell_1$-norm is $L$. At least half the columns have norm at most $e^{2L}$. If a proper $q$-coloring existed, one class would contain $s\ge N/(2q)$ of them. Its $2^s$ subset sums would be distinct, with total coordinate width at most $s e^{2L}$. Thus

$$2^s\le(1+s e^{2L}/D)^D.$$

The elementary bound $2^u\le1+Mu\Rightarrow u<2\log_2(2M)$, for $M\ge1$, gives $s<2D(1+2L/\log2)$, proving the strict first inequality in (3). Finally $L\le\log(t+1)\sum_{i\le t}1/(i+2)\le\log^2(t+2)$.

Consequently, for every fixed finite $q$, every $q$-coloring of the point set of a sufficiently large box has a nonzero vector

$$b\in W_t\cap\{-1,0,1\}^{X_t} \quad\text{whose nonzero coordinates have one actual point color.} \tag{4}$$

The quotient relation has genuinely ternary coefficients, not merely bounded rational denominators. Membership in $W_t$ follows integrally from (1); it does not bound coefficients of a line representation.

3. Exact missing colored lift

In a one-coordinate extension, let $C_{s,c}$ be the old lines of color $c$ in layer $s$, and $P_c$ the points whose new-axis line has color $c$. The exact extension criterion needs a nonzero vector

$$b\in\left(\bigcap_s\Sigma_1(C_{s,c})\right) \cap\{-1,0,1\}^{P_c}. \tag{5}$$

Equation (4) supplies a vector in the larger set $W_t\cap\{-1,0,1\}^{P_c}$. It does not supply (5). There are two separate gaps:

1. An integral preimage under the full line matrix need not be known to have every coefficient in $\{-1,0,1\}$.
2. Even a flat preimage in the full line family need not use the correct actual color, much less give one in each layer's correct color class.

The first obligation is proved without color constraints in at most three directions, by the integer graph-gauge theorem. It is not proved in arbitrary dimension. The second obligation remains even in settings where an unrestricted flat preimage exists.

The second gap is sharp. Existential colored lifting from common real spans holds in two directions, but is false in three: an explicit independent twelve-line class has a nonzero flat image on five points and no nonzero ternary sum on that point support. This rules out replacing (5) by real-span membership alone. A successful argument must use further structure from the joint coloring, not just independence of its individual classes.

Neither saturation nor the quotient coloring bound permits silently replacing $W_t$ by the intersection in (5). No general colored lifting theorem is asserted here.

The simple quotient identification and the finite counting argument are algebraic. No computation is used in (1)--(4). This is not an unbounded coloring theorem for the original line configuration.