The exact tensor coloring obligation

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The tensor extraction theorem supplies one uniform full-dissociation constant. This page isolates what is needed to turn it into a counterexample. All statements below are elementary; the chromatic-growth obligation remains unproved.

Layer criterion — proved

Throughout, $B$ is a nonempty finite set of nonzero vectors in a rational vector space, and simplex side sizes $p,m$ are integers at least three. The colorings may have empty color classes. For $C\subseteq B$, define

$$\Sigma(C)=\left\{\sum_{x\in C}\epsilon_xx: \epsilon_x\in\{-1,0,1\}\right\}.$$

A subset of $B\otimes S_p$ has layers $C_1,\ldots,C_p\subseteq B$. It is dissociated if and only if

1. each $C_j$ is dissociated; and
2. $\bigcap_{j=1}^p\Sigma(C_j)=\{0\}$.

Indeed a relation has one signed sum $u_j$ from each layer, and $\sum_j u_j\otimes s_j=0$ is equivalent to $u_1=\cdots=u_p$, because the sole rational relation of the simplex is $s_1+\cdots+s_p=0$. Common nonzero sums give a relation, while a common zero sum is trivial exactly when every layer is dissociated.

A finite separation parameter

Suppose $B$ has a proper dissociated $q$-coloring. Let $\tau_q(B)$ be the least positive integer $p$ for which there are proper $q$-colorings $f_1,\ldots,f_p$ of $B$ satisfying

$$\bigcap_{j=1}^p\Sigma(f_j^{-1}(c))=\{0\} \quad\text{for every }c\in[q]. \tag{1}$$

Set $\tau_q(B)=\infty$ if there are no such colorings. Repeating layers shows that if (1) holds for $p$, it holds for every larger $p$. The layer criterion gives the exact equivalence

$$\chi(B\otimes S_p)\leq q \quad\Longleftrightarrow\quad p\geq\tau_q(B). \tag{2}$$

Define the universal signed-span set

$$U_q(B)=\bigcap_{f\text{ proper }q\text{-coloring}} \ \bigcap_{c\in[q]}\Sigma(f^{-1}(c)).$$

Then $\tau_q(B)=\infty$ if and only if $U_q(B)\ne\{0\}$. To prove the nontrivial direction, take all finitely many proper colorings as layers; by permuting color names, the intersection for any one fixed color is exactly $U_q(B)$.

If $\chi(B)=q$ and $p\geq q+1$, then $q\leq\chi(B\otimes S_p)\leq q+1$. For the upper bound, use a fixed proper $q$-partition in every layer but choose its palette from $q+1$ colors so that each color is absent from at least one layer. Its signed-span intersection is then zero. Thus (2) tells precisely when a single factor raises the coloring number.

A concrete but unproved route to completion

Two elementary constraints on the threshold

Always $\tau_q(B)\geq3$ when $\chi(B)=q$. If two colorings witnessed zero intersections, their corresponding color classes would be disjoint and their unions dissociated. These $q$ unions cover every point twice, so discarding any one union still covers $B$, contradicting $\chi(B)=q$.

There is also a reset after a sufficiently large chromatic jump. If $\chi(B)=q$, $Y=B\otimes S_m$, $m\geq2q+1$, and $\chi(Y)=q+1$, then

$$\tau_{q+1}(Y)\leq2(q+1). \tag{3}$$

Use palette colorings putting the new color on a chosen $q$-set of blocks, and another coloring putting it on a disjoint $q$-set. Their signed spans intersect only at zero, since the union uses at most $m-1$ independent simplex directions. Repeat these two colorings with each of the $q+1$ cyclic permutations of the color names. For every color the intersection of all resulting signed spans is zero, proving (3). Thus immediate repetition of very large factors cannot by itself force successive chromatic jumps.

A finite benchmark is $8\leq\tau_2(S_3\otimes S_5)\leq10$, with exact proofs and bounded certificates in the tensor-layer test.

Critical cores give another restriction. If $\chi(B)=q$ and deleting either of two distinct non-antipodal elements $x,y$ lowers the coloring number, then $\tau_q(B)\leq2q$. Use a proper coloring having singleton class $\{x\}$, another having singleton class $\{y\}$, and all cyclic permutations of their color names. For each color the intersection is contained in $\{0,\pm x\}\cap\{0,\pm y\}=\{0\}$. Thus reducing to critical cores does not produce arbitrarily large thresholds cheaply.

The remaining quantitative obligation

The analytic extension loss for a tensor base $B$ and side size $p$ is at most $Q(B)/(2p^2)$, with $Q$ explicitly defined in the extraction theorem. To force a chromatic jump while keeping the analytic losses summable, one would like to choose

$$p<\tau_q(B),\qquad Q(B)/(2p^2)\leq\varepsilon_q, \qquad\sum_q\varepsilon_q<\rho(B_0).$$

No family providing these inequalities at all stages is known in these notes. In particular, neither $\tau_q(B)\gg\sqrt{Q(B)}$ nor unbounded coloring for the explicit side sizes in the extraction theorem has been proved.

Universal anchors allow arbitrarily large $p$ for one jump, but the anchor-gluing obstruction explains why they do not automatically propagate after that jump.