Twisted transversals, a scoped reset, and matching gadgets

---

1. The allowed twists

Given a finite $0/1$ configuration on rows $X$, take $d$ row-disjoint copies, indexed by $s\in[d]$. For arbitrary bijections $T_s:X\to X$, adjoin the disjoint transversal columns

$$n_x=\sum_{s=1}^d e_{(T_sx,s)}. \tag{1}$$

The existing root-polygon phase induction survives: correct each point $(T_sx,s)$ by the same random $Y_x$, where $Y_x^d$ is the prescribed new-term correction and $\mathbb EY_x=\cos(\pi/d)$. An old column of weight $w$ lies in one layer and meets $w$ distinct transversals, so its expected multiplier is exactly $\cos(\pi/d)^w$. Prescale its coefficient by that multiplier before optimizing old phases. The proof neither assumes coordinate-preserving permutations nor identical old phases. It is the mechanism behind the ordinary incidence-cover theorem.

2. Two witness colorings and the exact reset scope

Suppose the old configuration admits a proper $q$-coloring, and let $Y$ be (1)'s extension for $d\ge2$. Choose a nonempty proper set $S$ of layers and use $q+1$ colors as follows:

• Keep old color 1 in layers in $S$.
• Replace old color 1 by the new color $q+1$ outside $S$.
• Leave all other old colors unchanged.
• Give every new transversal color $q+1$.

This coloring is proper even with arbitrary twists in (1). A fresh-color relation, restricted to a layer in $S$, has only the transversal columns there. Their restrictions are a point basis, so every transversal coefficient vanishes. Remaining relations separate into proper old layers. All other colors use only old columns and are likewise proper.

Swap $S$ and its complement to obtain a second proper coloring. The two distinguished color-1 classes are supported in complementary row sets $U$ and $U^c$, so their rational spans intersect only at zero.

Hence any next untwisted pure-transversal extension with at least two layers remains $(q+1)$-colorable: put those witness colorings in two layers, arbitrary proper witness colorings elsewhere, and give every newest transversal color 1. Their required common signed sum is zero. In particular an actual jump $q\to q+1$ cannot be followed immediately by another jump under this untwisted operation.

For a next extension with twists $R_s$, the two relevant supports are instead $R_1^{-1}U$ and $R_2^{-1}U^c$. The same conclusion holds when $R_1^{-1}U=R_2^{-1}U$, for example when both twists preserve $U$. No unconditional arbitrary-successive-twist reset is proved. Different row permutations can destroy the disjoint-support argument. The initial witness construction survives arbitrary twists; its next-step consequence has this additional compatibility hypothesis.

There is also a size-conditioned reset: if the old configuration has n rows and the two new layer counts are d and D, then $nd<d^D$ ensures a $(q+1)$-coloring after both steps even with arbitrary twists. Choose the sparse distinguished support independently in each second-stage layer; the expected common intersection has size $nd/d^D<1$. This does not recover a sparse distinguished class after the second step, so no indefinitely repeatable fixed-palette reset is inferred.

3. Matching decorations retain a fixed extraction constant

There is a proved enlargement that prevents giving all new columns one fresh color. Start with the two point-basis columns on two rows. At stage $i$, take $d_i=i+2$ copies, add all transversals (1), and choose a perfect matching of their labels. For each matched pair $x,y$, also add the union column $n_x+n_y$. The number of row labels stays even, so a perfect matching always exists. Twists and matchings are prescribed as part of the construction, not selected after seeing a coloring.

The new columns, considered as frequencies in the independent transversal variables, are disjoint triples $\{e_x,e_y,e_x+e_y\}$. Each triple has positive-real Sidon parameter at least $1/2$. To see this directly, conjugate the entire third term inside the real part: replace $c_3z_xz_y$ by $\overline{c_3}\,\overline{z_x}\,\overline{z_y}$. The resulting three simplex characters have product one. Align their ideal phases, then divide the remaining product error, of angle at most $\pi$, equally among the three characters. Every real term retains at least $\cos(\pi/3)=1/2$ of its magnitude. Different matched pairs are optimized independently.

Now optimize the old coefficients after their attenuation as in Section 1. Choose the desired transversal values to optimize the new triple polynomial, then use independent roots to realize those values exactly. A column born at stage $i$ has weight $d_i$ or $2d_i$, receives at least $1/2$ of its coefficient magnitude when born, and subsequently has multiplier

$$\prod_{j>i}\cos(\pi/d_j)^w \ge e^{-\pi^2}\quad(w\le2d_i),$$

by the previously checked telescoping estimate. The initial basis columns retain at least $\tfrac12e^{-\pi^2/6}$, separating the first side 3 from the later sides. Thus every finite constructed configuration satisfies

$$\boxed{\rho\ge\tfrac12e^{-\pi^2}.} \tag{2}$$

One direct-sum application of Pisier's Theorem 2 gives a uniform full-dissociation extraction constant over all finite stages and choices. This uses a fixed matching gadget; no degree bound is allowed to grow silently with the desired number of colors.

4. A proved ternary three-color obstruction, not an iteration

Let $R_t$ and $n_t$ be the row and column counts. The construction has exactly

$$R_t=2\prod_{i\le t}d_i,\qquad \frac{n_t}{R_t}=1+\frac32\sum_{i\le t}\frac1{d_i}, \qquad \text{row degree}=1+2t. \tag{3}$$

Every old row is copied, and every new row belongs to one transversal and one matching union, proving the degree formula and the count recurrence.

The already proved variance-entropy bound, with coefficient alphabet $\{0,1\}$, implies that a proper two-coloring would satisfy

$$\frac{n_t}{R_t}\log2 \le\log\left(2\pi e\left[\frac{1+2t}{8}+\frac1{12}\right]\right). \tag{4}$$

The coefficient of $\log t$ on the left tends to $\tfrac32\log2>1$, whereas it is one on the right. Thus sufficiently large stages require at least three dissociated colors.

The matched triple itself prevents coloring every new column with one fresh color, so the pure-basis reset's proof is unavailable here. This is not evidence of a reusable chromatic jump: no four-color lower bound or arbitrary-palette amplification is proved. Equations (2)--(4) are the complete established positive conclusion of this gadget.

The constants, coefficient prescaling, counts, entropy conclusion, and compatible-twist reset scope passed independent internal review.