Full-column periodic circuit masks and bounded-period exclusion

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The fixed-symbol automaton theorem periodicizes an infinite proper coloring. The first result here prevents replacing its remaining ternary question by rational rank. The final result excludes arbitrary palettes of a prescribed bounded period, but does not control the period supplied by that automaton.

1. Thirteen full columns: a nonflat rational circuit

Fix $k\geq3$ and the ordinary translation-cover symbol

$$M(z)=\begin{pmatrix} 1&1&0&0\\ z^k&1&z&1\\0&1&1&0 \end{pmatrix}. \tag{1}$$

Its base columns are distinct nonzero zero-one vectors. Different sheet lifts of a fixed type are also distinct. A finite relation is described by Laurent polynomials $f_i(z)$. The first and third rows give

$$f_1=-Q,\quad f_2=Q,\quad f_3=-Q, \quad f_4=(z+z^k-1)Q. \tag{2}$$

Select sheets $\{0,1,k\}$ in types $1,2,3$, and $\{0,2,k+1,2k\}$ in type (4). A supported relation has $Q=a+bz+cz^k$, with

$$f_4=-a+(a-b)z+bz^2+(a-c)z^k+(b+c)z^{k+1}+cz^{2k}.$$

The missing sheets $1,k$ force $a=b=c$. The kernel on these thirteen columns is therefore one-dimensional, generated by

$$(1+z+z^k)(-1,1,-1,z+z^k-1), \quad (1+z+z^k)(z+z^k-1)=-1+z^2+2z^{k+1}+z^{2k}. \tag{3}$$

All thirteen coefficients are nonzero. The generator is primitive, with one coefficient (2) and all other magnitudes one. These columns are consequently a rational circuit and nevertheless fully dissociated: an integer relation is an integer multiple of this generator, and no nonzero multiple is ternary.

This uses all incident rows, not just a selected submatrix. The incident row times are

There are twelve rows and rank twelve.

2. Periodic masks, including a connected one

For any $L>2k$, repeat this thirteen-column set at translates $jL$, $j\in\mathbb Z$. Its incident rows lie in disjoint intervals $[jL,jL+2k]$. Every finite relation splits over the blocks, so the whole periodic mask is dissociated although it has rational dependencies. It also descends safely to cyclic cover degrees $mL$: its $m$ blocks remain row-disjoint. This assertion is not made for arbitrary moduli.

Connectivity of the mask does not by itself repair the inference. For $k=3$, instead take sheets $0,1,3\pmod6$ in types $1,2,3$, and $0,2,4\pmod6$ in type (4). Equation (2), with the forbidden fourth coordinate residues $1,3$, forces

$$Q(6j)=Q(6j+1)=Q(6j+3)=a_j.$$

The remaining coefficients are exactly

$$f_4(6j)=a_{j-1}-a_j,\qquad f_4(6j+2)=a_j,\qquad f_4(6j+4)=2a_j. \tag{4}$$

Every finitely supported rational sequence $a_j$ gives a rational relation. A ternary relation must have integer $a_j\in\{0,\pm1\}$ by its type-two coefficients, and (4) then forces every $a_j=0$. Adjacent individual circuit supports share their type-four column at sheet $6j$, so their incidence-support union is connected. Thus overlap through low-coefficient columns alone is insufficient.

Independent incidence variables have the same obstruction

The obstruction is not an artifact of specializing to one sheet coordinate. In the free abelian voltage group on eight incidences, write

$$M=\begin{pmatrix} a&b&0&0\\c&d&e&u\\0&f&g&0 \end{pmatrix},\qquad h=(-bug,aug,-afu,bcg-adg+aef).$$

These letters are independent Laurent variables. Put

$$X=\frac{bc}{ad},\qquad Y=\frac{ef}{dg}.$$

Then $h_4=adg(X+Y-1)$. The vector $h(1+X+Y)$ again has thirteen supported columns, with one coefficient two. A sub-supported kernel vector is $hQ$, where its second component forces $Q=\alpha+\beta X+\gamma Y$. The missing $X,Y$ coefficients in the fourth component force $\alpha=\beta=\gamma$. Independence of the exponent vectors of $X,Y$ ensures all displayed monomials are distinct. Thus this is a full-column height-two circuit in the generic cover.

Its full row support has twelve rows: the first and third rows each have three monomials, and the second has the six monomials $1,X,Y,X^2,XY,Y^2$ times one common factor. Deleting the height-two column gives a square matrix with determinant $\pm2$. Indeed, first- and third-row unit eliminations leave the three coefficients $\alpha,\beta,\gamma$ of $Q$. Eliminate the other three fourth-type columns at $1,X^2,Y^2$. The residual equations are

$$\beta-\alpha=0,\qquad \gamma-\alpha=0,\qquad \beta+\gamma=0,$$

whose matrix has determinant $\pm2$. All eliminations use unit pivots. This is a rational/binary rank discrepancy for full columns, not a minor obtained by discarding additional incident rows.

For each fixed row type, the exponent range in each of the eight incidence coordinates has width at most two. Translates of this circuit by $3\mathbb Z^8$ therefore have disjoint row supports. Their union is a periodic dissociated mask with rational dependencies. This generic version is author-recorded; the separate reviewer's check recorded below concerned the one-dimensional version.

3. An arbitrary bounded-period palette exclusion

Let a finite zero-one base have $n$ distinct nonzero column types and $r$ row types. Enumerate its incidences $e$, and give them voltages $g_e=2^{j(e)}$ with distinct nonnegative integers $j(e)$.

Theorem. Every proper $q$-coloring of its infinite translation cover which is periodic with period $\ell$ satisfies

$$\boxed{n\leq qr\ell.} \tag{5}$$

Proof. For every base type, choose a color occurring on one of its sheet residues. If $n>qr\ell$, some color is chosen on more than $r\ell$ distinct types. Choose one residue $a_v\in[0,\ell-1]$ of that color for each such type. Allow only coefficients of the form $z^{a_v}p_v(z^\ell)$.

Group equations by row type and residue $b\in[0,\ell-1]$. This gives a matrix over $\mathbb Q[Z,Z^{-1}]$, $Z=z^\ell$, with $r\ell$ rows and more columns. Its entry for incidence $e=(x,v)$, in the one row residue $b\equiv a_v+g_e\pmod\ell$, is

$$Z^{(a_v+g_e-b)/\ell}.$$

Crucially, each base type occurs only once among these columns. In a determinant expansion, different matchings use different subsets of original incidences. Their exponents differ by the difference of those incidence-voltage sums divided by $\ell$: the sums of selected column residues and selected row residues are fixed. Distinct subsets have distinct sums of binary powers. Hence every nonzero minor is a flat Laurent polynomial with coefficients in $\{0,\pm1\}$.

Let the matrix have rank $h$. Choose $h$ independent columns and rows, and one extra column. Alternating $h$-minors give a nonzero kernel vector on these $h+1$ columns. It annihilates every row, since the chosen rows span the row space over $\mathbb Q(Z)$. Its components are flat by the preceding argument. Returning from $Z$ to sheet labels produces a nonzero finite-support ternary relation, entirely in the chosen actual color. This contradicts properness and proves (5). $\square$

The same proof for independent incidence voltages in $\mathbb Z^E$ gives $n\leq qrs$ when the coloring has a period subgroup of finite index $s$. Unique matching incidence sums replace binary-power sums.

Exact limitation and checks

The automaton's upper bound on a possible period depends on the whole chosen symbol and is far larger than the lower bound (5) presently provides. No comparison excludes all possible periods. The circuit masks do not color the whole cover, and (5) does not prove unbounded coloring.

The thirteen-column circuit, its full rows, primitiveness, dissociation, and separated periodic/cyclic repetitions are author-recorded; a separate reviewer checked them. Statement (4) and the cofactor proof of (5) are author-recorded without a separate check. No computation or external theorem is used on this page.