Recurrent Hadamard gluing and anchor loss

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Standing and scope

This is a new recurrent, overlapping gluing design, outside the long private buffers in triangular gluing. It supplies an arbitrarily cheap chromatic jump from any fixed anchored source, but loses the next anchor at all sufficiently large Hadamard orders. The sufficient threshold depends on the source. Changing-source iterations below that threshold remain unexcluded; no successful iteration is proved. An independent internal audit checked the saturation, constants, full closure, scalar transfer, and diagonal-copy scope.

1. Saturated construction and analytic bound

Let $B\subset\Gamma\cong\mathbb Z^d$ be finite, nonzero, and without antipodal pairs, with $\chi(B)=q$ and universal nonzero $q$-anchor $w$. Write

$$M^2=\max_{b\in B}\|b\|_2^2,\qquad R=\max_{x\in\Sigma(B)}\|x\|_2^2\in\mathbb Z_{\geq1}.$$

For $r=2^k\geq2$, construct the Sylvester matrix recursively by $H_1=(1)$, $H_{2r}=\bigl(\begin{smallmatrix}H_r&H_r\\H_r&-H_r\end{smallmatrix}\bigr)$. Induction gives $H_rH_r^{\mathsf T}=rI$. Quotient $\Gamma^{2r}$ by

$$\Lambda=\{(a,H_r^{\mathsf T}a):a\in\Gamma^r\},$$

and let $G_r$ be the images of all source blocks. The identity half of $(I_r,H_r)\otimes I_d$ makes this lattice saturated. Every nonzero gluing word has at least three nonzero blocks: one occupied left block gives $r$ occupied right blocks; two or more occupied left blocks have at least one occupied right block, since $H_r$ is invertible. Hence the source blocks embed without new zero, equal, or antipodal labels. The entire right block family is independent in the quotient. Here independence of blocks means that their copies of $\Gamma$ embed as a direct sum; it does not mean that the source elements inside one block are rationally independent.

For the positive-real parameter $\rho$ of the phase argument,

$$\rho(G_r)\geq\rho(B)-\frac{\pi^2M^2r}{2(r+1)^2}. \tag{1}$$

Here is a direct verification of the saturated-basis issue. Put $A=(I_r,H_r)\otimes I_d$, with right inverse $P=A^{\mathsf T}/(r+1)$. Independently maximize every block polynomial. For each of the $rd$ basis-row phases, choose a real lift of the required correction which has mean zero and variance at most $\pi^2$: a phase $\theta\in[0,2\pi]$ is lifted to $\theta$ and $\theta-2\pi$ with probabilities making its mean zero. Choose these lifts independently. Applying $P$ to the lift vector solves every row constraint, because $AP=I$. These are all quotient constraints because the basis is integral and saturated.

The expected squared phase change of a frequency $b$ is bounded by $\pi^2\|P_v^{\mathsf T}b\|_2^2$, at most $\pi^2rM^2/(r+1)^2$ in any block. The mean-zero linear term cancels in $\operatorname{Re}(ze^{it})\geq\operatorname{Re}z -t\operatorname{Im}z-|z|t^2/2$. Summing coefficient-weighted losses and choosing a realization at least as good as its expectation proves (1).

Every gluing row with $a_i=w$ at one left block and zero elsewhere gives an obstruction to a $q$-coloring: each of its $\pm w$ entries has a signed representation in every restricted source color, and the blocks are disjoint. Thus $\chi(G_r)>q$.

2. Full-closure anchor loss at large order

If

$$2q(2dr+1)^R\exp\left(-\frac{r}{(2q+1)R}\right)<1, \tag{2}$$

then

$$\chi(G_r)=q+1,\qquad \tau_{q+1}(L(G_r))\leq2(q+1) \tag{3}$$

for every linear scalar encoding $L$ preserving ternary relations exactly. In particular no next-color universal anchor survives such an encoding. Condition (2) holds for every sufficiently large power of two, with $B$ fixed.

Counting every realizable gluing word

Any induced relation has block sums $(a,b)=(a,H_r^{\mathsf T}a)\in\Sigma(B)^{2r}$. Orthogonality gives $r\|a\|_2^2=\|b\|_2^2\leq rR$, hence $\|a\|_2^2\leq R$. There are at most $(2dr+1)^R$ possible integer vectors $a$: $\|a\|_1\leq\|a\|_2^2\leq R$, so at most $R$ signed coordinate steps, padded with zero steps, encode each one. A nonzero such word satisfies

$$|\operatorname{supp}_{\rm blocks}b| \geq\frac{\|b\|_2^2}{R} =\frac{r\|a\|_2^2}{R}\geq\frac rR.$$

Partition the right blocks randomly into $2q+1$ bins: one unused bin and bins $S_{t,c}$ for $t=1,2$, $c\in[q]$. A nonzero bounded word misses a given designated bin with probability at most $e^{-r/((2q+1)R)}$. The union bound and (2) give a partition in which every nonzero bounded word meets every designated bin. This covers all rationally induced ternary relations, not only the displayed Hadamard rows.

Two colorings with disjoint extra-color spans

Fix a proper source partition $B=P_1\sqcup\cdots\sqcup P_q$. For coloring $t$, use it on every block, but on right blocks in $S_{t,c}$, replace the color of $P_c$ by the new color $q+1$. A relation in an old color $c$ must vanish on $S_{t,c}$, contrary to the hitting property unless its block-sum word is zero. A zero word is excluded by source dissociation. The new color is also dissociated, since it lies in the independent right family of blocks, each with a dissociated source subset. Thus both are proper $(q+1)$-colorings.

Their new classes occupy disjoint right blocks, so their vector signed spans intersect only at zero. The assertion also survives exact scalar encoding: representations of a common scalar on these disjoint classes differ by a ternary relation; exactness makes their vector lifts equal, and disjoint right-coordinate spans make that lift zero. Take all $q+1$ cyclic relabelings of both colorings. For each color the intersection is then zero, proving (3). Restriction and the anchor obstruction give the exact chromatic number.

3. Fixed-order diagonal padding does not preserve the anchor

Repeat every column of $(I_r,H_r)$ $s$ times, equivalently replacing each source block by $s$ independent copies and gluing along the diagonal copy of $\Gamma$, which contains the diagonal anchor. This repeats columns, not rows; applying Section 1 to the entire group $\Gamma^s$ would be a different construction. The repeated row matrix still has an identity minor and is saturated. Its right inverse spreads the correction among the $s$ copies, giving

$$\rho(G_{r,s})\geq\rho(B)- \frac{\pi^2M^2r}{2s^2(r+1)^2}.$$

If $s\geq2q$, assign one copy of each right coordinate to each of $2q$ disjoint designated bins. Every nonzero right word is repeated in each such bin, so the hitting property holds without a probabilistic count. The preceding two-coloring construction gives $\chi(G_{r,s})=q+1$ and the same threshold bound (3).

The construction therefore does not provide a fixed-source cheap restart via large order or diagonal padding. It does not exclude all correlated gluing, nor changing-source orders below the quantified threshold (2).