Quadratic-loss primitive-anchor gluing

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Standing

The analytic lemma below is author-recorded; a separate reviewer checked it internally, which is not a tier. It improves the available estimate for a particular gluing operation. It does not supply iterated chromatic growth: a non-antipodal example below shows that even arbitrarily inexpensive boundary gluing can destroy every universal anchor.

Primitive normalization

Let $B$ be a finite set of nonzero elements of a free abelian group $\Gamma=\langle B\rangle_{\mathbb Z}$, with no antipodal pair. Suppose $\chi(B)=q$ and a nonzero $w\in\Gamma$ belongs to $\Sigma(C)$ for every class $C$ of every proper $q$-coloring. Write $w=ku$, where $u$ generates the rank-one group $\Gamma\cap\mathbb Qw$, and $k$ is a nonzero integer. Thus $u$ is primitive in $\Gamma$. Define

$$\lambda_B(u)=\min_{\ell(u)=1}\max_{b\in B}|\ell(b)|, \qquad \ell\in\operatorname{Hom}(\Gamma,\mathbb R). \tag{1}$$

The minimum exists: because $B$ spans $\Gamma\otimes\mathbb R$, $\max_B|\ell(b)|$ is a norm, and its sublevel sets are compact.

For $m\geq3$, form the torsion-free group

$$\Gamma_m=\Gamma^m/\mathbb Z(u_1+\cdots+u_m)$$

and let $G_m$ be the union of the images of $B$ in its $m$ summands. Primitivity of $u$ implies that the displayed quotient is free abelian. The copies of $B$ remain distinct and nonzero, with no new antipodal pairs: a relation between at most two summands cannot be a nonzero multiple of $u_1+\cdots+u_m$.

The primitive normalization is essential. Quotienting by $w_1+\cdots+w_m$ without saturation can leave torsion and is not the rational-vector gluing used for positive-integer realization. In particular, one cannot replace $\lambda_B(u)$ by $\lambda_B(w)$ in the bound below.

Analytic gluing lemma — proved

Use the positive-real Sidon parameter $\rho$ defined in the tensor extraction theorem. Then

$$\rho(G_m)\geq\rho(B)- \frac{\pi^2\lambda_B(u)^2}{2m^2}. \tag{2}$$

Indeed, the character tuples on $\Gamma_m$ are precisely $(g_1,\ldots,g_m)\in\widehat\Gamma^m$ such that $\prod_i u(g_i)=1$. Write an arbitrary polynomial on $G_m$ as $\sum_i f_i(g_i)$, and independently maximize $\operatorname{Re}f_i$ on each torus. The resulting value is at least $\rho(B)$ times the sum of the absolute values of all coefficients.

Choose $\theta\in[-\pi,\pi]$ with $e^{i\theta}=(\prod_i u(g_i))^{-1}$, and choose a minimizing $\ell$ in (1). Replace each $g_i$ by the character

$$g'_i(\gamma)=g_i(\gamma) \exp(i\theta\ell(\gamma)/m).$$

Their product constraint now holds because $\ell(u)=1$. At the independent maximum of $\operatorname{Re}f_i$, its derivative in the $\ell$ direction vanishes. Applying $\operatorname{Re}(ze^{it})\geq\operatorname{Re}z -t\operatorname{Im}z-|z|t^2/2$ term by term therefore bounds the total loss by

$$\frac{\theta^2}{2m^2}\sum_{i,b}|c_{i,b}|\ell(b)^2 \leq \frac{\pi^2\lambda_B(u)^2}{2m^2} \sum_{i,b}|c_{i,b}|.$$

This proves (2), without a randomized lattice lift.

For $m\geq q+1$, the coloring number is exactly $q+1$. Under a putative $q$-coloring, each block class signed-spans $w$, and their representations sum to zero in the quotient. Conversely, use $q+1$ colors, a proper $q$-palette in every block, and arrange that every color is absent from a block. Any proper collection of block summands embeds as a direct sum in the quotient, so every resulting class is dissociated.

Cheap boundary gluing need not propagate an anchor — proved

Let $A\subset\mathbb Z_{>0}$ have gcd one, maximum $M$, coloring number $q\geq2$, and a universal anchor $a\ne0$. The punctured Grow–Whicher example in anchor gluing supplies such an $A$, with $q=2$, $a=3$, and $M=253$.

For an integer $r\geq2$, put

$$B=\bigsqcup_{j=1}^r A e_j\subset\mathbb Z^r, \qquad w=a(e_1+\cdots+e_r),\qquad u=e_1+\cdots+e_r.$$

Direct sums preserve the coloring number and the hereditary dissociation constant. The vector $w$ is universal: restrict any proper $q$-coloring to each summand, represent $ae_j$ in the given color, and add the disjoint representations. Moreover

$$\rho(B)=\rho(A),\qquad \lambda_B(u)=M/r. \tag{3}$$

The first equality follows by independently maximizing the polynomials on the summands. For the second, write $\ell(e_j)=t_j$. The constraint is $\sum_jt_j=1$, while the objective is $M\max_j|t_j|$; its minimum $M/r$ is attained at $t_j=1/r$.

Apply the boundary gluing $m=q+1$ to $(B,w)$. Its $rm$ coordinate rays have exactly the one relation that their sum is zero. Consequently the output is exactly

$$G_m\cong A\otimes S_{r(q+1)}. \tag{4}$$

It has coloring number $q+1$, and (2) has loss $\pi^2M^2/[2r^2(q+1)^2]$, tending to zero as $r\to\infty$. Nevertheless $r(q+1)\geq q+2$, so the proved large-simplex obstruction says that it has no nonzero universal anchor for $(q+1)$-colorings, even after any exact ternary-relation-preserving scalar encoding.

Thus non-antipodality, hereditary extraction, and arbitrarily small analytic loss do not imply boundary-anchor propagation. This does not rule out a more structured scalar boundary construction; that narrower problem remains open here.