Version 3 · current · External review, round 3: correct

# Anchor for an unpinned field of oscillator cells: lowest eigenvector, long-range mark and weighing

- **Subproject:** 08-ilang-propagation
- **Package:** 05-anchor
- **Version:** v3
- **Mode:** external regeneration
- **Date:** 2026-10-25

## Changes from previous version

- Step 8, cubic torus: the limiting criterion for $a_{\rm bond}$ is now the exact one, $N/(D-H)\to0$ (image differences at the bodies over the direct-minus-image difference at the midpoint). It reduces to $N/D\to0$ only under $H/D\to0$, which families (A) and (B) supply. A new control, a pair far from a source next to the anchor, has $N/D\to0$ but $a_{\rm bond}\not\to0$; it is checked numerically (I1).
- Result 14, (5.27) in the twin: the unrestricted equivalence for $a_{\rm bond}$ is removed and the control is added (I1).
- Consistency check 1: the limit $\kappa_a\to0$ is separated from the criterion at $\kappa_a=0$. For linear grips the anchored lowest eigenvectors have no normalizable limit (I2).
- Code: `code/v3/check_anchor.py` gets the pair control of I1 (asymptotics.txt). The script was run again, and all numbers come from `code/v3/output/`.

## Response to verification

- I1: Accepted and fixed in Step 8 (cubic torus, after (5.27)) and in Result 14. With $N:=\Delta g_h(x_1)+\Delta g_h(x_2)$, $D:=\Delta g_s(\bar x)$, $H:=\Delta g_h(\bar x)$, the exact criterion is $a_{\rm bond}\to0$ iff $N/(D-H)\to0$. It is equivalent to $N/D\to0$ when also $H/D\to0$, not in general. Families (A) and (B) give $H/D\to0$, so their conclusions, (5.24) and the far-anchor theorem, are unchanged. The verifier's example ($h=0$, $x_s=e_x$, $x_{1,2}=(n^3,\pm n^4,0)$) is added as a control and checked on $\mathbb Z^3$ (asymptotics.txt). In it, $N/D\to0$, $N/(D-H)\to1$ and $a_{\rm bond}\to-(c_\phi-g_\infty(e_x))/c_\phi$.
- I2: Accepted and fixed in Consistency check 1. For $F_2\ne0$, the limit $\kappa_a\to0$ of (5.10) gives (5.5). For a linear grip, (5.10) holds at every $\kappa_a>0$, while at $\kappa_a=0$ Step 2 gives no lowest eigenvector. The anchored lowest eigenvectors lose their normalizable limit (mean shift $-\mu F_1/\kappa_a$, diverging spread of the uniform change), and their eigenvalue tends to $-\infty$. The criterion at $\kappa_a=0$ is that of Step 2, not the limit of (5.10).

## Setup and assumptions

**Description** (assumptions M8, M9, A11, A15; all from the question).
- *Field:* oscillator cells, one at each place $x$ of a ring ($D=1$) or cubic torus ($D=3$) of side $L\ge3$, $n=L^D$; contract the compression $C_M=\Pi H\Pi$ of $H=\tfrac12p^{\rm T}p+\tfrac12q^{\rm T}\kappa_h\Delta q$ ($\Delta$ the graph Laplacian, $\kappa_h>0$, no pinning), plus the terms below; $\Phi_x=q_x$. $P_x:=e_xe_x^{\rm T}$.
- *Anchor:* $\tfrac{\kappa_a}2q_h^2$, $\kappa_a>0$, at one place $h$.
- *Bodies:* held Model D bodies ($\tau=0$, no rest and clock term) at places $x_i$, in the eigenvector $\lvert\sigma_i\rangle\lvert f_{k_i}\rangle$ of $S$ with eigenvalue $\mu_i=\sigma_iM_{k_i}$, $M_k=m+2\pi k/(N\lambda_0)>0$; grip $\sum_x\lvert x\rangle\langle x\rvert\otimes S\otimes F^{(i)}(q_x)$, $F^{(i)}=F^{(i)}_0+F^{(i)}_1q+F^{(i)}_2q^2$, $F^{(i)}_1\ne0$ (class G2). Notation: $\alpha_i:=2\mu_iF_2^{(i)}$, and for $F_0^{(i)}\ne0$: $\hat b_i:=F_1^{(i)}/(2F_0^{(i)})$, $\hat\beta_i:=2F_2^{(i)}/F_1^{(i)}$. The square grip $4F_0F_2=F_1^2$ is $\hat\beta=\hat b$, i.e. $F=F_0(1+\hat\beta q)^2$.
- *Sources:* held objects coupled by $g_sq_{x_s}$.
- *State:* the total contract is diagonal in the place label of every held body and commutes with its $S$, so it is block diagonal in (place, $\sigma_i$, $k_i$). "Lowest eigenvector" means that of the block of the stated body labels (the reading of (2.5)); statements about the whole contract (all blocks) are marked as such.

**Inputs** (quoted in the question, from 08-ilang-propagation/02-grip-weight@v3, pinned field $K=\kappa_h\Delta+m_h^2\mathbb 1$, $G_0=(K^{-1})_{xx}$, $w_0=\langle q_x^2\rangle$):
- (2.5): a lowest eigenvector exists ($M\to\infty$) iff $1+2\sigma M_kF_2G_0>0$; otherwise the lowest eigenvalue of $C_M\to-\infty$.
- (2.18): for grips $F^{(i)}=F^{(i)}(0)\Phi_i(\beta q)$, at order $\mu^2$ and leading order in $\beta$: $\rho_{{\rm mark},i}=1$ and $\rho_{{\rm bond},i}=m_{\mathbf r}$ for every level and both branches iff $2\Phi_i''(0)=\Phi_i'(0)^2$ and $\Phi_1'(0)=\Phi_2'(0)$, i.e. square grips with a common $\hat\beta$.

**Definitions** (those of 02-grip-weight, applied with the anchor). *Cost* $C$: lowest eigenvalue with the body minus that without it, no sources. Replacing $F\to sF$: $C(s)=sC^{(1)}+s^2C^{\rm mark}+O(s^3)$. *Weight:* $W(x)$ is the $g_s$-linear part of the cost in the presence of the source, $W^{(1)},W^{\rm mark}$ its $s$ and $s^2$ coefficients; the weight is $W(x)-W(x')$. *Bond* $B=C_{12}-C_1-C_2$ (no sources) and its $g_s$-linear part $W_{\rm bond}$, both at order $s^2$. For two placements $x,x'$ of a body (or two placements of a pair related by a translation): $\Delta f:=f(x)-f(x')$, $\bar f:=\tfrac12(f(x)+f(x'))$, and
$$\rho_{\rm mark}:=\frac{\Delta W^{\rm mark}}{\Delta W^{(1)}}\,\frac{\bar C^{(1)}}{\bar C^{\rm mark}},\qquad\rho_{{\rm bond},i}:=\frac{\Delta W_{\rm bond}}{\bar B}\,\frac{\bar C_i^{(1)}}{\Delta W_i^{(1)}},\qquad m_{\mathbf r}:=\frac{\Delta\phi_a(x_1)+\Delta\phi_a(x_2)}{2\Delta\phi_a(\bar x)},$$
where $C_i^{(1)},W_i^{(1)}$ belong to one body with the grip and level of body $i$ at the midpoint $\bar x$ of the pair (a place), and $\phi_a$ is the made field (5.11). With place-independent costs these are the ratios of 02-grip-weight; with an anchor the costs depend on the place and are taken as means over the two placements. *Mutual cost* of two sources: the part $\propto g_1g_2$ of the lowest eigenvalue.

**Regime.** Exact statements hold for every $L\ge3$, $\kappa_h,\kappa_a>0$, $h$, on both graphs; asymptotic statements on the cubic torus take $L\to\infty$ first at fixed places. Places are in lattice units (separations are cell separations, M2). Standard mathematics used: Gaussian lowest eigenvectors of quadratic contracts, min-max, Schur complements, Sherman–Morrison and the matrix determinant lemma, M-matrices, lattice Green functions and their asymptotics.

## Derivation

### Step 1. The contract of a block and the limit $M\to\infty$

In the block of the bodies, $S\to\mu_i$ and $\lvert x\rangle\langle x\rvert\to\delta_{x,x_i}$, so the total contract acts on the field as
$$C_{\rm f}=\tfrac12p^{\rm T}p+\tfrac12q^{\rm T}Aq+b^{\rm T}q+c_0,\quad A=\kappa_h\Delta+\kappa_aP_h+\sum_i\alpha_iP_{x_i},\quad b=\sum_i\mu_iF_1^{(i)}e_{x_i}+\sum_sg_se_{x_s},\quad c_0=\sum_i\mu_iF_0^{(i)},\tag{5.1}$$
with $\kappa_a=0$ without an anchor. For $A>0$ the uncompressed $C_{\rm f}$ has a simple isolated lowest eigenvalue with a Gaussian eigenvector (standard):
$$E_0=c_0+\tfrac12{\rm Tr}\sqrt A-\tfrac12b^{\rm T}A^{-1}b,\qquad\langle q\rangle=-A^{-1}b,\qquad\langle q_xq_y\rangle-\langle q_x\rangle\langle q_y\rangle=\tfrac12(A^{-1/2})_{xy}.\tag{5.2}$$
If $A$ has a negative eigenvalue, $C_{\rm f}$ is unbounded below. If $A\ge0$ has a kernel vector $v$, the contract along $v$ is $\tfrac12p_v^2+(b\cdot v)q_v$: unbounded below if $b\cdot v\ne0$, otherwise a free mode whose spectrum is continuous at its bottom, with no eigenvector there. *Sense of $M\to\infty$:* the ranges of $\Pi$ grow with $M$ and their union (finite superpositions of oscillator levels) is a core of the uncompressed contract (standard for quadratic operators). By min-max the lowest eigenvalue of $C_M$ does not increase with $M$ and tends to the bottom of the spectrum; if that is an isolated simple eigenvalue, the lowest eigenvectors of $C_M$ converge in norm to it (a unit vector with cost within $\epsilon$ of $E_0$ has overlap$^2\ge1-\epsilon/\text{gap}$). Hence
$$\text{a lowest eigenvector exists }(M\to\infty)\iff A>0;\ \text{otherwise the lowest eigenvalue of }C_M\text{ tends to }-\infty\text{ or to the bottom of a continuum}.\tag{5.3}$$

### Step 2. Without an anchor (item 1)

$\Delta\ge0$ with kernel the uniform vector $\mathbf 1$ (connected graph): the only free direction is the uniform change. Let the bodies sit at $n$ distinct places (bodies at one place add their $\alpha$), $\mathcal A:={\rm diag}(\alpha_i)$. *Kron reduction* (Schur complement): $\min\{q^{\rm T}\kappa_h\Delta q:\ q_{x_i}=y_i\}=y^{\rm T}\mathcal L_ny$, attained at the harmonic extension, with $\mathcal L_n\ge0$ the reduced Laplacian of the body places, kernel ${\rm span}\{\mathbf 1\}$; $\mathcal L_1=0$, $\mathcal L_2=r_{12}^{-1}\left(\begin{smallmatrix}1&-1\\-1&1\end{smallmatrix}\right)$ with $r_{12}=r(x_1,x_2)$ of (5.8). So $q^{\rm T}A_0q\ge y^{\rm T}(\mathcal L_n+\mathcal A)y$ with equality at the extension, and for $y=0$, $q^{\rm T}A_0q=\kappa_hq^{\rm T}\Delta q>0$ unless $q=0$:
$$A_0>0\iff\mathcal L_n+\mathcal A>0.\tag{5.4}$$
*Members.* With $y=\mathbf 1$: necessary $\sum_i\mu_iF_2^{(i)}>0$; sufficient: all $\alpha_i\ge0$, one $>0$. One body:
$$\text{lowest eigenvector}\iff\mu F_2>0\iff\sigma={\rm sign}\,F_2\ (\text{every level }k);\quad\text{for }\sigma=-{\rm sign}\,F_2\text{ the lowest eigenvalue of }C_M\to-\infty.\tag{5.5}$$
Two bodies (trace and determinant of a $2\times2$ matrix):
$$A_0>0\iff\alpha_1+\alpha_2>0\ \text{and}\ \alpha_1\alpha_2+(\alpha_1+\alpha_2)/r_{12}>0.\tag{5.6}$$
If all $\alpha_i=0$ (no bodies, or only linear grips), $A_0=\kappa_h\Delta$ and $b\cdot\mathbf 1=\sum_i\mu_iF_1^{(i)}+\sum_sg_s$: the lowest eigenvalue tends to $-\infty$ if this is $\ne0$; otherwise the uniform change is free and there is no lowest eigenvector. In particular the field alone has none, so the reference eigenvalue of the cost of a body does not exist. *Whole contract:* the block with $\sigma_i=-{\rm sign}\,F_2^{(i)}$ for every body has all $\alpha_i\le0$, so without an anchor the whole contract has no lowest eigenvector, for any bodies of (G2).
*A grip as anchor.* A body with $\alpha=2\mu F_2>0$ adds $\tfrac\alpha2q_{x}^2$: an anchor of strength $\alpha$ at its own place (and a source $\mu F_1$). For one such body, summing the components of $(\kappa_h\Delta+\alpha P_x)v=e_y$ gives $\alpha v_x=1$, so $(A_0^{-1})_{xy}=1/\alpha$ for every $y$, and its $g_s$-linear cost $-\mu F_1g_s(A_0^{-1})_{xx_s}$ is
$$W(x)=-\frac{F_1g_s}{2F_2}\quad\text{for every }x,x_s:\qquad W(x)-W(x')=0.\tag{5.7}$$
A body that anchors the field itself has no weight; and $C(s)$ has no lowest eigenvector for $s\le0$, so $C^{(1)},C^{\rm mark}$ do not exist.

### Step 3. The anchored Green function

$K_a:=\kappa_h\Delta+\kappa_aP_h>0$ ($q^{\rm T}K_aq=0$ forces $q$ uniform and $q_h=0$). Let $g:=(\kappa_h\Delta)^+$, $g(x)=\frac1{\kappa_hL^D}\sum_{k\ne0}\frac{e^{ik\cdot x}}{\hat\Delta(k)}$, $\hat\Delta(k)=\sum_j(2-2\cos k_j)$, $k\in\frac{2\pi}L\mathbb Z_L^D$, and $r(x,y):=2[g(0)-g(x-y)]$ (the resistance distance of the springs $\kappa_h$). For $K_av=f$, summing components gives $\kappa_av_h=\sum_xf_x=:F$; then $\kappa_h\Delta v=f-Fe_h\perp\mathbf 1$, so $v=g*(f-Fe_h)+c\mathbf 1$ with $c$ fixed by $v_h=F/\kappa_a$. For $f=e_y$:
$$G_a(x,y):=(K_a^{-1})_{xy}=\frac1{\kappa_a}+g(x-y)-g(x-h)-g(y-h)+g(0)=\frac1{\kappa_a}+\frac12\bigl[r(x,h)+r(y,h)-r(x,y)\bigr],\tag{5.8}$$
exact for every $L$, $D$, $\kappa_h,\kappa_a>0$, $h$. $G_a(h,y)=1/\kappa_a$ for all $y$; $G_a>0$ entrywise ($K_a$ is an irreducible M-matrix). *Ring:* $g(x)=\frac1{\kappa_h}\bigl[\frac{L^2-1}{12L}-\frac{s(L-s)}{2L}\bigr]$ ($s$ the ring distance; zero sum, $\kappa_h\Delta g=\delta-\frac1L$), so $r=\frac{s(L-s)}{\kappa_hL}$. *Cubic torus:* as $L\to\infty$ at fixed arguments, $g\to g_\infty$, the Green function of $\kappa_h\Delta$ on $\mathbb Z^3$, with $g_\infty(0)=\gamma_3/\kappa_h$, $\gamma_3=0.2527310$, and $g_\infty(x)=\frac1{4\pi\kappa_h\lvert x\rvert}+O(\lvert x\rvert^{-3})$ (asymptotics.txt). For $\nu>0$, Sherman–Morrison with $\gamma_\nu:=(\kappa_h\Delta+\nu^2)^{-1}$ gives
$$R_{xy}(\nu):=\bigl[(K_a+\nu^2)^{-1}\bigr]_{xy}=\gamma_\nu(x-y)-\frac{\kappa_a\gamma_\nu(x-h)\gamma_\nu(y-h)}{1+\kappa_a\gamma_\nu(0)},\qquad R(0)=K_a^{-1};\tag{5.9}$$
$R_{xy}(\nu)>0$, and $R_{xx}(\nu)\le\min(\Gamma(x),\nu^{-2})$ decreases in $\nu$.

### Step 4. With an anchor: existence, made field, field constants (item 2)

*Existence.* $\min\{q^{\rm T}K_aq:\ q_{x_i}=y_i\}=y^{\rm T}\Gamma_n^{-1}y$ with $(\Gamma_n)_{ij}:=G_a(x_i,x_j)$ (standard); as in Step 2,
$$A>0\iff\Gamma_n^{-1}+\mathcal A>0;\qquad\text{one body: }1+2\mu F_2\Gamma(x)>0,\quad\Gamma(x):=G_a(x,x)=\frac1{\kappa_a}+r(x,h).\tag{5.10}$$
It holds in every block with all $\mu_iF_2^{(i)}\ge0$ (linear grips, both branches; no bodies). For $\mu F_2<0$ it is $2M_k\lvert F_2\rvert\Gamma(x)<1$: (2.5) with $G_0\to\Gamma(x)$. On the ring $\Gamma=\frac1{\kappa_a}+\frac{d(L-d)}{\kappa_hL}$ ($d$ the ring distance to $h$): the branch $\sigma=-{\rm sign}\,F_2$ loses its lowest eigenvector once $\frac{d(L-d)}{\kappa_hL}\ge\frac1{2M_k\lvert F_2\rvert}-\frac1{\kappa_a}$. On the cubic torus $\Gamma$ is bounded uniformly in $L$ and tends to $\Gamma_\infty-2g_\infty(x-h)<\Gamma_\infty:=\frac1{\kappa_a}+\frac{2\gamma_3}{\kappa_h}$, so for $L\to\infty$ the condition holds at every place iff $2M_k\lvert F_2\rvert\Gamma_\infty\le1$. For the whole contract with one body the condition is needed for the highest level and the branch with $\mu F_2<0$.

*Two bodies.* $\Gamma_n^{-1}+\mathcal A$ ($n=2$) is positive iff its trace and determinant are. Multiplying both by $\det\Gamma_n>0$, with $G_{12}:=G_a(x_1,x_2)$:
$$A>0\iff\bigl(1+\alpha_1\Gamma(x_1)\bigr)\bigl(1+\alpha_2\Gamma(x_2)\bigr)>\alpha_1\alpha_2G_{12}^2\ \text{and}\ \Gamma(x_1)+\Gamma(x_2)+(\alpha_1+\alpha_2)\bigl(\Gamma(x_1)\Gamma(x_2)-G_{12}^2\bigr)>0;\qquad\text{sufficient: }\bar a\,\bigl(\max_i\Gamma(x_i)+G_{12}\bigr)<1,\tag{5.26}$$
with $\bar a:=\max_i\max(0,-\alpha_i)$. The sufficient condition holds because $\Gamma_n^{-1}+\mathcal A\ge\lambda_{\max}(\Gamma_n)^{-1}-\bar a$, and $\lambda_{\max}(\Gamma_n)$ is at most the largest row sum ($G_{12}>0$). Two bodies that each meet the one-body condition need not meet (5.26). For $\alpha_1=\alpha_2=-a$ and $\Gamma(x_1)=\Gamma(x_2)=\Gamma$, the pair needs $a(\Gamma+G_{12})<1$, while each body alone needs only $a\Gamma<1$. On the cubic torus far from the anchor, $G_{12}\to c_\phi+g_\infty(\mathbf r)\ge c_\phi>0$ ((5.13)), so the gap persists at every separation.

Checked against $\lambda_{\min}(A)$, on two sizes of each graph (existence.txt):
- one body: three grips, two levels, both branches, three distances, with (5.5) and (5.6);
- random pairs of bodies: (5.10) and (5.26) agree with $\lambda_{\min}(A)$ in 40/40 cases;
- for each lattice, a pair with $\alpha_1=\alpha_2=-a$ and $1/\lambda_{\max}(\Gamma_n)<a<1/\max_i\Gamma(x_i)$: each body alone has a lowest eigenvector, but the pair has $\lambda_{\min}(A)<0$, as (5.26) predicts.

*Made field.* Without bodies, (5.2) with $b=g_se_{x_s}$:
$$\phi_a(x):=\langle q_x\rangle=-g_sG_a(x,x_s)=-g_s\Bigl[\frac1{\kappa_a}+\frac{r(x_s,h)}2\Bigr]-\frac{g_s}2\bigl[r(x,h)-r(x,x_s)\bigr].\tag{5.11}$$
It is a uniform shift (the anchored cell moves by $-g_s/\kappa_a$, plus half the resistance from source to anchor) plus a place-dependent part. Cubic torus, $L\to\infty$, with $c_\phi:=\frac1{\kappa_a}+\frac{\gamma_3}{\kappa_h}$: $\phi_a(x)=-g_s\bigl[c_\phi-g_\infty(x_s-h)\bigr]-g_s\bigl[g_\infty(x-x_s)-g_\infty(x-h)\bigr]$: the source and an image of opposite sign at the anchor, over a nonzero constant. Line (ring, $L\to\infty$): $\phi_a=-g_s[\frac1{\kappa_a}+\frac{\min(d_x,d_s)}{\kappa_h}]$ for $x,x_s$ on one side of $h$, $-g_s/\kappa_a$ on opposite sides.

*Field constants.* With $K_au_a=\omega_a^2u_a$ orthonormal, $\frac1{2\sqrt\lambda}=\frac1\pi\int_0^\infty\frac{d\nu}{\lambda+\nu^2}$ and $\frac1{\omega_a\omega_b(\omega_a+\omega_b)}=\frac2\pi\int_0^\infty\frac{d\nu}{(\omega_a^2+\nu^2)(\omega_b^2+\nu^2)}$, the constants that enter (Steps 6–7) are $\Gamma(x)$ (analogue of $G_0$), $G_a(x,y)$ (of $G(x-y)$) and
$$w(x):=\langle q_x^2\rangle=\tfrac12(K_a^{-1/2})_{xx}=\frac1\pi\int_0^\infty R_{xx}(\nu)\,d\nu,\qquad\Sigma(x,y):=\sum_{a,b}\frac{u_a(x)u_a(y)u_b(x)u_b(y)}{\omega_a\omega_b(\omega_a+\omega_b)}=\frac2\pi\int_0^\infty R_{xy}(\nu)^2d\nu,\tag{5.12}$$
the analogue of $w_0$ and the static correlation of $q_x^2,q_y^2$ (a term $\epsilon q_y^2$ changes $w(x)$ by $-\epsilon\Sigma(x,y)$). Against the distance $d$ to $h$ ($x-y$ fixed):
$$\begin{aligned}&\text{line }(L\to\infty):&&\Gamma=\tfrac1{\kappa_a}+\tfrac d{\kappa_h},\quad w=\tfrac{\ln d}{2\pi\sqrt{\kappa_h}}+O(1),\quad\Sigma(x,x)=\tfrac{2\ln2}{\pi\kappa_h^{3/2}}\,d\,(1+o(1));\\&\text{cubic }(L\to\infty):&&\Gamma=\Gamma_\infty-2g_\infty(x-h),\quad G_a(x,y)\xrightarrow{d\to\infty}c_\phi+g_\infty(x-y),\quad w-w_\infty,\ \Sigma-\Sigma_\infty=O(d^{-3}),\end{aligned}\tag{5.13}$$
with $w_\infty,\Sigma_\infty(x-y)$ the values (5.12) of the unanchored infinite lattice. *Line:* the anchor term of (5.9) makes $R_{xx}(\nu)\simeq\frac{1-e^{-2\nu d/\sqrt{\kappa_h}}}{2\nu\sqrt{\kappa_h}}$ for $d\gg1$ (up to $O(1/d)$ relative terms from $\kappa_a$); $\frac1\pi\int$ gives the logarithm, and $\frac2\pi\int$ gives $\frac{d}{\pi\kappa_h^{3/2}}\int_0^\infty\frac{(1-e^{-t})^2}{t^2}dt=\frac{2\ln2\,d}{\pi\kappa_h^{3/2}}$. From (5.9) on $\mathbb Z$ at $\kappa_h=1$, $\kappa_a=0.7$: $w(10d)-w(d)=0.3485,0.3645,0.3663$ for $d=10,100,1000$ against $\frac{\ln10}{2\pi}=0.3665$, and $\Sigma/d=0.4895,0.4463,0.4418,0.4413$ for $d=10,\dots,10^4$ against $\frac{2\ln2}\pi=0.4413$ (asymptotics.txt). *Cubic:* for each $\nu>0$ the torus resolvent tends to (5.9) on $\mathbb Z^3$, dominated by $\min(\Gamma_{\max},\nu^{-2})$, so (5.12) converges; the anchor term is $O(e^{-2\nu d/\sqrt{\kappa_h}}/d^2)$, whose $\nu$-integral is $O(d^{-3})$. At $\nu=0$ the zero mode of $\Delta$ keeps the anchor at every distance: $\Gamma\to\frac1{\kappa_a}+\frac{2\gamma_3}{\kappa_h}$, not the unanchored $g_\infty(0)=\gamma_3/\kappa_h$. The constants $w,\Sigma$ forget a distant anchor; $\Gamma,G_a$ do not.

### Step 5. Mutual cost of two sources (item 3)

By (5.2) with $A=K_a$, $b=g_1e_{x_1}+g_2e_{x_2}$:
$$\mathcal E_{12}=-g_1g_2\,G_a(x_1,x_2)=-g_1g_2\Bigl[\frac1{\kappa_a}+g(x_1-x_2)-g(x_1-h)-g(x_2-h)+g(0)\Bigr],\tag{5.14}$$
exact on both graphs for every $L$ (with held bodies, $G_a\to(A^{-1})_{x_1x_2}$); it is the constant $-g_1g_2/\kappa_a$ if one source sits at the anchor. Checked against the exact eigenvalue to 10 digits on rings $L=64,256$ and tori $L=16,32$ (mutual_cost.txt). *Cubic torus, $L\to\infty$ first*, all of $\lvert x_1-x_2\rvert,\lvert x_1-h\rvert,\lvert x_2-h\rvert\ge\ell\gg1$:
$$\mathcal E_{12}=-g_1g_2\,c_\phi-\frac{g_1g_2}{4\pi\kappa_h}\Bigl[\frac1{\lvert x_1-x_2\rvert}-\frac1{\lvert x_1-h\rvert}-\frac1{\lvert x_2-h\rvert}\Bigr]+O(\ell^{-3}).\tag{5.15}$$
With $\mathbf r=x_2-x_1$, midpoint $X$, $D_X=\lvert X-h\rvert$:
- *(a) $L\to\infty$, then $D_X\to\infty$ at fixed $\mathbf r$, then $\lvert\mathbf r\rvert\to\infty$:* $\mathcal E_{12}\to-g_1g_2[c_\phi+g_\infty(\mathbf r)]$, a fall as $-\frac{g_1g_2}{4\pi\kappa_h\lvert\mathbf r\rvert}$, the coefficient of the field without an anchor. Uniformly for $1\ll\lvert\mathbf r\rvert\ll D_X$ (the dipole parts of the two anchor terms cancel): $\mathcal E_{12}=-g_1g_2[c_\phi-\frac1{2\pi\kappa_hD_X}]-\frac{g_1g_2}{4\pi\kappa_h\lvert\mathbf r\rvert}[1+O(\lvert\mathbf r\rvert^3/D_X^3)+O(\lvert\mathbf r\rvert^{-2})]$.
- *(b) $L\to\infty$, then $\lvert\mathbf r\rvert\to\infty$ with $x_1$ fixed:* $\mathcal E_{12}=-g_1g_2[c_\phi-g_\infty(x_1-h)]-\frac{g_1g_2}{4\pi\kappa_h}\frac{\hat{\mathbf r}\cdot(x_1-h)}{\lvert\mathbf r\rvert^2}+O(\lvert\mathbf r\rvert^{-3})$: a fall as $\lvert\mathbf r\rvert^{-2}$ or faster, no long-range mark; no place dependence at all for $x_1=h$.
- *(c) $L\to\infty$, then $\lvert\mathbf r\rvert\to\infty$ at fixed $X$:* $\lvert x_i-h\rvert=\frac{\lvert\mathbf r\rvert}2(1+O(D_X/\lvert\mathbf r\rvert))$ and $\mathcal E_{12}=-g_1g_2c_\phi+\frac{3g_1g_2}{4\pi\kappa_h\lvert\mathbf r\rvert}[1+O(D_X^2/\lvert\mathbf r\rvert^2)+O(\lvert\mathbf r\rvert^{-2})]$: a fall as the inverse separation with $-3$ times the coefficient of (a).
- *$L$ not first:* (5.14) is a bounded function on the torus; the forms above need $\lvert\mathbf r\rvert,D_X\ll L$.

So the coefficient of the unanchored field is obtained when $\lvert\mathbf r\rvert/D_X\to0$ after $L\to\infty$, and not in the orders (b), (c). On $\mathbb Z^3$ the coefficient of $-\frac{g_1g_2}{4\pi\kappa_h\lvert\mathbf r\rvert}$ between consecutive separations is $+1.03,+1.00$ for $D_X=40$, $\lvert\mathbf r\rvert=4\to8\to16$ (a), and $-3.14,-3.03,-3.01$ for $X=h$, $\lvert\mathbf r\rvert=8\to16\to32\to64$ (c); for $x_1-h=3\hat e_x$, $4\pi\kappa_h\lvert\mathbf r\rvert^2[\mathcal E_{12}-\mathcal E_{12}(\infty)]=-2.20,-2.53,-2.74,-2.87$ for $\lvert\mathbf r\rvert=8,\dots,64$, approaching $-\hat{\mathbf r}\cdot(x_1-h)=-3$ (b) (asymptotics.txt). On the torus $L=32$ with $D_X=6$ it runs from $+1.08$ ($\lvert\mathbf r\rvert=2\to4$) through $\approx0$ to $-1.11$ ($12\to16$), the crossover from (a) to (c); with one source at $h$, $\mathcal E_{12}=-1/\kappa_a$ to 11 digits (mutual_cost.txt). *Ring (control):* from $r=\frac{s(L-s)}{\kappa_hL}$,
$$\mathcal E_{12}=-g_1g_2\Bigl[\frac1{\kappa_a}+\frac{d_1(L-d_1)+d_2(L-d_2)-s(L-s)}{2\kappa_hL}\Bigr]\xrightarrow{L\to\infty}\begin{cases}-g_1g_2\bigl[\frac1{\kappa_a}+\frac{\min(d_1,d_2)}{\kappa_h}\bigr]&\text{one side of }h,\\-g_1g_2/\kappa_a&\text{opposite sides}.\end{cases}\tag{5.16}$$
It is a quadratic polynomial in the distances at finite $L$, independent of the separation at fixed distance of the nearer source for $L\to\infty$, and grows with that distance: no part falls as the inverse separation in any order of the limits.

### Step 6. Cost and weight of one body (item 4)

One body at $x$: $A=K_a+\alpha P_x$. Determinant lemma, $\det(A+\nu^2)/\det(K_a+\nu^2)=1+\alpha R_{xx}(\nu)$, with $\int_0^\infty\ln\frac{\lambda'+\nu^2}{\lambda+\nu^2}d\nu=\pi(\sqrt{\lambda'}-\sqrt\lambda)$, gives $\tfrac12{\rm Tr}(\sqrt A-\sqrt{K_a})=\frac1{2\pi}\int_0^\infty\ln(1+\alpha R_{xx})d\nu$; Sherman–Morrison gives $(A^{-1})_{xy}=G_a(x,y)/(1+\alpha\Gamma(x))$. By (5.2), the cost and its $g_s$-linear part with a source (the trace does not depend on $g_s$) are, under (5.10),
$$C(x)=\mu F_0+\frac1{2\pi}\int_0^\infty\ln\bigl(1+2\mu F_2R_{xx}(\nu)\bigr)d\nu-\frac{\mu^2F_1^2\,\Gamma(x)}{2\bigl(1+2\mu F_2\Gamma(x)\bigr)},\qquad W(x)=\frac{\mu F_1\,\phi_a(x)}{1+2\mu F_2\Gamma(x)}.\tag{5.17}$$
Expanding in $s$ with (5.12):
$$C^{(1)}=\mu\bigl(F_0+F_2w(x)\bigr),\quad C^{\rm mark}=-\frac{\mu^2}2\bigl(F_1^2\Gamma(x)+F_2^2\Sigma(x,x)\bigr),\quad W^{(1)}=\mu F_1\phi_a(x),\quad W^{\rm mark}=-2\mu^2F_1F_2\,\Gamma(x)\,\phi_a(x).\tag{5.18}$$
All four are fixed by eigenvalues of blocks (M6). At leading order in the grip scale ($\hat\beta\hat bw,\ \hat\beta^2\Sigma/\Gamma\ll1$) they give, at every place, $W^{(1)}=2\hat b\,C^{(1)}\phi_a$ and $W^{\rm mark}=2\hat\beta\,C^{\rm mark}\phi_a$: each component's $g_s$-linear part is its own cost times the field at the body, with equal factors iff the grip is square. Beyond order $s^2$, (5.17) carries $u(x):=2M_k\lvert F_2\rvert\Gamma(x)$ and depends on the level.

### Step 7. Bond (item 4)

Two bodies at $x_1\ne x_2$, $G_{12}:=G_a(x_1,x_2)$, $\Sigma_{12}:=\Sigma(x_1,x_2)$. The $2\times2$ determinant lemma gives the trace part $\frac1{2\pi}\int\ln[(1+\alpha_1R_{11})(1+\alpha_2R_{22})-\alpha_1\alpha_2R_{12}^2]d\nu$, whose part depending on both bodies is $-\frac{\alpha_1\alpha_2}{2\pi}\int R_{12}^2d\nu=-\frac{\alpha_1\alpha_2}4\Sigma_{12}$ at order $s^2$. In $-\frac12b^{\rm T}A^{-1}b$, $A^{-1}=K_a^{-1}-\sum_i\alpha_iK_a^{-1}P_{x_i}K_a^{-1}+O(\alpha^2)$. At order $s^2$ and first order in $g_s$:
$$B=-\mu_1\mu_2\bigl[F_1^{(1)}F_1^{(2)}G_{12}+F_2^{(1)}F_2^{(2)}\Sigma_{12}\bigr],\qquad W_{\rm bond}=-2\mu_1\mu_2G_{12}\bigl[F_1^{(1)}F_2^{(2)}\phi_a(x_2)+F_2^{(1)}F_1^{(2)}\phi_a(x_1)\bigr],\tag{5.19}$$
i.e. $W_{\rm bond}=-\mu_1\mu_2F_1^{(1)}F_1^{(2)}G_{12}\Psi$ with $\Psi:=\hat\beta_1\phi_a(x_1)+\hat\beta_2\phi_a(x_2)$.

### Step 8. Weighing ratios, conditions and orders (item 4)

From (5.18), (5.19) and $\Delta(fg)=\bar f\Delta g+\bar g\Delta f$:
$$\rho_{\rm mark}=\frac{4F_2(F_0+F_2\bar w)\,\bar\Gamma}{F_1^2\bar\Gamma+F_2^2\bar\Sigma}\,(1+a_{\rm mark}),\qquad a_{\rm mark}:=\frac{\bar\phi_a\,\Delta\Gamma}{\bar\Gamma\,\Delta\phi_a},\tag{5.20}$$
$$\rho_{{\rm bond},i}=\frac{\bar G_{12}\,\Delta\Psi+\bar\Psi\,\Delta G_{12}}{\bar G_{12}+\frac14\hat\beta_1\hat\beta_2\bar\Sigma_{12}}\cdot\frac{1+\hat\beta_i\hat b_i\,\bar w(\bar x)}{2\hat b_i\,\Delta\phi_a(\bar x)},\tag{5.21}$$
($\bar w,\bar\Sigma,\bar\Gamma$ at the body places; for $F_0^{(i)}=0$ replace $\frac{2\hat b_i}{1+\hat\beta_i\hat b_i\bar w}$ by $\frac{F_1^{(i)}}{F_0^{(i)}+F_2^{(i)}\bar w}$). Both are exact at order $s^2$ for (G2) and independent of $\mu$: the same for every level and both branches.

*Coefficients and states.* For $\lvert s\rvert$ small, $A(s)=K_a+s\mathcal A>0$ for every grip, level and branch. So the $s$-coefficients of Steps 6–8 belong to an existing lowest eigenvector of the contract with $F\to sF$, and (5.20), (5.21) are defined for every member. They describe the lowest eigenvector of the stated contract ($s=1$) iff (5.10) holds at $s=1$ ((5.26) for a pair); then $A(s)=(1-s)K_a+sA(1)>0$ on all of $[0,1]$. $A$ increases with every $\alpha_i$, so a statement for every level up to $k_i$ and both branches needs (5.10) or (5.26) at $\alpha_i=-2M_{k_i}\lvert F_2^{(i)}\rvert$. Every statement below about the lowest eigenvector of the stated contract carries this condition.

*Numerical check* against the exact eigenvalue (weighing.txt; rings $L=32,64$ and tori $L=8,12$, three distances each, two levels, both branches):
- $C^{(1)},C^{\rm mark},W^{(1)},W^{\rm mark}$ agree with (5.18) to $3\times10^{-6}$ relative.
- $\rho_{\rm mark}$ agrees with (5.20) to $4\times10^{-6}$ relative for square, non-square, $8F_0F_2=F_1^2$ and linear grips. The linear grip gives $\rho_{\rm mark}=0$ within $10^{-12}$.
- $B$ and $W_{\rm bond}$ agree with (5.19) to $9\times10^{-5}$, and $\rho_{{\rm bond},i}$ with (5.21) to $1.4\times10^{-4}$ relative. This covers common square grips (five level and branch combinations), and common non-square, common $8F_0F_2=F_1^2$, common linear grips and a mixed (square, non-square) pair (two combinations each, giving each body both levels and both branches). The linear pair gives $\rho_{{\rm bond},i}=0$ within $10^{-8}$.
- The largest deviations come from the truncation of the $s$-stencil, largest on the ring at large $\Gamma$. Levels and branches agree within this accuracy.
- At $s=1$, $\lambda_{\min}$ of every pair block (156 cases) agrees with (5.26). On the tori every block of the check exists. On the rings some blocks do not exist at $s=1$, and there the ratios are formal coefficients only:
  - single blocks of the square grip with $\sigma=-1$: the upper level at $d=2$, both levels at $d=5,10$;
  - the single block of the grip $8F_0F_2=F_1^2$ with $\sigma=-1$, $k=1$, at $d=10$;
  - 22 of the 78 pair blocks, each containing a square-grip body with $\sigma=-1$.

*Excluded members:*
- $\Delta\phi_a=0$ (then $\Delta W^{(1)}=0$; on the line, (5.28) with $\Delta m=0$);
- $\Delta\phi_a(\bar x)=0$;
- $F_0+F_2\bar w=0$;
- $\bar G_{12}+\frac14\hat\beta_1\hat\beta_2\bar\Sigma_{12}=0$ (possible only for $\hat\beta_1\hat\beta_2<0$ outside $\varepsilon\ll1$).

$\bar C^{\rm mark}\ne0$ always ($\Gamma,\Sigma>0$).

*Controlled regime.* For $F_0^{(i)}\ne0$, let
$$\varepsilon^2:=\max_i\Bigl(\frac{\lvert F_2^{(i)}\rvert}{\lvert F_0^{(i)}\rvert}\bar w,\ \hat\beta_i^2\frac{\bar\Sigma}{\bar\Gamma},\ \lvert\hat\beta_1\hat\beta_2\rvert\frac{\bar\Sigma_{12}}{\bar G_{12}}\Bigr)\ll1 .$$
Every entry is nonnegative, $\lvert F_2\rvert/\lvert F_0\rvert=\lvert\hat\beta\hat b\rvert$, and each $\bar w$ is taken at the places where it enters. Then the factors $1+\hat\beta\hat b\bar w$, $1+\frac14\hat\beta^2\bar\Sigma/\bar\Gamma$ and $1+\frac14\hat\beta_1\hat\beta_2\bar\Sigma_{12}/\bar G_{12}$ of (5.20), (5.21) are $1+O(\varepsilon^2)$; the rest of (5.20), (5.21) is exact. With $\phi_j:=\phi_a(x_j)$ and $\bar\phi:=\phi_a(\bar x)$:
$$\rho_{\rm mark}=\frac{\hat\beta}{\hat b}(1+a_{\rm mark})\bigl(1+O(\varepsilon^2)\bigr),\qquad\rho_{{\rm bond},i}=\Bigl[\frac{\hat\beta_1\Delta\phi_1+\hat\beta_2\Delta\phi_2}{2\hat b_i\,\Delta\bar\phi}+\frac{(\hat\beta_1\bar\phi_1+\hat\beta_2\bar\phi_2)\,\Delta G_{12}}{2\hat b_i\,\bar G_{12}\,\Delta\bar\phi}\Bigr]\bigl(1+O(\varepsilon^2)\bigr).\tag{5.22}$$
*Grips with $F_0=0$* ($\hat b$ undefined). (5.20), (5.21) give exactly
$$\rho_{\rm mark}=\frac{\hat\beta^2\bar w\,(1+a_{\rm mark})}{1+\hat\beta^2\bar\Sigma/(4\bar\Gamma)},\qquad\rho_{{\rm bond},i}=\frac{\hat\beta_i\bar w(\bar x)}2\cdot\frac{\bar G_{12}\Delta\Psi+\bar\Psi\Delta G_{12}}{(\bar G_{12}+\frac14\hat\beta_1\hat\beta_2\bar\Sigma_{12})\,\Delta\bar\phi}.$$
In their weak-grip regime $\eta_0^2:=\max_j(\hat\beta_j^2\bar w,\hat\beta_j^2\bar\Sigma/\bar\Gamma,\lvert\hat\beta_1\hat\beta_2\rvert\bar\Sigma_{12}/\bar G_{12})\ll1$:
- $\rho_{\rm mark}=\hat\beta^2\bar w(1+a_{\rm mark})(1+O(\eta_0^2))$;
- $\rho_{{\rm bond},i}=\frac{\hat\beta_i\bar w(\bar x)}2\bigl[\frac{\hat\beta_1\Delta\phi_1+\hat\beta_2\Delta\phi_2}{\Delta\bar\phi}+\frac{(\hat\beta_1\bar\phi_1+\hat\beta_2\bar\phi_2)\Delta G_{12}}{\bar G_{12}\Delta\bar\phi}\bigr](1+O(\eta_0^2))$.

Both carry a factor $O(\eta_0^2)$, so at a fixed placement they vanish as $\eta_0\to0$. The first-order cost is then only $\mu F_2w$, and neither the mark nor a bond is weighed like it. Outside this regime there is no small parameter, and only (5.20), (5.21) apply.

For common square grips, $\rho_{\rm mark}=(1+a_{\rm mark})(1+O(\varepsilon^2))$ and $\rho_{{\rm bond},i}=(m_{\mathbf r}+a_{\rm bond})(1+O(\varepsilon^2))$, with $a_{\rm bond}:=\frac{(\bar\phi_1+\bar\phi_2)\Delta G_{12}}{2\bar G_{12}\Delta\bar\phi}$. For placements with $\Delta\Gamma=\Delta G_{12}=0$, (5.22) has the form of the pinned case and gives the conditions of (2.18) with $G_0\to\Gamma$, $w_0\to w$. In general the terms $a$ are the place dependence of the costs ($\Delta\Gamma$, $\Delta G_{12}$) times the uniform part of $\phi_a$ (5.11); the place-wise relations of Step 6 are not affected.

*Direct and image differences.* Let $\Delta g_h:=g(x-h)-g(x'-h)$ and $\Delta g_s:=g(x-x_s)-g(x'-x_s)$. By (5.8), (5.10), (5.11), exactly for every $L$ on both graphs, $\Delta\Gamma=-2\Delta g_h$ and $\Delta\phi_a=-g_s(\Delta g_s-\Delta g_h)$. For a pair translated by $\delta$, $\Delta G_{12}=-[\Delta g_h(x_1)+\Delta g_h(x_2)]$ and $\Delta\bar\phi=-g_s[\Delta g_s(\bar x)-\Delta g_h(\bar x)]$. Hence
$$a_{\rm mark}=\frac{2\bar\phi_a}{g_s\bar\Gamma}\,\frac{\Delta g_h}{\Delta g_s-\Delta g_h},\qquad a_{\rm bond}=\frac{\bar\phi_1+\bar\phi_2}{2g_s\bar G_{12}}\,\frac{\Delta g_h(x_1)+\Delta g_h(x_2)}{\Delta g_s(\bar x)-\Delta g_h(\bar x)}.\tag{5.27}$$
On the cubic torus both prefactors are bounded away from $0$ and $\infty$, uniformly in $L$ and in the places:
- $G_a\ge1/\kappa_a$ by the triangle inequality for $r$ in (5.8), so $-\phi_a/g_s\ge1/\kappa_a$ and $\Gamma,G_{12}\ge1/\kappa_a$;
- $G_a(x,y)\le\sqrt{\Gamma(x)\Gamma(y)}$, and $\Gamma$ is bounded uniformly in $L$ for $D=3$.

So after $L\to\infty$, with $N:=\Delta g_h(x_1)+\Delta g_h(x_2)$, $D:=\Delta g_s(\bar x)$ and $H:=\Delta g_h(\bar x)$:
- $a_{\rm mark}\to0$ iff $\Delta g_h/(\Delta g_s-\Delta g_h)\to0$, i.e. iff $\Delta g_h/\Delta g_s\to0$ with $\Delta g_s\ne0$ (for $t:=\Delta g_h/\Delta g_s$ the ratio is $t/(1-t)$);
- $a_{\rm bond}\to0$ iff $N/(D-H)\to0$. Here the numerator is not the image difference $H$ of the denominator. So this is equivalent to $N/D\to0$ when also $H/D\to0$ (then $N/(D-H)=(N/D)/(1-H/D)$), but not in general (pair control below).

Distance from the anchor alone does not suffice. Two sufficient families, both with $\Delta g_h/\Delta g_s\to0$, $H/D\to0$ and $N/D\to0$:
- *(A) Fixed relative placement.* Fix $x-x_s$ and $x'-x_s$ (for a pair, $x_j-x_s$ in both placements) with $\Delta g_s\ne0$ (resp. $\Delta g_s(\bar x)\ne0$), and move the configuration away from $h$ to distance $d$. Then $\Delta g_h=O(\lvert\delta\rvert d^{-2})$, so $a_{\rm mark},a_{\rm bond}=O(d^{-2})$.
- *(B) Non-cancellation.* Let $y:=\frac12(x+x')$ be the center of the comparison; for a pair of size $\lvert\mathbf r\rvert\ll R$, $y$ is the center of its two midpoints. Let $d:=\lvert y-h\rvert$, $R:=\lvert y-x_s\rvert$, $\hat e_h:=(y-h)/d$, $\hat e_s:=(y-x_s)/R$ and $\delta:=x-x'$. Require $\lvert\delta\cdot\hat e_s\rvert\ge c\lvert\delta\rvert$ with a fixed $c\in(0,1]$, and $1\ll R$, $\lvert\delta\rvert\ll R\ll d$.

In (B), the centered expansion of $g_\infty(x)=\frac1{4\pi\kappa_h\lvert x\rvert}+O(\lvert x\rvert^{-3})$ gives:
- $\Delta g_s=-\frac{\delta\cdot\hat e_s}{4\pi\kappa_hR^2}\bigl[1+O(\frac{\lvert\delta\rvert^2+1}{cR^2})\bigr]$;
- $\Delta g_h=-\frac{\delta\cdot\hat e_h}{4\pi\kappa_hd^2}+O(\lvert\delta\rvert(\lvert\delta\rvert^2+1)d^{-4})$, and $\Delta g_h(x_1)+\Delta g_h(x_2)=2\Delta g_h(y)+O(\lvert\delta\rvert(\lvert\mathbf r\rvert^2+\lvert\delta\rvert^2+1)d^{-4})$;
- $-\bar\phi_a/g_s=c_\phi+O(R^{-1})$, $\bar\Gamma=\Gamma_\infty+O(d^{-1})$, $\bar G_{12}=c_\phi+g_\infty(\mathbf r)+O(d^{-1})$.

With $\eta:=R^{-1}+\frac{\lvert\delta\rvert^2+1}{cR^2}+\frac{R^2}{cd^2}$, (5.27) gives
$$a_{\rm mark}=-\frac{2c_\phi}{\Gamma_\infty}\,\frac{R^2}{d^2}\,\frac{\delta\cdot\hat e_h}{\delta\cdot\hat e_s}+O\Bigl(\frac{R^2\eta}{cd^2}\Bigr),\qquad\frac{2c_\phi}{\Gamma_\infty}\in(1,2),\tag{5.23}$$
$$a_{\rm bond}=-\frac{2c_\phi}{c_\phi+g_\infty(\mathbf r)}\,\frac{R^2}{d^2}\,\frac{\delta\cdot\hat e_h}{\delta\cdot\hat e_s}+O\Bigl(\frac{R^2\eta}{cd^2}\Bigr),\tag{5.24}$$
so both are $O(R^2/(cd^2))$. $\kappa_a$ enters only through $c_\phi$ and $\Gamma_\infty$; $2c_\phi/\Gamma_\infty$ runs from 2 ($\kappa_a\to0$) to 1 ($\kappa_a\to\infty$).

*Numbers.* On $\mathbb Z^3$ with collinear placements ($c=1$; $R'=3,6$, $d=8,\dots,64$; asymptotics.txt), at $R'=3$ $a_{\rm mark}$ falls from $0.24$ to $4.6\times10^{-3}$ and $a_{\rm bond}$ from $0.25$ to $5.2\times10^{-3}$. Their ratios to the leading terms of (5.23), (5.24) reach $0.91$ ($R'=3$) and $0.97$ ($R'=6$) at $d=64$, within the $O(\eta)$ correction. On the tori of the check ($R'=1$, $d=1,2,3$), $a_{\rm mark}=0.92,0.47,0.23$ at $L=12$ (weighing.txt).

*Control: the mirror placement* $h=0$, $x=(d,0,0)$, $x'=(d+2,0,0)$, $x_s=(d+1,R,0)$. Here $\Delta g_s=0$ by reflection, while $\Delta\phi_a=g_s\Delta g_h\ne0$, so it is not an excluded member. (5.27) gives $a_{\rm mark}=-2\bar\phi_a/(g_s\bar\Gamma)\to2[c_\phi+g_\infty((1,R,0))]/\Gamma_\infty$ as $d\to\infty$. The placement has $\delta\cdot\hat e_s=0$, so it lies outside (B). On $\mathbb Z^3$ ($R=3$), $a_{\rm mark}=1.7634,1.7640,1.7645,1.7647$ for $d=8,16,32,64$, against the limit $1.7650$; for $R=6$ the limit is $1.7523$ (asymptotics.txt). At leading order in $\varepsilon$, common square grips there give $\rho_{\rm mark}\to2.77$, not 1. Hence, at leading order in $\varepsilon$, no grip gives $\rho_{\rm mark}=1$ at every comparison: $a_{\rm mark}$ tends to 0 in (A) and not at the mirror placement.

*Control: a pair far from a source next to the anchor.* $h=0$, $x_s=e_x$, $x_{1,2}=(d,\pm t,0)$ translated by $e_x$, with $d=n^3$, $t=n^4$, $n\to\infty$, so $\bar x=(d,0,0)$. With $A_*:=1/(4\pi\kappa_h)$ and $g_\infty$ of Step 3: $D\simeq A_*d^{-2}$, $D-H$ is the second difference of $g_\infty$ along $e_x$ at $\bar x$, $\simeq2A_*d^{-3}$, and $N\simeq2A_*d\,t^{-3}=2A_*d^{-3}$. So $N/D\to0$ and $H/D\to1$, but $N/(D-H)\to1$. As $t\to\infty$, $\bar\phi_j\to-g_s[c_\phi-g_\infty(e_x)]$ and $\bar G_{12}\to c_\phi$, so (5.27) gives $a_{\rm bond}\to-[c_\phi-g_\infty(e_x)]/c_\phi\ne0$, while $m_{\mathbf r}\simeq-d^3/(2t^3)\to0$. Here $\Delta\bar\phi=-g_s(D-H)\ne0$, and (5.26) holds for weak enough grips ($\Gamma$ and $G_{12}$ are bounded), so this is not an excluded member. The placement lies outside (A) (the $x_j-x_s$ are not fixed) and outside (B) ($\lvert\mathbf r\rvert\gg R\simeq d$). On $\mathbb Z^3$ for $n=2,\dots,6$ (asymptotics.txt): $N/(D-H)=0.703,0.862,0.918,0.946,0.962$ and $a_{\rm bond}=-0.669,-0.819,-0.872,-0.898,-0.913$, against the limit $-0.9488$. Over the same range $N/D$ falls from $0.16$ to $8.9\times10^{-3}$, and $\lvert m_{\mathbf r}\rvert$ from $0.017$ to $0.0020$ ($m_{\mathbf r}<0$). For common square grips, at leading order in $\varepsilon$, $\rho_{{\rm bond},i}-m_{\mathbf r}=a_{\rm bond}$ does not vanish there.

*Far from the anchor.* Take the order $L\to\infty$, then $d\to\infty$, within (A), or within (B) with $R/d\to0$ at fixed $c$. In the controlled regime ($\varepsilon\to0$ for $F_0^{(i)}\ne0$; grips with $F_0=0$ fail there), (5.22) reduces to the form of (2.18): $\rho_{\rm mark}=1$ and $\rho_{{\rm bond},i}=m_{\mathbf r}$, for every level and both branches, at order $\mu^2$ and leading order in the grip, iff the grips are square with a common $\hat\beta$. They then hold in the lowest eigenvector of the stated contract iff (5.10) holds for one body, and (5.26) for the pair, at $\alpha_i=-2M_{k_i}\lvert F_2^{(i)}\rvert$. For a pair this is stronger than the one-body conditions $2M_{k_i}\lvert F_2^{(i)}\rvert\Gamma(x_i)<1$. Far from the anchor, with equal $\alpha_i=-a$, it reads $a(\Gamma_\infty+c_\phi+g_\infty(\mathbf r))<1$. The first deviations are:
- at order $\hat\beta^0$: $a_{\rm mark},a_{\rm bond}$, of order $d^{-2}$ in (A) and $R^2/(cd^2)$ in (B);
- at the next order in the grip, for $a=0$: $\rho_{\rm mark}-1=\hat\beta^2(\bar w-\frac{\bar\Sigma}{4\bar\Gamma})+O(\hat\beta^4)$ and $\rho_{{\rm bond},i}-m_{\mathbf r}=m_{\mathbf r}\hat\beta^2(\bar w(\bar x)-\frac{\bar\Sigma_{12}}{4\bar G_{12}})+O(\hat\beta^4)$.

*Line (ring, $L\to\infty$ first).* Let $d_x$ be the distance of $x$ to $h$, and $m_x:=\min(d_x,d_s)$ for $x$ on the side of the source ($m_x:=0$ on the other side). Then (5.11), (5.13) give $\phi_a(x)=-g_s(\frac1{\kappa_a}+\frac{m_x}{\kappa_h})$ and $\Gamma(x)=\frac1{\kappa_a}+\frac{d_x}{\kappa_h}$. Also $G_{12}=\frac1{\kappa_a}+\frac{\min(d_1,d_2)}{\kappa_h}$ for two places on one side of $h$, and $\frac1{\kappa_a}$ across. Every comparison with $\Delta m\ne0$ ($\Delta m=0$ is the excluded $\Delta\phi_a=0$) has
$$a_{\rm mark}=\frac{(\kappa_h/\kappa_a+\bar m)\,\Delta d}{(\kappa_h/\kappa_a+\bar d)\,\Delta m}.\tag{5.28}$$
- *Both placements between the anchor and the source* ($m=d$ at both): $a_{\rm mark}=1$ exactly.
- *Straddling the source* ($d_{x'}<d_s<d_x$): $a_{\rm mark}-1=\frac{vP}{u(P+(u+v)/2)}>0$, with $P:=\kappa_h/\kappa_a+d_{x'}$, $u:=d_s-d_{x'}$, $v:=d_x-d_s$. Far from the anchor it tends to $\Delta d/\Delta m-1$; for example $x=x_s+1$, $x'=x_s-1$ gives $a_{\rm mark}\to2$.
- *Across the anchor* ($x'$ on the other side, $m_{x'}=0<m_x$): $a_{\rm mark}$ has the sign of $d_x-d_{x'}$.

The ring closed forms at $L=4096$ reproduce (5.28) in all three cases to $3\times10^{-3}$, the size of the finite-$L$ corrections (asymptotics.txt). For example, $x=21$, $x'=19$, $x_s=20$ gives $1.9531$ against $1.9533$. At leading order in $\varepsilon$, the grip with $\rho_{\rm mark}=1$, $\hat\beta/\hat b=1/(1+a_{\rm mark})$, therefore depends on the comparison, and no grip gives $\rho_{\rm mark}=1$ at every comparison on the line. A pair outside the range below needs (5.21) with the piecewise $\phi_a$ and $G_{12}$.

For comparisons in which both placements, and both bodies of a pair in both placements, lie between the anchor and the source on its side, $\phi_a=-g_s\Gamma$ and $G_{12}=\Gamma(x_{\rm near})$, so $a_{\rm mark}=1$ exactly and
$$\rho_{\rm mark}=\frac{2\hat\beta}{\hat b}\bigl(1+O(\varepsilon^2)\bigr),\qquad\rho_{{\rm bond},i}=\Bigl[\frac{\hat\beta_1+\hat\beta_2}{\hat b_i}+\frac{\hat\beta_{\rm far}\,s}{2\hat b_i(\kappa_h/\kappa_a+\bar d_{\rm near})}\Bigr]\bigl(1+O(\varepsilon^2)\bigr),\qquad m_{\mathbf r}=1,\tag{5.25}$$
with $s$ the separation, $d_{\rm near}$ the distance of the nearer body to $h$, and $\hat\beta_{\rm far}$ the grip of the farther body. In this range:
- square grips give $\rho_{\rm mark}=2$;
- $\rho_{\rm mark}=1$ at every such comparison iff $8F_0F_2=F_1^2$ ($\hat\beta=\hat b/2$);
- with a common such grip, $\rho_{{\rm bond},i}=1+\frac{s}{4(\kappa_h/\kappa_a+\bar d_{\rm near})}$, which tends to $m_{\mathbf r}$ only as $d_{\rm near}/s\to\infty$.

All three hold at leading order in $\varepsilon$, which here contains $\lvert F_2/F_0\rvert\bar w\propto\ln d$. On rings with $d=10$, $R'=3$: $a_{\rm mark}=0.53,0.81,0.96,0.990,0.9975$ for $L=32,64,256,1024,4096$ (asymptotics.txt); on the rings of the check the grip $8F_0F_2=F_1^2$ gives $\rho_{\rm mark}=1.008$ ($L=32$, $d=2$, $a_{\rm mark}=0.96$) and $0.937$ ($L=64$, $d=10$, $a_{\rm mark}=0.81$) (weighing.txt).

*Controls.*
- Linear grip ($F_2=0$): $W^{\rm mark}=W_{\rm bond}=0$ while $C^{\rm mark},B\ne0$, so $\rho_{\rm mark}=\rho_{{\rm bond},i}=0$ at every place, on both graphs.
- Grips with $F_0=0$: $\rho_{\rm mark},\rho_{{\rm bond},i}=O(\eta_0^2)$ in their weak-grip regime.
- Square grips, cubic torus: placements with $d\sim R$ ($a=O(1)$), the mirror placement at every $d$, and the pair far from a source next to the anchor ($a_{\rm bond}\not\to0$).
- Square grips, line: $\rho_{\rm mark}=2$ between the anchor and the source, $\rho_{\rm mark}>2$ straddling the source.

## Result

Class: oscillator cells on a ring or cubic torus of side $L\ge3$ without pinning, held Model D bodies with grips of class (G2) in a stated eigenvector of $S$, held sources; the block of the stated body labels; $M\to\infty$ first, in the sense of (5.3).

1. **(5.3)** *Criterion.* A lowest eigenvector exists iff $A>0$ in (5.1); otherwise the lowest eigenvalue of $C_M$ tends to $-\infty$ or to the bottom of a continuum.
2. **(5.4)–(5.6)** *Criterion, item 1 (no anchor).* A lowest eigenvector exists iff $\mathcal L_n+\mathcal A>0$; necessary: $\sum_i\mu_iF_2^{(i)}>0$. One body with $F_2\ne0$: iff $\sigma={\rm sign}\,F_2$, every level (otherwise the eigenvalue tends to $-\infty$). Two bodies: iff $\alpha_1+\alpha_2>0$ and $\alpha_1\alpha_2+(\alpha_1+\alpha_2)/r_{12}>0$. If no body has $F_2\ne0$ there is none ($-\infty$ if $\sum_i\mu_iF_1^{(i)}+\sum_sg_s\ne0$, a free uniform change otherwise); the field alone has none, so the cost of a body is undefined. The whole contract (all blocks) has none, for any bodies of (G2).
3. **(5.7)** *Negative.* A body with $\mu F_2>0$ (branch $\sigma={\rm sign}\,F_2$, every level) acts as an anchor of strength $2\mu F_2$ at its own place; alone, its $g_s$-linear cost is $-F_1g_s/(2F_2)$ at every place (no weight), and its cost has no expansion in $s$.
4. **(5.8), (5.9)** *Formula.* $G_a(x,y)=\frac1{\kappa_a}+\frac12[r(x,h)+r(y,h)-r(x,y)]$, exact for every $L$ on both graphs, with $G_a(h,\cdot)=1/\kappa_a$, ring $r=\frac{s(L-s)}{\kappa_hL}$; the resolvent (5.9).
5. **(5.10)** *Criterion, item 2.* With the anchor, a lowest eigenvector exists iff $\Gamma_n^{-1}+\mathcal A>0$; one body: iff $1+2\mu F_2\Gamma(x)>0$, $\Gamma(x)=\frac1{\kappa_a}+r(x,h)$, i.e. (2.5) with $G_0\to\Gamma(x)$. It holds whenever all $\mu_iF_2^{(i)}\ge0$. For $\mu F_2<0$ it fails on the ring once $\frac{d(L-d)}{\kappa_hL}\ge\frac1{2M_k\lvert F_2\rvert}-\frac1{\kappa_a}$, and on the cubic torus ($L\to\infty$) it holds at every place iff $2M_k\lvert F_2\rvert\Gamma_\infty\le1$, $\Gamma_\infty=\frac1{\kappa_a}+\frac{2\gamma_3}{\kappa_h}$.
   - **(5.26)** *Criterion.* Two bodies: a lowest eigenvector exists iff $(1+\alpha_1\Gamma(x_1))(1+\alpha_2\Gamma(x_2))>\alpha_1\alpha_2G_{12}^2$ and $\Gamma(x_1)+\Gamma(x_2)+(\alpha_1+\alpha_2)(\Gamma(x_1)\Gamma(x_2)-G_{12}^2)>0$. Sufficient: $\bar a(\max_i\Gamma(x_i)+G_{12})<1$. Two bodies that each meet the one-body condition need not meet it. With $\alpha_1=\alpha_2=-a$ and equal $\Gamma$, the pair needs $a(\Gamma+G_{12})<1$, each body alone only $a\Gamma<1$; on the cubic torus far from the anchor, $G_{12}\to c_\phi+g_\infty(\mathbf r)\ge c_\phi$ at every separation.
6. **(5.11)** *Formula.* Made field $\phi_a(x)=-g_sG_a(x,x_s)$: a uniform shift $-g_s[\frac1{\kappa_a}+\frac12r(x_s,h)]$ plus $-\frac{g_s}2[r(x,h)-r(x,x_s)]$. Cubic, $L\to\infty$: source plus an opposite image at the anchor, over the constant $-g_s[c_\phi-g_\infty(x_s-h)]$. Line: $-g_s[\frac1{\kappa_a}+\frac{\min(d_x,d_s)}{\kappa_h}]$ on one side of $h$, $-g_s/\kappa_a$ across.
7. **(5.12), (5.13)** *Formula.* Field constants $\Gamma$, $G_a$, $w$, $\Sigma$. Line ($L\to\infty$): $\Gamma=\frac1{\kappa_a}+\frac d{\kappa_h}$, $w=\frac{\ln d}{2\pi\sqrt{\kappa_h}}+O(1)$, $\Sigma(x,x)=\frac{2\ln2}{\pi\kappa_h^{3/2}}d(1+o(1))$. Cubic ($L\to\infty$, then $d\to\infty$): $\Gamma\to\Gamma_\infty$ and $G_a(x,y)\to c_\phi+g_\infty(x-y)$, which differ from the unanchored values; $w,\Sigma$ tend to the unanchored values with corrections $O(d^{-3})$.
8. **(5.14), (5.15)** *Formula and theorem, item 3.* $\mathcal E_{12}=-g_1g_2G_a(x_1,x_2)$, exact; constant if one source is at $h$. Cubic torus, $L\to\infty$ first: the image form (5.15), error $O(\ell^{-3})$. The separation dependence is $-\frac{g_1g_2}{4\pi\kappa_h\lvert\mathbf r\rvert}$, the coefficient of the unanchored field, when $\lvert\mathbf r\rvert/D_X\to0$ (order (a); uniformly for $1\ll\lvert\mathbf r\rvert\ll D_X$ with relative error $O(\lvert\mathbf r\rvert^3/D_X^3)+O(\lvert\mathbf r\rvert^{-2})$). With one source at a fixed place it falls as $\lvert\mathbf r\rvert^{-2}$ or faster (order (b)). At fixed midpoint it is $+\frac{3g_1g_2}{4\pi\kappa_h\lvert\mathbf r\rvert}$ (order (c)). $L\to\infty$ must come before both other limits.
9. **(5.16)** *Negative (control).* On the ring the mutual cost is a quadratic polynomial in the distances, for $L\to\infty$ independent of the separation at fixed distance of the nearer source: it falls as the inverse separation in no order of the limits.
10. **(5.17), (5.18)** *Formula, item 4.* Exact cost and $g_s$-linear cost of one body under (5.10). At order $s$ and $s^2$: $C^{(1)}=\mu(F_0+F_2w)$, $C^{\rm mark}=-\frac{\mu^2}2(F_1^2\Gamma+F_2^2\Sigma)$, $W^{(1)}=\mu F_1\phi_a$, $W^{\rm mark}=-2\mu^2F_1F_2\Gamma\phi_a$, at the body's place. At leading order in the grip, at every place: $W^{(1)}=2\hat bC^{(1)}\phi_a$ and $W^{\rm mark}=2\hat\beta C^{\rm mark}\phi_a$.
11. **(5.19)** *Formula.* Bond and its weight at order $s^2$ and first order in $g_s$: $B=-\mu_1\mu_2[F_1^{(1)}F_1^{(2)}G_{12}+F_2^{(1)}F_2^{(2)}\Sigma_{12}]$, $W_{\rm bond}=-\mu_1\mu_2F_1^{(1)}F_1^{(2)}G_{12}\Psi$.
12. **(5.20), (5.21)** *Formula.* $\rho_{\rm mark}$ and $\rho_{{\rm bond},i}$, exact at order $s^2$, independent of level and branch, with the excluded members of Step 8. As $s$-coefficients they are defined for every member. They refer to the lowest eigenvector of the stated contract iff (5.10) holds at $s=1$ ((5.26) for a pair), at $\alpha_i=-2M_{k_i}\lvert F_2^{(i)}\rvert$ when every level up to $k_i$ and both branches are claimed.
13. **(5.22)** *Criterion.* In the controlled regime, $F_0^{(i)}\ne0$ and $\varepsilon^2:=\max_i(\lvert F_2^{(i)}\rvert\bar w/\lvert F_0^{(i)}\rvert,\hat\beta_i^2\bar\Sigma/\bar\Gamma,\lvert\hat\beta_1\hat\beta_2\rvert\bar\Sigma_{12}/\bar G_{12})\ll1$: $\rho_{\rm mark}=\frac{\hat\beta}{\hat b}(1+a_{\rm mark})(1+O(\varepsilon^2))$, and $\rho_{{\rm bond},i}$ is the bracket of (5.22) times $1+O(\varepsilon^2)$. For placements with $\Delta\Gamma=\Delta G_{12}=0$ these are the conditions of (2.18) with $G_0\to\Gamma$, $w_0\to w$. Grips with $F_0=0$ give $\rho_{\rm mark},\rho_{{\rm bond},i}=O(\eta_0^2)$ in their weak-grip regime $\eta_0\ll1$: they do not weigh.
14. **(5.27), (5.23), (5.24)** *Formula and theorem.* Exactly, $a_{\rm mark}$ is the image difference $\Delta g_h$ over $\Delta g_s-\Delta g_h$ at the body, and $a_{\rm bond}$ is $N=\Delta g_h(x_1)+\Delta g_h(x_2)$ over $D-H=\Delta g_s(\bar x)-\Delta g_h(\bar x)$ at the midpoint, each times a prefactor bounded away from 0 and $\infty$ on the cubic torus. After $L\to\infty$, $a_{\rm mark}\to0$ iff $\Delta g_h/\Delta g_s\to0$ (with $\Delta g_s\ne0$), and $a_{\rm bond}\to0$ iff $N/(D-H)\to0$; the latter is equivalent to $N/D\to0$ when also $H/D\to0$, not in general. Distance from the anchor alone does not suffice. Two sufficient families, both with $H/D\to0$:
    - (A) fixed relative placement with $\Delta g_s\ne0$: $a=O(d^{-2})$;
    - (B) non-cancellation $\lvert\delta\cdot\hat e_s\rvert\ge c\lvert\delta\rvert$ at the center of the comparison, $1\ll R$, $\lvert\delta\rvert\ll R\ll d$: (5.23), (5.24), with $a=O(R^2/(cd^2))$.

    Take the order $L\to\infty$, then $d\to\infty$, within (A), or within (B) with $R/d\to0$ at fixed $c$, in the controlled regime. Then $\rho_{\rm mark}=1$ and $\rho_{{\rm bond},i}=m_{\mathbf r}$, for every level and both branches, at order $\mu^2$ and leading order in the grip, iff the grips are square with a common $\hat\beta$. They hold in the lowest eigenvector of the stated contract iff in addition (5.10) holds for one body, and (5.26) for the pair, at $\alpha_i=-2M_{k_i}\lvert F_2^{(i)}\rvert$; for a pair this is stronger than the one-body conditions. The first deviations are $a_{\rm mark},a_{\rm bond}$ and, at the next order in the grip, $\hat\beta^2(\bar w-\frac{\bar\Sigma}{4\bar\Gamma})$ and $m_{\mathbf r}\hat\beta^2(\bar w(\bar x)-\frac{\bar\Sigma_{12}}{4\bar G_{12}})$. *Controls:* at the mirror placement ($\Delta g_s=0$), $a_{\rm mark}\to2[c_\phi+g_\infty((1,R,0))]/\Gamma_\infty\ne0$ at every $d$. The pair with $x_s-h=e_x$, $x_{1,2}-h=(n^3,\pm n^4,0)$ has $N/D\to0$ but $a_{\rm bond}\to-[c_\phi-g_\infty(e_x)]/c_\phi\ne0$ as $n\to\infty$. At leading order in $\varepsilon$, no grip gives $\rho_{\rm mark}=1$ at every comparison.
15. **(5.28), (5.25)** *Formula, criterion and negative.* Line ($L\to\infty$ first): every comparison with $\Delta m\ne0$ has $a_{\rm mark}=\frac{(\kappa_h/\kappa_a+\bar m)\Delta d}{(\kappa_h/\kappa_a+\bar d)\Delta m}$. This equals 1 between the anchor and the source, is $>1$ straddling the source, and has the sign of $d_x-d_{x'}$ across the anchor. At leading order in $\varepsilon$, no grip gives $\rho_{\rm mark}=1$ at every comparison. For comparisons with both placements (both bodies of a pair in both placements) between the anchor and the source, at leading order in $\varepsilon$:
    - square grips give $\rho_{\rm mark}=2$;
    - $\rho_{\rm mark}=1$ iff $8F_0F_2=F_1^2$;
    - with that common grip, $\rho_{{\rm bond},i}=1+\frac{s}{4(\kappa_h/\kappa_a+\bar d_{\rm near})}$, which tends to $m_{\mathbf r}=1$ only for $d_{\rm near}/s\to\infty$.
16. *Control.* Linear grip: $\rho_{\rm mark}=\rho_{{\rm bond},i}=0$ at every place, every $L$, on both graphs. Grips with $F_0=0$: both ratios are $O(\eta_0^2)$ in their weak-grip regime.

## Consistency checks

1. *Limits of the anchor strength.* $\kappa_a\to\infty$: (5.8) becomes $g(x-y)-g(x-h)-g(y-h)+g(0)$, which vanishes for $x=h$: the cell at $h$ is held at $q_h=0$, and $\phi_a(h)=-g_s/\kappa_a\to0$. $\kappa_a\to0$ through positive values: $\Gamma=\frac1{\kappa_a}+r(x,h)\to\infty$. For $F_2\ne0$, (5.10) holds for all small $\kappa_a$ iff $\mu F_2>0$, the criterion (5.5) without an anchor. For a linear grip ($F_2=0$), (5.10) holds at every $\kappa_a>0$, while at $\kappa_a=0$ Step 2 gives no lowest eigenvector ($b\cdot\mathbf 1=\mu F_1\ne0$). The anchored lowest eigenvectors have no normalizable limit: their mean shift is $\langle q_y\rangle=-\mu F_1G_a(y,x)=-\mu F_1/\kappa_a+O(1)$, and the variance $\frac12(K_a^{-1/2})_{yy}$ diverges, because the lowest eigenvalue of $K_a$ is at most $\kappa_a/L^D$, with an eigenvector tending to the uniform one. Their eigenvalue $c_0+\frac12{\rm Tr}\sqrt{K_a}-\frac12\mu^2F_1^2\Gamma(x)\to-\infty$, as Step 2 states for $\kappa_a=0$. So the criterion at $\kappa_a=0$ is that of Step 2, not the limit of (5.10). Passed.
2. *Comparison with 02-grip-weight.* (5.10) is (2.5) with $G_0\to\Gamma(x)$; for $a_{\rm mark}=a_{\rm bond}=0$ (placements with $\Delta\Gamma=\Delta G_{12}=0$, or the far-anchor limit of the families (A), (B) of Step 8 on the cubic torus) (5.22) gives exactly the conditions of (2.18). Passed.
3. *Dimensions.* $[p]^2=$ cost $E$, $[q]^2=E^{-1}$; $[\kappa_h]=[\kappa_a]=E^2$, $[G_a]=[\Gamma]=E^{-2}$, $[g_s]=E^{3/2}$, $[\mu]=E$, $[F_1]=E^{1/2}$, $[F_2]=E$, $[\Sigma]=E^{-3}$: $g_1g_2G_a$, $\mu^2F_1^2\Gamma$, $\mu F_2w$ are costs; $\mu F_2\Gamma$, $a_{\rm mark}$, $\hat\beta^2w$, $\hat\beta^2\Sigma/\Gamma$ are pure numbers; $\ln d/\sqrt{\kappa_h}$ and $d/\kappa_h^{3/2}$ have the units of $w$ and $\Sigma$. Passed.

## Open issues

- The ratios use differences between placements with means of the place-dependent costs (the definitions of 02-grip-weight extended). The anchor terms $a_{\rm mark},a_{\rm bond}$ come from that extension, while the place-wise relation of Step 6 holds at every place. Which notion of weighing the programme adopts with an anchor is a choice for the main model.
- Finite-$L$ corrections to (5.15) are not derived in closed form (only $L\to\infty$ first; the tori of the check show them).
- $w_\infty$ and $\Sigma_\infty$ are not evaluated; the $O(1)$ and $o(1)$ terms of (5.13) on the line are not derived.
- Out of scope: formation of the mark with an anchor, moving bodies and inertia, cells with finitely many places, other anchors, typicality (item 6).

## Code

- `code/v3/check_anchor.py` (role: check). It computes the exact lowest eigenvalue (5.2) of the uncompressed quadratic contract and compares it with the closed forms.
  - Method: dense linear algebra ($\frac12{\rm Tr}\sqrt A$ from all eigenvalues) or sparse linear algebra. The Green function $g$ is computed independently by FFT, and $g_\infty$ by a Bessel integral. Small differences of $g_\infty$ (the pair control of (5.27)) come from one Bessel integral of the combined integrand.
  - Parameters: $\kappa_h=1$, $\kappa_a=0.7$, $M_k\in\{1,1.5\}$. The $s$-coefficients come from a five-point stencil (step 0.01). The $g_s$-linear part is exact (antisymmetric part in $g_s=\pm1$).

  Outputs:
  - `code/v3/output/existence.txt`: (i) $\lambda_{\min}(A)$ against (5.5), (5.6) and (5.10), and against (5.10), (5.26) for random pairs. For each lattice, a pair whose bodies meet the one-body condition while the pair violates (5.26). Rings $L=32,64$, tori $L=8,12$.
  - `code/v3/output/mutual_cost.txt`: (ii) $\mathcal E_{12}$ against (5.14), (5.15), (5.16) versus the separation, at midpoint distances 0 and 6 from the anchor. Rings $L=64,256$, tori $L=16,32$.
  - `code/v3/output/weighing.txt`: (iii) $C^{(1)},C^{\rm mark},W^{(1)},W^{\rm mark}$ against (5.18), and $\rho_{\rm mark}$ against (5.20), with the leading form (5.22), for square, non-square, $8F_0F_2=F_1^2$ and linear grips. Bonds $B$, $W_{\rm bond}$ against (5.19), and $\rho_{{\rm bond},i}$ against (5.21) and (5.22), for common square grips (five level and branch combinations), common non-square, $8F_0F_2=F_1^2$ and linear grips, and a mixed (square, non-square) pair (two combinations each). Existence at $s=1$ by (5.10), and for pairs $\lambda_{\min}$ against (5.26). Two levels, both branches, three distances; rings $L=32,64$, tori $L=8,12$; maximal deviations per lattice.
  - `code/v3/output/asymptotics.txt`: $\gamma_3$ and $g_\infty$; the orders of (5.15) on $\mathbb Z^3$; (5.23), (5.24) for collinear placements; the mirror placement of (5.27); the pair control of (5.27) ($N/D$, $H/D$, $N/(D-H)$, $m_{\mathbf r}$, $a_{\rm bond}$ for $n=2,\dots,6$); the line limit $a_{\rm mark}\to1$; (5.28) against the ring closed forms; (5.13) on the line.

## Methods used

- Gaussian lowest eigenvector of a quadratic contract; min-max and compressions
- Schur complement (Kron reduction), constrained quadratic minimum
- Pseudo-inverse of the graph Laplacian, resistance distance, M-matrices
- Sherman–Morrison formula and matrix determinant lemma; resolvent integral representations
- Lattice Green functions on tori and $\mathbb Z^3$ and their asymptotics; image form
- Perturbative expansion in the grip strength; numerical linear algebra (check)

Author: Ivan Honis · ivan.honis@ndot.io · License: CC BY 4.0 · Human–AI note