Version 3 · current

# Weighing of the own mark and of the bond against the grip, oscillator cells

- **Subproject:** 08-ilang-propagation
- **Package:** 02-grip-weight
- **Version:** v3
- **Mode:** external regeneration
- **Date:** 2026-10-24

## Changes from previous version

- Step 1, "Limit": for (G2) grips with (2.5) the form core now follows from the equivalence of the positive quadratic form norms. The false claim of a relative form bound below one is removed (I1).
- Step 3, after (2.5): the two-body condition is stated as the condition for the pair and both single-body references. The pair alone can have a lowest eigenvector without it (I4).
- Step 5, item (iv): replaced by a weighted Galerkin argument (Agmon bound in the number of quanta, uniform in $M$; strong convergence of the compressed reduced resolvents; form-norm convergence for third-order terms with a linear middle factor). Convergence is now claimed only for the coefficients through order $\theta^2$ (Gθ) and $\beta^3$ (GΦ), which are the ones the results use (I2).
- Step 5, "Where this eigenvalue is the lowest": the claim that the continued eigenvalue stops being the lowest is replaced by the absence of a neighbourhood that does not depend on $M$. The remainder orders are formal where $C$ has no lowest eigenvalue (I3).
- Step 2 and Step 6, midpoint factor for $d\gg\ell$: now given with the lattice decay vector $t^*(\hat d)$, and in the scaling regime with the lattice error $O(\lvert\mathbf r\rvert/\ell^3)$. The referral error is changed the same way (I6).
- Step 7(b), (2.16): the unfactored ratio, valid whenever $U\ne0$; the factored form only for $U_{\rm cl}\ne0$; "Reference and excluded members" updated (I5).
- Result: the entries (2.9)–(2.10), (2.11)–(2.13) and (2.15)–(2.16) are changed to match.
- Code (v3): `check_compression.py` adds the Taylor coefficients for an unbounded (Gθ) grip (I2). `check_g2_ratios.py` computes the exact bond ratio from the unfactored (2.16) and adds, on every graph, a pair of different grips with $U_{\rm cl}=0$ (I5). Both scripts were rerun.

## Response to verification

- I1: Accepted and fixed in Step 1, "Limit". For $K'>0$ the norms $q\cdot K'q$ and $q\cdot Kq$ are equivalent, the linear term is infinitesimally form-bounded, and the Hermite core is therefore a form core of $C$ for every stable (G2) grip. The relative-bound argument is kept only for bounded grips (bound 0).
- I2: Accepted and fixed in Step 5, item (iv). The missing weighted estimate is supplied: because $\Pi$ commutes with the number of quanta $N$, an Agmon estimate gives $\langle\Omega_M,e^{2\epsilon N}\Omega_M\rangle\le C$ uniformly in $M$. Together with strong convergence of the reduced resolvents and form-norm convergence of $R_MV\Omega_M$, the first- and second-order coefficients converge for every polynomially bounded grip, and so do third-order terms with a linear middle factor. The claim is limited to orders $\le\theta^2$ (Gθ) and $\le\beta^3$ (GΦ). Code (ii) adds an unbounded grip.
- I3: Accepted and fixed in Step 5, "Where this eigenvalue is the lowest", and in the Result entry (2.9)–(2.10).
- I4: Accepted and fixed in Step 3, after (2.5).
- I5: Accepted and fixed in Step 7(b), (2.16), and in "Reference and excluded members"; checked in Code (i) on a pair with $U_{\rm cl}=0$.
- I6: Accepted and fixed in Step 2 (decay vector), Step 6 (midpoint factor and referral) and the Result entry (2.11)–(2.13). The unrestricted form uses the lattice decay vector. The scaling-regime form carries the extra error $O(\lvert\mathbf r\rvert/\ell^3)$, which is small for $\lvert\mathbf r\rvert\ll\ell^3$ and so in the range $d\ll\ell^3$ of (2.4).

## Setup and assumptions

- **Field** (M8), as in the question: oscillator cells on $X_f$ (ring, square or cubic torus of side $L\ge3$, $N_f=L^D$ cells). With the graph Laplacian $\Delta$, $K:=\kappa_h\Delta+m_h^2\mathbb 1$ and $H=\tfrac12p\cdot p+\tfrac12q\cdot Kq$; normal modes $\omega_k^2=m_h^2+\kappa_h\sum_{j=1}^D4\sin^2(k_j/2)$, $k\in(2\pi/L)\mathbb Z_L^D$; pinning length $\ell:=\sqrt{\kappa_h}/m_h$. The $M$ places of a cell are the $M$ lowest number states of $\tfrac12(p_x^2+q_x^2)$, and at finite $M$ the whole total contract $C$ (field, bodies, source) is compressed, $C_M:=\Pi C\Pi$ (choice of this package).
- **Bodies** (M8, A15): held Model D bodies, $\tau=0$, no term of their own. The labels $(x_b,k,\sigma)$ are conserved ($S$ commutes with $C$), and in that sector $S$ acts as $\mu:=\sigma M_k$, $M_k=m+2\pi k/(N\lambda_0)>0$.
- **Grips.** (G2) as in the question. (G) as restricted by note (a) of the question:
  - (Gθ) $F=F(0)+\theta f$, $f(0)=0$, $f'(0)=1$, expanded in $\theta$ at fixed $f$;
  - (GΦ) $F=F(0)\Phi(\beta q)$, $F(0)\ne0$, $\Phi(0)=1$, $\Phi'(0)\ne0$, expanded in $\beta$ at fixed $\Phi$.

  Here $f$ and $\Phi$ are smooth, and each of their derivatives is bounded by a polynomial. Two bodies may carry different grips (note (b)). In (GΦ) they share $\beta$; this is no restriction, since $\Phi_i(\beta_iq)=\tilde\Phi_i(\beta q)$ with the fixed shape $\tilde\Phi_i(x)=\Phi_i(\beta_ix/\beta)$ at fixed ratios $\beta_i/\beta$.
- **Source** (M8, A15): $g_sq_{x_s}$, first order in $g_s$. **State** (M9, A11): as in the question; Steps 1 and 5 fix the sense of $M\to\infty$.
- **Definitions** of the question: cost, $C^{(1)}$, $C^{\rm mark}$, weight $W$, bond, $\rho_{\rm mark}$, and $\rho_{{\rm bond},i}$ with the reference of note (b).
- **Notation.** $\langle\cdot\rangle_0$ is the lowest eigenvector of $H$ (a Gaussian). $a_n:=\langle F^{(n)}(q_x)\rangle_0$ ($n\ge0$) is a Gaussian average of variance $w_0$; $a_n^{(i)}$ is the same for the grip of body $i$. $\Delta\phi:=\phi(x)-\phi(x')$.
- **Dimensions:** $[C]=E$, $[q]=E^{-1/2}$, $[K]=E^2$, $[F]=1$, $[\mu]=E$, $[g_s]=E^{3/2}$, $[\beta]=E^{1/2}$.
- **Standard mathematics** (allowed by the question): Gaussian lowest eigenvector of a positive quadratic contract; Sherman–Morrison, matrix determinant lemma; min-max; strong resolvent convergence of compressions (monotone convergence of forms); Rayleigh–Schrödinger perturbation theory for an isolated eigenvalue; Wick's theorem, the Mehler (Hermite) expansion, Gaussian integration by parts and the heat equation for Gaussian averages; lattice Fourier sums, their continuum limits and their saddle-point asymptotics.

## Derivation

### Step 1. Sector, limit $M\to\infty$, shift by the made field

In the sector of held bodies at $x_i$ with labels $(k_i,\sigma_i)$: $C=H+g_sq_{x_s}+\sum_i\mu_iF_i(q_{x_i})$.

*Limit.* $\operatorname{ran}\Pi_M$ grows with $M$. The union of these subspaces (finite sums of products of Hermite functions) is a form core of $H$. It is also a form core of $C$, with an equivalent form norm, in two cases:
- every bounded grip (relative form bound 0);
- every (G2) grip with (2.5), one body or two (Step 3). Here $K'>0$, so $cK\le K'\le c'K$ with $c,c'>0$ (positive matrices of finite size). The quadratic parts of $C$ and $H$ then have equivalent form norms, and the linear term $\mu F_1q_b$ is infinitesimally form-bounded. No relative bound below one is needed; such a bound fails for large $\mu F_2>0$.

In both cases min-max makes the lowest eigenvalue of $C_M$ nonincreasing in $M$, and it tends to the lowest eigenvalue of $C$ (Rayleigh–Ritz). For a (G2) grip that violates (2.5), the quadratic form of $C$ is continuous in the form norm of $H$ and unbounded below, and the union is dense in that norm. So the lowest eigenvalue of $C_M$ tends to $-\infty$. For (G), where $C$ need not be bounded below, the limit is taken coefficient by coefficient (Step 5). Statements about $C$ are statements about these limits (Code (ii)).

*Shift.* Let $\phi:=-g_sK^{-1}e_{x_s}$, $\phi(x)=-g_sG(x-x_s)$ (the made field), and $U:=\exp(-i\sum_x\phi(x)p_x)$. Then $U^\dagger q_xU=q_x+\phi(x)$, and with $K\phi=-g_se_{x_s}$:
$$
U^\dagger CU=H+\sum_i\mu_iF_i\bigl(q_{x_i}+\phi(x_i)\bigr)-\tfrac12g_s^2G(0).
\tag{2.1}
$$
So with the source, every eigenvalue is the one without the source for the shifted grips $F_{c_i}(q):=F_i(q+c_i)$, $c_i=\phi(x_i)$, minus a term that does not depend on the bodies. For one body, the eigenvalue $E[F]$ of $H+\mu F(q_{x_b})$ does not depend on $x_b$ (translations of the torus). The $s$-expansion commutes with the shift ($(sF)_c=sF_c$). Hence, at first order in $g_s$, for $X\in\{\text{cost},C^{(1)},C^{\rm mark}\}$:
$$
W_X(x)-W_X(x')=\partial_cX[F_c]\big|_{c=0}\,\Delta\phi,\qquad
\rho_{\rm mark}=\frac{\partial_c\ln C^{\rm mark}[F_c]}{\partial_c\ln C^{(1)}[F_c]}\Big|_{c=0}.
\tag{2.2}
$$
Every weight at first order in $g_s$ thus has the profile $\Delta\phi$, and $\rho_{\rm mark}$ does not depend on $x,x'$ or on the distance to the source.

*Fixed by the total contract.* $C^{(1)}=\frac{d}{ds}E_0\bigl(C_\emptyset+s(C_b-C_\emptyset)\bigr)|_{s=0}$, where $C_b,C_\emptyset$ are the total contracts with and without the body and $C_b-C_\emptyset=\mu F(q_{x_b})$ (the body has no term of its own). So the cost, $C^{(1)}$ and $C^{\rm mark}$ are fixed by the two total contracts, whatever pair term the single-body part $\mu F(0)$ is listed in (A5, M6). The lowest eigenvalue of the field alone drops out. By Hellmann–Feynman, $C^{(1)}=\mu\langle F\rangle_0=\mu a_0$ for every grip, so $\partial_c\ln C^{(1)}=a_1/a_0$ for every level and both branches.

### Step 2. Field constants and the made field

$$
\begin{aligned}
&G(x)=\tfrac1{N_f}\textstyle\sum_k e^{ik\cdot x}\omega_k^{-2},\quad G_0:=G(0),\quad w_0:=\langle q_x^2\rangle_0=\tfrac1{2N_f}\textstyle\sum_k\omega_k^{-1},\quad g_r(\nu):=[(K+\nu^2)^{-1}]_{x+r,x},\ g:=g_0,\\
&c_r(\tau):=\langle q_{x+r}e^{-(H-E_f)\tau}q_x\rangle_0=\tfrac1{2N_f}\textstyle\sum_k\cos(k\cdot r)\,e^{-\omega_k\tau}\omega_k^{-1},\quad J_n(r):=\int_0^\infty c_r(\tau)^n\,d\tau,\quad J_n:=J_n(0),\\
&J_1(r)=\tfrac12G(r),\quad \textstyle\int_0^\infty g_r(\nu)^2d\nu=2\pi J_2(r),\quad 0<c_r(\tau)\le c_0(\tau)\le w_0,\quad J_n(r)\le w_0^{\,n-1}J_1(r),\quad J_2\le\tfrac12w_0G_0.
\end{aligned}
\tag{2.3}
$$
Here $g_r(\nu)=2\int_0^\infty c_r(\tau)\cos\nu\tau\,d\tau$ and Parseval give the integral of $g_r^2$. $c_r>0$ because $c_r(\lvert\tau\rvert)$ is the kernel of $(-\partial_\tau^2+K)^{-1}$, whose semigroup preserves positivity ($K$ has nonpositive off-diagonal entries). $c_r\le c_0$ by Cauchy–Schwarz for $e^{-(H-E_f)\tau/2}q_{x+r}\Omega$ and $e^{-(H-E_f)\tau/2}q_x\Omega$, with translation invariance; $c_0$ is decreasing.

*Made field at large distances* ($L\to\infty$ first). In the scaling regime $\ell\gg1$ (weak pinning), the Fourier integral tends to its continuum form, with relative errors $O(\lvert d\rvert^{-2}+\ell^{-2})$ for $1\ll\lvert d\rvert\lesssim\ell$. For $\lvert d\rvert\gg\ell$ the decay rate carries an error $O(\ell^{-3})$ (see below), so the exponential forms hold for $\ell\ll\lvert d\rvert\ll\ell^3$:
$$
G(d)\simeq
\begin{cases}
\dfrac{e^{-\alpha\lvert d\rvert}}{2\kappa_h\sinh\alpha}\ \ (\cosh\alpha=1+\tfrac1{2\ell^2},\text{ exact on the chain at every }\ell)\ \simeq\dfrac{\ell}{2\kappa_h}e^{-\lvert d\rvert/\ell}, & D=1,\\[2mm]
\dfrac{K_0(\lvert d\rvert/\ell)}{2\pi\kappa_h}\to\dfrac{\ln(2\ell/\lvert d\rvert)-\gamma_E}{2\pi\kappa_h}\ (\lvert d\rvert\ll\ell),\ \ \dfrac{1}{2\pi\kappa_h}\sqrt{\dfrac{\pi\ell}{2\lvert d\rvert}}e^{-\lvert d\rvert/\ell}\ (\lvert d\rvert\gg\ell), & D=2,\\[2mm]
\dfrac{e^{-\lvert d\rvert/\ell}}{4\pi\kappa_h\lvert d\rvert}\to\dfrac1{4\pi\kappa_h\lvert d\rvert}\ (1\ll\lvert d\rvert\ll\ell), & D=3.
\end{cases}
\tag{2.4}
$$
At any $\ell$ and $\lvert d\rvert\gg\max(1,\ell)$, the saddle point of the lattice Fourier integral gives $G(d)\asymp\lvert d\rvert^{-(D-1)/2}e^{-\alpha(\hat d)\lvert d\rvert}$. The decay rate $\alpha(\hat n):=\max\{\hat n\cdot t:\sum_j4\sinh^2(t_j/2)\le\ell^{-2}\}$ depends on the direction for $D\ge2$. It is attained at the *decay vector* $t^*(\hat n)$, and $t^*(\hat d)=\nabla_d[\alpha(\hat d)\lvert d\rvert]$; the Hessian of $\alpha(\hat d)\lvert d\rvert$ is $O(\lvert t^*\rvert/\lvert d\rvert)$ (homogeneity of degree one). Along an axis $\cosh\alpha=1+1/(2\ell^2)$, as on the chain. For $\ell\gg1$, $\alpha(\hat n)=\ell^{-1}+O(\ell^{-3})$ and $t^*(\hat n)=\hat n/\ell+O(\ell^{-3})$, since $4\sinh^2(t/2)=t^2+O(t^4)$. Below the pinning length ($\ell\gg1$) the made field is long-range for $D=3$ ($1/\lvert d\rvert$), logarithmic for $D=2$ and linear for $D=1$; above it, it decays exponentially. On a finite torus $G$ is the sum over images; for $L\ll\ell$, $G(d)=1/(N_fm_h^2)+G_{\rm per}(d)+O(m_h^2)$, and the uniform part cancels in $\Delta\phi$.

### Step 3. (G2): existence and cost

With one body (no source), $C=H+\mu F_0+\mu F_1q_b+\mu F_2q_b^2$ is quadratic, with matrix $K':=K+a\,e_be_b^{T}$, $a:=2\mu F_2$. By the determinant lemma $\det K'=\det K\,(1+aG_0)$, and by interlacing at most one eigenvalue of $K'$ lies below the spectrum of $K$. So $K'>0$ iff $D_\mu:=1+aG_0>0$. For $D_\mu<0$, $C$ has a direction of negative curvature. For $D_\mu=0$, $K'$ has the null vector $K^{-1}e_b$, on which the linear term is $\mu F_1G_0\ne0$. In both cases $C$ is unbounded below. With Step 1:
$$
\text{a lowest eigenvector exists }(M\to\infty)\iff 1+2\sigma M_kF_2G_0>0;\ \text{otherwise the lowest eigenvalue of }C_M\to-\infty .
\tag{2.5}
$$
For both branches this reads $2M_k\lvert F_2\rvert G_0<1$.

For two bodies, the pair and both single bodies (all three enter the bond, Step 6) have lowest eigenvectors iff $D_1,D_2>0$ and $\Delta_2:=D_1D_2-a_1a_2G(\mathbf r)^2>0$. Given $D_1>0$, adding $a_2e_{x_2}e_{x_2}^T$ moves at most one eigenvalue below the spectrum, and $\det K'_{12}=\det K\,\Delta_2$ (determinant lemma). The pair alone needs only $K'_{12}>0$. That can hold with $D_1<0$ when $a_2>0$, since then $[(K+a_2e_{x_2}e_{x_2}^T)^{-1}]_{x_1x_1}<G_0$. Such pairs have no bond in the sense of the question, because $\text{cost}_1$ does not exist.

For $D_\mu>0$: $E=\mu F_0+\frac12\operatorname{Tr}\sqrt{K'}-\frac12\mu^2F_1^2(K'^{-1})_{bb}$, and $(K'^{-1})_{bb}=G_0/D_\mu$ by Sherman–Morrison. The zero-point part follows from three facts: $\sqrt A=\pi^{-1}\int_0^\infty dz\,z^{-1/2}A(A+z)^{-1}$; $\operatorname{Tr}[(K'+z)^{-1}-(K+z)^{-1}]=\partial_z\ln(1+a[(K+z)^{-1}]_{bb})$; and an integration by parts ($z=\nu^2$). Together they give $\frac12\operatorname{Tr}(\sqrt{K'}-\sqrt K)=Z(a):=\frac1{2\pi}\int_0^\infty\ln(1+a\,g(\nu))\,d\nu$; for one mode this is $\frac12(\sqrt{\omega^2+a}-\omega)$. $F\to sF$ maps $a\to sa$, and $Z'(0)=\frac1{2\pi}\int g\,d\nu=\frac12w_0$. Hence
$$
\text{cost}=\mu F_0-\frac{\mu^2F_1^2G_0}{2D_\mu}+Z(2\mu F_2),\qquad
C^{(1)}=\mu(F_0+F_2w_0)=\mu\langle F\rangle_0,\qquad
C^{\rm mark}=-\frac{\mu^2G_0}{2D_\mu}\bigl(F_1^2+\zeta\bigr),
\tag{2.6}
$$
with $\chi(y):=(y-\ln(1+y))/y^2=\int_0^1t\,dt/(1+ty)$ and
$$
\zeta:=\frac{4F_2^2}{\pi G_0}\int_0^\infty g(\nu)^2\,(1+2\mu F_2G_0)\,\chi\bigl(2\mu F_2g(\nu)\bigr)\,d\nu\ \ge0,\qquad
\zeta=\frac{4F_2^2J_2}{G_0}\bigl[1+O(\mu F_2G_0)\bigr].
\tag{2.7}
$$
$\zeta$ is the zero-point part of the mark beyond first order ($\zeta=-2D_\mu[Z(a)-\tfrac12aw_0]/(\mu^2G_0)$). Since $g(\nu)<G_0$ for $\nu>0$, $\partial_y[(1+yG_0)/(1+tyg)]=(G_0-tg)/(1+tyg)^2>0$. So, at fixed $F_2^2\neq0$, $\zeta$ is strictly increasing in $\sigma M_kF_2$. $C^{\rm mark}<0$ for every (G2) grip with (2.5).

### Step 4. (G2): weights

$F_c=F(\cdot+c)$ has $F_0\to F_0+F_1c+F_2c^2$, $F_1\to F_1+2F_2c$, $F_2\to F_2$. So $a$, $D_\mu$, $Z$, $\zeta$ do not depend on $c$: the zero-point part is not weighed. By (2.2) and (2.6):
$$
W(x)-W(x')=\frac{\mu F_1}{D_\mu}\Delta\phi,\qquad
\Delta W^{(1)}=\mu F_1\,\Delta\phi,\qquad
\Delta W^{\rm mark}=-\frac{2\mu^2F_1F_2G_0}{D_\mu}\Delta\phi .
\tag{2.8}
$$

### Step 5. (G): lowest orders

*Consequences of the class* (note (a)). For (Gθ) and (GΦ):
- (i) every $a_n$ is finite, and $\langle F^{(n)}(q_x+c)\rangle_0$ is smooth in $c$ with $\partial_ca_n=a_{n+1}$: the integrand is dominated by a polynomial times the Gaussian, so one may differentiate under the integral.
- (ii) Let $(q',q)$ be a Gaussian pair with variances $w_0$ and covariance $c$, $\lvert c\rvert\le w_0$, and let $g,h$ have polynomially bounded derivatives. Then $\operatorname{Cov}(g(q'),h(q))=\sum_{n\ge1}\langle g^{(n)}\rangle_0\langle h^{(n)}\rangle_0\,c^n/n!$, absolutely convergent. This is Mehler's formula; convergence follows from Parseval for the Hermite coefficients $\langle g^{(n)}\rangle_0w_0^{n/2}/\sqrt{n!}$ and Cauchy–Schwarz. Moreover $\partial_c\langle g(q')h(q)\rangle=\langle g'(q')h'(q)\rangle$ (Gaussian integration by parts).
- (iii) The matrix elements of $\Pi F(q_x)\Pi$ are finite.
- (iv) *Compressed coefficients.* Fix $g_s$. Let $H_g:=H+g_sq_{x_s}+\mu F(0)$, with lowest eigenvalue $E_g$ and eigenvector $\Omega$, and let $E_M$, $\Omega_M$ be those of $\Pi H_g\Pi$. Let $N:=\sum_xn_x$ be the total number of quanta of the cell oscillators $\frac12(p_x^2+q_x^2)$. $\Pi$ commutes with $N$.
  - *Ritz limits.* $E_M\downarrow E_g$ and $\Omega_M\to\Omega$ in $L^2$ (phase fixed), because the ranges exhaust a form core and $E_g$ is simple and isolated. By min-max the second eigenvalue of $\Pi H_g\Pi$ is $\ge E_g+m_h$, so its gap is $\ge m_h/2$ for large $M$.
  - *Weighted bound* (Agmon estimate in $N$). Write $H_g=\sum_{j=-2}^2H_j$, where $H_j$ raises $N$ by $j$. $H_{\pm2}$ are the terms $a_xa_y$, $a_x^\dagger a_y^\dagger$ with coefficient $\frac14(K-\mathbb 1)_{xy}$, and $H_{\pm1}$ is the source term. Then $e^{\epsilon N}H_ge^{-\epsilon N}=\sum_je^{j\epsilon}H_j$. Put $\varphi:=e^{\epsilon N}\Omega_M\in\operatorname{ran}\Pi$; it satisfies $e^{\epsilon N}\Pi H_g\Pi e^{-\epsilon N}\varphi=E_M\varphi$. The real part of the corresponding expectation value is
  $E_M\lVert\varphi\rVert^2=\langle\varphi,H_g\varphi\rangle+\sum_{j=1,2}(\cosh j\epsilon-1)\langle\varphi,(H_j+H_{-j})\varphi\rangle\ge(c_1-C\epsilon^2)\langle\varphi,N\varphi\rangle-C'\lVert\varphi\rVert^2$.
  Here $c_1>0$ comes from $H_g\ge c_1N-{\rm const}$ (since $H\ge\min(1,m_h^2)(N+N_f/2)$), $\lvert\langle\varphi,(H_j+H_{-j})\varphi\rangle\rvert\le C''\langle\varphi,(N+1)\varphi\rangle$, and $C'$ is uniform because $E_M$ is nonincreasing. For $C\epsilon^2\le c_1/2$, the sum of $e^{2\epsilon n}\lVert P_n\Omega_M\rVert^2$ over the levels $n\ge4C'/c_1$ of $N$ is bounded by its sum over the finitely many levels below. So $\langle\Omega_M,e^{2\epsilon N}\Omega_M\rangle\le C_\epsilon$ uniformly in $M$. Since $\langle q_b^{2j}\rangle\le C_j\langle(N+1)^j\rangle$, every polynomial moment of $q_b$ in $\Omega_M$ is bounded uniformly in $M$. With $\Omega_M\to\Omega$ and Cauchy–Schwarz, $P(q_b)\Omega_M\to P(q_b)\Omega$ in $L^2$ for every polynomially bounded $P$.
  - *Resolvents.* $e^{-(\Pi H_g\Pi-E_M)\tau}\Pi\to e^{-(H_g-E_g)\tau}$ strongly (monotone convergence of forms). The reduced resolvents $R_M=\int_0^\infty e^{-(\Pi H_g\Pi-E_M)\tau}Q_M\,d\tau$, where $Q_M$ projects onto $\operatorname{ran}\Pi\ominus\Omega_M$, are bounded by $2/m_h$. So they converge strongly (dominated convergence).
  - *Coefficients.* For polynomially bounded $V$, the first two Rayleigh–Schrödinger coefficients $\langle\Omega_M,V\Omega_M\rangle$ and $-\langle V\Omega_M,R_MV\Omega_M\rangle$ converge to those of $H_g+V$. Consider a third-order term $\langle\psi_M,V'\psi_M\rangle-E_{1,M}\lVert\psi_M\rVert^2$ with $\psi_M:=R_MV\Omega_M$ and a middle factor $V'$ linear in $q$. It converges too. Indeed $\psi_M\to\psi$ in $L^2$, and $\langle\psi_M,(H_g-E_M)\psi_M\rangle=\langle\psi_M,V\Omega_M\rangle\to\langle\psi,(H_g-E_g)\psi\rangle$. So $\psi_M\to\psi$ in the form norm, which bounds $q_b$.

  The coefficient of $\theta^n$ (Gθ) is the $n$-th order coefficient for $V=\mu f(q_b)$. At fixed $M$, $\Pi\Phi(\beta q_b)\Pi$ is smooth in $\beta$, with derivatives $\Pi\Phi^{(m)}(\beta q_b)q_b^m\Pi$. So the coefficient of $\beta^n$ (GΦ) is a sum of $k$-th order terms ($k\le n$) in the polynomials $P_m=\mu F(0)\Phi^{(m)}(0)q_b^m/m!$ with $\sum m=n$. For $n\le3$ the only third-order term has $m=(1,1,1)$. Hence, at fixed $g_s$, the Taylor coefficients of the eigenvalue of $C_M$ connected to the lowest one converge as $M\to\infty$ to the coefficients for $C$ through order $\theta^2$ (Gθ) and $\beta^3$ (GΦ). The same holds for two bodies, with $V=V_1+V_2$. These are all the orders that (2.9)–(2.14) and (2.17) use. Higher coefficients are not claimed to converge. The first order in $g_s$ is taken after this limit (Code (ii) checks a bounded and an unbounded grip).

*Second order.* At zero coupling the lowest eigenvalue of $H+\mu F(0)$ is nondegenerate, with gap $m_h$. Take $V=\mu(F-F(0))$ and $\delta F=F-\langle F\rangle_0$, and let $q(\tau)=e^{(H-E_f)\tau}qe^{-(H-E_f)\tau}$. Rayleigh–Schrödinger theory gives $E=E_f+\mu a_0-\mu^2\int_0^\infty\langle\delta F(q_b(\tau))\,\delta F(q_b(0))\rangle_0d\tau+\dots$, by $\sum_{n\ne0}\lvert V_{n0}\rvert^2/(E_n-E_0)=\int_0^\infty\langle\delta V e^{-(H-E_0)\tau}\delta V\rangle d\tau$. By Wick's theorem, $(q_{x+r}(\tau),q_x(0))$ is a Gaussian pair with covariance $c_r(\tau)$. So (ii) gives, with $S(\mathbf r):=\sum_{n\ge1}a_n^2J_n(\mathbf r)/n!$, $T(\mathbf r):=\sum_{n\ge1}a_na_{n+1}J_n(\mathbf r)/n!$, $S:=S(0)$, $T:=T(0)$:
$$
\text{cost}=\mu a_0-\mu^2S+R_3,\quad C^{(1)}=\mu a_0,\quad C^{\rm mark}=-\mu^2S+R_3,\quad S>0,\qquad R_3=O(\theta^3)\ \text{(Gθ)},\ O(\beta^4)\ \text{(GΦ)} .
\tag{2.9}
$$
$S>0$ because $J_n>0$ and $\delta F\ne0$ has a nonzero Hermite coefficient. Under the shift, $\partial_ca_n=a_{n+1}$ (i); by (2.2):
$$
\Delta W^{(1)}=\mu a_1\Delta\phi,\qquad \Delta W^{\rm mark}=-2\mu^2T\,\Delta\phi+R_3',\qquad R_3'=O(\theta^3)\ \text{(Gθ)},\ O(\beta^5)\ \text{(GΦ)}.
\tag{2.10}
$$

*Order counting for (GΦ).* Expand $V=\mu F(0)\sum_{m\ge1}\beta^m\Phi^{(m)}(0)q_b^m/m!$ inside each Rayleigh–Schrödinger term; every term is a finite Gaussian expectation by (i). The $k$-th order term has $k$ factors $V$, each at least of order $\beta$. Its part in which every factor is linear in $q_b$ vanishes for $k\ge3$, because for a linear grip the eigenvalue is exactly quadratic in the grip (Step 3 with $F_2=0$). So the $k$-th order term is $O(\beta^{k+1})$ for $k\ge3$. In its weight, $\partial_c$ replaces one factor $F$ by $F'=F(0)\beta\Phi'(\beta(q+c))$, whose constant part drops out for $k\ge2$. So the weight of the $k$-th order term is $O(\beta^{k+2})$ for $k\ge3$. This gives $R_3$ and $R_3'$ in (2.9)–(2.10): corrections of relative order $\beta^2$ to the second-order terms, which carry the sign of $\mu$ and so depend on the level and the branch. In (G2) they are the expansion of $1/D_\mu$ in $u:=2M_k\lvert F_2\rvert G_0$.

*Orders of the second-order terms in (GΦ).* Let $\varepsilon^2:=\beta^2w_0$. The heat equation $\partial_s\langle g\rangle_s=\frac12\langle g''\rangle_s$ for a centred Gaussian of variance $s$ gives $\langle\Phi^{(n)}(\beta q_x)\rangle_0=\Phi^{(n)}(0)+O(\varepsilon^2)$. By (ii), $T(\mathbf r)=\int_0^\infty h(c_{\mathbf r}(\tau))d\tau$ with $h(c):=\operatorname{Cov}(F'(q'),F(q))$. Here $h(0)=0$, $h'(0)=a_1a_2$, and $\lvert h''(c)\rvert=\lvert\langle F'''(q')F''(q)\rangle\rvert\le(\langle F'''^2\rangle_0\langle F''^2\rangle_0)^{1/2}$ by Cauchy–Schwarz. Hence $\lvert T(\mathbf r)-a_1a_2J_1(\mathbf r)\rvert\le\frac12(\langle F'''^2\rangle_0\langle F''^2\rangle_0)^{1/2}J_2(\mathbf r)$, and likewise $\lvert S(\mathbf r)-a_1^2J_1(\mathbf r)\rvert\le\frac12\langle F''^2\rangle_0J_2(\mathbf r)$. With $J_2(\mathbf r)\le w_0J_1(\mathbf r)$ (2.3) and $\langle F^{(n)}(q)^2\rangle_0=F(0)^2\beta^{2n}[\Phi^{(n)}(0)^2+O(\varepsilon^2)]$:
$$
a_0=F(0)[1+O(\varepsilon^2)],\quad a_n=F(0)\beta^n[\Phi^{(n)}(0)+O(\varepsilon^2)],\quad
S(\mathbf r)=F(0)^2\beta^2J_1(\mathbf r)[\Phi'(0)^2+O(\varepsilon^2)],\quad
T(\mathbf r)=F(0)^2\beta^3J_1(\mathbf r)[\Phi'(0)\Phi''(0)+O(\varepsilon^2)],
\tag{2.22}
$$
uniformly in $\mathbf r$. The same bounds hold for the two-grip sums $S_{12},T_{12},T_{21}$ of Step 6, with $\Phi_1,\Phi_2$. So in (GΦ) the mark first appears at order $\beta^2$ and, if $\Phi''(0)\ne0$, its weight at order $\beta^3$. In (Gθ), $a_n=\theta\langle f^{(n)}\rangle_0$ ($n\ge1$) gives $S=\theta^2S_f$ and $T=\theta^2T_f$, with $S_f:=\sum_n\langle f^{(n)}\rangle_0^2J_n/n!$ and $T_f:=\sum_n\langle f^{(n)}\rangle_0\langle f^{(n+1)}\rangle_0J_n/n!$. So the mark first appears at order $\theta^2$, and so does its weight if $T_f\ne0$.

*Mark weight vanishing at the lowest order* (note (c)):
$$
\text{(Gθ) with }T_f=0:\ \Delta W^{\rm mark}=O(\theta^3);\qquad \text{(GΦ) with }\Phi''(0)=0:\ \Delta W^{\rm mark}=O(\beta^5).
\tag{2.23}
$$
So $\theta^3$, resp. $\beta^5$, is the lowest order at which a nonzero mark weight can appear; its value is not computed. Examples: an odd $f$ (each product in $T_f$ contains an even derivative of $f$, which is odd and averages to zero), and $\Phi-1$ odd. For $\Phi''(0)=0$: $a_2=O(\beta^4)$ by the heat equation, so $a_1a_2J_1=O(\beta^5)$; the bound above is $O(\beta^5)$; and $R_3'=O(\beta^5)$. The mark itself stays at $\theta^2$, resp. $\beta^2$. Hence $\rho_{\rm mark}=O(1)$ for (Gθ) with $F(0)\ne0$, and $\rho_{\rm mark}=O(\beta^2)$ for (GΦ). For the linear grip the mark weight vanishes at every order (Step 4).

*Where this eigenvalue is the lowest.* Suppose $\mu F(0)(\Phi(x)-1)\ge-A-Bx^2$ for all $x$ (resp. $\mu\theta f(q)\ge-\lvert\theta\rvert(A+Bq^2)$), with constants $A,B\ge0$. Take (GΦ) and $\eta>0$. Splitting at $\lvert\beta q_b\rvert=\eta$ gives $V\ge-\lvert\mu F(0)\rvert c_\eta-\gamma q_b^2$, with $c_\eta:=\sup_{\lvert x\rvert\le\eta}\lvert\Phi(x)-1\rvert$ and $\gamma:=(A/\eta^2+B)\beta^2$. The quadratic contract $H-\gamma q_b^2$ has its two lowest eigenvalues at $E_f+O(\gamma)$ and $E_f+m_h+O(\gamma)$. By min-max, the second eigenvalue of $C-\mu F(0)$ is $\ge E_f+m_h-O(\gamma)-\lvert\mu F(0)\rvert c_\eta$, while the connected eigenvalue is $E_f+O(\beta^2)$. Choosing first $\eta$, then $\beta$ small makes the connected eigenvalue the only one below $E_f+\mu F(0)+m_h/2$, hence the lowest. For (Gθ) the same holds with $\gamma=\lvert\theta\rvert B$ and $\lvert\theta\rvert A$ in place of $\lvert\mu F(0)\rvert c_\eta$. If instead $\mu F(0)\Phi(x)$ (resp. $\mu\theta f(q)$) tends to $-\infty$ faster than $x^2$ (resp. $q^2$) along some direction, $C$ has no lowest eigenvalue, for every $\beta\ne0$ (resp. for that $\theta$). Hermite expansions converge in the Schwartz topology, and $\langle C\rangle$ is continuous in it. So the lowest eigenvalue of $C_M$ tends to $-\infty$ as $M\to\infty$ at every such fixed $\beta$ (resp. $\theta$). Hence no neighbourhood of $\beta=0$ (resp. $\theta=0$) that does not depend on $M$ keeps the lowest eigenvalue of $C_M$ near $E_f+\mu F(0)$. At each $M$, the eigenvalue of $C_M$ connected to the lowest one at $\beta=0$ is smooth in $\beta$. It may stay the lowest through avoided crossings; this is not determined here. (2.9)–(2.10) then state the limits of its Taylor coefficients ((iv)). Their remainders $O(\theta^3)$, $O(\beta^4)$, $O(\beta^5)$ are formal orders of the coefficient series: the remainder has no Taylor coefficient below that order. They are not remainder bounds for an eigenvalue of $C$, since $C$ has no lowest eigenvalue.

### Step 6. Bond

Two held bodies sit at $x_1$ and $x_2=x_1+\mathbf r$, with labels $\mu_i$, grips $F^{(i)}$ of the class (they may differ, note (b)) and no pair term between them. Define $U(\mathbf r):=\text{cost}_{12}-\text{cost}_1-\text{cost}_2$, fixed by four lowest eigenvalues; its $\mathbf r$-dependence is the bond. The cross term of second-order perturbation theory in $V_1+V_2$, with (ii) for the Gaussian pair $(q_{x_2}(\tau),q_{x_1}(0))$ of covariance $c_{\mathbf r}(\tau)$, gives
$$
U(\mathbf r)=-2\mu_1\mu_2S_{12}(\mathbf r)+O(\mu^3),\quad S_{12}(\mathbf r):=\sum_{n\ge1}\frac{a_n^{(1)}a_n^{(2)}J_n(\mathbf r)}{n!};\qquad \text{(G2): }U=-\mu_1\mu_2\bigl[F_1^{(1)}F_1^{(2)}G(\mathbf r)+4F_2^{(1)}F_2^{(2)}J_2(\mathbf r)\bigr]+O(\mu^3).
\tag{2.11}
$$
$O(\mu^3)$ is third order in the body–field coupling: $O(\theta^3)$ for (Gθ), relative $O(\beta^2)$ for (GΦ) (Step 5). For a common grip, $S_{12}=S(\mathbf r)>0$, so at this order $U<0$ for $\sigma_1=\sigma_2$ and $U>0$ for $\sigma_1=-\sigma_2$. For different grips $S_{12}(\mathbf r)$ can vanish. An example is (G2) with $F_1^{(1)}=F_1^{(2)}=b$, $F_2^{(1)}=-F_2^{(2)}=c$ and $c^2=b^2G(\mathbf r)/(4J_2(\mathbf r))$; it is stable once both grips are scaled down. The shift acts on each grip separately, $\partial_{c_i}a_n^{(i)}=a_{n+1}^{(i)}$. So
$$
\Delta W_{\rm bond}=-2\mu_1\mu_2\bigl[T_{12}(\mathbf r)\Delta\phi_1+T_{21}(\mathbf r)\Delta\phi_2\bigr]+O(\mu^3),\quad T_{12}:=\sum_{n\ge1}\frac{a_{n+1}^{(1)}a_n^{(2)}J_n(\mathbf r)}{n!},\ T_{21}:=\sum_{n\ge1}\frac{a_n^{(1)}a_{n+1}^{(2)}J_n(\mathbf r)}{n!},\qquad \Delta\phi_1+\Delta\phi_2=2m_{\mathbf r}\,\Delta\bar\phi,
\tag{2.12}
$$
where $\Delta\bar\phi:=\phi(\bar x)-\phi(\bar x')$ at the midpoints ($\bar x=x_1+\mathbf r/2$, $\mathbf r$ with even components) and $m_{\mathbf r}$ is the *midpoint factor*. For a common grip $T_{12}=T_{21}=T(\mathbf r)$ and $\Delta W_{\rm bond}=-4\mu_1\mu_2m_{\mathbf r}T(\mathbf r)\Delta\bar\phi$. Take a far reference ($\phi(\bar x')\to0$), $d:=\lvert\bar x-x_s\rvert\gg\lvert\mathbf r\rvert$ and $L\to\infty$ first. Then (2.4) and the expansion of $\phi(\bar x\pm\mathbf r/2)$ give:
- $D=3$, $1\ll d\ll\ell$: $m_{\mathbf r}-1=(3(\mathbf r\cdot\hat d)^2-\lvert\mathbf r\rvert^2)/(8d^2)+O(\lvert\mathbf r\rvert^2/(d\ell)+\lvert\mathbf r\rvert^4/d^4)$;
- $D=2$, $1\ll d\ll\ell$: $m_{\mathbf r}-1=(2(\mathbf r\cdot\hat d)^2-\lvert\mathbf r\rvert^2)/(8d^2[\ln(2\ell/d)-\gamma_E])+O(\lvert\mathbf r\rvert^2/\ell^2+\lvert\mathbf r\rvert^4/d^4)$;
- $D=1$, both bodies on one side of the source, any reference on that side: $m_r=\cosh(\alpha r/2)$ exactly;
- every $D$, any $\ell$, $d\gg\max(1,\ell)$: the lattice form of Step 2, with the decay vector $t^*=t^*(\hat d)$, gives $m_{\mathbf r}=\cosh(t^*\cdot\mathbf r/2)\bigl[1+O\bigl(\lvert\mathbf r\rvert/d+\lvert t^*\rvert\lvert\mathbf r\rvert^2/d\bigr)\bigr]$. Here the first-order terms of the prefactor $\lvert d\rvert^{-(D-1)/2}$ enter only with $\tanh(t^*\cdot\mathbf r/2)$, and the second-order term of the exponent is bounded by the Hessian of Step 2. For $D=1$, $t^*=\alpha$, and the form is exact (above).
- scaling regime $\ell\gg1$, $d\gg\ell$: with $t^*=\hat d/\ell+O(\ell^{-3})$, $m_{\mathbf r}=\cosh\bigl((\mathbf r\cdot\hat d)/(2\ell)\bigr)\bigl[1+O(\lvert\mathbf r\rvert/d+\lvert\mathbf r\rvert^2/(d\ell)+\lvert\mathbf r\rvert/\ell^3)\bigr]$. The last term is the lattice correction; it is small for $\lvert\mathbf r\rvert\ll\ell^3$, and so in the range $d\ll\ell^3$ of (2.4). For $\lvert\mathbf r\rvert\ll\ell$ this is $m_{\mathbf r}-1=(\mathbf r\cdot\hat d)^2/(8\ell^2)+O(\lvert\mathbf r\rvert^4/\ell^4+\lvert\mathbf r\rvert^2/(d\ell)+\lvert\mathbf r\rvert^2/\ell^4)$. For $\lvert\mathbf r\rvert\gtrsim\ell$, $m_{\mathbf r}$ grows exponentially in $\lvert t^*\cdot\mathbf r\rvert$.

If the pair is referred to $x_1$ instead of $\bar x$, the factor becomes $(\Delta\phi_1+\Delta\phi_2)/(2\Delta\phi_1)=1+(\mathbf r\cdot\nabla\phi)/(2\phi)+O(\lvert\mathbf r\rvert^2/\min(d,\ell)^2)$ for $\lvert\mathbf r\rvert\ll\min(d,\ell)$. That is a first-order error: $-(\mathbf r\cdot\hat d)/(2d)$ for $D=3$, $d\ll\ell$, and $-t^*(\hat d)\cdot\mathbf r/2$ for $d\gg\max(1,\ell)$ ($\simeq-(\mathbf r\cdot\hat d)/(2\ell)$ in the scaling regime).

*(G2) exactly.* The exact pair cost is the two-body form of Step 3, $U=U_{\rm cl}+Z_b$. Let $v_i:=\mu_iF_1^{(i)}$, $a_i:=2\mu_iF_2^{(i)}$, $D_i:=1+a_iG_0$ and $\eta_i:=F_2^{(i)}/F_1^{(i)}$. The shift changes only $v_i$, with $\partial_{c_i}v_i=2\eta_iv_i$, and leaves $K'$ unchanged. So
$$
U_{\rm cl}=-\tfrac12v^{T}\mathbb M v,\quad \mathbb M_{ij}:=(K'^{-1})_{x_ix_j}-\delta_{ij}\frac{G_0}{D_i},\quad
Z_b=\frac1{2\pi}\int_0^\infty\ln\Bigl[1-\frac{a_1a_2\,g_{\mathbf r}^2}{(1+a_1g)(1+a_2g)}\Bigr]d\nu,\quad
\Delta W_{\rm bond}=-2\sum_i\eta_i\,\Delta\phi_i\,v_i(\mathbb Mv)_i .
\tag{2.13}
$$
The zero-point part $Z_b$ of the bond is not weighed. With $\Delta_2>0$ (Step 3), $\mathbb M_{12}=G(\mathbf r)/\Delta_2$ and $\mathbb M_{ii}=-a_jG(\mathbf r)^2/(D_i\Delta_2)$ ($j\ne i$). For a common grip this gives $v^T\mathbb Mv=2\mu_1\mu_2F_1^2G(\mathbf r)[1-G(\mathbf r)F_2(\mu_1/D_1+\mu_2/D_2)]/\Delta_2$. The bracket is positive, because $\mu_iF_2/D_i<1/(2G_0)$ and $0<G(\mathbf r)\le G_0$. The integrand of $Z_b$ is $\ln(1-x)$ with $x$ of the sign of $a_1a_2=4\mu_1\mu_2F_2^2$. So for a common grip $U_{\rm cl}$ and $Z_b$ both have the sign of $-\mu_1\mu_2$, and $U\ne0$. For different grips $U$ can vanish (example above).

### Step 7. Weighing ratios and which grips weigh

*Reference and excluded members.* By note (b), $\rho_{{\rm bond},i}$ refers the bond weight per unit of bond, $\Delta W_{\rm bond}/U$, to $\Delta W^{(1)}/C^{(1)}=\Delta\bar\phi\,a_1^{(i)}/a_0^{(i)}$. This is the value for a single body with the grip of body $i$ at the midpoint (Step 1), the same for every level and branch. It differs from the pair's own first-order quantity $\sum_i\mu_ia_1^{(i)}\Delta\phi_i/\sum_i\mu_ia_0^{(i)}$, which is not used. Pairs with $\mu_1a_0^{(1)}+\mu_2a_0^{(2)}=0$ are therefore admitted. Excluded members (vanishing denominators):
- $a_0^{(i)}=\langle F^{(i)}\rangle_0=0$: $C^{(1)}=0$ for every level; in (G2), $F_0=-F_2w_0$;
- $a_1^{(i)}=\langle F^{(i)\prime}\rangle_0=0$: $C^{(1)}$ is not weighed; in (G2) this is excluded by $F_1\ne0$;
- $\Delta\bar\phi=0$;
- $U(\mathbf r)=0$: never for a common grip (Step 6; exactly in (G2), at order $\mu^2$ otherwise), but possible for different grips at some separations (at order $\mu^2$, where $S_{12}(\mathbf r)=0$).

$U_{\rm cl}=0$ with $U=Z_b\ne0$ is not excluded. It occurs for different (G2) grips with $a_1,a_2>0$: $\mathbb M$ is then indefinite, so $v^T\mathbb Mv=0$ for some $v\ne0$. There the unfactored form of (2.16) applies.

$C^{\rm mark}$ does not vanish: in (G2) $C^{\rm mark}<0$, and at order $\mu^2$, $C^{\rm mark}=-\mu^2S$ with $S>0$.

*(a) Order $\mu^2$, every pair of grips of the class.* From (2.2), (2.9)–(2.12):
$$
\rho_{{\rm mark},i}=\frac{2a_0^{(i)}T_i}{a_1^{(i)}S_i},\qquad
\rho_{{\rm bond},i}=\frac{a_0^{(i)}}{a_1^{(i)}}\,\frac{T_{12}(\mathbf r)\Delta\phi_1+T_{21}(\mathbf r)\Delta\phi_2}{S_{12}(\mathbf r)\,\Delta\bar\phi};\qquad
\text{common grip: }\rho_{\rm bond}=m_{\mathbf r}\,\rho(\mathbf r),\quad \rho(\mathbf r):=\frac{2a_0\,T(\mathbf r)}{a_1\,S(\mathbf r)} .
\tag{2.14}
$$
Here $T_i,S_i$ are $T,S$ for the grip of body $i$, and $\rho_{\rm mark}=\rho(0)$. $\mu$ cancels, so at this order both ratios are the same for every level and both branches (for the bond, for every pair of labels). $\rho_{\rm mark}$ does not depend on the distance to the source. $\rho_{{\rm bond},i}$ depends on it through $\Delta\phi_1/\Delta\bar\phi$ and $\Delta\phi_2/\Delta\bar\phi$; for a common grip only through $m_{\mathbf r}$ (Step 6), so that $\rho_{\rm bond}/m_{\mathbf r}=\rho(\mathbf r)$ does not.

*(b) (G2), exact in the grip.* (2.6)–(2.8) and (2.13) give
$$
\rho_{\rm mark}=\frac{4F_2\langle F\rangle_0}{F_1^2+\zeta(\sigma M_kF_2)},
\tag{2.15}
$$
$$
\begin{aligned}
&\rho_{{\rm bond},i}=\frac{\langle F^{(i)}\rangle_0}{F_1^{(i)}}\,\frac{\Delta W_{\rm bond}}{U\,\Delta\bar\phi}=-\frac{2\langle F^{(i)}\rangle_0}{F_1^{(i)}\,(U_{\rm cl}+Z_b)}\sum_j\eta_jv_j(\mathbb Mv)_j\frac{\Delta\phi_j}{\Delta\bar\phi}\quad(U\ne0);\\
&U_{\rm cl}\ne0:\ \rho_{{\rm bond},i}=\frac{4\langle F^{(i)}\rangle_0}{F_1^{(i)}\,(1+Z_b/U_{\rm cl})}\sum_j\eta_jw_j\frac{\Delta\phi_j}{\Delta\bar\phi},\quad w_j:=\frac{v_j(\mathbb Mv)_j}{v^T\mathbb Mv};\qquad
\text{common grip: }\rho_{\rm bond}=\frac{4F_2\langle F\rangle_0}{F_1^2+\zeta_b}\sum_jw_j\frac{\Delta\phi_j}{\Delta\bar\phi},\quad \zeta_b:=\frac{F_1^2Z_b}{U_{\rm cl}}=\frac{4F_2^2J_2(\mathbf r)}{G(\mathbf r)}\bigl[1+O(u)\bigr],
\end{aligned}
\tag{2.16}
$$
with $u:=2\max_iM_{k_i}\lvert F_2^{(i)}\rvert G_0$. The first line is (2.13) inserted into the reference of note (b), with $a_0^{(i)}/a_1^{(i)}=\langle F^{(i)}\rangle_0/F_1^{(i)}$. For $U_{\rm cl}\ne0$, $\Delta W_{\rm bond}=4U_{\rm cl}\sum_j\eta_jw_j\Delta\phi_j$ and $U=U_{\rm cl}(1+Z_b/U_{\rm cl})$ give the second line. For a common grip $U_{\rm cl}\ne0$ always (Step 6). Here $\sum_jw_j=1$. For a common grip and equal labels $w_j=\frac12$, and the sum is $m_{\mathbf r}$. In general $w_j=\frac12+O(uG(\mathbf r)/G_0)$ at fixed ratios of the grip coefficients. For a common grip and unequal labels the bond is therefore weighed at the place $\sum_jw_jx_j$, not at the midpoint. Referred to the midpoint, this adds $(w_1-\frac12)(\Delta\phi_1-\Delta\phi_2)/\Delta\bar\phi=O\bigl(u\,\tfrac{G(\mathbf r)}{G_0}\,\tfrac{\lvert\mathbf r\rvert}{\min(d,\ell)}\bigr)$ to the sum ($\lvert\mathbf r\rvert\ll\min(d,\ell)$), a label dependence beyond order $\mu^2$. Code (i) reproduces (2.15) and (2.16) from exact lowest eigenvalues, for common and for different grips, including a pair with $U_{\rm cl}=0$ (Code section).

*(c) Fixed shape (GΦ).* By (2.22), $a_0^{(i)}/a_1^{(i)}=[1+O(\varepsilon^2)]/(\beta\Phi_i'(0))$, and (2.14) gives, uniformly in $\mathbf r$ at order $\mu^2$,
$$
\rho_{{\rm mark},i}=\frac{2\Phi_i''(0)}{\Phi_i'(0)^2}+O(\varepsilon^2),\qquad
\rho_{{\rm bond},i}=\frac{\kappa_1\Delta\phi_1+\kappa_2\Delta\phi_2}{\Phi_i'(0)\,\Delta\bar\phi}+O(\varepsilon^2)\,\frac{\lvert\Delta\phi_1\rvert+\lvert\Delta\phi_2\rvert}{\lvert\Delta\bar\phi\rvert},\qquad \kappa_j:=\frac{\Phi_j''(0)}{\Phi_j'(0)} .
\tag{2.17}
$$
For a common grip, $\rho(\mathbf r)=2\Phi''(0)/\Phi'(0)^2+O(\varepsilon^2)=2F(0)F''(0)/F'(0)^2+O(\varepsilon^2)$ uniformly in $\mathbf r$. The higher orders of perturbation theory add relative corrections $O(\beta^2)$, which depend on the level and the branch (Step 5).

*(d) Which grips weigh.* By (2.17), at leading order in $\beta$:
$$
\rho_{{\rm mark},i}=1\ \text{and}\ \rho_{{\rm bond},i}=m_{\mathbf r}\ (i=1,2)\ \text{for every level and both branches}\iff 2\Phi_i''(0)=\Phi_i'(0)^2\ (i=1,2)\ \text{and}\ \Phi_1'(0)=\Phi_2'(0)\iff F^{(i)}(q)=F^{(i)}(0)(1+\hat\beta q)^2+O(q^3)\ \text{with a common}\ \hat\beta .
\tag{2.18}
$$
- *If:* $\kappa_j=\Phi'(0)/2$ for both bodies gives $\rho_{{\rm bond},i}=(\Delta\phi_1+\Delta\phi_2)/(2\Delta\bar\phi)=m_{\mathbf r}$.
- *Only if:* $\rho_{{\rm mark},i}=1$ is $2\Phi_i''(0)=\Phi_i'(0)^2$. Then $\rho_{{\rm bond},1}=m_{\mathbf r}$ reads $\Phi_2'(0)\Delta\phi_2=\Phi_1'(0)\Delta\phi_2$, so $\Phi_1'(0)=\Phi_2'(0)$ wherever $\Delta\phi_2\ne0$.
- *In terms of $F$:* this is the *square grip* $2F(0)F''(0)=F'(0)^2$, i.e. $(\sqrt{\lvert F\rvert})''(0)=0$, in (G2) $4F_0F_2=F_1^2$, with a common $\hat\beta:=F'(0)/(2F(0))=\beta\Phi'(0)/2$. The values $F^{(i)}(0)$ are free.
- *Range:* it holds at every distance from the source and every separation, on the three graphs, for all $L$, $\kappa_h$, $m_h>0$ in the regime $\varepsilon,u\ll1$ where the leading order holds. Since $m_{\mathbf r}=1+O(\lvert\mathbf r\rvert^2/\min(d,\ell)^2)$ for $\lvert\mathbf r\rvert\ll\min(d,\ell)$ (Step 6), $\rho_{\rm bond}=1$ there. At larger separations $\rho_{\rm bond}=m_{\mathbf r}\ne1$ is set by the shape of the made field, not by the grip.
- *First deviation otherwise:* $\rho_{{\rm mark},i}-1=(2\Phi_i''(0)-\Phi_i'(0)^2)/\Phi_i'(0)^2+O(\beta^2)$, the same at every distance and not vanishing as $\beta\to0$. For square grips with $\hat\beta_1\ne\hat\beta_2$, $\rho_{{\rm bond},i}=(\hat\beta_1\Delta\phi_1+\hat\beta_2\Delta\phi_2)/(2\hat\beta_i\Delta\bar\phi)+O(\beta^2)$, which is $(\hat\beta_1+\hat\beta_2)/(2\hat\beta_i)+O(\lvert\mathbf r\rvert/\min(d,\ell))$: its parameter is $\hat\beta_2/\hat\beta_1$, independent of the distance at this order.
- *Controls:* the linear grip ($\rho=0$); (G2) with $4F_0F_2\neq F_1^2$; square grips with $\hat\beta_1\ne\hat\beta_2$.

For (Gθ) see (2.20)(iii) and (2.23).

*(e) First deviation for a common square (G2) grip.* Insert $a_0=F_0(1+\beta^2w_0)$, $a_1=2F_0\beta$, $a_2=2F_0\beta^2$, $a_{n\ge3}=0$ (here $\beta=\hat\beta$) into (2.14), then use (2.3):
$$
\rho(\mathbf r)=\frac{1+\beta^2w_0}{1+\beta^2J_2(\mathbf r)/G(\mathbf r)},\qquad
\rho_{\rm mark}-1=\beta^2\Bigl(w_0-\frac{J_2}{G_0}\Bigr)+O(\varepsilon^4)\in\Bigl[\frac{\beta^2w_0}2,\beta^2w_0\Bigr)+O(\varepsilon^4),\qquad
\frac{\rho_{\rm bond}}{m_{\mathbf r}}-1=\beta^2\Bigl(w_0-\frac{J_2(\mathbf r)}{G(\mathbf r)}\Bigr)+O(\varepsilon^4).
\tag{2.19}
$$
The ratio $J_2(\mathbf r)/G(\mathbf r)$ decays exponentially for $\lvert\mathbf r\rvert\gg\ell$. In the scaling regime it is $\simeq1/(16\pi^2\sqrt{\kappa_h}\lvert\mathbf r\rvert^2)$ for $D=3$, $1\ll\lvert\mathbf r\rvert\ll\ell$, from the continuum $c_{\mathbf r}(\tau)\simeq[4\pi^2\sqrt{\kappa_h}(\lvert\mathbf r\rvert^2+\kappa_h\tau^2)]^{-1}$. So $\rho_{\rm bond}/m_{\mathbf r}-1\to\beta^2w_0$ at large separation. Both deviations in (2.19) are independent of the distance to the source; $\rho_{\rm bond}$ itself carries $m_{\mathbf r}$. Beyond order $\mu^2$, (2.15) depends on $(k,\sigma)$ only through $\zeta(\sigma M_kF_2)$, which brings a level and branch dependence of relative order $\varepsilon^2u$; the same holds for (2.16) with a common grip and equal labels. For unequal labels, (2.16) adds the referral term of (b). At $w_0=\zeta=0$ (tree level), (2.15) and (2.16) with a common grip and equal labels are exactly $1$ and $m_{\mathbf r}$ for the square grip, at every order in $u$. All deviations of the common square grip at equal labels therefore come from the zero-point spread of the field. Code (i) confirms (2.19) on rings and cubic tori (Code section).

*(f) What cannot be obtained.*
$$
\begin{aligned}
&\text{(i) no (G2) grip gives }\rho_{\rm mark}=1\text{ exactly for every level and both branches;}\\
&\text{(ii) at order }\mu^2\text{ no common (G2) grip gives }\rho_{\rm mark}=1\text{ and }\rho_{\rm bond}=m_{\mathbf r}\text{ at a separation with }J_2(\mathbf r)/G(\mathbf r)\ne J_2/G_0;\\
&\text{(iii) a (Gθ) grip with }F(0)\ne0,\ \langle f'\rangle_0\ne0,\ T_f\ne0\text{ has }\rho_{\rm mark}=\frac{2F(0)\,T_f}{\theta\,\langle f'\rangle_0\,S_f}+O(1),\ \lvert\rho_{\rm mark}\rvert\to\infty\ (\theta\to0).
\end{aligned}
\tag{2.20}
$$
- (i): for $F_2=0$, $\rho_{\rm mark}=0$. For $F_2\ne0$, $\zeta(\sigma M_kF_2)$ is strictly monotone in $\sigma M_k$ (Step 3), so the $2N$ values (2.15) are pairwise different ($\langle F\rangle_0\ne0$).
- (ii): the two conditions $F_1^2G_0+4F_2^2J_2=4F_2\langle F\rangle_0G_0$ and $F_1^2G(\mathbf r)+4F_2^2J_2(\mathbf r)=4F_2\langle F\rangle_0G(\mathbf r)$ imply $F_2^2\,(J_2/G_0-J_2(\mathbf r)/G(\mathbf r))=0$, and $F_2=0$ gives $\rho=0$. Since $J_2(\mathbf r)/G(\mathbf r)\to0$ for $\lvert\mathbf r\rvert/\ell\to\infty$ ($L\to\infty$ first) while $J_2/G_0>0$, a common grip weighs the mark and the bond like $C^{(1)}$ together only at leading order in $\varepsilon$.
- (iii): $T=\theta^2T_f$, $S=\theta^2S_f$, $a_0=F(0)+O(\theta)$ and $a_1=\theta\langle f'\rangle_0$ in (2.14), with $O(\theta)$ relative corrections to $T/S$. So a (Gθ) grip with $T_f\ne0$ never weighs its mark like $C^{(1)}$ at small $\theta$. For $T_f=0$, $\rho_{\rm mark}=O(1)$ by (2.23), and its value is decided at order $\theta^3$, which is not computed. (For $F(0)=0$, $a_0=\theta\langle f\rangle_0$, and (2.14) gives the finite value $2\langle f\rangle_0T_f/(\langle f'\rangle_0S_f)+O(\theta)$.)

*Dependence on $m_h$ and $L$.* The deviations are controlled by $\varepsilon^2=\beta^2w_0$ and $u$, and (2.5) requires $u<1$ for both branches.
- $L\to\infty$ at fixed $m_h>0$: $G_0$, $w_0$ and $J_n(\mathbf r)$ tend to finite lattice integrals in every $D$, and all results have limits.
- $m_h\to0$ at fixed $L$: $G_0\ge1/(N_fm_h^2)$ and $w_0\ge1/(2N_fm_h)$ diverge (the uniform mode), so $\varepsilon\to\infty$ and the expansions of Steps 5 and 7(c) are not uniform in $m_h$. For (G2):
$$
\begin{aligned}
&\text{linear grip: }\rho_{\rm mark}=\rho_{\rm bond}=0\ \text{at every }m_h,\quad \text{cost}=\mu F_0-\tfrac12\mu^2F_1^2G_0\to-\infty;\\
&F_2\ne0,\ \mu F_2<0:\ \text{no lowest eigenvector once }2M_k\lvert F_2\rvert G_0\ge1;\\
&\mu_iF_2^{(i)}>0:\ \text{cost finite},\ C^{(1)}=-C^{\rm mark}+O(1)=\mu F_2w_0+O(1)\to\infty,\ \rho_{\rm mark}=1+O(N_fm_h);\\
&\phantom{\mu_iF_2^{(i)}>0:\ }\rho_{\rm bond}\to0\ (\propto m_h)\ \text{for a common grip},\quad \lvert\rho_{{\rm bond},i}\rvert\to\infty\ (\propto1/m_h)\ \text{for different grips in general}.
\end{aligned}
\tag{2.24}
$$
Derivation. For $\mu_iF_2^{(i)}>0$, $K'$ has a finite positive limit, because $a_i>0$ pins the uniform mode. So the cost (2.6) and all two-body quantities (2.13) have finite limits, and $\Delta\phi$ converges (the uniform part cancels). For one body, $D_\mu\simeq aG_0$, so $\Delta W^{\rm mark}/\Delta W^{(1)}=-2\mu F_2G_0/D_\mu=-1+O(m_h^2)$. Moreover $C^{(1)}=\mu(F_0+F_2w_0)$ diverges while the cost stays finite, so $C^{(1)}/C^{\rm mark}=-1+O(1/w_0)$, and $\rho_{\rm mark}\to1$. The bond reference $a_1^{(i)}/a_0^{(i)}=F_1^{(i)}/\langle F^{(i)}\rangle_0$ tends to 0 like $1/w_0$. For a common grip, $v=(F_1/(2F_2))\,a$, and $K'_\infty\mathbf 1=\sum_ia_ie_{x_i}$ (with $K_\infty=\kappa_h\Delta$) gives $\mathbb M_\infty v=0$. So the bond weight (2.13) and $U_{\rm cl}$ vanish like $1/G_0\propto m_h^2$, while $U\to Z_b$; hence $\rho_{\rm bond}\propto m_h^2w_0\propto m_h$. For different grips ($\eta_1\ne\eta_2$), $v$ is not parallel to $a$, and $\Delta W_{\rm bond}/U$ has a nonzero limit in general. Code (i) follows the ratios on a ring down to $m_h=10^{-3}$. For (G) the limit is not determined: $\varepsilon\to\infty$.
- After $L\to\infty$, as $m_h\to0$: $G_0$ diverges for $D=1$ ($\propto\ell$) and $D=2$ ($\propto\ln\ell$) and tends to $0.2527/\kappa_h$ for $D=3$; $w_0$ diverges logarithmically for $D=1$ and stays finite for $D=2,3$.

### Step 8. Another eigenvector of the field

Take (G2) and the eigenvector $\lvert1\rangle$ with one quantum in the normal mode of $K'$ connected, as $s\to0$, to the uniform mode of $K$ (nondegenerate at fixed $L$). In its Fock eigenvectors the eigenvalues of the quadratic contract are $\mu F_0-\frac12J^TK'^{-1}J+\sum_j(n_j+\frac12)\Omega_j(K')$, with $\Omega_j^2$ the eigenvalues of $K'$. The source changes $J$, not $K'$, and the spectrum of $K'$ does not depend on the body's place. So the weights (2.8) and (2.13) are the same in every Fock eigenvector. In $\lvert1\rangle$, $C^{(1)}=\mu(F_0+F_2\langle q_b^2\rangle_1)$ with $\langle q_b^2\rangle_1=w_0+1/(N_fm_h)$ (Hellmann–Feynman). The mark gains the $c$-independent term $\Omega_0(a)-m_h-a/(2N_fm_h)=O(a^2)$, which is not weighed. Hence
$$
\rho_{\rm mark}\big|_{\lvert1\rangle}=\frac{4F_2\bigl(\langle F\rangle_0+F_2/(N_fm_h)\bigr)}{F_1^2+\zeta_1},\qquad \zeta_1-\zeta=O(F_2^2).
\tag{2.21}
$$
The answer (2.18) is unchanged. The deviation (2.19) changes, by $\beta^2/(N_fm_h)$ and by $(\zeta_1-\zeta)/F_1^2$. The first term vanishes as $L\to\infty$ and diverges as $m_h\to0$ at fixed $L$. Eigenvectors with $\beta^2\langle q_b^2\rangle_n\gtrsim1$ lose (2.18) even at leading order.

## Result

All results hold for held Model D bodies gripping a field of oscillator cells on a ring, square or cubic torus ($m_h>0$). The limit $M\to\infty$ is taken first (Steps 1 and 5), and all results are at first order in $g_s$. Class (G) is the one fixed by note (a) of the question: (Gθ) $F(0)+\theta f$ and (GΦ) $F(0)\Phi(\beta q)$ with polynomially bounded derivatives.

- **(2.1)–(2.2), theorem.** For every grip, the source enters the eigenvalues only by shifting the argument of each grip by the made field $\phi(x_i)=-g_sG(x_i-x_s)$. At first order in $g_s$, every weight is $\partial_cX[F_c]\,\Delta\phi$, and $\rho_{\rm mark}=\partial_c\ln C^{\rm mark}/\partial_c\ln C^{(1)}$, independent of the places and of the distance. The cost, $C^{(1)}=\mu\langle F\rangle_0$ and $C^{\rm mark}$ are fixed by the total contracts with and without the body.
- **(2.3)–(2.4), formula.** These are the field constants and the made field ($L\to\infty$ first). In the scaling regime $\ell=\sqrt{\kappa_h}/m_h\gg1$, for $1\ll\lvert d\rvert\ll\ell^3$, $\phi\propto G$ is linear, logarithmic or $1/\lvert d\rvert$ (for $D=1,2,3$) below $\ell$, and $\propto e^{-\lvert d\rvert/\ell}$ above it; the chain form $e^{-\alpha\lvert d\rvert}/(2\kappa_h\sinh\alpha)$ is exact at every $\ell$. At any $\ell$ and $\lvert d\rvert\gg\max(1,\ell)$, $G\asymp\lvert d\rvert^{-(D-1)/2}e^{-\alpha(\hat d)\lvert d\rvert}$, with the direction-dependent lattice rate $\alpha(\hat n)=\ell^{-1}+O(\ell^{-3})$.
- **(2.5), criterion.** For (G2), a lowest eigenvector exists iff $1+2\sigma M_kF_2G_0>0$. Otherwise the lowest eigenvalue of $C_M$ tends to $-\infty$.
- **(2.6)–(2.7), formula.** For (G2), exactly: cost $=\mu F_0-\mu^2F_1^2G_0/(2D_\mu)+Z(2\mu F_2)$, $C^{(1)}=\mu(F_0+F_2w_0)$, and $C^{\rm mark}=-\mu^2G_0(F_1^2+\zeta)/(2D_\mu)<0$. The zero-point term $\zeta\ge0$ is, for $F_2\ne0$, strictly monotone in $\sigma M_kF_2$.
- **(2.8), formula.** For (G2), exactly: $\Delta W=\mu F_1\Delta\phi/D_\mu$, $\Delta W^{(1)}=\mu F_1\Delta\phi$, $\Delta W^{\rm mark}=-2\mu^2F_1F_2G_0\Delta\phi/D_\mu$. The zero-point part is not weighed.
- **(2.9)–(2.10), formula.** For (G), $C^{(1)}=\mu a_0$ and $\Delta W^{(1)}=\mu a_1\Delta\phi$ hold exactly. Further, $C^{\rm mark}=-\mu^2S$ with $S=\sum_{n\ge1}a_n^2J_n/n!>0$, and $\Delta W^{\rm mark}=-2\mu^2T\Delta\phi$ with $T=\sum_{n\ge1}a_na_{n+1}J_n/n!$, up to $O(\theta^3)$ for (Gθ), and up to $O(\beta^4)$ resp. $O(\beta^5)$ (relative $O(\beta^2)$, level dependent) for (GΦ). The $a_n$ exist, $\partial_ca_n=a_{n+1}$, and the series converge absolutely. At fixed $g_s$, the Taylor coefficients of the compressed eigenvalue through order $\theta^2$ (Gθ), resp. $\beta^3$ (GΦ), converge to these values as $M\to\infty$; higher coefficients are not claimed. Where $C$ has no lowest eigenvalue, the remainder orders are formal orders of the coefficient series, not remainder bounds.
- **(2.22), formula.** For (GΦ), with $\varepsilon^2=\beta^2w_0$: $a_n=F(0)\beta^n[\Phi^{(n)}(0)+O(\varepsilon^2)]$, $S(\mathbf r)=F(0)^2\beta^2J_1(\mathbf r)[\Phi'(0)^2+O(\varepsilon^2)]$ and $T(\mathbf r)=F(0)^2\beta^3J_1(\mathbf r)[\Phi'(0)\Phi''(0)+O(\varepsilon^2)]$, uniformly in $\mathbf r$. The same holds for the two-grip sums.
- **(2.23), formula.** The mark weight vanishes at the lowest order for (Gθ) with $T_f=0$ (e.g. odd $f$) and for (GΦ) with $\Phi''(0)=0$ (e.g. $\Phi-1$ odd). A nonzero weight can first appear at order $\theta^3$, resp. $\beta^5$; its value is not computed. Then $\rho_{\rm mark}=O(1)$ (Gθ, $F(0)\ne0$), resp. $O(\beta^2)$.
- **(2.11)–(2.13), formula.** For two grips of the class at order $\mu_1\mu_2$:
  - The bond is $U=-2\mu_1\mu_2S_{12}(\mathbf r)$, with weight $-2\mu_1\mu_2[T_{12}(\mathbf r)\Delta\phi_1+T_{21}(\mathbf r)\Delta\phi_2]$.
  - For a common grip, $U$ has the sign of $-\mu_1\mu_2$; in (G2) this holds exactly. For different grips $U$ can vanish at some separations.
  - For (G2), exactly, $\Delta W_{\rm bond}=-2\sum_i\eta_i\Delta\phi_iv_i(\mathbb Mv)_i$, and the zero-point part $Z_b$ is not weighed.
  - The midpoint factor is $m_{\mathbf r}=(\Delta\phi_1+\Delta\phi_2)/(2\Delta\bar\phi)$, for a far reference, $d\gg\lvert\mathbf r\rvert$ and $L\to\infty$ first. It is $1+O(\lvert\mathbf r\rvert^2/d^2)$ for $1\ll d\ll\ell$ ($D=2,3$). For $d\gg\max(1,\ell)$ it is $\cosh(t^*\cdot\mathbf r/2)[1+O(\lvert\mathbf r\rvert/d+\lvert t^*\rvert\lvert\mathbf r\rvert^2/d)]$, with the lattice decay vector $t^*(\hat d)$; for $D=1$ it is exactly $\cosh(\alpha r/2)$. In the scaling regime, $t^*=\hat d/\ell+O(\ell^{-3})$ gives $\cosh((\mathbf r\cdot\hat d)/(2\ell))[1+O(\lvert\mathbf r\rvert/d+\lvert\mathbf r\rvert^2/(d\ell)+\lvert\mathbf r\rvert/\ell^3)]$, i.e. $1+(\mathbf r\cdot\hat d)^2/(8\ell^2)+O(\lvert\mathbf r\rvert^4/\ell^4+\lvert\mathbf r\rvert^2/(d\ell)+\lvert\mathbf r\rvert^2/\ell^4)$ for $\lvert\mathbf r\rvert\ll\ell$.
  - Referring the pair to a body instead of the midpoint gives an error of first order in $\lvert\mathbf r\rvert/\min(d,\ell)$.
- **(2.14), formula.** At order $\mu^2$, for every pair of grips of the class: $\rho_{{\rm mark},i}=2a_0^{(i)}T_i/(a_1^{(i)}S_i)$ and $\rho_{{\rm bond},i}=(a_0^{(i)}/a_1^{(i)})(T_{12}\Delta\phi_1+T_{21}\Delta\phi_2)/(S_{12}\Delta\bar\phi)$, with the single-body reference of note (b). For a common grip, $\rho_{\rm bond}=m_{\mathbf r}\rho(\mathbf r)$ with $\rho(\mathbf r)=2a_0T(\mathbf r)/(a_1S(\mathbf r))$. The ratios are the same for every level and both branches. $\rho_{\rm mark}$ and, for a common grip, $\rho_{\rm bond}/m_{\mathbf r}$ do not depend on the distance to the source; $\rho_{\rm bond}$ does, through $\Delta\phi_j/\Delta\bar\phi$. Excluded are $a_0^{(i)}=0$, $a_1^{(i)}=0$, $\Delta\bar\phi=0$ and separations with $U=0$ (different grips only).
- **(2.15)–(2.16), formula.** For (G2), exactly:
  - $\rho_{\rm mark}=4F_2\langle F\rangle_0/(F_1^2+\zeta)$.
  - Whenever $U\ne0$: $\rho_{{\rm bond},i}=-2\langle F^{(i)}\rangle_0\sum_j\eta_jv_j(\mathbb Mv)_j\Delta\phi_j/[F_1^{(i)}(U_{\rm cl}+Z_b)\Delta\bar\phi]$. For $U_{\rm cl}\ne0$ this equals $4\langle F^{(i)}\rangle_0\sum_j\eta_jw_j\Delta\phi_j/[F_1^{(i)}(1+Z_b/U_{\rm cl})\Delta\bar\phi]$. Pairs of different grips with $U_{\rm cl}=0\ne Z_b$ exist and are not excluded.
  - For a common grip $U_{\rm cl}\ne0$, and the bond is weighed at the place $\sum_jw_jx_j$. This is the midpoint for equal labels; otherwise it is shifted at relative order $uG(\mathbf r)/G_0$.
- **(2.17)–(2.18), criterion.** For (GΦ) at order $\mu^2$, uniformly in $\mathbf r$: $\rho_{{\rm mark},i}=2\Phi_i''(0)/\Phi_i'(0)^2+O(\varepsilon^2)$ and $\rho_{{\rm bond},i}=(\kappa_1\Delta\phi_1+\kappa_2\Delta\phi_2)/(\Phi_i'(0)\Delta\bar\phi)+O(\varepsilon^2)(\lvert\Delta\phi_1\rvert+\lvert\Delta\phi_2\rvert)/\lvert\Delta\bar\phi\rvert$, with $\kappa_j=\Phi_j''(0)/\Phi_j'(0)$; higher orders add relative $O(\beta^2)$.
  - Hence, at leading order in $\beta$, $\rho_{{\rm mark},i}=1$ and $\rho_{{\rm bond},i}=m_{\mathbf r}$ for both bodies, every level and both branches iff both grips are square, $2F(0)F''(0)=F'(0)^2$ (in (G2) $4F_0F_2=F_1^2$), with a common $\hat\beta=F'(0)/(2F(0))$.
  - This holds at every distance and separation, on all three graphs, and for all $L,\kappa_h,m_h>0$ with $\varepsilon,u\ll1$. $\rho_{\rm bond}=m_{\mathbf r}=1+O(\lvert\mathbf r\rvert^2/\min(d,\ell)^2)$ for $\lvert\mathbf r\rvert\ll\min(d,\ell)$.
  - Otherwise $\rho_{{\rm mark},i}-1=(2\Phi_i''(0)-\Phi_i'(0)^2)/\Phi_i'(0)^2$ does not vanish as $\beta\to0$. Square grips with $\hat\beta_1\ne\hat\beta_2$ have $\rho_{{\rm bond},i}\to(\hat\beta_1+\hat\beta_2)/(2\hat\beta_i)$ for $\lvert\mathbf r\rvert\ll\min(d,\ell)$.
- **(2.19), formula.** For a common square (G2) grip at order $\mu^2$:
  - $\rho_{\rm mark}-1=\beta^2(w_0-J_2/G_0)+O(\varepsilon^4)$, which lies in $[\beta^2w_0/2,\beta^2w_0)$ and is independent of the distance to the source.
  - $\rho_{\rm bond}/m_{\mathbf r}-1=\beta^2(w_0-J_2(\mathbf r)/G(\mathbf r))+O(\varepsilon^4)$, also independent of that distance, which tends to $\beta^2w_0$ at separations $\gg\ell$.
  - Beyond order $\mu^2$, level dependence enters at relative order $\varepsilon^2u$ for the mark and for the bond with equal labels. At tree level the square grip weighs the mark, and the bond with equal labels, exactly.
- **(2.20), negative.**
  - (i) No (G2) grip has $\rho_{\rm mark}=1$ exactly for every level and both branches.
  - (ii) At order $\mu^2$, no common (G2) grip has $\rho_{\rm mark}=1$ and $\rho_{\rm bond}=m_{\mathbf r}$ at any separation with $J_2(\mathbf r)/G(\mathbf r)\ne J_2/G_0$; this includes all separations $\lvert\mathbf r\rvert\gg\ell$ ($L\to\infty$ first).
  - (iii) A (Gθ) grip with $F(0)\ne0$, $\langle f'\rangle_0\ne0$ and $T_f\ne0$ has $\lvert\rho_{\rm mark}\rvert\to\infty$ as $\theta\to0$.
- **(2.21), formula.** In the (G2) eigenvector with one quantum in the uniform mode, the weights are unchanged, and (2.18) holds at leading order. The deviation changes by $\beta^2/(N_fm_h)$ and by an $O(F_2^2)$ zero-point term.
- **(2.24), formula.** As $m_h\to0$ at fixed $L$, for (G2):
  - The linear grip has $\rho_{\rm mark}=\rho_{\rm bond}=0$ at every $m_h$, while its cost diverges.
  - For $F_2\ne0$, the branch with $\mu F_2<0$ loses its lowest eigenvector.
  - In sectors with $\mu_iF_2^{(i)}>0$: the cost has a finite limit; $C^{(1)}$ and $C^{\rm mark}$ diverge as $\pm\mu F_2w_0$; $\rho_{\rm mark}=1+O(N_fm_h)\to1$; $\rho_{\rm bond}\to0$ for a common grip, while $\lvert\rho_{{\rm bond},i}\rvert\to\infty$ for different grips in general.
  - For (G) the expansion parameter $\varepsilon^2=\beta^2w_0$ diverges. As $L\to\infty$ at fixed $m_h>0$, all results have limits.

## Consistency checks

1. **Dimensions.** $[a_n]=E^{n/2}$, $[J_n]=E^{-n-1}$, $[G]=E^{-2}$, $[w_0]=E^{-1}$, $[\beta]=E^{1/2}$. So $\mu^2a_n^2J_n$ has dimension $E$; $\rho(\mathbf r)$, $\kappa_j/\Phi_i'(0)$, $\varepsilon^2=\beta^2w_0$, $\beta^2J_2/G_0$ and $u=2M_k\lvert F_2\rvert G_0$ are dimensionless. Passed.
2. **(G) against (G2) and the linear grip.** (2.9)–(2.12) with $a_1=F_1$, $a_2=2F_2$, $a_{n\ge3}=0$ give $C^{\rm mark}=-\mu^2(F_1^2G_0/2+2F_2^2J_2)$, $\Delta W^{\rm mark}=-2\mu^2F_1F_2G_0\Delta\phi$ and the two-grip bond weight $-2\mu_1\mu_2G(\mathbf r)[F_2^{(1)}F_1^{(2)}\Delta\phi_1+F_1^{(1)}F_2^{(2)}\Delta\phi_2]$. These are the order-$\mu^2$ terms of (2.6)–(2.8) and (2.13), with $\zeta\simeq4F_2^2J_2/G_0$. For $F_2=0$ all give $\rho=0$. Passed.
3. **Square grip at tree level.** Without the zero-point part, (2.1) and (2.6) give for $F=F_0(1+\beta q)^2$ the shifted cost $\mu F_0(1+\beta c)^2/(1+2\mu F_0\beta^2G_0)=\mu F(c)/D_\mu$. This is proportional to $F(c)$ for every $\mu$, so mark and $C^{(1)}$ are weighed alike, consistent with (2.15) at $w_0=\zeta=0$. Passed.

## Open issues

- (G) beyond the leading order in $\beta$: the level-dependent relative corrections $O(\beta^2)$ to (2.17)–(2.18) are bounded but computed only for (G2) (through $\zeta$ and $u$).
- The values of the first nonzero mark weights of (2.23) ($\theta^3$, $\beta^5$), and hence $\rho_{\rm mark}$ for (Gθ) with $T_f=0$, are not computed (not asked by note (c)).
- The limit $m_h\to0$ at fixed $L$ is determined only for (G2) grips.
- The fields of 06-ilang-source and 07-ilang-weight (03-gas-grip), $m_h=0$ with an anchor, and the formation speed are outside this package.
- Whether the square grip with a common $\hat\beta$, or the tuned grip $4F_2\langle F\rangle_0=F_1^2+\zeta$, is selected by anything in the language is not addressed (item 6, M10).

## Code

- `code/v3/check_g2_ratios.py` (check, item (i) of the question; cited in Step 7 and in (2.24)).
  - *Setup.* Rings $L=40,80$ and cubic tori $L=8,12$, with $\kappa_h=1$, $m_h=0.3$ ($\ell=3.3$). Levels $M_0=0.5$, $M_1=0.75$, both branches. Grips $(F_0,F_1,F_2)$: square $(1,0.6,0.09)$ ($\hat\beta=0.3$), generic $(1,0.6,0.15)$, linear $(1,0.6,0)$, and, in pairs only, square $(1,0.4,0.04)$ ($\hat\beta=0.2$). Bonds are computed for the three common grips and for the pairs square($0.3$)+square($0.2$) and square+generic.
  - *Method.* The exact lowest eigenvalue $\sum_is\mu_iF_0^{(i)}+\frac12\operatorname{Tr}\sqrt{K'}-\frac12J^TK'^{-1}J$ of the uncompressed contract is computed by dense linear algebra. $C^{(1)}$ and $W^{(1)}$ come from 4-point differences in $s$. Weights are the exact linear coefficient in $g_s$, $-J_b^TK'^{-1}e_{x_s}$; the eigenvalue is quadratic in $g_s$ and its trace part does not depend on $g_s$. Marks are evaluated at three distances from the source per graph. Bonds are evaluated at separations $2,4$, three midpoint distances and four label pairs: $(0,+)(0,+)$, $(1,-)(1,-)$, $(0,+)(1,+)$, $(0,+)(0,-)$. The reference of $\rho_{{\rm bond},i}$ is computed from a single body with the grip of body $i$ at the midpoint (note (b)). The exact formula compared is the unfactored (2.16). On every graph one further pair, labels $(0,+)(0,+)$, $\lvert\mathbf r\rvert=2$, has the square grip and the grip $(1,tF_1,0.09)$, with $t$ the smaller root of $v^T\mathbb Mv=0$, so that $U_{\rm cl}=0$ (I5).
  - *Results* (`output/g2_summary.json`, `output/g2_mark.csv`, `output/g2_bond.csv`):
    - $\lvert C^{(1)}-\mu\langle F\rangle_0\rvert\le3.4\cdot10^{-10}$.
    - $\rho_{\rm mark}$ agrees with (2.15) to $3.7\cdot10^{-9}$; its spread over distances is $\le1.7\cdot10^{-13}$.
    - $U$ agrees with $U_{\rm cl}+Z_b$ (2.13) to $8.7\cdot10^{-13}$. Both $\rho_{{\rm bond},i}$ agree with (2.16) to $1.2\cdot10^{-9}$, for common and for different grips.
    - Pair with $U_{\rm cl}=0$ ($\lvert U_{\rm cl}\rvert\le10^{-16}$; tuned $F_1=0.0213$ on the rings, $9.4\cdot10^{-4}$ and $7.0\cdot10^{-4}$ on the cubic tori): $U$ agrees with $Z_b$ to $6.7\cdot10^{-13}$, with $Z_b=-2.1\cdot10^{-4}$ (rings) and $-5.3\cdot10^{-7}$, $-3.7\cdot10^{-7}$ (cubic). Both $\rho_{{\rm bond},i}$ (from $97$ to $2.5\cdot10^6$, large because $U$ is small) agree with the unfactored (2.16) to a relative $1.8\cdot10^{-6}$. That is the accuracy of $U$ on the cubic tori; on the rings the agreement is $\le1.8\cdot10^{-10}$.
    - Linear grip: $\lvert\rho\rvert\le2.2\cdot10^{-13}$.
    - Square grip: $\rho_{\rm mark}-1\in[0.0357,0.0381]$ on the rings against $\beta^2(w_0-J_2/G_0)=0.0372$, and $\in[0.0154,0.0157]$ on the cubic tori against $0.0157$ ($L=8$) and $0.0155$ ($L=12$). The level spread is $\le0.0024$.
    - Common square grip, equal labels: $\rho_{\rm bond}/m_{\mathbf r}-1=0.0430,\,0.0452$ (rings, $\lvert\mathbf r\rvert=2,4$) against $0.0436,\,0.0454$ of (2.19); $0.02004,\,0.02014$ ($L=8$) and $0.01998,\,0.02012$ ($L=12$), the same as (2.19) to the digits shown. The values are the same at all midpoint distances, while $m_{\mathbf r}$ itself ranges from $0.23$ to $0.91$ on the cubic tori.
    - Generic grip: $\rho_{\rm mark}=1.742$–$1.760$ (rings) and $1.700$–$1.702$ (cubic) against $1.750$ and $1.701$ of (2.14), with leading value $4F_0F_2/F_1^2=1.667$.
    - Different square grips ($\hat\beta=0.3,0.2$), equal labels: $\rho_{{\rm bond},1}$ and $\rho_{{\rm bond},2}$ differ by the factor $1.46$–$1.48$, against $\hat\beta_1/\hat\beta_2=1.5$ at leading order. They lie within $5\,\%$ (rings, $\varepsilon^2=0.047$) and $2.1\,\%$ (cubic, $\varepsilon^2=0.020$) of the leading values $(\hat\beta_1\Delta\phi_1+\hat\beta_2\Delta\phi_2)/(2\hat\beta_i\Delta\bar\phi)$ of (2.18).
    - The order-$\mu^2$ formula (2.14) differs from the exact ratio by up to $0.11$ for a common grip, for unequal labels on the rings ($u\approx0.3$, $G(\mathbf r)/G_0\approx0.55$, $\lvert\mathbf r\rvert\sim\ell$). This is the referral term of Step 7(b). For different grips the difference is up to $0.042$.
  - *Limit $m_h\to0$* (`output/g2_mh_limit.csv`; ring $L=40$, $m_h=0.3,0.1,0.03,0.01,0.003,0.001$, sectors $\mu>0$):
    - $\rho_{\rm mark}$ agrees with (2.15) to $3.2\cdot10^{-7}$. For the generic grip it runs $1.744,1.771,1.782,1.693,1.449,1.220$ ($\mu=0.5$), towards 1, while $C^{(1)}$ grows from $0.54$ to $1.48$ and the cost settles at $0.27$.
    - The branch $\mu=-0.75$ has no lowest eigenvector from $m_h=0.1$ (generic) and $m_h=0.03$ (square) on.
    - Linear grip: $\rho=0$, cost $-1125$ and $-2531$ at $m_h=10^{-3}$.
    - Common generic grip, $r=2$: $\rho_{\rm bond}=1.86,1.79,1.61,0.73,0.11,0.021$, towards 0.
    - Square+generic: $\rho_{{\rm bond},1}=1.34,\dots,1.79,2.82$ and $\rho_{{\rm bond},2}=1.38,\dots,2.15,3.84$, growing. Its $U$ changes sign between $m_h=0.03$ and $0.01$ at equal labels, an instance of a vanishing two-grip bond profile.
  - *Outputs:* `output/g2_mark.csv`, `output/g2_bond.csv`, `output/g2_mh_limit.csv`, `output/g2_summary.json`.
- `code/v3/check_compression.py` (check, item (ii); cited in Steps 1 and 5).
  - *Setup.* Ring of three cells, $\kappa_h=m_h=1$ ($\omega_k=1,2,2$, $G_0=1/2$, $w_0=1/3$), $M=4,\dots,32$; sparse matrices, lowest eigenvalue by Lanczos.
  - *(G2)*, grip $(1,0.5,0.2)$ with $\mu=\pm0.75$. The lowest eigenvalue of $C_M$ decreases monotonically in $M$ to the exact value of (2.6). The error is $2.6\cdot10^{-2}$ at $M=4$, $1.0\cdot10^{-6}$ at $M=12$, $2\cdot10^{-13}$ at $M=24$ and $\le2.5\cdot10^{-14}$ at $M=32$; the costs converge likewise.
  - *(G)*, grip $F=1+\theta(\sin q+\frac12(1-\cos q))$ with $\mu=0.75$, class (Gθ), at $\theta=0.05,0.1,0.2,0.4$. All four lie in the range $2\lvert\theta\rvert\mu\sup\lvert f\rvert<m_h$ ($\sup\lvert f\rvert=1.618$) where the eigenvalue is the lowest.
    - *Reference.* The uncompressed lowest eigenvalue is computed independently on a sinc (Colbert–Miller) position grid in all three cells, with spacings $0.30$ ($35^3$ points) and $0.24$ ($47^3$ points). The two grids agree to $\le1.1\cdot10^{-14}$, and at $\theta=0$ they reproduce the exact value $3.25$ to $7\cdot10^{-15}$.
    - *Compressed eigenvalue.* The lowest eigenvalue of $C_M$ lies above the reference and decreases monotonically, down to round-off at $M=32$. The error is $2.3$–$2.5\cdot10^{-2}$ at $M=4$, $0.6$–$0.9\cdot10^{-6}$ at $M=12$, $0.5$–$4.6\cdot10^{-13}$ at $M=24$, and $\le1.0\cdot10^{-14}$ at $M=32$, for all four $\theta$.
    - *Taylor coefficients.* From Rayleigh–Schrödinger theory on the finite matrix, they converge to $\mu\langle f\rangle_0=0.0575694$ and $-\mu^2S/\theta^2=-0.1033310$ of (2.9). The errors are $4.6\cdot10^{-3}$ and $6\cdot10^{-4}$ at $M=4$, $\le8\cdot10^{-13}$ at $M=24$, and $\le3.1\cdot10^{-15}$ at $M=32$.
    - *Remainder.* The residual of the reference against the second-order expression, divided by $\theta^3$, is $0.01397,0.01401,0.01407,0.01414$ for the four $\theta$: the remainder of (2.9) is cubic in $\theta$.
  - *Unbounded (Gθ) grip* $f=q+q^2/2+q^3/6$, $\mu=0.75$ (Step 5 (iv)). Here $C$ is unbounded below for $\theta\ne0$. The first two Taylor coefficients of the compressed eigenvalue converge to $\mu\langle f\rangle_0=\mu w_0/2=0.125$ and $-\mu^2S=-\mu^2[(1+w_0/2)^2J_1+J_2/2+J_3/6]=-0.2032769$ of (2.9). Their errors are $1.0\cdot10^{-2}$ and $6.6\cdot10^{-3}$ at $M=4$, $6.8\cdot10^{-7}$ and $7.2\cdot10^{-7}$ at $M=12$, $1.1\cdot10^{-13}$ and $3.3\cdot10^{-13}$ at $M=24$, and $\le6.2\cdot10^{-15}$ at $M=32$.
  - *Output:* `output/compression.json`.

## Methods used

- Sector reduction, unitary shift by the made field
- Gaussian lowest eigenvector of a quadratic contract; Sherman–Morrison, matrix determinant lemma, interlacing
- Min-max for compressions; strong resolvent convergence of compressions; Agmon (weighted) estimate in the number of quanta
- Rayleigh–Schrödinger perturbation theory and order counting; Wick theorem, Mehler expansion, Gaussian integration by parts, heat equation for Gaussian averages; imaginary-$\lambda$ correlation functions
- Lattice Green functions on rings and tori, continuum and saddle-point asymptotics
- Numerical linear algebra (dense and sparse eigenvalues, finite differences); sinc position grid (Colbert–Miller)

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