Version 6 · current

# Inertia of a body with its own mark in a weakly pinned field of oscillator cells

- **Subproject:** 08-ilang-propagation
- **Package:** 06-mark-inertia
- **Version:** v6
- **Mode:** external regeneration
- **Date:** 2026-10-25

## Changes from previous version

- Step 8 (Odd part in $\sigma$, Heavy regime) and Result (6.10): the first relation of (6.10) is now stated for $L\gg\ell\ge1$. The other-branch term is given explicitly and bounded for every $L$, and its comparison with (iv) is shown to fail at finite $L$ when the uniform mode dominates. The heavy-regime part keeps every $L$ and $\ell$, now with this bound (verification I1).
- Step 6 (Rapidity identity) and Result (6.7): the relative error of the first relation of (6.7) is now $O(\ell^{-1}\ln\ell)$, from pairs with one soft and one lattice-order momentum (verification I2).
- Step 8 (Size of $R$): the heavy condition is stated as $\varepsilon_kR\gg1+r^2$. On the ring $R_2=\Theta(\ln^2\ell/c_0\ell)$, so the scale is $c_0\ell/\ln\ell$ for the linear grip and $c_0\ell/\ln^2\ell$ with a nonzero two-quantum part; the former is a sufficient condition for every grip (verification I3).
- Code: both scripts are copied unchanged from v5 and re-run into `code/v6/output/`; every number cited in Steps 3, 6, 7 and 10 is from this run.

## Response to verification

- I1: Accepted and fixed in Step 8 (Odd part in $\sigma$; Heavy regime) and Result (6.10). The other-branch term adds $\sigma\langle\!\langle(1+\cos Q_1)^2\rangle\!\rangle/(4\varepsilon_k\langle\!\langle1/E\rangle\!\rangle)$ to $\rho$; its smallness against (iv) needs $\langle\!\langle1\rangle\!\rangle=O(c_0^2\langle\!\langle\sin^2Q_1/E^2\rangle\!\rangle)$, which holds for $L\gg\ell\ge1$ but not when the uniform mode dominates at fixed $L$, as in the verifier's example ($\rho_{k,-}-\rho^{\rm e}_k=-m_h/\varepsilon_k$). The first relation is now restricted to $L\gg\ell\ge1$. The heavy part stays valid for every $L$: there the term is at most $2\Omega_{\max}/\varepsilon_k\ll1\ll\varepsilon_kR$.
- I2: Accepted and fixed in Step 6 (Rapidity identity) and Result (6.7): the error is now relative $O(\ell^{-1}\ln\ell)$. It is absorbed in $O(\delta_\ell)$, so (6.8) and (6.9) are unchanged.
- I3: Accepted and fixed in Step 8 (Size of $R$). The heavy condition is $\varepsilon_kR\gg1+r^2$. The scale $c_0\ell/\ln\ell$ is now given as the linear-grip scale and as a sufficient condition for every grip; with a nonzero two-quantum part the scale is $c_0\ell/\ln^2\ell$, up to grip-dependent constants.

## Setup and assumptions

- **Description** (assumptions, M8, M9, A11, A15), as in the question. Oscillator cells on the ring ($D=1$) or the cubic torus ($D=3$) of side $L$, $\Phi_y=q_y$, $c_0=\sqrt{\kappa_h}$, $\ell=c_0/m_h$. One Model D body with own contract $A$ only and the grip $\sum_x\lvert x\rangle\langle x\rvert\otimes S\otimes F(q_{x\mathbf e_1})$, $S=\sigma_3\otimes(m\mathbb 1_N+D)$, $F=F_0+F_1q+F_2q^2$, $F_0>0$, $F_1\neq0$; its chain runs over the cells $x\mathbf e_1$. State: the level $E_{k,\sigma}(P)$ connected to branch $\sigma$, clock level $k$, field in its lowest eigenvector. Speeds: $v_{\max}=\tau$ cells per unit $\lambda$ ((5.11) of 04-ilang-time/05-moving-clock with $k=1$), field speed $c_0$ ((4.17)), $r:=\tau/c_0$. The speeds are equal iff $r=1$, which is a condition on the description (M7).
- **Notation.** $\varepsilon_k:=F_0M_k$, with $M_k$ of (6.7) of 04-ilang-time/06-inertia-internal-energy, and $\bar F:=F_0+F_2w_0$. $\omega_k(p):=\sqrt{\varepsilon_k^2+\tau^2\sin^2p}$, $\theta_k(p):=\arctan(\tau\sin p/\varepsilon_k)$. Modes $q\in(2\pi/L)\mathbb Z_L^D$, $\Omega_q^2:=m_h^2+2\kappa_h\sum_{i=1}^D(1-\cos q_i)$, $c_q^2:=1/(2L^D\Omega_q)$, $\Omega_{\max}:=\sqrt{m_h^2+4D\kappa_h}$. Then the quantities of (2.6) are $G_0=\sum_q2c_q^2/\Omega_q$, $w_0=\sum_qc_q^2$, $\Gamma(t)=\sum_qc_q^2e^{-\Omega_qt}$ and $J_2=\sum_{q,q'}c_q^2c_{q'}^2/(\Omega_q+\Omega_{q'})$. The index $k$ is dropped where this is unambiguous. For a function $f(E,Q)$ we use the spectral sum of the grip and a ratio $R$ built from it:
$$
\langle\!\langle f\rangle\!\rangle:=F_1^2\sum_qc_q^2f(\Omega_q,q_1)+2F_2^2\sum_{q,q'}c_q^2c_{q'}^2f(\Omega_q+\Omega_{q'},q_1+q_1'),\qquad R:=\frac{\langle\!\langle\sin^2Q/E^2\rangle\!\rangle}{\langle\!\langle1/E\rangle\!\rangle}>0 .
\tag{6.1}
$$
- **Inputs** (as quoted in the question). From (2.6) of 08-ilang-propagation/02-grip-weight@v3, with $\mu=\sigma M_k$: $C^{(1)}=\mu\langle F\rangle_0$, $C^{\rm mark}=-\frac{\mu^2G_0}{2D_\mu}(F_1^2+\zeta)$, $\zeta=\frac{4F_2^2J_2}{G_0}[1+O(\mu F_2G_0)]$. From (2.8): $W(x)-W(x')=\mu F_1\Delta\phi/D_\mu$, with $D_\mu=1+2\mu F_2G_0$. Also (2.18) (the square grip $4F_0F_2=F_1^2$), (4.17), (5.11), and (6.7) of 04-ilang-time/06-inertia-internal-energy ($\omega_k''(0)=\tau^2/M_k$ for the bare body).
- **Order.** Expansion in $\delta F$ at fixed $F_0$, with $F_1$ and $F_2$ counted alike except in Step 9(b). It needs $M_k\lvert F_2\rvert G_0\ll1$ and $M_kF_1^2G_0\ll F_0$. On the ring $G_0\propto\ell$, so the statements at $\ell\gg1$ concern the coefficients; at fixed $\delta F$ they need $M_k(F_1^2+F_2^2/2c_0)\,\ell\ll c_0^2F_0$.
- **Limits.**
  - $M\to\infty$ first, at fixed $L$ and order by order in $\delta F$. The coefficients below are built from the lowest eigenvector of the field, its one- and two-quantum excitations and the matrix elements of $q_0$ and $q_0^2$, which are vectors and matrix elements with finitely many quanta. Those of $C_M$ tend to them as $M\to\infty$.
  - Then $L$ large. By Poisson summation, a lattice sum differs from its value at $L=\infty$ by $O(e^{-\kappa L})$ if its summand is analytic in the strip $\lvert\operatorname{Im}q_i\rvert<\kappa$.
    - *Held-field sums* ($G_0$, $w_0$, $J_2$, $\Gamma$): the only singularities are the zeros of $\Omega_q^2$, at distance $\simeq1/\ell$. They need $L\gg\ell$.
    - *Dressed sums* ((6.3) and its $P$-derivatives): $\omega$ and $\theta$ are also singular at $\lvert\operatorname{Im}q_1\rvert=\operatorname{arcsinh}(\varepsilon_k/\tau)$, and the denominators vanish at complex $q$. At $P=0$ their smallest value on the real Brillouin zone is the isolation gap $\Delta_k$ of Step 3. Near a quadratic minimum, $d(q)=\Delta_k+a_k(q_1-q_*)^2+\dots$, the zeros lie at distance $\simeq\sqrt{\Delta_k/a_k}$ from the real axis. So $\kappa_k\simeq\min\bigl(1/\ell,\ \operatorname{arcsinh}(\varepsilon_k/\tau),\ \sqrt{\Delta_k/a_k}\bigr)$, and the finite-$L$ statements for a state need $L\gg1/\kappa_k$.
    - This control is not uniform as $\Delta_k\to0$. Statements at large $L$ are therefore made for gaps bounded below. In Step 10 the lower-branch findings ((a)–(e) and the grid findings of (6.14), (6.15)) are for $\Delta_k>0.2\,m_h$. The light levels (finding (f), Part G, and the light-level member of (6.14)) reach upper-branch gaps down to $\Delta_k=0.022\,m_h$; at that gap the torus values at $L=128$ and $256$ differ by $0.067$ (finding (f)). $L$-convergence is checked at the smallest gaps of both kinds (Part G for the lower branch, $L=64$ against $128$; finding (f) for the light levels).
  - Then $\ell\gg1$, at fixed $\tau$, $c_0$ and $\varepsilon_k$.
- **Joint label.** At finite $L$ the allowed labels are $P\in(2\pi/L)\mathbb Z_L$. The expressions below are analytic in real $P$ and give the levels at these values. $E''(0)$ denotes the second derivative of this function.
- **Weights.** Those of (2.8), without corrections in $\tau$ (approximation allowed by the question).

## Derivation

### Step 1. Conserved labels and zeroth order

- **Clock level.** The clock operator $D$ (not the dimension) is diagonal in $\lvert f_k\rangle$, and every term of $C$ acts on the clock through $\mathbb 1_N$ or $D$. The clock level $k$ is therefore conserved, and $S=M_k\sigma_3$ in it.
- **Joint label and parity.** The joint translation (body $x\to x+1$, cells $y\to y+\mathbf e_1$) commutes with $A$, with the grip and with $H$, so $P$ is conserved. Reflect $x\to-x$ for the body and the cells together, and conjugate the branch by $\sigma_3$. This keeps $A$, $S$ and $H$ and maps $P\to-P$, so $E_{k,\sigma}(P)$ is even in $P$.
- **Body bands.** With $\lvert p\rangle=L^{-1/2}\sum_xe^{ipx}\lvert x\rangle$, $A=\sum_p\lvert p\rangle\langle p\rvert\otimes(-\tau\sin p\,\sigma_1)$. At $\delta F=0$, in level $k$, the body block is $\omega(p)(\cos\theta_p\,\sigma_3-\sin\theta_p\,\sigma_1)$. Its bands are $\sigma\omega(p)$, with eigenvectors $u_+=(\cos\frac{\theta_p}2,-\sin\frac{\theta_p}2)$ and $u_-=(\sin\frac{\theta_p}2,\cos\frac{\theta_p}2)$, and $\omega''(0)=\tau^2/\varepsilon_k$.
- **Field.** $H=E_f+\sum_q\Omega_qa_q^\dagger a_q$ and $q_y=\sum_qc_q(a_qe^{iq\cdot y}+a_q^\dagger e^{-iq\cdot y})$. The eigenvectors $\lvert n\rangle$ of $H$ have energies $E_f+E_n$ and momenta $Q_n$ along $\mathbf e_1$.
- **Unperturbed sector.** In the sector $P$ the eigenvectors at $\delta F=0$ are $\lvert P-Q_n,\sigma'\rangle\lvert n\rangle$, with energies $\sigma'\omega(P-Q_n)+E_f+E_n$.
- **Matrix elements of $\sigma_3$.** Multiplying out, with $\vartheta=\frac12(\theta_p+\theta_{p'})$: $\lvert\langle u_\sigma(p')\vert\sigma_3\vert u_\sigma(p)\rangle\rvert^2=\cos^2\vartheta$ and $\lvert\langle u_{-\sigma}(p')\vert\sigma_3\vert u_\sigma(p)\rangle\rvert^2=\sin^2\vartheta$.

### Step 2. The level at second order

By Wick's theorem, $q_y^2=w_0+{:}q_y^2{:}$. The grip is therefore $\bar F S+V'$, with $V'=\sum_x\lvert x\rangle\langle x\rvert\otimes S\otimes O_x$ and $O_x:=F_1q_x+F_2{:}q_x^2{:}$. The term $\bar FS$ replaces $\varepsilon_k$ by $M_k\bar F$ exactly.
- $O_x\lvert0\rangle$ has only one- and two-quantum components, and $\langle n\vert O_x\vert0\rangle=e^{-iQ_nx}\langle n\vert O_0\vert0\rangle$. Hence $\langle P-Q_n,\sigma';n\vert V'\vert P,\sigma;0\rangle=M_k\langle u_{\sigma'}(P-Q_n)\vert\sigma_3\vert u_\sigma(P)\rangle\langle n\vert O_0\vert0\rangle$.
- The weights are $\lvert\langle q\vert O_0\vert0\rangle\rvert^2=F_1^2c_q^2$ for one quantum. The two-quantum states of a pair of modes carry $2F_2^2c_q^2c_{q'}^2$, summed over ordered pairs; this follows from $\langle0\vert{:}q_0^2{:}\,e^{-(H-E_f)t}\,{:}q_0^2{:}\vert0\rangle=2\Gamma(t)^2$.
- Second-order perturbation theory for an isolated level (Step 3) then gives (6.2) and (6.3); for an upper branch that is not isolated they define its formal coefficients (Step 3). The energy denominators are taken at $F_0$; the difference is $O(\delta F^3)$. Both expressions hold on both graphs and for every $L$.
$$
E_{k,\sigma}(P)=E_f+\sigma\sqrt{M_k^2\bar F^2+\tau^2\sin^2P}+E^{(2)}_{k,\sigma}(P)+O(\delta F^3),
\tag{6.2}
$$
$$
E^{(2)}_{k,\sigma}(P)=-M_k^2\Bigl\langle\!\!\Bigl\langle\frac{\cos^2\vartheta}{E+\sigma[\omega(P-Q)-\omega(P)]}+\frac{\sin^2\vartheta}{E-\sigma[\omega(P-Q)+\omega(P)]}\Bigr\rangle\!\!\Bigr\rangle,\qquad\vartheta=\tfrac12[\theta(P)+\theta(P-Q)].
\tag{6.3}
$$
The first fraction has the intermediate body in its own branch, the second in the other branch.

### Step 3. Isolation and analyticity

Isolated means: separated, by a gap uniform in $L$, from all other states of the sector $(k,P)$, with any number of field quanta (question, note (a)).

**Spectrum of the sector at $\delta F=0$, $P=0$** (Step 1: energies $E_f+\sigma'\omega(Q_n)+E_n$).
- *Lower bound.* $\Omega$ is subadditive in the momentum: the map $q\mapsto(2\lvert\sin\frac{q_i}2\rvert)_i$ is componentwise subadditive, and $\sqrt{m_h^2+x^2}$ is increasing and subadditive. Hence a state with $n\ge1$ quanta and total momentum $\mathbf Q$ has $E_n\ge\Omega_{\mathbf Q}\ge\Omega_{(Q_1,0,0)}$. The states with $n\ge1$ quanta therefore lie at or above $E_f+b_k$ ($\sigma'=-1$) and $E_f+m_h+\varepsilon_k$ ($\sigma'=+1$), with $b_k:=\min_q[\Omega_q-\omega_k(q_1)]$. The minimum lies at $q_2=q_3=0$, so $b_k$ is the same function on both graphs, and $b_k\le m_h-\varepsilon_k$ (value at $q=0$).
- *Filling.* As $L\to\infty$, the lower-branch states with one quantum fill $E_f+[b_k,\Omega_{\max}-\varepsilon_k]$: $q\mapsto\Omega_q-\omega_k(q_1)$ is continuous on the connected Brillouin zone and equals $\Omega_{\max}-\varepsilon_k$ at $q=(\pi,\dots,\pi)$. States of $n$ pairs $(q,q')$ with $q_1'=-q_1$ fill $E_f-\varepsilon_k+[2nm_h,2n\Omega_{\max}]$. For $\ell^2\ge3/(4D)$, $\Omega_{\max}\ge2m_h$, so these intervals overlap, and the lower branch with $n\ge1$ quanta fills $E_f+[b_k,\infty)$.
- *Criterion.* The level $E_f+\sigma\varepsilon_k$ is isolated iff $\sigma\varepsilon_k<b_k$; otherwise it lies inside the continuum, or at its edge, as $L\to\infty$. The other zero-quantum level is at distance $2\varepsilon_k$. A strict gap persists for $P$ near 0 and small $\delta F$; the level is then analytic in $P$ and $\delta F$ (isolated eigenvalue of an analytic family).
- *Isolation gap.* $\Delta_k:=b_k-\sigma\varepsilon_k>0$ is the distance to the continuum. It is also the smallest denominator of (6.3) at $P=0$ (one quantum, own branch for $\sigma=-1$, other branch for $\sigma=+1$; two-quantum denominators are larger), so it sets a finite-size scale of its own (Setup, Limits).

**Consequences.**
- *Lower branch.* $b_k+\varepsilon_k=\min_q[\Omega_q-\tau^2\sin^2q_1/(\omega_k+\varepsilon_k)]$ increases with $\varepsilon_k$, so the lower branch is isolated iff $\varepsilon_k>\varepsilon_-(r)$, for a threshold $\varepsilon_-(r)\ge0$. Since $2(1-\cos q_1)\ge\sin^2q_1$, $\Omega_q>c_0\lvert\sin q_1\rvert$, while $\omega-\varepsilon_k=\tau^2\sin^2q_1/(\omega+\varepsilon_k)$ is at most $\tau\lvert\sin q_1\rvert$ and at most $\tau^2\lvert\sin q_1\rvert/2\varepsilon_k$. Hence $\varepsilon_-=0$ for $r\le1$, and $\tau^2\le2\varepsilon_kc_0$ suffices.
- *Upper branch.* $b_k-\varepsilon_k$ decreases with $\varepsilon_k$, so the upper branch is isolated iff $\varepsilon_k<\varepsilon_+(r)$, and $\varepsilon_+\le m_h/2$ by $b_k\le m_h-\varepsilon_k$. By Cauchy–Schwarz, $2\varepsilon_k+\tau\lvert\sin q_1\rvert\le\sqrt{(2\varepsilon_k/m_h)^2+r^2}\,\sqrt{m_h^2+c_0^2\sin^2q_1}$, so $(2\varepsilon_k/m_h)^2+r^2<1$ suffices. In a weakly pinned field, $m_h=c_0/\ell$, only *light levels*, $\varepsilon_k<c_0/2\ell$, have an isolated upper branch.
- *Formal coefficients.* Where the upper branch is not isolated, (6.3) is still finite and analytic in $P$ near 0 if its denominators do not vanish. This is condition (U): $q\mapsto\Omega_q-\varepsilon_k-\omega(q_1)$ and, if $F_2\ne0$, $(q,q')\mapsto\Omega_q+\Omega_{q'}-\varepsilon_k-\omega(q_1+q_1')$ have no zero on the Brillouin zone (sufficient: $\Omega_{\max}<\varepsilon_k$, or $\Omega_{\max}<2\varepsilon_k$ if $F_2=0$). Under (U) we call (6.2), (6.3) and all quantities built from them at $\sigma=+1$ *formal coefficients*. They are not coefficients of an isolated eigenvalue, and they are not an answer to items 2 and 3.
$$
(k,\sigma)\ \text{isolated}\iff b_k>\sigma\varepsilon_k;\qquad\sigma=-1:\ \varepsilon_k>\varepsilon_-(r),\ \ \varepsilon_-=0\ \text{for}\ r\le1\ (\text{and }\Leftarrow\tau^2\le2\varepsilon_kc_0);\qquad\sigma=+1:\ \varepsilon_k<\varepsilon_+(r)\le\tfrac{m_h}2\ \ (\Leftarrow(2\varepsilon_k/m_h)^2+r^2<1).
\tag{6.4}
$$
$\mathcal S$ denotes the set of isolated states $(k,\sigma)$ of a description. Model D has $N\ge2$ levels, so for $r\le1$, $\mathcal S$ contains the lower branches of at least two levels. The thresholds are computed in `code/v6/output/torus_mark_inertia.txt` (Part D). At $\ell=8$: $\varepsilon_+=m_h/2$ for $r\le0.5$, $\varepsilon_+=0.0620\,c_0$ at $r=0.75$, $0.0260\,c_0$ at $r=1$, and $\varepsilon_+=0$ for $r\ge1.25$. Further, $\varepsilon_-=0.117,\,0.346,\,0.985,\,2.877$ (in units of $c_0$) at $r=1.25,\,1.5,\,2,\,3$.

### Step 4. Cost

At $P=0$ the zero-quantum term of (6.2) is exactly $\sigma M_k\bar F$, and $\vartheta=-\theta(Q)/2$. For the expansion in $\tau^2$, use $\cos^2\vartheta=1-\tau^2\sin^2Q/4\varepsilon^2$, $\sin^2\vartheta=\tau^2\sin^2Q/4\varepsilon^2$, $\omega(Q)-\varepsilon=\tau^2\sin^2Q/2\varepsilon$ and $\omega(Q)+\varepsilon=2\varepsilon$, each up to $O(\tau^4)$. Combining the three resulting terms of first order in $\tau^2$ gives
$$
C^{(1)}_{k,\sigma}=\sigma M_k\bar F,\qquad C^{\rm mark}_{k,\sigma}=E^{(2)}_{k,\sigma}(0)=-M_k^2\Bigl(\frac{F_1^2G_0}2+2F_2^2J_2\Bigr)(1+\delta_C),\qquad\delta_C=\tau^2\frac{\langle\!\langle\sin^2Q/[E^2(E-2\sigma\varepsilon_k)]\rangle\!\rangle}{\langle\!\langle1/E\rangle\!\rangle}+O(\tau^4).
\tag{6.5}
$$
The prefactor $-M_k^2\langle\!\langle1/E\rangle\!\rangle$ is the held mark, i.e. the second-order part of (2.6). It is strictly negative because $F_1\ne0$.

### Step 5. Inertia: first-order part and mark

From (6.2), $E''(0)=\sigma\tau^2/(M_k\bar F)+E^{(2)\prime\prime}(0)$. The inverse of the first term is exactly linear in $\delta F$, and the second term is of second order. Expanding $I=1/E''$ gives (6.6), on both graphs and for every $L$, $\ell$ and $r>0$. At $\tau=0$ (held body) $E''=0$, and the inertia is not defined.
$$
I^{(1)}_{k,\sigma}=\frac{\sigma M_k\bar F}{\tau^2}=\frac{C^{(1)}_{k,\sigma}}{\tau^2},\qquad I^{\rm mark}_{k,\sigma}=-\frac{\varepsilon_k^2}{\tau^4}\,E^{(2)\prime\prime}_{k,\sigma}(0),\qquad\rho_{k,\sigma}:=\frac{\tau^2I^{\rm mark}_{k,\sigma}}{C^{\rm mark}_{k,\sigma}} .
\tag{6.6}
$$
Since $C^{(1)}\ne0$, $I^{(1)}=C^{(1)}/v^2$ forces $v=\tau$. A single $v$ for every isolated state therefore exists iff $\rho_{k,\sigma}=1$ on $\mathcal S$. For $\lvert\delta_C\rvert<1$, $C^{\rm mark}\ne0$ by (6.5), so $\rho$ has no vanishing denominator.

### Step 6. Ring, $\ell\gg1$: the scaling region

**Size of the sums.** On the ring $Q=q$. At $L=\infty$, $G_0=1/(m_h\sqrt{m_h^2+4\kappa_h})$, and $\langle\!\langle1/E\rangle\!\rangle\propto\ell$ is dominated by $\lvert q\rvert\lesssim1/\ell$, $E\sim m_h$. By contrast $\langle\!\langle1\rangle\!\rangle$ and $\langle\!\langle\sin^2Q/E^2\rangle\!\rangle$ grow at most like $\ln^2\ell$. Hence $R=O(\ln^2\ell/c_0\ell)$; its one-quantum part is $\frac2\pi\frac{\ln\ell}{c_0\ell}[1+O(1/\ln\ell)]$.

**Expansion of the own-branch fraction.** Let $g=\cos^2\vartheta$ and $\mathcal D=E+\sigma[\omega(P-Q)-\omega(P)]$. Then $(g/\mathcal D)''=g''/\mathcal D-2g'\mathcal D'/\mathcal D^2+2g\mathcal D'^2/\mathcal D^3-g\mathcal D''/\mathcal D^2$. At $P=0$:
- $g''=-2\tau^2/\varepsilon^2\,[1+O(Q^2)]$ and $g'=O(\tau^2Q/\varepsilon^2)$;
- $\mathcal D'=-\sigma\omega'(Q)=-\sigma\tau^2Q/\varepsilon\,[1+O(Q^2)]$;
- $\mathcal D''=\sigma[\omega''(Q)-\omega''(0)]=-2\sigma\tau^2\sin^2Q/\varepsilon\,[1+O(\tau^2/\varepsilon^2)]$;
- $\mathcal D=E\,[1+O(\tau^2Q^2/\varepsilon E)]$.

The four terms are therefore:
- (i) $-2\tau^2/\varepsilon^2E$;
- (ii) $O(\tau^4Q^2/\varepsilon^3E^2)$;
- (iii) $2\tau^4Q^2/\varepsilon^2E^3$;
- (iv) $2\sigma\tau^2\sin^2Q/\varepsilon E^2$.

In the scaling region, (ii), (iii) and (iv) are of relative size $\tau^2/\varepsilon c_0\ell$, $r^2$ and $\varepsilon/c_0\ell$ against (i). The other-branch fraction has $\sin^2\vartheta=\tau^2(2P-Q)^2/4\varepsilon^2\,[1+\dots]$ over $-2\sigma\varepsilon+O(E)$; it is of relative size $O(c_0/\varepsilon\ell)$.

**Rapidity identity.** In the scaling region $\Omega_q=\sqrt{m_h^2+c_0^2q^2}\,[1+O(q^2)]$. Substituting $q=(m_h/c_0)\sinh\alpha$ gives $\sum_qc_q^2f\to\int d\alpha\,f/4\pi c_0$, with $E=m_h\cosh\alpha$ and $c_0Q=m_h\sinh\alpha$. For pairs, put $\alpha=\mathcal A+a$ and $\beta=\mathcal A-a$. Then $E=M_a\cosh\mathcal A$ and $c_0Q=M_a\sinh\mathcal A$, with $M_a=2m_h\cosh a$ and $d\alpha\,d\beta=2\,d\mathcal A\,da$. In both cases $\langle\!\langle f\rangle\!\rangle=\int d\nu(M)\int d\mathcal A\,f(M\cosh\mathcal A,M\sinh\mathcal A)$, with a positive weight $d\nu$ independent of $\mathcal A$. Two integrals then give (6.7):
- $\int d\mathcal A\,\sinh^2\!\mathcal A/\cosh^3\!\mathcal A=\pi/2=\frac12\int d\mathcal A/\cosh\mathcal A$ gives the first relation;
- $\int d\alpha/\cosh\alpha=\pi$ and $\iint d\alpha\,d\beta/(\cosh\alpha+\cosh\beta)=\pi^2$ give $G_0=\ell/2c_0^2$ and $J_2=\ell/16c_0^3$ at leading order.

All these integrals converge in $\mathcal A$ and $a$, so the region $\lvert q\rvert\gg1/\ell$ enters only the corrections. For one quantum they are of relative order $\ell^{-1}$. In the two-quantum sums, pairs with one soft momentum ($\lvert q\rvert\lesssim1/\ell$) and one of lattice order deviate from the scaling form by a nonzero finite part per soft mode, and the sum over the soft partner adds a factor $\ln\ell$. Against the leading order $\ell$ this gives a relative $O(\ell^{-1}\ln\ell)$. For $D=1$:
$$
c_0^2\langle\!\langle Q^2/E^3\rangle\!\rangle=\tfrac12\langle\!\langle1/E\rangle\!\rangle\,[1+O(\ell^{-1}\ln\ell)],\qquad\langle\!\langle1/E\rangle\!\rangle=\frac{F_1^2G_0}2+2F_2^2J_2=\frac{\ell}{4c_0^2}\Bigl(F_1^2+\frac{F_2^2}{2c_0}\Bigr)[1+O(\ell^{-1}\ln\ell)] .
\tag{6.7}
$$

**Summing the terms.** Let $\delta_\ell:=(1+r^2+c_0/\varepsilon_k+\tau^2/\varepsilon_kc_0)\ln^2\ell/\ell$. Sum the four terms with the weights (6.1), keep (iv) explicitly, use (6.7), and use (6.5) with $\lvert\delta_C\rvert=O(\tau^2R/\varepsilon)$. This gives
$$
C^{\rm mark}_{k,\sigma}=-\frac{M_k^2\ell}{4c_0^2}\Bigl(F_1^2+\frac{F_2^2}{2c_0}\Bigr)[1+O(\delta_\ell)],\qquad\zeta=\frac{F_2^2}{2c_0}[1+O(\delta_\ell)],\qquad E^{(2)\prime\prime}_{k,\sigma}(0)=-\frac{2\tau^2}{\varepsilon_k^2}\,C^{\rm mark}_{k,\sigma}\Bigl[1-\frac{r^2}2-\sigma\varepsilon_kR+O(\delta_\ell)\Bigr] .
\tag{6.8}
$$
$\delta_\ell\ll1$ needs $\varepsilon_k\gg c_0\ln^2\ell/\ell>m_h/2$. In the range of (6.8) the upper branch is therefore never isolated (6.4), and its entries there are formal coefficients.

Code check (`code/v6/output/check_mark_inertia.txt`, part B), on rings with $L=8\ell$ at $\ell=512$:
- $G_0\cdot2c_0^2/\ell=1.00067$; the excess is the finite-$L$ term $\approx2e^{-8}$;
- $J_2\cdot16c_0^3/\ell=0.99938$;
- $C^{\rm mark}(\tau)/C^{\rm mark}(0)$ lies within $3.1\times10^{-3}$ of 1.

### Step 7. Inertia against cost on the ring

Inserting (6.8) into (6.6) gives (6.9). Since $\varepsilon_kR+\delta_\ell=O(\ln^2\ell/\ell)$, $\rho_{k,\sigma}\to2-r^2$ as $\ell\to\infty$. In this limit $\mathcal S$ consists of lower branches: those of all levels for $r\le1$, and of the levels with $\varepsilon_k>\varepsilon_-(r)$ otherwise (6.4). The statement on $v$ is for nonempty $\mathcal S$.
$$
\rho_{k,\sigma}=2-r^2-2\sigma\varepsilon_kR+O(\delta_\ell),\qquad\text{at leading order in }1/\ell:\ \ I^{\rm mark}_{k,-}=\frac{(2-r^2)\,C^{\rm mark}_{k,-}}{\tau^2}\ \Rightarrow\ \bigl[\exists\,v:\ I^{(1)}=C^{(1)}/v^2,\ I^{\rm mark}=C^{\rm mark}/v^2\ \text{on }\mathcal S\bigr]\iff r=1,\ v=\tau=c_0 .
\tag{6.9}
$$
- **First difference.** $I^{\rm mark}-C^{\rm mark}/\tau^2=(1-r^2)C^{\rm mark}/\tau^2$.
- **Mark alone, $r\neq1$.** For $r<\sqrt2$ the mark alone has $v_{\rm mark}^2=\tau^2/(2-r^2)$. At $r=\sqrt2$, $I^{\rm mark}=0$ at this order; for $r>\sqrt2$ it has the sign opposite to $C^{\rm mark}$.
- **Grip dependence.** The law holds for the $F_1^2$ part and the $F_2^2$ part separately. It is therefore independent of the grip, the linear grip ($F_2=0$) included.
- **Upper branch.** Its formal coefficients obey the same leading law.
- **Next order in $1/\ell$.** For the lower branch the term $+2\varepsilon_kR$ (from (iv)) grows with the level. The remainder $O(\delta_\ell)$ is of comparable order, and its difference between two levels is not controlled. So (6.9) does not decide whether two lower levels have different $\rho$ at finite $\ell$. They do at the tested members (code check below), and in the heavy regime of (6.10) for levels whose $\varepsilon_k$ differ by more than its relative remainder.
- **Code check** (part B, $\ell=512$, $r=1$). For the lower branch, the $F_1^2$ part gives $\rho_{1,-}=1.056$ and $1.072$ ($M_k=3,4$), and the $F_2^2$ part gives $1.211$ and $1.275$. With the term (iv) removed, the values are $1.00919$–$1.00920$ and $1.021$–$1.023$ for both levels (and for the formal upper-branch coefficients). For $r=0.5$ and $1.5$ the values approach $2-r^2$ in the same way.
- **Order of limits.** First $L\gg\ell$, then $\ell\to\infty$ at fixed $\varepsilon_k/c_0$ and $r$. Taking $\varepsilon_k/c_0\to\infty$ at fixed $\ell$ gives a different result (Step 8). Light levels ($\varepsilon_k<m_h/2$, the only ones with an isolated upper branch) lie outside this limit; they are treated in Step 10.

### Step 8. Heavy body: lower branch on both graphs

**Odd part in $\sigma$.** Let $\Delta=\omega(P-Q)-\omega(P)$ and $\Sigma=\omega(P-Q)+\omega(P)$. The part of (6.3) odd in $\sigma$ is $-M_k^2\langle\!\langle-\cos^2\vartheta\,\Delta/(E^2-\Delta^2)+\sin^2\vartheta\,\Sigma/(E^2-\Sigma^2)\rangle\!\rangle$. At $P=0$ its second derivative is dominated by $\cos^2\vartheta\,\Delta''/E^2$, with $\Delta''(0)=\omega''(Q)-\omega''(0)=-2\tau^2\sin^2Q_1/\varepsilon\,[1+O(\tau^2/\varepsilon^2)]$; this is the term (iv). The remaining odd terms are smaller:
- those with $(\cos^2\vartheta)''\Delta$ and $(\cos^2\vartheta)'\Delta'$, by $\tau^2/\varepsilon^2$;
- the one with $\Delta\Delta'^2/E^2$, by $r^2\tau^2/\varepsilon^2$;
- the other-branch term. With $\vartheta'(0)=\frac{\tau}{2\varepsilon}(1+\cos Q_1)[1+O(\tau^2/\varepsilon^2)]$ and $\Sigma=2\varepsilon[1+O(\tau^2/\varepsilon^2)]$, its odd part adds $\sigma\langle\!\langle(1+\cos Q_1)^2\rangle\!\rangle/(4\varepsilon\langle\!\langle1/E\rangle\!\rangle)\,[1+O((\Omega_{\max}^2+\tau^2)/\varepsilon^2)]$ to $\rho$. For every $L$ this is at most $\langle\!\langle1\rangle\!\rangle/(\varepsilon\langle\!\langle1/E\rangle\!\rangle)\le2\Omega_{\max}/\varepsilon$, since $E\le2\Omega_{\max}$. For $L\gg\ell\ge1$ the sums take their $L=\infty$ values, where $\langle\!\langle1\rangle\!\rangle=O(c_0^2\langle\!\langle\sin^2Q_1/E^2\rangle\!\rangle)$; the term is then smaller than (iv) by $c_0^2/\varepsilon^2$. This comparison is not uniform in $L$. At fixed $L$ and $m_h\to0$ the uniform mode $q=0$ dominates $\langle\!\langle1\rangle\!\rangle$ and $\langle\!\langle1/E\rangle\!\rangle$ but has $\sin Q_1=0$. Then $R=O(m_h^2)$, while this term gives $\rho_{k,-}-\rho^{\rm e}_k=-m_h/\varepsilon_k+O(m_h^2)$;
- the odd part of $\delta_C$ in $C^{\rm mark}$, by $(1+r^2)\tau^2/\varepsilon^2$, because the even part of $\rho$ is $O(1+r^2)$ for every $L$: (i) gives at most 2, and (iii) gives $r^2$ times a lattice ratio of order 1, using $\omega'(Q)=\tau^2\sin2Q_1/2\varepsilon$.

So, for $L\gg\ell\ge1$, the part of $\rho_{k,\sigma}$ odd in $\sigma$ is $-2\sigma\varepsilon_kR[1+O((c_0^2+\tau^2(1+r^2))/\varepsilon_k^2)]$, and the even part $\rho^{\rm e}_k$ (built from the lower branch and the formal upper coefficients) is $O(1+r^2)$. For the lower branch this gives the first part of (6.10). At finite $L$ the exact expressions (6.3) and (6.6) remain; only this estimate is restricted.

**Heavy regime.** If moreover $\varepsilon_kR\gg1+r^2$ and $\tau^2\ll\varepsilon_kc_0$, the term (iv), $2\sigma\tau^2\sin^2Q_1/\varepsilon_kE^2\,[1+O(\tau^2/\varepsilon_kc_0)]$, dominates $\rho$ itself, for every $L$. Indeed $E\ge c_0\lvert\sin Q_1\rvert$ (Step 3) and $E\ge m_h$ give $R\le1/c_0$ and $R\le1/m_h$, so $\varepsilon_kR\gg1$ implies $\varepsilon_k\gg c_0+m_h$. The other-branch term, at most $2\Omega_{\max}/\varepsilon_k\ll1$, is then inside the remainder. With (6.6) this gives the second part of (6.10). It holds on both graphs, for every $L$ and $\ell$, for an isolated lower branch (6.4); in the heavy regime it is isolated, since $\tau^2\le2\varepsilon_kc_0$.
$$
\rho_{k,-}=\rho^{\rm e}_k+2\varepsilon_kR\Bigl[1+O\Bigl(\frac{c_0^2+\tau^2(1+r^2)}{\varepsilon_k^2}\Bigr)\Bigr]\ \ (L\gg\ell\ge1),\ \ \rho^{\rm e}_k=O(1+r^2);\qquad\text{heavy: }\ I^{\rm mark}_{k,-}=I^{(1)}_{k,-}Y_k\Bigl[1+O\Bigl(\frac{1+r^2}{\varepsilon_kR},\frac{\tau^2}{\varepsilon_kc_0}\Bigr)\Bigr],\ \ Y_k:=2M_k^2\langle\!\langle\sin^2Q_1/E^2\rangle\!\rangle>0 .
\tag{6.10}
$$
**Consequences.**
- *Heavy regime.* $\rho_{k,-}=2\varepsilon_kR\gg1$ at leading order, for every grip. So no single $v$ exists at second order, at any $r$ ($r=1$ included) and at any finite $\ell$ and $L$. Moreover, at leading order $\rho_{k,-}$ is proportional to $\varepsilon_k$: two levels whose $\varepsilon_k$ differ by more than the relative remainder of (6.10) have different $\rho_{k,-}$.
- *Upper branch.* The same sums give the formal coefficient $\rho_{k,+}=\rho^{\rm e}_k-2\varepsilon_kR[\dots]$. For $\varepsilon_k\gg c_0$ the upper branch is not isolated (6.4).

**Size of $R$.** The heavy regime is defined by $\varepsilon_kR\gg1+r^2$ (with $\tau^2\ll\varepsilon_kc_0$); the scale of $\varepsilon_k$ this needs depends on the graph and the grip. Let $R_1$, $R_2$ be the ratio $R$ of the $F_1^2$ and the $F_2^2$ part; $R$ lies between them.
- On $D=3$ the sums are dominated by $\lvert q\rvert\sim1$, and $R$ tends to a finite limit of order $1/c_0$ as $\ell\to\infty$. The heavy regime is then $\varepsilon_k\gg(1+r^2)c_0$, up to a lattice number.
- On the ring ($L\gg\ell$), $R_1=\frac2\pi\frac{\ln\ell}{c_0\ell}[1+O(1/\ln\ell)]$ (Step 6). In the two-quantum numerator the soft pairs $1/\ell\ll\lvert q\rvert,\lvert q'\rvert\ll1$ contribute $\propto\iint dq\,dq'/\lvert qq'\rvert\sim\ln^2\ell$, with nonnegative summands, and Step 6 bounds it by $O(\ln^2\ell)$; so $R_2=\Theta(\ln^2\ell/c_0\ell)$. For the linear grip the heavy regime is therefore $\varepsilon_k\gg(1+r^2)c_0\ell/\ln\ell$. With a fixed nonzero two-quantum part ($F_2\ne0$ at fixed $\xi=F_2^2/c_0F_1^2$) it is reached already on the smaller scale $(1+r^2)c_0\ell/\ln^2\ell$, up to grip-dependent constants. Since $R\ge\min(R_1,R_2)=R_1$ at large $\ell$, $\varepsilon_k\gg(1+r^2)c_0\ell/\ln\ell$ suffices for every grip of (G2).

**$D=3$ otherwise.** The scaling argument of Step 6 fails on $D=3$. Outside the regime of (6.10), $\rho$ is given by the lattice sums of (6.3). They are evaluated in Step 10.

### Step 9. Weight over inertia

Let $\Delta W_{k,\sigma}=\sigma M_kF_1\Delta\phi/(1+2\sigma M_kF_2G_0)$ by (2.8), with $\Delta\phi\ne0$. Then $E''\Delta W=\Delta W/I$.

**(a) $F_1$ and $F_2$ counted alike.** $E''(0)=\sigma\tau^2/(M_k\bar F)+O(\delta F^2)$ and $\Delta W=\sigma M_kF_1\Delta\phi(1-2\sigma M_kF_2G_0)+O(\delta F^3)$. Hence, on both graphs and for every $L$, $\ell$ and $r$,
$$
E''_{k,\sigma}(0)\,\Delta W_{k,\sigma}=\frac{\tau^2F_1\Delta\phi}{F_0}\Bigl[1-\frac{F_2w_0}{F_0}-2\sigma M_kF_2G_0\Bigr]+O(\delta F^3).
\tag{6.11}
$$
In this counting the inertia of the mark enters only at $O(\delta F^3)$.
- **Linear grip.** For $F_2=0$ exactly, (6.11) is the same for every level and both branches, but only at this order. With $F_2=0$, $\Delta W$ is exactly $\sigma M_kF_1\Delta\phi$, and $E''$ is even in $F_1$. The first dependence on the level is therefore the term $-I^{\rm mark}/I^{(1)}$ of relative order $F_1^2$, i.e. (6.12) with $F_2=0$. On the ring, at leading order in $1/\ell$, it vanishes only at $r=\sqrt2$ (below). On the cubic torus it differs between any two isolated lower levels at the computed points (Step 10).

**(b) Grips of a fixed shape.** For grips $F=F_0\Phi(\beta q)$ of a fixed shape, as in (2.18), $F_1=O(\beta)$ and $F_2=O(\beta^2)$. The weight of the mark ($\propto F_1F_2$) and its inertia ($\propto F_1^3$ in $E''\Delta W$) are then of the same order $\beta^3$.
- $E''$ is even in $F_1$ (symmetry $q\to-q$, $p\to-p$ of $H$). Through relative order $\beta^2$ it therefore needs only the term $F_2w_0$ and the $F_1^2$ part of $E^{(2)}$; the $F_2^2$ part is $O(\beta^4)$.
- By (6.6), $\sigma M_kE^{(2)\prime\prime}(0)=-\sigma M_k\tau^4I^{\rm mark}/\varepsilon_k^2=-(\tau^2/F_0)\,I^{\rm mark}/I^{(1)}$ at this order. Hence, on both graphs and for every $L$ and $\ell$, with $I^{\rm mark}$ its $F_1^2$ part:
$$
E''_{k,\sigma}(0)\,\Delta W_{k,\sigma}=\frac{\tau^2F_1\Delta\phi}{F_0}\Bigl[1-\frac{F_2w_0}{F_0}-2\sigma M_kF_2G_0-\frac{I^{\rm mark}_{k,\sigma}}{I^{(1)}_{k,\sigma}}\Bigr][1+O(\beta^4)].
\tag{6.12}
$$
- **Ring, leading order in $1/\ell$.** By (6.9), $I^{\rm mark}/I^{(1)}=(2-r^2)C^{\rm mark}/C^{(1)}=-(2-r^2)\sigma M_kF_1^2G_0/2F_0$. The bracket is then $1-F_2w_0/F_0-\sigma M_kG_0[2F_2-(2-r^2)F_1^2/2F_0]$. If $\mathcal S$ (lower branches, Step 7) contains at least two levels, it is the same on $\mathcal S$ iff $(2-r^2)F_1^2=4F_0F_2$. The formal upper-branch coefficients obey the same condition. At $r=1$ this is the square grip of (2.18). The square grip satisfies it only at $r=1$, and the linear grip only at $r=\sqrt2$.
- **Heavy regime (both graphs, every finite $\ell$, lower branches).** By (6.10), with $R_1$ the ratio $R$ of the $F_1^2$ part, $-I^{\rm mark}_{k,-}/I^{(1)}_{k,-}=-M_k^2F_1^2G_0R_1$ at leading order. The bracket of (6.12) is then $1-F_2w_0/F_0+2M_kF_2G_0-M_k^2F_1^2G_0R_1$. This is a polynomial of degree two in $M_k$ with leading coefficient $-F_1^2G_0R_1\neq0$, while $F_2$ enters only through the terms of degree $\le1$. So, at this order, no grip of (G2) gives one value at three levels. For two levels $a,b$ exactly one grip does: $F_0F_2/F_1^2=R_1(\varepsilon_a+\varepsilon_b)/2$. The rest of the range on the cubic torus is treated in Step 10.

### Step 10. Cubic torus: items 2 and 3 from the lattice sums

**Two parts (exact).** Write (6.3) as $E^{(2)}_{k,\sigma}(P)=-M_k^2[F_1^2S_1(P)+F_2^2S_2(P)]$. Here $S_1$ and $S_2$ are its one- and two-quantum parts per unit $F_1^2$ and $F_2^2$; they depend only on $\varepsilon_k$, $\tau$, $\sigma$ and the field. Put $S_i:=S_i(0)$ and $\rho_i:=-(\varepsilon_k^2/\tau^2)S_i''(0)/S_i$, the ratio of (6.6) for part $i$. Define the grip ratios $\xi:=F_2^2/(c_0F_1^2)$ and $y:=F_0F_2/F_1^2$; $y=\frac14$ is the square grip and $y=0$ the linear grip. Finally put $T_{k,\sigma}:=\rho_1S_1$ and $t_{k,\sigma}:=T_{k,\sigma}/G_0$. Then (6.6) gives $I^{\rm mark}/I^{(1)}=-\sigma M_kF_1^2T_{k,\sigma}/F_0$ for the $F_1^2$ part, and with (6.12):
$$
\rho_{k,\sigma}=\frac{S_1\rho_1+\xi c_0S_2\rho_2}{S_1+\xi c_0S_2},\qquad\rho_{k,\sigma}=1\iff S_1(\rho_1-1)+\xi c_0S_2(\rho_2-1)=0,\qquad[\,\cdot\,]_{(6.12)}=1-\frac{F_2w_0}{F_0}-\sigma\frac{\varepsilon_kF_1^2G_0}{F_0^2}\,(2y-t_{k,\sigma}).
\tag{6.13}
$$
- *Singular cases.* Let $S_1,S_2>0$ (true at all computed points, finding (a) below). A grip of (G2), i.e. a finite $\xi\ge0$, gives $\rho_{k,\sigma}=1$ in exactly two cases:
  - $\rho_1=1$: then $\xi=0$, the linear grip, works (and every $\xi$ if also $\rho_2=1$);
  - $(\rho_1-1)(\rho_2-1)<0$: then exactly one grip works, $\xi=\xi^*_{k,\sigma}:=S_1(1-\rho_1)/[c_0S_2(\rho_2-1)]>0$.

  If $\rho_2=1\ne\rho_1$, then $\rho_{k,\sigma}\to1$ only as $\xi\to\infty$, i.e. $F_1\to0$, which (G2) excludes. In all other cases no grip works.
- *Item 3 reduced.* By (6.13), the bracket of (6.12) is the same on $\mathcal S$ iff $\sigma\varepsilon_k(2y-t_{k,\sigma})$ is. Let $\mathcal S$ consist of lower branches, and put $\varphi(\varepsilon):=\varepsilon\,t_{-}(\varepsilon)$.
  - Two levels $a\ne b$ need exactly $y=y_{ab}:=[\varphi(\varepsilon_b)-\varphi(\varepsilon_a)]/[2(\varepsilon_b-\varepsilon_a)]$.
  - Three levels need the points $(\varepsilon_k,\varphi(\varepsilon_k))$ on one straight line.
  - The linear grip needs $\varphi(\varepsilon_a)=\varphi(\varepsilon_b)$.
- *Check on the ring.* At leading order in $1/\ell$, $t=(2-r^2)/2$ by (6.9), and (6.13) gives back the condition of Step 9(b).

**Isolated states.** The thresholds at $\ell=8$ are those of Step 3. The grid below has $\varepsilon_k\ge0.1\,c_0>m_h/2$, so there $\mathcal S$ consists of lower branches. They are evaluated where $b_k+\varepsilon_k>0.2\,m_h$. Light levels are treated at the end of this step.

**Evaluation** (`code/v6/output/torus_mark_inertia.txt`).
- *Setting.* Units $c_0=F_0=1$.
  - Part D: torus $L=64$, $\ell=8$; $r\in\{0.1,0.25,0.5,0.75,1,1.25,1.5,2,3\}$; 71 geometric values of $\varepsilon_k\in[0.1,64]$, refined to 81 equidistant points across each window.
  - Part D2: $L=256$, $\ell=32$; $r\in\{0.1,0.5,1,1.5,2,3\}$; 36 values.
- *Method.* $S_1$ is an exact mode sum and $S_2$ an FFT convolution (Code). $S_i''(0)$ is the analytic second $P$-derivative of the summands, so no finite-difference step enters. On $L=12$ the sums agree with brute-force sums to $4\times10^{-14}$ and $2.5\times10^{-13}$ (Part A). Exact diagonalisation with at most two quanta on the torus $L=4$ reproduces the second-order coefficients of the cost and of $E''(0)$ from (6.2), (6.3) to $3.8\times10^{-9}$ (Part E).
- *Curvature and finite size at the smallest gaps* (Part G). The test points are the light levels at $f=0.9$ (gaps down to $0.022\,m_h$) and the lower-branch grid points nearest their threshold (gaps $0.279$–$0.455\,m_h$ at $\ell=8$, $m_h$ at $\ell=32$).
  - Richardson differences approach the analytic $\rho_1$, asymptotically like $h^4$. At the step $h_0$ of v3 they deviate by up to $15$ (ring) and $0.11$ (torus) at the light levels and by $8\times10^{-3}$ at the lower-branch points; at $h_0/4$, by $0.019$, $1.5\times10^{-4}$ and $3\times10^{-5}$. For $\rho_2$ the deviation is at most $6\times10^{-6}$.
  - At the lower-branch points, $L=64$ and $L=128$ ($\ell=8$) give $\rho_i$ within $1.6\times10^{-4}$.
- *Change with $\ell$* (Parts B, B2; $L=8\ell$ throughout), over the points of Part C, from $(L,\ell)=(64,8)$ to $(256,32)$:
  - for $\varepsilon_k\le4c_0$, $\rho_1$, $\rho_2$ and $t$ change by at most $0.093$, $0.012$ and $0.058$;
  - above, by at most $2.7\%$, $0.8\%$ and $2.8\%$;
  - nearer the lower threshold the change is larger: at $r=3$, $\varepsilon_k=4c_0$, $\rho_{1,-}=1.95$, $1.71$, $1.61$ at $\ell=8$, $16$, $32$.

**Findings at the computed points** (numerical: grid, refinement and roots; nothing is claimed between computed points).
- (a) At every computed point, $S_i(0)$ is at least its held value, so $S_1,S_2>0$.
- (b) $\rho_{1,-}$ and $\rho_{2,-}$ increase along the grid at every $r$. At $r=1$, $\ell=8$, they go from $0.28$ and $0.06$ at $\varepsilon_k=0.1\,c_0$ to $30.6$ and $16.8$ at $64\,c_0$.
- (c) *Windows.* By (6.13), a finite grip with $\rho_{k,-}=1$ needs $\rho_{1,-}=1$ or $(\rho_{1,-}-1)(\rho_{2,-}-1)<0$.
  - *Roots.* In each grid interval where $\rho_{i,-}-1$ changes sign, the root of $\rho_{i,-}(\varepsilon)=1$ is found by Brent's method on the lattice sums. Its estimated numerical error is below $2\times10^{-10}$: residual over slope, and for $\rho_2$ also the binning error, estimated from a lattice with half the energy step.
  - *Windows.* The windows are the intervals between consecutive roots (or the edge of the computed range) on which $(\rho_{1,-}-1)(\rho_{2,-}-1)<0$. At every computed point, grid and refinement (98–152 points per $r$), a finite grip exists iff the point lies in a window. Values in units of $c_0$, at $\ell=8$ with $\ell=32$ in brackets.
  - For $r\le2$ the window runs from the root of $\rho_{1,-}=1$, attained by the linear grip, to the root of $\rho_{2,-}=1$, not attained ($\xi^*\to\infty$):
    - $r=0.1$: $[0.9721,2.6683)$ $([0.9274,2.6406))$;
    - $r=1$: $[0.9123,2.6119)$ $([0.8957,2.5838))$;
    - $r=1.5$: $[0.8447,2.5342)$ $([0.8909,2.5053))$;
    - $r=2$: $[1.3385,2.4089)$ $([1.4187,2.3785))$.

    $\xi^*_{k,-}$ increases along the refined grid. The refined point $\varepsilon=1.434$ ($\ell=32$, $r=2$, $\xi^*=0.5056$) lies inside its window.
  - At $r=3$, $\rho_{2,-}>1$ at all computed points. The window runs from the edge of the computed range, $3.043$ ($3.337$), to the root of $\rho_{1,-}=1$ at $3.5649$ ($3.6997$), which the linear grip attains. $\xi^*$ decreases along it.
- (d) $\varphi$ increases along the grid at every $r$. Its discrete second differences are positive at all grid points for $r\le1$. For $r\ge1.25$ they are negative in a band just above the lower threshold: $[0.174,0.209]$ at $r=1.25$, $[0.438,0.634]$ ($[0.527,0.634]$) at $r=1.5$, $[1.209,1.749]$ ($[1.326,1.595]$) at $r=2$, and $[3.337,4.827]$ ($[4.014,4.827]$) at $r=3$.
- (e) *Large $\varepsilon_k$.* $R_1=0.232/c_0$ ($0.229/c_0$ at $\ell=32$). At $\varepsilon_k=64\,c_0$, $\rho_{1,-}-2\varepsilon_kR_1=0.75$–$0.88$ ($0.58$–$0.91$) and $\rho_{1,-}/(2\varepsilon_kR_1)=1.025$–$1.030$ ($1.020$–$1.031$), as (6.10) describes.
- (f) *Light levels* (Part F): $\ell=8$, $\varepsilon_k=f\varepsilon_+(r)$ with $f\in\{0.3,0.5,0.7,0.9\}$ and $r\in\{0.25,0.5,0.75,1\}$, so both branches are isolated.
  - On the ring ($L=2048$; $L=4096$ for the $F_1^2$ part, the same to four digits), $\rho_{1,+}\in[-65.2,-0.52]$, $\rho_{2,+}\in[-0.85,-0.13]$, and $\rho_{1,-},\rho_{2,-}\in[0.1065,0.7715]$.
  - On the torus ($L=128$, $256$), $\rho_{i,+}\in[-1.277,-0.0049]$ and $\rho_{i,-}\in[0.0049,0.1473]$. The two sizes agree to $0.003$, except at $r=1$, $f=0.9$ (gap $0.022\,m_h$), where $\rho_{1,+}=-1.277$ and $-1.210$.
  - So at every light-level point both parts of both branches have $\rho_i<1$, and by (6.13) no grip gives $\rho=1$ for either branch.

**Item 2.** By (6.6) a single $v$ must be $\tau$, and it needs $\rho_{k,\sigma}=1$ on $\mathcal S$ with one common grip. By (6.13) each isolated state then needs $\rho_1=1$ or $(\rho_1-1)(\rho_2-1)<0$.
- *Negative for the class, by members with margins.* Take $r=1$, where $\mathcal S$ contains the lower branches of all levels (6.4).
  - A description with a level at $\varepsilon_k=4c_0$: $\rho_{1,-}=2.6235$ and $\rho_{2,-}=1.4061$ at $\ell=8$ ($2.6102$ and $1.4175$ at $\ell=32$). So $\rho_{k,-}\ge1.40$ for every grip.
  - Two levels at $1.2\,c_0$ and $2\,c_0$, both in the window, need $\xi^*=1.96$ and $16.40$ ($2.105$ and $17.38$ at $\ell=32$).
  - Any description with a light level at a point of (f).
- *Members found on the grid.* One $v$ exists exactly when $\mathcal S$ is the lower branch of a single level with $\varepsilon_k$ in the window of (c) and $\xi=\xi^*_{k,-}$; then $v=\tau$.
  - Since $N\ge2$, this requires $r>1$ with all other levels at or below $\varepsilon_-(r)$; for $r\ge1.25$ there are no light levels.
  - For $r\le1$, $\mathcal S$ contains at least two lower levels. On the refined grid distinct points of the window have distinct $\xi^*$, and outside it no grip works, so no member with its levels on the grid has one $v$.
$$
D=3,\ \text{grid of Step 10}:\ \ \bigl[\exists\,v:\ I^{(1)}=C^{(1)}/v^2,\ I^{\rm mark}=C^{\rm mark}/v^2\ \text{on }\mathcal S\bigr]\iff\mathcal S=\{(k,-)\},\ \varepsilon_k\ \text{in the window},\ \xi=\xi^*_{k,-}\ (\text{then }v=\tau);\qquad r=1,\ \varepsilon_k=4c_0:\ \rho_{k,-}\ge1.40\ \text{for every grip}.
\tag{6.14}
$$
- The first difference is $I^{\rm mark}-C^{\rm mark}/\tau^2=(\rho_{k,\sigma}-1)\,C^{\rm mark}/\tau^2$.
- Order of limits: $L\gg\ell$, then $\ell$ large. Unlike on the ring, the sums have finite limits as $\ell\to\infty$; the findings are stated at $\ell=8$ and $\ell=32$.

**Item 3** ($\mathcal S$ without light levels; with them it is not evaluated).
- *Negative for the class, by a member with margins.* Take $r=1$ and the levels $\varepsilon_k=(1,2,3)\,c_0$: Model D with $m=d_1=c_0/F_0$, $N=3$, all three lower branches isolated. Then $y_{12}=0.563$ and $y_{23}=0.804$ at $\ell=8$ ($0.561$ and $0.798$ at $\ell=32$), so no grip of (G2) works.
- *Linear grip.* By (d), $y_{ab}>0$ for any two grid levels, so $y=0$ fails for any two isolated lower levels on the grid.
- *Members found on the grid.*
  - If $\mathcal S$ has two lower levels, exactly one grip works, $y=y_{ab}$. It depends on the levels and on $r$: at $r=1$, $y_{ab}=0.25,0.37,0.56,0.92,1.62$ for $(\varepsilon_a,\varepsilon_b)/c_0=(0.25,0.5),(0.5,1),(1,2),(2,4),(4,8)$, and $0.26,0.37,0.56,0.91,1.605$ at $\ell=32$ (direct evaluation at these levels).
  - If $\mathcal S$ has three or more lower levels at grid points where $\varphi$ has positive discrete second differences (all grid points for $r\le1$), no grip works, since the chord slopes of $\varphi$ then increase.
  - The square grip, which gives (6.12) on the ring at $r=1$, works only for pairs with $y_{ab}=\frac14$; it is not singled out.
$$
D=3:\ \ E''_{k,\sigma}(0)\,\Delta W_{k,\sigma}\ \text{the same on }\mathcal S\iff\sigma\varepsilon_k(2y-t_{k,\sigma})\ \text{the same on }\mathcal S;\qquad r=1,\ \varepsilon_k=(1,2,3)\,c_0:\ y_{12}=0.563\ne y_{23}=0.804\ \Rightarrow\ \text{no grip of (G2)} .
\tag{6.15}
$$

## Result

- **(6.2), (6.3)** [formula]. Class: ring and cubic torus, every $L$, $M\to\infty$ first, every isolated state (6.4), $P$ near 0. The level is $E_f+\sigma\sqrt{M_k^2\bar F^2+\tau^2\sin^2P}+E^{(2)}_{k,\sigma}(P)$ up to $O(\delta F^3)$, with $E^{(2)}$ the spectral sum (6.3) over one- and two-quantum states; it is even in $P$. For an upper branch that is not isolated, under (U), the same expressions are its formal coefficients.
- **(6.4)** [criterion]. Both graphs, $\ell^2\ge3/(4D)$, $L\to\infty$, $\delta F\to0$. $(k,\sigma)$ is separated by a gap from every sector of field excitations, and then analytic in $P$ near 0 and in $\delta F$, iff $b_k=\min_q[\Omega_q-\omega_k(q_1)]>\sigma\varepsilon_k$.
  - Lower branch: iff $\varepsilon_k>\varepsilon_-(r)$; $\varepsilon_-=0$ for $r\le1$, and $\tau^2\le2\varepsilon_kc_0$ suffices.
  - Upper branch: iff $\varepsilon_k<\varepsilon_+(r)\le m_h/2$, so only for light levels; $(2\varepsilon_k/m_h)^2+r^2<1$ suffices.
  - Otherwise the upper-branch coefficients are formal, finite under (U).
- **(6.5)** [formula]. Same class. $C^{(1)}_{k,\sigma}=\sigma M_k\bar F$ for every $\tau$. $C^{\rm mark}_{k,\sigma}=E^{(2)}_{k,\sigma}(0)=-M_k^2(F_1^2G_0/2+2F_2^2J_2)(1+\delta_C)$, with $\delta_C$ as in (6.5) at first order in $\tau^2$. The held mark is strictly negative.
- **(6.6)** [theorem]. Same class, $\tau>0$, exact at its order: $I^{(1)}_{k,\sigma}=C^{(1)}_{k,\sigma}/\tau^2$ for every isolated state and every $r$; $I^{\rm mark}=-\varepsilon_k^2E^{(2)\prime\prime}(0)/\tau^4$. A single $v$ for all isolated states exists iff $\rho_{k,\sigma}=1$ on $\mathcal S$, and then $v=\tau$.
- **(6.7)** [formula]. Ring, $L\gg\ell$: $c_0^2\langle\!\langle Q^2/E^3\rangle\!\rangle=\frac12\langle\!\langle1/E\rangle\!\rangle[1+O(\ell^{-1}\ln\ell)]$ and $\langle\!\langle1/E\rangle\!\rangle=\frac{\ell}{4c_0^2}(F_1^2+F_2^2/2c_0)[1+O(\ln\ell/\ell)]$.
- **(6.8)** [formula]. Ring, $L\gg\ell$, under (6.4) (for $\sigma=+1$: formal coefficients, since the upper branch is not isolated where $\delta_\ell\ll1$), up to relative $O(\delta_\ell)$: $C^{\rm mark}_{k,\sigma}=-\frac{M_k^2\ell}{4c_0^2}(F_1^2+\frac{F_2^2}{2c_0})$, $\zeta=F_2^2/2c_0$, and $E^{(2)\prime\prime}(0)=-\frac{2\tau^2}{\varepsilon_k^2}C^{\rm mark}[1-\frac{r^2}2-\sigma\varepsilon_kR]$, with $R=O(\ln^2\ell/c_0\ell)$.
- **(6.9)** [theorem]. Ring, $L\gg\ell$, every (G2) grip, every $r>0$, isolated lower branches (all levels for $r\le1$, levels with $\varepsilon_k>\varepsilon_-(r)$ otherwise). $\rho_{k,-}=2-r^2+2\varepsilon_kR+O(\delta_\ell)$, with $\varepsilon_kR=O((\varepsilon_k/c_0)\ln^2\ell/\ell)$. At leading order in $1/\ell$ (first $L\to\infty$, then $\ell\to\infty$ at fixed $\varepsilon_k/c_0$ and $r$, where $\mathcal S$ contains only lower branches), $I^{\rm mark}_{k,-}=(2-r^2)C^{\rm mark}_{k,-}/\tau^2$. Hence, for nonempty $\mathcal S$, one $v$ with $I^{(1)}=C^{(1)}/v^2$ and $I^{\rm mark}=C^{\rm mark}/v^2$ on $\mathcal S$ exists iff $r=1$, and then $v=\tau=c_0$. The first difference is $(1-r^2)C^{\rm mark}/\tau^2$. The formal upper-branch coefficients obey the same leading law.
- **(6.10)** [negative]. Both graphs, isolated lower branch with $\varepsilon_k\gg c_0$ and $\varepsilon_k^2\gg\tau^2(1+r^2)$. For $L\gg\ell\ge1$: $\rho_{k,-}=\rho^{\rm e}_k+2\varepsilon_kR\,[1+O((c_0^2+\tau^2(1+r^2))/\varepsilon_k^2)]$, with $\rho^{\rm e}_k=O(1+r^2)$; this estimate is not uniform at finite $L$ (uniform mode, Step 8). In the heavy regime (additionally $\varepsilon_kR\gg1+r^2$ and $\tau^2\ll\varepsilon_kc_0$), for every $L$ and $\ell$: $I^{\rm mark}_{k,-}=I^{(1)}_{k,-}Y_k$ and $\rho_{k,-}=2\varepsilon_kR\gg1$ at leading order. So no single $v$ exists at second order, at any $r$ ($r=1$ included) and at any finite $\ell$ and $L$. At leading order $\rho_{k,-}\propto\varepsilon_k$, so levels whose $\varepsilon_k$ differ by more than the relative remainder have different $\rho_{k,-}$.
- **(6.11)** [criterion]. Both graphs, every $L$, $\ell$, $r$, isolated states, at second order in $\delta F$ with $F_1$ and $F_2$ counted alike, with the weights (2.8): $E''\Delta W=\frac{\tau^2F_1\Delta\phi}{F_0}[1-F_2w_0/F_0-2\sigma M_kF_2G_0]$. At this order it is the same on $\mathcal S$ (with at least two states) iff $F_2=0$, independently of $r$. For $F_2=0$ the first dependence on the level comes from $I^{\rm mark}$ at relative order $F_1^2$, given by (6.12) with $F_2=0$. On the ring, at leading order in $1/\ell$, it vanishes only at $r=\sqrt2$. On the cubic torus it differs between any two isolated lower levels at the computed points (6.15).
- **(6.12)** [criterion]. Grips of fixed shape through relative order $\beta^2$, weights (2.8), both graphs, every $L$ and $\ell$, isolated states: $E''\Delta W=\frac{\tau^2F_1\Delta\phi}{F_0}[1-F_2w_0/F_0-2\sigma M_kF_2G_0-I^{\rm mark}/I^{(1)}]$.
  - Ring ($L\gg\ell$, leading order in $1/\ell$), $\mathcal S$ with at least two levels: the same on $\mathcal S$ iff $(2-r^2)F_1^2=4F_0F_2$. At $r=1$ this holds iff the grip is the square grip; for the linear grip it holds only at $r=\sqrt2$.
  - Heavy regime of (6.10), lower branches, at leading order: the bracket is a polynomial of degree two in $M_k$ with leading coefficient $-F_1^2G_0R_1\ne0$. No grip of (G2) makes it the same for three levels; for two levels $a,b$ exactly one grip does, $F_0F_2/F_1^2=R_1(\varepsilon_a+\varepsilon_b)/2$.
- **(6.13)** [formula]. Both graphs, every $L$ and $\ell$, isolated states. $\rho_{k,\sigma}$ is the mean of $\rho_1,\rho_2$ with weights $S_1$, $\xi c_0S_2$, and $\rho_{k,\sigma}=1$ iff $S_1(\rho_1-1)+\xi c_0S_2(\rho_2-1)=0$. For $S_1,S_2>0$ a grip of (G2) achieves this iff $\rho_1=1$ (with $\xi=0$, the linear grip; every $\xi$ if also $\rho_2=1$) or $(\rho_1-1)(\rho_2-1)<0$ (exactly one $\xi=\xi^*_{k,\sigma}>0$). If $\rho_2=1\ne\rho_1$ no grip does. The bracket of (6.12) is $1-F_2w_0/F_0-\sigma(\varepsilon_kF_1^2G_0/F_0^2)(2y-t_{k,\sigma})$.
- **(6.14)** [negative]. Cubic torus, every grip of (G2), isolated states. A single $v$ on $\mathcal S$ requires $\rho_{k,\sigma}=1$ on $\mathcal S$ with one grip, and then $v=\tau$ (exact).
  - Negative for the class, by members with margins at $r=1$: a level at $\varepsilon_k=4c_0$ has $\rho_{k,-}\ge1.40$ for every grip ($\ell=8$ and $32$, both parts). Two levels at $1.2\,c_0$ and $2\,c_0$ need different $\xi^*$. A light level at the points of Part F has $\rho<1$ for both branches and every grip.
  - Numerical finding at the computed points (grid and refinement; $L=8\ell$, $\ell=8$ and $32$, $r\in[0.1,3]$, $\varepsilon_k/c_0\in[0.1,64]$ with lower gap $>0.2\,m_h$, where $\mathcal S$ has only lower branches): one $v$ exists exactly when $\mathcal S$ is the lower branch of a single level, with $\varepsilon_k$ in the window of Step 10 (c) and $\xi=\xi^*_{k,-}$. The window ends are roots of $\rho_{1,-}=1$ and $\rho_{2,-}=1$, found to within $2\times10^{-10}$. At $r=1$, $\ell=8$ the window is $[0.9123,2.6119)\,c_0$; its lower end is attained by the linear grip, its upper end is not. This requires $r>1$. For the computed $r\le1$ ($r=1$ included) no member with its levels on the grid has one $v$.
  - The first difference is $(\rho_{k,\sigma}-1)C^{\rm mark}/\tau^2$.
- **(6.15)** [criterion]. Cubic torus, isolated states without light levels, relative order $\beta^2$, weights (2.8). $E''\Delta W$ is the same on $\mathcal S$ iff $\sigma\varepsilon_k(2y-t_{k,\sigma})$ is (exact); for lower branches, two levels need $y=y_{ab}$ and three need $(\varepsilon_k,\varphi(\varepsilon_k))$ collinear.
  - Negative for the class, by a member with margins: at $r=1$ the levels $(1,2,3)\,c_0$ give $y_{12}=0.563\ne y_{23}=0.804$ ($0.561\ne0.798$ at $\ell=32$), so no grip works.
  - Numerical finding at the grid points of (6.14): with two lower levels exactly one grip works, $y=y_{ab}$, depending on the levels and on $r$. With three or more lower levels where $\varphi$ is discretely convex (all grid points for $r\le1$) no grip works. The linear grip fails for any two lower levels, and the square grip is not singled out.

## Consistency checks

1. Held limit $\tau\to0$: $\theta=0$ and $\omega=\varepsilon_k$, so $E^{(2)}(0)=-M_k^2(F_1^2G_0/2+2F_2^2J_2)$. This is the second-order part of $C^{\rm mark}$ in (2.6), with $\mu^2=M_k^2$, $D_\mu=1+O(\delta F)$ and $\zeta=4F_2^2J_2/G_0$.
2. Bare body ($\delta F=0$, $F_0=1$): $E''(0)=\sigma\tau^2/M_k$, which is $\omega_k''(0)$ of (6.7) of 04-ilang-time/06-inertia-internal-energy (mirrored for $\sigma=-1$).
3. Dimensions:
   - $[\tau]=[c_0]=[m_h]=[M_k]=$ cost, $[q^2]=$ cost$^{-1}$, $[F_1^2]=[F_2^2/c_0]=$ cost, $[G_0]=$ cost$^{-2}$;
   - $C^{\rm mark}$ is a cost, $I$ and $C/\tau^2$ are cost$^{-1}$, $r$ is dimensionless;
   - $E''\Delta W$ and $\tau^2F_1\Delta\phi/F_0$ are cost$^2$.

## Open issues

- Cubic torus: the findings of (6.14) and (6.15) hold at the grid points of Step 10 ($L=64$, $\ell=8$ and $L=256$, $\ell=32$; $r\in[0.1,3]$; $\varepsilon_k/c_0\in[0.1,64]$; lower branch at least $0.2\,m_h$ above its threshold; light levels at the points of Part F). Nothing is claimed between grid points, for $r>3$, closer to the lower threshold, or in the limit $\ell\to\infty$. Near the threshold the values still change with $\ell$.
- Light levels: item 3 is not evaluated for descriptions with a light level, and item 2 only at the points of Part F ($\ell=8$).
- Finite size near threshold: the control of the dressed sums in $L$ is not uniform as the isolation gap $\Delta_k\to0$ (Setup, Limits). Lower branches with $\Delta_k\le0.2\,m_h$ are not evaluated. At the light level nearest its boundary ($\Delta_k=0.022\,m_h$), the torus values at $L=128$ and $256$ still differ by $0.067$.
- Upper branch outside $\mathcal S$: only its formal coefficients are given. How the embedded level behaves beyond second order (its decay into the continuum) is not computed.
- The weight of the moving body is taken from (2.8) without corrections in $\tau$; (6.11) and (6.12) rest on this.
- On the ring the expansion parameter grows like $\ell$, so (6.8)–(6.12) need $\delta F$ to shrink as $\ell$ grows. The regime $M_kF_1^2G_0\sim F_0$ is open.

## Code

- `code/v6/check_mark_inertia.py` (role: check). It is the v1 script, unchanged, run again; its output `code/v6/output/check_mark_inertia.txt` is identical to that of v1–v5.
- **Part A** (the check required by the question).
  - *Setting.* Ring of $L=32$ cells, $c_0=1$, $m_h=0.25$, $F_0=1$. A square grip ($F_1=0.3$, $F_2=0.0225$) and another grip ($F_1=0.3$, $F_2=-0.15$); levels $M_k=3,4$; both branches; $r=0.5,1,1.5$.
  - *Independent computation.* Exact diagonalisation of the total contract restricted to states with at most two field quanta, in the sector $P$ (dimension 1122). $E''(0)$ is the exact second derivative of the eigenvalue, from the eigen-decomposition. The coefficient of $s^2$ is fitted from $\delta F\to s\,\delta F$ with $s=\pm0.01,\pm0.02,\pm0.04$.
  - *Formula side.* (6.2), (6.3), (6.5) are evaluated as lattice sums.
  - *Results.* The second-order coefficients of $\mathrm{Cost}_{k,\sigma}$ and $E''_{k,\sigma}(0)$ agree to a maximal relative deviation of $2.26\times10^{-7}$. The first-order coefficients agree to $\le2.6\times10^{-8}$ absolute. The zeroth order reproduces $\sigma F_0M_k$ and $\sigma\tau^2/F_0M_k$.
  - At these parameters ($\varepsilon_k\ge3>m_h/2$) the upper branch is not isolated (6.4). Its rows are formal coefficients, which the restricted diagonalisation reproduces as well; the lower-branch rows are those of isolated states.
- **Part B.** Formula (6.3) on rings with $L=8\ell$, $\ell=16,\dots,512$. It reports $\rho$ of the $F_1^2$ and $F_2^2$ parts against $2-r^2$, the same values without the term (iv), the held sums against (6.7), and $C^{\rm mark}(\tau)/C^{\rm mark}(0)$; the $\sigma=+1$ rows are formal coefficients. These are the numbers quoted in Steps 6 and 7.
- `code/v6/torus_mark_inertia.py` (role: computation). It is the v4 script, unchanged, run again; its output `code/v6/output/torus_mark_inertia.txt` is identical to that of v4 and v5.
  - It evaluates the lattice sums of (6.3) for the $F_1^2$ and $F_2^2$ parts separately (units $c_0=F_0=1$, $\varepsilon_k=M_k$), at isolated states only, except in Part E.
  - Two-quantum sums: the one-quantum energies of each $q_1$ are binned with linear weights (step $10^{-3}$). The pair distribution per total $q_1$ is their convolution, computed by FFT, and the binning variance is removed to second order.
  - $S_i''(0)$: analytic second derivative in $P$ of each summand (of $\cos^2\vartheta$, $\sin^2\vartheta$ and the denominators, through $\omega'$, $\omega''$, $\theta'$, $\theta''$), summed with the same weights.
  - Crossings $\rho_{i,-}=1$: Brent's method in each sign-changing grid interval; error estimate from the residual, the slope, and for $\rho_2$ a lattice with energy step $5\times10^{-4}$. The windows are built from the roots and checked against all computed points.
  - Thresholds $\varepsilon_\pm(r)$ from $b_k$ at $L=\infty$.
  - *Part A:* mode-class and pair sums against brute-force sums on $L=12$.
  - *Parts B, B2:* convergence in $L$ and $\ell$, both parts, up to $(L,\ell)=(256,32)$.
  - *Part C:* table of $S_i(0)$ relative to the held value, $\rho_1$, $\rho_2$, $t$ and $\xi^*$ on $L=64$, $\ell=8$.
  - *Part D:* fine grid in $\varepsilon_k$ for nine values of $r$ on $L=64$, $\ell=8$. It gives the thresholds, monotonicity of $\rho_{i,-}$ at the grid points, the crossings $\rho_{i,-}=1$, the points with a finite $\xi^*_{k,-}$, monotonicity and discrete convexity of $\varphi$, $y_{ab}$, and the members with margins.
  - *Part D2:* Part D on $L=256$, $\ell=32$, both parts.
  - *Part E:* exact diagonalisation of the contract restricted to at most two field quanta on the torus $L=4$ (dimension 4290), against (6.2), (6.3) for both branches ($\sigma=+1$ formal).
  - *Part F:* light levels on the ring ($L=2048$, $4096$) and the torus ($L=128$, $256$), $\ell=8$.
  - *Part G:* at the light levels with $f=0.9$ and the lower-branch grid points nearest threshold, the analytic $\rho_i$ against Richardson differences with steps $h_0=\min(10^{-2},0.05\,\varepsilon_k/\tau)$, $h_0/2$, $h_0/4$, and $L=64$ against $L=128$ at $\ell=8$.
  These are the numbers quoted in Steps 3 and 10.

## Methods used

- Rayleigh–Schrödinger perturbation theory for an isolated eigenvalue in a sector of a conserved label; analyticity of isolated eigenvalues
- Fourier analysis on ring and torus; normal modes and Wick's theorem for the oscillator field
- Asymptotics of lattice sums dominated by a scaling region; rapidity substitution; Poisson summation
- Exact diagonalisation in a truncated Fock space; second derivative of an eigenvalue from its eigen-decomposition (code check)
- Spectrum of a sector: continuum edges from the subadditivity of the field dispersion; isolation by a uniform gap
- Numerical lattice sums on the cubic torus: FFT convolution of linearly binned energies with a variance correction, analytic differentiation of the summands in $P$ (checked against Richardson finite differences), Brent root-finding, convergence in $L$ and $\ell$; monotonicity and convexity at grid points
- Finite-size errors from the analyticity strip of the summands (Poisson summation), including the zeros of the energy denominators

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