Version 2 · earlier version; the current one is v6 · External review, round 1: major errors

# 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:** v2
- **Mode:** internal regeneration
- **Date:** 2026-10-25

## Changes from previous version

- New Step 10 and results (6.13)–(6.15): on the cubic torus, items 2 and 3 are now settled outside the large-$\varepsilon_k$ regime of (6.10), from the lattice sums of (6.3), which are evaluated numerically (`code/v2/torus_mark_inertia.py`); the size of $\varepsilon_kR$ where (U) holds is given (review I1).
- Step 8, "$D=3$ otherwise", and Step 9(b): now point to Step 10 (review I1).
- Step 9(a) and Result (6.11): for the linear grip, the first dependence on the level comes from $I^{\rm mark}$ at the next order, via (6.12) (review I2).
- Open issues: the item on the cubic torus is replaced by the limits of the numerical evaluation. Code: the v1 check is re-run unchanged in `code/v2/`, and the numbers of Steps 6 and 7 are cited from there.

## 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 $L\gg\ell$. The lattice sums then differ from their values at $L=\infty$ by terms exponentially small in $L/\ell$ (Poisson summation: the summands are analytic in a strip of width of order $1/\ell$).
  - 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 the isolated level (Step 3) then gives (6.2) and (6.3). 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

**Branch $\sigma=-1$.**
- *Sufficient condition.* Since $2(1-\cos q_1)\ge\sin^2q_1$, $\Omega_q^2\ge m_h^2+\kappa_h\sin^2q_1$. If $\tau^2\le2\varepsilon_kc_0$, then $\omega(q_1)-\varepsilon_k=\tau^2\sin^2q_1/(\omega+\varepsilon_k)\le\tau^2\lvert\sin q_1\rvert/2\varepsilon_k\le c_0\lvert\sin q_1\rvert<\Omega_q$, because $m_h>0$.
- *Several quanta.* $\Omega$ 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 $Q$ has $E_n\ge\Omega_Q$.
- *Conclusion.* At $\delta F=0$, $-\varepsilon_k$ is the lowest level of the sector $(k,P=0)$, separated by a gap. It stays isolated for $P$ near 0 and small $\delta F$, and is therefore analytic in $P$ and $\delta F$ (isolated eigenvalue of an analytic family).

**Branch $\sigma=+1$.**
- *Own-branch denominators.* They exceed $m_h-[\omega(P)-\varepsilon_k]>0$ for $P$ near 0.
- *Other-branch denominators.* They are $E_n-\omega(P-Q_n)-\omega(P)$. Condition (U): the continuous functions $q\mapsto\Omega_q-\varepsilon_k-\omega(q_1)$ 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. Under (U) these denominators are nonzero near $P=0$, uniformly in $L$. A sufficient condition is $\Omega_{\max}<\varepsilon_k$, or $\Omega_{\max}<2\varepsilon_k$ if $F_2=0$.
- *If one of these functions changes sign.* For $L\to\infty$ the level then lies inside a one- or two-quantum continuum of the other branch. It couples to that continuum with weight $\sin^2\vartheta$, which is nonzero on an open set ($\tau>0$), so it is not isolated at second order.
- *Several quanta.* States with three or more quanta of the other branch may be degenerate with the level; they enter at $O(\delta F^4)$.
$$
\sigma=-1:\ \ \Omega_q>\omega_k(q_1)-\varepsilon_k\ \forall q\ \ (\Leftarrow\ \tau^2\le2\varepsilon_kc_0);\qquad\sigma=+1:\ \ \text{(U)}\ \ (\Leftarrow\ \Omega_{\max}<\varepsilon_k,\ \text{or }\Omega_{\max}<2\varepsilon_k\text{ if }F_2=0).
\tag{6.4}
$$

### 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 level and both branches therefore exists iff $\rho_{k,\sigma}=1$ for all $k,\sigma$. 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 $D=1$:
$$
c_0^2\langle\!\langle Q^2/E^3\rangle\!\rangle=\tfrac12\langle\!\langle1/E\rangle\!\rangle\,[1+O(\ell^{-1})],\qquad\langle\!\langle1/E\rangle\!\rangle=\frac{F_1^2G_0}2+2F_2^2J_2=\frac{\ell}{4c_0^2}\Bigl(F_1^2+\frac{F_2^2}{2c_0}\Bigr)[1+O(\ell^{-1}\ln\ell)] .
\tag{6.7}
$$

**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
$$
C^{\rm mark}_{k,\sigma}=-\frac{M_k^2\ell}{4c_0^2}\Bigl(F_1^2+\frac{F_2^2}{2c_0}\Bigr)[1+O(\delta_\ell)],\qquad\zeta=\frac{F_2^2}{2c_0}[1+O(\delta_\ell)],\qquad E^{(2)\prime\prime}_{k,\sigma}(0)=-\frac{2\tau^2}{\varepsilon_k^2}\,C^{\rm mark}_{k,\sigma}\Bigl[1-\frac{r^2}2-\sigma\varepsilon_kR+O(\delta_\ell)\Bigr] .
\tag{6.8}
$$
Code check (`code/v2/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 $\ell\to\infty$.
$$
\rho_{k,\sigma}=2-r^2-2\sigma\varepsilon_kR+O(\delta_\ell),\qquad\text{at leading order in }1/\ell:\ \ I^{\rm mark}_{k,\sigma}=\frac{(2-r^2)\,C^{\rm mark}_{k,\sigma}}{\tau^2}\ \Rightarrow\ \bigl[\exists\,v:\ I^{(1)}=C^{(1)}/v^2,\ I^{\rm mark}=C^{\rm mark}/v^2\ \forall k,\sigma\bigr]\iff r=1,\ v=\tau=c_0 .
\tag{6.9}
$$
- **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.
- **Next order in $1/\ell$.** The term $-2\sigma\varepsilon_kR$ (from (iv)) is odd in $\sigma$ and grows with the level. By (6.10) it makes the two branches differ at every finite $\ell$ when $\varepsilon_k\gg c_0,\ \tau\sqrt{1+r^2}$.
- **Code check.** At $\ell=512$ and $r=1$, the $F_1^2$ part gives $\rho=0.962$ and $1.056$ ($M_k=3$, $\sigma=\pm1$). With the term (iv) removed, the values are $1.00920$ and $1.00919$ for both levels and both branches; for the $F_2^2$ part they are $1.021$–$1.023$. For $r=0.5$ and $1.5$ the values approach $2-r^2$ in the same way (part B).
- **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).

### Step 8. Branch asymmetry and heavy body: controls 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, by $c_0^2/\varepsilon^2$, since $\langle\!\langle1\rangle\!\rangle=O(c_0^2\langle\!\langle\sin^2Q/E^2\rangle\!\rangle)$;
- 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 $O(1+r^2)$: (i) gives at most 2, and (iii) gives $r^2$ times a lattice ratio of order 1, using $\omega'(Q)=\tau^2\sin2Q_1/2\varepsilon$.

**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. With (6.6) this gives the second part of (6.10). Both parts hold on both graphs, for every $L$ and $\ell$, under (6.4).
$$
\rho_{k,+}-\rho_{k,-}=-4\varepsilon_kR\Bigl[1+O\Bigl(\frac{c_0^2+\tau^2(1+r^2)}{\varepsilon_k^2}\Bigr)\Bigr];\qquad\text{heavy: }\ I^{\rm mark}_{k,\sigma}=I^{(1)}_{k,\sigma}Y_k\Bigl[1+O\Bigl(\frac{1+r^2}{\varepsilon_kR},\frac{\tau^2}{\varepsilon_kc_0}\Bigr)\Bigr],\ \ Y_k:=2M_k^2\langle\!\langle\sin^2Q_1/E^2\rangle\!\rangle>0 .
\tag{6.10}
$$
**Consequences.**
- *Branch asymmetry.* $R>0$. For $\varepsilon_k\gg c_0$ and $\varepsilon_k^2\gg\tau^2(1+r^2)$ the two branches therefore have different $\rho$, so $\rho_{k,\sigma}=1$ cannot hold for both. At second order no single $v$ exists then, at any finite $\ell$ and $L$ and at any $r$ ($r=1$ included). On the ring this difference vanishes as $\ell\to\infty$, consistently with (6.9).
- *Heavy regime.* $\rho_{k,\sigma}=-2\sigma\varepsilon_kR$ at leading order, so $\lvert\rho\rvert\gg1$, and $\rho$ is proportional to $\varepsilon_k$, so it depends on the level.

**Size of $R$.** 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 $\varepsilon_k\gg c_0$, up to a lattice number. On the ring it requires $\varepsilon_k\gg c_0\ell/\ln\ell$.

**$D=3$ 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, which also gives the size of $\varepsilon_kR$ where (U) holds.

### 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 does not vanish once two lower levels are isolated, at any computed $r$ (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 the same for every level and both branches iff $(2-r^2)F_1^2=4F_0F_2$. 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 $r=\sqrt2$.
- **$\varepsilon_k\gg c_0,\ \tau\sqrt{1+r^2}$ (both graphs, every finite $\ell$).** By (6.10), the part of $-I^{\rm mark}/I^{(1)}$ even in $\sigma$ is $-M_k^2G_0F_1^2R\,[1+O((c_0^2+\tau^2(1+r^2))/\varepsilon_k^2)]$, with $R$ of the $F_1^2$ part. All $F_2$ terms are odd in $\sigma$ or independent of the level. Take two levels whose values of $M_k^2$ differ by a relative amount larger than this error. Then no grip of (G2) makes the bracket the same for every level and both branches; this includes the heavy regime. 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.** 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):
$$
\rho_{k,\sigma}=\frac{S_1\rho_1+\xi c_0S_2\rho_2}{S_1+\xi c_0S_2},\qquad\rho_{k,\sigma}=1\iff\xi=\xi^*_{k,\sigma}:=\frac{S_1(1-\rho_1)}{c_0S_2(\rho_2-1)}\ge0,\qquad[\,\cdot\,]_{(6.12)}=1-\frac{F_2w_0}{F_0}-\sigma\frac{\varepsilon_kF_1^2G_0}{F_0^2}\,(2y-t_{k,\sigma}).
\tag{6.13}
$$
- Where $S_1,S_2>0$ (all computed points: each is at least $0.55$ times its held value, Part D), $\rho$ is a mean of $\rho_1$ and $\rho_2$ with positive weights. Hence some grip gives $\rho_{k,\sigma}=1$ iff 1 lies between $\rho_1$ and $\rho_2$.
- 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 quantity $\min_q[\Omega_q-\omega_k(q_1)+\varepsilon_k]$ of (6.4) increases with $\varepsilon_k$, since $\omega_k(q_1)-\varepsilon_k=\tau^2\sin^2q_1/(\omega_k+\varepsilon_k)$. In (U), the maxima of $\Omega_q-\omega_k(q_1)$ and of $\Omega_q+\Omega_{q'}-\omega_k(q_1+q_1')$ lie at $q=q'=(\pi,\pi,\pi)$, where $\omega_k=\varepsilon_k$. For $2\varepsilon_k>m_h$ the functions of (U) are negative at $q=0$. So, as $L\to\infty$, (U) holds iff $\Omega_{\max}<2\varepsilon_k$ for $F_2=0$, and iff $\Omega_{\max}<\varepsilon_k$ otherwise, with $\Omega_{\max}\approx2\sqrt3\,c_0$. The set $\mathcal S$ of isolated states $(k,\sigma)$ is therefore made of the lower branches of all levels above one threshold, together with the upper branches of all levels above another.

**Evaluation** (`code/v2/output/torus_mark_inertia.txt`).
- *Setting.* Units $c_0=F_0=1$. Torus $L=64$, $\ell=8$. Values $r\in\{0.1,0.25,0.5,0.75,1,1.25,1.5,2,3\}$ and 71 values of $\varepsilon_k\in[0.1,64]$. Lower branches are taken where their gap exceeds $m_h/2$, upper branches where (U) holds with a margin of $0.1$.
- *Method.* $S_1$ is an exact mode sum. $S_2$ is summed from histograms of pair energies, with the exact moments in each bin; this agrees with brute-force pair sums on $L=12$ to $4\times10^{-10}$ (Part A). Exact diagonalisation with at most two quanta on the torus $L=4$ reproduces the second-order coefficients of $\mathrm{Cost}$ and $E''(0)$ from (6.2), (6.3) to $9\times10^{-9}$, for both branches (Part E).
- *Convergence in $L$* (Part B). At $\ell=8$, $L=48$ and $L=64$ give the same $\rho_i$ to $6\times10^{-4}$.
- *Convergence in $\ell$* (Part B2), over the points of Part C. From $\ell=8$ to $\ell=32$ ($L=256$), $\rho_1$ and $t$ change by at most $0.093$ and $0.058$ for $\varepsilon_k\le4c_0$, and by at most $3.4\%$ above. From $\ell=6$ to $\ell=8$, $\rho_2$ changes by at most $0.008$ and $1.2\%$ respectively.
- *Check at $\ell=32$* (Part D2). The tests of Part D on the $F_1^2$ part are repeated at $\ell=32$, $L=256$, with the same outcomes.

**Findings** (Parts C, D, D2).
- (a) *Upper branch:* $\rho_{1,+}\le0.61$ and $\rho_{2,+}\le0.44$ (at $\ell=32$, $\rho_{1,+}\le0.63$).
- (b) *Lower branch.*
  - $\rho_{1,-}$ and $\rho_{2,-}$ increase strictly with $\varepsilon_k$ at every $r$. At $r=1$ they go from $0.28$ and $0.06$ at $\varepsilon_k=0.1\,c_0$ to $30.6$ and $16.8$ at $64\,c_0$.
  - Hence $\xi^*_{k,-}\ge0$ only in a window of $\varepsilon_k$: $[0.99,2.66]\,c_0$ at $r=0.1$, $[0.93,2.61]\,c_0$ at $r=1$, $[1.34,2.39]\,c_0$ at $r=2$, and $[3.34,3.56]\,c_0$ at $r=3$, whose lower end is the edge of the computed range.
  - Across the window $\xi^*_{k,-}$ is strictly monotone: increasing for $r\le2$, decreasing at $r=3$.
  - At $\ell=32$ the lower end of the window moves by at most $0.11\,c_0$ for $r\le2$, and by $0.3\,c_0$ at $r=3$.
- (c) *Both branches of one level:* for every grip, $\rho_{k,-}-\rho_{k,+}\ge0.85$ for $r\le2$, and $\ge0.44$ at $r=3$. For $r\le2$ the ratio $(\rho_{1,-}-\rho_{1,+})/(4\varepsilon_kR_1)$ is $0.54$–$0.74$ at $\varepsilon_k=1.92\,c_0$ (edge of (U)), and $0.9997$ at $64\,c_0$, as (6.10) predicts.
- (d) *Size of $\varepsilon_kR$:* $R_1=0.232/c_0$ and $R_2=0.127/c_0$ ($R_1=0.229/c_0$ at $\ell=32$).
  - At the edge of (U), $\varepsilon_kR_1=0.40$ for $F_2=0$ ($\varepsilon_k=\Omega_{\max}/2$). For $F_2\ne0$ ($\varepsilon_k=\Omega_{\max}$), $\varepsilon_kR_1=0.80$ and $\varepsilon_kR_2=0.44$.
  - Where (U) begins, $\varepsilon_kR<1$, so the heavy regime of (6.10) is not reached there, and the corrections to the first part of (6.10) are of order one.
- (e) *Item 3 functions.*
  - $\varphi(\varepsilon):=\varepsilon\,t_{-}(\varepsilon)$ increases strictly. It is strictly convex for $r\le1.25$. For $r\ge1.5$ it is strictly convex except in a band just above the edge of isolation of the lower branch: $\varepsilon\in[0.48,0.63]$ at $r=1.5$, $[1.33,1.75]$ at $r=2$, and $[3.66,4.83]$ at $r=3$ (narrower at $\ell=32$).
  - $\psi(\varepsilon):=\varepsilon(t_--t_+)$ increases strictly.
  - $y_{\rm b}:=(t_++t_-)/4$ lies in $[0.22,0.28]$ for $r\le2$, and changes sign at $r=3$.
- (f) *Dependence on $r$* (Part D). For $\varepsilon_k\ge2c_0$, $\rho_{1,-}$ and $\rho_{2,-}$ change by at most $0.08$ between $r=0.25$ and $r=2$; $\rho_{1,+}$ changes by up to $0.32$, near the edge of (U). Without a scaling region there is no counterpart of the term $-r^2$ of (6.9).

**Item 2.** By (6.6), a single $v$ must be $\tau$, and it needs $\rho_{k,\sigma}=1$ on all of $\mathcal S$.
- By (a), no grip gives this on an isolated upper branch.
- By (b), two isolated lower levels would need two different values of $\xi^*$.

Hence, for nonempty $\mathcal S$:
$$
D=3:\ \ \bigl[\exists\,v:\ I^{(1)}=C^{(1)}/v^2,\ I^{\rm mark}=C^{\rm mark}/v^2\ \text{on }\mathcal S\bigr]\iff\mathcal S=\{(k,-)\},\ \varepsilon_k\ \text{in the window of (b)},\ \xi=\xi^*_{k,-};\ \ \text{then }v=\tau .
\tag{6.14}
$$
- In particular, once two states are isolated, no grip and no computed $r$ ($r=1$ included) give one $v$.
- The first difference is $I^{\rm mark}-C^{\rm mark}/\tau^2=(\rho_{k,\sigma}-1)\,C^{\rm mark}/\tau^2$. For example, at $r=1$ and $\varepsilon_k=4c_0$: $\rho_{1,-}=2.62$, $\rho_{2,-}=1.41$, $\rho_{1,+}=-0.83$, $\rho_{2,+}=0.16$.
- Order of limits: $L\gg\ell$, then $\ell\to\infty$. Unlike on the ring, the sums have finite limits.

**Item 3.** By (6.13), the bracket of (6.12) is the same on $\mathcal S$ iff $\sigma\varepsilon_k(2y-t_{k,\sigma})$ is the same for all $(k,\sigma)\in\mathcal S$.
- *Both branches of a level $a$.* Adding the two conditions gives $y=y_{\rm b}(\varepsilon_a)$, and the common value is then $\psi(\varepsilon_a)/2$. For two such levels this needs $\psi(\varepsilon_a)=\psi(\varepsilon_b)$, which (e) excludes.
- *Lower branches of levels $a\ne b$.*
  - They give $y=y_{ab}:=[\varphi(\varepsilon_b)-\varphi(\varepsilon_a)]/[2(\varepsilon_b-\varepsilon_a)]$, which depends on the levels.
  - At $r=1$, $y_{ab}=0.25,0.37,0.56,0.92,1.62$ for $(\varepsilon_a,\varepsilon_b)/c_0=(0.25,0.5),(0.5,1),(1,2),(2,4),(4,8)$; at $\ell=32$ the values are $0.26,0.37,0.56,0.91,1.60$.
  - Three lower levels would have to lie on one straight line on the graph of $\varphi$, which strict convexity excludes.
- *Linear grip* ($y=0$).
  - Since $\varphi$ is increasing, $y_{ab}>0$, so the linear grip fails for any two lower levels.
  - Since $y_{\rm b}>0$ for $r\le2$, it also fails for both branches of one level.
$$
D=3:\ \ E''_{k,\sigma}(0)\,\Delta W_{k,\sigma}\ \text{the same on }\mathcal S\iff\sigma\varepsilon_k\,(2y-t_{k,\sigma})\ \text{the same on }\mathcal S;\ \ \text{no grip of (G2) if }\mathcal S\ \text{has both branches of two levels, or three lower levels where }\varphi\ \text{is strictly convex} .
\tag{6.15}
$$
If $\mathcal S$ has two states, exactly one value of $y$ works, and it depends on the levels and on $r$. The square grip, which gives (6.12) on the ring at $r=1$, is not singled out here.

## Result

- **(6.2), (6.3)** [formula]. Class: ring and cubic torus, every $L$, $M\to\infty$ first, every level and both branches under (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 $P$.
- **(6.4)** [criterion]. $\sigma=-1$: under $\Omega_q>\omega_k(q_1)-\varepsilon_k$ for all $q$ (implied by $\tau^2\le2\varepsilon_kc_0$), the level is isolated and analytic in $P$ and $\delta F$. $\sigma=+1$: under (U) (implied by $\Omega_{\max}<\varepsilon_k$, or by $\Omega_{\max}<2\varepsilon_k$ for $F_2=0$), the second-order coefficients are analytic in $P$ near 0. If a function of (U) changes sign, the level is not isolated at second order as $L\to\infty$.
- **(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 level, both branches and every $r$; $I^{\rm mark}=-\varepsilon_k^2E^{(2)\prime\prime}(0)/\tau^4$. A single $v$ exists iff $\rho_{k,\sigma}=1$ for all $k,\sigma$, and then $v=\tau$.
- **(6.7)** [formula]. Ring, $L\gg\ell$: $c_0^2\langle\!\langle Q^2/E^3\rangle\!\rangle=\frac12\langle\!\langle1/E\rangle\!\rangle[1+O(1/\ell)]$ 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), 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 level and both branches under (6.4), every $r>0$. $\rho_{k,\sigma}=2-r^2-2\sigma\varepsilon_kR+O(\delta_\ell)$, with $\varepsilon_kR=O((\varepsilon_k/c_0)\ln^2\ell/\ell)$. At leading order in $1/\ell$, $I^{\rm mark}=(2-r^2)C^{\rm mark}/\tau^2$. Hence one $v$ with $I^{(1)}=C^{(1)}/v^2$ and $I^{\rm mark}=C^{\rm mark}/v^2$ for every level and both branches exists iff $r=1$, and then $v=\tau=c_0$. The first difference is $(1-r^2)C^{\rm mark}/\tau^2$.
- **(6.10)** [negative]. Both graphs, every $L$ and $\ell$, under (6.4).
  - $\rho_{k,+}-\rho_{k,-}=-4\varepsilon_kR\,[1+O((c_0^2+\tau^2(1+r^2))/\varepsilon_k^2)]$. For $\varepsilon_k\gg c_0$ and $\varepsilon_k^2\gg\tau^2(1+r^2)$, no single $v$ exists at second order, at any $r$ ($r=1$ included) and at any finite $\ell$.
  - In the heavy regime (additionally $\varepsilon_kR\gg1+r^2$ and $\tau^2\ll\varepsilon_kc_0$): $I^{\rm mark}=I^{(1)}Y_k$ and $\rho_{k,\sigma}=-2\sigma\varepsilon_kR$ at leading order.
- **(6.11)** [criterion]. Both graphs, every $L$, $\ell$, $r$, 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 for every level and both branches 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 does not vanish once two lower levels are isolated, at any computed $r$, by (6.15).
- **(6.12)** [criterion]. Grips of fixed shape through relative order $\beta^2$, weights (2.8), both graphs, every $L$ and $\ell$: $E''\Delta W=\frac{\tau^2F_1\Delta\phi}{F_0}[1-F_2w_0/F_0-2\sigma M_kF_2G_0-I^{\rm mark}/I^{(1)}]$.
  - On the ring ($L\gg\ell$, leading order in $1/\ell$) it is the same for every level and both branches 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 $r=\sqrt2$.
  - For $\varepsilon_k\gg c_0,\ \tau\sqrt{1+r^2}$, with levels whose $M_k^2$ differ by more than the relative error of (6.10), no grip of (G2) gives this at any finite $\ell$, on either graph.
- **(6.13)** [formula]. Both graphs, every $L$ and $\ell$, under (6.4). $\rho_{k,\sigma}$ is the mean of $\rho_1,\rho_2$ with weights $S_1$, $\xi c_0S_2$, and $\rho_{k,\sigma}=1$ iff $\xi=\xi^*_{k,\sigma}\ge0$. 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, $L=64$, $\ell=8$ (with the convergence check of Step 10), every grip of (G2), $r\in[0.1,3]$, $\varepsilon_k/c_0\in[0.1,64]$ on the computed grid. $\mathcal S$ is the nonempty set of isolated states: lower branches where the minimum of (6.4) exceeds $m_h/2$, upper branches where $\Omega_{\max}<2\varepsilon_k$ ($F_2=0$) or $\Omega_{\max}<\varepsilon_k$ ($F_2\neq0$).
  - One $v$ with $I^{(1)}=C^{(1)}/v^2$ and $I^{\rm mark}=C^{\rm mark}/v^2$ on all of $\mathcal S$ exists iff $\mathcal S$ is the lower branch of a single level, with $\varepsilon_k$ in a window and $\xi=\xi^*_{k,-}$; then $v=\tau$. The window is about $[0.93,2.61]\,c_0$ at $r=1$.
  - In particular, no grip and no computed $r$ ($r=1$ included) give one $v$ once two states are isolated: $\rho_{k,+}<1$ for every grip, and $\xi^*_{k,-}$ is strictly monotone in $\varepsilon_k$.
  - The first difference is $(\rho_{k,\sigma}-1)C^{\rm mark}/\tau^2$. Where (U) begins, $\varepsilon_kR\le0.80$ (Step 10 (d)).
- **(6.15)** [criterion]. Same setting as (6.14). $E''\Delta W$ (to relative order $\beta^2$, weights (2.8)) is the same on $\mathcal S$ iff $\sigma\varepsilon_k(2y-t_{k,\sigma})$ is the same on $\mathcal S$.
  - No grip of (G2) achieves this if $\mathcal S$ contains both branches of two levels, or three lower levels in the range where $\varphi$ is strictly convex (all computed $\varepsilon_k$ for $r\le1.25$).
  - If $\mathcal S$ has two states, exactly one value of $y$ works, and it depends on the levels and on $r$.
  - The linear grip fails for any two lower levels, and for both branches of one level when $r\le2$. 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: (6.14) and (6.15) rest on lattice sums evaluated on a grid ($r\in[0.1,3]$, $\varepsilon_k/c_0\in[0.1,64]$, $L=64$, $\ell=8$, with the convergence check of Step 10). Between the grid points they rest on the monotonicity and convexity found on the grid, and above $64c_0$ on (6.10). No points were computed in the band where the gap of the lower branch is below $m_h/2$, or for $r>3$.
- Branch $\sigma=+1$ beyond second order: lower-branch states with three or more quanta can be degenerate with the level. Whether the exact level is isolated as $L\to\infty$ is not settled.
- 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/v2/check_mark_inertia.py` (role: check). It is the v1 script, unchanged, run again; its output `code/v2/output/check_mark_inertia.txt` is identical to that of v1.
- **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 $\sigma\tau^2/F_0M_k$.
- **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 $C^{\rm mark}(\tau)/C^{\rm mark}(0)$. These are the numbers quoted in Steps 6 and 7.
- `code/v2/torus_mark_inertia.py` (role: computation). Output: `code/v2/output/torus_mark_inertia.txt`. It evaluates the lattice sums of (6.3) on the cubic torus, for the $F_1^2$ and $F_2^2$ parts separately (units $c_0=F_0=1$, $\varepsilon_k=M_k$). The two-quantum sums use histograms of pair energies per total $q_1$, with the exact zeroth to second moments in each bin.
  - *Part A:* mode-class and histogram sums against brute-force sums on $L=12$.
  - *Parts B, B2:* convergence in $L$ and $\ell$.
  - *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 $r$: monotonicity of $\rho_{i,-}$, window and monotonicity of $\xi^*_{k,-}$, bounds on $\rho_{i,+}$, monotonicity and convexity of $\varphi$, monotonicity of $\psi$, $y_{\rm b}$, $y_{ab}$, and the dependence on $r$.
  - *Part D2:* the $F_1^2$ tests of Part D repeated at $\ell=32$, $L=256$.
  - *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.
  These are the numbers quoted in Step 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)
- Numerical lattice sums on the cubic torus: pair-energy histograms with exact bin moments, Richardson-extrapolated finite differences in $P$, convergence in $L$ and $\ell$; monotonicity and convexity tests on a grid

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