Version 1 · earlier version; the current one is v6 · Review of v1

# 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:** v1
- **Mode:** new
- **Date:** 2026-10-25

## 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/v1/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 two regimes above, $\rho$ is given by the lattice sums of (6.3), which are not evaluated here.

### 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 $O(\delta F^3)$.

**(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.

## 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 the weights (2.8): $E''\Delta W=\frac{\tau^2F_1\Delta\phi}{F_0}[1-F_2w_0/F_0-2\sigma M_kF_2G_0]$. This is the same for every level and both branches iff $F_2=0$, independently of $r$.
- **(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.

## 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 when $\varepsilon_k$ is not large against $c_0$ and $\tau\sqrt{1+r^2}$: $\rho_{k,\sigma}$ is given by the lattice sums of (6.3) and is not evaluated. Whether a single $v$ exists there is open.
- 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/v1/check_mark_inertia.py` (role: check). Output: `code/v1/output/check_mark_inertia.txt`.
- **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.

## 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)

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