Version 2 · current

# Cost on the background, and the inertia of a body

- **Subproject:** 03-ilang-space
- **Package:** 03-ilang-space/20-cost-and-mass
- **Version:** v2
- **Mode:** external regeneration
- **Date:** 2026-10-09

## Changes from previous version

- **Setup and assumptions, last bullet.** v1 claimed that all sums over $m$ on the infinite chain converge absolutely because of normalization alone. That does not control sums weighted by $m$ or $m^2$ (verification I1). v2 restricts the infinite chain to the starts of finite support stipulated in the question. It derives the moment bound $\sum_mm^2p_m(\lambda)\le(\lVert M\phi\rVert+2t\lambda)^2$ from (17.17), and uses Cauchy–Schwarz for the fixed-separation sums with weights up to $m^2$. This justifies the termwise differentiations, rearrangements and index shifts used for the chain.
- **Step 6, (20.13) and the sentence before it.** "for every state" and "for every start" now read "for every start of finite support (and every $\lambda$)". The tag (20.13) and its content are unchanged on this domain.
- **Methods used.** One keyword added for the moment bound.
- No other step, formula, tag or result has changed.

## Response to verification

- **I1:** Accepted and fixed in "Setup and assumptions" (last bullet) and in Step 6 ((20.13) and the sentence before it). The infinite chain is now restricted to starts of finite support. The bound $\lVert MU_\xi\phi\rVert\le\lVert M\phi\rVert+2t\lambda$ suggested by the verifier is derived from (17.17), including $\xi=0$. It gives a finite second moment, uniformly on bounded $\lambda$-intervals, for every $\lambda$. Fixed-separation sums are controlled by Cauchy–Schwarz. The claims "for every state" and "for every start" are restricted to these starts. No result changes.

## Setup and assumptions

- **Item 1.** Any description satisfying A1–A7; $b$ is the only instance of its type $T$, $A=\{b\}$. Inputs (2.1), (2.2): $\lvert\Psi\rangle=\sum_h\lvert h\rangle\otimes\lvert\psi_h\rangle$, $(V_A)_{hh'}=\langle\psi_{h'}\vert\psi_h\rangle$, $p_h=\lVert\psi_h\rVert^2$. Notation: $Q_h:=\lvert h\rangle\langle h\rvert\otimes\mathbb 1_{\bar A}$; for a set $R$ of places $Q_R:=\sum_{h\in R}Q_h$, $\bar Q:=\mathbb 1-Q_R$, $p_R:=\sum_{h\in R}p_h$.
- **Items 2–4.** Body–medium setting of the question: $C_{hh'}=t_{hh'}\mathbb 1_c$ ($h\neq h'$), $C_{hh}=t_{hh}\mathbb 1_c+K_h$; the cost edges between different places are the hopping edges $E$. Background vectors $u_h=(K_h-\epsilon_h)\chi$ (5.10), (17.2); they are fixed by the start and do not depend on $\lambda$. Notation: $\hat K_h:=C_{hh}-\Omega_h\mathbb 1_c=K_h-\epsilon_h$, so that $u_h=\hat K_h\chi$ and $\langle\chi\vert\hat K_h\vert\chi\rangle=0$; $\Delta_{hh'}:=u_h-u_{h'}$.
- **Inputs from 17**, used as quoted: (17.1), (17.4), (17.11)–(17.12), (17.15), (17.19), (17.22), and from Step 9 of 17 the decomposition $P_j\to\Pi_\xi$, $D_\xi=\xi M$, $U_\xi=e^{-i(C_b+\xi M)\lambda}$, $U_\xi^\dagger SU_\xi=e^{-i\xi\lambda}S$.
- $C$ does not depend on $\lambda$; $\lambda\ge0$. Items 1, 2 are exact for every $\lambda$; item 3 is at $\lambda=0$. **Infinite chain (item 4).** As stipulated in the question, only starts $\phi$ of finite support are used, so $\lVert M\phi\rVert<\infty$. With the decomposition $\psi_m=\sum_\xi(U_\xi\phi)_m\Pi_\xi\chi$ of Step 6(c), $p_m=\sum_\xi w_\xi\lvert(U_\xi\phi)_m\rvert^2$ (finitely many $\xi$, $\sum_\xi w_\xi=1$). By (17.17), with $\lVert S\rVert=\lVert S^\dagger\rVert=1$ and $\lvert1-e^{\mp i\xi\lambda}\rvert\le\lvert\xi\rvert\lambda$, $\lVert MU_\xi\phi\rVert=\lVert U_\xi^\dagger MU_\xi\phi\rVert\le\lVert M\phi\rVert+2t\lambda$, also for $\xi=0$. Hence $\sum_mm^2p_m(\lambda)\le(\lVert M\phi\rVert+2t\lambda)^2$. This sum is continuous in $\lambda$ (by (17.17), $U_\xi^\dagger MU_\xi-M$ is bounded and continuous in $\lambda$), so by Dini's theorem the series converges uniformly on bounded $\lambda$-intervals. For bounded $Y$ on $\mathcal H_c$, $\lvert\langle\psi_m\vert Y\vert\psi_n\rangle\rvert\le\lVert Y\rVert\sqrt{p_mp_n}$. By Cauchy–Schwarz, for a fixed separation $k=n-m$, $\sum_m(1+m^2)\sqrt{p_mp_{m+k}}\le\bigl(\sum_m(1+m^2)p_m\bigr)^{1/2}\bigl(\sum_n(1+2n^2+2k^2)p_n\bigr)^{1/2}$, and the tails are uniformly small. Every sum over $m$ used for the chain in Steps 1–4 and 6 has this form, with $\lvert k\rvert\le2$ and a weight of at most $m^2$ (one factor $m$ from $u_{h_m}=m\,e$, one from $C_{h_mh_m}=mX$). These sums therefore converge absolutely, locally uniformly in $\lambda$, so the termwise differentiations, rearrangements and index shifts are allowed. A uniform gradient on the chain means $t_{h_mh_m}=c+\kappa m$ with real $c,\kappa$. Up to $C\to C+c\mathbb 1$, $C$ then has the chain form with $X$ replaced by $X+\kappa\mathbb 1$, so the same bounds hold for every description that enters $a_0$ and $\mu$.

## Derivation

### Step 1. Block form and the cost identity (item 1a)

Inserting $\sum_hQ_h=\mathbb 1$ on both sides of $C$, and using $(\langle h\rvert\otimes\mathbb 1)\lvert\Psi\rangle=\lvert\psi_h\rangle$ (2.1) in (1.9):

$$
C=\sum_{h,h'}\lvert h\rangle\langle h'\rvert\otimes C_{hh'},\quad C_{h'h}=C_{hh'}^\dagger,\qquad
i\frac{\mathrm d}{\mathrm d\lambda}\lvert\psi_h\rangle=\sum_{h'}C_{hh'}\lvert\psi_{h'}\rangle .
\tag{20.1}
$$

In the setting of items 2–4 the second equation is (17.1). The definitions read $L_h=\langle\Psi\vert Q_hCQ_h\vert\Psi\rangle$ and $L_{hh'}=2\operatorname{Re}\langle\Psi\vert Q_hCQ_{h'}\vert\Psi\rangle$. Now $L=\sum_{h,h'}\langle\psi_h\vert C_{hh'}\vert\psi_{h'}\rangle$. The diagonal terms are the $L_h$, which are real because $C_{hh}$ is self-adjoint. For $h\neq h'$ the terms $(h,h')$ and $(h',h)$ are complex conjugates by $C_{h'h}=C_{hh'}^\dagger$, so together they give $L_{hh'}$. Hence, for every state and every $\lambda$,

$$
L=\sum_hL_h+\sum_{\{h,h'\}}L_{hh'},\qquad L_{hh'}=0\ \text{unless }\{h,h'\}\text{ is a cost edge}.
\tag{20.2}
$$

**Conventions.** (i) *Placement (A5):* $L_h$ and $L_{hh'}$ involve only the total $C$, $\Psi$ and $Q_h$. Moving a single-object term to another pair term changes the contract list but not $C$, so it changes nothing. (ii) *Common phase (1.2):* both are sesquilinear in $\Psi$, so they are invariant. (iii) *Phase split (1.6):* $\psi_h=a_h\lvert E_h\rangle$ is defined by (2.1) directly from $\Psi$, so no split enters. (iv) *Labels in $\bar A$ (M6):* a relabelling of identical-type objects in $\bar A$ acts as a permutation unitary $P$ on $\mathcal H_{\bar A}$, $\Psi\to P\Psi$, $C\to PCP^\dagger$. Since $b$ is unique of its type, $P$ commutes with $Q_h$, so both quantities are unchanged. Hence every site and edge cost is free of these conventions, and so is every aggregate of them on background points.

### Step 2. Shift $C\to C+c\mathbb 1$ (item 1b)

The shift adds $c\mathbb 1_{\bar A}$ to every $C_{hh}$ and leaves $C_{hh'}$ ($h\neq h'$) unchanged. The evolved state changes by $e^{-ic\lambda}$, which cancels in sesquilinear quantities. Hence

$$
L_h\to L_h+c\,p_h,\qquad L_{hh'}\to L_{hh'},\qquad L\to L+c .
\tag{20.3}
$$

Every edge cost is invariant; no single site cost is (unless $p_h=0$). A real combination $\sum_h\alpha_hL_h+\sum\beta_{hh'}L_{hh'}$, with coefficients that may depend on the state, is invariant iff $\sum_h\alpha_hp_h=0$. If the coefficients are state independent, this forces $\alpha_h=0$. Examples of invariant combinations:

$$
L_h-p_hL,\qquad L_R-p_RL,\qquad \frac{L_h}{p_h}-\frac{L_{h'}}{p_{h'}}\quad(p_h,p_{h'}>0).
\tag{20.4}
$$

### Step 3. Rate of change of $L_R$ (item 1c)

1. The algebra of Step 1, restricted to $R$, gives $L_R=\langle\Psi\vert Q_RCQ_R\vert\Psi\rangle$.
2. By (1.9), for a $\lambda$-independent operator $B$: $\frac{\mathrm d}{\mathrm d\lambda}\langle\Psi\vert B\vert\Psi\rangle=i\langle\Psi\vert[C,B]\vert\Psi\rangle$.
3. Split $C=Q_RCQ_R+Q_RC\bar Q+\bar QCQ_R+\bar QC\bar Q$. Since $Q_R\bar Q=0$, the outer block $\bar QC\bar Q$ commutes with $Q_RCQ_R$, and $[C,Q_RCQ_R]=\bar QCQ_RCQ_R-Q_RCQ_RC\bar Q$. The second operator is the adjoint of the first, so $i\langle[C,Q_RCQ_R]\rangle=2\operatorname{Im}\langle Q_RCQ_RC\bar Q\rangle$. Written in blocks:

$$
\frac{\mathrm dL_R}{\mathrm d\lambda}=\sum_{h\in R}\sum_{k\notin R}2\operatorname{Im}\langle\zeta^R_h\vert C_{hk}\vert\psi_k\rangle,
\qquad \lvert\zeta^R_h\rangle:=\sum_{h'\in R}C_{hh'}\lvert\psi_{h'}\rangle .
\tag{20.5}
$$

4. **Which blocks enter.** Every term contains exactly one boundary block $C_{hk}$ ($h\in R$, $k\notin R$), linearly. The interior blocks $C_{hh'}$ ($h,h'\in R$, diagonal blocks included) enter through $\zeta^R_h$. Exterior blocks $C_{kk'}$ ($k,k'\notin R$) do not enter. Hence **$L_R$ can change only through the boundary blocks**: if all of them vanish, $L_R$ is constant. The rate is not a function of the boundary blocks alone, however. It is weighted by the interior blocks and by the branch vectors on both sides, and through $\zeta^R_h$ it depends on $R$; it is not a sum of $R$-independent pair currents. In the setting of items 2–4 the boundary blocks are $t_{hk}\mathbb 1_c$, so cost crosses the boundary of $R$ only along hopping edges.
5. Under $C\to C+c\mathbb 1$, $\zeta^R_h\to\zeta^R_h+c\psi_h$. The rate therefore gains $c\sum_{h\in R,k\notin R}2\operatorname{Im}\langle\psi_h\vert C_{hk}\vert\psi_k\rangle=c\,\mathrm dp_R/\mathrm d\lambda$ (by (20.6) below), consistent with $L_R\to L_R+cp_R$.

### Step 4. Acceleration (item 2)

1. By (20.1), $\dot p_h=2\operatorname{Re}\langle\psi_h\vert\dot\psi_h\rangle$. The term with $C_{hh}$ drops out because it is real, so

$$
\frac{\mathrm dp_h}{\mathrm d\lambda}=\sum_{h'\neq h}J_{h'\to h},\qquad J_{h'\to h}:=2\operatorname{Im}\langle\psi_h\vert C_{hh'}\vert\psi_{h'}\rangle=-J_{h\to h'} .
\tag{20.6}
$$

   With $C_{hh'}=t_{hh'}\mathbb 1_c$ and (2.2) this is (17.4).
2. Since the $u_h$ do not depend on $\lambda$: $a=\sum_h\ddot p_hu_h=\sum_{h\neq h'}\dot J_{h'\to h}u_h=\sum_{\{h,h'\}}\dot J_{h'\to h}\Delta_{hh'}$, by antisymmetry.
3. By (20.1), $\langle\dot\psi_h\rvert=i\sum_k\langle\psi_k\rvert C_{kh}$, so $\frac{\mathrm d}{\mathrm d\lambda}\langle\psi_h\vert C_{hh'}\vert\psi_{h'}\rangle=i\sum_k\bigl[\langle\psi_k\vert C_{kh}C_{hh'}\vert\psi_{h'}\rangle-\langle\psi_h\vert C_{hh'}C_{h'k}\vert\psi_k\rangle\bigr]$. Separating $k=h$ in the first sum and $k=h'$ in the second gives the exact identity

$$
a(\lambda)=\sum_{\{h,h'\}\in E}\bigl[A_{hh'}+B_{hh'}\bigr]\Delta_{hh'},\qquad
\begin{aligned}
A_{hh'}&=2\operatorname{Re}\bigl[\langle\eta_h\vert C_{hh'}\vert\psi_{h'}\rangle-\langle\psi_h\vert C_{hh'}\vert\eta_{h'}\rangle\bigr],\quad \lvert\eta_h\rangle:=\textstyle\sum_{k\neq h}C_{hk}\lvert\psi_k\rangle,\\
B_{hh'}&=2\operatorname{Re}\langle\psi_h\vert C_{hh}C_{hh'}-C_{hh'}C_{h'h'}\vert\psi_{h'}\rangle .
\end{aligned}
\tag{20.7}
$$

   $A$ and $B$ are antisymmetric under $h\leftrightarrow h'$ and contain the factor $C_{hh'}$, so only cost edges contribute. The sum runs over $E$.
4. **Split by diagonal blocks.** The *hopping part* $a^{\rm hop}:=\sum_EA_{hh'}\Delta_{hh'}$ contains only off-diagonal blocks. The *cost part* $a^{\rm cost}:=\sum_EB_{hh'}\Delta_{hh'}$ contains each diagonal block once. With $C_{hh'}=t_{hh'}\mathbb 1_c$, $B_{hh'}=2\operatorname{Re}[t_{hh'}\langle\psi_h\vert C_{hh}-C_{h'h'}\vert\psi_{h'}\rangle]$, so diagonal blocks enter only through **differences $C_{hh}-C_{h'h'}$ across hopping edges**. Inserting $C_{hh}=\Omega_h\mathbb 1_c+\hat K_h$ and $L_{hh'}=2\operatorname{Re}[t_{hh'}\langle\psi_h\vert\psi_{h'}\rangle]$:

$$
B_{hh'}(\lambda)=L_{hh'}(\lambda)\,(\Omega_h-\Omega_{h'})+Z_{hh'}(\lambda),\qquad Z_{hh'}:=2\operatorname{Re}\bigl[t_{hh'}\langle\psi_h\vert\hat K_h-\hat K_{h'}\vert\psi_{h'}\rangle\bigr],
\tag{20.8}
$$

   and $A_{hh'}=2\operatorname{Re}[t_{hh'}(\langle\eta_h\vert\psi_{h'}\rangle-\langle\psi_h\vert\eta_{h'}\rangle)]$ with $\eta_h=\sum_{k\neq h}t_{hk}\psi_k$. The cost part thus has two terms: site-cost-per-weight differences weighted by edge costs, and a record term $Z$. Both parts also depend on all blocks implicitly, through $\psi_h(\lambda)$.
5. **Invariance.** $u_h=(C_{hh}-\Omega_h)\chi$ and $\Omega_h$ are functions of $C_{hh}$ and $\chi$, and (20.7) contains only blocks of the total $C$. Hence $a$ does not depend on the placement of single-object terms; in particular $K_h\to K_h+c_h$, $t_{hh}\to t_{hh}-c_h$ leaves $C_{hh}$, $u_h$, $a$ unchanged. Under $C\to C+c\mathbb 1$: $\psi_h(\lambda)\to e^{-ic\lambda}\psi_h(\lambda)$ cancels; $\eta_h$, $C_{hh}C_{hh'}-C_{hh'}C_{h'h'}$, $\Omega_h-\Omega_{h'}$, $\hat K_h$ and $u_h$ are unchanged. So $a$, and each of $A$, $L_{hh'}(\Omega_h-\Omega_{h'})$ and $Z$ separately, is unchanged. The phase split of the start sends $u_h\to e^{i\varphi}u_h$ and $a\to e^{i\varphi}a$, the common isometry of 17.

### Step 5. Inertia at $\lambda=0$ (item 3)

For the product start, $\psi_h(0)=\phi_h\chi$ and $\eta_h(0)=(\tau\phi)_h\chi$ with $(\tau\phi)_h:=\sum_{k\neq h}t_{hk}\phi_k$. Hence $L_{hh'}(0)=2\operatorname{Re}[t_{hh'}\phi_h^*\phi_{h'}]$, and $Z_{hh'}(0)=2\operatorname{Re}[t_{hh'}\phi_h^*\phi_{h'}]\,\langle\chi\vert\hat K_h-\hat K_{h'}\vert\chi\rangle=0$.

**(a)** By (20.7)–(20.8), for arbitrary $\Omega_h$:

$$
a(0)=a_0+\sum_{\{h,h'\}\in E}L_{hh'}(0)\,(\Omega_h-\Omega_{h'})\,\Delta_{hh'},\qquad
a_0=\sum_{\{h,h'\}\in E}2\operatorname{Re}\Bigl[t_{hh'}\bigl((\tau\phi)_h^*\phi_{h'}-\phi_h^*(\tau\phi)_{h'}\bigr)\Bigr]\Delta_{hh'} .
\tag{20.9}
$$

$a_0$ contains only the off-diagonal $t_{hh'}$, $\phi$ and the $u_h$, none of which changes when the $t_{hh}$ vary. Choosing $t_{hh}=\text{const}-\epsilon_h$ makes all $\Omega_h$ equal and removes the second term. So $a_0$ in (20.9) is the $a_0$ of the definition.

**(b)** For a uniform gradient, $\Omega_h-\Omega_{h'}=g(f,\Delta_{hh'})=-g(F,\Delta_{hh'})$. Then (20.9) gives

$$
a(0)=a_0+\mu F,\qquad \mu=-\sum_{\{h,h'\}\in E}L_{hh'}(0)\,\Delta_{hh'}\,g\bigl(\Delta_{hh'},\,\cdot\,\bigr) .
\tag{20.10}
$$

- *Existence and uniqueness.* Every $f\in\mathcal H_c$ is realized, by $t_{hh}=g(f,u_h)-\epsilon_h$ (real). So the defining identity, required for all uniform gradients, fixes $\mu$ on all of $\mathcal H_c$, and (20.10) satisfies it. $\mu F$ depends on $f$ only through the $g(f,\Delta_{hh'})$, so the freedom in $f$ for given $\Omega_h$ does not matter.
- *Symmetry.* $g(w,\mu w')=-\sum_EL_{hh'}(0)\,g(w,\Delta_{hh'})\,g(\Delta_{hh'},w')$ is symmetric, because $g$ is. So **$\mu$ is $g$-symmetric**.
- *Which cost.* **$\mu$ is a function of the edge costs $L_{hh'}(0)$ and the background differences $\Delta_{hh'}$ alone.** It is unchanged by any variation of the $t_{hh}$, which changes every site cost $L_h(0)=p_h\Omega_h$ with $p_h>0$ arbitrarily, and $L$ by (20.2). It is not a function of the site costs: in Step 6(b), $\theta=0$ and $\theta=\pi/2$ have the same $p_h$, the same $\Omega_h$, hence the same site costs, but different $\mu$. It is not a function of $L$ either: $\theta=0$ with $C$ and $\theta=\pi/2$ with $C+t\mathbb 1$ have the same $L(0)=\tfrac12\langle X\rangle+t$, while $\mu$ (invariant under the shift by (20.3)) differs. The edge costs are free of all conventions (Steps 1–2), so $\mu$ is as well, up to the common isometry $e^{i\varphi}$ of the phase split.
- *Positivity.* $g(w,\mu w)=-\sum_EL_{hh'}(0)\,g(\Delta_{hh'},w)^2$. Hence $\mu\ge0$ iff this form is non-negative. It is sufficient that $L_{hh'}(0)\le0$ on every edge with $\Delta_{hh'}\neq0$; edges inside one background point do not contribute. The condition is also necessary when the nonzero $\Delta_{hh'}$ are linearly independent over $\mathbb R$: take $w$ dual to one of them. If all $\Delta_{hh'}=s_{hh'}n$ are collinear, then $\mu\ge0$ iff $\sum_Es_{hh'}^2L_{hh'}(0)\le0$.

**(c)** Localized start $\phi=\lvert\beta\rangle$. One of $\phi_h,\phi_{h'}$ vanishes on every edge, so $L_{hh'}(0)=0$ and $\mu=0$. Further, $(\tau\phi)_h=t_{h\beta}$ for $h\neq\beta$ and $(\tau\phi)_\beta=0$. Only edges $\{h,\beta\}$ contribute to (20.9), with coefficient $2\operatorname{Re}[t_{h\beta}t_{h\beta}^*]$:

$$
a_0=2\sum_h\lvert t_{h\beta}\rvert^2(u_h-u_\beta)=2w_2,\qquad \mu=0,\qquad a(0)=2w_2\ \text{ for all }\Omega_h .
\tag{20.11}
$$

This agrees with (17.11)–(17.12): $\mathrm dv/\mathrm d\lambda(0)=2w_2$, and $w_2$ does not depend on the $t_{hh}$.

### Step 6. Chain with linear records (item 4)

Data: $t_{h_mh_{m\pm1}}=t$, all other $t_{hh'}=0$ (including $t_{hh}=0$), $\epsilon_{h_m}=m\langle X\rangle$, $\hat K_{h_m}=m(X-\langle X\rangle)$, $u_{h_m}=m\,e$, $\lVert e\rVert=\sigma_X$ (17.15). Orient every edge as $h=h_{m+1}$, $h'=h_m$, so that $\Delta=e$. Write $\psi_m:=\psi_{h_m}$, $p_m:=p_{h_m}$.

**(a)** $\Omega_{h_m}=t_{h_mh_m}+\epsilon_{h_m}=m\langle X\rangle$. A uniform gradient needs $g(f,(m-n)e)=(m-n)\langle X\rangle$, i.e. $g(f,e)=\langle X\rangle$. This is solved by $f=\langle X\rangle e/\sigma_X^2$, and every other solution differs from it by a vector $g$-orthogonal to $e$, which $\mu$ annihilates. So the gradient is uniform, and

$$
\Omega_{h_m}=m\langle X\rangle,\qquad F=-\frac{\langle X\rangle}{\sigma_X^2}\,e .
\tag{20.12}
$$

**The hopping part vanishes on the chain.** $\eta_m=t(\psi_{m+1}+\psi_{m-1})$, so $A_{h_{m+1}h_m}=2t^2\operatorname{Re}\bigl[\langle\psi_{m+2}\vert\psi_m\rangle-\langle\psi_{m+1}\vert\psi_{m-1}\rangle\bigr]+2t^2(p_m-p_{m+1})$. All $\Delta=e$, so $a^{\rm hop}=e\sum_mA_{h_{m+1}h_m}=0$ by an index shift, applied separately to each of the three absolutely convergent sums (Setup), for every start of finite support and every $\lambda$:

$$
a^{\rm hop}(\lambda)\equiv0,\qquad a_0=0\ \text{ for every start of finite support} .
\tag{20.13}
$$

**(b)** Start $\phi_{h_0}=2^{-1/2}$, $\phi_{h_1}=2^{-1/2}e^{i\theta}$. The only edge with $\phi_h\phi_{h'}\neq0$ is $\{h_0,h_1\}$. By (20.10), (20.13) and (20.12):

$$
L_{h_0h_1}(0)=t\cos\theta\ \ (\text{all other edge costs }0),\qquad \mu=-t\cos\theta\;e\,g(e,\,\cdot\,),\qquad a_0=0,\qquad a(0)=t\cos\theta\,\langle X\rangle\,e .
\tag{20.14}
$$

Here $a(0)=\mu F=-t\cos\theta\,e\,g(e,F)=t\cos\theta\,\langle X\rangle e$. The derivative of (17.22) at $\lambda=0$ is $t\sum_\xi w_\xi\,\xi\cos\theta\;e=t\cos\theta\,\langle X\rangle e$, using $\sum_\xi w_\xi\xi=\langle X\rangle$: **agreement**. Since $g(w,\mu w)=-t\cos\theta\,g(e,w)^2$ with $t>0$ and $e\neq0$, **$\mu$ is positive semidefinite iff $\cos\theta\le0$**. It is positive on the span of $e$ iff $\cos\theta<0$, and $\mu=0$ at $\cos\theta=0$.

Start $\lvert h_0\rangle\otimes\lvert\chi\rangle$: by (20.11), $\mu=0$ and $a_0=2t^2(u_{h_1}+u_{h_{-1}}-2u_{h_0})=0$, hence $a(0)=0$. By (17.19), $\bar x\equiv0$, so $a(0)=0$: **agreement**.

**(c)** $\lambda>0$, start of (b). $C(\mathbb 1\otimes\Pi_\xi)=(C_b+\xi M)\otimes\Pi_\xi$, so $\psi_m(\lambda)=\sum_\xi(U_\xi\phi)_m\,\Pi_\xi\chi$ (Step 9 of 17). For any function $q$ of $X$, using $\sum_m\varphi^*_{m+1}\varphi_m=\langle\varphi\vert S^\dagger\vert\varphi\rangle$, $U_\xi^\dagger S^\dagger U_\xi=e^{i\xi\lambda}S^\dagger$ and $\langle\phi\vert S^\dagger\vert\phi\rangle=\tfrac12e^{-i\theta}$:

$$
\sum_m2\operatorname{Re}\bigl[t\langle\psi_{m+1}\vert q(X)\vert\psi_m\rangle\bigr]=t\sum_\xi w_\xi\,q(\xi)\cos(\xi\lambda-\theta) .
$$

With $q=1$ this is the total edge cost $\mathcal T(\lambda):=\sum_mL_{h_mh_{m+1}}(\lambda)=t\sum_\xi w_\xi\cos(\xi\lambda-\theta)$. All $\Delta=e$, so the edge-cost expression (20.10) gives $\mu(\lambda)=-\mathcal T(\lambda)\,e\,g(e,\cdot)$ and $\mu(\lambda)F=\langle X\rangle\mathcal T(\lambda)\,e$. With $q=X-\langle X\rangle$ it gives $\sum_mZ_{h_{m+1}h_m}$. By (20.7), (20.8), (20.13), $a=\mu(\lambda)F+e\sum_mZ_{h_{m+1}h_m}$:

$$
a(\lambda)=t\sum_\xi w_\xi\,\xi\cos(\xi\lambda-\theta)\,e,\qquad
a(\lambda)-\mu(\lambda)F=t\sum_\xi w_\xi(\xi-\langle X\rangle)\cos(\xi\lambda-\theta)\,e=t\operatorname{Re}\Bigl[e^{-i\theta}\langle\chi\vert(X-\langle X\rangle)e^{iX\lambda}\vert\chi\rangle\Bigr]e .
\tag{20.15}
$$

$a(\lambda)$ is the derivative of (17.22). **$a(\lambda)=\mu(\lambda)F$ does not hold in general.** The difference is the record term, $t\langle\Psi\vert(S+S^\dagger)\otimes(X-\langle X\rangle)\vert\Psi\rangle\,e$: the correlation between hopping and record fluctuation, which vanishes at $\lambda=0$ because the start is a product. For small $\lambda$ it is $t\sigma_X^2\sin\theta\,\lambda\,e+O(\lambda^2)$, so it is nonzero whenever $\sin\theta\neq0$. If $\sin\theta=0$, the difference is $\pm t\sum_{\nu\ge0}c_\nu\cos(\nu\lambda)\,e$ with $c_\nu:=\sum_{\xi=\pm\nu}w_\xi(\xi-\langle X\rangle)$. Cosines of distinct frequencies are linearly independent, so the identity holds for all $\lambda$ iff $\sin\theta=0$ and $c_\nu=0$ for every $\nu\ge0$. This happens, for example, when the spectrum of $X$ in $\chi$ lies in $\{\nu,-\nu\}$.

## Result

- **(20.2)** $L=\sum_hL_h+\sum_{\{h,h'\}}L_{hh'}$ exactly. Site and edge costs are free of the placement of single-object terms, of the common phase, of the phase split and of relabelling in $\bar A$.
- **(20.3)–(20.4)** Under $C\to C+c\mathbb 1$: $L_h\to L_h+cp_h$, $L_{hh'}$ invariant. The invariant combinations are exactly those with $\sum_h\alpha_hp_h=0$, e.g. $L_h-p_hL$, $L_R-p_RL$, $L_h/p_h-L_{h'}/p_{h'}$.
- **(20.5)** $\mathrm dL_R/\mathrm d\lambda=\sum_{h\in R,k\notin R}2\operatorname{Im}\langle\zeta^R_h\vert C_{hk}\vert\psi_k\rangle$. Each term contains exactly one boundary block, exterior blocks do not enter, and $L_R$ is constant if all boundary blocks vanish.
- **(20.7)–(20.8)** $a=a^{\rm hop}+a^{\rm cost}$. The diagonal blocks enter explicitly only through $C_{hh}-C_{h'h'}$ on hopping edges, as $L_{hh'}(\Omega_h-\Omega_{h'})$ plus the record term $Z_{hh'}$. $a$ is independent of placement and invariant under $C\to C+c\mathbb 1$.
- **(20.9)–(20.10)** $a(0)=a_0+\sum_EL_{hh'}(0)(\Omega_h-\Omega_{h'})\Delta_{hh'}$ and $\mu=-\sum_EL_{hh'}(0)\,\Delta_{hh'}g(\Delta_{hh'},\cdot)$. $\mu$ is $g$-symmetric and a function of the edge costs and the background alone, not of the site costs or of $L$. It is positive semidefinite iff $-\sum_EL_{hh'}(0)g(\Delta_{hh'},w)^2\ge0$ for all $w$, which holds in particular if all edge costs are $\le0$. **The inertia is governed by the edge (hopping) cost of the body.**
- **(20.11)** Localized start: $\mu=0$ and $a_0=a(0)=2w_2$, in agreement with (17.11)–(17.12).
- **(20.12)–(20.15)** Chain: $\Omega_{h_m}=m\langle X\rangle$ and $F=-\langle X\rangle e/\sigma_X^2$. For the start of 4(b): $L_{h_0h_1}(0)=t\cos\theta$, $\mu=-t\cos\theta\,e\,g(e,\cdot)$, $a_0=0$, $a(0)=t\cos\theta\langle X\rangle e$ (agrees with (17.22)), and $\mu\ge0$ iff $\cos\theta\le0$. For the start $\lvert h_0\rangle$: $\mu=0$ and $a(0)=0$ (agrees with (17.19)). For $\lambda>0$, $a(\lambda)\neq\mu(\lambda)F$ in general, with the difference given in (20.15).

## Consistency checks

1. **Dimensions.** Let $[C]=\kappa$ and $[\lambda]=\kappa^{-1}$. Then $[u_h]=[\Omega_h]=[L]=\kappa$ and $[a]=\kappa^3$. In (20.9), $L\,\Omega\,\Delta$ and $t^2u$ are both $\kappa^3$. From $g(f,\Delta)=\Delta\Omega$, $f$ is dimensionless, so $[\mu]=\kappa^3$. In (20.14)–(20.15), $t\langle X\rangle e$ is $\kappa^3$. In (20.5), $C^2\psi^2$ is $\kappa^2=[L]/[\lambda]$.
2. **No hopping; equal records.** If $t_{hh'}=0$ for $h\neq h'$, then $E=\emptyset$, so $a\equiv0$ and $\mu=0$. By (20.5) and (20.6) every $L_R$ and every $p_h$ is constant, as it must be for block-diagonal $C$. If all $K_h$ are equal, all $u_h$ coincide and $\Delta=0$, so $a\equiv0$ and $\mu=0$, consistent with $\bar x$ being constant.
3. **Conservation of $L$.** For $R$ = all places, $\bar Q=0$ and (20.5) gives $\mathrm dL/\mathrm d\lambda=0$. The shift check of Step 3.5 reproduces (20.3).

## Open issues

- $\mu$ is defined at $\lambda=0$ only. 4(c) shows that its edge-cost extension $\mu(\lambda)$ misses the record term $Z$ for $\lambda>0$. Whether a convention-free inertia exists for $\lambda>0$ is not addressed here.
- For general (neither independent nor collinear) edge vectors $\Delta_{hh'}$, positivity of $\mu$ is stated as positivity of the quadratic form in (20.10), not as a condition on single edge costs.

## Methods used

- block decomposition with respect to place projectors
- derivative of expectation values via commutators (Heisenberg form)
- antisymmetric pair currents and telescoping sums
- spectral decomposition of a conserved medium operator
- real quadratic forms and positivity
- linear independence of trigonometric functions
- moment bounds via Cauchy–Schwarz and Dini's theorem (infinite chain)