Version 4 · current

# Formation of the made field: bound, conserved densities and oscillator cells

- **Subproject:** 08-ilang-propagation
- **Package:** 04-formation
- **Version:** v4
- **Mode:** external regeneration
- **Date:** 2026-10-24

## Changes from previous version

- Step 3, after (4.5): the uniformity of $c_A$ in $L$ is stated with its conditions: bounded range, bounded number and norms of the local terms that do not commute with $A_0$, and bounded $\lVert A_0\rVert$ (note (f)). Bounded support alone is no longer claimed to suffice (I1).
- Step 4, (4.8): the bounds $c_A\le\bar c_A$, $s_B\le\bar s_B$ uniform in $L$ and $s\ne0$ in the limit $L\to\infty$ are now assumptions of (4.8), and $C_\Phi$ is written with $\bar c_A$, $\bar s_B$ (I1, note (f)).
- Step 4, (4.8), Result: for $B=A$ the bound $\chi_A\le s_A/\lvert\mathbf q\rvert^2$ is now an assumption (note (f)). The sufficient condition through a summable remainder of the mark is removed (I2).
- Step 9(i): the precursor exponent of the control is reported as a finite-distance observation. Asymptotically, Step 6(b) gives $O(\rho^{-1/2})$ on that interval (I3).
- Step 9(i), Code: the torus made field is the infinite-lattice one plus exponentially small images, not equal to it. A size-refinement check bounds the images on every window. `oscillator_formation.py` was edited (docstring and check), and both scripts were run again in `code/v4/` (I4).

## Response to verification

- I1: Accepted and fixed in Steps 3 and 4. Bounded support of $A_0$ alone does not bound $c_A$ uniformly in $L$. Step 3 now gives the sufficient condition of note (f): uniformly bounded range, number and norms of the local terms of $C_F$ that do not commute with $A_0$, and $\lVert A_0\rVert$. (4.8) now assumes $c_A\le\bar c_A$, $s_B\le\bar s_B$ uniformly in $L$ and $s\ne0$ in the limit $L\to\infty$. $C_\Phi$ is built from these bounds and is independent of $L$.
- I2: Accepted and fixed in Step 4 and in the Result entry (4.8). The question's new note (f) asks item 2 for $B=A$ only under $\chi_A\le s_A/\lvert\mathbf q\rvert^2$. This bound is now an assumption of (4.8) for $B=A$, as for $B\ne A$. The sentence deriving it from a summable remainder of the mark is removed, and the case without it is out of scope.
- I3: Accepted and fixed in Step 9(i). On the ring the interval $0.485\le w\le0.8$ stays a fixed distance from the fold at $w=0.404$ and from $w=1$, so Step 6(b) gives $O(\rho^{-1/2})$ there. The nearly constant product with $\rho^{1/3}$ for $\rho=32,\dots,512$ is now reported as a finite-distance observation, not as an exponent.
- I4: Accepted and fixed in `oscillator_formation.py` and Step 9(i). By Poisson summation the torus made field is the infinite-lattice made field plus its images. Before the image front these are exponentially small (Step 7(i)), not zero. The docstring now says so, and a size-refinement check ($L'=32768$ on the ring, $L'=384$ on the torus) bounds the image contamination on every window.

## Setup and assumptions

**Class F** (assumption, M8). A field of finitely many objects with finitely many places each (A1) on a torus $X_f=\mathbb Z_L^D$, contract $C_F$ a sum of pair terms of finite range (A5), invariant under the translations: unitaries $U_y$ with $U_yC_FU_y^\dagger=C_F$, $U_yA_xU_y^\dagger=A_{x+y}$, $U_yB_xU_y^\dagger=B_{x+y}$, where $A_x$ (read) and $B_x$ (coupled to the source) are self-adjoint translates acting near $x$. The source sits at $x_s=0$ (translation).

**Source and start** (assumption, M8, M9, A11). The contract is $C_F+g_sB_0$ at every $\lambda\ge0$; at $\lambda=0$ the state is the lowest eigenvector $\Omega$ of $C_F$, assumed nondegenerate, eigenvalue $E_0$. Since $U_y\Omega$ is again a lowest eigenvector, $U_y\Omega=e^{ip_0\cdot y}\Omega$. Eigenvectors $\lvert n\rangle$ of $C_F$ are chosen with $U_y\lvert n\rangle=e^{ip_n\cdot y}\lvert n\rangle$; $\omega_n=E_n-E_0>0$ for $n\neq0$.

**Made field.** $\delta a(x,\lambda)=\partial_{g_s}\langle A_x\rangle_\lambda\vert_{g_s=0}$ (first order: the limit $g_s\to0$ of the coefficient is taken first); $\delta a_{\rm st}(x)$ the same in the lowest eigenvector of $C_F+g_sB_0$ (simple for small $\lvert g_s\rvert$). Label components $Y_{\mathbf q}=\sum_xe^{-i\mathbf q\cdot x}Y_x$ for operators, $\delta a_{\mathbf q}=\sum_xe^{-i\mathbf q\cdot x}\delta a(x)$, $\mathbf q\in(2\pi/L)\mathbb Z^D$; $Y_{\mathbf q}^\dagger=Y_{-\mathbf q}$.

**Inputs** (quoted in the question, with their settings). (8.10), (8.14), (8.15) of 07-ilang-weight/08-field-speed; (5.11) of 04-ilang-time/05-moving-clock, $v_{\max}=\tau\kappa\sigma_X$; (4.15) of 06-ilang-source/04-profile, $d_\Gamma(\pi(m),\pi(m'))=k\lvert m-m'\rvert$.

**Mathematics used** (M5): first-order (Duhamel) perturbation of $e^{-i(C_F+g_sB_0)\lambda}$, Cauchy–Schwarz, Fourier series on tori, Parseval and Poisson summation, closed forms of explicit Fourier integrals (Bessel functions), stationary phase (uniform in a parameter, with Airy functions at folds), contour shifts. All are applied to the stated operators; no field equation is used (M4).

## Derivation

### Step 1. First-order response in spectral form

With $\langle A_x\rangle_\lambda=\langle\Omega\vert e^{i(C_F+g_sB_0)\lambda}A_xe^{-i(C_F+g_sB_0)\lambda}\vert\Omega\rangle$, Duhamel's formula and $e^{-iC_F\lambda}\Omega=e^{-iE_0\lambda}\Omega$ give $\delta a(x,\lambda)=-i\int_0^\lambda\mathrm ds\,\langle\Omega\vert[A_x(s),B_0]\vert\Omega\rangle$, $A_x(s)=e^{iC_Fs}A_xe^{-iC_Fs}$. Inserting the eigenvectors (the $n=0$ terms cancel in the commutator) and integrating over $s$:

$$
\delta a(x,\lambda)=-\sum_{n\neq0}\Bigl[\langle\Omega\vert A_x\vert n\rangle\langle n\vert B_0\vert\Omega\rangle\,g_\lambda(\omega_n)+\langle\Omega\vert B_0\vert n\rangle\langle n\vert A_x\vert\Omega\rangle\,\overline{g_\lambda(\omega_n)}\Bigr],\qquad g_\lambda(\omega)=\frac{1-e^{-i\omega\lambda}}{\omega}.
\tag{4.1}
$$

The first-order change of the lowest eigenvector of $C_F+g_sB_0$ is $-\sum_{n\ne0}\lvert n\rangle\langle n\vert B_0\vert\Omega\rangle/\omega_n$, so $\delta a_{\rm st}$ is (4.1) with $g_\lambda(\omega)\to1/\omega$. Translation covariance gives $\langle\Omega\vert A_x\vert n\rangle=e^{i(p_0-p_n)\cdot x}\langle\Omega\vert A_0\vert n\rangle$ and $\langle n\vert B_x\vert\Omega\rangle=e^{i(p_n-p_0)\cdot x}\langle n\vert B_0\vert\Omega\rangle$. Summing over $x$, the first term of (4.1) survives only for $p_n=p_0-\mathbf q$, where $\langle n\vert B_0\vert\Omega\rangle=\lvert X_f\rvert^{-1}\langle n\vert B_{-\mathbf q}\vert\Omega\rangle$; the second only for $p_n=p_0+\mathbf q$. Hence

$$
\delta a_{\mathbf q}(\lambda)=-\frac1{\lvert X_f\rvert}\sum_{n\neq0}\Bigl[\langle\Omega\vert A_{\mathbf q}\vert n\rangle\langle n\vert B_{-\mathbf q}\vert\Omega\rangle\,g_\lambda(\omega_n)+\langle\Omega\vert B_{-\mathbf q}\vert n\rangle\langle n\vert A_{\mathbf q}\vert\Omega\rangle\,\overline{g_\lambda(\omega_n)}\Bigr],
\tag{4.2}
$$

and $\delta a^{\rm st}_{\mathbf q}$ is (4.2) with $g_\lambda,\overline{g_\lambda}\to1/\omega_n$.

### Step 2. The bound on formation (item 1)

Define, for the read operator and for any translate family $Y$,

$$
f_A(\mathbf q)=\frac{\langle\Omega\vert[A_{\mathbf q}^\dagger,[C_F,A_{\mathbf q}]]\vert\Omega\rangle}{\lvert X_f\rvert}=\frac1{\lvert X_f\rvert}\sum_n\omega_n\bigl(\lvert\langle n\vert A_{\mathbf q}\Omega\rangle\rvert^2+\lvert\langle n\vert A_{\mathbf q}^\dagger\Omega\rangle\rvert^2\bigr),\qquad
\chi_Y(\mathbf q)=\frac1{\lvert X_f\rvert}\sum_{n\neq0}\frac{\lvert\langle n\vert Y_{\mathbf q}\Omega\rangle\rvert^2+\lvert\langle n\vert Y_{\mathbf q}^\dagger\Omega\rangle\rvert^2}{\omega_n}.
\tag{4.3}
$$

The second form of $f_A$ follows by expanding the double commutator in the eigenvectors (sum rule). Both are $\ge0$, fixed by $C_F$, $A$ or $Y$, and $\Omega$, and unchanged by the common phase of $\Omega$ and by every replacement of contract terms that leaves $C_F$ unchanged (M6). By (4.2) with $B=A$, $\delta a^{\rm st}_{\mathbf q}=-\chi_A(\mathbf q)$: $\chi_Y$ is the label component of the static self-mark of $Y$.

Since $\lvert g_\lambda(\omega)\rvert=2\lvert\sin(\omega\lambda/2)\rvert/\omega\le\min(\lambda,2/\omega)$, bounding (4.2) termwise and applying Cauchy–Schwarz to the joint sum over $n$ and the two terms, with the weights $\omega_n^{\pm1/2}$ (for the bound $\lambda$) or $\omega_n^{-1/2}$ on both factors (for the bound $2/\omega$), gives for every $\mathbf q$, every $L$ and every $\lambda\ge0$

$$
\lvert\delta a_{\mathbf q}(\lambda)\rvert\le\min\Bigl\{\lambda\sqrt{f_A(\mathbf q)\,\chi_B(\mathbf q)}\,,\;2\sqrt{\chi_A(\mathbf q)\,\chi_B(\mathbf q)}\Bigr\},\qquad\text{for }B=A:\ \ \Bigl\lvert\frac{\delta a_{\mathbf q}(\lambda)}{\delta a^{\rm st}_{\mathbf q}}\Bigr\rvert\le\lambda\sqrt{\frac{f_A(\mathbf q)}{\chi_A(\mathbf q)}} .
\tag{4.4}
$$

*Highest eigenvector.* If $\Omega$ is the nondegenerate highest eigenvector, all $\omega_n<0$; Steps 1–2 hold with $\lvert\omega_n\rvert$, and (4.4) holds with $\lvert f_A\rvert$ and $\lvert\chi\rvert$, since the sum rule still equals $-\sum_n\lvert\omega_n\rvert(\dots)$. For an eigenvector inside the spectrum, $\sum_n\lvert\omega_n\rvert(\dots)$ is not a double commutator, and Step 3 does not apply.

### Step 3. Read operators that are conserved densities

Assumption (the question's item 2, M8): $N=\sum_xA_x$ commutes with $C_F$, and the terms of $C_F$ that do not commute with the $A_x$ have finite range. Then $[C_F,A_0]$ acts on the objects within a finite distance of $0$, and $R_A$, the largest $\lvert d\rvert$ with $[A_d,[C_F,A_0]]\ne0$, is finite. (For the field of 06-ilang-source, where $\Phi_x$ acts on every object, $R_A=1$ by the nearest-neighbour hopping.) Let

$$
F_A(d)=\langle\Omega\vert[A_d,[C_F,A_0]]\vert\Omega\rangle .
$$

Since $U_x[A_d,[C_F,A_0]]U_x^\dagger=[A_{x+d},[C_F,A_x]]$ and $U_x\Omega$ is $\Omega$ up to a phase, (4.3) gives $f_A(\mathbf q)=\sum_{\lvert d\rvert\le R_A}e^{i\mathbf q\cdot d}F_A(d)$, a finite sum. By the second form of (4.3), $f_A$ is real and even in $\mathbf q$ ($A_{\mathbf q}^\dagger=A_{-\mathbf q}$), and $f_A(0)=\lvert X_f\rvert^{-1}\langle\Omega\vert[N,[C_F,N]]\vert\Omega\rangle=0$. Hence $f_A(\mathbf q)=\sum_d{\rm Re}\,F_A(d)\,(\cos\mathbf q\cdot d-1)$. With $1-\cos\mathbf q\cdot d\le(\mathbf q\cdot d)^2/2\le\lvert\mathbf q\rvert^2\lvert d\rvert^2/2$ and $\lvert F_A(d)\rvert\le\lVert[A_d,[C_F,A_0]]\rVert\le2\lVert A_0\rVert\,\lVert[C_F,A_0]\rVert$:

$$
0\le f_A(\mathbf q)\le c_A\lvert\mathbf q\rvert^2,\qquad c_A=\tfrac12\sum_{0<\lvert d\rvert\le R_A}\lvert d\rvert^2\lvert F_A(d)\rvert\ \le\ \lVert A_0\rVert\,\lVert[C_F,A_0]\rVert\sum_{0<\lvert d\rvert\le R_A}\lvert d\rvert^2 ,
\tag{4.5}
$$

for every $L$. $c_A$ is fixed by $C_F$, $A$ and $\Omega$, and its bound by $C_F$ and $A$. Both depend on $C_F$ only through the total contract, so they are unchanged by every replacement of contract terms that leaves $C_F$ unchanged (M6). For the highest eigenvector, (4.5) holds with $\lvert f_A\rvert$.

*Uniformity in $L$.* Bounded support of $A_0$ alone does not bound $c_A$ uniformly in $L$, since the bound in (4.5) contains norms. Let $\Gamma_0$ be the set of terms of $C_F$ that do not commute with $A_0$; then $\lVert[C_F,A_0]\rVert\le2\lVert A_0\rVert\sum_{\gamma\in\Gamma_0}\lVert C_\gamma\rVert$. If $R_A$, the number and the norms of the terms in $\Gamma_0$, and $\lVert A_0\rVert$ are bounded uniformly in $L$ (the question's note (f)), then (4.5) gives $c_A\le\bar c_A$ with $\bar c_A$ independent of $L$. Where these norms grow with $L$, $c_A$ itself is used. An example is the field of 06-ilang-source at fixed density, where $\Phi_x$ acts on every object (Consistency check 2).

### Step 4. No formation within a linear $\lambda$ for conserved densities (item 2)

**Criterion** (stated before use; cubic torus, $D=3$). Let $\Phi\ge0$ be smooth on $\mathbb R^3$, not identically zero, supported in the shell $1\le\lvert y\rvert\le2$, and $\varphi_\ell(x)=\Phi(x/\ell)$. The made field at distance $\ell$ has formed in the mean to accuracy $\varepsilon\in(0,1)$ at $\lambda$ if

$$
\Bigl\lvert\sum_x\varphi_\ell(x)\bigl[\delta a(x,\lambda)-\delta a_{\rm st}(x)\bigr]\Bigr\rvert\ \le\ \varepsilon\,\Bigl\lvert\sum_x\varphi_\ell(x)\,\delta a_{\rm st}(x)\Bigr\rvert ;
\tag{4.19}
$$

it has formed from $\lambda_f$ on if (4.19) holds for every $\lambda\ge\lambda_f$, and $\lambda^\Phi_{\rm form}(\ell;\varepsilon)$ is the least such $\lambda_f$ (order: $g_s\to0$, $L\to\infty$, then $\ell\to\infty$). This is not the arrival of a first change. Suppose the made field has formed at every place of the shell $\ell\le\lvert x\rvert\le2\ell$, i.e. $\lvert\delta a-\delta a_{\rm st}\rvert\le\varepsilon\lvert\delta a_{\rm st}\rvert$ there, and $\delta a_{\rm st}$ has one sign there; then (4.19) holds. **Long-range mark:** in the limit $L\to\infty$, $\delta a_{\rm st}(x)=s/\lvert x\rvert+o(\lvert x\rvert^{-1})$, $s\neq0$. **Sources** (the question's notes (a), (f)): $\chi_B(\mathbf q)\le s_B/\lvert\mathbf q\rvert^2$ for $\mathbf q\ne0$, for $B\ne A$ and for $B=A$ ($\chi_A\le s_A/\lvert\mathbf q\rvert^2$). For $B=A$ this is an assumption, not derived from the long-range mark, and the case without the bound is out of scope by note (f). **Uniformity** (note (f)): $c_A\le\bar c_A$ and $s_B\le\bar s_B$ for all large $L$, with $\bar c_A,\bar s_B$ independent of $L$ (Step 3 gives a sufficient condition for $c_A$).

*Bound.* For real $\varphi$, Parseval gives $\sum_x\varphi(x)\delta a(x,\lambda)=\lvert X_f\rvert^{-1}\sum_{\mathbf q}\overline{\hat\varphi(\mathbf q)}\,\delta a_{\mathbf q}(\lambda)$. By (4.4) and (4.5), $\lvert\delta a_{\mathbf q}(\lambda)\rvert\le\lambda\sqrt{c_As_B}\le\lambda\sqrt{\bar c_A\bar s_B}$ for $\mathbf q\ne0$ and all large $L$, and $\delta a_{\mathbf 0}=0$ since $f_A(0)=0$. Poisson summation gives $\hat\varphi_\ell(\mathbf q)=\sum_{\mathbf n\in\mathbb Z^3}\ell^3\hat\Phi(\ell(\mathbf q+2\pi\mathbf n))$, so as $L\to\infty$, $\lvert X_f\rvert^{-1}\sum_{\mathbf q}\lvert\hat\varphi_\ell\rvert\to\int_{[-\pi,\pi]^3}\lvert\hat\varphi_\ell\rvert\,\mathrm d^3q/(2\pi)^3\le\lVert\hat\Phi\rVert_1=\int_{\mathbb R^3}\lvert\hat\Phi(\mathbf p)\rvert\,\mathrm d^3p/(2\pi)^3$, independent of $\ell$. Hence, in the limit $L\to\infty$, $\lvert\sum_x\varphi_\ell\,\delta a(\cdot,\lambda)\rvert\le\lambda\sqrt{\bar c_A\bar s_B}\,\lVert\hat\Phi\rVert_1$ for every $\lambda\ge0$. On the other side, formation in the mean at $\lambda$ requires $\lvert\sum_x\varphi_\ell\,\delta a\rvert\ge(1-\varepsilon)\lvert\sum_x\varphi_\ell\,\delta a_{\rm st}\rvert=(1-\varepsilon)\lvert s\rvert\ell^2\int\Phi(y)\lvert y\rvert^{-1}\mathrm d^3y\,(1+o(1))$. Therefore at every $\lambda$ at which (4.19) holds,

$$
\lambda\ \ge\ C_\Phi\,\ell^{2}\,(1+o(1)),\qquad C_\Phi=\frac{(1-\varepsilon)\,\lvert s\rvert\int\Phi(y)\lvert y\rvert^{-1}\mathrm d^3y}{\sqrt{\bar c_A\bar s_B}\;\lVert\hat\Phi\rVert_1}\qquad(\ell\to\infty),
\tag{4.8}
$$

and $\lambda^\Phi_{\rm form}(\ell;\varepsilon)\ge C_\Phi\ell^2(1+o(1))$. For every $\beta>0$, $\lambda^\Phi_{\rm form}(\ell;\varepsilon)>\beta\ell$ for all large $\ell$. **The long-range part of a conserved density cannot form within a $\lambda$ linear in the distance**, neither in the mean on a shell nor at every place of a shell; it needs a $\lambda$ at least quadratic in the distance. This holds for $B=A$ and for every source of note (a), each under the bounds of note (f), including fields of several kinds of objects: only the conservation of $N$, the locality of $A$, the bound on $\chi_B$ and the uniformity in $L$ were used. It also holds for the highest eigenvector, with $\lvert f_A\rvert$ and $\lvert\chi\rvert$ (Steps 2, 3). Sources with a stronger self-response, $B=A$ included, and cases without the uniform bounds are out of scope by notes (a) and (f). **Control:** oscillator cells with $A=B=q_x$; there $[q_x,[H,q_y]]=\delta_{xy}$, so $f_A=1$, $q_x$ is not the density of a conserved number, and (4.20) of Step 6(c) gives formation in the sense of (4.19) within a linear $\lambda$.

*Label form* (bounds on one component). The label component $\mathbf q$ has formed to accuracy $\varepsilon$ from $\lambda_f$ on if $\lvert\delta a_{\mathbf q}(\lambda)-\delta a^{\rm st}_{\mathbf q}\rvert\le\varepsilon\lvert\delta a^{\rm st}_{\mathbf q}\rvert$ for every $\lambda\ge\lambda_f$; $\lambda_{\rm form}(\mathbf q;\varepsilon)$ is the least such $\lambda_f$ ($L\to\infty$ at fixed $\mathbf q$, or $\lambda$ below the wrap time). This criterion concerns one component. A component can oscillate for ever while the made field at every distant place converges (oscillator cells, (4.10)), so it does not by itself constrain formation at a distance. With a long-range mark in label form, $\lvert\delta a^{\rm st}_{\mathbf q}\rvert\ge s_0/\lvert\mathbf q\rvert^2$ for $\lvert\mathbf q\rvert\le q_1$, formation of $\mathbf q$ requires $\lvert\delta a_{\mathbf q}(\lambda)\rvert\ge(1-\varepsilon)\lvert\delta a^{\rm st}_{\mathbf q}\rvert$. With (4.4) and (4.5), this gives, for every $B$,

$$
\lambda_{\rm form}(\mathbf q;\varepsilon)\ \ge\ \frac{(1-\varepsilon)\,\lvert\delta a^{\rm st}_{\mathbf q}\rvert}{\lvert\mathbf q\rvert\sqrt{c_A\,\chi_B(\mathbf q)}} ,
\tag{4.6}
$$

and, for $B=A$ ($\chi_B=\lvert\delta a^{\rm st}_{\mathbf q}\rvert$),

$$
\lambda_{\rm form}(\mathbf q;\varepsilon)\ \ge\ \frac{1-\varepsilon}{\lvert\mathbf q\rvert}\sqrt{\frac{\lvert\delta a^{\rm st}_{\mathbf q}\rvert}{c_A}}\ \ge\ \frac{1-\varepsilon}{\lvert\mathbf q\rvert^{2}}\sqrt{\frac{s_0}{c_A}}\qquad(\lvert\mathbf q\rvert\le q_1).
\tag{4.7}
$$

For $B\ne A$, formation of $\mathbf q$ within $\lambda\le\beta/\lvert\mathbf q\rvert$ requires the first relation below. For a source of note (a), (4.6) with $\lvert\mathbf q\rvert\sqrt{c_A\chi_B}\le\sqrt{c_As_B}$ gives the second:

$$
\chi_B(\mathbf q)\ \ge\ \frac{(1-\varepsilon)^2s_0^2}{\beta^2c_A}\,\frac1{\lvert\mathbf q\rvert^{4}};\qquad
\chi_B(\mathbf q)\le\frac{s_B}{\lvert\mathbf q\rvert^2}\ \Longrightarrow\ \lambda_{\rm form}(\mathbf q;\varepsilon)\ \ge\ \frac{(1-\varepsilon)\,s_0}{\sqrt{c_A s_B}}\,\frac1{\lvert\mathbf q\rvert^{2}}\quad(\lvert\mathbf q\rvert\le q_1).
\tag{4.9}
$$

So under (4.5) a label component of a long-range mark cannot form before a $\lambda\propto\lvert\mathbf q\rvert^{-2}$, for $B=A$ and for every source of note (a). For any $B$ it is excluded within $\lambda\le\beta/\lvert\mathbf q\rvert$ whenever $\lvert\mathbf q\rvert^4\chi_B(\mathbf q)\to0$. The highest eigenvector gives the same, with $\lvert f_A\rvert$ and $\lvert\chi\rvert$.

### Step 5. Oscillator cells: the made field and the criterion (item 3)

**Class** (M8; the question's item 3 with its notes (c) and (d)). $H=\tfrac12\sum_xp_x^2+\tfrac12\sum_{x,y}q_xK_{xy}q_y$ and $A_x=B_x=q_x$. $K$ has finite range and is invariant under the translations and under the permutations and reflections of the axes. $\hat K(\mathbf k)=m_h^2+\hat K_0(\mathbf k)$ with $\hat K_0(\mathbf k)=c_0^2\lvert\mathbf k\rvert^2+O(\lvert\mathbf k\rvert^4)$ (isotropic at second order by the cubic symmetry), and $m_h,c_0>0$. $\hat K$ is real and even, and $\Omega_{\mathbf k}=\hat K(\mathbf k)^{1/2}$ are the normal-mode frequencies. Two further conditions define the class:
- (c) a single soft point, $\hat K_0(\mathbf k)>0$ for $\mathbf k\ne0$;
- (d) on $\mathbf k\ne0$, at $m_h\to0$:
  - the phases $\rho k_1\pm\lambda\Omega_{\mathbf k}$ have finitely many caustic values of $\rho/\lambda$;
  - at each degenerate stationary point there is one degenerate direction, with a nonzero third derivative along it (a fold);
  - the critical points of $\Omega$ are nondegenerate.

Since $\Omega$ depends smoothly on $m_h^2$ on $\lvert\mathbf k\rvert\ge\delta$, (d) persists for small $m_h$. The other members of the oscillator class are out of scope by notes (c) and (d). Limit: $M\to\infty$ first. The lowest eigenvector and the first-order response of the compression $C_M$ then become those of $H$ (a stated limit, standard for compressions onto growing sets of low levels).

For $H+g_sq_0$, quadratic plus linear, the equations of the means are closed and exact: $\mathrm d\langle q\rangle/\mathrm d\lambda=\langle p\rangle$, $\mathrm d\langle p\rangle/\mathrm d\lambda=-K\langle q\rangle-g_se_0$. The ground state of $H$ is invariant under $(q,p)\to(-q,-p)$, so $\langle q\rangle=\langle p\rangle=0$ at $\lambda=0$, and $\langle q\rangle_\lambda=-g_sK^{-1}(1-\cos K^{1/2}\lambda)e_0$, exactly linear in $g_s$. The lowest eigenvector of $H+g_sq_0$ is the ground state displaced by $-g_sK^{-1}e_0$. Hence $\delta a_{\mathbf k}(\lambda)=-(1-\cos\Omega_{\mathbf k}\lambda)/\Omega_{\mathbf k}^2$, $\delta a^{\rm st}_{\mathbf k}=-1/\Omega_{\mathbf k}^2$, and for $L\to\infty$

$$
u(x,\lambda):=\delta a(x,\lambda)-\delta a_{\rm st}(x)=\int_{[-\pi,\pi]^D}\frac{\mathrm d^D k}{(2\pi)^D}\,e^{i\mathbf k\cdot x}\,\frac{\cos\Omega_{\mathbf k}\lambda}{\Omega_{\mathbf k}^2}.
\tag{4.10}
$$

(With (4.3): $f_A=1$, $\chi_A=\Omega_{\mathbf q}^{-2}$, and (4.4) reads $1-\cos\Omega_{\mathbf q}\lambda\le\lambda\Omega_{\mathbf q}$.)

**Criterion** (stated before use; places $\rho\mathbf e_1$, $\rho\in\mathbb N$, along an axis):

$$
\Delta_3(\rho,\lambda)=\frac{\delta a(\rho\mathbf e_1,\lambda)}{\delta a_{\rm st}(\rho\mathbf e_1)}-1\ \ (D=3),\qquad
\Delta_1(\rho,\lambda)=\frac{\nabla\delta a(\rho,\lambda)}{\nabla\delta a_{\rm st}(\rho)}-1,\ \ \nabla v(\rho)=v(\rho+1)-v(\rho)\ \ (D=1);
\tag{4.11}
$$

the made field at distance $\rho$ has formed to accuracy $\varepsilon\in(0,1)$ from $\lambda_f$ on if $\lvert\Delta_D(\rho,\lambda)\rvert\le\varepsilon$ for every $\lambda\ge\lambda_f$; $\lambda_{\rm form}(\rho;\varepsilon)$ is the least such $\lambda_f$. The first change arrives at $\lambda_{\rm arr}(\rho;\eta)$, the first $\lambda$ with $\lvert\Delta_D+1\rvert\ge\eta\rho^{-1/2}$ ($\eta>0$ fixed; a threshold that decreases with the distance, Step 7). In $D=3$ the static mark at $1\ll\rho\ll c_0/m_h$ is dominated by its inverse-distance part. In $D=1$ it is a constant plus a part linear in $\rho$, and the difference removes the constant, as the gradient does in (8.10). The square torus is out of scope by note (b) of the question.

### Step 6. Scaling limit and the formation speed (item 3)

**Order of limits.** $g_s\to0$, $M\to\infty$, $L\to\infty$ (or $\lambda$ below the wrap time), then $\rho\to\infty$ along $\mathbf e_1$ at fixed $w=c_0\lambda/\rho$ and fixed $\mu=m_h\rho/c_0$, then $\mu$ small. Thus $m_h=\mu c_0/\rho\to0$ with $\hat K_0$ fixed: the distances are large against the lattice spacing and a fixed fraction $\mu$ of the pinning length $c_0/m_h$.

*(a) Small labels*, $\lvert\mathbf k\rvert<\delta$. With $\mathbf k=\mathbf p/\rho$: $\Omega_{\mathbf k}\lambda=w\sqrt{\mu^2+p^2}\,(1+O(p^2/\rho^2))$ and $\Omega_{\mathbf k}^{-2}=\rho^2c_0^{-2}(\mu^2+p^2)^{-1}(1+O(p^2/\rho^2))$. So (4.10) and the static mark tend to $\rho^{2-D}c_0^{-2}$ times the integrals $\int\mathrm d^Dp\,(2\pi)^{-D}e^{ip_1}\cos(w\sqrt{\mu^2+p^2})/(\mu^2+p^2)$ at $w$ and at $w=0$. These are standard Fourier integrals of explicit functions (M5). In $D=3$ (value at unit distance) and in $D=1$ (derivative in the place at unit distance) they give the same ratio

$$
\Delta_\infty(w,\mu)=\begin{cases}-1, & w<1,\\[2pt] e^{\mu}-1-\mu e^{\mu}\,I_\mu\bigl(\mu\sqrt{w^2-1}\bigr), & w>1,\end{cases}\qquad I_\mu(Y)=\int_0^Y\frac{J_1(y)\,\mathrm dy}{\sqrt{y^2+\mu^2}} .
\tag{4.12}
$$

In $D=3$, $\partial_w$ of the first integral is $-(4\pi)^{-1}[\delta(w-1)-\theta(w-1)\,\mu J_1(\mu\sqrt{w^2-1})/\sqrt{w^2-1}]$, and its value at $w=0$ is $e^{-\mu}/(4\pi)$. In $D=1$, $\partial_w$ of the place derivative is $\tfrac12[\delta(w-1)-\theta(w-1)\,\mu J_1(\mu\sqrt{w^2-1})/\sqrt{w^2-1}]$, starting from $-e^{-\mu}/2$. The substitution $y=\mu\sqrt{w'^2-1}$ gives $I_\mu$. The lobes of $J_1$ between its zeros alternate in sign and decrease in weight against $1/\sqrt{y^2+\mu^2}$, so $0\le I_\mu(Y)\le I_\mu(j_{1,1})<1.1$ for all $Y\ge0$ (checked numerically for $10^{-3}\le\mu\le1$: $\min I_\mu=0$, $\max I_\mu\le1.092$). Write $\Delta_\infty=(e^\mu-1-\mu e^\mu)+\mu e^\mu(1-I_\mu)$. Then $0\le1-e^{\mu}(1-\mu)\le\mu^2e^\mu/2$ and $\lvert1-I_\mu\rvert\le1$ give

$$
\Delta_\infty(w,\mu)=-1\ \ (w<1),\qquad \lvert\Delta_\infty(w,\mu)\rvert\le\mu e^{\mu}\bigl(1+\tfrac{\mu}{2}\bigr)\ \ (w>1).
\tag{4.13}
$$

*(b) Uniform estimates.* Fix $\eta>0$. A smooth cutoff splits (4.10) (and its difference $\nabla$ in $D=1$) into $\lvert\mathbf k\rvert<2\delta$ and $\lvert\mathbf k\rvert>\delta$, with $\delta=\delta(\eta)$ small.

On $\lvert\mathbf k\rvert>\delta$ the amplitude $\Omega_{\mathbf k}^{-2}$ (times $e^{ik_1}-1$ for $\nabla$) is smooth in $\mathbf k$ and $m_h$. The phases are $\rho k_1\pm\lambda\Omega_{\mathbf k}=\lambda(vk_1\pm\Omega_{\mathbf k})$, $v=\rho/\lambda=c_0/w$, with stationary points where $\nabla\Omega_{\mathbf k}=\mp v\,\mathbf e_1$, transverse components included. Let $v_{\rm front}$ be the largest $\lvert\partial_{k_1}\Omega_{\mathbf k}\rvert$ at points with $\partial_{k_j}\Omega_{\mathbf k}=0$ for all $j\ge2$ (at $m_h\to0$).
- For $w\ge1+\eta$ the large parameter is $\lambda\ge(1+\eta)\rho/c_0$, and $v$ lies in the compact set $[0,c_0/(1+\eta)]$.
- For $w\le1-\eta$ the large parameter is $\rho$, and $\lambda/\rho$ lies in $[0,(1-\eta)/c_0]$.

By (d), on these parameter sets the stationary points are nondegenerate or folds, with finitely many caustic values, and at $v=0$ they are the critical points of $\Omega$, which are nondegenerate. Stationary phase uniform in the parameter, with Airy functions at the folds, bounds this part by $O(\lambda^{-D/2})$ off the caustic values and by $O(\lambda^{-(D-1)/2-1/3})$ everywhere. Relative to the static value ($\propto\rho^{-1}$, $D=3$) or gradient ($\propto\rho^0$, $D=1$) this is $O(\rho^{-1/2})$, and $O(\rho^{-1/3})$ near caustic values, uniformly on each range; for $w\ge1+\eta$ it decreases with $w$.

On $\delta\le\lvert\mathbf k\rvert<2\delta$ and down to the scaling region $\lvert\mathbf k\rvert=P/\rho$, $\lvert\nabla\Omega_{\mathbf k}\rvert=c_0(1+O(\delta^2)+O(\mu^2/P^2))$. So for $\delta$ small and $P$ large there is no stationary point at $\lvert w-1\rvert\ge\eta$, and this part is small by non-stationary phase. In the scaling region, (a) holds uniformly on compact sets of $w$ (dominated convergence). For large $w$, the only critical point of $\nu(\mathbf p)=\rho\Omega_{\mathbf p/\rho}/c_0$ is $\mathbf p=0$, where $\nu=\mu$ exactly and the Hessian is the continuum one up to $O(\rho^{-2})$. Stationary phase in $w$ therefore reproduces the large-$w$ behaviour of (a) with remainders uniform in $\rho$.

**Members** (short argument). The nearest-neighbour contract and the control (4.16) are separable: $\hat K_0=\sum_ig(k_i)$ with $g(k)=a(1-\cos k)+b(1-\cos2k)$, where $(a,b)=(2\kappa_h,0)$ for the nearest neighbour and $(14,-3)$ for the control. Here $g'=\sin k\,(a+4b\cos k)$ with $a+4b\cos k>0$ ($2\kappa_h$, resp. $14-12\cos k\ge2$), so on $[0,\pi]$ $g'$ vanishes only at $0$ and $\pi$. Moreover $g''(0)=a+4b=2c_0^2>0$ and $g''(\pi)=4b-a<0$. Hence:
- (i) The critical points of $\Omega$ away from $0$ have all $k_i\in\{0,\pi\}$, with the diagonal, nonzero Hessian $g''(k_i)/(2\Omega)$, so they are nondegenerate.
- (ii) $\partial_{k_j}\Omega=g'(k_j)/(2\Omega)=0$ ($j\ge2$) puts the stationary points on the branches $\mathbf k_\perp\in\{0,\pi\}^{D-1}$. There the Hessian of $\Omega$ is diagonal, with transverse entries $g''(k_j)/(2\Omega)\neq0$ and axial entry $\Omega_b''$, where $\Omega_b(k_1)=(m_h^2+c_b+g(k_1))^{1/2}$ and $c_b=\sum_{j\ge2}g(k_j)$. So there is at most one degenerate direction, $\mathbf e_1$.
- (iii) $\Omega_b''=N_b/(4\Omega_b^3)$ with $N_b=2g''\Omega_b^2-g'^2$, a trigonometric polynomial that is not identically zero, so it has finitely many zeros and finitely many caustic values. Since $N_b'=2g'''\Omega_b^2$, $\Omega_b'''=g'''/(2\Omega_b)$ at every zero. $N_b(\pi)=2g''(\pi)\Omega_b^2<0$, and $N_b(0)=2g''(0)c_b>0$ on the side branches, so the zeros lie in $(0,\pi)$.
  - Nearest neighbour: $g'''=-a\sin k\neq0$ on $(0,\pi)$.
  - Control: $g'''=\sin k\,(48\cos k-14)$ vanishes in $(0,\pi)$ only at $\cos k=7/24$, where $N_b>0$ on every branch.

So every zero of $N_b$ is simple, a fold, and both contracts are members, in $D=1$ and in $D=3$. For the nearest neighbour at $m_h=0$, $N_b=-a^2[\cos^2k-2\alpha_b\cos k+1]$ with $\alpha_b=1+c_b/a\in\{1,3,5\}$. The main branch ($\alpha_b=1$) has no zero for $k\ne0$, and the side branches have one zero each, at $\cos k=\alpha_b-\sqrt{\alpha_b^2-1}$, with caustic values $w=(\alpha_b+\sqrt{\alpha_b^2-1})^{1/2}=1+\sqrt2$ and $\sqrt2+\sqrt3$ ($D=3$). For the control the caustic values are $w=0.404$ (the front), and in $D=3$ also $0.757$ and $0.975$, all below 1 (`code/v4/output/oscillator_formation.json`, `fold_check`).

Near $w=1$ the stationary points approach the soft point $\mathbf k=0$, which is outside (d). There the $O(\lvert\mathbf k\rvert^4)$ term of $\hat K_0$ smooths the step of (4.12) over a window $\lvert c_0\lambda-\rho\rvert=O(\rho^{1/3})$, from labels $\lvert\mathbf k\rvert=O(\rho^{-1/3})$. For a dispersion that is concave near $0$ (the nearest neighbour), a wake of relative size $O(\rho^{-1/2}(w-1)^{-3/4})$ follows the window, from the stationary point at $\lvert\mathbf k\rvert\propto(w-1)^{1/2}$.

Hence, for every $K$ of the class, $\Delta_D(\rho,\rho w/c_0)\to\Delta_\infty(w,\mu)$ uniformly on $w\ge1+\eta$, and $\Delta_D\to-1$ uniformly on $0\le w\le1-\eta$, for every $\eta>0$. The lattice part contributes $O(\rho^{-1/3})$ at worst. With (4.13): for every $\varepsilon\in(0,1)$ and every $\mu$ with $\mu e^\mu(1+\mu/2)<\varepsilon$,

$$
\lim_{\rho\to\infty}\frac{c_0\,\lambda_{\rm form}(\rho;\varepsilon)}{\rho}=1,\qquad \lambda_{\rm form}(\rho;\varepsilon)-\frac{\rho}{c_0}=O(\rho^{1/3})\ \ \text{once the caustic wakes }O(\rho^{-1/3})\text{ are below }\varepsilon-\mu e^\mu(1+\tfrac{\mu}{2}) ,
\tag{4.14}
$$

on the ring ($D=1$) and on the cubic torus ($D=3$). The long-range part forms within a $\lambda$ linear in the distance, at the formation speed $v_{\rm form}=c_0$, independent of $\varepsilon$. For $\lambda<(1-\eta)\rho/c_0$ it has not formed ($\Delta_D\to-1$).

*(c) Control of item 2* (short argument; $D=3$; class (c), note (d) not needed). In the mean (4.19) only labels $\lvert\mathbf q\rvert\lesssim1/\ell$ enter. By Parseval, $\sum_x\varphi_\ell\,u(x,\lambda)=\int\mathrm d^3q\,(2\pi)^{-3}\,\overline{\hat\varphi_\ell(\mathbf q)}\cos(\Omega_{\mathbf q}\lambda)/\Omega_{\mathbf q}^2$, and on the zone $\hat\varphi_\ell(\mathbf q)=\ell^3\hat\Phi(\ell\mathbf q)+O(\ell^{-\infty})$. Labels with $\lvert\mathbf q\rvert\ge\delta$ contribute $O(\ell^{-\infty})$ uniformly in $\lambda$, since $\Omega_{\mathbf q}^{-2}\le1/\min_{\lvert\mathbf q\rvert\ge\delta}\hat K_0$ by (c). With $\mathbf q=\mathbf p/\ell$, the rest is $\ell^2c_0^{-2}\int\mathrm d^3p\,(2\pi)^{-3}\,\overline{\hat\Phi(\mathbf p)}\cos(w\nu(\mathbf p))/\nu(\mathbf p)^2$, with $w=c_0\lambda/\ell$, $\mu=m_h\ell/c_0$, and $\nu\to\sqrt{\mu^2+p^2}$ with all derivatives on bounded sets. This converges to its continuum form uniformly in $w\ge0$: by dominated convergence for bounded $w$, and for large $w$ by stationary phase at $\mathbf p=0$, where $\nu=\mu$ exactly. The continuum form is the mean of $\delta a_{\rm st}(x)\Delta_\infty(c_0\lambda/\lvert x\rvert,m_h\lvert x\rvert/c_0)$, by (a), which is isotropic. Hence

$$
\frac{\sum_x\varphi_\ell(\delta a-\delta a_{\rm st})}{\sum_x\varphi_\ell\,\delta a_{\rm st}}\ \longrightarrow\ S_\infty(w,\mu)=\frac{\int\Phi(y)\,e^{-\mu\lvert y\rvert}\lvert y\rvert^{-1}\,\Delta_\infty(w/\lvert y\rvert,\mu\lvert y\rvert)\,\mathrm d^3y}{\int\Phi(y)\,e^{-\mu\lvert y\rvert}\lvert y\rvert^{-1}\,\mathrm d^3y}\quad\text{uniformly in }w\ge0 .
\tag{4.20}
$$

On the shell $1\le\lvert y\rvert\le2$, $S_\infty=-1$ for $w<1$, and by (4.13) $\lvert S_\infty\rvert\le2\mu e^{2\mu}(1+\mu)$ for $w>2$. So for every $\varepsilon\in(0,1)$ with $2\mu e^{2\mu}(1+\mu)<\varepsilon$ and every $\eta>0$, $(1-\eta)\ell/c_0\le\lambda^\Phi_{\rm form}(\ell;\varepsilon)\le(2+\eta)\ell/c_0$ for all large $\ell$. In this member of a stated limit ($M\to\infty$), the long-range part forms in the mean within a $\lambda$ linear in the distance.

### Step 7. Formation speed, slope and front

Let $v_{\rm slope}=\sup_{\mathbf k}\lvert\nabla\hat K_0(\mathbf k)^{1/2}\rvert$, the largest slope of the normal-mode frequencies, and $v_{\rm proj}=\sup_{\mathbf k}\partial_{k_1}\hat K_0(\mathbf k)^{1/2}$ (both at $m_h\to0$, the regime of Step 6), with $v_{\rm front}$ as in Step 6(b). On the axis $\mathbf k=k_1\mathbf e_1$ the reflections $k_j\to-k_j$ give $\partial_{k_j}\Omega=0$ ($j\ge2$), and $\partial_{k_1}\Omega\to c_0$ as $k_1\to0^+$. A supremum over the stationary points of the axial ray is at most one over all $\mathbf k$, and $\partial_{k_1}\le\lvert\nabla\rvert$. Hence

$$
v_{\rm form}=c_0\ \le\ v_{\rm front}\ \le\ v_{\rm proj}\ \le\ v_{\rm slope}.
\tag{4.15}
$$

The inequalities hold for every $K$ of the class; $v_{\rm form}=c_0$ under the conditions of (4.14). $v_{\rm proj}$ ignores the transverse conditions. For separable $\hat K_0=\sum_ig(k_i)$ the stationary points lie on the branches of Step 6(b), where $\Omega_b'=g'/(2\Omega_b)$ is largest for $c_b=0$. So $v_{\rm front}=v_{\rm proj}=\max_kg'(k)/(2g(k)^{1/2})$.

**Arrival.**
- (i) For $\rho/\lambda>v_{\rm proj}$ the made field at $\rho\mathbf e_1$ is exponentially small in $\rho$. Its integrand $e^{ik_1\rho}(1-\cos\Omega_{\mathbf k}\lambda)/\Omega_{\mathbf k}^2$ is entire in $\hat K(\mathbf k)$, and $\lvert(1-\cos z)/z^2\rvert\le2e^{\lvert{\rm Im}\,z\rvert}$. Shifting $k_1\to k_1+i\sigma$ therefore bounds the made field by $2\lambda^2\exp[-\sigma(\rho-v_{\rm proj}\lambda)+O(\sigma^2\lambda)]$; the same holds for $\nabla$.
- (ii) For $c_0/v_{\rm front}<w\le1-\eta$ the lattice part has stationary points (Step 6(b)), and the small-label part has none and tends to its continuum value $0$. The made field there is a dispersive precursor of vanishing relative size: $\asymp\rho^{-1/3}$ at its fold edge $\lambda\simeq\rho/v_{\rm front}$ (when $v_{\rm front}>c_0$, a fold by (d) with nonzero amplitude $\Omega^{-2}$), and $O(\rho^{-1/2})$ behind it, with a constant that grows as $w\to1$.

With the threshold $\eta\rho^{-1/2}$ of (4.11) and $v_{\rm front}=v_{\rm proj}$, therefore $\lambda_{\rm arr}(\rho;\eta)v_{\rm front}/\rho\to1$. With a fixed threshold the precursor falls below it at large $\rho$, and by the uniform estimates of Step 6(b) the arrival then moves with $c_0$. For the nearest-neighbour contract, $\hat K_0=2\kappa_h\sum_i(1-\cos k_i)$ and $\lvert\nabla\hat K_0^{1/2}\rvert^2=\kappa_h\sum_i\sin^2\tfrac{k_i}2\cos^2\tfrac{k_i}2/\sum_i\sin^2\tfrac{k_i}2\le\kappa_h$. So $c_0=v_{\rm front}=v_{\rm proj}=v_{\rm slope}=\sqrt{\kappa_h}$: arrival and formation coincide, as in (8.15). **Control** (a member of the class with $c_0\ne v_{\rm slope}$):

$$
\hat K_0(\mathbf k)=\sum_{i}\bigl[14(1-\cos k_i)-3(1-\cos2k_i)\bigr]=\sum_i(1-\cos k_i)(8-6\cos k_i),\qquad c_0=1,\quad v_{\rm front}=v_{\rm proj}=v_{\rm slope}=2.475 .
\tag{4.16}
$$

It has range 2 along the axes, the symmetries of the class, $\hat K_0\ge2\sum_i(1-\cos k_i)$ (note (c)), and it is a member of note (d) (Step 6(b)). The speeds are computed by `code/v4/oscillator_formation.py`. There the precursor runs ahead of the formation by the factor 2.475, with vanishing relative amplitude.

### Step 8. Speeds of bodies and field (item 4)

Let $\ell_0$ be the recorded distance of one step of the body's chain ($\ell_0=\kappa\sigma_X$ in the setting of (5.11)); by (5.11) $v_{\max}=\tau\ell_0$. By the coupling map (4.15) of 06-ilang-source/04-profile, one chain step is $k$ field steps along the ring (along a lattice axis of the tori, a stated assumption, M8). So one field step is the recorded distance $\ell_0/k$, and the formation speed $c_0$ (field steps per unit $\lambda$) is $c_0\ell_0/k$. Hence

$$
\frac{v_{\rm form}}{v_{\max}}=\frac{c_0}{k\,\tau},\qquad v_{\rm form}=v_{\max}\iff c_0=k\tau\quad(\text{nearest neighbour: }\sqrt{\kappa_h}=k\tau),
\tag{4.17}
$$

independent of $\ell_0$; the front moves at $v_{\rm front}\ell_0/k\ge v_{\rm form}$. Equality of the two speeds is a condition on the description (M7).

### Step 9. Numerical check (code)

Criterion (4.11) with $\varepsilon=0.1$; all numbers are from `code/v4/output/`. Every computed formation value is a finite-window value $\lambda^{\rm win}_{\rm form}$: the last threshold crossing on $[0,\lambda_{\max}]$, before the first image front, with window end $w_{\max}=c_0\lambda_{\max}/\rho$. Horizon check: the same on $[0,\lambda_{\max}/2]$, and the largest $\lvert\Delta\rvert$ on $[\lambda_{\max}/2,\lambda_{\max}]$.

*(i) Oscillator cells* (`oscillator_formation.py`; $c_0=1$; ring $L=16384$, $m_h=5\cdot10^{-5}$; cubic torus $L=256$, $m_h=2.5\cdot10^{-4}$; so $\mu\le0.026$).
- **Windows.** On the ring, $w_{\max}$ falls from 893 ($\rho=16$) to 27.9 ($\rho=512$) for the nearest neighbour, and from 361 to 11.3 for the control. On the torus, $w_{\max}=23.3,\,11.6,\,7.8,\,3.9$ (nearest neighbour) and $9.4,\,4.7,\,3.1,\,1.6$ (control) for $\rho=8,16,24,48$.
- **Horizon check.** The half-window value equals the full-window value in every case except $\rho=48$ on the torus. On the second half of the window the largest $\lvert\Delta\rvert$ is at most 0.029 on the ring, and at most 0.086 on the torus for $\rho\le24$.
- **Images.** By Poisson summation the torus made field is the infinite-lattice made field plus its images at $x+nL$, $n\ne0$. Before the image front these are exponentially small (Step 7(i)), not zero. Size-refinement check: $\Delta_D$ recomputed on $L'=32768$ (ring) and $L'=384$ (torus), at every 20th $\lambda$ of the window, differs from its value on $L$ by at most $4\cdot10^{-10}$ on the ring, and on the torus by at most $1.1\cdot10^{-8}$ (nearest neighbour) and $6.1\cdot10^{-6}$ (control, $\rho=48$; at most $4\cdot10^{-13}$ for $\rho\le24$). At the sampled $\lambda$ the image contamination is therefore negligible against $\varepsilon=0.1$ on every window.
- **Control (4.16).** $c_0\lambda^{\rm win}_{\rm form}/\rho=1.266,\,1.162,\,1.089,\,1.053,\,1.031,\,1.018$ for $\rho=16,32,\dots,512$ ($D=1$), and $1.344,\,1.191,\,1.154,\,1.082$ for $\rho=8,16,24,48$ ($D=3$).
  - The arrival (threshold $0.1\rho^{-1/2}$) gives $v_{\rm front}\lambda_{\rm arr}/\rho$ from 0.928 to 0.985 ($D=1$) and from 0.743 to 0.882 ($D=3$).
  - The precursor amplitude is $\max\lvert\Delta_D+1\rvert$ on $1.2\,c_0/v_{\rm front}\le w\le0.8$. In $D=1$ this interval stays a fixed distance from the only caustic value $w=0.404$ (the fold at the front) and from $w=1$, so its stationary points are nondegenerate, and Step 6(b) gives $O(\rho^{-1/2})$ there asymptotically. The computed amplitude times $\rho^{1/3}$ lies between 0.66 and 0.72 for $\rho=32,\dots,512$. It decreases, but its near-constant product with $\rho^{1/3}$ is a finite-distance observation, not an asymptotic exponent. In $D=3$ it is $0.41,\,0.38,\,0.37,\,0.38$ for $\rho=8,\dots,48$ and is not yet decreasing, since $w=0.8$ is still within three front windows $\rho^{1/3}$ of the front.
- **Nearest neighbour.** $c_0\lambda^{\rm win}_{\rm form}/\rho=2.47,\,1.72,\,1.43,\,1.26,\,1.16,\,1.13$ ($D=1$, $\rho=16,\dots,512$), with $(\lambda_{\rm form}-\rho/c_0)/\rho^{1/3}$ between 6.4 and 9.3, bounded as (4.14) states.
  - In $D=3$: $4.22,\,3.64,\,3.72$ for $\rho=8,16,24$, and $3.14$ at $\rho=48$ ($w_{\max}=3.88$). These are set by the caustic wakes at $w=1+\sqrt2$ and $\sqrt2+\sqrt3$.
  - The largest $\lvert\Delta_3\rvert$ on $1.5\le w\le3$ falls as $0.238,\,0.197,\,0.178,\,0.130$. The ratio from $\rho=8$ to 48 is 0.546, against $6^{-1/3}=0.550$, consistent with the fold estimate $O(\rho^{-1/3})$ of Step 6(b) (the range contains the caustic value $1+\sqrt2$). It extrapolates below $\varepsilon=0.1$ at $\rho\approx10^2$.
- **Membership** (`fold_check`). The zeros of $N_b$ and the caustic values are those of Step 6(b), with $\Omega_b'''=-0.414,\,-0.318$ there (nearest neighbour, side branches) and $-2.35,\,-2.28,\,-1.83$ (control).

In short,

$$
\frac{c_0\lambda^{\rm win}_{\rm form}}{\rho}\Big\vert_{\rm control}=1.018\ (D=1,\ \rho=512),\ \ 1.082\ (D=3,\ \rho=48);\qquad \frac{c_0\lambda^{\rm win}_{\rm form}}{\rho}\Big\vert_{\rm n.n.}=1.13\ (D=1,\ \rho=512),\ \ 3.1\text{–}4.2\ (D=3,\ \rho\le48).
\tag{4.18}
$$

Static marks: the relative deviation from the closed or Bessel form is $1\cdot10^{-9}$ ($D=1$) and $4\cdot10^{-13}$ ($D=3$). $I_\mu$: $\min=0$, $\max\le1.092$, $\sup\lvert1-I_\mu\rvert=1$ for $\mu=10^{-3},\dots,1$.

*(ii) Field of 06-ilang-source* (`conserved_density_formation.py`; $t_f=1$, infinite lattice; the criterion of (8.10) in both dimensions, i.e. the ratio of the radial lattice differences along an axis; window $\lambda\le3\rho^2/(4\zeta_D)$).
- $\lambda^{\rm win}_{\rm form}/[\rho^2/(4\zeta_D(0.1))]=1.686,\,1.320,\,1.155,\,1.076$ for $\rho=4,8,16,32$ ($D=1$, $\zeta_1(0.1)=0.01555$), and $2.291,\,1.730,\,1.391,\,1.209,\,1.108$ for $\rho=4,8,16,32,64$ ($D=3$, $\zeta_3(0.1)=0.4068$). The excess over 1 halves with each doubling of $\rho$, an $O(1/\rho)$ lattice correction.
- The local exponents of $\lambda_{\rm form}$ in $\rho$ over the last two distances are 1.90 ($D=1$) and 1.87 ($D=3$).
- Horizon check: the half-window value equals the full-window value for $\rho\ge8$ ($D=1$) and $\rho\ge16$ ($D=3$).
- On the linear range $2\rho\le\lambda\le4\rho$ the largest $\lvert\Delta\rvert$ is 1.37 ($D=1$, $\rho=32$) and 3.60 ($D=3$, $\rho=64$): there is no linear formation, as (8.14) states and consistent with (4.8).

## Result

Class F and its source and start as stated (Ω the nondegenerate lowest eigenvector of $C_F$), first order in $g_s$ throughout.

- **(4.1), (4.2)** (formula; every $L$, every $\lambda\ge0$): the made field and its label components in the eigenvectors of $C_F$; the static mark is the same with $g_\lambda(\omega)\to1/\omega$.
- **(4.3)** (definition): the sum-rule weight $f_A(\mathbf q)$ and the static self-response $\chi_Y(\mathbf q)$, fixed by $C_F$, the read or source operator and Ω; $\delta a^{\rm st}_{\mathbf q}=-\chi_A(\mathbf q)$ for $B=A$.
- **(4.4)** (theorem, item 1): $\lvert\delta a_{\mathbf q}(\lambda)\rvert\le\min\{\lambda\sqrt{f_A\chi_B},\,2\sqrt{\chi_A\chi_B}\}$ for every $\mathbf q$, $L$ and $\lambda\ge0$; for $B=A$, $\lvert\delta a_{\mathbf q}(\lambda)/\delta a^{\rm st}_{\mathbf q}\rvert\le\lambda\sqrt{f_A/\chi_A}$. It holds for the nondegenerate highest eigenvector with $\lvert f_A\rvert,\lvert\chi\rvert$.
- **(4.5)** (theorem): if $N=\sum_xA_x$ commutes with $C_F$ and the terms that do not commute with the $A_x$ have finite range, then $0\le f_A(\mathbf q)\le c_A\lvert\mathbf q\rvert^2$ for every $L$, with $c_A=\tfrac12\sum_d\lvert d\rvert^2\lvert F_A(d)\rvert\le\lVert A_0\rVert\,\lVert[C_F,A_0]\rVert\sum_{0<\lvert d\rvert\le R_A}\lvert d\rvert^2$, fixed by $C_F$, $A$ and Ω (the bound by $C_F$ and $A$); for the highest eigenvector with $\lvert f_A\rvert$. $c_A\le\bar c_A$ uniformly in $L$ if $R_A$, the number and the norms of the terms of $C_F$ that do not commute with $A_0$, and $\lVert A_0\rVert$ are bounded uniformly in $L$ (note (f)).
- **(4.6)** (theorem, one label component): under (4.5), for every source operator $B$, formation of the label component $\mathbf q$ to accuracy $\varepsilon$ needs $\lambda\ge(1-\varepsilon)\lvert\delta a^{\rm st}_{\mathbf q}\rvert/(\lvert\mathbf q\rvert\sqrt{c_A\chi_B(\mathbf q)})$.
- **(4.7)** (negative, one label component, $B=A$): with $\lvert\delta a^{\rm st}_{\mathbf q}\rvert\ge s_0/\lvert\mathbf q\rvert^2$ ($\lvert\mathbf q\rvert\le q_1$), $\lambda_{\rm form}(\mathbf q;\varepsilon)\ge(1-\varepsilon)\sqrt{s_0/c_A}\,\lvert\mathbf q\rvert^{-2}$; also for the highest eigenvector.
- **(4.8)** (negative, item 2; cubic torus, order $g_s\to0$, $L\to\infty$, then $\ell\to\infty$): for a conserved density as in (4.5), with any source $B$, $B=A$ included, with $\chi_B\le s_B/\lvert\mathbf q\rvert^2$ (notes (a), (f); for $B=A$ an assumption, $\chi_A\le s_A/\lvert\mathbf q\rvert^2$; several kinds of objects included), with $c_A\le\bar c_A$ and $s_B\le\bar s_B$ uniformly in $L$ (note (f)), and with a long-range mark $\delta a_{\rm st}=s/\lvert x\rvert+o(\lvert x\rvert^{-1})$, $s\ne0$, in the limit $L\to\infty$: every $\lambda$ at which the made field at distance $\ell$ has formed in the mean (4.19) satisfies $\lambda\ge C_\Phi\ell^2(1+o(1))$, with $C_\Phi>0$ independent of $L$, given in (4.8). Hence for every $\beta>0$, $\lambda^\Phi_{\rm form}(\ell;\varepsilon)>\beta\ell$ for all large $\ell$, and the same holds for formation at every place of the shell: the long-range part cannot form within a $\lambda$ linear in the distance; also for the highest eigenvector. Control: (4.20).
- **(4.9)** (negative, one label component, $B\ne A$; under (4.5), with a long-range mark): formation of $\mathbf q$ within $\lambda\le\beta/\lvert\mathbf q\rvert$ requires $\chi_B(\mathbf q)\ge(1-\varepsilon)^2s_0^2/(\beta^2c_A\lvert\mathbf q\rvert^4)$; for $\chi_B\le s_B/\lvert\mathbf q\rvert^2$, $\lambda_{\rm form}(\mathbf q;\varepsilon)\ge(1-\varepsilon)s_0/(\sqrt{c_As_B}\,\lvert\mathbf q\rvert^2)$ for $\lvert\mathbf q\rvert\le q_1$; also for the highest eigenvector.
- **(4.10)** (formula, oscillator class, $M\to\infty$ then $L\to\infty$): exact made field $\delta a_{\mathbf k}(\lambda)=-(1-\cos\Omega_{\mathbf k}\lambda)/\Omega_{\mathbf k}^2$ and its deviation from the static mark.
- **(4.11)** (definition): the formation criterion along an axis (value ratio in $D=3$, difference ratio in $D=1$), $\lambda_{\rm form}(\rho;\varepsilon)$, and the arrival $\lambda_{\rm arr}(\rho;\eta)$ with the threshold $\eta\rho^{-1/2}$.
- **(4.12), (4.13)** (formula, theorem; scaling limit $\rho\to\infty$ at fixed $w=c_0\lambda/\rho$, $\mu=m_h\rho/c_0$): $\Delta_D\to\Delta_\infty(w,\mu)$, equal to $-1$ for $w<1$ and bounded by $\mu e^\mu(1+\mu/2)$ for $w>1$, in $D=1$ and $D=3$.
- **(4.14)** (theorem, item 3; ring ($D=1$) and cubic torus ($D=3$); the square torus is out of scope by note (b)): for every $K$ of the oscillator class with a single soft point (note (c)) whose part away from $\mathbf k=0$ has only folds as degenerate stationary points, finitely many caustic values and nondegenerate critical points of $\Omega$ (note (d)), for every $\varepsilon\in(0,1)$ and $\mu e^\mu(1+\mu/2)<\varepsilon$: $c_0\lambda_{\rm form}(\rho;\varepsilon)/\rho\to1$. The long-range part forms within a $\lambda$ linear in the distance, at the speed $c_0$, independent of $\varepsilon$. The lattice correction is $O(\rho^{1/3})$ in $\lambda$ once the caustic wakes, $O(\rho^{-1/3})$ relative, are below $\varepsilon-\mu e^\mu(1+\mu/2)$. The nearest-neighbour contract is a member (caustic values: none in $D=1$; $w=1+\sqrt2$ and $\sqrt2+\sqrt3$ in $D=3$), and so is the control (4.16) (caustic values $w=0.404$, and in $D=3$ also $0.757$, $0.975$). Order: $g_s\to0$, $M\to\infty$, $L\to\infty$, $\rho\to\infty$, then $\mu$ small.
- **(4.15)** (theorem): $c_0\le v_{\rm front}\le v_{\rm proj}\le v_{\rm slope}$ for every $K$ of the class. Under the conditions of (4.14), $v_{\rm form}=c_0$. The made field is exponentially small ahead of $\rho/\lambda=v_{\rm proj}$. Between $\rho/v_{\rm front}$ and $(1-\eta)\rho/c_0$, for every $\eta>0$, it is a precursor of relative size $O(\rho^{-1/3})$. With the threshold $\eta\rho^{-1/2}$ the first change arrives with $v_{\rm front}$ when $v_{\rm front}=v_{\rm proj}$ (every separable $\hat K_0$); with a fixed threshold it arrives with $c_0$. For the nearest-neighbour contract all four speeds equal $\sqrt{\kappa_h}$.
- **(4.16)** (example, control of item 3): $\hat K_0=\sum_i(1-\cos k_i)(8-6\cos k_i)$ is a member of the class (notes (c), (d)) with $c_0=1<v_{\rm front}=v_{\rm proj}=v_{\rm slope}=2.475$.
- **(4.17)** (formula, item 4; ring, or a lattice axis of the tori by assumption): $v_{\rm form}/v_{\max}=c_0/(k\tau)$; the two speeds are equal if and only if $c_0=k\tau$.
- **(4.18)** (numerical, Step 9, $\varepsilon=0.1$, finite-window values with their windows):
  - For the control (4.16), $c_0\lambda^{\rm win}_{\rm form}/\rho$ tends to 1 (1.018 at $\rho=512$, $D=1$; 1.082 at $\rho=48$, $D=3$, window $w\le1.57$), and $v_{\rm front}\lambda_{\rm arr}/\rho$ tends to 1 (0.985 at $\rho=512$, $D=1$).
  - For the nearest-neighbour $K$, $c_0\lambda^{\rm win}_{\rm form}/\rho=1.13$ at $\rho=512$ ($D=1$), and lies between 3.1 and 4.2 in $D=3$ at $\rho\le48$, set by caustic wakes of relative size $O(\rho^{-1/3})$ (Step 6(b)).
  - For the field of 06-ilang-source with the criterion of (8.10), $4\zeta_D\lambda^{\rm win}_{\rm form}/\rho^2$ tends to 1 (1.076 at $\rho=32$, $D=1$; 1.108 at $\rho=64$, $D=3$).
- **(4.19)** (definition): formation in the mean at distance $\ell$ on the shell $\ell\le\lvert x\rvert\le2\ell$ (cubic torus), with weight $\Phi(x/\ell)$, and $\lambda^\Phi_{\rm form}(\ell;\varepsilon)$; implied by formation at every place of the shell.
- **(4.20)** (example, control of item 2; oscillator cells, $D=3$, class (c), $M\to\infty$, $L\to\infty$, $\ell\to\infty$ at fixed $\mu=m_h\ell/c_0$): the mean deviation tends to $S_\infty(w,\mu)$ uniformly in $w=c_0\lambda/\ell$, with $S_\infty=-1$ for $w<1$ and $\lvert S_\infty\rvert\le2\mu e^{2\mu}(1+\mu)$ for $w>2$. So for $2\mu e^{2\mu}(1+\mu)<\varepsilon<1$, $\lambda^\Phi_{\rm form}(\ell;\varepsilon)$ lies between $(1-\eta)\ell/c_0$ and $(2+\eta)\ell/c_0$ for all large $\ell$: formation within a linear $\lambda$ under criterion (4.19).

## Consistency checks

1. **Dimensions.** $g_sB$ is a rate, so $\delta a$ per unit $g_s$ has the units $[A][B][\lambda]$; $f_A\sim[A]^2/[\lambda]$ and $\chi_B\sim[B]^2[\lambda]$, so both bounds of (4.4) and (4.6) are consistent, and so is (4.8), where $c_As_B\sim[A]^2[B]^2$ per unit $\lambda$ per squared distance. $c_0$ is in field steps per unit $\lambda$ and $\tau$ is a rate, so (4.17) is a pure number.
2. **(8.10), (8.14) as a special case.** For the field of 06-ilang-source with one field object ($n=1/\lvert X_f\rvert$) and $A=B=\Phi$, $F_A(0)=4Dnt_f$, $F_A(\pm\mathbf e_i)=-2nt_f$, so $f_A=2n\omega_{\mathbf q}$ and (4.5) holds with $c_A=2Dnt_f$. (4.2) gives one mode per label, $\delta a_{\mathbf q}(\lambda)/\delta a^{\rm st}_{\mathbf q}=1-\cos\omega_{\mathbf q}\lambda$, and (4.8) a $\lambda$ quadratic in the distance, as (8.10) and (8.14). The code reproduces $\lambda_{\rm form}\to\rho^2/(4t_f\zeta_D)$ with the criterion of (8.10) (Step 9). It gives $\zeta_1(0.1)=0.01555$, against $\pi\varepsilon^2/2=0.01571$, and $\zeta_3(0.1)=0.4068$, against the leading term $(3\sqrt{2\pi}\varepsilon/4)^{2/3}=0.3282$ of (8.10), within its correction $O(\varepsilon^{2/3})$ ($\varepsilon^{2/3}=0.22$).
3. **(8.15) as a special case.** For the nearest-neighbour contract in $D=3$ with the value criterion, $c_0=\sqrt{\kappa_h}$, $w=c$ and $\mu=m_Y\rho$. (4.13) gives $\lvert\Delta\rvert\le\mu e^\mu(1+\mu/2)$ for $c>1$, the same order $O(m_Y\rho)$ as the bound $8.4\,m_Y\rho$ of (8.15). The remainder $O(\rho^{-1/2})$ of nondegenerate stationary points matches $o(1)=O(\rho^{-1/2})$ for $1<c<2$, below the first caustic value $c=1+\sqrt2$.

## Open issues

- Item 2: (4.8) excludes linear formation in the mean on a shell and at every place of a shell. Formation at a single place, or on a set that is not a positive fraction of a shell, is excluded only for the field of 06-ilang-source ((8.14) and the code).
- Item 3: the front window $\lvert c_0\lambda-\rho\rvert=O(\rho^{1/3})$ near the soft point rests on the scaling of the labels $\lvert\mathbf k\rvert=O(\rho^{-1/3})$, not on note (d). The approach to $v_{\rm form}=c_0$ is slow when the front coincides with $c_0$ (nearest neighbour), and at the distances computed in $D=3$ it is still dominated by the caustic wakes.
- Item 3, arrival: for non-separable members with $v_{\rm front}<v_{\rm proj}$, the decreasing-threshold arrival speed is only bracketed by $v_{\rm front}$ and $v_{\rm proj}$.
- Numerics: the computed formation values are finite-window values; the all-future statement rests on the uniform estimates of Step 6(b).
- Item 1 for eigenvectors inside the spectrum: (4.4) holds with $\sum_n\lvert\omega_n\rvert(\dots)$ in place of $f_A$, but (4.5) does not apply.
- Item 4 on the tori rests on the assumption that the coupling map runs along a lattice axis.

## Code

- `code/v4/oscillator_formation.py` (check, items 3 and 4's speeds; edited from v3 for I4 of round 2: docstring and image check). It computes:
  - the exact made field (4.10) on a ring ($L=16384$) and a cubic torus ($L=256$), and the static mark on the infinite lattice by its Laplace representation, cross-checked against the closed or Bessel form for the nearest-neighbour $K$;
  - the image check: $\Delta_D$ again on a larger torus ($L'=32768$, $L'=384$) at every 20th $\lambda$ of the grid, and the largest $\lvert\Delta_L-\Delta_{L'}\rvert$ on each window;
  - criterion (4.11) with $\varepsilon=0.1$, as finite-window values on the full window and on the half window, plus the largest $\lvert\Delta\rvert$ on the second half of the window;
  - the arrival with the threshold $0.1\rho^{-1/2}$ and the precursor amplitude;
  - the speeds $c_0$, $v_{\rm front}$ and $v_{\rm slope}$;
  - the membership in note (d) (`fold_check`: zeros of $N_b$, $\Omega_b'''$ there, caustic values);
  - the bounds on $I_\mu$ used in (4.13).

  It runs for the nearest-neighbour $K$ and for the control (4.16). Output: `code/v4/output/oscillator_formation.json`.
- `code/v4/conserved_density_formation.py` (check, item 2; unchanged from v3, run again). It computes the field of 06-ilang-source with one field object and $A=B=\Phi$, on the infinite chain and the infinite cubic lattice, from the Bessel-function forms of (4.1). The criterion is that of (8.10) in both dimensions (ratio of the radial lattice differences), with $\varepsilon=0.1$, as finite-window values with a half-window check. It compares them with $\rho^2/(4t_f\zeta_D(\varepsilon))$, where $\zeta_D$ is the first zero of $\lvert F_D\rvert-\varepsilon$, $F_1=V$, $F_3=2\zeta V'-V$. Output: `code/v4/output/conserved_density_formation.json`.

## Methods used

- first-order (Duhamel) perturbation, spectral representation
- sum rule of a double commutator, Cauchy–Schwarz inequality
- translation covariance, Fourier series on tori, Parseval and Poisson summation
- Fourier integrals with Bessel functions, scaling limits
- stationary phase uniform in a parameter, folds and Airy functions, contour shift
- exact equations of the means for a quadratic contract
- numerical lattice sums and Bessel-function integrals (Python)

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