Version 1 · current · External review, round 1: minor issues

# Recording contract: exact witness data and leading-order witness geometry

- **Subproject:** 03-ilang-space
- **Package:** 05-recording-contract
- **Version:** v1
- **Mode:** new
- **Date:** 2026-10-08

## Setup and assumptions

- Two objects $a$, $c$ of different types, each the only instance of its type. There is no pair of objects of the same type, so (1.11) and Law 5 impose no constraint and $\mathcal H_{\rm adm}=\mathcal H=\mathcal H_{T(a)}\otimes\mathcal H_c$. Part $A=\{a\}$, companion $\bar A=\{c\}$. $a$ has $d_{T(a)}$ places (finite, A1), and $d_c=\dim\mathcal H_c$.
- Recording contract $C=\sum_h\lvert h\rangle\langle h\rvert\otimes K_h$ with $K_h=K_h^\dagger$ on $\mathcal H_c$, independent of $\lambda$ (A5). This is a single pair term of the form (1.10), and $C^\dagger=\sum_h\lvert h\rangle\langle h\rvert\otimes K_h^\dagger=C$.
- Initial state $\lvert\Psi(0)\rangle=\lvert\phi\rangle\otimes\lvert\chi\rangle$, $\lvert\phi\rangle=\sum_h\phi_h\lvert h\rangle$, $\phi_h\neq0$ for all $h$, $\lVert\phi\rVert=\lVert\chi\rVert=1$. $\langle X\rangle:=\langle\chi\vert X\vert\chi\rangle$ and $\operatorname{Var}_\chi X:=\langle X^2\rangle-\langle X\rangle^2$.
- $\lambda\ge0$ is only the ordering parameter of (1.9) (M2). The only approximation is the expansion $\lambda\to0^+$ in Steps 4 and 5. Every other statement is exact for all $\lambda\ge0$.
- Inputs as quoted: (2.1), (2.2), (2.11) from 02-view-content@v1, and (3.1), (3.2), (3.5) from 03-witness-distance@v1, with the notation $p_h$, $H_V$, $\lvert\psi_h\rangle$, $G_{hh'}$, $\approx$, $X_V$, $\alpha$ fixed there. The $\lambda$-dependence is written explicitly: $p_h(\lambda)$, $W(h,h';\lambda)$, $\alpha(h,h';\lambda)$.

## Derivation

### Step 1. Exact evolution

The projectors $\Pi_h:=\lvert h\rangle\langle h\rvert$ satisfy $\Pi_h\Pi_{h'}=\delta_{hh'}\Pi_h$ and $\sum_h\Pi_h=\mathbb 1$ (place basis orthonormal, A2). By induction $C^n=\sum_h\Pi_h\otimes K_h^n$ for $n\ge0$ ($n=0$: $\sum_h\Pi_h\otimes\mathbb 1=\mathbb 1$). Summing the exponential series, which converges in finite dimension, gives

$$
e^{-iC\lambda}=\sum_h\lvert h\rangle\langle h\rvert\otimes e^{-iK_h\lambda},
\tag{5.1}
$$

and so by (1.9)

$$
\lvert\Psi(\lambda)\rangle=\sum_h\phi_h\,\lvert h\rangle\otimes e^{-iK_h\lambda}\lvert\chi\rangle .
\tag{5.2}
$$

### Step 2. Weights, $H_V$ and branch states

Comparing (5.2) with (2.1) gives $\lvert\psi_h(\lambda)\rangle=\phi_h e^{-iK_h\lambda}\lvert\chi\rangle$. Since $e^{-iK_h\lambda}$ is unitary and $\lVert\chi\rVert=1$, (2.2) gives

$$
p_h(\lambda)=\lVert\psi_h(\lambda)\rVert^2=\lvert\phi_h\rvert^2,\qquad H_V=H_{T(a)}\quad\text{for every }\lambda\ge0 ,
\tag{5.3}
$$

because $\phi_h\neq0$ for every $h$. The state (5.2) has the form (1.6) with $a_h=\phi_h$ and the unit branch states

$$
\lvert E_h(\lambda)\rangle=e^{-iK_h\lambda}\lvert\chi\rangle,\qquad
\varrho_h(\lambda)=e^{-iK_h\lambda}\lvert\chi\rangle\langle\chi\rvert e^{iK_h\lambda}.
\tag{5.4}
$$

Another phase split in (1.6) multiplies $\lvert E_h\rangle$ by a unit factor and leaves $\varrho_h$ (3.1) unchanged.

### Step 3. View and witness data

From (5.4) and (2.2) with $\lvert\psi_h\rangle=\phi_h\lvert E_h\rangle$:

$$
G_{hh'}(\lambda)=\langle E_{h'}\vert E_h\rangle=\langle\chi\vert e^{iK_{h'}\lambda}e^{-iK_h\lambda}\vert\chi\rangle,
\qquad
\bigl(V_A(\lambda)\bigr)_{hh'}=\phi_h\phi_{h'}^{*}\,G_{hh'}(\lambda).
\tag{5.5}
$$

By (2.11), $W=\lvert G_{hh'}\rvert^2$:

$$
W(h,h';\lambda)=\bigl\lvert\langle\chi\vert e^{iK_{h'}\lambda}e^{-iK_h\lambda}\vert\chi\rangle\bigr\rvert^2 .
\tag{5.6}
$$

$W$ does not depend on $\phi$. It is also independent of the conventions listed in M6. The common phase (1.2) and the phase split between $\lvert\phi\rangle$ and $\lvert\chi\rangle$ only multiply $\lvert\chi\rangle$ by a unit factor, and this factor cancels in (5.6).

### Step 4. Small-$\lambda$ expansion of $W$

Let $A:=K_h$, $B:=K_{h'}$ and $D:=A-B$. In finite dimension both exponentials are convergent power series, so Taylor's theorem with remainder applies:

$$
e^{iB\lambda}e^{-iA\lambda}=\mathbb 1-i\lambda D-\frac{\lambda^2}{2}\bigl(A^2+B^2-2BA\bigr)+O(\lambda^3).
\tag{5.7}
$$

Taking the expectation in $\lvert\chi\rangle$ gives $G_{hh'}=1-i\lambda\langle D\rangle-\tfrac{\lambda^2}{2}\mu+O(\lambda^3)$ with $\mu:=\langle A^2\rangle+\langle B^2\rangle-2\langle BA\rangle$. Self-adjointness makes $\langle A^2\rangle,\langle B^2\rangle,\langle D\rangle$ real and gives $\langle AB\rangle=\langle BA\rangle^*$, so $\operatorname{Re}\mu=\langle A^2\rangle+\langle B^2\rangle-\langle AB\rangle-\langle BA\rangle=\langle D^2\rangle$. In $W=G_{hh'}G_{hh'}^*$ the first-order terms $\mp i\lambda\langle D\rangle$ cancel. The second order is $\lambda^2\langle D\rangle^2-\lambda^2\operatorname{Re}\mu$, so

$$
W(h,h';\lambda)=1-\lambda^2\operatorname{Var}_\chi(K_h-K_{h'})+O(\lambda^3).
\tag{5.8}
$$

Since $\lvert u_h\rangle-\lvert u_{h'}\rangle=(D-\langle D\rangle)\lvert\chi\rangle$ and $D-\langle D\rangle$ is self-adjoint,

$$
\operatorname{Var}_\chi(K_h-K_{h'})=\langle\chi\vert(D-\langle D\rangle)^2\vert\chi\rangle=\bigl\lVert u_h-u_{h'}\bigr\rVert^2 .
\tag{5.9}
$$

### Step 5. Limit of the witness angle

By (3.5), $\alpha(h,h';\lambda)=\arccos\sqrt{W(h,h';\lambda)}\in[0,\pi/2]$, so $\sin^2\alpha=1-W$. By (5.8) and (5.9) there are constants $c,\lambda_1>0$ such that $1-W=\lambda^2\lVert u_h-u_{h'}\rVert^2+R(\lambda)$ with $\lvert R(\lambda)\rvert\le c\lambda^3$ for $0<\lambda\le\lambda_1$. The radicand below equals $\sin^2\alpha/\lambda^2\ge0$, and $\sqrt{\cdot}$ is continuous on $[0,\infty)$, hence

$$
\frac{\sin\alpha}{\lambda}=\sqrt{\lVert u_h-u_{h'}\rVert^2+R(\lambda)/\lambda^2}\;\longrightarrow\;\lVert u_h-u_{h'}\rVert .
$$

In particular $\sin\alpha\to0$. On $[0,\pi/2]$ this forces $\alpha\to0$, and then $\alpha/\sin\alpha\to1$ (the ratio is set to $1$ where $\alpha=0$). Therefore

$$
\lim_{\lambda\to0^+}\frac{\alpha(h,h';\lambda)}{\lambda}=d_0(h,h')=\bigl\lVert u_h-u_{h'}\bigr\rVert,
\qquad \lvert u_h\rangle=\bigl(K_h-\langle K_h\rangle\bigr)\lvert\chi\rangle .
\tag{5.10}
$$

The argument needs no case distinction, so it covers $\operatorname{Var}_\chi(K_h-K_{h'})=0$. In that case $d_0=0$ and $\alpha=O(\lambda^{3/2})$.

### Step 6. Euclidean structure and dimension

$(\mathcal H_c,g)$ with $g(v,w):=\operatorname{Re}\langle v\vert w\rangle$ is a real Euclidean space of dimension $2d_c$ with $g(v,v)=\lVert v\rVert^2$ (standard).

(i) *Isometry.* By (5.10), $d_0(h,h')=\sqrt{g(u_h-u_{h'},u_h-u_{h'})}$ is the Euclidean distance of $u_h$ and $u_{h'}$. So $d_0$ is a pseudometric on $H_V$ with $d_0(h,h')=0\iff u_h=u_{h'}$. Define $h\sim_0h'$ iff $u_h=u_{h'}$. Then $[h]_0\mapsto u_h$ is a well-defined, distance-preserving bijection:

$$
\bigl(H_V/{\sim_0},\,d_0\bigr)\;\cong\;\bigl(\{u_h:h\in H_V\},\,\lVert\cdot\rVert\bigr)\subset(\mathcal H_c,g).
\tag{5.11}
$$

The conventions of M6 change $\lvert\chi\rangle$ only by a unit factor $e^{i\theta}$. Then $u_h\mapsto e^{i\theta}u_h$, which is a $g$-isometry. So the configuration is fixed up to a global isometry, and $d_0$ is invariant.

(ii) *Criterion.* $d_0(h,h')=0$ iff $D\lvert\chi\rangle=\langle D\rangle\lvert\chi\rangle$. Conversely, if $D\lvert\chi\rangle=z\lvert\chi\rangle$, then $z=\langle\chi\vert D\vert\chi\rangle=\langle D\rangle$. Hence

$$
d_0(h,h')=0\iff(K_h-K_{h'})\lvert\chi\rangle\in\mathbb C\lvert\chi\rangle .
\tag{5.12}
$$

(iii) *Orthogonality.*

$$
\langle\chi\vert u_h\rangle=\langle K_h\rangle-\langle K_h\rangle\langle\chi\vert\chi\rangle=0 ,
\tag{5.13}
$$

so every $u_h$ lies in $\chi^\perp$. This is a complex subspace of dimension $d_c-1$, that is, a real subspace of dimension $2d_c-2$: the $g$-orthogonal complement of $\lvert\chi\rangle$ and $i\lvert\chi\rangle$.

(iv) *Dimension.* Let $\dim_0:=\dim_{\mathbb R}\operatorname{aff}_{\mathbb R}\{u_h\}$. There are at most $d_{T(a)}$ points, so their affine span has dimension at most $d_{T(a)}-1$. By (5.13) the span lies in the linear subspace $\chi^\perp$. Hence

$$
\dim_0\le\min\bigl(d_{T(a)}-1,\;2d_c-2\bigr).
\tag{5.14}
$$

The bound is attained. For any $w\in\chi^\perp$, the operator $K=\lvert w\rangle\langle\chi\rvert+\lvert\chi\rangle\langle w\rvert$ is self-adjoint with $K\lvert\chi\rangle=w$ and $\langle K\rangle=0$, so $u=w$. Let $m$ be the right side of (5.14). Choose $u=0$ for one place, a real basis of an $m$-dimensional real subspace of $\chi^\perp$ for $m$ further places, and $u=0$ for the remaining places. This gives $\dim_0=m$.

### Step 7. Example (a): line

In Steps 7 and 8 the imaginary unit is written $\mathrm i$, because $i$ and $j$ are indices. Since $\langle K_{h_i}\rangle=i\langle X\rangle$, we get $u_{h_i}=i\,u_X$ with $\lvert u_X\rangle=(X-\langle X\rangle)\lvert\chi\rangle$, and $\lVert u_X\rVert^2=\operatorname{Var}_\chi X=\sigma_X^2$ by the computation in (5.9). By (5.10),

$$
d_0(h_i,h_j)=\lvert i-j\rvert\,\sigma_X .
\tag{5.15}
$$

Since $\sigma_X>0$, $u_X\neq0$, and the points $i\,u_X$ are distinct and equally spaced on the line $\mathbb R u_X$. So $\sim_0$ is trivial, $(H_V,d_0)$ is isometric to $\sigma_X\{1,\dots,n\}\subset\mathbb R$, and $\dim_0=1$ for $n\ge2$ ($\dim_0=0$ for $n=1$).

$K_{h_i}=iX$ and $K_{h_j}=jX$ are functions of the same $X$, so they commute and $e^{\mathrm i jX\lambda}e^{-\mathrm i iX\lambda}=e^{-\mathrm i(i-j)\lambda X}$. By (5.6), exactly for all $\lambda\ge0$,

$$
W(h_i,h_j;\lambda)=F\bigl((i-j)\lambda\bigr),\qquad F(s):=\bigl\lvert\langle\chi\vert e^{-\mathrm isX}\vert\chi\rangle\bigr\rvert^2 .
\tag{5.16}
$$

$F(-s)=F(s)$, because $\langle\chi\vert e^{\mathrm isX}\vert\chi\rangle=\langle\chi\vert e^{-\mathrm isX}\vert\chi\rangle^*$. So $W$ depends only on $\lvert i-j\rvert\lambda$. As a cross-check, (5.8) gives $F(s)=1-s^2\sigma_X^2+O(s^3)$, consistent with (5.15).

### Step 8. Example (b): plane

Linearity gives $u_{h_{(i,j)}}=i\,u_X+j\,u_Y$, with $\lVert u_X\rVert=\sigma_X$, $\lVert u_Y\rVert=\sigma_Y:=\sqrt{\operatorname{Var}_\chi Y}$ and $g(u_X,u_Y)=\operatorname{Cov}_\chi(X,Y)=\tfrac12\langle XY+YX\rangle-\langle X\rangle\langle Y\rangle$. Here $\sigma_X,\sigma_Y\ge0$, and positivity is not assumed. With $\Delta i:=i-i'$ and $\Delta j:=j-j'$, (5.10) gives

$$
d_0\bigl(h_{(i,j)},h_{(i',j')}\bigr)=\sqrt{\sigma_X^2(\Delta i)^2+2\operatorname{Cov}_\chi(X,Y)\,\Delta i\,\Delta j+\sigma_Y^2(\Delta j)^2}\, .
\tag{5.17}
$$

*Square lattice.* Suppose $d_0=\sigma\sqrt{(\Delta i)^2+(\Delta j)^2}$ for all pairs, with $\sigma>0$. Since $n\ge2$, the differences $(\Delta i,\Delta j)=(1,0),(0,1),(1,1)$ all occur, from $(2,1),(1,2),(2,2)$ against $(1,1)$. Then (5.17) gives $\sigma_X^2=\sigma^2$, $\sigma_Y^2=\sigma^2$ and $\sigma_X^2+2\operatorname{Cov}_\chi(X,Y)+\sigma_Y^2=2\sigma^2$. The converse is immediate from (5.17). Hence

$$
d_0=\sigma\sqrt{(\Delta i)^2+(\Delta j)^2}\ \text{ for all pairs}
\iff
\sigma_X=\sigma_Y=\sigma>0\ \text{ and }\ \operatorname{Cov}_\chi(X,Y)=0 .
\tag{5.18}
$$

Equivalently, $u_X$ and $u_Y$ are $g$-orthogonal and have equal non-zero length. Then $(i,j)\mapsto u_{h_{(i,j)}}$ is an isometry from the square lattice $\sigma\{1,\dots,n\}^2\subset\mathbb R^2$ onto $\{u_h\}$.

*Dimension.* Each difference $u_{h_{(i,j)}}-u_{h_{(1,1)}}=(i-1)u_X+(j-1)u_Y$ is a combination of $u_X$ and $u_Y$, and both occur as differences ($n\ge2$). So the affine span is $u_{h_{(1,1)}}+\operatorname{span}_{\mathbb R}\{u_X,u_Y\}$, and $\dim_0$ is the rank of the real Gram matrix $\bigl(\begin{smallmatrix}\sigma_X^2&\operatorname{Cov}\\ \operatorname{Cov}&\sigma_Y^2\end{smallmatrix}\bigr)$. Cauchy–Schwarz gives $\lvert\operatorname{Cov}_\chi(X,Y)\rvert\le\sigma_X\sigma_Y$, and

$$
\dim_0=2\iff\lvert\operatorname{Cov}_\chi(X,Y)\rvert<\sigma_X\sigma_Y .
\tag{5.19}
$$

$\dim_0=1$ if $\lvert\operatorname{Cov}_\chi\rvert=\sigma_X\sigma_Y$ and $\sigma_X^2+\sigma_Y^2>0$, and $\dim_0=0$ if $\sigma_X=\sigma_Y=0$. The square-lattice condition (5.18) implies (5.19). Only real independence matters, so (5.18) can already hold for $d_c=2$, where (5.14) allows $\dim_0\le2$. Example: $\lvert\chi\rangle=\lvert0\rangle$, $X=\lvert0\rangle\langle1\rvert+\lvert1\rangle\langle0\rvert$, $Y=-\mathrm i\lvert0\rangle\langle1\rvert+\mathrm i\lvert1\rangle\langle0\rvert$. Then $u_X=\lvert1\rangle$, $u_Y=\mathrm i\lvert1\rangle$, $\sigma_X=\sigma_Y=1$ and $\operatorname{Cov}_\chi=0$.

## Result

1. Exact evolution: $e^{-iC\lambda}$ (5.1), $\lvert\Psi(\lambda)\rangle$ (5.2), $p_h(\lambda)=\lvert\phi_h\rvert^2$ with $H_V=H_{T(a)}$ (5.3), branch states (5.4), view (5.5), and $W(h,h';\lambda)$ as expected (5.6).
2. Small $\lambda$: $W=1-\lambda^2\operatorname{Var}_\chi(K_h-K_{h'})+O(\lambda^3)$ (5.8), with $\operatorname{Var}_\chi(K_h-K_{h'})=\lVert u_h-u_{h'}\rVert^2$ (5.9), and $\alpha/\lambda\to d_0=\lVert u_h-u_{h'}\rVert$ including the zero-variance case (5.10).
3. Geometry: $(H_V/{\sim_0},d_0)$ is isometric to $\{u_h\}\subset(\mathcal H_c,\operatorname{Re}\langle\cdot\vert\cdot\rangle)$ (5.11); $d_0=0$ iff $(K_h-K_{h'})\lvert\chi\rangle\propto\lvert\chi\rangle$ (5.12); $\langle\chi\vert u_h\rangle=0$ (5.13); $\dim_0\le\min(d_{T(a)}-1,\,2d_c-2)$, and the bound is attained (5.14).
4. Line: $d_0=\lvert i-j\rvert\sigma_X$, $\dim_0=1$ (5.15), and exactly $W=F((i-j)\lambda)$ with $F$ even (5.16). Plane: $d_0$ (5.17); square lattice iff $\sigma_X=\sigma_Y>0$ and $\operatorname{Cov}_\chi(X,Y)=0$ (5.18); $\dim_0=2$ iff $\lvert\operatorname{Cov}_\chi(X,Y)\rvert<\sigma_X\sigma_Y$ (5.19).

No discrepancy with the expected result.

## Consistency checks

1. **$\lambda=0$.** (5.2) gives $\lvert\phi\rangle\otimes\lvert\chi\rangle$, and (5.6) gives $W(h,h';0)=\lvert\langle\chi\vert\chi\rangle\rvert^2=1$ for all pairs. So all places are $\approx$-equivalent and $X_V$ is a single point.
2. **All $K_h=K$.** $e^{iK\lambda}e^{-iK\lambda}=\mathbb 1$, so $W\equiv1$ by (5.6). All $u_h$ are equal, so $d_0\equiv0$ and $\dim_0=0$, consistent with (5.8) at zero variance.
3. **$K_h\lvert\chi\rangle=k_h\lvert\chi\rangle$ for all $h$.** (5.4) gives $\lvert E_h\rangle=e^{-ik_h\lambda}\lvert\chi\rangle$, so $W=\lvert e^{-i(k_h-k_{h'})\lambda}\rvert^2\equiv1$ and $u_h=(k_h-k_h)\lvert\chi\rangle=0$, consistent with (5.12) and (5.13).

## Open issues

- $d_0(h,h')=0$ gives only $1-W=o(\lambda^2)$. It does not imply $h\approx h'$ at $\lambda>0$ when $K_h$ and $K_{h'}$ do not commute, because the higher orders of (5.7) contain products like $BA$. Conversely, $W=1$ at isolated $\lambda$ (revivals, for example $F(s_0)=1$ in (5.16) for integer spectrum of $X$) does not imply $d_0=0$. So $\sim_0$ and $\approx$ at finite $\lambda$ are different identifications. Their relation at finite $\lambda$ is out of scope.
- In (b), the exact finite-$\lambda$ form of $W$ is not derived. It depends only on $(\Delta i\,\lambda,\Delta j\,\lambda)$ when $[X,Y]=0$, by the argument of Step 7, but in general not otherwise. Large $\lambda$ is out of scope.

## Methods used

- Functional calculus for block-diagonal (controlled) operators; matrix exponential
- Taylor expansion with remainder of matrix exponentials
- Witness-rule inputs: view via partial trace (2.2), witness overlap (2.11), angle (3.5)
- Realification of a complex Hilbert space; Euclidean distance, affine span, Gram matrix, Cauchy–Schwarz