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

# Loop data of the witness geometry in a recording medium

- **Subproject:** 03-ilang-space
- **Package:** 10-loop-data
- **Version:** v1
- **Mode:** new
- **Date:** 2026-10-08

## Setup and assumptions

**Setting** (question.md, from 05). Objects $a$ and $c$ of different types, each the only instance of its type; part $A=\{a\}$; recording contract $C=\sum_h\lvert h\rangle\langle h\rvert\otimes K_h$; initial state $\lvert\phi\rangle\otimes\lvert\chi\rangle$ with $\phi_h\neq0$ for every $h$. Each $K_h$ is self-adjoint on $\mathcal H_c$ (because $C=C^\dagger$) and independent of $\lambda$ (A5). $\langle X\rangle:=\langle\chi\vert X\vert\chi\rangle$, $\langle\chi\vert\chi\rangle=1$.

**Inputs** (as quoted in question.md). From 02: $G_{hh'}=\langle E_{h'}\vert E_h\rangle$, $W(h_1,\dots,h_k)=G_{h_1h_2}\cdots G_{h_kh_1}$, rephasing invariance (2.9), (2.10), and $W(h,h')=\lvert G_{hh'}\rvert^2$ (2.11). From 05: $G_{hh'}(\lambda)=\langle\chi\vert e^{\mathrm iK_{h'}\lambda}e^{-\mathrm iK_h\lambda}\vert\chi\rangle$ (5.5); $d_0(h,h')=\lVert u_h-u_{h'}\rVert$ with $\lvert u_h\rangle=(K_h-\langle K_h\rangle)\lvert\chi\rangle$ (5.10); the isometry (5.11); $\langle\chi\vert u_h\rangle=0$ (5.13). From 06: $g_1,g_2$ of (6.8) at $B_{\beta_0}=0$, i.e. $M=K_h$, $M'=K_{h'}$, $D=K_h-K_{h'}$.

**Notation.** On $\mathcal H_c$ (question.md):
$$
g(v,w)=\operatorname{Re}\langle v\vert w\rangle,\qquad
\omega(v,w)=\operatorname{Im}\langle v\vert w\rangle,\qquad
Jv=\mathrm i\,v,\qquad \lVert v\rVert^2=g(v,v)=\langle v\vert v\rangle .
\tag{10.1}
$$
$g$ and $\omega$ are real-bilinear; $g$ is symmetric and $\omega$ antisymmetric, since $\langle w\vert v\rangle=\langle v\vert w\rangle^*$. Throughout, $\mathrm i$ is the imaginary unit. In Steps 1–6 the loop index is $j$, taken cyclically ($h_{k+1}:=h_1$), and $u_j:=u_{h_j}$. In Steps 7–8, $i,j$ are lattice indices and the loop index is $m$. Local abbreviation: $\Delta_{hh'}:=\langle K_h\rangle-\langle K_{h'}\rangle\in\mathbb R$.

**Branch and range of validity.** $\Phi$ takes values in $(-\pi,\pi]$, and $\operatorname{Arg}$, $\operatorname{Log}$ are principal. Each $G_{hh'}(\lambda)$ is an entire function of $\lambda$ with $G_{hh'}(0)=1$ (finite dimension, A1). Hence for a given loop there is $\lambda_0>0$ such that for $\lvert\lambda\rvert<\lambda_0$ every $\operatorname{Re}G_{h_jh_{j+1}}>0$ and $\sum_j\lvert\operatorname{Arg}G_{h_jh_{j+1}}\rvert<\pi$. There $W\neq0$, every $\operatorname{Log}G_{h_jh_{j+1}}$ is analytic, and
$$
\Phi(h_1,\dots,h_k;\lambda)=\sum_{j=1}^k\operatorname{Arg}G_{h_jh_{j+1}}(\lambda)\qquad(\lvert\lambda\rvert<\lambda_0).
\tag{10.2}
$$
Every $O(\lambda^n)$ refers to $\lambda\to0$ in this range. Negative $\lambda$ is used only in Step 6(b), as a formal parameter of (1.9). Step 1 and (10.11) hold for every $\lambda$. $\Phi$ is a datum of the view by (2.9)–(2.10); it is not read as a field, flux or curvature (M1, M4).

## Derivation

### Step 1. The modulus is pairwise data (item 1a)

Since $G_{h'h}=\langle E_h\vert E_{h'}\rangle=G_{hh'}^*$, (2.11) reads $W(h_j,h_{j+1})=G_{h_jh_{j+1}}G_{h_{j+1}h_j}=\lvert G_{h_jh_{j+1}}\rvert^2$. Taking the squared modulus of the product that defines $W$,
$$
\lvert W(h_1,\dots,h_k)\rvert^2=\prod_{j=1}^k\lvert G_{h_jh_{j+1}}\rvert^2=\prod_{j=1}^kW(h_j,h_{j+1}) .
\tag{10.3}
$$
No expansion is used, so (10.3) holds exactly for every $\lambda$. Beyond the pairwise data, $W$ carries only its phase $\Phi$.

### Step 2. A single overlap to second order (item 1b)

(i) Expanding both exponentials in (5.5),
$$e^{\mathrm iK_{h'}\lambda}e^{-\mathrm iK_h\lambda}=\mathbb 1-\mathrm i\lambda D+\lambda^2\bigl(K_{h'}K_h-\tfrac12K_h^2-\tfrac12K_{h'}^2\bigr)+O(\lambda^3),$$
and $K_{h'}K_h-\tfrac12K_h^2-\tfrac12K_{h'}^2=-\tfrac12D^2-\tfrac12[K_h,K_{h'}]$. Taking $\langle\cdot\rangle$ gives $g_1=-\mathrm i\langle D\rangle$ and $g_2=-\frac12\bigl(\langle D^2\rangle+\langle[K_h,K_{h'}]\rangle\bigr)$, in agreement with (6.8) at $B_{\beta_0}=0$.

(ii) By (5.10), $K_h\lvert\chi\rangle=\lvert u_h\rangle+\langle K_h\rangle\lvert\chi\rangle$, and $\langle\chi\vert u_h\rangle=0$ by (5.13). Since $K_h=K_h^\dagger$, $\langle K_hK_{h'}\rangle=\langle K_h\chi\vert K_{h'}\chi\rangle=\langle u_h\vert u_{h'}\rangle+\langle K_h\rangle\langle K_{h'}\rangle$. Also, $D\lvert\chi\rangle=(u_h-u_{h'})+\Delta_{hh'}\lvert\chi\rangle$ is an orthogonal decomposition. With (10.1) and (5.10), it follows that
$$
\langle[K_h,K_{h'}]\rangle=\langle u_h\vert u_{h'}\rangle-\langle u_{h'}\vert u_h\rangle=2\mathrm i\,\omega(u_h,u_{h'}),
\qquad
\langle D^2\rangle=\lVert D\chi\rVert^2=d_0(h,h')^2+\Delta_{hh'}^2 .
\tag{10.4}
$$

(iii) Inserting (10.4) and $\langle D\rangle=\Delta_{hh'}$ gives
$$
G_{hh'}(\lambda)=1-\mathrm i\lambda\Delta_{hh'}-\frac{\lambda^2}{2}\Bigl(d_0(h,h')^2+\Delta_{hh'}^2+2\mathrm i\,\omega(u_h,u_{h'})\Bigr)+O(\lambda^3).
\tag{10.5}
$$

(iv) Use $\operatorname{Log}(1+x)=x-\tfrac12x^2+O(x^3)$ with $x^2=-\Delta_{hh'}^2\lambda^2+O(\lambda^3)$. The $\Delta_{hh'}^2$ terms cancel:
$$
\operatorname{Log}G_{hh'}(\lambda)=-\mathrm i\lambda\Delta_{hh'}-\lambda^2\Bigl(\tfrac12d_0(h,h')^2+\mathrm i\,\omega(u_h,u_{h'})\Bigr)+O(\lambda^3).
\tag{10.6}
$$
The real part is $\ln\lvert G_{hh'}\rvert=-\frac12\lambda^2d_0(h,h')^2+O(\lambda^3)$. The imaginary part, rewritten with the antisymmetry of $\omega$, is the phase:
$$
\operatorname{Arg}G_{hh'}(\lambda)=-\lambda\bigl(\langle K_h\rangle-\langle K_{h'}\rangle\bigr)+\lambda^2\,\omega(u_{h'},u_h)+O(\lambda^3).
\tag{10.7}
$$
The phase of a single overlap depends on the phase convention of the branch states fixed in (5.5) (M6). Only cyclic sums of (10.7) are data (2.9).

### Step 3. The loop formula (item 1c)

Sum (10.7) around the loop, using (10.2). The first-order terms telescope, $\sum_j(\langle K_{h_j}\rangle-\langle K_{h_{j+1}}\rangle)=0$, and $\sum_j\omega(u_{j+1},u_j)=-\sum_j\omega(u_j,u_{j+1})$. Hence
$$
\Phi(h_1,\dots,h_k;\lambda)=-\lambda^2\sum_{j=1}^k\omega\bigl(u_{h_j},u_{h_{j+1}}\bigr)+O(\lambda^3),\qquad k\ge3 .
\tag{10.8}
$$
The sum is translation invariant. For every $x\in\mathcal H_c$, with $S=\sum_ju_j=\sum_ju_{j+1}$ and $\omega(x,x)=0$,
$$
\sum_j\omega(u_j+x,u_{j+1}+x)=\sum_j\omega(u_j,u_{j+1})+\omega(x,S)+\omega(S,x)=\sum_j\omega(u_j,u_{j+1}).
\tag{10.9}
$$
For $k=3$, take $x=-u_1$. The terms with a zero argument drop out, and
$$
\Phi(h_1,h_2,h_3;\lambda)=-\lambda^2\,\omega\bigl(u_{h_2}-u_{h_1},\,u_{h_3}-u_{h_1}\bigr)+O(\lambda^3).
\tag{10.10}
$$

### Step 4. Reversal (item 1d)

With $G_{h'h}=G_{hh'}^*$ (Step 1), for every $\lambda$,
$$
W(h_k,\dots,h_1)=\prod_{j=1}^kG_{h_{j+1}h_j}=\prod_{j=1}^kG_{h_jh_{j+1}}^*=W(h_1,\dots,h_k)^* .
\tag{10.11}
$$
Hence $\Phi(h_k,\dots,h_1)=-\Phi(h_1,\dots,h_k)$. This holds exactly for $\lvert\lambda\rvert<\lambda_0$, by (10.2) and $\operatorname{Arg}z^*=-\operatorname{Arg}z$ for $\operatorname{Re}z>0$, and modulo $2\pi$ wherever $W\neq0$. At order $\lambda^2$ it is the antisymmetry of $\omega$ in (10.8).

### Step 5. Relation to the metric (item 2)

(i) From (10.1), $g(Jv,w)=\operatorname{Re}\bigl(-\mathrm i\langle v\vert w\rangle\bigr)=\operatorname{Im}\langle v\vert w\rangle$, $g(Jv,Jw)=\operatorname{Re}\langle v\vert w\rangle$ and $J^2v=-v$. Thus
$$
\omega(v,w)=g(Jv,w),\qquad g(Jv,Jw)=g(v,w),\qquad J^2=-\mathbb 1 .
\tag{10.12}
$$

(ii) *Bound.* Cauchy–Schwarz for $g$, together with $\lVert Jv\rVert=\lVert v\rVert$, gives $\lvert\omega(v,w)\rvert\le\lVert v\rVert\,\lVert w\rVert$. Take $v=u_{h_2}-u_{h_1}$ and $w=u_{h_3}-u_{h_1}$. By (10.10) and (5.10),
$$
\lvert\Phi(h_1,h_2,h_3;\lambda)\rvert\le\lambda^2\,d_0(h_1,h_2)\,d_0(h_1,h_3)+O(\lambda^3).
\tag{10.13}
$$

(iii) *Saturation.* Cauchy–Schwarz is an equality iff $Jv$ and $w$ are real-linearly dependent. So the leading term saturates (10.13) iff
$$
u_{h_3}-u_{h_1}\in\mathbb R\,J\bigl(u_{h_2}-u_{h_1}\bigr)\qquad\text{or}\qquad u_{h_2}=u_{h_1} .
\tag{10.14}
$$
For $w=tJv$ the leading term is $-\lambda^2t\lVert v\rVert^2$. Since $g(Jv,v)=\omega(v,v)=0$, condition (10.14) splits into two parts. First, the two edges at $h_1$ are $g$-orthogonal. This is a condition on $d_0$ alone, because by polarization $g(v,w)=\frac12\bigl[d_0(h_1,h_2)^2+d_0(h_1,h_3)^2-d_0(h_2,h_3)^2\bigr]$. Second, $w$ lies in the complex line $\mathbb Cv=\operatorname{span}_{\mathbb R}\{v,Jv\}$, which is a condition that involves $J$.

(iv) *In what sense the leading loop data are fixed.* By (10.8), (10.9) and (10.12), the leading datum $\Phi^{(2)}:=-\sum_jg(Ju_j,u_{j+1})$ depends only on the differences of the record vectors in $(\mathcal H_c,g)$, which realize $d_0$ by (5.11), and on $J$. The metric $d_0$ fixes this configuration up to an isometry $x\mapsto Rx+b$ of $(\mathcal H_c,g)$ (a finite subset of a Euclidean space is fixed by its distances up to an ambient isometry). By (10.12), an orthogonal $R$ with $RJ=JR$ (a unitary map) leaves every $\Phi^{(2)}$ unchanged, and one with $RJ=-JR$ (an antiunitary map) reverses every sign. So, at leading order, the loop data are fixed by the $d_0$-configuration together with its position relative to $J$, up to unitary motions. By itself, $d_0$ fixes only $g$ on differences, and it bounds $\Phi^{(2)}$ by (10.13). It does not fix the loop data:
- *Sign.* In a basis in which $\chi$ is real, replace every $K_h$ by its entrywise conjugate. The new records are again self-adjoint and have the same $\langle K_h\rangle$. Then $u_h\mapsto u_h^*$, every $d_0$ is unchanged, and $\omega(u_h^*,u_{h'}^*)=\operatorname{Im}\langle u_h\vert u_{h'}\rangle^*=-\omega(u_h,u_{h'})$. Every $\Phi^{(2)}$ changes sign.
- *Magnitude, $d_c\ge3$.* Take orthonormal $e_1,e_2\perp\chi$. A record $K=\lvert u\rangle\langle\chi\rvert+\lvert\chi\rangle\langle u\rvert$ with $u\perp\chi$ has $\langle K\rangle=0$ and record vector $u$. The triangles with record vectors $(0,e_1,\mathrm ie_1)$ and $(0,e_1,e_2)$ have the same distances $1,1,\sqrt2$. By (10.10), however, the first has $\Phi=-\lambda^2+O(\lambda^3)$ and the second $\Phi=O(\lambda^3)$.

For $d_c=2$, all $u_h$ lie in the complex line $\chi^\perp$ (5.13). This is a real plane on which $J$ acts as a rotation by $\pi/2$. On it, $\lvert g(Jv,w)\rvert$ is twice the Euclidean area of the triangle, which $d_0$ fixes. Only the sign of $\Phi^{(2)}$ then needs $J$.

### Step 6. When the $\lambda^2$ term vanishes (item 3)

(a) *Commuting records.* If $[K_{h_j},K_{h_{j+1}}]=0$ for all $j$ (in particular, if the records commute pairwise), then (10.4) gives $\omega(u_j,u_{j+1})=0$, and the $\lambda^2$ term of (10.8) vanishes.

(b) *Real records.* Let $\chi$ be real and every $K_h$ real symmetric in some basis. Then $u_h=(K_h-\langle K_h\rangle)\chi$ is real, so $\langle u_h\vert u_{h'}\rangle\in\mathbb R$ and $\omega(u_h,u_{h'})=0$: the $\lambda^2$ term vanishes. Moreover, for real $K$ and real $\lambda$ the entrywise conjugate of $e^{-\mathrm iK\lambda}$ is $e^{\mathrm iK\lambda}$. Conjugating (5.5) entrywise therefore gives $G_{hh'}(\lambda)^*=\chi^{\mathsf T}e^{-\mathrm iK_{h'}\lambda}e^{\mathrm iK_h\lambda}\chi=G_{hh'}(-\lambda)$, and
$$
W(h_1,\dots,h_k;-\lambda)=W(h_1,\dots,h_k;\lambda)^* .
\tag{10.15}
$$
By (10.2), $\Phi(-\lambda)=-\Phi(\lambda)$ for $\lvert\lambda\rvert<\lambda_0$, where $\Phi$ is real-analytic. Its Taylor series therefore has only odd powers, and with (10.8), $\Phi=O(\lambda^3)$.

(c) *Collinear record vectors.* Let $u_{h_j}=x+t_je$ with $x,e\in\mathcal H_c$ and $t_j\in\mathbb R$; the line need not pass through $0$. Applying (10.9) with $-x$ gives $\sum_j\omega(u_j,u_{j+1})=\sum_jt_jt_{j+1}\,\omega(e,e)=0$.

### Step 7. Plane (item 4a)

Here $\chi=\lvert0\rangle$, $X\lvert0\rangle=\lvert1\rangle$, $Y\lvert0\rangle=\mathrm i\lvert1\rangle$ and $\langle0\vert X\vert0\rangle=\langle0\vert Y\vert0\rangle=0$. For $K_{h_{(i,j)}}=iX+jY$ with $i,j\in\mathbb Z$, we get $\langle K_{h_{(i,j)}}\rangle=0$ and
$$
u_{h_{(i,j)}}=(i+\mathrm i\,j)\lvert1\rangle,\qquad
\omega\bigl((i+\mathrm i\,j)\lvert1\rangle,(i'+\mathrm i\,j')\lvert1\rangle\bigr)=\operatorname{Im}\bigl[(i-\mathrm i\,j)(i'+\mathrm i\,j')\bigr]=ij'-ji' ,
\tag{10.16}
$$
and the second identity holds for all real $i,j,i',j'$. Also, $d_0(h_{(i,j)},h_{(i',j')})=\sqrt{(i-i')^2+(j-j')^2}$ and $J\bigl((i+\mathrm ij)\lvert1\rangle\bigr)=(-j+\mathrm ii)\lvert1\rangle$, so $J$ acts on the labels as the rotation $(i,j)\mapsto(-j,i)$. Take a lattice triangle with vertices $p_m=(i_m,j_m)$, $m=1,2,3$. Since $u_{h_{p}}$ is linear in $p$, we have $u_{h_{p_2}}-u_{h_{p_1}}=\bigl((i_2-i_1)+\mathrm i(j_2-j_1)\bigr)\lvert1\rangle$, and likewise for $p_3$. Then (10.10) and (10.16) give
$$
\Phi(h_{p_1},h_{p_2},h_{p_3};\lambda)=-\lambda^2\bigl[(i_2-i_1)(j_3-j_1)-(j_2-j_1)(i_3-i_1)\bigr]+O(\lambda^3)=-2\lambda^2A+O(\lambda^3),
\tag{10.17}
$$
where $A=\frac12\det(p_2-p_1,\,p_3-p_1)$ is the signed area in the coordinates $(i,j)$. $A$ is positive when $p_1\to p_2\to p_3$ runs counterclockwise, with $i$ as the first axis and $j$ as the second. $A$ is computed from the labels by which the description specifies the records. By (5.11) and (10.16), these labels are isometric to the record vectors in the plane $\chi^\perp$, and the orientation is the one fixed by $J$. Nothing further is attributed to $A$.

### Step 8. Line (item 4b)

For $K_{h_i}=iX$ the records commute, so $e^{\mathrm ii'X\lambda}e^{-\mathrm iiX\lambda}=e^{-\mathrm i(i-i')X\lambda}$. With $e^{-\mathrm itX}=\cos t\,\mathbb 1-\mathrm i\sin t\,X$ and $\langle0\vert X\vert0\rangle=0$,
$$
G_{h_ih_{i'}}(\lambda)=\cos\bigl((i-i')\lambda\bigr),\qquad
W(h_{i_1},\dots,h_{i_k};\lambda)=\prod_{m=1}^k\cos\bigl((i_m-i_{m+1})\lambda\bigr).
\tag{10.18}
$$
$W$ is real and positive whenever $\lvert\lambda\rvert\max_m\lvert i_m-i_{m+1}\rvert<\pi/2$. There $\Phi=0$ identically, so in particular its $\lambda^2$ term vanishes. This agrees with Step 6, all three cases of which apply: (a) the records commute; (b) $X$ is real symmetric and $\chi$ is real; (c) $u_{h_i}=i\lvert1\rangle\in\mathbb R\lvert1\rangle$.

## Result

1. Exact modulus identity (10.3); single-overlap phase (10.7); loop formula (10.8), with the triangle form (10.10); reversal (10.11).
2. $\omega=g(J\cdot,\cdot)$, $J$ is $g$-orthogonal and $J^2=-\mathbb 1$ (10.12). Bound (10.13); saturation iff (10.14). At leading order the loop data are fixed by the $d_0$-configuration together with $J$, up to unitary motions. They are not fixed by $d_0$ alone: $d_0$ never fixes the sign, and for $d_c\ge3$ it does not fix the magnitude either (Step 5(iv)).
3. The $\lambda^2$ term vanishes for commuting records, for real records, and for collinear record vectors (Step 6). For real records, $W(-\lambda)=W(\lambda)^*$ (10.15) and $\Phi$ is odd.
4. Plane: $\Phi=-2\lambda^2A+O(\lambda^3)$ (10.17). Line: $\Phi\equiv0$ for small $\lambda$ (10.18).

Every formula expected in question.md is confirmed; I found no discrepancy in sign or factor.

## Consistency checks

1. **$\lambda=0$ and dimensions.** At $\lambda=0$, $E_h=\chi$ for every $h$, so every $G=1$, $W=1$ and $\Phi=0$. This is consistent with (10.3) and with (10.8), which has no $\lambda^0$ or $\lambda^1$ term. In (5.5), $\lambda K_h$ is dimensionless, and $u_h$ has the dimension of $K_h$. Hence $\lambda\Delta_{hh'}$, $\lambda^2d_0^2$, $\lambda^2\omega(u,u')$ and both sides of (10.13) are dimensionless, as phases and log-moduli must be.
2. **Unit lattice triangle, exactly.** Take $(0,0)\to(1,0)\to(0,1)$, i.e. $K=0,X,Y$. Then $G_{h_1h_2}=\langle0\vert e^{\mathrm iX\lambda}\vert0\rangle=\cos\lambda$, $G_{h_3h_1}=\langle0\vert e^{-\mathrm iY\lambda}\vert0\rangle=\cos\lambda$, and $G_{h_2h_3}=\langle0\vert e^{\mathrm iY\lambda}e^{-\mathrm iX\lambda}\vert0\rangle=\cos^2\lambda-\mathrm i\sin^2\lambda$. So $W=\cos^2\lambda\,(\cos^2\lambda-\mathrm i\sin^2\lambda)$ and $\Phi=-\arctan(\tan^2\lambda)=-\lambda^2+O(\lambda^4)$. This agrees with (10.17) for $A=\frac12$, and with (10.7), since $\lambda^2\omega(\mathrm i\lvert1\rangle,\lvert1\rangle)=-\lambda^2$. Also, $\lvert W\rvert^2=\cos^4\lambda\,(\cos^4\lambda+\sin^4\lambda)=W(h_1,h_2)W(h_2,h_3)W(h_3,h_1)$, exactly as (10.3) requires.
3. **Saturation in the plane.** $J$ acts on the labels as the rotation by $+\pi/2$ (Step 7). So (10.14) says that $p_3-p_1$ is perpendicular to $p_2-p_1$, i.e. that the triangle has a right angle at $p_1$. In the Euclidean plane, $2\lvert A\rvert=\lvert p_2-p_1\rvert\,\lvert p_3-p_1\rvert\,\lvert\sin\theta_1\rvert$, which saturates (10.13) under exactly this condition. The unit triangle has its right angle at $(0,0)$ and gives $\lvert\Phi\rvert=\lambda^2=\lambda^2\cdot1\cdot1$.

## Open issues

- $\lambda_0$ depends on the loop. No statement is made at finite $\lambda$, nor beyond order $\lambda^2$, except the oddness for real records (out of scope).
- For commuting but non-real records, only the $\lambda^2$ term is shown to vanish. Whether the $\lambda^3$ term vanishes is not examined.
- (10.7) is the phase of a single overlap in the branch-state convention of (5.5). Only its cyclic sums are convention-free.

## Methods used

- Taylor expansion of matrix exponentials; principal logarithm of a near-unit analytic function
- Decomposition of a Hermitian inner product into a Euclidean form $g$, a symplectic form $\omega$ and a complex structure $J$
- Cauchy–Schwarz inequality and its equality case; polarization identity
- Complex conjugation symmetry for real operators
- Shoelace (determinant) formula for the signed area of a triangle