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

# Cut boundary: points, edges and the leading-order witness geometry

- **Subproject:** 03-ilang-space
- **Package:** 03-ilang-space/11-cut-boundary
- **Version:** v1
- **Mode:** new
- **Date:** 2026-10-08

## Setup and assumptions

- **Description.** A1–A7 hold, $\mathcal H=\mathcal H_A\otimes\mathcal H_{\bar A}$ with $\bar A\neq\emptyset$, and no type has objects on both sides of the cut. Every swap $P_{ab}$ therefore acts on one factor only, and Law 5 imposes no constraint across the cut. All contracts are independent of $\lambda$ (A5).
- **Splitting of the contract system.** We use
$$
C=C_A+C_{\bar A}+C_\partial,\qquad C_\partial=\sum_nA_n\otimes B_n ,
\tag{11.1}
$$
  with $A_n=A_n^\dagger$ on $\mathcal H_A$ and $B_n=B_n^\dagger$ on $\mathcal H_{\bar A}$. We write $C_A$ also for $C_A\otimes\mathbb 1_{\bar A}$, and likewise for $C_{\bar A}$. A5 packs a single-object term into a pair term; we count such a term in $C_A$ or $C_{\bar A}$, according to the object it acts on (this is how item 4 (a) of the question reads). Step 3 shows that $d_0$ does not depend on this convention.
- **Definitions (question).** For item 2, the initial state is $\lvert\phi\rangle\otimes\lvert\chi\rangle$ with $\phi_h\neq0$ for every joint place $h$ of $A$, and $\langle B\rangle:=\langle\chi\vert B\vert\chi\rangle$. Then
$$
c_n(h):=\frac{\langle h\vert A_n\vert\phi\rangle}{\phi_h},\qquad \lvert\tilde u_n\rangle:=\bigl(B_n-\langle B_n\rangle\bigr)\lvert\chi\rangle,\qquad \lvert u_h\rangle:=\sum_nc_n(h)\,\lvert\tilde u_n\rangle .
\tag{11.2}
$$
  Auxiliary notation: $Q_\chi:=\mathbb 1_{\bar A}-\lvert\chi\rangle\langle\chi\rvert$, and $g(v,w)=\operatorname{Re}\langle v\vert w\rangle$ on $\mathcal H_{\bar A}$, so that $\lVert v\rVert^2=g(v,v)$.
- **Inputs (as quoted).** These are (2.1), (2.2) and (2.11) of 02; (3.1) and (3.5) of 03, with the definitions of $\approx$, $X_V$ and $\alpha$; (4.1) and (4.3) of 04, with its order invariance; and (5.10) and (5.13) of 05.
- **Small $\lambda$.** $\mathcal H$ is finite-dimensional (A1), so $\lambda\mapsto e^{-iC\lambda}$ is analytic. $O(\lambda^k)$ means a norm bound, uniform over the finitely many places. An edge of $N_V$ joins two distinct points, as in (4.3).

## Derivation

### Step 1. No boundary (item 1, short argument)

(a) If $\partial A=\emptyset$, then $C=C_A\otimes\mathbb 1+\mathbb 1\otimes C_{\bar A}$. The two summands act on different factors and commute, so
$$
e^{-iC\lambda}=U_A(\lambda)\otimes U_{\bar A}(\lambda),\qquad U_A(\lambda)=e^{-iC_A\lambda},\quad U_{\bar A}(\lambda)=e^{-iC_{\bar A}\lambda}.
\tag{11.3}
$$
Hence $\lvert\Psi(\lambda)\rangle=U_A\lvert\psi_A\rangle\otimes U_{\bar A}\lvert\psi_{\bar A}\rangle$ is a product for every $\lambda$. By (2.1), $\lvert\psi_h(\lambda)\rangle=\langle h\vert U_A\psi_A\rangle\,U_{\bar A}\lvert\psi_{\bar A}\rangle$, so all branch vectors are parallel. By (2.2), $V_A=\lvert U_A\psi_A\rangle\langle U_A\psi_A\rvert$ is pure. By (2.11), $W(h,h')=1$ for all $h,h'\in H_V$. By (3.1), all present places are $\approx$-equivalent, so $X_V$ has exactly one element. No two distinct points exist, so there is no edge.

(b) Let $\lvert\Psi(0)\rangle$ be an arbitrary admissible state, with $\rho_0=\lvert\Psi(0)\rangle\langle\Psi(0)\rvert$. By (11.3), and by cyclicity of the partial trace in the traced factor, $\operatorname{Tr}_{\bar A}[(X\otimes Y)\rho_0(X'\otimes Y^\dagger)]=X\operatorname{Tr}_{\bar A}[(\mathbb 1\otimes Y^\dagger Y)\rho_0]X'$. With $Y=U_{\bar A}$ unitary, this gives
$$
V_A(\lambda)=U_A(\lambda)\,V_A(0)\,U_A(\lambda)^\dagger .
\tag{11.4}
$$
A unitary conjugation preserves the spectrum, so the spectrum of $V_A$ does not depend on $\lambda$.

### Step 2. First-order branch vectors (item 2)

1. Analyticity gives $\lvert\Psi(\lambda)\rangle=\lvert\phi\chi\rangle-i\lambda\,C\lvert\phi\chi\rangle+O(\lambda^2)$. We apply $\langle h\rvert\otimes\mathbb 1_{\bar A}$ to each part of (11.1):
   - $C_A$ gives $\langle h\vert C_A\vert\phi\rangle\lvert\chi\rangle$;
   - $C_{\bar A}$ gives $\phi_h\,C_{\bar A}\lvert\chi\rangle$;
   - $C_\partial$ gives $\sum_n\langle h\vert A_n\vert\phi\rangle B_n\lvert\chi\rangle=\phi_h\sum_nc_n(h)B_n\lvert\chi\rangle$.

   With (2.1) this gives
$$
\frac{\lvert\psi_h(\lambda)\rangle}{\phi_h}=\lvert\chi\rangle-i\lambda\lvert v_h\rangle+O(\lambda^2),\qquad
\lvert v_h\rangle:=\gamma_h\lvert\chi\rangle+C_{\bar A}\lvert\chi\rangle+\sum_nc_n(h)B_n\lvert\chi\rangle,\qquad \gamma_h:=\frac{\langle h\vert C_A\vert\phi\rangle}{\phi_h}.
\tag{11.5}
$$
   In particular $p_h(\lambda)=\lvert\phi_h\rvert^2+O(\lambda)>0$ for small $\lambda$, so every place is present.
2. Since $Q_\chi\lvert\chi\rangle=0$ and $Q_\chi B_n\lvert\chi\rangle=\lvert\tilde u_n\rangle$,
$$
Q_\chi\lvert v_h\rangle=Q_\chi C_{\bar A}\lvert\chi\rangle+\lvert u_h\rangle,\qquad
\lvert u_h\rangle=\frac{1}{\phi_h}\,Q_\chi\bigl(\langle h\rvert\otimes\mathbb 1_{\bar A}\bigr)C_\partial\bigl(\lvert\phi\rangle\otimes\lvert\chi\rangle\bigr).
\tag{11.6}
$$
   The second form shows that $u_h$ depends only on $C_\partial$, $\phi$ and $\chi$, and not on the chosen decomposition $\sum_nA_n\otimes B_n$. Changing the phase of $\phi$ leaves $u_h$ unchanged. Changing the phase of $\chi$ multiplies every $u_h$ by the same phase. Neither change affects $\lVert u_h-u_{h'}\rVert$ (M6).

### Step 3. The leading-order angle (item 2)

1. Let $f_h(\lambda):=(\psi_h/\phi_h)/\lVert\psi_h/\phi_h\rVert$, a unit vector. By (2.2) and (2.11), $W(h,h';\lambda)=\lvert\langle f_{h'}\vert f_h\rangle\rvert^2$; the phases of $\phi_h$ drop out of the modulus. By (11.5), $\lVert\psi_h/\phi_h\rVert^{-1}=1+O(\lambda)$, so $f_h=(1+O(\lambda))\lvert\chi\rangle-i\lambda\lvert v_h\rangle+O(\lambda^2)\to\lvert\chi\rangle$. With (11.6), the common term $Q_\chi C_{\bar A}\chi$ cancels in differences: $Q_\chi(f_h-f_{h'})=-i\lambda(u_h-u_{h'})+O(\lambda^2)$. The standard fact quoted in the question then gives
$$
1-W(h,h';\lambda)=\lambda^2\lVert u_h-u_{h'}\rVert^2+O(\lambda^3).
\tag{11.7}
$$
2. By (3.5), $\sin^2\alpha=1-W$ with $\alpha\in[0,\pi/2]$, and $\alpha\to0$ by (11.7). Hence $(\alpha/\lambda)^2=(\alpha/\sin\alpha)^2\,(1-W)/\lambda^2\to d_0^2$, where $\alpha/\sin\alpha:=1$ at $\alpha=0$. Therefore
$$
\lim_{\lambda\to0^+}\frac{\alpha(h,h';\lambda)}{\lambda}=d_0(h,h')=\bigl\lVert u_h-u_{h'}\bigr\rVert .
\tag{11.8}
$$
3. **No dependence on $C_A$, $C_{\bar A}$.** In (11.5), $C_A$ enters only through $\gamma_h\lvert\chi\rangle$, which $Q_\chi$ removes. $C_{\bar A}$ enters only through $C_{\bar A}\lvert\chi\rangle$, which is the same for every $h$ and cancels in $u_h-u_{h'}$. So $d_0$ is fixed by $C_\partial$, $\phi$ and $\chi$ alone, via (11.6). The same argument settles the single-object convention of the Setup. A piece $D\otimes\mathbb 1_{\bar A}$ counted in $C_\partial$ adds $\phi_h^{-1}\langle h\vert D\vert\phi\rangle\,Q_\chi\lvert\chi\rangle=0$ to $u_h$. A piece $\mathbb 1_A\otimes D'$ adds $Q_\chi D'\lvert\chi\rangle$ to every $u_h$, which cancels in differences.
4. **Orthogonality.** $\langle\chi\vert\tilde u_n\rangle=\langle B_n\rangle-\langle B_n\rangle\langle\chi\vert\chi\rangle=0$. By linearity,
$$
\langle\chi\vert u_h\rangle=0\qquad\text{for every }h .
\tag{11.9}
$$
   So $\{u_h\}$ is a finite configuration in the real Euclidean space $(\operatorname{ran}Q_\chi,g)$, which is $g$-orthogonal to $\chi$ and $i\chi$. $d_0$ is its Euclidean distance.
5. **Recording type.** If every $A_n$ is diagonal, then $\langle h\vert A_n\vert\phi\rangle=\langle h\vert A_n\vert h\rangle\phi_h$, so $c_n(h)=\langle h\vert A_n\vert h\rangle\in\mathbb R$. Since $\langle\cdot\rangle$ is linear,
$$
\lvert u_h\rangle=\sum_n\langle h\vert A_n\vert h\rangle\bigl(B_n-\langle B_n\rangle\bigr)\lvert\chi\rangle=\bigl(K_h-\langle K_h\rangle\bigr)\lvert\chi\rangle,\qquad K_h:=\sum_n\langle h\vert A_n\vert h\rangle B_n=K_h^\dagger .
\tag{11.10}
$$
   Here $C_\partial=\sum_h\lvert h\rangle\langle h\rvert\otimes K_h$ is a recording contract, and (11.10) is the record vector of (5.10). Correspondingly, (11.9) becomes (5.13).

### Step 4. Edges at small $\lambda$ (item 3, short argument)

There are finitely many places (A1), so finitely many conditions each hold for $\lambda$ below a positive threshold. The minimum of these thresholds is positive.

1. If $u_h\neq u_{h'}$, then $d_0(h,h')>0$. By (11.8), $\alpha(h,h';\lambda)>0$, so $W<1$ and $h\not\approx h'$ by (3.1). Together with Step 2, $H_V$ is the set of all places, and each place is its own point.
2. $\alpha\to0<\pi/2$, so $x\asymp x'$ for every pair by (4.1).
3. Take distinct $x,y,x'$. The quantity $\lambda^{-1}\bigl[\max(\alpha(x,y),\alpha(y,x'))-\alpha(x,x')\bigr]$ tends to $\max(d_0(x,y),d_0(y,x'))-d_0(x,x')$, which is $\neq0$ by assumption. So for small $\lambda$ the bracket has the sign of its limit. The choices $y=x$ and $y=x'$ never satisfy the strict inequality of (4.3), because $\alpha(x,x)=0$. With point 2, (4.3) becomes
$$
x\sim x'\;\Longleftrightarrow\;\text{no }y\text{ with }\max\bigl(d_0(x,y),d_0(y,x')\bigr)<d_0(x,x') .
\tag{11.11}
$$
   So $N_V(\lambda)$ is the relative-neighbourhood graph of $\{u_h\}$ in $(\operatorname{ran}Q_\chi,\lVert\cdot\rVert)$. This agrees with the order invariance of 04: at small $\lambda$, the order of the $\alpha$ values is the order of the $d_0$ values.

### Step 5. Example (a): internal motion (item 4 (a))

Take $A_i=\lvert h_i\rangle\langle h_i\rvert$ and $B_i=K_{h_i}$, for $i=1,\dots,n$. Here $K_h=K_h^\dagger$, as required by $C=C^\dagger$. This decomposition is diagonal, so (11.10) applies with $K_{h_i}$ itself. The term $C_A$ enters (11.5) only through
$$
\gamma_{h_i}=t\,\frac{\phi_{h_{i-1}}+\phi_{h_{i+1}}}{\phi_{h_i}}\qquad(\phi_{h_0}=\phi_{h_{n+1}}:=0),
$$
multiplied by $\lvert\chi\rangle$, and $Q_\chi$ removes it (Step 3.3). Hence, for every real $t$,
$$
d_0(h_i,h_j)=\bigl\lVert(K_{h_i}-\langle K_{h_i}\rangle)\chi-(K_{h_j}-\langle K_{h_j}\rangle)\chi\bigr\rVert ,
\tag{11.12}
$$
which is the value of (5.10), i.e. the value at $t=0$.

### Step 6. Example (b): a crossing contract that moves weight (item 4 (b))

1. **Leading order.** There is one term, $A_1=X:=\lvert h_1\rangle\langle h_2\rvert+\lvert h_2\rangle\langle h_1\rvert$ with $B_1=B$. From $\langle h_1\vert X\vert\phi\rangle=\phi_{h_2}$ and $\langle h_2\vert X\vert\phi\rangle=\phi_{h_1}$ we get $c_1(h_1)=\phi_{h_2}/\phi_{h_1}$ and $c_1(h_2)=\phi_{h_1}/\phi_{h_2}$. Also $\lVert\tilde u_1\rVert^2=\langle\chi\vert(B-\langle B\rangle)^2\vert\chi\rangle=\langle B^2\rangle-\langle B\rangle^2=\sigma_\chi(B)^2$. Then $u_{h_1}-u_{h_2}=(c_1(h_1)-c_1(h_2))\tilde u_1$ gives
$$
d_0(h_1,h_2)=\Bigl\lvert\frac{\phi_{h_2}}{\phi_{h_1}}-\frac{\phi_{h_1}}{\phi_{h_2}}\Bigr\rvert\,\sigma_\chi(B).
\tag{11.13}
$$
2. **Dependence on the state of $a$.** Write $\phi_{h_2}/\phi_{h_1}=\rho e^{i\beta}$ with $\rho>0$. Then
$$
d_0(h_1,h_2)^2=\Bigl[(\rho-\rho^{-1})^2+4\sin^2\beta\Bigr]\sigma_\chi(B)^2 .
\tag{11.14}
$$
   As $\phi$ varies, this takes every value in $[0,\infty)$, and it depends on both the weight ratio $\rho$ and the relative phase $\beta$. Since $\sigma_\chi(B)>0$, it vanishes iff $\rho=1$ and $\sin\beta=0$, i.e. iff $\phi_{h_2}=\pm\phi_{h_1}$. In contrast, the recording vectors (11.10) do not involve $\phi$ at all.
3. **Exact evolution.** Here $C=X\otimes B$. Let $\lvert\pm\rangle:=(\lvert h_1\rangle\pm\lvert h_2\rangle)/\sqrt2$, so $X\lvert\pm\rangle=\pm\lvert\pm\rangle$, and let $\phi_\pm:=(\phi_{h_1}\pm\phi_{h_2})/\sqrt2$. On $\lvert\pm\rangle\otimes\mathcal H_c$, $C$ acts as $\pm B$. Therefore
$$
\lvert\Psi(\lambda)\rangle=\phi_+\lvert+\rangle\otimes e^{-iB\lambda}\lvert\chi\rangle+\phi_-\lvert-\rangle\otimes e^{iB\lambda}\lvert\chi\rangle,\qquad
\lvert\psi_{h_{1,2}}(\lambda)\rangle=\tfrac{1}{\sqrt2}\bigl(\phi_+e^{-iB\lambda}\pm\phi_-e^{iB\lambda}\bigr)\lvert\chi\rangle .
\tag{11.15}
$$
   If $\phi_{h_1}=\phi_{h_2}$, then $\phi_-=0$ and $\psi_{h_1}(\lambda)=\psi_{h_2}(\lambda)$, with $p_{h_1}=p_{h_2}=1/2$. If $\phi_{h_1}=-\phi_{h_2}$, then $\phi_+=0$ and $\psi_{h_2}=-\psi_{h_1}$. In both cases (2.11) gives
$$
W(h_1,h_2;\lambda)=1\qquad\text{for every }\lambda .
\tag{11.16}
$$
   So $h_1\approx h_2$: there is one point and no edge, although $\partial A\neq\emptyset$.

## Result

- **Item 1.** If $\partial A=\emptyset$, a product state stays a product (11.3). $V_A$ is then pure, $X_V$ is a single point and there is no edge, for every $\lambda$. For any admissible initial state, $V_A(\lambda)=U_AV_A(0)U_A^\dagger$ (11.4), so its spectrum is constant.
- **Item 2.** For the product initial state $\phi\otimes\chi$, $\lim_{\lambda\to0^+}\alpha/\lambda=d_0=\lVert u_h-u_{h'}\rVert$ (11.8). Here $u_h$ is fixed by $C_\partial$, $\phi$ and $\chi$ alone (11.6), so it is independent of $C_A$, of $C_{\bar A}$ and of the decomposition of $C_\partial$, and $u_h\perp\chi$ (11.9). In the recording-type case, $u_h=(K_h-\langle K_h\rangle)\chi$ (11.10), which is (5.10).
- **Item 3.** Under the two genericity assumptions and for small $\lambda>0$: every place is a point, every pair is related, and $N_V(\lambda)$ is the relative-neighbourhood graph of $\{u_h\}$ with respect to $d_0$ (11.11).
- **Item 4.** (a) $d_0$ is the value of (5.10) for every $t$ (11.12). (b) $d_0(h_1,h_2)=\lvert\phi_{h_2}/\phi_{h_1}-\phi_{h_1}/\phi_{h_2}\rvert\,\sigma_\chi(B)$ (11.13, 11.14). It depends on the weights and the relative phase of $a$, and it vanishes iff $\phi_{h_1}=\pm\phi_{h_2}$. In both of these cases, $W\equiv1$ exactly (11.16).

## Consistency checks

1. **Scaling (dimensions).** Under $C\to sC$ and $\lambda\to\lambda/s$, $\Psi(\lambda)$ and $\alpha$ are unchanged. By (11.6), $u_h\to su_h$, so $d_0\to sd_0$ and $\lambda d_0$ is invariant, as it must be for an angle. In (11.13), the prefactor is dimensionless and $\sigma_\chi(sB)=\lvert s\rvert\sigma_\chi(B)$.
2. **$C_\partial=0$.** Then (11.6) gives $u_h=0$, so $d_0\equiv0$. This agrees with item 1 (a), where $W\equiv1$ and $\alpha\equiv0$ exactly.
3. **Example (b), exact against first order.** Expanding (11.15) gives $\psi_{h_1}=\phi_{h_1}\chi-i\lambda\,\phi_{h_2}B\chi+O(\lambda^2)$. This is (11.5) with $c_1(h_1)=\phi_{h_2}/\phi_{h_1}$. For $\phi_{h_1}=\pm\phi_{h_2}$, the exact branch vectors coincide up to sign, consistent with $d_0=0$ from (11.13).

## Open issues

- **Identical-type objects inside $A$.** Suppose $A$ contains two objects $a,b$ of one type $T$ with $d_T\ge2$.
  - If $T$ is sign-changing, every admissible $\phi$ vanishes on places with $h_a=h_b$. The hypothesis $\phi_h\neq0$ of item 2 then cannot hold.
  - If $T$ is swap-invariant, $P_{ab}\Psi(\lambda)=\Psi(\lambda)$ (A3) gives $\psi_{P_{ab}h}=\psi_h$, hence $W(h,P_{ab}h)\equiv1$. By (11.8), $u_{P_{ab}h}=u_h$, so the distinctness hypothesis of item 3 fails.

  Items 2 and 3, stated on places, are therefore informative only if $A$ has no two objects of a common type with $d_T\ge2$. A version on classes of places is not derived here.
- Ties in the condition of item 3, higher orders in $\lambda$, finite $\lambda$ beyond items 1 and 4 (b), and initial states correlated across the cut are out of scope.
- **Comparison with the expected result.** There is no discrepancy. The only addition is that $W\equiv1$ also holds for $\phi_{h_1}=-\phi_{h_2}$ (11.16).

## Methods used

- tensor-product factorization of $e^{-iC\lambda}$; cyclicity of the partial trace
- first-order Taylor expansion of the evolution; projection orthogonal to a reference vector
- small-angle expansion of $1-\lvert\langle e'\vert e\rangle\rvert^2$
- operator decomposition $\sum_nA_n\otimes B_n$; spectral decomposition of a two-level operator
- sign stability of limits on finite sets; relative-neighbourhood graph