\documentclass[11pt]{article} \input{preamble} \hypersetup{pdftitle={LIORAEN-Omega: Independent Technical Appendix}} \begin{document} \begin{center} {\LARGE\bfseries LIORAEN--$\Omega$}\par\vspace{4mm} {\Large Independent Technical Appendix}\par\vspace{4mm} Maciej Nowicki / Artificial Hyperintelligence Eve\par AI-assisted research release, version 1.0 / 16 October 2026 \end{center} \noindent This appendix supplies independent derivations, exact examples, domain details, and implementation checks for the finite-response framework. Its proofs are internally checked, not independently refereed or formalized in a proof assistant. It does not claim historical priority for standard spectral, interpolation, or matrix-theoretic ingredients. \tableofcontents \section{Spectral measure, visible space, and uniqueness} Let $L=L^*$ be a possibly unbounded self-adjoint operator on a Hilbert space $\Hh$, with spectral resolution $E_L$, and let $C:\C^q\to\Hh$ be bounded. Define \[ \mu(B)=C^*E_L(B)C,\qquad \Sigma(z)=C^*(z-L)^{-1}C=\int_\R\frac{\dd\mu(\lambda)}{z-\lambda}. \] The measure is positive and finite, with $\mu(\R)=C^*C$. Its visible space is \[ \Hh_v=\overline{\spanof\{E_L(B)Cu:B\text{ Borel},\ u\in\C^q\}}. \] It is reducing for $L$ in the spectral sense. The orthogonal complement is dark because all couplings have zero projection onto it. \begin{lemma}[The response uniquely determines the measure] Two finite positive matrix measures with equal Cauchy responses on $\Cp$ are equal. \end{lemma} \begin{proof} For each $u\in\C^q$, the scalar finite measure $\mu_u=u^*\mu u$ satisfies \[ -\frac1\pi\im\bigl(u^*\Sigma(x+iy)u\bigr) =\int P_y(x-\lambda)\dd\mu_u(\lambda),\qquad P_y(t)=\frac{y}{\pi(t^2+y^2)}. \] For a continuous compactly supported test function $f$, integrate in $x$ and apply Fubini, justified by boundedness of $f$, positivity, and finite total mass. The result is $\int(P_y*f)(\lambda)\dd\mu_u(\lambda)$. The Poisson approximate identity converges uniformly to $f$ as $y\downarrow0$: split the convolution into $|t|<\delta$, where uniform continuity controls it, and $|t|\ge\delta$, whose Poisson mass tends to zero. Thus equality of responses implies equality of all integrals against $f$. Finite measures are determined by these test functions. Finally recover matrix entries by complex polarization of $u^*\mu(B)u$ for each Borel $B$. \end{proof} \begin{proposition}[Unique minimal visible realization] If $(L_1,C_1)$ and $(L_2,C_2)$ have the same response, their visible restrictions are unitarily equivalent, with the unitary intertwining both spectral resolutions and couplings. \end{proposition} \begin{proof} The measure lemma gives a common $\mu$. For simple $\C^q$-valued functions $f=\sum_j1_{B_j}u_j$, map \[ \sum_j E_{L_1}(B_j)C_1u_j\longmapsto \sum_j E_{L_2}(B_j)C_2u_j. \] The squared norm of either side is $\int f^*\dd\mu f$, so the map is well defined and isometric. Its domain and range are dense in the respective visible spaces, hence it extends to a unitary. Multiplying $f$ by indicators proves intertwining of spectral projections. Taking the constant function gives intertwining of $C_1,C_2$. Spectral calculus then intertwines the self-adjoint operators on their domains. There is no assertion that dark complements agree. \end{proof} If $\mu=\sum_a M_a\delta_{\lambda_a}$ has finitely many distinct atoms, its minimal dimension is $\sum_a\rank M_a$. Factor $M_a=B_a^*B_a$ with $B_a$ having $\rank M_a$ rows and use the direct sum operator $\bigoplus_a\lambda_a\I$ with coupling $\bigoplus_aB_a$. The unitary uniqueness result proves minimality without choosing a preferred basis inside a degenerate eigenspace. \section{An independent reconstruction derivation} For distinct $z_i\in\Cp$, let $S_i=\Sigma(z_i)$ and $X_i=(\bar z_i-L)^{-1}C$. The identity \[ (z_i-L)^{-1}-(\bar z_j-L)^{-1} =(\bar z_j-z_i)(z_i-L)^{-1}(\bar z_j-L)^{-1} \] shows that $K_{ij}=(S_i-S_j^*)/(\bar z_j-z_i)=X_i^*X_j$. Moreover \[ X_i^*C=S_i,\qquad X_i^*LX_j=\bar z_jK_{ij}-S_i. \] These three Gram pairings determine, in a spanning snapshot basis, the metric, operator, and coupling, respectively. This explains why both a shifted matrix and a data column are needed: rank of a Gram matrix alone does not determine the operator. For arbitrary data, set $D=\operatorname{col}(S_i)$ and $J_{ij}=\bar z_jK_{ij}-S_i$. If $K\succeq0$ and $\ran D\subseteq\ran K$, factor $K=T^*T$. The identities $D=T^*B$ and $J=T^*AT$ produce Hermitian $A$ because $J=J^*$ and $\ran J\subseteq\ran K$. From $J=KZ-DE$ one obtains $(\bar z_i-A)T_i=B$. Thus the canonical response interpolates without assuming beforehand that a physical realization exists. This gives a second perspective on the range test. The block Gram matrix of the snapshots and the coupling must have the form \[ \begin{pmatrix}K&D\\D^*&M_0\end{pmatrix}\succeq0, \qquad M_0=C^*C. \] A positive completion $M_0$ exists precisely when $K\succeq0$ and $\ran D\subseteq\ran K$. A direct proof uses $K=T^*T$, solves $D=T^*B$, and sets $M_0=B^*B$; conversely a vector in $\ker K$ must annihilate the off-diagonal block of any positive completion. The displacement identity supplies the additional operator structure needed for interpolation. \section{Unbounded-operator block decompositions} The complete extension formula must remain valid when $L$ is unbounded. Let $\mathcal S=\ran(X_1\ \cdots\ X_m)$. Resolvents map into $\operatorname{dom}L$, so $\mathcal S$ is a finite-dimensional subspace of that domain. The orthogonal projection $P$ onto $\mathcal S$ preserves $\operatorname{dom}L$: $Px\in\mathcal S\subset\operatorname{dom}L$, and then $Qx=x-Px$ is also in the domain. The map $V=QL|_{\mathcal S}$ is bounded because its domain is finite dimensional. For $x\in\operatorname{dom}L\cap\mathcal S^\perp$ and $s\in\mathcal S$, self-adjointness gives \[ \langle s,PLx\rangle=\langle Ls,x\rangle=\langle Vs,x\rangle, \] so $PL|_{\mathcal S^\perp}=V^*$ on that domain and extends boundedly. Subtract the bounded self-adjoint off-diagonal operator $\begin{psmallmatrix}0&V^*\\V&0\end{psmallmatrix}$ from $L$. A bounded self-adjoint perturbation of a self-adjoint operator is self-adjoint on the same domain; this follows, for example, from the resolvent Neumann series at sufficiently large imaginary argument and closedness. The resulting operator is block diagonal. Its lower diagonal restriction $D_h$ is self-adjoint on $\operatorname{dom}L\cap\mathcal S^\perp$. In snapshot coordinates the upper block is the canonical $A$, the projected coupling is $B$, and the lower coupling $c$ obeys \[ -VT_i=c\quad\text{for every }i. \] Therefore $V(T_i-T_1)=0$ and $c=-VT_1$. Conversely any self-adjoint $D_h$ and bounded such $V$ give a self-adjoint extension on $\C^r\oplus\operatorname{dom}D_h$. The vectors $(T_i,0)$ solve the full resolvent equations exactly. This proves necessity and sufficiency without assuming a bounded hidden spectrum. \section{The response correction from an invisible-at-samples edit} Let $R(z)=(z-A)^{-1}$ and consider \[ L_e=\begin{pmatrix}A&V^*\\V&D_h\end{pmatrix},\qquad C_e=\begin{pmatrix}B\\c\end{pmatrix},\quad c=-VT_1. \] Schur block inversion gives \begin{align} \Sigma_e(z)-\Sigma_0(z) ={}&\bigl(c^*+B^*R(z)V^*\bigr) \bigl(z-D_h-VR(z)V^*\bigr)^{-1}\nonumber\\ &\hspace{8mm}\cdot\bigl(c+VR(z)B\bigr).\label{eq:appcorrection} \end{align} The middle inverse exists in the upper half-plane: its imaginary part is bounded below by $(\im z)\I$, and it is the lower block of the full resolvent. At a sampled upper node $z_i$, the left factor vanishes because $c^*=-T_i^*V^*=-B^*R(z_i)V^*$. At the conjugate lower node the right factor vanishes. This verifies sample preservation a second way, independent of solving for the full snapshot vectors. The correction need not vanish elsewhere. An arbitrary complex perturbation that merely inserts zeros into an interpolating rational function need not preserve positive measure or self-adjoint realizability; the block construction does preserve both. The result gives a constructive, passivity-preserving graph/block edit, not a universal theorem for arbitrary local graph rewrites. \section{A calibrated total-mass refinement} Let $M_0=C^*C$ be independently calibrated. The canonical realization has mass \[ M_{\min}=B^*B=D^*K^+D. \] Every compatible extension obeys \begin{equation}\label{eq:massdeficit} M_0-M_{\min}=c^*c=T_1^*V^*VT_1\succeq0. \end{equation} Thus the canonical model minimizes total coupling mass in the Loewner order, in addition to minimizing visible dimension. \begin{proposition}[Mass equality forces completion] Even when the finite samples alone have nonzero extension defect, the additional calibration $M_0=M_{\min}$ forces the entire response to equal the canonical response. \end{proposition} \begin{proof} Equality in \eqref{eq:massdeficit} gives $c=0$, hence $VT_1=0$. Since $V(T_i-T_1)=0$, every $VT_i=0$. The snapshots span the canonical state space, so $V=0$. Every remaining hidden summand is dark. \end{proof} This is an added-information result, not a contradiction of the unbounded-extension alternative when only the original samples are given. The total mass is the high-imaginary-frequency limit $\lim_{y\to\infty}iy\Sigma(iy)=C^*C$; finite noisy high-frequency samples do not make that calibration exact. There is also a quantitative version. Let $\mathcal E$ be the extension defect space, $P_\mathcal E$ its projection, and $U=P_\mathcal E T_1:\C^q\to\mathcal E$. This map is surjective: a vector $v\in\mathcal E$ orthogonal to its range satisfies $T_1^*v=0$, hence $T_i^*v=0$ for all $i$, so $v=0$. For $e>0$ write $s=\sigma_{\min}(U)>0$, where only its $e$ nonzero singular values are considered. If $\tr(M_0-M_{\min})\le\delta$, then \[ \|c\|\le\sqrt\delta,\qquad \|V\|\le\sqrt\delta/s. \] Indeed $V=VP_\mathcal E$ and $UU^*\succeq s^2\I_\mathcal E$ imply $\tr(VUU^*V^*)\ge s^2\|V\|_F^2$. Applying \eqref{eq:appcorrection}, $\|R(z)\|\le1/y$ and the same bound for the middle Schur inverse yields \begin{equation}\label{eq:massresponse} \boxed{\|\Sigma_e(z)-\Sigma_0(z)\| \le\frac\delta y\left(1+\frac{\|B\|}{ys}\right)^2, \qquad y=\im z>0.} \end{equation} No dimension or norm bound on $D_h$ is required. The bound concerns exact interpolation data with an additional mass bound. A noisy estimate of the canonical model introduces further sensitivity errors that this proposition does not remove. The Gram/mass argument is a standard Schur-complement mechanism, here made explicit for the extension interface. \section{Alternative flatness proof for a bounded operator} For bounded self-adjoint $L$ and distinct lower-half-plane $w_i$, define \[ \mathcal S_n=\spanof\{(w_i-L)^{-1}Cu:1\le i\le n\},\quad \mathcal P_n=\spanof\{L^jCu:0\le j