\section{Finite precision certificates with analytic feedback}\label{sec:feedback}\label{ff:section}

Let $Q_1(t),\ldots,Q_q(t)$ be $D\times D$ real or complex matrices, real analytic in a real parameter near zero. Fix $p_i>0$ with $\sum_i p_i=1$. At each step independently choose $I+Q_i(t)$ or $I-Q_i(t)$ with probabilities $p_i/2$. For the ordered product $V_N$, put
\begin{equation}\label{ff:setting}
 \Phi_t(X)=\sum_i p_iQ_i(t)^*XQ_i(t),\qquad
 G_N(t)=\mathbb E(V_N^*V_N)=(\operatorname{Id}+\Phi_t)^N(I).
\end{equation}
Throughout this section $\Phi_0$ is nilpotent; no marginality assumption is imposed at $t\ne0$. Unsubscripted state-matrix norms are operator norms. Write
\[
 C_k(t)=\Phi_t^k(I),\qquad c_k(t)=\tr C_k(t),\qquad
 C_h(0)\ne0,\quad C_{h+1}(0)=0.
\]
Thus $h\ge0$ is the baseline height. All $C_k$ are positive semidefinite, and
\begin{equation}\label{ff:positive}
 G_N=\sum_{k=0}^N\binom Nk C_k,\qquad
 c_k=\sum_{|w|=k}p_w\norm{Q_w}_F^2.
\end{equation}
Here $Q_w$ ranges over ordered length-$k$ products. Positivity implies that $C_{h+1}(0)=0$ is equivalent to the vanishing of all baseline words of that length. Their image spans form a strictly descending chain until zero, so $h\le D-1$.

Set $M=D^2$ and take the characteristic polynomial on the $M$-dimensional matrix space:
\begin{equation}\label{ff:characteristic}
 \det(\lambda\operatorname{Id}-\Phi_t)
 =\lambda^M+a_1(t)\lambda^{M-1}+\cdots+a_M(t).
\end{equation}
For a scalar analytic germ, $\operatorname{ord}_0$ denotes its vanishing order, with value infinity for the zero germ. Define
\begin{equation}\label{ff:certificate}
 v_k=\operatorname{ord}_0c_k,\qquad
 b=\min_{1\le j\le M}\frac{\operatorname{ord}_0a_j}{j},\qquad
 \boxed{\displaystyle
 \beta=\min\left\{b,\ \min_{h<k<M}\frac{v_k}{k-h}\right\}.}
\end{equation}
Infinite terms are omitted and empty minima are infinity. The certificate uses only a characteristic polynomial and $M$ initial moments. Every finite candidate is positive. Equation~\eqref{ff:positive} shows that $v_k$ is even and retains every cancellation inside each ordered word.

\subsection{A sharp frontier from finite data}

The characteristic rescaling below is the classical first Newton slope for an eigenvalue perturbation problem; see, for example, \cite{Moro}. We include the elementary estimate needed here.

\begin{lemma}[Spectral order]\label{ff:spectral}
If $b<\infty$, then $\rho(\Phi_t)\asymp t^b$ as $t\downarrow0$. More precisely, with
\[
 a_j^{[b]}=\lim_{t\downarrow0}t^{-bj}a_j(t),\qquad
 P_b(z)=z^M+\sum_{j=1}^M a_j^{[b]}z^{M-j},
\]
one has $\rho(\Phi_t)/t^b\to r_b>0$, where $r_b$ is the largest modulus of a root of $P_b$. If $b=\infty$, then $\Phi_t^M=0$ identically near zero.
\end{lemma}
\begin{proof}
For any matrix of spectral radius $r$, its characteristic coefficients satisfy
$|a_j|\le\binom Mj r^j$. Conversely, if $\eta=\max_j|a_j|^{1/j}$, the characteristic equation rules out a root of modulus greater than $2\eta$. These inequalities give $\rho(\Phi_t)\asymp t^b$. Alternatively, substituting $\lambda=t^bz$ in \eqref{ff:characteristic} gives monic polynomials converging to $P_b\ne z^M$. Continuity of their root multisets gives the stated positive limit. When all $a_j$ vanish, Cayley--Hamilton gives $\Phi_t^M=0$.
\end{proof}

\begin{theorem}[Complete algebraic-horizon frontier]\label{ff:frontier}
If $\beta<\infty$, there are $c,C,A,t_0>0$ such that for every integer $N\ge0$ and $0<t\le t_0$,
\begin{equation}\label{ff:global}
 c(N+1)^h\le\norm{G_N(t)}
 \le C(N+1)^h\exp\{ANt^\beta\}.
\end{equation}
For each $\alpha>0$, at $N=\lfloor t^{-\alpha}\rfloor$,
\begin{equation}\label{ff:powerfrontier}
 \frac{\norm{G_N(t)}}{(N+1)^h}=O(1)
 \quad\Longleftrightarrow\quad \alpha\le\beta.
\end{equation}
For $\alpha>\beta$ this ratio tends to infinity. If $\beta=\infty$, then $C_k(t)=0$ for all $k>h$ and $\norm{G_N(t)}\asymp(N+1)^h$ uniformly at every horizon.
\end{theorem}
\begin{proof}
Assume $\beta<\infty$. The definition of $\beta$ bounds the finite initial sequence
$u_k(t)=t^{-\beta(k-h)}c_k(t)$, $0\le k<M$, and every coefficient $a_j(t)t^{-\beta j}$. Cayley--Hamilton gives
\[
 u_k=-\sum_{j=1}^M\bigl(a_jt^{-\beta j}\bigr)u_{k-j},\qquad k\ge M.
\]
Choose $A\ge1$ with $\sum_j\sup|a_jt^{-\beta j}|A^{-j}\le1$ and enlarge $K$ to cover the initial sequence. Induction proves, uniformly in $k$ and $t$,
\begin{equation}\label{ff:recurrencebound}
 c_k(t)\le KA^kt^{\beta(k-h)}.
\end{equation}
For $k\le h$ also use the sharper bound $c_k(t)\le K$. The low terms in \eqref{ff:positive} contribute $O((N+1)^h)$. For $k=h+j>h$, the remaining trace is at most
\[
 KA^hN^h\sum_{j\ge1}\frac{(ANt^\beta)^j}{(h+j)!}
 \le C(N+1)^h e^{ANt^\beta}.
\]
Since $\norm G\le\tr G$ for $G\succeq0$, this proves the upper bound. Continuity gives $c_h(t)\ge c_0>0$. Its positive binomial term and $\tr G\le D\norm G$ prove the lower bound for large $N$; $G_N\succeq I$ covers the finitely many remaining $N$.

The upper bound gives \eqref{ff:powerfrontier} for $\alpha\le\beta$, including equality. If a finite initial moment attains $\beta$, its term divided by $N^h$ is bounded below by a positive multiple of $N^{k-h}t^{v_k}$, which diverges for $\alpha>\beta$. In the other case $\beta=b<\infty$. A positive-map Perron eigenmatrix $R_t\succeq0$, normalized by $\norm{R_t}=1$, satisfies $R_t\preceq I$ and $\Phi_t(R_t)=\rho(\Phi_t)R_t$. Hence
\begin{equation}\label{ff:Perronlower}
 G_N(t)\succeq(1+\rho(\Phi_t))^N R_t,
 \qquad\norm{G_N(t)}\ge(1+\rho(\Phi_t))^N.
\end{equation}
For $\alpha>b$ Lemma~\ref{ff:spectral} makes this exponential dominate $N^h$. The positive Perron fact follows by regularizing $\Phi$ with $\epsilon\tr(\cdot)I$, finding a fixed point on the trace-one positive section, identifying its eigenvalue with the spectral radius by the order-unit norm, and passing to the limit. No regular choice of eigenvectors is required.

Finally, $\beta=\infty$ makes every characteristic coefficient and every initial $C_k$, $h<k<M$, identically zero. The recurrence then gives $C_k=0$ for all $k>h$, proving the last assertion.
\end{proof}

\begin{proposition}[The infinite word invariant is finite]\label{ff:words}
The same certificate satisfies
\begin{equation}\label{ff:infinite}
 \boxed{\displaystyle\beta=\inf_{k>h}\frac{\operatorname{ord}_0c_k}{k-h}.}
\end{equation}
Consequently $h$ and $\beta$ are independent of the particular fixed strictly positive label probabilities.
\end{proposition}
\begin{proof}
If $\beta=\infty$, Theorem~\ref{ff:frontier} makes both sides infinity. Otherwise bound~\eqref{ff:recurrencebound} makes the infimum at least $\beta$. It is at most every finite initial candidate. If $b<\infty$, \eqref{ff:Perronlower} applied to individual powers also gives $c_k(t)\ge\rho(\Phi_t)^k$, hence $v_k\le bk$. Letting $k\to\infty$ bounds the infimum by $b$. If $b=\infty$, Cayley--Hamilton truncates the sequence and the finite initial candidates suffice. Finally, positive word weights cannot cancel a squared leading word coefficient in \eqref{ff:positive}; they change neither its order nor the baseline height.
\end{proof}

\subsection{The full polynomial, ODE, and exponential phase diagram}

For $v_k<\infty$ let $A_k=\lim_{t\downarrow0}t^{-v_k}C_k(t)\succeq0$, which is nonzero. For $0<\alpha\le b$, with every finite $\alpha$ permitted when $b=\infty$, put
\begin{equation}\label{ff:gamma}
 \gamma_\alpha=\max_{0\le k<M,\,v_k<\infty}(\alpha k-v_k),\qquad
 \mathcal F_\alpha=\{k<M:\alpha k-v_k=\gamma_\alpha\}.
\end{equation}

\begin{theorem}[Finite crossover and the spectral transition]\label{ff:phases}
For $0<\alpha\le b$, the limits
\[
 L_k^{[\alpha]}=\lim_{t\downarrow0}t^{\gamma_\alpha-\alpha k}C_k(t),\qquad
 a_j^{[\alpha]}=\lim_{t\downarrow0}t^{-\alpha j}a_j(t)
\]
exist for all $k\ge0$. The entire matrix function
\begin{equation}\label{ff:entire}
 E_\alpha(s)=\sum_{k=0}^\infty\frac{s^k}{k!}L_k^{[\alpha]}
\end{equation}
is determined by the finite constant-coefficient ODE
\begin{equation}\label{ff:ODE}
 E_\alpha^{(M)}+\sum_{j=1}^Ma_j^{[\alpha]}E_\alpha^{(M-j)}=0,
 \qquad E_\alpha^{(k)}(0)=L_k^{[\alpha]},\quad 0\le k<M.
\end{equation}
Uniformly for $s$ in compact subsets of $(0,\infty)$,
\begin{equation}\label{ff:crossover}
 t^{\gamma_\alpha}G_{\lfloor st^{-\alpha}\rfloor}(t)\longrightarrow E_\alpha(s)\succeq0,
 \qquad E_\alpha(s)\ne0.
\end{equation}
The three algebraic-horizon regimes are as follows:
\begin{enumerate}
\item If $0<\alpha<b$, then every $a_j^{[\alpha]}=0$ and
\begin{equation}\label{ff:polynomial}
 E_\alpha(s)=\sum_{k\in\mathcal F_\alpha}\frac{s^k}{k!}A_k.
\end{equation}
\item At $\alpha=b<\infty$, the correct normalization is
\begin{equation}\label{ff:gammab}
 \gamma_b=\max_{k<M,\,v_k<\infty}(bk-v_k)\ge bh.
\end{equation}
It may be strictly greater than $bh$. The crossover is the entire ODE solution $E_b$, which may contain exponential terms; such terms need not be excited by its initial data.
\item If $b<\alpha<\infty$, then
\begin{equation}\label{ff:exponential}
 t^{\alpha-b}\log\norm{G_{\lfloor st^{-\alpha}\rfloor}(t)}\longrightarrow sr_b,
\end{equation}
uniformly on those compact sets, with $r_b$ from Lemma~\ref{ff:spectral}.
\end{enumerate}
\end{theorem}
\begin{proof}
For $\alpha\le b$, normalize $B_k=t^{\gamma_\alpha-\alpha k}C_k$. Its $M$ initial matrices are bounded and have limits by analyticity and \eqref{ff:gamma}. The scaled characteristic coefficients $a_jt^{-\alpha j}$ are bounded and convergent. The same Cayley--Hamilton induction as above gives $\norm{B_k}\le KA^k$ uniformly, and the finite limiting recurrence gives every $L_k^{[\alpha]}$. For $N=\lfloor st^{-\alpha}\rfloor$,
\[
 t^{\gamma_\alpha}\binom Nk C_k
 =\bigl(\binom Nk t^{\alpha k}\bigr)B_k
 \longrightarrow\frac{s^k}{k!}L_k^{[\alpha]}.
\]
On $0<s\le S$ its norm is at most $K(AS)^k/k!$. Dominated convergence proves \eqref{ff:crossover} and its uniformity. The limiting recurrence is exactly \eqref{ff:ODE}. At least one initial limiting matrix is PSD and nonzero, so its positive series is nonzero for $s>0$. When $\alpha<b$ all limiting recurrence coefficients vanish, so the series truncates and gives \eqref{ff:polynomial}. At $\alpha=b$ this argument instead gives \eqref{ff:gammab} and the stated ODE.

For $\alpha>b$, apply a unitary Schur triangularization to the bounded superoperator family $T_t=\operatorname{Id}+\Phi_t$ in the Hilbert--Schmidt norm. Its diagonal part has norm at most $1+\rho(\Phi_t)$, and its strictly upper triangular part is uniformly bounded. A product containing $M$ strictly upper triangular factors, even with diagonal factors inserted, is zero. Expanding $T_t^N$ therefore gives
\[
 \norm{G_N(t)}\le C(N+1)^{M-1}(1+\rho(\Phi_t))^N.
\]
Together with \eqref{ff:Perronlower}, this bounds the difference between $\log\norm{G_N}$ and $N\log(1+\rho(\Phi_t))$ between zero and $O(\log(N+1))$. Lemma~\ref{ff:spectral} then proves \eqref{ff:exponential}.
\end{proof}

At the preservation frontier $\alpha=\beta<\infty$, one has $\gamma_\beta=\beta h$. Write $L_k=L_k^{[\beta]}$ and $E=E_\beta$. Then $L_k=0$ for $k<h$, $L_h=C_h(0)\ne0$, and
\begin{equation}\label{ff:normalizedcritical}
 \frac{\norm{G_{\lfloor st^{-\beta}\rfloor}(t)}}{\lfloor st^{-\beta}\rfloor^h}
 \longrightarrow\frac{\norm{E(s)}}{s^h}.
\end{equation}
For $\alpha<\beta$, the high-order remainder in the proof of \eqref{ff:global} tends to zero after division by $N^h$, yielding $N^{-h}G_N(t)\to C_h(0)/h!$.

For an explicit feedback transition, take the single generator $Q(t)=E_{12}+tE_{21}$. Then $Q(t)^2=tI$, $h=1$, and $b=\beta=1$. For $0<t\le1$ the exact moment is
\[
 G_N=A_N I+B_NQ(t)^*Q(t),\qquad
 A_N=\frac{(1+t)^N+(1-t)^N}{2},\quad
 B_N=\frac{(1+t)^N-(1-t)^N}{2t}.
\]
Since $Q(t)^*Q(t)=\operatorname{diag}(t^2,1)$, its norm is $A_N+B_N$. At $N=\lfloor s/t\rfloor$,
\[
 tG_N(t)\longrightarrow\sinh(s)E_{22},\qquad
 \frac{\norm{G_N(t)}}{N}\longrightarrow\frac{\sinh(s)}s.
\]
Here the finite ODE reduces to $E''-E=0$, with $E(0)=0$ and $E'(0)=E_{22}$; the exponential feedback mode survives the critical normalization.

\subsection{When the same frontier controls arbitrary horizons}

\begin{theorem}[A finite test for the full-sequence law]\label{ff:arbitrary}
Assume $\beta<\infty$. The following are equivalent:
\begin{enumerate}
\item For every $t_j\to0^+$ and $N_j\to\infty$,
\[
 \frac{\norm{G_{N_j}(t_j)}}{(N_j+1)^h}=O(1)
 \quad\Longleftrightarrow\quad N_jt_j^\beta=O(1).
\]
\item Some $L_k$ with $k>h$ is nonzero.
\item Some matrix in the finite block $L_{h+1},\ldots,L_{h+M}$ is nonzero.
\end{enumerate}
The finite block is computed from the ODE data in \eqref{ff:ODE}; no additional analytic moments are needed.
\end{theorem}
\begin{proof}
If $L_k\ne0$ for $k>h$, then $v_k=\beta(k-h)$. The positive $k$th binomial term forces bounded $Nt^\beta$ whenever the normalized moment is bounded. The converse implication is \eqref{ff:global}.

If all higher $L_k$ vanish, $E(s)=s^hL_h/h!$. Choose $S_j\to\infty$ and then $t_j\downarrow0$ sufficiently small that, at $N_j=\lfloor S_jt_j^{-\beta}\rfloor$, the normalized matrix $G_{N_j}(t_j)/N_j^h$ differs from $L_h/h!$ by at most one in norm. This is possible by \eqref{ff:normalizedcritical} and its matrix version at each fixed $s=S_j$. These moments stay bounded while $N_jt_j^\beta\to\infty$. Finally, $M$ consecutive zero terms beyond $h$ propagate forever through the order-$M$ recurrence, proving the finite-block equivalence.
\end{proof}

The qualification is necessary. For an integer $r\ge1$, take the single generator
\[
 Q(t)=E_{12}\oplus[t^r].
\]
Then $h=1$, $b=\beta=2r$, but exactly
\[
 \norm{G_N(t)}=\max\{N+1,(1+t^{2r})^N\},\qquad E(s)=sE_{22}.
\]
At $N=\lfloor rt^{-2r}\log(1/t)\rfloor$, one has $Nt^\beta\to\infty$, while the feedback term is asymptotic to $t^{-r}$ and remains smaller than $N$. The normalized moment is bounded. Thus the algebraic frontier is exact even when a later logarithmic window is invisible to its leading critical normalization.

\subsection{Adaptive labels, resolvents, and exact computation}

\begin{theorem}[Uniformity under adapted positive mixing]\label{ff:adaptive}
Allow the next label probabilities to depend on all preceding labels and signs, subject to fixed bounds $0<p_-\le p_{i,r}\le p_+$. Conditional on the chosen label and past, the next sign is fair. Keep the matrices $Q_i(t)$ fixed. For finite $\beta$, \eqref{ff:global} holds uniformly over these policies; for $\beta=\infty$, the uniform bound is $\norm{G_N^{\rm policy}}\asymp(N+1)^h$. At $N=\lfloor t^{-\alpha}\rfloor$ the normalized moments are uniformly bounded exactly for $\alpha\le\beta$. For $\alpha>\beta$ they diverge uniformly over all admissible policies. Exact crossover coefficients remain statements about stationary reference probabilities.
\end{theorem}
\begin{proof}
Define fixed positive maps
\[
 T_\pm(X)=X+p_\pm\sum_iQ_i^*XQ_i.
\]
Backward induction on the remaining horizon gives
\[
 T_-^N(I)\preceq G_N^{\rm policy}\preceq T_+^N(I).
\]
Indeed, conjugate the fixed lower and upper continuation matrices by the next branch, average its conditionally fair sign, and apply the probability bounds; positivity preserves the inequalities. Each comparison map is a stationary uniform-label model after rescaling all $Q_i$ by $\sqrt{qp_\pm}$. Its baseline height and word orders are unchanged. Proposition~\ref{ff:words} gives the same $\beta$ for both maps. Apply Theorem~\ref{ff:frontier} separately to these stationary bounds. This avoids comparing infinitely many word weights by a single constant.
\end{proof}

There is also a finite-prefix comparison valid without analyticity. For any fixed positive map $A$ on $D\times D$ matrices, put $r=\rho(A)$ and
\[
 S_N=\sum_{k=0}^{M-1}\binom Nk\tr A^k(I).
\]
For each $L\ge0$ there is $K_{M,L}$, independent of $A$, such that
\begin{equation}\label{ff:resolvent}
 S_N\le\tr\bigl[(\operatorname{Id}+A)^N(I)\bigr]\le K_{M,L}S_N
 \qquad\text{whenever }Nr\le L.
\end{equation}
To see this, write $R(z)=\tr\bigl[(\operatorname{Id}-zA)^{-1}(I)\bigr]=\sum c_kz^k$. Its denominator is $P(z)=\sum a_jz^j$, with $a_0=1$, and its numerator has degree less than $M$, with coefficients $\sum_{i=0}^ja_ic_{j-i}$. Since $|a_i|\le\binom Mi r^i$, for $zr\le1/2$,
\[
 R(z)\le3^M\sum_{k<M}z^kc_k.
\]
Take $z=cN$ with $c=[2\max(1,L)]^{-1}$ and use $\binom Nk\le e^{1/c}(cN)^k$. For $N\ge M$, $N^k\le k!M^k\binom Nk$ for $k<M$; for $N<M$ the finite prefix is already exact. Positivity proves the lower bound. Keeping binomial coefficients in \eqref{ff:resolvent} is essential for constants independent of an arbitrary unbounded nonnormal family at small $N$.

For polynomial $Q_i$ and rational coefficients and probabilities, the superoperator matrix, its characteristic polynomial, and $C_0,\ldots,C_{M-1}$ can be computed by exact polynomial matrix arithmetic. Their leading coefficients determine $b$, $\beta$, every Newton face, and the finite ODE data. Recurrence computes the additional block in Theorem~\ref{ff:arbitrary}. No exponential enumeration of words is required. These are arithmetic statements for explicit polynomial input; an arbitrary analytic oracle need not permit finite detection of a zero germ.

The results concern a fixed finite dimension near a nilpotent centered baseline. The polynomial and exponential phase limits refer to stationary matrices and probabilities; the adapted theorem preserves the frontier and its uniform bounds. Positive graded finite quotients of genuine gapped lifts form an exactly marginal subclass. General feedback matrices in this section are a broader calibration model and are not asserted to be gapped-algebroid lifts.