\section{A sharp resolution hierarchy for holonomy-forced completion}
\label{sec:resolution-hierarchy}

The exact auxiliary rank can jump even when a flat connection changes by
an arbitrarily small amount. We now compute the full finite-probe resolution
law of such a jump. This section uses a specified observation experiment;
its sample counts are not claims about an unrelated calibration experiment.

Fix an integer $d\ge2$ and use the standard basis $e_1,\ldots,e_d$ of
$\mathbb R^d$. For $1\le r\le d$, define
\[
 N_r=\sum_{i=d-r+1}^{d-1}E_{i,i+1},\qquad v=e_d,
 \qquad T_{\epsilon,r}=\exp(-\epsilon N_r).
\]
An empty sum is zero. On the same five-dimensional base
$M=S^1_t\times\mathbb R^4_x$, with $t$ of period one, consider
\[
 \nabla_r=d+\epsilon N_r\,dt,\qquad
 B=-x^1v\,dt\wedge dx^2\wedge dx^3\wedge dx^4.
\]
The connection is flat because its matrix is constant and $dt\wedge dt=0$.
For every $r$ it has exactly the same globally defined curvature datum
\[
 F=d_{\nabla_r}B
   =v\,dt\wedge dx^1\wedge dx^2\wedge dx^3\wedge dx^4.
 \label{eq:hierarchy-common-F}
\]
Indeed the connection term on $B$ vanishes, while the sign from commuting
$dx^1$ past $dt$ cancels the minus sign in $B$. Thus $F$ has pointwise
image of dimension one, independently of $r$ and $\epsilon$.

For $\epsilon\ne0$, the smallest parallel subbundle containing $F$ has
fibre $\operatorname{span}\{e_{d-r+1},\ldots,e_d\}$ and rank $r$.
To see this, parallel transport once around the circle is $T_{\epsilon,r}$.
An invariant subspace for this invertible operator is invariant for its
inverse and for the finite polynomial
\[
 \log T_{\epsilon,r}=-\epsilon N_r.
\]
It must therefore contain every $N_r^mv$, $0\le m<r$, once it contains
$v$. These $r$ vectors are independent, and their span is parallel.
The minimum parallel auxiliary rank in the completion theorem is consequently
$r$. At $\epsilon=0$ it is one for all $r$.

\subsection{Exact loop matrices and every singular scale}
Define a finite probe matrix using the $d$ loop counts $j=0,\ldots,d-1$:
\[
 K^{(r)}_\epsilon
   =\bigl[T_{\epsilon,r}^{\,j}v\bigr]_{j=0}^{d-1}
   =\left[\sum_{m=0}^{r-1}\frac{(-j\epsilon)^m}{m!}e_{d-m}
      \right]_{j=0}^{d-1}.
\]
Write $K_\epsilon=K^{(d)}_\epsilon$.

\begin{theorem}[Exact determinant and singular-value hierarchy]
\label{thm:loop-singular-scales}
For every real $\epsilon$,
\[
 \det K_\epsilon=\epsilon^{d(d-1)/2}.
\]
There are constants $0<c_d\le C_d<\infty$, depending only on $d$, such
that for $0<|\epsilon|\le1$ and singular values in decreasing order,
\[
 c_d|\epsilon|^{j-1}
 \le \sigma_j(K_\epsilon)
 \le C_d|\epsilon|^{j-1},\qquad 1\le j\le d.
\]
Hence, for $0\le r<d$,
\[
 \inf_{\operatorname{rank}L\le r}\|K_\epsilon-L\|_{\rm op}
   =\sigma_{r+1}(K_\epsilon)\asymp_d|\epsilon|^r,
\]
and
\[
 \inf_{\operatorname{rank}L\le r}\|K_\epsilon-L\|_F
   =\left(\sum_{j=r+1}^d\sigma_j(K_\epsilon)^2\right)^{1/2}
   \asymp_d|\epsilon|^r.
\]
\end{theorem}

\begin{proof}
Let $R$ reverse the coordinate rows and put
\[
 D_\epsilon=\operatorname{diag}(1,\epsilon,\ldots,\epsilon^{d-1}),
 \qquad V_{m,j}=\frac{(-j)^m}{m!},\quad 0\le m,j<d.
\]
Direct evaluation gives $K_\epsilon=RD_\epsilon V$. The Vandermonde
formula and $\prod_{0\le i<j<d}(j-i)=\prod_{m=0}^{d-1}m!$ imply
\[
 \det V=(-1)^{d(d-1)/2}=\det R.
\]
The two signs cancel, proving the determinant identity. Multiplication by
an invertible fixed matrix satisfies
\[
 \sigma_j(D_\epsilon)\,\sigma_{\min}(V)
 \le\sigma_j(D_\epsilon V)
 \le\sigma_j(D_\epsilon)\,\sigma_{\max}(V).
\]
For $|\epsilon|\le1$, the ordered singular values of $D_\epsilon$ are
$1,|\epsilon|,\ldots,|\epsilon|^{d-1}$. This proves the displayed bounds
with $c_d=\sigma_{\min}(V)$ and $C_d=\sigma_{\max}(V)$.
Explicit, possibly weaker constants are
\[
 c_d=\|V^{-1}\|_F^{-1},\qquad C_d=\|V\|_F.
\]
The optimal rank-approximation formulas follow by truncating an SVD;
the matching lower bounds follow by the min--max characterization of
singular values. Finally the largest term in the discarded tail is of
order $|\epsilon|^{2r}$ and at most $d-r$ such terms occur.
\end{proof}

This is a fixed-$d$ theorem. Its constants are not asserted to be uniform
as the required auxiliary dimension grows.

\begin{proposition}[Sharp deterministic certification radius]
For a real matrix $Y$, the closed operator-norm ball of radius $\eta$
around $Y$ excludes every matrix of rank at most $r$ if and only if
$\eta<\sigma_{r+1}(Y)$. In particular the certification radius at the
exact centre $K_\epsilon$ for full rank is
$\sigma_d(K_\epsilon)\asymp_d|\epsilon|^{d-1}$.

If the centre is itself corrupted by unknown operator-norm error at most
$\eta$, uniform distinction between $K_\epsilon$ and all rank-at-most-$r$
matrices is possible exactly when
\[
 2\eta<\sigma_{r+1}(K_\epsilon).
\]
\end{proposition}

\begin{proof}
The first claim is the operator-norm distance formula in the preceding
theorem, including attainment. For the second, the two radius-$\eta$
observation sets are disjoint when their centres have distance greater
than $2\eta$. At distance at most $2\eta$, the midpoint between
$K_\epsilon$ and an attaining rank-$r$ approximant is compatible with
both hypotheses. Conversely, if the distance is greater than $2\eta$,
the two closed observation sets are disjoint and a nearest-set rule
distinguishes them.
\end{proof}

\subsection{A statistical experiment with an exact resource frontier}
Suppose $n$ independent real $d\times d$ response matrices are observed, each
equal to an unknown mean matrix $M$ plus independent $N(0,\sigma^2)$
noise in every entry. Here $d$ is fixed, and observing a whole matrix is
one observation by definition. Its columns are the responses to the single
specified seed $v$ at loop counts $0,\ldots,d-1$. These are not observations
of the complete parallel-transport operators on all possible input vectors.
This is a fixed loop-response experiment, not a binary-window experiment
and not arbitrary linear sensing.

\begin{theorem}[Detecting every additional required direction]
\label{thm:geometric-gaussian-hierarchy}
For $1\le r<d$, the consecutive genuine geometric models above satisfy
\[
 \|K^{(r+1)}_\epsilon-K^{(r)}_\epsilon\|_F^2
 =\frac{|\epsilon|^{2r}}{(r!)^2}\sum_{j=0}^{d-1}j^{2r}.
\]
Their $n$-observation Kullback--Leibler divergence is exactly this quantity
times $n/(2\sigma^2)$. The optimal sum of their two testing errors is
\[
 2\Phi\!\left(-\frac{\sqrt n\,|\epsilon|^r}{2\sigma r!}
                   \sqrt{\sum_{j=0}^{d-1}j^{2r}}\right),
\]
where $\Phi$ is the standard normal distribution function. Consequently,
for a sequence $0<|\epsilon_n|\le1$, consistently distinguishing the
models of auxiliary ranks $r$ and $r+1$ is possible if and only if
$n|\epsilon_n|^{2r}/\sigma^2\to\infty$.
\end{theorem}

\begin{proof}
The two probe matrices differ in the single row $e_{d-r}$, whose entries
are $(-j\epsilon)^r/r!$. This proves the squared Frobenius norm formula.
Vectorizing the observations gives two Gaussian location laws with common
covariance $\sigma^2 I$. Their KL divergence is squared mean distance
divided by $2\sigma^2$, and independent repetitions multiply it by $n$.
Projection onto the mean-difference direction gives the optimal likelihood
ratio test, with threshold at the midpoint. Both errors are the same normal
tail, giving the exact formula and its consistency criterion.
\end{proof}

For clarity, the final step $d-1$ versus $d$ compares two flat connections
on the same base with the same exact rank-one five-form $F$. The lower-rank
challenger is therefore geometrically realizable; the resource obstruction
does not arise merely by admitting arbitrary matrices.

\begin{theorem}[Certification against every rank-deficient challenger]
\label{thm:gaussian-full-rank-frontier}
In the same experiment let $\mathcal R_{n,\epsilon}$ be the infimum,
over all tests, of the sum of the error at $M=K_\epsilon$ and the worst
error over $\operatorname{rank}M\le d-1$. Put $s=\sigma_d(K_\epsilon)$.
Then
\[
 2\Phi\!\left(-\frac{\sqrt n\,s}{2\sigma}\right)
 \le\mathcal R_{n,\epsilon}
 \le\min\!\left\{1,\frac{8d^2\sigma^2}{ns^2}\right\}.
\]
In particular, for fixed $d$ and $0<|\epsilon_n|\le1$,
\[
 \mathcal R_{n,\epsilon_n}\longrightarrow0
 \quad\Longleftrightarrow\quad
 \frac{n|\epsilon_n|^{2(d-1)}}{\sigma^2}\longrightarrow\infty.
\]
\end{theorem}

\begin{proof}
An SVD rank-$d-1$ truncation is at Frobenius distance $s$ from
$K_\epsilon$. The optimal two-Gaussian sum-error against this one
challenger gives the lower bound.
For the upper bound let $\bar Y$ be the sample mean and declare full rank
when $\sigma_d(\bar Y)>s/2$. Singular values are 1-Lipschitz in operator
norm. Under either hypothesis a testing error therefore implies
$\|\bar Y-M\|_F\ge s/2$. Its squared expectation is
$d^2\sigma^2/n$, so Markov's inequality bounds each error by
$4d^2\sigma^2/(ns^2)$. A constant test always has sum-error one.
The singular-scale theorem proves sufficiency of the stated criterion.
If that criterion fails to diverge, a subsequence has bounded
$ns^2/\sigma^2$, and the normal-tail lower bound remains positive,
proving necessity.
\end{proof}

Testing only $\epsilon=0$ against a fully specified nonzero Jordan model
is a different experiment: its first new row is already of order
$\epsilon$. The cost $|\epsilon|^{-2(d-1)}$ concerns the last required
direction and the possibility of a rank-$d-1$ challenger. It must not be
assigned to every possible question about the connection.

\subsection{The explicit loop-length tradeoff}
The response restriction is material. If probe $j$ instead traverses $jL$
loops for an integer $L\ge1$, its matrix is exactly
\[
 K^{(r)}_{\epsilon,L}=K^{(r)}_{L\epsilon}.
\]
All consecutive-model squared distances therefore replace
$|\epsilon|^{2r}$ by $|L\epsilon|^{2r}$. For
$0<|L\epsilon|\le1$, the full-rank singular resolution is
$\asymp_d|L\epsilon|^{d-1}$, and the stated Gaussian consistency
threshold becomes $n|L\epsilon|^{2(d-1)}/\sigma^2\to\infty$.
Thus loops of length on the order of $1/|\epsilon|$ can remove the
small-amplitude penalty in this idealized fixed-noise response model.
Their traversal cost, any noise accumulated along a loop, and experiments
with additional seed vectors require separate analysis. No resource lower
bound uniform over free unbounded loops or free additional inputs is claimed.