| \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. |
|
|
