Datasets:

PureOne
/

AEQORIA-Sharp-Holonomy-Observation

\chapter{An exact adaptive testing law}
\section{Two geometric hypotheses and one noise contract}
We now compare the product-limit connection, denoted by hypothesis $\theta=0$,
with the known connection parameter $0<\lambda<1$, denoted by $\theta=1$.
The soul and the marked unit vector are the same in both experiments. Each
query chooses a based loop of length at most $L$, resets the vector, and
returns a scalar value
\begin{equation}\label{eq:gaussianmodel}
 Y=\begin{cases}
 1+\sigma Z,&\theta=0,\\
 1-d_\lambda(\gamma)+\sigma Z,&\theta=1,
 \end{cases}
 \qquad Z\sim N(0,1),\quad\sigma>0.
\end{equation}
Every shot has independent fresh noise. The loop can be selected adaptively
from previous outcomes and hypothesis-independent private randomization.
The number of shots $n$ is fixed in advance, and every shot has the same
individual length budget. Only these scalar outcomes enter the decision.

Let $D=D_\lambda(L)$. For all permitted loops,
$0\le d_\lambda(\gamma)\le D$. By the compactness result, an absolutely
continuous maximizing loop exists. The question is whether adaptation can
use differently oriented loops to beat repetition of this maximizer.

\section{A single strongest experiment simulates every weaker one}
Consider one scalar observation $Z_*$ with variance $\sigma^2$ and means
$1$ under the null and $1-D$ under the alternative. Given a desired gap
$0\le d\le D$, put $a=d/D$ and generate
\begin{equation}\label{eq:garbling}
 Y=1+a(Z_*-1)+\sigma\sqrt{1-a^2}\,\xi,
 \qquad \xi\sim N(0,1)
\end{equation}
independently. Under either hypothesis, this has variance $\sigma^2$.
Its two means are $1$ and $1-d$. The transformation does not use the unknown
hypothesis. Hence a strongest observation can simulate any permitted weaker
observation exactly.

This is an explicit comparison kernel in the sense of Blackwell's
comparison of experiments \cite{Blackwell}. The general comparison framework
is established; the point here is that the geometric control set reduces
to the one scalar $D$, with no unproved claim about the usefulness of
adaptation.

\begin{theorem}[No adaptive advantage for this testing experiment]
\label{thm:gaussianexact}
Under \eqref{eq:gaussianmodel}, the minimum equal-prior error probability
over all fixed-$n$ adaptive strategies is
\begin{equation}\label{eq:exactrisk}
 \risk_n(\lambda,L,\sigma)
 =\PhiN\left(-\frac{\sqrt n D_\lambda(L)}{2\sigma}\right),
\end{equation}
where $\PhiN$ is the standard normal cumulative distribution function.
Repeating a maximally separating loop attains the minimum.
\end{theorem}
\begin{proof}
Take $n$ independent strongest observations. To simulate any adaptive
strategy, proceed in order. Its first loop choice depends only on its private
random seed. Use \eqref{eq:garbling} with the gap of that loop to generate
the first simulated response. The strategy then chooses its second loop
from this simulated history. Apply \eqref{eq:garbling} to the second fresh
strongest observation with the newly selected gap. Continue recursively.
Conditioned on each simulated history, the next observation has exactly
the law prescribed by the original adaptive experiment under either
hypothesis. Induction proves equality of the complete transcript laws.
Any decision based on that transcript can therefore be simulated from the
$n$ strongest observations. Its risk cannot be smaller than their optimal
risk.

For repeated strongest observations, the sample mean is Gaussian with
variance $\sigma^2/n$ and means $1$ and $1-D$. The likelihood-ratio rule
with equal priors thresholds at $1-D/2$. Its two error probabilities are
equal and both are $\PhiN(-\sqrt nD/(2\sigma))$. This proves the lower
bound and its attainment. If only smooth paths are allowed, smooth
approximations to a maximizing absolutely continuous loop give the same
infimal risk by continuity.
\end{proof}

\section{The sharp length--noise--sample constant}
Fix a target risk $0<\alpha<1/2$ and write
$z_{1-\alpha}=\PhiN^{-1}(1-\alpha)>0$. The exact minimum number of shots
is
\begin{equation}\label{eq:sampleexact}
 n_\alpha=\left\lceil
 \frac{4\sigma^2z_{1-\alpha}^2}{D_\lambda(L)^2}
 \right\rceil.
\end{equation}
Combining the sharp scalar law with this formula gives
\begin{theorem}[Sharp short-loop statistical cost]\label{thm:samplessharp}
For fixed $d\ge3$ and $0<\alpha<1/2$,
\begin{equation}\label{eq:samplesasymptotic}
 n_\alpha=
 \frac{4096\pi^4\sigma^2z_{1-\alpha}^2}{\coup^4L^8}
 \bigl(1+O_d(L)\bigr)+O(1),
\end{equation}
uniformly for $0<\lambda<1$ as $L\downarrow0$. The final $O(1)$ is
only the integer rounding error. Repetition of the explicit circle writer
attains the leading constant.
\end{theorem}
\begin{proof}
Square the uniform expansion
$D=\coup^2L^4/(32\pi^2)(1+O_d(L))$ and invert it for sufficiently small
$L$. The coefficient in \eqref{eq:sampleexact} becomes
$4(32\pi^2)^2=4096\pi^4$. The circle writer has a gap whose ratio to
$D$ tends uniformly to one, so replacing the maximizing loop by this
constructive loop changes the real-valued sample requirement only by a
relative $1+O_d(L)$ factor.
\end{proof}

The exponent eight is not an arbitrary difficulty rating. A short loop
produces displacement of order $\coup L^2$. A scalar self-echo deficit
squares it, producing order $\coup^2L^4$. Independent Gaussian mean-shift
information then squares that deficit again. These are three distinct
operations; eliminating one by changing the observation contract changes
the statistical exponent.

\section{A finite-budget impossibility statement}
For every fixed $n,L,\sigma>0$,
\begin{equation}\label{eq:indistinguishable}
 \lim_{\lambda\downarrow0}\risk_n(\lambda,L,\sigma)=\tfrac12.
\end{equation}
Indeed, the global bound gives $D\le\coup^2L^4/8\to0$. Nevertheless,
every alternative with positive $\lambda$ has full $\SO(d)$ normal
holonomy and the same soul metric. Thus no decision rule in this experiment
can uniformly distinguish full normal holonomy from trivial transport at a
fixed finite sample and loop budget.

For a simple explicit estimate, the elementary bound
$\Phi(-x)\ge1/2-x/\sqrt{2\pi}$ gives
\[
 \risk_n\ge\max\left\{0,\frac12-
 \frac{\sqrt n\,\coup^2L^4}{16\sqrt{2\pi}\,\sigma}\right\}.
\]
This estimate is not sharp at finite parameters; \eqref{eq:exactrisk} is
the exact statement in terms of the geometric scalar extremum.

\section{Where adaptation can matter again}
The domination theorem is deliberately narrow. It compares two known
hypotheses with a common fixed variance, a fixed number of samples, and an
individual cost bound on every query. It does not address an unknown
orientation parameter, an unknown connection strength, variable shot costs,
sequential stopping, a persistent unreset state, or a detector with
loop-dependent variance. In those models the simulation kernel may cease
to be available or may consume a different resource.

For example, when $\lambda$ is unknown and the loop maximizing the gap
depends on it, a policy may first estimate the parameter and then choose a
better loop. That is a composite inference problem, not a counterexample
to Theorem~\ref{thm:gaussianexact}. Likewise, a coordinate-valued detector
can measure a signed displacement of order $\coup L^2$, whereas the
scalar detector here sees its square. Such a detector can have a different
sample law because it receives more information per shot.

\section{A comparison with deterministic tomography}
The statistical cost in \eqref{eq:samplesasymptotic} concerns distinguishing
a specified connection from the product limit. The inverse radius in
\eqref{eq:optimalminimax} concerns reconstructing an arbitrary skew target
in a calibrated frame. Their error metrics, unknown objects, and resource
counts differ. Their common dependence on $D_\lambda(L)$ is a structural
bridge, not a claim that the two experiments are interchangeable.

In particular, replacing the random error by an adversarial error of
arbitrarily small but fixed magnitude can make some weak targets
indistinguishable indefinitely. Repetition helps independent random noise;
it need not help a systematic calibration bias. The manuscript therefore
keeps the deterministic, Gaussian, and finite-radius error models in
separate theorems.

	\chapter{An exact adaptive testing law}
	\section{Two geometric hypotheses and one noise contract}
	We now compare the product-limit connection, denoted by hypothesis $\theta=0$,
	with the known connection parameter $0<\lambda<1$, denoted by $\theta=1$.
	The soul and the marked unit vector are the same in both experiments. Each
	query chooses a based loop of length at most $L$, resets the vector, and
	returns a scalar value
	\begin{equation}\label{eq:gaussianmodel}
	Y=\begin{cases}
	1+\sigma Z,&\theta=0,\\
	1-d_\lambda(\gamma)+\sigma Z,&\theta=1,
	\end{cases}
	\qquad Z\sim N(0,1),\quad\sigma>0.
	\end{equation}
	Every shot has independent fresh noise. The loop can be selected adaptively
	from previous outcomes and hypothesis-independent private randomization.
	The number of shots $n$ is fixed in advance, and every shot has the same
	individual length budget. Only these scalar outcomes enter the decision.

	Let $D=D_\lambda(L)$. For all permitted loops,
	$0\le d_\lambda(\gamma)\le D$. By the compactness result, an absolutely
	continuous maximizing loop exists. The question is whether adaptation can
	use differently oriented loops to beat repetition of this maximizer.

	\section{A single strongest experiment simulates every weaker one}
	Consider one scalar observation $Z_*$ with variance $\sigma^2$ and means
	$1$ under the null and $1-D$ under the alternative. Given a desired gap
	$0\le d\le D$, put $a=d/D$ and generate
	\begin{equation}\label{eq:garbling}
	Y=1+a(Z_*-1)+\sigma\sqrt{1-a^2}\,\xi,
	\qquad \xi\sim N(0,1)
	\end{equation}
	independently. Under either hypothesis, this has variance $\sigma^2$.
	Its two means are $1$ and $1-d$. The transformation does not use the unknown
	hypothesis. Hence a strongest observation can simulate any permitted weaker
	observation exactly.

	This is an explicit comparison kernel in the sense of Blackwell's
	comparison of experiments \cite{Blackwell}. The general comparison framework
	is established; the point here is that the geometric control set reduces
	to the one scalar $D$, with no unproved claim about the usefulness of
	adaptation.

	\begin{theorem}[No adaptive advantage for this testing experiment]
	\label{thm:gaussianexact}
	Under \eqref{eq:gaussianmodel}, the minimum equal-prior error probability
	over all fixed-$n$ adaptive strategies is
	\begin{equation}\label{eq:exactrisk}
	\risk_n(\lambda,L,\sigma)
	=\PhiN\left(-\frac{\sqrt n D_\lambda(L)}{2\sigma}\right),
	\end{equation}
	where $\PhiN$ is the standard normal cumulative distribution function.
	Repeating a maximally separating loop attains the minimum.
	\end{theorem}
	\begin{proof}
	Take $n$ independent strongest observations. To simulate any adaptive
	strategy, proceed in order. Its first loop choice depends only on its private
	random seed. Use \eqref{eq:garbling} with the gap of that loop to generate
	the first simulated response. The strategy then chooses its second loop
	from this simulated history. Apply \eqref{eq:garbling} to the second fresh
	strongest observation with the newly selected gap. Continue recursively.
	Conditioned on each simulated history, the next observation has exactly
	the law prescribed by the original adaptive experiment under either
	hypothesis. Induction proves equality of the complete transcript laws.
	Any decision based on that transcript can therefore be simulated from the
	$n$ strongest observations. Its risk cannot be smaller than their optimal
	risk.

	For repeated strongest observations, the sample mean is Gaussian with
	variance $\sigma^2/n$ and means $1$ and $1-D$. The likelihood-ratio rule
	with equal priors thresholds at $1-D/2$. Its two error probabilities are
	equal and both are $\PhiN(-\sqrt nD/(2\sigma))$. This proves the lower
	bound and its attainment. If only smooth paths are allowed, smooth
	approximations to a maximizing absolutely continuous loop give the same
	infimal risk by continuity.
	\end{proof}

	\section{The sharp length--noise--sample constant}
	Fix a target risk $0<\alpha<1/2$ and write
	$z_{1-\alpha}=\PhiN^{-1}(1-\alpha)>0$. The exact minimum number of shots
	is
	\begin{equation}\label{eq:sampleexact}
	n_\alpha=\left\lceil
	\frac{4\sigma^2z_{1-\alpha}^2}{D_\lambda(L)^2}
	\right\rceil.
	\end{equation}
	Combining the sharp scalar law with this formula gives
	\begin{theorem}[Sharp short-loop statistical cost]\label{thm:samplessharp}
	For fixed $d\ge3$ and $0<\alpha<1/2$,
	\begin{equation}\label{eq:samplesasymptotic}
	n_\alpha=
	\frac{4096\pi^4\sigma^2z_{1-\alpha}^2}{\coup^4L^8}
	\bigl(1+O_d(L)\bigr)+O(1),
	\end{equation}
	uniformly for $0<\lambda<1$ as $L\downarrow0$. The final $O(1)$ is
	only the integer rounding error. Repetition of the explicit circle writer
	attains the leading constant.
	\end{theorem}
	\begin{proof}
	Square the uniform expansion
	$D=\coup^2L^4/(32\pi^2)(1+O_d(L))$ and invert it for sufficiently small
	$L$. The coefficient in \eqref{eq:sampleexact} becomes
	$4(32\pi^2)^2=4096\pi^4$. The circle writer has a gap whose ratio to
	$D$ tends uniformly to one, so replacing the maximizing loop by this
	constructive loop changes the real-valued sample requirement only by a
	relative $1+O_d(L)$ factor.
	\end{proof}

	The exponent eight is not an arbitrary difficulty rating. A short loop
	produces displacement of order $\coup L^2$. A scalar self-echo deficit
	squares it, producing order $\coup^2L^4$. Independent Gaussian mean-shift
	information then squares that deficit again. These are three distinct
	operations; eliminating one by changing the observation contract changes
	the statistical exponent.

	\section{A finite-budget impossibility statement}
	For every fixed $n,L,\sigma>0$,
	\begin{equation}\label{eq:indistinguishable}
	\lim_{\lambda\downarrow0}\risk_n(\lambda,L,\sigma)=\tfrac12.
	\end{equation}
	Indeed, the global bound gives $D\le\coup^2L^4/8\to0$. Nevertheless,
	every alternative with positive $\lambda$ has full $\SO(d)$ normal
	holonomy and the same soul metric. Thus no decision rule in this experiment
	can uniformly distinguish full normal holonomy from trivial transport at a
	fixed finite sample and loop budget.

	For a simple explicit estimate, the elementary bound
	$\Phi(-x)\ge1/2-x/\sqrt{2\pi}$ gives
	\[
	\risk_n\ge\max\left\{0,\frac12-
	\frac{\sqrt n\,\coup^2L^4}{16\sqrt{2\pi}\,\sigma}\right\}.
	\]
	This estimate is not sharp at finite parameters; \eqref{eq:exactrisk} is
	the exact statement in terms of the geometric scalar extremum.

	\section{Where adaptation can matter again}
	The domination theorem is deliberately narrow. It compares two known
	hypotheses with a common fixed variance, a fixed number of samples, and an
	individual cost bound on every query. It does not address an unknown
	orientation parameter, an unknown connection strength, variable shot costs,
	sequential stopping, a persistent unreset state, or a detector with
	loop-dependent variance. In those models the simulation kernel may cease
	to be available or may consume a different resource.

	For example, when $\lambda$ is unknown and the loop maximizing the gap
	depends on it, a policy may first estimate the parameter and then choose a
	better loop. That is a composite inference problem, not a counterexample
	to Theorem~\ref{thm:gaussianexact}. Likewise, a coordinate-valued detector
	can measure a signed displacement of order $\coup L^2$, whereas the
	scalar detector here sees its square. Such a detector can have a different
	sample law because it receives more information per shot.

	\section{A comparison with deterministic tomography}
	The statistical cost in \eqref{eq:samplesasymptotic} concerns distinguishing
	a specified connection from the product limit. The inverse radius in
	\eqref{eq:optimalminimax} concerns reconstructing an arbitrary skew target
	in a calibrated frame. Their error metrics, unknown objects, and resource
	counts differ. Their common dependence on $D_\lambda(L)$ is a structural
	bridge, not a claim that the two experiments are interchangeable.

	In particular, replacing the random error by an adversarial error of
	arbitrarily small but fixed magnitude can make some weak targets
	indistinguishable indefinitely. Repetition helps independent random noise;
	it need not help a systematic calibration bias. The manuscript therefore
	keeps the deterministic, Gaussian, and finite-radius error models in
	separate theorems.