\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.