\chapter{One scalar controls the optimal observation resources}
\section{Attainment and symmetry of the loop problem}
Before using the cap results, we justify two features of the loop extremum:
attainment in the absolutely continuous class and invariance under rotations
fixing the marked ray.

\begin{lemma}[Compactness of bounded-length transport]\label{lem:compact}
For fixed $L$, $\lambda$, and $d$, the set
$\{H_\lambda(\gamma):\ell(\gamma)\le L\}$ is compact. In particular,
$D_\lambda(L)$ is attained by an absolutely continuous based loop.
\end{lemma}
\begin{proof}
Reparametrize each loop to constant speed on $[0,1]$, allowing stationary
pieces, so that $\norm{X(t)}_Q\le L$ almost everywhere. The base equation
is $\dot\gamma=\gamma X$ and the transport equation is
$\dot U=-\lambda XU$. Both state curves are uniformly Lipschitz in a compact
matrix group. A sequence therefore has uniformly convergent subsequences for
both states. The controls have a weakly convergent subsequence in $L^2$.

In the integral equations, products of a uniformly convergent state with a
weakly convergent control pass to the limit: separate the product into the
uniform-state error times a bounded control, and the weak-control term against
a fixed continuous state. Thus the limit solves both state equations. The
base endpoints remain the identity. Lower semicontinuity of the length
integral, or weak closedness of its convex sublevel set, preserves the budget.
The limiting endpoint is therefore an allowed holonomy. Compactness and
continuity of the scalar deficit prove attainment.
\end{proof}

Smooth or piecewise smooth controls approximate the same endpoints without
changing the suprema. If a small budget excess appears in an approximation,
first approximate a slightly shorter near-optimal loop and then take a limit.
For the exact attainment assertions we retain the absolutely continuous
class, avoiding an unnecessary smooth-maximizer claim.

For any $P\in\SO(d)$, the conjugated loop
$\gamma^P(t)=P\gamma(t)P^{-1}$ has the same length and satisfies
\begin{equation}\label{eq:conjugationsym}
 H_\lambda(\gamma^P)=PH_\lambda(\gamma)P^{-1}.
\end{equation}
This follows directly from the transport ODE. If $Pu=u$, its output ray is
$PH_\lambda(\gamma)u$. The stabilizer of $u$ acts transitively on each
latitude sphere around $u$. Therefore a single extremal output ray supplies
all the vertices needed for a simplex on its latitude.

\section{The exact scalar-to-matrix bridge}
Set
\begin{equation}\label{eq:alphamax}
 \alpha_\lambda(L)=\arccos(1-D_\lambda(L)).
\end{equation}
Every allowed output ray belongs to the spherical cap of radius
$\alpha_\lambda(L)$. We do not need a claim that every point in the cap is
reachable. The extremal latitude and its symmetry orbit are sufficient.

\begin{theorem}[Exact resource bridge]\label{thm:bridge}
Suppose
\begin{equation}\label{eq:bridgeregime}
 0<D_\lambda(L)\le1-\frac1{\sqrt d}.
\end{equation}
Then
\begin{equation}\label{eq:bridge}
 \Gamma_\lambda(L)=\Omega_\lambda(L)
 =\frac d{d-1}\left(2D_\lambda(L)-D_\lambda(L)^2\right).
\end{equation}
In addition,
\begin{align}
 \sup\det(V^{\mathsf T}V)
 &=d(1-D_\lambda(L))^2
 \left[\frac d{d-1}(2D_\lambda(L)-D_\lambda(L)^2)\right]^{d-1},
 \label{eq:detbridge}\\
 \inf\tr\bigl((V^{\mathsf T}V)^{-1}\bigr)
 &=\frac1{d(1-D_\lambda(L))^2}
   +\frac{(d-1)^2}{d(2D_\lambda(L)-D_\lambda(L)^2)}.
 \label{eq:inversebridge}
\end{align}
All extrema are over $d$ preparation loops of individual length at most $L$;
the inverse-trace infimum is over invertible frames.
\end{theorem}
\begin{proof}
Let $\alpha=\alpha_\lambda(L)$. Condition \eqref{eq:bridgeregime} is exactly
the cap regime of Chapter~5. Those cap theorems provide all the upper or lower
bounds.

Choose a maximizing loop from Lemma~\ref{lem:compact}. Its output can be
written $\cos\alpha\,u+\sin\alpha\,z$ for a unit $z\perp u$.
For every vertex $z_i$ of a regular simplex in $u^\perp$, choose a rotation
$P_i$ fixing $u$ and taking $z$ to $z_i$. The conjugated loops from
\eqref{eq:conjugationsym} have the same budget and produce the simplex-rim
frame. They attain all the cap optima simultaneously. Finally use
$\cos\alpha=1-D_\lambda(L)$ and
$\sin^2\alpha=2D_\lambda(L)-D_\lambda(L)^2$.
\end{proof}

This theorem is exact even though the finite-$L$ value of $D_\lambda(L)$ has
not been reduced to a closed expression. It is a reduction of several
optimization problems to one scalar extremal. Knowing that scalar, or
bounding it above and below, gives the corresponding optimal reconstruction
and uncertainty quantities without solving separate matrix-design problems.

\section{Sharp local constants}
Since $\coup\le1/4$, the global bound \eqref{eq:globaldeficit} ensures the
cap regime for all sufficiently small $L$, uniformly in the parameter.
Combining Theorems~\ref{thm:scalarsharp} and~\ref{thm:bridge} yields
\begin{corollary}[Sharp Gram and exterior constants]\label{cor:sharpframe}
Uniformly for $0<\lambda<1$,
\begin{equation}\label{eq:sharpframe}
 \Gamma_\lambda(L)=\Omega_\lambda(L)
 =\frac d{16\pi^2(d-1)}\coup^2L^4\bigl(1+O_d(L)\bigr).
\end{equation}
The optimal determinant has asymptotic
\begin{equation}\label{eq:sharpdet}
 \sup\det(V^{\mathsf T}V)
 =d\left(\frac d{d-1}\right)^{d-1}
   \left(\frac{\coup L^2}{4\pi}\right)^{2(d-1)}
   \bigl(1+O_d(L)\bigr).
\end{equation}
\end{corollary}
\begin{proof}
The scalar deficit is order $\coup^2L^4$, so its square is lower order
uniformly in the parameter. Substitute its sharp expansion into the exact
bridge formulas. The dimension is fixed in the $O_d$ notation.
\end{proof}

The determinant exponent $4(d-1)$ is not an information capacity in bits. It
is the exponent of a squared Euclidean volume in the prepared frame. A coding
problem would require a specified parameter class, decoding norm, and code
contract. Here the volume is used only as a design and conditioning invariant.

\section{A fixed-soul obstruction to uniform reconstruction}
\begin{theorem}[No positive uniform frame floor]\label{thm:no-uniform}
Fix $d\ge3$ and $L>0$. In the family of Chapter~2,
\begin{equation}\label{eq:nouniform}
 \inf_{0<\lambda<1}\Gamma_\lambda(L)=0.
\end{equation}
Thus a positive lower bound for optimal one-ray conditioning cannot depend
only on the isometry type of the soul, its dimension, the normal dimension,
the full normal holonomy group, and the loop budget.
\end{theorem}
\begin{proof}
For any prepared ray and any unit $z\perp u$,
\[
 |\ip z{H_\lambda(\gamma)u}|\le
 \norm{H_\lambda(\gamma)-\Id}_\op\le\frac{\coup L^2}{2}.
\]
More directly, the total transverse norm has this same upper bound. The
transverse trace argument gives, for every frame,
\[
 \lambda_{\min}(V^{\mathsf T}V)
 \le\frac d{4(d-1)}\coup^2L^4.
\]
The right side tends to zero with $\lambda$, independently of the choice of
loops. Every positive parameter has the same fixed soul and full holonomy by
Chapter~2. Taking the supremum over loop choices does not change the bound.
\end{proof}

The conclusion persists if one additionally asks for uniform local curvature
bounds and noncollapse on a prescribed compact tube around the soul: the
metrics converge smoothly there to the product metric. It does not refute
bounds involving a quantitative lower nondegeneracy condition on normal
curvature. Such a condition is precisely what the parameter $\coup$ measures
in this example.

\section{The operational meaning of the bridge}
The same scalar extremum has three distinct interpretations.
The quantity $\sqrt{2D_\lambda(L)}$ is the largest one-ray displacement.
The polynomial $\frac d{d-1}(2D_\lambda(L)-D_\lambda(L)^2)$ gives the
optimal least Gram eigenvalue and least skew-sensor singular value in the cap
regime. Its square $D_\lambda(L)^2$ controls the Gaussian mean-shift
information per scalar observation. These powers arise from orthogonality,
frame geometry, and statistical likelihood, respectively. They describe
separate resources rather than interchangeable meanings of one quantity.