\chapter{Optimal frames inside a spherical cap}
\section{The finite-dimensional problem}
The geometry has reduced the reachable rays to a small angular region around
$u$. Before returning to loops, we solve the resulting frame-design problem
exactly. This part uses ordinary Euclidean linear algebra. The appearance of
a regular simplex is consistent with classical tight-frame theory
\cite{benedetto-fickus}; the proof below supplies the cap constraint and all
constants needed here.

Fix $0<\alpha\le\arccos(1/\sqrt d)$ and unit vectors
$v_1,\ldots,v_d$ with
\begin{equation}\label{eq:cap}
 \ip u{v_i}\ge\cos\alpha.
\end{equation}
Let $V=[v_i]$, $S=VV^{\mathsf T}$, and $G_V=V^{\mathsf T}V$.
The two positive semidefinite matrices have the same eigenvalues and trace
$d$. Write
\begin{equation}\label{eq:capblock}
 S=\begin{pmatrix}a&q^{\mathsf T}\\q&B\end{pmatrix}
 \quad\text{relative to }\R u\oplus u^\perp.
\end{equation}
Then
\begin{equation}\label{eq:captraces}
 a=\sum_i\ip u{v_i}^2\ge d\cos^2\alpha,\qquad
 \tr B=d-a\le d\sin^2\alpha.
\end{equation}
Put
\begin{equation}\label{eq:capc}
 c_\alpha=\frac d{d-1}\sin^2\alpha.
\end{equation}

\section{The least-eigenvalue optimum}
\begin{theorem}[Cap E-optimality and rigidity]\label{thm:capE}
Under \eqref{eq:cap},
\begin{equation}\label{eq:capE}
 \lambda_{\min}(G_V)\le c_\alpha.
\end{equation}
The bound is attained by
$v_i=\cos\alpha\,u+\sin\alpha\,z_i$, where the $z_i$ form a regular
simplex in $u^\perp$. If $0<\alpha<\arccos(1/\sqrt d)$, every maximizing
frame has this form, up to an orthogonal transformation fixing $u$ and a
permutation of its columns.
\end{theorem}
\begin{proof}
The trace of $S$ on the $(d-1)$-dimensional subspace $u^\perp$ is at most
$d\sin^2\alpha$. Its average Rayleigh quotient is therefore at most
$c_\alpha$. The least Rayleigh quotient on the entire space cannot exceed
this average. This proves the inequality.

For the displayed simplex frame, the eigenvalues are
$d\cos^2\alpha$ and $c_\alpha$ with multiplicity $d-1$.
The cap threshold ensures $d\cos^2\alpha\ge c_\alpha$, proving
attainment.

For rigidity, equality implies
$S-c_\alpha\Id\succeq0$ and
$\tr(P_{u^\perp}(S-c_\alpha\Id))=0$.
A positive semidefinite matrix with zero trace on a subspace vanishes on that
subspace, including its cross block. Hence $B=c_\alpha\Id$ and $q=0$.
Equality in \eqref{eq:captraces} forces every vector to be on the cap rim:
$\ip u{v_i}=\cos\alpha$. Write
$v_i=\cos\alpha\,u+\sin\alpha\,z_i$. The equations $q=0$ and
$B=c_\alpha\Id$ imply $\sum_i z_i=0$ and
$\sum_i z_iz_i^{\mathsf T}=d\Id/(d-1)$ on $u^\perp$.
The Gram matrix of the $z_i$ has kernel containing $\mathbf1$, rank $d-1$,
and all nonzero eigenvalues $d/(d-1)$. It is consequently
$\frac d{d-1}(\Id-\mathbf1\mathbf1^{\mathsf T}/d)$, which is exactly the
regular-simplex Gram matrix. Gram equality gives the stated orthogonal
uniqueness.
\end{proof}

\section{Determinant and inverse-trace optimality}
The same frame optimizes two further criteria. These are useful when total
uncertainty volume or average linear reconstruction error is the objective,
rather than the worst direction.

\begin{theorem}[Cap D- and A-optimality]\label{thm:capDA}
Under the cap assumption,
\begin{equation}\label{eq:capD}
 \det G_V\le d\cos^2\alpha\left(
           \frac d{d-1}\sin^2\alpha\right)^{d-1}.
\end{equation}
For invertible $V$,
\begin{equation}\label{eq:capA}
 \tr(G_V^{-1})\ge
 \frac1{d\cos^2\alpha}+\frac{(d-1)^2}{d\sin^2\alpha}.
\end{equation}
The simplex-rim frame attains both bounds. In the strict cap regime their
maximizers or minimizers have the same rigidity as in Theorem~\ref{thm:capE}.
\end{theorem}
\begin{proof}
If $S$ is singular the determinant statement is immediate. Otherwise the
Schur complement gives
\[
 \det S=a\det(B-qq^{\mathsf T}/a)
 \le a\left(\frac{d-a}{d-1}\right)^{d-1}.
\]
The inequality uses the arithmetic--geometric mean of the positive eigenvalues
and discards the nonnegative quantity $|q|^2/a$ from the trace. The function
on the right is decreasing for $a\ge1$. Since
$a\ge d\cos^2\alpha\ge1$, its maximum occurs at the smallest permitted
$a$, which gives \eqref{eq:capD}.

For the inverse trace, the block inverse or Schur-complement inequalities give
\[
 \tr(S^{-1})\ge\frac1a+\tr(B^{-1})
 \ge\frac1a+\frac{(d-1)^2}{d-a}.
\]
The last function is increasing for $a\ge1$. Substituting
$a=d\cos^2\alpha$ proves \eqref{eq:capA}. Equality in either argument
requires $q=0$, scalar $B$, and all vectors on the cap rim; the rigidity
argument of the preceding theorem then applies.
\end{proof}

\section{The exterior optimum}
The scalar cap radius also controls a different inverse map. On skew matrices,
use the isometric identification with $\Lambda^2\R^d$ induced by the
orthonormal basis of elementary plane rotations.

\begin{lemma}[Singular values of skew congruence]\label{lem:wedgeSVD}
If $V$ has singular values $s_1\ge\cdots\ge s_d\ge0$, then the singular
values of $K\mapsto V^{\mathsf T}KV$ on $\so(d)$ are precisely
\begin{equation}\label{eq:wedgeSVD}
 \{s_is_j:1\le i<j\le d\}.
\end{equation}
\end{lemma}
\begin{proof}
Take an ordinary singular value decomposition
$V=U\diag(s_i)W^{\mathsf T}$. Orthogonal conjugation on either side is an
isometry of the skew-matrix space. It is enough to consider diagonal $V$.
An elementary skew matrix in the $i,j$ plane is then multiplied by $s_is_j$.
These elementary matrices form an orthonormal basis for $\norm\cdot_\wedge$.
\end{proof}

\begin{theorem}[Cap-optimal skew sensing]\label{thm:capexterior}
For $d\ge3$ and vectors satisfying \eqref{eq:cap},
\begin{equation}\label{eq:capexterior}
 \sigma_{\min}(\sensor_V)\le c_\alpha.
\end{equation}
The simplex-rim frame attains the bound.
\end{theorem}
\begin{proof}
Let $\mu_1\ge\cdots\ge\mu_d$ be the eigenvalues of $S$. Since
$\mu_1\ge a\ge d\cos^2\alpha$,
$\sum_{i=2}^d\mu_i\le d\sin^2\alpha$.
The geometric mean of the two smallest of these $d-1$ values is at most their
full average. Thus
\[
 \sigma_{\min}(\sensor_V)=\sqrt{\mu_{d-1}\mu_d}
 \le\frac1{d-1}\sum_{i=2}^d\mu_i\le c_\alpha.
\]
The simplex frame has $d-1$ equal small eigenvalues $c_\alpha$ and one
larger eigenvalue. Their pair products therefore attain the bound.
\end{proof}

The restriction $d\ge3$ is meaningful: two weak transverse directions are
needed for the least pair product to have order $\sin^2\alpha$. In dimension
two there is only one skew degree of freedom and the product involves one
large and one small singular value instead.

\section{A quantitative near-equality statement}
The equality configuration is not merely a possible design. Near optimality
already forces most of its covariance structure.

\begin{proposition}[Near-optimal cap covariance]\label{prop:nearcap}
Suppose \eqref{eq:cap} holds and
$\lambda_{\min}(S)\ge(1-\eps)c_\alpha$, with $0\le\eps<1$.
Then
\begin{align}
 \sum_i\bigl(\sin^2\alpha-|P_{u^\perp}v_i|^2\bigr)
       &\le d\eps\sin^2\alpha,\label{eq:near-rim}\\
 \norm{B-c_\alpha\Id}_\op
       &\le\max(1,d-2)\eps c_\alpha,\label{eq:near-tight}\\
 |q|^2&\le\bigl(a-(1-\eps)c_\alpha\bigr)
                    (d-1)\eps c_\alpha.\label{eq:near-balance}
\end{align}
\end{proposition}
\begin{proof}
The lower spectral bound gives
$\tr B\ge(d-1)(1-\eps)c_\alpha$. Compare this with
$\tr B\le(d-1)c_\alpha$ to obtain \eqref{eq:near-rim}. Every eigenvalue
of $B$ is at least $(1-\eps)c_\alpha$, and the trace bound forces each
to be at most $(1+(d-2)\eps)c_\alpha$, proving
\eqref{eq:near-tight}. Finally set $m=(1-\eps)c_\alpha$.
The positive semidefinite block matrix $S-m\Id$ has lower block
$B-m\Id$ with trace at most $(d-1)\eps c_\alpha$.
Its two-by-two scalar compressions imply
$|q|^2\le(a-m)\norm{B-m\Id}_\op$, and the norm is at most that trace.
\end{proof}

These bounds quantify concentration near the rim, near isotropy in the
transverse space, and near vanishing of the longitudinal--transverse cross
covariance. They do not by themselves give a dimension-free perturbation
bound for every individual labeled simplex vertex. Such a stronger conclusion
would require an additional alignment argument and a declared norm.