% 226_Addendum_WheelCommutator.tex
% Skill: gap-closure v1.1  |  Outcome: Case A  |  Date: 2026-05-22
%
% Computes [D^2_{B^4}, Delta_{S^3}] in the Gegenbauer basis exactly.
% Closes the "well-posed calculation" left open in the HUP Saturation
% research file (Option B, open item 3).

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\title{Addendum 226: Wheel Commutator $[D^2_{B^4},\,\Delta_{S^3}]$ in the Gegenbauer Basis}
\author{TOE Corpus}
\date{2026-05-22}

\begin{document}
\maketitle

\textbf{Skill:} gap-closure v1.1 \quad|\quad \textbf{Outcome:} Case A (closed) \quad|\quad \textbf{Date:} 2026-05-22

\section{Setup and Prior Work}

\textbf{Operators (TOE Paper 18, Wheel definition):}
\begin{itemize}
  \item $D^2_{B^4} \equiv \text{Fork}$ — biharmonic operator on $B^4$
  \item $\Delta_{S^3} \equiv \text{Weld}$ — Laplace-Beltrami on $S^3$
\end{itemize}

\textbf{Open item being closed:} HUP Saturation research file (2026-05-22),
Open Item~3:
\begin{quote}
\emph{``The correct statement of HUP for the Wheel\ldots is $[D^2_{B^4},\Delta_{S^3}]$,
not $\Delta(\alpha^{-1})\cdot\Delta(\text{conjugate})$.  Writing out this
commutator in terms of the layer fractions is a well-posed calculation and
should be done.''}
\end{quote}

\textbf{Basis:} Gegenbauer-Chebyshev $\{U_0,U_1,U_2,U_3\}$ on $S^3$, where
$U_n(\nu)$ satisfies
\[
  \Delta_{S^3}\,U_n = -n(n+2)\,U_n, \qquad \mu_n := -n(n+2).
\]
The Hopf coordinate $x = \cos^2\!\beta \in [0,1]$ and $\nu = 2x-1$ relate
the $S^3$ zonal function calculus to the TOE density polynomial.
In monomials: $U_0=1$, $U_1=4x-2$, $U_2=16x^2-16x+3$,
$U_3=64x^3-96x^2+40x-4$.

\textbf{Density polynomial (P02, eq.~1.4):}
$\rho(x) = 16\pi^3 x^3 + 3\pi^2 x^2 + 2\pi x$,\ \ $\Omega = \int_0^1\!\rho\,dx
= 4\pi^3+\pi^2+\pi$.

\textbf{Key prior result (A225, Thm.~1):} the Gegenbauer coefficients of $\rho$ are
\[
  a_3 = \frac{\pi^3}{4} =: \Omega_0 \quad\text{(exactly)}, \qquad a_2 \approx 48.36,
  \qquad a_1 \approx 117.49, \qquad a_0 \approx 120.92.
\]

\section{Biharmonic $D^2_{B^4}$ in the Monomial Basis}

In $B^4$ with radial coordinate $r$ and $x = r^2$, the Laplacian on radial
functions is $\Delta_{B^4} f = 4x f''(x) + 8f'(x)$.  The biharmonic is

\[
  D^2_{B^4}\,f(x)
  = \Delta_{B^4}^2\,f
  = 16x^2 f''''(x) + 96x f'''(x) + 96 f''(x).
\]

Acting on monomials $x^n$:

\begin{align*}
  D^2_{B^4}(1) &= 0, \\
  D^2_{B^4}(x) &= 0, \\
  D^2_{B^4}(x^2) &= 192, \\
  D^2_{B^4}(x^3) &= 1152\,x.
\end{align*}

The matrix in $\{1,x,x^2,x^3\}$ (row = output, column = input) is therefore
\[
  [D^2_{B^4}]_{\text{mono}} =
  \begin{pmatrix}
    0 & 0 & 192 & 0 \\
    0 & 0 & 0 & 1152 \\
    0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0
  \end{pmatrix}.
\]

\section{Weld $\Delta_{S^3}$ in the Monomial Basis}

For $x \in [0,1]$ with $\nu = 2x-1$, the zonal Laplacian is
$\Delta_{S^3} f(x) = x(1-x)f''(x) - \tfrac{3}{2}(2x-1)f'(x)$.
Acting on $x^n$:
\[
  \Delta_{S^3}(x^n)
  = n\!\left(n+\tfrac{1}{2}\right)x^{n-1} - n(n+2)\,x^n.
\]
Matrix:
\[
  [\Delta_{S^3}]_{\text{mono}} =
  \begin{pmatrix}
    0 & \frac{3}{2} & 0 & 0 \\
    0 & -3 & 5 & 0 \\
    0 & 0 & -8 & \frac{21}{2} \\
    0 & 0 & 0 & -15
  \end{pmatrix}.
\]
\textit{Verification:} $\Delta_{S^3}(4x-2) = 4(\tfrac{3}{2}-3x) = 6-12x = -3\cdot U_1$ ($\checkmark$, eigenvalue~$-3$); $\Delta_{S^3}(16x^2-16x+3) = -8\cdot U_2$ ($\checkmark$, eigenvalue~$-8$).

\section{Main Result: Commutator in the Gegenbauer Basis}

\begin{theorem}[Wheel Commutator — Spectral Gap Formula]\label{thm:commutator}
In the basis $\{U_0,U_1,U_2,U_3\}$, in which $\Delta_{S^3} =
\operatorname{diag}(0,-3,-8,-15)$:
\begin{align*}
  D^2_{B^4}(U_2) &= 3072\,U_0,\\
  D^2_{B^4}(U_3) &= 18432\,(U_0 + U_1),\\
  D^2_{B^4}(U_0) &= D^2_{B^4}(U_1) = 0.
\end{align*}
Consequently the commutator satisfies
\begin{equation}\label{eq:spectral-gap}
  [D^2_{B^4},\Delta_{S^3}]_{ij}
  = [D^2_{B^4}]_{ij}\,(\mu_j - \mu_i),
  \qquad \mu_n = -n(n+2),
\end{equation}
giving the non-zero entries:
\[
  [D^2_{B^4},\Delta_{S^3}] =
  \begin{pmatrix}
    0 & 0 & -24576 & -276480 \\
    0 & 0 & 0 & -221184 \\
    0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0
  \end{pmatrix}
\]
where the columns $U_2$ and $U_3$ carry the only non-trivial action.
\end{theorem}

\begin{proof}
$D^2_{B^4}(U_3)$: From the monomial formula,
$D^2_{B^4}(64x^3 - 96x^2 + 40x - 4) = 64\cdot 1152\,x - 96\cdot 192
= 73728\,x - 18432$.  Using $x = (U_1+2)/4$:
$73728\,x - 18432 = 18432(U_1+2) - 18432 = 18432(U_0+U_1)$.
All other entries follow by direct computation or the diagonal identity $[A,B]_{ij} = A_{ij}(\lambda_j - \lambda_i)$ for $B$ diagonal.
\end{proof}

\begin{remark}[Equal-weight mixing]\label{rem:mixing}
The identity $D^2_{B^4}(U_3) = 18432(U_0 + U_1)$ means the biharmonic maps
the $n=3$ (bulk-sourced) mode to \emph{equal amplitudes} on the scalar ($n=0$)
and dipole ($n=1$) modes.  The integer $18432 = 2^{11}\!\cdot 3^2$ factors as
$64\cdot 1152/4$, where $64$ is the leading coefficient of $U_3$ and $1152$
is the biharmonic eigenvalue of $x^3$.
\end{remark}

\section{Layer Descent Theorem}

\begin{theorem}[Layer Descent]\label{thm:descent}
The commutator $[D^2_{B^4},\Delta_{S^3}]$ is strictly lower-triangular in the
Gegenbauer basis: it maps modes $U_n$ to lower modes $U_m$ with $m < n$ only.
In particular, its action on $\rho$ satisfies
\[
  [D^2_{B^4},\Delta_{S^3}]\,\rho
  \;\in\;\operatorname{span}\{U_0,\,U_1\} \;=\; \operatorname{span}\{1,\,x\},
\]
independently of the $\pi$-coefficients of $\rho$.  Explicitly:
\begin{align}
  [D^2_{B^4},\Delta_{S^3}]\,\rho
  &= \bigl(4608\pi^2(\pi-1)\bigr)\cdot 1
     + \bigl(-221184\pi^3\bigr)\cdot x \notag\\
  &= -221184\,\Omega_0\cdot(U_0 + U_1)
     - 24576\,a_2\cdot U_0 + (\text{correction from }a_2)\label{eq:descent}
\end{align}
where $\Omega_0 = \pi^3/4 = a_3$ is the kernel constant (A225, Thm.~1).
\end{theorem}

\begin{proof}
The support statement follows from $[D^2,\Delta]_{ij} = 0$ for $i \geq 2$ (all
rows below the second are zero in Theorem~\ref{thm:commutator}).  The explicit
formula uses $[D^2,\Delta]\rho = \sum_n a_n [D^2,\Delta](U_n)$ with
$a_3 = \Omega_0$ (A225), giving:
\begin{align*}
  [D^2,\Delta]\rho
  &= a_2(-24576\,U_0) + a_3(-276480\,U_0 - 221184\,U_1)\\
  &= (-24576\,a_2 - 276480\,\Omega_0)\,U_0 - 221184\,\Omega_0\,U_1.
\end{align*}
Converting back to monomial form yields $4608\pi^2(\pi-1) - 221184\pi^3 x$.
\end{proof}

\begin{remark}[Interpretation]
The density $\rho$ has support in modes $\{U_1,U_2,U_3\}$; mode $U_0$ carries
normalization.  The commutator ``burns off'' the highest two modes (boundary
$U_2$ and bulk $U_3$), projecting them onto the lowest two ($U_0,U_1$).
This is the quantitative version of P224's observation that Fork (biharmonic
evolution in the bulk) and Weld (spectral propagation on the boundary) do not
commute: their failure to commute is localised to the two highest-layer modes.
\end{remark}

\section{Implications for the HUP Path to $\Delta$}

The Robertson uncertainty relation applied to states of $\rho$:
\[
  (\Delta D^2_{B^4})(\Delta\Delta_{S^3}) \;\geq\;
  \tfrac{1}{2}\bigl|\langle[D^2_{B^4},\Delta_{S^3}]\rangle\bigr|.
\]

Numerical evaluation under the Haar measure ($\langle\,\cdot\,\rangle_\rho$):
\begin{alignat*}{2}
  \langle D^2_{B^4}\rangle_\rho &\approx 4.59\times10^5,\quad&
  \langle \Delta_{S^3}\rangle_\rho &\approx -8.01\times10^2,\\
  \langle[D^2,\Delta]\rangle_\rho &\approx -5.34\times10^6,\quad&
  \Delta &= \Omega - \alpha^{-1}_{\rm exp} \approx 3.05\times10^{-4}.
\end{alignat*}
The ratio
\[
  \frac{|\langle[D^2,\Delta]\rangle_\rho|}
  {2|\langle D^2\rangle_\rho|\,|\langle\Delta_{S^3}\rangle_\rho|}
  \approx 7.3\times10^{-3}
\]
is far from unity (the HUP saturation value), and the commutator expectation is
$\sim10^{10}$ times larger than $\Delta$, so no simple evaluation of the Haar
expectation yields $\Delta$ algebraically.

\textbf{Why the HUP path is not closed here:} the boundary eigenstate at
$\nu^* = 2x^*-1$ (the eigentrajectory fixed point, A224/A225 open item~1) has
not been computed in closed form.  The Robertson bound \emph{could} produce a
near-cancellation at $\nu^*$ if the boundary state lies near the null space of
$[D^2,\Delta]$; this requires $\nu^*$ explicitly and is deferred.

\section{Summary and Open Items}

\paragraph{Closed (this addendum).}
The commutator $[D^2_{B^4},\Delta_{S^3}]$ is computed exactly in the Gegenbauer
basis.  The spectral-gap formula (eq.~\ref{eq:spectral-gap}) gives all entries
as integers.  The layer-descent theorem shows the commutator maps $\rho$ into
$\operatorname{span}\{1,x\}$ exactly, with the leading coefficient $-221184\,\Omega_0\cdot U_1$
being a direct product of the kernel constant $\Omega_0 = a_3$ (A225) and the
integer spectral-gap factor $221184 = 18432\times12$.

\paragraph{Open item 1.}
The boundary eigenstate $|\psi_{\rm bnd}\rangle$ requires $\nu^* = 2x^*-1$ in
closed form (A224/A225 open item~1, OP-alpha blocker).  Once $\nu^*$ is known,
the Robertson ratio
$|\langle\psi_{\rm bnd}|[D^2,\Delta]|\psi_{\rm bnd}\rangle| /
(2\,\Delta D^2\cdot\Delta\Delta_{S^3})$
can be evaluated.

\paragraph{Open item 2.}
The integer $18432 = D^2_{B^4}(U_3)$ coefficient encodes equal-weight mixing of
$U_0$ and $U_1$ (Remark~\ref{rem:mixing}).  Whether this
$U_0\!:\!U_1 = 1\!:\!1$ ratio is forced by the Hopf fibration structure (i.e.\ by
the $\pi/\Omega$ layer fraction at the edge) or is a numerical coincidence is
unresolved.

\paragraph{Open item 3.}
The monomial ratio of commutator output terms:
\[
  \frac{\text{linear coefficient}}{\text{constant coefficient}}
  = \frac{-221184\pi^3}{4608\pi^2(\pi-1)}
  = \frac{-48\pi}{\pi-1}
  \approx -70.41.
\]
The factor $48 = |U_3\text{ leading coeff}|/2 + \ldots$\ is purely combinatorial
here, but its appearance alongside $\pi/(\pi-1)$ (a ratio of consecutive TOE
layers) may carry geometric meaning in the Hopf stratification.

\bigskip
\noindent\textit{Cross-references:} P18 (Wheel/\^{O} identification), A224
(bulk-boundary stratification), A225 (Peter-Weyl expansion, $a_3=\Omega_0$),
HUP Saturation research file (2026-05-22).

\end{document}
