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\hypersetup{pdftitle={Boundary Reconstruction, Volume-Preserving Transport, and an Invariant Cubic},pdfauthor={Prepared for Matthew Alan Newman},pdfsubject={A self-contained mathematical proof}}
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\title{\textbf{Boundary Reconstruction,\\Volume-Preserving Transport,\\and an Invariant Cubic}\\[7pt]
\large A self-contained mathematical proof}
\author{Prepared for Matthew Alan Newman}
\date{21 September 2026}

\begin{document}
\maketitle
\thispagestyle{empty}

\begin{abstract}
An explicit twelve-point icosahedral rule reconstructs every harmonic
polynomial of degree at most two from its boundary values. We prove that an
invertible golden-ratio matrix transports this decoder from a ball to an
ellipsoid, together with its elliptic differential operator and boundary
measure. The matrix preserves three-dimensional volume and the harmonic cubic
\(J(x,y,z)=z(x^2+xy-y^2)\). An explicit orthogonal change of coordinates
identifies this invariant with \(\sqrt5\,XYZ\). On an aligned icosahedral grid,
the invariant is invisible to the twelve samples, and its square obstructs
degree-six cubature. One additional, explicitly specified boundary value
recovers the coefficient of this cubic. We obtain a complete thirteen-sample
decoder on a ten-dimensional field space, prove its compatibility with
transport, and show how an exact integer
quotient--remainder record preserves finitely encoded sampling data.
All reconstruction statements are restricted to their stated field spaces.
\end{abstract}

\section{Definitions and statement of the combined theorem}

Work over the real numbers. Put
\begin{equation}\label{eq:constants}
\varphi=\frac{1+\sqrt5}{2},\qquad
\varphi'=1-\varphi,\qquad
\tau=\frac1{\varphi+\pi}.
\end{equation}
Thus \(\varphi+\varphi'=1\), \(\varphi\varphi'=-1\),
\(\varphi^2=\varphi+1\), and \(\tau>0\). Define
\begin{equation}\label{eq:maps}
M=\begin{pmatrix}0&1\\1&1\end{pmatrix},\qquad
A=\begin{pmatrix}0&\tau&0\\\tau&\tau&0\\0&0&-\tau^{-2}\end{pmatrix},\qquad
\Psi(\xi)=A\xi.
\end{equation}
For \(w=(x,y)\in\R^2\) and \(\xi=(x,y,z)\in\R^3\), set
\begin{equation}\label{eq:invariants}
\normform(w)=x^2+xy-y^2,\qquad J(\xi)=z\normform(w).
\end{equation}
Let \(\ball=\{\xi:\|\xi\|\le1\}\), \(\sphere=\partial\ball\), and let
\(\sigma\) be normalized spherical area measure, so that
\(\dd\sigma=\dd\Omega/(4\pi)\) and \(\sigma(\sphere)=1\).
For a nonnegative integer \(m\), \(\mathcal P_{\le m}\) denotes the real
polynomials on \(\R^3\) of total degree at most \(m\). Define
\begin{equation}\label{eq:hspace}
\HH=\{u\in\mathcal P_{\le2}:\Delta u=0\}.
\end{equation}

Put \(r=\sqrt{1+\varphi^2}\). Let \(v_1,\ldots,v_{12}\) enumerate
\begin{equation}\label{eq:standardnodes}
\mathcal V=\frac1r\bigl\{(0,\pm1,\pm\varphi),
 (\pm1,\pm\varphi,0),(\pm\varphi,0,\pm1)\bigr\},
\end{equation}
where the two signs in each group are independent. Let
\begin{equation}\label{eq:orthogonal}
r_+=\sqrt{1+\varphi^2},\qquad r_-=\sqrt{1+(\varphi')^2},\qquad
R=\begin{pmatrix}
1/r_+&\varphi/r_+&0\\
1/r_-&\varphi'/r_-&0\\
0&0&1
\end{pmatrix}.
\end{equation}
The aligned nodes and the additional point are
\begin{equation}\label{eq:alignednodes}
n_i=R^Tv_i\quad(1\le i\le12),\qquad
p_*=R^T\frac{(1,1,1)^T}{\sqrt3}.
\end{equation}
Define the ellipsoid, transported nodes, and diffusion matrix by
\begin{equation}\label{eq:ellipsoid}
E=A\ball,\qquad q_i=An_i,\qquad q_*=Ap_*,\qquad
D=AA^T=\begin{pmatrix}
\tau^2&\tau^2&0\\\tau^2&2\tau^2&0\\0&0&\tau^{-4}
\end{pmatrix}.
\end{equation}
The transported quadratic space and its cubic extension are
\begin{equation}\label{eq:transportspace}
\mathcal H_A=\{u\circ A^{-1}:u\in\HH\},\qquad
\mathcal F_A=\mathcal H_A\oplus\R J.
\end{equation}
For \(y,q\in\R^3\), write
\begin{equation}\label{eq:transportkernel}
K_A(y,q)=1+3y^TD^{-1}q+
\frac52\left(3(y^TD^{-1}q)^2-y^TD^{-1}y\right).
\end{equation}

\begin{theorem}[Combined reconstruction and transport]\label{thm:main}
The preceding definitions have the following consequences.
\begin{enumerate}[label=\textup{(\roman*)}]
\item The map \(\Psi\) is invertible, preserves volume, and satisfies
\(J\circ\Psi=J\). The space \(\mathcal H_A\) is the nine-dimensional space
of polynomials \(v\) of degree at most two satisfying
\(\nabla\cdot(D\nabla v)=0\).
\item For \(v\in\mathcal H_A\),
\begin{equation}\label{eq:mainquadratic}
v(y)=\frac1{12}\sum_{i=1}^{12}v(q_i)K_A(y,q_i)
\qquad(y\in\R^3).
\end{equation}
\item The space \(\mathcal F_A\) has dimension ten. Moreover,
\(J(q_i)=0\) for every \(i\), whereas
\begin{equation}\label{eq:extraJ}
J(q_*)=\frac{\sqrt5}{3\sqrt3}\ne0.
\end{equation}
For every \(f\in\mathcal F_A\), define
\(\widehat v(y)=\frac1{12}\sum_i f(q_i)K_A(y,q_i)\). Then
\begin{equation}\label{eq:main13}
\boxed{\displaystyle
f(y)=\widehat v(y)+\frac{J(y)}{J(q_*)}
\bigl(f(q_*)-\widehat v(q_*)\bigr).}
\end{equation}
Thus the thirteen stated boundary measurements uniquely determine \(f\).
The twelve aligned measurements alone do not.
\item Under \(\nu=A_*\sigma\), the image of normalized spherical measure,
the twelve transported nodes have degree-five cubature exactness. No choice
of real weights on those twelve nodes has degree-six exactness for \(\nu\).
\end{enumerate}
\end{theorem}

The following sections prove every assertion of Theorem~\ref{thm:main} and
then establish exact recovery of finitely encoded field records.

\section{The spherical sampling rule}

\begin{lemma}[Spherical moments]\label{lem:moments}
For distinct coordinate indices \(j,k\), normalized spherical integration
satisfies
\begin{equation}\label{eq:moments}
\int_\sphere \xi_j^2\dd\sigma=\frac13,\qquad
\int_\sphere \xi_j^4\dd\sigma=\frac15,\qquad
\int_\sphere \xi_j^2\xi_k^2\dd\sigma=\frac1{15},\qquad
\int_\sphere x^2y^2z^2\dd\sigma=\frac1{105}.
\end{equation}
Any monomial with an odd exponent in at least one coordinate has integral
zero. In particular,
\begin{equation}\label{eq:tensormoment}
\int_\sphere \xi_i\xi_j\xi_k\xi_l\dd\sigma
=\frac{\delta_{ij}\delta_{kl}+\delta_{ik}\delta_{jl}
+\delta_{il}\delta_{jk}}{15}.
\end{equation}
\end{lemma}
\begin{proof}
Reflection in a coordinate plane changes the sign of an odd monomial and
preserves \(\sigma\), which proves its zero integral. Spherical coordinates
give, for every integer \(m\ge0\),
\[
\int_\sphere z^{2m}\dd\sigma
=\frac12\int_{-1}^1 t^{2m}\dd t=\frac1{2m+1}.
\]
Coordinate symmetry gives the same formula for \(x\) and \(y\). From
\((x^2+y^2+z^2)^2=1\) on \(\sphere\), symmetry yields
\(1=3/5+6\int x^2y^2\dd\sigma\); hence the mixed fourth moment is \(1/15\).
The tensor formula follows by considering the possible even exponent
patterns: four equal indices give \(1/5\), two pairs give \(1/15\), and all
other patterns have an odd exponent.

Next,
\[
\frac15=\int x^4(x^2+y^2+z^2)\dd\sigma
=\frac17+2\int x^4y^2\dd\sigma,
\]
so \(\int x^4y^2\dd\sigma=1/35\). Expanding
\((x^2+y^2+z^2)^3=1\) now gives
\[
1=\frac37+\frac{18}{35}+6\int x^2y^2z^2\dd\sigma.
\]
The final moment in~\eqref{eq:moments} follows.
\end{proof}

\begin{lemma}[Degree-five icosahedral cubature]\label{lem:cubature}
For every \(p\in\mathcal P_{\le5}\),
\begin{equation}\label{eq:cubature}
\frac1{12}\sum_{i=1}^{12}p(v_i)=\int_\sphere p\dd\sigma.
\end{equation}
The same assertion holds after any fixed orthogonal change of the nodes.
\end{lemma}
\begin{proof}
The node set is invariant under independent coordinate sign changes and
cyclic coordinate permutations. Thus its average annihilates every monomial
with an odd coordinate exponent. For degree at most five, the remaining
monomials have total degree zero, two, or four. Since
\[
r^2=\varphi+2,\qquad r^4=5\varphi^2,\qquad
1+\varphi^4=3\varphi^2,
\]
direct averaging on~\eqref{eq:standardnodes} gives
\[
\frac1{12}\sum_i(v_i)_1^2=
\frac{1+\varphi^2}{3r^2}=\frac13,\qquad
\frac1{12}\sum_i(v_i)_1^4=
\frac{1+\varphi^4}{3r^4}=\frac15,
\]
and
\[
\frac1{12}\sum_i(v_i)_1^2(v_i)_2^2
=\frac{\varphi^2}{3r^4}=\frac1{15}.
\]
Together with the constant moment, these are all required even moments,
up to coordinate permutation. Lemma~\ref{lem:moments} proves equality with
spherical integration, and linearity proves~\eqref{eq:cubature}.

If \(O\) is orthogonal, then \(p\circ O\) has the same degree as \(p\), and
\(\sigma\) is orthogonally invariant. Applying~\eqref{eq:cubature} to
\(p\circ O\) proves the last assertion.
\end{proof}

\begin{proposition}[The twelve-sample quadratic decoder]\label{prop:decoder}
Let \(n_i\) be any orthogonal image of the nodes in~\eqref{eq:standardnodes}.
Every \(u\in\HH\) has the unique form
\begin{equation}\label{eq:quadraticform}
u(\xi)=a+b^T\xi+\xi^TC\xi,
\qquad C=C^T,\quad\tr C=0.
\end{equation}
The space \(\HH\) has dimension nine, and
\begin{equation}\label{eq:sphericaldecoder}
u(\xi)=\frac1{12}\sum_{i=1}^{12}u(n_i)
\left[1+3\xi^Tn_i+
\frac52\bigl(3(\xi^Tn_i)^2-\|\xi\|^2\bigr)\right]
\end{equation}
for every \(\xi\in\R^3\).
\end{proposition}
\begin{proof}
Every quadratic polynomial admits a unique representation with a symmetric
quadratic matrix. Its Laplacian is \(2\tr C\), so harmonicity is equivalent
to the displayed trace condition. The dimensions are \(1\), \(3\), and
\(6-1=5\), respectively.

Write \(\avg{g}=\frac1{12}\sum_i g(n_i)\). Products used below have degree
at most four, so Lemma~\ref{lem:cubature} permits their exact integration.
The second and fourth moments in Lemma~\ref{lem:moments} give
\begin{equation}\label{eq:coeffrecovery}
\avg{u}=a,\qquad
\avg{u(n)n}=\frac b3,\qquad
\avg{u(n)(nn^T-I/3)}=\frac{2C}{15}.
\end{equation}
For the last identity, the constant contribution vanishes, the linear
contribution has odd degree, and the quadratic contribution has entries
\[
\sum_{k,l}C_{kl}\int n_kn_ln_in_j\dd\sigma
-\frac{\delta_{ij}}3\sum_{k,l}C_{kl}\int n_kn_l\dd\sigma
=\frac{2C_{ij}}{15},
\]
because \(C=C^T\) and \(\tr C=0\).
Substitution of~\eqref{eq:coeffrecovery} into~\eqref{eq:quadraticform}
gives~\eqref{eq:sphericaldecoder}. Its validity for every \(\xi\) also
proves uniqueness from the twelve samples.
\end{proof}

\section{The reversible map and its invariant}

\begin{proposition}[Volume and cubic preservation]\label{prop:invariance}
The matrix \(A\) is invertible with determinant one. It satisfies
\begin{equation}\label{eq:invariance}
J(A\xi)=J(\xi),\qquad
A^{-1}=\begin{pmatrix}
-\tau^{-1}&\tau^{-1}&0\\
\tau^{-1}&0&0\\
0&0&-\tau^2
\end{pmatrix}.
\end{equation}
\end{proposition}
\begin{proof}
Since \(\det M=-1\),
\(\det A=\det(\tau M)(-\tau^{-2})=(-\tau^2)(-\tau^{-2})=1\).
Multiplication verifies the stated inverse. Linear change of variables
therefore shows that \(\Psi\) preserves three-dimensional Lebesgue volume.
Direct calculation gives
\[
\normform(M(x,y))=\normform(y,x+y)
=y^2+y(x+y)-(x+y)^2=-\normform(x,y).
\]
By homogeneity, \(\normform(\tau Mw)=-\tau^2\normform(w)\), so
\[
J(A(w,z))=(-\tau^{-2}z)(-\tau^2\normform(w))=J(w,z).
\]
\end{proof}

\section{Transport of the equation, decoder, and measure}

\begin{theorem}[Ellipsoidal reconstruction]\label{thm:transport}
For the matrix \(A\), ellipsoid \(E\), and nodes in~\eqref{eq:ellipsoid},
\begin{equation}\label{eq:ellipsoidset}
E=\{y\in\R^3:y^TD^{-1}y\le1\},\qquad q_i\in\partial E.
\end{equation}
For every \(u\in C^2(\R^3)\), with
\(v=u\circ A^{-1}\),
\begin{equation}\label{eq:operatortransport}
\nabla_y\cdot(D\nabla_yv)(y)=(\Delta u)(A^{-1}y).
\end{equation}
In particular,
\begin{equation}\label{eq:Hchar}
\mathcal H_A=\{v\in\mathcal P_{\le2}:\nabla\cdot(D\nabla v)=0\},
\end{equation}
and \(\dim\mathcal H_A=9\). Its reconstruction formula
is~\eqref{eq:mainquadratic}.
\end{theorem}
\begin{proof}
The matrix \(D\) is symmetric positive definite and
\(D^{-1}=A^{-T}A^{-1}\). Hence
\(y^TD^{-1}y=\|A^{-1}y\|^2\), which proves~\eqref{eq:ellipsoidset}.
The chain rule gives
\[
\nabla_yv(y)=A^{-T}\nabla u(A^{-1}y),\qquad
\operatorname{Hess}_y v(y)
=A^{-T}\operatorname{Hess}u(A^{-1}y)A^{-1}.
\]
Since \(D\) is constant, cyclicity of trace gives
\begin{align*}
\nabla_y\cdot(D\nabla_yv)(y)
&=\tr\!\left(AA^T A^{-T}\operatorname{Hess}u(A^{-1}y)A^{-1}\right)\\
&=\tr\bigl(\operatorname{Hess}u(A^{-1}y)\bigr)
=(\Delta u)(A^{-1}y).
\end{align*}
The inverse substitution \(u=v\circ A\) proves both directions
of~\eqref{eq:Hchar}. Composition by an invertible linear map preserves
polynomial degree and gives a vector-space isomorphism, so the dimension is
nine.

For \(v=u\circ A^{-1}\), the samples satisfy \(v(q_i)=u(n_i)\). Apply
Proposition~\ref{prop:decoder} at \(\xi=A^{-1}y\). As
\[
(A^{-1}y)^Tn_i
=(A^{-1}y)^T(A^{-1}q_i)=y^TD^{-1}q_i,
\qquad \|A^{-1}y\|^2=y^TD^{-1}y,
\]
the result is precisely~\eqref{eq:mainquadratic}.
\end{proof}

\begin{remark}[Ordinary harmonicity is not generally transported]
The harmonic polynomial \(u(x,y,z)=x^2-y^2\) becomes
\[
(u\circ A^{-1})(y_1,y_2,y_3)
=\tau^{-2}(y_2^2-2y_1y_2).
\]
Its ordinary Laplacian is \(2\tau^{-2}\), while its operator
\(\nabla\cdot(D\nabla\,\cdot\,)\) is zero. Thus the operator in
Theorem~\ref{thm:transport} is part of the transported specification.
\end{remark}

\begin{proposition}[The transported boundary measure]\label{prop:measure}
Let \(\nu=A_*\sigma\), so that
\(\int_{\partial E}g\dd\nu=\int_\sphere g(An)\dd\sigma(n)\).
For every \(p\in\mathcal P_{\le5}\),
\begin{equation}\label{eq:ellipsoidcubature}
\frac1{12}\sum_{i=1}^{12}p(q_i)=\int_{\partial E}p\dd\nu.
\end{equation}
If \(\dd S_E\) is Euclidean surface area on \(\partial E\), then at
\(q=An\),
\begin{equation}\label{eq:measurejacobian}
\dd S_E(q)=\|A^{-T}n\|\dd\Omega(n),\qquad
\dd\nu(q)=\frac{\dd S_E(q)}{4\pi\|D^{-1}q\|}.
\end{equation}
\end{proposition}
\begin{proof}
The polynomial \(p\circ A\) has degree at most five, so
Lemma~\ref{lem:cubature} proves~\eqref{eq:ellipsoidcubature}.
For tangent vectors \(a,b\) to the sphere,
\[
(Aa)\mathbin{\times}(Ab)=\det(A)A^{-T}(a\mathbin{\times}b).
\]
Taking lengths in a local surface parametrization and using \(\det A=1\)
gives the first surface Jacobian. Since
\(A^{-T}n=A^{-T}A^{-1}q=D^{-1}q\), the definition
\(\dd\sigma=\dd\Omega/(4\pi)\) gives the second identity.
The Jacobian is nonconstant: at the first and second coordinate unit
vectors its values are \(\sqrt2\,\tau^{-1}\) and \(\tau^{-1}\),
respectively. Thus \(\nu\) differs from normalized Euclidean surface area
on \(\partial E\).
\end{proof}

\section{The aligned invariant and the sampling obstruction}

\begin{lemma}[An orthogonal product representation]\label{lem:alignment}
The matrix \(R\) in~\eqref{eq:orthogonal} is orthogonal. With
\((X,Y,Z)^T=R(x,y,z)^T\),
\begin{equation}\label{eq:Jproduct}
J(x,y,z)=\sqrt5\,XYZ.
\end{equation}
Moreover,
\begin{equation}\label{eq:Jharmonic}
\Delta J=0,\qquad \nabla\cdot(D\nabla J)=0,
\qquad J(n_i)=J(q_i)=0,
\qquad J(p_*)=J(q_*)=\frac{\sqrt5}{3\sqrt3}.
\end{equation}
\end{lemma}
\begin{proof}
The first two rows of \(R\) have unit length and their inner product is
\((1+\varphi\varphi')/(r_+r_-)=0\). The third row is orthogonal to them.
Thus \(RR^T=I\). Also
\[
r_+^2r_-^2=(\varphi+2)(3-\varphi)=5,
\qquad r_+r_-=\sqrt5,
\]
and
\[
(x+\varphi y)(x+\varphi' y)=x^2+xy-y^2.
\]
These equations prove~\eqref{eq:Jproduct}.
Orthogonality also shows that all \(n_i\) and \(p_*\) lie on \(\sphere\).

Direct differentiation gives
\(\Delta J=2z-2z+0=0\).
Proposition~\ref{prop:invariance} also implies \(J\circ A^{-1}=J\), so
Theorem~\ref{thm:transport}, applied to \(u=J\), gives
\(\nabla\cdot(D\nabla J)=0\). This use of
\eqref{eq:operatortransport} is valid for a cubic because the operator
identity holds for every \(C^2\) function.

For \(n_i=R^Tv_i\), one of the coordinates of \(Rn_i=v_i\) is zero.
Equation~\eqref{eq:Jproduct} therefore gives \(J(n_i)=0\).
At \(p_*\), the three transformed coordinates are all \(1/\sqrt3\), so
\(J(p_*)=\sqrt5/(3\sqrt3)\). Invariance under \(A\) proves the corresponding
statements for \(q_i\) and \(q_*\).
\end{proof}

\begin{proposition}[A common reconstruction and cubature obstruction]\label{prop:obstruction}
The twelve aligned samples cannot determine an arbitrary harmonic
polynomial of degree at most three. Furthermore,
\begin{equation}\label{eq:Jmoment}
\int_\sphere J^2\dd\sigma
=\int_{\partial E}J^2\dd\nu=\frac1{21}.
\end{equation}
No real weights on the twelve nodes \(n_i\) give degree-six cubature for
\(\sigma\), and no real weights on \(q_i\) give degree-six cubature for
\(\nu\). In both cases degree five is attained by weights \(1/12\).
\end{proposition}
\begin{proof}
The nonzero harmonic cubic \(J\) and the zero polynomial have the same
values on all twelve aligned nodes. This proves the first assertion; the
same argument applies to the transported nodes.

Orthogonal invariance, Lemma~\ref{lem:alignment}, and
Lemma~\ref{lem:moments} give
\[
\int_\sphere J^2\dd\sigma
=5\int_\sphere X^2Y^2Z^2\dd\sigma=\frac5{105}=\frac1{21}.
\]
Since \(J(An)=J(n)\), pushforward of the measure gives the second integral
in~\eqref{eq:Jmoment}. For any real weights \(a_i\),
\(\sum_i a_iJ(n_i)^2=\sum_i a_iJ(q_i)^2=0\).
As \(J^2\) has degree six and its integral is nonzero, degree-six exactness
is impossible. Degree-five exactness follows from
Lemma~\ref{lem:cubature} and Proposition~\ref{prop:measure}.
\end{proof}

\begin{remark}[Alignment and time sampling]
The assertion \(J(n_i)=0\) uses the explicitly aligned nodes
\eqref{eq:alignednodes}; it is not asserted for an arbitrary orientation
of an icosahedron. For these aligned nodes,
\(J(A^kn_i)=0\) for every integer \(k\). Thus measurements of this invariant
at the transported copies of the same twelve nodes remain zero at every
step.
\end{remark}

\section{The thirteenth measurement and completion of the main proof}

\begin{theorem}[Exact recovery of the invariant coefficient]\label{thm:13}
Let \(f\in\mathcal F_A\). Its thirteen samples at
\(q_1,\ldots,q_{12},q_*\) uniquely determine \(f\) by
\eqref{eq:main13}. One additional measurement with \(J\ne0\) is necessary
and sufficient to augment these fixed twelve aligned samples for this
field space.
\end{theorem}
\begin{proof}
Because \(J\) is a nonzero cubic, it is not in the quadratic space
\(\mathcal H_A\). Hence the sum in~\eqref{eq:transportspace} is direct and
\(\dim\mathcal F_A=9+1=10\). There are unique \(v\in\mathcal H_A\) and
\(\beta\in\R\) with \(f=v+\beta J\).

By Lemma~\ref{lem:alignment}, \(f(q_i)=v(q_i)\) for \(1\le i\le12\).
Theorem~\ref{thm:transport} therefore gives
\[
\widehat v(y)=\frac1{12}\sum_i f(q_i)K_A(y,q_i)=v(y).
\]
Evaluation at \(q_*\) now yields
\[
\beta=\frac{f(q_*)-\widehat v(q_*)}{J(q_*)},
\]
whose denominator is the nonzero number in~\eqref{eq:extraJ}. Substituting
this value into \(f=v+\beta J\) proves~\eqref{eq:main13}.

If two fields have all thirteen values in common, their quadratic parts
coincide by the twelve-sample formula, and their cubic coefficients coincide
by the last equation. This proves uniqueness. Without an additional
measurement, adding any multiple of \(J\) leaves all twelve values
unchanged. A point with \(J\ne0\) distinguishes these multiples, proving
necessity and sufficiency for augmentation of this fixed sampling set.
\end{proof}

\begin{corollary}[Commutation of transport and reconstruction]\label{cor:commuting}
Let \(\mathcal F=\HH\oplus\R J\), and define
\(\mathcal U_Af=f\circ A^{-1}\). Let \(\mathcal S\) evaluate fields at
\(n_1,\ldots,n_{12},p_*\), and let \(\mathcal S_A\) evaluate fields at
\(q_1,\ldots,q_{12},q_*\). Let \(\mathcal D\) and \(\mathcal D_A\) be their
reconstruction maps, with \(\mathcal D_A\) given
by~\eqref{eq:main13}. Then
\begin{equation}\label{eq:commuting}
\mathcal S_A\mathcal U_A=\mathcal S,\qquad
\mathcal D_A\mathcal S_A=\id_{\mathcal F_A},\qquad
\mathcal D_A\mathcal S_A\mathcal U_A=\mathcal U_A.
\end{equation}
In particular, on admissible sample vectors,
\(\mathcal D_A=\mathcal U_A\mathcal D\).
\end{corollary}
\begin{proof}
On the original ball, Proposition~\ref{prop:decoder} recovers the quadratic
part from the twelve aligned values. Evaluation at \(p_*\), where
\(J(p_*)\ne0\), then recovers the cubic coefficient by the same subtraction
and division as in Theorem~\ref{thm:13}. This defines \(\mathcal D\) and
proves \(\mathcal D\mathcal S=\id_{\mathcal F}\).

The identity \(J\circ A^{-1}=J\) shows that \(\mathcal U_A\) maps
\(\mathcal F\) bijectively onto \(\mathcal F_A\).
At every transported sampling point, \((f\circ A^{-1})(Ap)=f(p)\), proving
the first identity. The second is Theorem~\ref{thm:13}; their composition
gives the third. The same pointwise argument together with uniqueness gives
the final assertion on sample vectors arising from these field spaces.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:main}]
Proposition~\ref{prop:invariance} proves volume and invariant preservation.
Theorem~\ref{thm:transport} proves the characterization, dimension, and
decoder of \(\mathcal H_A\). Lemma~\ref{lem:alignment} and
Theorem~\ref{thm:13} prove the ten-dimensional extension and its sampling
formula. Proposition~\ref{prop:measure} gives degree-five exactness for
\(\nu\), and Proposition~\ref{prop:obstruction} excludes degree six for
any real weights. These are all four assertions.
\end{proof}

\section{Exact integer records and information accounting}

\begin{theorem}[Quotient--remainder reversibility]\label{thm:ledger}
Let \(Q,S\) be integers with \(0<Q<S\). For \(d\in\Nzero\), define
\begin{equation}\label{eq:ledger}
F(d)=\left\lfloor\frac{Qd}{S}\right\rfloor,
\qquad \rho(d)=Qd-SF(d).
\end{equation}
The map \(\mathcal L(d)=(F(d),\rho(d))\) is a bijection from
\(\Nzero\) onto
\begin{equation}\label{eq:ledgerimage}
\mathcal A_{Q,S}=\{(a,r)\in\Nzero\times\mathbb Z:
0\le r<S,\; Sa+r\equiv0\pmod Q\}.
\end{equation}
Its inverse is
\begin{equation}\label{eq:ledgerinverse}
\mathcal L^{-1}(a,r)=\frac{Sa+r}{Q}.
\end{equation}
An ordered finite remainder sequence and the terminal quotient reconstruct
every preceding iterate exactly.
\end{theorem}
\begin{proof}
Euclidean division gives \(Qd=SF(d)+\rho(d)\) with \(0\le\rho(d)<S\), so
\(\mathcal L(d)\in\mathcal A_{Q,S}\) and~\eqref{eq:ledgerinverse}
recovers \(d\). Conversely, for \((a,r)\in\mathcal A_{Q,S}\), the formula
defines a nonnegative integer \(d\). Then
\(Qd/S=a+r/S\), whose floor is \(a\); its remainder is \(r\).
This proves both directions of the bijection.

For a trajectory \(d_{k+1}=F(d_k)\), the identity
\(d_k=(Sd_{k+1}+\rho_k)/Q\) reconstructs its preceding term. Backward
induction reconstructs the whole finite trajectory from its terminal value
and ordered remainder list. Since \(F(d)<d\) for positive integers \(d\),
every forward trajectory reaches zero in finitely many steps. A proposed
ledger is valid only when the reconstructed values are nonnegative integers
and all remainder bounds and identities hold.
\end{proof}

\begin{corollary}[Exact Shannon identity]\label{cor:entropy}
If \(\mathsf D\) is a nonnegative integer-valued random variable with
finite Shannon entropy, measured in bits, then
\begin{equation}\label{eq:entropy}
H(\mathsf D)=H(F(\mathsf D))+H(\rho(\mathsf D)\mid F(\mathsf D)).
\end{equation}
\end{corollary}
\begin{proof}
The bijection in Theorem~\ref{thm:ledger} reassigns the probabilities of
\(\mathsf D\) to distinct pairs without changing them. Thus
\(H(\mathsf D)=H(F(\mathsf D),\rho(\mathsf D))\). The discrete entropy chain
rule gives~\eqref{eq:entropy}. It can also be obtained directly by writing
the pair probability as the product of its first-coordinate marginal and
the conditional probability of its second coordinate, and expanding the
logarithm in the entropy sum.
\end{proof}

\begin{theorem}[Composition with finite field records]\label{thm:recordcomposition}
Fix the construction parameters and sample labels used in
Theorem~\ref{thm:13}. Let \(\mathfrak F\subseteq\mathcal F_A\) be a class
whose exact thirteen-sample records admit an integer encoding
\(\operatorname{enc}\) and a decoding function \(\operatorname{dec}\)
with
\begin{equation}\label{eq:encodingassumption}
\operatorname{dec}(\operatorname{enc}(\mathcal S_A f))
=\mathcal S_A f\qquad(f\in\mathfrak F).
\end{equation}
Then quotient--remainder storage followed by reversal and field
reconstruction recovers every field in \(\mathfrak F\) exactly:
\begin{equation}\label{eq:fullcomposition}
\mathcal D_A\circ\operatorname{dec}\circ\mathcal L^{-1}
\circ\mathcal L\circ\operatorname{enc}\circ\mathcal S_A
=\id_{\mathfrak F}.
\end{equation}
\end{theorem}
\begin{proof}
For any \(f\in\mathfrak F\), first cancel
\(\mathcal L^{-1}\circ\mathcal L\) by Theorem~\ref{thm:ledger}. Next use
\eqref{eq:encodingassumption} to recover \(\mathcal S_Af\). Finally,
\(\mathcal D_A\mathcal S_Af=f\) by Theorem~\ref{thm:13}. This gives exactly
\eqref{eq:fullcomposition}.
\end{proof}

The finite-encoding hypothesis is explicit. For example, any finite byte
record \(b=(b_0,\ldots,b_{L-1})\), including a length-zero record, has the
injective integer encoding
\begin{equation}\label{eq:byteencoding}
\iota(b)=256^L+\sum_{j=0}^{L-1}b_j256^{L-1-j},\qquad 0\le b_j<256.
\end{equation}
The base-256 representation consists of a leading digit \(1\) followed by
the \(L\) bytes. Its leading digit therefore retains length and any zero
bytes. If those bytes specify exact sample values in a declared notation,
decoding recovers those values. No assertion is made that arbitrary real
sample vectors have finite exact encodings.

\section{Scope of the proved result}

The reconstructed field belongs to the explicitly stated ten-dimensional
space \(\mathcal F_A\). The cubic mode is identified by the specified
additional sample; the twelve aligned values alone leave its coefficient
undetermined. The differential operator is transported with the domain,
and the cubature measure is \(A_*\sigma\). Ordinary Euclidean surface area
has the separate Jacobian~\eqref{eq:measurejacobian}. Integer reversal
recovers the exact encoded record under~\eqref{eq:encodingassumption}.

Within these hypotheses, the combined identity~\eqref{eq:fullcomposition}
and the commuting relations~\eqref{eq:commuting} prove compatibility of
boundary reconstruction, reversible geometric transport, and exact record
recovery. Every sampling formula in this note is a polynomial identity,
so its validity on the stated domain follows without a numerical
approximation or a discretized PDE solution.

\end{document}
