# Supplementary proofs for the 21 September 2026 boundary-reconstruction note

Revision: 22 September 2026. These proofs supplement the original nine-page
PDF; they are not revisions to its bytes or an additional source paper.

Use the definitions of the original note. In particular,

\[
\varphi=\frac{1+\sqrt5}{2},\quad
\varphi'=1-\varphi,\quad
\tau=\frac1{\varphi+\pi},\quad
M=\begin{pmatrix}0&1\\1&1\end{pmatrix},\quad
A=\begin{pmatrix}0&\tau&0\\\tau&\tau&0\\0&0&-\tau^{-2}\end{pmatrix}.
\]

Write \(\Psi(\xi)=A\xi\) and \(J(x,y,z)=z(x^2+xy-y^2)\).
Let \(B=\{x\in\mathbb R^3:|x|\le1\}\), \(E=AB\), and

\[
\mathcal H_{\le2}=\{u\in\mathbb R[x,y,z]:\deg u\le2,\ \Delta u=0\},\qquad
\mathcal H_A=\{u\circ A^{-1}:u\in\mathcal H_{\le2}\},\qquad
\mathcal F_A=\mathcal H_A\oplus\mathbb RJ.
\]

The dimensions are nine and ten, respectively. The direct sum follows because
every member of \(\mathcal H_A\) has degree at most two, whereas \(J\) is a
nonzero homogeneous cubic.

## Two-step compatibility

Set \(K=(2M-I)/\sqrt5\). Then

\[
K^2=I,\qquad M^2=\tfrac32 I+\tfrac{\sqrt5}{2}K.
\]

Indeed, \(M^2=M+I\) gives both identities by substitution. Since
\(\varphi^{-1}=\varphi-1\),

\[
\cosh(2\log\varphi)=\frac{\varphi^2+\varphi^{-2}}{2}=\frac32,\qquad
\sinh(2\log\varphi)=\frac{\varphi^2-\varphi^{-2}}{2}=\frac{\sqrt5}{2}.
\]

The exponential series, grouped into even and odd powers of \(K\), therefore
yields

\[
M^2=e^{2(\log\varphi)K}.
\]

For \(\kappa>0\), define

\[
\Phi_{\kappa,\eta}(w,z)=(\kappa e^{\eta K}w,\kappa^{-2}z).
\]

Applying \(\Psi\) twice proves

\[
\boxed{\Psi^2=\Phi_{\tau^2,\,2\log\varphi}.}
\]

The map \(\Psi\) itself is outside this family: its planar block has
determinant \(-\tau^2\), whereas \(\det(\kappa e^{\eta K})=\kappa^2>0\) because
\(\operatorname{tr}K=0\). This proves the two-step correspondence without
identifying an odd step with a member of the positive-scale family.

## Aligned nodes and the additional sample

Define the orthogonal matrix

\[
R=\begin{pmatrix}
1/\sqrt{1+\varphi^2}&\varphi/\sqrt{1+\varphi^2}&0\\
1/\sqrt{1+(\varphi')^2}&\varphi'/\sqrt{1+(\varphi')^2}&0\\
0&0&1
\end{pmatrix}.
\]

Let \(v_i\) be the twelve points

\[
\frac1{\sqrt{1+\varphi^2}}
\bigl\{(0,\pm1,\pm\varphi),(\pm1,\pm\varphi,0),(\pm\varphi,0,\pm1)\bigr\},
\]

with independent signs, and put \(n_i=R^T v_i\), \(q_i=A n_i\). Choose one
representative \(a_j\) from each of the six antipodal pairs among the \(n_i\).
For distinct pairs,

\[
|a_i\cdot a_j|=\frac1{\sqrt5},\qquad |a_j|=1.
\]

These identities follow directly from the listed vertex coordinates and are
unchanged by the orthogonal map \(R^T\).

In coordinates \((X,Y,Z)=R(x,y,z)\), factorization gives \(J=\sqrt5\,XYZ\).
Thus \(J(n_i)=0\). The original note proves \(J\circ A=J\), so \(J(q_i)=0\) as
well. Put

\[
p_*=R^T\frac{(1,1,1)^T}{\sqrt3},\qquad q_*=A p_*.
\]

Then

\[
J(q_*)=J(p_*)=\frac{\sqrt5}{3\sqrt3}\ne0.
\]

## Even interpolants

For \(j=1,\ldots,6\), define the polynomial on all of \(\mathbb R^3\)

\[
\boxed{E_j(x)=\frac16+\frac54\left((x\cdot a_j)^2-\frac{|x|^2}{3}\right).}
\]

Its Laplacian is zero, since \(\Delta(x\cdot a_j)^2=2\) and \(\Delta|x|^2=6\).
It is even, and substitution gives

\[
E_j(\pm a_i)=\delta_{ij}.
\]

The even subspace of \(\mathcal H_{\le2}\) consists of a constant and a
trace-free quadratic form, and has dimension \(1+5=6\). The six interpolation
identities make the \(E_j\) linearly independent; consequently they form a
basis for that subspace. In particular, for every even \(e\in\mathcal H_{\le2}\),

\[
e(x)=\sum_{j=1}^6 e(a_j)E_j(x).
\]

## Independence of three axes

For any three distinct antipodal axes, the Gram matrix has diagonal entries
one and off-diagonal entries of magnitude \(1/\sqrt5\). If those entries are
\(u,v,w\), its determinant is

\[
1-u^2-v^2-w^2+2uvw=\frac25\pm\frac{2}{5\sqrt5}>0.
\]

Thus every such triple spans \(\mathbb R^3\).

## Nine-node criterion

A set of nine of the twelve nodes uniquely determines every
\(u\in\mathcal H_{\le2}\) if and only if it meets all six antipodal pairs.

To prove sufficiency, decompose uniquely

\[
u(x)=e(x)+b\cdot x,
\]

where \(e\) is even and harmonic of degree at most two, and \(b\in\mathbb R^3\).
A nine-node set meeting all six pairs contains three complete pairs and three
single representatives. On each complete pair,

\[
u(a_j)-u(-a_j)=2b\cdot a_j.
\]

The three-axis independence just proved determines \(b\). Subtracting
\(b\cdot x\) from the remaining measurements gives all six even pair-values, so
the interpolation formula determines \(e\).

For necessity, if pair \(j\) is absent, the nonzero polynomial \(E_j\) vanishes
on every selected node. Uniqueness fails. A nine-node selection cannot omit two
pairs, so every failure consists of four complete pairs, one singleton, and one
omitted pair. The four complete pairs still determine the odd part. Five even
pair-values are known, leaving exactly \(\mathbb R E_j\) as the kernel on
\(\mathcal H_{\le2}\). Its rank is therefore eight.

There are

\[
\binom63 2^3=160
\]

successful selections: choose the three complete pairs and then one sign from
each remaining pair. There are

\[
6\cdot5\cdot2=60
\]

failures: choose the omitted pair, the singleton pair, and its sign. These
counts total \(\binom{12}{9}=220\).

## Transported ranks and the surviving kernel

Define

\[
\widetilde E_j(y)=E_j(A^{-1}y).
\]

For a set \(I\) of node indices, let \(S_I:\mathcal F_A\to\mathbb R^{|I|}\) send
\(f\) to \((f(q_i))_{i\in I}\); let \(S_{I,*}\) append \(f(q_*)\). Pullback by
\(A^{-1}\) is an isomorphism between the quadratic field spaces and preserves
the evaluation ranks.

On all twelve nodes, the quadratic part is determined and \(J\) is invisible.
Hence

\[
\ker S_{12}=\mathbb RJ,\qquad \operatorname{rank}S_{12}=9.
\]

Adding \(q_*\), where \(J(q_*)\ne0\), gives a zero kernel and rank ten. For a
successful nine-node selection, the same argument gives rank ten after adding
\(q_*\).

For a failing nine-node selection omitting pair \(j\),

\[
\ker S_I=\operatorname{span}\{\widetilde E_j,J\}.
\]

An element \(\alpha\widetilde E_j+\beta J\) also vanishes at \(q_*\) precisely
when

\[
\alpha E_j(p_*)+\beta J(q_*)=0.
\]

Since \(J(q_*)\ne0\), solving this single equation proves

\[
\boxed{
\ker S_{I,*}
=
\operatorname{span}\left\{
\widetilde E_j-\frac{E_j(p_*)}{J(q_*)}J
\right\}.
}
\]

This polynomial is nonzero: at either transported vertex of the omitted pair
its value is one. Thus the kernel has dimension one and the rank is nine.

For the specified \(p_*\), the vertex coordinates give

\[
(p_*\cdot a_j)^2=\frac13\pm\frac{2}{3\sqrt5},\qquad
E_j(p_*)=\frac{1\pm\sqrt5}{6}\ne0.
\]

The surviving mode is therefore a mixture of an even quadratic mode and the
cubic mode; the extra sample has removed one of the two formerly invisible
directions.

Ten scalar evaluations are the minimum possible for injectivity on the
ten-dimensional vector space \(\mathcal F_A\), by rank-nullity. A successful
nine-node selection together with \(q_*\) attains that number. The
thirteen-sample construction retains the symmetric twelve-node decoder and adds
one evaluation; it is not a minimality assertion for that symmetric formula or
a claim about unrestricted cubic fields.

## Computed spherical moment

With normalized spherical measure \(\sigma=d\Omega/(4\pi)\),

\[
J^2=x^4z^2+2x^3yz^2-x^2y^2z^2-2xy^3z^2+y^4z^2.
\]

The odd terms have zero integral. Using the normalized spherical monomial
moments,

\[
\int x^4z^2\,d\sigma=\int y^4z^2\,d\sigma=\frac1{35},\qquad
\int x^2y^2z^2\,d\sigma=\frac1{105},
\]

therefore gives

\[
\boxed{\int J^2\,d\sigma=\frac1{35}-\frac1{105}+\frac1{35}=\frac1{21}.}
\]

The corrected checker evaluates this polynomial integral using exact rational
arithmetic. Its remaining NumPy checks are numerical corroboration of selected
claims. The polynomial arguments above and in the original note supply the
written proofs.