Boundary reconstruction, volume-preserving transport, and an invariant cubic

Work with $\tau=1/(\varphi+\pi)$. This is $1/\Omega_0$, not the contraction $\lambda=\varphi/\Omega_0$. The matrix

$A=\begin{pmatrix}0&\tau&0\\\tau&\tau&0\\0&0&-\tau^{-2}\end{pmatrix}$

has determinant one, preserves volume, and satisfies $J\circ A=J$ for $J(x,y,z)=z(x^2+xy-y^2)$. After an orthogonal alignment $J=\sqrt5\,XYZ$. The twelve aligned icosahedral nodes lie on the kernel of $J$; the extra point $p_*=R^T(1,1,1)^T/\sqrt3$ does not, and $J(q_*)=\sqrt5/(3\sqrt3)$.

On the nine-dimensional space of degree-at-most-two harmonics, the equal-weight twelve-node rule reconstructs every field, and $A$ transports that decoder to the ellipsoid with operator $\nabla\cdot(D\nabla\cdot)$ and measure $\nu=A_*\sigma$. Ordinary harmonicity is not transported. Degree five is exact for $\nu$; no real weights on those twelve nodes have degree six, because $J^2$ integrates to $1/21$ and vanishes at the nodes.

On the ten-dimensional space $\mathcal F_A=\mathcal H_A\oplus\mathbb R J$, the boxed thirteen-sample formula recovers every field. The twelve aligned samples alone do not. The original PDF is the proof.

Let $S_{12}$ evaluate fields in $\mathcal F_A$ at the twelve transported nodes. Then

$\ker S_{12}=\mathbb R J,\qquad \operatorname{rank} S_{12}=9,\qquad \ker S_{13}=\{0\},\qquad \operatorname{rank} S_{13}=10.$

Thirteen measurements describe this particular decoder, not the minimum possible sample count. Nine suitably chosen original rows, together with the extra point, already give ten independent measurements. The symmetric twelve-node formula does not require selecting a spanning subset. Why a positive degree-five rule makes sampling injective on $\mathcal H_{\le 2}$, and why $J$ and $J^2$ are different obstructions, is recorded in the exact field interface. That note does not change this decoder.

A nine-node selection determines every quadratic harmonic field if and only if it includes at least one node from every antipodal pair. The even interpolants $E_j$ satisfy $E_j(\pm a_i)=\delta_{ij}$. Omit a pair, and that $E_j$ is a nonzero kernel vector. Meet every pair, and three complete-pair differences recover the linear part because every triple of distinct icosahedral axes has Gram determinant $1-3/5\pm 2/(5\sqrt5)>0$.

Successful selections: $\binom63 2^3=160$. Failing selections: $6\cdot10=60$, rank eight on $\mathcal H_{\le2}$.

One such successful selection is written out as an explicit minimal ten-sample decoder: the three complete pairs $\pm a_1,\pm a_2,\pm a_3$, the three singles $a_4,a_5,a_6$, and $q_*$. Those labels are the note’s aligned coordinates, not this site’s browser node numbers.

After a failing selection that omits pair $j$, write $\widetilde E_j(y)=E_j(A^{-1}y)$ with

$E_j(x)=\tfrac16+\tfrac54\bigl((x\cdot a_j)^2-|x|^2/3\bigr).$

The nine measurements then have kernel $\operatorname{span}\{\widetilde E_j,J\}$. The extra sample imposes one equation. Since $J(q_*)\ne0$,

$\ker S_{I,*}=\operatorname{span}\Bigl\{\widetilde E_j-\frac{E_j(p_*)}{J(q_*)}J\Bigr\}.$

The first term is a function of transported coordinates. The numerator remains $E_j(p_*)$ because $A^{-1}q_*=p_*$. For the specified point, $J(q_*)=\sqrt5/(3\sqrt3)$, both nonzero. Neither the pure even mode nor the pure cubic remains invisible; one linear combination does.

With $K=(2M-I)/\sqrt5$,

$\Psi^2=\Phi_{\tau^2,\,2\log\varphi}.$

This is a compatibility lemma with the family on /extension. $\Psi$ itself is not a $\Phi_{c,\eta}$: the planar block has determinant $-\tau^2$, and the odd step flips the sign of $z$. The identity lives in REMARKS.md; the extension page points here.

• Boundary_Reconstruction_Transport_Formal_Proof.pdf — 344,230 bytes, SHA-256 a8c9e39e29f283272b3c7dc2933c7dad1a02bc2651ce0341508317de3d295571Copy
• Boundary_Reconstruction_Transport_Formal_Proof.tex — 27,553 bytes, SHA-256 fcaadf4b55fc24fae1c8cfb1d8839d22a95eea27d3124fbe19002c50cce6c89cCopy. Chat paste pasted_tex_as_received.txt (27,552 bytes, SHA-256 7db0d779b8e975fcf9a02f37d6d0895beedaf7f9049b5272afe7e6e3e04a3ba4Copy) omitted the leading backslash of \documentclass.
• REMARKS.md — 7,927 bytes, SHA-256 9c217a5e88f1ff04d0d8ec4bb5f4bf9862640258bb1112f5a2d774f72e12fe9bCopy. Two-step identity, interpolants, 160/60 count, transported surviving kernel, exact $J^2$ mean.
• verify_boundary_reconstruction.py — revision 2026-09-22.r2, 10,661 bytes, SHA-256 ce3e5ea29ca8de60e27e45164903065f340e54826935f88045b0ed715b5e3efeCopy. Supersedes acc774f6….
• HOSTING.md, README.md, ENVIRONMENT.json, SHA256SUMS.txt

Python 3.10.21, NumPy 2.2.6, captured 2026-09-22T12:35:49Z. Both commands below exit 0 with empty stderr and identical stdout (897 bytes, SHA-256 60ad47fa29f0c7ada869555a393e497411f9c8b08344b2850b71c84a840890b1Copy). That transcript is also produced by the superseded vacuous checker; ENVIRONMENT.json binds it to program hash ce3e5ea29ca8de60e27e45164903065f340e54826935f88045b0ed715b5e3efeCopy. TOTAL 30 PASS

python3 verify_boundary_reconstruction.py
python3 -O verify_boundary_reconstruction.py

Transcripts: this_archive_verify.stdout.txt, this_archive_verify.optimized.stdout.txt.

Temporary copies of the checker, not the accepted source, with expected mean $1/22$ and with the $xyz$ coefficient of $J$ changed from 1 to 2, each exit 1 in both normal and -O modes with FAIL J2_spherical_mean_one_twenty_first and no TOTAL 30 PASS.

Not an additional source paper, not peer review, not a proof-assistant certificate, not a joining theorem of Papers I–VII, not a 13-point degree-six cubature, not a blowup result, and not an enlargement of Paper IV, Paper VI, paper 8, or /extension. Alignment is required for $J(n_i)=0$. Integer reversal recovers only finitely encoded records. Related: geometry, cubature, reproduce.