Explicit ten-sample reconstruction

Let $\varphi=(1+\sqrt5)/2$, $\varphi'=1-\varphi$, $r=\sqrt{1+\varphi^2}$, and $r_-=\sqrt{1+(\varphi')^2}$. The orthogonal alignment is

Write $\xi=Rx$ and $\widehat f(\xi)=f(R^{\mathsf T}\xi)$. Then $RR^{\mathsf T}=I$ and $J(x)=\sqrt5\,XYZ$. A claim that $\det R=1$ is not needed.

The six unit axes in aligned coordinates are $a_1=r^{-1}(0,1,\varphi)$, $a_2=r^{-1}(1,\varphi,0)$, $a_3=r^{-1}(\varphi,0,1)$, $a_4=r^{-1}(0,1,-\varphi)$, $a_5=r^{-1}(1,-\varphi,0)$, $a_6=r^{-1}(\varphi,0,-1)$. They satisfy $|a_j|=1$ and $(a_i\cdot a_j)^2=1/5$ for $i\neq j$.

The nine icosahedral samples are at $a_1,-a_1,a_2,-a_2,a_3,-a_3,a_4,a_5,a_6$. The tenth is at $\xi_*=(1,1,1)/\sqrt3$. Physical points are $p=R^{\mathsf T}\xi$. All ten lie on the unit sphere. Record the values in the fixed order $v_{1,+},v_{1,-},v_{2,+},v_{2,-},v_{3,+},v_{3,-},v_4,v_5,v_6,v_*$. At the first nine points, $XYZ=0$. At the extra point, $J(p_*)=\sqrt5/(3\sqrt3)\neq 0$.

Every $u\in\mathcal H_{\le 2}$ decomposes as $\widehat u(\xi)=e(\xi)+\ell\cdot\xi$, with $e$ even and harmonic of degree at most two. The cubic contributes zero at every icosahedral sample. The paired differences $d_j=(v_{j,+}-v_{j,-})/2$ for $j=1,2,3$ satisfy $d_j=\ell\cdot a_j$. With

the inverse is explicit:

Set $m_j=(v_{j,+}+v_{j,-})/2$ for the three complete pairs, and $m_j=v_j-\ell\cdot a_j$ for the three singles. Then $m_j=e(a_j)$. The even cardinals

are harmonic and satisfy $E_j(\pm a_i)=\delta_{ij}$. They form a basis of the six-dimensional even space, so

Equivalently, $c_0=(1/6)\sum m_j$ and $C=(5/4)\sum m_j(a_ja_j^{\mathsf T}-I/3)$, with $\operatorname{tr}C=0$.

The extra sample gives $v_*=U(\xi_*)+\beta\sqrt5/(3\sqrt3)$, hence

In physical coordinates,

The extra-point values simplify to $E_j(\xi_*)=\varphi/3$ for $j=1,2,3$ and $\varphi'/3$ for $j=4,5,6$, so

Put $w(\xi)=\xi-\sum_{k=4}^6 a_k E_k(\xi)$ and $t(\xi)=B^{-\mathsf T}w(\xi)$. The nine quadratic cardinals are $L_{j,\pm}=(E_j\pm t_j)/2$ for $j=1,2,3$ and $L_k=E_k$ for $k=4,5,6$. Then $U=\sum_{i\in\mathcal I} v_i L_i$. With $C_*(\xi)=3\sqrt3\,XYZ$, $W_*=C_*$, and $W_i=L_i-L_i(\xi_*)C_*$,

These ten functions lie in $\mathcal F$ and satisfy $W_i(\xi_j)=\delta_{ij}$ at the ten sample points, including the additional point.

This callout is an explanatory supplement. It is not part of the hosted Markdown file. Let $p_1,\ldots,p_{10}$ be the ordered physical sample points, and let $\mathcal W_i(x)=W_i(Rx)$ be the physical-coordinate cardinal functions.

The cardinal formula is the decoder $S^{-1}$. For the invariant matrix $A$ in the transport section, set $\mathcal F_A=\{f\circ A^{-1}:f\in\mathcal F\}$ and $U_A f=f\circ A^{-1}$. Let $S_A$ evaluate $\mathcal F_A$ at $Ap_1,\ldots,Ap_{10}$. Then

Here $D_A$ is the decoding operator. The elliptic diffusion matrix, when displayed, is $AA^{\mathsf T}$, not $D$.

The derivation is constructive: paired differences determine $\ell$, the six even values determine $e$, and the additional sample determines $\beta$. The ten cardinal functions prove that the evaluation map is an isomorphism onto $\mathbb R^{10}$.

The space has dimension 10. Any collection of fewer than ten scalar point evaluations has a nonzero kernel, so ten is minimal for scalar point sampling on this space. Every vector of ten real readings produces exactly one field in the model. These ten readings contain no surplus linear consistency equation.

The existing thirteen-sample arrangement has rank ten and therefore three independent linear consistency equations. It can detect sample vectors that violate those equations. That is not an error-correction guarantee.

Let $g(y)=f(A^{-1}y)$ with

The physical points $q_i=Ap_i$, including $q_*=Ap_*$, lie on the boundary of $AB^3$ and retain the values $g(q_i)=v_i$. Because $J(Ax)=J(x)$,

The quadratic part lies in the transported space $\mathcal H_A$, with operator $\nabla\cdot(AA^{\mathsf T}\nabla)$. The complete field lies in $\mathcal H_A\oplus\mathbb R J$. Composition with the invertible map preserves the evaluation ranks and the ten-sample minimality result. The general transport statement is the written argument. The checker’s two rational executions, $\tau=1/2$ and $\tau=2/3$, are finite supporting checks.

For $f=2+3x+x^2-y^2+4J$ at $x_0=(1/5,-3/10,1/3)$,

The boundary-field tool still uses its twelve- and thirteen-sample calculation. This page does not add a ten-sample mode to that tool.

The two supplied transcripts are byte-identical. Their normal and optimized names record provenance, not different stdout. The author record names Python 3.12.14. This archive replayed the same checker with Python 3.10.21 at 2026-09-23T23:53:35Z. Both processes exited 0, stderr was empty, and both stdout files match a6e04971faf30e665f00fc1b60a16488dddce315a83d715291b5298031268be0Copy.

Fresh evidence, kept separate from the supplied transcripts: execution.json, normal.stdout.txt, optimized.stdout.txt.

Commands: python3 verify_ten_sample.py and python3 -O verify_ten_sample.py. Standard library only. The thirteen groups include all 100 cardinal identities, all ten polynomial reconstructions, agreement of the staged and closed decoders, the alignment, the worked example, and the two finite transport checks.

Not an additional counted source paper. Not peer review. Not independent third-party reproduction. Not arbitrary cubic or quintic reconstruction. Not a new positive cubature rule. Not an enlargement of Paper VI, Paper VII, or paper 8. Frozen G₆ is unchanged. Two rational transport executions do not prove the identity for every $\tau>0$; that statement is the written argument.