# Explicit ten-sample reconstruction with an invariant cubic

23 September 2026. A calculation on the field space already specified in the conversation. No website changes.

**Result.** Nine specified icosahedral samples and one additional boundary sample reconstruct every field in
\[
\mathcal F=\mathcal H_{\le2}\oplus\mathbb RJ,
\qquad J(x,y,z)=z(x^2+xy-y^2).
\]
The formulas below give the decoder directly. Ten scalar point evaluations are minimal on this ten-dimensional space. The same measurements transport to the ellipsoid under the stated invariant map.

## 1. Coordinates and the ten points

Let
\[
\varphi=\frac{1+\sqrt5}{2},\quad \varphi'=1-\varphi,
\quad r=\sqrt{1+\varphi^2},\quad r_-=\sqrt{1+(\varphi')^2},
\]
and define the orthogonal matrix
\[
R=\begin{pmatrix}
1/r&\varphi/r&0\\
1/r_-&\varphi'/r_-&0\\
0&0&1
\end{pmatrix}.
\]
Write \(\xi=(X,Y,Z)^\mathsf T=Rx\), and set \(\widehat f(\xi)=f(R^\mathsf T\xi)\). Since \(1+\varphi\varphi'=0\) and \(rr_-=\sqrt5\),
\[
RR^\mathsf T=I,\qquad J(x)=\sqrt5\,XYZ.
\]
This is an orthogonal alignment; a claim that \(R\) has determinant positive one is not needed.

Use the following six unit axis representatives in aligned coordinates:
\[
\begin{aligned}
a_1&=r^{-1}(0,1,\varphi),&a_2&=r^{-1}(1,\varphi,0),&a_3&=r^{-1}(\varphi,0,1),\\
a_4&=r^{-1}(0,1,-\varphi),&a_5&=r^{-1}(1,-\varphi,0),&a_6&=r^{-1}(\varphi,0,-1).
\end{aligned}
\]
The twelve points \(\{\pm a_j:1\le j\le6\}\) form the standard unit icosahedron. Direct calculation gives
\[
\|a_j\|=1,\qquad (a_i\cdot a_j)^2=\frac15\quad(i\ne j).
\]

Take the nine icosahedral samples at
\[
\boxed{a_1,-a_1,a_2,-a_2,a_3,-a_3,a_4,a_5,a_6}
\]
and the tenth sample at
\[
\xi_*=(1,1,1)/\sqrt3.
\]
The physical sample points are \(p_{j,\pm}=\pm R^\mathsf Ta_j\) for the three complete pairs, \(p_j=R^\mathsf Ta_j\) for the three singles, and \(p_*=R^\mathsf T\xi_*\). All ten points lie on the unit sphere.

Record the values in this fixed order:
\[
v_{1,+},v_{1,-},v_{2,+},v_{2,-},v_{3,+},v_{3,-},v_4,v_5,v_6,v_*.
\]
Here \(v_{j,\pm}=\widehat f(\pm a_j)\), \(v_j=\widehat f(a_j)\), and \(v_*=\widehat f(\xi_*)\). These are coordinate-defined labels, not an assertion about the browser tool's numeric node ordering.

At the first nine points, \(XYZ=0\). At the extra point,
\[
J(p_*)=\sqrt5\,X_*Y_*Z_*=\frac{\sqrt5}{3\sqrt3}\ne0.
\]

## 2. Recover the linear part

Every \(u\in\mathcal H_{\le2}\) has a unique decomposition
\[
\widehat u(\xi)=e(\xi)+\ell\cdot\xi,
\]
where \(e\) is an even harmonic polynomial of degree at most two, and \(\ell\in\mathbb R^3\). The cubic contributes zero to every icosahedral sample.

Set
\[
d_j=\frac{v_{j,+}-v_{j,-}}2,\qquad j=1,2,3.
\]
Then \(d_j=\ell\cdot a_j\). Thus, with
\[
B=\frac1r\begin{pmatrix}0&1&\varphi\\1&\varphi&0\\\varphi&0&1\end{pmatrix},
\qquad d=(d_1,d_2,d_3)^\mathsf T,
\]
we have \(B\ell=d\). Its determinant is
\[
\det B=-\frac{1+\varphi^3}{r^3}=-\frac{2\varphi^2}{r^3}\ne0.
\]
Consequently,
\[
\boxed{
\ell=\frac{r}{2\varphi^2}
\begin{pmatrix}
-\varphi&1&\varphi^2\\
1&\varphi^2&-\varphi\\
\varphi^2&-\varphi&1
\end{pmatrix}
\begin{pmatrix}d_1\\d_2\\d_3\end{pmatrix}.}
\tag{1}
\]
This is an explicit inverse, with no numerical matrix solve.

## 3. Recover the even part

For the complete pairs, define
\[
m_j=\frac{v_{j,+}+v_{j,-}}2\quad(j=1,2,3).
\]
For the singles, define
\[
m_j=v_j-\ell\cdot a_j\quad(j=4,5,6).
\]
All six numbers are now \(m_j=e(a_j)\).

Define
\[
\boxed{E_j(\xi)=\frac16+\frac54
\left((\xi\cdot a_j)^2-\frac{\|\xi\|^2}{3}\right).}
\tag{2}
\]
Each \(E_j\) is even and harmonic, because
\[
\Delta(\xi\cdot a_j)^2=2\|a_j\|^2=2,
\qquad \Delta\frac{\|\xi\|^2}{3}=2.
\]
Moreover, the inner-product identities give
\[
E_j(\pm a_i)=\delta_{ij}.
\]
The space of even harmonics of degree at most two has dimension six: one constant and five traceless quadratic components. The six functions \(E_j\) are linearly independent by their evaluation values, and hence form a basis. It follows that
\[
\boxed{U(\xi)=\ell\cdot\xi+\sum_{j=1}^6m_jE_j(\xi)=\widehat u(\xi).}
\tag{3}
\]
This proves that the nine chosen icosahedral values determine the entire harmonic quadratic part.

For an explicit coefficient representation,
\[
U(\xi)=c_0+\ell\cdot\xi+\xi^\mathsf TC\xi,
\]
where
\[
c_0=\frac16\sum_{j=1}^6m_j,
\qquad
C=\frac54\sum_{j=1}^6m_j\left(a_ja_j^\mathsf T-\frac13I\right),
\qquad \operatorname{tr}C=0.
\]

## 4. Recover the cubic and the complete field

At the additional point,
\[
v_*=U(\xi_*)+\beta\frac{\sqrt5}{3\sqrt3}.
\]
Therefore
\[
\boxed{\beta=\frac{3\sqrt3}{\sqrt5}\bigl(v_*-U(\xi_*)\bigr),}
\tag{4}
\]
and
\[
\boxed{\widehat f(\xi)=U(\xi)+3\sqrt3\,XYZ\bigl(v_*-U(\xi_*)\bigr).}
\tag{5}
\]
Equivalently, in the original coordinates,
\[
\boxed{f(x)=U(Rx)+\frac{J(x)}{J(p_*)}\bigl(v_*-U(\xi_*)\bigr).}
\tag{6}
\]

The extra-point evaluation can be simplified without substituting all coordinates:
\[
E_j(\xi_*)=
\begin{cases}
\varphi/3,&j=1,2,3,\\
\varphi'/3,&j=4,5,6.
\end{cases}
\]
Thus
\[
\boxed{U(\xi_*)=\frac{\ell_X+\ell_Y+\ell_Z}{\sqrt3}
+\frac\varphi3(m_1+m_2+m_3)
+\frac{\varphi'}3(m_4+m_5+m_6).}
\tag{7}
\]
For example, for the first three axes,
\((\xi_*\cdot a_j)^2=(1+2/\sqrt5)/3\); for the other three it is \((1-2/\sqrt5)/3\). Substitution into (2) proves (7).

## 5. A single cardinal reconstruction formula

The staged calculation above is already an explicit decoder. It also yields ten cardinal functions.

Put
\[
w(\xi)=\xi-\sum_{k=4}^6a_kE_k(\xi),
\qquad t(\xi)=B^{-\mathsf T}w(\xi).
\]
Define the nine quadratic functions
\[
L_{j,\pm}(\xi)=\frac12\bigl(E_j(\xi)\pm t_j(\xi)\bigr)
\quad(j=1,2,3),
\qquad L_k(\xi)=E_k(\xi)\quad(k=4,5,6).
\]
Writing \(\mathcal I=\{(1,+),(1,-),(2,+),(2,-),(3,+),(3,-),4,5,6\}\), equation (3) becomes
\[
U(\xi)=\sum_{i\in\mathcal I}v_iL_i(\xi).
\]
Indeed, the linear contribution after correcting the single samples is
\(\ell\cdot w=d^\mathsf TB^{-\mathsf T}w=d\cdot t\), which gives exactly the paired coefficients above.

Set
\[
C_*(\xi)=3\sqrt3\,XYZ,
\quad W_*(\xi)=C_*(\xi),
\quad W_i(\xi)=L_i(\xi)-L_i(\xi_*)C_*(\xi).
\]
Then
\[
\boxed{\widehat f(\xi)=\sum_{i\in\mathcal I}v_iW_i(\xi)+v_*W_*(\xi).}
\tag{8}
\]
The ten functions lie in the declared ten-dimensional field space and satisfy
\[
W_i(\xi_j)=\delta_{ij}
\]
at the ten sample points, including the additional point. Equation (8) is the requested ten-sample interpolation formula.

## 6. Uniqueness, minimality, and retained consistency checks

The derivation shows uniqueness constructively: the paired differences determine \(\ell\); the six even values determine \(e\); the additional sample determines \(\beta\). Equivalently, the ten cardinal functions prove that the evaluation map is an isomorphism onto \(\mathbb R^{10}\).

The field space has dimension \(9+1=10\), since the nonzero cubic \(J\) cannot belong to the degree-at-most-two subspace. Any collection of fewer than ten scalar point evaluations is a linear map from a ten-dimensional space to a space of smaller dimension. Rank-nullity gives a nonzero kernel. Such measurements cannot identify every field. Thus ten is minimal for scalar point sampling on this space.

This reduction has a practical cost. Every vector of ten real readings produces exactly one field in the model; these ten readings alone contain no surplus linear consistency equation. The original thirteen-sample arrangement has rank ten and therefore three independent linear consistency equations. It can detect sample vectors that violate those equations. This statement does not assert an error-correction guarantee for the thirteen samples.

## 7. Transport to the ellipsoid

Let
\[
A=\begin{pmatrix}0&\tau&0\\\tau&\tau&0\\0&0&-\tau^{-2}\end{pmatrix},
\qquad \tau>0,
\]
and let \(g(y)=f(A^{-1}y)\). The physical measurement points become \(q_i=Ap_i\), including \(q_*=Ap_*\), and lie on the boundary of \(AB^3\). They retain their values:
\[
g(q_i)=f(p_i)=v_i.
\]
The invariant identity \(J(Ax)=J(x)\) gives \(J(A^{-1}y)=J(y)\), and hence
\[
\boxed{g(y)=U(RA^{-1}y)+\frac{J(y)}{J(q_*)}
\bigl(v_*-U(\xi_*)\bigr),\qquad J(q_*)=\frac{\sqrt5}{3\sqrt3}.}
\tag{9}
\]
The quadratic part lies in the transported space \(\mathcal H_A\), with operator \(\nabla\cdot(AA^\mathsf T\nabla)\). The complete field lies in \(\mathcal H_A\oplus\mathbb RJ\). Composition with the invertible map preserves the evaluation ranks and the ten-sample minimality result.

## 8. Worked example and executable verification

For the physical-coordinate field
\[
f(x,y,z)=2+3x+x^2-y^2+4z(x^2+xy-y^2),
\]
the decoder returns \(\beta=4\). At \(x_0=(1/5,-3/10,1/3)\),
\[
J(x_0)=-\frac{11}{300},\qquad
u(x_0)=\frac{51}{20},\qquad
\boxed{f(x_0)=\frac{721}{300}.}
\]

`verify_ten_sample.py` checks the formulas with rational coefficients in
\[
\mathbb Q(\sqrt5,r,\sqrt3),\qquad r^2=(5+\sqrt5)/2.
\]
It requires only the Python standard library. The equality decisions are exact and use explicit exceptions, including under optimized execution. It checks both directions of the matrix inverse, all even-cardinal identities, all 100 ten-point cardinal identities, all ten polynomial reconstruction identities, agreement of the staged and closed cardinal decoders, the alignment, the worked example, two exact finite transport cases, and rejection of an incorrect sample count.

Run:

```sh
python verify_ten_sample.py
python -O verify_ten_sample.py
```

The general transport result and minimality are established by the written arguments above; the two rational transport executions are finite supporting checks. `verification.txt` records the normal execution, and `verification_optimized.txt` records the optimized execution.

**Recorded execution.** Python 3.12.14; normal and optimized executions both reported `TOTAL 13 PASS`. Their outputs are byte-identical (1186 bytes).

- Checker SHA-256: `cc850bda469cb33cc9d37c307dceede7e1b4e5ede002957c3ff7497c7a388c0f`.
- Each execution transcript SHA-256: `a6e04971faf30e665f00fc1b60a16488dddce315a83d715291b5298031268be0`.