SIGILLUM — Forty-State Spectral Computer

SIGILLUM · forty-state spectral wheel

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OPEN LIVE INSTRUMENT

Forty-state wheel. Affine map $F_s(j)=3j+s\pmod{40}$. Orbit projector $\Pi$. Laplacian contraction $T_\varepsilon^n\to\Pi$.

The Dee plate below remains provenance, not the machine.

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SIGILLUM · FORTY-STATE SPECTRAL COMPUTER

How does a finite wheel compute a kernel?

A working digital Sigillum Dei Aemeth: the affine map $F_s(j)=3j+s\pmod{40}$, exact orbit projector $\Pi$, Laplacian contraction $T_\varepsilon^n\to\Pi$, and the 47-projector Enochian heptagram.

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Sigillum Dei Aemeth — historical plate (provenance, not the machine)

SIGILLUM · live spectral wheel · City portal instrument

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LIVE INTERFACE

The instrument above runs in place. Wheel, Heptagram, and Proof tabs. Select a node, change offset $s$, run Orbit ($P\cdot e_j$) or Contract ($T_\varepsilon^n\to\Pi$). Certificates update with the operator.

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Research mandate

SIGILLUM is the Mini AI Labs geometric-computation instrument for the forty-state wheel. It sits with The Cube — E47 Geometric Computation as a selected system: The Cube interrogates the 125-state carrier; SIGILLUM interrogates the 40-state seal.

$$ V=\{0,\dots,39\}\;\longrightarrow\;F_s(j)=3j+s\pmod{40}\;\longrightarrow\;P_s\;\longrightarrow\;\Pi=(I+P+P^2+P^3)/4\;\longrightarrow\;T_\varepsilon^n\to\Pi $$

Mathematical spine

Carrier: forty indexed positions $V=\{0,1,\dots,39\}$.
Map: $F_s(j)=3j+s\pmod{40}$. Because $\gcd(3,40)=1$, $F_s$ is a bijection and $P_s$ is a permutation matrix.
Period: $3^4\equiv 1\pmod{40}$, so $P^4=I$.
Orbit projector: $\Pi=(I+P+P^2+P^3)/4$, $\Pi^2=\Pi=\Pi^\dagger$, rank equal to the number of cycles.
Laplacian: $L=I-(P+P^T)/2$, self-adjoint, $L\Pi=0$.
Contraction: $\varepsilon=2/3$, $T_\varepsilon=I-\varepsilon L$. Positive modes contract by $1/3$ and $-1/3$, so $T_\varepsilon^n\to\Pi$.
Heptagram: 25 Seniors ($E_6$, $\lambda=6$) + 22 Keys/Aethyrs ($E_{30}$, $\lambda=30$) = 47 projectors. Displayed as a typed catalogue, not as a historical identity with $\ker K$.

Machine certificates

For $s=1$: ten 4-cycles, $\operatorname{rank}\Pi=10$, $\sigma(L)=\{0^{(10)},1^{(20)},2^{(10)}\}$.

For $s=2$: cycle type $1,1,2,2,2,4^{8}$, $\operatorname{rank}\Pi=13$, $\sigma(L)=\{0^{(13)},1^{(16)},2^{(11)}\}$.