Q3_self_dual_vertex_count
plain-language theorem explainer
The binary cube in three dimensions has exactly eight vertices, matching the self-dual property where vertex count equals the number of 3-cells. Researchers deriving the Standard Model gauge group and fermion content from the Q3 cube in Recognition Science would cite this counting result. The proof is a one-line normalization that applies the definition of vertex count directly.
Claim. Let $V(D)$ denote the number of vertices of the $D$-dimensional binary cube. Then $V(3) = 2^3$.
background
The SpectralEmergence module starts from the forced spatial dimension D=3 (T8) to obtain the binary cube Q3 with 8 vertices. The referenced definition states that V(D) equals 2^D, the vertex count of the hypercube. This supplies the 8-dimensional vertex space on which the automorphism group B3 = S3 ⋉ (Z/2Z)^3 acts, decomposing into the color, weak, hypercharge, and conjugate sectors whose dimensions sum to 8.
proof idea
One-line wrapper that applies norm_num to the definition of V, confirming the numerical identity for D=3.
why it matters
This equality anchors the vertex space for the B3 decomposition that produces the SU(3) x SU(2) x U(1) gauge content and the 48 chiral fermion states. It directly implements the self-dual combinatorics noted in the declaration and connects T8 (D=3) to the 8-tick octave and the |Aut(Q3)| = 48 count used for fermionic degrees of freedom. The module treats it as the starting point for deriving three generations from face pairs and the phi-ladder mass formula.
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