Schläfli classification of regular polytopes
schlafli_classification
Submitter: Kim Morrison.
Notes: §42 of Knill's 'Some Fundamental Theorems in Mathematics'. Schläfli's dimension-by-dimension enumeration of the regular convex polytopes: p_3 = 5 (Euclid XIII — the five Platonic solids), p_4 = 6 (Schläfli's six regular 4-polytopes), and p_d = 3 for every d ≥ 5 (regular simplex, hypercube, cross-polytope). Companion to the broader Platonic classification problem, which additionally records p_2 = ∞. Mathlib has convexHull, extremePoints, IsExposed, vectorSpan, AffineIsometryEquiv, Set.encard but no convex-polytope datatype, no face lattice, no regular-polytope concept, and none of the classification counts. The Challenge ships ~1.5 pages of helper defs (ConvexPolytope, dim, IsFace, Flag, isSymmetry, IsRegular, Similar, regularPolytopes, regularSimilar, platonicCount).
Source: L. Schläfli, *Theorie der vielfachen Kontinuität* (1852, posthumously published 1901; first proof that p_3 = 5, p_4 = 6, and p_d = 3 for d ≥ 5). Listed as §42 in O. Knill, *Some Fundamental Theorems in Mathematics* (https://people.math.harvard.edu/~knill/graphgeometry/papers/fundamental.pdf).
Informal solution: p_3 = 5: any regular 3-polytope has Schläfli symbol {p, q} with p, q ≥ 3 and 1/p + 1/q > 1/2 (the spherical-angle-sum constraint at each vertex), giving exactly (3,3), (4,3), (3,4), (5,3), (3,5) — the five Platonic solids. p_4 = 6: any regular 4-polytope {p, q, r} satisfies the analogous spherical inequality, giving (3,3,3), (4,3,3), (3,3,4), (3,4,3), (5,3,3), (3,3,5) — Schläfli's six. p_d ≥ 5 = 3: in dim ≥ 5 the only solutions to the iterated angle constraint are the regular simplex {3, …, 3, 3}, hypercube {4, 3, …, 3}, and cross-polytope {3, …, 3, 4}. Existence of each candidate is by explicit construction (vertex coordinates); uniqueness within Schläfli class is by induction on dimension via vertex figures.
theorem schlafli_classification :
platonicCount 3 = 5 ∧
platonicCount 4 = 6 ∧
∀ d, 5 ≤ d → platonicCount d = 3 := ⊢ platonicCount 3 = 5 ∧ platonicCount 4 = 6 ∧ ∀ (d : ℕ), 5 ≤ d → platonicCount d = 3
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