The Kepler Conjecture 2025, 46 - Local Optimality
A contravening decomposition star D with the set of sphere packing vectors within a radius of 2t0 U(D), and assuming U(D) has 12 elements, and there being a bijection φ between U(D) and kissing arrangement Ufcc of 12 tangent unit balls in the fcc configuration, or a bijection with Uhcp constructed similarly. v, w ∈ U(D), |w - v| ≤ 2t0 iff |φ(w) - φ(v)| = 2. The proximity graph of U(D) is th same as the contact graph of Ufcc or Uhcp. It follows that σ(D) ≤ 8pt. Equality holds iff U is either Ufcc or Uhcp. This is the local optimality Theorem.
A quad cluster scores at most 0. It's exactly 0 if its corners have height 2, forming a square with sides 2. Other standard clusters have strictly negative scores. A constant pt is equal to σ(S) with S a regular tetrahedron with edges of length 2. pt = 4arctan(√2/5) - π/3 ≈ 0.05537. A quasi-regular tetrahedron S satisfies σ(S) ≤ 1pt. Equality, again, only for edge length 2. A simpliex S with faces having exclusively circumradii of at most √2 and not a quasi-regular tetrahedron or quarter, then s-vor(S) < 0.
A triangle T with circumradius < √2 and edges that stay clear of a barrier in B, then T doesn't overlap any barrier in B. T = {u, v, w} are a set of vertices and circumradius < √2, and one of the edges {v, w} passing through a barrier b in B, then {v, w} has length between 2t0 and √8. The vertex u is a vertex of b. One of the endpoints of the vertex forms a 4-tuple simplex in Q with the barrier. A vertex of height of at most 2t0 such that 0, v1, v2, v3 are distinct vertices, which are outside Q0, and {0, v2, v3} is a barrier, then the open cone at origin over B(0, √2) ∩ B(v1, √2) doesn't meet the closed cone C at origin over the other two vertices. Define an enlarged set of simplices Q0' as the set of simplices S with a vertex at origin so that either S ∈ Q0 or S is a simplex with a vertex at origin, and with circumradius < √2 so that no edge passes through a barrier. The simplices in Q0' don't overlap one another.
For a vertex v of height at most 2.36, define C(v) as the cone at origin generated by B(0, √2) ∩ B(v1, √2). C'(v) is a subset of C(v) with x ∈ C(v), so that for x in C(v), x is closer to 0 than v, part of B(0, √2), and doesn't lie in the cone over any simplex in Q0. Non-zero vertices u with face {0, u, v} is a barrier or has circumradius √2, and none of the edges of this face passes through a barrier, then x and v lie in the same half-space bounded by the plane perp. to {0, u, v}, passing through 0 and the circumcenter of them. For every 4-simplex in Q0 including the 0-vertex, the segment of x and one vertex doesn't cross through the cone C spanned with the other two vertices and the origin.
If x ∈ VC(0), 0 < |x| ≤ 1.18, and a point at distance 1.18 from 0 along the ray (0, x) not in VC(0), and if x is not in the cone over any simplex of Q0, then there is some v with x ∈ C'(v), |v| ≤ 2.36. If |u| ≤ 2.36 and |v| ≤ 2.36, C'(u), C'(v) don't overlap for u ≠ v. If Ω(0) ∩ Ω(v), which is the face of the Voronoi cell of Ω(0) associated with the vertex v. If F'(v) is part of F(v) ∩ B(0, 1.18), outside the cone over any simplex in Q0, and H(v) the closure of the union of segments from the origin to points of F'(v). If C''(v) is the cone at origin, spanned by B(0, 1.18) ∩ B(v, 1.18), then the closure C'(v) ∩ C''(v) is exactly H(v). For x ∈ VC(0), 0 < |x| ≤ 1.18 the point 1.18 away from 0 along the ray (0, x) is not in VC(0), and if x is not in the cone over any simplex of Q0, then ∃ v: x∈ C'(v), and |v| ≤ 2.36.