Pith. sign in
theorem

flatAngleT13_eq

proved
show as:
module
IndisputableMonolith.Gravity.Analysis.ReggeHinge4DStarKernel13
domain
Gravity
line
166 · github
papers citing
none yet

plain-language theorem explainer

The flat dihedral angle on every type-(1,3) star simplex equals π/3. Anyone assembling the six-simplex Freudenthal star deficit for the absolute masks {0,1,15} cites this identity. The proof is a one-line wrapper of the elementary evaluation arccos(1/2)=π/3.

Claim. The flat angle attached to the type-$(1,3)$ hinge equals $\pi/3$: $\arccos(1/2)=\pi/3$.

background

The module builds the Regge 4D type-(1,3) periodic-lattice star deficit kernel. The hinge is the triangle with absolute masks ${0,e_0,e_0+e_1+e_2+e_3}={0,1,15}$ (difference masks $(1,14)$, local flat squared lengths $(1,3,4)$). Exactly six Kuhn simplices in the origin unit cube contain the hinge; each carries the same flat Gram data after reordering so the hinge occupies slots $(0,1,2)$.

The shared flat cosine is $1/2$, obtained from the Gram projection of the local squared-edge vector t13FlatSqEdges. The angle itself is therefore defined as $\arccos(1/2)$. The elementary identity $\arccos(1/2)=\pi/3$ is already proved upstream as arccos_one_half via $\cos(\pi/3)=1/2$ and the range restriction of arccos.

proof idea

One-line wrapper: the definition of the flat angle is exactly $\arccos(1/2)$, so the claim is definitionally identical to the already-proved lemma arccos_one_half. No further rewriting is required.

why it matters

This identity is the arithmetic step inside the flatness gate. Downstream, star_flat_angle_sum_two_pi rewrites the six-fold star sum as $6\cdot(\pi/3)=2\pi$ by a single simp that invokes the present equality, then closes by ring. That gate is deliverable A.3 of the module: the type-(1,3) star is exactly flat, so the deficit vanishes on the lattice background.

The result sits inside the QG full-theory campaign after the type-(1,1) seed orbit and the 15-class orbit classification. It does not yet transport to the complementary type $(3,1)$, nor does it assemble the full flat Hessian or recover Einstein–Hilbert weights. It is a pure geometric evaluation needed before the signed deficit-class kernel on classes $(1,3,5,7,9,11,13)$ can be stated.

Switch to Lean above to see the machine-checked source, dependencies, and usage graph.