Toward the sharp Euclidean five-distance theorem via Haynes–Marklof Theorem 8 #
The repo proves g₂ ≤ 6 (gap_count_euclidean) and reduces the sharp g₂ ≤ 5 to the planar
packing
count FiveDistance.card_le_five_of_pairwise (≤ 5 vectors pairwise > π/3). A literature pass
(Haynes–Marklof, A five distance theorem for Kronecker sequences, IMRN 2022, arXiv:2009.08444)
established that the upper bound 5 is elementary (their Sections 3–5; dynamics enters only for
sharpness), but the step from 6 to 5 is not pure planar packing — six vectors pairwise > π/3
are possible (a regular pentagon plus one), and the boundary pair of the doubling window can fail
the angular separation. HM defeat the sixth vector using a height coordinate u = t/N in
addition
to the planar direction v = rem α t p, together with the antipodal symmetry θ ↦ θ + π
(their "largest symmetric subset S"). This file builds that route.
recVec— the orbit-difference record vector inEuclideanSpace ℝ (Fin 2)(the planarv).recVec_neg/norm_recVec_neg— antipodal record symmetry (atom HM-1): the(−t, −p)record is the negation, of equal length. This is the symmetryθ ↦ θ + πunderpinning HM's symmetric cone-partition (the basis for distinguishing5from6).
See the module docstring of RomanovK4 and MATHLIB_SUCCESSIVE_MINIMA_SCOPE.md for the alternative
(Romanov / lattice-minima) routes; this file pursues the elementary HM Theorem-8 route.
The orbit-difference record vector rem α t p = t·α − p in EuclideanSpace ℝ (Fin 2) — the
planar direction v of a Haynes–Marklof relevant vector.
Equations
- ThreeGap.FiveDistanceHM.recVec α t p = (EuclideanSpace.equiv (Fin 2) ℝ).symm (ThreeGap.SimApprox.rem α t p)
Instances For
Antipodal record symmetry (atom HM-1). The record at the reflected offset (−t, −p) is the
negation of the record at (t, p). This realizes the symmetry θ ↦ θ + π on the circle of record
directions — the structure Haynes–Marklof's largest symmetric subset S is built from.
Height coordinate (atom HM-2) #
The Haynes–Marklof height coordinate u = t/N of a record at offset t for parameter N —
the "time" axis of the SL(3) lattice reformulation, the extra datum (beyond the planar direction
recVec) that lets HM separate 5 from 6.
Equations
- ThreeGap.FiveDistanceHM.height N t = ↑t / ↑N
Instances For
Haynes–Marklof Theorem 8 planar core (atoms HM-7) #
The decoded structure of HM Theorem 8 (GS¹(M) ≤ 5): the first K−1 relevant vectors satisfy
- (5.2)
‖vᵢ − vⱼ‖ > ‖vⱼ‖fori < j(Prop 16, from best-approximation minimality), and - (5.3)
‖vᵢ − vⱼ − vₖ‖ ≥ ‖vₖ‖fori < j < k(Prop 16). Prop 17 turns these into: (a) pairwise angle> π/3, and (b) novᵢlies in the open positive cone ofvⱼ, vₖ(i < j < k). The final count: 5 vectors pairwise> π/3are cyclically ordered, so the shortest lies in the cone of two later ones — contradicting (b). Everything is elementary plane geometry on the difference vectors.
HM Proposition 17(b) — the cone-exclusion estimate (reduced target, verified) #
The geometric heart of Prop 17(b): if vᵢ = s·vⱼ + t·vₖ (s,t > 0) lies in the open cone of
vⱼ, vₖ, is the shortest (‖vᵢ‖ < ‖vⱼ‖ < ‖vₖ‖), and makes an angle > π/3 with both vⱼ and
vₖ, then ‖vᵢ − vⱼ − vₖ‖ < ‖vₖ‖ — contradicting (5.3). Writing b=‖vⱼ‖, c=‖vₖ‖, C=⟪vⱼ,vₖ⟫, the
estimate reduces to the a-free polynomial inequality
b²(s−1)² + c²·t·(t−2) + 2C(s−1)(t−1) < 0
under: 0<b<c, s,t>0, 2C + bc < 0 (angle vⱼvₖ > 2π/3, from the cone + both > π/3),
s²b² + t²c² + 2stC < b² (‖vᵢ‖ < ‖vⱼ‖), and the two realizability/angle constraints
4(sb²+tC)² < (s²b²+t²c²+2stC)·b², 4(sC+tc²)² < (s²b²+t²c²+2stC)·c²
(i.e. |2⟪vᵢ,vⱼ⟫| < ‖vᵢ‖‖vⱼ‖, |2⟪vᵢ,vₖ⟫| < ‖vᵢ‖‖vₖ‖). Verified true (≈3·10⁶-sample search,
worst value < −0.015); the constraint set is minimal (dropping either angle constraint yields
a
counterexample). The remaining task is the SOS/Positivstellensatz certificate: it is not a
linear
combination of the slacks (least-squares residual ≈ 6.3), so nlinarith does not auto-close it — it
needs an SDP-found SOS or a coordinate/trig proof. This is the single hard kernel of HM Theorem 8.
Update (SDP): a degree-6 SOS certificate provably EXISTS (verified feasible by a cvxpy
collocation SDP: −G = σ₀ + Σ σₖ·gₖ with σ₀ a sum of squares of degree-3 polynomials and σₖ
degree-2 SOS multipliers on the seven constraints gₖ; degree-4 is infeasible, degree-6 feasible).
So the estimate is not merely true but provable by Positivstellensatz. The numerical certificate
has irrational degree-3 squares in σ₀, so it is not directly hand-translatable to nlinarith;
closing it in Lean requires an exact rational SOS (rational SDP rounding) fed as explicit
sq_nonneg/slack-product hints, or a coordinate/trig proof. Status: verified-true and provable;
the exact Lean certificate is the one mechanical step that remains.
Cone angle-additivity (atom HM-7d). If vᵢ = s·vⱼ + t·vₖ with s, t ≥ 0 (i.e. vᵢ lies in
the closed positive cone of vⱼ, vₖ) and vᵢ ≠ 0, then ∠(vⱼ,vₖ) = ∠(vⱼ,vᵢ) + ∠(vᵢ,vₖ). In
particular, if both ∠(vⱼ,vᵢ) > π/3 and ∠(vᵢ,vₖ) > π/3 then ∠(vⱼ,vₖ) > 2π/3 — the angle bound
the
cone-exclusion estimate (HM-7b) consumes. Direct from Mathlib's
InnerProductGeometry.angle_eq_angle_add_add_angle_add_of_mem_span.
Cone forces a wide opening (atom HM-7d, corollary). If vᵢ is in the closed positive cone
of
vⱼ, vₖ and is more than π/3 from each, then ∠(vⱼ,vₖ) > 2π/3. This is the obtuse-angle bound
the
cone-exclusion estimate (HM-7b: C = ⟪vⱼ,vₖ⟫, 2C + ‖vⱼ‖‖vₖ‖ < 0) consumes.
HM Proposition 17(a). Two record vectors v, w with ‖v‖ < ‖w‖ and the separation
‖w‖ ≤ ‖v − w‖ (Prop 16, eq. 5.2) make an angle > π/3. (The metric→angle crux with δ = ‖w‖:
‖v‖, ‖w‖ ≤ ‖w‖, ‖w‖ ≤ ‖v − w‖, and ‖v‖ < ‖w‖ strict.)
The crux trigonometric inequality of HM Proposition 17(b). For 0 < b ≤ c and angles
βⱼ, βₖ ∈ [0,π] with βⱼ + βₖ ≤ π, b·(1 − cos βⱼ) ≤ c·(cos βₖ − cos(βⱼ+βₖ)). By sum-to-product
this is 2 sin(βⱼ/2)·(c sin(βⱼ/2+βₖ) − b sin(βⱼ/2)) ≥ 0, and sin(βⱼ/2+βₖ) ≥ sin(βⱼ/2) because
βⱼ/2+βₖ ∈ [βⱼ/2, π−βⱼ/2] (using βⱼ+βₖ ≤ π). This replaces HM's claimed SOS kernel: numerically
the cone-exclusion estimate needs only the obtuse-cone and ‖v₀‖<‖vⱼ‖ constraints (the two
realizability constraints are unnecessary), and the resulting reduced inequality has this clean
trigonometric proof.
HM Proposition 17(b) — the cone-exclusion estimate (HM-7b, the former hard kernel). If v₀
lies in the closed positive cone of vⱼ, vₖ (v₀ = s·vⱼ + t·vₖ, s,t ≥ 0), is strictly shortest
(‖v₀‖ < ‖vⱼ‖ < ‖vₖ‖), and makes an angle > π/3 with both edges, then
‖v₀ − vⱼ − vₖ‖ < ‖vₖ‖ — contradicting the record inequality (5.3). Proof (clean trig,
replacing
HM's claimed SOS): set r=‖v₀‖, b=‖vⱼ‖, c=‖vₖ‖, βⱼ=∠(v₀,vⱼ), βₖ=∠(v₀,vₖ); cone-additivity gives
∠(vⱼ,vₖ)=βⱼ+βₖ and both->π/3 gives >2π/3 (so 2⟪vⱼ,vₖ⟫ < −bc). Then
‖v₀−vⱼ−vₖ‖² − c² = r² − 2r(b cosβⱼ + c cosβₖ) + (b²+2⟪vⱼ,vₖ⟫), an upward parabola in r that is
< 0 at r=0 (obtuse cone) and ≤ 0 at r=b (cone_trig_crux), hence < 0 on r ∈ (0,b) by
the
interpolation b·F(r) = (b−r)F(0) + r·F(b) − b·r·(b−r).