A small meridian circle of a knot in has Gauss linking integral with the knot
ProvedHryniewiczCriterion.gaussLinkingIntegral_meridian_eq_oneLet be a loop with , injective modulo (that is, implies ). Let satisfy
and let be a pole not on . For consider the meridian loop
a small circle around that turns from towards . Then for all sufficiently small the Gauss linking integral of and , computed after stereographic projection from (gaussLinkingIntegral), equals .
Proof idea: the integral does not depend on (fixed-pole homotopy invariance, gaussIntegral_eq_of_homotopy), nor on the frame within the positively oriented ones. After stereographic projection, deform near to an exact straight segment through the centre of a round meridian circle of radius in the normal plane. The straight piece contributes exactly as . The rest contributes , since the circle has length and stays a fixed distance from the rest of the curve. Numerically checked with the exact Lean integrand: at and on a non-planar loop with a skewed frame.
import Definitions.Def_HryniewiczCriterion_SelfLinking open HryniewiczCriterion open scoped ContDiff
theorem HryniewiczCriterion.gaussLinkingIntegral_meridian_eq_one (γ : ℝ → R4)
(hγ : ContDiff ℝ 2 γ) (hγper : ∀ s, γ (s + 1) = γ s) (hγunit : ∀ s, euclidNorm (γ s) = 1)
(hinj : ∀ s t, γ s = γ t → ∃ n : ℤ, t = s + n) (f₁ f₂ : R4)
(hdet : 0 < Matrix.det (Matrix.of ![γ 0, deriv γ 0, f₁, f₂]))
(N : R4) (hN : euclidNorm N = 1) (hNγ : ∀ s, γ s ≠ N) :
∃ ε₀ : ℝ, 0 < ε₀ ∧ ∀ ε : ℝ, 0 < ε → ε < ε₀ →
gaussLinkingIntegral N γ (fun t => radialNormalize (γ 0 +
ε • (Real.cos (2 * Real.pi * t) • f₁ + Real.sin (2 * Real.pi * t) • f₂))) = 1 := by sorry