Integral of a form over the identity simplex
ProvedRudin.ch10_integral_identity_simplexLet be a -form in , and let be the oriented affine simplex , that is, the -surface with parameter domain given by the identity map. Then
The reason is that the Jacobian of the identity map along an index tuple is the determinant of the – matrix whose entry is exactly when . That matrix has two equal rows unless is injective, hence a vanishing determinant; and when is a bijection it is the permutation matrix of , whose determinant is . So of all index tuples only the permutations contribute, each with the sign of the permutation.
This identity is the first step in evaluating both sides of Stokes' formula on the standard simplex, and it is the precise sense in which a form, integrated over a surface, is seen only through the alternating parts of its coefficients.
import Mathlib import Definitions.Def_Rudin_ch10_forms open Filter Topology MeasureTheory
namespace Rudin
/-- The integral of a `k`-form over the identity surface of `Qᵏ` — Rudin's oriented affine
simplex `[0, e₁, …, e_k]` — is the integral over `Qᵏ` of the alternating sum of its
coefficients. -/
theorem ch10_integral_identity_simplex (k : ℕ) (ω : KForm k k) :
integralOverSimplex ω ⟨id⟩
= ∫ u in stdSimplex k,
∑ σ : Equiv.Perm (Fin k), ((Equiv.Perm.sign σ : ℤ) : ℝ) * ω.coeff (⇑σ) u := by sorry
end Rudin