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Spatial Poincaré lemma: integrate a spatially closed Hessian on R4\mathbb{R}^4R4

Proved
NewtonCartan.exists_potential_of_spatial_F

by Ethan Cheung Wang · Oct 21, 2026 · Mathlib 0df444a (Lean v4.33.1)

newton-cartanpoincare-lemmatrautman

Let FadF^a{}_dFad​ be smooth on R4\mathbb{R}^4R4 with F0d=Fa0=0F^0{}_d=F^a{}_0=0F0d​=Fa0​=0, Fij=FjiF^i{}_j=F^j{}_iFij​=Fji​, and

∂kFij=∂jFik(i,j,k spatial).\partial_k F^i{}_j=\partial_j F^i{}_k\qquad(i,j,k\text{ spatial}).∂k​Fij​=∂j​Fik​(i,j,k spatial).

Then there exists a smooth potential φ\varphiφ such that

∂j(∇iφ)=Fij(i,j spatial).\partial_j(\nabla^i\varphi)=F^i{}_j\qquad(i,j\text{ spatial}).∂j​(∇iφ)=Fij​(i,j spatial).

The construction integrates the spatial Hessian on each constant-ttt slice (global Poincaré lemma on R3\mathbb{R}^3R3) and assembles the slices smoothly in ttt.

Formalization Note. Corrects the retired exists_potential_of_closed_F.

Preamble
import Definitions.Def_NewtonCartan_defs

open scoped ContDiff
Formal statement
namespace NewtonCartan

/-- If $F^a{}_d$ is smooth, vanishes on temporal indices, is symmetric, and is closed as a Hessian in
spatial indices, then there is smooth $\varphi$ with $\partial_j\partial_i\varphi = F^i{}_j$ for spatial $i,j$
(equivalently $\partial_j(\nabla^i\varphi)=F^i{}_j$). -/
theorem exists_potential_of_spatial_F (F : E → Fin 4 → Fin 4 → ℝ)
    (hF : ContDiff ℝ ∞ F)
    (hrow : ∀ x d, F x 0 d = 0) (hcol : ∀ x a, F x a 0 = 0)
    (hsym : ∀ x i j, F x i j = F x j i)
    (hclosed : ∀ x i j k, i ≠ 0 → j ≠ 0 → k ≠ 0 →
      pd (fun y => F y i j) k x = pd (fun y => F y i k) j x) :
    ∃ φ : E → ℝ, ContDiff ℝ ∞ φ ∧
      ∀ x i j, i ≠ 0 → j ≠ 0 →
        pd (fun y => gradUp φ y i) j x = F x i j := by sorry

end NewtonCartan
Source
Malament Prop. 4.2.5 Poincaré step; Schwartz Thm. 2.29

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