Trautman is spatially closed as a Hessian
ProvedNewtonCartan.riem00_spatial_closed_Fcurvaturenewton-cartantrautman
Assume a classical connection on satisfies CurvConds for smooth and that its Riemann tensor has the Trautman form
Then is smooth and satisfies , , , the spatial closed-Hessian identity
and .
Temporal closedness need not hold: yields with . Spatial closedness is the integrability needed to recover slicewise.
Formalization Note. Corrects the retired riem00_is_closed_F. Uses classical_christoffel_form.
Preamble
import Definitions.Def_NewtonCartan_defs open scoped ContDiff
Formal statement
namespace NewtonCartan
/-- Under CurvConds and the Trautman $t_b$-form, $F^a{}_d := R^a{}_{00d}$ is smooth, vanishes on
temporal indices, is symmetric, and is closed as a Hessian in the *spatial* indices; its trace is $4\pi\rho$.
(Temporal closedness $\partial_0 F^i{}_j = \partial_j F^i{}_0$ can fail, e.g. $\Gamma^i{}_{00}=x^i f(t)$.) -/
theorem riem00_spatial_closed_F (Γt : Conn) (ρ : E → ℝ)
(hΓt : IsClassical Γt) (hρ : ContDiff ℝ ∞ ρ) (hcurv : CurvConds Γt ρ)
(hform : ∀ x a b c d,
riem Γt x a b c d =
tt b * (tt c * riem Γt x a 0 0 d - tt d * riem Γt x a 0 0 c)) :
let F : E → Fin 4 → Fin 4 → ℝ := fun x a d => riem Γt x a 0 0 d
ContDiff ℝ ∞ F ∧
(∀ x d, F x 0 d = 0) ∧ (∀ x a, F x a 0 = 0) ∧
(∀ x i j, F x i j = F x j i) ∧
(∀ 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) ∧
(∀ x, ∑ i : Fin 4, F x i i = 4 * Real.pi * ρ x) := by sorry
end NewtonCartanSource
Malament Prop. 4.2.5 / Weatherall Prop. 2.2