The metric is covariantly constant

Proved

EinsteinFieldEquationsD.metric_covariantly_constant

by Lucas · Oct 9, 2026 · Mathlib 0df444a (Lean v4.33.1)

differential-geometryeinstein-field-equationsgeneral-relativity

Let $g$ be a smooth, symmetric, nondegenerate metric on an open set $U\subseteq\mathbb{R}^D$ . Then $g$ is covariantly constant for its Levi-Civita connection:

\nabla_\lambda g_{\mu\nu}=\partial_\lambda g_{\mu\nu}-\Gamma^\sigma_{\lambda\mu}g_{\sigma\nu}-\Gamma^\sigma_{\lambda\nu}g_{\mu\sigma}=0\quad\text{on }U.

Preamble

import Mathlib
import Definitions.Def_EinsteinFieldEquationsD_Defs

Formal statement

namespace EinsteinFieldEquationsD

theorem metric_covariantly_constant {D : ℕ} (g : Tensor2 D) (U : Set (Coord D)) (hg : IsMetricOn g U) :
    ∀ x ∈ U, ∀ l a b : Fin D, covDeriv2 g g x l a b = 0 := by sorry

end EinsteinFieldEquationsD

Source

Wikipedia, "Einstein field equations", revision oldid=1374672993, https://en.wikipedia.org/w/index.php?title=Einstein_field_equations&oldid=1374672993; section 'Features — Conservation of energy and momentum', derivation box

Read-back

What the Lean code literally says, in plain math · Aristotle (Harmonic) - same agent as the drafter; non-blind

Non-blind read-back — not independent testimony. This read-back was written by the same agent (Aristotle, by Harmonic) that drafted the Lean statements it describes, with full knowledge of the source and of the intended meaning. It is not the output of a blind, independent auditor, and no reviewer should treat it as such. Compare it against the Lean code yourself.

Here $D$ is an arbitrary natural number, points are $x \in \mathbb{R}^D$ , $g$ is any function assigning to each point a real $D\times D$ matrix $(g_{ab}(x))$ , and $U \subseteq \mathbb{R}^D$ is any set. The hypothesis that $g$ is a metric on $U$ means: $U$ is open; every component $x \mapsto g_{ab}(x)$ is $C^\infty$ on $U$ ; $g_{ab}(x) = g_{ba}(x)$ for $x \in U$ ; and $\det g(x) \neq 0$ for $x \in U$ . No signature (Lorentzian or otherwise) is imposed. Outside $U$ , $g$ is unconstrained.

Claim. For every $x\in U$ and all $l,a,b$ ,

\partial_l g_{ab}(x)-\sum_s\big(\Gamma^s_{la}(x)\,g_{sb}(x)+\Gamma^s_{lb}(x)\,g_{as}(x)\big)=0,

with $\Gamma^c_{ab}=\tfrac12\sum_s g^{cs}(\partial_a g_{sb}+\partial_b g_{sa}-\partial_s g_{ab})$ and $g^{cs}$ the matrix inverse of $g(x)$ . This is the covariant derivative of the 2-tensor field $g$ itself, as defined in the definitions file.

Human review

Endorsed by Shuze Chen · Oct 10, 2026
Confirmed by the moderator at approval.
Endorsed by Lucas · Oct 10, 2026
Confirmed by the mission captain (proposal self-audit).

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