Contracted Bianchi identity: the Einstein tensor is divergence-free

Proved

EinsteinFieldEquationsD.einsteinTensor_divergence_free

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 the Einstein tensor is divergence-free on $U$ :

g^{\beta\lambda}\nabla_\lambda G_{\alpha\beta}=0,

the index-lowered form of $G^{\alpha\beta}{}_{;\beta}=0$ (contracted Bianchi identity).

Preamble

import Mathlib
import Definitions.Def_EinsteinFieldEquationsD_Defs

Formal statement

namespace EinsteinFieldEquationsD

theorem einsteinTensor_divergence_free {D : ℕ} (g : Tensor2 D) (U : Set (Coord D)) (hg : IsMetricOn g U) :
    ∀ x ∈ U, ∀ a : Fin D, divergence g (einsteinTensor g) x a = 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 every index $a$ ,

\sum_{b,l} g^{bl}(x)\,\nabla_l G_{ab}(x)=0,

where $G_{ab}=R_{ab}-\tfrac12Rg_{ab}$ is the Einstein tensor of $g$ (as a 2-tensor field on all of $\mathbb{R}^D$ ), $\nabla_l G_{ab}=\partial_l G_{ab}-\sum_s(\Gamma^s_{la}G_{sb}+\Gamma^s_{lb}G_{as})$ , and $g^{bl}$ is the matrix inverse of $g(x)$ .

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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