TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

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Second Bianchi identity for a bundle connection

Statement

Let $\nabla^{End}$ be the connection induced on $End (E)$ . For an $End (E)$ -valued $k$ -form $A$ , define its covariant exterior derivative by alternating covariant differentiation:

$(d^{\nabla} A) (X_{0}, \dots, X_{k}) = \sum_{a} (- 1)^{a} \nabla_{X_{a}}^{End} (A (X_{0}, \dots, {\hat{X}}_{a}, \dots, X_{k})) + \sum_{a < b} (- 1)^{a + b} A ([X_{a}, X_{b}], X_{0}, \dots, {\hat{X}}_{a}, \dots, {\hat{X}}_{b}, \dots, X_{k}) .$

If $Ω$ is the curvature two-form, then

$d^{\nabla} Ω = 0.$

Facts & Assumptions

[F1]

In a local frame the curvature matrix obeys $Ω = d ω + ω \land ω$ , with matrix factors in the displayed order. Curvature two-form structure equation.

[F2]

The induced Hom connection satisfies $(\nabla_{X} A) (s) = \nabla_{X} (A (s)) - A (\nabla_{X} s)$ . Product connection on tensor and hom bundles.

[F3]

Induced covariant differentiation preserves exterior powers and is a degree-zero derivation. Induced connection on exterior powers is a degree zero derivation.

[F4]

The ordinary exterior derivative is a degree-one graded derivation. The exterior derivative is a graded derivation.

[F5]

The ordinary exterior derivative satisfies $d^{2} = 0$ . The exterior derivative squares to zero.

Proof

Given: A local frame $e$ with connection matrix $ω$ and curvature matrix $Ω$ .

1.1F2F3algebra

From [F2], the local matrix of the End $(E)$ connection is the commutator action $\nabla_{X}^{End} A = X (A) + ω (X) A - A ω (X)$ . Alternating this formula as in the definition of $d^{\nabla}$ , with [F3] ensuring the alternating degrees are preserved, gives for an End $(E)$ -valued $k$ -form $A$ the local identity $d^{\nabla} A = d A + ω \land A - (- 1)^{k} A \land ω$ . In particular, $d^{\nabla} Ω = d Ω + ω \land Ω - Ω \land ω$ .

2.1F1F4F5step 1.1algebra∎

Substitute [F1] into step 1.1 and apply the graded Leibniz rule: $d^{\nabla} Ω = d^{2} ω + d ω \land ω - ω \land d ω + ω \land d ω + ω \land ω \land ω - d ω \land ω - ω \land ω \land ω = 0$ by [F5] and associativity of matrix/wedge multiplication. Since this holds in every local frame, it is the intrinsic identity $d^{\nabla} Ω = 0$ .

Second Bianchi identity for a bundle connection

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources