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TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14
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Curvature two-form structure equation

Statement

Let e=(e1,…,er) be a local frame, let ω=(ωij) be its connection matrix, and let Ω=(Ωij) be the matrix of the curvature two-form, defined by R∇(X,Y)ej=Ωij(X,Y)ei. Then

Ω=dω+ω∧ω,

where multiplication order is fixed by

(ω∧ω)ij=∑kωik∧ωkj.

Facts & Assumptions

[F1]

Bundle curvature is an End⁡(E)-valued two-form. Vector-bundle curvature is an endomorphism-valued two-form.

[F2]

In the supplied frame, ∇Xej=ωij(X)ei. Connection one form in a local frame.

[F3]

For s=eu, ∇Xs=e(Xu+ω(X)u). Local coordinate formula for a bundle connection.

[F4]

The exterior derivative of a local coordinate expansion differentiates its scalar coefficients. The local coordinate formula for the exterior derivative.

[F5]

The wedge product is the pointwise alternating product of forms. The wedge product of differential forms.

Proof

Given: A local frame e, a coordinate chart on its domain, and coordinate fields ∂a,∂b.

1.1F1F2F3algebra

Since coordinate fields commute, expand the defining curvature commutator on ej with [F2]–[F3]. The coefficient of ei is ∂a(ωij(∂b))−∂b(ωij(∂a))+∑k(ωik(∂a)ωkj(∂b)−ωik(∂b)ωkj(∂a)).

2.1F1F4F5step 1.1algebra∎

By [F4], the first two terms in step 1.1 are dωij(∂a,∂b); by [F5], the sum is ∑k(ωik∧ωkj)(∂a,∂b) in precisely the stated matrix order. Both sides are two-forms by [F1], so equality on every coordinate-frame pair proves Ωij=dωij+∑kωik∧ωkj.

Depends on

Dependency tree · 2 levels

Sources