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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Codazzi equation for a Riemannian submanifold

Statement

Assume ${A C}_{ω}$ . For tangent fields on an embedded Riemannian submanifold, define

$(\nabla I I) (X; Y, Z) : = \nabla_{X}^{⊥} I I (Y, Z) - I I (\nabla_{X}^{M} Y, Z) - I I (Y, \nabla_{X}^{M} Z) .$

Then the Codazzi equation is

$(\overline{R} (X, Y) Z)^{⊥} = (\nabla I I) (X; Y, Z) - (\nabla I I) (Y; X, Z) .$

The curvature convention is $\overline{R} (X, Y) Z = {\overline{\nabla}}_{X} {\overline{\nabla}}_{Y} Z - {\overline{\nabla}}_{Y} {\overline{\nabla}}_{X} Z - {\overline{\nabla}}_{[X, Y]} Z$ . The choice hypothesis is inherited exactly through the smooth tangent and normal projection constructions.

Facts & Assumptions

Given: Countable choice, an embedded Riemannian submanifold, and tangent fields $X, Y, Z$ .

[F1]

The Gauss decomposition is ${\overline{\nabla}}_{X} Y = \nabla_{X}^{M} Y + I I (X, Y)$ . Induced connection and second fundamental form.

[F2]

The normal component of the ambient derivative of a normal field is $\nabla^{⊥}$ . Normal connection.

[F3]

The second fundamental form is a smooth normal-valued two-tensor. The second fundamental form is a symmetric normal-bundle-valued two-tensor.

[F4]

The induced connection is torsion free. The induced connection is Levi–Civita.

[F5]

Curvature is the bracket-corrected commutator with the sign used in the Statement. Curvature of an affine connection.

Proof

technique · direct

1.1

Apply [F1] to the tangent field $\nabla_{Y}^{M} Z$ and [F2] to the normal field $I I (Y, Z)$ . Taking normal components gives $({\overline{\nabla}}_{X} {\overline{\nabla}}_{Y} Z)^{⊥} = I I (X, \nabla_{Y}^{M} Z) + \nabla_{X}^{⊥} I I (Y, Z) .$ The same formula with $X$ and $Y$ interchanged also holds, while [F1] gives $({\overline{\nabla}}_{[X, Y]} Z)^{⊥} = I I ([X, Y], Z)$ .

F1F2F3algebra

2.1

Substitute step 1.1 into [F5]: $(\overline{R} (X, Y) Z)^{⊥} = \nabla_{X}^{⊥} I I (Y, Z) - \nabla_{Y}^{⊥} I I (X, Z) + I I (X, \nabla_{Y}^{M} Z) - I I (Y, \nabla_{X}^{M} Z) - I I ([X, Y], Z) .$

F5step 1.1algebra

3.1

By [F4], $[X, Y] = \nabla_{X}^{M} Y - \nabla_{Y}^{M} X$ . Replace the bracket term in step 2.1 and regroup the first, third, and fifth terms and then the remaining terms according to the definition in the Statement. The result is precisely $(\nabla I I) (X; Y, Z) - (\nabla I I) (Y; X, Z)$ .

F3F4step 2.1algebra

4.1

On an empty or zero-dimensional submanifold every term is the unique zero section. In dimension one the skew pair $(X, Y)$ forces both sides to vanish; normal rank zero also makes both sides zero. The tensorial formula applies at boundary points, and positive definiteness supplies its projections. The stated ${A C}_{ω}$ is inherited through [F1]–[F4], with no new selection.

F1F2F3F4step 1.1step 2.1step 3.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Chuu-Lian Terng, Lecture Notes on Curves and Surfaces in R^3 and Riemannian Geometry (standard reference, not scraped)