Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14
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Algebraic symmetries of the Riemann tensor

Statement

This item assumes ACω, namely countable choice. In the propagated dependency chain, that assumption is required through First Bianchi identity; after those interfaces are fixed, the remaining local or finite argument makes no additional countable-family choice.

The Riemann curvature four-tensor satisfies, for all vector fields X,Y,Z,W,

Rm(X,Y,Z,W)=Rm(Y,X,Z,W),

Rm(X,Y,Z,W)=Rm(X,Y,W,Z),

Rm(X,Y,Z,W)=Rm(Z,W,X,Y),

and

Rm(X,Y,Z,W)+Rm(Y,Z,X,W)+Rm(Z,X,Y,W)=0.

These are respectively first-pair skewness, last-pair skewness, pair interchange, and the cyclic first-Bianchi symmetry.

Facts & Assumptions

[A1]

ACω is countable choice and is required here through First Bianchi identity; after those supplied interfaces are fixed, the remaining local or finite calculation makes no additional countable-family choice.

[F1]

Rm(X,Y,Z,W)=g(R(X,Y)Z,W). Riemann curvature four-tensor.

[F2]

Curvature is skew in its first two arguments. Curvature is skew in its first two arguments.

[F3]

Curvature obeys the cyclic first Bianchi identity. First Bianchi identity.

[F4]

The Levi–Civita connection is metric compatible. Levi civita connection.

Proof

Given: ACω, smooth vector fields X,Y,Z,W and the Levi–Civita connection.

1.1

Combining [F1] with [F2] gives first-pair skewness, and pairing [F3] with W gives the displayed cyclic identity for Rm.

A1F1F2F3
1.2

Metric compatibility [F4] expands the scalar identity XYg(Z,W)YXg(Z,W)[X,Y]g(Z,W)=0. The mixed terms g(YZ,XW) and g(XZ,YW) cancel in pairs, leaving g(R(X,Y)Z,W)+g(Z,R(X,Y)W)=0. Symmetry of g and [F1] give last-pair skewness.

F1F4algebra
2.1

Write the cyclic identity from step 1.1 for the four ordered triples (X,Y,Z;W), (Y,Z,W;X), (Z,W,X;Y), and (W,X,Y;Z) and add them. Last-pair skewness from step 1.2 cancels the eight terms whose first pair is respectively (X,Y), (Y,Z), (Z,W), or (W,X). The four remaining terms, simplified with both pair skews, give 2Rm(Y,W,X,Z)2Rm(X,Z,Y,W)=0. Renaming (X,Z,Y,W) as an arbitrary quadruple yields pair interchange.

step 1.1step 1.2algebra

Depends on

Used by

Dependency tree · two levels

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Sources