LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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Three-dimensional curvature is determined by Ricci curvature

Physical scope

Pure mathematics; a finite-dimensional Lorentz-signature tensor identity, with no empirical claim.

Statement

Let $V$ be a three-dimensional real vector space with Lorentzian inner product $g$ of signature $(-, +, +)$ . Let $R_{a b c d}$ be an algebraic curvature tensor: it is antisymmetric in $a, b$ and in $c, d$ , symmetric under exchange of the pairs, and satisfies $R_{a b c d} + R_{a c d b} + R_{a d b c} = 0$ . Define ${Ric}_{b d} = g^{a c} R_{a b c d}$ and $Scal = g^{b d} {Ric}_{b d}$ . Then

$R_{a b c d} = g_{a c} {Ric}_{b d} - g_{a d} {Ric}_{b c} - g_{b c} {Ric}_{a d} + g_{b d} {Ric}_{a c} - \frac{1}{2} Scal (g_{a c} g_{b d} - g_{a d} g_{b c}) .$

In particular, $Ric = 0$ if and only if $R_{a b c d} = 0$ .

Facts & Assumptions

Given: The finite-dimensional vector space, metric, and algebraic curvature tensor in the statement.

[A1]

The metric is nondegenerate with an orthonormal basis in which $g = diag (- 1, 1, 1)$ ; the inverse $g^{a b}$ is used for contractions.

Proof

1.1givenA1algebra

Define $S_{a b c d}$ to be the right-hand side of the displayed formula. Symmetry of $g$ and $Ric$ gives antisymmetry of $S$ in each index pair and symmetry under exchanging the pairs. Substitution into $S_{a b c d} + S_{a c d b} + S_{a d b c}$ cancels the Ricci and scalar terms pairwise, so $S$ satisfies the first Bianchi identity and is an algebraic curvature tensor. Contracting the first and third indices gives $g^{a c} S_{a b c d} = 3 {Ric}_{b d} - {Ric}_{b d} - {Ric}_{b d} + Scal g_{b d} - \frac{1}{2} Scal (3 g_{b d} - g_{b d}) = {Ric}_{b d} .$ Thus $W : = R - S$ is an algebraic curvature tensor with zero Ricci contraction.

2.1A1step 1.1algebra

In the basis of [A1], an algebraic curvature tensor is represented by a symmetric $3 \times 3$ matrix on the basis bivectors $e_{0} \land e_{1}, e_{0} \land e_{2}, e_{1} \land e_{2}$ . Write its diagonal entries as $x = W_{0101}$ , $y = W_{0202}$ , and $z = W_{1212}$ . The three diagonal zero-Ricci equations are $x + y = 0$ , $- x + z = 0$ , and $- y + z = 0$ . They imply $2 x = 0$ , hence $x = y = z = 0$ .

3.1step 1.1step 2.1algebra∎

The three off-diagonal zero-Ricci equations are $Ric (W)_{01} = W_{0212} = 0$ , $Ric (W)_{02} = - W_{0112} = 0$ , and $Ric (W)_{12} = - W_{0102} = 0$ . They set the three off-diagonal entries of the same symmetric bivector matrix to zero. Therefore $W = 0$ , so $R = S$ and the displayed identity holds. If $Ric = 0$ , then its scalar contraction is zero and the formula gives $R = 0$ ; conversely $R = 0$ implies its Ricci contraction vanishes.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Steven Carlip, Quantum Gravity in 2+1 Dimensions: The Case of a Closed Universe