ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6.1-sol)

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.

Worked component signs of the oriented volume form

Example

Worked computation of the components of the oriented volume form in an inertial frame with $ϵ (e_{0}, e_{1}, e_{2}, e_{3}) = + 1$ . The covariant components are the permutation signs: $ϵ_{0123} = + 1$ ; $ϵ_{0213} = - 1$ , the sign of a single transposition; $ϵ_{2301} = + 1$ , the product of the two transpositions $(0 2)$ and $(1 3)$ ; $ϵ_{3210} = + 1$ , the product of the two transpositions $(0 3)$ and $(1 2)$ ; and $ϵ_{a b c d} = 0$ whenever two of the indices coincide. Raising all indices with the diagonal inverse metric gives $ϵ^{0123} = η^{00} η^{11} η^{22} η^{33} ϵ_{0123} = - 1$ . For the spatial symbol, $ϵ_{123} = + 1$ , $ϵ_{213} = - 1$ , $ϵ_{312} = + 1$ , and $ϵ_{i j k} = 0$ if two spatial indices coincide.

Facts & Assumptions

Given: An inertial frame $(o; e_{0}, e_{1}, e_{2}, e_{3})$ of the model and the volume form $ϵ$ of The oriented volume form epsilon and the permutation symbols of an inertial frame.

[L1]

$ϵ (v_{0}, v_{1}, v_{2}, v_{3}) = \det [v_{j}^{a}]$ , where the $j$ -th column lists the coordinates of $v_{j}$ in the inertial frame; consequently $ϵ_{a b c d} = ϵ (e_{a}, e_{b}, e_{c}, e_{d})$ is $+ 1$ on $(0, 1, 2, 3)$ , the sign of the permutation for a permutation tuple, and $0$ when two indices repeat. The inverse metric is $η^{a b} = diag (- 1, 1, 1, 1)^{a b}$ and $ϵ^{a b c d} = η^{a a^{'}} η^{b b^{'}} η^{c c^{'}} η^{d d^{'}} ϵ_{a^{'} b^{'} c^{'} d^{'}}$ ; the spatial symbol is defined by $ϵ_{123} = + 1$ together with alternation (The oriented volume form epsilon and the permutation symbols of an inertial frame).

[L2]

The determinant is the unique column-multilinear, alternating, normalized function of the columns; in particular a matrix with two equal columns has determinant $0$ (The determinant is the unique normalized alternating multilinear function on the columns).

[L3]

For $A = (a_{i j})$ , $\det A = \sum_{σ \in S_{n}} sgn (σ) \prod_{i < n} a_{σ (i), i}$ ; on a permutation matrix the sum has exactly one nonzero term, of value $sgn (σ)$ (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[L4]

The form $ϵ$ and the value $ϵ (e_{0}, e_{1}, e_{2}, e_{3}) = 1$ are the same in every inertial frame (The oriented volume form is independent of the positively oriented inertial frame).

Verification

technique · direct determinant computations on permutation matrices

1.1L1L4algebra

The coordinate matrix of the frame tuple $(e_{0}, e_{1}, e_{2}, e_{3})$ is the identity matrix, whose determinant is $+ 1$ ; hence $ϵ_{0123} = + 1$ by [L1]; this value is frame independent by [L4].

1.2L1L3algebra

The tuple $(e_{0}, e_{2}, e_{1}, e_{3})$ has coordinate matrix equal to the permutation matrix of the transposition $(1 2)$ ; its determinant is the single nonzero term of [L3], namely $sgn (1 2) = - 1$ , so $ϵ_{0213} = - 1$ .

1.3L1L3algebra

The tuple $(e_{2}, e_{3}, e_{0}, e_{1})$ has coordinate matrix the permutation matrix of $σ = (0 2) (1 3)$ , a product of two transpositions, so $sgn σ = (- 1)^{2} = + 1$ and $ϵ_{2301} = + 1$ by [L1] and [L3].

1.4L1L3algebra

The tuple $(e_{3}, e_{2}, e_{1}, e_{0})$ has coordinate matrix the permutation matrix of $σ = (0 3) (1 2)$ , again a product of two transpositions, so $ϵ_{3210} = + 1$ by [L1] and [L3].

1.5L1L2algebra

If two of the indices $a, b, c, d$ coincide, then two columns of the coordinate matrix $[e_{a}, e_{b}, e_{c}, e_{d}]$ coincide, so its determinant is $0$ by [L2]; hence $ϵ_{a b c d} = 0$ for every tuple with a repeated index.

2.1L1L2L3algebra

For the spatial symbol: the coordinate matrix of $(e_{1}, e_{2}, e_{3})$ in the spatial ordered basis $(e_{1}, e_{2}, e_{3})$ is the identity, giving $ϵ_{123} = + 1$ ; the tuple $(e_{2}, e_{1}, e_{3})$ is a transposition, giving $ϵ_{213} = - 1$ ; and $(e_{3}, e_{1}, e_{2})$ corresponds to the $3$ -cycle $(1 3 2)$ , an even permutation, giving $ϵ_{312} = + 1$ ; repeated spatial indices give $0$ by the same column-coincidence argument as step 1.5.

3.1step 1.1L1L4algebra∎

Raising the indices with the diagonal inverse metric, exactly one term of the sum in [L1] survives for $(0, 1, 2, 3)$ , and $ϵ^{0123} = η^{00} η^{11} η^{22} η^{33} ϵ_{0123} = (- 1) (+ 1) (+ 1) (+ 1) (+ 1) = - 1$ ; this is the value in every inertial frame, since both $ϵ$ and the diagonal inverse metric are frame independent by [L1] and [L4].

Depends on

Dependency tree · 2 levels

Sources

S. M. Carroll, Lecture Notes on General Relativity, arXiv:gr-qc/9712019v1 (1997), Ch. 1 pp. 2-7 (PDF pp. 9-14)