ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-11-04 (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.

Matrix form of the vector-square decomposition

Example

Represent $8_{v} \otimes 8_{v}$ by real $8 \times 8$ matrices via $e_{i} \otimes e_{j} \mapsto E_{i j}$ . The Spin(8) action is $A \mapsto g A g^{T}$ , where $g$ is the orthogonal vector action. Every such matrix has the unique decomposition $A = \frac{tr A}{8} I_{8} + \frac{A - A^{T}}{2} + (\frac{A + A^{T}}{2} - \frac{tr A}{8} I_{8}) .$ The three terms are respectively the scalar line, the skew-symmetric two-form module $28$ , and the symmetric trace-free module $35_{v}$ from the Spin(8) tensor-product theorem.

Example references

Spin(8) vector and half-spin modules

Spin(8) tensor-product decompositions

Facts & Assumptions

Given: $8_{v} = R^{8}$ with its standard Euclidean inner product and the Spin(8) vector action.

[F1]

$8_{v}$ is the standard real vector representation of Spin(8) on oriented Euclidean $R^{8}$ (Spin(8) vector and half-spin modules).

[F2]

$8_{v} \otimes 8_{v} ≅ 1 \oplus 28 \oplus 35_{v}$ , with $28 = Λ^{2} (8_{v})$ and $35_{v}$ the trace-free part of ${Sym}^{2} (8_{v})$ (Spin(8) tensor-product decompositions).

[A1]

The Axiom of Choice is assumed because the Spin(8) tensor-product theorem used in [F2] depends on it (The Axiom of Choice).

Verification

1.1givenF1construct

The tensors $e_{i} \otimes e_{j}$ form a basis of $8_{v} \otimes 8_{v}$ , and the assignment $e_{i} \otimes e_{j} \mapsto E_{i j}$ sends it bijectively to the matrix-unit basis, so it is a linear isomorphism. Under the diagonal action, $(g e_{i}) \otimes (g e_{j})$ maps to $g E_{i j} g^{T}$ ; by linearity the induced tensor action is $A \mapsto g A g^{T}$ .

2.1step 1.1algebra

The first matrix term is a scalar multiple of $I_{8}$ ; the second is skew-symmetric; the third is symmetric and has trace $tr A - tr A = 0$ . Their sum is $A$ by expanding the symmetric and skew parts. A matrix that is both symmetric and skew-symmetric is zero, and a scalar matrix with zero trace is zero, so this decomposition is direct and unique over $R$ .

3.1F1F2step 2.1algebra∎

Orthogonal conjugation fixes $I_{8}$ and the trace, preserves transpose, and therefore preserves each of the three summands. Their dimensions are $1$ , $8 \cdot 7 / 2 = 28$ , and $8 \cdot 9 / 2 - 1 = 35$ respectively. By [F2] these are exactly the scalar, antisymmetric, and symmetric trace-free modules in the theorem.

Matrix form of the vector-square decomposition

Example

Example references

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources