ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pass

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.

A mixed-sign even lift has norm minus one

Example

Let $V = R^{1, 1}$ with orthogonal basis $e_{+}, e_{-}$ satisfying $Q (e_{+}) = 1$ and $Q (e_{-}) = - 1$ . The even Pin element $E = e_{+} e_{-}$ has $N (E) = - 1$ and maps to $- I \in S O (1, 1)$ , which is outside $S O_{0} (1, 1)$ . It belongs to full-even Spin and not to norm-one Spin.

Facts & Assumptions

Given: The signature $(+, -)$ Clifford relation and the conventions for full-even and norm-one Spin.

[F1]

For a product of anisotropic vectors, $N (u_{1} \dots u_{k}) = \prod_{j} (- Q (u_{j}))$ (Normalized Pin products and Clifford norm).

[F2]

Full-even Spin maps onto all $S O (1, 1)$ (Real Pin and full-even Spin images, kernels, and definite compactness).

[F3]

In mixed signature, $N = 1$ on full-even lifts exactly over $S O_{0} (p, q)$ (The even Clifford norm detects the identity component in mixed signature).

Verification

technique · direct

1.1givenF1algebra

The Clifford relations give $e_{-} e_{+} = - e_{+} e_{-}$ , so $E^{2} = e_{+} e_{-} e_{+} e_{-} = - e_{+}^{2} e_{-}^{2} = 1$ . It is a product of two normalized vectors and therefore belongs to full-even Spin. By [F1], $N (E) = (- Q (e_{+})) (- Q (e_{-})) = (- 1) (1) = - 1$ .

2.1F2F3step 1.1algebra∎

The reflection from $e_{+}$ negates $e_{+}$ and fixes $e_{-}$ ; the reflection from $e_{-}$ fixes $e_{+}$ and negates $e_{-}$ . Their product is $ρ_{E} = - I$ , with determinant $+ 1$ . Its polar block factor is $K_{+} = - 1$ on the positive line and $K_{-} = - 1$ on the negative line, so [F3] puts it outside $S O_{0} (1, 1)$ . Hence the full-even lift $E$ is excluded from norm-one Spin although it lies over $S O (1, 1)$ , as asserted.

Depends on

Used by

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Dependency tree · two levels

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Sources

Eckhard Meinrenken, Lie groups and Clifford algebras, Chapter 2 §4.1 (standard reference, not scraped)
Jean Gallier, Clifford Algebras, Clifford Groups, and a Generalization of the Quaternions: The Pin and Spin Groups, §1.9 (standard reference, not scraped)