TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-08-29

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 real linear isomorphism preserves or reverses orientation according to the sign of its determinant

Statement

Let $T : V \to V$ be a linear isomorphism of an oriented real vector space with $\dim V = n \geq 1$ . If $\det T > 0$ , then $T$ carries every positively oriented basis to a positively oriented basis; if $\det T < 0$ , then $T$ carries every positively oriented basis to a negatively oriented one. Thus the sign of $\det T$ is exactly whether $T$ preserves or reverses the orientation.

Facts & Assumptions

Given: A linear isomorphism $T$ of an oriented $n$ -dimensional real vector space, $n \geq 1$ , and a positively oriented ordered basis $B$ .

[L1]

An orientation is an equivalence class of ordered bases, with two bases equivalent exactly when the determinant of their change-of-basis isomorphism is positive; a nonzero real determinant is positive or negative (Orientation of a finite-dimensional real vector space).

[L2]

The top exterior power acts by the determinant: $Λ^{n} T = \det (T) \cdot id$ (On $Λ^{n} V$ , the induced map $Λ^{n} T$ is multiplication by $\det T$ ).

Proof

technique · direct

1.1

Fix a positively oriented ordered basis $B = (b_{1}, \dots, b_{n})$ ; the transported list $T (B) = (T (b_{1}), \dots, T (b_{n}))$ is again an ordered basis because $T$ is an isomorphism, and the change-of-basis map from $B$ to $T (B)$ is $T$ itself.

givenL1

1.2

By [L1], $T (B)$ is positively oriented exactly when $\det T > 0$ , and negatively oriented exactly when $\det T < 0$ ; both cases are exhaustive because $T$ invertible has $\det T \neq 0$ .

2.1

The top exterior action of [L2] records the same sign: $T (b_{1}) \land \dots \land T (b_{n}) = Λ^{n} T (b_{1} \land \dots \land b_{n}) = \det (T) b_{1} \land \dots \land b_{n}$ , so the transported basis belongs to the class of $B$ precisely when the scalar is positive.

L2step 1.1

3.1

Steps 1.2 and 2.1 prove both the preservation and the reversal claims, and the orientation class of the transported basis is independent of the positively oriented basis chosen.

step 1.2step 2.1∎

Depends on

Used by

Reversing orientation negates the Hodge star while keeping the metric fixed Example
FALSE: an inner product determines an orientation False statement

Dependency tree · two levels

8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Reyer Sjamaar, Manifolds and Differential Forms, §8.2 (standard reference, not scraped)