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Orientation of a finite-dimensional real vector space
Definition
Let be a finite-dimensional real vector space (Finite-dimensional vector space, and its dimension ; infinite-dimensional means having no finite basis). Two ordered bases of (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis) are declared equivalent when the determinant of the unique change-of-basis linear isomorphism carrying to satisfies , where the determinant is The determinant of an endomorphism of a finite-dimensional vector space: its matrix determinant in an ordered basis in positive dimension, and on the zero space and positivity is the order of the real field The reals form a totally ordered field.
This is an equivalence relation. The identity change of basis has determinant . If through , then through , and . If through , then by For same-sized finite square matrices over a commutative ring, .
An orientation of is an equivalence class of ordered bases under this relation; a basis in the chosen class is positively oriented for that orientation. When there are exactly two orientations: fixing one ordered basis , every other basis has or by the trichotomy of The reals form a totally ordered field. If , replacing the sole basis vector of by its negative produces a basis with determinant ; if , interchanging two entries of does the same. Thus both classes occur. When the only ordered basis is the empty one and there is exactly one orientation.
Remarks
Orientations depend only on the real vector-space structure; no inner product or basis preference enters the definition.
Depends on
- Finite-dimensional vector space, and its dimension $\dim_F V$; infinite-dimensional means having no finite basis
- Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis
- The determinant of an endomorphism of a finite-dimensional vector space: its matrix determinant in an ordered basis in positive dimension, and $1$ on the zero space
- For same-sized finite square matrices over a commutative ring, $\det(AB)=\det(A)\det(B)$
- The reals form a totally ordered field
Used by
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Sources
- Reyer Sjamaar, Manifolds and Differential Forms, §8.2 (standard reference, not scraped)