Timeline for Vector spaces without natural bases
Current License: CC BY-SA 2.5
6 events
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| Nov 23, 2023 at 18:41 | comment | added | étale-cohomology | But if you define Euclidean n-space as $\mathbb{R}^n$, then the "canonical basis" would be the standard basis (so Euclidean space cannot be $\mathbb{R}^n$; at best it's isomorphic to it, with a nontrivial space of isomorphisms). | |
| Nov 23, 2023 at 18:40 | comment | added | étale-cohomology | Do not affine spaces and affine manifolds formalize this very idea? | |
| Aug 6, 2010 at 16:31 | comment | added | Gabriel Ebner | @Vectornaut I learned it from Beutelspacher's Projective Geometry (German; English translation available). To actually construct the vector space you need to pick a hyperplane of points at infinity, and an origin. The scalings centered at the origin correspond to elements of the skew field; the translations form the underlying vector space of the affine space (the complement of the hyperplane). (E.g. you can define a translation as an automorphism that has all points at infinity as fixed points, and is invariant on all lines through a point at infinity (the direction of the translation)). | |
| Aug 5, 2010 at 16:58 | comment | added | Vectornaut | @Gabriel Ebner: Cool! What kind of textbook would cover this stuff? How could I figure out which module generates, say, the real projective plane? | |
| Aug 4, 2010 at 1:06 | comment | added | Gabriel Ebner | Indeed, every Desarguesian projective geometry (i.e. a set with an incidence relation satisifying the axioms for a projective space together with Desargues' theorem) is generated by a vector space over a skew field. (This skew field is a field if Pappus' theorem holds) Furthermore there is a isomorphism that turns any given basis of that vector space into another basis, so all bases are equally natural, provided that you do not distinguish isomorphic geometries. | |
| Aug 3, 2010 at 19:31 | history | answered | Vectornaut | CC BY-SA 2.5 |