Dear Colleagues, The artificial choice of unit of length should indeed be avoided in fundamental considerations since, among other things, it trivializes the relation between length & area, etc. Over a given rig R, a Euclidean space E seems to be 1) a torsor over an R-module V where V is equipped with 2) an isomorphism V->Hom(V, L) 3) where L is an invertible R-module. The fact that L itself has no given rig structure can be compared with the general idea of metric (TAC Reprints 1) as valued in a monoidal category (which has one covariant binary operation, NOT two.) The R-modules R, L, L@L, etc, may be non-isomorphic, and there may even be other invertibles corresponding to time, force, etc. However, this Picard group will have all its elements of order two (i.e.,each invertible module will have its own R-valued pairing) if R is real in the sense that a sum of several squares is invertible if one of them is (as shown a few years ago by Steve Schanuel). That result is for the category of abstract sets, destroying my hope that the free abelian group on three generators might occur for a suitable spatial topos of affine algebraic geometry. Of course, a unit of M, where M is an invertible module, should be an isomorphism R->M, but in general such isomorphisms exist only locally. To see nontrivial invertible modules, look at classical arithmetic or complex analysis, or in the present geometric vein, look at rigs and R-modules not in abstract sets, but in more cohesive or variable toposes. Garrett Birkhoff recommended the topos of G-sets for a certain group G of homogeneities, obtaining (in a rather tautological way) the result that the group of dimensional analysis occurs as a Picard group. Looking forward to your thoughts Bill Quoting Steve Vickers <s.j.vickers@cs.bham.ac.uk>:
Top-down:
A Euclidean space is a torsor for a Euclidean inner product space E.
A torsor, or principal homogeneous space, is (in this instance) a generalized metric space with vector distances in place of real distances so as to make the triangle inequality an equality, with d(x,y) = -d(y,x) and d(x,y) = 0 iff x = y. Like metric spaces torsors have no origin. E supplies the distances. I put "Euclidean" as a modifier for "inner product space" to connote the liberalization of morphisms
Vaughan Pratt wrote: thereof
to preserving inner product only up to a constant factor (as opposed to the presumed default of on the nose). This liberalization accommodates scaling, and can be considered as forgetting the scale. That and torsors for forgetting the origin makes this approach "top-down."
Dear Vaughan,
I too thought about torsors, but couldn't see round the problem that they fix a unit length. (Suppose E is an inner product space and X a torsor for it, then for any x, y in X there is a unique v in E taking x to y, and so the length of v gives the distance from x to y.) Does "liberalizing the morphisms" in the way you suggest really do the trick? That seems to require a new notion of torsor, and I can't see how it would work technically.
Regards,
Steve.