On 12/16/09, Dan Asimov <dasimov@earthlink.net> wrote:
Fred wrote:
<< . . . the rings involved are isomorphic.
I'll denote this
\H ~= Cl(0,2)
. . .
The quaternions have a natural order-3 automorphism i -> j -> k -> i, which Cl(0,2) lacks. This causes both computational inconvenience and conceptual confusion, when neophytes attmept to impose a symmetry which doesn't exist.
. . .
OK, so do I have it right: You're saying the quaternions and Cl(0,2) are isomorphic as rings, but unlike the quaternions, Cl(0,2) has no natural order-3 automorphism ?
(But Cl(0,2) must have *some* order-3 isomorphism, even if not "natural", by virtue of being isomorphic to H.)
Yes --- as a ring (skew field, even) --- but not as a graded algebra, which turns out in practice of central importance to Clifford algebra techniques. [Not that every practitioner seems fully to appreciate this: there's a substantial community promoting an abomination called "paravectors" --- sums of scalars and vectors --- I mean, dash it all, these bounders are not even versors! Harrumph --- I digress ...]
I'm not sure I'd call i -> j -> k -> i a "natural" isomorphism of H, because its naturalness depends on a chosen basis for H. (More precisely, a choice of oriented 2-subspace of pure quaternions.)
The automorphism group of H is SO(3), so without such a choice, any 120-degree rotation in SO(3) would be equally natural. I think.
Agreed --- the point I'm trying to make here is just that the notational visibility of this particular symmetry encourages people to assume a cod non-homomorphism \H -> Cl(0,3): i -> \u, j -> \v, k -> \w ?? WFL