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.) 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. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele