Hello all,
I've been studying the octionions... off and on...
Lately, I've been experimenting with some of the various pictorial representations of the octonion multiplication table and how each relates to the possible bases for the octonions...
This pictorial representation of the octonion multiplication table is my favorite so far...
where the arrows define the positive direction around the Hamilton triangle... so, if you multiply two distinct basis elements in one of the seven Hamilton triangles in a counter-clockwise direction their product is positive... clockwise, negative.
The general rule is usually expressed:
It seems that when the octonions are represented in Clifford Algebra they're either represented as paravectors in , or 1-vectors in . The variation in the adjustment necessary to the Clifford product to represent the octonion product is relatively minor.
I also notice that many authors build up the octonions through the Cayley-Dickson doubling process:
which yields to this equivalent pictorial representation of the octonion multiplication table:
The complex numbers have two Clifford representations... Cl<sub>0,1</sub> where i is a 1-vector, and Cl<sub>2</sub><sup>+</sup> where i is a 2-vector. The quaternions also have two Clfford representations... {1,i,j,ij} in Cl<sub>0,2</sub> where i, j are 1-vectors and ij = k is a 2-vector, and {1,i,j,k} in Cl<sub>3</sub><sup>+</sup> where i,j,k are all 2-vectors.
I've not seen a Clifford Algebra representation that appears to more directly correspond to the basis of the octonions as represented above, i.e. O = {1,i,j,k,l,il,jl,kl}... Similar to the representation we have of the quaternions for {1,i,j,ij} in Cl<sub>0,2</sub> where i, j are 1-vectors and ij = k is a 2-vector. Some representation where il, jl, and kl are composed of some Clifford elements that are i, j, k, l...?
Has anyone seen any such Clifford representation of the octionions? Can such a Clifford representation be constructed?