Like the Clifford algebras these are characterised by what they square to, 'C' squares to -1, 'Do' squares to +1 and 'Du' squares to 0 so we can build up more and more complex algebras by combining these in different ways using the Kronecker product, this will allow us to generate all of the Clifford algebras.
For example we could generate CC as follows: first we take C given by:
We want to combine this with another copy of 'C', this requires another independent dimension so this time we will use j instead if i:
We combine these using the Kronecker product, by taking all combinations of i and j, that is: 1, i, j, i^j. Then put in 4×4 array going across the table (red) and down (blue) which gives:
We can simplify this by using the individual tables to give us the following identities:
- j^i = - i^j
- i^i = -1
- j^j = -1
This is exactly the same as G 0,2,0 which is a geometric algebra based on two dimensions which both square to -ve, shown here:
Its not surprising the table is the same (with only notation differences) as is was derived in the same way.
But here is where the division algebras diverge from geometric algebra, in geometric algebra terms like i^i are called bivectors and remain in that form as they can't be simplified any further, in division algebras like this we convert the bivector terms into vectors like this:
If we only did this at the end of the calculation it would not matter since we have already shown that:
CC = H = G 0,2,0
But we are going on to build up further algebras from this and since the vector base k has slightly different properties to the bivector base i^i then the algebras are going to diverge more and more as we build up more complex algebras.
For instance, if we add another multiplication by C to give octonions,
CCC = O
This can cause non-associativity because theoperator involves converting bivector bases to vector bases and vector×bivector may involve a sign change from bivector×vector. So it may be that:
(CC)C ≠ C(CC) ?
This may become even more ambiguous for 16ions, do we define them as (CC)(CC) or C(CC)C ?
We can calculate the result as follows:
The terms of CC are 1,i,j,k and we will multiply this by a third complex number, say,
which gives the product:
again we convert bivectors to vectors as follows:
i =e1, j=e2, k=e3, m=e4, i^m=e5, j^m=e6, k^m=e7
which is the multiplication table for octonions.
I have not yet worked out the table when the terms are combined in a different order, my plan is to adapt the program here to do this. Such a program would become more necessary as we build really big algebras (for example E8).