I don't understand any of the physics, but this is an interesting math idea
that I'd never encountered before.
I think this refers to the tensor product of the four real division algebras:
R (x) C (x) H (x) O
(where R = reals, C = complexes, H = quaternions, O = octonions, and (x) denotes
tensor product over the real numbers.
Tensor products are one of the two basic ways to combine algebras over a ring.
But if I recall how I learned about this in a course called 18.26, I think
everything was commutative. (Which the quaternions and octonions aren't, but
I'm not sure it matters.)
I will attempt to unwind the definition of tensor product of algebras, since
I'd like to know what this weird thing is.
*****-----until further notice, R means any commutative ring-----*****
Let's say A and B are algebras over a commutative ring R. Then their tensor
product A (x) B will be the quotient of the free algebra over R generated
by the elements
{a_j (x) b_j | a_j in A, b_j in B},
i.e.,
A*B = {Sum r_j (a_j (x) b_j), a_j in A, b_j in B, r_j in R, Sum is finite}.
To get the tensor product we need to factor out by the subring Z generated by
elements of the form
((ra)(x)b) - (a(x)(rb)), r in R, a in A, b in B
and the like that give the tensor product its bilinearity.
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*****-----From now on R is just the reals again-----*****
Dimensions of tensor products multiply, so the dimension of
R (x) C (x) H (x) O
where (x) is always understood as tensor with respect to R=reals, is
1 x 2 x 4 x 8 = 64.
Hmm, weird. Anyone care to explain (anything about) the physics?
—Dan
Henry Baker wrote:
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Combined as $latex \mathbb{R}\otimes\mathbb{C}\otimes\mathbb{H}\otimes\mathbb{O}$, the four number systems form a 64-dimensional abstract space.
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