FrediFizzx wrote:Well, good luck with it, Joy. I used to like John Baez but now I see he is just another person that can't think for himself or didn't want to take the time to. I've been through that math forwards and backwards doing simulations and computer verifications of it. There is nothing wrong with it. Who would have thought? 7-spheres constructed from Euclidean primitives! It is probably the way Nature does it instead of via pure octonions.
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Fred, take a look at the multiplication table, Figure 1, in Joy's RSOS paper. In fact, there are two tables, one for lambda = +1 and one for lambda = -1. These are the multiplication tables for 8 basis vectors of, in principle, two real vector spaces K+ and K-. But K- and K+ are the same linear space! The spans, i.e., the sets of all real linear combinations of each 8 basis elements are the same. Let me call it K. The space K is closed under multiplication (everything *in* either table is +/- one of the 8 basis elements of either space; so all linear combinations, multiplied together, also form a linear combination). The space has a zero and a unit. It's a unitary, non-commutative, algebra, but it is associative, which is nice [that has to be checked, somehow]. In fact, the space K is (or is isomorphic to) Cl(0,3)(R), as Joy himself does remark, since, as one can see directly, it *is* the even subalgebra of Cl(4, 0). The associativity of K follows from the associativity of Cl(4, 0). It is a trivial matter to verify this isomorphism with, e.g. GAviewer. It is moreover a well known fact from the theory of classification of real Clifford algebras. [The octonions are not associative; all Clifford algebras, by definition, are].
Now let me define M = lambda I_3 e_infty. M is an element of K. It's square is +1, which can be read off the multiplication table. So we have M^2 = 1 and hence M^2 - 1 = 0.
Therefore (M - 1)(M+1) = 0.
Suppose we could find a norm ||.|| making K a normed unitary algebra. By definition, this would mean that the norm ||.|| *is* a norm, which furthermore satisfies ||AB|| = ||A||.||B||, ||A|| = 0 if and only if A = 0, ||a A|| = |a| ||A||, ||1|| = 1.
Joy claims that he has been able to define such a norm on his algebra K.
Now define A = M - 1, B = M + 1. These are two elements of K and neither is equal to zero. We have AB = 0. One says that A and B are zero divsors. Neither is zero but their product is. It follows that ||A||.||B|| = ||0|| = 0. Therefore, either ||A|| = 0, or ||B|| = 0. Therefore, either A = 0, or B = 0. Therefore, either M = 1 or M = -1 ... a contradiction.
You can check all this really easily with any decent Geometric Algebra computer algebra program, e.g., GAViewer. I did that, myself, several times. Very educative!
A division algebra is a unitary algebra with a norm making it a normed vector space and such that the norm is multiplicative and such that every element except 0 has a multiplicative inverse. In a division algebra, there are no zero divisors. A and B just discussed are zero divisors. If such a space were associative and if B had a multiplicative inverse, then from AB = 0 we would get A = A B B{-1} = 0 B^{-1} =0.
A famous theorem states that the *only* normed unitary division algebras are R, C, H and O (reals, complex numbers, quaternions, octonions). Christian claims he has found a norm on his algebra, which makes it a normed algebra, and which satisfies ||AB||= ||A||.||B||. But then his algebra cannot have zero divisors. But it does have zero divisors, in particular: (M - 1)(M + 1) = 0.
You can read about all this here
https://en.wikipedia.org/wiki/Hurwitz%27s_theorem_(composition_algebras) and here
https://math.ucr.edu/home/baez/octonions/node2.html
