Having the table of ordered permutation triplet basis product rules for Right Octonion Algebra in the body of my essay, it can be used to visualize my algebraic variance/invariance sieve, and demonstrate that all product terms for any number of Octonion algebraic element products will fall into an algebraically invariant set, or one of 14 algebraically variant sets.
To make sure everyone gets the drift on the ordered permutation triplet basis product rule, going cyclically left to right, the consecutive product of two basis elements is + the third element, and commuting the product order going cyclically right to left, the consecutive product of two basis elements is - the third element. For any three unlike non-scalar basis elements there are two possible definitions, starting with one order then forming another rule exchanging any two elements which is fully equivalent to ordering them in the opposite direction. This singular rule change has the effect of changing the sign on all 6 basis product pairs. Very important that it is just a sign change.
It will not be of any importance that I use a table for Right Octonions and not Left Octonions, nor that the table is relative to algebra R0. Any other choice will sieve the same product terms into the same sets, the only difference will be the relative signs within the variant sets.
I will do a proof by example using a sequence of basis element products describing the product history of the final resultant basis element. The product e1 * e2 is ruled by the ordered permutation triplet including the set {e1 e2 e3}, using curly braces to not imply a sign rule quite yet. For R0, R1, R2 and R3 the sign rule is (e1 e2 e3) and for R4, R5, R6 and R7 the rule is the opposite: (e3 e2 e1). The correspondence between algebra enumerations and triplet enumerations is no accident, they were purposefully enumerated to make it such. Anyway, for the former 4 the result is +e3, and for the latter 4 it is -e3. This would be a simple example of an algebraically variant product term; it can change signs when a change of algebra definition is made.
Now take the e3 result and multiply on the left by e2, effectively doing e2 * (e1 * e2). This is based on the same triplet rule {e1 e2 e3}, so any algebra change induced negation will be done twice, meaning for every possible Octonion Algebra choice, the result will always be +e1. This is an example of a non-trivial (not singularly defined e0 * en, ej * ej etc.) algebraically invariant product term.
If instead we did e5 * (e1 * e2), the second product would be using the rule for {e6 e5 e3} so the sign on the final result basis element e6 would be dependent on both {e1 e2 e3} and {e6 e5 e3} rules, and they can change in different ways for specific changes in algebra. Relative to R0, which has all +1 values in its column and results in +e6, the e6 sign for some other Right algebra will be determined by the product of the row {e1 e2 e3} and {e6 e5 e3} values for that algebra's column. If the row product is +1, the result in that algebra will be the same as in R0: +e6, and if the row product is -1, the result for that algebra will have opposite sign as R0 indicates, or -e6. These column products are precisely the compositions mentioned in my essay. {e1 e2 e3} and {e6 e5 e3} have e3 in common, and the only other triplet including e3 is {e7 e4 e3} which indeed is the resultant row from the composition operation. So we can look at this row and see for R0, R3, R4 and R7 the final result for e5 * (e1 * e2) will be +e6, and for R1, R2, R5 and R6 the result will be -e6. Doing the same row composition on our first example above, the composition ends up on the all +1 row, which is where every algebraic invariant product term will end up, independent of the number of products. If along the way we needed to multiply by the scalar e0 or the very same current result basis, these rules are singularly defined, so one must stay on the current row.
You can see now 7 algebraically variant rows to land on, and might be questioning how I get to my claim of twice this number. We must bring in the Left Octonion Algebras into the discussion, or more precisely the anti-automorphism Right to Left morph which negates all seven triplet rules. If you ended up on a particular variant row through the application of an even number of variant products, changing every one would not change the final result. However, if you did an odd number of variant products, the anti-automorphism would change the result sign. Thus the variant count parity doubles the 7 to 14. My V+{abc} is even count, and V-{abc} is odd count. Since the composition rule is closed, any number of basis products may be done with comparable results.
Now, what happens in each of these variant sets is that for any legitimate Octonion Algebra definition change, every product term in a set will either change sign, or not change sign. My "Law of Octonion Algebraic Invariance" states any observable described by Octonion Algebra must be an algebraically invariant form. The corollary to this is any algebraically variant form is not observable. If we individually add/subtract per sign all product terms in a variant set and force a zero result for each set, a mixed bag of variant and invariant product terms now becomes fully invariant since +0 = -0. These are my "Homogeneous Equations of Algebraic Constraint". It is easy to believe experimentation will not show everything that needs to be seen. I am thinking these equations of constraint are extremely important.
Octonion Algebra is not talking softly here, it is shouting!
