I read that a couple times. I found it unsatisfying because you are dropping terms that happen to have a coefficient of zero without a clear explanation of how those terms are even well formed in the first place.
I suppose this could be answered if you explicitly stated what set the geometric product acted on. As a guess, and from skimming Wikipedia, it’s the direct sum of scalars, vectors, bivectors, etc, up through n-vectors. So 2 + x∧y + 3x∧y∧z is a valid output. And it’s probably straightforward to show that the geometric product is actually defined on this space.
Ok I understand your question better, might update the article. Like the other commenter said, you can keep applying the geometric product more than once. To find what happens you can just work with the basis vectors.
So, you basically have three case:
x (xx) = x (1) = x
-> a vector
x (xy) = x (x.y + x^y) = x (x.y) + xxy = x (x.y) + y
-> a vector
x (yz) = x (y.z + x^y) = x (y.z) + xyz
-> a vector + a trivector. This only happens if the three vectors are independent, which can never be the case for -ava
I suppose this could be answered if you explicitly stated what set the geometric product acted on. As a guess, and from skimming Wikipedia, it’s the direct sum of scalars, vectors, bivectors, etc, up through n-vectors. So 2 + x∧y + 3x∧y∧z is a valid output. And it’s probably straightforward to show that the geometric product is actually defined on this space.
(Hi Marc!)