Proof that geometric product is associative

Question

Geometric product has nice property since it is a ring and it is associative to multiplication, which is not the case for vector cross product. But besides it is an axiom for geometric product, in the process of actually defining geometric product in a constructive way, is there a proof that it is indeed satisfy the associativity? i.e., the geometric product of a blade $A_r$ and blade $B_s$ by grade expansion of $$ A_rB_s = \langle A_rB_s\rangle_{|r-s|} + ... + \langle A_rB_s\rangle_{r+s}$$ is this associative?

Answer 1

If you defined the geometric product "from the bottom" on basis elements, then it follows from the fact the product is defined to be associative on the basis elements (See proposition 1 pg 4 in this text by Jacobson for a proof that associativity on basis elements is sufficient for associativity of the ring.) (I think this is basically what you're asking when you're talking about the grade decomposition.)

If you are defining it from the top down as a quotient of the tensor algebra on $V$, then it is associative because the tensor algebra is associative.

Answer 2

There are two approaches:

1. Write down a list of axioms for geometric algebra. Associativity probably needs to be on the list. Advantage: One can get down to the business of using the algebra right away. Disadvantage: How does one know that the list of axioms does not hide an inconsistency?

2. For most people "it has been proved consistent" is a good enough answer to the question above. But some will want to see a construction of GA and proofs of its properties, including associativity of the geometric product. There are several proofs. My own is the topic of the paper An elementary construction of the geometric algebra, Adv. Appl. Clif. Alg. 12, 1-6 (2002). A somewhat improved version is available at my website http://faculty.luther.edu/~macdonal/ . The paper cites other proofs.