Axiomatic introduction of the geometric product (Clifford Algebra)

Question

While reading "Introduction to Clifford Algebra" by John Denker (html version or pdf version), I have trouble to understand the generalization of the geometric product from vectors to general clifs. In particular I dont see where equation (24) comes from. That equations states that from $A:=γ_1∧γ_2$ and $B:=(γ_2 + γ_3)∧(γ_4 + γ_1)$ should follow the geometric product: $$AB = 1 + γ_1γ_4 + γ_2γ_3 + γ_1 γ_2 γ_3 γ_4.$$

It is unclear to me what terms like $γ_1γ_4$ mean and where the overall expression comes from.

Answer 1

Using the identity for vectors $v,w$ that $vw=v\cdot w + v\wedge w$ and the corollary that $vw=v\wedge w=-wv$ if $w\perp v$, you get

$$(\gamma_1\wedge\gamma_2)((\gamma_2 + \gamma_3)\wedge(\gamma_4 + \gamma_1))\\ =\gamma_1\gamma_2((\gamma_2+\gamma_3)(\gamma_4+\gamma_1)-(\gamma_2+\gamma_3)\cdot(\gamma_4+\gamma_1))\\ \gamma_1\gamma_2(\gamma_2\gamma_4+\gamma_2\gamma_1+\gamma_3\gamma_4+\gamma_3\gamma_1-0)\\ =\gamma_1\gamma_2\gamma_2\gamma_4+\gamma_1\gamma_2\gamma_2\gamma_1+\gamma_1\gamma_2\gamma_3\gamma_4+\gamma_1\gamma_2\gamma_3\gamma_1\\ =\gamma_1\gamma_4+1+\gamma_1\gamma_2\gamma_3\gamma_4+\gamma_2\gamma_3$$

Answer 2

If $\gamma_1$ and $\gamma_2$ are vectors then $$ \gamma_1 \gamma_2= \gamma_1 \ast \gamma_2 + \gamma_1 \wedge \gamma_2 $$ that is the geometric product is decomposed as a sum of a scalar and wedge product. The object $ \gamma_1 \wedge \gamma_2$ is a directed plane element, i.e. a bivector, while $\gamma_1 \ast \gamma_2$ is a number.