Squaring a vector using geometric algebra I'm doing research involving clifford algebra and I'm having difficulty understanding this one axiom: $a^2  = g(a,a)$. It states that this is the square of a vector and dividing the original vector by this can give an inverse vector. I was wondering how I should go about calculating this.
 A: If $a^2  = g(a,a)\neq 0$, then, $\frac{1}{g(a,a)}a$ is clearly the inverse of $a$.
As for "understanding" the axiom, it is simply one of the defining features of a Clifford algebra. The quadratic form associated with the bilinear form is encoded into the product by design. We do this because we can do useful things with it.
A: I don't understand what it is you don't understand. 
Given a vector space $V$ with a symmetric non-degenerate bilinear form, $B$, you define its Clifford algebra as:
$$
\operatorname{Cl}(V,B) = \bigotimes V\,/\, I\left(u\otimes v+v\otimes u-2B(u,v)1_{\otimes V}\,\left.\right\rvert u,v\in V\,\right).
$$
Where $\otimes V$ is the tensor algebra of $V$ and $I(\cdots)$ is the ideal generated by these elementes. This implies that in the Clifford algebra (in the quotient),
$$
uv + vu = 2B(u,v)1_{\operatorname{Cl}}.
$$
There isn't anything to calculate here, this is only a "symbolic" construction. If, for example, you were working with some representation of the Clifford algebra (like the gamma matrices in the Dirac equation), the product is represented  as the matrix product.
