I'm reading the Oersted Medal Lecture by David Hestenes to improve my understanding of Geometric Algebra and its applications in Physics. I understand he does not start from a mathematical "clean slate", but I don't care for that. I want to understand what he's saying and what I can do with this geometric algebra.
On page 10 he introduces the unit bivector i. I understand (I think) what unit vectors are: multiply by a scalar and get a scaled directional line. But a bivector is a(n oriented) parallellogram (plane). So if I multiply the unit bivector i with a scalar, I get a scaled parallellogram?
The bivector "i" is the Hestenes thing which corresponds to what is normally called dx wedge dy. This is an antisymmetric 2-tensor with components 1 and -1 at the x,y and y,x positions, and it is represented by a little area square in the x-y plane. This is a differential form.
What Hestenes does to produce geometric algebra is to multiply every vector index by gamma matrices, and these gamma matrices make an antisymmetric algebra with the following law:
$$ \gamma_i \gamma_j + \gamma_j \gamma_i = 2g_{ij}$$
Where g is the metric tensor, usually $g_{ij}=\delta_{ij}$ for 3 dimensional (or Euclidean) geometric algebra. In 3d, the $\gamma$ matrices have a standard representation with the Pauli matrices.
This means that when Hestenes is talking about the unit vector, he isn't thinking of it like a unit vector, but as what other people would call the slash of the unit vector, which is the $gamma$ matrices dotted with this unit vector, or in this case, just $\gamma_x$. The $\gamma$ matrices square to 1, and anticommute. So when he multiplies two unit vectors in geometric algebra, he gets
$$ \gamma_x \gamma_y = {1\over 2} (\gamma_x \gamma_y - \gamma_y\gamma_x)=\sigma_{xy}$$
Where the last equality is a definition: $\sigma_{ij}$ is defined as the antisymmetric product of $\gamma$ matrices:
$$ 2\sigma_{ij} = \gamma_i \gamma_j - \gamma_j \gamma_i $$
(most authors omit the factor of 2 on the left hand side). This definition of $\sigma$ is redundant off the diagonal, by antisymmetry of $\gamma$ multiplication. It's just defined this way to make sure that $\sigma_{ii}$ is zero, not 1.
So for Hestense, the unit two-form $dx\wedge dy$ is contracted with sigma, so it is just $\sigma_{xy}$. This means that it has the property that it squares to -1. You can see this by squaring and anticommuting the $\gamma$ matrices.
Hestene's approach hides the gamma matrices. The upside is that this gives you a quick intuition for $\gamma$ algebra without cumbersome notation for the $\gamma$ matrices or explicit matrix representations. The downside is that you don't learn symmetric tensors, and the notation is wildly different from what everyone else uses, with insufficient payoff (as far as I can see) to make up for it. Keeping the gammas explicit makes GA readable to me, but Hestenes prefers to hide them. It's no big deal to translate.