A new product – Geometric Algebra

After having defined the elements of geometric algebra as combinations of its bases, we are curious to see how the product of two vectors is expressed. Let us assume that we are in three-dimensional space but for the sake of computation the two vectors lie in the plane $e_1 e_2$ :

$\pmb{a} = a_1 e_1 + a_2 e_2$
$\pmb{b} = b_1 e_1 + b_2 e_2$

$\pmb{ab} = a_1 e_1 b_1 e_1 + a_1 e_1 b_2 e_2 + a_2 e_2 b_1 e_1 + a_2 e_2 b_2 e_2 =$
$= (a_1 b_1 + a_2 b_2) + (a_1 b_2 - a_2 b_1) \: e_1 e_2$

The first term sounds familiar: we immediately see that it is the scalar product of the two vectors.
The second term, on the other hand, reminds us of something we have already seen, but not quite the same: it looks like a disguised vector product. In fact the vector product between $\pmb{a}$ and $\pmb{b}$ , if we want to express it by means of the coordinates of the vectors, is

$\pmb{a} \times \pmb{b} = (a_1 b_2 - a_2 b_1)\: e_3$

where the $e_3$ versor reminds us that the resulting vector is perpendicular to the plane defined by the factors and is oriented with the right-hand rule (right-handed basis).

In other words, the second term has the same magnitude as the vector product, but is not imagined in a perpendicular plane, but lies in the plane of the two factors. It’s a bivector, generated by the geometrical extension of the factors (wedge).
At this point we could write the complete expression of the geometric product between vectors:

$\pmb{ab} &=& \pmb{a} \cdot \pmb{b} + \pmb{a} \wedge \pmb{b}$

if we really like the “old” vector product, we need to transform the base bivector $e_1 e_2$ into the base $e_3$ . This result is obtained with the duality operator, as we will see later.
The notable result is $\pmb{a} \wedge \pmb{b} = - I \: (\pmb{a} \times \pmb{b})$ or the reverse expression: $\pmb{a} \times \pmb{b} = I \: (\pmb{a} \wedge \pmb{b})$

And let’s say that this result is surprising: in geometric algebra the product of two vectors generates a number composed of a scalar and a bivector and therefore reminds us a complex number! In fact, he has the scalar product as real part and the module of the vector product as imaginary part.

The geometric product in the calculations behaves quite well, but it has a crucial peculiarity:

but we can’t even say that it is anticommutative, because the geometric product is composed of two parts: one that commutes (the scalar product) and the other that anticommutates (the external product).

Exploiting this property, it is useful to extract the two components from the geometric product: it will be very useful in the calculations.

$\begin{equation} \pmb{a} \cdot \pmb{b} = \frac {1}{2} (\pmb{ab} + \pmb{ba})$
$\begin{equation} \pmb{a} \wedge \pmb{b} = \frac {1}{2} (\pmb{ab} - \pmb{ba})$

A final definition: the external product of a certain number of linearly independent vectors is called blade. It is certainly a multivector, but a multivector is not necessarily a blade. For example, in four dimensions the multivector $e_0e_1 + e_2e_3$ is not a blade because it cannot be expressed as an external product of vectors.
If the vectors are orthogonal, it can be said that the blade is the geometric product of $k$ non-zero vectors.

$(AB)C = A(BC)$	is associative
$\lambda A = A \lambda$	scalar multiplication is commutative
$A(B+C) = AB+AC$	is distributive wrt the sum
In general $AB \neq BA$	it’s NOT commutative!