On "The Inner Products of Geometric Algebra" (Leo Dorst), page 39, equation 2.3, the outer product between multivector $A_r$ and $B_s$, of grades $r$, $s$, is defined as:
$$A_r \wedge B_s = \sum_{r,s} \left \langle \left \langle A \right \rangle _r \left \langle B \right \rangle _s \right \rangle _{r+s}$$
I have doubts about how to calculate it, so I need confirmation of what I think it means.
I interpret that to calculate the product, it should be taken only the highest degree blades of each multivector, calculate the geometric product, and discard from the result anything that is not of degree $r+s$, like on this example:
$A_r=1+e_1+e_{12}+2*e_{23}$
$B_s=2+e_1+e_4$
$r=2, s=1$
$\left \langle A_r \right \rangle _r =e_{12}+2*e_{23}$ (not exactly $\left \langle A \right \rangle _r$)
$\left \langle B_s \right \rangle _s =e_1+e_4$ (not exactly $\left \langle B \right \rangle _s$)
$A_r * B_s =-1*e_2 + 2*e_{123} + 1*e_{124} + 2*e_{234}$
$r+s=3$
$A_r \wedge B_s = 2*e_{123} + 1*e_{124} + 2*e_{234}$
I made this table of blade products (without the sign on the result). The $1$ on the binary numbers represent the presence or absence of a basis vector $e_i$ on each blade.
What I deduce from the definition, is that the outer product only multiplies blades with no basis vector in common. The rest get discarded, so, when calculating the wedge product, it should be possible to avoid multiplying blades with common basis vectors, and only the ones marked in green are worth calculating (It looks like they form a Sierpinski triangle).
Is this correct? Is the most general definition of wedge product?
Best Answer
I'd write, for multivectors $ A, B $ $$A \wedge B = \sum_{r,s} \left \langle \left \langle A \right \rangle _r \left \langle B \right \rangle _s \right \rangle _{r+s}$$ or for $ A_r, B_s $, k-vectors of grades $ k = r, s $ respectively $$ A_r \wedge B_s = {\left\langle{{A_r B_s}}\right\rangle}_{{r+s}}.$$ i.e.: for the general outer product of two multivectors, you take the outer products of all the respective grades, but if you are operating on k-vectors (multivectors with only a single grade $k$), of blades (which are, by definition, products of perpendicular vectors, so they can be written as outer products), then the outer product is just the grade selection of the sum of the respective grades.
I'd rewrite your example as $$\begin{aligned} A &= 1 + e_1 + e_{12} + 2 e_{23} \\ B &= 2 + e_1 + e_4,\end{aligned}$$ not using any $r,s$ suffixes here, as these are not k-vectors of any specific grade. For these multivectors, you can select out all the specific grades, which are $$\begin{aligned} A_0 &= 1 \\ A_1 &= e_1 \\ A_2 &= e_{12} + 2 e_{23} \\ B_0 &= 2 \\ B_1 &= e_1 + e_4.\end{aligned}$$ Now you can form the outer products of all these single-grade components $$\begin{aligned}A \wedge B &=\left( { A_0 \wedge B_0 } \right)+ \left( { A_0 \wedge B_1 + A_1 \wedge B_0 } \right)+ \left( { A_2 \wedge B_0 + A_1 \wedge B_1 } \right)+ \left( { A_2 \wedge B_1 } \right) \\ &=\left( { 2 } \right) + \left( { e_1 + e_4 + 2 e_1 } \right) + \left( { 2 e_{12} + 4 e_{23} + e_1 \wedge \left( { e_1 + e_4 } \right) } \right) + \left( { e_{124} + 2 e_{231} + 2 e_{234} } \right) \\ &= \left( { 2 } \right) + \left( { 3 e_1 + e_4 } \right) + \left( { 2 e_{12} + 4 e_{23} + e_{14} } \right) + \left( { e_{124} + 2 e_{123} + 2 e_{234} } \right).\end{aligned}$$ Here, I've used braces to group the respective grade $0,1,2,3$ components of this general outer product.
If you look closer at the cited paper, you'll see that it isn't $A_r \wedge B_s$ on the LHS, but $A \wedge B$. The sum on the RHS is over all grades $r$ of $A$ and all grades $s$ of $B$ (not the max grade as mentioned in the comment.)
In practice, this abstract wedge product is not often used. Instead, wedge products are most often taken between k-vectors that have a single grade (k), in which case, the $r,s$ indexes are the max (and only) grades, and there is no sum.