2
$\begingroup$

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.

enter image description here

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ This adds to my confusion. Then the outer (wedge) product result in not only r+s grade blades? I thought that r,s only represents the highest grades. $\endgroup$
    – Colim
    Commented Aug 12, 2022 at 17:14
  • $\begingroup$ I've amended my answer to try to clarify. $\endgroup$ Commented Aug 12, 2022 at 21:17
  • $\begingroup$ I thought that $A_r$ meant any multivector with, at most, blades or r grade $\endgroup$
    – Colim
    Commented Aug 12, 2022 at 22:32
  • $\begingroup$ $ \left \langle \left \langle A \right \rangle _r \left \langle B \right \rangle _s \right \rangle _{r+s}$ means that only the $r+s$ blades are taken from the product of r-blades with s-blades. Then the product of blades that share a common basis vector will have it eliminated, and will never reach the $r+s$ grade. The table shows in green all the blades that satisfy $rblade*sblade=[r+s]blade$. Is then correct to only consider the products marked in green, when calculating the wedge product? $\endgroup$
    – Colim
    Commented Aug 12, 2022 at 22:37
  • $\begingroup$ Usually when one writes $A_r$, what is meant is $A_r = {\langle A \rangle}_r$. This is the grade $r$ component of $A$. $\endgroup$ Commented Aug 13, 2022 at 2:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .