Orthogonal rewriting of multi-vectors

Question

Use Orthogonal rewriting of multi-vectors to verify $(a \land b \land c)$$(d \land e)$ where a,b,c,d,e are vectors in a dimension of at least 5?
Note: by definition the grade $\vert r-s \vert$ vector of $A_r$ $B_s$ is $A_r$ $\cdot$ $B_s$

I know that you can use the geometric product of an r-vector($A_r$ i.e., a vector of grade r) and a s-vector ($B_s$) is a sum of multi-vectors whose lowest grade is $\vert r-s \vert$ and highest grade is $\vert r+s \vert$. I need help in determining the how to verify the multi-vectors. I am new to multi-vectors

Answer 1

As with one of your commenters, I don't know what you mean by orthogaonal rewriting, nor verify.

I can only guess that by verify you mean to expand the product in terms of it's component grades. That expansion takes the form

$$\begin{aligned}\left( { a \wedge b \wedge c } \right) \left( { d \wedge e } \right)&=\left( { a \wedge b \wedge c } \right) \cdot \left( { d \wedge e } \right) \\ &+{\left\langle{{ \left( { a \wedge b \wedge c } \right) \left( { d \wedge e } \right) }}\right\rangle}_{3} \\ &+a \wedge b \wedge c \wedge d \wedge e \end{aligned}$$

The first term is a vector and expands as

$$\begin{aligned}\left( { a \wedge b \wedge c } \right) \cdot \left( { d \wedge e } \right)&=\left( { \left( { a \wedge b \wedge c } \right) \cdot d } \right) \cdot e \\ &=\left( { a \wedge b (c \cdot d) +b \wedge c (a \cdot d) +c \wedge a (b \cdot d) } \right) \cdot e \\ &= a (b \cdot e) (c \cdot d) +b (c \cdot e) (a \cdot d) +c (a \cdot e) (b \cdot d) \\ &-(a\cdot e) b (c \cdot d) -(b\cdot e) c (a \cdot d) -(c\cdot e) a (b \cdot d) \\ &= a \left( { (b \cdot e) (c \cdot d) -(c\cdot e) (b \cdot d) } \right) \\ &+b \left( { (c \cdot e) (a \cdot d) -(a\cdot e) (c \cdot d) } \right) \\ &+c \left( { (a \cdot e) (b \cdot d) -(b\cdot e) (a \cdot d) } \right).\end{aligned}$$

The second term is a trivector and expands as

$$\begin{aligned}2 {\left\langle{{ \left( { a \wedge b \wedge c } \right) \left( { d \wedge e } \right) }}\right\rangle}_{3}&={\left\langle{{ \left( { a \wedge b \wedge c } \right) (d e - e d)}}\right\rangle}_{3} \\ &={\left\langle{{ \left( { a \wedge b \wedge c } \right) d e -\left( { a \wedge b \wedge c } \right) e d}}\right\rangle}_{3} \\ &={\left\langle{{ \left( { a \wedge b \wedge c \wedge d } \right) e +\left( { \left( { a \wedge b \wedge c } \right) \cdot d } \right) e }}\right\rangle}_{3} \\ &-{\left\langle{{ \left( { a \wedge b \wedge c \wedge e } \right) d+\left( {\left( { a \wedge b \wedge c } \right) \cdot e } \right) d}}\right\rangle}_{3} \\ &= \left( { a \wedge b \wedge c \wedge d } \right) \cdot e + \left( { \left( { a \wedge b \wedge c } \right) \cdot d } \right) \wedge e \\ &- \left( { a \wedge b \wedge c \wedge e } \right) \cdot d- \left( {\left( { a \wedge b \wedge c } \right) \cdot e } \right) \wedge d \\ &= \left( { a \wedge b \wedge c } \right) d \cdot e - \left( { a \wedge b \wedge d } \right) c \cdot e \\ &+ \left( { a \wedge c \wedge d } \right) b \cdot e - \left( { b \wedge c \wedge d } \right) a \cdot e \\ &+ \left( { a \wedge b \wedge e } \right) c \cdot d - \left( { a \wedge c \wedge e } \right) b \cdot d + \left( { b \wedge c \wedge e } \right) a \cdot d \\ &- \left( { a \wedge b \wedge c } \right) e \cdot d + \left( { a \wedge b \wedge e } \right) c \cdot d \\ &- \left( { a \wedge c \wedge e } \right) b \cdot d + \left( { b \wedge c \wedge e } \right) a \cdot d \\ &- \left( { a \wedge b \wedge d } \right) c \cdot e + \left( { a \wedge c \wedge d } \right) b \cdot e - \left( { b \wedge c \wedge d } \right) a \cdot e \\ &= - 2 \left( { a \wedge b \wedge d } \right) c \cdot e + 2 \left( { a \wedge c \wedge d } \right) b \cdot e \\ &- 2 \left( { b \wedge c \wedge d } \right) a \cdot e + 2 \left( { a \wedge b \wedge e } \right) c \cdot d \\ &- 2 \left( { a \wedge c \wedge e } \right) b \cdot d + 2 \left( { b \wedge c \wedge e } \right) a \cdot d.\end{aligned}$$

Putting all the pieces together, assuming no algebra errors (this is really a task better suited for a computer algebra system), the multivector product expands as

$$\begin{aligned}\left( { a \wedge b \wedge c } \right) \left( { d \wedge e } \right)&= a \left( { (b \cdot e) (c \cdot d) -(c\cdot e) (b \cdot d) } \right) \\ &+b \left( { (c \cdot e) (a \cdot d) -(a\cdot e) (c \cdot d) } \right) \\ &+c \left( { (a \cdot e) (b \cdot d) -(b\cdot e) (a \cdot d) } \right) \\ &- \left( { a \wedge b \wedge d } \right) c \cdot e + \left( { a \wedge c \wedge d } \right) b \cdot e \\ &- \left( { b \wedge c \wedge d } \right) a \cdot e + \left( { a \wedge b \wedge e } \right) c \cdot d \\ &- \left( { a \wedge c \wedge e } \right) b \cdot d + \left( { b \wedge c \wedge e } \right) a \cdot d \\ &+a \wedge b \wedge c \wedge d \wedge e \end{aligned}$$

Answer 2

For computer algebra method look at the bac_cab section of -

http://github.com/brombo/galgebra/blob/master/examples/LaTeX/latex_check.pdf

The galgebra modules (python) are at -

http://github.com/brombo/galgebra

If you are interested read the doc first.