Applying the Triple Product in n-dimension

[rachbomb]

I would like to show that (cxb).a = (axc).b in Rⁿ where x denotes the cross product and . denotes the dot product.

Since the cross product is only defined in R¹, R³, and R⁷, my inclination is to prove the above equation in cases (case one being a,b,c are vectors in R¹, etc). However, this seems a bit tedious, particularly for R⁷.

I am familiar with the Triple Product in R³, but am unsure if it applies in Rⁿ, and if so how to prove it. If so, this seems like a much quicker and more concise proof.

Please help! Any assistance is greatly appreciated. Thanks!

 

[Simon_Tyler]

The cross-product in 7 dimensions is not unique - since there is a 5-dimensional subspace orthogonal to any 2 linearly independent vectors. (There is a nontrivial binary cross product only in 3 and 7 dimensions.)

Let's use the choice of cross product given by Lounesto (http://en.wikipedia.org/wiki/Seven-dimensional_cross_product#Coordinate_expressions").
Given the basis e_i where i=1,...,7 modulo 7, the cross product is the antisymmetric and cyclic extension of $$e_i\times e_{i+1}=e_{i+3}$$ i.e.
$$\begin{align*} e_i\times e_{i+1}&=e_{i+3}=-e_{i+1}\times e_i \\ e_{i+3}\times e_i&=e_{i+1}=-e_i\times e_{i+3} \\ e_{i+1}\times e_{i+3}&=e_i=-e_{i+3}\times e_{i+1} \end{align*}$$
This can be written using Kronecker deltas
$$e_i\times e_j=\Delta_{ijk}e_k =\left(\delta _{i,j+1,k+3}-\delta _{i+1,j,k+3}+\delta _{i+3,j,k+1} -\delta _{i,j+3,k+1}+\delta _{i+1,j+3,k}-\delta _{i+3,j+1,k}\right)e_k$$
where, by construction, the symbol $$\Delta_{ijk}$$ is antisymmetric under exchange of indices and symmetric under cyclic permutations.

The scalar triple product is then
$$(x\times y)\cdot z = (x_iy_j\Delta_{ijk}e_k)\cdot(z_le_l) = x_iy_jz_k\Delta_{ijk}$$
from which it's clear that it the scalar triple product inherits all of the symmetry properties of $$\Delta$$.

Using the above choice of cross product, the explicit scalar triple product is
$$\begin{align*} (x\times y)\cdot z &= \left(x_6 y_2+x_4 y_3-x_3 y_4+x_7 y_5-x_2 y_6-x_5 y_7\right) z_1 +\left(-x_6 y_1+x_7 y_3+x_5 y_4-x_4 y_5+x_1 y_6-x_3 y_7\right) z_2\\ &+\left(-x_4 y_1-x_7 y_2+x_1 y_4+x_6 y_5-x_5 y_6+x_2 y_7\right) z_3 +\left(x_3 y_1-x_5 y_2-x_1 y_3+x_2 y_5+x_7 y_6-x_6 y_7\right) z_4\\ &+\left(-x_7 y_1+x_4 y_2-x_6 y_3-x_2 y_4+x_3 y_6+x_1 y_7\right) z_5 +\left(x_2 y_1-x_1 y_2+x_5 y_3-x_7 y_4-x_3 y_5+x_4 y_7\right) z_6\\ &+\left(x_5 y_1+x_3 y_2-x_2 y_3+x_6 y_4-x_1 y_5-x_4 y_6\right) z_7 \end{align}$$
with which you can also check the symmetries.

 

[BruceG]

This may be not the answer you are looking for, but you can do a proper generalisation to Rⁿ by moving over to the wedge-product of 1-forms.

Then the proof follows immediately for the assocoativity and antisymmetry of the wedge product:

a^b^c = -a^c^b = +c^a^b

Notes:
1. The special algebraic form of the wedge product in 1,3,7, I guess, follows from the existence of normed division algebras in 2,4,8 dims (namely C,H,O).

2. To relate back to (axb).c in 3-d you have to appreciate that in 3-d a 3-form a^b^c is dual to a 0-form (i.e. scalar) thus a^b^c and (axb).c both represent the volume spanned by a,b,c.