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*}<br /> e_i\times e_{i+1}&amp;=e_{i+3}=-e_{i+1}\times e_i \\<br /> e_{i+3}\times e_i&amp;=e_{i+1}=-e_i\times e_{i+3} \\<br /> e_{i+1}\times e_{i+3}&amp;=e_i=-e_{i+3}\times e_{i+1}<br /> \end{align*}<br />
This can be written using Kronecker deltas
<br /> e_i\times e_j=\Delta_{ijk}e_k<br /> =\left(\delta _{i,j+1,k+3}-\delta _{i+1,j,k+3}+\delta _{i+3,j,k+1}<br /> -\delta _{i,j+3,k+1}+\delta _{i+1,j+3,k}-\delta _{i+3,j+1,k}\right)e_k<br />
where, by construction, the symbol \Delta_{ijk} is antisymmetric under exchange of indices and symmetric under cyclic permutations.

The scalar triple product is then
<br /> (x\times y)\cdot z = (x_iy_j\Delta_{ijk}e_k)\cdot(z_le_l) = x_iy_jz_k\Delta_{ijk}<br />
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
<br /> \begin{align*}<br /> (x\times y)\cdot z &amp;=<br /> \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<br /> +\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\\<br /> &amp;+\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<br /> +\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\\<br /> &amp;+\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<br /> +\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\\<br /> &amp;+\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<br /> \end{align}<br />
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.