# Applying the Triple Product in n-dimension

• rachbomb
A volume is invariant under any permutation of a,b,c, i.e. they are equivalent.In summary, the cross-product and scalar triple product have specific algebraic forms in dimensions 1,3,7 and can be related to the wedge product in higher dimensions using the antisymmetry and associativity properties.
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.

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 ei 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.

Last edited by a moderator:
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.

## 1. What is the Triple Product in n-dimension?

The Triple Product in n-dimension is a mathematical operation that involves the cross product and dot product of three vectors in n-dimensional space. It is used to determine the volume of a parallelepiped, which is a three-dimensional shape with six faces.

## 2. How is the Triple Product calculated in n-dimension?

The Triple Product in n-dimension is calculated by taking the dot product of one vector with the cross product of the other two vectors. This can be represented by the formula:
(a x b) ⋅ c = |a| |b| |c| sin(θ), where a, b, and c are the three vectors and θ is the angle between a and b.

## 3. What is the significance of the Triple Product in n-dimension?

The Triple Product in n-dimension is significant because it can be used to calculate the volume of a parallelepiped in n-dimensional space. It is also used in physics and engineering to solve problems involving forces and torque.

## 4. Can the Triple Product be used in any dimension?

Yes, the Triple Product can be used in any dimension. It is not limited to three-dimensional space, although it is most commonly used in three dimensions.

## 5. What are some real-world applications of the Triple Product in n-dimension?

The Triple Product in n-dimension has various applications in fields such as physics, engineering, and computer graphics. It is used to calculate the volume of a parallelepiped in three-dimensional space, determine the moment of inertia of an object, and solve problems involving forces and torque. It is also used in computer graphics to calculate the lighting and shading of 3D objects.

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