Seven-dimensional Cross Product - Coordinate Expressions

Coordinate Expressions

To define a particular cross product, an orthonormal basis {e_j} may be selected and a multiplication table provided that determines all the products {e_i × e_j}. One possible multiplication table is described in the Example section, but it is not unique. Unlike three dimensions, there are many tables because every pair of unit vectors is perpendicular to five other unit vectors, allowing many choices for each cross product.

Once we have established a multiplication table, it is then applied to general vectors x and y by expressing x and y in terms of the basis and expanding x × y through bilinearity.

Lounesto's multiplication table
×	e₁	e₂	e₃	e₄	e₅	e₆	e₇
e₁	0	e₄	e₇	−e₂	e₆	−e₅	−e₃
e₂	−e₄	0	e₅	e₁	−e₃	e₇	−e₆
e₃	−e₇	−e₅	0	e₆	e₂	−e₄	e₁
e₄	e₂	−e₁	−e₆	0	e₇	e₃	−e₅
e₅	−e₆	e₃	−e₂	−e₇	0	e₁	e₄
e₆	e₅	−e₇	e₄	−e₃	−e₁	0	e₂
e₇	e₃	e₆	−e₁	e₅	−e₄	−e₂	0

Using e₁ to e₇ for the basis vectors a different multiplication table from the one in the Introduction, leading to a different cross product, is given with anticommutativity by

More compactly this rule can be written as

with i = 1...7 modulo 7 and the indices i, i + 1 and i + 3 allowed to permute evenly. Together with anticommutativity this generates the product. This rule directly produces the two diagonals immediately adjacent to the diagonal of zeros in the table. Also, from an identity in the subsection on consequences,

which produces diagonals further out, and so on.

The e_j component of cross product x × y is given by selecting all occurrences of e_j in the table and collecting the corresponding components of x from the left column and of y from the top row. The result is:

$\begin{align}\mathbf{x} \times \mathbf{y} = (x_2y_4 - x_4y_2 + x_3y_7 - x_7y_3 + x_5y_6 - x_6y_5)\,&\mathbf{e}_1 \\ {}+ (x_3y_5 - x_5y_3 + x_4y_1 - x_1y_4 + x_6y_7 - x_7y_6)\,&\mathbf {e}_2 \\ {}+ (x_4y_6 - x_6y_4 + x_5y_2 - x_2y_5 + x_7y_1 - x_1y_7)\,&\mathbf{e}_3 \\ {}+ (x_5y_7 - x_7y_5 + x_6y_3 - x_3y_6 + x_1y_2 - x_2y_1)\,&\mathbf{e}_4 \\ {}+ (x_6y_1 - x_1y_6 + x_7y_4 - x_4y_7 + x_2y_3 - x_3y_2)\,&\mathbf{e}_5 \\ {}+ (x_7y_2 - x_2y_7 + x_1y_5 - x_5y_1 + x_3y_4 - x_4y_3)\,&\mathbf{e}_6 \\ {}+ (x_1y_3 - x_3y_1 + x_2y_6 - x_6y_2 + x_4y_5 - x_5y_4)\,&\mathbf{e}_7. \\ \end{align}$

As the cross product is bilinear the operator x×– can be written as a matrix, which takes the form

$T_{\mathbf x} = \begin{bmatrix} 0 & -x_4 & -x_7 & x_2 & -x_6 & x_5 & x_3 \\ x_4 & 0 & -x_5 & -x_1 & x_3 & -x_7 & x_6 \\ x_7 & x_5 & 0 & -x_6 & -x_2 & x_4 & -x_1 \\ -x_2 & x_1 & x_6 & 0 & -x_7 & -x_3 & x_5 \\ x_6 & -x_3 & x_2 & x_7 & 0 & -x_1 & -x_4 \\ -x_5 & x_7 & -x_4 & x_3 & x_1 & 0 & -x_2 \\ -x_3 & -x_6 & x_1 & -x_5 & x_4 & x_2 & 0 \end{bmatrix}.$

The cross product is then given by

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