Vector product in DD dimensions

It is often1 said that the vector product (“×\times”, cross product) cannot be generalized to other than three dimensions. But it depends a bit on how one defines the vector product. If we forgo the requirement that is must be a binary operation (i.e. the product of two vectors) but instead require D−1D - 1 vectors for the DD dimensional vector product, it generalizes quite naturally: u⃗=×(v⃗1,v⃗2,...,v⃗D−1)=v⃗1×v⃗2×...,v⃗D−1=∣e⃗1e⃗2...e⃗Dv1,1v⃗1,2...v1,Dv2,1v⃗2,2...v2,D⋮⋮⋱⋮vD−1,1v⃗D−1,2...vD−1,D∣.\vec u = \times(\vec v_1, \vec v_2, ... , \vec v_{D-1}) = \vec{v}_1\times\vec v_2 \times ..., \vec v_{D-1} = \begin{vmatrix} \vec e_1 & \vec e_2 & ... & \vec e_D\\ v_{1,1} & \vec v_{1,2} & ... & v_{1,D}\\ v_{2,1} & \vec v_{2,2} & ... & v_{2,D}\\ \vdots & \vdots & \ddots & \vdots\\ v_{D-1,1} & \vec v_{D-1,2} & ... & v_{D-1,D}\end{vmatrix}. By the properties of determinants, it remains to be distributive over addition (i.e. it is a linear, product-style operator), u⃗=0⃗\vec u = \vec 0 if any two vectors in the product are parallel, and u⃗\vec u is perpendicular to every argument of the product. Using this definition, we can use it for any dimension larger than one. For example, the vector product of V⃗1=2e⃗1+1e⃗2\vec V_1 = 2\vec e_1 + 1\vec e_2 in 2 dimensions is U⃗\vec U, U⃗=×(V⃗1)=∣e⃗1e⃗221∣=1e⃗1−2e⃗2.\vec U = \times(\vec V_1) = \begin{vmatrix}\vec e_1 & \vec e_2\\ 2& 1\end{vmatrix} = 1\vec{e}_1 - 2\vec{e}_2.

  1. It is not often a topic of discussion, really.↩︎

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