[ww520]

The second term is not a cross product. It's the exterior product (or bivector). Cross product only works in 3D. Exterior product can work in any higher dimension.

A cross product of two 3D vectors is another vector, perpendicular to the plane containing the two vectors. The exterior product is a 2-vector (thus bivector) that sweeps the parallelogram between the two vectors; it's on the plane containing the two vectors. In 3D, the cross product vector is perpendicular to bivector plane.

[mjcohen]

No. In n dimensions a cross product of n-1 n-vectors is the determinant with the top row being the basis elements (e_1, e_2, ..., e_n) and the next n-1 rows being the n-1 vectors. The result is a n-vector orthogonal to the n-1 vectors.

In 2 dimensions it is i j x y

In 3 dimensions it is

i j k x1 y1 z1 x2 y2 z2

And so on.

[itishappy]

That's not a particularly fruitful generalization. The traditional definition begins:

> The cross product on a Euclidean space V is a bilinear map from V × V to V.

This is required for some of the cross product's useful properties such as:

  |a x b| = |a| |b| sin(theta)

This only works in 0, 1, 3, and 7 dimensions.

[lupire]

The exterior product is the "correct" version of the cross product. I'm OK with the good guys seizing the name.

Cross product works in dimension 2^k-1 for k in 0,1,2,3. (real / singleton, complex / duonion, quaternion, octonion) It is trivial in 0D and 1D, defined uniquely in 3D, and unique up to some arbitrary negations in 7D.

https://en.m.wikipedia.org/wiki/Seven-dimensional_cross_prod...