Cross product of 4D Vectors I am trying to understand how to do the cross product of two four-dimensional vectors. From what I understood it's not possible, unless the vectors are of the form
$\mathbb R^3 \times \{0\}$ or $\{0\} \times \mathbb R^3$.
So for example, if I have  $u = (1, \frac12, 0, 0)$, $v = (0, 2, 1, 0)$, what is the cross product $w = u \times v$?
 A: Four-dimensional Euclidean space does not have a binary cross product. (If it did, you could use it to define a five-dimensional division algebra, which isn't possible.) However, if you choose a three-dimensional subspace of $\mathbb{R}^4$, then that subspace inherits the inner product, which uniquely determines a cross product (up to choice of orientation).
You seem to be talking about $\mathbb{R}^3\times\{0\}$ as a 3D subspace of $\mathbb{R}^4$, in which case to calculate the cross product of two vectors (in this 3D subspace) you simply ignore the fourth coordinate (which is $0$) and do the calculation with the first three coordinates.
There is a ternary cross product on $\mathbb{R}^4$ in which you can compute a vector perpendicular to three given ones, with size and orientation based on the parallelotope generated by the three vectors (instead of a parallelogram as with two vectors). This can be calculated with differential forms if one was so inclined.
A: Cross products are only defined in $\mathbb{R}^3$ and $\mathbb{R}^7$, because of the direct relationship of the cross-product to quaternions and octonions respectively.
$\mathbb{R}$ (the reals), $\mathbb{C}$ (the complex numbers), quaternions and octonions are the only examples of division algebras that exist. Moving from  $\mathbb{C}$ to the quaternions, you lose commutativity; moving from the quaternions to the octonions, you lose associativity.
If the reason why you are trying to take the cross product of two four-dimensional vectors is because you want to find another vector that is orthogonal to both vectors, then you could consider this suggestion, although you need to use three vectors in $\mathbb{R}^4$ to find a fourth vector that is orthogonal to all three.
A: Being a vector, the vector product has a magnitude and direction.
For the magnitude, you can take the usual definition, i.e., $$||w||=||u||\cdot||v||\cdot \sin(u,v),$$ with $\cos(u,v) = u\cdot v / (||u||\cdot ||v||)$ and $\sin^2 + \cos^2 = 1$.
For its direction, you could impose the usual $w \perp u$ and $w \perp v$, i.e., $w \cdot u = 0$ and $w \cdot v = 0$.   
Presumably, an extra condition is required to define $w$ uniquely in the fourth dimension.
