Exterior product and four vectors?

Question

Quite a straight forward question but something I don't really understand. I have a mediocre grasp on the concept of exterior products in $3$ dimensions. However, is the product defined for two $4$-vectors? If so how is it defined explicitly given two four vectors?

Answer 1

Here is a coordinate description of the exterior product of two vectors in $\mathbb{R}^4$: if $e_i$ denotes the $i$th basis vector, then any $v \in \mathbb{R}^4$ has coordinates $(a^1,a^2 ,a^3,a^4)$, meaning that $v = \sum a^ie_i$. The defining characteristic of the exterior product is that it is bilinear and antisymmetric, so if $w = \sum b^i e_i$ is another vector we must have

$$v \wedge w = \sum_i a^ie_i \wedge \sum_j b^je_j = \sum_{i,j}a^ib^je_i\wedge e_j,$$

and since $e_i \wedge e_j = -e_j \wedge e_i$ this amounts to

$$v\wedge w = (a^1b^2 - a^2b^1)(e_1 \wedge e_2) + \cdots + (a^3b^4 - a^4b^3)(e^3 \wedge e^4)$$

as in Ted Shifrin's answer.

Answer 2

Just as with the cross product of two vectors in $\Bbb R^3$, you want to compute things this way. Take the matrix whose columns are the two vectors $a$ and $b$: $$\begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \\ a_3 & b_3 \\ a_4 & b_4 \end{bmatrix}.$$ Now compute its $\binom 42=6$ $2\times 2$ minors: \begin{align*} x_{12} &= a_1b_2-a_2b_1 \\ x_{13} &= a_1b_3-a_3b_1 \\ x_{14} &= a_1b_4-a_4b_1 \\ x_{23} &= a_2b_3-a_3b_2 \\ x_{24} &= a_2b_4-a_4b_2 \\ x_{34} &= a_3b_4-a_4b_3. \end{align*} These are the standard coordinates of the $2$-vector $a\wedge b$.