# 'Cross product' for 3 vectors in 4D

so I understood that for the cross product to have meaning in $n$-dimensions, one needs $n-1$ vectors. I tried reading about it, but I couldn't find any good resources on exterior algebra. So my question is if there is a simple way to define the cross product in $4$D for three vectors (or just application of the generalization for this specific case?), or if anyone has a good resource on exterior algebra?

• Try Winitzki: sites.google.com/site/winitzki/linalg Nov 13, 2017 at 7:18
• What do you know about the Levi-Civita tensor? Nov 13, 2017 at 7:19
• Just extend the determinant version of the cross product en.wikipedia.org/wiki/Cross_product#Matrix_notation to 4-by-4 matrices. Nov 13, 2017 at 7:23
Yes, you can generalize using the matrix determinant, e.g. $$\det\left(\begin{array}{llll}\mathbf{i}&x_1&y_1&z_1\\\mathbf{j}&x_2&y_2&z_2\\\mathbf{k}&x_3&y_3&z_3\\\mathbf{l}&x_4&y_4&z_4\\\end{array}\right).$$
• There are $n-1$ vectors, i.e. $x=(x_1,x_2,x_3)$, $y=(y_1,y_2,y_3)$, and $z=(z_1,z_2,z_3)$. The $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ vectors are the base vectors. You can extend this to any dimension by way of the determinant, which is defined for any $n\times n$ matrices. Nov 13, 2017 at 14:46
• @Yoav2000 you can put the base vectors vertically or horizontally, and the vectors you're attempting to "cross" must be placed vertically or horizontally accordingly. What's more you can put them on any row or column. If the base vectors are on an even row or column then the result is negated, so it alternates between $\pm$. Don't forget to +1 if you liked the answer. Dec 11, 2017 at 8:54
• Correction to the above earlier comment: There are $n-1$ vectors, i.e. $x=(x_1,x_2,x_3,x_4),y=(y_1,y_2,y_3,y_4)$, and $z=(z_1,z_2,z_3,z_4)$. The $\textbf{i}$, $\textbf{j}$, $\textbf{k}$, and $\textbf{l}$ vectors... so there are $n-1=4-1=3$ vectors. Dec 11, 2017 at 9:02