3

There are many MSE posts about how to define a cross product in $\mathbb{R^4}$. It is impossible to define a cross product of two vectors in $\mathbb{R^4}$, since there are infinitely many directions perpendicular to those two vectors, and we don't know which direction to choose. However, If we are given THREE vectors $A,B,C$, it is possible to find a unique direction perpendicular to this three vectors, if $A,B,C$ are independent. However, finding this perpendicular vector involves solving a system of equations.

So my question is：can we define a Quasi Cross Product $\{A,B,C\}$ on $\mathbb{R^4}$, so that we can find a direction perpendicular to $A,B,C$ without solving a system of equations?

Ma Joad
  • 7,420

3 Answers3

6

You have the "same" determinant formula. If $\vec{a} = (a_1,a_2,a_3,a_4)$, similarly for $\vec{b}$ and $\vec{c}$, then $$\vec{a}\times\vec{b}\times \vec{c} = \begin{vmatrix} \vec{e}_1 & \vec{e}_2 & \vec{e}_3 & \vec{e}_4 \\ a_1 & a_2 & a_3 & a_4 \\ b_1 & b_2 & b_3 & b_4 \\ c_1 & c_2 & c_3 & c_4\end{vmatrix},$$where $(\vec{e}_1,\ldots,\vec{e}_4)$ is the standard basis for $\Bbb R^4$. This does not require solving a system. Example: $$(1,1,0,0)\times (0,1,1,0) \times (0,0,1,1) = \begin{vmatrix} \vec{e}_1 & \vec{e}_2 & \vec{e}_3 & \vec{e}_4 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1\end{vmatrix} = (1,-1,1,-1).$$

Ivo Terek
  • 77,665
  • 1
    An interested person could check this for determinant $\vec x = \vec a \times \vec b \times \vec c$ by computing the three dot products $\vec x \circ \vec a$, $\vec x \circ \vec b$, $\vec x \circ \vec c$ and ensuring they are all zero. – DanielV Dec 02 '18 at 08:36
1

Yes, it is the Hodge dual of the wegde product. Given three vectors $A=(a_1,a_2,a_3,a_4)$, $B=(b_1,b_2,b_3,b_4)$, and $C=(c_1,c_2,c_3,c_4)$, the triple-product is defined as $$\{A,B,C\}= (\det M_1, \det M_2, \det M_3, \det M_4)$$ with $$ M_1 = \begin{pmatrix} a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ a_4 & b_4 & c_4 \end{pmatrix} $$ where the other matrices $M_{2}$, $M_3$, $M_4$ are obtained by cyclically permuting the indices. Note that the product is antisymmetric with respect to exchanging any of its arguments.

Fabian
  • 23,360
1

The short answer is yes. One way is to take the formal determinant $$\left|\begin{matrix}e_1&e_2&e_3&e_4\\ a_1&a_2&a_3&a_4\\ b_1&b_2&b_3&b_4\\ c_1&c_2&c_3&c_4\\ \end{matrix}\right|$$ where $e_1,\ldots,e_4$ are the standard unit vectors, and $a=\sum a_ie_i$ etc., are the three vectors.

Or you can rephrase this in terms of exterior powers and the Hodge star operator.

All this works in $n$ dimensions too.

Angina Seng
  • 158,341