Operations between vectors include the dot product and the cross product (also known as the vector product or outer product). Among them, the dot product is relatively simple and easily generalizes to high dimensions. However, the cross product is different; generally speaking, it is considered a concept specific to three-dimensional space. The difficulty in generalizing the cross product lies in its multiple interpretations. If we view vectors and their cross products within the framework of tensors (which belongs to the study of differential forms), then cross products for vectors in dimensions higher than three do not exist. But if we simply view the cross product as a tool to “generate a vector orthogonal to given vectors,” then it can be generalized to higher dimensions. The generalization technique is quite ingenious and very similar to the cross product in three-dimensional space.
Reviewing Three-Dimensional Space
To generalize the 3D cross product, it is helpful to first review its origins. The cross product originated from quaternion multiplication, but from a functional perspective, we want to construct a vector \boldsymbol{w}=(w_1,w_2,w_3) such that it is perpendicular (orthogonal) to two known non-collinear vectors \boldsymbol{u}=(u_1,u_2,u_3) and \boldsymbol{v}=(v_1,v_2,v_3). From the perspective of universality, we also hope that the constructed vector has no “singularities”; for this reason, we use only multiplication for the construction. As for the geometric meaning of the cross product, that is a later consideration; after all, the primary goal must be achieved first.
Here, in order to construct such a vector, determinants play a huge role! We consider the determinant: \begin{vmatrix} u_1 & v_1 & w_1 \\ u_2 & v_2 & w_2 \\ u_3 & v_3 & w_3 \end{vmatrix}
From the properties of determinants, we know that if (w_1,w_2,w_3)=(u_1,u_2,u_3), then the determinant is 0. Expanding it gives: u_1 \begin{vmatrix} u_2 & v_2\\u_3 &v_3\end{vmatrix}-u_2 \begin{vmatrix} u_1 & v_1\\u_3 &v_3\end{vmatrix}+u_3 \begin{vmatrix} u_1 & v_1\\u_2 &v_2\end{vmatrix}=0
Similarly, if (w_1,w_2,w_3)=(v_1,v_2,v_3), then the determinant is also 0, and expanding it yields: v_1 \begin{vmatrix} u_2 & v_2\\u_3 &v_3\end{vmatrix}-v_2 \begin{vmatrix} u_1 & v_1\\u_3 &v_3\end{vmatrix}+v_3 \begin{vmatrix} u_1 & v_1\\u_2 &v_2\end{vmatrix}=0
Thus, the vector \left(\begin{vmatrix} u_2 & v_2\\u_3 &v_3\end{vmatrix},-\begin{vmatrix} u_1 & v_1\\u_3 &v_3\end{vmatrix}, \begin{vmatrix} u_1 & v_1\\u_2 &v_2\end{vmatrix}\right) is automatically perpendicular to vectors \boldsymbol{u}=(u_1,u_2,u_3) and \boldsymbol{v}=(v_1,v_2,v_3). The above expression is exactly how we currently define the vector cross product, providing a glimpse into its origin.
In theoretical analysis, we generally write the cross product as: \boldsymbol{u}\times\boldsymbol{v}=\begin{vmatrix} u_1 & v_1 & \boldsymbol{e}_1 \\ u_2 & v_2 & \boldsymbol{e}_2 \\ u_3 & v_3 & \boldsymbol{e}_3 \end{vmatrix} where \boldsymbol{e}_1,\boldsymbol{e}_2,\boldsymbol{e}_3 are the basis vectors of the three-dimensional space.
High-Dimensional Cross Product
With the foundation above, defining the cross product in higher-dimensional spaces is not difficult. For example, for three vectors in four-dimensional space \boldsymbol{x}=(x_1,x_2,x_3,x_4), \boldsymbol{y}=(y_1,y_2,y_3,y_4), \boldsymbol{z}=(z_1,z_2,z_3,z_4), their cross product can be defined as: Cross(\boldsymbol{x},\boldsymbol{y},\boldsymbol{z})=\begin{vmatrix} x_1 & y_1 & z_1 &\boldsymbol{e}_1 \\ x_2 & y_2 & z_2 &\boldsymbol{e}_2 \\ x_3 & y_3 & z_3 &\boldsymbol{e}_3 \\ x_4 & y_4 & z_4 &\boldsymbol{e}_4 \end{vmatrix} Vectors in even higher-dimensional spaces can be defined similarly.
Now let us consider the geometric meaning of its magnitude. Let Cross(\boldsymbol{x},\boldsymbol{y},\boldsymbol{z})=\boldsymbol{w}=(\omega_1,\omega_2,\omega_3,\omega_4), where: \begin{aligned} \omega_1=\begin{vmatrix} x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \\ x_4 & y_4 & z_4 \end{vmatrix}, \omega_2=-\begin{vmatrix} x_1 & y_1 & z_1 \\ x_3 & y_3 & z_3 \\ x_4 & y_4 & z_4 \end{vmatrix}&, \\ \omega_3=\begin{vmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_4 & y_4 & z_4 \end{vmatrix}, \omega_4=-\begin{vmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{vmatrix}&. \end{aligned}
Then the determinant \begin{vmatrix} x_1 & y_1 & z_1 & \omega_1 \\ x_2 & y_2 & z_2 & \omega_2 \\ x_3 & y_3 & z_3 & \omega_3 \\ x_4 & y_4 & z_4 & \omega_4 \end{vmatrix} represents the four-dimensional hypervolume of the parallelotope formed by \boldsymbol{x}, \boldsymbol{y}, \boldsymbol{z}, \boldsymbol{w}. Furthermore, since \boldsymbol{w} is perpendicular to the other three vectors, this parallelotope is actually a “four-dimensional right prism” (imagine a 3D right prism) in 4D space. Its “base” is the 3D parallelepiped formed by \boldsymbol{x}, \boldsymbol{y}, \boldsymbol{z}. Based on the analogy “volume = base area \times height,” the aforementioned determinant should equal the “volume of the parallelepiped multiplied by the magnitude of \boldsymbol{w}.” When the magnitude is 1, the hypervolume of the 4D prism is numerically equal to the volume of the 3D parallelepiped. Consequently, the determinant \begin{vmatrix} x_1 & y_1 & z_1 & \frac{\omega_1}{|\boldsymbol{w}|} \\ x_2 & y_2 & z_2 & \frac{\omega_2}{|\boldsymbol{w}|} \\ x_3 & y_3 & z_3 & \frac{\omega_3}{|\boldsymbol{w}|} \\ x_4 & y_4 & z_4 & \frac{\omega_4}{|\boldsymbol{w}|} \end{vmatrix} is equal to the volume of the 3D parallelepiped formed by \boldsymbol{x}, \boldsymbol{y}, \boldsymbol{z}. Expanding it along the last column yields |\boldsymbol{w}|. This is the geometric meaning of the magnitude, which is analogous to the cross product in three-dimensional space. The geometric meaning of the cross product and its magnitude in even higher dimensions can be analogized accordingly: in an n-dimensional space, (n-1) linearly independent vectors can define a cross product, and its magnitude is the (n-1)-dimensional hypervolume of the body formed by the original (n-1) vectors.
Looking back, we find that this result even holds for 2D space. Constructing a cross product from a single vector (a,b), i.e., a vector perpendicular to it, requires only considering the determinant: \begin{vmatrix} a & \boldsymbol{e}_1 \\ b & \boldsymbol{e}_2 \\ \end{vmatrix} = -b \boldsymbol{e}_1 + a \boldsymbol{e}_2 = (-b, a)
Evidently, the cross product provides the most natural description for generating perpendicular vectors.
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