Cross Products in High-Dimensional Space and Their Geometric Meaning
Operations between vectors include the dot product and the cross product (also called the vector product or exterior product). The dot product is relatively simple and generalizes easily to higher dimensions; the cross product, however, is different — generally speaking, it is a peculiarly three-dimensional object. The difficulty in generalizing the cross product lies in its multiple layers of meaning. If we view vectors and their cross products through the lens of tensors (which belongs to the realm of differential forms), then cross products of vectors in dimensions higher than three simply do not exist. But if we instead think of the cross product merely as "a tool that generates, from two vectors, a third vector orthogonal to both," then the cross product can indeed be generalized to higher dimensions — and the technique for doing so is remarkably elegant, bearing a close resemblance to the three-dimensional cross product.
Revisiting Three-Dimensional Space
To generalize the three-dimensional cross product, it helps first to revisit where the three-dimensional cross product comes from. The cross product originates from quaternion multiplication, but from a purely goal-oriented point of view, what we want is to construct a vector $\boldsymbol{w}=(w_1,w_2,w_3)$ that is perpendicular (orthogonal) to two given, non-collinear vectors $\boldsymbol{u}=(u_1,u_2,u_3),\boldsymbol{v}=(v_1,v_2,v_3)$. For the sake of generality, we also want the constructed vector to be free of any "singularities," so we will build it using multiplication alone. As for the geometric meaning of the cross product, that can wait — let's first achieve the basic goal. more
Here, in order to construct such a vector, the determinant plays a crucial role! 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 zero. Expanding it gives:
$$u_1 \begin{vmatrix} u_2 & v_2\\u_3 &v_3\end{vmatrix}-u_2 \begin{vmatrix} u_1 & v_1\\u_3 &v_3\end{vmatrix}+u_3 \begin{vmatrix} u_1 & v_1\\u_2 &v_2\end{vmatrix}=0$$
Likewise, if $(w_1,w_2,w_3)=(v_1,v_2,v_3)$, the determinant is also zero, and expanding it gives
$$v_1 \begin{vmatrix} u_2 & v_2\\u_3 &v_3\end{vmatrix}-v_2 \begin{vmatrix} u_1 & v_1\\u_3 &v_3\end{vmatrix}+v_3 \begin{vmatrix} u_1 & v_1\\u_2 &v_2\end{vmatrix}=0$$
From this we see that 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 the vector $\boldsymbol{u}=(u_1,u_2,u_3),\boldsymbol{v}=(v_1,v_2,v_3)$. And this is exactly the vector cross product as we currently define it — so we can see clearly where the cross product comes from.
In theoretical analysis, we usually 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 $\boldsymbol{e}_1,\boldsymbol{e}_2,\boldsymbol{e}_3$ is the basis of three-dimensional space.
Cross Products in Higher Dimensions
With the groundwork above, defining a cross product in higher-dimensional space is no longer difficult. For example, given three vectors $\boldsymbol{x}=(x_1,x_2,x_3,x_4),\boldsymbol{y}=(y_1,y_2,y_3,y_4),\boldsymbol{z}=(z_1,z_2,z_3,z_4)$ in four-dimensional space, their cross product can be defined as
$$Cross(\boldsymbol{x},\boldsymbol{y},\boldsymbol{z})=\begin{vmatrix} x_1 & y_1 & z_1 &\boldsymbol{e}_1 \\ x_2 & y_2 & z_2 &\boldsymbol{e}_2 \\ x_3 & y_3 & z_3 &\boldsymbol{e}_3 \\ x_4 & y_4 & z_4 &\boldsymbol{e}_4 \end{vmatrix}$$
Cross products of vectors in even higher-dimensional spaces can be defined analogously.
Now let's consider the geometric meaning of its norm. Write $Cross(\boldsymbol{x},\boldsymbol{y},\boldsymbol{z})=\boldsymbol{w}=(\omega_1,\omega_2,\omega_3,\omega_4)$, 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 parallelepiped-like "four-dimensional body" spanned by $\boldsymbol{x},\boldsymbol{y},\boldsymbol{z},\boldsymbol{w}$. Moreover, since $\boldsymbol{w}$ is perpendicular to the other three vectors, this four-dimensional parallelepiped is essentially a "four-dimensional right prism" (think of a three-dimensional right prism), whose "base" is the three-dimensional parallelepiped spanned by $\boldsymbol{x},\boldsymbol{y},\boldsymbol{z}$. By analogy with "volume = base area times height," the above determinant should equal "the volume of the parallelepiped times the norm of $\boldsymbol{w}$." When the norm equals 1, the hypervolume of the four-dimensional prism is numerically equal to the volume of the three-dimensional parallelepiped. Thus 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 three-dimensional parallelepiped spanned by $\boldsymbol{x},\boldsymbol{y},\boldsymbol{z}$; expanding it along the last column then gives $|\boldsymbol{w}|$. This is the geometric meaning of the norm, and it parallels the three-dimensional cross product. The geometric meaning of the cross product and its norm in even higher dimensions can be understood by the same analogy: in n-dimensional space, a cross product can be defined for (n-1) linearly independent vectors, and its norm is exactly the hypervolume of the (n-1)-dimensional body spanned by those original (n-1) vectors.
Looking back, we can see that this result even holds in two-dimensional space. To construct a cross product from a single vector $(a,b)$ — that is, to construct a vector perpendicular to it — we only need to consider the determinant
$$\begin{vmatrix} a &\boldsymbol{e}_1 \\ b &\boldsymbol{e}_2 \\ \end{vmatrix}=-b \boldsymbol{e}_1+a\boldsymbol{e}_2=(-b,a)$$
As we can see, the cross product offers the most natural way to describe the generation of a perpendicular vector.
Translated automatically with claude-sonnet-5; all equations are reproduced verbatim from the source. Copyright remains with the original author.