Angle between two planes in four dimensions

Suppose I have two planes defined in 4D space, either in terms of vectors spanning the planes, $X = t_1 A_1 + t_2 B_2$ and $X = t_3 A_3 + t_4 B_4$ (where $X$, $A$'s, and $B$'s are vectors with four elements and $t$'s are scalars), or in terms of null space, $[C_1; D_1] X = 0$ and $[C_2; D_2] X = 0$ (where the matrices are $2$ x $4$ and $X$ has four elements).

I understand that these two planes generally intersect in just a single point (unless the matrix $[C_1; D_1; C_2; D_2]$ is rank deficient). But is it meaningful to ask what the angle is between the planes? If so, how would it be computed? There is an explicit formula for the 3D case: simply the angle between the normals to the planes. Is there no equivalent explicit expression for the 4D case?

Solution 1:

Choose orthonormal bases $(E_1, E_2)$ and $(F_1, F_2)$ for the two planes. Then, the (hyper)volume of the parallelepiped spanned by the $E_1, E_2, F_1, F_2$ on the one hand is $\sin \theta$ where $\theta$ is the angle between the two planes, and on the other hand is (the absolute value of) the determinant of the matrix given by adjoining the four vectors:

$|\det[E_1 \, E_2 \, F_1 \, F_2]|$.

This quantity is independent of the choices of orthogonal bases, and in fact, we can even take the bases of the planes to be any that span parallelograms of unit area, so we don't need to produce an orthogonal basis.

If we start out with any bases $(E_i)$ or $(F_i)$, say ones that span areas $\lambda$ and $\mu$, we can always normalize them by rescaling vectors, but we might as well build this into our equation: The bases $(\lambda^{-1} E_1, E_2)$ and $(\mu^{-1} F_1, F_2)$ both span parallelograms of unit area, so for general bases the hypervolume is

$\sin \theta = |\det[\lambda^{-1} E_1 \, E_2 \, \mu^{-1} F_1 \, F_2]| = \frac{|\det[E_1 \, E_2 \, F_1 \, F_2]|}{\lambda \mu}$,

which we might write as

$\color{red}{\sin \theta = \dfrac{|\det[E_1 \, E_2 \, F_1 \, F_2]|}{|E_1 \wedge E_2| |F_1 \wedge F_2|}}$,

where $|G_1 \wedge G_2|$ denotes the area of the parallelograms spanned by $G_1, G_2$.

This generalizes readily to formulas the angle between $k$-planes and $(n - k)$-planes in vector spaces of dimension $n$ (try this for the familiar situation $n = 2$, $k = 1$), and with just a little more work to finding the angle between $k$- and $l$- planes in vector spaces of dimension more than $k + l$.

Remark This leaves the matter of computing explicitly the areas $|G_1 \wedge G_2|$ of the parallelograms the bases defime. The area $A$ of the parallelogram defined by vectors $H_1 = (x_1, y_1), H_2 = (x_2, y_2)$ in the plane is

$A = \left\vert\det \left(\begin{array}{cc} x_1 & x_2 \\ y_1 & y_2\end{array}\right)\right\vert = |x_1 y_2 - x_2 y_1|$, and its square is $A^2 = (x_1 y_2 - x_2 y_1)^2$, which we can rewrite as

$A^2 = [(x, y) \cdot (x, y)] [(x', y') \cdot (x', y')] - [(x, y) \cdot (x', y')]^2 = (H_1 \cdot H_1) (H_2 \cdot H_2) - (H_1 \cdot H_2)^2$.

Now, the formula $A^2 = (H_1 \cdot H_1) (H_2 \cdot H_2) - (H_1 \cdot H_2)^2$ doesn't depend on coordinates, it just uses the Euclidean structure (namely the dot product $\cdot\,$), so it works just as well for computing the areas of the parallelograms in our original problem, that is, we may write

$|G_1 \wedge G_2|^2 = (G_1 \cdot G_1) (G_2 \cdot G_2) - (G_1 \cdot G_2)^2$

and then take square roots if we like.