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?

  • $\begingroup$ Yes; just as in lower dimensions, it's the smallest angle between two vectors, one in the first plane, one in the second. $\endgroup$ Aug 27 '14 at 13:34
  • $\begingroup$ So how would that be computed? Would I have to conduct a minimization? $\endgroup$
    – 8bar
    Aug 27 '14 at 13:35
  • $\begingroup$ You could conduct a minimization in terms of your parameters $A_1, B_2, A_3, B_4$, which would solve the general case. $\endgroup$ Aug 27 '14 at 13:41
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    $\begingroup$ I guess one minor simplification would be that min invcos(X1.X2/(norm(x1) norm(x2))) would be equivalent to max X1.X2/(norm(x1) norm(x2)) $\endgroup$
    – 8bar
    Aug 27 '14 at 13:51
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    $\begingroup$ Yes, that's right, also notice that changing vector length doesn't affect angles, so you might as well restrict to unit length vectors, and now you're just maximizing $X_1 \cdot X_2$. There's a circle of each of these in each plane, and we can parameterize these circles in terms of your $A$'s and $B$'s. We might as well assume your bases $(A_i, B_{i + 1})$, $i = 1, 3$ for the planes are orthonormal, in which case we can parameterize the circle of unit vectors by $r_i(\theta) := A_i \cos \theta + B_i \sin \theta$. $\endgroup$ Aug 27 '14 at 13:53

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.

  • $\begingroup$ Wonderful @Travis! I obviously can't use the cross product to find the area of the parallelogram in 4D, but after a little searching, I found that the area of the parallelogram spanned by G1 and G2 is simply the square root of G1.G1 G2.G2 - (G1.G2)^2. $\endgroup$
    – 8bar
    Aug 28 '14 at 10:49
  • $\begingroup$ @8bar, yes, the formula gives the sine of the angle, not the cosine (I was thinking of the usual dot product formula). $\endgroup$ Aug 28 '14 at 11:07
  • $\begingroup$ @8bar Yes, you're quite correct, I've added an explanation as a remark to my answer. Notice that the area $|x_1 y_2 - x_2 y_1|$ of the parallelogram in the ($(x, y)$-)plane is the length of the cross product of the vectors regarded as living in the $xy$-plane inside $xyz$-space, namely, $(x_1, y_1, 0), (x_2, y_2, 0)$, which gives rise in $3$ dimensions to the identity $(V \cdot V)(W \cdot W) = (V \cdot W)^2 + (V \times W) \cdot (V \times W)$. So the formula you point out for the area of the parallelogram does use cross product in a way, even though it applies in dimensions other than $3$. $\endgroup$ Aug 28 '14 at 11:34
  • $\begingroup$ @Travis said, "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." So does this generalize to the the angle between a plane and a line in 4D as well? I have two vectors spanning the plane and one vector spanning the line, so the matrix in the numerator isn't square. $\endgroup$
    – 8bar
    Aug 29 '14 at 12:36
  • $\begingroup$ Yes it does, this is the "a little more work". The vectors in a basis $(U_1, U_2)$ of the plane and a vector $V$ spanning the line together span a (3d) parallelepiped, and it will have volume $|U_1 \wedge U_2||V| \sin \theta$, where $\theta$ is the angle between the two subspaces. $\endgroup$ Aug 29 '14 at 12:41

I wonder whether we should distinguish following two cases when defining angle between two planes (not intersecting) in dimension four.

Case one is when both planes $K$, $L$ belong to the same complex structure on $\mathbb R^4$. In such case for each vector $v$ in $K$ angle between $v$ and $L$ is the same. (The angle between vector $v$ and plane $L$ is defined as minimum angle between vectors $v$ and $w$ for $w$ in $L$). The examples are $ K=<[1,0,0,0], [0,1,0,0]>, L_1=<[1,1,1,1]/2, [-1,1,-1,1]/2>$ $ L_2:=<[t,0,1,-s]/2,[0,t,s,1]/2> $. In this definition $t=\dfrac{\sqrt 5-1}{2}$ and $s=\dfrac{\sqrt 5+1}{2}$.

Case two is when both planes $K$, $L$ do not belong to the same complex structure. In this case for $v$ in $K$ the resulting angle between $v$ and $L$ can vary depending on choice of $v$. Examples are $K=<[1,0,0,0], [0,1,0,0]>, L=<[1,1,1,1]/2, [0,1,-1,0]/\sqrt 2 >$, $L_2=<[1,1,1,1]/2, [0,1,-s, t]/2>$. How to define angle in this case ? We may define it as minimum of angles between vectors $v$ and $w$ for $v$ in $K$ and $w$ in $L$. We may also define it as determinant of matrix $[v_1,v_2,w_1,w_2]$ where $v_1,v_2$ is orthonormal basis of $K$ and $w_1, w_2$ is orthonormal basis of $L$.

Interesting question is to ask about product of reflections in the planes. By reflection in the plane I mean matrix which reverse vectors in the plane and fix vectors perpendicular to the plane. If the angle between planes is $\dfrac{2 \pi}{k}$ then we expect that product of reflections has order $k$.

Of course I do not settle things in my answer. I rather asked another questions here in order to expand horizons $\begin {matrix}'''\\{..}\\{\smile} \end {matrix}$


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