# Finding angles between planes in 4D space.

Suppose I have two intersecting planes in a four dimensional Euclidean space. It seems to me that there are two angles between these planes. If the two planes intersect in a line then one of those angles is zero. If the two angles are non-zero then the planes intersect in a point. If one plane is the WX plane and the other the YZ plane then the two angles are both 90 degrees.

The best I can do is this. Each plane is defined by two orthonormal basis vectors. Project a basis vector onto the other plane. Then the angle is the arccos of the length of the projection. Do this with each basis vector to get the two angles. However I believe that the result depends on the choice of basis vectors.

What I need could be to find a unit vector in a plane so that it's projection onto the other plane is of maximum length. I don't know how to do that. I suppose find an expression for the length of the projection, take the derivative, and find where it is zero. Is there an easier way?

• Because in 4D there is no unique normal. Jul 7, 2022 at 2:31

Suppose $$\Pi_1$$ and $$\Pi_2$$ are two oriented subspaces of a Euclidean inner product space $$U$$ of the same dimension which intersect in a codimension $$1$$ subspace. This subspace can be afforded an orthonormal basis, which we can extend to ordered orthonormal bases of $$\Pi_1$$ and $$\Pi_2$$ by adding unit vectors $$v_1$$ and $$v_2$$. (Keep in mind the order and direction of basis vectors must agree with the orientations.) The angle $$\phi=\angle v_1v_2$$, defined by $$\cos\phi=v_1\cdot v_2$$, is the dihedral angle between the two subspaces. You can start with an arbitrary pair of ordered orthonormal bases for $$\Pi_1$$ and $$\Pi_2$$ and construct the bases we need using Gram-Schmidt. If you don't care about orientation, you can solve $$\cos\phi=\pm v_1\cdot v_2$$ for the acute or convex angle $$\phi$$ you want.

It is possible to go another route and generalize. The orthogonal projection $$\pi_2:U\to\Pi_2$$ may be restricted to $$\Pi_1$$, and it has a "hypervolume distortion factor" $$\lambda$$. That is, the effect of $$\pi_2$$ on any region $$D\subset\Pi_1$$ with finite volume $$\mathrm{vol}(D)$$ is to first shrink it along the $$v_1$$-axis by a factor of $$\lambda$$ (and flip it if $$\lambda<0$$), then rotate $$\Pi_1$$'s $$v_1$$-axis through $$U$$ to $$\Pi_2$$'s $$v_2$$-axis, to get a region with hypervolume $$\mathrm{vol}(\pi_2(D))=\lambda\,\mathrm{vol}(D)$$. This is just like how determinants work on a fixed space. This definition of $$\lambda=\cos\phi$$ applies to any two equal-dimension subspaces $$\Pi_1$$ and $$\Pi_2$$, regardless of their intersection's (co)dimension.

If $$\{v_1,\cdots,v_d\}$$ is an oriented orthonormal basis for $$\Pi_1$$, then we can form the matrix $$V_1=(v_1~\cdots~v_d)$$ and similarly $$V_2$$ for $$\Pi_2$$. The Grammian matrix is $$G=V_2^T V_1$$ and the Grammian determinant $$\det G=\det(V_2^T V_1)$$ is the volume distortion factor $$\lambda$$ and then we can define $$\phi$$ by $$\lambda=\cos\phi$$.

I think one way to see this works is to verify both $$\det G$$ and $$\lambda$$ "transform" the same way (with some kind of column operations involving both $$V_1$$ and $$V_2$$, which represent geometric shearing aka transvection) and are equal in a nice set of "base cases." Or, perhaps we can extend $$\pi_2$$ to a linear operator on $$\Pi_1+\Pi_2$$ which has the same determinant as $$\lambda$$. Or, you can view $$V_2^TV_1$$ as a matrix whose columns represent the parallelotope you get by projecting the unit cube of $$\Pi_1$$ (the parallelotope generated by $$V_1$$) to $$\Pi_2$$ and using $$V_2$$ as an ordered basis; then the determinant is well-known to measure the hypervolume.

Also note if we represent $$V_1$$ by $$\omega_1=v_1\wedge\cdots\wedge v_d$$ and $$V_2$$ by $$\omega_2$$ in the exterior power $$\Lambda^dU$$, then this is the inherited inner product $$\cos\phi=\langle\omega_1,\omega_2\rangle$$.

We can generalize even further to cases where $$\dim\Pi_1\ne\dim\Pi_2$$. In these cases, the Grammian matrix $$G=V_2^TV_1$$ is rectangular but not square, so we can't take its determinant. Plus, $$\cos\phi$$ being signed ($$\pm$$) doesn't really make sense because all embeddings of one space into another are related by ambient isometries (as opposted to when $$\dim\Pi_1=\dim\Pi_2$$, there are two classes corresponding to preserving or reversing orientations). However, $$\lambda^2=\cos^2\phi$$ makes sense, and this equals $$\det(G^TG)$$ where $$G=V_2^TV_1$$. This way we can measure angles between subspaces of different dimension. Note, though, linear dependence implies when projecting a larger dimensions subspace to a smaller dimension one, this determinant will be $$0$$, because smaller-dimensional volume is considered zero as a higher-dimensional volume.

• $$\mathrm{vol}(\pi_2(D))=\lambda\,\pi_1(D)$$ I feel like something is off, isnt it? Jul 11, 2022 at 20:19
• @Filippo Fixed. Jul 12, 2022 at 2:12
• Thank you. I still don't get it though: How can $\lambda$ be negative if$$\mathrm{vol}(\pi_2(D))=\lambda\,\mathrm{vol}(D)$$for all $D$? I am obviously assuming that vol is a (non-negative) measure, is that my mistake? Jul 12, 2022 at 7:10
• @Filippo Yes - this is signed volume, which is positive or negative depending on orientation. Jul 12, 2022 at 8:01
• I am impressed. Thank you for the insight and +1. Jul 12, 2022 at 8:11