Orthogonally intersecting geodesics on the surface of a four-dimensional hypersphere

A multidimensional maze. Say you have a single dimension A which can have values 0,1,2,3. Now say you have a second dimension B which can also have these values. A simple 2-D planar maze could be written as a series of 2-digit numbers, AB, tracking one's path through the maze. 00, 10, 20, 21, 22, 23. In at 00 and out at 23. You could do the same with a 3-D maze with a series of 3-digit numbers, ABC. For 4-D, ABCD. And so on.

Now, put the values 0,1,2,3 on a circle. And intersect the circles at right angles, so that you have three perpendicular geodesics on a sphere: ABC. You will then have for A, 0123; for B, 0123; and for C, 0123. They will intersect at 6 points on the sphere, and there will be no common "origin" but rather the central figure is the spherical triangle formed by ABC with each vector running perpendicular and counterclockwise on the surface of the sphere (right-hand rule). So where A starts and equals 0, B=1. Where B starts at 0, C=1. Where C starts at 0, A=1. At each of these intersections the excluded parameter (C, A, and B respectively) is designated N for null. So you have (along A), 01N, 1N0, 2,3,N, 3N2, and then you come back to where you started, at 01N. The same applies along B: N01, 01N, N23, 23N, and back to N01. You can do the same along C. And although I refer to a "central figure" there are really 8 equal spherical triangles on the surface of the sphere. Note that because of their being on the surface of a sphere, each vertex of the triangle is a right angle.

Now I'll just briefly digress and mention that the central spherical triangle can be oriented clockwise, instead, creating a sphere that is nonsuperimposable with respect to our first sphere.

Now if you take a 4-D hypersphere, you get the same kind of series, but with 6 circular geodesics: A, B, C, D, E, and F. The intersections are in 4-space and are therefore harder to imagine. The central figure is a spherical tetrahedron, whose surfaces represent the positively curved spherical triangles that appear at the intersection of 4 spheres of equal volume. Note carefully that, because of the curvature of 4-space, each vertex of the tetrahedron is composed of 3 intersecting geodesics that are at right angles to one another. In choosing the directions of the 6 vectors you can pick arbitrarily for each one. But be aware of all of the possible permutations, and of those permutations that are equivalent (superimposable) when the spherical tetrahedron is rotated.

Think of a hexagon ABCDE and label each point according to one of the two possible directions for each vector, + or -. So you can have ++++++ or +-++++ or +--+++ and so on. But because of the ring formation many of these are equivalent. If you remember the positions of two side groups on a benzene ring, ortho, meta and para, you have: all +, all -, 1+, 1-, ortho +, ortho-, meta+, meta-, para+ and para-. That's it. Let's take for our discussion below the case of all of the geodesics/vectors going in the + direction. Understand that what we call the + direction is arbitrary for each vector. But once we establish these, we have to be consistent in naming the arrangements as above (e.g. ortho-, para+, etc.) as relative to how the "++++++" vectors are arranged. Again we are talking about six geodesics in 4-space forming a hyperspherical tetrahedron, each geodesic having a positive and negative direction, and we are preparing to be able to name all of the intersections, in order, along any of the six geodesics, A, B, C, D, E or F.

We start at each vertex of the tetrahedron, which lies on the surface of the hypersphere. Note that each edge of the tetrahedron extends geodesically along the surface of the hypersphere in opposite directions and meets again at its antipode on the opposite side of the hypersphere. Thus the surface volume of the hypersphere is entirely divided into adjacent hyperspherical tetrahedrons. This is analogous to the spherical triangles on the surface of the sphere above.

We proceed in the positive direction with each geodesic, for example A, and at each intersection label the value of A starting with zero; we do the same with B, C, D, and E. Each intersection will have 3 of the geodesics possessing a defined value, and three undefined. Those which do not exist at a given intersection are labeled N as in the simpler case of the sphere. So a given vertex label might have the form of 0N15NN, or 34NNN1, and so on, referring to the values ABCDEF. On my (arbitrary) "central" tetrahedron, A, B, and C radiate orthogonally away from point 000NNN in a positive direction. D and E radiate orthogonally away from point 1NN00N in a positive direction. F radiates from poin N1N1N0. C, E and F intersect at NN1N11. So what are the remaining intersections? You can deduce them from the naming scheme. But how are they placed with respect to one another? How many adjacent tetrahedrons are there on the surface of the hypersphere?