I can’t stop thinking about the 600-cell:
It’s a ‘Platonic solid in 4 dimensions’ with 600 tetrahedral faces and 120 vertices. One reason I like it is that you can think of these vertices as forming a group: a double cover of the rotational symmetry group of the icosahedron. Another reason is that it’s a halfway house between the icosahedron and the lattice. I explained all this in my last post here:
I wrote that post as a spinoff of an article I was writing for the Newsletter of the London Mathematical Society, which had a deadline attached to it. Now I should be writing something else, for another deadline. But somehow deadlines strongly demotivate me—they make me want to do anything else. So I’ve been continuing to think about the 600-cell. I posed some puzzles about it in the comments to my last post, and they led me to some interesting thoughts, which I feel like explaining. But they’re not quite solidified, so right now I just want to give a fairly concrete picture of the 600-cell, or at least its vertices.
This will be a much less demanding post than the last one—and correspondingly less rewarding. Remember the basic idea:
Points in the 3-sphere can be seen as quaternions of norm 1, and these form a group that double covers The vertices of the 600-cell are the points of a subgroup that double covers the rotational symmetry group of the icosahedron. This group is the famous binary icosahedral group.
Thus, we can name the vertices of the 600-cell by rotations of the icosahedron—as long as we remember to distinguish between a rotation by and a rotation by Let’s do it!
• 0° (1 of these). We can take the identity rotation as our chosen ‘favorite’ vertex of the 600-cell.
• 72° (12 of these). The nearest neighbors of our chosen vertex correspond to the rotations by the smallest angles that are symmetries of the icosahedron; these correspond to taking any of its 12 vertices and giving it a 1/5 turn clockwise.
• 120° (20 of these). The next nearest neighbors correspond to taking one of the 20 faces of the icosahedron and giving it a 1/3 turn clockwise.
• 144° (12 of these). These correspond to taking one of the vertices of the icosahedron and giving it a 2/5 turn clockwise.
• 180° (30 of these). These correspond to taking one of the edges and giving it a 1/2 turn clockwise. (Note that since we’re working in the double cover rather than giving one edge a half turn clockwise counts as different than giving the opposite edge a half turn clockwise.)
• 216° (12 of these). These correspond to taking one of the vertices of the icosahedron and giving it a 3/5 turn clockwise. (Again, this counts as different than rotating the opposite vertex by a 2/5 turn clockwise.)
• 240° (20 of these). These correspond to taking one of the faces of the icosahedron and giving it a 2/3 turn clockwise. (Again, this counts as different than rotating the opposite vertex by a 1/3 turn clockwise.)
• 288° (12 of these). These correspond to taking any of the vertices and giving it a 4/5 turn clockwise.
• 360° (1 of these). This corresponds to a full turn in any direction.
Good! We need a total of 120 vertices.
This calculation also shows that if we move a hyperplane through the 3-sphere, which hits our favorite vertex the moment it touches the 3-sphere, it will give the following slices of the 600-cell:
• Slice 1: a point (our favorite vertex),
• Slice 2: an icosahedron (whose vertices are the 12 nearest neighbors of our favorite vertex),
• Slice 3: a dodecahedron (whose vertices are the 20 next-nearest neighbors),
• Slice 4: an icosahedron (the 12 third-nearest neighbors),
• Slice 5: an icosidodecahedron (the 30 fourth-nearest neighbors),
• Slice 6: an icosahedron (the 12 fifth-nearest neighbors),
• Slice 7: a dodecahedron (the 20 sixth-nearest neighbors),
• Slice 8: an icosahedron (the 12 seventh-nearest neighbors),
• Slice 9: a point (the vertex opposite our favorite).
Here’s a picture drawn by J. Gregory Moxness, illustrating this:
Note that there are 9 slices. Each corresponds to a different conjugacy class in the group These in turn correspond to the dots in the extended Dynkin diagram of which has the usual 8 dots and one more.
The usual Dynkin diagram has ‘legs’ of lengths and
The three legs correspond to conjugacy classes in that map to rotational symmetries of an icosahedron that preserve a vertex (5 conjugacy classes), an edge (2 conjugacy classes), and a (3 conjugacy classes)… not counting the element That last element gives the extra dot in the extended Dynkin diagram.
You can click on an image to see its source. The shiny ball-and-strut picture of the 120-cell was made by Tom Ruen using Robert Webb’s Stella software and placed on Wikicommons. The picture of slices of the 120-cell was drawn by J. Gregory Moxness and placed on Wikicommons under a Creative Commons Attribution-Share Alike 4.0 International license.