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B Is there an easy way to traverse a 4-polytope?

  1. Dec 1, 2016 #1
    I'm running into a math problem in a story I'm writing that I can't figure out how to solve, nor can I find a solution on google (probably just don't know what to search for.)

    I'm writing a story, where a character gets lost in a "cave" of sorts. It's more in a holodeck type thing, so geometry acts strangely. The cave is laid out such that it's a https://en.wikipedia.org/wiki/600-cell. Each cell is a chamber and each face represents a tunnel connecting them.

    I'm trying to figure out how to traverse the thing in a way that I can keep track of. Not only do I need to know what cell I'm in and what the neighbors are, but also orientation, since depending on how you enter a chamber, it's walls and floors might not be where I expect them.
  2. jcsd
  3. Dec 1, 2016 #2
    Interesting question! I will give a few first impressions. Each triangular face is shared by just 2 tetrahedral cells; each edge is shared by 5 tetrahedra; and each vertex is shared by 20 tetrahedra. There are a number of ways to visualize how the 600 tetrahedral cells are arranged.

    One of these goes as follows. consider an icosahedron. For each of its 20 triangles, the 600-cell has a *ring* of 30 terahedra with each of these tetrahedra touching each of the two adjacent ones in the ring along one of its two opposite edges. Any two of these 20 rings link each other in the 600-cell. (The entire 600-cell is, like any ordinary regular polytope, a topological sphere of the appropriate dimension — in this case it is a 3-sphere.)

    Despite this arrangement of 20 rings according to how the triangles of an icosahedron are arranged, you cannot find an actual icosahedron among the 1200 triangular faces of the 600-cell such that through each face of that icosahedron passes just one of the 20 rings of 30 tetrahedra each. (There are, however, plenty of icoshedra among those 1200 triangular faces: Each of the 120 vertices of the 600-cell belongs to 20 tetrahedra, and their 20 faces that lie opposite to that vertex form a perfect regular icoshedron.

    So, taking any one way of viewing the 600-cell as 20 rings of 30 tetrahedra each (there are many!), if you'd like to give a label to each of the 20 faces of that icosahedron (say using the 20 letters A through T), and also choose a "marked" tetrahedron on each of those 20 rings of 30, then each tetrahedron of the 600-cell would naturally have a label consisting of one letter, and one number between 1 and 30. That would also help with the orientation within each tetrahedron, but I'll let you take over from here.
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