# I How to show that a 4-polytope is a tiling of the 3-sphere

1. Jul 1, 2017

### t_r_theta_phi

Is there a way to figure out if a 4-polytope is a tiling of the 3-sphere based on only the number of vertices, edges, faces, and cells?

Here is a specific example. Say that there are two "polytopes" - one is a tesseract, and one is a disjoint union of two tesseracts. Both will have an Euler characteristic of 0, but single tesseract is homeomorphic to a 3-sphere and the pair of tesseracts is not. Is there a way to figure out this fact based only on the number of vertices, edges, faces, and cells in each polytope?

In three dimensions the analogous problem could be easily solved because the single polyhedron would have an Euler characteristic of 2 and the disjoint union would have a characteristic of 4.

I'm not quite sure my terminology is correct but I gave it a try.

Last edited: Jul 2, 2017
2. Jul 2, 2017

### lavinia

Not sure if this helps.

Consider a tiling of a circle as a triangle - 3 vertices, 3 edges (Euler characteristic zero). Two disjoint triangles have six vertices and six edges. Subdivide the triangle by adding a vertex in the interior of each line segment. This is a second tiling of the circle. It also has six vertices and six edges.

Last edited: Jul 2, 2017
3. Jul 2, 2017

### t_r_theta_phi

I see. So 6 vertices and 6 edges can correspond to different structures, so it is not possible to differentiate between the them based on numbers of elements alone.

As you mentioned in your example, for the triangular tiling there exists a tiling with double the number of elements that also tiles the circle. However, as far as I know, it is not like that with my tesseract example. 16 vertices, 32 edges, 24 square faces, and 8 cubic cells is a valid tiling of the 3-sphere, but 32 vertices, 64 edges, 48 square faces, and 16 cubic cells is not. If it were, wouldn't it correspond to a new regular polytope? In other words, since both have the same type of elements, what is it about the number of elements in the tesseract that makes it a tiling of the 3-sphere? I have a feeling that this is much more complicated than I originally thought.

The reason I am asking this is because I found a way to find the ratios of the numbers of each type of element in a regular 4-polytope given its Schlafli symbol. The ratios can be found by knowing how many of each type of element must connect to every other type of element, and then correcting for overcounting. I am wondering how this ratio corresponds to the actual number of elements in the polytope. Wouldn't the extra condition be that the polytope must correspond to a tiling of the 3-sphere?

Last edited: Jul 2, 2017
4. Jul 3, 2017

### lavinia

I don't know anything about regular polytopes so without some research I can't comment. If you want to explain then we can try to figure it out together.

Last edited: Jul 3, 2017
5. Jul 5, 2017

### t_r_theta_phi

For any given polytope, imagine we know the information about its structure but not any of the actual numbers of vertices, edges, faces, etc. Let's say we want to find the number of type A elements (could be vertices, cells, anything) in terms of the number of type B elements. This is given by:

A = (# of A connected to each B) * B / ( # of B connected to each A)

The denominator corrects for overcounting.

Consider a cube for example. There are 3 E connected to each V, and 2 V connected to each E. Therefore

E = 3/2 V

Using this same method we can find

F = 3/4 V

Using the identity V - E + F = 2,

V = 8, E = 12, and F = 6.

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For the 4-dimensional tesseract we can use the same method to find

E = 2 V, F = 3/2 V, and C = 1/2 V

However, Euler's identity V - E + F - C = 0 holds no matter what V actually is. The actual numbers of elements cannot be found from this method alone.

There are many topological spaces that have an Euler characteristic of 0, so I am guessing that the missing condition is that the polytope must correspond to a tiling of the 3-sphere.

Last edited: Jul 5, 2017