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## Main Question or Discussion Point

Now, let's consider mapping the three dimensions of space onto the "surface" of a 3-sphere. Like Non-Euclidean de Sitter geometry.

The surface of a sphere is positionally symmetrical. However, the positional symmetry of the surface being mapped to is distinct from the positional symmetry of the space being mapped.

We could represent the space in terms of a co-ordinate system, and likewise for its positional symmetry.

A Cartesian co-ordinate system has a high degree of positional symmetry, so we consider a three-dimensional Cartesian grid, equivalent to a honeycomb of cubes. Each cube being surrounded by and making contact with 26 cubes.

I think a problem arises when we consider all the Platonic polyhedroids/polychorons. There's only one made with cubes, the tesseract.

Because there doesn't appear to be a magical number of cubes which space can be divided into. And so, if the three dimensions of space can be mapped to a 3-sphere, I daresay we might expect there to be an infinite number of cubic Platonic polyhedroids.

And a tesseract has a different configuration as well, each of its cubes only makes contact with 6 cubes, not 26.

For the cubic honeycomb, each edge is shared by 4 cubes, and each vertex by 8 cubes, yielding for 8 cubes, 24 edges and 8 vertices.

A tesseract, on the other hand, has 32 edges and 16 vertices.

Now you might be confused by the mentions of "non-Euclidean" and "Cartesian", however, I've already mentioned a distinction between the two positional symmetries.

Also, there's still only one cubic Platonic polyhedroid.

All of this appears to imply to me that it is illogical to map the three dimensions of space onto a 3-sphere.

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