# Can one "chop" up S^3 into little cubes?

1. Mar 17, 2015

### Spinnor

Say I live in a very large, fixed radius, three-sphere. Could a mathematician pick a very large set of points at the right locations in my three-sphere such that when each point was connected to their nearest 6 neighbors with short line segments that locally I would have a nearly rectangular 3 dimensional grid that would also be uniform in the sense that from any point the grid would look the same, lengths of line segments all the same and angles between nearest line segments nearly or exactly 90 degrees?

Would be some kind of translation in-variance?

Would this be in effect "chopping" up a three-sphere into little cubes? Is there a smallest number of cubes that a three-sphere can be chopped into or do we only get cubes when the number of cubes is very large?

Thanks for any help!

2. Mar 17, 2015

### Ben Niehoff

90 degree corners will probably not work, but if all you want to do is divide $S^3$ into hexahedra, then I think the answer is yes. The answer to the analogous question for $S^2$ (dividing it into quadrilaterals) is yes, as follows:

Consider a unit cube, and a sphere circumscribing it, such that they intersect in the 8 corners of the cube. Now project the edges of the cube up onto the sphere. The sphere is now divided into 6 squares, which are quadrilaterals of equal side lengths and equal angles (which will, in this case, be 120 degrees). If you want to divide the sphere into smaller "squares", then put a grid on the faces of your cube, and project that up as well. But the smaller quadrilaterals will not have equal side lengths and equal angles.

You can imagine a similar process should work for $S^3$; just start with a hypercube and its circumscribing 3-sphere.