ae wrote:The way the vertices of 24-cell are two sets of axis-aligned vertices which are permutations of (\pm1,0,0,0) and diagonal vertices of (\pm 1/2, \pm 1/2, \pm 1/2, \pm 1/2) is very similar to the structure of rhombic dodecahedron in 3D. The vertices of rhombic dodecahedron are at permutations of (\pm 1, 0, 0) and (\pm 1/2, \pm 1/2, \pm 1/2).
Correct. In fact, the 24-cell is in many ways the 4D analogue of the rhombic dodecahedron (see below). It is a happy coincidence that in 4D, it also happens to have regular facets.
Correct. It might interest you to know that there are exactly
two ways of doing this. This actually corresponds with the vertex-first projection of a 4D cube into 3D: four of the cells project onto one possible arrangement of the four parallelopipeds, and the other four cells project onto the other possible arrangement.
Now, the other way of constructing a rhombic dodecahedron is to take two 3D cubes, cut one of them into 6 square pyramids, and stick the pyramids onto the faces of the other cube. If you do the analogous thing in 4D: cut a 4D cube into 8 cubical pyramids and stick them onto the facets of another 4D cube, you end up with the 24-cell.
I am wondering if there is any way of constructing a 24-cell with some finite number of four dimensional parallelepipeds.
Any ideas?![/code]
I'm not familiar enough with 5D to be able to answer this either way, but my intuition suggests that this doesn't work. The main reason is that in 3D, the faces of the rhombic dodecahedron are trapezoids---distorted squares, if you will, which correspond with projections of the square ridges of a 4D cube. Unfortunately, in 4D, the 24-cell has octahedral facets, which are not parallelopipeds, and so cannot correspond with projected 5D cube ridges.
I know Wendy has mentioned before that in 4D and above, there are actually (at least) 2 possible analogues of the rhombic dodecahedron: one is shape resulting from cutting an n-cube into pyramids and attaching them to another n-cube, and the other is the maximal projection of an (n+1)-cube into n-space. I'm not sure what the second shape in 4D is called, though. Its cells would have to be parallelopipeds, so it would be neither regular nor uniform (Archimedean). Maybe Wendy can help us with this one. :-)