In the euclidean space, the points belonging to a regular octahedron are those that satisfy the inequalities
where . These eight inequalities can be divided into two groups of four according to the number (even or odd) of negative signs they contain. For example, the inequalities
all have one or three negative signs and the points satisfying these form a tetrahedron. The other four inequalities correspond to the dual tetrahedron of the first, which shows that the intersection of two regular dual tetrahedra form a regular octahedron. Moreover, the vertices of the two tetrahedra can be seen as the eight vertices of a cube.
I am wondering if there exists a similar relationship between regular polytopes in four dimensions. As it is another case of a regular cross-polytope, the hexadecachoron (or 16-cell) is defined by the sixteen inequalities
If one were to take the eight inequalities containing an odd number of negative signs, say
which 4-polytope would be obtained ? I doubt it would be a regular 5-cell, since (obviously) the number of cells and the number of hyperplanes don't add up. Besides, the intersection of the two 4-polytopes corresponding to the two sets of eight inequalities should technically correspond to the 16-cell.
The tesseract, having eight cells, could be a candidate, but I have been unable to show that these eight inequalities define one (or any other 4-polytope). Any ideas?