In the euclidean space, the points $(x,y,z)$ belonging to a regular octahedron are those that satisfy the inequalities
$$
\pm x\pm y \pm z \leq a
$$
where $a \geq 0$. 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
\begin{align}
x-y+z\leq a\\
-x+y+z \leq a\\
x+y-z\leq a\\
-x-y-z\leq a
\end{align}
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
$$
\pm x\pm y \pm z \pm w \leq a.
$$
If one were to take the eight inequalities containing an odd number of negative signs, say
\begin{align}
x-y-z-w\leq a\\
-x+y-z-w\leq a\\
-x-y+z-w \leq a\\
-x-y-z+w\leq a\\
x+y+z-w \leq a\\
x+y-z+w \leq a\\
x-y+z+w \leq a\\
-x+y+z+w \leq a\\
\end{align}
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 ?
Edit: I've discovered just now that 16-cells are four dimensional demihypercubes (see https://en.wikipedia.org/wiki/Demihypercube), and so they are analogous to tetrahedra in that two of them can be combined to obtain the 16 vertices of a tesseract. I am still interested to know what type of polytope corresponds to the eight inequalities above, however.
Best Answer
This is indeed a tesseract. Look at them inequalities: they define eight hyperplanes, which split in four pairs of parallel hyperplanes, which in turn are all perpendicular to each other.
This is a part of a broader picture, which has many things in common with the 3D case and many things different from it.
Let's look closer. Here is the 3D case:
Now the 4D case:
So it goes.