I'm having trouble identifying a 4-polytope from its system of inequalities

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.

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:

1. By throwing away half of the vertices of a 3D cube, we get a demicube which happens to be a tetrahedron.
2. No faces of the cube survive; the faces of the tetrahedron are located in different planes.
3. This can be done in two ways. By doing both, we get two tetrahedra. Their union is the Kepler's stella octangula, their convex hull is the original cube, and their intersection is an octahedron.
4. The symmetry of the tetrahedron is a subset (subgroup) of that of the cube.
5. By looking at a tetrahedron, we may reconstruct the unique original cube of which it is a demicube.

Now the 4D case:

1. By throwing away half of the vertices of a tesseract (4D cube), we get a demitesseract which happens to be an orthoplex (cross-polytope).
2. The hyperfaces (cells) of the hypercube are not lost entirely, but "trimmed" to form some of those of the new figure.
3. The procedure can be done in two ways, leading to two different orthoplexes.
4. The symmetry group of the orthoplex is the same as that of the tesseract (they aren's called duals for nothing, after all), but when positioned like this, only a part of their symmetry elements is common to both.
5. By looking at an orthoplex, we may reconstruct the original tesseract using your procedure or otherwise, but it can be done in two ways.
6. By applying both procedures repeatedly, we arrive at the following magnificent construction. Imagine three sets of 8 vertices each, say: $$(\pm1,\pm1,\pm1,\pm1)\text{ with odd number of -1's}\\ (\pm1,\pm1,\pm1,\pm1)\text{ with even number of -1's}\\ (\pm2,0,0,0)\text{ and the permutations}$$ Each set defines an orthoplex. Each two sets together define a tesseract. Each tesseract has two inscribed orthoplexes, and each orthoplex is inscribed in two tesseracts. The convex hull of everything is a 24-cell. The intersection of everything is a smaller 24-cell, sitting in dual orientation to the first one.

So it goes.

• This is fascinating ! Thank you for your very articulate answer. I did not even think of looking at the inequalities more closely, having in mind the 3D case where no planes defined by the four inequalities are parallel or perpendicular to one another. If I might ask : taking instead the eight inequalities with an even number of -1's in my original post yields another tesseract. What can be said of their union ? Is it another 4-polytope ? – Croisillon Jun 25 '20 at 1:33
• Their union is some stellated body, much like stella octangula, only bigger, 4D, and without a name. – Ivan Neretin Jun 25 '20 at 5:25
• Not only the hull of those 3 sets above would result in the 24-cell, each individual hull results in a 16-cell, while each pair results in a tesseract. Accordingly there is also a compound of 3 16-cells (the stellated 24-cell) and a compound of 3 tesseracts (the great 24-cell). Cf. bendwavy.org/klitzing/incmats/stico.htm resp. bendwavy.org/klitzing/incmats/gico.htm – Dr. Richard Klitzing Jun 27 '20 at 21:47