Polytope constructed from $2^{D-1}$ intersecting linear inequalities

Question

I have a polytope that has come up in my research, and I am not a geometer so I'm unsure how to characterise it. I will outline its construction.

In $D$ dimensions, we consider the set of $x$'s that satisfy: $$L_0 \leq c_d^\top x \leq L_1, d=1,\dots,2^{D-1},$$ for given $c_d \in \mathbb{R}^D$ and $L_0, L_1 \in \mathbb{R}$.

In one dimension we have an interval. In two dimensions we have a parallelogram, and in three dimensions I believe the shape is some kind of octahedron. Below are some graphs that illustrate the 2D and 3D cases.

Can someone identify what shape this forms as we increase the dimension? Is there a neat characterisation?

Answer 1

When restricting further to all unit edges, then what obviously belongs here are all the orthoplexes (hyperoctahedra) x3o...3o4o, which are well-known regular polytopes, existing within every dimension. But even then this wouldn't be all: Obviously within 3D there also would be the truncated tetrahedron x3x3o, which only is a non-regular uniform polyhedron.

How could the latter observation be generalized into higher dimensions, if your restriction still has to be applied?

First we observe that the orthoplex, when oriented simplex first, can be represented as some "lace prism" or "segmentotope", i.e. a polytope, which uses 2 parallel facets (bases - which here are a pair of dually arranged simplices) and the remainder then are lacing facets, which simply interpolate any (d-k)-face of the top base with the accordingly placed k-face of the bottom base. Thence we might write x3o...3o4o (n nodes) = x3o...3o || o3o...3x (n-1 nodes each).

Then we observe that the former occurance of the truncated tetrahedron within this latter description would serve alike: x3x3o || o3x3x indeed is a non-uniform scaliform polychoron (4D polytope), the facets of which are 2 truncated tetrahedra (bases) + 6 tetrahedra (lacing pairs of orthogonal edges) + 8 triangular cupolae (being laced between triangles and hexagons, pointing either up or down) = 16 facets again.

The same applies within 5D too: the orthoplex here could be written as x3o3o3o || o3o3o3x. But both, x3x3o3o || o3x3o3x and x3x3x3o || o3x3x3x, have 32 facets as well.

And this applies within any higher dimension alike: just exchange any ...o..... || ...o..... node pair by an ...x..... || ...x..... node pair. Thereby proving that all its facets still come in parallel pairs: It is obvious for the orthoplexes themselves. And those above replacements are nothing but partial Stott expansions thereof: in fact those lacing faces, which are described by the complementary subsymbols of any ...x..... || ...x..... pair, has just been pulled radially out orthogonally to the axis of the stack of those segmentotopes.

(Sure, your quest was a bit broader, then allowing for different edge sizes as well. I.e. all facet-parallel variations of the above described polytopes would be allowed then too.)

--- rk