Face lattice of n-simplex is an (n+1)-cube

[quickfur]

Hi all,

Just dropping to mention a curious fact that I discovered today: the Hasse diagram of the face lattice of an n-simplex is isomorphic to the edges and vertices of an (n+1)-hypercube.

For example, given a triangle ABC, we have the set ABC, then the edges, AB, AC, and BC; and then the vertices, A, B, and C, and finally the empty set. If you draw the Hasse diagram, you'll see that there are 8 nodes, corresponding with 8 vertices, and 12 subset relationships, corresponding with 12 edges.

The case of the triangle by itself is unremarkable, and may be attributed to coincidence... but if you draw the Hasse diagram of the face lattice of a tetrahedron, you'll see that it's isomorphic to the edges and vertices of a tetracube.

In general, an n-simplex's face lattice gives rise to a Hasse diagram that is isomorphic to an (n+1)-cube. This is a rather remarkable fact, and quite unexpected to me. What do people think about this strange connection?

(Some explanations for those who are unfamiliar with face lattices: to understand a what a lattice is, wrt to polytopes, consider the set V of the vertices of, say, a cube. We may represent each face as a 4-element subset of V, containing those vertices that define the face, and we may represent each edge as a 2-element subset of V, and each vertex as a singleton set containing the vertex. Now, add the empty set to the mix, and you have a collection of subsets of V, including V itself (which represents the entire cube). Now, to make the Hasse diagram of this structure, represent each of these subsets of V as a node, and connect two nodes A and B if A is a subset of B. But omit the edges between vertices and V, since it's implied (the subset relation is transitive). This gives you a graph, the Hasse diagram with a number of nodes and some edges connecting them, describing the face lattice of the cube.)

Now, I've also wondered what the Hasse diagram of a square may represent, and I get a graph that looks like the dual of a square antiprism. I haven't tried to draw the Hasse diagram of a cube yet, but it might turn out to be an interesting shape.

[wendy]

The Hasse diagram for any polytope is the same as its antitegum, and likewise, an antitegum is the hasse diagram of either of its bases (figure or dual).

Antitegums are quiet interesting, really. Every surtope of an antitegum is an antitegum. A prism-product of antitegums is itself an antitegum, (being the Hasse diagram for the pyramid product of the bases). Since the cube is the prism-product of line-segments (the hasse-diagram for a point), it is itself the hasse-diagram for the pyramid-product of points (ie a simplex).

In the simplest sense, imagine you have an open pyramid, with a cubic base. Onto this, you allow it to intersect with an open pyramid on an octahedral base, so that the axis of both pyramids are the same, and the apices point inwards (eg <----> ). When the cube and octahedron are in dual position, you get a hasse diagram for either the cube or the octahedron.

The octahedron end looks like a set of 8 cubes at a corner, but bent backwards somewhat. The cube end consist of six square antitegums.

[Marek14]

For n-simplex, the total number of elements is equal to 2^(n+1), because it has n+1 vertices and every of its elements can be uniquely described as a combination of those vertices. Also, in the diagram, each element will be connected to all those that differ from it in the presence or absence of exactly one vertex. So it follows that it's diagram has to be isomorphic to (n+1)-hypercube.

For other shapes, you can probably imagine the diagram as lace tower - each of its element types will transform in vertices in one layer of the tower, and we want all edges to be between two adjacent layers. Thus follows that square will be square deltahedron, and generally any n-gon will transform in n-gonal deltahedron, dual to n-gonal antiprism (note that triangle transforms into cube, which is dual to triangular antiprism, a.k.a. octahedron).

Let's look in three dimensions. One thing that seems clear is that a polyhedron will be always assigned the same diagram as its dual. Let's look at the cube.

We can construct the 4D polychoron easily:

1. Take "cube node" for full cube, and put it in [0,0,0,0]
2. Take six "face nodes", and put them in centres of the faces, shifted in 4th dimension.
3. Take twelve "edge nodes", and put them in centres of the edges, shifted once again.
4. Take eight "vertex nodes", and put them in vertices, once again shifted.
5. Take "empty set node", and put it at the very end.

Each layer can be changed in 4-position or in size.

o4o3o||o4o3x||o4x3o||x4o3o||o4o3o

o3o||o3x||x3o||o3o is isomorphic to cube, while o4o||o4x||x4o||o4o is isomorphic to square deltahedron. Thus it seems that the polychoron is composed of cubes and square deltahedra.