# $3\mathrm D$ Projection Of $4\mathrm D$ Polyhedron

Can someone identify this shape?

I think it is a $3\mathrm D$ projection of $4\mathrm D$ polyhedron. The body in the center seems to be a truncated octahedron, so as the body in the middle. The outer one is a snub cube, I think.

• Can someone help me to determine whether it is indeed a $3\mathrm D$ projecttion of a $4\mathrm D$ polyhedra?
• If it is, what is the $4\mathrm D$ polyhedron and what is the projection centered at?
• Do you have a vertex count? That might help identify it. – Joseph O'Rourke Jun 8 '15 at 12:02
• All I have is this singe image, unfortunately =( – Vlad Jun 8 '15 at 12:53
• The 'inner' facet looks like a truncated octahedron/permutohedron, and the outer face appears to be the same, but with pyramids on each of the hexagonal faces. (you can see squares in top right/front, and bottom left/front, as well as, at the very top, what appears to be a hexagonal prism). It looks like a permutohedral prism (i.e., line segment times $3$-permutohedron times a line segment, plus those pyramids on the 'outer' face) – pjs36 Jun 8 '15 at 15:18
• But keep in mind, I don't think that's how $4$-polytopes work; at least not their Schlegel diagrams: we can't just attach pyramids to facets and end up with another Schlegel diagram, in general. That was just my initial observation. – pjs36 Jun 8 '15 at 16:07

## 1 Answer

As I start writing, I'm not convinced it's even a $4$-polytope. It probably is, but I'll at least attempt a decomposition into $3$-faces and hopefully compute the Euler characteristic. I don't believe it's a well-known polytope, and I think the 'outer' facet is actually a truncated octahedron with hexagonal pyramids attached to each square face (there are several vertices of degree $6$, and the snub cube doesn't have any of those).

I definitely believe the inner facet is a truncated octahedron (I'll call any of these $P^3$, for the $3$-dimensional permutohedron; also I keep writing 'icosahedron' instead of 'octahedron'). Also, please excuse the right side of my highlighted edges, something seems a little off.

Now, when it comes to those "joining" facets, things are a little more complicated, but not too much so. Essentially, we almost have prisms coming off each face of the central $P^3$, but not quite. For the square faces of $P^3$, we have

two cubes, depicted here with red and blue edges, sharing the purple square $2$-face. Let's start counting faces of each dimension now.

We get $12$ vertices for each of the $6$ square faces of the central $P^3$, for a total of $72$ with only a few missing.

Each cube duo contributes $5 \cdot 4 = 20$ edges, for $120$ so far.

Assuming the purple squares are legitimate $2$-faces (I'm not sold on this), we have $6 + 5 = 11$ faces of dimension $2$, for $66$ so far, and finally $2 \cdot 6$ facets, plus the central $P^3$, for $13$ facets.

The final sort of facets are a bit odd

as they seem to be an 'inner' hexagonal prism (red edges) and a hexagonal prism with hexagonal pyramid 'attached' at a hexagonal $2$-face (blue edges), with each of these facets also sharing a purple hexagonal $2$-face.

There are $8$ of these facet pairs (one for each hexagon in the central $P^3$), bringing our total up to 29 (plus the outer facet), for $30$ total facets.

These facet pairs contribute only $1$ vertex not yet accounted for, and we have $72 + 8 = 80$ vertices by my count.

Now, many of the edges of these facets are contained in things listed so far. It'll be helpful to consider the $12$ green edges of the central $P^3$, each will give us $3$ edges yet uncounted, plus $6$ edges (from the apex of the pyramid) for each of the $8$ hexagonal faces of the central $P^3$. This is $12 \cdot 3 + 6 \cdot 8 = 84$ new edges, for a total whopping count of $120 + 84 = 204$ edges.

Finally, for $2$-faces, each of the $12$ green edges of the central $P^3$ gives $2$ new square $2$-faces. Each of the $8$ hexagonal faces of the central $P^3$ gives $2$ hexagonal $2$-faces (one in the central $P^3$, one with purple edges in the picture) and $6$ triangular $2$-faces, for a total of $12\cdot 2 + 8\cdot 8 = 88$ new $2$-faces, and a grand total of $66 + 88 = 154$ faces of dimension $2$.

Thus, it seems like the $f$-vector is $(80, 204, 154, 30)$, and the alleged Euler characteristic is indeed $80 - 204 + 154 - 30 = 0$, as it should be for any $4$-polytope.

• And just to be clear, I mostly wanted to confirm that it's probably a (Schlegel diagram of a) polytope by making sure the alternating sum of face counts sums to $0$, like it had better, for a $4$-polytope. Schlegel diagrams are indeed a kind of projection of a polytope down a dimension. In the process, we learned what kind of facets comprise our polytope, but I'm still not convinced it's a well-known polytope. It's like we stuck two permutohedral prisms together but those degree $6$ vertices are weird. – pjs36 Jun 13 '15 at 1:51
• But, I'm now solidly in the "not a polytope" camp, due to the degree $6$ vertices. Each is only a vertex in a single $3$-face, and that's not OK. For a polytope as an intersection of half-spaces, the vertices are where several half-spaces intersect. Those degree $6$ vertices simply don't show up this way. But if we threw them away, it's much more likely to be a projection of a $4$-polytope. – pjs36 Jun 13 '15 at 16:15
• (+1): Nice analysis! Looks to me like a cylinder over a truncated octahedron, with the hexagons at one end "capped off" with pyramids, just as you say. (The model can also be viewed as the $1$-skeleton a closed polytope homeomorphic to a three-sphere by "capping off" with the truncated octahedron of your first photo, and by "capping off" the entire outside, viewing the "hexagonal pyramids" as triangular subdivisions of the "outer" hexagonal faces, but that interpretation is "not a polytope" due to the subdivided hexagons.) – Andrew D. Hwang Jun 17 '15 at 2:02
• Thanks @AndrewD.Hwang! I've been kicking around the idea of adding another answer, on why we don't seem to have a polytope here, but it's hard to get the argument just right. I like your other approach with the subdivided hexagons, it's good to have a 2nd opinion! – pjs36 Jun 18 '15 at 23:37
• @pjs36 Thank you, your answer clarifies a lot for me! – Vlad Jul 9 '15 at 23:26