Vertex-first projections of hypercubes

A few days ago I was answering a question posted on math.stackexchange.com. It was asked what would be the next polytope in the following sequence "Hexagon,Rhombic Dodecahedron,???"?

One possible answer for this question goes along the following lines: Both the hexagon and the rhombic dodecahedron are vertex-first projections of cubes. The hexagon is the projection of the 3-cube, and the rhombic dodecahedron is the projection of the 4-cube. So the next polytope in this sequence would be the vertex-first projection of the 5d cube. (I consider the vertex-first projection that aligns two opposing cube vertices along the normal vector of the projection hyperplane.)

Vertex-first projection of the 3-cube.

Vertex-first projection of the 3-cube.

When answering this question I did a quick web search and tried to find this polytope. To my surprise I couldn't find anything about it and it seems that this polytope is not well know. To get an intuition I wanted to compute it. However this was somehow tricky so I decided to think about this later.

Today I finally found the time to get back to this question. My plan was to use polymake to compute the polytope. The first difficulties came with installing polymake. Fortunately, there is a pre-compiled package for MacOS. I did not succeed to install polymake under Lion, but the Mountain Lion version on my Laptop worked out of the box. Polymake is written in Perl and to use it I had to get into Perl basics first. Moreover, it seems that polymake is not the best tool to compute a vertex-first projection.  That's why I computed the coordinates in Mathematica (there is a very handy function called RotationTransform which makes it easy to get the cubes into the right position). As a first test I computed the vertex-first projection of the 4-cube, and after some tweaking I got the rhombic dodecahedron shown in the picture.

Vertex-first projection of the 4-cube.

Vertex-first projection of the 4-cube.

The next step was the projection of the 5-cube. This was a bit trickier, since the coordinates of the projection are not rationals (they were rationals in the 4-cube projection). So instead of projecting the all the vertices of the 5-cube, I was projecting only the vertices . The projected vertices are with

Then I computed with polymake the zonotope spanned by the segments by the points and the origin. Clearly this gives the desired polytope but I did not had any problems with precision. Here is what I got:

Vertex-first projection of the 5-cube.

Vertex-first projection of the 5-cube.

The computed polytope has 30 vertices, 70 edges, 60 2d-faces, and 20 3d-faces. I was surprised that I couldn't find the polytope on Wikipedia. The construction seems very natural,  and  the way it is built out of cubes is very neat. Does it have a name or does it appear somewhere else in a different context? This would be interesting to know.

Since it is really hard to understand the Schlegel diagram from a static picture I recorded a small animation that shows the rotation of the Schlegel diagram.

The 5d analogue (projection of the 6d cube) has the f-vector (62, 180, 210, 120, 30).

2 thoughts on “Vertex-first projections of hypercubes

  1. Permalink  ⋅ Reply

    Paco Santos

    March 4, 2013 at 12:43pm

    I have just posted the following on stackexchange:

    "Hi, I got here via A. Schulz's blog entry. As he says, you are looking for the projection of the (d+1)-dimensional hypercube (the (d+1)-cube for short) along a diagonal. But, actually, any generic projection of the (d+1)-cube will give (combinatorially) the same polytope. By "generic" I just mean that the direction of projection is not parallel to any coordinate hyperplane.

    A different description is that your polytope is the zonotope obtained as the Minkowski sum of any set of d+1 generic vectors in R^d, where generic means that no d of them lie in a hyperplane."

    In particular, for computations one can use (e.g.), the standard basis vectors plus the vector (1,...,1).

    • Permalink  ⋅ Reply

      Andre

      March 4, 2013 at 6:28pm

      Dear Paco,
      you are completely right that in this case the combinatorial description of the polytope is the same no matter how you project down (avoiding degeneracies). However, I was also interested in the actual geometric vertex-first realization, so I missed that almost all projections give combinatorically the same polytope. In fact, I was using the zonotope interpretation to get rid of the perturbations introduced by rounding the reals in the 5d cube projection.

      Thanks for reading my blog,

      André

Leave a Reply