# Name this polytope

I was wondering what people call a certain type of shape. It is the shape formed by an orthogonal projection of a hypercube along one of its longest diagonals.

In other words, fill in the missing entries:

n-simplex : Triangle ->      Tetrahedon        ->  5-cell    ...
n-cube   :  Square  ->         Cube           -> Tesseract  ...
?      :  Hexagon -> Rhomibic Dodecahedron  ->     ?      ...


IMPORTANT: I'm looking for the name of the class of shapes, not just the 4D analogue.

-
It seems to me your arrows should point in the opposite direction---from right to left. –  Michael Hardy Dec 10 '12 at 3:53
@MichaelHardy The arrows represent the well known coolness ordering of polytopes, read "X -> Y" as "Y is cooler than X". I think I get your intuition, there are increasing freedoms the way I drew them. e.g. the tetrahedron uniquely specifies the triangle, but not vice-versa. –  Lucas Dec 10 '12 at 4:16

What you are looking for is the vertex-frist projection of a hypercube. Apparently it seems that the 4d analogue has no special name.

Edit: I computed the polytope. Read my blog post about it. Here is a teaser:

-

Is there any chance that you are looking for the name of a {3,4,3} regular convex polychoron? It is called 24-cell (or icositetrachoron, octaplex, octacube polyoctahedron).

24-cell can be derived as a rectified 16-cell (different from 16-cell).

Other related polychrons (uniform, but not necessary regular and convex):

Or if you are looking for a {5,3,3} regular and convex 4D polytope, it is called 120-cell (or hecatonicosachoron).

Here is a table I found on Wikipedia that describes the polytope families. Hope you find it useful.

-
That isn't what I'm asking, sorry :( Possibly it is the dual (in some sense) of the 24-cell. –  Lucas Dec 18 '12 at 11:55
Also, I want the name of a whole class. –  Lucas Dec 18 '12 at 11:55

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 $\mathbb R^d$, where generic means that no $d$ of them lie in a hyperplane.

-
Can you please add a link to A. Schultz's blog entry? –  MJD Mar 4 '13 at 14:14
The blog at least seems to be Aaron Schultz's, but I could not find the specific post. –  MJD Mar 4 '13 at 14:17
It is not Aaron, but André Schultz. I did not add the link since he put the link in his answer, above (math.stackexchange.com/a/261323/64942). Anyway, the link is: ongraphs.de/blog/2013/02/vertex-first-projection-of-hypercubes –  Paco Santos Mar 4 '13 at 20:41