Are there any names yet for this kind of polytope?

52 Views Asked by At

Are there any names yet for this kind of polytope (either the 3D or 4D instance): A half of the hypercube $[-1,1]^n$ cut by the hyperplane $\sum_{i=1}^n x_i = 0$? The 3D instance is a polyhedron with $1+3+3=7$ vertices and $3+3+1=7$ faces, whose spheric polar projection is like a triangular support embedded in the regular hexagon. The 4D instance is a polytope with $1+4+6=11$ vertices and $4+4+1=9$ cells, whose spheric polar projection is like a tetrahedral support embedded in an octahedron.

1

There are 1 best solutions below

0
On

I don't think that those polytopes will have any other name than the one provided by / refering to its construction: "vertex-first half of (hyper)cube". This especially is because those polytopes do have quite a different structure for even versus odd dimensions.

Consider a vertex-first $nD$ hypercube with unit edge size $x$. Then the various vertex layers form exactly the following sequence of (multi)rectified simplices (then of edge size $q = \sqrt2\cdot x$), where each Coxeter-Dynkin diagram of the respective layer has $n-1$ nodes and there are $n+1$ such layers:

$$\begin{array}{c l} o3o3o3...o & (point)\\ q3o3o3...o & (simplex)\\ o3q3o3...o & (rectified\ simplex)\\ o3o3q3...o & (birectified\ simplex)\\ \vdots\\ o3o3o3...q & (dual\ simplex)\\ o3o3o3...o & (point)\\ \end{array}$$

Whenever $n$ is even then $n+1$ will be odd and thence there is an according central vertex layer of the form $o3o3...o3q3o...3o3o$.

However when $n$ is odd then there will be an even number of vertex layers, or, told differently, there will be an odd number of segments inbetween. The central one then will be an alterprism with top base $o3o3...o3q3o3o...3o3o$ and bottom base $o3o3...o3o3q3o...3o3o$. Thus the requested midsection would dissect right this alterprism midwise in turn (parallel to its bases). This midsection too can be given by its Coxeter-Dynkin diagram as $o3o3...o3y3y3o...3o3o$, where the respective edge size clearly is $y=q/2$, because the lacing triangles thereof likewise get dissected midwise.

--- rk