## What is a step prism?

For general discussion of polyhedra, not necessarily Stella-specific.
Nikayama NDos
Posts: 2
Joined: Sat Feb 16, 2013 6:06 am

### What is a step prism?

It's outstanding that all 4D dice with 5+ cells is possible, "gyrochora" being the general method.
Stella4D tells gyrochora are dual of step prisms, which can be created by Stella4D.
But what is a step prism? I can't find any reference to it in Google or DBpia.

robertw
Posts: 568
Joined: Thu Jan 10, 2008 6:47 am
Location: Melbourne, Australia
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### Re: What is a step prism?

It was so long ago I worked on this that I really don't remember how they're defined, and possibly didn't understand it fully then! If you understand what gyrochora are, then these are their duals, so should have the dual properties.

I asked around though, and got a reply from Dr. Richard Klitzing (although I can't say I really followed it, but also haven't spent too long trying).
When considering step prisms, first think about a purely n-gonal multiprism {n} x {n} x {n} x .... x {n}, say k times. That one clearly would be a polytope living within 2k dimensions and having n^k vertices. In fact, its vertices thus can be coded by all the k-tuples of (independently chosen) numbers from 1 to n.

The next thing is to select a specific subset of just n vertices therefrom, i.e. some specific k-tuples only, and then consider the convex hull of those n points. For a typical selection this would still result in a 2k dimensional polytope.

The way of selection then is where the name step derives from. Moreover it assures the full dimensionality. You just will have to specify k-1 bijective maps f_j from the set of numbers 1,..,n onto itself (i.e. one such permutation for each j=2,...,k) - for instance say f_j(i) = m*i mod n, for some (j dependent) number m - which step the vertices of the first n-gon in a pre-defined distinct way through the other ones. That is, you just select those n specific k-tuples, which start by 1,…,n respectively and follow in the remaining k-1 entries each the globally chosen pattern: (a,b,c,...,k) = (id, f_2, f_3, ..., f_k)(a).

Nikayama NDos
Posts: 2
Joined: Sat Feb 16, 2013 6:06 am

### Re: What is a step prism?

robertw wrote:
Fri Aug 10, 2018 2:30 am
When considering step prisms, first think about a purely n-gonal multiprism {n} x {n} x {n} x .... x {n}, say k times. That one clearly would be a polytope living within 2k dimensions and having n^k vertices. In fact, its vertices thus can be coded by all the k-tuples of (independently chosen) numbers from 1 to n.

The next thing is to select a specific subset of just n vertices therefrom, i.e. some specific k-tuples only, and then consider the convex hull of those n points. For a typical selection this would still result in a 2k dimensional polytope.

The way of selection then is where the name step derives from. Moreover it assures the full dimensionality. You just will have to specify k-1 bijective maps f_j from the set of numbers 1,..,n onto itself (i.e. one such permutation for each j=2,...,k) - for instance say f_j(i) = m*i mod n, for some (j dependent) number m - which step the vertices of the first n-gon in a pre-defined distinct way through the other ones. That is, you just select those n specific k-tuples, which start by 1,…,n respectively and follow in the remaining k-1 entries each the globally chosen pattern: (a,b,c,...,k) = (id, f_2, f_3, ..., f_k)(a).
(Shouldn't the vertices be encoded by those of from 0 to n-1?)

I see. So n-m step prism is made by selecting n vertices out of n-n duoprism, one each of each n-gonal layers. f_j selects the vertices at even distance, so the resulting step prism is vertex-transitive. (Hence its dual, the gyrochoron, is cell-transitive.)

What is the symmetry of the n-m step prism? I guess it's [n,2]⁺ in Coxeter notation, is it?

robertw
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### Re: What is a step prism?

Here's some further discussion from Dr. Richard Klitzing:
Let's reduce to 4D. Then you'd start with an (n,n)-duoprism. That one could be represented by its set of square faces, which altogether form the surface of a torus. When cut open you'll get an n x n grid of squares.

Code: Select all

``````+---+---+---+---
|   |   |   |
+---+---+---+---
|   |   |   |
+---+---+---+---
|   |   |   |
+---+---+---+---
|   |   |   |  ``````
You could imagine that the cells of that duoprism, the individual n-prisms would be represented behind the columns respectively before the rows. Esp. the n-gons would be represented here as the vertical respectively horizontal edge sequences.

As I told yesterday you can number the vertices of those n-gons. Thus any vertex here is represented by a pair of numbers, the position number wrt. the vertical and wrt. the horizontal edge sequence. This in fact is just the same as for the indices of matrix elements M=(m_ij)_ij.

The issue for step prisms now is to select a subset of n vertices from this set of n^2 vertices. In order to be still full dimensional, it is required to use one vertex from every horizontal and every vertical edge sequence each (and, for sure, n being large enough). As such this is like applying the numbers 1 to n repeatedly on that grid in a sudoku like manner.

Code: Select all

``````4---3---2---1---
|   |   |   |
1---4---3---2---
|   |   |   |
2---1---4---3---
|   |   |   |
3---2---1---4---
|   |   |   |``````

And then selecting the subset of vertices with any single specific number. The step prism of consideration would be the convex hull of those n vertices. In fact, the given rule tells you how the starting vertex, chosen on the leftmost vertical line, steps through the grid while moving on towards the right.

In order to be full dimensional in our setup of 4D, n surely has to be 5 at least (because the smallest 4D figure is the 4-simplex, having 5 vertices). Moreover this setup of step prisms shows that there are polytopes with any individual number of vertices within any (even) dimensional space, starting from the number of vertices of the according dimensional simplex, onwards.