Permutohedron Facets

Question

The permutohedron P4 is the 3d truncated octahedron. Its facets consist of 6 hexagons and 8 squares.

The permutohedron P5 is a 4d polytope with 30 3d facets: there are 10 truncated octahedra, and 20 hexagonal prisms (I believe).

The permutohedron P6 is a 5d polytope with 62 4d facets: there are 12 P5 polytopes, 30 truncated octadedral prisms, and 20 polytopes with 36 vertices. These latter seem to be 4d polytopes with 36 vertices and 12 3d facets, each of which is a hexagonal prism. Is that correct?

Answer 1

The permutatopes $P(n)$ are just the omnitruncated $n-1$ dimensional simplices. Accordingly these are represented by the Coxeter-Dynkin diagrams $x3x3x3...3x3x$ with $n-1$ nodes and in here each node symbol $x$ represents a ringed node (in a typewriter friendly way).

As is known for Coxeter Dynkin diagrams, the according facets are obtained by omission of any possible single node each (as long as those do not become degenerate). I.e. in here we have $$\begin{array}{ccl}.\ x3x3...3x3x & = & P(0)\times P(n-1)\\ x\ .\ x3...3x3x & = & P(1)\times P(n-2)\\ ... & = & P(k)\times P(n-k-1)\\ x3x3x3...3x\ . & = & P(n-1)\times P(0)\end{array}$$

The respective counts each are provided by the $n$-th row of the Pascal triangle (excluding the extremal 1 each), e.g. the truncated octahedron $P(4)=x3x3x$ has $choose(4,1)=4$ hexagons $x3x\ .$, $choose(4,2)=6$ squares $x\ . \ x$, and $choose(4,3)=4$ more hexagons $.\ x3x$.

As for your question about the 5D $P(6)$ this esp. provides $choose(6,1)=6\ P(5)$s, $choose(6,2)=15\ P(4)$-prisms, $choose(6,3)=20\ P(3)$-duoprisms, further $choose(6,4)=15\ P(4)$-prisms, and further $choose(6,5)=6\ P(5)$s. Here $P(3)$ clearly is the regular hexagon. Thus those 36 mentioned vertices of yours indeed divide as $6\times 6$ within these duoprisms.

--- rk

Answer 2

From @pjs36 's answer to this linked question: Nice embedding of the permutohedron of order $n$ in ${\mathbb R}^{n-1}$

[The] $k$-dimensional faces of the permutohedron $P_n$ are in bijection with chains of subsets of $[n] = \{1, 2, \ldots, n\}$ of length $n - k$, if we include $[n]$ in the chain (but never $\varnothing$).

...

Because of this description, we can also pick out facets (faces of codimension $1$) rather easily: they're in bijection with proper, nonempty subsets of $[n]$.

I think there is enough information in that answer to understand the structure of the facets. There is an example explaining the square facets of $P_4$.