TITLE: Nice embedding of the permutohedron of order $n$ in ${\mathbb R}^{n-1}$
QUESTION [7 upvotes]: The permutohedron $P_n$ of order $n$ ($n\geqslant 2$) is the convex hull of the points $P_\pi=(\pi(1),\dots,\pi(n))$ where $\pi$ ranges over all permutations of $\{1,2,\dots,n\}$. Obviously, since these points are all in the plane with equation $x_1+\dots+x_n=\frac{n(n+1)}{2}$, the polytope $P_n$ has dimension at most $n-1$. If I am not wrong, its dimension is exactly $n-1$. 
Is it possible to find a "nice" embedding of $P_n$ in ${\mathbb R}^{n-1}$? "Nice" means here a simple description of the vertices. Since I am mostly interested in the combinatorics of polytopes, my question can be formulated as: is there a $(n-1)$-polytope in ${\mathbb R}^{n-1}$ with a description of vertices is as simple as possible and that is combinatorially equaivalent to $P_n$?

REPLY [3 votes]: There is a very simple formula. If $v(\sigma)\in\Bbb R^d$ is the vertex that corresponds to the permutation $\sigma:\{1,...,d+1\}\to\{1,...,d+1\}$, then you can choose its coordinates as
$$v_i(\sigma)=\sigma(d+1)+\sigma(i),\quad\text{for $i\in\{1,...,d\}$}.$$

If you want a standard permutahedron, centered at the origin and with all edges of the same length, then you can use
$$v_i(\sigma) = \frac{d+2}2 -\frac{\sigma(d+1)}{\sqrt{d+1}}-\sigma(i)\Big(1-\frac1{\sqrt{d+1}}\Big).$$
All edges are of length $$\sqrt2\Big(1-\frac1{\sqrt{d+1}}\Big).$$

If you use the Mathematica code with the second formula
vert[d_] := Table[
   (d+2)/2-s[[d+1]]/Sqrt[d+1]-s[[i]](1-1/Sqrt[d+1]),
{s,Permutations[Range[d+1]]}, {i,1,d}]

then vert[1] gives you the points $1/2-1/\sqrt2$ and $-1/2+1/\sqrt2$.
And if you use ConvexHullMesh to plot vert[2] and vert[3] you find