Are there any graphic portrayals of von Neumann polytopes in low dimensions?

2$\begingroup$ Is it the same as the Birkhoff polytope? $\endgroup$ – Igor Khavkine Sep 17 '14 at 19:03

2$\begingroup$ The BirkoffvonNeumann polytope (if that's what you mean) lives in $\mathbb{R}^{n^2}$, so it is difficult to picture for $n>1$. I think for $n=2$, it is a segment in $\mathbb{R}^4$. $\endgroup$ – Joseph O'Rourke Sep 17 '14 at 19:11

14$\begingroup$ As a general good practice, I recommend including a definition of the terminology needed to understand a question (specially if the question is very short, as is the case here). $\endgroup$ – André Henriques Sep 17 '14 at 22:29
One can get some intuition for what this polytope $\mathcal{B}_n$ looks like from the fact that the face lattice of $\mathcal{B}_n$ is isomorphic to the set of all subsets $S$ of $[n]\times[n]$ (where $[n]=\lbrace 1,2,\dots,n\rbrace$), ordered by inclusion, such that $S$ is the support of a doublystochastic matrix, together with the empty face. For instance, when $n=3$ the 3dimensional faces correspond to removing a single element from $[3]\times [3]$ (so there are nine of them), the 2faces correspond to removing two elements, no two in the same row or column (so 18 of them), the 1faces (edges) from removing three elements, no two in the same row or column (six of them), or by choosing an element $(i,j)\in [3]\times[3]$ and removing the two elements in the same row and the two elements in the same column as $(i,j)$ (so nine of them). The 0faces (vertices) are of course the supports of the six permutation matrices. The $f$vector is thus $(6, 15,18,9)$. A graphical portrayal of this 4dimensional polytope would be messy.
This answer is meant to supplement the second part of Richard Stanley's answer.
Given a natural number $n$, the von Neumann (or Birkhoff) polytope is defined as the convex hull of $n \times n$ permutation matrices. Its ambient space is therefore $\mathbb{R}^{n^2}$ as Joseph O'Rourke points out in his comment. However, its dimension is $(n1)^2$. So for $n = 3$ you get a $4$dimensional polytope whose Schlegel diagram looks as follows:
The vertices I used are: $0 = (1, 0, 0, 1)$, $1 = (0, 0, 0, 1)$, $2 = (0, 0, 1, 0)$, $3 = (0, 1, 1, 0)$, $4 = (0, 1, 0, 0)$, $5 = (1, 0, 0, 0)$ (sorry for the awkward notation, it was chosen by Polymake).
The set of doubly stochastic matrices is a subset of the set of nonnegative $n \times n$ matrices whose sum of entries is $n$.
These matrices form an $(n^21)$dimensional simplex in $\mathbb{R}^{n^2}$, whose vertices are the $n^2$ matrices (written as row vectors of length $n^2$) $(n,0,0,...,0)$, $(0,n,0,...,0)$, ..., $(0,0,... 0,n)$.
The vertices of the Birkhoff polytope are the arithmetic means of $n$ of these  suitably chosen  matrices.
For $n=2$ the Birkhoff polytope lies within a tetrahedron, between the midpoints of the edges that connect points $1 = (2, 0, 0, 0)$ and $4 = (0, 0, 0, 2)$ and points $2 = (0, 2, 0, 0)$ and $3 = (0, 0, 2, 0)$.