Permutohedron and triangulation of cube via Eulerian numbers

Question

The $h$-vector of the (simplicial complex given by the boundary of the polytope dual to the) permutohedron is the sequence of Eulerian numbers $A(n,k)=\#\{w\in S_n\colon \mathrm{des}(w)=k\}$.

Example: The (dual of the) 2-dimensional permutohedron is a hexagon, with 6 1-cells, 6 0-cells, and 1 -1-cell. So we compute the $h$-vector by $$\begin{array}{c c} (1,6,6) &= 1 \times (1,2,1)\\ & +4 \times (0,1,1) \\ &+1 \times (0,0,1) \end{array}$$ so that its $h$-vector is $(1,4,1)$.

Meanwhile the $h^*$-vector of the unit hypercube is also given by the Eulerian numbers. This means that the $h$-vector of any unimodular triangulation of the hypercube is the Eulerian numbers, padded with two trailing 0's.

Example: Looking at

we see that a unimodular triangulation of the cube has 6 tetrahedra, 18 triangles, 19 edges, and 8 vertices. So we compute the $h$-vector of this triangulation by $$\begin{array}{c c} (1,8,19,18,6) &= 1 \times (1,4,6,4,1)\\ & +4 \times (0,1,3,3,1) \\ &+1 \times (0,0,1,2,1) \\ &+0 \times (0,0,0,1,1) \\ &+0 \times (0,0,0,0,1) \end{array}$$ so that the $h$-vector is $(1,4,1,0,0)$.

Note: There is a pretty canonical unimodular triangulation of the hypercube whose maximal simplices are $0 \leq x_{w_1} \leq x_{w_2} \leq \cdots \leq x_{w_n} \leq 1$ for all permutations $w = w_1\cdots w_n$.

Question: Is there some deeper geometric connection between a triangulation of the hypercube and the permuotohedron suggested by this numerology?

In terms of what an answer could be: I believe if $\Delta$ is a $(d-1)$-dimensional simplicial complex with $h$-vector $(c_0,\ldots,c_d)$, then $\Delta \ast \mathrm{pt}$ (the cone over $\Delta$, i.e., the result of joining $\Delta$ with a single vertex) has $h$-vector $(c_0,\ldots,c_d,0)$. The simplest thing would be if the triangulation of the cube were the result of coning the permutohedron twice. So...

Question, hopeful version: Does double cone over the permutohedron give a unimodular triangulation of the hypercube?

Answer 1

The canonical unimodular triangulation of the cube $\mathcal{C}_n$ is a double cone over the subcomplex $\Gamma$ whose facets are its $(n-2)$-dimensional faces $F$ on the boundary of $\mathcal{C}_n$ not containing the origin or its antipode. Such faces $F$ are defined by $$ 0=x_{w(1)}\leq x_{w(2)}\leq\cdots\leq x_{w(n)}=1, \qquad (*) $$ where $w\in S_n$. Moreover, the faces of $F$ are obtained by converting some of the inequalities in (*) to equalities. This is exactly dual to the description of the faces of the permutohedron, so $\Gamma$ is indeed the geometric realization of an abstract simplicial complex that is isomorphic to the boundary of the dual permutohedron, as suggested in my earlier comment.

Answer 2

One context in which this relationship between the hypercube and the permutahedron appears is as follows. The space of weights of the Lie algebra $\mathfrak{sl}_n$ is naturally identified with $\mathbb R^n/(1,\dots,1)$. Under this identification vertices of the unit hypercube (other than $(0,\dots0)$ and $(1,\dots,1)$) project precisely into all Weyl group translates of fundamental weights. These are the primitive vectors of the braid fan (Weyl fan) and their convex hull is dual to the permutahedron.

In other words, the orthogonal projection along $(1,\dots,1)$ takes the hypercube to a dual of the permutahedron and identifies the subcomplex in Richard's answer with the boundary of said dual polytope. It also provides a bijection between the $n!$ maximal simplices of the mentioned triangulation and the $n!$ Weyl chambers.

This can be generalized by replacing the hypercube by an arbitrary order polytope and the permutahedron by its sum with a certain cone.