Are descents in alternating subgroup counted by $h$-vector? Consider the alternating subgroup $A_n$ of the symmetric group $S_n$ (or in general any Coxeter Group). Is there a simplicial complex whose $h$-vector $h_i$ equals the number of elements of $A_n$ with $i$ descents? Note that the $h$-vector of the Coxeter Complex of $S_n$ counts descent in $S_n$.
I found this paper where "Coxeter-like complexes" for the alternating subgroup was defined (but these complexes depend on the choice of a distinguished generator). I am not sure if one can compute $h$-vectors of these complexes or show that they count descents. Any help regarding this will be appreciated. 
 A: The descents of the alternating group do not give the $h$-vector of the simplicial complex from the paper you linked. This is easy to see whenever $n$ is $0$ or $1 \bmod 4$, because then $w_0$ is an even permutation with $n-1$ descents, but the simplicial complex only has dimension $n-2$. 
As a concrete example, when $n=4$, the number of even permutations with $k$ descents is $(1,5,5,1)$, for $k=0$, $1$, $2$ and $3$. This would be the $h$-vector of a simplicial complex with $8$ vertices, $18$ edges and $12$ triangles and (if we are assuming it is shellable) the homology of $S^2$. There are many such. One concrete way to achieve this is take a cube and divide each face along a diagonal into two triangles.
The complex of Brenti, Reiner and Roichman, in this example, is $1$-dimensional. It is a subcomplex of the $1$-skeleton of the Coxeter complex. Which subcomplex it is depends on a choice of simple generator of $S_4$. It is either the $1$-skeleton of a cube, or else it is the graph obtained by taking the $1$-skeleton of a tetrahedron and subdividing each edge in half.  See figure $2.1$.
