What are midway sections of simplices?

Question

This is a (slightly modified) crosspost.

Subsequent edit - coordinates are changed to obtain simpler expressions; the existing answer is not affected.

There is a family of convex polytopes: $P_n$ is $n$-dimensional and has $(m+1)^2$ vertices for $n=2m$ and $(m+1)^2+1$ vertex for $n=2m+1$.

I define $P_n$ to be the section of the simplex in $\mathbb R^{n+1}$ with vertices $$ (\underbrace{-1,...,-1}_i,\underbrace{1,...,1}_{n+1-i}), $$ $i=0,1,...,n+1$ with the "midway" hyperplane $x_0+...+x_n=0$. For example, $P_1$ is just the segment in $\mathbb R^2$ with vertices $(-1,1)$ and $(0,0)$,

while $P_2$ it is the quadrangle in $\mathbb R^3$ with vertices $(-1,0,1)$, $(-1,\frac12,\frac12)$, $(-\frac12,-\frac12,1)$ and $(0,0,0)$ (not a square, neither a parallelogram or trapezoid, rather something of a kite).

In general, $P_n$ is the convex hull of the points $$ (\underbrace{-1,...,-1}_i,\underbrace{r,...,r}_{n+1-i-j},\underbrace{1,...,1}_j) $$ with $0\leqslant i,j\leqslant\frac n2$ and $r=\frac{i-j}{n+1-i-j}$, and one more point $(-1,..,-1,1,...,1)$ with $i=j=\frac{n+1}2$ for odd $n$. The number of such points is $m^2$ for $n=2m$ and $m^2+1$ for $n=2m+1$.

In fact some experimenting suggests that $P_{2m+1}$ is a pyramid - namely, removing that extra point $(-1,...,-1,1,...,1)$ gives a single facet of $P_{2m+1}$. In fact the base of the perpendicular from $(-1,...,-1,1,...,1)$ to this facet is another vertex $(-1,...,-1,0,0,1,...,1)$. Whether this facet is combinatorially equivalent to $P_{2m}$ I don't know. It is certainly not isometric to $P_{2m}$.

Is there some nicer description? Say, a more tidy realization of the same combinatorial type? For example, can one easily find numbers of faces of all dimensions in $P_n$? What about the dual polytope?

Update

Using the Sage code from the @MoritzFirsching's answer one checks that $P_{2m}$ is in fact the Cartesian square of an $m$-simplex. How to prove this?

Answer 1

Too long for a comment:

The $f$-vector of the polytopes in question appear to be:

P_1 (1, 2, 1)
P_2 (1, 4, 4, 1)
P_3 (1, 5, 8, 5, 1)
P_4 (1, 9, 18, 15, 6, 1)
P_5 (1, 10, 27, 33, 21, 7, 1)
P_6 (1, 16, 48, 68, 56, 28, 8, 1)
P_7 (1, 17, 64, 116, 124, 84, 36, 9, 1)
P_8 (1, 25, 100, 200, 250, 210, 120, 45, 10, 1)
P_9 (1, 26, 125, 300, 450, 460, 330, 165, 55, 11, 1)
P_10 (1, 36, 180, 465, 780, 922, 792, 495, 220, 66, 12, 1)
P_11 (1, 37, 216, 645, 1245, 1702, 1714, 1287, 715, 286, 78, 13, 1)
P_12 (1, 49, 294, 931, 1960, 2989, 3430, 3003, 2002, 1001, 364, 91, 14, 1)
P_13 (1, 50, 343, 1225, 2891, 4949, 6419, 6433, 5005, 3003, 1365, 455, 105, 15, 1)
P_14 (1, 64, 448, 1680, 4256, 7952, 11424, 12868, 11440, 8008, 4368, 1820, 560, 120, 16, 1)
P_15 (1, 65, 512, 2128, 5936, 12208, 19376, 24292, 24308, 19448, 12376, 6188, 2380, 680, 136, 17, 1)

So the number of vertices appears to be either $(n-1)^2 + 1$ or $n^2$ depending on even and odd $n$.

For even $n$, all facets of $P_n$ are combinatorially isomorphic.

For odd $n$, all but one of the facets of $P_n$ are combinatorially isomorphic.

We can also check that it seems that for odd $n$, $P_n$ isomorphic to a pyramid over $P_{n- 1}$. This was suggested in a comment by @მამუკა ჯიბლაძე. Here's some sage code to do that:

def get_P(n):
    S = Polyhedron([0]*(n + 1 - k) + [1]*k for k in range(n + 2))
    return Polyhedron(ieqs=S.inequalities_list(), eqns=[[-(n + 1)/2] + [1]*(n + 1)])

for n in range(1, 24, 2):
    assert get_P(n-1).pyramid().is_combinatorially_isomorphic(get_P(n))