# What are midway sections of simplices?

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?

• The combinatorial automorphism group of $P_{2m}$ seems to be isomorphic to the automorphism group of the complete bipartite Graph $K_{m+1, m+1}$, a group with $2((m+1)!)^2$ elements. I'm starting to think there is an easy realization of, perhaps as the join of two simplices of dimension $m$ or something similar.. Dec 16 '20 at 8:35
• Interesting! There seem to be two simplicial faces, one on vertices with $i=0$ and another on vertices with $j=0$. It would be great if $P_{2m}$ is their join. Dec 16 '20 at 11:05
• But join does not have $m^2$ vertices, would it? It would have $m^2$ edges... Dec 16 '20 at 11:18
• @MoritzFirsching Could it be the Cartesian product of two simplices?? But then it would have $m^2$ facets while you computed that it has only $2m+2$ facets... Dec 16 '20 at 12:13
• @MoritzFirsching In fact [(polytopes.simplex(n)*polytopes.simplex(n)).is_combinatorially_isomorphic(get_P(2*n)) for n in range(14)] evaluates to True throughout! I was wrong about facets, it seems to fit. So it "only" remains to prove it Dec 16 '20 at 15:58

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))


• Fantastic! By all faces you mean all faces of the same dimension, or only all facets (faces of codimension one)? Could you also check whether that one distinguished face of $P_n$ is combinatorially isomorphic to $P_{n-1}$? Dec 15 '20 at 13:50
• I meant to write "facets" not "faces". And "yes" $P_{n-1}$ is isomorphic to the extraordinary facets of $P_n$ for odd $n$. Dec 15 '20 at 13:57
• Presumably $P_{2m+1}$ is a pyramid with base $P_{2m}$, hence facets of $P_{2m+1}$ are pyramids over facets of $P_{2m}$ Dec 15 '20 at 14:12