3
$\begingroup$

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 $(0,0)$,

enter image description here

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).

enter image description here

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?

$\endgroup$
8
  • 1
    $\begingroup$ 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.. $\endgroup$ Dec 16 '20 at 8:35
  • $\begingroup$ 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. $\endgroup$ Dec 16 '20 at 11:05
  • $\begingroup$ But join does not have $m^2$ vertices, would it? It would have $m^2$ edges... $\endgroup$ Dec 16 '20 at 11:18
  • $\begingroup$ @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... $\endgroup$ Dec 16 '20 at 12:13
  • 1
    $\begingroup$ @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 $\endgroup$ Dec 16 '20 at 15:58
2
$\begingroup$

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))

    
$\endgroup$
3
  • $\begingroup$ 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 $P_{n-1}$? $\endgroup$ Dec 15 '20 at 13:50
  • 1
    $\begingroup$ I meant to write "facets" not "faces". And "yes" $P_{n-1}$ is isomorphic to the extraordinary facets of $P_n$ for odd $n$. $\endgroup$ Dec 15 '20 at 13:57
  • 1
    $\begingroup$ Presumably $P_{2m+1}$ is a pyramid with base $P_{2m}$, hence facets of $P_{2m+1}$ are pyramids over facets of $P_{2m}$ $\endgroup$ Dec 15 '20 at 14:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.