Existence of 5-d centrally symmetric self-dual polytope

Question

Does there exist a 5-dimensional centrally symmetric self-dual polytope?

Answer 1

I recently discovered a 5-dimensional centrally symmetric, self-dual polytope, and I am not aware of anyone else discussing it, so I decided to ask & answer this question in the affirmative. I phrased the question this way because the 4-dimensional 24-cell is famously regular, centrally symmetric and self-dual. If nothing else, this answer might be useful for cocktail party banter: “I was reading the other day about a 5-dimensional polytope that’s both centrally symmetric and self-dual. Of course, it’s not regular.”

The tridiagonal Birkhoff polytope $TB_{d+1}$ is a d-polytope. It has been noted (footnote 1) that $TB_{d+1}$ is combinatorially equivalent to the Fibonacci d-polytope (of order 2) $FP_d$ (2); i.e., the 0/1 d-polytope whose vertices have no consecutive pair of 1s. (Hereinafter just called $FP_d$.) This equivalence results from the fact that a tridiagonal permutation matrix is determined by its superdiagonal.

I call the subject polytope $JCS_5$. Its construction begins by embedding $FP_4$ in the hyperplane $x_5$ = 1. So, the first 8 vertices are:

$v_1$: (0, 0, 0, 0, 1)

$v_2$: (1, 0, 0, 0, 1)

$v_3$: (1, 0, 1, 0, 1)

$v_4$: (0, 1, 0, 0, 1)

$v_5$: (1, 0, 0, 1, 1)

$v_6$: (0, 0, 1, 0, 1)

$v_7$: (0, 1, 0, 1, 1)

$v_8$: (0, 0, 0, 1, 1)

Then, let $v_{8+i}$ = -$v_i$ for i = 1 to 8. By construction, the convex hull of this vertex set is a centrally symmetric polytope, and the 16 vertices identified are polytope vertices.

The first 8 facets of the form $f_i$ $\cdot$ x = 1 are:

$f_1$: (0, 0, 0, 0, 1)

$f_2$: (-2, -2, 0, 0, 1)

$f_3$: (-2, 0, 0, 0, 1)

$f_4$: (0, -2, 0, 0, 1)

$f_5$: (0, -2, -2, 0, 1)

$f_6$: (0, 0, -2, 0, 1)

$f_7$: (0, 0, 0, -2, 1)

$f_8$: (0, 0, -2, -2, 1)

The last 8 facets are given by $f_{8+i}$ = -$f_i$ for i = 1 to 8.

The ordering of the vertices and facets was chosen to produce a nicely transposable facet-vertex incidence matrix M:

By applying the following permutation to the columns of $M$:

(2 3 6 8 7 4) (10 11 14 16 15 12);

And the following permutation to the rows:

(2 6 7) (3 8 4) (10 14 15) (11 16 12);

You obtain the transpose of $M$; $M^T$. Per (2), this implies that $JCS_5$ is self-dual.

The f-vector of $JCS_5$ is (16, 64, 98, 64, 16). Eight of the 16 facets of $JCS_5$ are of type $FP_4$ (facets 1, 4, 5, 6, 9, 11, 12 and 13), and the other eight are of a type I call $J_4$. The seven facets of $FP_4$ are two triangular prisms, four pyramids, and a tetrahedron. The nine facets of $J_4$ are an octahedron, four pyramids and four tetrahedra. $J_4$ is the combinatorial type called the ‘special 0/1 polytope P’ in (3).

$J_d$ is constructed by beginning with $FP_{d-1}$ (having $F_{d+1}$ vertices, where $F_i$ is the Fibonacci sequence) embedded in $x_d$ = 1. This polytope has four facets of the type $FP_{d-2}$ (having $F_d$ vertices). (These facets are given by $x_1$ = 0, $x_{d-1}$ = 0, $x_1$ + $x_2$ = 1, and $x_{d-2}$ + $x_{d-1}$ = 1.) The inverses of the vertices of one of these facets are added, and the convex hull taken, obtaining $J_d$, a d-polytope with $F_{d+2}$ vertices.

Based on low dimensional results, it appears that:

• For d $\ge$ 4, $J_d$ has a facet of type $JCS_{d-1}$, which in turn has facets of type $J_{d-2}$.
• Also, $J_d$ has facets of type $J_{d-1}$.
• $JCS_d$ has 4(d – 1) facets, and $J_d$ has 4d – 7 facets.

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(1) da Fonseca, C.M., “Some open questions on the Fibonacci polytope”, Linear and Multilinear Algebra, Volume 59, Issue 3, March 2011

(2) Tiwary, H.R., Elbassioniy, K., “On the Complexity of Checking Self-duality of Polytopes and its Relations to Vertex Enumeration and Graph Isomorphism”, SCG ’08, June 9-11, 2008

(3) Baumeister, B.; Haase, C.; Nill, B.; Paffenholz, A.; “On permuation polytopes”, Adv. Math. 222 (2009) pp. 431-452

Answer 2

The simplex is the only regular self dual polytope in dim $\ge 5$, see e.g. wikipedia Did you mean this?