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A convex polytope $P\subset\Bbb R^d$ is centrally symmetric if $-P=P$. It is self-dual (or better, self-polar?) if its polar dual $P^\circ$ is congruent to $P$, that is, there is a map $X\in\mathrm O(\smash{\Bbb R^d})$ with $\smash{P^\circ}=XP$.

Question: Are there centrally symmetric self-dual polytopes in dimension $d>4$?

Such exist in dimension $d=2$ and $d=4$:

  • for $d=2$ we have the regular 2n-gons,
  • for $d=4$ we have the regular 24-cell.
M. Winter
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  • Had you check this text? It does not seem to be about centrally symmetrical polytopes, but about self-dual. https://arxiv.org/pdf/1902.00784 – Fedor Petrov Dec 31 '20 at 12:25
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    @FedorPetrov Yes. The paper mainly deals with the case $P^\circ=-P$ (then $P$ cannot be centrally symmetric). The two mentioned constructions, pyramids and joins, do not yield centrally symmetric polytopes either. The add-and-cut construction works only if we already have a self-polar polytope in dimension $d>4$. – M. Winter Dec 31 '20 at 12:29
  • what is $X$ for a 24-cell? – Fedor Petrov Dec 31 '20 at 13:26
  • @FedorPetrov I guess I could compute the matrix explicitly. This would take me some time. Why do you want to know? Do you doubt its existence? – M. Winter Dec 31 '20 at 13:29
  • I do not doubt, but it could (theoretically) somehow suggest other examples. – Fedor Petrov Dec 31 '20 at 13:36
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    @FedorPetrov Let $P$ be the convex hull of the $24$ units in the ring of half integer quaternions. Let $Q$ be the convex hull of the $24$ integer quaternions with norm $2$. Then $P$ and $\tfrac{1}{\sqrt{2}} Q$ are dual polytopes, each of which is isomorphic to the $24$-cell. Multiplication (either left or right) by $\tfrac{1+i}{\sqrt{2}}$ is an isomorphism from $P$ to $\tfrac{1}{\sqrt{2}} Q$. – David E Speyer Dec 31 '20 at 14:08
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    I think that, with the standard inner product, the actual statement should be that $\sqrt[4]{2} P$ and $\tfrac{1}{\sqrt[4]{2}} Q$ are dual; the isomorphism is still given by multiplication by $\tfrac{1+i}{\sqrt{2}}$. – David E Speyer Dec 31 '20 at 16:25

2 Answers2

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There are centrally symmetric self-dual polytopes in every dimension. This follows from Proposition 3.9 in Reisner, S., Certain Banach spaces associated with graphs and CL-spaces with 1- unconditional bases, J. Lond. Math. Soc., II. Ser. 43, No. 1, 137-148 (1991). ZBL0757.46030.

Moreover, in dimension $\geqslant 3$ the matrix $X$ can be chosen to be a permutation matrix.

Here is an example in dimension $3^d$ for every $d$. Start with Sztencel-Zaramba polytope $P$. This is the unit ball for the norm on $\mathbf{R}^3$ $$ \|(x,y,z)\| = \max \left( |y|+|z|, |x|+\frac 12 |z| \right)$$ whose dual norm satisfies $$ \|(x,y,z)\|_* = \|(z,y,x)\|. $$ We may now define inductively a sequence $\|\cdot\|_d$, which is norm on $\mathbf{R}^{3^d}$ (identified with $\mathbf{R}^{3^{d-1}}\times\mathbf{R}^{3^{d-1}}\times\mathbf{R}^{3^{d-1}}$). Chose $\|\cdot\|_1$ to be above norm, and use the recursive formula $$ \|(x,y,z)\|_{d+1} = \|( \|x\|_d ,\|y\|_d , \|z\|_d )\|_1 .$$ One checks by induction that there is a permutation matrix which maps the unit ball onto its polar.

To visualize the polytope $P$ you may use the Sage code

p1 = Polyhedron(vertices=[[0,1,1],[0,1,-1],[0,-1,1],[0,-1,-1],[1,0,1/2],[1,0,-1/2],[-1,0,1/2],[-1,0,-1/2]])
p1.projection().plot()
Guillaume Aubrun
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  • This is amazing! Thank your for this answer. I had already mentally excluded the possibility that I could have overlooked an example in dimension three. – M. Winter Jan 07 '21 at 13:43
  • Allowing for nonconvex polytopes, there are even uniform examples in (at least) 7D and 8D. – Daniel Sebald May 13 '21 at 13:35
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I stumbled upon another way to construct many self-dual (actually self-polar) centrally symmetric polytopes in higher dimensions.

Start from a graph $G$ on vertex set $[d]:=\{1,...,d\}$ and consider the following polytope (where $e_i$ is the $i$-th unit normal vector):

$$\mathcal P(G):=\mathrm{conv}\Big\{\sum_{i\in I} \pm e_i\;\Big\vert\;\text{$I\subseteq[d]$ induces a complete subgraph in $G$}\Big\}.$$

The $\pm$ means to consider all $2^{|I|}$ sums with all combinations of $+$ and $-$ so that the resulting polytope is centrally symmetric, in fact, coordinate symmetric (i.e. invariant under reflection at any coordinate hyperplane). This is also known as an unconditional polytope in the literature.

By a result due to Lovász, $G$ is a perfect graph if and only if $(\mathcal P(G))^\circ=\mathcal P(\overline G)$, where $\overline G$ denotes the complement of $G$. Thus, if $G$ is perfect and self-complementary, we constructed a centrally symmetric self-dual (actually, self-polar) polytope in dimension $d$. The smallest (non-trivial) example is a 4-dimensional polytope $\mathcal P(P_3)$, where $P_3$ is the path of length three (with four vertices). Considering this list of small self-complementary graphs there is exactly one further polytope in dimension $d=5$ based on the bull graph. According to this answer there seem to exist many self-complementary perfect graphs. But note that self-complementary graphs require $d\equiv 0,1\pmod 4$, and so this technique cannot construct polytopes in dimension $2,3,6,7,...$.

Some references

  • L. Lovász. "Normal hypergraphs and the perfect graph conjecture"
  • F. Kohl, M. Olsen, R. Sanyal. "Unconditional reflexive polytopes"
M. Winter
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