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

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.
• Had you check this text? It does not seem to be about centrally symmetrical polytopes, but about self-dual. arxiv.org/pdf/1902.00784 – Fedor Petrov Dec 31 '20 at 12:25
• @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 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
• I think that, with the standard inner product, the actual statement should be that $\sqrt{2} P$ and $\tfrac{1}{\sqrt{2}} Q$ are dual; the isomorphism is still given by multiplication by $\tfrac{1+i}{\sqrt{2}}$. – David E Speyer Dec 31 '20 at 16:25

## 1 Answer

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()

• 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 7 at 13:43