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.



Are there centrally-symmetric self-dual polytopes in dimension $d> 4$?
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**](http://dx.doi.org/10.1112/jlms/s2-43.1.137), J. Lond. Math. Soc., II. Ser. 43, No. 1, 137-148 (1991). [ZBL0757.46030](https://zbmath.org/?q=an:0757.46030).


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()

```

