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.

I do not know of any others.

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.

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.

I do not know of any others.

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.

I do not know of any others.