TITLE: Geometric realization of combinatorial self-duality in polytopes
QUESTION [11 upvotes]: Let's say I have a combinatorially self-dual polytope $P\subseteq\Bbb R^d$, i.e., its face lattice is isomorphic to its dual (you reverse the direction of the lattice order).

Question: Is it always possible to realize $P$ geometrically, so that $P$ and its polar
  $$P^\circ := \{x\in\Bbb R^d  \mid \langle x,s\rangle \le 1 \text{ for all $s\in P$}\}$$
  are geometrically related in some sense, e.g. via orthogonal, linear, affine or projective transformations? In other words: so that they are geometrically self-dual?

I would prefer to have a realization of the duality under the weakest possible transformations (i.e. orthogonal), but I wonder whether more general are necessary.

REPLY [3 votes]: Alathea Jensen defines "self-polar":

Self-polar polytopes are convex polytopes that are equal to an orthogonal transformation of their polar sets.

and writes some interesting things about self-polar polytopes here:
https://arxiv.org/abs/1902.00784
Your questions are partially answered there:

For all two-dimensional polytopes (a.k.a. polygons) we have (obviously): All self-dual polytopes have self-polar realizations.
For all 3-dimensional polytopes the answer is a we have (less obviously, using Koebe-Andreev-Thurston): All self-dual polytopes have self-polar realizations. (This is Theorem 4.6)
For arbitrary dimensions this question is still open:
"Question 9.2. Does every self-dual polytope have a self-polar realization?"

The paper does not consider affine or projective transformations in general, but does consider the notion "negatively self-polar".
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