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).



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> **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*?
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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.



Geometric realization of combinatorial self-duality in polytopes
Alathea Jensen defines "self-polar":



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


