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I just learned the existence of Gil Kalai's $3^{d}$ conjecture, which according to Wikipedia, is proven for $d$ at most $4$. It states that every $d$ dimensional polytope with central symmetry has at least $3^{d}$ "faces" (vertices, edges, true faces and so on, including the polytope itself but not the empty set). As Wikipedia gives the example of the parallelogram and the cube and itsdual the octahedron, I came to think that maybe there could be a combinatorial interpretation of this conjecture in terms of the symmetry group of an "orthogonalized" version of the polytope or of its dual, that is a $d$-dimensional parallelotope, which would have at least $(\mathbb{Z}/2)^{d}$ as isometry group and which would thus entail the central symmetry property.

The number $3^{d}$ would then emerge from taking $0$, $1$ or $2$ elements of each copy of $(\mathbb{Z}/2)$ and combining them to get a subgroup of the isometry group of a "face". It would then remain to show that any given "flexible" polytope or the dual thereof can be continuously deformed into such a parallelotope (necessarily of same $d$-dimensional volume, as Lebesgue measure) with corners cut out by hyperplanes in such a way that the central symmetry is preserved (a repetition of such "CS cuts" being possibly needed to recover the considered polytope).

Has such an approach been considered so far to tackle this problem? If so, can you give some reference?

Of course, Gil himself is more than welcome to express his feeling about these ideas as well as his original motivation in formulating this conjecture.

Edit: it seems the notion of "vertex figure" as defined in https://en.m.wikipedia.org/wiki/Vertex_figure corresponds to the idea of CS cut I had in mind, but I'm not sure it helps much. Maybe we should first classify uniform polytopes one can obtain that way from hypercubes.

Second edit after Pietro Majer's comment and my response thereto: if we consider the class $CS_{d}$ of $d$-dimensional centrosymmetric polytopes, and $CS=\cup_{d>0}CS_{d}$, we can define automorphisms $\varphi$ of $CS$ as continuous bijections thereof such that for any $(A,B)\in CS^{2},\varphi(A\times B)=\varphi(A)\times\varphi(B)$, so that $P_{\varphi(A\times B)}=P_{\varphi(A)}P_{\varphi(B)}$, leading to ring homorphisms of $\mathbb{Z}[X]$ preserving face counting polynomials of centrosymmetric polynomials, and thus an interpretation of the group of automorphisms of $CS$ preserving a given face counting polynomial $P_{A}$ of a centrosymmetric polytope $A$ as the Galois group thereof. The central symmetry requirement implies the existence of an order $2$ element of this group, possibly realizing the wish to make $(\mathbb{Z}/2\mathbb{Z})^{d}$ appear as a subgroup of the isometry group of $A$.

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    $\begingroup$ More exactly, the idea is to show that a $d$-dimensional polytope with central symmetry can be deformed into a convex polytope having $(\mathbb{Z}/2)^{d}$ as a subgroup of its isometry group, as $3^{d}$ is a lower bound. $\endgroup$
    – Sylvain JULIEN
    Commented Aug 24, 2017 at 19:18
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    $\begingroup$ Dear Sylvain. one intruiging aspect of the conjecture is that equality holds for all "Hammer polytopes" namely polytopes obtained from an interval by applying a sequence of operations of two types: 1) Cartesian product 2) Moving from a polytope to its polar So testing your approach on Hammer polytopes would be a first step to try. $\endgroup$
    – Gil Kalai
    Commented Aug 25, 2017 at 5:58
  • $\begingroup$ Dear Gil, there's a little ambiguity in the notion of polytope. In your original conjecture, is a polytope supposed to be convex? If so, I think it should not be too hard to prove that any polytope can be obtained from a right parallelotope applying CS cuts (maybe "truncations" should be preferred, here the issue I face is purely linguistic). As CS cuts can only increase the total number of "faces" (this is to be proven rigorously but shouldn't be too much of an issue), your conjectured lower bound would be recovered rather easily. $\endgroup$
    – Sylvain JULIEN
    Commented Aug 25, 2017 at 10:05
  • $\begingroup$ It seems that for a $ d $-dimensional parallelotope the number of $ k $ -dimensional "faces" is given by the integer $ \binom{d}{k}2^{d-k}1^k $ appearing in the development of $ (2+1)^{d} $, while the number of $ k $-dimensional "faces"of its polar is given by $ \binom{d}{k}1^{d-k}2^k $ appearing in the development of $ (1+2)^{d} $. I guess the 2 stands for a pair of symmetric points and 1 for the center of symmetry. Has this been proven ? $\endgroup$
    – Sylvain JULIEN
    Commented Feb 23, 2018 at 9:20
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    $\begingroup$ Dear Sylvain, if you attach to each polytope $A$ the “counting faces polynomial” $P_A$ whose k-th degree coefficient is the number of k-dimensional "faces" then $P_{A\times B}=P_AP_B$ $\endgroup$
    – Pietro Majer
    Commented Aug 4, 2021 at 21:22

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