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Let $P\subset\Bbb R^d$ be a vertex-transitive polytope aka. an orbit polytope. Can there be a matrix $T\in\mathrm{SO}(\Bbb R^d)$ that commutes with all symmetries in $\mathrm{Aut}(P)\subset\mathrm O(\Bbb R^d)$?

Probably one approach to the question is as follows: can there be a vertex-transitive polytope $P\subset\Bbb R^d$ for which $\mathrm{Aut}(P)$ is (real) irreducible, but not absolutely irreducible (that is, not irreducible over $\Bbb C$).

Vertex-transitivity is necessary for all these questions. For example, there is a polytope (not vertex-transitive) whose symmetry group is a finite subgroup of $\mathrm{SO}(\Bbb R^2)$, which is real irreducible, but reducible over $\Bbb C$. Since $\mathrm{SO}(\Bbb R^2)$ is commutative, every element of that group would then commute with $\mathrm{Aut}(P)$. It is known that most commutative groups cannot be symmetry groups of vertex-transitive polytopes (only exceptions are elementary 2-abelian groups).

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  • $\begingroup$ I assume $T$ should not be a scalar multiple of the identity? By Schur's Lemma, there is such a $T$ in the general linear group if and only if the representation of $\mathrm{Aut}(P)$ on $\mathbb{R}^d$ is reducible. Is it known when this happens? $\endgroup$
    – Mark Wildon
    Commented Nov 1, 2019 at 10:28
  • $\begingroup$ @Mark Yes, it should not be scalar. But I phrased it as $T\in\mathrm{SO}(\Bbb R^d)$ to ensure this. And I am not aware of any such classification. For that matter, is there a classification of real irreducible, but not absolutely irreducible matrix groups $\Gamma\subset\mathrm{O}(\Bbb R^d)$? $\endgroup$
    – M. Winter
    Commented Nov 1, 2019 at 10:32
  • $\begingroup$ Can't you just take a rectange (centered at the origin) in $\mathbb{R}^2$ with two different side length. Then the automorphism group is $\mathbb{Z}/2\times \mathbb{Z}/2$, and minus the identity is an element in $SO_2$ that commutes with all symmetries? $\endgroup$
    – HenrikRüping
    Commented Nov 2, 2019 at 8:44
  • $\begingroup$ @HenrikRüping You are right! My intentions were to exclude all scalar transformations by restricting to $T\in\mathrm{SO}(\Bbb R^d)$, but obviously this does not work in even dimensions. I have to rethink the formulation of my question. $\endgroup$
    – M. Winter
    Commented Nov 2, 2019 at 10:33

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There are examples of vertex-transitive polytopes with irreducible symmetry groups for which there are still a non-scalar transformations $T\in\mathrm O(\Bbb R^d)$ that commutes with all the symmetries of the polytopes.

Fix a group $G$ with the following properties:

  • $G$ is neither abelian nor generalized dicyclic.
  • the centralizer of every irreducible representation of $G$ is isomorphic to either $\Bbb C$ or $\Bbb H$.

Examples are contained among the finite subgroups of $\mathrm U(n)$ (for $n\ge 2$) or $\mathrm{SU}(2)$. The first property ensures that $G$ is not among the groups that are excluded as symmetry groups of vertex-transitive polytopes as determined in [1]. So there is a vertex-transitive polytope $P$ whose automorphism group can be considered as a matrix group representation of $G$. By the second assumption, the centralizer of this matrix group containst then more than just scalar matrices.

[1] E. Friese, F. Ladisch. "Classification of Affine Symmetry Groups of Orbit Polytopes"

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