Suppose I have a (finite, real, central, essential) hyperplane arrangement $\mathcal{H}$ such that all regions "have the same shape": for any two regions $R,R'$, there is an orthogonal transformation taking $R$ to $R'$ (these transformations are not required to do anything nice to the rest of the arrangement). Is $\mathcal{H}$ necessarily a reflection arrangement?
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asked Nov 11, 2020 at 20:00
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1$\begingroup$ This is more restricted than what you ask, but in this preprint I prove that if the arrangement is transitive on the regions, then it is a reflection arrangement (though it is proven via a dual formulation with zonotopes, see also this question). You probably know that an arrangement with congruent (aka. same-shaped) regions must be simplicial. As far as I know, no classification of simplicial arrangements exists beyond 3D (and I am not sure whether 3D is settled). $\endgroup$– M. WinterNov 12, 2020 at 15:32
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This is a known open problem (for isometric regions), which, as far as I know, is still not settled.
The dimension 3 case was proved affirmatively in https://arxiv.org/abs/1501.05991, where also some history of the question is outlined. I am not aware of any progress since.
answered Nov 12, 2020 at 10:50