Timeline for Are orbit polytopes of rotation subgroup of Coxeter group combinatorially equivalent?
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15 events
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| Jul 16, 2020 at 15:47 | comment | added | Grant B. | @M.Winter Yes, I think we are used to polytopes which use a subset of the defining inequalities of permutahedra rather than a subset of vertices, since these tend to be better behaved. | |
| Jul 16, 2020 at 7:58 | comment | added | M. Winter | @GrantB. Oh I got it, I was under the impression that the reflection groups are too nice to have non-generic dependencies inside a Weyl chamber. I was aware of the linked question but have not interpreted the answer in that way. Interesting. | |
| Jul 16, 2020 at 1:21 | comment | added | Grant B. | @M.Winter I agree that the problem reduces to showing that the face lattice of $P(G^+;v)$ is independent of $v$ for $v$ in a given Weyl chamber. However this statement is already not true for all subgroups of $G$, as the linked post indicates. The problem is that points in the interior of a Weyl chamber are simply not always generic in the sense of "no extra affine dependencies". There can be and are non-generic affine dependencies in non-degenerate permutahedra, they just aren't among vertices comprising a face of the polytope. | |
| Jul 15, 2020 at 21:28 | comment | added | M. Winter | @GrantB. There is certainly something special about $G^+$ that cannot be generalized to other subgroups of $G$. For example, there are no non-generic dependencies as long as $v$ stays in the same Weyl chamber. We know this because there are no non-generic dependencies for $G$. Now, $P(G^+;v)$ has points in exactly half of the Weyl chambers and so we know that for $v$ in any of these we get the same combinatorial type. But putting $v$ in one of the other chambers gives you the mirrored polytope, still combinatorially equivalent to the previous one. Might this already be the subtlety? | |
| Jul 15, 2020 at 16:42 | comment | added | Grant B. | @M.Winter I think it's more subtle than that. The statement is not true for every subgroup of $G$; the answer to the OP's recent question gives a counterexample for a cyclic subgroup (a "non-generic" affine dependency). The paper I mentioned above discusses the problem of characterizing subgroups with combinatorial orbit polytopes, but to my knowledge not much is known. | |
| Jul 14, 2020 at 12:18 | comment | added | M. Winter | My heuristic argument would be the following: the reason that all the $P(G;v)$ have the same combinatorial type is that there is no generic $v$ (i.e. $v$ not fixed by any non-trivial $g$) so that $G\cdot v$ has non-generic affine dependencies (because such must occur when transitioning from one combinatorial type to another). This obviously translates to no non-generic affine dependencies between the points in $G^+\cdot v$ (since the notion of "generic $v$" stays the same). This can probably be made precise. | |
| Jul 13, 2020 at 23:47 | comment | added | Bob | Great; thank you | |
| Jul 13, 2020 at 23:38 | comment | added | LSpice | @GrantB.'s reference: Cruickshank and Kelly - Rearrangement inequalities and the alternahedron (MSN). | |
| Jul 13, 2020 at 20:39 | comment | added | Grant B. | The second question is answered in the positive for the symmetric group in Cruickshank, J., Kelly, S. Rearrangement Inequalities and the Alternahedron. | |
| Jul 13, 2020 at 20:21 | history | edited | Bob |
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| Jul 13, 2020 at 18:49 | history | edited | Bob | CC BY-SA 4.0 |
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| Jul 13, 2020 at 18:00 | answer | added | Nathan Reading | timeline score: 2 | |
| Jul 13, 2020 at 16:54 | comment | added | Johannes Hahn | It depends on what exactly you mean by "alternation". It is the case that $P(G^+;v)$ has exactly half as many vertices as $P(G;v)$. So in this sense $P(G^+;v)$ is obtained from $P(G;v)$ by "skipping every second vertex". Note that $Gv$ is indeed the set of vertices of the polytope $P(G;v)$, because all points in the orbit lie on a sphere, so that none is a convex combination of any others. | |
| Jul 13, 2020 at 14:23 | history | edited | Bob | CC BY-SA 4.0 |
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| Jul 13, 2020 at 13:56 | history | asked | Bob | CC BY-SA 4.0 |