Suppose that $G\subset O(d)$ is a finite reflection (finite Coxeter) group. For any $v\in \mathbb{R}^d$ which is not fixed by any non-trivial $g\in G$, one can consider the orbit polytope (Coxeter) permutahedra \begin{equation} P(G;v)=Conv (G\cdot v) \end{equation} given by the orbit.

Now consider $G^+\subset SO(d)$, the index-two rotation subgroup of $G$. Again one can consider the orbit polytope \begin{equation} P(G^+;v)=Conv(G^+\cdot v) \end{equation} for $v$ as above (i.e., not fixed by any non-trivial $g$ from the original group). Is it necessarily the case that $P(G^+;v)$ is just obtained by $P(G;v)$ by "alternation"?

If $v_1,v_2\in \mathbb{R}^d$ are not fixed by any $g\in G$, it can be shown that $P(G;v_1)$ and $P(G;v_2)$ are combinatorially equivalent. Must the same be true for $P(G^+;v_1)$ and $P(G^+;v_2)$ as well?

This would definitely seem to be the case for a number of examples (e.g., for $G=A_2\times A_2\times A_2$, for which $P(G;v)$ is a box, $P(G^+;v)$ is a tetrahedra).

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    $\begingroup$ 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. $\endgroup$ – Johannes Hahn Jul 13 '20 at 16:54
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    $\begingroup$ The second question is answered in the positive for the symmetric group in Cruickshank, J., Kelly, S. Rearrangement Inequalities and the Alternahedron. $\endgroup$ – Grant B. Jul 13 '20 at 20:39
  • $\begingroup$ @GrantB.'s reference: Cruickshank and Kelly - Rearrangement inequalities and the alternahedron (MSN). $\endgroup$ – LSpice Jul 13 '20 at 23:38
  • $\begingroup$ Great; thank you $\endgroup$ – Bob Jul 13 '20 at 23:47
  • $\begingroup$ 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. $\endgroup$ – M. Winter Jul 14 '20 at 12:18

Let me try to give a rigorous definition to "by alternation" such that the answer to your first question is "yes".

Given a polytope $P$ whose vertex-edge graph is bipartite, one might say that there are two polytope $Q$ is obtained from $P$ "by alternation" if there is a bipartition of the graph such that $Q$ is the convex hull of one block of the bipartition.

Under this definition, yes, your graph $P(G^+,v)$ is obtained from $P(G,v)$ by alternation. The requirement that $v$ is not fixed by any (nontrivial) element of $G$ is equivalent to the requirement that $v$ is not fixed by any reflection, or equivalently, not contained in any reflecting hyperplane. (This is standard...See for example Section 1.12 of Humphreys "Reflection Groups and Coxeter groups".) The reflecting hyperplanes cut the ambient space into simplicial cones, so we're just choosing $v$ in the interior of one of the cones. Then the orbit of $v$ contains exactly one point in each of the cones, and this gives a bijection between the orbit and the elements of the group.

The cones define a fan structure on the ambient space (i.e. any two cones intersect each other in faces), and this fan is the normal fan of $P(G,v)$. If two maximal cones in this fan are adjacent (i.e. share a codimension-1 face), then they are related by a reflection in $G$, so exactly one of the corresponding group elements is in $G^+$. Thus the vertices of $P(G^+,v)$ are one block of a bipartition of the vertices of $P(G,v)$.

  • $\begingroup$ That's great; thanks. Does it follow that if $v_1$ and $v_2$ are not fixed by any (non-identity) $g\in G$, then $P(G^+,v_1)$ and $P(G^+,v_2)$ have to combinatorially equivalent? $\endgroup$ – Bob Jul 13 '20 at 18:38
  • $\begingroup$ I'm not sure if it follows from the "by alternation" fact, but I suspect it is true. I also suspect that this is known and that someone has written it down. But I have not thought carefully about it to be sure. $\endgroup$ – Nathan Reading Jul 13 '20 at 19:30
  • $\begingroup$ Thanks, I should probably put a "reference request" tag on this. I'm pretty certain its true in $\mathbb{R}^3$. $\endgroup$ – Bob Jul 13 '20 at 20:20

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