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A Wythoffian polytope $P\subset\Bbb R^d$ is an orbit polytope of a finite reflection group, that is,

$$P:=\mathrm{Orb}(\Gamma,x):=\mathrm{conv}\{Tx\mid T\in\Gamma\},$$

where $\Gamma$ is a finite reflection group, and $x\in\Bbb R^d$.

Question: Is it true, that if $e$ is an edge of $P$, then there is a reflection $T\in\Gamma$ that fixes $e$ set-wise but flips its orientation?

I am pretty sure that this is true, but I cannot find a concise proof from first principles. I would be most happy with a short self-contained argument (possibly using some well-known properties of reflection groups).


Partial proof

Suppose that the generator $x\in\Bbb R^d$ was chosen from the interior of a Weyl chamber of $\Gamma$ (the resulting polytope is sometimes called a $\Gamma$-permutahedron or omnitruncated uniform polytope). Since $\Gamma$ acts regularly on the Weyl chambers, every Weyl chamber contains a single vertex. In particular, if $e$ is an edge of $P$, its end vertices are in different chambers and the edge must cross a bounding reflection hyperplane of the chamber. Since reflection on this hyperplane is a symmetry of $P$, the edge must be perpendicular to the hyperplane and the reflcetion on it must flip its orientation.

My hope is that this can be extended to an argument for general placements of $x$. Each Wythoffian polytope can be obtained as a "limit" (in some sense) of such $\Gamma$-permutahedra, and maybe this sufficec to prove the statement. However, I was not able to make a clear case for why the edges of the resulting polytope are "limits" of edges of the $\Gamma$-permutahedra.

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Suppose $W$ is a fundamental domain of $\Gamma$, and $v \in W$ is a seed vertex.

By vertex transitivity, every edge of $P$ is $\gamma e$ for some edge $e$ incident to $v$ and some $\gamma \in \Gamma$, and if $e$ is bisected by a reflection $T \in \Gamma$, then $\gamma T \gamma^{-1}$ is a reflection bisecting $\gamma e$. In fact, every edge is the image of an edge leaving $v$ and passing through $W$, by the nature of the fundamental domain (possibly along part of the boundary of $W$.)

The other endpoint of $e$ cannot be in $W$, not even on its boundary, because each point of $P$ is the image of exactly one point in the fundamental region. (Wikipedia cites Hall, Brian C. (2015), Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, Proposition 8.29 for this fact.)

At the point where the edge $e$ leaves $W$, it passes through a bounding reflection of $W$. This reflection must carry $e$ onto some edge of $P$, and since it fixes one point in $e$, it must carry $e$ to itself.

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  • $\begingroup$ Thank you for your answer. How is it clear that an edge is not connecting two vertices in the same fundamental region? Sure, two vertices cannot live in the interior of the same region, but what about the boundary? $\endgroup$ – M. Winter Mar 18 at 11:04
  • $\begingroup$ Because each point of $P$ is the image of exactly one point in the fundamental region. (Wikipedia cites Hall, Brian C. (2015), Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, Proposition 8.29 for this.) $\endgroup$ – Nick Matteo Mar 18 at 16:05
  • $\begingroup$ That was the crucial missing ingredient for me. Thank you very much, especially for this explicit reference! $\endgroup$ – M. Winter Mar 18 at 16:07
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Isn't that exactly the definition of being Wythoffian?

I.e. by virtue of kaleidoscopical construction the edges occur only as convex hulls of a seed point (vertex) and its mirror image wrt. to some mirror out of the reflection group. In fact one which occurs as boundary of the fundamental domain. Thence this very mirror fixes this very edge setwise, but reverses its orientation.

--- rk

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  • $\begingroup$ It is intuitively clear to me, but I have a hard time formalizing this using the given definition. You say: "edges occur only as convex hulls of a seed point (vertex) and its mirror image wrt. some mirror". I agree, but how can I see that exactly these pairs of vertices form the edges, and no others. It is easy for when the seed point is in the interior of a Weyl chamber. The other cases might follow via some limit argument, but I was not able to formalize this rigorously. I disagree that this is "exactly the definition of being Wythoffian", at least not as I stated it. $\endgroup$ – M. Winter Feb 5 at 0:37
  • $\begingroup$ To me "a Wythoffian polytope" is just one that follows the kaleidoscopical construction device, as given by Wythoff. $\endgroup$ – Dr. Richard Klitzing Feb 5 at 6:24
  • $\begingroup$ To start with you solely have to consider those mirrors, which are the walls of your Weyl chamber (fundamental region) and the seed point. On each of those mirrors your seed point gets reflected into an image point. Either such an individual image point together with the seed point provides an edge where the corresponding mirror is midwise perpendicular, or the seed point already was coincident to the respective mirror and no edge will occur here. The remainder then is just the reflection of this local configuration all around the polytope. Thus again all edge images have such a mirror image. $\endgroup$ – Dr. Richard Klitzing Feb 5 at 6:31
  • $\begingroup$ I disagree that this is obvious and I would prefer a formal argument. What is a support vector of the claimed edge? And how to see that the other vertex pairs have no support vectors? $\endgroup$ – M. Winter Feb 5 at 11:26
  • $\begingroup$ Each of the original vertex pairs (seed point and image) are defined by the respective mirror hyperplane, i.e. the normal of that thence would be the asked for supporting vector (when being re-attached to the seed point). - Still quite obvious and direct, no mystics in here. $\endgroup$ – Dr. Richard Klitzing Feb 5 at 19:32

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