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Given two origin symmetric convex polytopes $P_1$ and $P_2$ (that is $P_i=-P_i$) with the same edge-graph, but potentially of different dimensions and combinatorial types. Let $\phi: G_{P_1}\to G_{P_2}$ be an isomorphism between their edge-graphs.

Question: Does for each vertex $v\in P_1$ hold $\phi(-v)=-\phi(v)$?

Inuitively, I am asking whether the edge-graph already determines which vertices form an antipodal pair.

The following example shows that for a vertex (black) its antipodal vertex (white) is not necessarily a vertex of maximal graph-distance (gray).


This question is a more precise formulation of this older question.

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    $\begingroup$ Basic observation -- for zonotopes (Minkowski sums of line segments) -- the answer is yes -- in a $2n$-vertex zonotope, the distance between $v$ and $w$ is $n$ if and only $v$ and $w$ are antipodal. $\endgroup$ Commented Feb 9, 2023 at 19:39
  • $\begingroup$ @DavidESpeyer: does your observation generalize to belt polytopes (= normal fan is a hyperplane arrangement)? See also mathoverflow.net/questions/377438 $\endgroup$ Commented Feb 10, 2023 at 1:00
  • $\begingroup$ I thought every belt polytope was combinatorially equivalent to a zonotope, which would imply that the answer is "yes". Am I wrong about that? $\endgroup$ Commented Feb 10, 2023 at 1:22
  • $\begingroup$ @DavidESpeyer To every hyperplane arrangement you get a zonotope with this arrangement as its normal fan (take the Minkowski sum of the line segments spanned by the normal vectors). So your comment should apply. $\endgroup$
    – M. Winter
    Commented Feb 10, 2023 at 11:54
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    $\begingroup$ @DimaPasechnik It's true for zonotopes, but not in general. Think of a bipyramid, any two vertices have graph-distance at most two. I added another example in the post. $\endgroup$
    – M. Winter
    Commented Feb 13, 2023 at 10:42

2 Answers 2

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$\def\RR{\mathbb{R}}$ The answer is no! I will give an example of a centrally symmetric polytope in $\RR^4$ with $12$ vertices where there is a symmetry of the edge graph interchanging two non-antipodal vertices, and fixing the other ten vertices.

Define the function $f : \RR \to \RR^4$ by $$f(\theta) = (\cos \theta, \sin \theta, \cos (3 \theta), \sin (3 \theta)).$$ Observe that $f(\theta+\pi) = -f(\theta)$, so the convex hull of $f(\theta_1)$, $f(\theta_2)$, ..., $f(\theta_n)$, $f(\theta_1+\pi)$, $f(\theta_2+\pi)$, ..., $f(\theta_n+\pi)$ is always a centrally symmetric polytope. I'll write $P(\theta_1, \theta_2, \ldots, \theta_n)$ for the convex hull of $f(\theta_1)$, $f(\theta_2)$, ..., $f(\theta_n)$, $f(\theta_1+\pi)$, $f(\theta_2+\pi)$, ..., $f(\theta_n+\pi)$.

We need two lemmas:

Lemma 1: Let $|\theta_1 - \theta_2| < 2 \pi/3$ and let $\theta_3$, $\theta_4$, ..., $\theta_n$ be any other angles. Then $(f(\theta_1), f(\theta_2))$ is an edge of $P(\theta_1, \theta_2, \theta_3, \theta_4, \ldots, \theta_n)$.

Lemma 2: Let $0 < \alpha < \beta < \pi/2$. Then there is $\delta>0$ (dependent on $\alpha$ and $\beta$) such that, for $|\gamma-\pi/2| < \delta$, the line segement $(f(\gamma), f(- \gamma))$ is NOT an edge of $P(-\gamma, -\beta, - \alpha, \alpha, \beta, \gamma)$.

Once we have these lemmas, our construction will be to choose $0 < \alpha < \beta < \pi/6$ and then $\gamma$ extremely close to $\pi/2$. Our polytope will be $P(-\gamma, - \beta, - \alpha, \alpha, \beta, \gamma)$. If we have chosen $\gamma$ close enough to $\pi/2$, then the above lemmas guarantee that both $f(\gamma)$ and $f(\pi - \gamma)$ will NOT neighbor $f(- \gamma)$ and $f(-\pi+\gamma)$, but will neighbor the other eight vertices (and each other). So switching $f(\gamma)$ and $f(\pi-\gamma)$ will be an symmetry of the edge graph which does not preserve the antipodal pairing.

We now prove the lemmas.

Proof of Lemma 1: Rotating the circle, we may assume that $\theta_1 = - \theta_2$ and $0 < \theta_1 < \pi/3$. Put $a = \cos \theta_1 > 1/2$ and consider the function $g(\theta) = 3 a^2 \cos \theta - \cos^3 \theta$. Basic calculus shows that this is maximized at $\theta = \pm \theta_1$. (We need that $a>1/2$ in order to make sure that the value at $\theta_1$, namely $2 a^3$, beats the other local maximum at $\pi$, namely $1-3a^2$.) Expanding $\cos^3 \theta = (3/4) \cos \theta + (1/4) \cos (3 \theta)$, we have $g(\theta) = (3 a^2-3/4) \cos \theta - (1/4) \cos (3 \theta)$. Writing $(x_1, x_2, x_3, x_4)$ for the coordinates on $\RR^4$, the linear functional $(3a^2 - 3/4) x_1 - (1/4) x_3$ is larger at $f(\pm \theta_1)$ than at any other $f(\theta)$, so $((f(\theta_1), f(-\theta_1))$ is an edge of $P(\theta_1, -\theta_1, \theta_3, \theta_4, \ldots, \theta_n)$ as desired. $\square$

Proof of Lemma 2: It is enough to show that some point on the line segment from $f(\gamma)$ to $f(- \gamma)$ is in the convex hull of $f(\pm \alpha)$, $f(\pm \beta)$, $f(\pi \pm \alpha)$ and $f(\pi \pm \beta)$. Putting $h(\theta) = ((f(\theta)+f(-\theta))/2$, we will show that $h(\gamma)$ is in the convex hull of $h(\alpha)$, $h(\beta)$, $h(\pi-\alpha)$ and $h(\pi - \beta)$. Explicitly, $h(\theta) = (\cos \theta, 0, \cos (3 \theta), 0)$, so all of these points are in $2$ dimensions. The points $h(\alpha)$, $h(\beta)$, $h(\pi-\alpha)$ and $h(\pi - \beta)$ are the vertices of a parallelogram with center at $(0,0)$. As $\gamma$ approaches $\pi/2$, the point $h(\gamma)$ approaches $(0,0)$ so, for $\gamma$ close enough to $\pi/2$, the point $h(\gamma)$ will be inside this parallelogram. $\square$

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  • $\begingroup$ You prove that there is a symmetry that doesn't preserve antipodal pairs, which indeed answers the OP. It isn't what your second sentence says though (which is satisfied by my cube example in a comment to Dima's answer that does not answer the OP). I'm only questioning your wording. $\endgroup$ Commented Feb 21, 2023 at 7:35
  • $\begingroup$ Wait, the second sentence says " I will give an example of a centrally symmetric polytope in $\mathbb{R}^4$ with $12$ vertices where there is a symmetry of the edge graph interchanging two non-antipodal vertices." This is right; the symmetry interchanges $f(\gamma)$ and $f(\pi-\gamma)$, which are not antipodal, and fixes the others. (I'm going to add "and fixes the other ten vertices" for clarity, though.) $\endgroup$ Commented Feb 21, 2023 at 11:07
  • $\begingroup$ Yes, that clarifies it, thanks. $\endgroup$ Commented Feb 21, 2023 at 13:39
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An extended comment:

A graph-theoretic reformulation might be as follows. Can a graph $\Gamma=(V,E)$, $2n=|V|$, with two different antipodal structures of class size 2, be the edge graph of a polytope?

Here an antipodal structure of class size 2 is an equivalence relation on $V$ with all classes $\{v_k,v'_k\}$ of size 2, such that none of the classes $\{v_k,v'_k\}$ is an edge, and the involutory permutation $\phi:=(v_1,v'_1)(v_2,v'_2)...(v_n,v'_n)$ is an automorphism of $\Gamma$.

If you have two such antipodal structures, $\phi_1$ and $\phi_2$, they together generate a dihedral subgroup $H$ in the automorphism group of $\Gamma$. (That's what two involutions generate.) Then notice that $H$ itself acts on these antipodal structures. It seems to lead somewhere...

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    $\begingroup$ Consider a cube in its standard orientation. One antipodal structure pairs each vertex with its most distant vertex. Another pairs each vertex on the top face with the vertex diagonally opposite on the top face, and similarly for the bottom face. $\endgroup$ Commented Feb 15, 2023 at 8:51
  • $\begingroup$ right - except that the latter cannot be realised by a centrally symmetric polytope - because the centre cannot be at the same time on the top and bottom faces. (basically, one pins down the location of the centre, once there is a quadrangle in the graph preserved by $\phi$.) Hmm, but how to make this kind of argument work in general, I don't know. $\endgroup$ Commented Feb 15, 2023 at 13:29

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