Number of Zonotope Edges Parallel to Generator

Question

Suppose we have a zonotope $Z$ that is the Minkowski sum of line segments $U_1+\dots +U_n$. All the edges of $Z$ are parallel to some $U_i$. Is it also true that the number of edges parallel to $U_i$ equals the number of edges parallel to $U_j$ for every $j\neq i$? Surely this question has been asked and probably answered before. I looked through a page of zonohedra and verified that they all answer the question true.

https://www.ics.uci.edu/~eppstein/junkyard/ukraine/ukraine.html

It is known that $Z$ is a zonotope iff $Z$ is the projection of a cube from $\mathbb{R}^n$. Of course, a cube in $\mathbb{R}^n$ is the sum of $n$ orthonormal vectors and has the same number of edges parallel to each of its generators. I suspect there may be a good induction proof down this way by adding one generator, $v$, at a time and but haven't found one. The new generator divides the old zonotope into two pieces. The edges parallel to $v$ will stretch between the two halves.

Any help is appreciated! Thanks! Aaron

Answer 1

Partial answer.

This is true as long as no three of the $n$ vector summands are coplanar. In that case all the two dimensional faces are parallelograms. For each summand you build a zone of parallelograms by following an edge across to its opposite edge. Each pair of zones meets twice. That means there are $2(n-1)$ edges in each direction.

The truncated octahedron and the other examples in the page you link mean that this condition is not necessary. But something is, since the hexagonal prism fails, as @MWinter points out.

Answer 2

The thing is more or less obvious as long as the set of your spanning vectors $\{U_i\}_{i\in I\ }$ is such that none of those is being contained within the spanned cone of the others. For then all edges will be exterior to the produced zonotope.

Now consider the simple case of 3 coplanar vectors. Any 2 of them would span a rhomb and the effect by the respective third is to expand that into an hexagon. So again your proposition is true.

--- rk

Answer 3

That is not the case.

Consider a $2n$-gonal prism with $n\ge 3$ (with top and bottom face being regular polygons). It is a zonotope.

Then there are $2n$ edges connecting the top and the bottom face, all parallel. But choose an edge in the top or bottom face: it is parallel to only three other edges.

Answer 4

Following up Ethan's answer.

Let $S\subset \mathbb{R}^d$ be finite and generate the zonotope $Z$. By Ethan's argument, $Z$ will have the same number of edges per generator if all 2D faces are parallelograms, i.e. no three elements of $S$ are coplanar.

Seeing coplanarity in $S$ as degenerate, we can still decompose the boundary of $Z$ into parallelograms. This is done by "parallelogramizing" all the 2D faces. Now, count the number of edges per generator by distinguishing between legit edges (the 1D faces of $Z$) and illegit edges (the parallelogramizing segments inside a 2D face). We seek a condition * that determines when $Z$ has the same number of legit edges per generator.

We know $Z$ has the same number of edges per generator if we include the illegit ones. Hence, * must hold iff $Z$ has the same number of illegit edges per generator, so let's count those. Say a generator $s\in S$ is parallel to a 2D face $F\subset Z$. Suppose $F$ is a $2n$-gon. Each pair of generators parallel to $F$ determines a parallelogram in the parallelogramization of $F$, so there are $n-1$ parallelograms in the $s$-zone of $F$, meaning $n$ total edges parallel to $s$, $2$ of which are legit, leaving $n-2$ illegit edges.

The total number of illegit edges parallel to $s$ is then the total "surplus" of all the 2D faces in the $s$-zone, which counts how many generators after $2$, i.e. hexagons are surplus 1, octagons are surplus 2, etc. This turns into the following condition:

Let $\mathcal{H}$ be the set of all 2D subspaces in $\mathbb{R}^d$. The zonotope generated by $S$ will satisfy * iff the total nullity,

$$ T(s):=\sum_{s\in H\in\mathcal{H}} \mathrm{nullity}(H\cap S) $$

does not depend on $s$. As expected, it's a sort of balancing condition.