Polytope with indegree-increasing property.

Question

I have a question about a simple polytope. I am worried that my question would be inappropriate for mathoverflow. So I am sorry that I am ignorant of combinatorics.

Let $\mathcal{P}$ be a simple convex polytope of dimension $n$ embedded in $\mathbb{R}^n$. Then any vector $v \in \mathbb{R}$ defines a height function $h_v$ of $\mathcal{P}$ such that $$ h_v(x) = \langle v, x \rangle$$ where $\langle \cdot, \cdot \rangle$ is the usual inner product in $\mathbb{R}^n$. Assume that $v$ is chosen not to be perpendicular to any face of $\mathcal{P}$. Then $h_v$ gives an orientation on each edge of $\mathcal{P}$ such that $$ \overrightarrow{pq} \Leftrightarrow h_v(p) < h_v(q)$$ for every edge $\overline{pq}$ connecting $p$ and $q$. With this orientation, we can define an index $\mathrm{ind}(p)$ (or, indegree of $p$ as a digraph) as a number of edges which end at p. And we call $h_v$ is `index-increasing' if $$\mathrm{ind}(p) < \mathrm{ind}(q) ~\mathrm{ implies} ~h_v(p) < h_v(q)$$.

Allowing to deform $\mathcal{P}$ continuously (preserving combinatorial structure and convexity, but a normal fan can vary), is it always possible to find a deformation of $\mathcal{P}$ and a vector $v \in \mathbb{R}^n$ such that $h_v$ is index-increasing?

I would really appriciate if you could give me any comment. Thank you.

Answer 1

This answer benefited from a discussion with I. Izmestiev.

Consider a $3$-dimensional simple polytope $P$. The indegrees of the vertices can only be $0$, $1$, $2$ or $3$. The lowest vertex has indegree $0$, the highest vertex has indegree $3$, and the other vertices have indegree $1$ or $2$. For an index-increasing realization of $P$, there is a plane that separates the indegree-$\le 1$ vertices and the indegree-$\ge 2$ vertices.

Moreover, every face of $P$ contains a vertex of indegree $\le 1$ (its lowest vertex) and a vertex of indegree $\ge 2$ (its highest vertex). Therefore, the existence of the required realization means that there must be a plane that cuts all the faces.

From this point of view, it becomes much clearer that index-increasing realization is very unlikely to exist even in dimension $3$. Counter-examples can be easily constructed, for example, by taking truncations.

Answer 2

This is not an answer, but rather a request for clarification with an image. It seems to me much depends on what you mean by "a deformation of $\mathcal{P}$." You want to reorient and deform $\cal P$ so that the indegree increases monotonically with height. An example that might be problematical is a polytope with a mixture of high- and low-degree vertices, uniformly distributed around the surface. An example in $\mathbb{R}^3$ is the triakis icosahedron, which has 20 degree-3 vertices and 12 degree-10 vertices. It is convex, but its structure is more easily understood in this nonconvex version:

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                  ^{(Image from Wikipedia page.)}

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Perhaps you could clarify by describing a deformation and a $v$ that makes $h_v$ index-increasing for the convex version of the triakis icosahedron.