Average degree of graph from polyhedral complex

Question

Consider a graph constructed as follows. We begin with a pure polyhedral complex $C$ of dimension $d$ in $\mathbb{R}^d$ -- for our purposes this is just a finite collection $\{P_1, \dots, P_k\}$ of distinct $d$-dimensional convex polytopes ("cells") in $\mathbb{R}^d$ such that for any $P_i$ and $P_j$, the intersection $P_i \cap P_j$ is a face of both $P_i$ and $P_j$ (possibly empty). Now we define the "cell graph" $G(C)$ to be the graph with vertices $P_1, \dots, P_k$, and an edge between $P_i$ and $P_j$ if their intersection has dimension $d-1$ (i.e. their intersection is a facet of both). Let's say a graph is $d$-cellular if it is the cell graph of some pure polyhedral complex of dimension $d$.

My question is:

What is the best upper bound on the average degree of a $d$-cellular graph (if there is one)?

Note that when $d = 2$, a $2$-cellular graph is necessarily planar, and thus has average degree $< 6$, and indeed there are $2$-cellular graphs which have average degree arbitrarily close to $6$: for example we can take the cell graph of a large bounded restriction of the regular hexagonal tiling of $\mathbb{R}^2$. I would guess that a potential upper bound for the case $d=3$ is $14$, asymptotically achieved by the Bitruncated cubic honeycomb (pictured below), but it's not immediately clear to me that there's any upper bound on the average degree for fixed $d \geq 3$.

Edit: As @quarague points out, the Triakis truncated tetrahedral honeycomb has a higher limit average degree of $16$.

Answer 1

There's no upper bound when $d \ge 3$.

If all we wanted was to have some 3D objects touch each other with unbounded degree, there would be a straightforward method: lay down a bunch of long north-to-south pipes parallel to each other, then lay down a bunch of long east-to-west pipes on top of them. Every north-to-south pipe touches every east-to-west pipe, and so degree can be made arbitrarily large.

For this problem, we want to adjust the pipes so that they're polyhedra, and that the points of contact are facets. Moreover (and this is the challenge) we want to make the adjustment so that all the polyhedra are convex.

First, let's bend the pipes so that convexity will be easy to achieve. Take the surface $z = x^2-y^2$ and lay down pipes on this surface: pipes with a fixed $x$ (in blue in the diagram below) are just below the surface, and pipes with a fixed $y$ (in red in the diagram below) are just above the surface.

Now I'm going to handwave a bit because there's no good way to draw pictures of the result, though I'll try anyway. To get the polyhedral complex, we do the following:

1. At each contact point between a red pipe and a blue pipe, put down a polygon where that red pipe and blue pipe are supposed to "meet".
2. Replace each red pipe by the convex hull of all the polygons where it meets blue pipes. Do the same (vice versa) for each blue pipe.

Here's a picture, though it's not as helpful:

The idea is that because all the contact points of a red pipe are in convex position bending upward, taking the convex hull will give us a bunch of points that stay above the surface $z = x^2-y^2$; for blue convex hulls, we'll get a bunch of points that stay below that surface, so we don't create any inconvenient intersections. Taking convex hulls won't interfere with the contact polygons between the red objects and the blue objects.