Representing a Hyperbolic Polyhedral Net in ℝ3

Question

This question is inspired by the classic board game Wiz-War. Suppose we have a polyhedral net of 5 squares, with the labeled edges connected in the resulting polyhedron. This layout makes every square touch every other square.

Illustration:

    +-A-+
    |   |
    B   C
    |   |
+-B-+---+-C-+
|   |   |   |
D   |   |   D
|   |   |   |
+-E-+---+-F-+
    |   |
    E   F
    |   |
    +-A-+

• I believe that the Euler characteristic (3 = 8? vertices - 10 edges + 5 faces) of this prevents it from being convex.
• By my calculations it is Schlafli {4,5} which means it must be hyperbolic.

Given this, it appears the correct representation is a hyperbolic tiling, specifically Order-5 Square Tiling

My question is, provided I am correct in all of the above, is there a way to relax the constraints of the definition of a regular closed polyhedron to represent this in ℝ³ while maintaining as many properties of regularity as possible (face, edge, & vertex transitivity)? Forgive my inability to state these elegantly, but I mean things such as:

• Concavity
• Compound polygonal faces
• Faces that are not polygons
• Faces that stretch into the 3rd dimension
• Fundamental polygon join directionality manipulation
• Manifold-related trickery (Mobius strip, Klein bottle, Gabriel's horn, etc...)

Answer 1

There are many ways to glue the edges you labelled together, depending on whether you preserve or reverse orientation. This has an effect on the resulting manifold.

Suppose you glue the edges together in an oriented manner, like this:

Then you'll get an orientable surface, with $5$ faces, $10$ edges, and $5$ vertices (note that all "outer" vertices are completely identified by the gluing, so they form a single vertex of order $8$, while the inner vertices have order $3$ each). The resulting structure is orientable with Euler characteristic $0$, so it is topologically a torus. Note that this cannot be realized as a geometric polyhedron with flat faces, since among other weirdness it has some edges connecting a vertex to itself.

Alternatively, you could glue the edges together such that orientation is always reversed:

This results in $5$ faces, $10$ edges, but only $2$ vertices. The resulting Euler characteristic is $-3$.

Using other variations of gluing together the edges as you've indicated, you can produce any nonorientable surface with Euler characteristic from $1$ to $-4$.

In most cases, the abstract polyhedron you get has vertices of different orders, but there are some cases where each vertex has the same order: in one case (the second case I described) you get two vertices of order $10$, and in three cases you get a single vertex which has order $20$. Their universal covers would be $\{4,10\}$ and $\{4,20\}$ respectively.

EDIT:

The universal cover of the orientable gluing looks like this:

It's neither face-transitive nor vertex-transitive. Note that all quadrilaterals displayed here are squares, so the tiling is not flat, but it's still approximately Euclidean.