Is it possible to construct a seamless, boundless, finite surface solely with heptagons?

Question

Note, the motivation for this question essentially comes from game design. I was wondering if it's possible and/or if it even makes sense to have a playing field that is both hyperbolic (the area you can reach is more than $\pi r^2$ for a given travel distance $r$) and toroidal (going past the boundary relocates you to the other side of the world).

Is it possible to stitch together more than two heptagons such that:

• there are a finite number of heptagons,
• the heptagons are joined edge-to-edge,
• there are at least three heptagons around each vertex,
• and every edge is shared by exactly two distinct heptagons?

Alternatively, four hexagons around each vertex would work, or any configuration where the tiling would necessarily be hyperbolic (like these).

My intuition is being pretty unhelpful in this case since hyperbolic and toroidal seem contradictory, but I can't immediately think any reason that definitely establishes whether a hyperbolic and toroidal field is possible or not.

Answer 1

There is indeed a compact hyperbolic surface $S$ with a heptagon tiling as you describe, such that exactly 3 heptagons meet around every vertex. Actually there are infinitely many examples of such surfaces, but the most famous example is the Klein quartic, consisting of 24 heptagons. A sculpture of this surface was made by Helaman Fergusun who entitled it The Eightfold Way (a pun on the name of Murray Gell-man's particle theory). There is an entire book of articles about this surface, also entitled The Eightfold Way.

Here's a very brief description of the Klein quartic (the terminology comes from Klein's original construction using algebraic geometry, but I am going to describe a different construction, which more closely highlights features that answer your question).

One starts with a tiling of the hyperbolic plane by regular heptagons with angles of $120^\circ$, meeting 3 around each vertex. The symmetry group of this tiling is the $(2,3,7)$ triangle group, a member of the family of Coxeter groups; one can see that the $120^\circ$ regular heptagon subdivides into $14$ triangles with angles $\pi/2,\pi/3,\pi/7$.

There is a general theorem about symmetry groups of the hyperbolic plane having a compact fundamental domain, saying that every such group has a finite index torsion free subgroup, in fact it has infinitely many such subgroups, whose indices can be arbitrarily large. The quotient of the hyperbolic plane by this subgroup is always a hyperbolic surface.

In the case of the $(2,3,7)$ triangle group, one can find a particular subgroup of index $24 \cdot 14 = 336$ which is torsion free. The quotient of the heptagon tiled hyperbolic plane under this subgroup is the heptagon tiled Klein quartic.

Answer 2

Gauss-Bonnet prevents a hyperbolic torus without cone points. If you're willing to use a cone point, then you can even have squares. If you're willing to use octagons, then you can construct genus-2 hyperbolic surfaces.

Answer 3

no answer just a long comment or my 2 cents (meaning I am not very knowledgable on the intersection between hyperbolic geometry and topology )

I find hyperbolic and finite allready difficult to combine. Let alone combining Hyperbolic and torodial.

You can read about it in "The shape of space, second edition" by Jeffrey Weeks, (ISBN 978-0-8247-0709-5) but to be honnest I did not understand a lot of it, especially not when trying to combine topology with hyperbolic geometry.

Maybe have a look at hyperouge http://steamcommunity.com/app/342610 a game played on a hyperbolic plane (but not torodial nor finite).

Also was thinking why not play it on a real torus (the inside of a torus is negativly curved so has a bit of hyperbolic geometry about it)

or play it on an pseudosphere (https://en.wikipedia.org/wiki/Pseudosphere ) or dini's surface ( https://en.wikipedia.org/wiki/Dini%27s_surface ) but I am wondering on how this would work .

hope this helps ( a bit)