# Smallest tile to tessellate the hyperbolic plane

Is it known what the smallest tile (in terms of area) that can tessellate the hyperbolic plane is? In particular, it should tessellate the plane by itself.

I think it will be a Triangle group, but I'm not sure.

(In spherical geometry, the answer is that there is no smallest tile, because you can make bipyramids with arbitrarily small faces. I don't think this will be the case with the hyperbolic plane though.)

• Do you require your tiles to have geodesic edges? – Neal Jan 26 at 14:13
• @Neal: The tilings Ian Agol and I mentioned can be done with geodesic edges. – Douglas Zare Jan 26 at 16:20
• @Neal I do not. – PyRulez Jan 26 at 17:18
• Since it hasn't been mentioned: there's a uniform lower bound on covolumes of discrete subgroups of $\mathrm{PGL}_2(\mathbf{R})$ (Siegel 1945). As a consequence, in contrast to answers below, there is a uniform positive lower bound on the area of tiles, when one restricts to tilings whose isometry group acts transitively on tiles. (In Agol's picture, the oriented isometry group is cyclic, generated by the loxodromic element represented in the upper half-plane by the homothety $z\mapsto 2z$.) – YCor Jan 29 at 16:14
• By the way what if one requires a tiling whose isometry group contains a free subgroup? – YCor Jan 29 at 16:17

The tilings mentioned by Ian Agol are related to an action of a Baumslag-Solitar group $\{ a,b \bigg| b^{-1}a^2b=a \}$ on the hyperbolic plane. They have arbitrarily small area, but diameter uniformly bounded away from $0$. It is possible to tesselate the hyperbolic plane with a single tile with arbitrarily small diameter, too. Let there be $n$ arcs on top and $n+1$ arcs on the bottom. As $n \to \infty$ the distance between the top and bottom goes to $0$. The region is naturally related to a (non-faithful) action of a Baumslag-Solitar group $\{a,b \bigg|b^{-1}a^{n+1}b=a^n\}$ on the hyperbolic plane.

• I wonder which tile that can tile the hyperbolic plane isohedrally has the smallest diameter. – PyRulez Jan 26 at 7:20

# Binary Tiling

In fact, one can tile the hyperbolic plane with arbitrarily small tiles. There is a tiling of the hyperbolic plane (apparently due to Boroczky) by pentagons.

The horizontal edges are horocycles in the upper half-space model of the hyperbolic plane, and the vertical lines geodesics. The edge at the top of each tile is half the length of the one at the bottom. One can make these arbitrarily thin, and hence have arbitrarily small area.

• Cool! Out of curiosity, do you know what the dual tiling looks like? – PyRulez Jan 26 at 0:27
• @PyRulez : not sure what you mean by dual tiling? – Ian Agol Jan 26 at 1:14
• Abstractly the dual would be a tiling with a vertex for each face of the original tiling and facets corresponding to the vertices of the original tiling; in the case of a tiling by regular polygons this is straightforward, but explicitly defining and constructing the dual in the irregular case seems substantially less trivial. The first method that comes to mind is a Vornoi construction, but it's not clear to me that that actually produces a topologically correct tiling. – Steven Stadnicki Jan 26 at 1:32
• Oh, I see. It will have triangles and quadrilaterals, since the vertices have degrees 3 and 4. – Ian Agol Jan 26 at 2:21
• I haven't found an electronic copy of Boroczky's paper, but a reference (and pictures) may be found here: jstor.org/stable/4145214 Some authors, such as Goodman-Strauss and Margulis-Mozes do not cite this paper, so it seems to be unknown even to some experts. zbmath.org/?q=an:03612055 – Ian Agol Jan 26 at 7:50