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Section 10.4 Tiling the Poincaré Disk

Having considered tiling in Euclidean geometry (Section 10.1) and elliptic geometry (Sections 10.2–10.3), we now consider hyperbolic geometry.

Repeating the argument in Section 10.2, but using the fact that angle sums for hyperbolic triangles must be less than \(\pi\text{,}\) we now obtain

\begin{equation} k \gt \frac{2n}{n-2}\tag{10.4.1} \end{equation}

where as usual \(n\) denotes the number of sides of the polygon and \(k\) denotes the number of polygons that meet at a poing. This condition is easy to satisfy! All we have to do is make our tiles big enough that their interior angles are small, so that we can fit many of them together at a point! There are indeed infinitely many hyperbolic tilings, several of which are illustrated in Figure 10.4.1.

Figure 10.4.1. Some partial tilings of the hyperbolic plane, using regular polygonal tiles with \(n\) sides, \(k\) of which meet at a point.

In order to construct such tilings, it is helpful to know the lengths of the sides. We can again use a trigonometric identity to answer this question.

Many identities from spherical geometry, such as the Law of Cosines, continue to hold in hyperbolic geometry, with appropriate sign changes and/or conversion between spherical ahd hyperbolic trigonometric functions. Somewhat surprisingly, although the Supplemental Law of Cosines is normally derived on a sphere using the polar property of elliptic geometry, this analogy is still valid in this case. In hyperbolic geometry, the Supplemental Law of Cosines takes the form

\begin{equation} \cosh\left(\frac{c}{r}\right) = \frac{\cos(C)+\cos(A)\cos(B)}{\sin(A)\sin(B)}\tag{10.4.2} \end{equation}

where \(A\text{,}\) \(B\text{,}\) \(C\) denote the angles of the triangle, with \(C\) opposite side \(c\text{,}\) and where \(r\) now denotes the “radius” of the underlying hyperbolic model, usually assumed to be \(1\text{.}\) As in Section 10.2, we have broken our \(n\)-sided polygon into \(n\) central triangles, with central angle \(C=\frac{2\pi}{n}\text{,}\) and remaining angles equal to half the interior angle of the polygon, so \(A=B=\frac12\left(\frac{2\pi}{k}\right)\text{.}\) Thus,

\begin{equation} \frac{c}{r} = \arccosh\left( \frac{\cos\frac{2\pi}{n}+\cos^2\frac{\pi}{k}}{\sin^2\frac{\pi}{k}} \right) .\tag{10.4.3} \end{equation}

Similarly, the “radius” \(\rho\) of each tile is

\begin{equation} \frac{\rho}{r} = \arccosh\left( \frac{\cos\frac{\pi}{k}+\cos\frac{\pi}{k}\,\cos\frac{2\pi}{n}} {\sin\frac{\pi}{k}\,\sin\frac{2\pi}{n}} \right) .\tag{10.4.4} \end{equation}