Realizing Right-Angled Hyperbolic Pentagons in the Upper Half Plane

From abstract side-length data to Möbius-placed coordinates.

In a previous note we described the moduli space of right angled hyperbolic pentagons, which are uniquely determined by a pair of adjacent sides $a,b$ which must satisfy $\sinh a\sinh b>1$. In fact, we worked out the remaining side lengths $c,d,e$ in terms of these two:

$\cosh d = \sinh a\sinh b$

$\sinh c = \frac{\cosh a}{\sqrt{\sinh^2a\sinh^2b-1}}$

$\cosh e =\frac{\cosh a\sinh b}{\sqrt{\sinh^2a\sinh^2 b-1}}$

Having this description in hand, we take it one step further and for each point in moduli space, construct an explicit such pentagon in the upper half plane model, where geodesics are faithfully represented by their endpoints, as an unordered pair in $\overline{\mathbb{R}}:=\mathbb{R}\cup\{\infty\}$. Having this allows one to compute right angled pentagon reflection groups, and draw beautiful images of their corresponding tilings:

Note we will allow ourselves to use all side lengths $a-e$ throughout the calculations, as we know from above this can easily be recast in terms of $a,b$ alone. For each $a,b$ only need to compute one representative right angled hexagon, and so are free to use the isometries of the hyperbolic plane to simplify the problem. Indeed, using homogeneity we can take one vertex $v$ of the hexagon to $i$, and then use isotropy to rotate it so one of the sides through $i$ is the unit circle. If we take this to be the side $b$, then

$\gamma_b \mapsto \{0,\infty\}$

But, as $H$ is right angled, the other side through $v=i$ must be perpendicular to the vertical - and thus, is the unit circle. This determines $\gamma_a$:

$\gamma_a\mapsto \{-1,1\}$

As the hyperbolic length of the vertical side is $b$ and its first endpoint is at $i$, the second endpoint must lie at $e^b i$ as

$$\begin{align} b&=\mathrm{length}(\gamma_b)\\ &=\int_1^? \frac{dt}{t}\\ &=\ln|t|\Big|_1^?\\ &=\ln(?) \end{align}$$

Because this geodesic also intersects the vertical at a right angle, it is a circle centered at zero. And as it intersects the $y$ axis at height $e^b$ it must intersect the $x$ axis at $\pm e^b$:

$\gamma_c\mapsto (-e^b,e^b)$

Now we skip to computing the side $e$. This lies on the other end of $a$ from the vertical geodesic, and so is the result of simply translating the vertical geodesic along the unit circle by distance $b$. This is accomplished by the Möbius transformation

$T=\begin{pmatrix}\cosh\frac{a}{2}&\sinh\frac{a}{2}\\\sinh\frac{a}{2}&\cosh\frac{a}{2}\end{pmatrix}$

Applying this to the endpoints $\{0,\infty\}$ of the vertical geodesic yields

$$\begin{align*} T_{0} &= \begin{pmatrix} \cosh \left( \frac{a}{2} \right) & \sinh \left( \frac{a}{2} \right) \\ \sinh \left( \frac{a}{2} \right) & \cosh \left( \frac{a}{2} \right) \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} \sinh \left( \frac{a}{2} \right) \\ \cosh \left( \frac{a}{2} \right) \end{pmatrix} \rightarrow \tanh \left( \frac{a}{2} \right) \\ T_{\infty} &= \begin{pmatrix} \cosh \left( \frac{a}{2} \right) & \sinh \left( \frac{a}{2} \right) \\ \sinh \left( \frac{a}{2} \right) & \cosh \left( \frac{a}{2} \right) \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} \cosh \left( \frac{a}{2} \right) \\ \sinh \left( \frac{a}{2} \right) \end{pmatrix} \rightarrow \coth \left( \frac{a}{2} \right) \end{align*}$$

Thus the geodesic $e$ is simply

$\gamma_e\mapsto \left(\tanh\frac{a}{2},\coth\frac{a}{2}\right)$

Finally we compute the geodesic $d$. This is also a translation of the vertical geodesic, but not along the unit circle, instead along $\gamma_c$, a circle of radius $e^b$ in our model. We form the translation along this geodesic by conjugation: if $V_b\colon p\mapsto e^b p$ is the translation of length $b$ along the vertical geodesic, and $H_c$ be the translation by distance $c$ along the horizontal unit circle geodesic. Then the Möbius transformation $M$ we seek is

$M = V_bH_cV_b^{-1}$

We are interested in the image of the vertical geodesic’s endpoints under this isometry. As $V_b$ is translation along the vertical geodesic, it (and hence it’s inverse as well) fixes these endpoints, so

$$\begin{align} M.\{0,\infty\}&=V_bH_cV_b^{-1}.\{0,\infty\}\\ &= V_bH_c \{0,\infty\} \end{align}$$

Using what we’ve previously learned about translation of $\{0,\infty\}$ along the unit circle, we see

$$\begin{align} V_bH_c.\{0,\infty\} &= V_b.\left\{\tanh \frac{c}{2},\coth\frac{c}{2}\right\}\\ &=\left\{e^b\tanh \frac{c}{2}, e^b\coth\frac{c}{2}\right\} \end{align}$$

Thus we have it,

$\gamma_d\mapsto \left\{e^b\tanh \frac{c}{2}, e^b\coth\frac{c}{2}\right\}$

And that’s it! We’ll box off the result for easy finding:

The same pentagon in the disk

The half plane is where the computation wants to happen — geodesics are pairs of real numbers and the isometries are $2\times2$ matrices — but it is not where one wants to look at the answer. The realization is lopsided: $\gamma_a$ and $\gamma_b$ pile up near $i$ while $\gamma_d$ runs out towards $e^b\coth\frac{c}{2}$, far off to the right. So let us carry it to the Poincaré disk by the Cayley transform

$\psi(z)=\frac{iz+1}{z+i}$

which takes $i$ to the center and $\overline{\mathbb{R}}$ to the unit circle, rotated so that $\infty$ lands due north. Being a Möbius transformation it takes geodesics to geodesics and preserves angles, so we lose nothing by moving there.

It is tempting to chase the ten endpoints around the boundary circle, but there is a better bookkeeping. In the disk a geodesic is an arc of a circle meeting the boundary at right angles, and a circle with center $C$ and radius $r$ is orthogonal to the unit circle exactly when

$|C|^2=1+r^2$

So the center alone determines the geodesic — the radius is $\sqrt{|C|^2-1}$ — and we may as well name a geodesic by its center $C$. Better still, expanding $|C_1-C_2|^2=r_1^2+r_2^2$ with that relation, two of them are perpendicular precisely when

$C_1\cdot C_2=1$

an ordinary dot product. Right angles, which is all our pentagon is made of, have become linear.

Now the sides fall out almost without computation. $\gamma_a$ and $\gamma_b$ pass through $i$, which is the center of the disk, so they are diameters — the horizontal and the vertical one. That is our whole normalization made visible: the vertex $v$ is at the center, and the two sides through it are two perpendicular straight lines.

For $\gamma_e$: it is perpendicular to the diameter $\gamma_a$, so its center lies on that axis, and $|C|^2=1+r^2$ pins the rest down. The same argument on the other axis gives $\gamma_c$. And $\gamma_d$ is perpendicular to both $\gamma_c$ and $\gamma_e$, so its center is read straight off the two dot products — one coordinate each.

The last radius is the moduli condition in disguise. Clearing denominators,

$\tanh^2a+\tanh^2b-1=\frac{\sinh^2a\sinh^2b-1}{\cosh^2a\,\cosh^2b}$

so $\gamma_d$ is a real circle exactly when $\sinh a\sinh b>1$. The pentagon closes up precisely when the fifth side has somewhere to be.

The realization is now concrete enough to draw with. In the next note we turn it into a shader that renders the whole tiling, one pixel at a time.