Note: Hyperbolic Dodecahedra

I just had reason to calculate the size of a regular right angled dodecahedron again, a quantity I know I’ve calculated several times before, but could not find my old notes. To prevent this happening again, I’ll record the calculation here.

We’ll proceed in a little more generality, and consider an arbitrary regular hyperbolic dodecahedron, with dihedral angles $\theta$. In the end, we will specialize and give the measurements for three dodecahedra: the right angled one, Seifert & Weber’s, and the idea regular dodecahedron. Our goal will be to calculate several quantities:

• $f$, the distance from the center to the center of a face
• $v$, the distance from the center to a vertex
• $e$, the distance from the center to an edge’s midpoint
• $s$, the side-length

PICTURE

The trick is to get the computation down a dimension into the hyperbolic plane. Call the center of the dodecahedron $O$, and choose two adjacent faces meeting in an edge. Slice the dodecahedron with the hyperbolic plane passing through the center and the midpoint of each of the faces. This plane also passes through the edge-midpoint, so in our slice we have the following figure:

PICTURE

The angle $\delta$ at the center is the same as the Euclidean angle between two adjacent face centers of a dodecahedron (as one can see by looking in the tangent space to $O$, and thinking about the geodesics heading out to face centers). This is readily computable if we recall some properties of a Euclidean dodecahedron of side length $a$ (where $\phi$ is the golden ratio)

• The outer radius (origin to vertex) is $r_{\mathrm{out}}=\frac{\sqrt{3}}{2}\phi a$
• The mid-radius (origin to edge center) is $r_\mathrm{mid}=\frac{\phi^2}{2}a$
• The in-radius (origin to face center) is $r_\mathrm{in}=\frac{\phi^2}{2\sqrt{3-\phi}}$

Below is an image of the corresponding slice of a Euclidean dodecahedron, where we can read of the trigonometry of $\delta$: PICTURE

$$\cos\frac{\delta}{2} = \frac{r_\mathrm{in}}{r_\mathrm{mid}}=\frac{1}{\sqrt{3-\phi}}$$ $$\implies \sin\frac{\delta}{2}=\sqrt{\frac{2-\phi}{3-\phi}},\hspace{0.5cm}\tan\frac{\delta}{2}=\sqrt{2-\phi}$$

However, these simplify in $\mathbb{Q}[\phi]$ which will help us going forward:

$$\cos\frac\delta 2 = \sqrt{\frac{\phi}{\sqrt{5}}}\hspace{1cm}\sin\frac\delta 2=\sqrt{\frac{1}{\phi\sqrt{5}}}\hspace{1cm}\tan\frac\delta 2=\frac{1}{\phi}$$

Subdividing the quadrilateral above by a geodesic from $O$ to the edge center gives two congruent right triangles, and we can proceed by hyperbolic trigonometry.

PICTURE

All three angles of the triangle are known, so we can immediately read off relations for the side lengths:

$$\cosh f = \frac{\cos\frac\theta 2}{\sin\frac\delta 2}=\hspace{1cm}\cosh e = \cot\frac{\delta}{2}\cot\frac{\theta}{2}$$

$$f=\operatorname{acosh}\left(\sqrt{\phi\sqrt{5}}\cos\frac\theta 2\right)$$ $$e=\operatorname{acosh}\left(\phi\cot\frac\theta 2\right)$$

To compute $v$ and $s$ we need to slice our dodecahedron differently, with a slice passing through two vertices and the center $O$. In this slice, the two sides eminating from $O$ are both length $v$, and the remaining is an actual edge of the dodecahedron, with side length $s$. The angle $\eta$ at $O$ agrees with the corresponding angle for a Euclidean dodecahedron, which we compute (as above) from its side length and radii:

PICTURE

$$\cos\frac{\eta}{2} = \frac{r_\mathrm{mid}}{r_\mathrm{out}}=\frac{\phi}{\sqrt{3}}$$

From this we find simplified forms of the trigonometric values in $\mathbb{Q}[\phi]$:

$$\cos\frac\eta 2 = \frac{\phi}{\sqrt{3}}\hspace{1cm}\sin\frac\eta 2=\frac{1}{\phi\sqrt{3}}\hspace{1cm}\tan\frac\eta 2=\frac{1}{\phi^2}$$

Back in the hyperbolic case, a similar subdivision into two right triangles completes our work

PICTURE

$$\tan\frac\eta 2 = \frac{\tanh\frac s 2}{\sinh e}\hspace{1cm}\cos\frac\eta 2 = \frac{\tanh e}{\tanh v}$$

Computing $\sinh e, \tanh e$ explicitly in terms of known quantities:

$$\sinh e =\sqrt{\cosh^2 e-1}=\sqrt{\phi^2\cot^2\frac{\theta}{2}-1}=\sqrt{\frac{\phi^2}{\tan^2\frac\theta 2}-1}$$ $$\tanh e=\sqrt{1-\operatorname{sech}^2 e}=\sqrt{1-\frac{1}{\phi^2\cot^2\frac\theta 2}}=\sqrt{1-\frac{\tan^2\frac\theta 2}{\phi^2}}$$

Using this, we can solve the previous relations for $v$

$$\tanh v = \frac{\tanh e}{\cos\frac\eta 2}=\frac{\sqrt{1-\frac{\tan^2\frac\theta 2}{\phi^2}}}{\frac{\phi}{\sqrt{3}}}$$ $$\implies v = \operatorname{atanh}\left(\frac{\sqrt{3}}{\phi}\sqrt{1-\frac{\tan^2\frac\theta 2}{\phi^2}}\right)$$

and $s$:

$$\tanh\frac{s}{2}=\tan\frac\eta 2 \sinh e = \frac{1}{\phi^2}\sqrt{\frac{\phi^2}{\tan^2\frac\theta 2}-1}$$ $$\implies s=2\operatorname{atanh}\left(\frac{1}{\phi^2}\sqrt{\frac{\phi^2}{\tan^2\frac\theta 2}-1}\right)$$

Right Angled Coxeter Dodecahedron

When the dihedral angles are right, $\theta = \pi/2$ so plugging in gives

$$f=\operatorname{acosh}\left(\sqrt{\phi\sqrt{5}}\frac{1}{\sqrt{2}}\right)\approx 0.808461$$ $$e=\operatorname{acosh}\left(\phi\right)\approx 1.06128$$ $$v=\operatorname{atanh}\left(\frac{\sqrt{3}}{\phi}\sqrt{1-\frac{1}{\phi^2}}\right)\approx 1.22646$$ $$s = 2\operatorname{atanh}\left( \frac{\sqrt{\phi^2-1}}{\phi^2}\right)\approx 0.530638$$

We record simplified versions for future reference:

Seifert Weber Dodecahdron

Seifert weber dodecahedral space is built by identifying opposite faces of a dodecahedron with a $3/10$ twist. Such a gluing pairs up edges in collections of five, meaning in the universal cover there are five dodecahedra around each edge. Realizing this in the hyperbolic metric requires a regular dodecahedron with all dihedral angles $2\pi/5$. The trigonometry of $\pi/5$ involves even more copies of the golden ratio:

$$\cos\frac\pi 5=\frac\phi 2\hspace{0.5cm}\sin\frac\pi 5=\frac{1}{2}\sqrt{\frac{\sqrt{5}}{\phi}}\hspace{1cm}\tan\frac\pi 5=\frac{1}{\phi}\sqrt{\frac{\sqrt{5}}{\phi}}$$

Using these,

$$f=\operatorname{acosh}\left(\sqrt{\phi\sqrt{5}}\frac\phi 2\right)\approx 0.996384$$ $$e=\operatorname{acosh}\left(\phi\cdot\phi\sqrt{\frac{\phi}{\sqrt{5}}}\right)\approx 0.952797$$ $$v=\operatorname{atanh}\left(\frac{\sqrt{3}}{\phi}\sqrt{1-\frac{1}{\phi^2}\frac{\sqrt{5}}{\phi^3}}\right)\approx 1.90285$$ $$s = 2\operatorname{atanh}\left(\frac{1}{\phi^2}\sqrt{\phi^2\frac{\phi^3}{\sqrt{5}}-1}\right)\approx 1.99277$$

Again we record for future reference

Ideal Dodecahedron

To make a hyperbolic dodecahedron with all vertices idea we cannot follow the procedure above and start with the dihedral angle. Instead we are assuming that $v$ is infinite. In terms of our formulas, this means $\tanh v=1$, which lets us find $e$:

$$\frac{\phi}{\sqrt{3}}=\cos\frac\eta 2 = \frac{\tanh e}{\tanh v}=\tanh e$$ $$\implies e=\operatorname{atanh}\frac{\phi}{\sqrt{3}}$$

We can then use the fact that we know $e$ to solve the dihedral angle $\theta$ itself:

$$\cosh e = \frac{1}{\sqrt{1-\tanh^2 e}}=\frac{1}{\sqrt{1-\frac{\phi^2}{3}}}=\phi\sqrt{3}$$ $$\cosh e =\phi\cot\frac{\theta}{2}$$ $$\implies \tan\frac{\theta}{2}=\frac{1}{\sqrt{3}}$$

This is incredibly nice: a tangent of $1/\sqrt{3}$ means an angle of $\pi/6$ so $\theta=\pi/3$, the dihedral angle of $60$ degrees. Knowing the dihedral angle, we can find the radius to the face centers using the original formula

$$\cosh f = \sqrt{\phi\sqrt{5}}\cos\frac\theta 2=\sqrt{\phi\sqrt{5}}\frac{\sqrt{3}}{2}$$

Collected together: