Regular Polytopes in the Poincare Ball

Bounding spheres for every regular hyperbolic polytope, and the eight honeycombs that come out of them.

We want to draw honeycombs of hyperbolic space by Coxeter polytopes, and as before that means writing a polytope down: not knowing that one exists, but producing the actual numbers.

A previous note did this in the plane. We take the same two choices - regular polytopes, centred at the origin, in the Poincare ball - for the same reason. Hyperplanes of the ball are spheres orthogonal to the boundary, or flat disks through the centre, and centring the polytope kills the second case: no face passes through the origin, so every face is an honest sphere.

Almost all of that argument carries over, because none of it was two-dimensional: it was about two spheres, and two spheres span a plane in any dimension and meet each other the same way inside it. What changes is the dual. In the plane, saying that the exterior points form the dual polygon was saying very little, a $p$-gon’s dual being a $p$-gon. Here a dodecahedron is bounded by spheres centred on an icosahedron.

A Plane is a Point Outside the Ball

The correspondence is the same, and it is worth one picture before we lean on it.

A sphere of centre $c$ and radius $r$ meets the unit sphere at right angles exactly when $r^2+1=\|c\|^2$, by Pythagoras on the two radii and the line of centres - nothing in that argument mentions dimension. So the radius is not a second choice: $r=\sqrt{\|c\|^2-1}$, which is the length of the tangent from $c$ to the ball.

In the plane the two tangents from $c$ touched the circle at two points, and those were the ends of the geodesic. Here the tangent lines from $c$ form a cone, and it touches the sphere along a whole circle - which is exactly where the sphere centred at $c$ meets the boundary, and so is the ideal boundary of the plane that sphere carries.

So a hyperbolic polytope centred at the origin is a finite set of points outside the ball, one per face, and everything that follows is about locating that set.

The Exterior Points are a Copy of the Dual

Let $P$ be a regular polytope with dihedral angle $\theta$, centred at the origin, and let $c_1,\dots,c_N$ be the exterior points of its faces.

The argument that placed them in the plane was about the origin, and it is unchanged here. The isometries of $\HH^3$ fixing the centre of the ball are exactly the Euclidean rotations and reflections about it, so the symmetry group of $P$ acts on the exterior points as an ordinary Euclidean symmetry group. It carries faces to faces transitively, so the $c_i$ form a single orbit and all lie at one distance $\rho$ from the origin; and the reflection in the plane through the origin and a face’s centre fixes that face, hence its sphere, hence its centre, so each $c_i$ lies along the direction $\hat u_i$ of its own face’s centre.

Those directions are exactly the vertex directions of the dual polytope. So the exterior points are a Euclidean copy of $P^*$ at circumradius $\rho$: a cube is bounded by spheres centred on an octahedron, a dodecahedron by spheres centred on an icosahedron, a tetrahedron by spheres centred on another tetrahedron.

The directions are settled before we start, then, and the only freedom left is how far out along them the points sit. That single number is $\rho$, and finding it is the rest of the problem.

The Scale

To find it we need the dihedral angle in a form we can compute with, and the dihedral angle is a two-dimensional measurement: two faces of $P$ meet along an edge, and $\theta$ is the angle between them measured in a plane cutting that edge at right angles. So we should cut.

Take the plane $\Pi$ through the origin spanned by two adjacent face directions $\hat u_i$ and $\hat u_j$. Three things happen at once, and together they put us back in the previous note’s picture exactly:

• $\Pi$ passes through the centre of the ball, so it meets the ball in a flat unit disk - and a plane through the centre is a hyperbolic plane, so that disk is a copy of the Poincare disk.
• $\Pi$ contains both sphere centres $\rho\hat u_i$ and $\rho\hat u_j$, so it cuts each face sphere in a full circle, and each such circle is orthogonal to the boundary of the disk because its sphere was orthogonal to the boundary of the ball.
• The two spheres cross on a circle, and $\Pi$ meets that circle. At such a point $\Pi$ contains both radii, so it contains both tangent planes’ normals, and the angle between the two circles in $\Pi$ is the angle between the two spheres. No correction: the slice measures $\theta$ itself.

So we are looking at two circles in a Poincare disk, meeting at $\theta$, each orthogonal to the boundary. That is the previous note’s configuration, and it needs the previous note’s relation: the law of cosines on the triangle made of the two radii and the line of centres, remembering that its apex angle is $\chi=\pi-\theta$ and not $\theta$, since $\theta$ is measured between the tangents and each radius is perpendicular to its own. Substituting $\cos\chi=-\cos\theta$,

$r_1^2+r_2^2+2r_1r_2\cos\theta = d^2$

Now fill in what the slice supplies. Both circles have the same radius, and orthogonality makes it $\sqrt{\rho^2-1}$. Their centres are two adjacent vertices of the dual at circumradius $\rho$, so the only thing still missing is how far apart two adjacent directions are. Write $\delta$ for the angle they subtend at the origin - it is a fact about the dual alone, and the one place a particular solid will enter this calculation. Then the distance between the centres is the dual’s edge,

$d = \rho\sqrt{2-2\cos\delta}$

Putting those in leaves one equation in the single unknown $\rho$,

$2(\rho^2-1)(1+\cos\theta) = \rho^2(2-2\cos\delta)$

and collecting the $\rho^2$ terms finishes it:

This is the plane’s theorem word for word, with $\delta=2\pi/p$ replaced by whatever angle the dual happens to have. Nothing in the derivation ever knew the dimension; the slice is what let us borrow it whole.

The Five Solids

Only one number in the theorem knows which polytope we mean, and it is $\delta$. So we need, for each Platonic solid, the directions of its face centres and the angle between two adjacent ones. Those directions are the dual’s vertices, and the duals are Platonic solids again, so it is a short list of coordinates:

solid	face directions $\hat u_i$ (before normalising)
tetrahedron	$(1,1,1)$ and the three with two signs flipped
cube	$(\pm1,0,0)$ and cyclic
octahedron	$(\pm1,\pm1,\pm1)$
dodecahedron	$(0,\pm1,\pm\phi)$ and cyclic
icosahedron	$(\pm1,\pm1,\pm1)$ together with $(0,\pm\tfrac1\phi,\pm\phi)$ and cyclic

Taking the dot product of two adjacent ones gives $\cos\delta$, and $\sigma=\sqrt{2-2\cos\delta}$ is the corresponding chord. For the dodecahedron, say, two adjacent icosahedral directions are $(0,1,\phi)$ and $(0,-1,\phi)$, whose dot product is $\phi^2-1=\phi$ against a norm of $1+\phi^2=2+\phi$, so $\cos\delta = \phi/(2+\phi) = 1/\sqrt5$. The whole table:

solid	dual	$\cos\delta$	$\delta$	$\sigma$	$4/\sigma^2$
tetrahedron	tetrahedron	$-1/3$	$109.47^\circ$	$2\sqrt{2/3}$	$3/2$
cube	octahedron	$0$	$90^\circ$	$\sqrt2$	$2$
octahedron	cube	$1/3$	$70.53^\circ$	$2/\sqrt3$	$3$
dodecahedron	icosahedron	$1/\sqrt5$	$63.43^\circ$	$2/\sqrt{2+\phi}$	$2+\phi$
icosahedron	dodecahedron	$\sqrt5/3$	$41.81^\circ$	$2/(\sqrt3\,\phi)$	$3\phi^2$

The last column is the one that does the work. Written in terms of $\sigma$ the theorem becomes

$\frac{1}{\rho^2-1} = \frac{4}{\sigma^2}\cos^2\frac\theta2 - 1$

so the five families differ in exactly one constant: $3/2$, $2$, $3$, $2+\phi$, $3\phi^2$.

The cube is the case to look at, since $\sigma=\sqrt2$ collapses it as far as it will go: $4/\sigma^2=2$, and $2\cos^2\tfrac\theta2-1$ is just $\cos\theta$. The bounding spheres of a hyperbolic cube have radius $1/\sqrt{\cos\theta}$.

Where Each Family Lives

Each formula is only good on an interval of $\theta$, and the two ends are quite different from each other - and both differ from the plane, where the family always ran from an ideal polygon at $\theta=0$ up to the Euclidean angle.

The Euclidean end. The denominator $\cos\theta+\cos\delta$ is positive exactly when $\theta+\delta<\pi$, so the family stops at

$\theta_{\mathrm{euc}} = \pi - \delta$

where $\rho$ runs off to infinity: the spheres have flattened into planes and the solid has shrunk to a point. That angle is the dihedral angle of the Euclidean solid, and not by coincidence - the dihedral angle between two faces is supplementary to the angle between their outward normals, which is exactly $\delta$. Nor is the coincidence surprising once one sees why: a very small region of $\HH^3$ is very nearly Euclidean, so as the solid shrinks its angles must approach the Euclidean ones. That is what the family runs out of - not room, but size.

For the cube $\theta_\mathrm{euc}=\pi/2$, so there is no right-angled hyperbolic cube. Push the dihedral angle up towards a right angle and the cube shrinks away in front of you, reaching nothing at the same moment it would have reached $90^\circ$.

The ideal end. Going the other way the solid grows and its vertices reach the sphere at infinity, but not at $\theta=0$. Look at a vertex where $k$ faces meet: a small sphere around it cuts the solid in a spherical $k$-gon whose angles are the dihedral angles. A spherical polygon has angle sum greater than a flat one, and it degenerates precisely when the two agree, $k\theta=(k-2)\pi$:

$\theta_{\mathrm{ideal}} = \frac{(k-2)\pi}{k}$

So the ideal end depends only on the vertex degree, and the solid runs out of room while its angles are still substantial. The tetrahedron, cube and dodecahedron are all trivalent and turn ideal at $\pi/3$; the octahedron is $4$-valent, at $\pi/2$; the icosahedron is $5$-valent, at $3\pi/5$. Below those the vertices have passed beyond infinity and there is no polytope in $\HH^3$ any more.

solid	$k$	ideal $\theta$	Euclidean $\theta$
tetrahedron	3	$60^\circ$	$70.53^\circ$
cube	3	$60^\circ$	$90^\circ$
octahedron	4	$90^\circ$	$109.47^\circ$
dodecahedron	3	$60^\circ$	$116.57^\circ$
icosahedron	5	$108^\circ$	$138.19^\circ$

Those intervals are short. The cube has $30^\circ$ to live in and the tetrahedron barely $10^\circ$, which is already a hint that not much is going to fit inside them.

Which Ones Tile

A polytope tiles $\HH^3$ face-to-face exactly when a whole number of copies close up around each edge. If $r$ of them do, they share the $2\pi$ of angle around that edge equally, so

$\theta = \frac{2\pi}{r}$

and the honeycomb is the regular one with Schlafli symbol $\{p,q,r\}$, where $\{p,q\}$ is the cell. When it happens, the theorem hands over the coordinates at once: with $\theta=2\pi/r$,

$\rho=\sqrt{\frac{1+\cos\frac{2\pi}r}{\cos\frac{2\pi}r+\cos\delta}}$

and the cell is the part of the ball outside the spheres of radius $\sqrt{\rho^2-1}$ centred at $\rho\hat u_i$, one for each face direction in the table above. That is the whole recipe; everything left is bookkeeping about which $r$ are allowed.

Since the only angles at which anything tiles are $2\pi/r$, and each solid lives on one interval, the classification is a matter of which of those angles land in which interval.

Eight crossings, and no more are possible:

The four on the top row are the four compact regular honeycombs of hyperbolic $3$-space, and it is worth appreciating how little went into producing them: one formula for a radius, two endpoints for an interval, and a count of which $2\pi/r$ fit inside.

Two of these deserve a remark. $\{5,3,5\}$ has $\theta=2\pi/5$, five dodecahedra around each edge - the universal cover of Seifert-Weber dodecahedral space, whose gluing identifies opposite faces with a $3/10$ twist. And $\{5,3,4\}$ is the right-angled dodecahedron, the only member of any of these families with a right dihedral angle at all, which is why it turns up so often.

Which of Them Are Reflection Groups

Reflection in a face plane is a symmetry of $\{p,q,r\}$ for every $r$. What the parity of $r$ decides is whether the cell is a Coxeter polytope, meaning all its dihedral angles are $\pi/m$ for a whole number $m$. Since $\theta=2\pi/r$, that happens exactly when $r$ is even, which picks out five of the eight:

$\{3,3,6\}\quad \{4,3,6\}\quad \{3,4,4\}\quad \{5,3,4\}\quad \{5,3,6\}$

Of these only $\{5,3,4\}$ has compact cells, so the right-angled dodecahedron is the only compact regular polytope in $\HH^3$ whose reflections generate a discrete group.

The three left out - $\{4,3,5\}$, $\{5,3,5\}$ and $\{3,5,3\}$ - are no less real as honeycombs; their cells simply are not chambers of a reflection group, having a dihedral angle of $2\pi/5$ or $2\pi/3$ that no mirror pair produces. Seifert-Weber is the case to hold onto: five dodecahedra close up around an edge with a $3/10$ twist, and the cells fit whether or not any mirror performs that twist.

Generalising Further

The dictionary and the angle relation used here are completely general - they say nothing about regularity, and hold for any collection of planes at all. What regularity bought was that a single $\sigma$ and a single $\theta$ reduced everything to one equation with one unknown.

Without it the problem becomes a genuine system: one equation $\langle n_i,n_j\rangle=-\cos\theta_{ij}$ per adjacent pair, with the unit conditions $\langle n_i,n_i\rangle=1$, to be solved for the poles. That is the Gram matrix picture of a Coxeter polytope, and whether a solution exists is Andreev’s theorem rather than a range check. A note for another day.