Regular Polytopes in the Poincaré Ball

Getting the faces of every regular hyperbolic polytope from euclidean constructions.

A previous note constructed regular polygons in the hyperbolic plane: working in the Poincaré disk, each side is a circle orthogonal to the boundary circle, and one formula locates all $p$ of them at once. Beautifully, essentially the same construction works one dimension up, with regular polytopes in the Poincaré ball.

Almost everything carries over, because almost nothing in the argument was really two-dimensional. But the answers come out quite differently. In the plane, a regular $p$-gon could have any angle smaller than the Euclidean value, and infinitely many regular tilings came out. Here the possible dihedral angles of each solid fill out only a short interval, and when we go looking for honeycombs, the count is finite: there are exactly eight.

A Plane is a Point Outside the Ball

Hyperplanes of the ball model come in two kinds, just like geodesics of the disk did: flat disks through the center, and spheres meeting the boundary at right angles. And the correspondence with exterior points works the same way. A sphere of center $c$ and radius $r$ meets the unit sphere at right angles exactly when

$r^2+1=\|c\|^2$

by the Pythagorean theorem on the two radii and the segment joining the centers, the same computation as in the plane: everything happens inside the plane spanned by those three segments, and that plane doesn’t care what dimension it sits in. So again the radius isn’t a second piece of data, $r=\sqrt{\|c\|^2-1}$ is the length of a tangent from $c$ to the ball, and a plane of $\HH^3$ not through the center is named by a single point $c$ outside the ball.

In the disk, the tangent segment from $c$ swept around $c$ to trace out the geodesic circle. Here the tangent lines from $c$ form a whole cone, touching the unit sphere along a circle, and that circle is exactly where the sphere centered at $c$ crosses the boundary: the ideal boundary of the hyperbolic plane that $c$ names.

As before, we’ll center our polytope at the origin so that no face passes through the center, and then the polytope is a finite set of points outside the ball, one per face.

The Exterior Points are a Copy of the Dual

Now take a regular polytope $P$ in $\HH^3$: all faces congruent regular polygons, all dihedral angles equal, its symmetry group taking any face to any other. Just as in the plane, the symmetry group is finite and so fixes a point, the center of $P$, and we move that center to the origin of the ball. Then no face passes through the origin, and each face’s sphere has its own exterior point $c_i$.

The disk model’s gift carries over too: the isometries of $\HH^3$ fixing the center of the ball are exactly the Euclidean rotations and reflections about it. So the symmetry group of $P$ acts on the points $c_1,\dots,c_N$ as a group of ordinary Euclidean symmetries. It takes faces to faces transitively, so it takes the $c_i$ to one another transitively: they form a single orbit, and in particular they all sit at one common distance $\rho$ from the origin. And each $c_i$ lies on the axis through its own face’s center, since the rotations about that axis fix the face and therefore fix $c_i$.

But the face-center directions $\hat u_i$ of a polytope are exactly the vertex directions of its dual. So the exterior points are a Euclidean copy of the dual polytope $P^*$, scaled to circumradius $\rho$: a cube is bounded by spheres centered on an octahedron, a dodecahedron by spheres centered on an icosahedron, a tetrahedron by spheres centered on another tetrahedron. This is where one dimension up gets more interesting than the plane, where the dual of a $p$-gon was just another $p$-gon.

So the directions $\hat u_i$ are settled before we start: they’re a fact about which solid we chose. The only freedom left is how far out along them the points sit, and that single number is $\rho$. Finding it is the rest of the problem.

The Scale

To find $\rho$ we need an equation relating it to the dihedral angle $\theta$, and the dihedral angle is fundamentally a two-dimensional measurement: two faces meet along an edge, and $\theta$ is measured in a plane cutting that edge at right angles. So we should cut.

Take two adjacent faces of $P$. Slice the whole picture with the Euclidean plane $\Pi$ through the center of the ball and the centers of their two spheres.

Why does this work? Passing through the ball’s center, $\Pi$ meets the ball in a flat unit disk: a copy of the Poincaré disk. Passing through the sphere centers, it meets each sphere in a great circle, so it shows the true angle at which the two spheres cross, and each circle is orthogonal to the boundary of the disk since its sphere was orthogonal to the boundary of the ball. And since the ball model is conformal, just like the disk one dimension down, that crossing angle is the dihedral angle $\theta$.

So inside the slice we’re looking at the previous note’s picture exactly: two circles in a Poincaré disk, each orthogonal to the boundary, crossing at $\theta$.

The previous note’s relation applies as it stands. The law of cosines on the triangle of the two radii and the segment joining the centers, remembering that the apex angle is the supplement $\chi=\pi-\theta$ and not $\theta$ itself, gives

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

Now fill in what the slice supplies. Both circles have the same radius, and orthogonality with the boundary makes it $r=\sqrt{\rho^2-1}$. Their centers are two adjacent vertices of the dual polytope at circumradius $\rho$, so the only missing ingredient is the angle between two adjacent directions. Write $\delta$ for that angle: it’s a fact about the dual solid alone, and it’s the only place a particular choice of solid enters this calculation. Then the two centers sit at distance

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

apart, and substituting everything into the angle relation leaves one equation in the single unknown $\rho$:

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

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 solid serves up. The slice is what let us borrow it whole: the derivation only ever looked at two spheres at a time, and two crossing spheres, cut through their centers, are just two crossing circles.

The Five Solids

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

solid	face directions $\hat u_i$ (before normalizing)
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

Now dot two adjacent ones together. For the dodecahedron, say, two adjacent icosahedral directions are $(0,1,\phi)$ and $(0,-1,\phi)$: their dot product is $\phi^2-1=\phi$, each has squared norm $1+\phi^2=2+\phi$, and so

$\cos\delta=\frac{\phi}{2+\phi}=\frac{1}{\sqrt5}$

Doing this for all five solids fills out the table:

solid	dual	$\cos\delta$	$\delta$
tetrahedron	tetrahedron	$-1/3$	$109.47^\circ$
cube	octahedron	$0$	$90^\circ$
octahedron	cube	$1/3$	$70.53^\circ$
dodecahedron	icosahedron	$1/\sqrt5$	$63.43^\circ$
icosahedron	dodecahedron	$\sqrt5/3$	$41.81^\circ$

The cube’s row is the friendliest: $\cos\delta=0$ makes the bounding radius

$r^2=\rho^2-1=\frac{1-\cos\delta}{\cos\theta+\cos\delta}=\frac{1}{\cos\theta}$

so a hyperbolic cube of dihedral angle $\theta$ is cut out by six spheres of radius $1/\sqrt{\cos\theta}$, centered on an octahedron.

Where the Families End

Each solid’s formula is only good on an interval of dihedral angles, and the two ends of that interval are different in kind: one is a limit of the edges, the other of the vertices. Both also differ from the plane, where every family ran from an ideal polygon at $\theta=0$ all the way 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 flatten into planes and the solid shrinks to a point, exactly as the polygons did. And this endpoint is a familiar number: the dihedral angle between two faces is supplementary to the angle between their outward normals, and the angle between adjacent normals is $\delta$, so $\pi-\delta$ is the dihedral angle of the Euclidean solid. Of course it is: a very small region of $\HH^3$ is very nearly Euclidean, so as the solid shrinks its angles must approach the Euclidean ones.

For the cube, $\theta_{\mathrm{euc}}=\pi/2$. So there is no right-angled hyperbolic cube: push the dihedral angle up toward a right angle and the cube shrinks away in front of you, vanishing at the same moment it would have reached $90^\circ$.

The Ideal End

Going the other way the solid grows, and this end is stranger. The number we control is the dihedral angle, measured along an edge. But a polytope also has corners, and it turns out the corners give out first: the vertices escape to the sphere at infinity while the dihedral angle is still substantial, nowhere near $0$. And this, too, is something we can just compute: we have every face of the polytope as an explicit sphere, so we can solve for the vertices and watch them leave.

Where is a vertex? On its axis, first of all: the rotations of $P$ about a vertex direction $\hat v$ fix that vertex, so the vertex is a point $t\hat v$ for some $t>0$, exactly the argument that put each $c_i$ on its face’s axis. And the vertex lies on each of the spheres bounding its adjacent faces. So let $\hat u$ be an adjacent face direction and intersect the axis with that sphere:

$\|t\hat v-\rho\hat u\|^2=\rho^2-1$

which multiplies out to a quadratic in $t$,

$t^2-2t\rho\cos\gamma+1=0,\qquad t=\rho\cos\gamma\pm\sqrt{\rho^2\cos^2\gamma-1}$

where $\gamma$ is the angle between a vertex direction and an adjacent face direction. Like $\delta$, it’s a constant of the solid, read off the coordinate table, and it’s the second and last place the particular solid enters. (It’s also the same constant for a solid and its dual, since dualizing swaps face directions with vertex directions and so leaves the angle between them alone.)

This quadratic knows everything about the vertex. Its two roots multiply to $1$, so the axis crosses the sphere at a pair of points that are inverses across the unit sphere, one inside the ball and one outside: that’s the orthogonality of the face sphere and the boundary sphere, showing up one last time. The vertex is the inner root. And the vertex exists exactly as long as the roots are real, so the discriminant runs the show:

$\rho\cos\gamma>1:\ \text{a vertex inside the ball},\qquad \rho\cos\gamma=1:\ \text{both roots at }t=1$

At $\rho\cos\gamma=1$ the two inverse points collide, and the only points that are their own inverses lie on the unit sphere itself: the vertex has arrived at infinity. (Below that the roots go complex and the axis misses the sphere entirely, but our families will end before we ever need this case.)

So the family turns ideal at $\rho_{\mathrm{ideal}}=1/\cos\gamma$, and the main formula converts this to the angle. For the cube, $\hat v=\tfrac{1}{\sqrt3}(1,1,1)$ and $\hat u=(1,0,0)$ give $\cos\gamma=1/\sqrt3$, so $\rho^2_{\mathrm{ideal}}=3$, and solving $\frac{1+\cos\theta}{\cos\theta}=3$ gives $\cos\theta=\tfrac12$: the cube’s vertices reach infinity at $\theta_{\mathrm{ideal}}=60^\circ$, well before its dihedral angle would have run out at $90^\circ$. The other solids go the same way:

solid	$\rho^2_{\mathrm{ideal}}$	ideal $\theta$	Euclidean $\theta$
tetrahedron	$9$	$60^\circ$	$70.53^\circ$
cube	$3$	$60^\circ$	$90^\circ$
octahedron	$3$	$90^\circ$	$109.47^\circ$
dodecahedron	$15-6\sqrt5$	$60^\circ$	$116.57^\circ$
icosahedron	$15-6\sqrt5$	$108^\circ$	$138.19^\circ$

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

Which Ones Tile

A polytope tiles $\HH^3$ face-to-face when copies of it fit together around every edge and every vertex. The edges give the numerical condition. If $r$ copies meet around an edge, they share the $2\pi$ of dihedral angle equally, so

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

and the candidate honeycomb has Schläfli symbol $\{p,q,r\}$, with $\{p,q\}$ the cell. Each solid lives on one short interval of dihedral angles, so the search is a finite check of which angles $2\pi/r$ land in which interval, and running through the table there are exactly eight hits:

What about the vertices? Around a vertex, each cell presents a corner, and the corners fit together in a pattern of their own: $q$ faces of the cell meet at each of its vertices and $r$ cells wrap around each edge, so the arrangement of cells around a vertex is described by the symbol $\{q,r\}$, the vertex figure. And now look at what the eight candidates serve up. The compact four have vertex figures $\{3,5\}$, $\{3,4\}$, $\{3,5\}$ and $\{5,3\}$: the icosahedron, the octahedron, and the dodecahedron. Platonic solids, that is, patterns that wrap a sphere, which is exactly what must surround an ordinary point. The ideal four have vertex figures $\{3,6\}$, $\{3,6\}$, $\{4,4\}$ and $\{3,6\}$: the triangular and square tilings of the Euclidean plane, wrapping a vertex at infinity, whose cross-section is flat¹. The edge condition and the vertex condition pass together, and the split between them is the same split the discriminant found.

So let’s draw all eight. With $\theta=2\pi/r$ the theorem hands over each cell explicitly,

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

the part of the ball outside the spheres of radius $\sqrt{\rho^2-1}$ centered at $\rho\hat u_i$:

Two of the eight deserve a remark. $\{5,3,5\}$, with five dodecahedra around each edge, is the universal cover of the Seifert–Weber dodecahedral space, whose gluing identifies opposite faces of one dodecahedron with a $3/10$ turn. And $\{5,3,4\}$ is the right-angled dodecahedron: the only right dihedral angle anywhere in these five families, which is a large part of why it shows up so often in hyperbolic geometry.

Which Are Fundamental Domains of Reflection Groups

Reflection in a face plane is a symmetry of $\{p,q,r\}$ for every $r$. The finer question is whether the cell is a fundamental domain for the group its own face reflections generate: whether the reflected copies of the cell tile without the group ever folding the cell onto itself. For that, the cell must be a Coxeter polytope, with every dihedral angle of the form $\pi/m$: two mirrors meeting at $2\pi/r$ for odd $r$ generate additional reflections cutting through the cell itself, so the true fundamental domain would be something smaller. Since $\theta=2\pi/r=\pi/(r/2)$, the cell is Coxeter exactly when $r$ is even, which picks out five of the eight:

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

Only one of these has compact cells. So the right-angled dodecahedron of $\{5,3,4\}$ is the only compact regular polytope in $\HH^3$ that’s the chamber of a reflection group. The odd three, $\{4,3,5\}$, $\{5,3,5\}$ and $\{3,5,3\}$, exist all the same: five dodecahedra genuinely close up around each edge of $\{5,3,5\}$ whether or not any mirror could have put them there.

Generalizing Further

Nothing in the sphere dictionary or the angle relation required regularity: they hold for any collection of planes whatsoever. What regularity did was collapse the unknowns, one $\theta$, one $\delta$, one $\rho$, until the whole configuration hung on a single equation.

Drop the symmetry and the problem becomes a genuine system. In the language of the hyperboloid remark, each face has a pole $n_i$ with $\langle n_i,n_i\rangle=1$, each prescribed dihedral angle is an equation $\langle n_i,n_j\rangle=-\cos\theta_{ij}$, and the poles are the unknowns. This is the Gram matrix picture of a hyperbolic polytope, and whether the system has a solution is no longer a range check: it’s Andreev’s theorem. A note for another day.

Footnotes

1. That these local conditions really do assemble into a global honeycomb is Poincaré’s polyhedron theorem, which we won’t prove here: this note’s business is producing the cells, and the theorem’s hypotheses are exactly what our formula lets us check. ↩