1.1 Colours

Let $P\subset {\mathbb {X}}^n$ be a right-angled finite polytope in some space $ {\mathbb {X}}^n = {\mathbb {H}}^n, {\mathbb {R}}^n$ or $ {\mathbb {S}}^n$ . We always suppose that P has finite volume. When $ {\mathbb {X}}^n = {\mathbb {H}}^n$ , the polytope P may have both finite and ideal vertices. We can interpret P as an orbifold $P= {\mathbb {X}}^n /\Gamma $ , where $\Gamma $ is the right-angled Coxeter group generated by the reflections $r_F$ along the facets F of P. A presentation for $\Gamma $ is

$$ \begin{align*}\langle\ r_F\ |\ r_F^2, [r_F, r_{F'}]\ \rangle,\end{align*} $$

where F varies among the facets of P and $F,F'$ among the pairs of adjacent facets.

A c-colouring of P is the assignment of a colour (taken from some fixed set of c elements) to each facet of P, such that incident facets have distinct colours. We generally use $\{1,\ldots , c\}$ as a palette of colours and suppose that every colour is painted on at least one facet.

Let $e_1,\ldots , e_c$ be the canonical basis of the ${\mathbb {Z}}_2$ -vector space ${\mathbb {Z}}_2^c$ . A colouring on P induces a homomorphism $\Gamma \to {\mathbb {Z}}_2^c$ that sends $r_F$ of to $e_j$ , where j is the colour of F. One verifies that the kernel $\Gamma ' \triangleleft \Gamma $ acts freely on $ {\mathbb {X}}^n$ , and, hence, we get a manifold $M= {\mathbb {X}}^n/{\Gamma '}$ that orbifold-covers $P= {\mathbb {X}}^n/\Gamma $ with degree $2^c$ .

The manifold $M= {\mathbb {X}}^n/\Gamma '$ is hyperbolic, flat or elliptic, according to the model $ {\mathbb {X}}^n$ , and is tessellated into $2^c$ copies of P. Geometrically, we may see M as constructed by mirroring P iteratively along facets sharing the same colours $1,\ldots , c$ .

More precisely, we can describe the tessellation of M into $2^c$ copies of P as follows. For every vector $v\in {\mathbb {Z}}_2^c$ , we denote by $P_v$ an identical copy of P. We identify each facet F of $P_v$ via the identity map with the same facet of $P_{v+e_j}$ , where j is the colour of F. This gives the tessellation of M.

We say that two colourings on P are isomorphic if they induce the same partition of facets, possibly after acting by some isometry of P. Isomorphic colourings yield isometric manifolds M.

As an example, we can always colour P by assigning distinct colours to distinct facets. In this case, c equals the number of facets of P and $\Gamma \to {\mathbb {Z}}_2^c$ is just the abelianisation homomorphism. With this choice, the resulting manifold M can be quite big and often intractable (especially in higher dimension $n>3$ ), so it is often preferable to work with a small number of colours. Another fundamental reason for rejecting this inefficient colouring will be given below in Corollary 13.

Here are some more interesting examples:

• • The Euclidean n-cube has a unique n-colouring up to isomorphisms, where opposite facets are coloured with the same colour. This colouring produces a flat torus tessellated into $2^n$ cubes. The case $n=2$ is shown in Figure 1 and is easily generalised to any n. More generally, we will prove below that any colouring on the n-cube produces a flat torus.

  Figure 1

  A square P with two colours (left). The flat manifold M is constructed by taking four copies of P and identifying the edges as shown (centre). We get a flat square torus (right).

  

• • The right-angled spherical n-simplex has a unique colouring up to isomorphisms: It has $n+1$ colours, and it produces the spherical manifold $S^n$ with its standard tessellation into $2^{n+1}$ right-angled simplexes.

• • Every ideal hyperbolic polygon is right-angled in a vacuous sense (it has no finite vertices) and can be 1-coloured! Indeed, the edges are pairwise disjoint and, hence, can all be given the same colour. The construction produces the double of the polygon, a hyperbolic punctured sphere.

• • Every right-angled hyperbolic hexagon can be 2-coloured, and the result is the double of a geodesic pair-of-pants, that is a genus-2 hyperbolic surface, tessellated into four hexagons.

• • The ideal octahedron in $ {\mathbb {H}}^3$ has a unique 2-colouring up to isomorphisms. The colouring produces a cusped hyperbolic 3-manifold which is the complement of the minimally twisted chain link with six components shown in Figure 2 (see [Reference Kolpakov and Martelli26] for more details).

  Figure 2

  The minimally twisted chain link with six components.

  

• • The ideal 24-cell in $ {\mathbb {H}}^4$ has a unique 3-colouring, that produces a hyperbolic 4-manifold with 24 cusps with 3-torus sections (see [Reference Kolpakov and Martelli26]).

1.2 Cusp sections

When $P\subset {\mathbb {H}}^n$ has some ideal vertex, the resulting manifold M has some cusps, and there is a simple and straightforward procedure to derive its shape directly from the combinatorics of P and its colouring, that we now explain.

Let v be an ideal vertex of $P\subset {\mathbb {H}}^n$ . The link of v in P is by definition the intersection of P with a small horosphere centered at v. It is a right-angled Euclidean $(n-1)$ -parallelepiped C. We use the letter C because a parallelepiped is combinatorially a cube, and, in fact, it will also be isometric to a cube in all the cases that are of interest here.

The parallelepiped C inherits a colouring from that of P: it suffices to assign to every $(n-2)$ -facet of C the colour of the $(n-1)$ -facet of P that contains it. The induced colouring on C generates an abstract compact flat $(n-1)$ -manifold N that orbifold-covers C by the procedure explained above. The manifold N is tessellated into $2^{c'}$ copies of C, where $c'\leq c$ is the number of colours of C.

By construction, the preimage of C in M consists of some copies of N. The number of copies is equal to $2^h$ , where $h=c-c'$ is the number of colours in $\{1,\ldots , c\}$ that are not assigned to any facet incident to v, that is that are not assigned to any facet of C. The preimage of C in M consists of $2^c = 2^h \cdot 2^{c'}$ copies of C in total.

Summing up: There are $2^h$ cusps in M lying above v, each with section N derived directly from C and its induced colouring. Here are some examples:

• • If $P\subset {\mathbb {H}}^2$ is a 1-coloured ideal polygon, the link C at each ideal vertex v is a 1-coloured 1-cube (that is, a segment). Here, $h=0$ , the preimage of C is a circle and there is one cusp above each v. The punctured sphere M has one cusp above each ideal vertex of P.

• • If $P\subset {\mathbb {H}}^2$ is a 2-coloured ideal triangle, there are two types of ideal vertices. Two ideal vertices have a 2-coloured 1-cube as a link C, while the third ideal vertex has a 1-coloured 1-cube C. We have $h=0$ for the first two ideal vertices, and $h=1$ for the third. Therefore, the counterimage of C consists of one circle for each of the first two ideal vertices and two circles for the third. The manifold M has four cusps overall, two above the first two vertices and two above the third. It is a four-punctured sphere tessellated into four copies of P (see Figure 3).

  Figure 3

  When P is an ideal triangle with one or two colours, the manifold M is a sphere with three or four punctures, respectively.

  

• • If P is a 2-coloured ideal octahedron, it has six ideal vertices, and the link of each is a 2-coloured square C. We have $h=0$ on each ideal vertex, so the counterimage of C in M is a unique torus. The hyperbolic 3-manifold M has six cusps overall, one above each ideal vertex of P. As already stated, M is the complement of the link in Figure 2.

• • If P is a 3-coloured ideal 24-cell in $ {\mathbb {H}}^4$ , it has 24 ideal vertices, and the section of each is a 3-coloured 3-cube C (see [Reference Kolpakov and Martelli26]). We have $h=0$ on each ideal vertex, so the counterimage of C is a single 3-torus. The hyperbolic 4-manifold M has 24 toric cusps, one above each ideal vertex of P.

1.3 The Euclidean parallelepiped

One basic example is the Euclidean right-angled n-parallelepiped

$$ \begin{align*}C = [0,l_1] \times \cdots \times [0,l_n] \subset {\mathbb{R}}^n.\end{align*} $$

Fix a c-colouring of C. Only opposite facets are disjoint and, hence, may share the same colour. Therefore, we have $n \leq c \leq 2n$ , there are $2n-c$ pairs of opposite facets with the same colour and the remaining $2(c-n)$ facets with distinct colours. Let M be the flat manifold produced by the c-colouring of C.

Proof. Recall that $M = {\mathbb {R}}^n/\Gamma '$ , where $\Gamma $ is the reflection group of C and $\Gamma '\triangleleft \Gamma $ is the kernel of the map $\Gamma \to {\mathbb {Z}}_2^c$ induced by the colouring.

Let $r_{i,1}$ and $r_{i,2}$ be the reflections along the opposite facets of C that are orthogonal to the i-th axis, for $i=1,\ldots , n$ . The composition $r_{i,1}r_{i,2}$ is a translation along the axis of distance $2l_i$ . If the facets share the same colour, we have $r_{i,1}r_{i,2} \in \Gamma '$ , while if they do not, we have $r_{i,1}r_{i,2}r_{i,1}r_{i,2} \in \Gamma '$ . This shows that

$$ \begin{align*}a_1l_1 {\mathbb{Z}} \times \cdots \times a_nl_n {\mathbb{Z}} < \Gamma',\end{align*} $$

where $a_i$ equals 2 or 4 depending on whether the i-th pair of opposite facets share the same colour. These two subgroups have the same index in $\Gamma $ since

$$ \begin{align*}2^{2n-c} \cdot 4^{c-n} = 2^c.\end{align*} $$

Therefore, $\Gamma ' = a_1l_1 {\mathbb {Z}} \times \cdots \times a_nl_n {\mathbb {Z}}$ and M is as stated.

The proof also shows that M is tessellated into $2^{2n-c} \cdot 4^{c-n} = 2^c$ copies of C. The two extreme cases are the following: If $c=n$ , then M is tessellated into $2^c$ copies of C, while if $c=2n$ , then M is tessellated into $4^c$ copies.

A cusp in a hyperbolic n-manifold is toric if its section is a flat $(n-1)$ -torus. We summarise our discussion as follows.

1.4 A program in Sage

We have written a general program in Sage, available from [Reference Martelli39], that may be used to study a coloured right-angled polytope P and the resulting manifold M. The program takes as an input the incidence graph of the facets of P and their colouring, and produces as an output some information on P and, more importantly, on M. It calculates, in particular, the Betti numbers of M via the formula stated in [Reference Choi and Park9, Theorem 1.1], also explained in [Reference Ferrari, Kolpakov and Slavich15, Section 2.2], and the number of cusps of M using Corollary 8.

1.5 The right-angled hyperbolic polytopes

We refer to the excellent papers [Reference Everitt, Ratcliffe and Tschantz14] and [Reference Potyagailo and Vinberg30] for an introduction to the sequence of right-angled hyperbolic polytopes $P^3,\ldots , P^8$ . These have many beautiful properties that we now briefly summarise.

Each $P^n \subset {\mathbb {H}}^n$ is a finite volume right-angled polytope with both finite and ideal vertices. The link of a finite or ideal vertex is a right-angled spherical $(n-1)$ -simplex or a Euclidean $(n-1)$ -cube, respectively. The numbers of facets, ideal vertices and finite vertices of $P^n$ are listed in Table 2, together with the isometry group of $P^n$ and its order. The isometry group acts transitively on the facets, so, in particular, these are all isometric: In fact, every facet of $P^n$ is isometric to $P^{n-1}$ when $n\geq 4$ . The quotient of $P^n$ by its isometry group is a simplex.

Table 2

The number of facets, ideal vertices and finite vertices of $P^n$ , the isometry group $ {\mathrm {Isom}}(P^n)$ expressed as a Weyl group and its order $| {\mathrm {Isom}}(P^n)|$ , and the dual Euclidean polytope.

1.6 Euler characteristic

Recall that the orbifold Euler characteristic of a hyperbolic right-angled polyhedron P is zero in odd dimension, while in even dimension it can be calculated via the simple formula

$$ \begin{align*}\chi(P) = \sum_{i=0}^n (-1)^i \frac{f_i}{2^{n-i}},\end{align*} $$

where $f_i$ is the number of i-dimensional faces of P, including P itself (so $f_n=1$ ). Only real vertices (not the ideal ones) contribute to $f_0$ . From this formula, we deduce the well-known [Reference Everitt, Ratcliffe and Tschantz14] values

$$ \begin{align*}\chi(P^4) = 1/16, \qquad \chi(P^6) = -1/8, \qquad \chi(P^8) = 17/2.\end{align*} $$

In even dimension, the Euler characteristic and the volume are roughly the same thing, up to a constant that will be recalled below.

1.7 The dual Gosset polytopes

Combinatorially, the polytopes $P^n$ are dual to the Gosset polytopes listed in the last column of Table 2 and discovered by Gosset [Reference Gosset16] in 1900 (see [Reference Everitt, Ratcliffe and Tschantz14]). Every Gosset polytope is a Euclidean polytope with regular facets, whose isometry group (which is the same as $ {\mathrm {Isom}}(P^n)$ ) acts transitively on the vertices. The regular facets of the Gosset polytope are of two types: some $(n-1)$ -simplexes (dual to the real vertices of $P^n$ ) and some $(n-1)$ -octahedra (dual to the ideal vertices of $P^n$ ). A k-octahedron, here, is the regular polytope dual to the k-cube (sometimes also called k-orthoplex).

We will describe a colouring of $P^n$ as a colouring of the vertices of the dual Gosset polytope, where we require, of course, that two vertices adjacent connected by an edge must have distinct colours (so only the 1-skeleton of the dual Gosset polytope is important at this stage). We would like to find some colouring with a reasonably small number of colours, and possibly a high degree of symmetry: We are confident that some natural choices should arise from the exceptional properties of $P^n$ and their dual Gosset polytopes, and this is indeed the case as we will see.

We now analyse the polyhedra $P^3, \ldots , P^8$ individually. For each $P^n$ , we define a colouring and study the resulting hyperbolic manifold $M^n$ .

1.8 The manifold $M^3$

The hyperbolic polyhedron $P^3\subset {\mathbb {H}}^3$ is the right-angled double pyramid with triangular base shown in Figure 4. The three vertices of the triangular base are ideal, while the two remaining vertices are real. Each of the six faces F of $P^3$ is a triangle with a right-angled real vertex and two ideal vertices.

Figure 4

The polyhedron $P^3$ is a right-angled bipyramid with three ideal vertices along the horizontal plane and two real ones (top and bottom in the figure).

The dual Gosset polytope is a triangular prism, whose faces are two base equilateral triangles and three lateral squares. Its 1-skeleton is shown schematically in Figure 5. It can be coloured with three colours in a unique way (shown in the figure) up to isomorphism. Therefore, $P^3$ has a unique 3-colouring up to isomorphism. The polyhedron cannot be coloured with less than three colours.

Figure 5

The 1-skeleton of the triangular prism has a unique 3-colouring up to isomorphism, shown here.

We equip $P^3$ with this 3-colouring. This produces a hyperbolic 3-manifold $M^3$ , tessellated into $2^3=8$ copies of $P^3$ .

The link of each ideal vertex v of $P^3$ is a square C, that is dual to a square face of the Gosset prism. We see from Figure 5 that C is 3-coloured: two opposite edges of C have distinct colours, and the other two opposite edges have the same colour. By Corollary 8, the counterimage of C consists of a single (because $2^{3-3}=1$ ) torus cusp section in $M^3$ . The hyperbolic manifold $M^3$ has therefore three cusps, one above each vertex v of P.

Using Sage, we have calculated the Betti numbers of $M^3$ :

$$ \begin{align*}b_0 = 1, \qquad b_1 = 3, \qquad b_2 = 2.\end{align*} $$

We get of course, $\chi (M^3)=0$ .

1.9 The manifold $M^4$

The hyperbolic polytope $P^4\subset {\mathbb {H}}^4$ is fully described in [Reference Potyagailo and Vinberg30, Reference Ratcliffe and Tschantz32], and we refer to these sources for more details. It has 10 facets, each isometric to $P^3$ . It has also five real vertices and five ideal vertices.

The dual Gosset polytope $0_{21}$ is the 4-dimensional rectified simplex. That is, it is the convex hull of the midpoints of the 10 edges of a regular 4-dimensional simplex. Its 10 vertices may be seen in $ {\mathbb {R}}^5$ as the points obtained by permuting the coordinates of $(0,0,0,1,1)$ . Two such vertices are adjacent if they differ only in two coordinates. The Gosset polytope $0_{21}$ has 10 facets; of these, five are regular tetrahedra (created by the rectification) dual to the finite vertices of $P^4$ , and five are regular octahedra (the rectified facets of the original regular 4-simplex) dual to the ideal vertices of $P^4$ .

A convenient orthogonal plane projection of the 1-skeleton of $0_{21}$ is shown in Figure 6. We assign to $0_{21}$ , and, hence, to $P^4$ , the 5-colouring depicted in Figure 7. This produces a hyperbolic 4-manifold $M^4$ , tessellated into $2^5 = 32$ copies of $P^4$ . We have $\chi (M^4) = 32/16 = 2$ .

Figure 6

The orthogonal projection of the 1-skeleton of the rectified simplex $0_{21}$ on the plane in $ {\mathbb {R}}^5$ generated by $(1, \epsilon , -\epsilon , -1, 0)$ and its cyclic permutations, where $\epsilon =(\sqrt 5 -1)/2 = 2\cos (2\pi /5)$ is the positive root of $\epsilon ^2 + \epsilon -1$ . The image of the vertex $(0,0,0,1,1)$ is indicated as $00011$ , and so on. Some edges are superposed along the projection, so two vertices that are connected by an edge on the plane projection may not be so in $0_{21}$ . To clarify this ambiguity, we have chosen a blue vertex and painted in red the six vertices adjacent to it, in two cases (all the other cases are obtained by rotation).

Figure 7

A 5-colouring of the 1-skeleton of $0_{21}$ and hence of $P^4$ .

The polytope $P^4$ has five ideal vertices $v_1,\ldots , v_5$ . Each $v_i$ is dual to the octahedral facet of $0_{21}$ contained in the coordinate hyperplane $x_i=0$ , whose six vertices in Figure 6 are precisely those with $x_i=0$ . The case $i=1$ is shown in Figure 8. We can see on the figure that the octahedron is 5-coloured. The other four octahedra are obtained from this one by rotating the plane projection diagram, and they are also 5-coloured.

Figure 8

An octahedral facet of $P^4$ . This is a subgraph of the 1-skeleton in Figure 7. Some edges are superposed.

We have discovered that the link of each ideal vertex of $P^4$ is a 5-coloured cube C. By Corollary 8, the counterimage of C consists of a single (since $2^{5-5}=1$ ) toric cusp section in $M^4$ . Therefore, the hyperbolic manifold $M^4$ has five cusps overall, one above each ideal vertex of $P^4$ .

Using Sage, we have calculated the Betti numbers of $M^4$ :

$$ \begin{align*}b_0 = 1, \qquad b_1 = 5, \qquad b_2 = 10, \qquad b_3 = 4.\end{align*} $$

We get $\chi (M^4)=2$ again.

1.10 The manifold $M^5$

The hyperbolic polytope $P^5\subset {\mathbb {H}}^5$ is fully described in [Reference Potyagailo and Vinberg30, Reference Ratcliffe and Tschantz33], and we refer to these sources for more details. It has 16 facets, each isometric to $P^4$ . It also has 16 real vertices and 10 ideal vertices. Every real vertex is opposed to a facet.

The dual Gosset polytope $1_{21}$ has 16 vertices. We can represent these in $ {\mathbb {R}}^5$ as the vertices $(\pm 1, \pm 1, \pm 1, \pm 1, \pm 1)$ with an odd number of minus signs. Two vertices are connected by an edge if they differ only in two coordinates. The Gosset polytope $1_{21}$ has 26 facets; of these, 16 are regular 4-simplexes dual to the finite vertices of $P^5$ , and 10 are regular 4-octahedra dual to the ideal vertices of $P^5$ .

A convenient planar projection of its 1-skeleton is shown in Figure 9. We assign to $1_{21}$ , and, hence, to $P^5$ , the 8-colouring depicted in Figure 10. This produces a hyperbolic 5-manifold $M^5$ tessellated into $2^8 = 256$ copies of $P^5$ .

Figure 9

The orthogonal projection of the 1-skeleton of $1_{21}$ on the plane spanned by the vectors $(\sqrt 2, \sqrt 2, 2-\sqrt 2, 2-\sqrt 2, 0)$ and $(2-\sqrt 2, \sqrt 2-2, \sqrt 2, -\sqrt 2, 0)$ . The string $\pm \pm \pm \pm \pm $ indicates the projection of the vertex $(\pm 1, \pm 1, \pm 1, \pm 1, \pm 1)$ . Some edges are superposed along the projection, so two vertices that are connected by an edge on the plane projection may not be so in $1_{21}$ . To clarify visually, for this ambiguity, we have chosen a blue vertex and painted in red the 10 vertices adjacent to it, in two cases (all the other cases are obtained by rotation).

Figure 10

The chosen colouring for $P^5$ .

The polytope $P^5$ has 10 ideal vertices. Each ideal vertex is dual to a 4-octahedral facet of $1_{21}$ contained in a hyperplane $x_i = \pm 1$ . We deduce then that there are two types of 4-octahedral facets, depicted in Figure 11. Eight facets are of the left type, and two of the right type (all obtained by rotating the graphs shown in the figure). The vertices of the facets of the first type inherit an 8-colouring, while those of the facets of the second type inherit a 4-colouring.

Figure 11

The ten 4-octahedral facets of $1_{21}$ are of two types. Eight are obtained by rotating the type shown on the left, and two by rotating the type shown on the right. These are subgraphs of the 1-skeleton in Figure 10. Some edges are superposed.

We have discovered that there are eight ideal vertices of the first type and two ideal vertices of the second type in $P^5$ . The link of an ideal vertex of the first type of $P^5$ is an 8-coloured 4-cube C, while the link of an ideal vertex of the second type is a 4-coloured 4-cube C. Note that four and eight are precisely the minimum and maximum number of colours on a 4-cube. By Corollary 8, the counterimage of C consists of a single (since $2^{8-8}=1$ ) toric cusp section in the first case, and of $2^{8-4} = 2^4 = 16$ toric cusp sections in the second case. Therefore, the hyperbolic manifold $M^4$ has $8\cdot 1 + 2\cdot 16 = 40$ cusps overall. The first eight cusps lie above the eight vertices of the first type, and the remaining 32 cusps lie above the two vertices of the second type, distributed as 16 above each.

Using Sage, we have calculated the Betti numbers of $M^5$ :

$$ \begin{align*}b_0 = 1, \qquad b_1 = 24, \qquad b_2 = 120, \qquad b_3 = 136, \qquad b_4 = 39.\end{align*} $$

We get of course, $\chi (M^5)=0$ .

1.11 The manifold $M^6$

The hyperbolic polytope $P^6\subset {\mathbb {H}}^6$ is fully described in [Reference Everitt, Ratcliffe and Tschantz14, Reference Potyagailo and Vinberg30], and we refer to these sources for more details. It has 27 facets, each isometric to $P^5$ . It also has 72 finite vertices and 27 ideal vertices. Every ideal vertex is opposed to a facet.

The dual Gosset polytope $2_{21}$ has 27 vertices. We can represent them in the affine hyperspace of $ {\mathbb {R}}^7$ of equation $x_1 + \cdots + x_6 - 3x_7 = -1$ , as the vertices

$$ \begin{align*}(-1,0,0,0,0,0,0), \quad (1,1,0,0,0,0,1), \quad (0,1,1,1,1,1,2)\end{align*} $$

and all the other vertices obtained from these by permuting the first six coordinates, so we get $6+15+6=27$ vertices in total (see [Reference Everitt, Ratcliffe and Tschantz14, Table 2]. Two vertices are connected by an edge if their Lorentzian product in $ {\mathbb {R}}^7$ with signature $(++++++-)$ is zero. The Gosset polytope $2_{21}$ has 99 facets; of these, 72 are regular 5-simplexes dual to the finite vertices of $P^6$ , and 27 are regular 5-octahedra dual to the ideal vertices of $P^6$ .

Both $P^6$ and $2_{21}$ have many remarkable properties. To mention one, the 1-skeleton of $2_{21}$ is the configuration graph of the 27 lines in a general cubic surface (see [Reference Coxeter11]). A planar projection of the 1-skeleton of $2_{21}$ taken from [Reference Coxeter11] is shown in Figure 12. In the figure, we see that there are nine lines that intersect in the centre, containing each three mutually nonadjacent vertices. This suggests that the polytope may have a nice 9-colouring.

Figure 12

An orthogonal projection of the 1-skeleton of $2_{21}$ on the plane. Some edges are superposed. There are nine lines intersecting in the centre of the figure, each line containing three vertices that are mutually non incident.

Inspired by the figure, we describe a 9-colouring for $2_{21}$ . The three vertices

$$ \begin{align*}(-1,0,0,0,0,0,0), \quad (1,1,0,0,0,0,1), \quad (1,0,1,1,1,1,2)\end{align*} $$

are mutually non connected by any edge since their Lorentzian products are not zero. We assign them the colour 1. If we permute cyclically the first six entries of these three vertices, we get five more triplets of mutually non connected vertices, and we assign them the colours $2,\ldots , 6$ . Finally, we assign the colours $7,8,9$ to the following remaining triplets of mutually disjoint vertices:

$$ \begin{gather*} (1,0,1,0,0,0,1), \quad (0,1,0,0,1,0,1), \quad (0,0,0,1,0,1,1); \\ (0,1,0,1,0,0,1), \quad (0,0,1,0,0,1,1), \quad (1,0,0,0,1,0,1); \\ (0,0,1,0,1,0,1), \quad (1,0,0,1,0,0,1), \quad (0,1,0,0,0,1,1). \end{gather*} $$

We equip $P^6$ with this 9-colouring. Each triple of facets with the same colour is called a triplet. The colouring produces a hyperbolic 6-manifold $M^6$ , tessellated into $2^9 = 512$ copies of $P^6$ . We have $\chi (M^6) = -512/8 = -64$ .

The polytope $P^6$ has 27 ideal vertices. Using our program in Sage [Reference Martelli39], we discover that the link of each of the 27 ideal vertices of $P^6$ is a 9-coloured 5-cube C. We show one explicit example. Every facet F of $P^6$ is opposite to an ideal vertex, which is incident precisely to those facets that are not incident to F. Correspondingly, every vertex v in $2_{21}$ is opposite to a 5-octahedral facet, whose vertices are precisely those that are not connected to v. The 5-octahedral facet opposite to the vertex $v=(-1,0,0,0,0,0,0)$ has the following vertices:

$$ \begin{gather*} (1,1,0,0,0,0,1),\! (1,0,1,0,0,0,1),\! (1,0,0,1,0,0,1),\! (1,0,0,0,1,0,1),\! (1,0,0,0,0,1,1), \\ (1,0,1,1,1,1,2),\! (1,1,0,1,1,1,2),\! (1,1,1,0,1,1,2),\! (1,1,1,1,0,1,2),\! (1,1,1,1,1,0,2). \end{gather*} $$

The two vertices that lie in the same column are not connected. Their colours are

$$ \begin{gather*} 1,\ 7,\ 9,\ 8,\ 6, \\ 1,\ 2,\ 3,\ 4,\ 5. \end{gather*} $$

All the nine colours are present. As we said above, using our Sage program, we discover that a similar configuration holds at every vertex v. Therefore, by Corollary 8, the counterimage of C consists of a single (since $2^{9-9}=1$ ) toric cusp section. We deduce finally that the hyperbolic manifold $M^6$ has 27 cusps, one above each vertex of $P^6$ . The Betti numbers of $M^6$ , calculated by our program, are:

$$ \begin{align*}b_0 = 1, \qquad b_1 = 18, \qquad b_2 = 183, \qquad b_3 = 411, \qquad b_4 = 207, \qquad b_5 = 26.\end{align*} $$

We get $\chi (M^6)=-64$ again.

1.12 The manifold $M^7$

The hyperbolic polytope $P^7 \subset {\mathbb {H}}^7$ is described in [Reference Everitt, Ratcliffe and Tschantz14, Reference Potyagailo and Vinberg30]. It has 56 facets, each isometric to $P^6$ . It also has 576 finite vertices and 126 ideal vertices.

The dual Gosset polytope $3_{21}$ has 56 vertices. We will discover below that the 56 vertices can be partitioned into 14 sets of four mutually disjoint vertices, called quartets. This partition will be induced from a colouring of $P^8$ , that is in turn easily described using octonions. The precise description of the partition is given below in Section 1.14.

We equip $P^7$ with the 14-colouring induced by the partition into 14 quartets. This produces a hyperbolic 7-manifold $M^7$ , tessellated into $2^{14}=16,384$ copies of $P^7$ .

The polytope $P^7$ has 126 ideal vertices. Using our Sage program, we discover that, similarly as with $P^5$ , there are two types of ideal vertices with respect to the chosen 14-colouring of $P^7$ . The first type consists of 112 vertices, and the second type only 14. The link of an ideal vertex of the first type is a 12-coloured 6-cube C, while the link of an ideal vertex of the second type is a 6-coloured 6-cube. Note that, as with $P_5$ , the numbers 6 and 12 are the minimum and maximum number of colours in a 6-cube. From Corollary 8, we deduce that $M^7$ has $14 \cdot 2^{14-6} + 112 \cdot 2^{14-12} = 4,032$ cusps overall.

Using Sage, we have calculated the Betti numbers of $M^7$ :

$$ \begin{align*}b_0 = 1, \ b_1 = 182, \ b_2 = 6321, \ b_3 = 41300, \ b_4 = 55139, \ b_5 = 24010, \ b_6 = 4031.\end{align*} $$

We get of course, $\chi (M^7)=0$ .

1.13 The manifold $M^8$

The hyperbolic polytope $P^8\subset {\mathbb {H}}^8$ is described in [Reference Everitt, Ratcliffe and Tschantz14, Reference Potyagailo and Vinberg30]. It has 240 facets, each isometric to $P^7$ . It also has 17,280 finite vertices and 2,160 ideal vertices.

The dual Gosset polytope $4_{21}$ has 240 vertices. This beautiful albeit complicated polytope can be described elegantly using octonions, much in the same way as the 4-dimensional 24-cell may be defined using quaternions. This viewpoint is crucial in this paper, so we introduce it carefully.

A 3-colouring for the 24-cell

To warm up, we start by recalling that the 24 vertices of the 24-cell are the quaternions

$$ \begin{align*}\pm 1, \ \pm i,\ \pm j,\ \pm k,\ \tfrac 12 (\pm 1 \pm i \pm j \pm k).\end{align*} $$

Two such vertices are adjacent along an edge if and only if their Euclidean scalar product is $\frac 12$ (we identify the quaternions space with the Euclidean $ {\mathbb {R}}^4$ , as usual). Every vertex is adjacent to eight other vertices.

We can assign three colours to the 24 vertices, by subdividing them into three sets of eight vertices each, that we call octets. These are:

1. (1) $ \pm 1, \pm i, \pm j, \pm k$ ;

2. (2) the elements $\tfrac 12 (\pm 1 \pm i \pm j \pm k) $ with an even number of minus signs;

3. (3) the elements $\tfrac 12 (\pm 1 \pm i \pm j \pm k) $ with an odd number of minus signs.

The scalar product of two vertices lying in the same octet is an integer, so it is never $\frac 12$ . Therefore, this indeed defines a 3-colouring of the vertices of the 24-cell. Since the dual of a 24-cell is another 24-cell, we also get a 3-colouring of the facets of the dual 24-cell. This colouring was heavily employed in [Reference Kolpakov and Martelli26].

Here is an algebraic description of this 3-colouring that will be useful below. The 24 vertices of the 24-cell described above form a group called the binary tetrahedral group. The eight elements $ \pm 1, \pm i, \pm j, \pm k$ form a normal subgroup of index 3, called the quaternion group and indicated with the symbol $Q_8$ . The octets are just the three lateral classes of $Q_8$ .

Octonions

We now turn to the Gosset polytope $4_{21}$ and the octonions. For a nice introduction to the subject, we recommend [Reference Baez5]. We describe an octonion as a linear combination of $1, e_1, e_2, \ldots , e_7$ . We have $e_i^2 = -1$ , and the multiplication of two distinct elements, $e_i$ and $e_j$ , is beautifully described by the Fano plane shown in Figure 13. The Fano plane is the projective plane over ${\mathbb {Z}}_2$ , and it contains seven points and seven oriented lines: every line is a cyclically ordered triple of points as in the figure. For every $i\neq j$ , we have $e_ie_j = \pm e_k$ , where $e_k$ is the third vertex in the unique line containing $e_i$ and $e_j$ , and the sign is positive if and only if the line is cyclically oriented like $e_i\to e_j \to e_k$ . So, for instance, $e_1e_2 = e_4$ and $e_1e_6 = -e_5$ . In general, we get

$$ \begin{align*}e_{n} e_{n+1} = e_{n+3},\end{align*} $$

where the subscripts run modulo 7. The product is neither commutative nor associative: for every $i,j,k$ , we have

$$ \begin{align*}(e_ie_j)e_k = \pm e_i(e_je_k),\end{align*} $$

where the sign is $+1$ if and only if $e_i,e_j,e_k$ belong to the same line in the Fano plane (which is always the case, if $i,j,k$ are not distinct).

Figure 13

The Fano plane. The circle should be interpreted as a line.

A 15-colouring for the Gosset polytope $4_{21}$

The 240 vertices of the Gosset polytope $4_{21}$ are the octonions

$$ \begin{gather*} \pm 1,\ \pm e_1,\ \pm e_2, \ \pm e_3, \ \pm e_4, \ \pm e_5, \ \pm e_6, \ \pm e_7, \\ \tfrac 12(\pm 1 \pm e_{n} \pm e_{n+1} \pm e_{n+3}), \quad \tfrac 12(\pm e_{n+2} \pm e_{n+4} \pm e_{n+5} \pm e_{n+6}), \end{gather*} $$

where n runs modulo 7. Although we will not need this information, we mention that these are (up to rescaling) precisely the 240 nontrivial elements of smallest norm in the $E_8$ lattice.

We have 16 elements of type $\pm 1$ or $\pm e_i$ . Each line l in the Fano plane contains three vertices $e_n, e_{n+1}, e_{n+3}$ and determines 16 elements of type $\tfrac 12(\pm 1 \pm e_{n} \pm e_{n+1} \pm e_{n+3})$ and 16 elements of type $\tfrac 12(\pm e_{n+2} \pm e_{n+4} \pm e_{n+5} \pm e_{n+6})$ , so we indeed get $16 + 7 \cdot 16 + 7 \cdot 16 = 15 \cdot 16 = 240$ vertices overall. Two vertices of $4_{21}$ are connected by an edge if and only if their Euclidean scalar product is $\frac 12$ . One can check that every vertex is adjacent to 56 other vertices (its link is dual to $P^7$ that has 56 facets).

Similarly to what we did with the 24-cell, we can assign a 15-colouring to $4_{21}$ by subdividing the 240 vertices into 15 sets of 16 elements each; we call each such set a hextet. The hextets are:

1. (1) $\pm 1,\pm e_1, \pm e_2, \pm e_3, \pm e_4, \pm e_5, \pm e_6, \pm e_7$ ;

2. (2) the elements $\tfrac 12(\pm 1 \pm e_{n} \pm e_{n+1} \pm e_{n+3})$ and $\tfrac 12(\pm e_{n+2} \pm e_{n+4} \pm e_{n+5} \pm e_{n+6})$ with an even number of minus signs;

3. (3) the elements $\tfrac 12(\pm 1 \pm e_{n} \pm e_{n+1} \pm e_{n+3})$ and $\tfrac 12(\pm e_{n+2} \pm e_{n+4} \pm e_{n+5} \pm e_{n+6})$ with an odd number of minus signs.

The hextets of type (2) and (3) depend on the choice of n modulo 7. So we get $1+2\cdot 7 = 15$ hextets overall. One can check that the scalar product of two vertices lying in the same hextet is always an integer, so it is never $\frac 12$ . Therefore, we can assign the same colour to all the 16 members of a given hextet, and, hence, obtain a 15-colouring for $4_{21}$ as promised.

Algebraic description

There is an algebraic interpretation for the 15-colouring of $4_{21}$ analogous to that for the 3-colouring of the 24-cell. We warn the reader that some caution is needed when passing from quaternions to octonions: first, the product of octonions is notoriously nonassociative; second, contrary to a common mistake (see [Reference Conway and Smith10, Chapter 9] for a discussion), and as proved by Coxeter [Reference Coxeter12], the 240 vertices of $4_{21}$ are not closed under multiplication! Indeed, the product of the two vertices

$$ \begin{align*}\frac 12 (1+e_1+e_3+e_7) \cdot \frac 12(1+e_1+e_2+e_4) = \frac 12(e_1 + e_3+ e_4 + e_6)\end{align*} $$

is not a vertex. We could fix this via a single reflection that transforms the 240 vertices into a multiplicatively closed set (this is explained in [Reference Conway and Smith10, Section 9.2]), thus obtaining another isometric description of $4_{21}$ , but we do not really need this here, so we just keep them as they are. The only thing that we need here is that the 240 octonions are closed under left multiplication by each of the 16 elements in the hextet $S=\{\pm 1, \pm e_i\}$ , a fact that can be verified easily. The set S is closed under multiplication, but it is not a group since it is not associative. One can also verify that the left multiplication by each element of S preserves each hextet, and that this ‘action’ of S is free and transitive, in the sense that for very pair of distinct elements in a hextet, there is a unique element of S that sends the first to the second by left-multiplication.

Summing up, the 15 hextets that we have constructed are the orbits of the action of S by left-multiplication on the set of 240 vertices of $4_{21}$ . This is analogous to the 3-colouring of the 24-cell, where the three octets are the orbits of the action of the quaternion group $Q_8$ by left multiplication.

The manifold $M^8$

We equip $P^8$ with the 15-colouring just defined. This produces a hyperbolic 8-manifold $M^8$ , tessellated into $2^{15} = 32,768$ copies of $P^8$ . We have $\chi (M^8) = 2^{15}\cdot 17/2 = 27,8528$ .

The polytope $P^8$ has 2,160 ideal vertices. Using our Sage program, we discover a phenomenon that was already present with $P^5$ and $P^7$ . The ideal vertices are of two types: the first type contains 1,920 of them, and the second type 240. The link of a vertex of the first type is a 14-coloured 7-cube, while the link of a vertex of the second type is a 7-coloured 7-cube. As with $P^5$ and $P^7$ , we note that 7 and 14 are the minimum and maximum possible number of colours in a 7-cube. From Corollary 8, we deduce that $M^8$ has $240\cdot 2^{15-7} + 1,920 \cdot 2^{15-14} = 65,280$ cusps.

Using Sage, we have calculated the Betti numbers of $M^8$ :

$$ \begin{gather*} b_0 = 1,\quad b_1 = 365, \quad b_2 = 33,670, \quad b_3 = 583,290, \\ b_4 = 1,783,226, \quad b_5 = 1,346,030, \quad b_6 = 456,595, \quad b_7 = 65,279. \end{gather*} $$

We get $\chi (M^8)=278,528$ again.

1.14 Back to the polytope $P^7$

The polytope $P^7$ is a facet of $P^8$ . We think of $P^7$ as the facet dual to the vertex 1 of $4_{21}$ . As we already said, we equip $P^7$ with the colouring induced by the 15-colouring of $P^8$ just introduced.

We study this inherited colouring of $P^7$ . We think of $3_{21}$ as the link figure of the vertex $1$ of $4_{21}$ . The vertices of $4_{21}$ adjacent to $1$ are precisely those of the form

$$ \begin{align*}\tfrac 12(1 \pm e_{n} \pm e_{n+1} \pm e_{n+3}),\end{align*} $$

where n runs modulo 7. So we get $7\cdot 8 = 56$ vertices, as required. These vertices are contained in the hyperplane $x_0 = \frac 12$ , and their convex hull is $3_{21}$ . Two such vertices are connected by an edge in $3_{21}$ if and only if their scalar product is $\frac 12$ .

The 15-colouring of $4_{21}$ induces a 14-colouring of $3_{21}$ that partitions the 56 vertices into 14 sets of four vertices each, that we call quartets. Each quartet consists of the vertices $\frac 12(1 \pm e_n \pm e_{n+1} \pm e_{n+3})$ that share the same n and the same parity of the minus signs.

1.15 Volumes

We have constructed a colouring on each polytope $P^3, \ldots , P^8$ , and, hence, obtained a list of manifolds $M^3,\ldots , M^8$ . Table 3 summarises the colouring type of each polytope.

Table 3

The colouring type of each $P^3,\ldots , P^8$ .

The volumes of the hyperbolic manifolds $M^3, \ldots , M^8$ are listed in Table 4. In even dimension $n=2m$ , we have used the Gauss-Bonnet formula

$$ \begin{align*} {\mathrm{Vol}}(P) = (-2\pi)^m / (n-1)!! \cdot \chi(P).\end{align*} $$

In odd dimension, we have

$$ \begin{align*} {\mathrm{Vol}}(P^3) = L(2)\sim 0.91, \quad {\mathrm{Vol}}(P^5) = 7 \zeta(3) / 8 \sim 1.05, \quad {\mathrm{Vol}}(P^7) = 8L(4) \sim 7.92.\end{align*} $$

The symbols $\zeta $ and L indicate the Riemann and Dirichlet functions (see [Reference Everitt, Ratcliffe and Tschantz14, Reference Ratcliffe and Tschantz31]).

Table 4

The volume, the Euler characteristic and the number of cusps of each hyperbolic n-manifold $M^n$ .

1.16 The chosen colourings are all minimal

Although we will not need it, we mention the following fact.

1.17 The last nonzero Betti number

The Betti numbers $b_i$ and the number c of cusps of each $M^n$ were listed in Table 1. In all the cases, we have $b_{n-1} = c-1$ . Since $M^n$ is the interior of a compact manifold with c boundary components, in general, we must have $b_{n-1} \geq c-1$ . Therefore, here, the Betti number $b_{n-1}$ is as small as possible, given the number c of cusps.

2.1 States

Let $P \subset {\mathbb {X}}^n$ be a right-angled polytope in some space $ {\mathbb {X}}^n = {\mathbb {H}}^n, {\mathbb {R}}^n$ or $S^n$ . Following [Reference Jankiewicz, Norin and Wise18], a state is a partition of the facets of P into two subsets, that we denote as I (in) and O (out). Every facet thus inherits a status I or O.

Let P be equipped with a colouring with c colours. This induces a free action of ${\mathbb {Z}}_2^c$ on the set of all the states of P, in the following way. For every $j\in \{1,\ldots , c\}$ , the basis element $e_j$ acts by reversing the I/O status of every facet of P coloured by j, while leaving the status of the other facets unaffected. The action is free, hence, each orbit consists of $2^c$ distinct states.

2.2 Diagonal maps

As discovered in [Reference Jankiewicz, Norin and Wise18], the choice of a colouring and a state for a right-angled polytope P induce both a manifold M and a diagonal map $M \to S^1$ . (The construction of [Reference Jankiewicz, Norin and Wise18] is actually more general than this, but this interpretation is enough here.) Shortly:

We explain how this works. We already know how a colouring on P produces a manifold M, so it remains to explain how a state induces a map $f\colon M\to S^1$ .

The manifold M is tessellated into the $2^c$ polytopes $P_v$ with varying $v\in {\mathbb {Z}}_2^c$ . Since these are right-angled, the tessellation is dual to a cube complex C with $2^c$ vertices. We work in the piecewise-linear category (see [Reference Rourke and Sanderson34] for an introduction) and think of C as piecewise-linearly embedded inside M. If P has some ideal vertices (as it will be the case with all the polytopes $P^n\subset {\mathbb {H}}^n$ that we consider here), the complement $M \setminus C$ consists of open cusps, so there is a deformation retraction $r\colon M \to C$ . The cube complex C is a spine of M.

We indicate the vertex of C dual to $P_v$ simply as v, so the vertices of C are identified with ${\mathbb {Z}}_2^c$ . Here, v stands both for a vector of ${\mathbb {Z}}_2^c$ and a vertex of C.

The edges of C are dual to the facets of the tessellation: an edge of C connects v and $v+e_j$ if the dual facet F is coloured as j. So, in particular, there are k distinct edges connecting v to $v+e_j$ , where k is the number of facets in P coloured with j. In all the colourings that we have chosen for the polytopes $P^n$ , the number k does not depend on the colour j. The 1-skeleton of C for $P=P^3$ is shown in Figure 14.

Figure 14

The 1-skeleton of the dual cubulation C for $M^3$ , tessellated into eight polyhedra $P_v^3$ . The polyhedron $P^3$ is 3-coloured, and each colour is painted on two faces. The vertices of C are identified with ${\mathbb {Z}}_2^3$ . There are two edges connecting v and $v+e_j$ corresponding to the two faces in $P_v$ with the same colour j, for each $j=1,2,3$ .

Let now s be a fixed state for P. The state s induces an orientation on all the edges of C, in the simplest possible way: consider an edge connecting v and $v+e_j$ , where the j-th component of v is zero, that is $v_j=0$ . The edge is dual to some facet of the tessellation that is a precise identical copy of a facet F of P. If the status of F is O, we orient the edge outward, that is from v to $v+e_j$ , while if it is I, we orient it inward, from $v+e_j$ to v.

By construction, this orientation is coherent, that is on every square of C (and, hence, on any k-cube), the orientations of two opposite edges match as in Figure 15. This crucial fact allows us to apply the Bestvina-Brady theory [Reference Bestvina and Brady7]. We identify every k-cube of C with the standard k-cube $[0,1]^k \subset {\mathbb {R}}^k$ , so that the orientations on the edges of C match with the orientations of the axis in $ {\mathbb {R}}^k$ . The diagonal map on the standard k-cube is

$$ \begin{align*}[0,1]^k \longrightarrow S^1 = {\mathbb{R}}/{{\mathbb{Z}}}, \qquad x \longmapsto x_1+ \cdots + x_k.\end{align*} $$

The diagonal maps on the k-cubes of C match to give a well-defined continuous piecewise-linear map $C \to S^1$ . By precomposing it with the deformation retraction $r\colon M \to C$ , we finally get a diagonal map

$$ \begin{align*}f\colon M \to S^1.\end{align*} $$

This is the main protagonist of our construction. The diagonal map induces a homomorphism $f_* \colon \pi _1(M) \to \pi _1(S^1) = {\mathbb {Z}}$ . A dichotomy arises here:

Figure 15

Every square (and, hence, every k-cube) of the cubulation has its opposite edges oriented coherently as shown here.

The case (1) is not so interesting: All the examples that we construct here on the hyperbolic manifolds $M^n$ will be of type (2). In (2), since $\mathrm {Im} f_* = 2 {\mathbb {Z}}$ , one may decide to replace f with a lift along a degree-2 covering of $S^1$ to get a surjective $f_*$ .

This inefficient colouring is therefore of no use here.

Example 14. For the 3-coloured $P^3$ , we will choose the following state: For every pair of faces with the same colour, assign I to one face and O to the other (the choice of which face gets I and which face gets O will not affect the result much, as we will see). The resulting 1-skeleton of C is then oriented as in Figure 16. By Proposition 12, the homomorphism $f_*$ is not trivial.

Figure 16

We assign a state to $P^3$ , where faces with the same colours have opposite status. We get the orientation of the 1-skeleton of C shown here.

2.3 States and orbits

Let a right-angled $P\subset {\mathbb {X}}^n$ be equipped with a colouring and a state s. These determine a diagonal map $f\colon M \to S^1$ as explained above. We now would like to study f and how it depends on s. A powerful machinery is already available for this task and is described by Bestvina and Brady in [Reference Bestvina and Brady7].

We call s the initial state. Recall that s induces a coherent orientation of the edges of the dual cubulation C. It also induces a state on every polytope $P_v$ of the tessellation, as follows: Every facet F of $P_v$ is dual to an edge e of C, and, hence, inherits a transverse orientation from that of e. We assign the status O or I to F according to whether the transverse orientation points outwards or inward with respect to $P_v$ . It is easy to see that the state induced on $P_v$ is precisely $v(s)$ , the result of the action of v on the initial state s (as described in Section 2.1).

Summing up, the polyhedron $P_0$ has the initial state s, while $P_v$ inherits the state $v(s)$ for each $v\in {\mathbb {Z}}_2^c$ . The following proposition says that the states that lie in the same orbits produce equivalent diagonal maps.

2.4 Ascending and descending links

Let a right-angled $P\subset {\mathbb {X}}^n$ be equipped with a state s. Let Q be a Euclidean polytope combinatorially dual to P. When $P=P^3, \ldots , P^8$ , of course Q is a Gosset polytope. The state s induces a dual state on Q, that is the assignment of a status I or O to each vertex of Q.

If we remove the interiors of the $(n-1)$ -octahedral facets from $\partial Q$ (that correspond to the ideal vertices of P), we are left with a flag simplicial complex $\dot Q$ . This holds because P is right-angled, and, hence, simple; as a consequence, every face of Q is a simplex, except the $(n-1)$ -octahedral facets dual to the ideal vertices of P.

Following [Reference Bestvina and Brady7], we define the ascending link (respectively, descending link) as the subcomplex of $\dot Q$ generated by the vertices with status O (respectively, I). Since $\dot Q$ is a flag complex, these subcomplexes are determined by their 1-skeleton.

Let now P be equipped with both a colouring and a state. We get a manifold M and a diagonal map $f\colon M \to S^1$ . For every vertex $v\in {\mathbb {Z}}_2^c$ of the dual cubulation C, the link of v in C is precisely the simplicial complex $ {\mathrm {link}}(v) = \dot Q$ , and it inherits the state $v(s)$ of $P_v$ . The status of a vertex of $ {\mathrm {link}}(v)$ is I or O according to whether the corresponding oriented edge of C points inward or outward with respect to v. The ascending and descending links at v are denoted, respectively, as $ {\mathrm {link}}_\uparrow (v)$ and $ {\mathrm {link}}_\downarrow (v)$ , and they are disjoint subcomplexes of $ {\mathrm {link}}(v)$ .

The diagonal map $f\colon M \to S^1$ induces a homomorphism $f_*\colon \pi _1(M) \to {\mathbb {Z}}$ . We are interested in its kernel H.

2.5 Legal states

Following [Reference Jankiewicz, Norin and Wise18], a state s on P is legal if the ascending and descending links that it determines on the dual flag simplicial complex $\dot Q$ are both connected. The group ${\mathbb {Z}}_2^c$ acts on the set of all states of P, and an orbit is legal if it consists only of legal states. As noted in [Reference Jankiewicz, Norin and Wise18], Theorem 16 implies the following.

The chase of a legal orbit is the combinatorial game introduced in [Reference Jankiewicz, Norin and Wise18]. After introducing the rules of the game, the authors exhibited some legal orbits on two remarkable right-angled polytopes in $ {\mathbb {H}}^4$ , namely, the ideal 24-cell and the compact right-angled 120-cell [Reference Jankiewicz, Norin and Wise18], so providing the first algebraically fibring hyperbolic 4-manifolds. Here, we play with the right-angled polytopes $P^n$ and find some legal orbits on all of them. More than that, we find some even better kind of orbits in the cases $n=3,7,8$ , that we call 1-legal.

2.6 1-legal states

We extend the nomenclature of [Reference Jankiewicz, Norin and Wise18] by saying that a state s is 1-legal if its ascending and descending links are both simply connected. An orbit is 1-legal if it consists only of 1-legal state. Here is a consequence of Theorem 16.

2.7 The Euler characteristic check

In the following pages, we will double count the Euler characteristic of our manifolds as a safety check. If a colouring and a state on P produce a manifold M and a diagonal function $f\colon M \to S^1$ , we always have

(1) $$ \begin{align} \chi(M) = \sum_{v \in {\mathbb{Z}}_2^c} \big(1-\chi( {\mathrm{link}}_{\uparrow}(v)) \big). \end{align} $$

The same formula holds with the descending link $ {\mathrm {link}}_{\downarrow }(v)$ . We say that the integer $1-\chi ( {\mathrm {link}}_\uparrow (v))$ is the contribution of v to the Euler characteristic of M. Note that a contractible ascending link contributes with zero, while a k-sphere contributes with $(-1)^{k+1}$ .

We now construct a legal orbit on each individual polytope $P^3, \ldots , P^8$ . We have used a code written with Sage to analyse all these cases; both the code and the resulting data are available from [Reference Martelli39].

2.8 A 1-legal orbit for $P^3$

In the 3-colouring of $P^3$ , the facets are partitioned into three pairs. For every pair, we assign the status O to one facet and I to the other, arbitrarily. The orbit of this state s is independent of this choice and consists precisely of all the $2^3=8$ states that can be constructed in this way.

By direct inspection, we find that the eight states reduce to two up to isomorphism, and they are shown in Figure 17. The ascending and descending links are both either a triangle or two segments connected along an endpoint. In both cases, they are contractible. One can, in fact, verify that the conditions of [Reference Battista and Martelli6, Theorem 15] are satisfied, and, hence, the diagonal map $f\colon M^3 \to S^1$ can be smoothened to become a fibration (we will not need this here).

Figure 17

We exhibit a state by colouring the vertices in black and white, with black (white) corresponding to the status I (O). There are only two states in the orbit of $P^3$ up to isomorphism, and in both cases, the ascending and descending links are contractible: they are a triangle and two segments joined along an endpoint.

The ascending and descending links are simply connected, and, hence, the orbit is 1-legal. By Corollary 18, the kernel H of $f_*\colon \pi _1(M^3) \to {\mathbb {Z}}$ is finitely presented: It is the fundamental group of the surface fibre of the fibration $f\colon M^3 \to S^1$ .

The formula (1) holds since $\chi (M^3)=0$ and each contractible link contributes with zero to the sum.

2.9 A legal orbit for $P^4$

In the 5-colouring of $P^4$ , the facets are partitioned into five pairs. As in the previous case, we assign the statuses O and I arbitrarily to each pair. The orbit consists of all the states that assign distinct statuses to each pair. We get $2^5=32$ states.

By direct inspection, we find that these states reduce to four up to isomorphism, depicted in Figure 18. As shown in the figure, the ascending and descending links are always connected, so the orbit is legal. However, we note that the orbit is not 1-legal, since in the first case, both the ascending and descending links are circles. The first case occurs only in two of the 32 states.

Figure 18

We exhibit a state by colouring the vertices in black and white, with black (white) corresponding to the status I (O). There are only four states in the orbit of $P^4$ up to isomorphism. We show here the descending link, generated by the black vertices. In the first case, we get a circle, while in the other cases, we always get a contractible complex made of three triangles, two triangles and one tetrahedron and one triangle, respectively. The ascending links are of the same types.

In fact, one can verify that the ascending and descending links in the first case form a Hopf link in $S^3$ , if considered in the boundary of the Gosset polytope, and that the conditions of [Reference Battista and Martelli6, Theorem 15] are satisfied, so the diagonal function $f\colon M^4 \to S^1$ can be smoothened to a circle-valued Morse function with two index-2 critical points. This is the best that we can get in dimension 4, since no fibrations may occur on an even-dimensional hyperbolic manifold (we will not need these facts here, for more details see [Reference Battista and Martelli6]).

By Corollary 17, the kernel H of $f_*\colon \pi _1(M^4) \to {\mathbb {Z}}$ is finitely generated. The formula (1) holds since $\chi (M^4)=2$ and the two states of the first kind contribute each with 1, while all the others contribute with 0.

2.10 A legal orbit for $P^5$

In the 8-colouring for $P^5$ , the facets are partitioned into eight pairs. As in the previous cases, we assign the statuses O and I arbitrarily to each pair. The orbit consists of all the states that assign different statuses to each pair. We get $2^8 = 256$ states.

Each state produces a pair of ascending and descending links. Since these are flag simplicial complexes, they are determined by their 1-skeleta. Either using our program with Sage or by direct inspection, we find that the resulting 512 graphs reduce to only seven up to isomorphism. These seven graphs are those generated by the black vertices in Figure 19.

Figure 19

Every ascending or descending link for $P^5$ is isomorphic to one of the seven descending links shown here.

We can check by hand (or with our Sage program) that the first four graphs in the figure generate a contractible simplicial complex. The remaining three generate a simplicial complex that collapses, respectively, to $S^2$ , a wedge of three circles $\vee _3 S^1$ and $S^3$ . The complexes that collapse to $S^2$ and $\vee _3 S^1$ are shown in Figure 20. The complex that collapses to $S^3$ is actually homeomorphic to $S^3$ , and it is the boundary of a 4-octahedron, decomposed into 16 tetrahedra. It corresponds dually to an ideal vertex of $P^5$ .

Figure 20

Two simplicial complexes that collapse to $S^2$ and $\vee _3 S^1$ . The first is the boundary of an octahedron with two tetrahedra attached to a pair of opposite faces. The second consists of two tetrahedra and four edges joining them.

In all the cases, the simplcial complex is connected, so the orbit is legal. It is not always simply connected, so the orbit is not 1-legal. By Corollary 17, the kernel H of $f_*\colon \pi _1(M^5) \to {\mathbb {Z}}$ is finitely generated.

The formula (1) holds since $\chi (M^5)=0$ , and using Sage, we find that we get 32 occurrences of $S^2$ , eight of $\vee _3 S^1$ and eight of $S^3$ . Their contributions to the Euler characteristic are $32\cdot (-1) + 8\cdot 3 + 8 \cdot 1 = 0$ .

2.11 A legal orbit for $P^6$

In the 9-colouring for $P^6$ , the 27 facets are partitioned into nine triplets. As opposite to the previous cases, there does not seem to be a natural choice of a state here. However, a brute computer search shows that there are many legal orbits for $P^6$ . For instance, we may take s as the state where the first vertex in each triple listed in Section 1.11 is O and the remaining two are I. By using our Sage program, we find that the orbit of this state is legal. By Corollary 17, the kernel H of $f_*\colon \pi _1(M^6) \to {\mathbb {Z}}$ is finitely generated.

As we said, a computer search shows that many initial states s yield legal orbits. We could not find a single 1-legal orbit, but, admittedly, we have not checked all the possible initial states.

2.12 A 1-legal orbit for $P^7$

In the 14-colouring for $P^7$ , the 56 facets are partitioned into 14 quartets. We see $P^7$ as the facet of $P^8$ dual to the vertex $1$ in $4_{21}$ . We will define below a state for $P^8$ , and this will induce one s for $P^7$ in the obvious way: every facet of $P^7$ inherits the status of the adjacent facet in $P^8$ .

The state s inherited in this way turns out to be balanced with respect to the colouring, in the following sense: There is an even number $2m$ of facets sharing the same colour, and precisely half of them m are given the status I, and the other half m the status O. If a state is balanced, then every other state in the orbit is also balanced. The states that we have chosen for $P^3, P^4$ and $P^5$ are balanced: For these polyhedra, we have $m=1$ , and there was, in fact, a unique orbit of balanced states. Here, $m=2$ , so in each quartet, two facets receive the status I and two the status O.

The orbit of s consists of $2^{14}$ states, each contributing with an ascending and descending link. Using Sage, we are pleased to discover that the resulting $2^{15} = 32,768$ graphs reduce to only 106 up to isomorphism (this is probably due to the many symmetries of s that are inherited from $P^8$ ).

Using Sage, we also see that all the simplicial complexes generated by the 106 graphs are simply connected. Therefore, the orbit is 1-legal. With Sage, we have also checked (1). All the data can be found in [Reference Martelli39]. By Corollary 18, the kernel H of $f_*\colon \pi _1(M^7) \to {\mathbb {Z}}$ is finitely presented.

2.13 A 1-legal orbit for $P^8$

In the 15-colouring for $P^8$ , the 240 facets are partitioned into 15 hextets. How can we find a good initial state s for $P^8$ ? The numbers are overwhelming: the polytope $P^8$ has 240 facets, so there are $2^{240}$ possible states to choose from. Each orbit consists of $2^{15}$ distinct states, and we would like to find one orbit where all these $2^{15}$ states are legal, or even better, 1-legal. A brute force computer search is out of reach.

To define a legal state, we take inspiration from the 24-cell sitting inside quaternions space, since this has some strong analogies with the Gosset polytope $4_{21}$ sitting in octonions space as we already noticed above. We have already exploited this analogy when we fixed a convenient colouring for $P^8$ , and we do it again now to design a convenient state.

A state for the 24-cell

A nice legal state for the 24-cell was constructed in [Reference Battista and Martelli6] as follows. Recall that its 24 vertices are divided into three octets: these are $ \pm 1, \pm i, \pm j, \pm k$ , the elements $\tfrac 12 (\pm 1 \pm i \pm j \pm k) $ with an even number of minus signs and those with an odd number of minus signs.

Consider the group $G=\{\pm 1, \pm i\}$ and its action on the 24 vertices by left-multiplication. We can verify easily that each octet is invariant by this action, and is subdivided into two orbits of four elements each. We assign arbitrarily the status I to one orbit, and O to the other. The resulting state s is balanced (see the definition above), and also legal, as it was, in fact, already observed in [Reference Jankiewicz, Norin and Wise18]. The ascending and descending links are each homeomorphic to a G-invariant annulus as in Figure 21, so they are connected but not simply connected (the state is not 1-legal). The two G-invariant annuli form altogether a banded Hopf link in $S^3$ .

Figure 21

The ascending and descending links are both an annulus decomposed into 12 triangles, and altogether, they form two annuli that collapse onto a Hopf link in $S^3$ . The figure (taken from [Reference Battista and Martelli6]) shows the vertices of the 24-cell, with their 3-colouring (Blue / Red / Yellow) and their state (the vertices with a O status have an additional black circle). Only some edges of the 24-cell are shown for the sake of clarity: more edges should be added that connect each yellow vertex and its eight neighbours.

The orbit of s along the action of ${\mathbb {Z}}_2^3$ consists of the $2^3$ states obtained from s by reversing the I/O status of some octet. The geometry of the 24-cell is so extraordinary that these $2^3$ states are all isomorphic [Reference Battista and Martelli6]. In particular, the orbit is legal (but it is not 1-legal). The choice of which orbit is I and which is O inside each octet is irrelevant, since different choices lead to the same orbit.

A state for $4_{21}$

We now try to mimic the above construction for $4_{21}$ . The 240 vertices are partitioned into 15 hextets, that is $\pm 1,\pm e_1, \pm e_2, \pm e_3, \pm e_4, \pm e_5, \pm e_6, \pm e_7$ , the elements $\tfrac 12(\pm 1 \pm e_{n} \pm e_{n+1} \pm e_{n+3})$ and $\tfrac 12(\pm e_{n+2} \pm e_{n+4} \pm e_{n+5} \pm e_{n+6})$ with an even number of minus signs, and those with an odd number of minus signs, with the integer n varying modulo 7.

It is now natural to consider the quaternion group $Q = \{ \pm 1, \pm e_1, \pm e_2, \pm e_4\}$ and its ‘action’ on the 240 vertices of $4_{21}$ by left-multiplication. This is the analogue of the group $G=\{\pm 1, \pm i\}$ defined above, roughly because taking quaternions inside octonions looks like taking complex numbers inside quaternions. However, this is not really a group action because octonions are not associative, and, hence, we may have that $e_1(e_2(x)) \neq (e_1e_2)(x)$ . Therefore, some caution is needed.

We already know that the set $S=\{\pm 1, \pm e_1, \ldots , \pm e_7\}$ acts freely and transitively by left-multiplication on every hextet (that is, for every pair of elements in the hextet, there is a unique element in S whose left-multiplication sends the first to the second). We pick the following 15 base elements, one inside each hextet:

$$ \begin{align*}1,\quad 1+e_n+e_{n+1}+e_{n+3},\quad -1+e_n+e_{n+1}+e_{n+3},\end{align*} $$

where n runs modulo 7. We consider inside each hextet the eight elements obtained by left-multiplying the base element by the elements of Q. We assign to these eight elements the status O, and the status I to the remaining eight of the hextet. We have defined a balanced state s. The orbit consists of $2^{15}$ balanced states.

Remark 19. By analogy with the 24-cell, it would be tempting to guess that the $2^{15}$ states in the orbit are all isomorphic, and maybe that the ascending and descending links are homotopic to two copies of $S^3$ linked in $S^7$ . This is, however, impossible by a Euler characteristic argument, due to the fact that the 24-cell has $\chi = 1$ , while $\chi (P^8) = 17/2$ is much bigger. In general, one should not push the analogies too far: The situation is intrinsically more complicated here. We will come back to this point below.

Using our Sage code, we have determined the ascending and descending links of each of the $2^{15} = 32,768$ states. Note that each graph can have up to 240 vertices, and we have $2^{16} = 65,536$ graphs to analyse. Luckily, these graphs reduce up to isomorphism to only 185. This is probably due to the extraordinary symmetries of both the colouring and the state that we have chosen for $P^8$ . Our Sage program says that each of the 185 simplicial complexes generated by these graphs is connected and simply connected. Therefore, the orbit is 1-legal. By Corollary 18, the kernel H of $f_*\colon \pi _1(M^8) \to {\mathbb {Z}}$ is finitely presented.

Remark 20. We also checked (1). Both sides give (quite reassuringly) the same number $278,528$ . The formula (1) also explains a fact we alluded to in Remark 19. Since $\chi (P^8) = 17/2$ , the average contribution to the Euler characteristic of an ascending link is $17/2$ , which is a relatively big number if compared to the Euler characteristic of the other polytopes considered above. Referring to Remark 19, we note that it is certainly impossible that all the ascending links be $S^3$ , since their individual contribution would be 1. The ascending links that we find with Sage are indeed quite complicated (they are typically some wedges of spheres of various dimensions $k\geq 2$ ), much more than those discovered with $P^3,\ldots , P^7$ . They can be found in [Reference Martelli39].

2.14 The restriction of $f$ to the cusps

In our analysis, we have briefly mentioned the fact that when $n=3,4$ , the chosen orbits satisfy the conditions of [Reference Battista and Martelli6, Theorem 15] and, hence, $f\colon M^n \to S^1$ can be smoothened to become a perfect smooth circle-valued Morse function (for $n=3$ , this is a fibration).

One may wonder if this is also the case when $n\geq 5$ , and, indeed, this was our hope at the beginning of our study: It would be extremely interesting to find a fibration on an odd-dimensional hyperbolic manifold of dimension 5 or 7. We show that this is not the case, for any possible choice of initial state, a serious obstruction being the restriction of f to the cusps of $M^n$ .

After writing a first draft of this paper, we found a fibration for $M^5$ with a more elaborated construction that overcomes this problem (see [Reference Italiano, Martelli and Migliorini17]).

2.15 The geometrically infinite coverings

For every $n=3, \ldots , 8$ , the kernel of $f_*\colon \pi _1(M^n) \to \pi _1(S^1) = {\mathbb {Z}}$ is a normal subgroup $H\triangleleft \pi _1(M^n)$ of infinite index. We summarise our discoveries.

The limit set is the whole sphere because H is normal in $\pi _1(M^n)$ and $M^n$ has finite volume. In particular, the hyperbolic n-manifold

$$ \begin{align*}\widetilde M^n = {\mathbb{H}}^n/H\end{align*} $$

that covers $M^n$ is geometrically infinite. The dimension $n=4$ was investigated in [Reference Battista and Martelli6]. Here, we are particularly interested in the dimensions $n=5,\ldots , 8$ .

Recall that a group H is of type $\mathrm {F}_m$ if there exists a $K(H,1)$ with finite m-skeleton [Reference Bestvina and Brady7]. If H is $\mathrm {F}_m$ , the Betti number $b_m(H)$ is obviously finite.