A small cusped hyperbolic 4-manifold

By gluing some copies of a polytope of Kerckhoff and Storm's, we build the smallest known orientable hyperbolic 4-manifold that is not commensurable with the ideal 24-cell or the ideal rectified simplex. It is cusped and arithemtic, and has twice the minimal volume.


Introduction
There is a natural interest in hyperbolic manifolds of low volume, and this note addresses dimension four; see the survey [17].
Hyperbolic manifolds in the paper are understood to be complete and of finite volume.Two manifolds (or orbifolds) are commensurable if they are both finitely covered by a third manifold (or orbifold).Recall also that the generalised Gauss-Bonnet formula relates the volume of a hyperbolic 4-manifold M to its Euler characteristic χ(M ) as follows: The smallest known closed hyperbolic 4-manifolds are non-orientable and have χ = 8 [3,13].These and the few other explicit examples of closed manifolds that we could find in the literature [4,6,14] are tessellated by 2π/k-angled 120-cells for k = 3, 4, 5, and are arithmetic and commensurable [9].
We find here another commensurability class of low-volume hyperbolic 4-manifolds: Theorem.There exists an orientable, cusped, arithmetic, hyperbolic 4-manifold M with χ(M ) = 2, not belonging to any of the commensurability classes above.
The manifold M is commensurable with a Coxeter polytope Q belonging to a continuous family of hyperbolic 4-polytopes discovered in 2010 by Kerckhoff and Storm [10].Notably, classes (1) and (3) are represented by other Coxeter polytopes of the family, and the examples in (4) are hybrids of manifolds in (3) and (2).Our method applies to two additional Coxeter polytopes of the family, but giving less interesting examples from the viewpoint of this paper (see Section 2.4).
We build M explicitly, by gluing together some copies of a bigger polytope P of Kerckhoff and Storm's (Q is a quotient of P ).The results in [18,24,23] are obtained in the same spirit, by gluing polytopes of the Kerckhoff-Storm family as well.
The polytope P has volume (2/5) • (4π 2 /3) and octahedral symmetry.It has a few 2-faces with dihedral angle 2π/3, while the remaining ones are right angled.These are the main features of P that will be exploited.
2010 Mathematics Subject Classification.57M50.The author has been partially supported by the SNSF "Ambizione" grant PZ00P2-193559.As a first step, we "kill" in a natural way the 2π/3 angles by gluing in pairs some facets of 5 copies of P , to get a particularly symmetric hyperbolic manifold X with right-angled corners.These objects have been fruitfully used in four-dimensional hyperbolic geometry in the very last years [16,19,20,1].
A study of X and its symmetries then allows us to build M as follows.The facets of X are hyperbolic 3-manifolds with geodesic boundary (the corners), and are divided in two types.The corners, some punctured surfaces, are always the intersection of two facets of different type.We then close X up via two commuting, fixed-point free, isometric involutions of ∂X as gluing maps: one for each type of facet.
By construction, M is commensurable with the orbifold Q. Being Q arithmetic, Maclachlan's work [15] allows us to distinguish its commensurability class.One similarly gets a few non-orientable manifolds with χ = 2 in the same class (Remark 2.4).
Structure of the paper.The polytopes Q and P are introduced in Section 1, while the manifolds X and M are built in Section 2.
Acknowledgements.I am grateful to Leone Slavich for several discussions on the topic.

The polytopes
We introduce here two polytopes from [10]: the Coxeter polytope Q in Section 1.1, and the bigger polytope P in Section 1.2.The commensurability class of Q is distinguished in Proposition 1.1.
We refer the reader to the book [27] for the general theory of hyperbolic Coxeter polytopes and reflection groups, including arithmeticity.
Recall that the path graph with n − 1 vertices is a Coxeter diagram for the symmetric group S n [27, Table 1].In the following sections, we will be interested in the edge I of Q corresponding to the reflection subgroup (1) and its vertices V, V ′ ∈ I, called top and bottom vertices, corresponding to the reflection groups (2) We conclude the section distinguishing the commensurability class of Q.
Proof.By Maclachlan's work [15], the ramification set of the quadratic form associated to Γ is a commensurability invariant of arithmetic reflection groups.We compute this set as explained in [8].
For a quick description of the computation without proofs, one may consult [18,Section 4.5].Up to isometry of H 4 in its hyperboloid model, the bounding hyperplanes of the polytope Q ⊂ H 4 (coherently oriented) are dually represented by these spacelike vectors of R1,4 : (see [10,Section 4] for i 0 , i 1 , i 2 , and [10, Table 3] with t = t 3 = √ 7/7 for the remaining vectors).The Gram matrix of the corresponding vectors of unit Minkowski norm is: For some basis, a matrix representing the associated quadratic form over Q is: .
1.2.The bigger polytope.Recall Formula (1).By reflecting the Coxeter polytope Q around its edge I, we get a bigger polytope We resume here all the needed information on P , referring to [10,18] for proofs and further details (see especially [18,Proposition 3.16]).
A remarkable property of P is its antipodal symmetry.We call antipodal map the inversion a ∈ Isom(P ) through the centre of P (the midpoint of the edge I of Q).
The polytope P has 22 facets, partitioned up to symmetry of P into 3 sets: (E) the extremal facets, divided into top and bottom facets By definition, the names of the facets agree on whether they are contained in one of the two half-spaces bounded by H 3 ("top/bottom" and "upper/lower") or not ("central").The group G I preserves the two half-spaces, while a exchanges them.
With this conventions, all is well defined up to co-orientation of H 3 ("top/bottom" and "upper/lower") and permutation of the facet indices, thanks to the following lemma.Let A n < S n denote the alternating subgroup.
Lemma 1.2.The action of S 4 on the facet indices is induced by an isomorphism Moreover, the quotient P/Isom(P ) is isometric to the orbifold Q/Isom(Q).
Proof.The same argument of [24, Proposition 2.4] applies, replacing the words "upper tetrahedral facet" with "link of the top vertex V " (see also [18,Lemma 4.15]).The key point is that the unique non-trivial symmetry of Q (an order-two rotation corresponding to the reflection along the edge {i 0 , c} of the diagram in Figure 1) writes as the composition of a with an element of G I .Some facets of P are drawn in Figure 2, and some vertex links in Figure 3. From these two pictures and the respective captions, one recovers the whole combinatorics and geometry of P (like for instance the adjacency graph of the facets) by means of Lemma 1.2.We record here a few consequences that will be relevant for us.First, the top (resp.bottom) facets intersect each other with angle 2π/3, and share the top (resp.bottom . Second, the central, resp.half-height, facets are pairwise disjoint.Third, if two non-isometric facets intersect, then they are orthogonal.Fourth, the antipodal map acts on the central facets as follows: (3) a(C i,j ) = C k,l for all distinct i, j, k, l.

The construction
In this section we build the manifold M by gluing some copies of P .Recall that the top and bottom vertices of P have as link the spherical tetrahedron with dihedral angles 2π/3.The latter tiles S 3 in the regular honeycomb combinatorially equivalent to the triangulation of the boundary of the 4-simplex.Therefore we need at least 5 copies of P to build a manifold, and we will see that 5 is enough.
We proceed as follows.In Section 2.1, we build a manifold with corners X by gluing in pairs the extremal facets of 5 copies of P .In Section 2.2, we study X and its symmetries.In Section 2.3, we close X up by pairing the remaining facets via two symmetries of X, and conclude our proof.Finally, in Section 2.4, we give some additional information on how to apply the construction to two more polytopes of Kerckhoff and Storm's.
Before beginning with the construction, let us introduce here some terminology.A hyperbolic nmanifold with "pure" right-angled corners is a hyperbolic manifold X with boundary, locally modelled on the intersection in H n of two closed half-spaces bounded by orthogonal hyperplanes.The boundary ∂X is naturally stratified into maximal connected, totally geodesic, submanifolds: the (n − 1)dimensional facets have (possibly empty) totally geodesic boundary, while the (n − 2)-dimensional corners have no boundary.We say that ∂X is bicolourable if there are two unions of facets A and B such that ∂X = A ∪ B and the corners of X are the connected components of A ∩ B. This implies that each corner is contained in exactly two facets: one in A, and one in B. In particular, the facets of X are isometrically embedded hyperbolic manifolds with totally geodesic boundary (made of the corners of X).

Extremal gluing.
We build here X by gluing isometrically in pairs all the extremal facets of 5 copies of P : the top ones first and the bottom ones after.There is in fact a unique way to perform the first gluing, and a preferred one for the second.
Recall Formula (2).By reflecting the smaller Coxeter polytope Q ⊂ P around its top vertex Since G I < G V with index 5 = 5!/4!, the polytope P contains P and is tesselleted by 5 isometric copies of P .These are glued altogether around V (the centre of P ), in the pattern of the triangulation of S 3 induced by the boundary of the 4-simplex.
Note that we can alternatively think of P as built by pairing isometrically all the top facets of 5 abstract copies of P as described in Figure 4. We now build X from P by pairing all the bottom facets Figure 4.The complete graph K 5 , with its nodes and half-edges labelled to remind the "more abstract" construction of X and P , which follows here.To build X (resp.P ), we glue the extremal (resp.top) facets of 5 abstract copies P 0 , . . ., P 4 of P as follows.Consider the edge of K 5 joining the nodes i and j, where i < j.If i = 0 (blue edge), glue the facets E j and E ′ j (resp.the facet E j ) of P 0 to the corresponding facets of P j via the map induced by the identity of P .If instead i = 0 (black edge), glue the facets E j and E ′ j (resp.the facet E j ) of P i to the facets E i and E ′ i (resp.the facet E i ) of P j via the map induced by the reflection (ij) ∈ S 4 = G I of P .
. A schematic picture of the "more concrete" construction of X from P = P ∪ P 1 ∪ . . .∪ of these 5 copies of P as follows.Let ϕ : F → G be a pairing map between two top facets of two copies P x and P y of P .Then we want to glue a x (F ) to a y (G) via a y • ϕ • a x , where a z denotes the antipodal map of P z .This is done in Figure 4.
More concretely, let r i be the reflection along the top face E i of P , and define we glue the facet E ′ i of P to the facet r i (E ′ i ) of P i via r i , and, for i = j, the facet r i (E ′ j ) of P i to the facet r j (E ′ i ) of P j via the reflection r ij ∈ G I corresponding to (ij) ∈ S 4 (recall Lemma 1.2 and see Figure 5).

2.2.
The manifold with corners.We study here the space X just constructed.
Let H and C denote the union of the copies of the half-height and central facets of P in X, respectively.These facets are the unpaired ones.
Proposition 2.1.The resulting complex X is an orientable hyperbolic 4-manifold with pure rightangled corners and boundary ∂X = H ∪ C bicoloured by H and C.
Proof.We analyse the effect of the gluing on the vertex links of the copies of P along their extremal facets.
We refer to Figures 3 and 4. By construction, the gluing graphs of the type-k vertex links correspond to the subgraphs of K 5 spanned by 6 − k nodes (repeated twice to deal with the antipodal vertices).Precisely, the type-1 links are glued so as to tessellate two copies of S 3 (thus giving points in the interior of X) like the boundary of the 4-simplex, the type-2 links tessellate some copies of a halfsphere of S 3 (giving internal points of the facets of X), and the type-3 links tessellate some copies of a right-angled bigon of S 3 (giving points of the corners of X).The links of the ideal vertices are glued to form flat 3-manifolds with right-angled corners, stratum-preserving homeomorphic to [0, 1] 2 -bundles over S 1 (giving the cusps of X).
We have shown that the vertex links glue to form spherical or flat 3-manifolds with pure right-angled corners, so X is a hyperbolic 4-manifold with pure right-angled corners.It is orientable, being obtained from P via orientation-reversing gluing maps.
By construction ∂X = H ∪ C and H and C are unions of facets.The previous analysis also shows that the corners are contained in the intersection of copies of a half-height and a central facet of P (see the gluing of the type-3 vertex links).Therefore H and C define a bicolouring of ∂X, and this concludes the proof.
It is not difficult to show that H consists of two disjoint facets (an "upper" one and a "lower" one), while C is a facet.We omit the proof since this fact is not needed.
Recall now Lemma 1.2 and Formulas ( 1) and ( 2).In the sequel, we shall think of S 4 ⊂ S 5 as the stabiliser of 0 in the permutation group S 5 of {0, . . ., 4}.Lemma 2.2.Every symmetry of P is the restriction of an isometry of X.
Proof.This is true for a thanks for the "a-equivariance" of the gluing maps producing X, and for each σ ∈ S 4 ⊂ S 5 ∼ = Aut(K 5 ) since it induces a permutation (σ itself!) of the half-edge labels of the graph K 5 in Figure 4.
Since moreover G V ∼ = S 5 permutes the 5 copies of P in X, we can write: In the following section, we will implicitly use Lemma 2.2 and adopt the above convention.

2.3.
Half-height and central gluing.We are finally ready to build M from X.
Let r ij ∈ Isom(X) correspond to the reflection (ij) ∈ S 4 ⊂ Isom(P ), and recall that ∂X = H ∪ C. Now pick X, glue H to itself via h = a • r 12 and C to itself via c = a • r 34 , and call M the resulting space.
Proof.Observe that h, c ∈ Isom(P ) are both the composition of an inversion through a point with a reflection through a hyperplane containing the point, so their fixed-point set is a line.In both cases, this line joins two ideal vertices of P : , respectively, as it is easily checked by Lemma 1.2 and Formula (3).Therefore the fixed-point sets of h, c ∈ Isom(X) are contained in the interior of P ⊂ X.
Since H and C define a bicolouring of ∂X (Proposition 2.1) and the gluing maps h and c are two distinct, commuting, fixed-point free, isometric involutions of ∂X, it follows that M is a hyperbolic manifold: near the points corresponding to the internal points of the facets of X because h and c are fixed-point free isometric involutions of ∂X, and near the points corresponding to the corners of X because moreover h and c are distinct and commute (so each corner cycle has length 4 and trivial return map).
The manifold M is orientable because h and c are orientation reversing by Lemma 1.2 (recall that X is orientable by Proposition 2.1).
Finally, note that the gluing maps used to build M from the copies of P are induced by symmetries of P .Therefore M covers the orbifold P/Isom(P ) ∼ = Q/Isom(Q) (recall Lemma 1.2), and so M and Q are commensurable.
We conclude the paper with some additional information.
Remark 2.4.One builds a few non-orientable manifolds with χ = 2 in the same commensurability class, just by choosing as gluing maps h and c other pairs of distinct, commuting, isometric involutions of X with no fixed point on ∂X, without requiring they both reverse the orientation.Any two among a, a • r 12 , a • r 34 and a • r 12 • r 34 work.
The main difference is that now the pentagonal faces of P have dihedral angle π/2 (resp.2π/5) in place of 2π/3, therefore the link of the top vertex V tessellates S 3 like the boundary of the regular 16-cell (resp.600-cell) R in place of the simplex ∆.So, to build X, one has to glue 16 (resp.600) copies of P in place of 5.This time S 4 acts on P ⊂ X like a facet stabiliser in Isom(R), in place of Isom(∆) ∼ = S 5 .
The analogous construction gives an orientable manifold M with χ = 10 (resp.482).Moreover, in this case, M has an orientation-preserving isometric involution ι without fixed points, so the quotient M/ ι is a twice-smaller orientable manifold.One gets ι by composing the map induced by a with that induced by the antipodal map of R (a symmetry that ∆ does not enjoy!).

Figure 1 .
Figure 1.The Coxeter diagram of the reflection group Γ of Q.If two nodes are joined by a thin, thick, or dashed edge, then the two corresponding bounding hyperplanes meet with angle π/3, are tangent at infinity, are ultraparallel, respectively.There is no edge joining two nodes if the corresponding hyperplanes are orthogonal.
Figure 2. The top, upper, and central facets E 4 , H 4 , and C 12 of P .The ideal vertices are drawn as white dots, while the top vertex V and the type-2 and type-3 vertices are in blue, green and black, respectively.The black, red, and yellow edges have dihedral angle π/2, 2π/3, and arcos(−1/3), respectively.The red pentagons are 2π/3-angled ridges of P , while the other 2-faces in the picture are right angled.

1 Figure 3 .
Figure 3.The vertex links of P up to symmetry: three spherical tetrahedra for the finite vertices (from left to right, of type 1, 2 and 3), and a Euclidean parallelepiped for the ideal vertices.The black edges are right angled, while the red ones are 2π/3 angled.
tessellated by the copies of the facet t, resp.b, of Q, where E ′ i = a(E i ); (H) the half-height facets, divided into upper and lower facetsH 1 , H 2 , H 3 , H 4 and H ′ 1 , H ′ 2 , H ′ 3 , H ′ 4 ,tessellated by the copies of the facet u, resp.ℓ, of Q, where H ′ i = a(H i ); (C) the central facets C 12 , C 13 , C 14 , C 23 , C 24 , C 34 , tessellated by the copies of the facet e of Q, where C ij is the unique such facet of P that intersects the top facets E i and E j .The ideal vertices of P all lie in the ideal boundary of a hyperplane H 3 ⊂ H 4 [18, Proposition 3.19].