HHYPERBOLIC THREE-MANIFOLDSTHAT EMBED GEODESICALLY
BRUNO MARTELLI
Abstract.
We prove that every complete finite-volume hyperbolic 3-manifold M that may be tessellated into right-angled regular polyhedra (dodecahe-dra or ideal octahedra) embeds geodesically in a complete finite-volume con-nected orientable hyperbolic 4-manifold W , which is also tessellated into right-angled regular polytopes (120-cells and ideal 24-cells). If M is connected, thenVol( W ) < Vol( M ).This applies for instance to the Whitehead and the Borromean links com-plements. As a consequence, the Borromean link complement bounds geomet-rically a hyperbolic 4-manifold. Introduction
In this note we addess the following question.
Question 1.
Given a complete finite-volume hyperbolic n -manifold M , is there aconnected complete finite-volume orientable hyperbolic ( n + 1)-manifold W thatcontains M as a geodesic hypersurface?If the answer is positive, we say that M embeds geodesically . We note that W is assumed to be connected and orientable, but it makes perfectly sense to askQuestion 1 for disconnected and/or non-orientable hyperbolic manifolds M .Embedding geodesically a given hyperbolic manifold M is not a trivial task:among the uncountably many connected orientable hyperbolic surfaces, only count-ably many embed geodesically, and they form a dense subset of Teichm¨uller space,as proved by Fujii and Soma [5].In dimension n (cid:62) M which does not em-bed geodesically. On the other hand, only few explicit hyperbolic finite-volume3-manifolds M are known to embed, so the question is still wide open.The main object of this paper is to provide new examples. Recall that there areprecisely two right-angled hyperbolic platonic solids: the (non-compact) ideal octa-hedron and the (compact) right-angled dodecahedron. Inspired by [4], we say thata complete (possibly disconnected) finite-volume hyperbolic 3-manifold is dodeca-hedral or octahedral if it may be tessellated into right-angled dodecahedra or idealoctahedra, respectively. Dodecahedral manifolds are compact, while octahedralones have cusps.For instance, the Whitehead and Borromean link complements (see Fig. 1) areoctahedral manifolds and Thurston’s first famous examples of closed hyperbolicmanifolds fibering over S are dodecahedral, see [17]. We prove here the following. a r X i v : . [ m a t h . G T ] M a r BRUNO MARTELLI
Figure 1.
The Borromean rings are hyperbolic. Their comple-ment M is octahedral and hence embeds geodesically in a finite-volume hyperbolic four-manifold. Theorem 2.
Every (possibly disconnected) dodecahedral or octahedral hyperbolicthree-manifold M embeds geodesically in some connected complete finite-volumeorientable hyperbolic four-manifold W . In particular, the Whithehead and Borromean link complements embed geodesi-cally. We note that M may be disconnected: for instance, we can embed multiplecopies of both the Whitehead and Borromean link complements disjointly in a singleconnected complete finite-volume hyperbolic four-manifold W .We can also estimate the volume of W in terms of that of M . For simplicity werestrict ourselves to connected manifolds M . Theorem 3. If M is connected, there is a W with Vol( W ) ≤ · · Vol( M ) , Vol( W ) ≤ · Vol( M ) in the dodecahedral and octahedral case respectively. In our construction the hyperbolic four-manifold W is tessellated into four-dimensional right-angled hyperbolic regular polytopes, the and ,whose facets are octahedra and dodecahedra. Some of the techniques we use areborrowed from [8]. We prove Theorems 2 and 3 in Section 1 and then make furthercomments in Section 2. Acknowledgements.
The author warmly thanks Steven Tschantz for drawing thepictures in Fig. 5 and 6. The pictures in Fig. 1 and Fig. 4 are taken from WikipediaCommons: that in Fig. 1 lies in the Public Domain, those in Fig. 4 were producedusing the software
Stella by its author [20].1.
Manifolds with right-angled corners
We prove here Theorems 2 and 3.
YPERBOLIC THREE-MANIFOLDS THAT EMBED GEODESICALLY 3
MW' MW
Figure 2.
We place a right-angled 24- or 120-cell above each oc-tahedron or dodecahedron in M to get W (cid:48) , then we double it toget a manifold with corner W containing M in its interior.1.1. Manifolds with right-angled corners.
We now generalize both hyperbolicmanifolds and right-angled polyhedra in a single notion.We visualize hyperbolic space via the disc model D n and define P ⊂ D n as theintersection of D n with the positive sector x , . . . , x n (cid:62)
0. A hyperbolic manifoldwith (right-angled) corners is a topological n -manifold M with an atlas in P andtransition maps that are restrictions of isometries.The boundary ∂M is stratified into vertices, edges, . . . , and facets. Distinct strataof the same dimension meet at right-angles. Examples of such M are hyperbolicmanifolds with geodesic boundary and right-angled polytopes.A simple albeit crucial property is that if we glue two hyperbolic manifoldswith corners M and M along two isometric facets, the result is a new hyperbolicmanifold with corners.We prove here the following. Proposition 4.
Every octahedral or dodecahedral hyperbolic 3-manifold M em-beds geodesically in the interior of a connected, complete, finite-volume orientablehyperbolic four-manifold W with corners.Proof. We first consider the case M is orientable. The manifold M is tessellatedinto right-angled octahedra or dodecahedra: by placing a right-angled 24-cell or120-cell “above” each octahedron or dodecahedron of the tessellation, we obtain ahyperbolic four-manifold W (cid:48) with corners, whose boundary ∂W (cid:48) contains M as aconnected component. By doubling W (cid:48) along M we get a hyperbolic manifold withcorner W containing M in its interior (see Fig. 2).If M is connected, then W also is and we are done. Otherwise, each component M i of M is contained in a component W i of W , for i = 1 , . . . , k . We note thatevery 24-cell or 120-cell has a facet opposite to that contained in M which is stilla facet of W , hence ∂W i contains at least one octahedral or dodecahedral facet f i for each i = 1 , . . . , k .We only need to connect all these facets f i altogether. To do this, we observethat both the 24- and the 120-cell contain (at least) three pairwise non-incidentdistinct facets, and by doubling the polytope multiple times along one of these weget a bigger right-angled (not regular) polytope P with disjoint facets g , . . . , g k that are isometric to the f i . We attach P to the disconnected W by identifying f i BRUNO MARTELLI
M W MW MW P Figure 3. If M is not connected, we use another right-angledpolytope P to connect all the components W i .to g i for every i = 1 , . . . , k and thus get a connected manifold with corners that westill name W (see Fig. 3).Finally, if a component M i is non-orientable, we consider its orientable doublecovering ˜ M i and construct a ˜ W i containing ˜ M i as above. The orientation-reversingdeck involution ι : ˜ M i → ˜ M i extends uniquely to an orientation- preserving involu-tion ι : ˜ W i → ˜ W i that exchanges the two sides of ˜ W i \ ˜ M i and we set W i = ˜ W i / ι . (cid:3) Remark 5. If M is connected and tessellated into n right-angled dodecahedra orideal octahedra, the proof of Proposition 4 produces a W that is tessellated into2 n right-angled 120-cells or ideal 24-cells.1.2. Colorings.
We now show how to promote a hyperbolic manifold with cornersto a manifold without corners. We first note that a manifold with corners W isnaturally a hyperbolic orbifold, and as such it has a finite cover that is a hyperbolicmanifold by Selberg’s Lemma. However, we want a manifold cover that still contains M , and this is not guaranteed.We now construct some explicit manifold finite covers, following the coloring technique used by various authors in similar contexts, see for instance Vesnin[18, 19], Davis and Januszkiewicz [3], Izmestiev [6], and Kolpakov, Martelli, andTschantz [8].Let W be a hyperbolic manifold with corners. Consider a finite set S = { , . . . , k } of colors. Let a coloring λ of W be the assignment of a color c ∈ S at every facetof W , such that adjacent facets have different colors. In particular, the n facetsincident to a vertex all have distinct colors, hence if ∂W contains vertices we musthave k (cid:62) n . We suppose for simplicity that all the k colors in S are used by λ (ifnot, just take a smaller k ). Proposition 6.
Let W be a connected orientable hyperbolic manifold with corners.If W has a coloring λ with k colors, there is a connected orientable hyperbolic YPERBOLIC THREE-MANIFOLDS THAT EMBED GEODESICALLY 5
Figure 4.
The 120- and 24-cells. The picture shows the tessel-lation of S into 120 or 24 regular polyhedra and its layers: thefirst layer is the unbounded face containing ∞ in the picture, theadjacencies between the second and third layers can be seen in thefigure with little effort. manifold ˜ W that is tessellated into k identical copies of W . The induced map ˜ W → W is well-defined and is an orbifold cover of degree k .Proof. We construct ˜ W as follows. We consider the Z -vector space Z k , with canon-ical basis e , . . . , e k and finite cardinality 2 k . For every vector v ∈ Z k we define acopy vW of W , so that we get 2 k disjoint identical copies overall. For a facet F of W , we indicate by vF the corresponding facet in vW .For every v ∈ Z k and every facet F of W , we identify vF with ( v + e c ) F where c is the color of F . Since adjacent facets have distinct colors, one sees easily thatthe result of this gluing is a genuine hyperbolic n -dimensional manifold ˜ W withoutcorners. If W is connected and orientable, then ˜ W is. The manifold ˜ W is tessellatedinto 2 k copies of W and we get an orbifold covering ˜ W → W of degree 2 k . (cid:3) We can now refine Proposition 4 by estimating the number of colors needed for W . The number 43 is probably not optimal, but the important point is that it doesnot depend on M . Lemma 7. If M is connected, the boundary ∂W of the manifold W constructed inProposition 4 can be colored with at most or colors, depending on whether M is dodecahedral or octahedral.Proof. The proof depends heavily on the combinatoric of the 120- and 24-cells,shown in Fig. 4. We may suppose to simplify notations that M is orientable: in thenon-orientable case the proof is just the same. Let us first consider the case where M decomposes into dodecahedra. BRUNO MARTELLI
Figure 5.
A hyperbolic right-angled polyhedron Q obtained byattaching 8 right-angled dodecahedra to the same vertex v , which isthe barycenter of Q . The polyhedron Q has 42 faces: 24 pentagons,12 hexagons, and 6 octagons.A right-angled 120-cell Z is attached to each right-angled dodecahedron D in M to form W (cid:48) , and then W is the double of W (cid:48) along M . Starting from D as a “northpole”, and ending to its opposite “south pole”, the 120 facets of Z decompose intonine spherical layers consisting of 1, 12, 20, 12, 30, 12, 20, 12, and 1 dodecahedra.The 12 facets in the second layer are incident to D and are identified to thesecond-layer facets of the adjacent 120-cells (attached to the dodecahedra in M incident to D ). All the facets in the higher layers form the boundary ∂W . Notehowever that ∂W does not consist simply of dodecahedra, because some of them areglued together with dihedral angle π to form more complicate facets, and we nowneed to control this phenomenon. To understand the problem, imagine the lowerdimensional situation where one attaches right-angled dodecahedra to a surface thattessellates into right-angled regular pentagons. One checks easily that in this casethe faces of ∂W are right-angled pentagons and octagons, the latter partitionedinto four pentagons. The situation here is slightly more complicate but analogous.Fig. 4-(left) shows that a second-layer facet is incident to 5 third-layer facets and1 fourth-layer one. Each third-layered facet F is incident to three second-layeredones, all incident to a single vertex v . The facet F is attached to three similar third-layered facets of other 120-cells, which are also incident to v , and by repeating thiswe get that F is contained in a facet Q of ∂W consisting of eight dodecahedra, allsharing the same vertex v which lies in the center of Q . The polyhedron Q is shownin Fig. 5 and has 42 faces: 24 pentagons, 12 hexagons, and 6 octagons. YPERBOLIC THREE-MANIFOLDS THAT EMBED GEODESICALLY 7
Figure 6.
A hyperbolic right-angled polyhedron R obtained bymirroring a right-angled dodecahedron along a face.Every fourth-layered facet F is incident to a single second-layered one, and ishence attached to a single fourth-layered one in some adjacent 120-cell: the twoform altogether the polyhedron R shown in Fig. 6, which has 17 faces: 12 pentagonsand 5 hexagons. Summing up, the boundary ∂W is tessellated into polyhedra ofthree types Q , R , and D , with 42, 17, and 12 faces respectively. The adjacencygraph of these facets is a graph where every vertex has valence ≤
42, and hence itcan be colored with at most 43 colors. (Every finite graph with valence bounded by k can be ( k + 1)-colored, simply by ordering the vertices and then assigning themcolors in a sequence.) The proof is complete.We now turn to the non-compact octahedral case, which is a bit different. Aright-angled ideal 24-cell is attached to every ideal regular octahedron O in M toform W (cid:48) . The 24-cell layers into 1, 8, 6, 8, 1 octahedra, and it has the remarkableproperty that it can be colored with three colors, say yellow, red, and blue, eachassigned to 8 faces. The 8 octahedra in the odd layers have the same color, sayyellow. Each even layer contains four red and four blue octahedra.As above, the 8 octahedra in the second layer are identified to second-layeroctahedra in the adjacent 24-cells. Fig. 4-(right) shows that each second-layeroctahedron is incident to 3 third-layer (yellow) octahedra and to 1 (non-yellow)fourth-layer one.Every third-layer yellow octahedron is adjacent to four second-layer octahedra,and is therefore attached to four third-layer yellow octahedra in adjacent 24-cells.In contrast to the 120-cell case, these four adjacent octahedra may not intersecteach other; each is adjacent to three more yellow octahedra, so that the facet R in ∂W containing them may consist of an arbitrarily big number of octahedra! BRUNO MARTELLI
Luckily, all these octahedra are yellow, so we color R in yellow. These big yellowfacets in ∂W are pairwise not adjacent, so we need only to color the rest with othercolors.The remaining fourth-layer octahedra are just paired to adjacent ones producingmore facets. Each such facet F is obtained by doubling a right-angled octahedronalong a face, and it is easily seen to have 11 faces: 8 ideal triangles and three idealquadrilaterals. One such facet F does not have a natural color, and it is adjacentto 5 yellow facets and 6 more uncolored facets isometric to F . The adjacencygraph of the uncolored facets in ∂W has valence 6 and hence can be colored with 7(non-yellow) colors. Therefore ∂W can be colored with at most 8 colors overall. (cid:3) Proof of the main result.
We can now prove Theorems 2 and 3.
Proof of Theorem 2 . By Proposition 4 the manifold M embeds geodesically in theinterior of a connected complete orientable finite-volume hyperbolic four-manifold W with corners. By Proposition 6 there is a connected complete orientable finite-volume hyperbolic manifold ˜ W containing W and hence M geodesically. (cid:3) Proof of Theorem 3 . If M is connected, then W is tessellated into 2 n right-angled120-cells (ideal 24-cells), see Remark 5, and ∂W colors with at most 43 (8) colorsby Lemma 7. By Proposition 6 the manifold ˜ W is tessellated into 2 · n = 2 n (2 · n = 2 n ) right-angled 120-cells (ideal 24-cells).The volumes of the right-angled dodecahedron D , ideal octahedron O , 120-cell Z , and ideal 24-cell C , areVol( D ) = 4 . . . . , Vol( O ) = 3 . . . . , Vol( Z ) = 343 π = 111 . . . . , Vol( C ) = 43 π = 13 . . . . which gives Vol( Z )Vol( D ) ≤ , Vol( C )Vol( O ) ≤ . . This implies the result since 3 . · ≤ (cid:3) One can probably prove that a connected dodecaheral or octahedral three-manifold embeds geodesically also using the subgroup separability property forright-angled polytopes stated in [1, Theorem 3.1]. The disconnected case and thebounds on the volumes however do not seem to follow easily from such arguments.2.
Comments
A related interesting question, already studied in the literature, is the following:
Question 8.
Given a complete finite-volume orientable hyperbolic n -manifold M ,is there a complete finite-volume orientable hyperbolic ( n + 1)-manifold W withgeodesic boundary isometric to M ?Here both M and W can be disconnected. If the answer is positive, we usuallysay that M bounds geometrically . Theorem 2 has the following corollaries. Givena manifold M , we denote by 2 M the disconnected manifold that consists of twocopies of M . YPERBOLIC THREE-MANIFOLDS THAT EMBED GEODESICALLY 9
Corollary 9.
Let M be an octahedral or dodecahedral orientable manifold. Themanifold M bounds geometrically.Proof. Theorem 2 says that there is an orientable W containing M geodesically.By cutting W along M we get an orientable hyperbolic manifold with geodesicboundary 2 M . (cid:3) Corollary 10.
Let M be a connected octahedral or dodecahedral orientable mani-fold. If M has a fixed-point free orientation-reversing involution ι , then M boundsgeometrically.Proof. The manifold M embeds geodesically in an orientable W . By cutting W along M we get a W (cid:48) whose geodesic boundary consists of two copies of M . Wecan kill one boundary component by quotienting it with ι , and the result is anorientable W (cid:48)(cid:48) with geodesic boundary M . (cid:3) Corollary 11.
The Borromean rings complement M bounds geometrically.Proof. As one can check with SnapPy, the manifold M is the orientable double coverof the non-orientable octahedral manifold m N , in the Callahan –Hildebrand – Weeks cusped census [2]. Therefore M has such an involution ι . (cid:3) Hyperbolic manifolds that bound geometrically exist in all dimensions [11] andthe first three-dimensional examples are contained in [12, 13]. More examples werethen constructed in [8] with techniques similar to the ones used here. An explicitlink complement was produced in [15].The Borromean rings complement is tessellated into two ideal octahedra andhas volume 7.32772 . . .
When the first version of this paper appeared, this was thesmallest hyperbolic three-manifold known to bound geometrically: shortly after,it was shown by Slavich [16] that the figure-eight knot complement also boundsgeometrically, using the rectified simplex that was employed previously in [9]. Thefigure-eight knot complement is (together with its sibling) the smallest cusped ori-entable hyperbolic three-manifold and has volume 2 . . . . To the best of ourknowledge, the smallest compact one known has volume 68.8992 . . . and is tessel-lated into 16 right-angled dodecahedra, see [8].If M bounds geometrically a manifold W , then of course it also embeds geodesi-cally in the double of W . Although we suspect that the latter notion is strongerthan the former, we do not know a single example of hyperbolic orientable manifoldthat embeds geodesically but does not bound geometrically.A fundamental result of Long and Reid [10] shows in fact that “most” closedhyperbolic 3-manifolds do not bound geodesically: a closed hyperbolic 3-manifold M that bounds geometrically must have integral η -invariant η ( M ) ∈ Z . Note that η ( M ) = − η ( M ) and hence a mirrorable 3-manifold (ie one admitting an orientation-reversing isometry) has vanishing η -invariant, coherently with Corollary 10. We askthe following. Question 12.
Is there a dodecahedral 3-manifold with non-integral η -invariant? One such manifold M would embed geodesically, but would not bound. In factTheorem 2 could suggest that many hyperbolic 3-manifolds embed geodesically,whereas as we said only few of them bound.A natural question is whether Theorem 2 extends to the whole commensurabilityclasses of dodecahedral and octahedral manifolds: note that a manifold in thesecommensurability class need not to decompose into right-angle dodecahedra oroctahedra.Finally, we note that, from a more four-dimensional perspective, Theorem 2shows that 24-cell and 120-cell manifolds form a big set, rich enough to con-tain geodesically all octahedral and dodecahedral manifolds. Some of these four-dimensional hyperbolic manifolds were constructed in [14, 7]. References [1]
I. Agol – D. D. Long – A. W. Reid , The Bianchi groups are separable on geometricallyfinite subgroups , Ann. Math. (2001), 599–621.[2]
P. J. Callahan – M. V. Hildebrand – J. R. Weeks , A census of cusped hyperbolic -manifolds , Math. Comp. (1999), 321-332.[3] M. Davis – T. Januszkiewicz , Convex polytopes, Coxeter orbifolds and torus actions , DukeMath. J. (1991), 417–451.[4] E. Fominykh – S. Garoufalidis – M. Goerner – V. Tarkaev – A. Vesnin , A census oftetrahedral hyperbolic manifolds , arXiv:1502.00383 [5] M. Fujii – T. Soma , Totally geodesic boundaries are dense in the moduli space , J. Math.Soc. Japan (1997), 589–601.[6] I.V. Izmestiev , Three-dimensional manifolds defined by coloring a simple polytope , Math.Notes (2001), 340–346.[7] A. Kolpakov – B. Martelli , Hyperbolic four-manifolds with one cusp , Geom. & Funct.Anal. (2013), 1903–1933.[8] A. Kolpakov – B. Martelli – S. Tschantz , Some hyperbolic three-manifolds that boundgeometrically , Proc. Amer. Math. Soc. (2015), 4103–4111.[9]
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