Infinitely many virtual geometric triangulations
IINFINITELY MANY VIRTUAL GEOMETRIC TRIANGULATIONS
DAVID FUTER, EMILY HAMILTON, AND NEIL R. HOFFMAN
Abstract.
We prove that every cusped hyperbolic 3–manifold has a finite cover admitting in-finitely many geometric ideal triangulations. Furthermore, every long Dehn filling of one cuspin this cover admits infinitely many geometric ideal triangulations. This cover is constructed inseveral stages, using results about separability of peripheral subgroups and their double cosets,in addition to a new conjugacy separability theorem that may be of independent interest. Theinfinite sequence of geometric triangulations is supported in a geometric submanifold associatedto one cusp, and can be organized into an infinite trivalent tree of Pachner moves. Introduction
A hyperbolic 3–manifold M is called cusped if it is noncompact and has finite volume. Everycusped 3–manifold M admits a topological ideal triangulation: that is, a decomposition into finitelymany tetrahedra whose vertices have been removed, with faces identified in pairs by affine maps. A geometric ideal triangulation is a stronger notion, where each tetrahedron is isometric to the convexhull of 4 non-coplanar points in H , and where the tetrahedra are glued by isometry to give thecomplete hyperbolic metric on M . See Definition 2.2 for precise details. The focus of this paper ison geometric triangulations.The presence of a geometric triangulation makes the geometry of M much more accessible to bothpractical and theoretical study. On the practical side, geometric ideal triangulations are central to theworkings of the computer program SnapPy [7] that computes hyperbolic structures and rigorouslyverifies their geometric properties [20]. On the theoretical side, Thurston’s original proof of thehyperbolic Dehn filling theorem implicitly assumed the 3–manifold at hand admits a geometrictriangulation [29]. Similarly, Neumann and Zagier’s work on volume assumes that the complementof some closed geodesic in M admits a geometric triangulation [23].Despite the importance of geometric triangulations, the first part of the following conjecture hasbeen open for multiple decades. Conjecture 1.1.
Let M be a (finite volume) cusped hyperbolic –manifold. Then (1) (Folklore) M admits at least one geometric ideal triangulation. (2) M admits infinitely many geometric ideal triangulations. In the 1980s, Conjecture 1.1.(1) was widely believed to follow from the work of Epstein and Penner[10]. More precisely, the community believed that a geometric ideal polyhedral decomposition of M can always be subdivided to give a geometric ideal triangulation. It took time to realize thata naive refinement of the Epstein–Penner cell decomposition does not suffice; see the discussion ofconing in Section 2 for a description of some of the challenges. To our knowledge, the first recordof Conjecture 1.1.(1) in the literature is by Petronio [25, Conjecture 2.3], in 2000. See also Petronioand Porti for a useful account of the history [26].By contrast, Conjecture 1.1.(2) is new. We propose this tantalizing strengthening of the originalconjecture because searching for infinite and flexible sequences of geometric triangulations mightprovide a pathway to finding at least one. Indeed, our main result can be interpreted as a proof ofconcept that such a pathway exists in the context of finite covers and Dehn filling. Date : February 26, 2021.2020
Mathematics Subject Classification. a r X i v : . [ m a t h . G T ] F e b DAVID FUTER, EMILY HAMILTON, AND NEIL R. HOFFMAN
Passing to covers makes both parts of Conjecture 1.1 more amenable. Toward Part (1) of theConjecture, Luo, Schleimer and Tillmann showed that every cusped hyperbolic manifold M has afinite cover that supports a geometric triangulation [22]. We recall their proof strategy in Section 2,and incorporate several of their ideas in the proof of our theorems. Our main result, in the directionof Conjecture 1.1.(2), is the following. Theorem 1.2.
Let M be a cusped hyperbolic –manifold and A ⊂ M a horocusp. Then there is afinite cover (cid:99) M → M , such that A lifts to a cusp (cid:98) A ⊂ (cid:99) M , with the following properties: • (cid:99) M admits infinitely many geometric ideal triangulations. • For every sufficiently long slope s on ∂ (cid:98) A , the Dehn filling (cid:99) M ( s ) admits infinitely manygeometric ideal triangulations. Part of the interest of Theorem 1.2 comes from the fact that direct constructions of geometricideal triangulations are only known in special classes of manifolds. For example, Gu´eritaud provedthat certain well-studied triangulations of hyperbolic once-punctured torus bundles are geometric[13]. Futer extended Gu´eritaud’s method to hyperbolic 2–bridge link complements [13, Appendix].Gu´eritaud and Schleimer proved that if M is a generic multi-cusped hyperbolic manifold, thenlong Dehn fillings of M will admit geometric triangulations [15]. Ham and Purcell found geometricideal triangulations of highly twisted link complements, by adapting Gu´eritaud and Schleimer’sconstruction to some especially nice triangulations of fully augmented links [16].There have also been attacks on Conjecture 1.1.(1) that have attempted to subdivide a geo-metric polyhedral decomposition into geometric ideal tetrahedra. Wada, Yamashita, and Yoshida[30], building on work of Yoshida [32], described a sufficient condition on the dual 1–skeleton of apolyhedral decomposition to make such a subdivision possible. Sirotkina proved that a subdivisionis always possible if each 3–cell has at most six faces [28]. Goerner proved that a subdivision isalways possible if each 3–cell is a (not necessarily regular) ideal dodecahedron [12]. Champanerkar,Kofman, and Purcell have constructed interesting examples of link complements admitting a decom-position into regular ideal bipyramids, which can then be subdivided into geometric ideal tetrahedra[5, Theorem 3.5].To our knowledge, there is only one prior paper constructing infinitely many geometric triangula-tions on the same hyperbolic manifold. Dadd and Duan showed that the figure–8 knot complement,which decomposes into two regular ideal tetrahedra, supports infinitely many geometric triangu-lations [9]. Their proof strategy is very delicate, in that it does not extend to the figure–8 sistermanifold, which also decomposes into two regular ideal tetrahedra.Given a cusped manifold M , the topological Pachner graph of M is the graph whose vertices areisotopy classes of (topological) ideal triangulations, with edges corresponding to 2–3 moves and theirinverses. (See Definition 3.5 for the definition of a 2–3 move, and Figure 3 for an illustration.) The geometric Pachner graph of M is the induced subgraph whose vertices are geometric ideal triangu-lations. The infinitely many geometric triangulations found by Dadd and Duan [9] are organizedin the form of an infinite ray in a single component of the geometric Pachner graph of the figure–8knot complement. In a generic situation, the infinitely many geometric triangulations constructedin Theorem 1.2 contain an even greater amount of structure. Theorem 1.3.
Let M be a cusped hyperbolic –manifold containing a non-rectangular cusp. Thenthere exists a finite cover (cid:99) M → M such that the geometric Pachner graph of (cid:99) M contains a subgraphhomeomorphic to an infinite trivalent tree. The hypothesis on a non-rectangular cusp can be explained as follows. As we describe in Section 2,every non-compact end of M has the form A ∼ = T × [0 , ∞ ), where T is a torus endowed with aEuclidean metric that is well-defined up to similarity. We say that A is rectangular if the Euclideanmetric on T admits a rectangular fundamental domain, and non-rectangular otherwise. By the workof Nimershiem [24], the Euclidean structures on cusp tori of hyperbolic 3–manifolds form a dense NFINITELY MANY VIRTUAL GEOMETRIC TRIANGULATIONS 3 subset of the moduli space of M ( T ). Since rectangular tori represent a codimension-one slice of M ( T ), one can say that a generic cusped 3–manifold satisfies the hypotheses of Theorem 1.3.The infinite trivalent tree mentioned in Theorem 1.3 can be identified with the dual 1–skeletonof the Farey graph. See Definition 3.9 and Figure 4 for a review of the Farey graph; in brief, itsvertices correspond to slopes , or simple closed curves on a torus, and to rational numbers in RP .The branches of the trivalent tree of Theorem 1.3 limit to every point of RP . In particular, theinfinite sequence of geometric triangulations that we will construct can be chosen to approach anyrational or irrational foliation on a cusp torus of (cid:99) M . Manifolds with rectangular cusps satisfy aslightly weaker version of Theorem 1.3; see Remark 6.12 for details.1.1. Proof strategy.
Next, we outline the main ideas in the proofs of Theorems 1.2 and 1.3. Bothproofs use the same initial setup and general strategy. Since having a non-rectangular cusp simplifiesthe argument considerably, Theorem 1.3 will be proved first.Let M be a cusped hyperbolic 3–manifold. We will obtain geometric triangulations by subdivid-ing the canonical (Epstein–Penner) polyhedral decompositions of covers of M . Section 2 reviews theEpstein–Penner construction [10], emphasizing the way in which the canonical polyhedral decompo-sition P depends on the choice of neighborhoods of the cusps. That section also reviews the processof subdivision via coning and lays out a sufficient condition (involving an order on the cusps) thatensures P can be subdivided into geometric ideal tetrahedra. See Lemma 2.9, which is essentiallydue to Luo, Schleimer, and Tillmann [22], for details.In Section 3, we describe a particular feature of the canonical polyhedral decomposition P thatoccurs in the “generic” scenario when a manifold M has multiple cusps, one cusp A is chosen to besufficiently small, and there is a unique shortest path from A to the other cusps. In this situation,Gu´eritaud and Schleimer [15] show the canonical polyhedral decomposition P has only one or twocells poking into this cusp A . These cells fit together to form a submanifold called a drilled ananas (see Definition 3.3). In Lemma 3.6, we show that a drilled ananas admits an infinite sequence ofgeometric ideal triangulations. When A is a non-rectangular cusp, these triangulations are arrangedin a trivalent tree of 2–3 moves, as described in Theorem 1.3.To build covers of M satisfying the above-mentioned conditions, we will need to separate certainsubgroups and subsets of π ( M ) from group elements that cause undesired coincidences. Section 4reviews several key definitions and results about separability that are needed for our purposes. Thestrongest result that is needed for the proof of Theorem 1.3 is Theorem 4.4, due to Hamilton, Wilton,and Zalesskii [18], which provides separability of double cosets of peripheral subgroups.With this background in hand, we can begin to construct covers. Assuming that M has a non-rectangular cusp, Section 5 produces a sequence of finite covers (cid:99) M → ˚ M → M , with increasinglystrong properties. In particular, ˚ M contains a drilled ananas, while (cid:99) M has a polyhedral decomposi-tion (cid:98) P that can be subdivided via coning. It will follow that (cid:99) M admits an infinite trivalent tree ofgeometric ideal triangulations, establishing Theorem 1.3.1.2. New separability tools.
To prove Theorem 1.2, which handles hyperbolic manifolds withrectangular cusps and provides an additional conclusion about Dehn fillings, we need stronger sep-arability tools than what was previously available in the literature. The following new result maybe of independent interest. In the theorem statement, a peripheral subgroup of Γ = π ( M ) is asubgroup coming from the inclusion of a cusp. Theorem 1.4 (Conjugacy separation of peripheral cosets) . Let M = H / Γ be a cusped hyperbolic –manifold. Let H and K be (maximal) peripheral subgroups of Γ corresponding to distinct cuspsof M . Let g ∈ Γ be an element such that K is disjoint from every conjugate of gH . Then thereis a homomorphism ϕ : Γ → G , where G is a finite group, such that ϕ ( K ) is disjoint from everyconjugate of ϕ ( gH ) . Theorem 1.4 has the following topological interpretation. A maximal peripheral subgroup H ⊂ Γis the stabilizer of a horoball (cid:101) B ⊂ H . Given g ∈ Γ (cid:114) H , the coset gH is the set of all elements of DAVID FUTER, EMILY HAMILTON, AND NEIL R. HOFFMAN
Γ that move (cid:101) B to g (cid:101) B . Connecting these two horoballs is a geodesic arc (cid:101) β that projects to an arc β ⊂ M . We wish to find a finite cover (cid:99) M → M where the cusp corresponding to K lifts, and where every preimage of β connects distinct cusps. Theorem 1.4 provides such a cover, corresponding tothe subgroup (cid:98) Γ = ϕ − ◦ ϕ ( K ) that contains K but excludes every conjugate of gH .Several precursors of Theorem 1.4 appear in the recent literature on 3–manifold groups. Given aperipheral subgroup K and a single element g ∈ Γ that is disjoint from every conjugate of K , it isstraightforward to find a finite quotient that witnesses this disjointness [18, Lemma 4.5]. Given non-conjugate subgroups H and K , Chagas and Zalesskii find a finite quotient of Γ where their imagesare not conjugate [4]. Given a pair of non-conjugate peripheral subgroups H and K , Wilton andZalesskii use an argument of Hamilton to construct a finite quotient ϕ : Γ → G , such that non-trivialelements of ϕ ( H ) and ϕ ( K ) always lie in distinct conjugacy classes [31, Lemma 4.6].It is worth recalling the proof of the last result. First, take a hyperbolic Dehn filling M ( s )corresponding to a quotient Γ → Γ( s ), where K stays parabolic but non-trivial elements of H becomeloxodromic. In particular, the quotient of H is represented by loxodromic matrices of trace not equalto 2. Then, take a congruence quotient of the matrix group Γ( s ) in a matrix group over a finite ring(see Definition 4.6), where the traces of these loxodromic matrices can still be distinguished from2. Our contribution to this narrative is that we achieve even stronger separability for non-conjugateparabolic subgroups H and K , separating the image of K from the image of every conjugate of gH .The proof of Theorem 1.4 appears in Section 4, and uses a similar two-step method: first constructan appropriate Dehn filling, and then analyze the congruence quotients related to the Dehn filling. Aspart of this analysis, we apply tools from algebraic number theory, including a theorem of Hamilton[17, Corollary 2.5] (restated below as Proposition 4.9), to control the traces of an entire coset gH .Using the separability Theorem 1.4, we prove Theorem 1.2 in Section 6. If ˚ M is a cover of M containing a drilled ananas ˚ N , as above, we use the topological interpretation of Theorem 1.4 toconstruct two additional covers ( M → M → ˚ M where the ananas ˚ N lifts but most edges of thepolyhedral decomposition ( P connect distinct cusps. In particular, ( M has a drilled ananas ( N and apolyhedral decomposition ( P that can be subdivided into ideal tetrahedra via coning, which impliesinfinitely many geometric triangulations. Then, we build a cover (cid:99) M → ( M where the drilled ananas ( N has two distinct lifts. One of these lifts supports infinitely many geometric triangulations, whilethe other gets filled to obtain the Dehn filling conclusion of the theorem. The opening paragraphsof Section 6 outline this construction in much greater detail.1.3. Acknowledgements:
We thank Jessica Purcell, Saul Schleimer, and Henry Segerman for help-ful discussions about triangulations. We thank Ian Agol and Matthew Stover for enlightening dis-cussions about separating peripheral double cosets. Henry Wilton, who fielded our questions atseveral crucial points, deserves particular gratitude.This project was conceived when the first author visited Oklahoma State University in November2019, and reached a mature state when the first and third authors visited the University of Arkansasfor the Redbud Topology Conference in March 2020. We thank both universities for their hospitality.From there the collaboration proceeded over Zoom, with the second author joining at this virtualstage. The Redbud conference stands out in our memories as a last hurrah of in-person discussionand collaboration, for a significant time to come.During this project, Futer was partially supported by NSF grant DMS–1907708, while Hoffmanwas partially supported by Simons Foundation grant
Triangulations and polyhedral decompositions
This section reviews some standard definitions about hyperbolic manifolds and their polyhedraldecompositions and triangulations. Then, it proves Lemma 2.9 and Corollary 2.10, which will beour main ways to obtain a geometric triangulation from a polyhedral decomposition.
NFINITELY MANY VIRTUAL GEOMETRIC TRIANGULATIONS 5
For the remainder of this paper, the symbol M denotes a cusped, orientable hyperbolic 3–manifold.(Since Theorems 1.2 and 1.3 construct finite covers, no generality is lost in assuming that M isorientable.) We will use (cid:102) M to denote the universal cover of M , which is isometric to H . Otherdecorations, such as (cid:99) M and M , denote finite-sheeted covers of M . Definition 2.1.
Let M be a cusped hyperbolic manifold, and let (cid:98) f : (cid:99) M → M be a finite cover. Let A ⊂ M be an embedded submanifold. We say that A lifts to (cid:99) M if the inclusion map ι : A (cid:44) → M liftsto an inclusion (cid:98) ι : A (cid:44) → (cid:99) M . In this case, the image (cid:98) A = (cid:98) ι ( A ) is called a lift of A . The lift (cid:98) A formsonly one component of (cid:98) f − ( A ), and covers A with degree one.We remark that a lift is distinct from a path-lift. If γ ⊂ M is a (parametrized) closed curve basedat x , then γ always has a path-lift (cid:98) γ starting at any preimage (cid:98) x ∈ (cid:98) f − ( x ). This path-lift is only alift if it returns to (cid:98) x . Definition 2.2. A geometric ideal polyhedron P is the convex hull in H of n ≥ ∂ H . If n = 4, the polyhedron is called a geometric ideal tetrahedron , and its isometry classis determined by the cross-ratio of its 4 vertices. The polyhedron P and its boundary ∂P inherit anorientation from the embedding P (cid:44) → H .An ideal polyhedron P is called an ideal pyramid if P contains an ideal vertex v (called the apex )and a unique face F not incident to v (called the base ). It follows that every edge of P either belongsto ∂F (in which case it is called a base edge ) or connects v to a vertex of F (in which case it is calleda lateral edge ). Every pyramid is either an ideal tetrahedron, or has a unique choice of apex andbase.A geometric ideal polyhedral decomposition P is a decomposition of M into geometric ideal poly-hedra, glued together by orientation-reversing isometries along their boundary faces. The cusps of M are therefore in bijection with the equivalence classes of ideal vertices in P . If all the cells areideal tetrahedra, the decomposition P is called a geometric ideal triangulation , and denoted T . Thepreimage of P in a cover (cid:99) M → M is denoted (cid:98) P , and similarly for other decorations. Convention 2.3.
All triangulations and polyhedral decompositions described below are presumedto be geometric, unless specified otherwise. While there is a rich theory of topological ideal triangu-lations of 3–manifolds, sometimes endowed with extra data, our focus in this paper is on geometry.
Definition 2.4. A (closed) horocusp A is the quotient of a closed horoball in H by a discrete group G of parabolic isometries, where G ∼ = Z × Z . Topologically, A is homeomorphic to T × [0 , ∞ ) and ∂A is isometric to a flat torus. The interior of A is called an open horocusp .If M = H / Γ is a finite-volume hyperbolic 3–manifold, an open horocusp in M is an embeddednoncompact end that is isometric to an open horocusp. A (closed) horocusp in M is the closure ofan open horocusp in M . In particular, a horocusp A ⊂ M is homeomorphic to T × [0 , ∞ ) with afinite number of points of tangency on T × { } identified in pairs.A horocusp collection in M is a union of closed horocusps A , . . . , A n containing all the noncom-pact ends of M , such that the interiors of the A i are pairwise disjoint.For a hyperbolic 3–manifold M = H / Γ, we typically work with (cid:102) M = H in the upper half-spacemodel. The preimage of a horocusp collection in M is a collection of (closed) horoballs in H withdisjoint interiors, called a packing . When we mention a horoball (cid:101) A in H in this context, we implicitlyassume that (cid:101) A is one of the horoballs in the packing, meaning (cid:101) A covers one of the specified horocuspsof M . We further conjugate Γ in Isom( H ) ∼ = PSL(2 , C ) so that ∞ is a parabolic fixed point of Γ,which means that a horoball (cid:101) A about ∞ occurs in the packing. All other horoballs in the packingare tangent to points of C . The packing horoballs with largest Euclidean diameter (equivalently, thehoroballs closest to (cid:101) A ) are called full-sized . Definition 2.5.
Let M be a cusped hyperbolic 3–manifold with horocusp collection A , . . . , A n .An orthogeodesic is an immersed geodesic segment γ that begins at ∂A i and ends at ∂A j , such that DAVID FUTER, EMILY HAMILTON, AND NEIL R. HOFFMAN
Figure 1.
The construction of a canonical polyhedral decomposition in a cuspedhyperbolic surface. We have a horoball packing of H (black) and the universalcover (cid:101) Σ of the cut locus Σ (red). Dual to (cid:101)
Σ is a canonical triangulation (cid:101) P (blue).Every vertex (cid:101) v ∈ (cid:101) Σ is the center of a ball (grey) tangent to a maximal collection ofhoroballs that contain the vertices of the cell of (cid:101) P dual to (cid:101) v . γ is orthogonal to ∂A i and ∂A j at the respective endpoints. The case A i = A j is permitted. If A i is tangent to A j , then a point of tangency is considered an orthogeodesic of length 0. We note thatan orthogeodesic is necessarily the shortest path in its homotopy class.In a similar fashion, an orthogeodesic in H is the shortest path between a pair of disjoint horoballs (cid:101) A, (cid:101) A (cid:48) . This path is necessarily a geodesic segment that is orthogonal to ∂A and ∂A (cid:48) .A collection of horocusps in a hyperbolic manifold M determines a canonical decomposition of M into polyhedra, as follows. Definition 2.6.
Let M be a cusped hyperbolic 3–manifold, endowed with a horocusp collection A , . . . , A n . The Ford–Voronoi domain
F ⊂ M consists of all points of M that have a unique shortestpath to the union of the A i . The complement Σ = M (cid:114) F , called the cut locus , is a 2–dimensionalcell complex consisting of finitely many totally geodesic polygons. The combinatorial dual of Σ isdenoted P and called the canonical polyhedral decomposition determined by ( M, A , . . . , A n ). Thispolyhedral decomposition has one geodesic edge for each polygonal face of Σ, one totally geodesic2–cell for each edge of Σ, and one 3–cell for each vertex of Σ. The edges of P are bi-infinite extensionsof orthogeodesics between the cusps. See Figure 1 for a 2–dimensional example.The top-dimensional cells of P can be characterized as follows. By construction, every 3–cell P ⊂ P is dual to a vertex v ∈ Σ. There is a metric ball D centered at v , which is tangent to somenumber of horocusps (corresponding to the ideal vertices of P ), and disjoint from their interiors.Furthermore, the collection of cusps tangent to D is maximal with respect to inclusion.In the context of closed surfaces, the construction of the canonical decomposition P dates backto the work of Voronoi and Delaunay in the early 20th century. Epstein and Penner [10] gave acharacterization of P using convexity in the hyperboloid model of H . As a consequence, P issometimes called the Delaunay or Epstein–Penner decomposition of M .The canonical polyhedral decomposition P determined by a choice of horocusps is always geo-metric. Thus every cusped hyperbolic 3–manifold admits a geometric polyhedral decomposition.Furthermore, one may attempt to subdivide the polyhedra of P into tetrahedra by coning . Definition 2.7.
Let P be a (geometric) ideal polyhedron, and let v be an ideal vertex of P . The coning of P from v is the decomposition of P into (geometric) ideal pyramids whose apex is v andwhose bases are the polygonal faces of P not incident to v . If P is an ideal pyramid and w is avertex of the base of P , the coning of P from w results in ideal tetrahedra, because every face of P not incident to w is an ideal triangle. NFINITELY MANY VIRTUAL GEOMETRIC TRIANGULATIONS 7
Definition 2.8.
Let P be a (geometric) ideal polyhedral decomposition of M , and let V denotethe set of cusps of M . Consider a strict partial order ≺ on V . Observe that ≺ imposes a (strict)partial order on the vertices of any polyhedron P ⊂ P , because vertices of P map to cusps of M .The coning of P induced by ≺ is the following subdivision: if P has a unique ≺ –minimal vertex v ,then P is coned from v ; otherwise, P is not subdivided at all.The iterated coning of P induced by ≺ is the following two-step procedure. First, cone P from itsunique ≺ –smallest vertex (if such a vertex exists), which either leaves P unchanged or decomposesit into pyramids. Second, cone each pyramid from the unique ≺ –smallest vertex of its base (if sucha vertex exists). Lemma 2.9.
Let M be a cusped hyperbolic –manifold, and ≺ a strict partial order on the set ofcusps of M . Let P be a (geometric) ideal decomposition of M , with the property that every polyhedron P ⊂ P has a unique ≺ –minimal vertex v P . Then the iterated coning of P induced by ≺ produces awell-defined subdivision of P into geometric ideal pyramids.Furthermore, if ≺ gives a total order of the vertices of every polyhedron, then the iterated coningof P produces a geometric ideal triangulation.Proof. Suppose P and P (cid:48) are polyhedra of P that are identified along a face F . We need to checkthat the iterated coning of P and P (cid:48) induces the same subdivision of the face F . We show this byconsidering two cases. Case 1: F does not have a unique ≺ –minimal vertex. In this case, we claim that F will not besubdivided at all. For, polyhedron P will be coned from its unique minimal vertex v P , which is notcontained in F by hypothesis. This produces a collection of pyramids, with F a base of one of thepyramids. Since F does not have a unique minimal vertex, the second stage of iterated coning doesnot subdivide F at all. An identical argument applies to P (cid:48) , proving the claim. Case 2: F has a unique ≺ –minimal vertex w . In this case, we claim that F will be subdividedby coning from w . If w = v P is minimal in all of P , then P will be subdivided into pyramids byconing from w , hence F will be also. Otherwise, if w (cid:54) = v P , then P will be subdivided into pyramidsby coning from v P , and F will be the base of one of these pyramids. At the second stage of theiterated coning, the pyramid in P containing F will be coned from the minimal vertex of F , namely w . An identical argument applies to P (cid:48) , proving the claim.Finally, observe that if ≺ gives a total order of the vertices of each cell, then we must be in Case 2:every face F has a minimal vertex. Thus every face is subdivided into triangles, and every pyramidis subdivided into geometric ideal tetrahedra. (cid:3) The “furthermore” statement of Lemma 2.9 was previously observed by Luo, Schleimer, andTillmann [22, Lemma 7]. They also used the separability of peripheral subgroups (Proposition 4.3)to show that every cusped 3–manifold M with a geometric polyhedral decomposition P has a finitecover (cid:99) M such that any order on the cusps of (cid:99) M imposes a total order on the vertices of eachpolyhedron of (cid:98) P . Compare Lemma 5.4 below. Consequently, (cid:99) M has a geometric ideal triangulation.In fact, a total order is not necessary to produce a geometric triangulation: Corollary 2.10.
Let M be a cusped hyperbolic –manifold, and ≺ a strict partial order on theset of cusps of M . Let P be a (geometric) ideal decomposition of M , with the property that everypolyhedron P ⊂ P has a unique ≺ –minimal vertex v P . Let P (cid:48) be the pyramidal refinement of P guaranteed by Lemma 2.9. Then every choice of diagonals in the non-triangular faces of P (cid:48) leads toa decomposition of P (cid:48) into geometric ideal tetrahedra.Proof. Following Lemma 2.9, let P (cid:48) be the subdivision into ideal pyramids coming from the iteratedconing of P induced by ≺ . Then every non-triangular 2–cell F ⊂ P (cid:48) must be the base of exactlytwo pyramids. Thus the non-tetrahedral pyramids of P (cid:48) are glued in pairs, with each pair forminga bipyramid that is joined to other cells along ideal triangles only. This means we have completefreedom to choose diagonals of every non-triangular 2–cell F , subdividing the two pyramids adjacent DAVID FUTER, EMILY HAMILTON, AND NEIL R. HOFFMAN to F into tetrahedra, without impacting the choices anywhere else in the manifold. Every such choiceproduces a subdivision of P (cid:48) into geometric ideal tetrahedra. (cid:3) An infinite tree of triangulations
In the last section, we described a construction of Luo, Schleimer, and Tillmann for decomposingan ideal polyhedral decomposition P into a geometric ideal triangulation. Having one geometrictriangulation is clearly a prerequisite to having infinitely many. In this section, we describe aparticular geometric feature called a drilled ananas (see Definition 3.3) that admits an infinitesequence of geometric triangulations. By embedding a drilled ananas inside a triangulation of acusped manifold M , we obtain an infinite sequence of ideal triangulations of M . See Lemma 3.6 forthe construction of an infinite sequence of triangulations, and Proposition 3.10 for a more refineddescription of an infinite trivalent tree of triangulations.Consider a polyhedral decomposition P of M . Let A ⊂ M be a horocusp, chosen small enoughthat for every polyhedron P ⊂ P , the intersection P ∩ A consists of neighborhoods of ideal vertices.Then P induces a decomposition of the torus ∂A into Euclidean polygons, which truncate the idealvertices of polyhedra of P . We call this decomposition the cusp cellulation of ∂A , and denote it C ( A ). If P is the canonical polyhedral decomposition (determined by some choice of cusps), then C ( A ) satisfies the Delaunay condition : the vertices of every polygon can be inscribed on a circle,where the interior of the circle does not contain any other vertices.In the following proposition, A is a horocusp of M . Let A t be a sub-horocusp of A such that d ( ∂A, ∂A t ) = t . A particular feature occurs when t becomes sufficiently large. Proposition 3.1 (Gu´eritaud–Schleimer [15]) . Let M be a cusped hyperbolic –manifold, endowedwith a choice of horocusps A, B , . . . , B n for n ≥ . Assume that an orthogeodesic α from A to B is the unique shortest path from A to ∪ nj =1 B j . Then, for every sufficiently small sub-horocusp A t ⊂ A , the canonical decomposition P determined by A t , B , . . . , B n contains a unique edge from A t to ∪ nj =1 B j . This edge is the bi-infinite extension of α .Furthermore, there are one or two –cells of P that meet the cusp A t . Each such –cell is anideal pyramid with an apex at A t and all lateral edges identified to α . If A is a rectangular cusp,then the single –cell meeting A t is a rectangular pyramid and the induced cellulation of ∂A t is arectangle. Otherwise, if A is a non-rectangular cusp, then the two –cells meeting A t are isometricideal tetrahedra, and the induced cellulation of ∂A t consists of two isometric, acute triangles. This result is due to Gu´eritaud and Schleimer [15, Section 4.1], and appears in the form of adiscussion with the explicit hypothesis that the cusp torus ∂A is not rectangular. For completeness,we reproduce an expanded version of their proof. Proof.
Set M = H / Γ, where we view H in the upper half-space model. As described in Section 2,we conjugate Γ in Isom( H ) so that a horoball (cid:101) A about ∞ covers the horocusp A . Then K =Stab Γ ( ∞ ) ∼ = Z can be identified with π ( A ) ⊂ Γ. Every preimage of B j is a horoball tangent to apoint of C .By hypothesis, there is a unique shortest orthogeodesic from A to ∪ nj =1 B j , which leads from A to B . Extend α to be a bi-infinite geodesic. After shrinking A by an appropriate distance, we mayfurther assume that α is the unique shortest orthogeodesic from A to A ∪ ( ∪ nj =1 B j ). Consequently,there is a choice of horoball (cid:101) B covering B , such that all of the full-sized horoballs tangent to pointsof C are in the K –orbit of (cid:101) B .Consider a ball D ⊂ H that rests on the collection of horoballs tangent to points of C (seeFigure 2). If the Euclidean diameter of D is sufficiently large, then D will only touch the full-sizedhoroballs. Furthermore, if we shrink A by a sufficiently large distance t ≥
0, producing a sub-horocusp A t whose preimage horoball (cid:101) A t ⊂ (cid:101) A is at sufficient Euclidean height, then D will also bedisjoint from (cid:101) A t . Inflating D to a maximal (hyperbolic) radius produces a ball D + that is tangentto (cid:101) A t and some number of horoballs from the orbit K · (cid:101) B , and is disjoint from all other horoballs. NFINITELY MANY VIRTUAL GEOMETRIC TRIANGULATIONS 9
Figure 2.
A ball D (shown in red) resting on four full-sized horoballs (blue) andtangent to a horoball about ∞ (green). In the setting of Proposition 3.1, all of thefull-sized horoballs are in the same orbit of K ∼ = Z . It follows that the ball D canbe tangent to four full-sized horoballs at once if and only if K has a rectangularfundamental domain, as in this figure. The center of D corresponds to an idealrectangular pyramid (dotted).Observe that D + is tangent to either 3 or 4 horoballs in K · (cid:101) B , and that the case of 4 occurs preciselywhen ∂A t = ∂ (cid:101) A t /K has a rectangular fundamental domain.Now, consider the cut locus Σ t corresponding to the horocusp collection A t , B , . . . , B n . Let (cid:101) Σ t be the preimage of Σ t in H , and consider the polyhedral decomposition P = P t dual to Σ t .By Definition 2.6, every vertex v ∈ (cid:101) Σ t is equidistant from some collection of horoballs (which ismaximal with respect to inclusion), and conversely every point equidistant from a maximal collectionof horoballs is a vertex of (cid:101) Σ t . In particular, the hyperbolic center of the ball D + constructed inthe previous paragraph must be a vertex w ∈ (cid:101) Σ t . By Definition 2.6, this vertex w is dual to apolyhedron P w in the canonical decomposition P , whose ideal vertices lie in the horoballs tangentto D + . Recall that D + is tangent to (cid:101) A t and either 3 or 4 full-sized horoballs in K · (cid:101) B .If there are 3 full-sized horoballs tangent to D + , then P w is an ideal tetrahedron. Furthermore,up to the action of K = Stab Γ ( ∞ ), there must be exactly one other ideal tetrahedron in P thatintersects (cid:101) A t . In this case, the induced cusp cellulation of ∂A t consists of two isometric triangles.These triangles must be acute: otherwise, the circle that circumscribes the 3 vertices of an obtusetriangle would contain the fourth vertex of the parallelogram in its interior, violating the Delaunaycondition. Since the two triangles of ∂A t are isometric, the two tetrahedra meeting A t are alsoisometric. Furthermore, each of the two tetrahedra has 3 edges identified to α , corresponding to thefact that the cusp cellulation has a single vertex at α ∩ ∂A t .If there are 4 full-sized horoballs tangent to D + , as in Figure 2, then the polyhedron P w is anideal rectangular-based pyramid. In this case, P w is the only cell of P meeting A t , and the inducedcusp cellulation of A t is a single rectangle with a vertex at α ∩ ∂A t . Consequently, all lateral edgesof P w are identified to α . (cid:3) Remark 3.2.
Akiyoshi [1] proved that the canonical polyhedral decomposition P t determined bythe horocusp collection A t , B , . . . , B n must stabilize as t → ∞ . Thus, for sufficiently large t , there is a stable geometric decomposition P = P t independent of t . This is our motivation for droppingthe superscript t .In the polyhedral decomposition P that occurs in Proposition 3.1, the cells that enter the specialcusp A t fit together to form a geometric object that we call a drilled ananas . Definition 3.3. A drilled ananas is a 3–manifold N homeomorphic to T × [0 , ∞ ) (cid:114) { x } , where x ∈ T × { } , and endowed with a complete hyperbolic metric with the following properties. Forsome y >
0, the non-compact end T × [ y, ∞ ) ⊂ N is isometric to a horocusp, such that eachcross-section T × { y (cid:48) } for y (cid:48) > y is a horotorus. The boundary ∂N = T × { } (cid:114) { x } is made upof two totally geodesic ideal triangles, with vertices at x . These two triangles are glued by isometryalong their edges to form a standard two-triangle triangulation of a once-punctured torus, withshearing and bending allowed along the edges of this boundary triangulation. If the base of thedrilled ananas is comprised of a single totally geodesic ideal rectangle, then we make a choice ofdiagonal to decompose N into two ideal tetrahedra.The horocusp T × [ y, ∞ ) ⊂ N is called the cusp of N , while a regular neighborhood of x ∈ T ×{ } is called the thorn of N .Here are a few notes on terminology and past usage. The term thorn was coined by Baker andCooper [2]. A drilled ananas is a special case of a topological ideal polyhedron in the work of Gu´eritaud[14], and a slightly less special case of an ideal torihedron in the work of Champanerkar, Kofman,and Purcell [5, Definition 2.1]. Both of these generalizations capture the idea of placing a hyperbolicstructure on a 3–manifold endowed with a polyhedral graph on its boundary. Gu´eritaud coined theterm ananas (French for pineapple ) to describe a topological polyhedron with the topology of a solidtorus. (Compare the definition of a filled ananas immediately above Claim 6.10.) The object N inDefinition 3.3 can be obtained by removing the core of a solid torus, hence drilled ananas .With the above definition, the conclusion of Proposition 3.1 can be rephrased as follows. Corollary 3.4.
Let M be a cusped hyperbolic manifold satisfying the hypotheses of Proposition 3.1,and let P = P t be the polyhedral decomposition produced by that proposition. Let N ⊂ M be thesubmanifold obtained by gluing together all cells of P that have an ideal vertex in cusp A , alongtheir shared faces. Then N is a drilled ananas comprised of two acute ideal tetrahedra or one idealrectangular pyramid. Furthermore, N is convex, with an angle less than π at every edge of P ∩ ∂N .Proof. The conclusion that N is a drilled ananas is immediate from Proposition 3.1 and Defini-tion 3.3. It remains to check that N is convex.By Proposition 3.1, each cell of P comprising N is an ideal pyramid with a base along ∂N . If N contains a single rectangular pyramid, we subdivide it into two isometric ideal tetrahedra T, T (cid:48) bychoosing a diagonal along ∂N . Otherwise, N already consists of two isometric, acute-angled idealtetrahedra T, T (cid:48) . The three lateral faces of T are glued to the three lateral faces of T (cid:48) , and all lateraledges of T, T (cid:48) are identified to the single geodesic α that connects the thorn of N to the cusp of N .Let θ , θ , θ be the dihedral angles of T at the edges identified to α . Since T and T (cid:48) are necessarilyisometric, these are also the dihedral angles of T (cid:48) at the edges identified to α . See Figure 3(I). Thenthe three internal angles along the edges of ∂N are 2 θ , θ , θ .If T and T (cid:48) are cells of P , then Proposition 3.1 says that all of their angles are acute. Thus2 θ i < π for every i , hence ∂N is locally convex at every edge on its boundary. Otherwise, if T and T (cid:48) were created by subdividing an ideal pyramid, the cusp A ⊂ N is rectangular, hence theEuclidean triangles truncating the tips of T and T (cid:48) have a right angle θ and two acute angles θ , θ .Consequently, the internal angles along ∂N are 2 θ = π and 2 θ , θ < π . In either case, ∂N islocally convex, hence N is convex. (cid:3) A key feature of a drilled ananas is that it is made up of two tetrahedra glued along three faces.Each of these three faces supports a local move, called a 2–3 move.
NFINITELY MANY VIRTUAL GEOMETRIC TRIANGULATIONS 11 (i)
Gluing two ideal tetrahedra to form adrilled ananas N i . (ii) In a 2–3–move, the two tetrahedraglued along a vertical face, shown in part I,are replaced by three tetrahedra that sharean edge dual to that face (dashed). Theresult appears in part III. (iii)
After a 2–3 move, the original ananas N i decomposes into three ideal tetrahedra. (iv) The new ananas N i +1 is the union ofthe two tetrahedra in N i which are incidentto ∞ . The next 2–3 move will be alonga face that is exterior to the fundamentaldomain shown here. Figure 3.
The inductive step of Lemma 3.6.
Definition 3.5.
Let T be a topological ideal triangulation of a 3–manifold N , possibly with bound-ary. Let T and T (cid:48) be distinct tetrahedra in T that are glued together along a face F , and observethat T ∪ F T (cid:48) is a bipyramid. A topological – move replaces T ∪ T (cid:48) with three tetrahedra gluedtogether along a central edge dual to F , while all other tetrahedra of T remain the same. Theresulting triangulation is denoted T . We say that a 2–3 move is geometric if T and T are bothgeometric triangulations. Assuming T is geometric, a 2–3 move will be geometric whenever thebipyramid T ∪ F T (cid:48) is strictly convex.In Figure 3, tetrahedra T and T (cid:48) are shown in panel (I), with face F a darker shade of blue. Thedual edge to F is dashed in panel (II), and the resulting three tetrahedra are shown in panel (III). Lemma 3.6.
A drilled ananas N admits an infinite sequence of geometric triangulations, connectedby geometric – moves. The inductive construction that proves the lemma is illustrated in Figure 3.
Proof.
Let N be a drilled ananas. By definition, ∂N is subdivided into two totally geodesic idealtriangles. Consequently, N itself can be subdivided into two geometric ideal tetrahedra T, T (cid:48) byconing those triangles to the non-compact end at ∞ . As in Proposition 3.1, we think of each of T, T (cid:48) as a triangular pyramid with a base along ∂N . The three lateral faces of T are glued to the threelateral faces of T (cid:48) , and all lateral edges of T, T (cid:48) are identified to a single geodesic α .As in Corollary 3.4, let θ , θ , θ be the dihedral angles of T and T (cid:48) at the edges identified to α .See Figure 3(I). Since these angles are positive and θ + θ + θ = π , two of the three angles must be strictly less than π/
2. Then the three internal angles along the edges of ∂N are 2 θ , θ , θ . Atleast two of these angles (say, 2 θ and 2 θ ) are strictly less than π .We will prove the lemma by induction. Set N = N . The key inductive claim is: Claim 3.7.
Let N i be a drilled ananas with a two-tetrahedron geometric triangulation. Let β i ⊂ ∂N i be a boundary edge such that the internal angle at β i is less than π . Then performing a diagonalexchange on β i results in a geometric ideal tetrahedron ∆ i and a new sub-ananas N i +1 ⊂ N i , where N i = N i +1 ∪ ∆ i and N i +1 again has a two-tetrahedron geometric triangulation. The proof of the claim is almost immediate. Removing β i from a triangulation of ∂N i results ina quadrilateral. Let β (cid:48) i be the opposite diagonal of this quadrilateral. Observe that the geodesicrepresentative of β (cid:48) i lies strictly inside N i , because N i is locally convex at β i . The join of β i , β (cid:48) i is a(geometric) ideal tetrahedron ∆ i . Removing the interior of ∆ i from N i produces a sub-ananas N i +1 whose boundary is pleated along β (cid:48) i and the two remaining edges of ∂N i . (cid:54) Observe that the two-tetrahedron triangulation of N i gives rise to a three-tetrahedron triangula-tion of N i = ∆ i ∪ N i +1 , in a geometric 2–3 move. See Figure 3. We can now apply this move to N i +1 , and so on, resulting in an infinite sequence of geometric triangulations of N = N . (cid:3) Corollary 3.8.
Suppose N is a drilled ananas, and f : (cid:98) N → N is a finite cover. Then (cid:98) N admitsan infinite sequence of f –equivariant geometric triangulations. The corollary follows immediately from Lemma 3.6 because every triangulation of N lifts to atriangulation of (cid:98) N . We also remark that any finite cover (cid:98) N → N is regular and has abelian deckgroup, because π ( N ) ∼ = Z is abelian.3.1. Connections to the Farey complex.
We can now describe some additional structure in theset of triangulations of a drilled ananas N . Definition 3.9.
The
Farey complex F is a simplicial complex whose vertices are isotopy classes ofarcs on a torus T based at a marked point x , and whose edges correspond to arcs that are disjoint(except at x ). The vertices of F are commonly identified with the rational points QP ⊂ RP , asfollows. Endow T with a standard Euclidean metric, with fundamental domain a unit square. Thenevery loop in T based at x can be pulled tight to a Euclidean geodesic of some well-defined slope Q ∪ {∞} . Conversely, every rational slope defines a unique isotopy class of arc from x to x .Triangles in F correspond to (isotopy classes of) one-vertex triangulations of T with the vertexat x , or equivalently to ideal triangulations of T (cid:114) { x } . The dual 1–skeleton of F is a trivalent tree,with every edge of the dual tree corresponding to a diagonal exchange. See Figure 4.Using Definition 3.9, we can state the following stronger formulation of Lemma 3.6. Proposition 3.10.
Let N be a drilled ananas with a geometric triangulation consisting of two acute-angled tetrahedra. Then N admits an infinite trivalent tree of geometric – moves, with vertices ofthe tree in natural bijection with triangles of the Farey complex F .Proof. The proof amounts to adding some book-keeping to the proof of Lemma 3.6. Let N = N ,subdivided into ideal tetrahedra T, T (cid:48) . By hypothesis, all dihedral angles of T and T (cid:48) are acute.This triangulation of N defines an induced cellulation of the boundary of the horocusp A ⊂ N . Infact, this cusp cellulation is a one-vertex triangulation: the one vertex is the intersection between ∂A and the single edge α ⊂ N into cusp A , while the two triangles are the cross-sections ∂A ∩ T and ∂A ∩ T (cid:48) . Let δ be this triangulation of ∂A . Fix a framing of ∂A so that the three slopes occurringin τ are 0 /
1, 1 /
0, and 1 /
1, as in the central triangle of Figure 4.Observe that any one-vertex triangulation of ∂A defines an ideal triangulation of ∂N , by projectingoutward from the cusp. In the opposite direction, any ideal triangulation of the boundary ∂N i forsome sub-ananas N i ⊂ N defines a one-vertex cusp triangulation. We will pass freely between thetwo viewpoints. NFINITELY MANY VIRTUAL GEOMETRIC TRIANGULATIONS 131001 / − / Figure 4.
The Farey complex F . Edges of the dual tree correspond to diagonalexchanges in a torus with one marked point. Every non-backtracking path in thedual tree, starting from the central triangle, can be realized via geometric 2–3 movesin a drilled ananas. Figure from Ham and Purcell [16, Figure 3.1].Now, let β be a boundary edge of N . In order to carry out the construction of Claim 3.7, theinternal angle of β needs to be less than π . But since all dihedral angles of T and T (cid:48) are acute, wemay choose β at will. Now, the new sub-ananas N ⊂ N , constructed as in Claim 3.7, will inducea new cusp cellulation τ of ∂A , which differs from τ via a diagonal exchange. Thus we may choose τ to be any one of the three triangulations adjacent to τ in Figure 4.Continuing inductively, suppose that we have constructed the sub-ananas N i ⊂ N i − ⊂ . . . ⊂ N ,and that ∂N i has ideal triangulation τ i . Then, as noted in the proof of Lemma 3.6, there are twoedges of ∂N i that have interior angle less than π . The third edge necessarily has angle more than π ;it is the edge β (cid:48) i − that was just created in constructing N i . (See Figure 3(IV), with indices shiftedby 1.) Thus we may choose β i to be any edge of ∂N i other than β (cid:48) i − , which means the new cusptriangulation τ i +1 is allowed to be any of the two neighbors of τ i that are distinct from τ i − . Insummary, the path τ , τ , . . . of cusp cellulations associated with N , N , . . . is allowed to be anynon-backtracking path starting from τ . (cid:3) In any path path τ , τ , . . . of cusp cellulations constructed in the above proof, the slopes of edgesapproach some limiting value in RP , and the edges themselves approach a foliation with the limitingslope. Thus the geometric retriangulations of N can be chosen to limit to any foliation of the torus.The strategy for proving Theorem 1.2 and Theorem 1.3 can now come into view. Given a cusped3–manifold M , we will find a cover ˚ M with a polyhedral decomposition ˚ P as in Proposition 3.1,where two ideal tetrahedra fit together to form a drilled ananas. A further cover (cid:99) M ensures thatthe other cells of (cid:98) P can be subdivided into ideal tetrahedra as well. Producing these covers requirestools from subgroup separability, described in the next section.4. Separability
This section begins with a review of some standard results about separable subsets and subgroupsin a group G . The main content of the section is a proof of Theorem 1.4. Definition 4.1.
Let G be a group. The profinite topology on G is the topology whose basic opensets are cosets of finite-index normal subgroups. Since every coset of a finite-index subgroup H (cid:67) G is the complement of finitely many other cosets of H , the basic open sets are also closed.A subset S ⊂ G is called separable if it is closed in the profinite topology on G .The following characterization is standard. Lemma 4.2.
Let G be a group and S ⊂ G a subset. The following are equivalent: (1) S is separable.(2) For every element g ∈ G (cid:114) S , there is a homomorphism ϕ : G → F , where F is a finite groupand ϕ ( g ) / ∈ ϕ ( S ) .Proof. The set S is closed if and only if every g ∈ G (cid:114) S is contained in a basic open set disjointfrom S . But, by Definition 4.1, a basic open set containing g is precisely the preimage of an elementunder a homomorphism to a finite group. (cid:3) If M is a cusped hyperbolic 3–manifold, a subgroup of π ( M ) coming from the inclusion of ahorocusp A is called peripheral . This peripheral subgroup of π ( M ) is also maximal abelian : it isnot contained in any larger abelian subgroup.We will need to separate peripheral subgroups and their double cosets. The following separabilityresult has been known since the 1980s, if not earlier. See e.g. Long [21, page 484] for a proof. Proposition 4.3.
Let M = H / Γ be a cusped hyperbolic –manifold. Then maximal abelian sub-groups of Γ are separable. In particular, peripheral subgroups of Γ are separable. (cid:3) We will also need to separate peripheral double cosets, using the following theorem of Hamilton,Wilton, and Zalesskii [18, Theorem 1.4].
Theorem 4.4 (Hamilton–Wilton–Zalesskii [18]) . Let M = H / Γ be a finite-volume hyperbolic –manifold. Let H and K be abelian subgroups of Γ . Then, for every g ∈ Γ , the double coset HgK = { hgk : h ∈ H, k ∈ K } is separable in Γ . (cid:3) Theorem 4.4 is the strongest separability tool needed in the proof of Theorem 1.3. The readerwho is mainly interested in that result is invited to proceed directly to Section 5.4.1.
Algebraic tools for separability.
The separability results that we use in this paper, includingTheorem 4.4 and Theorem 1.4, are proved using tools from algebraic number theory. To set up theproof of Theorem 1.4, we review some needed definitions and results.A number field is a finite field extension of Q . An extremely useful connection between numberfields and 3–manifolds comes from the following result. Theorem 4.5 (Thurston [29], Bass [3], Culler–Shalen [6, 8]) . Let M = H / Γ be a cusped hyperbolic –manifold. Then (1) Γ ⊂ PSL(2 , C ) can be lifted to SL(2 , C ) . (2) Γ can be conjugated in SL(2 , C ) to lie in SL(2 , R ) ⊂ SL(2 , k ) , where R is a finitely generatedsubring of a number field k . (cid:3) Conclusion (1) was observed by Thurston [29, page 98] and carefully written down by Culler andShalen [8, Proposition 3.1.1], with a simplified proof by Culler [6, Corollary 2.2]. Conclusion (2) wasobserved by Thurston [29, Proposition 6.7.4] as an algebraic consequence of Mostow rigidity, andindependently proved by Bass [3].The arithmetic data associated to a ring R can be used to construct finite quotients of both ringsand groups. The construction uses the following notions. Definition 4.6. If k is a number field, let O k denote the ring of integers of k . If p is a non-zeroprime ideal of O k , we let k p denote the p -adic completion of k and O k p the ring of integers of k p .The ring O k p has a unique maximal ideal. The quotient of O k p by this maximal ideal is a finite fieldcalled the residue class field of O k p . The quotient map is called the residue class field map withrespect to p . If R is a finitely generated ring in a number field k , then R ⊂ O k p for all but finitelymany primes p of O k . Restricting the residue class field map to R yields a finite quotient of R .Now, suppose Γ ⊂ SL(2 , R ), where R is a ring. Let I ⊂ R be an ideal, such that S = R/I isfinite. Then the homomorphism SL(2 , R ) → SL(2 , S ) NFINITELY MANY VIRTUAL GEOMETRIC TRIANGULATIONS 15 where coefficients are reduced modulo I is called a congruence quotient of PSL(2 , R ). We also callthe composition ϕ : Γ (cid:44) → SL(2 , R ) → SL(2 , S )a congruence quotient of Γ. (This is slightly abusive, because Γ may fail to surject SL(2 , S ).)We now describe three algebraic results that will be used in the proof of Theorem 1.4. The firstof these is [18, Theorem 2.6]. Proposition 4.7 (Hamilton–Wilton–Zalesskii [18]) . Let R be a finitely generated ring in a numberfield k . By fixing a Q embedding of k into C , we may view k ⊂ C . Let ω ∈ R , and set Z ω = { m + nω | m, n ∈ Z } and Q ω = { m + nω | m, n ∈ Q } . If y ∈ R − Q ω , then there exist a finite ring S and a ring homomorphism ρ : R → S such that ρ ( y ) / ∈ ρ ( Z ω ) . (cid:3) The following technical lemma is a variant of Proposition 4.7.
Lemma 4.8.
Let R be a finitely generated ring in a number field k . By fixing a Q embedding of k into C , we may view k ⊂ C . Let ω ∈ R − R . Then there is an infinite collection Ω of primesof Q , such that for each prime p ∈ Ω , there exist a finite field F p of characteristic p and a ringhomomorphism η p : R → F p , such that { , η p ( ω ) } is linearly independent over F p .Consequently, η p has the following property. Consider an element y ∗ ∈ Q ω = { m + nω | m, n ∈ Q } .Express y ∗ in lowest terms: y ∗ = m ∗ + n ∗ ωv ∗ where m ∗ , n ∗ ∈ Z and v ∗ ∈ N . If η p ( y ∗ ) = η p ( m + nω ) for some m, n ∈ Z , then v ∗ m ≡ m ∗ (mod p ) and v ∗ n ≡ n ∗ (mod p ) .Proof. Since ω ∈ C − R , the set { , ω } is linearly independent over Q . By a standard argumentused in the proof of Theorem 4.4 (compare [18, page 278]), we can preserve this property in a finitequotient. We include the details for completeness. Let L denote the normal closure of k over Q andlet ¯ τ ∈ Gal( L/ Q ) represent complex conjugation. Since ω ∈ C − R , we know ω is not fixed by ¯ τ .By the Tchebotarev Density Theorem, there exist infinitely many primes p of Q with unramifiedextension p in L such that ¯ τ is the Frobenius automorphism for p /p . After eliminating a finite setof primes, if necessary, we may assume that R ⊂ O L p , where O L p denotes the ring of integers in the p -adic field L p . Given such a prime p , let F p denote the residue class field of O L p and let F p denotethe finite field of p elements. Let η p be the composition of the inclusion map of R into O L p with theresidue map: η p : R (cid:44) → O L p → F p . Since ¯ τ is the Frobenius automorphism of L/ Q with respect to p /p , Gal( L p / Q p ) = (cid:104) ¯ τ (cid:48) (cid:105) where ¯ τ (cid:48) = ¯ τ on L . Since ¯ τ ( ω ) (cid:54) = ω , it follows that ω / ∈ Q p . The Galois group of F p / F p is also induced by ¯ τ . Thisimplies that η p ( ω ) / ∈ F p and therefore, the set { , η p ( ω ) } is linearly independent over F p .Next, suppose that η p ( y ∗ ) = η p ( m + nω ) for some m, n ∈ Z . Then η p ( m ∗ + n ∗ ω ) = η p ( v ∗ y ∗ ) = η p ( v ∗ m + v ∗ nω ) . The above equality can be rewritten as η p ( v ∗ m − m ∗ )+ η p ( v ∗ n − n ∗ ) η p ( ω ) = 0. Since the set { , η p ( ω ) } is linearly independent over F p , it follows that v ∗ m ≡ m ∗ (mod p ) and v ∗ n ≡ n ∗ (mod p ). (cid:3) The third preliminary algebraic result is a combination of [17, Theorem 2.3 and Corollary 2.5].
Proposition 4.9 (Hamilton [17]) . Let R be a finitely generated ring in a number field k , let δ be anon-zero element of R that is not a root of unity, and let x , x , . . . , x j be non-zero elements of R .Then, for every sufficiently large integer q , there exist a non-zero prime ideal p of O k , a finite field F p , and and a ring homomorphism σ : R → F p such that R ⊂ O k p , the multiplicative order of σ ( δ ) is equal to q , and σ ( x i ) (cid:54) = 0 , for each ≤ i ≤ j . The field F p is the residue class field of O k p andthe map σ is the restriction to R of the residue class field map with respect to p . (cid:3) Conjugacy separation of peripheral cosets.
We can now prove Theorem 1.4.
Proof of Theorem 1.4.
Let H and K be maximal parabolic subgroups of Γ corresponding to distinctcusps of M , and let g ∈ Γ be an element such that K is disjoint from every conjugate of gH . Inparticular, this implies g / ∈ H . Fix a non-trivial element h ∈ H . Since H is a maximal abeliansubgroup of Γ, and g / ∈ H , the commutator [ g, h ] = gh g − h − is nontrivial. Let A be the cusp of M corresponding to K , let B be the cusp corresponding to H , and let C , . . . , C (cid:96) be the remainingcusps. We will leave the cusp A unfilled, and will fill the remaining cusps. So long as a tuple of slopes s on B, C , . . . , C (cid:96) avoids finitely many slopes on each cusp, the Dehn filled manifold M ( s ) will behyperbolic. Thus, by Thurston’s hyperbolic Dehn surgery theorem, we can choose generators h and h of H that can be completed to tuples s and s where M ( s ) and M ( s ) are both hyperbolic.Then, for j = 1 ,
2, the fundamental group π ( M ( s j )) has a discrete, faithful representation to a groupof isometries Γ( s j ) ⊂ PSL(2 , C ). By Theorem 4.5, we view Γ( s j ) as a subgroup of SL(2 , C ). Let ψ j : Γ → Γ( s j ) be the quotient homomorphism induced by the inclusion M (cid:44) → M ( s j ). By choosingsufficiently long Dehn fillings, we may assume that ψ ([ g, h ]) and ψ ([ g, h ]) are non-trivial. Thesechoices ensure the following properties for j ∈ { , } : • ψ j ( K ) is a parabolic subgroup of Γ( s j ), • ψ j ( H ) is a loxodromic subgroup of Γ( s j ), • ψ j ( g ) / ∈ ψ j ( H ).Before working with the Dehn filled manifolds, we examine the coset gH in Γ. By Theorem 4.5,we can conjugate Γ to lie in SL(2 , k ) for some number field k . After possibly expanding k , and thenconjugating Γ in SL(2 , k ), we may assume that g = (cid:18) a bc d (cid:19) , h = ± (cid:18) (cid:19) , and h = ± (cid:18) ω (cid:19) , for a fixed element ω ∈ C − R . Note that the traces of h and h are determined by the lift toSL(2 , C ), and might not coincide. Thus an arbitrary element gh m h n ∈ gH can be expressed as gh m h n = ± (cid:18) a bc d (cid:19) (cid:18) m + nω (cid:19) = ± (cid:18) a a ( m + nω ) + bc c ( m + nω ) + d (cid:19) , (4.1)where the sign ± depends on the traces of h , h and the parity of m, n .As above, set Z ω = { m + nω | m, n ∈ Z } and Q ω = { m + nω | m, n ∈ Q } . Then gH ⊂ (cid:26) ± (cid:18) a ax + bc cx + d (cid:19) (cid:12)(cid:12)(cid:12) x ∈ Z ω (cid:27) . Since g / ∈ H , and H is a maximal parabolic subgroup of Γ, we have c (cid:54) = 0. Thus gh m h n ∈ gH isparabolic if and only if tr( gh m h n ) = ± ( a + d + c ( m + nω )) ∈ {± } . Solving for x = m + nω ∈ Z ω ,let y + = y +1 = 2 − a − dc and y − = y − = − − a − dc . Then the coset gH contains a parabolic element if and only if { y + , y − } ∩ Z ω (cid:54) = ∅ . We will abusenotation slightly by thinking of the subscripts as either symbols ( ± ) or numbers ( ± y + and y − , corresponding to trace +2 and trace −
2, are the “problem elements” thatwe will need to track throughout the proof. In particular, let R ⊂ k be the ring generated by thecoefficients of the generators of Γ. By expanding R if necessary, we may assume that c − ∈ R , whichimplies y ± ∈ R .From here, the proof proceeds as follows. For each number i ∈ {± } , we will construct a homo-morphism ϕ i : Γ → G i , where G i is a finite group. Each ϕ i will be either a congruence quotient of Γ,or the product of a congruence quotient of Γ and a congruence quotient of Γ( s j ) for some j . Thenwe will package these homomorphisms together to obtain ϕ = ϕ − × ϕ + : Γ −→ G = G − × G + . NFINITELY MANY VIRTUAL GEOMETRIC TRIANGULATIONS 17
In particular, for every γ ∈ Γ, the image ϕ ( γ ) is a tuple of matrices, each with coefficients in a finitering, and each with a well-defined trace. To complete the proof, we will see that for every h ∈ H and every (cid:96) ∈ K , some coordinate of ϕ ( gh ) differs in trace from the same coordinate of ϕ ( (cid:96) ). Thiswill imply that ϕ ( gh ) cannot be conjugate to ϕ ( (cid:96) ).For each i ∈ {± } , the definition of ϕ i : Γ → G i depends on whether y i belongs to Q ω . If y i / ∈ Q ω ,we argue as follows. Claim 4.10.
Suppose y i ∈ R − Q ω . Then there is a ring homomorphism ρ i : R → S i , where S i is afinite ring, such that the following holds. For every pair ( m, n ) ∈ Z , we have ρ i ( m + nω ) (cid:54) = ρ i ( y i ) . By Proposition 4.7, there exist a finite ring S i and a ring homomorphism ρ i : R → S i such that ρ i ( y i ) / ∈ ρ i ( Z ω ). By the definition of Z ω , this means ρ i ( y i ) (cid:54) = ρ i ( m + nω ) for any ( m, n ). (cid:54) Using ρ i , we define a congruence quotient ϕ i : Γ (cid:44) → SL(2 , R ) → G i = SL(2 , S i ) if y i / ∈ Q ω , (4.2)completing the definition of G i and ϕ i in this case.Alternately, if y i ∈ Q ω , Lemma 4.8 provides an infinite set of primes Ω, such that for each p ∈ Ωthere is an associated finite field F p and ring homomorphism η p : R → F p . The central claim in thiscase is the following. Claim 4.11.
Suppose y i ∈ Q ω . Then there is a choice of Dehn filling quotient ψ j : Γ → Γ( s j ) where the coefficients of Γ( s j ) lie in a finitely generated ring T j , a prime number p ∈ Ω , and a ringhomomorphism σ p,i : T j → E i , where E i is a finite field, such that the following holds. For everypair ( m, n ) ∈ Z , we have η p ( m + nω ) (cid:54) = η p ( y i ) or σ p,i ◦ tr ◦ ψ j ( gh m h n ) (cid:54) = ± . In fact, the desired homomorphism σ p,i exists for all sufficiently large p ∈ Ω. However, we willonly need σ p,i for one p ∈ Ω.We momentarily postpone the proof of Claim 4.11 to describe the construction of ϕ i . The ringhomomorphism η p defines a congruence quotient ν : Γ (cid:44) → SL(2 , R ) → SL(2 , F p ) . Similarly, σ p,i defines a homomorphism ν i factoring through a congruence quotient: ν i : Γ ψ j −→ ψ j (Γ) = Γ( s j ) (cid:44) → SL(2 , T j ) → SL(2 , E i ) . We can now define ϕ i = ν × ν i : Γ −→ G i = SL(2 , F p ) × SL(2 , E i ) if y i ∈ Q ω , (4.3)completing the definition of G i and ϕ i in this case. Proof of Claim 4.11.
We begin by specifying the choice of Dehn filling quotient ψ or ψ . Write y i = ( m i + n i ω ) /v i in lowest terms, as in Lemma 4.8. If v i = 1, then we set j = 2 and work with theDehn filling ψ j = ψ : Γ → Γ( s ) for concreteness (although ψ would also work). Assuming v i (cid:54) = 1,we have either v i (cid:45) m i or v i (cid:45) n i . If v i (cid:45) m i , then we set j = 2 and select the Dehn filling M ( s ) andthe quotient map ψ : Γ → Γ( s ). Then ψ ( H ) is an infinite cyclic loxodromic subgroup of Γ( s )generated by ψ ( h ). Consequently, ψ ( h m h n ) = ψ ( h m ), where v i m (cid:54)≡ m i (mod v i ) because v i (cid:45) m i . Similarly, if v i (cid:45) n i , then we set j = 1 and select the Dehn filling M ( s ) and the quotient map ψ : Γ → Γ( s ). This has the effect that ψ ( h m h n ) = ψ ( h n ), where v i n (cid:54)≡ n i (mod v i ) because v i (cid:45) n i . In either case, the above non-congruences will be used in the endgame of the proof of the claim.Because the arguments for m i and n i are entirely parallel, and differ only by a substitution ofsymbols, we assume without loss of generality that v i (cid:45) m i , hence j = 2 and we have the Dehn fillingquotient ψ : Γ → Γ( s ).By Theorem 4.5, we can conjugate Γ( s ) in SL(2 , C ) such that Γ( s ) ⊂ SL(2 , T ), where T = T is a finitely generated ring in a number field. Moreover, we may assume that ψ ( g ) = (cid:18) r st u (cid:19) and ψ ( h ) = (cid:18) λ λ − (cid:19) , for some r, s, t, u, λ ∈ C with | λ | (cid:54) = 1. Then ψ ( gH ) = (cid:26)(cid:18) r st u (cid:19) (cid:18) λ m λ − m (cid:19) (cid:12)(cid:12)(cid:12) m ∈ Z (cid:27) = (cid:26)(cid:18) rλ m sλ − m tλ m uλ − m (cid:19) (cid:12)(cid:12)(cid:12) m ∈ Z (cid:27) , and tr ◦ ψ ( gh m h n ) = rλ m + uλ − m . In particular, ψ ( gH ) contains a parabolic matrix if and only if rλ m + uλ − m = ± m , which is true if and only if λ m is a root of rx ± x + u . Let ζ = 1 + √ − rur , ξ = 1 − √ − rur be the roots of rx − x + u . Then {− ζ, − ξ } are the roots of rx + 2 x + u . By expanding T = T ,if necessary, we may assume that { ζ, ξ } ⊂ T . Since ψ ( g ) does not commute with ψ ( h ), we have s (cid:54) = 0 and t (cid:54) = 0. Therefore, ru (cid:54) = 1, which implies that ζ (cid:54) = ξ .To prove the claim, we consider two cases. Case 1: λ m i / ∈ { ζ v i , ξ v i } .By Proposition 4.9, for all sufficiently large primes p ∈ Ω, there exist a finite field E i and a ringhomomorphism σ p,i : T → E i , such that σ p,i ( ζ − ξ ) (cid:54) = 0 , σ p,i ( λ m i − ζ v i ) (cid:54) = 0 , σ p,i ( λ m i − ξ v i ) (cid:54) = 0 , and the multiplicative order of σ p,i ( λ ) is equal to 2 p .Suppose for a contradiction that there exist m, n ∈ Z such that η p ( m + nω ) = η p ( y i ) and σ p,i ◦ tr ◦ ψ ( gh m h n ) = σ p,i ( rλ m + uλ − m ) = ± . Then Lemma 4.8 implies v i m ≡ m i (mod p ) ⇒ v i m ≡ m i (mod 2 p ) . If σ p,i ( rλ m + uλ − m ) = 2, then σ p,i ( λ m ) is a root of f ( x ) = σ p,i ( r ) x − x + σ p,i ( u ) over E i . Sincethe two distinct roots of f over E i are equal to σ p,i ( ζ ) and σ p,i ( ξ ), we have σ p,i ( λ m ) = σ p,i ( ζ ) or σ p,i ( λ m ) = σ p,i ( ξ ). Similarly, if σ p,i ( rλ m + uλ − m ) = −
2, then σ p,i ( λ m ) = σ p,i ( − ζ ) or σ p,i ( λ m ) = σ p,i ( − ξ ). In either case, σ p,i ( λ v i m ) = σ p,i ( ζ v i ) or σ p,i ( λ v i m ) = σ p,i ( ξ v i ) . Since 2 v i m ≡ m i (mod 2 p ) and the multiplicative order of σ p,i ( λ ) is equal to 2 p , this implies that σ p,i ( λ m i ) = σ p,i ( ζ v i ) or σ p,i ( λ m i ) = σ p,i ( ξ v i ) , a contradiction. Case 2: λ m i ∈ { ζ v i , ξ v i } .Without loss of generality, we may assume that λ m i = ζ v i . First, suppose that v i = 1. Thismeans that y i = m i + n i ω ∈ Z ω , hence gh m i h n i ∈ gH has trace ± λ m i = ζ v i and v i = 1, λ m i ∈ { ζ, − ζ } . Therefore, ψ ( gh m i h n i ) = ψ ( gh m i ) is also a parabolicelement in Γ( s ). This is a contradiction, since the only parabolic elements of Γ that remain parabolicafter the Dehn filling lie in conjugates of K , and the coset gH is disjoint from every conjugate of K .We conclude that v i > λ v i m i (cid:54) = ξ v i . By Proposition 4.9, for all sufficiently large primes p ∈ Ω, NFINITELY MANY VIRTUAL GEOMETRIC TRIANGULATIONS 19 there exist a finite fleld E i and a ring homomorphism σ p,i : T → E i , such that σ p,i ( ζ − ξ ) (cid:54) = 0 , σ p,i ( λ v i m i − ξ v i ) (cid:54) = 0 , and the multiplicative order of σ p,i ( λ ) is equal to 2 v i p . Suppose for a contradiction that there exist m, n ∈ Z such that η p ( m + nω ) = η p ( y i ) and σ p,i ◦ tr ◦ ψ ( gh m h n ) = σ p,i ( rλ m + uλ − m ) = ± . Then Lemma 4.8 implies v i m ≡ m i (mod p ) ⇒ v i m ≡ v i m i (mod 2 v i p ) . Since σ p,i ( rλ m + uλ − m ) = ±
2, we have σ p,i ( λ m ) = σ p,i ( ± ζ ) or σ p,i ( λ m ) = σ p,i ( ± ξ ). If σ p,i ( λ m ) = σ p,i ( ± ζ ), then σ p,i ( λ v i m ) = σ p,i ( ζ v i ) = σ p,i ( λ m i ) . Since the multiplicative order of σ p,i ( λ ) is divisible by 2 v i , we have2 v i m ≡ m i (mod 2 v i ) , contradicting the fact that v i (cid:45) m i . If σ p,i ( λ m ) = σ p,i ( ± ξ ), then, since the multiplicative order of σ p,i ( λ ) is equal to 2 v i p and 2 v i m ≡ v i m i (mod 2 v i p ), we obtain σ p,i ( λ v i m i ) = σ p,i ( λ v i m ) = σ p,i ( ξ v i ) , which contradicts our assumption in choosing σ p,i .Finally, assume that λ v i m i = ξ v i . By Proposition 4.9, there exist a finite field E i and a ringhomomorphism σ p,i : T → E i , such that σ p,i ( ζ − ξ ) (cid:54) = 0, and the multiplicative order of σ p,i ( λ )is divisible by 2 v i . Suppose that there exist m ∈ Z such that σ p,i ( rλ m + uλ − m ) = ±
2. Then σ p,i ( λ m ) = σ p,i ( ± ζ ) or σ p,i ( λ m ) = σ p,i ( ± ξ ). By the argument above, σ p,i ( λ m ) (cid:54) = σ p,i ( ± ζ ). If σ p,i ( λ m ) = σ p,i ( ± ξ ), then σ p,i ( λ v i m ) = σ p,i ( ξ v i ) = σ p,i ( λ v i m i ) . Since the multiplicative order of σ p,i ( λ ) is divisible by 2 v i , this implies that2 v i m ≡ v i m i (mod 2 v i ) . But this contradicts the fact that v i (cid:45) m i , completing the proof of Claim 4.11. (cid:54) We can now complete the proof of the theorem. For each i ∈ {± } , we have defined a homo-morphism ϕ i : Γ → G i , using Equation (4.2) if y i / ∈ Q ω and Equation (4.3) if y i ∈ Q ω . Now,define ϕ = ϕ − × ϕ +1 : Γ −→ G = G − × G +1 . We need to show that ϕ ( K ) is disjoint from every conjugate of ϕ ( gH ).Consider an arbitrary element gh m h n ∈ gH , and suppose for a contradiction that ϕ ( gh m h n ) isconjugate to ϕ ( (cid:96) ) for some (cid:96) ∈ K . Since (cid:96) is parabolic, we know tr( (cid:96) ) ∈ {± } . Recalling the generalform for gh m h n in Equation (4.1), define a number (cid:15) = (cid:15) ( (cid:96), m, n ) ∈ {± } so thattr( gh m h n ) = (cid:15) · tr( (cid:96) )2 · ( a + d + c ( m + nω )) . In other words, (cid:15) = 1 when tr( h m h n ) = tr( (cid:96) ), and (cid:15) = − ϕ (cid:15) of ϕ to obtain a contradiction.If y (cid:15) / ∈ Q ω , then ϕ (cid:15) ( (cid:96) ) ∈ SL(2 , S (cid:15) ). Since ϕ (cid:15) ( gh m h n ) is conjugate to ϕ (cid:15) ( (cid:96) ), we havetr ◦ ϕ (cid:15) ( gh m h n ) = ρ (cid:15) ◦ tr( gh m h n ) = ρ (cid:15) (cid:16) (cid:15) tr( (cid:96) )2 ( a + d + c ( m + nω )) (cid:17) = ρ (cid:15) (cid:0) tr( (cid:96) ) (cid:1) = tr ◦ ϕ (cid:15) ( (cid:96) ) . Since ρ (cid:15) is a ring homomorphism and (cid:15) tr( (cid:96) )2 ∈ {± } is a unit, we may rearrange terms to obtain ρ (cid:15) ( a + d + c ( m + nω )) = ρ (cid:15) (cid:0) (cid:15) (cid:1) ⇒ ρ (cid:15) ( c ( m + nω )) = ρ (cid:15) (cid:0) (cid:15) − a − d (cid:1) . But then ρ (cid:15) ( m + nω ) = ρ (cid:15) ( y (cid:15) ), contradicting Claim 4.10. If y (cid:15) ∈ Q ω , then ϕ (cid:15) ( (cid:96) ) = ( ν ( (cid:96) ) , ν (cid:15) ( (cid:96) )) ∈ SL(2 , F p ) × SL(2 , E p,(cid:15) ), a product of two matrices. Since ν ( gh m h n ) is conjugate to ν ( (cid:96) ), we obtaintr ◦ ν ( gh m h n ) = η p ◦ tr( gh m h n ) = η p (cid:16) (cid:15) tr( (cid:96) )2 ( a + d + c ( m + nω )) (cid:17) = η p (cid:0) tr( (cid:96) ) (cid:1) = tr ◦ ν ( (cid:96) ) . Then the same rearrangement of terms as before gives η p ( m + nω ) = η p ( y (cid:15) ). Meanwhile, since ν (cid:15) ( gh m h n ) is conjugate to ν (cid:15) ( (cid:96) ), we obtain σ p,(cid:15) ◦ tr ◦ ψ j ( gh m h n ) = tr ◦ ν (cid:15) ( gh m h n ) = tr ◦ ν (cid:15) ( (cid:96) ) = ± , contradicting Claim 4.11. In either case, the proof is complete. (cid:3) Manifolds with non-rectangular cusps
In this section, we prove Theorem 1.3. We begin with a cusped, hyperbolic 3–manifold M containing a horocusp A . We will construct a sequence of finite covers (cid:99) M → ˚ M → M , with anincreasingly strong sequence of properties. The final cover (cid:99) M will have infinitely many geometricideal triangulations, implying Theorem 1.3. The construction proceeds in four steps.Step 1. Construct a cover ˚ M → M where A lifts to a horocusp ˚ A that has a unique shortest pathto the other cusps. This is accomplished in Lemma 5.2.Step 2. Shrink ˚ A ⊂ ˚ M to a small sub-horocusp ˚ A t . By Proposition 3.1 and Corollary 3.4, thecanonical cell decomposition ˚ P determined by ˚ A t and the other cusps contains an embeddeddrilled ananas ˚ N consisting of one or two cells of ˚ P .Step 3. Construct a cover (cid:99) M → ˚ M where every polyhedron in the lifted polyhedral decomposition (cid:98) P has vertices at distinct cusps. This is accomplished in Lemma 5.4.Step 4. Now, (cid:98) P can be subdivided into geometric ideal tetrahedra by Lemma 2.9, and furthermore (cid:98) P contains a cover (cid:98) N of the original ananas ˚ N . If the cusp A was non-rectangular, theinduced triangulation of (cid:98) N is equivariant with respect to the cover of ˚ N . To conclude theproof, we use Lemma 3.6 to find infinitely many ideal triangulations of (cid:98) N , hence of (cid:99) M .Step 1 builds covers using Theorem 4.4, whereas Step 3 uses Proposition 4.3. Steps 2 and 4construct and subdivide polyhedral decompositions, but do not build any covers. The hypothesisthat A is a non-rectangular cusp is used only in Step 4. See Remark 5.5 for a detailed descriptionof how this hypothesis is used.The following basic lemma will be used to apply the results of Section 4. Lemma 5.1.
Let M = H / Γ be a cusped hyperbolic manifold. Let (cid:101) B, (cid:101) B (cid:48) ⊂ H be horoballs thatcover the same cusp in M , and let g ∈ Γ be an isometry such that g ( (cid:101) B ) = (cid:101) B (cid:48) . Then the set of all elements of Γ taking (cid:101) B to (cid:101) B (cid:48) is of the form S = Stab Γ ( (cid:101) B (cid:48) ) g Stab Γ ( (cid:101) B ) = g Stab Γ ( (cid:101) B ) = Stab Γ ( (cid:101) B (cid:48) ) g, both a left coset and a right coset of peripheral subgroups.Proof. Let h ∈ S . Then h differs from g by pre-composition with some element s ∈ Stab Γ ( (cid:101) B ) andpost-composition with some element s (cid:48) ∈ Stab Γ ( (cid:101) B (cid:48) ). In other words, S = Stab Γ ( (cid:101) B (cid:48) ) g Stab Γ ( (cid:101) B ) = (cid:8) s (cid:48) · g · s | s (cid:48) ∈ Stab Γ ( (cid:101) B (cid:48) ) , s ∈ Stab Γ ( (cid:101) B ) (cid:9) , proving the first equality of the lemma. Now, observe that Stab Γ ( (cid:101) B (cid:48) ) = g Stab Γ ( (cid:101) B ) g − . Thus S = g Stab Γ ( (cid:101) B ) g − · g Stab Γ ( (cid:101) B ) = g Stab Γ ( (cid:101) B ) , proving the second equality of the lemma. The final equality is proved similarly. (cid:3) NFINITELY MANY VIRTUAL GEOMETRIC TRIANGULATIONS 21
Now, let M be a cusped hyperbolic manifold containing a collection of disjoint, closed horocusps.Let A be one of the horocusps. Let γ , . . . , γ n be the set of all orthogeodesics below a certain length L that connect A to the union of the other cusps. This set is finite for any L , and non-empty when L is sufficiently large.Step 1 of the proof is accomplished by the following lemma. Lemma 5.2.
Let M = H / Γ be a cusped hyperbolic manifold and A ⊂ M a horocusp. For i =1 , . . . , n , let γ i be an orthogeodesic from ∂A to some horocusp of M . Assume that γ i (cid:54) = γ j for i (cid:54) = j .Then there is a finite cover ˚ f : ˚ M → M , where A lifts to a horocusp ˚ A ⊂ ˚ M . Furthermore, thepath-lifts ˚ γ , . . . , ˚ γ n that start at ˚ A lead to horocusps of ˚ M that are distinct from one another andfrom ˚ A . Recall from Definition 2.1 that a lift ˚ A must cover A with degree one. Thus, for each i , the path γ i has exactly one path-lift ˚ γ i starting at ˚ A . Proof.
Conjugate Γ in (cid:102) M = H so that one preimage of A is a horoball (cid:101) A about ∞ . The subgroup K = Stab Γ ( (cid:101) A ) ∼ = Z can be identified with π ( A ). Choose a fundamental domain D ⊂ ∂ (cid:101) A for theaction of K . For each i , let (cid:101) γ i be a path-lift of γ i to H , whose initial point lies in D . Let (cid:101) B i be thehoroball at the forward endpoint of (cid:101) γ i . By construction, the orthogeodesics (cid:101) γ , . . . , (cid:101) γ n lie in distinct K –orbits, hence the horoballs (cid:101) B , . . . , (cid:101) B n do also. In addition, each (cid:101) B i is disjoint from (cid:101) A .For every pair ( i, j ) with 1 ≤ i < j ≤ n , let S ij ⊂ Γ be the set of all deck transformations thatmap (cid:101) B j to (cid:101) B i . This set may be empty (this will be the case if (cid:101) B i and (cid:101) B j cover distinct cusps of M ).Otherwise, let g ij ∈ S ij be an arbitrary element and observe that by Lemma 5.1, S ij = g ij Stab Γ ( (cid:101) B j ).Let T ij be the set of all deck transformations that map (cid:101) B j to any horoball in the K –orbit of (cid:101) B i . If S ij (cid:54) = ∅ , we have T ij = KS ij = Kg ij Stab Γ ( (cid:101) B j ), a double coset of peripheral subgroups of Γ.In a similar fashion, let S j ⊂ Γ be the set of all deck transformations that map (cid:101) B j to (cid:101) A .If S j (cid:54) = ∅ , Lemma 5.1 says that S j = g j Stab Γ ( (cid:101) B j ) for an arbitrary element g j ∈ S j . Let T j = KS j . Finally, define T = (cid:91) ≤ i Let P be an ideal polyhedral decomposition of a cusped hyperbolic 3–manifold M .A diagonal of P is a bi-infinite geodesic β that is contained in some cell of P . A diagonal β is called returning if its endpoints are in the same horocusp of M .Step 3 of the proof is to apply the following result due to Luo, Schleimer, and Tillmann [22,Lemmas 8 and 9]. Lemma 5.4. Let ˚ M be a cusped hyperbolic –manifold with a polyhedral decomposition ˚ P . Thenthere is a finite regular cover (cid:98) f : (cid:99) M → ˚ M , such that the lifted polyhedral decomposition (cid:98) P has noreturning diagonals.Proof. This is proved in [22, Lemmas 8 and 9], using a fairly straightforward application of Propo-sition 4.3. (cid:3) We can now prove Theorem 1.3: a cusped hyperbolic 3–manifold M containing a non-rectangularcusp A has a cover (cid:99) M with infinitely many geometric ideal triangulations, organized in a trivalenttree. We stress that each edge of this tree represents a sequence of n geometric Pachner moves asdescribed by Corollary 3.8, where n is the degree of the local cover (cid:98) N → ˚ N of the ananas ˚ N ⊂ ˚ M . Proof of Theorem 1.3. Let M be a cusped hyperbolic 3–manifold, and let A ⊂ M be a non-rectangular cusp. Consider the sequence of finite covers (cid:99) M (cid:98) f −→ ˚ M ˚ f −→ M constructed in Lemma 5.2 and Lemma 5.4. In particular, ˚ M has a polyhedral decomposition ˚ P suchthat two acute-angled ideal tetrahedra of ˚ P fit together to form a drilled ananas ˚ N . Consequently, ∂ ˚ N consists of two triangular faces of ˚ P .Let (cid:98) P be the polyhedral decomposition of (cid:99) M obtained by pulling back ˚ P . Then (cid:98) P contains asubmanifold (cid:98) N that covers ˚ N . By Lemma 5.4, (cid:98) P has no returning diagonals, hence the vertices ofevery polyhedron P ⊂ (cid:98) P are mapped to distinct cusps of (cid:99) M .Choose an ordering ≺ on the cusps of (cid:99) M . Then, for every polyhedron P ⊂ (cid:98) P , we get a totalordering of the vertices of P . Thus, by Lemma 2.9, the iterated coning induced by ≺ subdivides (cid:98) P into geometric ideal tetrahedra. By construction, (cid:98) N already consists of tetrahedra, so does not needto be subdivided. Now, by Corollary 3.8, the initial geometric triangulation of (cid:98) N (which comes fromlifting the two-tetrahedron triangulation of ˚ N ) is the start of an infinite sequence of geometric idealtriangulations.Since A is non-rectangular, the triangulation of the drilled ananas ˚ N consists of two acute-angledtetrahedra. Thus every path in the trivalent tree of geometric triangulations of ˚ N that was describedin Proposition 3.10 lifts to a path of geometric triangulations of (cid:98) N , hence to a path of geometrictriangulations of (cid:99) M . (cid:3) Remark 5.5. If A is a rectangular cusp of M , the above proof of Theorem 1.3 still constructs covers (cid:99) M → ˚ M → M , where (cid:99) M contains a submanifold (cid:98) N that covers the drilled ananas ˚ N . However, this NFINITELY MANY VIRTUAL GEOMETRIC TRIANGULATIONS 23 time (cid:98) N consists of rectangular pyramids that need to be subdivided into tetrahedra. The subdivisionimposed by ordering the cusps of (cid:99) M may impose different choices of diagonals on the rectangles of ∂ (cid:98) N , which would obstruct the triangulation of (cid:98) N from being equivariant with respect to the cover (cid:98) N → ˚ N . This means we cannot apply Corollary 3.8 to obtain an infinite sequence of triangulations.The issue of equivariance does not arise if (cid:98) N is a lift of ˚ N .In the next section, we will deploy Theorem 1.4 to construct a cover (cid:99) M where (cid:98) N is indeed a lift,enabling us to handle rectangular cusps. This will have the additional benefit that each edge of thetrivalent tree will represent a single geometric 2–3 move, as in Proposition 3.10.6. Rectangular cusps and Dehn fillings In this section, we prove Theorem 1.2, which extends Theorem 1.3 to manifolds with rectangularcusps, and also provides infinitely many geometric triangulations of long Dehn fillings of (cid:99) M . Theproof begins in the same way as Steps 1 and 2 of the four-step outline described at the start ofSection 5. In particular, we will find a cover ˚ M → M that contains a drilled ananas ˚ N . The mainchallenge, as mentioned in Remark 5.5, is to build further covers where ˚ N continues to lift but(most) returning diagonals stop being returning. Before outlining how to do this, we introduce adefinition and a motivating example. Definition 6.1. Let M = H / Γ be a cusped hyperbolic 3–manifold with a distinguished horocusp A and a polyhedral decomposition P . Let (cid:101) A ⊂ H be a horoball covering A , and let (cid:101) P be the liftedpolyhedral decomposition. An (cid:101) A –parabolic diagonal of (cid:101) P is a bi-infinite geodesic (cid:101) β contained in acell of (cid:101) P , whose ends are in horoballs (cid:101) C and (cid:101) C (cid:48) such that there is a parabolic isometry g ∈ Stab Γ ( (cid:101) A )with g ( (cid:101) C ) = (cid:101) C (cid:48) .An A –parabolic diagonal of P is the projection β ⊂ M of an (cid:101) A –parabolic diagonal of (cid:101) P , forsome horoball (cid:101) A covering A . We remark that the choice of (cid:101) A is immaterial: for any other horoball (cid:101) A (cid:48) covering A , there will be some (cid:101) A (cid:48) –parabolic lift of β . On the other hand, the choice of P canaffect the collection of A –parabolic diagonals, because it affects the set of bi-infinite geodesics thatare diagonals of P in the first place. We also remark that an A –parabolic diagonal is necessarilyreturning, according to Definition 5.3.A horocusp C ⊂ M is called A –problematic (relative to P ) if there is an A –parabolic diagonal β of P whose endpoints lie in C . Example 6.2. Suppose, as in the conclusion of Corollary 3.4, that M = H / Γ contains a drilledananas N that is obtained by gluing one or two cells of P . Let A ⊂ M be a horocusp containingthe cusp of N , and let B ⊂ M be a horocusp containing the thorn of N . Every geodesic β ⊂ ∂N must have its endpoints in the single thorn. Consequently, β must be an A –parabolic diagonal of P ,because every lift of β to H must have its endpoints in horoballs that are permuted by Stab Γ ( (cid:101) A )for an appropriate horoball (cid:101) A covering A . Thus, by Definition 6.1, B is an A –problematic cusp.Now, suppose that M = H / Γ is some finite cover of M where A lifts to A . Since A lifts,or equivalently Stab Γ ( (cid:101) A ) = Stab Γ ( (cid:101) A ), every A –parabolic diagonal β of P lifts to an A –parabolicdiagonal β of the lifted polyhedral decomposition P . In particular, some preimage of β continues tobe a returning diagonal in M .The gist of the following outline is that Example 6.2 is a worst-case scenario that can be isolatedand handled. With Theorem 1.4 and with enough care, all diagonals that are not in the boundaryof a drilled ananas eventually lift to be non-returning, while the ananas continues to lift.Now, let M be a cusped hyperbolic 3–manifold containing a horocusp A . We will take thefollowing sequence of steps:Step 1. Describe a criterion on the distance from A that any A –problematic cusp must satisfy. Thisis accomplished in Lemma 6.3. Step 2. Using the criterion of Lemma 6.3, find a finite cover ˚ M → M , where A lifts to ˚ A . This cover˚ M contains a polyhedral decomposition ˚ P and a drilled ananas ˚ N with its cusp in ˚ A andits thorn in ˚ B , such that ˚ B is the only ˚ A –problematic cusp of ˚ M . In fact, all ˚ A –parabolicdiagonals of ˚ P lie in ∂ ˚ N . See Lemma 6.4 for details.Step 3. Using Theorem 1.4, find a finite cover M → ˚ M , where all of the above features hold (inparticular, ˚ N lifts to N ), and in addition every polyhedron of P has some vertex in ahorocusp other than B , the thorn of N . See Lemma 6.5 for details.Step 4. Using Theorem 1.4 again, find a finite cover ( M → M , where all of the above features hold(in particular, N lifts to ( N ), and in addition all returning diagonals have their endpoints incusps that cover B . This means there are very few returning diagonals, and the partial orderargument of Corollary 2.10 suffices to find a geometric triangulation ( T that is compatiblewith infinitely many geometric triangulations of ( M . See Lemma 6.6 for details.Step 5. Using H ( ( M ), find a double cover (cid:99) M → ( M where ( N has two distinct lifts, called (cid:98) N and (cid:98) N (cid:48) . See Lemma 6.7 for details. We will replace one lift (cid:98) N with a triangulated solid torusto perform a long Dehn filling (cid:99) M ( s ), while using the other lift (cid:98) N (cid:48) to obtain infinitely manygeometric triangulations of (cid:99) M ( s ).We now proceed to carry out these steps in detail. In the following lemma, M ( T ) is the modulispace of unit-area flat tori, and R + M ( T ) is the moduli space of flat tori of any area. Lemma 6.3. There is a function L : R + × R + M ( T ) → R + such that the following holds for everymulti-cusped hyperbolic –manifold M .Suppose that M contains a horocusp collection A, B , . . . , B k and that α is an orthogeodesic from A to B . Then, in the canonical polyhedral decomposition P determined by A, B , . . . , B k , any A –problematic horocusp B j must satisfy d ( A, B j ) < L = L (len( α ) , ∂A ) .Furthermore, if A is replaced by a sub-horocusp A t for some t > , then the distance bound L isreplaced by L + t . In symbols, L (len( α )+ t, ∂A t ) = L (len( α ) , ∂A ) + t. One particular consequence of Lemma 6.3 is that the length bound L only depends on the horo-cusps A and B . Although varying the sizes of B , . . . , B k may have the effect of changing thepolyhedral decomposition P , thereby changing the collection of A –parabolic diagonals of P , theconclusion of the lemma still holds for the same L . Proof. Write M = H / Γ, and conjugate Γ so that A is covered by a horoball (cid:101) A about ∞ . Then α has a lift (cid:101) α that leads from (cid:101) A to a horoball (cid:101) B covering B . A further conjugation, preserving thepoint ∞ , ensures that (cid:101) B has Euclidean diameter exactly 1. Then, setting (cid:96) = len( α ) = len( (cid:101) α ), itfollows that ∂ (cid:101) A lies at Euclidean height e (cid:96) . Let K = Stab Γ ( (cid:101) A ). Note that the length (cid:96) ∈ R + and theEuclidean metric ∂A ∈ R + M ( T ) determine the orbit of horoballs K (cid:101) B up to Euclidean isometry(equivalently, up to a hyperbolic isometry stabilizing ∂ (cid:101) A ).Suppose that (cid:101) β ⊂ H is an (cid:101) A –parabolic diagonal of (cid:101) P . Let (cid:101) C, (cid:101) C (cid:48) be the horoballs at the endpointsof (cid:101) β , and let (cid:101) P be a polyhedron containing (cid:101) β . Then, as in Definition 2.6, (cid:101) P contains the center of ametric ball D that is tangent to the horoballs about its vertices, including (cid:101) C and (cid:101) C (cid:48) , and is disjointfrom all other horoballs in the packing.Let h denote the Euclidean diameter of (cid:101) C , which is equal to the Euclidean diameter of (cid:101) C (cid:48) because (cid:101) C (cid:48) ∈ K (cid:101) C . Then d ( (cid:101) A, (cid:101) C ) = (cid:96) − log h . We will see that when h (cid:28) 1, or equivalently d ( (cid:101) A, (cid:101) C ) (cid:29) D lead to a contradiction.Let w be the shortest Euclidean translation length (along C = ∂ H (cid:114) {∞} ) of any element of K . Thus the Euclidean distance between the centers of (cid:101) C and (cid:101) C (cid:48) is at least w . Since D must betangent to (cid:101) C and (cid:101) C (cid:48) but disjoint from C , its Euclidean diameter is bounded below by a function of NFINITELY MANY VIRTUAL GEOMETRIC TRIANGULATIONS 25 w and h that grows without bound as h → 0. On the other hand, a ball of large Euclidean diameterwhose lowest point is below Euclidean height h must intersect one of the diameter 1 horoballs inthe K –orbit of (cid:101) B . Compare to Figure 2. Thus every sufficiently small value h (cid:28) L on d ( (cid:101) A, (cid:101) C ) = (cid:96) − log h . Observe that L depends onlyon the lattice of horoballs K (cid:101) B , hence on (cid:96) = len( α ) and the Euclidean metric on ∂A . Thus we maywrite L = L (len( α ) , ∂A ).To prove the “furthermore,” suppose that we replace A by a sub-horocusp A t . This has the effectof replacing (cid:96) by (cid:96) + t and replacing (cid:101) A by a horoball (cid:101) A t at Euclidean height e (cid:96) + t . Then the Euclideanlength w and the lattice of horoballs K (cid:101) B both remain the same, hence the same value of h (cid:28) d ( (cid:101) A, (cid:101) C ) has just increased by t . Thus replacing A by A t has the effect of replacing L by L + t . (cid:3) We can now begin constructing covers of a cusped hyperbolic 3–manifold M containing a horocusp A . In the following lemma, corresponding to Step 2, we build a cover ˚ M → M that supportsa polyhedral decomposition ˚ P that contains a drilled ananas ˚ N with its cusp in ˚ A , such that all˚ A –parabolic diagonals of ˚ P lie in ∂ ˚ N . Lemma 6.4. Let M be a cusped hyperbolic –manifold and A ⊂ M a horocusp. Then there is afinite cover ˚ f : ˚ M → M such that the following hold: • ˚ M contains a horocusp collection ˚ A, ˚ B = ˚ B , . . . , ˚ B k , where k ≥ and ˚ A ⊂ ˚ M is a lift of A . • There is an orthogeodesic ˚ α from ˚ A to ˚ B that is the unique shortest path from ˚ A to ∪ ˚ B j . • For large t > , the polyhedral decomposition ˚ P = ˚ P t determined by ˚ A t , ˚ B , . . . , ˚ B k containsa drilled ananas ˚ N , built out of one or two pyramids whose lateral edges are identified to ˚ α ,with its cusp in ˚ A and its thorn in ˚ B . • For every horoball (cid:101) A covering ˚ A , there is a corresponding preimage (cid:101) N of ˚ N , such that all (cid:101) A –parabolic diagonals of (cid:101) P lie in ∂ (cid:101) N . The main point in Lemma 6.4 is the last bullet, as it provides a partial converse to Example 6.2. Proof. Choose a collection of disjoint horocusps, containing A . Relative to this collection of horo-cusps, let α be a shortest orthogeodesic in M that starts at A . Let L = L (len( α ) , ∂A ) be the boundproduced by Lemma 6.3. Let S = { γ , . . . , γ n } be a set of n ≥ ∂A ,containing all the orthogeodesics that have length at most L . We set γ = α . By Lemma 5.2, thereis a finite cover ˚ M → M , where A lifts to a cusp ˚ A , and where the γ i ∈ S have path-lifts ˚ γ , . . . , ˚ γ n that start on ∂ ˚ A and lead to cusps that are distinct from one another and from ˚ A . Let ˚ B i be thehorocusp at the endpoint of ˚ γ i .Let ˚ B = ˚ B be the horocusp at the end of ˚ α = ˚ γ . We keep ˚ B fixed, but shrink each of ˚ B , . . . ˚ B n to ensure that d ( ˚ A, ˚ B i ) ≥ L for i ≥ , . . . n . Any other horocusp of ˚ M , labeled B i for i = n +1 , . . . , k ,must already satisfy d ( ˚ A, ˚ B i ) ≥ L . In particular, ˚ α is the unique shortest path from ˚ A to any cuspof ˚ M .Now, ˚ M and its collection of horocusps satisfies the hypotheses of Proposition 3.1. Thus, forsufficiently large t , we may replace ˚ A with ˚ A t and build a canonical polyhedral decomposition˚ P = ˚ P t that contains one or two ideal pyramids with a vertex in ˚ A and with their lateral edgesglued to ˚ α . By Corollary 3.4, these cells glue up to form an embedded, convex drilled ananas ˚ N ⊂ ˚ M .By construction, ˚ N has its cusp in ˚ A and its thorn in ˚ B .By Lemma 6.3, any ˚ A –problematic cusp ˚ B i in the polyhedral decomposition ˚ P must satisfy d ( ˚ A, ˚ B i ) < L or equivalently d ( ˚ A t , ˚ B i ) < L + t . The only horocusp satisfying these hypotheses is˚ B = ˚ B . Thus any ˚ A –parabolic diagonal must have its endpoints in ˚ B .Let ˚ β be an ˚ A –parabolic diagonal of ˚ P = ˚ P t . In the universal cover H , let (cid:101) A be a horoballcovering ˚ A , and let (cid:101) N be the preimage of ˚ N containing (cid:101) A t . The ˚ A –parabolic diagonal ˚ β lifts to an (cid:101) A –parabolic diagonal (cid:101) β , whose endpoints must be in horoballs (cid:101) B, (cid:101) B (cid:48) such that d ( (cid:101) A, (cid:101) B ) < L . By construction, ˚ α is the only orthogeodesic from ˚ A to ˚ B with length less than L . Therefore, (cid:101) B and (cid:101) B (cid:48) must be full-sized horoballs connected to (cid:101) A by lifts of ˚ α . Since the endpoints of (cid:101) β are in theideal vertices of (cid:101) N , and (cid:101) N is convex, it follows that (cid:101) β ⊂ (cid:101) N . In particular, ˚ β must lie in one of thepolyhedra comprising ˚ N .To complete the proof, recall that ˚ N consists of either two ideal tetrahedra or one rectangular-based ideal pyramid. In either case, we think of the constituent cells as pyramids with bases along ∂ ˚ N and lateral edges glued to ˚ α . Any diagonal in an ideal pyramid is either a lateral edge orcontained in the base. Since the ˚ A –parabolic diagonal ˚ β has both of its endpoints in ˚ B , it cannotbe a lateral edge, and must be contained in the base of the ambient pyramid. Thus ˚ β ⊂ ∂ ˚ N and (cid:101) β ⊂ ∂ (cid:101) N , as claimed. (cid:3) The next lemma, corresponding to Step 3 of the outline, is our first use of Theorem 1.4. Roughlyspeaking, the lemma says that there is a cover M → ˚ M where the drilled ananas ˚ N lifts, and wherereturning diagonals are controlled to a significant degree. Lemma 6.5. Let ˚ M = H / ˚Γ be a cusped hyperbolic –manifold containing a distinguished horocusp ˚ A and a drilled ananas ˚ N whose cusp is at ˚ A and whose thorn is in horocusp ˚ B . Suppose that ˚ P isa polyhedral decomposition of ˚ M , with the following property: for every horoball (cid:101) A covering ˚ A , thereis a corresponding preimage (cid:101) N of ˚ N , such that all (cid:101) A –parabolic diagonals of (cid:101) P lie in ∂ (cid:101) N . Then thereis a finite cover f : M → ˚ M such that the following hold: • ˚ A lifts to a distinguished cusp A . • ˚ N lifts to a drilled ananas N whose cusp is at A and whose thorn is in horocusp B . • Every A –parabolic diagonal of P has its endpoints in B . • Every polyhedron P ⊂ P has a vertex in some horocusp other than B .Proof. Let (cid:101) A be a horoball covering ˚ A , and let K = Stab ˚Γ ( (cid:101) A ). Let (cid:101) N ⊂ H be the preimage of ˚ N containing (cid:101) A . Then there is a horoball (cid:101) B covering ˚ B , such that all ideal vertices of (cid:101) N lie in (cid:101) A ∪ K (cid:101) B .By hypothesis, all (cid:101) A –parabolic diagonals of (cid:101) P lie in ∂ (cid:101) N and have their endpoints in horoballs of K (cid:101) B . Let ˚ P , . . . , ˚ P n be the polyhedra of ˚ P that have all of their vertices in ˚ B . (If no such polyhedraexist, we may simply set M = ˚ M and let f be the identity map.) For each ˚ P i , let ˚ β i be an edgethat is not ˚ A –parabolic. Such an edge must exist, because all ˚ A –parabolic edges belong to ∂ ˚ N .For each ˚ β i , choose a preimage (cid:101) β i ⊂ H . The ends of (cid:101) β i lie in horoballs (cid:101) B i , (cid:101) B (cid:48) i , which mustcover ˚ B because both endpoints of ˚ β i are in ˚ B . Thus there is an isometry g i ∈ ˚Γ = π ( ˚ M ) suchthat g i ( (cid:101) B i ) = (cid:101) B (cid:48) i . By Lemma 5.1, the set of all isometries in ˚Γ taking (cid:101) B i to (cid:101) B (cid:48) i is a left coset g i Stab ˚Γ ( (cid:101) B i ). Since ˚ β i is not ˚ A –parabolic, the coset g i Stab ˚Γ ( (cid:101) B i ) is disjoint from all ˚Γ–conjugates of K = Stab ˚Γ ( (cid:101) A ). Equivalently, K is disjoint from all ˚Γ–conjugates of g i Stab ˚Γ ( (cid:101) B i ).By Theorem 1.4, there is a homomorphism ϕ i : ˚Γ → G i , where G i is a finite group, such that ϕ i ( K ) is disjoint from all G i –conjugates of ϕ i (cid:0) g i Stab ˚Γ ( (cid:101) B i ) (cid:1) . We can now consider the producthomomorphism ϕ = ( ϕ , . . . , ϕ n ) : ˚Γ −→ G = G × · · · × G n . Then, for each i , the image ϕ ( K ) is disjoint from all G –conjugates of ϕ (cid:0) g i Stab ˚Γ ( (cid:101) B i ) (cid:1) .Now, let Γ = ϕ − ◦ ϕ ( K ), and let M = H / Γ. We get a covering map f : M → ˚ M . Then, byconstruction, every ˚Γ–conjugate of g i Stab ˚Γ ( (cid:101) B i ) is disjoint from Γ. Since K = Stab ˚Γ ( (cid:101) A ) = Stab Γ ( (cid:101) A ),the horocusp the horocusp ˚ A ⊂ ˚ M lifts to a horocusp A = (cid:101) A/K ⊂ M . Similarly, ˚ N lifts to a drilledananas N = (cid:101) N /K ⊂ M . The thorn of N is in the horocusp B that is covered by (cid:101) B , hence B covers˚ B . This proves the first two bullets in the lemma.For the next bullet, let γ be an A –parabolic diagonal in P . Since H → M is a regular cover, wemay choose a lift (cid:101) γ that is an (cid:101) A –parabolic diagonal in (cid:101) P . Recall that all (cid:101) A –parabolic diagonals in (cid:101) P lie in ∂ (cid:101) N and have their endpoints in horoballs of K (cid:101) B . Since K ⊂ Γ, it follows that γ ⊂ ∂N hasits endpoints in B , as desired. NFINITELY MANY VIRTUAL GEOMETRIC TRIANGULATIONS 27 To prove the remaining conclusion, let P be a polyhedron of P . If P has an ideal vertex in somecusp that does not belong to f − ( ˚ B ), then certainly P has a vertex that is not in B . Otherwise, P = P i is a lift of some ˚ P i . Let β i ⊂ P i be a lift of ˚ β i ⊂ ˚ P i . We claim that the endpoints of β i arein distinct cusps of M , and in particular one endpoint is not in B .Let (cid:101) γ i ⊂ H be an arbitrary preimage of β i , and let (cid:101) C i , (cid:101) C (cid:48) i be horoballs in the packing containingthe ends of (cid:101) γ i . Since (cid:101) γ i and (cid:101) β i both cover ˚ β i ⊂ ˚ M , there is an isometry h i ∈ ˚Γ such that h i ( (cid:101) γ i ) = (cid:101) β i ,which implies h i ( (cid:101) C i ) = (cid:101) B i and h i ( (cid:101) C (cid:48) i ) = (cid:101) B (cid:48) i . Thus the set of all isometries in ˚Γ taking (cid:101) C i to (cid:101) C (cid:48) i canbe written as h − i · g i Stab ˚Γ ( (cid:101) B i ) · h i . By construction, this conjugate of g i Stab ˚Γ ( (cid:101) B i ) is disjoint fromΓ. Thus (cid:101) C i and (cid:101) C (cid:48) i lie in distinct Γ–orbits, which means that the endpoints of β i are in distinctcusps. This proves the claim and the lemma. (cid:3) The next lemma, corresponding to Step 4 of the outline, builds a cover ( M with even strongerrestrictions on the returning diagonals of the polyhedral decomposition ( P . Lemma 6.6. Let M = H / Γ be a cusped hyperbolic –manifold containing a distinguished horocusp A and a horocusp B . Suppose that P is a polyhedral decomposition of M , such that every A –parabolicdiagonal of P has its endpoints in B . Then there is a finite cover ( f : ( M → M where A lifts to adistinguished cusp ( A , such that every returning diagonal of ( P has its endpoints in ( f − ( B ) .Proof. Let R = { β , . . . , β n } be the set of returning diagonals of P whose endpoints are not in B . This set is finite because P has finitely many polyhedra, and each polyhedron contains finitelymany diagonals. By hypothesis, every diagonal β i ∈ R is not A –parabolic. Assume that R (cid:54) = ∅ , asotherwise we may simply take ( M = M .Let (cid:101) A be a horoball covering A , and let K = Stab Γ ( (cid:101) A ). For each β i , choose a preimage (cid:101) β i ⊂ H .The ends of (cid:101) β i lie in horoballs (cid:101) B i , (cid:101) B (cid:48) i , which cover the same cusp of M because β i is a returningdiagonal. Thus there is an isometry g i ∈ Γ = π ( M ) such that g i ( (cid:101) B i ) = (cid:101) B (cid:48) i . By Lemma 5.1, the setof all isometries in Γ taking (cid:101) B i to (cid:101) B (cid:48) i is a left coset g i Stab Γ ( (cid:101) B i ). Since β i is not A –parabolic, thecoset g i Stab Γ ( (cid:101) B i ) is disjoint from all Γ–conjugates of K = Stab Γ ( (cid:101) A ).As in the proof of Lemma 6.5, we use Theorem 1.4 to find a finite-index subgroup ( Γ ⊂ Γ thatcontains K and is disjoint from every Γ–conjugate of g i Stab Γ ( (cid:101) B i ). Let ( M = H / ( Γ. Since K ⊂ ( Γ,the horocusp A ⊂ M lifts to a horocusp ( A ⊂ ( M . As in the proof of Lemma 6.5, the disjointnessof ( Γ and all Γ–conjugates of g i Stab Γ ( (cid:101) B i ) implies that every lift ( β i of β i has endpoints in distinctcusps, and is not a returning diagonal. Thus any returning diagonal in ( M must be the preimage ofa returning diagonal in M whose endpoints are in B , hence it has endpoints in ( f − ( B ). (cid:3) Following Lemma 6.6, the manifold ( M has so few returning diagonals that it is possible to imposea partial order ≺ on the cusps of ( M such that every polyhedron ( P ⊂ ( P has a unique lowest vertex.Using Corollary 2.10, we can refine ( P to a geometric triangulation T , and apply Lemma 3.6 to buildinfinitely many geometric triangulations of ( M . See Claim 6.8 below for details.To find geometric triangulations Dehn fillings, we need to take one more cover. Lemma 6.7. Let ( M = H / ( Γ be a cusped hyperbolic –manifold containing at least three cusps anda distinguished horocusp ( A . Then there is a double cover (cid:98) f : (cid:99) M → ( M , where ( A has two distinct lifts.Proof. Since ( M is the interior of a compact 3–manifold with at least three boundary tori, the “halflives, half dies” lemma [19, Lemma 3.5] implies that H ( ( M ) has a Z n direct summand for n ≥ 3. Let G be the subgroup of H ( ( M ) induced by the inclusion ( A → ( M . Since G has rank at most 2, theremust be a primitive, infinite-order homology class h ∈ H ( ( M ) such that (cid:104) h (cid:105) is a direct summand that is linearly independent from G . Thus we may define a projection π h : H ( ( M ) → (cid:104) h (cid:105) such that G ⊂ ker( π h ).Now, consider the sequence of surjective homomorphisms ( Γ = π ( ( M ) ab −−→ H ( ( M ) π h −−→ (cid:104) h (cid:105) ∼ = Z −→ Z / Z , where ab is abelianization. Let (cid:98) ϕ : ( Γ → Z / Z be the composition, and let (cid:98) Γ = ker( (cid:98) ϕ ). By construc-tion, G ⊂ ker( π h ), hence π ( ( A ) ⊂ ker( (cid:98) ϕ ) = (cid:98) Γ. By the lifting criterion, ( A lifts to (cid:99) M = H / (cid:98) Γ. Since (cid:98) f : (cid:99) M → ( M is a regular cover of degree 2, there must be two distinct lifts. (cid:3) We can now prove our main result, Theorem 1.2. Proof of Theorem 1.2. Let M be a cusped hyperbolic 3–manifold containing a horocusp A . Considerthe sequence of finite covers (cid:99) M (cid:98) f −→ ( M ( f −→ M f −→ ˚ M ˚ f −→ M constructed in the preceding lemmas. The cusp A ⊂ M lifts along each covering map. Recall thatby Lemma 6.4, ˚ M has at least three cusps, hence ( M does also, and we may indeed apply Lemma 6.7to construct (cid:98) f .By Lemma 6.4, ˚ M has a polyhedral decomposition ˚ P such that one or two ideal cells of ˚ P fittogether to form a drilled ananas ˚ N that deformation retracts to ˚ A . The 1–skeleton of ˚ P decomposes ∂ ˚ N into two ideal triangles or one ideal rectangle. Furthermore, since ˚ A lifts to A ⊂ M and ( A ⊂ ( M ,the drilled ananas ˚ N lifts to N ⊂ M and ( N ⊂ ( M . Claim 6.8. The polyhedral decomposition ( P of ( M can be refined to a geometric ideal triangulation ( T . If ∂ ( N is a single ideal rectangle, then we may choose either diagonal of this rectangle to be anedge in ( T . Finally, ( T is the start of an infinite sequence of geometric triangulations of ( M . Recall, from Lemma 6.5, that the drilled ananas N ⊂ M has its thorn in a horocusp B , such thatevery polyhedron P ⊂ P has a vertex in some horocusp apart from B . Thus, in the lifted polyhedraldecomposition ( P of ( M , every polyhedron ( P must have at least one vertex in a horocusp that is notin ( f − ( B ). We call the cusps of ( f − ( B ) blue . By Lemma 6.6, all returning diagonals of ( P havetheir endpoints in blue cusps.Let V be the set of cusps of ( M . We impose a partial order ≺ on V as follows: the non-blue cuspsare totally ordered in some fashion; the blue cusps are pairwise incomparable; and ( C ≺ ( B for everyblue cusp ( B and non-blue cusp ( C . Since every polyhedron P ⊂ ( P has at least one non-blue idealvertex, and the non-blue vertices of P are totally ordered below the blue ones, it follows that P hasa unique ≺ –minimal vertex. Thus, by Lemma 2.9, the iterated coning of P induced by ≺ produces ( P (cid:48) , a well-defined subdivision of ( P into geometric ideal pyramids. By Corollary 2.10, any choice ofdiagonals in the non-triangular faces of ( P (cid:48) produces a geometric ideal triangulation ( T .By construction, the thorn of ( N is in a blue cusp of ( f − ( B ). Thus the partial order ≺ does notimpose any ordering on the ideal vertices of ∂ ( N . If this boundary is a single ideal rectangle, theconing induced by ≺ in Lemma 2.9 does not subdivide it, and we may choose our preferred diagonalto subdivide ( N into two ideal tetrahedra. By Lemma 3.6, ( N admits infinitely many geometric idealtriangulations, hence ( M does also. This proves the claim. (cid:54) Next, we lift the triangulation ( T of ( M to a geometric triangulation (cid:98) T of (cid:99) M . Then the infinitesequence of geometric triangulations of ( M lifts to an infinite sequence of geometric triangulations of (cid:99) M , as claimed in the statement of the theorem. NFINITELY MANY VIRTUAL GEOMETRIC TRIANGULATIONS 29 By Lemma 6.7, the horocusp ( A ⊂ ( M has two distinct lifts to (cid:99) M , which we call (cid:98) A and (cid:98) A (cid:48) .Consequently, the drilled ananas ( N ⊃ ( A also has two distinct lifts to (cid:99) M , namely (cid:98) N ⊃ (cid:98) A and (cid:98) N (cid:48) ⊃ (cid:98) A (cid:48) . The two distinct lifts of ( A and ( N play distinct roles in the Dehn filling argument. Claim 6.9. For all but finitely many choices of slope s on ∂ (cid:98) A , the following hold: • The Dehn filled manifold (cid:99) M ( s ) has a hyperbolic structure where the core curve γ of the filledsolid torus is isotopic to a closed geodesic γ s . • The ideal tetrahedra of (cid:98) T remain geometric in the hyperbolic metric on (cid:99) M ( s ) . Each idealvertex of a tetrahedron of (cid:98) T that used to enter cusp (cid:98) A now spins about the geodesic γ s . • With an appropriate choice of diagonals in Claim 6.8, the union of the two tetrahedra in (cid:98) N has convex boundary in (cid:99) M ( s ) . The first two bullets in the claim follow from Thurston’s hyperbolic Dehn surgery theorem [29,Chapter 4]. For a sufficiently long slope s , the hyperbolic metric on (cid:99) M ( s ) is obtained via anarbitrarily small deformation of the metric on (cid:99) M . Thus, for every ideal tetrahedron T ⊂ (cid:98) T , asufficiently small deformation of the metric will keep T geometric and positively oriented. As in [29,Section 4.4], the two tips of ideal tetrahedra that enter (cid:98) A will now spin about the core geodesic γ s .For the last bullet of the claim, suppose first that the cusp A ⊂ M is non-rectangular (hence,so are its lifts). Then Corollary 3.4 implies that the original drilled ananas ˚ N consists of acutetetrahedra, and ∂ ˚ N is strictly convex at all three of its edges. The same properties are preserved inthe lift (cid:98) N and are still preserved in (cid:98) N ( s ) after a sufficiently small deformation of the metric.Next, suppose that A ⊂ M is rectangular. Then Corollary 3.4 implies that the original drilledananas ˚ N has convex boundary, with interior angles strictly less than π at the two edges of ∂ ˚ N ∩ ˚ P ,and an angle of π along the (arbitrary) diagonal of the ideal rectangle. The same properties remaintrue in the lift (cid:98) N ⊂ (cid:99) M . When we deform the metric on (cid:99) M to obtain (cid:99) M ( s ), the interior angle of π may become π + (cid:15) for small (cid:15) , violating convexity, but then the opposite choice of diagonal on ∂ (cid:98) N will have interior angle π − (cid:15) . Thus an appropriate choice of diagonal in Claim 6.8 keeps ∂ (cid:98) N convexin (cid:99) M ( s ). (cid:54) To construct geometric triangulations of (cid:99) M ( s ), we need to introduce the solid-torus analogueof a drilled ananas. A filled ananas is a 3–manifold X homeomorphic to a solid torus with oneboundary point removed, and endowed with a complete hyperbolic metric with the following prop-erties. The boundary ∂X is made up of two totally geodesic ideal triangles, with vertices at theremoved point. These two ideal triangles are glued by isometry along their edges to form a standardtwo-triangle triangulation of a once-punctured torus, with shearing and bending allowed along theedges. Furthermore, X is subdivided into geometric ideal tetrahedra. Claim 6.10. For all but finitely many choices of slope s on ∂ (cid:98) A , the hyperbolic manifold (cid:99) M ( s ) has a geometric triangulation (cid:98) T ( s ) with the following properties. Finitely many tetrahedra of (cid:98) T ( s ) fit together to form a filled ananas X ( s ) . Furthermore, the restriction of (cid:98) T ( s ) to the complement (cid:99) M ( s ) (cid:114) X ( s ) is combinatorially isomorphic to the restriction of (cid:98) T to the complement (cid:99) M (cid:114) (cid:98) N . This statement is due to Gu´eritaud and Schleimer, and closely resembles [15, Theorem 1]. As-suming that (cid:98) T is the canonical triangulation of (cid:99) M with respect to some choice of horocusps, theyconstruct the triangulated filled ananas X ( s ) and endow it with a geometric structure isometric tothe completion of the two spun tetrahedra of (cid:98) N mentioned in Claim 6.9. Then they replace (cid:98) N with X ( s ), and prove that the resulting triangulation (cid:98) T ( s ) is the canonical triangulation of (cid:99) M ( s ). Infact, the construction of X ( s ), which occurs in [15, Section 2] and is encapsulated in [15, Corollary16], only uses the hypotheses that (cid:98) T is geometric and that (cid:98) N remains convex in (cid:99) M ( s ). See also [16,Corollary 4.18]. The combinatorial structure of the triangulation of X ( s ) is closely guided by thecombinatorics of the Farey graph F and the continued fraction expansion of the filling slope s , while the hyperbolic metric on X ( s ) is constructed using Casson and Rivin’s work on angle structures andvolume optimization [27, 11]. In particular, the canonicity of (cid:98) T is not needed in the proof that (cid:98) T ( s )is geometric. (cid:54) To complete the proof of the theorem, we have Claim 6.11. For all but finitely many choices of slope s on ∂ (cid:98) A , the Dehn filled manifold (cid:99) M ( s ) hasan infinite sequence of geometric triangulations connected by geometric – moves. Observe that by Claim 6.10, the geometric triangulation (cid:98) T ( s ) agrees with (cid:98) T on the drilled ananas (cid:98) N (cid:48) ⊂ (cid:99) M ( s ). By Lemma 3.6, this two-tetrahedron geometric triangulation of (cid:98) N (cid:48) is the start of aninfinite sequence of geometric triangulations connected by geometric 2–3 moves. (cid:54) (cid:3) The following remark states a version of Theorem 1.3 for manifolds with rectangular cusps. Inthe statement, a 4 – move is a local move on triangulations, which takes an octahedron that hasbeen decomposed into 4 tetrahedra along one of its three internal diagonals and replaces it with adecomposition into 4 tetrahedra along a different internal diagonal. The move is called a geometric – move if both decompositions are into geometric ideal tetrahedra. Remark 6.12. If A ⊂ M is a rectangular cusp, the cover ( M contains an infinite trivalent tree ofgeometric ideal triangulations, where one edge of the tree is a geometric 4–4 move and the remainingedges are geometric 2–3 moves. This can be seen as follows. In Claim 6.8 of the above proof, thedrilled ananas ( N consists of an ideal rectangular pyramid ( P ⊂ ( P . After the pyramidal decompositioninduced by ≺ , the 3–cell ( P (cid:48) glued to ( P is also an ideal rectangular pyramid. The two choices ofdiagonal for the shared face of ( P ∩ ( P (cid:48) lead to ideal triangulations that differ by a geometric 4–4move. Each of these choices can serve in Lemma 3.6 as the starting configuration in an infinitesequence of geometric ideal triangulations. In the proof of Proposition 3.10, the dual tree of theFarey complex splits in half along the edge − : half of the tree is reachable by geometric 2–3moves if we choose the diagonal , and the other half is reachable if we choose the diagonal − . Theunion of these halves is an infinite trivalent tree of geometric 2–3 moves, with one edge replaced bya 4–4 move.We close the paper by pointing out that the figure–8 knot complement M does not contain adrilled ananas, because M has only one cusp. 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