Bull. Soc. math. France 130 (3), 2002, p. 457–491

HYPERIDEAL POLYHEDRA IN HYPERBOLIC 3-SPACE

by Xiliang Bao & Francis Bonahon

Abstract.—Ahyperidealpolyhedronisanon-compactpolyhedroninthehyperbolic 3-space 3 which, in the projective model for 3 3,isjusttheintersectionof ⊂ 3 with a projective polyhedron whose vertices are all outside 3 and whose edges all meet 3 .Weclassifyhyperidealpolyhedra,uptoisometriesof 3 ,intermsoftheir combinatorial type and of their dihedral angles. Resum´ e´ (Poly`edres hyperid´eaux de l’espace hyperbolique de dimension 3) Un poly`edre hyperid´eal est un poly`edre non-compact de l’espace hyperbolique 3 de dimension 3 qui, dans le mod`ele projectif pour 3 3,estsimplementl’intersection ⊂ de 3 avec un poly`edre projectif dont les sommets sont tous en dehors de 3 et dont toutes les arˆetes rencontrent 3 .Nousclassifionscespoly`edres hyperid´eaux, `aisom´etrie de 3 pr`es, en fonction de leur type combinatoire et de leurs angles di´edraux.

Consider a compact convex polyhedron P ,intersectionoffinitelymanyhalf- spaces in one of the three n-dimensional homogeneous spaces, namely the eu- clidean space En,thesphereSn or the hyperbolic space Hn.TheboundaryofP inherits a natural cell decomposition, coming from the faces of the polyhedron.

Texte re¸cu le 26 juin 2001, accept´ele8d´ecembre 2001 Xiliang Bao,DepartmentofMathematics,UniversityofSouthernCalifornia,LosAngeles CA 90089-1113 (USA) Computer Associates, 9740 Scranton Road # 200, San Diego, • CA 92121-1745 (USA) E-mail : [email protected] • Francis Bonahon,DepartmentofMathematics,UniversityofSouthernCalifornia,Los Angeles CA 90089-1113, (USA) E-mail : [email protected] • Url : http://math.usc.edu/~fbonahon 2000 Mathematics Subject Classification. — 51M09. Key words and phrases. — Hyperbolic space, polyhedron, ideal polyhedron, hyperideal. This work was partially supported by NSF grants DMS-9504282 and DMS-9803445 from the US National Science Foundation.

BULLETIN DE LA SOCIET´ EMATH´ EMATIQUE´ DE FRANCE 0037-9484/2002/457/$ 5.00 c Soci´et´eMath´ematique de France ! 458 BAO (X.) & BONAHON (F.)

Along each (n 2)-face e,wecanmeasuretheinternaldihedralangleαe ]0,π[ between the two− (n 1)-faces meeting along e. A natural question then∈ arises: − If we are given an (n 1)-dimensional cell complex X with a weight αe ]0,π[ attached to each (n − 2)-dimensional cell e,isthereaconvexpolyhedron∈ P in En, Sn or Hn whose− boundary has the combinatorial structure of this cell complex X,andsuchthatαe is the dihedral angle of P along the face e? An explicit computation provides a full answer in the simplest case where X is the boundary of the n-simplex. The solution involves the signatures of var- ious minors of the symmetric n n-matrix whose ij-entry is +1 if i = j and × is cos αeij ,whereeij is the edge joining the i-th vertex to the j-th vertex, if i−= j;see[6],[17].Inparticular,theanswerisexpressedintermsofthesigns $ of polynomials in cos αeij .Sincethisconditionontheanglesαe is not that easy to handle, one can expect the general case to be quite intractable, and this indeed seems to be the case. In general, the main technical difficulty is to control the combinatorics as one deforms the polyhedron P .Atypicalproblemoccurswhenap-dimensional face becomes (p 1)-dimensional, for instance when two vertices collide so that a1-dimensionalfaceshrinkstoonepoint.− One way to bypass this technical difficulty is to impose additional restric- tions which will prevent such vertex collisions and face collapses. For instance, one can require that all dihedral angles αe are acute, namely lie in the inter- 1 val ]0, 2 π]. In this context, Coxeter [6] proved that every acute angled compact convex polyhedron in the euclidean space En is an orthogonal product of eu- clidean simplices, possibly lower dimensional; this reduces the problem to the case of euclidean simplices, which we already discussed. Similarly, Coxeter also proved that every acute angled convex polyhedron in the sphere Sn is a simplex. The situation is more complex in the hyperbolic space Hn but, when n =3, Andreev was able to classify all acute angled compact convex polyhedra in H3 in terms of their combinatorics and their dihedral angles [2]. In hyperbolic space, another approach to prevent vertex collisions is to put these vertices infinitely apart, by considering (non-compact finite volume) ideal polyhedra,whereallverticessitonthesphereatinfinity∂ Hn of Hn.In[11], Rivin classifies all ideal polyhedra in H3 in terms of their combinatorics∞ and of their dihedral angles. The case of acute angled polyhedra with some vertices at infinity had been earlier considered by Andreev [3]. In this paper, we propose to go one step further by considering polyhedra in H3 whose vertices are ‘beyond infinity’, and which we call hyperideal polyhedra. These are best described in Klein’s projective model for H3.Recallthat,in this model, H3 is identified to the open unit ball in R3 RP3,thatgeodesics of H3 then correspond to the intersection of straight lines⊂ of R3 with H3,and that totally geodesic planes in H3 are the intersection of linear planes with H3. In this projective model H3 RP3,ahyperideal polyhedron is defined as the ⊂

tome 130 – 2002 – no 3 HYPERIDEAL POLYHEDRA IN HYPERBOLIC 3-SPACE 459 intersection P of H3 with a compact convex polyhedron P Proj of RP3 with the following properties: 1) Every vertex of P Proj is located outside of H3; 2) Every edge of P Proj meets H3. Note that we allow vertices of P Proj to be located on the unit sphere ∂ H3 bounding H3,sothathyperidealpolyhedraincludeidealpolyhedraasaspecial∞ case. From now on, we will restrict attention to the dimension n =3.Following the standard low-dimensional terminology, we will call vertex any 0-dimensional face or cell, an edge will be a 1-dimensional face or cell, and we will reserve the word face for any 2-dimensional face or cell. To describe the combinatorics of a hyperideal polyhedron P ,itisconvenient to consider the dual graph Γofthecelldecompositionof∂P,namelythegraph whose vertices correspond to the (2-dimensional) faces of P ,andwheretwo vertices v and v" are connected by an edge exactly when the corresponding faces f and f " of P have an edge in common. Note that Γis also the dual graph of the projective polyhedron P Proj associated to P . The graph Γmust be planar,inthesensethatitcanbeembeddedinthe sphere S2.Inaddition,Γis3-connected in the sense that it cannot be discon- nected or reduced to a single point by removing 0, 1 or 2 vertices and their adjacent edges; this easily follows from the fact that two distinct faces of P Proj can only meet along the empty set, one vertex or one edge. A famous theorem of Steinitz states that a graph is the dual graph of a convex polyhedron in R3 if and only if it is planar and 3-connected; see [8]. A classical consequence of 3- connectedness is that the embedding of Γin S2 is unique up to homeomorphism of S2;seeforinstance[9, 32]. In particular, it intrinsically makes sense to talk of the components of S2 § Γ. Note that these components of S2 Γnaturally correspond to the vertices− of P Proj. − The results are simpler to state if, instead of the internal dihedral angle αe of the polyhedron P along the edge e,weconsidertheexternal dihedral an- gle θ = π α ]0,π[. e − e ∈

Theorem 1.—Let Γ be a 3-connected planar graph with a weight θe ]0,π[ attached to each edge e of Γ.ThereexistsahyperidealpolyhedronP in H∈3 with dual graph isomorphic to Γ and with external dihedral angle θe along the edge corresponding to the edge e of Γ if and only if the following two conditions are satisfied:

1) For every closed curve γ embedded in Γ and passing through the edges e1, e ,...,e of Γ, n θ 2π with equality possible only if γ is the boundary 2 n i=1 ei ≥ of a component of S2 Γ; ! − 2) For every arc γ embedded in Γ,passingthroughtheedgese1, e2,...,en of Γ,joiningtwodistinctverticesv1 and v2 which are in the closure of the same

BULLETIN DE LA SOCIET´ EMATH´ EMATIQUE´ DE FRANCE 460 BAO (X.) & BONAHON (F.) component A of S2 Γ but such that γ is not contained in the boundary of A, n − i=1 θei >π. Proj ! In addition, for the projective polyhedron P associated to P ,avertexof P Proj is located on the sphere at infinity ∂ H3 if and only if equality holds in Condition 1 for the boundary of the corresponding∞ component of S2 Γ. − Note that Theorem 1 generalizes Rivin’s existence result for ideal polyhe- dra [11].

Theorem 2.—The hyperideal polyhedron P in Theorem 1 is unique up to isometry of H3.

Theorem 2 was proved by Rivin [10], [11] for ideal polyhedra, and by Rivin and Hodgson [12] (if we use the truncated polyhedra discussed in 1) for the other extreme, namely for hyperideal polyhedra with no vertex on the§ sphere at infinity. Even in these cases, one could argue that our proof is a little simpler, as it is based on relatively simple infinitesimal lemmas followed by covering space argument, as opposed to the more delicate global argument of Lemma 4.11 of [12]. However, the main point of Theorem 2 is that it is a key ingredient for the proof of Theorem 1, justifying once again the heuristic principle that “uniqueness implies existence”. Theorem 2 is the reason why we introduced Condition 2 in the definition of hyperideal polyhedra, as it fails for general polyhedra without vertices in H3. Our proof of Theorems 1 and 2 is based on the continuity method pioneered by Aleksandrov [1] and further exploited in [2] and [12]. We first use an implicit function theorem, proved through a variation of Cauchy’s celebrated rigidity theorem for euclidean polyhedra [5], to show that a hyperideal polyhedron is locally determined by its combinatorial type and its dihedral angles. We then go from local to global by a covering argument. Although the generalization of the results of [11] from ideal polyhedra to hyperideal polyhedra is of interest by itself, the real motivation for this work was to provide a proof of the classification of ideal polyhedra which locally controls the combinatorics of the polyhedra involved. Ideal polyhedra play an important role in 3-dimensional geometry, as they can be used as building blocks to construct hyperbolic 3-manifolds through the use of ideal triangula- tions, possibly not locally finite. To study deformations of a hyperbolic metric on a 3-manifold, it is therefore useful to have a good classification of the defor- mations of ideal polyhedra within a given combinatorial type. Unfortunately, Rivin’s argument in [11] is indirect. He first uses the classification of compact hyperbolic polyhedra by their dual polyhedra [12], where one completely looses control of the combinatorics, and he extends it to ideal polyhedra by passing to the limit as the vertices go to infinity; he then observes that for ideal polyhedra the dual polyhedron does determine the combinatorics of the ideal polyhedron.

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In this regard, the local characterization of hyperideal polyhedra by their di- hedral angles provided by our Theorem 11, which is already the key technical step in this paper, may be its most useful result for applications. The paper [10] provides a different approach to a local control of ideal polyhe- dra through their combinatorics and dihedral angles. The reader is also referred to [13], [14] for the consideration of other rigidity properties of polyhedra in hyperbolic 3-space. It may also be of interest that Theorems 1 and 2 can be translated into purely euclidean (or at least projective) statements. Indeed, they provide a classification of hyperideal projective polyhedra P Proj modulo the action of the group PO(3, 1) consisting of those projective transformations of RP3 that respect the unit sphere S2 R3 RP3.Thisisparticularlyremarkablewhen one notices that, for an edge⊂ e ⊂of P = H3 P Proj,thehyperbolicdihedral ∩ 2 angle θe of P is equal to the euclidean angle between the two circles Π S 2 ∩ and Π" S at their intersection points, where ΠandΠ " are the two euclidean planes respectively∩ containing the two faces of P that meet along e.Byduality, Theorems 1 and 2 also classify convex projective polyhedra whose faces all meet the unit sphere S2 R3 RP3 but whose edges are all disjoint from the closed ball H3 S2,modulotheactionofPO(3⊂ ⊂ , 1). Theorems∪ 1 and 2 for the somewhat simpler case of strictly ideal polyhedra, where all vertices of P Proj are outside of the closure of H3,appearedin[4]. The final draft of this paper was essentially completed while the second author was visiting the Institut des Hautes Etudes´ Scientifiques, which he would like to thank for its productive hospitality. The authors are also grateful to the referee for several suggestions of improvement of the exposition, including a simplification of the proof of Proposition 6.

1. Hyperideal polyhedra We first recall a few basic facts about the projective model for H3 (see for instance [16]). Consider the symmetric bilinear form B (X ,X ,X ,X ), (Y ,Y ,Y ,Y ) = X Y + X Y + X Y + X Y 0 1 2 3 0 1 2 3 − 0 0 1 1 2 2 3 3 on R4".IntheprojectivespaceRP3,weconsidertheimage# H3 of the set of those X R4 with B(X, X) < 0. For the standard embedding of R3 in RP3, ∈ 3 3 defined by associating to (x1,x2,x3) R the point of RP with homogeneous ∈ 3 coordinates (1,x1,x2,x3), the subset H just corresponds to the open unit ball in R3. The projection R4 RP3 induces a diffeomorphism between H3 and the set H of those X =(X→,X ,X ,X ) R4 with B(X, X)= 1andX > 0. 0 1 2 3 ∈ − 0 The tangent space TXH of H at X is equal to the B-orthogonal of X and,

BULLETIN DE LA SOCIET´ EMATH´ EMATIQUE´ DE FRANCE 462 BAO (X.) & BONAHON (F.)

since B(X, X) < 0andB has signature (3, 1), the restriction of B to TXS is H3 therefore positive definite. The corresponding riemannian metric on H ∼= is exactly the hyperbolic metric of the projective model for H3. The group O(3, 1) of linear B-isometries of R4 induces an action of PO(3, 1) = O(3, 1)/ Id on H3 which respects the metric of H3.The group PO(3, 1) is actually{± } equal to the whole isometry group of H3.Note that PO(3, 1) is naturally isomorphic to the subgroup of O(3, 1) consisting of those elements that respect H. In this model, the geodesics of H3 are exactly the non-empty intersections of H3 with projective lines of RP3 or, equivalently, with straight lines of R3 RP3. Similarly, the (totally geodesic) hyperbolic planes in H3 correspond to the⊂ non- empty intersections of H3 with projective planes of RP3 or, equivalently, with affine planes of R3.Wewilloccasionallyusethepropertythat,giventwo 3 projective planes Πand Π" whose intersection meets H ,thehyperbolicdihedral 3 3 angle between the hyperbolic planes Π H and Π" H along the geodesic 3 ∩ ∩ 3 Π Π" H is exactly equal to the euclidean angle between the circles Π ∂ H ∩ ∩ 3 ∩ ∞ 3 and Π" ∂ H at their two intersection points in the sphere at infinity ∂ H bounding∩ H∞3 in R3. ∞ One of the most valuable features of the projective model for H3 is its du- ality properties. Given a k-dimensional projective subspace & RP3,with ⊂ 0 k 2, the B-orthogonal L⊥ of the (k +1)-dimensionallinearsubspaceL of≤R4 ≤corresponding to & is a (3 k)-dimensional linear subspace of R4,and − 3 therefore defines a (2 k)-dimensional projective space &⊥ of RP . − 3 3 In particular, if x is a point which is not in the closure of H in RP , x⊥ is aplanewhichmustintersectH3,sinceotherwiseB would be positive definite 3 or degenerate. By elementary linear algebra, if y x⊥ H and if l denotes the projective line passing through x and y,thegeodesic∈ ∩l H3 is orthogonal 3 3 ∩ to the hyperbolic plane x⊥ H for the metric of H .Similarly,ifz belongs to ∩ 3 3 the intersection of x⊥ with the sphere ∂ H bounding H ,thentheprojective line joining x to z cannot meet H3,andisthereforetangentto∞ ∂ H3.Thislast ∞ point gives us a very geometric way to construct x⊥:Drawalltheprojective 3 lines passing through x and tangent to ∂ H ;thenx⊥ is the projective plane which intersects ∂ H3 along the circle formed∞ by all the points of tangency. ∞ 3 When x is a point of the sphere ∂ H ,thedualplanex⊥ is just the plane tangent to ∂ H3 at x.Thiscan,forinstance,beseenbycontinuityfromthe∞ previous case.∞ Conversely, if Πis a hyperbolic plane in H3,thepointx such that Π= 3 3 3 x⊥ H is necessarily outside of the closure of H in RP ,forsignaturereasons. In particular,∩ the line L in R4 corresponding to x RP3 contains exactly two vectors X R4 with B(X, X)=+1.If,inaddition,∈ Πisendowedwith atransverseorientation,thisdefinesatransverseorientationforthelinear∈ 4 subspace L⊥ R .Bydefinition,theunit normal vector of the transversely ⊂

tome 130 – 2002 – no 3 HYPERIDEAL POLYHEDRA IN HYPERBOLIC 3-SPACE 463 oriented hyperbolic plane Πis the vector X L with B(X, X)=+1andwhich ∈ crosses L⊥ in the direction of this transverse orientation. Now, consider a hyperideal polyhedron P in H3,asdefinedintheintroduc- tion. Recall that this means that P = P Proj H3,whereP Proj is a convex projective polyhedron in RP3 whose vertices are∩ all outside of H3 and whose edges all meet H3 ∂ H3.AvertexofP Proj which is on the sphere at infinity ∂ H3 is said to be∪ideal∞ ;otherwise,itisstrictly hyperideal. ∞At this point, it may be useful to remind the reader that a convex projective polyhedron in RP3 is the closure of a component of the complement of finitely many projective planes in RP3. To avoid degenerate cases, we require in addi- tion that a convex projective polyhedron does not contain any projective line or, equivalently, that it is disjoint from at least one projective plane in RP3, so that is is homeomorphic to a closed 3-ball. Since every edge of P Proj meets H3,sodoeseveryfaceanditfollowsthat the projective polyhedron P Proj is completely determined by P = P Proj H3. For future reference, we note the following easy property. ∩

Lemma 3.—For any two distinct vertices v and v" of the projective polyhedron Proj P associated to the hyperideal polyhedron P ,thelinesegmentjoiningv to v" in P Proj meets H3. Proof.—ModifyingP with an element of PO(3, 1) if necessary, we can assume that P Proj R3 RP3 without loss of generality. Then, consider the radial projection R⊂3 ⊂v S2 to the unit sphere centered at v.Theimageϕ(H3) is a spherical open−{ } disk→ contained in a hemisphere of S2,andϕ(P )isaconvex spherical polygon whose vertices are contained in ϕ(H3). It follows that ϕ(P ) 3 3 is contained in ϕ(H ), by convexity. In particular, ϕ(v")isinϕ(H ), and it 3 follows that the line segment vv" must meet H . Consider a strictly hyperideal vertex v of P Proj.Amongthetwoclosed 3 3 hyperbolic half-spaces delimited by the hyperbolic plane v⊥ H in H ,letHv ∩ Proj be the one with the following property: For every other vertex v" = v of P , 3 $ Proj the (unique) oriented line segment k of RP which goes from v to v" in P 3 exits Hv at its intersection point with v⊥ H .(Tomakesurethatthisreally makes sense, it may be useful to modify P Proj∩ by an element of PO(3, 1) so that 3 3 3 it is contained in R RP and to use the geometric description of v⊥ H by lines passing through⊂ v and tangent to ∂ H3.) In other words, the∩ half- 3 ∞ space Hv is on the same side of x⊥ H as v with respect to the other vertices of P Proj. ∩

Proj Lemma 4.—For any two distinct hyperideal vertices v, v" of P ,theasso- H3 ciated hyperbolic half-spaces Hv and Hv! in are disjoint.

Proof.—BydefinitionofHv and Hv! ,itclearlysufficestoshowthatthetwo 3 3 3 hyperbolic planes v⊥ H and v"⊥ H are disjoint. In RP ,theintersection ∩ ∩ BULLETIN DE LA SOCIET´ EMATH´ EMATIQUE´ DE FRANCE 464 BAO (X.) & BONAHON (F.)

of v⊥ and v"⊥ is the line (vv")⊥ dual to the line vv" passing through v and v". 3 By Lemma 3, the line vv" meets H and its dual line (vv")⊥ is consequently 3 3 disjoint from the closure of H .Itfollowsthatthetwohyperbolicplanesv⊥ H 3 ∩ and v"⊥ H are disjoint, and therefore that H and H are also disjoint. ∩ v v! Proj If v is an ideal vertex of P ,wechooseasmallhoroballHv centered at v. 3 3 3 3 In the projective model H R RP , Hv is just the intersection with H of a closed euclidean ball which⊂ is contained⊂ in H3 ∂ H3 and tangent to ∂ H3 at v.Thereareofcoursemanypossiblechoicesforsuchahoroball,butwe∪ ∞ ∞ choose it so that all Hv thus associated to strictly hyperideal and ideal vertices are pairwise disjoint. By definition, the truncated polyhedron P Trun associated to the hyperideal polyhedron P is obtained by removing from P its intersection with the interior Trun of each such Hv.NotethatP is completely determined if all the vertices Proj of P are strictly hyperideal, and defined modulo choice of the horoballs Hv associated to the ideal vertices in the general case. This truncated polyhedron P Trun will play an important rˆole in our argu- ments. By construction, P Trun is compact. Its faces are of two types. The first type of face is the intersection of a face of P with P Trun .Thesecondtypeof Trun face is the intersection of P with one of the ∂Hv.Thissecondtypecan itself be subdivided in two subtypes, according to whether ∂Hv is a horosphere, when the vertex v is ideal, or a hyperbolic plane, when the vertex v is strictly hyperideal.

2. Necessary conditions Proposition 5.—Let Γ be the dual graph of a hyperideal polyhedron P in H3 and, for every edge e of Γ,letθe ]0,π[ be the external dihedral angle of P at the edge corresponding to e.Foreveryclosedcurve∈ γ embedded in Γ and n passing through the edges e1, e2,...,en of Γ,then i=1 θei 2π with equality if only if γ is the boundary of a component of S2 Γ corresponding≥ to an ideal vertex of P Proj. −!

Proof.—Countingindicesmodulon,letvi be the vertex of Γthat is between 3 ei 1 and ei in γ,andletfi be the face of P corresponding to vi.LetΠi H be the− hyperbolic plane containing f ,andletH H3 be the half-space delimited⊂ i i ⊂ by Πi and containing P .LetP " be the intersection of the Hi.Theboundary ∂P" is an annulus, and is decomposed as a union of infinite strips Πi P ",insuch 3 ∩ awaythatΠ P " and Π P " meet along the geodesic of H containing the i ∩ i+1 ∩ edge of P corresponding to ei. 3 3 3 The closure of P " in H ∂ H meets ∂ H along two topological disks ∪ ∞ ∞ D1 and D2,oneofwhichwillbereducedtoasinglepointif(andonlyif)γ is the boundary of a component of S2 Γwhichcorrespondstoanidealvertex − tome 130 – 2002 – no 3 HYPERIDEAL POLYHEDRA IN HYPERBOLIC 3-SPACE 465

j j j j of P .ThediskDj is the union of circle arcs k1, k2,...,kn,whereeachki is 3 contained in the circle Ci ∂ H bounding the hyperbolic plane Hi.Note j⊂ ∞ j that, at the point where ki meets ki+1,theeuclideanexternalanglebetween these two arcs is equal to the angle between the circles Ci and Ci+1,namelyto the hyperbolic external dihedral angle θei of P along the edge ei.Thisholds j j even when ki or ki+1 is reduced to a point, provided we use the convention j that the tangent line at a point of ki is equal to the tangent line of Ci at the same point. First consider the case where neither one of the two disks D1 and D2 is reduced to a point. Let ϕ : ∂ H3 R2 be defined by stereographic projection ∞ → from a point in the interior of D2.Becausethestereographicprojectionsends circle to circle and because this stereographic projection was performed from 1 apointintheinteriorofD2,thearcsϕ(ki )formingtheboundaryofϕ(D1) are circle arcs whose curvature vectors point away from ϕ(D1). In particular, outside of the corners of ∂ϕ(D1), the geodesic curvature κ of ∂ϕ(D1)isnegative. Also, because the stereographic projection preserves angles, the angle between 1 1 1 1 ϕ(ki )andϕ(ki+1)isequaltotheanglebetweenki and ki+1,namelytoθei . If we apply to ϕ(D1) the Gauss-Bonnet formula (also known in this case as n n the Theorem of the Turning Tangents), 2π = κ + θe < θe ∂ϕ(D1) i=1 i i=1 i where the inequality is strict because of our assumption that D is not reduced $ ! 1 ! to a single point. This proves the expected result in this case. If one of D1 and D2,sayD2,isreducedtoapoint,consideragainthe H3 R2 1 stereographic projection ϕ : ∂ from this point of D2.Thesidesϕ(ki ) ∞ → of the polygon ϕ(D1)arenowstraightarcs,andtheGaussBonnetformulagives n 2π = i=1 θei in this case. Proposition! 6.—Let Γ be the dual graph of a hyperideal polyhedron P in H3 and, for every edge e of Γ,letθe ]0,π[ be the dihedral angle of P at the edge corresponding to e.Foreveryarc∈γ embedded in Γ,passingthroughtheedges e1, e2,...,en of Γ,andjoiningtwodistinctverticesv1 and vn+1 which are in the closure of the same component A of S2 Γ but such that γ is not contained n − in the boundary of A,then i=1 θei >π. Proof.—Itclearlysufficestorestrictattentiontothose! γ which meet the closure of A only at their two end points. Let v be the vertex of P Proj corre- sponding to A. If the vertex v is on the sphere at infinity ∂ H3 then, by Proposition 5, ∞ we can join the two end points of γ by an arc γ" in ∂A crossing edges en+1, n+p ..., en+p such that i=n+1 θei π.ApplyingProposition5tothesimple ≤ n n+p closed curve γ γ",weobtainthat θ θ π 2π π = π. ∪ ! i=1 ei ≥ i=1 ei − ≥ − In addition, equality can occur only when γ γ" is the boundary of a component 2 ! ∪ ! A" of S Γ, which is different from A since γ is not contained in the boundary − of A.Inthiscase,γ" is contained in ∂A ∂A" and therefore consists of a single ∩ BULLETIN DE LA SOCIET´ EMATH´ EMATIQUE´ DE FRANCE 466 BAO (X.) & BONAHON (F.)

n+p edge en+1,sothat i=n+1 θei = θen+1 <π.Asaconsequence,theinequality is always strict. Now consider the! other case, where the vertex v is strictly hyperideal. Let ρ be the hyperbolic reflection across the plane v⊥,andletP " be the union Proj P ρ(P ). Because v⊥ separates v from the other vertices of P and is ∪ orthogonal to the edges and faces of P that it meets, P " is a hyperideal poly- hedron. Each edge of P " is, either an edge of P that does not meet v⊥,orthe image under ρ of an edge of P that does not meet v⊥,oranedgeofP that meets v⊥.Similarly,afaceofP " is, either a face of P that does not meet v⊥, or the image under ρ of a face of P that does not meet v⊥,ortheintersection of a face of P that meets v⊥ with its image under ρ.Asaconsequence,the dual graph Γ" of P " is obtained by gluing together two copies of Γalong the two images of ∂A. Because of our assumption that γ meets ∂A only at its end points, the images of γ in each of the two copies of Γform a closed curve γ" embedded in Γ".ApplyingProposition5totheclosedcurveγ" in the graph Γ" dual to P ", n we conclude that 2 i=1 θei > 2π,whichconcludestheproof. Lemma 7.—Let P!be a convex cone with non-empty interior in H3,inter- section of finitely many half-spaces each containing the point x H3 in its ∈ boundary. If e1,...,en are the edges of P and if θe ]0,π[ denotes the n ∈ external dihedral angle of P along the edge e,then i=1 θei < 2π. 1H3 H3 Proof.—IntheunittangentsphereTx of at! x,considerthesetQ of those vectors which point towards P .ThenQ is a convex polygon with geodesic 1H3 sides in the sphere Tx .Inaddition,theexternalangleofQ at each of its corners is equal to the external dihedral angle of P at the corresponding edge. The inequality then follows by application of the Gauss-Bonnet formula to Q.

3. Spaces of polyhedra If Γis a planar 3-connected graph, let denote the space of hyperideal PΓ polyhedra P whose dual graph ΓP is identified to Γ. More precisely, an element of Γ is a pair consisting of a hyperideal polyhedron% P and of an isomorphism Γ P Γ between Γand the dual graph Γ of P . → P P Because% working with Γoften leads to confusing terminology (vertices of Γ correspond to faces of the corresponding polyhedra, while vertices of the projec- tive polyhedra correspond to components of S2 Γ, etc. . . ), it is convenient to fix − ahyperidealpolyhedronP ,withassociatedprojectivepolyhedronP Proj. 0 ∈ PΓ 0 We will see in 6that is always non-empty, but what is important here is § PΓ that the combinatorics of P0 %can be abstractly described in terms of Γ. In Proj particular, the set F of faces% of P0 and P0 is naturally identified to the set of

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Proj vertices of Γ, and the set V of vertices of P0 is naturally identified to the set of components of S2 Γ. Recall that, because Γis 3-connected, its embedding in S2 is unique up to− homeomorphism of S2 (see for instance [9, 32]), so that § these two sets F and V only depend on Γ. For every P Γ with associated projective polyhedron P Proj,thesetsoffacesandverticesof∈ P P Proj are now naturally identified to F and V . % We endow with the topology induced by the embedding Φ: (RP3)V PΓ PΓ → that associates to P Γ the vertices of its associated projective poly- hedron P Proj.% ∈ P % The group PO(3, 1) has% a natural action on .Considerthequotientspace PΓ Γ = Γ/PO(3, 1). P P % Lemma 8.—The space is Hausdorff. % PΓ Proof.—Weneedtoshowthat,whenevertwosequencesP and g n ∈ PΓ n ∈ PO(3, 1) are such that P converges to some P and g P converges to n ∈ PΓ n n some Q ,thereisag PO(3, 1) such that gP = Q. % ∈ PΓ ∈ Pick four distinct faces f , f , f , f F of P Proj% in such a way that f is 0 1 2 3 ∈ 0 0 adjacent to% the faces f1, f2 and f3.Foreachi,considerthehyperbolicplane that contains the face of P corresponding to fi,transverselyorientedwiththe outward boundary orientation from P ,andletX R4 be its unit normal i ∈ vector, as defined in 1. A consequence of the choice of the faces f0, f1, f2, f3 § RP3 is that the intersection of the four projective planes X0⊥, X1⊥, X2⊥, X3⊥ in is empty. It follows that the vectors X0, X1, X2, X3 are linearly independent, 4 and form a basis X0,X1,X2,X3 for R . { } 4 Similarly, the polyhedron Q Γ provides a basis Y0,Y1,Y2,Y3 for R , where Y R4 is the unit normal∈ P vector to the hyperbolic{ plane which} con- i ∈ tains the face of Q corresponding to%fi,transverselyorientedwiththeoutward boundary orientation from Q,andthepolyhedronPn Γ gives a similar n n n n ∈ P basis X0 ,X1 ,X2 ,X2 . { } Proj Since Pn converges to P for the topology of Γ,eachvertexof% Pn converges to the corresponding vertex of P Proj,andeachofitsfacesconse-P Proj n quently converges to the corresponding face of P %.Inparticular,eachXi R4 n converges to Xi in .Similarly,sincegnPn converges to Q in Γ,eachgnXi 4 P converges to Yi.IfAn denotes the matrix of gn from R with the basis X ,X ,X ,X to R4 with the basis Y ,Y ,Y ,Y ,itfollowsthat% A { 0 1 2 3} { 0 1 2 3} n converges to the identity matrix. As a consequence, gn converges to a linear isomorphism g of R4,whichmustbeinO(3, 1) since this subgroup is closed in the linear group. Therefore, g induces an element of PO(3, 1) which, by continuity, sends P to Q as required.

To analyze the local topology of Γ and Γ,subdividethefacesofthe Proj P P model projective polyhedron P0 into triangles by adding a few edges (and % BULLETIN DE LA SOCIET´ EMATH´ EMATIQUE´ DE FRANCE 468 BAO (X.) & BONAHON (F.)

Proj Proj no vertex). Let P0 denote P0 with this new cell decomposition of its Proj Proj boundary. The set V of vertices of P0 is the same as that of P0 ,however Proj Proj the set of edges of&P is now E E",whereE is the set of ‘old’ edges of P 0 ∪ 0 and E" is the set of ‘new’ edges introduced& to subdivide its faces into triangles. Let x (RP3)V &be a point which is close to the vertex set Φ(P ) (RP3)V ∈ Proj ∈ of a polyhedron P Γ.Fora(triangle)facef of P0 ,withverticesv1, v2 ∈ P RP3 and v3,thecorrespondingcoordinatesxv1 , xv2 , xv3 of x are the vertices 3 ∈ of a triangle tf in RP%,uniquelydeterminedifwerequirethat& tf is close to the triangle bounded by the corresponding vertices in a face of P (beware that three non-collinear points are the vertices of 4 distinct triangles in RP3). As f Proj ranges over all faces of P0 ,theunionofthetf forms the boundary of a 3 Proj polyhedron Px in RP .ThispolyhedronPx is a perturbation of P .In Proj contrast to P , Px is& not necessarily convex and some of its vertices may 3 be inside H .However,forx sufficiently close to Φ(P ), the polyhedron Px is embedded and each of its edges meets H3. Proj If e E E" is an edge of P , consider the corresponding edge of P , ∈ ∪ 0 x and let Θe(x) ] π,π[ R be the external dihedral angle of the hyperbolic ∈ −3 ⊂ polyhedron Px H along this edge.& Considering all such edges e E E", ∩ E E! ∈ ∪3 V this defines% a map Θ:U R ∪ on a neighborhood U of Φ(P )in(RP ) . → We now use ΘtocharacterizetheintersectionofU with the image of the 3 V embedding Φ: Γ % (RP ) . %P → 3 V Lemma 9.—For% every P Γ,thereisaneighborhoodU of Φ(P ) in (RP ) ∈ P E E! such that, for the function Θ:U R ∪ defined above, U Φ Γ consists of those x U with the following% → properties: ∩ P ∈ " # % % (i) Θ (x)=0for every ‘new’ edge e E"; e ∈ n (ii) for every vertex v V , i=1 Θei (x) 2π where e1,...,en E are the % Proj ∈ ≥ ∈ edges of P0 that contain the vertex v. ! % Proof.—ChooseU small enough that Θe(x) > 0forevery‘old’edgee E and every x U.Ifx U satisfies Conditions (i) and (ii), this property∈ of ∈ ∈ old edges and Condition (i) guarantee that% Px is convex and, if we erase those Proj edges where ∂Px is flat, has the same combinatorics as P0 .ByLemma7, 3 Condition (ii) implies that all the vertices of Px lie outside of H .Sinceallthe 3 3 edges of Px already meet H ,thisshowsthatPx H is a hyperideal polyhedron, whose image in (RP3)V is equal to x. ∩ Conversely, Conditions (i) and (ii) are clearly necessary for x U to be ∈ in Φ ,bydefinitionofΘandbyProposition5. PΓ " # The% key technical step in% the proof of the Rigidity Theorem 2 is the following infinitesimal rigidity result, whose proof will occupy the next section.

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Proposition 10 (Infinitesimal Rigidity Lemma). — For P Γ,letΘ: E E! ∈ P U R ∪ be the map defined as above on a neighborhood U of Φ(P ) → in (RP3)V .Then,thekernelofthetangentmapofΘ at Φ(P ) is% equal to% the image of the tangent map at Id PO(3, 1) of the map PO(3, 1) (RP3)V de- ∈ → fined by g gΦ(P ).Inotherwords,anyinfinitesimaldeformationof% Φ(P ) in 3 V ,→ (RP ) which infinitesimally respects the dihedral angles Θe for all e E E" must come from composition by an infinitesimal element of PO(3, 1).∈ ∪ % Assuming Proposition 10, we are now ready to determine the local type of = /PO(3, 1). PΓ PΓ The map ΘisdefinedonaneighborhoodofΦ( Γ), and we can consider % E E! P the composition Θ Φ: R ∪ .Propositions5,6andLemma9impose ◦ PΓ → constraints on% the image of Θ Φ. Namely, its image% is contained in K 0 ◦ Γ × ⊂ RE RE! where% the subset% K RE consists of those θ RE which satisfy × Γ ⊂ ∈ the following conditions: %

0) For every e E,thecoordinateθe of θ corresponding to e is in the interval ]0,π[; ∈ 1) For every closed curve γ embedded in Γand passing through the edges e , e ,...,e of Γ, n θ 2π with equality allowed only when γ is the 1 2 n i=1 ei ≥ boundary of a face of S2 Γ. ! − 2) For every arc γ embedded in Γ, passing through the edges e1, e2,..., en of Γ, and joining two distinct vertices which are in the closure of the same component of S2 Γbutsuchthatγ is not contained in the boundary of that n − component, i=1 θei >π. RE RE! The map !Θ Φ: Γ KΓ 0 is invariant under the action ◦ P → × ⊂ × of PO(3, 1) on ,andthereforeinducesamapΘ: K on = PΓ PΓ → Γ PΓ Γ/PO(3, 1).% % P % Theorem% 11 (Local characterization of hyperideal polyhedra by their dihedral angles) The map Θ: K is a local homeomorphism. PΓ → Γ Proof (assuming Proposition 10).—ConsiderahyperidealpolyhedronP ∈ Γ,anditsclassin Γ which, to avoid cumbersome notation, we will denote by theP same letter P P .WewanttoshowthatΘrestrictstoahomeomorphism ∈PΓ from% a neighborhood of P in Γ to a neighborhood of Θ(P )inKΓ. P E E! Consider the restriction Θ:U R ∪ of ΘtoanopenneighborhoodU of RP3 V → Proj Φ(P )in( ) .BecauseallthefacesofthecelldecompositionofP0 are triangles, a counting argument% shows that #E%+#E" =3#V +6,where#X denotes the cardinal of X.Inparticular,thedifferencebetweenthedimensions& E E! of the domain and of the range of Θ:U R ∪ is equal to 6. → ´ ´ ´ BULLETIN DE LA SOCIETEMATHEMATIQUE% DE FRANCE 470 BAO (X.) & BONAHON (F.)

The map PO(3, 1) (RP3)V defined by g gΦ(P )hasaninjectivetangent map at Id. This can→ easily be seen by picking,,→ as in the proof of Lemma 8, four faces f0, f1, f2, f3 F such that the unit normal vectors X0, X1, X2, 4 ∈ Proj X2 R of the planes containing the corresponding faces of P are linearly independent;∈ an element of the kernel of the tangent map infinitesimally fixes Proj the vertices of P ,thereforeinfinitesimallyfixestheXi,andconsequently must be trivial. It follows that the image of the tangent map of PO(3, 1) (RP3)V at Id has dimension 6. → Combining this computation of dimensions with Proposition 10, we conclude E E! that the differential of Θ:U R ∪ at Φ(P )issurjective.Inparticular, E E! → Θ(U)isopeninR ∪ if we choose U small enough. The Submersion Theorem% shows that we can choose the neighborhood U of Φ%(P )sothatitdecomposesasU = U " U "" in such a way that Θcorresponds ∼ × to the composition of the projection U " U "" U " and of a diffeomorphism × → U " Θ(U). Let (u" ,u"") U " U "" correspond to Φ(P ) U.Forevery% u" U ", → 0 0 ∈ × ∈ ∈ the map g g(u",u"")immersesaneighborhoodW of Id in PO(3, 1) in the ,→ 0 u! slice u%" U "" U since Θ g = Θ. By the above dimension computations, { }× ⊂ ◦ PO(3, 1) and U "" both have dimension 6, and we conclude that we can choose the neighborhood U and the identification% % U = U " U "" so that each slice u" U "" ∼ × { }× is of the form W (u",u0""), where W is a fixed neighborhood of Id in PO(3, 1). We claim that, if U and W are chosen small enough, two elements of U will be in the same PO(3, 1)-orbit if and only if they belong to the same slice 3 V u" U "" U.Itclearlysufficestoshowthat,ifx (RP ) and g PO(3, 1) {are} such× that⊂ x and gx are both close to Φ(P ), then∈ g must be close∈ to Id in PO(3, 1). For this, pick again four faces f0, f1, f2, f3 F such that the unit 4 ∈ normal vectors X0, X1, X2, X2 R of the planes containing the corresponding faces of P Proj are linearly independent.∈ Then, as in the proof of Lemma 8, each gXi is close to Xi,anditfollowsthatg is close to Id in PO(3, 1). Therefore, if we choose U small enough, two points (u",u""), (v",v"") U " ∈ × U "" ∼= U are in the same PO(3, 1)-orbit if and only if they are in the same slice u" U "",namelyifandonlyifv" = u",namelyifandonlyiftheyhave { }× the same image under Θ. In other words, the restriction of ΘtoU induces a homeomorphism between U/PO(3, 1) and Θ(U). By Lemma 9, the image% of U Φ Γ is equal to Θ(U) % (KΓ 0 ). It ∩ P % ∩ ×{ } follows that Θinducesahomeomorphismbetween" # U Φ Γ /PO(3, 1) and % %∩ P Θ(U) (KΓ 0 ). Since Φis a PO(3, 1)-equivariant homeomorphism" # between ∩ ×%{ } % Γ and Φ Γ ,weconcludethatΘ Φinducesahomeomorphismbetween P% 1 P ◦ 1 Φ− (U)/PO(3" #, 1) and Θ(U) (KΓ 0 ). Note that Φ− (U)/PO(3, 1) is an open % % ∩ ×{% } neighborhood of P in Γ,andthatΘ(U) (KΓ 0 )isanopenneighborhood P ∩ ×{E } E! of Θ Φ(P )inKΓ %0 since Θ(U)isopeninR R . ◦ ×{ } % × % % tome 130 – 2002 – no 3 HYPERIDEAL POLYHEDRA IN HYPERBOLIC 3-SPACE 471

This proves that the map Θ: Γ KΓ restricts to a homeomorphism from aneighborhoodofP in to a neighborhoodP → of Θ(P )inK . PΓ Γ

4. The Infinitesimal Rigidity Lemma This section is devoted to the proof of the Infinitesimal Rigidity Lemma of Proposition 10. The proof follows the general lines of the famous arguments of Cauchy [5] on the deformations of convex polyhedra in E3 (see also [15][7]), as adapted to hyperbolic polyhedra by Andreev [2] and Rivin [12]. For polyhedra which are strictly hyperideal, namely whose vertices are all outside of the clo- sure of H3,thisproofisessentiallycontainedin[12, 4] if we use the truncated polyhedron introduced in 1. § § E E! Recall that we have a map Θ:U R ∪ ,definedonaneighborhoodU of Φ(P )in(RP3)V ,whichtox U associates→ the hyperbolic external dihedral ∈ angles of the polyhedron Px whose% vertices are the coordinates of x and whose Proj RP3 V combinatorial type is that of P0 .WealsohaveamapG :PO(3, 1) ( ) defined by g gΦ(P ). We want to show that the kernel of the tangent→ map ,→ 3 V E E! T Θ:T (RP ) R &∪ is equal to the image of Φ(P ) Φ(P ) → T G : T PO(3, 1) T (RP3)V . % Id Id −→ Φ(P ) Letx ˙ T (RP3)V be a vector which is contained in the kernel of T Θ. ∈ Φ(P ) Φ(P ) We want to show thatx ˙ is in the image of TIdG,namelyistangenttoan infinitesimal deformation of Φ(P )byaninfinitesimalelementofPO(3, 1). % Lemma 12.—If, for a given v V ,thevertexofP Proj corresponding to v ∈ 3 3 is located on the sphere at infinity ∂ H RP ,thenthecoordinatevectorx˙ v of x˙ corresponding to v is tangent to∞∂ H⊂3 in RP3. ∞ Proj Proof.—Lete , e ,...,e E E" be the edges of P that are adjacent to 1 2 n ∈ ∪ 0 the vertex v.Letϕ : U R be the function defined by ϕ(x)= n Θ (x). → i=1 ei Note that the vectorx ˙ is contained in the kernel of the& tangent map TΦ(P )ϕ, ! since it is contained in the kernel of TΦ(P )Θ. % By Proposition 5 and Lemma 7, ϕ(x)=2π whenever the coordinate xv of x U (RP3)V corresponding to v belongs% to ∂ H3.Itfollowsthatthe ∈ ⊂ ∞ 3 V kernel of TΦ(P )ϕ contains all those vectorsy ˙ TΦ(P )(RP ) whose coordinate 3 ∈ y˙v is tangent to ∂ H .Thelemmawillbeprovedifweshowthatthekernelof ∞ 3 TΦ(P )ϕ consists only of thesey ˙ withy ˙v tangent to ∂ H .Forthis,itsuffices ∞ to show that TΦ(P )ϕ is non-trivial. Modifying P by an element of PO(3, 1) if necessary, we can arrange that the projective polyhedron P Proj is contained in R3 RP3 and contains the origin 0 H3 R3 in its interior. ⊂ ∈Let t⊂ xt U (RP3)V , t ] ε,ε[, be a curve such that x0 =Φ(P ), ,→ ∈ ⊂ t ∈ − 0 such that the coordinate x is equal to x for every v" = v and every t,and v! v! $ BULLETIN DE LA SOCIET´ EMATH´ EMATIQUE´ DE FRANCE 472 BAO (X.) & BONAHON (F.)

t RP3 such that the curve t xv is tangent to the outer unit normal vector of ∂ H3 R3 RP3 ,→at t =0.If˙∈ x0 denotes the tangent vector of t xt at ∞ ⊂ ⊂ ,→ 0 t =0,wewillusealittlebitofeuclideangeometrytoestimateTΦ(P )ϕ(˙x )= t dϕ(x )/dt t=0. | t 3 V Consider the projective polyhedron Pxt associated to x (RP ) ,namely t Proj∈ with vertex set x and with the combinatorial type of P0 .Fort 0, the H3 0 ≥ H3 intersection of Pxt with ∂ has one component At which is near xv ∂ . ∞ 3 ∈ ∞ This component At is a polygon on the sphere ∂ H ,reducedtoapointwhen& ∞ 3 t =0.Rememberthat,ifΠandΠ" are two projective planes which meet H , 3 3 the hyperbolic dihedral angle between the hyperbolic planes Π H and Π" H 3∩ ∩ 3 is equal to the euclidean angle between the two circles Π ∂ H and Π" ∂ H , for the euclidean metric on the sphere ∂ H3 R3.ApplyingtheGauss-Bonnet∩ ∞ ∩ ∞ ∞ ⊂ formula to At,weconcludethat n ϕ(xt)= Θ (xt)=2π κ area(A ) ei − − t i=1 ∂At ' ( % 3 where κ is the geodesic curvature of the boundary ∂At in ∂ H . ∞ Let us differentiate in t at t =0.Theareatermisboundedbyaquantityof order 2 in t,andthereforedoesnotcontributetothederivative.Becausewe arranged that the point 0 is in the interior of P ,thecurvatureκ is negative bounded away from 0 on each side of At.Ontheotherhand,becausethe 0 0 componentx ˙ v of the tangent vectorx ˙ is equal to the outer unit normal vector 3 of ∂ H ,thelengthof∂At has a positive derivative with respect to t at t =0. ∞ 0 t It follows that TΦ(P )ϕ(˙x )=dϕ(x )/dt t=0 is strictly positive. (Note that we | do not need to worry about the derivative of κ since ∂A0 has length 0.) This proves that T(p,x)ϕ is non-trivial. As a consequence, its kernel consists 3 V 3 of those vectorsy ˙ TΦ(P )(RP ) whose coordinatey ˙v is tangent to ∂ H . Sincex ˙ belongs to this∈ kernel, this completes the proof of Lemma 12. ∞

Let t xt, t ] ε,ε[, be a smooth curve in (RP3)V such that x0 =Φ(P ) t ,→ ∈ − and dx /dt t=0 =˙x.ByLemma12,wecanchoosethiscurvesothat,whenever | 0 0 H3 the coordinate xv of x =Φ(P )isinthesphere∂ ,thecorresponding t t H3 ∞ coordinate xv of x remains in ∂ for every t.Inparticular,fort sufficiently t t ∞ H3 RP3 small, all coordinates xv of x are in the complement of in . t 3 V Again, let Pxt be the projective polyhedron associated to x (RP ) , t Proj∈ namely with vertex set x and with the combinatorial type of P0 . Trun As in 1, consider the truncated polyhedron P t obtained from P t by § x x chopping offits intersection with the interiors of disjoint half-spaces& and t t horoballs Hv where, for each v V , Hv is a half-space delimited by the hyper- H3 t ∈ t t bolic plane (xv)⊥ if the vertex xv of Pxt is strictly hyperideal and Hv is ∩ t Trun ahoroballcenteredatxv if this vertex is ideal. From its construction, Pxt inherits a natural polyhedron structure where its faces are of three types:

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3 (i) the intersection of a face of Pxt with the complement in H of the interiors t of all the Hv;suchafaceisapolygoncontainedinahyperbolicplane,whose sides are either geodesic or horocyclic, and whose internal angles are all equal 1 to 2 π. t (ii) the intersection of Pxt with the hyperbolic plane ∂Hv associated to a t strictly hyperideal vertex xv;suchafaceisahyperbolicpolygonwithgeodesic sides and, at each of its vertices, its internal angle is equal to the dihedral angle 3 of the corresponding edge of Pxt H . ∩ t (iii) the intersection of Pxt with the horosphere ∂Hv associated to an ideal t vertex xv;suchafaceisaeuclideanpolygonwithgeodesicsidesand,ateach of its vertices, its internal angle is equal to the internal dihedral angle of the 3 corresponding edge of P t H . x ∩ Trun Recall that in the construction of Pxt we have a degree of freedom in the t choice of the horospherical Hv associated to the ideal vertices of Pxt .Wewill use this freedom to impose an additional condition on the faces of type (iii). t By adjusting the height of the corresponding horospheres ∂Hv,wecanalways arrange that the area of each face of type (iii) is constant, independent of t. Trun Note that the polyhedra Pxt all have the combinatorial type of the polyhe- Trun Proj dron P0 obtained by truncating the polyhedron P0 (which, as a reminder, Proj was defined by subdividing the faces of P0 into triangles). & Proj& Trun Consider now the unsubdivided polyhedron P0 ,andtruncateittoP0 . Trun Trun Each vertex v of P0 is also a vertex of P0 ,andisthereforeassociatedto t Trun avertexv of Pxt . Trun For each edge e of P0 ,lookatitsendvertices& v1, v2,andconsiderthe t t t t Trun hyperbolic distance d(v1,v2)betweenthecorrespondingverticesv1, v2 of Pxt . Label the edge e by the symbol +, 0 or according to whether the derivative t t − of d(v1,v2)att =0ispositive,zeroornegative. Our goal is to show that all the edges are actually labelled by 0. This will require a few preparatory lemmas. If Q is a polygon with each of its edges labelled by a symbol +, 0 or ,define − the number of sign changes of ∂Q as the minimum number of vertices v1,...,vn which one needs to remove from ∂Q so that the edges in each component of ∂Q v1,...,vn have the same sign, namely are either all in +, 0 or all in ,−0 {.Notethatthenumberofsignchangesisalwayseven.} { } {− } Trun We will first show that, for each face f of P0 ,eitheralltheedgesoff are labelled by 0, or it admits at least 4 sign changes in its boundary. In other words, there cannot be 0 or 2 sign changes in the boundary of f,unlessallthe edges of f are labelled by 0. We begin with faces of Type (iii), contained in horospheres. The argument uses the following euclidean geometry lemma, which is an infinitesimal and simpler version of Lemma M3 of [15] (a result attributed there to A.D. Alek- sandrov).

BULLETIN DE LA SOCIET´ EMATH´ EMATIQUE´ DE FRANCE 474 BAO (X.) & BONAHON (F.)

2 Lemma 13.—In the euclidean plane R ,letQt, t ] ε,ε[,beadifferentiable family of strictly convex polygons with straight sides.∈ − Suppose that, at t =0, the derivative of the angle of Qt at each of its vertices is equal to 0,andthat the derivative of the area of Qt is also equal to 0.LabeleachedgeofQ0 by the symbol +, 0,or according to whether the derivative of its length at t =0is positive, zero or− negative. Then, either all edges are labelled by 0,orthereare at least 4 sign changes in ∂Q0.

Here, “strictly convex” means that the internal angle of Qt at each of its vertices is strictly between 0 and π.

Proof.—(Compare[15,LemmaM3].) Suppose, in search of a contradic- tion, that there are exactly 2 sign changes. We can then index the edges t t t t t t of Qt as e1,...,eq,eq+1,...,er,er+1 = e1,goingcounterclockwiseinthisorder t around ∂Qt,insuchawaythatei is labelled by + or 0 if 1 i q,and t t ≤ ≤t by or 0 if q +1 i r.Letvi be the vertex intersection of ei and ei+1,and −t t ≤ ≤ t set v0 = vr for consistency of the notation. Finally, let Ti be the unit tangent t vector of the edge ei,orientedbythecounterclockwiseorientationof∂Qt,and let &t be the length of et.Then,for1 i

Proof.—Theface Q corresponds to a face Q" of the projective polyhe- Proj dron P0 .However,theverticesofPxt associated to the vertices of Q" are not necessarily coplanar any more. Nevertheless, because the vector x˙ T (RP3)V tangent to t xt is contained in the kernel of the tangent ∈ Φ(P ) ,→ map TΦ(P )Θ, they are infinitesimally coplanar in the following sense: we can

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3 find a projective plane Πt in RP such that, for every vertex v of Q",the t distance from the vertex xv of Pxt corresponding to v to the plane Πt is 0 and has derivative 0 at t =0,foranarbitraryriemannianmetriconRP3.In addition, we can choose the plane Πt so that it depends differentiably on t.For instance, we can take Πt to be the plane passing through three predetermined such vertices of Pxt . t For each vertex v of Q",chooseapointyv Πt,dependingdifferentiably 0 0 t∈ t on t,suchthatyv = xv and the distance from yv to xv has derivative 0 at t =0. t H3 In addition, we can choose yv so that it belongs to the sphere at infinity ∂ t 0 ∞t whenever xv does, namely whenever xv does by our choice of the curve t x 3 V ,→ ∈ (RP ) .LetQt be the convex projective polygon in Πt whose vertices are the t points yv associated to the vertices v of Q",andwhichisuniquelydetermined by the property that it depends continuously on t and that Q0 = Px0 H0 Proj ∩ (namely Q0 is the face of P = Px0 corresponding to Q"). Note that Qt RP2 is a hyperideal polygon in the projective plane Ht ∼= with respect to the H3 H2 hyperbolic plane Ht ∼= ,fort sufficiently small. In the hyperbolic∩ plane H H3,letQTrun be the truncated polygon associ- t ∩ t ated to the hyperideal polygon Qt.Notethatthereisanaturalidentification Trun Trun between the vertices of Qt and those of the face Q of P0 .Becausethe RP3 t distance (for an arbitrary riemannian metric on )fromthevertexyv of Qt t to the corresponding vertex xv of Pxt is infinitesimally 0, the hyperbolic dis- Trun tance between two vertices of Qt is infinitesimally equal to the hyperbolic Trun distance between the corresponding vertices of Pxt .Inparticular,ifwela- Trun bel the edges of Q0 = Q as above Lemma 13, the labelling is the same as ∼ Trun that induced on the face Q by our original labelling of P0 .Lemma16then follows from Lemma 15.

Finally, we consider faces of Type (ii) of P Trun ,containedinhyperbolic planes which are dual to strictly hyperideal vertices of P Proj.

2 Lemma 17.—In the hyperbolic plane H ,letQt, t ] ε,ε[,beadifferentiable family of compact strictly convex polygons with geodesic∈ − sides. Suppose that, at t =0,thederivativeoftheangleofQt at each of its vertices is equal to 0.Label each edge of Q0 by the symbol +, 0 or according to whether the derivative of its length at t =0is positive, zero or− negative. Then, either all edges are labelled by 0,orthereareatleast4 sign changes in ∂Q0.

Proof.—TheproofisessentiallythesameasthatofLemma15(withno horocyclic sides). The reader familiar with [12] will also recognize here an infinitesimal (and simpler) version of [12, Lemma 4.11], which provided the inspiration for our proof of Lemma 15.

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Trun Lemma 18.—For every face Q of P0 which corresponds to a face of Trun type (ii) of the truncated polyhedron P0 ,eitheralltheedgesof∂Q are labelled by 0,orthereareatleast4 sign changes in ∂Q. Proof.—TheproofisidenticaltothatofLemma16,usingLemma17instead of Lemma 15. We now conclude with a purely combinatorial argument.

Trun Lemma 19.—Label each edge of P0 by a sign +, 0 or in such a way Trun − that, for each face Q of P0 ,eitheralltheedgesofQ are labelled by 0 or there are at least 4 sign changes in ∂Q.Then,alltheedgesarelabelledby0. Proof.—ThisisCauchy’sfamouscombinatoriallemma[5];seealso[15, Lemma T]. Since it is fairly simple, we include a proof for the sake of completeness. Trun Let G be the 1-skeleton of the dual cell decomposition of ∂P0 .Note that G is a combinatorial graph, in the sense that an edge has two distinct end vertices and no two edges have the same end vertices. Label each edge Trun of G with the label of the dual edge of P0 .Thesignconditionforthefaces Trun of P0 translates to a similar sign condition for each vertex v of G:either all the edges adjacent to v are labelled by 0, or one encounters at last 4 sign changes& as one goes around v. Lemma 19 then follows from the following statement. Sublemma 20.—Let G be a graph embedded in S2 with each edge of G labelled by a sign +, 0 or in such a way that, for each vertex v of G,eitherallthe edges adjacent to v−are labelled by 0,oroneencountersatleast4 sign changes as we go around v.Then,alltheedgesofG are labelled by 0. Proof of Sublemma 20.—SupposethatthereisagraphG for which the prop- erty fails. Erasing from G all the edges labelled by 0 (which does not disturb the sign change condition) and restricting attention to a connected component of the remaining graph, we can assume that all edges are labelled by + or , and that G is connected and non-empty. Adding edges labelled by + if neces-− sary (which again does not disturb the sign change condition), we can assume that the complement of G in S2 consists of triangles. For i =0,2,letFi be the number of triangles in the complement of G with i sign changes in their boundary. Then, if V and E respectively denote the number of vertices and edges of G,thehypothesisthatthereareatleast4 sign changes at each vertex implies that 4V 2F2.Sincewearrangedthat the complement of G consists of triangles, counting≤ edge sides in two different ways gives 3(F0 + F2)=2E,whiletheEulercharacteristicequationshows that V E + F0 + F2 =2.Combiningthesethreerelations,weobtainthat F +4 −0, a contradiction. 0 ≤ tome 130 – 2002 – no 3 HYPERIDEAL POLYHEDRA IN HYPERBOLIC 3-SPACE 481

This completes the proof of Lemma 19. We are now ready to complete the proof of the Infinitesimal Rigidity Lemma of Proposition 10. Trun Proof of Proposition 10.—ByLemmas14,16and18,alltheedgesofP0 Trun are labelled by 0. This means that if, for each edge e of P0 ,weconsiderthe t t t Trun geodesic arc e joining the vertices v1, v2 of Pxt corresponding to the end t vertices v1, v2 of e,thelengthofe has derivative 0 at t =0.Inaddition,if Trun t t two edges e1 and e2 of P0 meet at the vertex v,theanglebetweene1 and e2 t t at v also has derivative 0 at t =0,becausethetangentvector˙x =dx /dt t=0 | is contained in the kernel of the tangent map TΦ(P )Θ(comparethepolygons Qt which we used in the proofs of Lemmas 14, 16 and 18). Finally, we already Trun saw in the proof of Lemma 16 that, for every face Q of% P0 that corresponds Proj Trun to a face of P0 ,theverticesofPxt corresponding to the vertices of Q are Trun infinitesimally coplanar near t =0.If,amongthesefacesQ of P0 corre- Proj sponding to faces of P0 ,wesuccessivelygofromonetotheotherbycrossing one edge at a time, we conclude that there exists an isometry gt PO(3, 1) of 3 ∈t H ,dependingdifferentiablyont and with g0 =Id,suchthatdgt(v )/dt t=0 =0 t Trun Trun | for every vertex v of Pxt corresponding to a vertex v of P0 . Note that the vertices of the projective polyhedron Pxt ,namelythecoordi- t Trun nates of x ,canbecompletelyrecoveredfromtheverticesofPxt that corre- Trun spond to vertices of P0 ,asintersectionofprojectivelinespassingthrough t these points. It follows that dgt(x )/dt t=0 =0.Weconcludethat | t 0 t 0=dgt(x )/dt t=0 =dgt(x )/dt t=0 +dx /dt t=0, | | |

=dgt Φ(P ) /dt t=0 +˙x | 0 t since g0 =Id,x =Φ(P ), and dx /dt t=0" =˙x#. | 1 In particular, the vectorx ˙ = dgt(Φ(P ))/dt t=0 =dgt− (Φ(P ))/dt t=0 1 − | | is tangent to the curve t gt− (Φ(P )) at t =0.Asaconsequence,˙x is in the image of the tangent,→ map at Id of the map PO(3, 1) (RP3)V defined by g g(Φ(P )). → This,→ is exactly what we needed to complete the proof of Proposition 10.

5. Convergence of hyperideal polyhedra E We showed in Theorem 11 that the map Θ: Γ KΓ R ,whichtoa hyperideal polyhedron associates its dihedral angles,P is→ a local⊂ homeomorphism. This section is devoted to proving that it is a covering map. Recall that K was defined as the set of those θ RE such that: Γ ∈ 0) For every e E,thecoordinateθe of θ corresponding to e is in the interval ]0,π[; ∈

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1) For every closed curve γ embedded in Γand passing through the edges n e1, e2,...,en of Γ, i=1 θei 2π with equality allowed only when γ is the boundary of a face of S2 Γ. ≤ ! − 2) For every arc γ embedded in Γ, passing through the edges e1, e2,..., en of Γ, and joining two distinct vertices which are in the closure of the same component of S2 Γbutsuchthatγ is not contained in the boundary of that n − component, i=1 θei <π. Proposition! 21.—The map Θ: K is proper. PΓ → Γ Proof.—SinceKΓ is locally compact and Γ is metrizable, it suffices to prove the following: If P , n N,isasequencesuchthatP Θ(P )convergesto n ∈PΓ ∈ n some θ∞ K ,thenthereisasubsequenceP , k N,whichconvergestoa ∈ Γ nk ∈ polyhedron P Γ. Consider such∞ ∈ aP sequence P , n N.Asusual,weusethesame n ∈PΓ ∈ symbol Pn Γ to denote a hyperideal polyhedron representing the class P ,andwelet∈ P P Proj be the projective polyhedron associated to the n ∈PΓ n polyhedron P .Let% xn =Φ(P ) (RP3)V be the vertex set of P Proj.Pick n n ∈ n three consecutive vertices v1, v2, v3 V on the boundary of a face f of the Proj ∈ model polyhedron P0 .AftermodifyingPn by an element of PO(3, 1) if n n Proj necessary, we can arrange that the corresponding vertices xv1 and xv2 of Pn sit on the closure of the axis R 0 R3 in RP3 while the vertex xn is in ×{ }⊂ v3 the closure of the plane 0 R2 in RP3.BycompactnessofRP3,wecanalso { }× n 3 V assume after passing to a subsequence that x converges to x∞ (RP ) . n Proj ∈ For each face f F ,thecorrespondingfacef of Pn is contained in a n∈ RP3 projective plane Πf of .Afterpassingtoasubsequence,wecanassume n that the plane Πf converges to a plane Πf∞ for each f F .Byconstruction, H3 RP3 ∈ the limit plane Πf∞ meets the closure of in .Inparticular,eitherit meets H3 or it is tangent to the sphere at infinity ∂ H3. ∞ n Similarly, for an edge e E joining two vertices v, v" V ,lete be the Proj∈ ∈ corresponding edge of Pn . Among the two line segments joining the vertices xn and xn of P Proj in RP3, en is also the one that meets H3.Asn tends to , v v! n en converges to a line segment e which joins x to x and which meets the∞ ∞ v∞ v∞! closure of H3 (after passing to a subsequence if by any chance x = x ). v∞ v∞! Note that the limit edge e∞ could conceivably be reduced to a single point. We first prove that this does not happen. Lemma 22.—There is no edge e E such that the corresponding limit edge ∈ e∞ is reduced to a single point. Proof of Lemma 22.—Supposeotherwise,andthatthereisapointp RP3 ∈ such that Ep = e E; e∞ = p is non-empty. Note that p must be on the sphere at infinity{∂ ∈H3.Indeed,itmustbeintheclosureof} H3 since each en meets H3,butcannotbein∞ H3 since the end points of en are outside H3.

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Consider the vertices v1, v2 and v3 which we selected at the beginning of the proof of Proposition 21. The edge e12 joining v1 to v2 cannot be in Ep, 3 since the corresponding line segment e∞ contains the diameter R 0 H . 12 ×{ }∩ The edge e23 joining v2 to v3 cannot be in Ep either, because its end points xv∞2 and xv∞3 are distinct since they are respectively contained in the closures of R 0 and 0 R2,whicharedisjoint.Intheboundaryofthemodel ×{ } Proj{ }× polyhedron P0 ,letKp be the union of all the edges e Ep.Bytheabove ∈ Proj remark and because e12 and e23 are contained in the same face of P0 ,the Proj interiors of e12 and e23 are contained in the same component of ∂P0 Kp. Proj − Let U be a regular neighborhood of Kp in ∂P0 ,andletγ be a component of ∂U which is contained in the same component of ∂PProj K as the interiors 0 − p of e12 and e23. Let e , e ,...,e , e = e E be the edges of P Proj traversed by γ,in 1 2 m m+1 1 ∈ 0 this order. Note that γ defines a closed curve γΓ in the dual graph Γ, crossing the edges corresponding to e1, e2,...,em.Thisclosedcurveisnotnecessarily embedded, but it is immersed in Γin the sense that each ei is different from ei+1. Also, by construction, each limit edge ei∞ has one end point equal to p,butis not reduced to this point. Let f1, f2,...,fm, fm+1 = f1 be the faces met by γ,sothatγ crosses the edge ei between the faces fi and fi+1.Byconstruction,eachofthelimit R3 R3 planes Πf∞i passes through p.In ,theeuclideanplaneΠf∞i bounds ∩ n apreferredeuclideanhalf-spaceHf∞i ,namelythelimitofthehalfspaceHfi n 3 3 bounded by Π R and containing the polyhedron Pn H .Theintersec- fi ∩ ⊂ tion C of these half-spaces H is a convex cone with vertex p,containingthe f∞i limit edges ei∞ in its boundary. In addition, every extremal half-line of C must coincide with one of the ei∞ near p. The intersection of C with a small horosphere S H3 centered at p is a ⊂ polygon QS,possiblynon-compactand/orreducedtoasubsetofageodesic of S.NotethatthepolygonQS is convex for the euclidean metric of S induced by the hyperbolic metric of H3. We split the analysis into several cases. H3 Case 1. None of the limit edges ei∞ is tangent to ∂ . H3 ∞ In particular, the planes Πf∞i all meet and define hyperbolic planes 3 3 n 3 H Π∞.Asn tends to ,thehyperbolicplaneH Π converges to H Π∞. ∩ fi ∞ ∩ fi ∩ fi The dihedral angle between two such consecutive planes H3 Πn and H3 Πn ∩ fi ∩ fi+1 n n RE is equal to the coordinate θei of θ =Θ(Pn) corresponding to ei E.Be- n ∈ E n ∈ cause θ converges to a point θ∞ K ]0,π[ ,thelimitθ∞ of θ is different ∈ Γ ⊂ ei ei from 0 and π.ItfollowsthatthetwohyperbolicplanesH3 Πn and H3 Πn ∩ fi ∩ fi+1 have non-trivial intersection, and that their dihedral angles along this intersec- tion geodesic is equal to θe∞i .Notethatthehyperbolicdihedralanglebetween the hyperbolic planes H3 Πn and H3 Πn is also equal to the euclidean ∩ fi ∩ fi+1

BULLETIN DE LA SOCIET´ EMATH´ EMATIQUE´ DE FRANCE 484 BAO (X.) & BONAHON (F.) angle between the two lines S Πn and S Πn in the horosphere S.Itfollows ∩ fi ∩ fi+1 that the polygon QS = C S is compact and, as in the proof of Proposition 5, m ∩ that i=1 θe∞i =2π. Because of the hypothesis that θ∞ satisfies Condition 1 in the definition ! of KΓ,thisimpliesthattheclosedcurveγΓ in Γdefined by γ is embedded, and 2 bounds a component of S Γcorrespondingtoavertexv4 V .Itfollowsthat Proj− ∈ Proj acomponentW of ∂P0 γ contains only one vertex of P0 ,namelyv4. This W cannot be a component− of the interior of the regular neighborhood U of Kp since, by definition of Kp,eachofthesecomponentscontainsatleastone edge of P Proj.Therefore,W must be a component of ∂PProj U and meets 0 0 − the interiors of the edges e12 and e23.Asaconsequence,thevertexv4 in W must be equal to v2,andv1 and v3 must be in Kp.However,thiswouldimply that xv∞1 = xv∞3 = p,contradictingthefactthatxv∞1 and xv∞3 belong to disjoint subsets of RP3,namelytherespectiveclosuresofR 0 and 0 R2.We ×{ } { }× consequently reach a contradiction in the case where none of the limit edges ei∞ is tangent to ∂ H3. ∞ H3 Case 2. Some but not all limit edges ei∞ are tangent to ∂ , and all H3 ∞ those ei∞ which are tangent to ∂ are equal. ∞ H3 Assume that ei∞ = ei∞+1 = = ej∞ are tangent to ∂ but that ei∞ 1 and ··· ∞ − e∞ are both different from e∞ = e∞.Then,fori

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As a consequence, the faces f and f share this vertex v V .Inparticular, i j+1 ∈ the end points of γΓ" (namely the vertices of Γrespectively corresponding to f and f )arecontainedintheclosureofthesamecomponentA of S2 Γ, i j+1 − namely the component corresponding to v V .Sinceθ∞ satisfies Condition 2 i 1 ∈ − in the definition of KΓ and since k=j+1 θe∞j = π,weconcludethatγΓ" is contained in the boundary of A.ButthisimpliesthattheclosedcurveγΓ is equal to the boundary of the component! A of S2 Γ. We already saw in Case 1 that this is impossible, as this leads to a contradiction.− H3 Case 3. Some but not all limit edges ei∞ are tangent to ∂ , and at least H3 ∞ two of the ei∞ which are tangent to ∂ are different. ∞ The convex polygon QS = S C in the horosphere S is non-compact since H3 ∩ some ei∞ is tangent to ∂ .Ontheotherhand,thehypothesisthatthere ∞ H3 exists an ej∞ which is not tangent to ∂ provides a corner of ∂QS where ∞ the external dihedral angle is at least θe∞j > 0. It follows that the boundary ∂QS is non-empty and connected. There consequently exists indices i1, i2 such H3 that the limit edge ei∞ is tangent to ∂ if and only if i1 i i2,counting indices modulo m. ∞ ≤ ≤ H3 If ei∞ and ei∞+1 are tangent to ∂ ,eitherei∞ = ei∞+1,orei∞ and ei∞+1 are collinear and point in opposite directions∞ at p.Indeed,thisimmediatelyfollows n by consideration of the possible limits for the face fi+1,usingthefactthatit is a convex polygon and that each of its edges must meet H3.(Therearetwo H3 cases, according to whether Πf∞ is tangent to ∂ or not). If ei∞ and ei∞+1 i+1 ∞ pointed in opposite directions at p,thiswouldprovideageodesiclinewhichis completely contained in the convex polyhedron QS,whichisincompatiblewith the fact that ∂QS has a corner with strictly positive external dihedral angle. Therefore ei∞ = ei∞+1 for all i1 i

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Otherwise, counting indices modulo m,therearei, j such that ei∞ = ei∞+1 = = ej∞ but such that ei∞ 1 and ej∞+1 are both different from ei∞ = ej∞.Then, ···the oriented edges e ,with− i k j have the same terminal end point v, k ≤ ≤ which is distinct from the terminal end point v" of ei 1 and from the terminal − end point v"" of ej+1.Thetwoverticesv" and v"" are actually equal; otherwise, Proj one would get a contradiction from the limit of the convex triangle Tn in Pn Proj whose vertices are the vertices of Pn corresponding to v, v" and v"",using 3 the fact that by Lemma 3 each edge of Tn must meet H . We conclude from this analysis that the terminal end point of each ei with 1 i m must be equal to v or v" = v"". ≤ ≤ If v and v" are joined by an edge e E,thenγ must be the boundary of a ∈ regular neighborhood of e,by3-connectednessofΓ.Inparticular,theedgee12 must be equal to e or to one of the e with 1 i m,bydefinitionofγ. i ≤ ≤ However, the limit edge e∞ joins the limit points x∞ and x∞,andiscontained v v! H3 R H3 in a line tangent to ∂ at p.Sincee12∞ contains the diameter 0 , ∞ ×{ }∩ it follows that e12 cannot be equal to e.Similarly,e12 cannot be equal to one H3 of the ei as ei∞ is tangent to ∂ at p by definition of γ. ∞ Finally, if v and v" are not joined by an edge, then the ei are the only edges whose interior is contained in the same component of ∂PProj K as γ.This 0 − p again contradicts the definition of γ since e12 cannot be equal to any of these ei, as above. As a consequence, we reach a contradiction in all cases. This completes the proof of Lemma 22.

H3 Lemma 23.—No plane Πf∞ is tangent to ∂ . ∞ 3 Proof of Lemma 23.—SupposethatΠ∞ is tangent to ∂ H .Then,because f ∞ no limit edge e∞ is reduced to a single point, by Lemma 22, the consideration n Proj of the possible degenerations of the face f of Pn shows the following: There are two edges e1 and e2 of the face f,meetingatavertexv,suchthatthelimit H3 point xv∞ is on the sphere ∂ and such that the limit edges e1∞ and e2∞ point ∞ in opposite directions at xv∞. By consideration of the convex cone C delimited by the limit planes Π∞ f ! associated to the faces f " F that contain v,oneconcludesasintheproofof ∈ Lemma 22 that there exists a third edge e3 containing v such that e3∞ is equal to e1∞ and e2∞.However,thisprovidesafacef0,containingv,suchthatthe n H3 convex polygon f0 converges to a line segment that is tangent to ∂ at one ∞ of its end points. This is not possible if no limit edge e∞ is reduced to a point. This proves Lemma 23.

For an edge e E,wealreadyknowbyLemma22thatthelimitedgee∞ ∈ is not reduced to a single point. If f and f " F are the two faces separated ∈ 3 3 by e,Lemma23showsthatthehyperbolicplanesΠ∞ H and Π∞ H are f ∩ f ! ∩

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n n H3 non-trivial. Since the dihedral angle θe between the hyperbolic planes Πf n 3 ∩ and Π H converges to the coordinate θ∞ =0, π of the limit point θ∞ K , f ! e Γ ∩ 3 3 $ ∈ it follows that Π∞ H and Π∞ H have a non-trivial intersection, and make f ∩ f ! ∩ dihedral angle of θe∞ along this intersection. In particular, the limit edge e∞ meets H3. n n For a face f F ,theplaneΠf containing the corresponding face f of Proj ∈ H3 Pn converges to a plane Πf∞ which meets and, for each edge e of f,the n n 3 edge e of f converges to an edge e∞ which meets H but whose end points H3 n are outside of .Itfollowsthatf converges to a convex polygon f ∞ Πf∞ which is really 2-dimensional and has the same combinatorics as f. ⊂ As a consequence, as f ranges over all the elements of F ,theunionofthe polygons f forms a polyhedral sphere S which has the same combinatorial ∞Proj ∞ type as ∂P0 .Inaddition,S is locally convex since the dihedral angle θe∞ ∞ of S along the edge e∞ corresponding to e E is in ]0,π[. Let∞ P Proj RP3 be bounded by the sphere∈S ,orientedbyitsidentification Proj∞ ⊂ Proj ∞ to ∂P0 .ToshowthatP is a projective convex polyhedron, we only have to show that it does not contain∞ any projective line. (Remember that we had included this condition in the definition of convex polyhedra in RP3). However, by local convexity, the only way this could fail is if P Proj was projectively equivalent to an infinite prism in R3 RP3,inwhichcasethedualgraphof∞ the cell decomposition of ∂PProj = S ⊂would be homeomorphic to a circle. But this dual graph is Γby construction,∞ ∞ and cannot be a circle by 3-connectedness. Therefore, P Proj is a convex projective polyhedron. We already noted that its vertices are∞ all outside H3,andthatitsedgesallmeetH3.Therefore P = P Proj H3 is a hyperideal polyhedron. By construction, the vertices ∞ Proj∞ ∩ Proj of P are the limit of the vertices of Pn ,anditfollowsthatP is the ∞ ∞ limit of the sequence Pn in Γ. This concludes the proofP of Proposition 21.

If we retrace back the proof of Proposition 21, and in particular the proof of Lemma 22, we can summarize it as follows: If the sequence Pn Γ, n N, admits no converging subsequence then, possibly after passing to a∈ subsequence,P ∈ it necessarily degenerates according to one of the following patterns:

1) one of the dihedral angles of Pn converges to 0 or π; 2) Pn develops a very long and thin ‘neck’ around a very short simple closed curve in ∂Pn which corresponds to a fixed curve in the model polyhedron ∂P0; Trun 3) the truncated polyhedron Pn develops a very long and thin ‘neck’ Trun around a very short simple closed curve in ∂Pn which corresponds to a fixed Trun Trun curve in ∂P0 and crosses exactly one of the new faces of ∂Pn associated Proj to vertices of P0 . An immediate corollary of Proposition 21 is the following.

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Corollary 24.—The map Θ: K is a covering map. PΓ → Γ Proof.—ThemapΘisalocalhomeomorphismbyTheorem11,anditis proper by Proposition 21. Since KΓ is locally compact, this implies that Θis acoveringmap.

6. Realizing large angles To realize hyperideal polyhedra with large external dihedral angles, namely with small internal dihedral angles, we will rely on Andreev’s classification of compact hyperbolic polyhedra with acute internal dihedral angles [2]. We first state this classification with our current terminology. Recall that the complete graph Kn is the graph with n vertices where any pair of distinct vertices are joined by one edge. Let Kn− be obtained from Kn by removing one arbitrary edge.

Theorem 25 (see [2]). — Let Γ be a planar 3-connected graph, different from 1 the graphs K and K−.Letaweightθ [ π,π[ be attached to each edge e of Γ. 4 5 e ∈ 2 There exists a compact polyhedron P in H3 with dual graph isomorphic to Γ and with external dihedral angle θe at the edge corresponding to the edge e of Γ if and only if the following three conditions are satisfied: 2 1) every component of S Γ contains exactly three edges e1, e2 and e3 of 3 − Γ,and i=1 θei < 2π; 2) for every simple closed curve γ embedded in Γ and passing through ex- ! 3 actly three edges e1, e2 and e3, i=1 θei > 2π unless γ is the boundary of a component of S2 Γ; 3) for every simple− closed curve!γ embedded in Γ and passing through exactly 4 S2 four edges e1, e2, e3 and e4, i=1 θei > 2π unless γ bounds in adiamond made up of two components of S2 Γ and of one edge of Γ separating them. ! − In addition, if P exists, it is unique up to isometry of H3.

Andreev also provides a similar result for the graph K5−,forwhichthereisan additional condition. Note that K5− is the dual graph of a prism with triangular basis. The case of the graph K4,whichisthedualgraphofatetrahedron,is treated in [17]. However, we will not need to consider these two graphs. Consider a strictly hyperideal polyhedron P0 Γ,anditsassociatedtrun- Trun Trun ∈PTrun cated polyhedron P0 .ThedualgraphΓ of P0 is obtained from Γas follows: For each component A of S2 Γ, add to Γa vertex v and join by an − A edge this new vertex vA to each vertex of Γlocated in the boundary of A.In particular, ΓTrun can be described purely in terms of Γ. Also, note that each Trun Trun edge of Γ ΓcorrespondstoanedgeofP0 where the external dihedral − 1 angle is equal to 2 π.

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Proposition 26.—Let Γ be a planar 3-connected graph with at least 4 ver- tices, and with a weight θ ] 2 π,π[ attached to each edge e of Γ.Thenthere e ∈ 3 exists a strictly hyperideal polyhedron P in H3 with dual graph isomorphic to Γ and with external dihedral angle θe at the edge corresponding to the edge e of Γ. In addition, P is unique up to isometry of H3. Proof.—LetΓTrun be the graph associated to Γas above. Let us associate to Trun 2 each edge e of Γ the weight θe [ 3 π,π[ofthedataofProposition26ife Trun ∈ 1 Trun is in Γ Γ ,andtheweightθe = 2 π otherwise. We first show that Γ with these⊂ edge weights satisfy the conditions of Theorem 25. By construction, ΓTrun is planar. One easily checks that it is 3-connected. Also, Γhas at least 4 vertices and S2 Γhasatleast4components,by3- connectedness. It follows that ΓTrun has− at least 8 vertices, and in particular is neither K4 nor K5−. 2 Trun Every component of S Γ is a triangle bounded by two edges e1 and e2 Trun − 1 of Γ Γandoneedgee3 of Γ. Then θe1 = θe2 = 2 π and θe3 <π,sothat 3 − i=1 θei < 2π.Asaconsequence,Condition1ofTheorem25issatisfied. Let γ be a simple closed curve embedded in ΓTrun which crosses exactly ! Trun three edges e1, e2 and e3.Ifatleastonevertexofγ is in Γ Γ, it easily follows from the 3-connectedness of Γthat γ is the boundary of− a component 2 Trun of S Γ .Otherwise,eachei is also an edge of Γbecause the edges of Trun − Trun Γ ΓalljoinavertexofΓtoavertexofΓ Γ. It follows that each θei − 2 3 − is greater than 3 π,andthereforethat i=1 θei > 2π.Therefore,Condition2 of Theorem 25 is satisfied. Finally, let γ be a simple closed curve! embedded in ΓTrun which crosses exactly four edges e1, e2, e3 and e4.Itcancontainatmosttwoverticesof ΓTrun Γ, since no two vertices of ΓTrun ΓareadjacentinΓTrun .Ifγ contains two vertices− of ΓTrun Γ, it again follows− from the 3-connectedness of Γthat γ bounds a diamond− made up of the union of two components of S2 ΓTrun and of one edge of Γ. If γ contains exactly one vertex of ΓTrun Γ,− say the Trun − vertex separating e1 from e2,thene1 and e2 are both in Γ Γande3 and 1 2 − 2 e4 are edges of Γ, so that θe1 = θe2 = 2 π, θe3 > 3 π and θe4 > 3 π;itfollows that 4 θ > 7 π>2π.Finally,ifnovertexofγ is in ΓTrun Γ, then each i=1 ei 3 − e is in Γ, so that θ > 2 π;consequently 4 θ > 8 π>2π in this case. It i ! ei 3 i=1 ei 3 follows that Condition 3 of Theorem 25 is satisfied. ! 3 By Theorem 25, there consequently exists a compact polyhedron P " in H Trun whose dual graph is isomorphic to Γ and with external dihedral angle θe at the edge corresponding to the edge e of ΓTrun . Trun The faces of P " correspond to vertices of Γ ,andthereforeareoftwotypes: those which correspond to vertices of Γ, and those which correspond to vertices Trun of Γ Γ. Let f1, f2,...,fn be the faces of P " that correspond to vertices − 3 Proj of Γ, and letΠ i RP be the projective plane containing fi.LetP be the ⊂ 3 n closure of the component of RP Π that contains the interior of P ". − i=1 i ´ ´ ´ BULLETIN DE LA SOCIETEMATHEMATIQUE, DE FRANCE 490 BAO (X.) & BONAHON (F.)

At this point, we know that the dual graph of the cell decomposition of Proj ∂P contains Γand has no additional vertex. If f is a face of P " which corresponds to a vertex of ΓTrun Γ, namely to a component of S2 Γ, and − − if fi1 , fi2 ,...,fik are the faces of P " adjacent to f,thefactthatthefij are orthogonal to f implies that the projective planes Πij all pass through the point 3 3 3 Π⊥ RP H ∂ H dual to the projective plane Πcontaining f.Itfollows that∈ the dual− graph∪ ∞ of ∂PProj has no additional edges, and is equal to Γ. Incidentally, this shows that P Proj is really a convex polyhedron in RP3, namely satisfies the additional condition that it contains no projective line. Otherwise, it would be projectively equivalent to the closure of an infinite prism in R3 RP3 and its dual graph Γwould be homeomorphic to a circle, which is excluded⊂ by 3-connectedness. This also shows that the hyperbolic polyhedron P = P Proj H3 is strictly ∩ hyperideal. By construction of P ",theexternaldihedralangleofP along the edge corresponding to the edge e of Γis equal to θe. Conversely, the uniqueness of P follows from the uniqueness of P " provided by Theorem 25.

7. Proof of Theorems 1 and 2 We are now ready to complete the proof of Theorems 1 and 2, by considering again the map Θ: K RE ,whichtoahyperidealpolyhedronassociates PΓ → Γ ⊂ its dihedral angles. Indeed, by definition of KΓ,Theorem1isequivalenttothe property that Θis surjective, whereas Theorem 2 is equivalent to the fact that Θ is injective. Both statements are simultaneously provided by the following result.

Proposition 27.—The map Θ: K is a homeomorphism. PΓ → Γ Proof.—WeprovedinCorollary24thatΘ: Γ KΓ is a covering map. E P → By definition, KΓ is convex in R ,andinparticularitissimplyconnected. It follows that the covering is trivial. Now, Proposition 26 shows that, for 2 E 1 every θ ] 3 π,π[ ,Θ− (θ)consistsofexactlyonepoint.Thisprovesthatthe covering∈ map Θ: K is actually a homeomorphism. PΓ → Γ

BIBLIOGRAPHY [1] Alexandrow (A.D.) – Konvexe Polyeder,Akademie-Verlag,Berlin, 1958. [2] Andreev (E.M.) – On Convex Polyhedra in Lobaˇcevski˘ıSpaces(Rus- sian),Mat.Sbornik,t.81 (123) (1970), pp. 445–478; English transl., Math. USSR Sb. t. 10 (1970), pp. 413–440.

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[3] , On Convex Polyhedra of Finite Volume in Lobaˇcevski˘ıSpaces (Russian),Mat.Sbornik,t.83 (125) (1970), pp. 256–260; English transl., Math. USSR Sb. t. 12 (1970), pp. 255–259. [4] Bao (X.) – Hyperideal Polyhedra in Hyperbolic 3-Space, Doctoral disser- tation,UniversityofSouthernCalifornia,LosAngeles,1998. [5] Cauchy (A.) – Sur les polygones et les poly`edres,J.Ecole´ Polytechnique, t. 9 (1813), pp. 87–98. [6] Coxeter (H.S.M.) – On Complexes with Transitive Groups of Automor- phisms,Ann.ofMath.,t.35 (1934), pp. 588–621. [7] Cromwell (P.R.) – Polyhedra,CambridgeUniversityPress,1997. [8] Grunbaum¨ (B.) – Convex Polytopes,IntersciencePublishers,JohnWiley &SonsInc.,NewYork,1967. [9] van Lint (J.H.) & Wilson (R.M.) – ACourseinCombinatorics,Cam- bridge University Press, 1992. [10] Rivin (I.) – Euclidean Structures on Simplicial Surfaces and Hyperbolic Volume,Ann.ofMath.,t.139 (1994), pp. 553–580. [11] , ACharacterizationofIdealPolyhedrainHyperbolic3-Space,Ann. of Math., t. 143 (1996), pp. 51–70. [12] Rivin (I.) & Hodgson (C.D.) – ACharacterizationofCompactConvex Polyhedra in Hyperbolic 3-Space,Invent.Math.,t.111 (1993), pp. 77–111; Corrigendum, Invent. Math. t. 117 (1994), pp. 359. [13] Schlenker (J.-M.) – M´etriques sur les poly`edres hyperboliques convexes, J. Diff. Geom., t. 48 (1998), pp. 323–405. [14] , Dihedral Angles of Convex Polyhedra,DiscreteComp.Geom., t. 23 (2000), pp. 409–417. [15] Stoker (J.J.) – Geometrical Problems Concerning Polyhedra in the Large,Comm.PureAppl.Math.,t.21 (1958), pp. 119–168. [16] Thurston (W.P.) – Three-Dimensional Geometry and Topology, (S. Levy, ed.),PrincetonMath.Series,vol.35,PrincetonUniversityPress, 1997. [17] Vinberg (E.B.)` – Discrete groups generated by reflections in Lobaˇcevski˘ı spaces,Mat.Sbornik,t.72 (114) (1967), pp. 471–488; English transl., Math. USSR Sb. t. 1 (1967), pp. 429–444; Correction, Mat. Sbornik t. 73 (115) (1967), pp. 303.

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