]]> 2002 American Mathematical Society Transactions of the American Mathematical Society Trans. Amer. Math. Soc. tran 354 06 20020204 December 8, 2000 October 5, 2001 February 4, 2002 2369 2369-2397 29 S0002-9947-02-02955-0 10.1090/S0002-9947-02-02955-0 0002-9947 1088-6850 English 57Mxx 2000 http://www.ams.org/jourcgi/jour-getitem?pii=S0002-9947-02-02955-0 Introduction sec:intro The twisted face-pairing construction , introduced in CFP3 , takes as input an arbitrary orientation-reversing face-pairing of a faceted 3-ball P (see definitions in Section 2), and returns an infinite, parametrized family of orientation-reversing face-pairings (Q,) for which every quotient Q is a closed, orientable 3-manifold. Even when the original face-pairing (P,) is very simple, or even trivial, the associated face-pairings (Q,) are almost always interesting and nontrivial, yet have relatively simple face-pairing descriptions, and have fundamental groups whose natural presentations are balanced with long relators. Thus twisted face-pairing 3-manifolds supply a wonderful and rich field of exploration for classroom use, and supply an incomparable source for those of us who study 3-manifolds in terms of the asymptotic combinatorial properties of their universal covers. Anyone who has sought relatively simple face-pairing descriptions of interesting 3-manifolds will recognize the difficulties resolved by this construction. We review the construction here. Given an orientation-reversing face-pairing on a faceted 3-ball P and a face f of P , let f denote the cellular homeomorphism by which identifies f with its paired face. Let be the equivalence relation on the set of edges of P generated by e f(e) when e is an edge of f . We call the resulting equivalence classes edge cycles . For each edge cycle e , let e be the number of edges in e and let m e be a positive integer. We call m e the multiplier of e and denote the function e m e by mul. We obtain a cell complex Q Q(, mul ) from P by subdividing every edge e into e m e subedges. Given a face f of Q , we construct an orientation-preserving cellular homeomorphism f:ff that takes each vertex v of f to the vertex of f that follows v in the orientation on f induced from the orientation on Q . We then obtain a face-pairing on Q by letting f f f , and we denote by M M(, mul ) the quotient complex Q . The fundamental theorem of twisted face-pairings is that M is always a 3-manifold. This was proved in CFP3 by an Euler characteristic argument. Our first goal in this paper will be to give a conceptual geometric proof of the same result. We do so by showing that the quotient complex has just one vertex and that its link is isomorphic with the faceted sphere dual to (Q) (Theorem 3.1). (Actually, when we are careful about orientations, we shall see that this isomorphism reverses orientations. We shall discuss orientations a bit more in the next paragraph.) We call M a twisted face-pairing 3-manifold . Note that the direction in which is twisted in forming the new face-pairing is determined by the orientation of Q , which is determined by the original given orientation on P . But P (and Q ) have two orientations, so that could have been defined so as to twist in the opposite direction. For our purposes, the natural way to encode this opposite twist is to denote by P and by Q the faceted 3-balls P and Q with the opposite orientation. Then (Q , ) denotes the twisted face-pairing resulting from twists in the opposite direction. Our main result in Section 4 is the proof of the highly nonobvious fact that Q and Q give dual cell structures of the same manifold. This material is crucial for the further development of the theory that will appear in papers CFP4 and CFP5 . In Section 5, we show, among other things, that there is an epimorphism 1(M) 1(P ) , where 1(P ) is taken to mean the fundamental group of P with vertices removed. The homology groups of M and P are compared in Section 6. We conclude the paper in Section 7 with examples. The examples include manifolds from five of the eight 3-dimensional geometries. In CFP5 we use a more general cellulation of P to give a twisted face-pairing 3-manifold with geometry S 2 . We know of no twisted face-pairing 3-manifold which has one of the product geometries 3 or 2 . This paper lays the groundwork for further papers, including CFP4 , CFP5 , and CFP6 . We say that a faceted 3-ball P is ample if it satisfies the following conditions: enumerate i) if two distinct faces of P intersect, then they intersect in a vertex or an edge; ii) if three distinct faces of P intersect pairwise, then their common intersection is a vertex; iii) no face of P is a triangle. enumerate In CFP4 we show that if P is an ample faceted 3-ball, is an orientation-reversing face-pairing on P , and mul is a multiplier function for , then 1(M(, mul )) is Gromov hyperbolic with space at infinity a 2-sphere. Hence one can use this procedure to easily construct 3-manifolds with Gromov hyperbolic fundamental groups. It is hoped that these will be useful test examples for our approach (see, for example, , CFP1 , CFP2 and CS ) to Thurston's Hyperbolization Conjecture. Indeed, the search for such examples was the primary motivation for the development of this twisted face-pairing construction. In CFP5 we describe twisted face-pairing 3-manifolds in terms of Heegaard diagrams and Dehn surgeries on framed links. Suppose we are given a faceted 3-ball P with 2n faces and an orientation-reversing face-pairing on P . Let E 1,,E m be the edge cycles. Let S be a closed orientable surface of genus n , and let 1,, n be a family of nontrivial, pairwise disjoint, simple closed curves on S whose union does not disconnect S . Then there is a family 1,, m of pairwise disjoint simple closed curves on S with the following property. For each i 1,,m , let i be one of the two Dehn twists on S about i , chosen so that the directions in which we twist are consistent. Given a multiplier function mul for , let mul 1 mul (E 1) m mul (E m) . We prove that S and the two families of curves 1,, n and mul ( 1),, mul ( n) form a Heegaard diagram for M(, mul ) . The surgery description comes naturally from this. From P and we get a diagram of a link in S 3 with components 1,, n, 1,, m . Then M(, mul ) is obtained by Dehn surgery from this link with framing 0 for each i and framing 1 mul (E j) plus the blackboard framing for each j . From the point of view of producing a single manifold M , there is no need to have a multiplier function. For suppose we are given a faceted 3-ball P with orientation-reversing face-pairing and multiplier function mul . Let P' be the faceted 3-ball obtained from P by subdividing each edge e of P into m e subedges. Then induces a face-pairing ' on P' , and M(, mul ) M(', mul ') , where mul ' is the constant multiplier function which assigns 1 to each edge cycle of P' . So by replacing P by P' one could assume that each multiplier is 1 . However, there are advantages to allowing and using multiplier functions. If P is an ample faceted 3-ball with orientation-reversing face-pairing and multiplier function mul , then the associated faceted 3-ball P' will not be ample if any multiplier is greater than one. So by switching to P' one would not be able to conclude from CFP4 that 1(M) is Gromov hyperbolic. Furthermore, as discussed in the previous paragraph, multipliers are inverses of framings of link components and by using multipliers one can often give unified treatments of infinite families of examples. Primary definitions sec:defns We begin with a cellulation of a 3-ball. We have in mind a polyhedron in that word's original meaning. Since the word polyhedron has come to mean something much more general, we instead use the expression faceted 3-ball. So we take a faceted 3-ball to be a cellulation of a 3-ball with exactly one 3-cell. We next must make the term cellulation precise. There is not an obvious class of cell complexes on the 2-sphere for us to use. We might work with more general cell complexes, but we find the class of cell complexes which we choose here to be interesting and convenient, so we proceed as follows. A faceted 3-ball P is an oriented regular CW complex such that P is a closed 3-ball and P has a single 3-cell. The regularity condition means that for each open cell the prescribed homeomorphism of an open Euclidean ball to that cell extends to a homeomorphism of the closed Euclidean ball to the closed cell. If P is a faceted 3-ball, a face-pairing on P consists of the following. First, the faces of P are paired: for every face f of P there exists a face f -1 f of P such that (f -1 ) -1 f . Second, for every face f of P there exists a cellular homeomorphism f:f f -1 called a face-pairing map such that f -1 -1 f . We furthermore impose a compatibility condition on the face-pairing maps two paragraphs below. Set f:f is a face of P . We say that is orientation-reversing if and only if every face-pairing map reverses orientation. Suppose P is a faceted 3-ball and is an orientation-reversing face-pairing on P . Let be the least equivalence relation on the set of edges of P such that, if f is a face of P and e is an edge of f , then e f(e) . The equivalence classes of are called edge cycles . We draw diagrams of edge cycles as follows. Let e 1,,e j be the distinct edges of an edge cycle E so that for every i 1,,j there exists a face f i of P with e if i and e i 1 f i (e i) , where the indices are taken modulo j . We draw the following diagram: equation lin:diagram e 1 f 1 e 2 f 2 f j-1 e j f j e 1. equation For each edge cycle E , let E be the number of edges in E and let m E be a positive integer. We call E the length of the edge cycle E . The function mul : edge cycles defined by Em E is called the multiplier function , and m E is called the multiplier of E . As on page 123 of T2 , to ensure that the quotient space P is a cell complex, we require that our face-pairing maps satisfy the following face-pairing compatibility condition . Every edge cycle diagram as in line lin:diagram in effect represents a composition of functions, these functions being face-pairing maps restricted to edges. We require that the composition of functions arising from every edge cycle diagram be the identity map. Hence for the edge cycle diagram in lin:diagram we require that f j e j f 2 e 2 f 1 e 1 be the identity map on e 1 . As in Problem 3.2.10 T2 , it follows that the action of the face-pairing maps on vertices determines P up to homeomorphism. Now construct a faceted 3-ball Q Q(, mul ) from P by just subdividing the edges of P as follows. Let e be an edge of P , and let E be the edge cycle of e . We subdivide e into Em E subedges. The face-pairing compatibility condition implies that we may choose this subdivision in an -invariant way: if v is a vertex of Q contained in a face f of P , then f(v) is also a vertex of Q . Every vertex of P determines a vertex of Q , called an original vertex of Q . Similarly, every edge of P determines a subcomplex of Q , called an original edge of Q . The vertices of Q which are not original vertices are called new vertices . figure tran2955el-fig-1 The link of x in Q . fig:lkq figure figure h tran2955el-fig-2 The link of in M . fig:lkm figure We orient Q so that the identity map P Q is orientation preserving. The orientations on P and Q induce orientations on each face f of P or Q , and these in turn determine orientations on the boundary f of each face f . The definition of faceted 3-ball assures that given a face f of P or Q and a vertex x of f , the notion of following vertex relative to f is meaningful: it is the vertex y of f adjacent to x in the direction of the orientation of f . We say that x is the vertex preceding y relative to f . Similarly, we have preceding and following edges and original edges relative to f . The edges of f in a given original edge of f are ordered relative to f . Because we obtain Q by subdividing P in an -invariant way, the face-pairing on P naturally determines a face-pairing, which we continue to call , on Q . For each face f of Q , let f be an orientation-preserving cellular homeomorphism of f which takes each vertex of f to its following vertex. We assume that these maps are defined -invariantly; that is, for each face f of Q we have f -1 f f f -1 . Let be the face-pairing on Q with the same pairing of faces as for the face-pairing , and with f f f for each face f of Q . We assume that the maps f are defined so that satisfies the face-pairing compatibility condition. Then is a face-pairing on Q called the twisted face-pairing . We refer to P as the model faceted 3-ball and we refer to as the model face-pairing from which Q and are constructed. Define M M(, mul ) to be the cell complex Q consisting of orbits of points of Q under . We show in Theorem thm:funthm that M is always a manifold. figure tran2955el-fig-3 Two faces of S . fig:faces figure figure tran2955el-fig-4 The simple closed curve . fig:gamma figure By a dual of a faceted 3-ball P we mean a faceted 3-ball for which there exists an orientation-reversing cellular homeomorphism to P . Given a faceted 3-ball P , we let P denote a faceted 3-ball dual to P . Given P , , mul , and Q as above, we fix an orientation-reversing cellular homeomorphism :Q Q . Notions such as original vertex and edge extend to Q . Let x be a vertex of Q , and denote by the image of x in M . Then link (x,Q) is canonically isomorphic to a face f(x) of link (,M) . If f is a face of Q containing x , then f determines an edge e x(f) of link (,M) . Similarly, if e is an edge of Q containing x , then e determines a vertex v x(e) of link (,M) . See Figures fig:lkq and fig:lkm . We use the clockwise orientation of faces in all figures. Fundamental theorem sec:funthmsec The following is the fundamental theorem of twisted face-pairing. We use the notation and terminology of the previous section. thm thm:funthm Let P be a faceted 3-ball, let be an orientation-reversing face-pairing on P , and let mul be a multiplier function for . Then the cell complex M M(, mul ) is an orientable closed 3-dimensional manifold. Furthermore, M has just one vertex, u , and the dual of link (u,M) is isomorphic to Q as oriented 2-complexes. This isomorphism is determined by the map :Q Q ; every face of link (u,M) is canonically isomorphic to link (x,Q) for some unique vertex x of Q and the barycenter of this face corresponds to the vertex (x) of Q . thm proof Our argument proceeds by first proving the last two sentences of the theorem. Once this is done, it follows that the link of every vertex of M is a 2-sphere, and so M is a manifold by a well-known theorem. For example, see 59, Theorem 1 ST or Prop. 3.2.7 T2 . Let f be a face of Q , let a be an original vertex of f , and let h be the edge of f preceding a . We investigate the star S of v a(h) in link (,M) . For this, let x 1 f(a) and let h 1 f(h) . It follows that 1) x 1 , 2) v x 1 (h 1) v a(h) and 3) e x 1 (f -1 ) e a(f) . See Figure fig:faces . Thus f(a) and f(x 1) are faces of S which share the edge e x 1 (f -1 ) e a(f) which contains v a(h) . Furthermore, because h is the edge of f preceding a , if B 1 is a barycenter of f(a) and B 2 is a barycenter of f(x 1) and is a simple closed curve from v a(h) to B 1 to B 2 to v a(h) as shown in Figure fig:gamma , then is oriented in the direction opposite to the orientation of the faces of Q . (Recall that all figures are drawn using the clockwise orientation for the faces of Q .) We use Figure fig:defg to continue the investigation of S . Let a 1 a , let b 1 be the original vertex of f following a and let e 1 be the original edge of f following a , so that a 1 and b 1 are the endpoints of e 1 . Let f 1 f . We maintain the meaning of x 1 and h 1 in the previous paragraph. The edge cycle E of the edge of P corresponding to e 1 is depicted in Figure fig:defg . Fix k 1,, Em E-1 for the rest of this paragraph. Suppose that a k , b k , e k , f k , x k , and h k are defined. Define a k 1 , b k 1 , e k 1 , and f k 1 so that a k 1 f k (a k) , b k 1 f k (b k) , e k 1 f k (e k) and f k 1 is the unique face of Q other than f -1 k which contains e k 1 . Note that a k , b k , e k and f k are periodic with period E . Define x k 1 and h k 1 so that x k 1 f k 1 (x k) and h k 1 f k 1 (h k) . Then h k is the k th edge of e k 1 relative to f k 1 and h Em E is the last edge of e 1 relative to f 1 . In particular, x Em E b 1 . Just as in the previous paragraph, we see that f(x k) and f(x k 1 ) are faces of S which share the edge e x k (f k 1 ) which contains v a(h) . See Figure fig:linkpart . Since the vertices x 1,,x Em E-1 have valence 2, the faces f(x 1),,f(x Em E-1 ) are digons. figure tran2955el-fig-5 Defining a k , b k , e k , f k , x k , and h k . fig:defg figure figure tran2955el-fig-6 The part of the link of the vertex of M corresponding to the original edge of f from a 1 to b 1 . fig:linkpart figure We now repeat the discussion of the previous three paragraphs, replacing h by h Em E and replacing a 1 by b 1 . Continuing, we see that the dual of the link of v a(h) in S is canonically isomorphic to f . Furthermore, the discussion of the simple closed curve in Figure fig:gamma shows that as we traverse f in the positive direction, we traverse the dual of the link of v a(h) in S in the opposite direction. Since the conclusions of the previous paragraph hold for every face of Q , it easily follows that M has just one vertex and all of the assertions of Theorem thm:funthm are true. This proves Theorem thm:funthm . proof We call M a twisted face-pairing 3-manifold . We wish to label and direct every face of P and Q . This direction of faces is not to be confused with the orientation of faces induced by the orientation of P and Q . For this, suppose that Q has n pairs of faces, and let F 1,,F n be representatives of these pairs, so that the faces of Q are F 1 1 ,,F n 1 . Let f be a face of Q . Then there exists a unique i 1,,n such that f F i, F -1 i . We label f with i . Suppose f F i . In this case we say that f is directed outward . By this we have in mind a vector normal to f pointing out of Q . Similarly, if f F -1 i , then we say that f is directed inward , and we have in mind a vector normal to f pointing into Q . We label and direct the faces of P so that the canonical correspondence between the faces of P and Q respects labels and directions. If we label a face of P or Q in a figure with i' , then that face has label i and is directed inward; a face of P or Q in a figure with label simply i is understood to be directed outward. If f is a face of P with label i and directed outward, then we allow ourselves to write i instead of f . We next direct the edges of Q and label each of these directed edges with an element of 1,,n as follows. Let e be an edge of Q . Let x and y be the vertices of e . Theorem thm:funthm implies that x and y determine faces f x and f y of the link of the vertex of M and e determines an edge common to f x and f y . As in Figure fig:faces , viewed from f x , this common edge corresponds to a face f of Q and viewed from f y , this common edge corresponds to f -1 . There exists a unique i 1,,n such that f F i,F -1 i . We label e with i . If f F i , then we direct e from x to y , and if f F -1 i , then we direct e from y to x . ex tetrahedron We give an example to illustrate the construction. Let the model faceted 3-ball P be a tetrahedron with vertices A , B , C , and D as in Figure fig:tet1p . The number 1 which appears in one corner of the face ABC indicates that face ABC has label 1 and is directed outward. The 1' which appears in one corner of face ABD indicates that face ABD has label 1 and is directed inward. The situation is analogous for faces ACD and BCD . The directed circular arcs around 1 , 1' , 2 and 2' simply remind us that, as always, faces in figures are oriented clockwise. We proceed as in Section 2 of CFP3 . We next define face-pairing maps. Face-pairing map 1 maps face ABC to face ABD fixing AB , and face-pairing map 2 maps face ACD to face BCD fixing CD . We write 1:( array ccc A B C A B D array ), 2:( array ccc A C D B C D array ), and 1 1 , 2 1 . The edge cycles for have diagrams as follows: equation 2 gather AB 1 AB, BC 1 BD 2 AD 1 AC 2 BC, lin:edgecycle CD 2 CD. gather figure tran2955el-fig-7 The complex P for the tetrahedron example. fig:tet1p figure At present we consider the simplest case and choose each of the edge cycle multipliers to be 1 . Then AB and CD are not subdivided in passing from P to Q , but BC , BD , AD , and AC are each subdivided into four subedges. Figure fig:tet1q shows the faceted 3-ball Q . We label the new vertices of Q arbitrarily. figure tran2955el-fig-8 The complex Q for the tetrahedron example. fig:tet1q figure Figure fig:tet1fogr shows the link of the vertex of M . For each vertex v of Q , there is a face f(v) in this link. The vertices of f(v) are parametrized by the edges of Q incident to v ; in Figure fig:tet1fogr we label the corners of f(v) by the other vertices incident to those edges. The edges of the link correspond to faces of Q , and are labeled 1 or 2 and given a transverse direction accordingly. figure tran2955el-fig-9 The link of the vertex of M . fig:tet1fogr figure Figure fig:tet1bata shows the faceted 3-ball Q dual to Q . The edge labels and directions of Q and the vertex labels of Q are induced by the isomorphism between Q and the dual of the link of the vertex of M . figure tran2955el-fig-10 The complex Q with edge labels and directions. fig:tet1bata figure ex Duality sec:dual We now wish to construct a twisted face-pairing 3-manifold from Q . It is clear that we may view Q as a subdivision of P . Furthermore, there exists a face-pairing on P corresponding to with multiplier function mul corresponding to mul so that Q is gotten from P , and mul in the same way that Q is gotten from P , and mul. We let be the twisted face-pairing on Q determined by and mul . Note that because the map :QQ reverses orientation, is not the conjugation'' of by ; the face-pairing twists in the opposite direction. The following theorem motivates us to consider . thm thm:resp The edge labels and directions of Q are respected by . thm proof To prove Theorem thm:resp we further investigate the edge direction and labeling of Q . We return to the setting of paragraph 4 of the proof of Theorem thm:funthm . With Figures fig:defg and fig:linkpart in mind, we find it useful to augment Figure fig:defg with diagonals as in Figure fig:diags . We represent (e 1) as the diagonal from a 1 to x 1 to x 2 , , ending at x Em E b 1 . For every i 1,, Em E the edge of (e 1) from x i-1 to x i (from a 1 to x 1 when i 1 ) is labeled with j , where f i F j,F -1 j and this edge is directed toward x i if f i F j and this edge is directed away from x i if f i F -1 j . Just as we view f -1 E as being left of e 1 and f 1 as being right of e 1 , we view (f 1) as being above (e 1) and we view (f -1 E ) as being below (e 1) . Analogous discussions hold for (e 2),,(e E ) . figure tran2955el-fig-11 Drawing the (e i) 's as diagonals. fig:diags figure Let i 1,, E . It is now clear that applying f i to a vertex of e i other than b i moves that vertex diagonally one unit right and up in Figure fig:diags . Likewise, applying f i to a vertex of (e i) other than a i moves that vertex down one unit in Figure fig:diags . Furthermore, just as f 1 (b 1) is the vertex below b 1 , it is also true that f 1 (a 1) is the vertex below a 1 . Theorem thm:resp follows easily. proof Theorem thm:funthm implies that Q and determine a manifold M with one vertex the dual of whose link is isomorphic to Q as oriented 2-complexes. We direct the faces of Q and label them with 1,,n so that :QQ preserves directions and labels of faces. We direct the edges of Q and label them with 1,,n in accordance with the isomorphism between Q and the dual of the link of the vertex of M . thm How to label and direct the edges of Q and Q thm:howto Let e be an edge of P . Let f be a face of P which contains e , and let E be the edge cycle of which contains e . Let F 1 1 ,,F n 1 be the faces of P so that F k is labeled with k and directed outward for k 1,,n . Let e 1 i 1 a 1 e 2 i 2 a 2 i j-1 a j-1 e j i j a j e 1, be a diagram of E , where e kF i k a k and a k 1 for k 1,,j , e 1 e and F i 1 a 1 f . Let k 1,,j . Then, relative to the orientation of f , the (j-k 1) th new edge of e in Q has label i k and its direction agrees with the orientation of f exactly when a k -1 . Likewise, relative to the orientation of ((f)) , the (j-k 1) th new edge of (e) in Q has label i k and its direction agrees with the orientation of ((f)) exactly when a k -1 . The above description is complete if mul (E) 1 . In general, there are mul (E) copies of the above pattern. thm proof As in the proof of Theorem thm:resp , this follows easily from Figure fig:diags . proof ex We continue with the tetrahedron example of Example 3.2. Just as Figure fig:tet1fogr shows the link of the vertex of M , Figure fig:tet1bagr shows the link of the vertex of M . Just as we label and direct the edges of Q in Figure fig:tet1bata , we label and direct the edges of Q in Figure fig:tet1fota . We can verify that the edge directions and labels of Q are correct by using either Figure fig:tet1bagr or (with much less effort) Theorem thm:howto and lin:edgecycle . ex figure tran2955el-fig-12 The link of the vertex of M . fig:tet1bagr figure figure tran2955el-fig-13 The complex Q with edge labels and directions. fig:tet1fota figure thm thm:edgelabel Two edges of Q map to the same edge of M if and only if they have the same label. thm proof Theorem thm:resp implies that if two edges of Q map to the same edge of M , then they have the same label. On the other hand, the proof of the main theorem of CFP3 shows that the number of edges of M equals the number of faces of M . In other words, the number of edges of M equals the number of edge labels of Q . Theorem thm:edgelabel follows. proof cor cor:induce The directions and labels of the faces and edges of Q induce directions and labels of the faces and edges of M so that M has exactly one face and edge with label i for every i 1,,n . cor proof The assertion for faces is clear, and the assertion for edges follows from Theorem thm:resp and Theorem thm:edgelabel . proof Our next goal is to construct cell complexes M and M which subdivide M and M , respectively, and to show that there exists a cellular homeomorphism from M to M . For this we introduce the notion of a dual cap subdivision . The terminology dual cap subdivision derives from the word dual as in dual complex and the word cap in the meaning of intersection. The dual cap subdivision of a complex is gotten by intersecting'' that complex with its dual complex''. Formally, we next define the dual cap subdivision of a regular CW complex X with dimension at most 3. We proceed by induction on the dimension of X . Suppose that our given complex is an edge e , consisting as usual of two 0-cells and one 1-cell. Choose a barycenter for e , and subdivide e into two edges by means of its barycenter. This determines the dual cap subdivision e of e . In turn this determines the dual cap subdivision X of X if X has dimension 1. Now suppose that our given complex is a face f , consisting of one 2-cell whose boundary is a 1-complex. We have (f) . Choose a barycenter for f . We join the barycenter of f with an edge to the barycenter of every edge of f . This determines the dual cap subdivision f of f , which subdivides f into quadrilaterals, one for every vertex of f . In turn this determines the dual cap subdivision X of X if X has dimension 2. Finally, suppose that our given complex is a 3-cell c whose boundary is a 2-complex. We have (c) . Choose a barycenter for c . We join the barycenter of c with an edge to the barycenter of every face of c . To complete our description of c , we describe the faces of the subdivision of c whose interiors lie in the interior of c . Let e be an edge of c . Let f and g be the faces of c which contain e . We have defined edges of c joining the barycenter of e to the barycenter of f , joining the barycenter of f to the barycenter of c , joining the barycenter of c to the barycenter of g and joining the barycenter of g to the barycenter of e . These four edges of c determine a face of c whose interior lies in the interior of c . Every face of the subdivision of c whose interior lies in the interior of c has this form. This determines the dual cap subdivision c of c . In turn this determines the dual cap subdivision X of X if X has dimension 3. In this paragraph we discuss the structure of the 3-cells which occur in the dual cap subdivision of a 3-cell. For 2-cells the situation is clear: the dual cap subdivision of a 2-cell is a union of quadrilaterals, one for every vertex of the given 2-cell. Figure fig:2cell shows a vertex v of a 2-cell f and parts of the two edges of f which contain v , drawn as thick line segments. The dots indicate that part of each edge is not shown. The vertex u is the barycenter of f , and the other two vertices are barycenters of the edges of f which contain v . We see one of the quadrilaterals in the dual cap subdivision of f . Figure fig:3cell is a 3-dimensional version of Figure fig:2cell . In Figure fig:3cell v is a vertex of a 3-cell c , and the thick line segments which contain v are parts of the edges of c which contain v . Now u is the barycenter of c . The 3-cell of the dual cap subdivision of c which contains v might be called an alternating suspension; we take an octagon, cone every other vertex of the octagon to v and cone the remaining vertices of the octagon to u . In general, suppose that v has valence k . Then we cone every other vertex of a 2 k -gon to v , and we cone the remaining vertices of the 2 k -gon to u . Both u and v have valence k in the resulting 3-cell, and every other vertex has valence 3. figure tran2955el-fig-14 One 2-cell in the dual cap subdivision of a 2-cell. fig:2cell figure figure tran2955el-fig-15 One 3-cell in the dual cap subdivision of a 3-cell. fig:3cell figure Having defined the notion of dual cap subdivision, we construct a dual cap subdivision Q of Q which is -invariant. The face-pairing on Q then induces what might be called a face-pairing on Q . (It is not quite a face-pairing as defined only because Q has more than one 3-cell.) We define the dual cap subdivision of M to be the cell complex M of orbits of points of Q under . We likewise have analogous statements involving Q , and M . Let C 1,,C k be the closed 3-cells of Q . For every i 1,,k let A i be a cell complex isomorphic to C i so that A 1,,A k are mutually disjoint. Let A be the disjoint union of A 1,,A k . The face-pairing on Q induces in a straightforward way what might be called a face-pairing on A . (This is more general than a face-pairing as defined because A has more than one 3-cell, these 3-cells are mutually disjoint and only those faces corresponding to faces in the boundary of Q are paired with other faces.) The proof of Theorem thm:funthm can be viewed as showing that the cell complex which consists of orbits of points of A under is isomorphic to Q . Thus we may view A 1,,A k as being cell complexes isomorphic to the 3-cells of Q . Now the face-pairing on Q induces a face-pairing on A . By symmetry, the complex which consists of orbits of points of A under is isomorphic to Q . It follows that if we begin with A , perform the identifications determined by and then perform the identifications determined by , we obtain a cell complex isomorphic to M . So M is isomorphic to the cell complex gotten from A by performing the identifications determined by both and . Likewise M is isomorphic to the cell complex gotten from A by performing the identifications determined by both and . This proves that there exists a cellular homeomorphism from M to M . Because there exists a cellular homeomorphism from M to M , we may view M and M as giving cellular decompositions of the same space: we take M M . When we do this we see that there exists a duality between the cells of M and the cells of M . The vertex of one corresponds to the 3-cell of the other. Every edge of one corresponds to a face of the other. In fact, Corollary cor:induce shows that the face and edge labels and directions of Q and Q determine face and edge labels and directions for M and M . The edge of M labeled with i 1,,n passes through the face of M labeled with i , and the edge of M labeled with i 1,,n passes through the face of M labeled with i . Furthermore, directions of edges agree with directions of faces. We summarize this paragraph in the following theorem. thm thm:duality The manifolds M and M are homeomorphic. We may identify them so that the complexes M and M are dual to each other in a way which respects directions and labels of edges and faces. thm cor cor:ccorccw Let P be an unoriented faceted 3-ball embedded in 3 . Suppose given an orientation-reversing face-pairing on P with a multiplier function. From this we can construct two twisted face-pairings; one twisted face-pairing is gotten by always twisting clockwise and one twisted face-pairing is gotten by always twisting counterclockwise. Then the manifolds gotten from these two twisted face-pairings are homeomorphic. cor proof This is essentially equivalent to saying that M and M are homeomorphic. proof The duality between M and M motivates us to say that objects such as , , M ,are dual to the objects , , M , ,. Let be the universal covering cell complex of M , and let be the universal covering cell complex of M . We assume that M M , and hence that . The edge and face directions and labels of M and M lift to and . Just as we have a duality between M and M , we have a duality between and . Let G be the fundamental group of M . We fix an action of G on . It is clear that G acts cellularly on both and and that G preserves edge and face labels and directions. We fix a vertex of , and we call the base vertex of . Likewise we fix a vertex of , and we call the base vertex of . Let C be the closed 3-cell of dual to . There exists a canonical cellular map from Q to C . Let F 1 1 ,,F n 1 denote the faces of Q as before, and let 1 1 ,, n 1 denote their images in C under the canonical map from Q to C . Define n elements of G as follows. Let i 1,,n . Let g i be the unique element of G such that i C g i C ; it follows that g i is the unique element of G such that -1 i C g -1 i C . We similarly define elements g 1 ,,g n of G using the 3-cell of dual to . thm thm:present Let F 1 1 ,,F n 1 be the faces of Q as usual. Then the elements g 1,,g n generate G , and a defining set of relators for them is determined by the closed edge paths of Q arising from the boundaries of (F 1),,(F n) . thm proof This is standard; for example, see 46 and 62 ST . Let p:Q M be the quotient map. The group G is isomorphic to the fundamental group of the spine p(Q ) , and the presentation given by Theorem thm:present is the standard presentation of the fundamental group of the surface complex p(Q ) . proof We refer to g 1,,g n as the geometric generating set of G . ex We continue with the tetrahedron example of Example 3.2. Let x 1,,x n form a basis of a free group. We consider the diagram of Q in Figure fig:tet1bata . Beginning with edge AB and proceeding in the counterclockwise direction, the labels and directions of the edges of the face ABC give the relator x 1 x 1 x 2x 1x 2 x 1 x 2 x 1x 2. Beginning with edge CD and proceeding in the counterclockwise direction, the labels and directions of the edges of face ACD give the relator x 2 x 2 x 1x 2x 1 x 2 x 1 x 2x 1. We conclude that Gx 1,x 2:x 1 x 1 x 2x 1x 2 x 1 x 2 x 1x 2, x 2 x 2 x 1x 2x 1 x 2 x 1 x 2x 1. Hence H 1(M) G G,G is the trivial group, and so M is a homology sphere. We obtain a dual presentation for G by using Q instead of Q : Gx 1,x 2:x 1 x 1 x 2 x 1x 2x 1 x 2x 1x 2, x 2 x 2 x 1 x 2x 1x 2 x 1x 2x 1. ex Returning to the general case, we let g denote the Cayley graph of G with respect to g 1,,g n (the subscript g stands for geometric). thm thm:cayley The 1-skeleton of with its labels, directions and base vertex is canonically isomorphic to g . thm proof After we map the vertex of g represented by the identity element of G to , Theorem thm:present easily implies that there exists a unique G -equivariant cellular isomorphism from g to the 1-skeleton of . This proves Theorem thm:cayley . proof The model face-pairing pseudomanifold N and the group H sec:orb Suppose given a faceted 3-ball P with an orientation-reversing face-pairing . Define N N() to be the cell complex consisting of the orbits of points of P under . We call N the model face-pairing pseudomanifold . Our next goal is to define a group H , which is closely related to the fundamental group of N . Let F 1 1 ,,F n 1 be the faces of P , and let i:F iF i -1 for i 1,,n be face-pairing maps as usual. Let x 1,,x n form a basis of a free group F . Let E be an edge cycle of . Suppose that E has the diagram e 1 i 1 a 1 e 2 i 2 a 2 i j-1 a j-1 e j i j a j e 1, where e 1,,e j are the edges of E and a 1,,a j 1 . To E we associate the element R E x i 1 a 1 x i j a j F . The element R E is defined only up to inversion and cyclic permutation, namely, it is defined only up to choosing a diagram for E . We define the group H to be given by the presentation H x 1,,x n: R E:E is an edge cycle of . thm thm:hom The map from g 1,,g n to x 1,,x n given by g i x i for i 1,,n determines a surjective group homomorphism from G to H . thm proof Theorems thm:howto and thm:present show that G is given by a presentation whose relators are products of cyclic permutations of the relators of H and their inverses. Theorem thm:hom follows. proof thm thm:hom2 Let N 0 be the open manifold gotten from N by deleting its vertices. Then H 1(N 0) . The inclusion N 0N gives a surjective group homomorphism 1(N 0) 1(N) which is an isomorphism if and only if N is a manifold. thm proof Let p:PN be the quotient map. By construction N has a natural cell structure induced from the cell structure of P . Construct a dual cap subdivision P of P which is -invariant. Let v be the barycenter of the 3-cell of P , and denote the barycenter of a face f of P by b(f) . Let K be the surface complex in N defined as follows. There is exactly one vertex p(v) . For each face f of P , let e f be the edge in P joining v and b(f) . For each pair f,f of paired faces in P , there is an edge p(e f)p(e f ) in K . For each edge e of P , there is a face in K which consists of the union of the images under p of the faces in P which contain v and the barycenter of an edge in the edge cycle of e . That is, the vertex of K is dual to the 3-cell of N , the edges of K are dual to the faces of N , and the faces of K are dual to the edges of N . Then H 1(K) (see 46 ST ). Since K is a strong deformation retract of N 0 , we have 1(K) 1(N 0) . If N is a manifold, then 1(N 0) 1(N) . If N is not a manifold, then some vertex of N has a link F which is not simply connected. Let and be simple closed curves in F which intersect transversely in a single point. It follows easily from duality that the homology classes of and cannot both be 0 in H 1(N 0) . Hence the image of 1(F) in 1(N 0) is nontrivial, and so by van Kampen's theorem 1(N) is a proper quotient of 1(N 0) . proof cor cor:hom3 There exists a surjective group homomorphism from 1(M) to 1(N) . cor proof This follows from Theorem thm:hom and Theorem thm:hom2 . proof Homology of M and N sec:homsec Since M is a 3-dimensional cell complex, to M there is associated a chain complex C(M) with boundary operator of the form 0 C 3(M) 3 C 2(M) 2 C 1(M) 1 C 0(M) 0. Likewise we have for N a chain complex C(N) with boundary operator of the form 0 C 3(N) 3 C 2(N) 2 C 1(N) 1 C 0(N) 0. There exists a chain map :C(N) C(M) with i:C i(N) C i(M) for i 0,1,2,3 defined as follows. The map 3 maps the 3-cell of N to the 3-cell of M . There exists a canonical bijection between the 2-cells of N and the 2-cells of M , and 2 maps every 2-cell of N to the corresponding 2-cell of M . The complex M has only one 0-cell, and 0 maps every 0-cell of N to the 0-cell of M . Finally, we discuss the map 1 . We identify the edges of N with the edge cycles E 1,,E m of . These edge cycles E 1,,E m form a -basis of C 1(N) . Let e 1,,e n be the edges of M , so that e i is labeled with i . We view e 1,,e n as forming a -basis of C 1(M) . Let i 1,,n , and let j 1,,m . Let x i and R E j be as in the definition of H in Section sec:orb . Define ij to be the sum of the exponents of x i in R E j . We set 1(E j) mul (E j) i 1 n ij e i for j 1,,m . It remains to orient e i and E j . The direction of e i determines the orientation of e i . Finally, the choice of edge cycle diagram in defining R E j determines an orientation of E j for which it is possible to prove using Theorem thm:howto that 0 , 1 , 2 , 3 give a chain map . It is easy to see that we have the following commutative diagram of Abelian groups and group homomorphisms, where the vertical sequences of maps are exact. array ccccccccccc 0 0 0 0 ker ( 1) ker ( 0) 0 0 C 3(N) C 2(N) C 1(N) C 0(N) 0 0 C 3(M) C 2(M) C 1(M) C 0(M) 0 0 0 0 array The long exact homology sequence associated to this diagram is equation eq:long array rcl 0 H 3(N) H 3(M) 0 H 2(N) 10 pt H 2(M) K H 1(N) H 1(M), array equation where K is the kernel of the homomorphism from ker ( 1) to ker ( 0) . Given a finite cell complex X , let b i(X) denote its i th Betti number for every nonnegative integer i . Statements 4 and 5 are the main statements of the following theorem. The other statements are included for completeness and because they are easy. thm 1 thm thm:homology We have the following statements. enumerate H 0(M) H 0(N) H 3(M) H 3(N) . b 2(N) b 2(M) . The groups K , H 2(M) and H 2(N) are free Abelian groups. b 1(M) rank (H H,H ) . We have b 1(N) b 1(M) , and equality holds if and only if N is a manifold. The image of the map in eq:long from H 1(N) to H 1(M) lies in the torsion subgroup of H 1(M) . enumerate thm proof Statement 1 is clear. Statement 2 is clear because eq:long shows that H 2(N) injects into H 2(M) . Statement 3 is also clear after noting that both 3:C 3(N) C 2(N) and 3:C 3(M) C 2(M) are zero maps. Now consider statement 4. Because 1:C 1(M) C 0(M) is the zero map, H 1(M) C 1(M) im ( 2) . We next investigate 2:C 2(M)C 1(M) . Let F 1,,F n be the usual faces of Q , now viewed as forming a -basis of C 2(M) . As usual, the orientation of Q determines orientations of F 1,,F n . Let be the matrix of 2 with respect to the ordered bases (-F 1,,-F n) and (e 1,,e n) . Let be the matrix of 1 with respect to the ordered bases (E 1,,E m) and (e 1,,e n) . We view the columns of and as vectors in n . Because H 1(M) C 1(M) im ( 2) , it follows that b 1(M) is the codimension of the -column space of . The definitions of H and 1 easily imply that rank (H H,H ) is the codimension of the -column space of . Hence to prove statement 4, it suffices to prove that the column space of equals the column space of . In this paragraph we prove that the column space of equals the column space of . Let j 1,,n . Using Theorem thm:howto for the first equation, it is not difficult to see that array rcl 2(-F j) k 1 m jk 1(E k) k 1 m jk mul (E k) i 1 n ik e i 10pt k 1 m mul (E k) i 1 n ik jk e i. array This implies that k 1 m mul (E k) k , where k is the nn matrix whose ij -entry is ik jk for k 1,,m . We prove for k 1,,m that the symmetric matrix k is positive semidefinite as follows. Let x (x 1,,x n) t be a column vector in n . Then the i th entry of the vector kx is j 1 n ik jk x j , and so equation 1 equation eq:psd array rcl x t kx i 1 nx i j 1 n ik jk x j i 1 n ik x i j 1 n jk x j 10pt ( i 1 n ik x i) 2 0. array equation This proves that k is positive semidefinite for k 1,,m . Hence is a positive semidefinite symmetric matrix because it is a positive linear combination of positive semidefinite symmetric matrices. It follows that the null space of is the set of all column vectors x in n such that 0 x t x k 1 m mul (E k)x t k x . Because k is positive semidefinite for k 1,,m , we have that k 1 m mul (E k)x t k x 0 if and only if x t k x 0 for every k 1,,m . In other words, the null space of is the intersection of the null spaces of 1,, m . But eq:psd shows that the null space of k is the orthogonal complement of the line spanned by the column ( 1k ,, nk ) t for k 1,,m . Because is symmetric, it follows that the orthogonal complement of the column space of equals the orthogonal complement of the column space of , and so the column space of equals the column space of . This proves statement 4. The inequality of statement 5 follows from Corollary cor:hom3 . If N is a manifold, then Theorem thm:hom2 and statement 4 imply that b 1(N) b 1(M) . Finally, a classical theorem on Euler characteristics, which is proved in Section 3 of CFP3 , implies that (N) 0 . In other words, b 2(N) b 1(N) . On the other hand, if b 1(N) b 1(M) , then b 1(N) b 1(M) b 2(M) b 2(N) , the second equality coming from Poincare duality and the inequality coming from statement 2. It follows that if b 1(N) b 1(M) , then b 1(N) b 2(N) , that is, (N) 0 . As in Section 3 of CFP3 , we have (N) 0 if and only if N is a manifold. Thus b 1(N) b 1(M) if and only if N is a manifold. Statement 6 follows from the fact that the column space of equals the column space of : given x C 1(N) there exists a positive integer a such that a 1(x) im ( 2) , that is, the homology class of x in H 1(N) maps to a homology class in H 1(M) whose order divides a . This proves Theorem thm:homology . proof thm 1 cor cor:betti Let P be a model faceted 3-ball, and let P' be a model faceted 3-ball gotten from P by subdividing P . Let M be the twisted face-pairing manifold gotten from P and some multiplier function, and let M' be the twisted face-pairing manifold gotten from P' and some multiplier function. Then b 1(M) b 1(M') . In particular, b 1(M) is independent of the choice of edge cycle multipliers. cor proof From statement 4 of Theorem thm:homology , it follows that b 1(M) is the rank of H H,H . Theorem thm:hom2 states that H is isomorphic to 1(N 0) . It is clear that subdividing P does not change 1(N 0) . Corollary cor:betti follows. proof Examples sec:examples In this section we give several examples of twisted face-pairing 3-manifolds. ex lens spaces ex:lens Let P be a faceted 3-ball with exactly two faces. For convenience we assume that P is the unit ball in 3 and that the faces of P are the northern and southern hemispheres. For any orientation-reversing face-pairing and multiplier function mul , M(, mul ) is a lens space (and hence has geometry S 3 ). ex By varying the numbers of sides of the faces, the face-pairing , and the multiplier function, one can realize many lens spaces as twisted face-pairing 3-manifolds. In CFP6 we give a different construction of lens spaces as twisted face-pairing 3-manifolds and show that every lens space is a twisted face-pairing 3-manifold. ex lunes ex:lunes Let P be a faceted 3-ball in which each face is a digon. Then P has exactly two vertices. For convenience we assume that P is the unit ball in 3 , the vertices of P are (0,0,1) , the edges of P are arcs of great circles, and all of the face-pairing maps are isometries. ex Let F be the intersection of P with the xy -plane. We call F the equatorial disk of P . Then F has a cell structure whose vertices are the intersections of F with the edges of P and whose edges are the intersections of F with the faces of P . Furthermore, every face-pairing on P restricts to an edge pairing on F . Conversely, given an edge pairing on a disk F in which no edge is paired with itself, one can construct a faceted 3-ball P with all faces digons and with an orientation-reversing face-pairing which restricts on the equator of P to an edge pairing of the equatorial disk of P which is equivalent to the given edge pairing on F . Since the face-pairing maps are orientation reversing, this can be done in a unique way up to cellular equivalence. Hence we can describe these examples by giving a disk F with edge pairing. For convenience we label the vertices of F by , , and label the corresponding edges of P by e , e , ,. As a simple example, let F be the disk with vertices (1,0,0) , (0,1,0) , (-1,0,0) , and (0,-1,0) , and with edge pairing the antipodal map. Then P has four lunes, and is the antipodal map. See Figure fig:pfootbal . We label and direct the faces of P as in Figure fig:pfootbal . The two edge cycles of have diagrams e 1 e 2 e and e 2 e 1 e . Choose each of the multipliers to be 1 . Then each edge of P is subdivided into two edges in Q , so each of the four faces of Q is a quadrilateral. See Figure fig:qfootbal . figure tran2955el-fig-16 The model faceted 3-ball P and its equatorial disk F . fig:pfootbal figure figure tran2955el-fig-17 The faceted 3-ball Q . fig:qfootbal figure Next we prepare to construct the universal cover of M . To begin this, take a regular octahedron in 3 with (0,0,1) as a pair of opposite vertices. For future convenience we assume that each of the octahedron's edges in the xy -plane is parallel to either the x -axis or the y -axis. Now remove the edges in the xy -plane. Instead of having eight triangles as faces, we now have four quadrilaterals as faces. This cell complex Q' is isomorphic to Q . The midpoints of the eight edges of Q' are the vertices of an inscribed cube. The horizontal edges of this cube are on the boundary of Q' . Now deform Q' , without changing its cell structure, by flattening the parts of Q' above and below the cube to the cube. Next deform Q' further by flattening the sides of Q' toward the sides of this cube. Then Q' , and hence Q , becomes a cube-with-fins'', which we denote by C . See Figure fig:cubefins , where the edges of C are drawn in bold. figure tran2955el-fig-18 The cube-with-fins C . fig:cubefins figure Figure fig:cubes shows a pair of nested cubes in 3 . By adding arcs of diagonals joining corresponding vertices of the two nested cubes, one gets a decomposition of the larger cube into seven 3-cells, each of which is cellularly isomorphic to a cube. By switching from 3 to S 3 , one gets a cellular decomposition of S 3 into eight cubes. figure tran2955el-fig-19 Decomposing S 3 into eight cubes. fig:cubes figure figure tran2955el-fig-20 The 1-skeleton of . fig:oct figure In this paragraph we construct and show that the 1-skeleton of is isomorphic to the 1-complex in Figure fig:oct , which has one vertex at infinity. For this we begin with our cube-with-fins C , which is cellularly isomorphic to Q . Although C is a cube-with-fins and not a cube, we view C as corresponding to the central cube in Figure fig:cubes . We label and direct the four faces of C so that the face which meets the positive x -axis has label 1 and is directed outward, the face which meets the negative x -axis has label 1 and is directed inward, the face which meets the positive y -axis has label 2 and is directed outward, and the face which meets the negative y -axis has label 2 and is directed inward. The central vertex in Figure fig:oct is dual to C and the four edges in Figure fig:oct which contain the central vertex are dual to the four faces of C in a way which respects labels and directions. Now we take a copy of C and attach the face of the copy with label 1 directed inward to the face of C with label 1 directed outward according to our twisted face-pairing. The vertex on the positive x -axis in Figure fig:oct is dual to this copy of C . It is easy to check that the four edges in Figure fig:oct which meet the positive x -axis are dual to the four faces of this copy of C in a way which respects labels and directions. We attach three more copies of C to C in the same way. We now have five cubes-with-fins corresponding to five of the cells in Figure fig:cubes . Next it is not difficult to see using Figure fig:oct that we can take another copy of C , place it above C and attach each of its four faces according to our twisted face-pairing to the top face of one of the four copies of C which we attached to C . We likewise take a copy of C , place it below C and attach each of its four faces according to our twisted face-pairing to the bottom face of one of the four copies of C which we attached to C . Our seven cubes-with-fins are the 3-cells in a cellular decomposition of a closed 3-ball. These cells are arranged like the cells in Figure fig:cubes , but our complex of cubes-with-fins is not the complex in Figure fig:cubes . Finally, it is not difficult to see that we can attach one more copy of C to our complex of cubes-with-fins according to our twisted face-pairing to obtain a cellular decomposition of S 3 . Thus we have just constructed . We see that is homeomorphic to S 3 and that the 1-skeleton of is isomorphic to the 1-complex in Figure fig:oct . In this paragraph we identify G 1(M) . We get a faceted 3-ball Q dual to Q by reflecting Q in the xy -plane. This and Theorem thm:howto show that we may take Q Q as oriented, directed and labeled complexes. Theorem thm:present implies that with respect to our geometric generating set g 1,g 2 , we have Gx 1,x 2:x 1x 2x 1x 2,x 2x 1x 2x 1, where g 1x 1 and g 2x 2 . It is easy to see that there exists a surjective group homomorphism from the latter group to the quaternion group 1,i,j,k of order 8 with x 1i and x 2j . The previous paragraph implies that G 8 , and so G is isomorphic to the quaternion group of order 8. As assured by Theorem thm:cayley , we see that Figure fig:oct gives a Cayley graph for G . Now consider using the same faceted 3-ball P with four lunes, but using as face-pairing the orientation-reversing maps which take each face to the opposite face and fix the vertices (0,0,1) . There is a single edge cycle. Given a positive integer m , let mul be the multiplier function which assigns m to this edge cycle. Using the surgery description in CFP5 , it is easy to see that M(, mul ) is obtained by Dehn surgery on the Borromean rings with framings 0 , 0 , and 1 m . When m 1 this is the Heisenberg manifold (the prototypical Nil manifold). In CFP6 we will show that every twisted face-pairing manifold which is obtained from a model faceted 3-ball whose faces are all digons is a Seifert fibered manifold whose base surface is the orbit space of its equatorial disk with edge pairing. In the above case of the antipodal map on a square, the quotient surface is the projective plane, which is consistent with the fact that our twisted face-pairing manifold has the geometry of S 3 . In the above case of the square with edge pairing which takes each edge to the opposite edge by translation, the quotient surface is the torus, which is consistent with the fact that our twisted face-pairing manifold is the Heisenberg manifold. ex ex:pact Let P be the unit ball in 3 , decomposed into a faceted 3-ball as follows. See Figure fig:pacman . There are three vertices, A (-1,0,0) , B (0,1,0) , and C (0,-1,0) . The three arcs AB , BC , and AC on the equator are edges of P . There are two edges n AB and n AC in the northern hemisphere; n AB joins A and B and n AC joins A and C . The images s AB and s AC of n AB and n AC under the reflection in the xy -plane are also edges. These are all of the edges. There are six faces in P ; there are two digons and a triangle in the northern hemisphere and there are two digons and a triangle in the southern hemisphere. Each face is paired with its reflection in the xy -plane, and each of the face-pairing maps is given by reflection in the xy -plane. We label and direct the faces of P as in Figure fig:pacman . ex figure tran2955el-fig-21 The northern hemisphere of P . fig:pacman figure figure tran2955el-fig-22 The northern hemispheres of Q and Q . fig:qpacman figure The edge cycles of the model face-pairing are AB , BC , AC , n AB ,s AB , and n AC ,s AC . Figure fig:qpacman shows the northern hemispheres of Q and Q (gotten from Q by reflection in the xy -plane) for the multiplier function which assigns 1 to every edge cycle. In general, given a multiplier function mul , let p mul ( AB ) , q mul ( AC ) , r mul ( BC ) , s mul ( n AB ,s AB ) , and t mul ( n AC ,s AC ) . Then Theorem thm:present shows that, with respect to our geometric generating set g 1,g 2,g 3 , G x 1,x 2,x 3: (x 1) p (x 3 -1 x 1) s,(x 2) q (x 3 -1 x 2) t, (x 3) r (x 2 -1 x 3) t (x 1 -1 x 3) s, where g 1x 1 , g 2x 2 and g 3x 3 . First suppose s t 1 . Then g 3 g 1 p 1 , g 3 g 2 q 1 , and 1 g 3 r g 2 -1 g 3 g 1 -1 g 3 g 2 -1 g 3g 1g 3 r 1 . This implies that g 2 g 1 (p 1)(r 1) p , g 1 p 1 g 1 (q 1)((p 1)(r 1) p) , and G ((p 1)(q 1)(r 1) pq-1) . Using the results of CFP5 , it can be shown that M is homeomorphic to the lens space L((p 1)(q 1)(r 1) pq-1,qr 2q r 1) . For general values of p , q , r , s , and t , Gx 1,x 3:(x 1) p((x 3) -1 x 1) s a,b x 2,x 3: (x 2) q((x 3) -1 x 2) t, where in the left homomorphism ax 3 and b(x 1) p(x 3) r , and in the right homomorphism ax 3 and b(x 2) -q . Note that because g 1 p (g 1g 3) s , it follows that g 1g 3 commutes with g 1 p . Hence g 3 commutes with g 1 p . It follows that the image of a in G commutes with the image of b , and so the above amalgamation is along an Abelian group. If p q s t 2 , then G is a Solv group for every choice of r . Using the results of CFP5 , it can be shown that if p q s t 2 , then M is a Solv manifold for every choice of r . ex cube ex:cube Let P be a cube in 3 with center at the origin. Pair each face of P with its opposite face, and let each face-pairing map be the antipodal map. ex Then P is the simplest example of an ample faceted 3-ball, and with this choice of model face-pairing each edge cycle has length two. We choose each multiplier to be 1 . Then Q is obtained from the cube P by subdividing each edge of P into two edges. Let M M(, mul ) , and let G 1(M) . We know from CFP4 that G is Gromov hyperbolic. According to SnapPea , M is hyperbolic. The volume of M is about 5.3335 , and the shortest geodesic in M has length about 1.2659 . Since the face-pairing is preserved by the group of symmetries of the cube, M has that group as a subgroup of its group of symmetries. According to SnapPea, this group of order 48 is the symmetry group of M . figure tran2955el-fig-23 A surgery description for M(, mul ) . fig:tetsurg1 figure figure tran2955el-fig-24 A second surgery description for M(, mul ) . fig:tetsurg2 figure ex hexahedron ex:hex For this example, let P be a hexahedron in 3 which is the union of two regular tetrahedra along a common face. We assume that P is centered at the origin and that its three vertices of valence 4 are in the xy -plane. Let be the model face-pairing for which each face is paired to its image under reflection in the xy -plane and each face-pairing map is reflection in the xy -plane. ex The three edges in the xy -plane are in edge cycles of length 1 , and each of the other three edge cycles has length 2 . We begin by choosing a multiplier function mul so that the multiplier is 2 if the edge cycle has length 1 and is 1 if the edge cycle has length 2 . So each edge of P is subdivided into two edges in Q , and each of the six faces of Q is a hexagon. By SnapPea, M(, mul ) is a hyperbolic 3-manifold with volume approximately 1.8319 . Now let mul ' be the constant multiplier function with each multiplier 2 . Again, by SnapPea, M(, mul ') is a hyperbolic 3-manifold with volume approximately 5.1379 and symmetry group of order 48 . ex We return to the tetrahedron example of Example 3.2. In Section sec:funthmsec we choose for multiplier function the constant function 1 , and in Example 4.9 we show that for this choice of multiplier function M is a homology sphere. More generally, suppose m is a positive integer and define the multiplier function mul by mul ( AB ) 1 , mul ( CD ) m , and mul ( BC,BD,AD,AC ) 1 . Using the surgery description in CFP5 , one sees that M(, mul ) is given by Dehn surgery on the framed link in S 3 shown in Figure fig:tetsurg1 . Fenn-Rourke moves on the two components which link the components with framing 0 and give an equivalent surgery description in Figure fig:tetsurg2 . More Fenn-Rourke moves reduce the framed link of Figure fig:tetsurg2 to the figure eight knot with framing -m , which is equivalent to the figure eight knot with framing m . Hence M(, mul ) is obtained from the (m,1) Dehn filling on the figure eight knot complement. At the end of Section 6 of CFP3 we related M(, mul ) to the figure eight knot in a rather different way using partial twisted face-pairings. If m 1 , then M(, mul ) is the Brieskorn homology sphere (2,3,7) , which has the geometry of the universal cover of PSL(2,) . According to Theorem 4.7 T1 , M(, mul ) is hyperbolic if m5 . ex