Minimal triangulations for an infinite family of lens spaces
aa r X i v : . [ m a t h . G T ] M a y Minimal triangulations for an infinite family of lens spaces
William Jaco, Hyam Rubinstein and Stephan Tillmann
Abstract
The notion of a layered triangulation of a lens space was defined by Jaco and Ru-binstein in [6], and, unless the lens space is L ( , ) , a layered triangulation with the minimalnumber of tetrahedra was shown to be unique and termed its minimal layered triangulation .This paper proves that for each n ≥ , the minimal layered triangulation of the lens space L ( n , ) is its unique minimal triangulation. More generally, the minimal triangulations (andhence the complexity) are determined for an infinite family of lens spaces containing the lensspaces L ( n , ) . AMS Classification
Keywords
Given a closed, irreducible 3–manifold, its complexity is the minimum number oftetrahedra in a (pseudo–simplicial) triangulation of the manifold. This number agreeswith the complexity defined by Matveev [7] unless the manifold is S , IR P or L ( , ) . Matveev’s complexity of these three manifolds is zero, but the respective minimumnumbers of tetrahedra are one, two and two. It follows from the definition that thecomplexity is known for all closed manifolds which appear in certain computer gen-erated censuses. Moreover, the question of determining the complexity of a givenclosed 3-manifold has an algorithmic solution which uses the solution to the
Home-omorphism Problem for 3–Manifolds and, in general, is impractical. It has been anopen problem to determine the complexity for an infinite family of closed manifolds.Two sided asymptopic bounds are given for two families of closed hyperbolic mani-folds by Matveev, Petronio and Vesnin [9] using hyperbolic volume. Other two-sidedbounds using homology groups have been obtained by Matveev [7], [8]. This paperdetermines minimal triangulations (and hence the complexity) for an infinite family ofclosed, irreducible 3–manifolds, which includes the following three families:1 heorem 1
The minimal layered triangulation of the lens space L ( n , ) , n ≥ , isits unique minimal triangulation. The complexity of L ( n , ) is therefore 2 n − . Theorem 2
The minimal layered triangulation of L (( s + )( t + ) + , t + ) , where t > s > , s odd and t even, is its unique minimal triangulation. The complexity of L (( s + )( t + ) + , t + ) istherefore s + t . Theorem 3
The minimal layered triangulation of L (( s + )( t + ) + , t + ) , where t > s > , s even and t odd, is its unique minimal triangulation. The complexity of L (( s + )( t + ) + , t + ) istherefore s + t . Layered triangulations of lens spaces and minimal layered triangulations of lens spacesare defined and studied in [6]. A brief review of layered triangulations of the solidtorus and lens spaces is given in Section 2. The above results imply that the followingconjecture holds for infinite families of lens spaces:
Conjecture 4 [6] A minimal layered triangulation of a lens space is a minimal tri-angulation; moreover, it is the unique minimal triangulation except for S , IR P and L ( , ) .In particular, given L ( p , q ) , where ( p , q ) = , p > q > p > , the conjecturedcomplexity is E ( p , q ) − , where E ( p , q ) is the number of steps needed in the Eu-clidean algorithm (viewed as a subtraction algorithm rather than a division algorithm)to transform the unordered tuple ( p , q ) to the unordered tuple ( , ) . Equivalently, E ( p , q ) is the sum, (cid:229) n i , of the “partial denominators” in the continued fractions ex-pansion of p / q : pq = n + n + n + n +
1. . . = [ n , n , n , n , . . . ] . There is an analogous conjecture in terms of special spines due to Matveev [7] givingthe same conjectured complexity. The minimal known special spine of L ( p , q ) is dualto the minimal layered triangulation. Anisov [1, 2] has recently found a geometriccharacterisation of the former as the cut locus of a generic point when L ( p , q ) , q = , is imbued with its standard spherical structure. In the case q = , the cut locus isnot a special spine, but perturbing either the metric or the cut locus slightly gives theminimal known special spine. 2he above conjectures were made in the 1990s, and this paper gives the first infinitefamily of examples which satisfy them. The new ingredients used to address the con-jectures are the following.First, we study minimal and 0–efficient triangulations of closed, orientable 3–manifoldsvia the edges of lowest degree. Any such triangulation contains at least two edges ofdegree at most five, and the neighbourhood of an edge of degree no more than five isdescribed in Section 3.Second, assume that M is a closed, orientable 3–manifold which admits a non-trivialhomomorphism j : p ( M ) → ZZ . Each (oriented) edge, e , in a one-vertex triangula-tion of M represents an element, [ e ] , of p ( M ) and is termed j –even if j [ e ] = j –odd otherwise. We exhibit a relationship between the number of j –even edges, thenumber of tetrahedra and the Euler characteristic of a canonical normal surface dual to j in Section 4.Third, we analyse in Section 5 how layered triangulations of solid tori which sit assubcomplexes of minimal and 0–efficient triangulations can intersect.As an application of these results, we prove the following in Section 6: Theorem 5
Alens spacewithevenfundamental group satisfies Conjecture 4ifithasa minimal layered triangulation such that there are no more j –odd edges than thereare j –even edges.It is then shown that this implies that a lens space never has more j –even edges than j –odd edges, and that a lens space satisfies this condition if it is contained in one ofthe families given in Theorems 1, 2 or 3. Moreover, a complete description of the setof all lens spaces to which Theorem 5 applies is given in Figure 6.The first author is partially supported by NSF Grant DMS-0505609 and the Grayce B.Kerr Foundation. The second and third authors are partially supported under the Aus-tralian Research Council’s Discovery funding scheme (project number DP0664276). This section establishes notation and basic definitions, and describes a family of lay-ered triangulations of the solid torus, denoted S k = { , k + , k + } , as well as theminimal layered triangulation of the lens space L ( k + , ) , k ≥ . A consequence ofTheorem 1 concerning minimal triangulations of the solid torus subject to a specifiedtriangulation of the boundary is also given (see Corollary 6).3 .1 Triangulations
The notation of [5, 6] will be used in this paper. A triangulation, T , of a 3–manifold M consists of a union of pairwise disjoint 3–simplices, e D , a set of face pairings, F , and a natural quotient map p : e D → e D / F = M . Since the quotient map is injectiveon the interior of each 3–simplex, we will refer to the image of a 3–simplex in M as a tetrahedron and to its faces, edges and vertices with respect to the pre-image.Similarly for images of 2– and 1–simplices, which will be referred to as faces and edges in M . For edge e , the number of pairwise distinct 1–simplices in p − ( e ) istermed its degree , denoted d ( e ) . If an edge is contained in ¶ M , then it is termed a boundary edge ; otherwise it is an interior edge . A triangulation of the closed, orientable, connected 3–manifold M is termed minimal if M has no other triangulation with fewer tetrahedra. A triangulation of M is termed if the only embedded, normal 2–spheres are vertex linking. It is shown bythe first two authors in [5] that (1) the existence of a 0–efficient triangulation impliesthat M is irreducible and M = IR P , (2) a minimal triangulation is 0–efficient unless M = IR P or L ( , ) , and (3) a 0–efficient triangulation has a single vertex unless M = S and the triangulation has precisely two vertices. These facts will be usedimplicitly. Layering along a boundary edge is defined in [6] and illustrated in Figure 1(a). Namely,suppose M is a 3–manifold, T ¶ is a triangulation of ¶ M , and e is an edge in T ¶ which is incident to two distinct faces. We say the 3–simplex s is layered along the(boundary) edge e if two faces of s are paired, “without a twist,” with the two facesof T ¶ incident with e . The resulting 3–manifold is homeomorphic with M . If T ¶ isthe restriction of a triangulation T of M to ¶ M , then we get a new triangulation of M and denote it T ∪ e s .Starting point for a layered triangulation of a solid torus is the one-tetrahedron triangu-lation of the solid torus shown in Figure 1(b), where the two back faces are identifiedas indicated by the edge labels. One can then layer on any of the three boundary edges(see [6]), giving a layered triangulation of the solid torus with two tetrahedra. Induc-tively, a layered triangulation of a solid torus with k tetrahedra is obtained by layering4Sfrag replacements M D e e e (a) Layering on a boundary edge PSfrag replacements M D e e e (b) The solid torus S Figure 1: Layered triangulation of the solid toruson a boundary edge of a layered triangulation with k − k tetrahedra has one vertex, 2 k + k + univalent edge .Given a solid torus with a one-vertex triangulation on its boundary, it follows froman Euler characteristic argument that this triangulation has two faces meeting alongthree edges. There is a unique set of unordered nonnegative integers, { p , q , p + q } ,associated with the triangulation on the boundary and determined from the geometricintersection of the meridional slope of the solid torus with the three edges of the tri-angulation. Two such triangulations with numbers { p , q , p + q } and { p ′ , q ′ , p ′ + q ′ } can be carried to each other via a homeomorphism of the solid torus if and only if thetwo sets of numbers are identical. There are three possible ways to identify the twofaces in the boundary, each giving a lens space. These identifications can be thoughtof as “folding along an edge in the boundary”, which is also referred to as “closingthe book” with an edge as the binding. Folding along the edge p gives the lens space L ( q + p , q ) ; along the edge q the lens space L ( p + q , p ) ; and along the edge p + q the lens space L ( | p − q | , p ) . If the solid torus is triangulated, then after identificationone obtains a triangulation of the lens space. If the triangulation of the solid torusis a layered triangulation of the solid torus, then the induced triangulation of the lensspace is termed a layered triangulation of the lens space. A layered triangulation ofa lens space having the the minimal number of tetrahedra is called a minimal layeredtriangulation of the lens space. It is shown in [6], that —except for the lens space L ( , ) — a lens space has a unique minimal layered triangulation. It is conjectured,see Conjecture 4, that a minimal layered triangulation is minimal.There is a companion conjecture for triangulations of the solid torus. If T ¶ is a trian-gulation on the boundary of a compact 3–manifold M , then a triangulation T of M having all its vertices in ¶ M and agreeing with T ¶ on ¶ M is called an extension of ¶ to M . A one-vertex triangulation on the boundary of a solid torus can be extendedto a layered triangulation of the solid torus. There are infinitely many ways to extendthe triangulation on the boundary; however, it is shown in [6] that there is a unique lay-ered extension having the minimal number of tetrahedra; termed the minimal layeredextension. The companion conjecture to that given above is that the minimal layeredextension is the minimal extension; a first infinite family satisfying this companionconjecture is given in Corollary 6. { k + , k + , } We now describe the combinatorics of a layered triangulation of the solid torus extend-ing the { k + , k + , } triangulation on the boundary of the solid torus. It is shownbelow that this is the minimal layered extension. This family of layered triangulationscan be characterized by the fact that there is a unique interior edge of degree three andall other interior edges have degree four. Denote by S the triangulation of the solidtorus with one tetrahedron, D , with edges labelled and oriented as shown in Figure1(b). A 3–simplex, D , is layered on edge e ; there is a unique edge, labelled e , which is not identified with any of e , e , e . The edge e is oriented such that its inci-dent faces give the relation e = e + e . Denote the resulting triangulated solid torusby S . Inductively, D k is layered on e k ; the new edge is labelled e k + with relation e k + = e + e k + . The resulting triangulated solid torus is denoted S k . Identify the fundamental group of the solid torus with ZZ . Analysing the meridian discin the triangulation consisting of just D , and then taking into account the relationsgiven from faces, one obtains: [ e k ] = k ∈ ZZ , with the given orientation conventions for edges. The edges are accordingly termed odd or even . When k = , we have d ( e ) = , d ( e ) = d ( e ) = d ( e ) = . Afterlayering k tetrahedra, k ≥ , the degrees of edges are as follows:edge e e e to e k e k + e k + degree 2 k + ( k + , ) In the above construction, at stage k , one can choose not to add D k + , but instead tofold along the edge e k + , identifying the two faces and identifying the edge e k + with − e . As noted above, one obtains a layered triangulation of the lens space L ( k + , ) k + . This is in fact the minimal layered triangulation of L ( k + , ) as can be deduced from the continued fraction expansion. This also impliesthat the above extension of { k + , k + , } is the minimal layered extension. Theprevious discussion shows that there are k + k ≥
3: edge e e e to e k e k + degree 2 k + k = , there are two edges of degree three and one of degree six, when k = , there is one of degree four and one of degree two.This triangulation of L ( k + , ) is denoted L = L k . If k is odd, then having e k + identified with − e , each of the k + L k can be termed odd or even with-out ambiguity and with consistency with their designation as “odd” or “even” in S k . Letting n = k + , we have L ( k + , ) = L ( n , ) , where n ≥ . In the sequel, we write L ( k + , ) = L ( n , ) to indicate that n ≥ k is an odd, positive integer.Theorem 1 states that L k is the unique minimal triangulation of L ( k + , ) , k ≥ Corollary 6
The minimal layered extension of the triangulations { k + , k + , } , k ≥ , on the boundary ofthe solid torus is theunique minimalextension. Proof
Let k ≥ E , of { k + , k + , } which is not the minimal layered extension. Folding along the edge k + L ( k + , ) with k or fewer tetrahedra. Since L k is the uniqueminimal triangulation of L ( k + , ) , it follows that E contains precisely k tetrahedraand that it is obtained by splitting L k open along a face. Now in L k there are preciselytwo faces along which one can split open to obtain the minimal layered extension of { k + , k + , } , hence not E . Splitting open along any other face gives boundary apinched 2–sphere, hence again not E . We thus arrive at a contradiction.If k is even, assume that there is an extension of { k + , k + , } with k or fewer tetra-hedra which is not the minimal layered extension. One may layer another tetrahedronon this to get an extension of { ( k + ) + , ( k + ) + , } with k + Given any 1–vertex triangulation of a closed 3–manifold, let E denote the numberof edges, E i denote the number of edges of degree i , and T denote the number of7etrahedra. Then E = T + , and hence 6 = (cid:229) ( − i ) E i . It follows that there are at leasttwo edges of degree at most five. The new contribution in this section is an analysisof edges of degree four and five in minimal and 0–efficient triangulations. For edgesof degree at most three, this has been done previously by the first two authors. Forconvenience, Proposition 6.3 of [5] is stated (with minor corrections) as Propositions 8and 9 in Subsection 3.1. A new proof is given for the case of degree three edges whichgeneralises to degrees four and five.As a simple application of below characterisation of low degree edges, the readermay find pleasure in determining all closed, orientable, connected and irreducible 3–manifolds having a minimal and 0–efficient triangulation with at most three tetrahedra.
Lemma 7
Suppose theclosed, orientable, connected 3–manifold M hasa0–efficienttriangulation. Ifthere isaface in M which isadunce hat, then M = S . Proof
The face which is a dunce hat is bounded by a single edge. Since a dunce hat iscontractible, the edge bounds an immersed disc in M . Dehn’s lemma (as stated in [4],I.6) implies that the edge bounds an embedded disc. Now [5], Proposition 5.3, impliesthat M = S . Proposition 8 (Edges of degree one or two) [5] A minimal and 0–efficient triangu-lation T of the closed, orientable, connected and irreducible 3–manifold M has(1) no edge oforder one unless M = S , and(2) no edge oforder twounless M = L ( , ) or L ( , ) . Proposition 9 (Edges of degree three) [5] A minimal and 0–efficient triangulation T of the closed, orientable, connected and irreducible 3–manifold M has no edge oforder three unless either(3a) T contains asingle tetrahedron and M = L ( , ) ; or(3b) T contains twotetrahedra and M = L ( , ) or L ( , ) ; or(3c) T contains, as an embedded subcomplex, the two tetrahedron, layered trian-gulation S = { , , } of the solid torus. Moreover, T contains at least threetetrahedra and every edge of degree three iscontained insuch asubcomplex.Moreover, the triangulations in (3b) are obtained by identifying the boundary faces of S appropriately. 8 roof Assume that e is an edge of degree three in T . Let e D e ⊂ e D be the set of all3–simplices containing a pre-image of e . Then 1 ≤ | e D e | ≤ . If | e D e | = , then the preimage of a small loop around e must meet all four faces ofthe single tetrahedron in e D e . It follows that e D = e D e . Analysing the possibilities gives M = L ( , ) and T is the a one-tetrahedron triangulation thereof. We are thus in case(3a).If | e D e | = , then one of the tetrahedra in e D e , denoted e s , contains exactly one pre-image of e , and the other, denoted e s , contains exactly two. The faces f , f of e s meeting in the pre-image of e have to be identified with faces f , f respectively of e s . First assume that f and f meet in a pre-image of e . Since d ( e ) = , one of f , f must contain another pre-image of e . But this implies that d ( e ) > M is closedand | e D e | = . It follows that f and f cannot meet in a pre-image of e . Similarly, ifthe pre-images of e in the two faces meet in a vertex, then we get a contradiction to d ( e ) = . Hence the pre-images form a pair of opposite edges and the remaining faces, f , f , of e s must be identified since | e D e | = . The face pairings f ↔ f , f ↔ f , f ↔ f are uniquely determined by this information and the fact that M is orientable.The resulting identification space, X , is equivalent to the layered triangulation S ofthe solid torus. The projection p : e D → M gives rise to a natural map p ′ : X → M suchthat the restriction of p to e s ∪ e s factors through p ′ . If p ′ is an embedding, then we are in case (3c). Hence assume that there are furtheridentifications between e s and e s . If there is a pairing between the remaining faces of e s , then e D = e D e . Analysing thethree possibilities (recalling that M is orientable) gives either of the following cases:(1) Folding along the univalent edge gives IR P . However, the triangulation is not0–efficient.(2) Folding along the edge of degree five gives L ( , ) . (3) Folding along the edge of degree three gives L ( , ) . These are the possibilities stated in case (3b).Now assume there is no further identification of faces but of edges between e s and e s . Note that there are three distinct edges in the boundary of X . If precisely two ofthem are identified, then either of the following cases applies:(4) The triangulation T contains a face which is a cone. Then [5], Corollary 5.4,implies that M = S . But a minimal triangulation of S contains a single tetra-hedron, contradicting | e D | > .
95) The boundary, F , of a regular neighbourhood of p ( e s ∪ e s ) in M is an embed-ded 2–sphere and a barrier surface (see [5]). Hence F shrinks to a stable surfacein the complement of p ( e s ∪ e s ) . Since every normal 2–sphere in a 0–efficienttriangulation is vertex linking and the complement of p ( e s ∪ e s ) cannot containa vertex linking surface, F bounds a 3–ball in the complement of p ( e s ∪ e s ) . There is a homotopy of M taking this ball to a disc which extends to a homotopyidentifying the two free faces of p ( e s ∪ e s ) , hence giving rise to a triangulationof M with fewer tetrahedra than T . This contradicts minimality of T . Now assume that all three edges in the boundary of X are identified. Then either aface is a dunce hat and Lemma 7 yields a contradiction as in (4), or the argument in(5) applies.If | e D e | = , then e is contained in three distinct tetrahedra and a 3 → T is minimal. The analysis of degree four and five edges is done more coarsely in order to reduce thenumber of cases to be considered. Some of the notions in the preceding proof are firstformalised. Given an edge, e , of degree n in a minimal and 0–efficient triangulation T of the closed, orientable, connected and irreducible 3–manifold M , assume thatit is contained in precisely k pairwise distinct tetrahedra, s , ..., s k . Then 1 ≤ k ≤ n and there is a triangulated complex, X = X n ; k , having a triangulation with k tetrahedracontaining an interior edge of degree n , denoted e , and a map p e : X → M taking e → e with the property that p | { e s ,..., e s k } : e D → M factors through p e . A complex X asabove is maximal if there is no other such complex X ′ with the property that p : { e s , ..., e s k } → X → M factors as p : { e s , ..., e s k } → X ′ → X → M . If X n ; k is maximal, then ( X n ; k , e ) is said to be a model for e . Proposition 10 (Edges of degree four) Assume the minimal and 0–efficient trian-gulation T of the closed, orientable, connected and irreducible 3–manifold M has anedgeofdegreefour,denoted e . Thenthemodel, ( X k , e ) , for e isoneofthefollowing:(4a) If k = , then X = X or X , where X and X are one-tetrahedrontriangulations of S and L ( , ) respectively that contain a(necessarily unique)edge of degree four, e . Inparticular, M = S or L ( , ) . S S e (a) X ∼ = S PSfrag replacements S S e (b) X ∼ = L ( , ) PSfrag replacements S S e (c) X ∼ = L ( , ) PSfrag replacements S S e (d) X ∼ = S / Q PSfrag replacements S S e (e) X ∼ = solid torus PSfrag replacements S S e (f) X ∼ = solid torus PSfrag replacements S S e (g) X ∼ = solid torus PSfrag replacements S S e (h) X ∼ = Figure 2: Degree four edges in minimal and 0–efficient triangulations(4b) If k = , then X = X , X or X , where X is a two-tetrahedron trian-gulation of L ( , ) with e theuniqueedgewhichhasdegreetwowithrespect toeachtetrahedron; X isatwo-tetrahedron triangulation of S / Q with e eitherofitsthree degreefouredges; and X = { , , } isthetriangulated solid torus S withanother tetrahedron layered onthe boundary edge ofdegree three with e theunique edge ofdegree four;(4c) If k = , then X = X or X , where X is S withtwotetrahedraattachedto its boundary faces such that the degree 2 boundary edge becomes an interioredge of degree four, e ; and X is a solid torus obtained by first identifying apair of opposite edges of a tetrahedron, and then layering two tetrahedra on theresulting edge making itaninterior edge of degree four, e . (4d) If k = , then X is an octahedron triangulated with four tetrahedra with e their common intersection.The maximal complexes are shown in Figure 2: To obtain X and X , identifythe two front faces and identify the two back faces as indicated by the arrows; toobtain X identify the two front faces of the first 3–simplex with the front faces of11hesecond, and thenthebackfaces ofeachtetrahedron; for X identify thetwofrontfaces of the first 3–simplex with the front faces of the second, and likewise with theback faces; to obtain X , identify the front faces of the first3–simplex withthe frontfaces ofthe second, and the back faces ofthe second with thefront faces of the third. Proof
We only have to show that the stated list of maximal complexes is correct. Themain arguments are as in the proof for Proposition 9, and we will therefore not give alldetails. As above, assume that e is an edge of degree four in M , and let e D e ⊂ e D be theset of all 3–simplices containing a pre-image of e . Then 1 ≤ | e D e | ≤ . If | e D e | = , then the preimage of a small loop around e must meet all four faces ofthe single tetrahedron in e D e . It follows that e D = e D e . Analysing the possibilities gives M = S or L ( , ) and T is a one-tetrahedron triangulation containing a (necessarilyunique) edge of degree four. We are thus in case (4a).If | e D e | = , denote the tetrahedra in e D e by e s and e s . Without loss of generality, e s contains precisely one or two pre-images of e . In the first case, analysing the possible face pairings yields the one tetrahedron solidtorus with another tetrahedron layered on the degree 3 boundary edge, thus having aunique interior edge of degree four. This is X , and there may be further identificationsfrom other face pairings in F . In the second case, first assume that the two pre-images of e are contained in a face of e s . This forces e s to also have a face containing two pre-images of e , and these twofaces are identified. Now there are two other faces of e s containing a pre-image of e . If they are identified, then a face is a cone. Hence both are identified with faces of e s . There is a unique way to do this under the condition that e has degree four. But thenthe remaining free face of each e s i is a cone or a dunce hat. In either case, this forces M = S , contradicting minimality.Hence assume that the two pre-images of e are not contained in a face of either e s nor e s . Then the triangulation contains precisely two tetrahedra. First assume thattwo faces of e s are identified. Then the same is true for e s . Analysing all possibleidentifications gives (up to combinatorial equivalence) the triangulation X with thespecified marked edge of degree four. The triangulation can be viewed as two solid toriidentified along their boundary, and the corresponding lens space identified as L ( , ) . Next, assume that no two faces of e s are identified. This yields the triangulation X of with a marked edge. However, all three edges are combinatorially equivalent. Toidentify the manifold, notice that for each edge, there is a normal surface made upof two quadrilaterals, one in each tetrahedron, which don’t meet that edge. This is anembedded Klein bottle and a one-sided Heegaard surface for the manifold. Using [10],the manifold can now be identified as S / Q . | e D e | = , denote the tetrahedra in e D e by e s , e s and e s . Without loss of generality, e s contains precisely two pre-images of e . First assume that there is no face pairingbetween faces of e s and e s . Then the two faces incident with the pre-image of e in e s i , i = , , are identified with faces of e s . Analysing the possibilities gives X , andthere may be further identifications from other face pairings in F . Next assume thatthere is a pairing between faces of e s and e s containing a pre-image of e . Then thereis a unique such pairing, and the remainder of the argument is as in the proof of (3c),giving X . If | e D e | = , then the only possibility is the octahedron with possible identificationsalong its boundary.The simplicial maps from X to a minimal, 0–efficient triangulation of a closed,orientable 3–manifold are X → L ( , ) and X → L ( , ) = X . The triangulation X of L ( , ) contains three edges of degree four; two of them have neighbourhoodsmodelled on X . Thus, we have classified all minimal, 0–efficient triangulations withat most two tetrahedra containing an edge of degree four.It is also true that S → L ( , ) , but S has no interior edge of degree four, and hencedoes not appear in the above list. Proposition 11 (Edges of degree five) Assume theminimaland 0–efficient triangu-lation T of the closed, orientable, connected and irreducible 3–manifold M has anedgeofdegreefive,denoted e . Thenthemodel, ( X k , e ) , for e isoneofthefollowing:(5a) If k = , then X is a one-tetrahedron triangulation of S which contains a(necessarily unique) edge ofdegree five, e . In particular, M = S . (5b) If k = , then X = X or X , where X isatwo-tetrahedron triangulationof L ( , ) with e either of the two edges of degree five; and X is a two-tetrahedron triangulation of L ( , ) with e the unique edge of degree five.(5c) If k = , then X = X , X or X , where X is S with two tetrahedraattached toitsboundary facessuchthatthedegree3boundary edgebecomesaninterior edge of degree five, e ; X = { , , } is X = { , , } with anothertetrahedron layered on the degree 4 boundary edge, giving a unique interioredge of degree five, e ; and X is a solid torus with a unique interior edge ofdegree five, e , obtained from a 3-tetrahedron triangulated prism by identifyingtwoboundary squares. 13Sfrag replacements X S e (a) X ∼ = S PSfrag replacements X S e (b) X ∼ = L ( , ) PSfrag replacements X S e (c) X ∼ = L ( , ) PSfrag replacements X S e (d) X ∼ = solid torus PSfrag replacements X S e (e) X ∼ = solid torus PSfrag replacements X S e (f) X ∼ = solid torus PSfrag replacements X S e (g) X ∼ = solid torus PSfrag replacements X S e (h) X ∼ = solid torus PSfrag replacements X S e (i) X ∼ = Figure 3: Degree five edges in minimal and 0–efficient triangulations(5d) If k = , then X = X or X , where X is S with three tetrahedra at-tached to its boundary faces such that the degree 2 boundary edge becomes aninterioredgeofdegreefive, e ; and X isasolidtorusobtained byidentifying apairofoppositeedgesofatetrahedron, andthenlayeringonetetrahedron ontheresulting edge, and attaching twotetrahedra to the remaining boundary faces tocreate aunique interior edge ofdegree five, e . (5e) If k = , then X , is a ball triangulated with five tetrahedra such that theirintersection isaunique interior edge of degree five, e . Themaximal complexes are shown inFigure 3.
Proof
The proof follows the same line of argument as the previous two propositions’proofs. We only highlight the main points that are different. Firstly, in the case where | e D e | = L ( , ) , minimality does not imply 0–efficiency. The latter property can be verified by computing the set of all connectednormal surfaces of Euler characteristic equal to two.14igure 4: Types of tetrahedra and normal discs of the dual surfaceIn the case where | e D e | = , there is a subcase where e s contains one pre-image of e , and e s and e s contain two each. If two pre-images are contained on a commonface, then one obtains X . Otherwise there are two subsubcases. First, one assumesthat each of e s and e s has two of its faces identified. Then each is equivalent to S , and these two subcomplexes must meet in a face. One thus obtains a pinched2–sphere made up of the two faces of e s which meet e s and e s respectively. Thisis not possible due to minimality. Hence assume that precisely one of e s and e s iscombinatorially equivalent to S . Analysing the possibilities gives X . Last, assumethat none of e s and e s has two of its faces identified. Analysing all possible gluingsof the remaining faces, one obtains a 3–tetrahedron complex whose boundary eitherconsists of two faces which form a pinched 2–sphere, or one of whose boundary facesis a cone or a dunce hat. In either case, one obtains a contradiction. ZZ –cohomology classes Throughout this section, let T be an arbitrary 1–vertex triangulation of the closed3–manifold M , and j : p ( M ) → ZZ be a non–trivial homomorphism. Additionalhypotheses will be stated. A colouring of edges arising from j is introduced and acanonical normal surface dual to j is determined. This yields a combinatorial con-straint on the triangulation, which is then specialised to a family of lens spaces. Each edge, e , is given a fixed orientation, and hence represents an element [ e ] ∈ p ( M ) . If j [ e ] = , the edge is termed j –even, otherwise it is termed j –odd. This terminol-ogy is independent of the chosen orientation for e . Faces in the triangulation giverelations between loops represented by edges. It follows that a tetrahedron falls intoone of the following categories, which are illustrated in Figure 4:15ype 1: A pair of opposite edges are j –even, all others are j –odd.Type 2: The three edges incident to a vertex are j –odd, all others are j –even.Type 3: All edges are j –even.It follows from the classification of the tetrahedra in T that, if j is non-trivial, thenone obtains a unique normal surface, S j ( T ) , with respect to T by introducing asingle vertex on each j –odd edge. This surface is disjoint from the tetrahedra of type3; it meets each tetrahedron of type 2 in a single triangle meeting all j –odd edges;and each tetrahedron of type 1 in a single quadrilateral dual to the j –even edges.Moreover, S j ( T ) is dual to the ZZ –cohomology class represented by j . The set-up and notation of the previous subsection is continued. Let A ( T ) = number of tetrahedra of type 1, B ( T ) = number of tetrahedra of type 2, C ( T ) = number of tetrahedra of type 3, o ( T ) = number of j –odd edges, e ( T ) = number of j –even edges,˜ e ( T ) = number of pre-images of j –even edges in e D . Assume that T contains T ( T ) tetrahedra. Then T ( T ) = A ( T ) + B ( T ) + C ( T ) . For the remainder of this subsection, we will write A = A ( T ) , etc. The complex K in M spanned by all j –even edges in T is homotopy equivalent to the complement, N , of a regular neighbourhood of S j in M . We therefore have:2 c ( S j ) = c ( ¶ N ) = c ( N ) = c ( K ) . The Euler characteristic of K can be computed from the combinatorial data. We have: c ( K ) = − e + B + C − C , and hence: 2 C + B = e − + c ( S j ) . (4.1)So: ˜ e = A + B + C = A + B + C + e − + c ( S j ) ≤ T − + c ( S j ) + e . (4.2)16 emma 12 Let M beaclosed,orientable, irreducible3–manifoldwhichisnothome-omorphic to one of IR P , L ( , ) , and j : p ( M ) → ZZ be a non–trivial homomor-phism. Suppose that T is a minimal triangulation with T tetrahedra, and let S j bethe canonical normal surface dual to j . Then e ≥ − T − c ( S j ) + ¥ (cid:229) j = ( j − ) e j , (4.3)where e i is thenumber of j –even edges ofdegree i . Proof
First note that the existence of j implies that M = S , L ( , ) . It now followsfrom [5], Theorem 6.1, that T has a single vertex; hence S j and e i are defined.Moreover, [5], Proposition 6.3 (see also Proposition 8), implies that the smallest degreeof an edge in T is three. One has ˜ e = (cid:229) i e i and e = (cid:229) e i . Putting this into (4.2) givesthe desired inequality.
Let M = L ( n , q ) with a minimal triangulation, T , and assume M = IR P or L ( , ) . There is a unique non–trivial homomorphism j : p M → ZZ . It is known throughwork by Bredon and Wood [3] and the second author [10] that, up to isotopy, thereis a unique incompressible non-orientable surface S in L ( n , q ) . The surface S j ( T ) must be non-orientable since it is dual to a non-trivial ZZ –cohomology class and thereis no non-trivial ZZ–cohomology class. It therefore compresses to the surface S , so c ( S ) ≥ c ( S j ( T )) . The smallest degree of an edge in T is three. Denote by L n , q the minimal layeredtriangulation of M = L ( n , q ) . Then Theorem 8.2 of [6] implies that c ( S j ( T )) ≤ c ( S ) = c ( S j ( L n , q )) = − e ( L n , q ) . We also have that the number of tetrahedra in T satisfies T ( T ) ≤ e ( L n , q ) + o ( L n , q ) − . Hence 4 − T ( T ) − c ( S j ( T )) ≥ + e ( L n , q ) − o ( L n , q ) . (4.4)Now assuming that e ( L n , q ) ≥ o ( L n , q ) , inequality (4.3) becomes: e ≥ + ¥ (cid:229) j = ( j − ) e j . (4.5)We will show that this inequality forces T = L n , q as well as e ( L n , q ) = o ( L n , q ) . L ( n , q ) having minimal layered triangulations satisfy-ing e ( L n , q ) ≥ o ( L n , q ) is non-empty: an edge in the triangulation L n , = L k of L ( n , ) = L ( k + , ) is j –even if and only if it is even; and hence odd if and only ifit is j –odd. Whence e ( L k ) = o ( L k ) . Throughout this section, assume that M is an irreducible, orientable, connected 3–manifold with a fixed triangulation, T . Further assumptions will be stated. We provesome general facts about layered solid tori and maximal layered solid tori in such atriangulation.
Definition 13 (Layered solid torus) A layered solid torus with respect to T in M isa subcomplex in M which is combinatorially equivalent to a layered solid torus. Anyreference to T is suppressed when T is fixed. The edges of the layered solid torus T are termed as follows: If there is more than one tetrahedron, then there is a uniqueedge which has been layered on first, termed the base-edge (of T ). The edges in theboundary of T are called boundary edges (of T ) and all other edges (including thebase-edge) are termed interior edges (of T ). For every edge in M , we will also refer to its degree as its M –degree, and its degreewith respect to the layered solid torus T in M is called its T –degree. The T –degreeof an edge not contained in T is zero. The unique edge of T –degree one is termed a univalent edge (for T ) . Clearly, T –degree and M –degree agree for all interior edgesof T . Lemma 14
If the intersection of two layered solid tori consists of two faces, then M is alens space withlayered triangulation. Proof
If two layered solid tori, T and T , meet in precisely two faces, then thesemust be their respective boundary faces. Assume that the identification of the facesextends to a homeomorphism of the boundary tori. Whence M is a lens space. Iftetrahedron s in T meets T , then s is layered on a boundary edge of T and T ∪ s is a layered solid torus unless there are further identifications of the two faces of s notmeeting T . But this forces s = T and M is a lens space with layered triangulation.Otherwise add s to T and subtract it from T ; the result now follows inductively.18ence assume that the identification of the faces does not extend to a homeomorphismof the boundary tori. Examining the resulting face pairings (using the fact that M isorientable), one observes that an edge folds back onto itself, which is not possible. Lemma 15
If two layered solid tori share a tetrahedron, then either one contains theother or M isalens space withlayered triangulation. Proof
Assume that one is not contained in the other, and denote the layered solid toriby T and T . Then the closure of T \ ( T ∩ T ) is non-empty and a layered solid torus.It meets T in precisely two faces and hence Lemma 14 yields the result. Lemma 16
If the triangulation is minimal and 0–efficient, then the intersection oftwolayered solid tori cannot consist ofasingle face.
Proof If T and T meet in a single face, then the free faces give rise to a pinched2–sphere in M . The boundary of a regular neighbourhood of T ∪ T in M is an em-bedded 2–sphere. As in the proof of Proposition 9, Case (5), a barrier argument givesa contradiction to minimality. Lemma 17
If the triangulation is minimal and 0–efficient, then the intersection oftwolayered solid tori cannot consists ofthree edges.
Proof If T and T meet in precisely three edges, then the boundary of a regularneighbourhood of T ∪ T in M consists of two embedded 2–spheres. As in the abovelemma, this can be evacuated, resulting in a smaller triangulation. Lemma 18
Ifthetriangulation isminimaland0–efficient,andtheintersection oftwolayered solid toriconsists oftwoedges, then M isalensspace and thetriangulation isaminimal layered triangulation. Proof If T and T meet in two edges, they form a spine for the boundary of eachsolid torus. The boundary of a regular neighbourhood of T ∪ T in M is an embed-ded 2–sphere. As above, this 2–sphere shrinks to a point in the complement of T ∪ T . Hence there is a homotopy in M identifying the boundaries of T and T . If this homo-topy identifies the remaining edges, then one obtains a new triangulation having fewertetrahedra. Hence assume that the homotopy does not identify the remaining edges.Then a triangulation is obtained by inserting a single tetrahedron. The complement of T ∪ T consists therefore of exactly one tetrahedron and the triangulation is a minimallayered triangulation. 19 .2 Maximal layered solid tori Definition 19 (Maximal layered solid torus) A layered solid torus is a maximal lay-ered solid torus with respect to T in M if it is not strictly contained in any otherlayered solid torus in M . A layered triangulation of a lens space contains precisely two maximal layered solidtori. The lemmata of the previous section directly imply the following:
Lemma 20
Assume that the triangulation is minimal and 0–efficient. If M is not alens space with layered triangulation, then the intersection of two distinct maximallayered solid tori in M consists ofatmost asingle edge. Lemma 21
Assume that the triangulation is minimal and 0–efficient, and supposethat M contains alayeredsolidtorus, T , madeupofatleasttwotetrahedra andhavingaboundary edge, e , which has degree four in M . Theneither(1) T isnot amaximal layered solid torus in M ; or(2) e istheunivalent edgefor T anditiscontained infourdistinct tetrahedra in M ;or(3) M is alens space withminimal layered triangulation. Proof
We apply Proposition 10 to identify the complex X k on which the neighbour-hood of e is modelled. Since T contains at least two tetrahedra, this rules out X , X , X and X . If the neighbourhood is modelled on X , then M is decomposed into T and the one-tetrahedron solid torus, S . Hence M is a lens space with layered triangulation. Byassumption, this must be a minimal layered triangulation.If the neighbourhood is modelled on X , then T meets a one-tetrahedron solid torus, S , in a face. This is not possible due to Lemma 16.If the neighbourhood is modelled on X , then either one or two tetrahedra of X aremapped to T . If one is mapped to T , then T cannot be maximal. If two are mapped to T , then either T is not maximal, or, as in the proof of Proposition 9, one observes thatsince the triangulation is minimal with at least two tetrahedra and M is irreducible,there is a face pairing between the remaining two free faces of the third tetrahedronwhich implies that M is a lens space with layered triangulation.The remaining possibility is X , which implies that e is contained in four distincttetrahedra. This forces e to be the univalent edge for T . Lemma 22
Assume that the triangulation contains a single vertex and that all edgeloops are coloured using a homomorphism j : p ( M ) → ZZ . Then all tetrahedra in alayered solid torus in M are either oftype one ortype three, but not both. Proof
The colouring of the layered solid torus is uniquely determined by the imageof the longitude under j ; the result follows from the description of the layering pro-cedure. Definition 23 (Types of layered solid tori) Assume that the triangulation contains asingle vertex and that all edge loops are coloured using a homomorphism j : p ( M ) → ZZ . A layered solid torus containing a tetrahedron of type one (respectively three) isaccordingly termed of type one (respectively three).
Lemma 24
Assumethat T isminimaland 0–efficient andthat M isatoroidal. Thenevery torus which is normal with respect to T bounds a solid torus in M on at leastone side. Proof
Let T be a normal torus in M . Since M is atoroidal, there is a compressiondisc for T . Denote the 2–sphere resulting from the compression by S . Since M isirreducible, S bounds a ball, B . If T is not contained in B , then a solid torus withboundary T is obtained by attaching a handle to B in M . Hence assume that T is contained in B . Then S can be pushed off T , and T is a barriersurface for S . Since T is 0–efficient, S either shrinks to a 2–sphere embedded in atetrahedron or to a vertex linking 2–sphere. It follows that M = S . Alexander’s torustheorem now implies that T bounds a solid torus on at least one side. Lemma 25
Assume that T is minimal and 0–efficient and that M isatoroidal. Sup-pose twomaximallayered solid torimeetinprecisely oneedge. Thentheboundary ofasmall regular neighbourhood of their union bounds asolid torus in M. Proof
Denote the two maximal layered solid tori by T , T , and the common edge by e . Then the boundary of a small regular neighborhood N of T ∪ T is a (topological)torus and a barrier surface. Hence, either ¶ N is isotopic to a normal surface or M \ N is a solid torus. In the second case, we are done, and in the first case Lemma 24 givesthe conclusion. 21 emma 26 Assume that T is minimal and 0–efficient and that M isatoroidal. Sup-pose theedges arecoloured byahomomorphism j : p ( M ) → ZZ , andthattwopair-wise distinct maximal layered solid tori of type one meet in an j –even edge. Theneither M isalensspacewithlayeredtriangulation, or M admitsaSeifertfibrationwithat least twoand at mostthree exceptional fibres. Proof
Denote the two maximal layered solid tori by T , T , and the common j –even edge by e . If T ∩ T properly contains e , then M is a lens space with layeredtriangulation according to Lemma 20.Hence assume T ∩ T = { e } . Let N be a small regular neighbourhood of T ∪ T . Then ¶ N is an embedded torus in M , and either N or M \ N is a solid torus according toLemma 25. Since T and T are of type one, e is not a longitude of either of them.It follows that N cannot be a solid torus. Whence M \ N is a solid torus. If e isnot homotopic to a longitude of M \ N , then we have a Seifert fibration of M withthree exceptional fibres; otherwise we have a Seifert fibration with two exceptionalfibres. Lemma 27
Assume that T is minimal and 0–efficient and that M isatoroidal. Sup-pose the edges are coloured by a homomorphism j : p ( M ) → ZZ , and that threepairwise distinct maximal layered solid tori meet in an j –even edge. Then at leastone of them isof type three unless M admits a Seifert fibration withthree exceptionalfibres. Proof
Denote the three maximal layered solid tori by T , T , T , and the common j –even edge by e . If T i ∩ T j strictly contains e for i = j , then M is a lens space withlayered triangulation according to Lemma 20. But then there are no three pairwisedistinct maximal layered solid tori. Hence M is not a lens space with layered triangu-lation.Assume that T and T are of type one. As in the proof of Lemma 26, let N be asmall regular neighbourhood of T ∪ T . Then M \ N is a solid torus, but N is not asolid torus as e is not a longitude of T or T . Assume that M does not admit a Seifertfibration with three exceptional fibres. Then e is homotopic to a longitude of M \ N and it follows that j restricted to M \ N is trivial. If T has e as its longitude, then it isof type three. Otherwise, the longitude of T is homotopic into M \ N . But this impliesthat j restricted to T is trivial which again implies that T is of type three. Lemma 28
Assume that T is minimal and 0–efficient and that M is a lens space.Suppose the edges arecoloured by ahomomorphism j : p ( M ) → ZZ . Iftwodistinct typeonemaximallayered solidtorimeetinan j –evenedge,thentheseare the only maximal layered solid tori oftype one inthe triangulation.22 roof Denote the maximal layered solid tori by T , T , and the common j –even edgeby e . If T ∩ T strictly contains e , then M is a lens space with layered triangulation,and hence T and T are the only maximal layered solid tori of type one in the trian-gulation. Hence assume T ∩ T = { e } . Let N be a small regular neighborhood of T ∪ T . As in the argument above, theclosure of M \ N is a solid torus but N is not a solid torus. However, since M isassumed to be a lens space, e must be a longitude of M \ N . Since e is j –even, j restricted to e is trivial and thus j restricted to M \ N is trivial. Now assume that T isa maximal layered solid torus of type one which is distinct from T , T . Since T is oftype one, its longitude is j –odd. But it is clearly homotopic into M \ N , contradictingthe fact that j restricted to M \ N is trivial. Whence T and T are the only maximallayered solid tori of type one in the triangulation. Theorem 5
A lens space with even fundamental group satisfies Conjecture 4 if it hasa minimal layered triangulation such that there are no more j –odd edges than thereare j –even edges. Proof
Let M be a lens space with even fundamental group and j : p ( M ) → ZZ bethe unique non-trivial homomorphism. Suppose that the minimal layered triangulationhas no more j –odd edges than j –even edges.If M = L ( , ) , then L is a one-tetrahedron triangulation satisfying the hypothesisand we only need to prove the uniqueness statement. This follows by enumerating allsingle tetrahedron triangulations of closed, orientable 3–manifolds.If M = IR P , then the minimal layered triangulation has two tetrahedra, one j –oddedge and one j –even edge. It is well-known that this is a minimal triangulation (asfollows, for instance, from the above method of enumeration), but that there is anotherminimal triangulation with two vertices coming from the standard lens space represen-tation.Hence we may assume that M = IR P or L ( , ) . In this case, it is shown in Subsec-tion 4.3 that we have the following inequality for j –even edges: e ≥ + ¥ (cid:229) j = ( j − ) e j . (6.1)23y way of contradiction, suppose that the given minimal triangulation is not the min-imal layered triangulation. Each edge of degree three is j –even, so (6.1) implies thatthere are at least two edges of degree three. Since the triangulation contains at leastthree tetrahedra, each edge of degree three, e , is the base edge of a layered solid torussubcomplex isomorphic to S . This subcomplex is contained in a unique maximallayered solid torus, T ( e ) . Conversely, if a maximal layered solid torus, T , contains anedge, e , of degree three, then e is unique and we write e = e ( T ) . Since we assume that the given minimal triangulation of M is not the minimal layeredtriangulation, it follows from Lemma 20 that any two distinct maximal layered solidtori share at most an edge. We seek a contradiction guided by inequality (6.1).The proof, a basic counting argument, is organised as follows. The set of all edges ofdegree three, Y , is divided into a number of pairwise disjoint subsets. These subsetsare bijectively pared with a set of pairwise distinct j –even edges. If a subset of Y contains h pairwise distinct edges of degree three and the associated j –even edge hasdegree d , then we say that there is an associated deficit of h − d + h ≤ d − , anda gain of h − d + h > d − . Then (6.1) implies that the total gain is at least two.Let e ∈ Y such that T ( e ) is of type three. Then S ∼ = T ⊆ T ( e ) . Denote e thelongitude of T . This is a j –even boundary edge with T –degree 5 and M –degree5 + m for some m ≥ . The total number of maximal layered solid tori in M meetingin e is bounded above by m + , since no two meet in a face. Hence the maximallayered solid tori containing a degree three edge and meeting in the j –even edge e contribute at most m + to the left hand side of (6.1). The contribution of e to the righthand side is d ( e ) − = + m . One therefore obtains a deficit since m + ≤ m + . To e associate the set, Y , of all degree three edges e ′ such that T ( e ′ ) contains e . We now proceed inductively. Let e be an edge of degree three such that T ( e ) is oftype three and e is not contained in the collection of subsets Y , ..., Y i − of Y . ThenT ( e ) contains a subcomplex isomorphic to S , whose longitude, e i , cannot be any ofthe edges e , ..., e i − . Consider the set Y i of all degree three edges e ′ such that T ( e ′ ) contains e i and e ′ is not contained in any of Y , ..., Y i − . Then the above calculationshows that there is a deficit associated to e i and Y i . It follows that there must also be a maximal layered solid torus of type one, T , whichcontains an edge of degree three not in ∪ Y i . Let e be the unique j –even boundaryedge of T . We observe:(1) Suppose T is a maximal layered solid torus of type one which meets T in e . Lemma 28 implies that T , T are the only maximal layered solid tori of typeone. If d ( e ) ≥ , then e can be associated with { e ( T ) , e ( T ) } and either givesa deficit or a gain of + . In either case, (6.1) cannot be satisfied since Y = ∪ Y i S { e ( T ) , e ( T ) } . Hence d ( e ) = . − − − − + + Figure 5: Edge flip: Labels indicate the changes in edge degree(2) If T contains a j –even interior edge, e T , of degree at least five, then e T can beassociated with { e ( T ) } and gives a deficit.(3) If each j –even interior edge of T has degree less than five, but d ( e ) ≥ , thenit can be associated with { e ( T ) } and gives a deficit.Let X be the set of all degree three edges such that there is an associated maximallayered solid torus of type one which satisfies (2) or (3) above. To the collection ∪ j Y j S X we have an associated deficit, and that its complement, Z = Y \ ( ∪ j Y j S X ) , contains precisely the degree three edges with the property that the associated maximallayered solid torus has all j –even interior edges of degree at most four and has a j –even boundary edge of degree four. If Z is empty, then (6.1) is not satisfied.Let T be the maximal layered solid torus associated to an element of Z . It followsfrom Lemma 21 that its j –even boundary edge of degree four, e , is its univalent edge,and that it is contained in four distinct tetrahedra in the triangulation. The argumentproceeds by replacing the four tetrahedra around e by a different constellation of fourtetrahedra using an appropriate edge flip , see Figure 5. The resulting triangulation T ′ is also minimal and not a layered triangulation since it contains an edge of degree fourcontained in four distinct tetrahedra; there are associated sets Y ′ , Y ′ i , X ′ and Z ′ . Wewill show that either Z ′ is a proper subset of Z , or that T ′ contains fewer tetrahedraof type three than T . It then follows inductively that there is a minimal triangulation T ′′ which is not the minimal layered triangulation and with Z ′′ = /0; thus giving acontradiction.It remains to describe the re-triangulation process. There are three different cases toconsider; they are listed below using the types of tetrahedra ordered cyclically aroundthe e , starting with T . The complex formed by the four tetrahedra around e is alsotermed an octahedron (even though it may not be embedded), and the set of all fouredges “linking” e its equator. 251,1,1,1): Denote the maximal layered solid tori containing e by T and T . (We mayhave T = T . ) Doing any edge flip replaces the octahedron with four tetrahedra oftype two meeting in an j –even edge of degree four. The j –even edges along theequator have their degrees increased by one. Since no maximal layered solid toruscontains a tetrahedron of type two, it follows that each maximal layered solid toruswith respect to T is an maximal layered solid torus with respect to T ′ except for T and T ; whence Y \ { e ( T ) , e ( T ) } ⊂ Y ′ . If T is isomorphic to S , then after theflip, the degree three edge becomes a degree four edge; so e ( T ) / ∈ Y ′ , which impliese ( T ) / ∈ Z ′ . Otherwise e ( T ) ∈ Y ′ , but T ( e ( T )) with respect to T ′ has an j –evenboundary edge of degree five. Whence e ( T ) ∈ X ′ , and so e ( T ) / ∈ Z ′ . Similarly for T . Since no j –even boundary edge has its degree decreased by the flip, and no othermaximal layered solid tori are affected, we have Y ′ j = Y j for each j and X ⊆ X ′ . Thisproves that Z ′ is a proper subset of Z in this case.(1,2,2,1) or (1,1,2,2): We have that T is the unique maximal layered solid torus con-taining e . Do the unique edge flip such that the two j –even edges not contained onthe equator stay of the same degree. This replaces the octahedron by four tetrahedraof type one meeting in an odd edge of degree four. The main line of the argumentis as above, showing that e ( T ) / ∈ Z ′ . If every other maximal layered solid torus withrespect to T is a maximal layered solid torus with respect to T ′ , then we are done.Hence assume that some maximal layered solid torus with respect to T ′ , T ′ , is ob-tained from a maximal layered solid torus with respect to T , T , by layering onto itone of the (new) four tetrahedra of type one. Note that two of these tetrahedra cannotbe contained in T ′ for otherwise T ′ meets another layered solid torus in a single face.Hence there is at most one such tetrahedron, and at most one such maximal layeredsolid torus which has been extended. Now also note that the j –even boundary edgeof T ′ has degree one more than the j –even boundary edge of T . Thus, if e ( T ) ∈ X , then e ( T ′ ) ∈ X ′ , and if e ( T ) ∈ Z , then also e ( T ′ ) ∈ X ′ . This proves that we again have Y ′ j = Y j for each j , X ⊆ X ′ , and Z ′ is a proper subset of Z . (1,2,3,2): In this case, doing any edge flip replaces this with a (1,1,2,2)–octahedronaround an odd edge of degree four, thus reducing the number of tetrahedra of typethree. Corollary 29
Let L ( n , q ) bealensspacewithevenfundamentalgroup,andminimallayered triangulation L n , q . Supposethattheedgeshavebeencoloured usingthenon-trivial homomorphism j : p ( L ( n , q )) → ZZ . Then e ( L n , q ) ≤ o ( L n , q ) . Proof
Assume that e ( L n , q ) > o ( L n , q ) . Then Theorem 5 implies that L n , q is theunique minimal triangulation. It follows from the proof that L ( n , q ) = IR P or L ( , ) , e ( L n , q ) = o ( L n , q ) . Hence (4.4) implies4 − T ( L n , q ) − c ( S j ( L n , q )) ≥ + e ( L n , q ) − o ( L n , q ) > . Using (4.3), this gives e > , which is not possible in a layered triangulation. Corollary 30
Let L ( n , q ) be a lens space with even fundamental group, and nothomeomorphic to IR P or L ( , ) . Denote the minimal layered triangulation L n , q , and suppose that the edges have been coloured using the non-trivial homomorphism j : p ( L ( n , q )) → ZZ . Thefollowing are equivalent:(1) e ( L n , q ) ≥ o ( L n , q ) ;(2) e ( L n , q ) = o ( L n , q ) ;(3) there are precisely two j –even edges of degree three, and all other j –evenedges areof degree four.Moreover,theonlylayeredtriangulation of L n , q satisfying (3)istheminimallayeredtriangulation. Proof
We have ( ) ⇔ ( ) due to the preceding corollary. We have ( ) ⇒ ( ) since(6.1) holds with e = . Hence assume that we are given an arbitrary layered triangulation T of L ( n , q ) withprecisely two j –even edges of degree three, and all other j –even edges are of degreefour. We will show that this implies that there are as many j –even as j –odd edges.Each tetrahedron in T is of type one and we may describe T starting from T ∼ = S = { , , } with e and e j –odd, and e j –even. Since T contains two edgesof degree three, there is a tetrahedron layered on e ; the resulting layered solid torus T ∼ = S = { , , } has univalent edge the j –even edge, e . Hence in T , there isa tetrahedron layered on either of the j –odd boundary edges e or e of T , givinga layered solid torus T which is either isomorphic to { , , } or { , , } with j –odd univalent edge. The even boundary edge e has T –degree three; hence either T is obtained from folding along e or by layering on e . The construction continuesinductively. Since we alternate between layering on j –even and j –odd edges, andin the end fold along an j –even edge of degree three, it follows that the resultingtriangulation has as many j –odd edges as j –even edges. This, in particular, proves ( ) ⇒ ( ) since the minimal layered triangulation must be constructed in this way.To prove the last statement, notice that the above procedure describes the sub-tree ofthe L –graph of [6] shown in Figure 6, and that any layered triangulation satisfying (3)corresponds to a unique path without looping or backtracking in this sub-tree. Henceit is a minimal layered triangulation. 27Sfrag replacements / / / / / / / / / / /
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16 7 / / p / qq / p + q p / p + q Figure 6: A complete description of the set of all lens spaces to which Theorem 5 ap-plies: Shown is part of the L –graph of [6]. The 0–cells are in bijective correspondencewith one–vertex triangulations on the boundary of the solid torus. The unique pathfrom p / q to 1 / { p , q , p + q } when q > p > . The minimal layered triangu-lation of a lens space satisfies the hypothesis of Theorem 5 if and only if it is obtainedby folding the minimal layered extension of { p , q , p + q } along q , where p / q = / L ( p + q , p ) . Remark 31
It is shown in [6] that the minimal layered extension of { p , q , p + q } contains E ( q , p ) − q > p > . This implies that upon foldingalong edge q , the minimal layered triangulation of L ( p + q , p ) contains E ( q , p ) − E ( p + q , p ) = E ( q , p ) + . As noted earlier, each of the lens spaces L ( n , ) contains as many j –odd edges as j –even edges, and hence satisfies Conjecture 4. To recognise more families that satisfythis condition, we study layered solid tori that are isomorphic to S m for some m ≥ . efinition 32 ( S –maximal layered solid torus) A layered solid torus is an S –maximal layered solid torus with respect to T in M if it is isomorphic to S m forsome m ≥ S h for some h > m . If T has more than two tetrahedra, then any edge of degree three is contained in an S –maximal layered solid torus. Moreover, this S –maximal layered solid torus hasthe edge of degree three as its base-edge and all other interior edges have degree four.To begin with, we analyse the result of identifying two layered solid tori, T ∼ = S t and T ∼ = S s , t ≥ s ≥ , along their boundaries. The identification of the solid toriis uniquely determined by a pairing of the boundary edges, and hence there are sixpossibilities: ( s + , s + , ) ↔ ( t + , t + , ) : This gives L ( s + t + , ) with minimal layered trian-gulation (as can be seem from the continued fraction expansion), and if s + t > , thereare precisely two S –maximal layered solid tori which meet in s + t − ( s + , s + , ) ↔ ( t + , , t + ) : This gives L (( s + )( t + ) + , t + ) with minimallayered triangulation. If s ≥ , there are precisely two S –maximal layered solid toriwhich meet in one tetrahedron; one is isomorphic to S s , the other to S t + . If s = , then there is at most one S –maximal layered solid torus. ( s + , s + , ) ↔ ( , t + , t + ) : This gives L (( s + )( t + ) + , t + ) with minimallayered triangulation. If s ≥ , there are precisely two S –maximal layered solid toriwhich meet in one tetrahedron; one is isomorphic to S s + , the other to S t . If s = , then there is at most one S –maximal layered solid torus. ( s + , s + , ) ↔ ( t + , t + , ) : This gives L ( t − s , ) , and there is an edge of degreetwo. The triangulation is not a minimal layered triangulation (since neither t − s = t + s = t − s = t + s = ( s + , s + , ) ↔ ( t + , , t + ) : This gives L ( s ( t + ) + t , t + ) and there is anedge of degree two. An algebraic argument shows again that the triangulation is nota minimal layered triangulation unless s = t = M = L ( , ) . In particular, thetriangulation does not contain an S –maximal layered solid torus. ( s + , s + , ) ↔ ( , t + , t + ) : This gives L (( s + )( t + ) + t + , t + ) with aminimal layered triangulation. If s ≥ , there are precisely two S –maximal layeredsolid tori which meet in two faces. Lemma 33
Iftwodistinct S –maximallayered solidtorimeetinprecisely twofacesthen M is a lens space with layered triangulation. If it is the minimal layered triangu-lation, then M ishomeomorphic to L (( s + )( t + ) + t + , t + ) , where t ≥ s ≥ . roof It follows from Lemma 14 that M is a lens space with layered triangulation.An S –maximal layered solid torus contains at least two tetrahedra. The lemma nowfollows directly from the stated possibilities.Note that if M as in the above lemma has even fundamental group, then s + ↔ ↔ t + s and t are odd; hence the number of tetrahedra is even. Inparticular, the number of edges in the triangulation is odd, so the number of j –oddedges is greater than the number of j –even edges.Recall that L ( p , q ) = L ( p , q ) if and only if p = p and q q ± ≡ ± ( p ) . Thefollowing two results give Theorems 1, 2 and 3.
Proposition 34
If two distinct S –maximal layered solid tori share a tetrahedron,then M is a lens space with minimal layered triangulation. Moreover, M is containedin one ofthe following, mutually exclusive families:(1) L ( s + t + , ) , t ≥ s ≥ L (( s + )( t + ) + , t + ) , where t > s > , (3) L (( s + )( t + ) + , t + ) , where t > s > , (4) L (( t + )( t + ) + , t + ) = L (( t + )( t + ) + , t + ) where t ≥ . Proof
Since the S –maximal layered solid tori are distinct, it follows from Lemma 15that M is a lens space. Moreover, M can be described as the union of two layered solidtori, T and T , along their boundary, where T ∼ = S s , and T ∼ = S t , with s , t ≥ . Without loss of generality, assume t ≥ s ≥ . The above possibilities show that, underthe assumption that there are two distinct S –maximal layered solid tori, M is one ofthe cases listed in (1)–(4). The equality in (4) follows since ( t + )( t + ) ≡ − ( t + )( t + ) + . Also note that a lens space as in (4) with t = S –maximal layered solidtorus. It remains to show that the cases are mutually exclusive.Since the minimal layered triangulation is unique, it follows from the description ofintersections of S –maximal layered solid tori that the lens spaces in the first list donot appear in the second or third unless s + t − = , whence M = L ( , ) . This doesnot appear in (2) or (3). It remains to show that the possibilities in (2) and (3) arecomplete and exclusive.Let t ≥ s ≥ t ≥ s ≥ . The minimal layered triangulations of L (( s + )( t + ) + , t + ) and L (( s + )( t + ) + , t + ) are characterised by the fact that thereare two S –ma-la-so-tos meeting in one tetrahedron. It follows that ( s = s + t + = t ) or ( s = t and t + = s + s = t ≥ s = t ≥ s , L (( s + )( t + ) + , t + ) = L (( s + )( t + ) + , t + ) only if s = s + t + = t . Since the fundamental groups must have the sameorder, we have t = s + , giving t + = t = s + = s ≤ t ; a contradiction. Corollary 35 If M is one of the lens spaces listed in (1)–(4) of Proposition 34 andhasevenfundamental group,thenthereareasmany j –evenedgesasthereare j –oddedges, where j : p ( M ) → Z istheunique non-trivial homomorphism. Moreover, M is either in(1) and s + t isodd;(2) and s isodd and t iseven;(3) and s iseven and t is odd. Proof
First note that every lens space in (4) has odd fundamental group. We give theargument for the third case; the others are analogous. Since the gluing is ( s + , s + , ) ↔ ( t + , , t + ) to give L (( s + )( t + ) + , t + ) , and we want to have evenfundamental group, the paring 1 ↔ t + t odd, and s + ↔ s even.Even edges in the S –layered solid tori correspond to j –even edges in the lens space;similarly for odd edges. Since the gluing identifies one pair of even edges and twopairs of odd edges, we have12 ( s + ) + ( t + ) − = ( s + t + ) j –even edges in L (( s + )( t + ) + , t + ) , and12 ( s + ) + ( t + ) − = ( s + t + ) j –odd edges in L (( s + )( t + ) + , t + ) . Any other example to which Theorem 5 applies contains two S –maximal layeredsolid tori which meet in precisely two edges, one edge or not at all. Every examplecan be constructed as in the proof of Corollary 30. For instance, take s ≥ S s = { , s + , s + } , then in turn layer on 1 , s + , s + , s + , s + , s + , s + s + . This yields L ( s + , s + ) with minimallayered triangulation having s + s + s + = [ , , , , s + ] , and hence E ( s + , s + ) − = ( s + ) − = s + . eferences [1] Sergei Anisov: Cut loci in lens manifolds.
C. R. Math. Acad. Sci. Paris 342 (2006), no.8, 595–600.[2] Sergei Anisov:
Geometrical spines of lens manifolds.
J. London Math. Soc. (2) 74(2006), no. 3, 799–816.[3] Glen E. Bredon and John W. Wood:
Non-orientable surfaces in orientable 3–manifolds.
Invent. Math., 7:83110, 1969.[4] William Jaco:
Lectures on three-manifold topology , CBMS Regional Conference Se-ries in Mathematics, 43, American Mathematical Society, Providence, R.I., 1980.[5] William Jaco and J. Hyam Rubinstein: , Jour-nal of Differential Geometry (2003), no. 1, 61–168.[6] William Jaco and J. Hyam Rubinstein: Layered-triangulations of 3–manifolds ,arXiv:math/0603601.[7] Sergei V. Matveev:
Complexity theory of three-dimensional manifolds , Acta Appl.Math. 19 (1990), no. 2, 101–130.[8] Sergei V. Matveev:
Recognition and tabulation of three-dimensional manifolds , (Rus-sian) Dokl. Akad. Nauk 400 (2005), no. 1, 26–28.[9] Sergei V. Matveev, Carlo Petronio and Andrei Vesnin:
Two-sided aymptotic bounds forthe complexity of some closed hyperbolic three-manifolds , to appear in the Journal ofthe Australian Mathematical Society, arXiv:math/0602372v1.[10] J. Hyam Rubinstein:
One-sided Heegaard splittings of 3–manifolds , Pacific J. Math. 76(1978), no. 1, 185–200.DepartmentofMathematics,OklahomaStateUniversity,Stillwater,OK74078-1058,USADepartmentofMathematicsandStatistics,TheUniversityofMelbourne,VIC3010,AustraliaDepartmentofMathematicsandStatistics,TheUniversityofMelbourne,VIC3010,AustraliaEmail: [email protected] [email protected]@[email protected] [email protected]@ms.unimelb.edu.au