aa r X i v : . [ m a t h . G T ] J un The Thurston norm via Normal Surfaces
Daryl Cooper and Stephan Tillmann
To Bill Thurston on the occasion of his sixtieth birthday
Abstract
Given a triangulation of a closed, oriented, irreducible, atoroidal 3–manifold everyoriented, incompressible surface may be isotoped into normal position relative to the triangu-lation. Such a normal oriented surface is then encoded by non-negative integer weights, 14 foreach 3–simplex, that describe how many copies of each oriented normal disc type there are.The Euler characteristic and homology class are both linear functions of the weights. There isa convex polytope in the space of weights, defined by linear equations given by the combina-torics of the triangulation, whose image under the homology map is the unit ball, B , of theThurston norm.Applications of this approach include (1) an algorithm to compute B and hence the Thurstonnorm of any homology class, (2) an explicit exponential bound on the number of vertices of B in terms of the number of simplices in the triangulation, (3) an algorithm to determine thefibred faces of B and hence an algorithm to decide whether a 3–manifold fibres over the circle. AMS Classification
Keywords
This work was inspired by a desire to understand the topological significance of thefaces of the unit ball of the Thurston norm. The main result of this paper implies thatthe unit ball of the Thurston norm for a closed, orientable, irreducible, atoroidal three-manifold is the projection under a linear map of a certain polyhedron in transverselyoriented normal surface space. This polyhedron can be computed from a triangulationusing linear algebra. In order to make this paper accessible to a wide audience we donot assume that the reader is familiar with the theory of normal surfaces. To keep thepaper short we have considered only the case that M is closed and orientable and haveintroduced the bare minimum of the theory of transversely oriented normal surfacesfor this application. To facilitate a quick overview, we first state most definitions andresults leading up to the main result (Theorem 5) as well as the applications, and fillthe gaps at the end of the paper. A more general treatment is to be found in [1], andreferences to normal surface theory can be found in [3].This work was partially supported by NSF grant DMS-0405963. The authors thankAndrew Casson and Bus Jaco for helpful comments on this project.1 ransversely oriented normal surfaces An arc a in a 2–simplex D is a normal arc if a ∩ ¶ D = ¶a , and in addition the endpoints of a are on distinct edges of D . A transversely oriented normal arc ( a , n a ) is a normal arc a in D together with a transverse orientation n a to a in D . Two such arcsare equivalent if there is a homeomorphism of D to itself whichpreserves each edge of D and takes one arc with its transverseorientation to the other. There are 6 equivalence classes.A disc D in a 3–simplex t is a normal disc if D ∩ ¶t = ¶ D is a union of normal arcs, notwo of which lie in the same face of t . The disc D is called a triangle if the boundaryconsists of three normal arcs; otherwise the boundary has four normal arcs and D iscalled a quad. A transversely oriented normal disc ( D , n D ) is a normal disc D together with a trans-verse orientation n D . Two such discs are equivalentif there is a homeomorphism of t to itself whichpreserves each face of t and takes takes one discwith its transverse orientation to the other. Thereare 14 equivalence classes. The boundary of atransversely oriented normal disc is a collection of transversely oriented normal arcs;three for a triangle and four for a quad. Each transversely oriented normal arc is con-tained in the boundary of exactly one equivalence class of transversely oriented normaltriangles and one of transversely oriented normal quads in t . Unless stated otherwise, M denotes a closed, oriented, irreducible 3–manifold and T a triangulation of M with t M which are more general than a simplicial triangulation. The interior of every simplex s in T , of every dimension, is required to be embedded in M , but there may be self-identifications on the boundary of s . A triangulation is if every normal2–sphere is vertex linking , i.e. it bounds a ball contained in a small neighbourhood ofa vertex. Jaco and Rubinstein [3] showed that if M is not homeomorphic to IR P or L ( , ) , then any minimal triangulation of M is 0–efficient. A minimal triangulation of M has the property that no triangulation of M contains fewer 3–simplices.The transversely oriented normal disc space ND n ( T ) is the real vector space of di-mension 14 t with a basis consisting of the equivalence classes of the transverselyoriented normal discs in each 3–simplex of T . A properly embedded surface S ⊂ M is a normal surface if the intersection of S with every 3–simplex of T is a collectionof pairwise disjoint normal discs. If the normal surface S has a transverse orienta-tion, n S , then it determines a unique point x n ( S , n S ) ∈ ND n ( T ) , where the coefficient2f ( D , n D ) is the number of transversely oriented normal discs of that type in S . Thefollowing result is routine:
Proposition 1 (Incompressible is isotopic to normal) Every closed, incompressiblesurfaceinaclosed,irreducible,triangulated 3-manifoldisisotopictoanormalsurface.There is a linear subspace NS n ( T ) ⊂ ND n ( T ) defined by the matching equations. There is one matching equation for each equivalence class of transversely orientednormal arc ( a , n a ) in each 2–simplex D of T . The two sides of D in M are labelled + and − arbitrarily. The matching equation is t − + q − = t + + q + . Here t − (respectively q − ) is the coefficient of the equivalence class of transversely ori-ented normal triangles (respectively quads) on the − side of D which contains ( a , n a ) in its boundary. Similarly for the + side. This equation expresses that there are thesame number of transversely oriented normal discs containing ( a , n a ) in their bound-ary on either side of D . It follows that if ( S , n S ) is a transversely oriented normalsurface in M , then x n ( S , n S ) ∈ NS n ( T ) . Proposition 2 (Branched immersions) Every non-zero point in NS n ( T ) with non-negativeintegralcoordinatesisrepresentedbyatransverselyorientednormalbranchedimmersion of a closed orientable surface. Conversely, a transversely oriented normalbranched immersion of a closed orientable surface, ( f , n f ) : S → M , determines aunique point, x n ( f , n f ) , in NS n ( T ) . The normal surface space is NS n + ( T ) = NS n ( T ) ∩ [ , ¥ ) t . It is a cone on the com-pact convex polytope C = NS n + ( T ) ∩ V , where V is the affine subspace consisting ofall points whose coordinates sum to one. Whence C is the intersection of a simplexof dimension 14 t − projective solution space and its extreme points are termed vertices . Given a transversely oriented normal sur-face, we would like to know its Euler characteristic as well as which homology classit represents. This information is given by the linear maps described in the followingtwo results. Lemma 3 ( c ∗ = c for immersions) There is a linear map c ∗ : NS n ( T ) → R withthe property that if ( f , n f ) : S → M is a transversely oriented normal (unbranched)immersion, then c ∗ ( x n ( f , n f )) = c ( S ) . The equality in the above lemma does not hold for branched immersions.3f f : S → M is a normal immersion of a closed oriented surface S , then the orientationon M determines an induced transverse orientation for the immersion, n f . Conversely,given a transversely oriented normal branched immersion of a closed surface S , thetransverse orientation and the orientation on M determine an orientation of S . Thisobservation leads to the following:
Proposition 4 (Homology map) There isasurjective homomorphism h : NS n ( T ) → H ( M ; R ) , called the homology map, with the following property: If ( f , n f ) : S → M is a trans-versely oriented normal branched immersion, then f ∗ ([ S ]) = h ( x n ( f , n f )) , where S isgiven the induced orientation.A point x ∈ NS n + ( T ) is admissible if at most one quad type appears in each 3-simplex(though both orientations are allowed). Two points x , y ∈ NS n + ( T ) are compatible if x + y is admissible. Theorem 5 (Thurston norm via normal surfaces) Let M be a closed, orientable, ir-reducible, atoroidal 3–manifold with simplicial or 0–efficient triangulation T . Let C be the projective solution space and B be the convex hull ofthe finite setof points (cid:26) v | c ∗ ( v ) | : v is avertex of C which isadmissible and satisfies c ∗ ( v ) < (cid:27) . Then h ( B ) is the unit ball, B , of the Thurston norm on H ( M ; IR ) . In particular, B has at most 2 t vertices. Algorithm 6 (Fibred faces) Let M be a closed, orientable, irreducible, atoroidal 3–manifold. The following is an algorithm to determine the fibred faces of B and henceto determine whether M fibres over the circle:(0) We may assume that M is given via a (simplicial or 0–efficient) triangulation.(1) Compute the unit norm ball of the Thurston norm using the above theorem. (2)Determine the top-dimensional faces. (3) For each top dimensional face F of B : (3a)Let ˆ F be the barycentre of F and let a ∈ H ( M ; Z ) be the smallest integral multipleof ˆ F . (3b) Find an embedded, norm minimising, transversely oriented normal surface S without sphere or torus components representing a . (3c) Use Haken’s algorithm tocheck whether M \ S is homeomorphic to S × [ , ] ; see [4]. Then F is a fibred face ifand only if Step (3c) yields an affirmative answer. Moreover, M fibres over the circleif and only if at least one face is fibred. 4his paper gives a self-contained proof of Theorem 5 for a 0–efficient triangulation. Inthe case of a simplicial triangulation, work by Tollefson and Wang [7] is used to proveTheorem 5. An alternative approach to construct B is given in [7], Algorithm 5.9, andused by Schleimer [5] to obtain an algorithm to determine whether a 3–manifold fibresover the circle. The algorithm to compute B given here is practical in the sense thatthe authors hope to implement it on a computer. Proofs, neat position and algebraically aspherical solutions
Proof of Proposition 2
A non-zero point in NS n ( T ) with non-negative integral co-ordinates determines an abstract collection of normal discs. Since the integers satisfythe matching equations, we may identify the edges of these discs in pairs to producea closed surface S such that there is a map of S into M mapping each disc in S toa normal disc in M . This map is an immersion on the complement of the set, V , ofvertices of S , and thus a branched immersion. The transverse orientations on the discsmatch along edges, so this map extends to a map of S × [ − , ] into M which is animmersion on the complement of V × [ − , ] . The map of S therefore has a transverseorientation and, since M is orientable, it follows that S is orientable. The conversedirection is obvious. Proof of Lemma 3 A corner of a 3-simplex is a small neighborhood of an edge.The degree of a 1–simplex e in T is the number of corners of 3-simplices in T which contain e . If all these 3-simplices are embedded this is just the number of3-simplices containing e . We define c ∗ on each transversely oriented normal disc ( D , n D ) as follows. For each corner v ∈ ¶ D let d ( v ) denote the degree of the 1–simplex in T that contains v . Then c ∗ ( D , n D ) = − ( ¶ D ) + (cid:229) v ∈ ¶ D d ( v ) , where ( ¶ D ) is the number of sides of ¶ D (three for a triangle and four for a quad).A normal immersion of a surface S gives it a cell decomposition into normal trianglesand quads. Every 1–cell appears in two 2–cells, and every 0–cell v appears in d ( v ) S using this cell decomposition. Proof of Proposition 4
Denote by C ( M ) the 2–dimensional singular chain group of M with coefficients R . A homomorphism h : ND n ( T ) → C ( M ) is first constructed.Suppose ( D , n D ) is a transversely oriented normal disc in some 3-simplex of T . Theorientation of M and the transverse orientation n D determine an orientation m D of D .5f D is a normal triangle then ( D , m D ) may be regarded as a singular 2–simplex. If D isa normal quad, we add a diagonal to subdivide D into two triangles and obtain the sumof two singular 2-simplices. This defines h on a basis and we extend linearly. It is easyto see that if x ∈ NS n ( T ) then h ( x ) is a singular 2–cycle. See [1] for details. We define h ( x ) = [ h ( x )] . It is also easy to see that if ( f , n f ) : S → M is a transversely orientedbranched normal immersion then f ∗ ([ S ]) = h ( x n ( f , n f )) . The map h is surjective sinceevery homology class is represented by an embedded, closed, oriented surface whichmay be isotoped to be normal.We obtain the theory of normal surfaces if we forget transverse orientations. Thus ND ( T ) is the vector space of dimension 7 t with basis the equivalence classes ofnormal discs in each 3-simplex of T . There is one un-oriented matching equationfor each equivalence class of normal arc in each 2-simplex in T . The solutions ofthe matching equations give a linear subspace NS ( T ) ⊂ ND ( T ) , and NS + ( T ) = NS ( T ) ∩ [ , ¥ ) t . A point in NS + ( T ) is admissible if in each 3-simplex the coeffi-cient of at most one quad type is non-zero.We now the describe the geometric sum of two embedded normal surfaces. Suppose D , D ′ are two normal discs in a 3-simplex t which intersect transversely in an arc a . After cutting D and D ′ along a there are two possible ways to cross-join to obtaintwo disjoint discs. There is exactly one way which yields disjoint normal discs unless D and D ′ are quads of different types. In that case there is no cross join which givesdisjoint normal discs. The direction of the cross join is determined by the way thenormal arcs intersect in the boundary. Now suppose S and F are two normal surfaces.We may isotope them so they intersect transversely and, for every 3–simplex t ∈ T , the intersection of a normal disc in S ∩ t with one in F ∩ t is either empty or a singlearc. If S and F are compatible , i.e. the sum of their normal coordinates is admissible,then there is a unique way to cut and cross join S and F along S ∩ F to obtain a new(possibly not connected) normal surface denoted S + F . The pair ( S , F ) is disc reduced if no curve of intersection in S ∩ F bounds a disc inboth S and F . An innermost discs argument shows that S and F can be replaced bynormal surfaces S ′ and F ′ homeomorphic to (but possibly not normally isotopic to) S and F respectively such that ( S ′ , F ′ ) is disc reduced and S ′ + F ′ is normally isotopicto S + F . An ambient isotopy of M is a normal isotopy if it preserves every cell in thetriangulation T . The following is well known:
Proposition 7 (Admissible implies unique embedded normal surface) Suppose M isa closed, triangulated 3-manifold. There is a one–to–one correspondence betweenadmissible non-negative integral points in NS ( T ) and normal isotopy classes of em-bedded normalsurfacesin M . Furthermore,additionofcompatiblepointscorrespondsto geometric sumof compatible normal surfaces.6 roof
Assume for simplicity that every simplex of T is embedded. The general caseonly involves more words. A point x ∈ NS + ( T ) determines a collection of normaldiscs in each 3-simplex of T . These in turn determine a collection of points on each1-simplex of T . These points determine a collection of normal arcs in each 2-simplexof T . Up to normal isotopy there is a unique way to embed these arcs in each 2-simplex. Thus up to normal isotopy x determines a unique set of disjoint simple closedcurves, C ( x , t ) , in the boundary of each 3-simplex t ∈ T . Up to normal isotopy thereis a unique set of disjoint discs in t with boundary C ( x , t ) . These discs are normaldiscs if and only if each simple closed curve in C ( x , t ) meets each face of t at mostonce. This happens if and only if x is admissible. If x is admissible these normaldiscs are identified along their boundaries to produce a normal surface S which is thusunique up to normal isotopy.The corresponding result in the transversely oriented setting is weaker: Proposition 8 (Admissibleimpliesimmersedtransversely oriented normal)If M isaclosed, orientable, triangulated 3–manifold, then every admissible non-negative inte-gralpointin NS n ( T ) isthecoordinate ofatransversely oriented(unbranched) normalimmersion ofaclosed oriented surface. Proof
A non-negative admissible integral point x n ∈ NS n ( T ) determines a collec-tion of transversely oriented normal discs in each 3-simplex. We embed each of thesediscs in the simplex in a canonical neat position . The boundary arcs of different discsmay intersect but only in the interior of 2-simplices in T , and canonical means thatthe normal discs in adjacent 3-simplices have boundary arcs that match pairwise withmatching transverse orientations to give a transversely oriented immersed normal sur-face realising x n . First we describe neat position.
Suppose ( a , n a ) is a transversely oriented normalarc in a 2-simplex D . The transverse orientation is thought of as a function on thecomponents of D \ a which sends one component to + − . Denote the former by C + and let s denotethe maximal sub-simplex of D contained in C + . There are6 such simplices and they are in bijective correspondencewith the equivalence classes of ( a , n a ) . The arc is called short if s is a 0–simplex. Otherwise s is a 1–simplex andthe arc is called long. A family of transversely oriented arcsis in neat position if each arc is straight, two arcs intersectonly in the interior of D , and the only arcs which intersect are long arcs of differenttypes. 7 family of transversely oriented normal discs in a 3-simplex t is in neat position ifthe intersection of the boundary of the family with each 2-simplex D ⊂ ¶t is in neatposition and each normal disc D is the cone on ¶ D from the barycentre of the verticesof ¶ D . Thus D is determined by the intersection points of ¶ D with the 1-skeleton T . Hence in neat position every triangle is flat. A transversely oriented normal triangle is small if the boundary consists of three short arcs and otherwise the boundary consistsof three long arcs and the triangle is called large.
It follows that in neat position, twotriangles intersect if and only if they are both large and of different un-oriented types.Suppose ( D , n D ) is a transversely oriented normal disc in a 3-simplex t . As above,a component of t \ D is labelled C + using n D . Let s ( D , n D ) denote the maximalsub-simplex of t contained in C + . There are 14 such simplices and they are in 1-1correspondence with the equivalence classes of transversely oriented normal discs. Ifa family of discs is in neat position, then, after an ambient isotopy, every disc ( D , n D ) in the family is in a small neighborhood of s ( D , n D ) . A triangle is small if s is a 0-simplex and large if s is a 2–simplex. For a quad s is always a 1–simplex and D will be chosen to be a long thin rectangle very close to s . In neat position, if two discs intersect then they are either two large triangles ofdifferent types, or a quad ( D , n D ) and a large triangle which meets s ( D , n D ) . Suppose we are given a collection of transversely oriented normal discs in a 3–simplex t which is admissible: only one quad type appears but both transverse orientations areallowed. We claim that up to normal isotopy there is a unique way to position thesediscs in neat position in t . For existence, it suffices to show that it is possible to place in neat position one copyof each transversely oriented normal triangle in t together with two normal quadswhich differ only in transverse orientation. For,having done this, one may take closely spacedparallel copies of these discs. The figure showshow to place the discs. First place the large tri-angles in neat position. Next place each trans-versely oriented quad in a small neighborhoodof the 1–simplex of s it determines, so that it isdisjoint from the large triangles it can be madedisjoint from. Finally place the small triangles in small neighbourhoods of the verticesdisjoint form all other discs. This proves existence.Uniqueness follows from the fact that the family of neat arcs in each 2-simplex D of s is determined up to normal isotopy by their endpoints, and the fact that neat discsare determined by their boundaries. This completes the claim.8e now describe canonical neat position for admissible solutions. A vector x n in NS + ( T ) determines, for each 1–simplex e of T , a finite number of transverselyoriented points on e . We first position these points on e in neat position which justmeans close to the end of e determined by the transverse orientation on that point,and equally spaced apart. The vector x n also determines a family of transverselyoriented normal arcs types in each 2-simplex of T . By the previous discussion, thereis a unique way to place these normal arcs in neat position with endpoints the chosenneatly positioned points. Since x n is admissible, there is then a unique way to placethe transversely oriented normal disc types determined by x n in each 3–simplex of T into neat position so the boundary is the set of neatly positioned arcs. This is thecanonical neat position.Since M is orientable and the immersed surface constructed above is transversely ori-entable it is also orientable.An orientable surface S is aspherical if no connected component of S is a sphere. Wewish to have an algebraic criterion for deciding whether a transversely oriented normalsurface is aspherical depending only on its transversely oriented normal coordinates.There is a partial order ≥ on ND n ( T ) , where x ≥ y if and only if every coordinate of x is greater than or equal to the corresponding coordinate of y . Definition
A point x ∈ NS n + ( T ) is algebraically aspherical if for all y ∈ NS n + ( T ) such that x ≥ y , we have c ∗ ( y ) ≤ . Lemma 9 (Properties of aspherical elements) The following properties follow di-rectly fromthe definition.(1) If x ∈ NS n + ( T ) isalgebraically aspherical, c ∗ ( x ) ≤ . (2) If x = (cid:229) x i isalgebraically aspherical and x i ∈ NS n + ( T ) foreach i , theneach x i isalgebraically aspherical.(3) If x ∈ NS n + ( T ) isalgebraically aspherical, so is a x for any a > . Proposition 10 (embedded aspherical in 0–efficient ⇒ algebraically aspherical) Let M be a closed, oriented 3–manifold with 0–efficient triangulation T . If ( F , n F ) isan embedded transversely oriented normal surface in M and no component of F is asphere, then x n ( F , n F ) isadmissible and algebraically aspherical. Proof
Since F is embedded x n = x n ( F , n F ) is admissible. Forgetting transverse ori-entations gives a linear map j : NS n ( T ) → NS ( T ) , and we write x = j ( x n ) . Thetopology of the embedded normal surface F is determined (up to normal isotopy) by9 . If x n is not algebraically aspherical, we will show that there is an integer n > nx > y , where y = x ( S ) for some embedded normal sphere S . Then nx = ( nx − y ) + y , where nx , ( nx − y ) , y ∈ NS + ( T ) are compatible admissible non-negative integral so-lutions to the unoriented matching equations. By Proposition 7 for each admissible z ∈ NS + ( T ) there is an embedded normal surface F ( z ) in M with normal coordinates z , and this surface is unique up to normal isotopy. Thus F ( y ) is normally isotopic to S . Since T is 0–efficient S is a vertex linking sphere and so may be isotoped into a smallneighborhood of that vertex disjoint from F ( nx − y ) . Let n · F denote an embeddednormal surface consisting of n parallel disjoint copies of the normal surface F . Theuniqueness part of Proposition 7 implies F ( nx ) = n · F . The above equation implies n · F is normal-isotopic to the geometric sum of F ( nx − x ( S )) and S . Since these sur-faces are disjoint, their geometric sum equals their disjoint union. Thus n · F containsa sphere normally isotopic to S , which is a contradiction.It remains to prove that if x n is not algebraically aspherical, then S exists. If x n is notalgebraically aspherical, then there is a n ∈ NS n + ( T ) with x n ≥ a n and c ∗ ( a n ) > . Wemay assume a n has non-negative rational coordinates. Then there is an integer n > b n = na n has integer coordinates and nx n ≥ b n . Now c ∗ ( b n ) = n c ∗ ( a n ) > . Since x n is admissible, b n is admissible and hence b = j ( b n ) ∈ NS + ( T ) is admissi-ble. The map c ∗ factors through a linear map c ∗ un : ND ( T ) → R with the property thatif A is an embedded normal surface, then c ∗ un ( x ( A )) = c ( A ) . So c ∗ ( b n ) = c ∗ un ( j ( b n )) and since F ( b ) is an embedded normal surface, c ( F ( b )) = c ∗ un ( b ) = c ∗ ( b n ) . Hence c ( F ( b )) > . Thus F ( b ) contains a component S with c ( S ) > . Since S ⊂ F ( b ) are normal surfaces it follows that b ≥ x ( S ) . If S = R P then, since M is orientable,a small regular neighborhood of S is bounded by a 2–sphere S ′ , and x ( S ′ ) = x ( S ) . Since T is 0–efficient, S ′ is a vertex linking sphere. But this implies that x ( S ) is notan integral solution, giving a contradiction. Thus S is a sphere and nx ≥ b ≥ x ( S ) . If S is a closed surface, then − c − ( S ) is the sum of the Euler characteristic over allconnected components of S with negative Euler characteristic (see [6]). If c ∈ H ( M ) , then the Thurston norm of c is || c || = inf S : [ S ]= c c − ( S ) . If M is irreducible, discarding sphere components from S does not change [ S ] or c − ( S ) and so the infimum can be taken over surfaces containing no spheres.A transversely oriented normal surface ( F , n F ) in M is called a taut normal surface ifit is aspherical and c ( F ) = −|| h ( x n ( F , n F )) || , and it is least weight taut if F minimises | F ∩ T ( ) | over all taut normal surfaces representing h ( x n ( F , n F )) . The following is are-formulation of (part of) Lemma 3.2 of Tollefson and Wang [7].10 emma 11 (Tollefson–Wang [7]) Let M beaclosed,oriented3–manifoldwithsim-plicial triangulation T . Let ( F , n F ) be least weight taut. Let G , H be orientablenormal surfaces such that mx ( F ) = x ( G ) + x ( H ) for some positive integer m and thepair ( G , H ) is disc reduced. Then the transverse orientation of F induces uniquetransverse orientations n G and n H on G and H respectively such that x n ( F , n F ) = x n ( G , n G ) + x n ( H , n H ) and ( G , n G ) and ( H , n H ) are least weight taut. Proposition 12 (least weight taut ⇒ algebraically aspherical) Let M be a closed,oriented, irreducible 3–manifold with simplicial triangulation T . If ( F , n F ) is a leastweight taut normal surface in M , then x n ( F , n F ) is admissible and algebraically as-pherical. Proof
The proof of the previous proposition is modified as follows. First note that if ( F ( nx − y ) , F ( y )) is not disc reduced, then there is y ′ such that ( F ( nx − y ′ ) , F ( y ′ )) isdisc reduced and c ( F ( y ′ )) = c ( F ( y )) = . Lemma 11 implies that F ( y ′ ) can be trans-versely oriented to give a least weight taut normal surface. Since F ( y ′ ) = /0 , it followsthat a sphere represents a non-trivial homology class contradicting the assumption that M is irreducible. Proof of Theorem 5
We claim that || h ( y ) || ≤ | c ∗ ( y ) | for all y ∈ cone ( B ) , the cone of B over the origin. It then follows that h ( B ) ⊆ B since B = cone ( B ) ∩ c ∗ ( − ) . First weprove the claim in the special case when y = v is a vertex of B . Since the inequalitiesdefining C have integral coefficients, v has rational coordinates. Let t > x = t · v is an integral point. Since B ⊂ NS n + ( T ) , x is non-negative.Since v is admissible, x is admissible. By Proposition 8 there is a transversely oriented(unbranched) normal immersion ( f , n f ) : S → M such that x n ( f , n f ) = x , and Propo-sition 4 gives h ( x ) = f ∗ ([ S ]) ∈ H ( M ) . Since S is (unbranched) immersed, Lemma 3yields c ∗ ( x ) = c ( S ) . If S is not connected, then x is the sum over all components S i ⊂ S of x ( S i ) ∈ cone ( C ) . Since v is an extreme point of C it is not a non-trivialconvex combination of points in C . However x n ( f , n s ) = (cid:229) i x n ( f | S i , n f | S i ) . Thus x n ( f , n s ) is a multiple of x n ( f | S i , n f | S i ) for every component S i of S . By minimal-ity of t it follows that S is connected. Now c ∗ ( x ) = t c ∗ ( v ) < S is not a sphere,hence − c − ( S ) = c ( S ) = c ∗ ( x ) . Gabai [2] showed that the embedded norm equals thesingular norm, and so || h ( x ) || ≤ c − ( S ) = | c ( S ) | = | c ∗ ( x ) | . This proves the claim inthe special case.To prove the claim in the general case, assume y = (cid:229) l i v i with l i ≥ , where each v i is a vertex of B . Then h ( y ) = (cid:229) l i h ( v i ) so || h ( y ) || ≤ (cid:229) l i || h ( v i ) || . Now c ∗ ( y ) = c ∗ ( (cid:229) l i v i ) = (cid:229) l i c ∗ ( v i ) . Since c ∗ ( v i ) < | c ∗ ( y ) | = (cid:229) l i | c ∗ ( v i ) | . By the spe-cial case, || h ( v i ) || ≤ | c ∗ ( v i ) | , and it follows that || h ( y ) || ≤ | c ∗ ( y ) | , proving the claim.11o prove the reverse containment, a rational point in b ∈ ¶ B can be expressed as b = [ S ] / | c ( S ) | for some norm-minimising, transversely oriented embedded surface ( S , n S ) no component of which is a sphere or a torus. By Proposition 1 S can beisotoped into normal position. The ray through p = x n ( S , n S ) intersects C at a point x . Since p is admissible, x is admissible. Lemma 3 gives | c ∗ ( p ) | = | c ( S ) | = || [ S ] || . Since S is embedded and aspherical, Proposition 10 implies that p is algebraicallyaspherical if T is 0–efficient. Otherwise, it follows from [7], Lemma 2.1, that ( S , n S ) may be chosen to be least weight taut in which case Proposition 12 implies that p is algebraically aspherical. Hence x is algebraically aspherical by Lemma 9(3). Wecan express x as a convex linear combination of some of the vertices of C . Then x = (cid:229) t i · x i , where each x i is a vertex of C and 0 < t i ≤ . Parts (2) and (3) of Lemma 9imply that each x i is algebraically aspherical, and therefore by Lemma 9(1) c ∗ ( x i ) ≤ . Moreover, x i ≤ x therefore x i is admissible. If c ∗ ( x i ) = , then the smallest integralmultiple of x i is the normal coordinate of a connected, immersed surface of zero Eulercharacteristic, giving h ( x i ) = M is atoroidal.Let ˆ x = (cid:229) t i · ˆ x i , where ˆ x i = x i if c ∗ ( x i ) < x i = h ( ˆ x ) = h ( x ) , c ∗ ( x ) = c ∗ ( ˆ x ) and each non-zero ˆ x i is in cone ( B ) . Since cone ( B ) is convex,ˆ x ∈ cone ( B ) . It follows that ˆ x / | c ∗ ( ˆ x ) | ∈ cone ( B ) ∩ c ∗ ( − ) = B . Now [ S ] = h ( p ) and h ( ˆ x ) / | c ∗ ( ˆ x ) | = h ( x ) / | c ∗ ( x ) | = h ( p ) / | c ∗ ( p ) | = [ S ] / | c ( S ) | = b ∈ ¶ B . Hence h ( B ) contains all the rational points in ¶ B . The set of rational points in ¶ B is dense because the Thurston norm takes integral values on integral points. Thus ¶ B ⊆ h ( B ) . Since h is linear and B is convex, B ⊆ h ( B ) . Each vertex of C = D t − ∩ NS n ( T ) lies in a unique minimal sub-simplex of D . Thesub-simplices of D t − = V ∩ [ , ¥ ) t correspond to coordinate subspaces of R t . Thus a vertex of ¶ C is uniquely determined by which normal coordinates are zero.There are 14 t coordinates, so C has at most 2 t vertices. Every vertex of the unit ballof the Thurston norm ball, B , is the image of a point on a ray through a vertex of C . This gives the claimed bound on the number of vertices of B . Proof of Algorithm 6
Only part (3b) needs to be explained. Let B ≤ be the convexhull of the finite set of all points v | c ∗ ( v ) | , where v is a vertex of C which is admissibleand satisfies c ∗ ( v ) ≤ . Define Q = cone ( B ≤ ) ∩ { x : h ( x ) = a } ∩ { x : c ∗ ( x ) = −|| a ||} . The total weight of a point in Q is the sum of its coordinates. Given an integer w > P ( w ) be the finite (possibly empty) set consisting of all admissible integralpoints in Q of total weight at most w . The algorithm to construct S is to increase w until one of the points in P ( w ) is found, by the following procedure, to be the12oordinate of an embedded transversely oriented normal surface without sphere ortorus components. Given x n ∈ P ( w ) , construct the unique embedded normal surface F with un-oriented normal coordinate x = j ( x n ) . Discard x n if some component of F is a torus, a sphere or 1–sided. Otherwise check whether the components of F can betransversely oriented to yield an embedded transversely oriented normal surface withcoordinate x . We now prove this algorithm terminates. Let F be a norm-minimising oriented surfacewithout sphere or torus components and with [ F ] = a ∈ H ( M ) , so || a || = − c ( F ) . By Proposition 1 we may assume F is a transversely oriented normal surface andset p = x n ( F , n F ) . Then p ∈ Q and p ∈ P ( w ) for w sufficiently large. Hence thealgorithm will construct F or a surface with the same properties of lower weight. Remark 13 If M is a closed, oriented 3–manifold, then h ( B ≤ ) is the unit ball of theThurston norm — this is a non-compact polytope if M is not atoroidal. References [1] Daryl Cooper and Stephan Tillmann:
Transversely Oriented Normal Surfaces , inpreparation.[2] David Gabai:
Foliations and the Topology of 3–Manifolds,
Journal of Differential Ge-ometry (1983), pp. 445-503.[3] William Jaco and J. Hyam Rubinstein: , Jour-nal of Differential Geometry (2003), no. 1, 61–168.[4] Sergei Matveev: Algorithmic topology and classification of 3-manifolds.
Algorithmsand Computation in Mathematics, 9. Springer-Verlag, Berlin, 2003.[5] Saul Schleimer:
Almost normal Heegaard splittings , PhD thesis, University of Califor-nia at Berkeley, PhD 2001.[6] William P. Thurston:
A norm for the homology of 3–manifolds , Mem. Amer. Math.Soc. 59 (1986), no. 339, i–vi and 99–130.[7] Jeffrey L. Tollefson and Ningyi Wang:
Taut normal surfaces , Topology 35 (1996),55–75.DepartmentofMathematics,UniversityofCaliforniaSantaBarbara,CA93106,USADepartmentofMathematicsandStatistics,TheUniversityofMelbourne,VIC3010,AustraliaEmail: [email protected] [email protected]@math.ucsb.edu [email protected]