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Cyclic surgery, degrees of maps of character curves, andvolume rigidity for hyperbolic manifolds.

Nathan M. Dunfield ⋆

Department of Mathematics, University of Chicago, 5734 S. University Ave., Chicago, IL 60637, USA.email: [email protected]

January 2, 2018

1 Introduction

This paper proves the following conjecture of Boyer and Zhang: If a small hyperbolicknot in a homotopy sphere has a non-trivial cyclic surgery sloper, then it has an incom-pressible surface with non-integer boundary slope strictly betweenr − 1 andr + 1. Istate this result below as Theorem 4.1, after giving the background needed to understandit. Corollary 1.1 of Theorem 4.1 is that any small knot which has only integer bound-ary slopes has Property P. The proof of Theorem 4.1 also gives information about thediameter of the set of boundary slopes of a hyperbolic knot.

The proof of Theorem 4.1 uses a new theorem about the PSL2C character varietyof the exterior,M , of the knot. This result, which should be of independent interest,is given below as Theorem 3.1. It says that for certain components of the charactervariety ofM , the map on character varieties induced by∂M → M is a birationalisomorphism onto its image. The proof of Theorem 3.1 depends on a fancy versionof Mostow rigidity due to Gromov, Thurston, and Goldman. The connection betweenTheorem 3.1 and Theorem 4.1 is the techniques introduced by Culler and Shalen whichconnect the topology ofM with its PSL2C character variety.

I will begin with the background needed for Theorem 4.1. LetK be aknot in acompact, closed, 3-manifoldΣ, that is, a tame embeddingS1 → Σ. The exterior ofK, M , isΣ minus an open regular neighborhood ofK. SoM is a compact 3-manifoldwhose boundary is a torus. Supposeγ is a simple closed curve in∂M . We can createa closed manifoldMγ from M by taking a solid torus and gluing its boundary to∂Min such a way thatγ bounds a disc in the solid torus (Mγ depends only on the isotopyclass ofγ). The new manifoldMγ is called aDehn fillingof M or aDehn surgeryon

⋆ This work was partially supported by an NSF Graduate Fellowship.

http://arxiv.org/abs/math/9802022v3

2 Nathan M. Dunfield

K. Recently, many people have studied what kinds of manifolds arise when you do this,especially in the case whereK is a knot inS3 (see the surveys [Gor,Lue]).

From now on, letΣ be a homotopy sphere. An interesting question is: Are thereany non-trivial Dehn surgeries onK such that the resulting manifold is a homotopysphere? Or more generally, where the resulting manifold has cyclic fundamentalgroup?A knot is said to haveProperty Pif no non-trivial Dehn surgery on it yields a homotopysphere. By the Cyclic Surgery Theorem [CGLS], for fixedK there is at most oneγother than the meridian for whichMγ is a homotopy sphere. Gordon and Luecke [GL]have shown that non-trivial Dehn surgery on a knot inS3 never yieldsS3, and so thePoincare Conjecture would imply that every knot has Property P. On the other hand, thereare plenty of examples of knots with non-trivial cyclic surgeries. Fintushel and Stern[FS] discovered that the(−2, 3, 7) pretzel knot has two non-trivial surgeries where theresulting manifold is a lens space (see also [BMZ]). More examples of knots with non-trivial cyclic surgeries are given in [Lue,BZ2]. This famed pretzel knot was also the firstexample found to have a non-integer boundary slope (I will define what boundary slopesare in a moment). Boyer and Zhang conjectured that there was a general connectionbetween cyclic surgeries and non-integer boundary slopes.

A properly embedded surface(F, ∂F ) → (M,∂M) will be called incompressibleif π1(F ) → π1(M) is injective and not a 2-sphere bounding a ball. An incompressiblesurfaceF which is not boundary parallel is calledessential. Isotopy classes of orientedsimple closed curves in the torus∂M are in bijective correspondence with primitiveelements ofH1(∂M,Z); unoriented isotopy classes with pairs of primitive elements{γ,−γ}. If (α, β) is a basis forH1(∂M,Z), theslopeof γ = aα+bβ with respect to thisbasis isab ∈ Q∪{∞}. Note that the slope of an unoriented isotopy class is well defined,and gives a bijection between unoriented isotopy classes of simple closed curves andQ∪{∞}. If ∂F is non-empty, it consists of disjoint simple closed curves in∂M . Thesecurves must all be parallel, and so correspond to the same pair of primitive elements{γ,−γ} in H1(∂M,Z). These are called theboundary classesof F , and their slopetheboundary slopeof F . Hatcher [Hat] has shown that the number of boundary slopesis always finite. AsM is the exterior of a knot in a homotopy sphere, fix a meridian-longitude basis(µ, λ), forH1(∂M); all slopes will be with respect to this basis. A slopeis integral if it is inZ. Many knots have only integral boundary slopes, e.g. all 2-bridgeknots [HT]. If the Dehn fillingMγ has cyclic fundamental group, then the slope ofγ iscalled acyclic surgery slope. A manifold issmall if it does not contain aclosed, non-boundary parallel, incompressible surface. A knot is small if its exterior is.A knot ishyperbolicif the interior ofM admits a complete hyperbolic metric of finite volume.Finally, here is the theorem which was conjectured by Boyer and Zhang in [BZ2],andimproves their partial results in this direction:

Theorem 4.1SupposeK is a small hyperbolic knot in a homotopy sphere which has anon-trivial cyclic surgery sloper. Then there is an essential surface in the exterior ofKwhose boundary slope is non-integral and lies in(r − 1, r + 1).

Property P, degrees of maps of character curves, and volume rigidity 3

Note that the Cyclic Surgery Theorem [CGLS] shows that a cyclic slope for a knot ina homotopy sphere is always integral. The(−2, 3, 7) pretzel satisfies the hypotheses ofthis theorem, which explains why it has a non-integral boundary slope. Theorem 4.1remains true for some knot exteriors in manifolds with non-trivial cyclic fundamentalgroup, see Section 4. A corollary of the theorem is a condition which implies that a knothas Property P:

Corollary 1.1 If a small knot in a homotopy sphere has only integral boundary slopes,then it has Property P.

This follows as a small non-hyperbolic knot is a torus knot, and surgeries on these havebeen classified [Mos].

I should mention that there are small hyperbolic knots which have non-integral bound-ary slopes but no cyclic surgeries. Boyer and Zhang gave examples in [BZ2], and onecanalso use the proof of Proposition 2.2 of [HO] to give examples of small Montesinos knotswith non-integral boundary slopes of absolute value less than 2. So it is not possible touse Corollary 1.1 to show that all small knots have Property P.

The ideas of the proof of Theorem 4.1 can also be used to give information about thediameter,d, of the set of boundary slopes. In [CS3], Culler and Shalen showed that forany knot the diameterd ≥ 2. For a hyperbolic knot, I show that ifd = 2 then the greatestand least slopes are not integral (see Section 5 for a detailed statement).

Let me change course and state the theorem about character varieties that underliesthe proof of Theorem 4.1, and which I hope will be of use in other situations. LetMbe a finite-volume hyperbolic 3-manifold with one cusp. LetX(M) denote the PSL2Ccharacter variety ofM . Basically,X(M) is the set of representations ofπ1(M) intoPSL2C mod conjugacy, and is an algebraic variety overC (for details, see Section 2).Culler and Shalen introduced a way of getting essential surfaces fromX(M), which hasbeen very useful in proving theorems about Dehn surgery. I will postpone explaininghow this works until the next section, but this is what connects the next theorem with thepreceding. LetX0 denote an irreducible component ofX(M) which contains the con-jugacy class of a discrete faithful representation. SinceM has one cusp, the (complex)dimension ofX0 is 1. The inclusioni: ∂M → M induces a map on character varietiesi∗: X0 → X(∂M). I will prove:

Theorem 3.1The mapi∗: X0 → X(∂M) is a birational isomorphism onto its image.

The conclusion of this theorem doesnot always hold for other components ofX(M).The key to the proof of Theorem 3.1 is:

Volume Rigidity Theorem 6.1 (Gromov-Thurston-Goldman) SupposeN is a com-pact, closed, hyperbolic 3-manifold. Ifρ: π1(N) → PSL2C is a representation withvol(ρ) = vol(N), thenρ is discrete and faithful.

If N is a closed manifold, the volume of a representationρ: π1(N) → PSL2C is definedas follows. Choose any smooth equivariant mapf : N → H3. The formf∗(VolH3) de-scends to a form onN . The volumevol(ρ) is the absolute value of the integral of this


form overN .The volume is independent off as any two such maps are equivariantlyhomotopic (see Section 2.5). Goldman [Gol] noticed that one can prove Theorem 6.1 inessentially the same way as the Strict Version of Gromov’s Theorem givenby Thurstonin his lecture notes [Thu]. As this section of Thurston’s notes remains unpublished, Iwill include a proof of Theorem 6.1. (Theorem 6.1 is now known to hold for all con-nected semisimple Lie groups, where in the definition of volumeH3 is replaced by theappropriate symmetric space. The cases not done in [Gol] are covered by [Cor1]and[Cor2]).

The Volume Rigidity Theorem is connected to Theorem 3.1 by a theorem which is areinterpretation of some of the results of [CCGLS]. It shows, roughly, that thevolumeof a representationρ depends only on the image of its conjugacy class under the mapi∗: X(M) → X(∂M) (for a precise statement, see Section 2.6).

Let me end the introduction with an outline of the proof of Theorem 4.1. The proofgoes by contradiction, and so suppose thatK has a cyclic surgery slope but no non-integer boundary slopes. A simple algebraic argument determines exactly what the im-age ofi∗: X0 → X(∂M) is. Theorem 3.1 says thatX0 is essentially the same asi∗(X0),and soX0 is now known. Reading off fromX0 information about the number of bound-ary components of a certain essential surface contained inM leads to a contradiction.

The rest of this paper is organized as follows: In Section 2 I will review thefactsabout character varieties that I will need later, and also prove some needed lemmas.Sections 3 and 4 are devoted to the proofs of Theorems 3.1 and 4.1 respectively. Section5 discusses the proof of Theorem 4.1 in the context of the norm introduced in [CGLS],and gives an application to the question of the diameter of the set of boundary slopes.Finally, Section 6 gives a proof of the Volume Rigidity Theorem.

I would like to thank Peter Shalen for his encouragement and innumerable usefulconversations, as well as telling me about Proposition 2.3. I thank Andrew Przeworskiand Marc Culler for their comments on an earlier draft of this paper. I also thank thereferee for very useful comments and suggestions, and especially for pointing out anerror in the original statement of Theorem 5.3.

2 Character varieties

2.1 Basics

In this section, I will review facts about character varieties that will be needed later. Thebasic references are the first chapter of [CGLS] and [CS1], as well as the expositoryarticle [Sha]. The case of PSL2C-, as opposed to SL2C-, character varieties, is treated in[BZ3].

If M is a topological space with finitely generated fundamental group, denote byR(M) the set of representations ofπ1(M) into PSL2C. This set has a natural struc-ture as an affine complex algebraic variety. Now PSL2C acts onR(M) by conjugationof representations. LetX(M) denote the quotient space (strictly speaking, the algebro-geometric quotient), andt: R(M) → X(M) the quotient map. The affine varietyX(M)


is called the character variety ofR(M), as two representations inR(M) map to thesame point ofX(M) if and only if they have the same character. If two representa-tions have the same character and one of them is irreducible, then they are conjugate.Moreover, for an irreducible componentX of X(M) which contains the character ofan irreducible representation, characters of reducible representations forma subvarietyof strictly smaller dimension. So usually just think ofX(M) as representations modconjugacy. A character is called discrete, faithful, or whatever if allrepresentations withthat character are discrete, faithful, or whatever. For eachγ in π1(M) there is a regularfunctionfγ onX(M) such that ifχ in X(M) is the character of a representationρ thenfγ(χ) = tr(ρ(γ))2 − 4. This is well defined as the trace of an element of PSL2C is de-fined up to sign. A map between spacesf : M → N gives a mapf∗: X(N) → X(M)via composition withf∗: π1(M) → π1(N). In particular,i: ∂M → M induces a mapi∗: X(M) → X(∂M).

We can also consider SL2C- rather than PSL2C- representations, and everythingworks the same way. LetR(M) denote the variety of representations ofπ1(M) intoSL2C, andX(M) the associated character variety. In this case, there are also regularfunctionstrγ on X(M) defined bytrγ(χ) = tr(ρ(γ)). The natural mapX(M) →X(M) is finite to 1, though it may not be onto, as not all representations into PSL2C liftto ones into SL2C.

WhenM is a finite-volume hyperbolic manifold with one cusp, I will denote byX0

a component ofX(M) which contains the character of a discrete faithful representation(noteX0 may not be unique, see Section 2.7). By [Cul], a torsion free discrete faithfulrepresentation lifts to SL2C and there is a componentX0 of X(M) which coversX0.Said another way, every representation inX0 lifts to one into SL2(C).

2.2 Associated actions on trees

Culler and Shalen discovered a way to construct essential surfaces fromX(M) or X(M).In this and the next section, I will explain their method for an irreducible curveX inX(M) which contains the character of an irreducible representation. LetY be a smoothprojective model forX andp: Y → X the associated birational isomorphism. The fi-nite set of pointsY \ p−1(X) are called ideal points. Culler and Shalen showed how toassociate to each ideal point an action ofπ1(M) on a simplicial tree as follows: Choosea curveR ⊂ R(M) so that the Zariski closure oft(R) isX (I will say more about howto chooseR later in Lemma 2.2). IfC(X) denotes the field of rational functions onX,thent induces an inclusion of fieldsC(X) → C(R). The pointy determines a valuationv on C(X) wherev(f) is the order off at y (zeros getting positive valuation). Thereis a valuationw onC(R) which extendsv in the sense that there is a positive integerdso thatw(f) = d · v(f) for all f ∈ C(X). Associated to the fieldC(R) and the valua-tion w is its Bruhat-Tits tree, which I will denoteTy. This is a simplicial tree on whichSL2C(R) acts by isometeries. The translation length of an elementA of SL2C(R) act-ing onTy is min(0,−2w(tr(A))) (Proposition II.3.15 of [MS]). There is a tautological


representationP : π1(M) → SL2C[R], whereC[R] ⊂ C(R) is the ring of regular func-tions, defined byP (γ) = A whereA is the matrix of regular functions so that at anyrepresentationρ ∈ R , A(ρ) = ρ(γ). ComposingP and the action of SL2C(R) on Tygives the promised action ofπ1(M) on a tree.

In order to prove Theorem 4.1 it will be necessary to know the relationship betweenthe valuationsv andw exactly. I will show:

Proposition 2.2For each ideal pointy in Y there is a simplicial treeTy on whichπ1(M)acts so that the translation length ofγ ∈ π1(M) acting onTy is twice the degree of poleof trγ at y.

Because of the formula for translation length above, to prove Proposition 2.2 it suf-fices to show we can chooseR so thatd = 1 (you may wish to skip ahead until thisbecomes crucial in the proof of Theorem 4.1). LetZ be a smooth projective model forR,andt: Z → Y be the map induced byt: R→ X. The mapt is a holomorphic branchedcovering of Riemann surfaces. If there is az ∈ t−1(y) such thatt is not branched atz,takew to be the valuation determined by order atz, and thend = 1. Thus Proposition 2.2will be proved by:

Lemma 2.2Lety be a fixed ideal point ofX. Then there is a curveR ⊂ R(M) such thatthe induced mapt: Z → Y has the property that for anyz in t−1(y), t is not branchedat z.

Proof By Corollary 1.4.5 of [CS1] there is anα ∈ π1(M) so thattrα has a pole aty.Sincetrα is non-constant there is an irreducible characterχ in X such thattrα(χ) 6=2. Let ρ be a representation with characterχ. By Proposition 1.5.1 of [CS1] there isa β ∈ π1(M) such thatρ restricted to the subgroup〈α, β〉 generated byα andβ isirreducible, or equivalentlytr[α,β](ρ) 6= 2. Considering the regular functiontr[α,β], wesee that all but finitely many charactersχ ∈ X are irreducible when restricted to〈α, β〉.There are two cases, depending on whethertrβ is constant or not. Supposetrβ is non-constant. Then there is a representationρ0 whose character is inX so thatρ0 restrictedto 〈α, β〉 is irreducible andρ0(α) andρ0(β) are hyperbolic elements of SL2C. Thenρ0restricted to〈α, βnαβ−n〉 is irreducible for largen. This is because the fixed point setof ρ0(βnαβ−n) is ρ0(βn)(fix(α)), which is disjoint fromfix(α) for largen. So there isa conjugateγ of α such thatρ0 restricted to〈α, γ〉 is irreducible. Again, all but finitelymany characters ofX are irreducible when restricted to〈α, γ〉. Let V be the subvarietyof R(M) consisting of representationsρ such that

ρ(α) =

(a 10 1/a

), and ρ(γ) =

(a 0s 1/a

), (1)

for somea, s ∈ C, with a 6= 0. Any representationρwhich is irreducible when restrictedto 〈α, γ〉 and whereρ(α) is not parabolic is conjugate to exactly two representations inV . For a conjugate of such aρ lying in V , there are two choices fora, and oncea isfixed, s is determined bytr(ρ(αγ)). Any two conjugates ofρ which agree on〈α, γ〉


are actually equal, as the stabilizer under conjugation of any irreducible representationis {I,−I}. Thus all but finitely many points ofX are in t(V ), so we can choose anirreducible curveR ⊂ V so thatt(R) is dense inX. The mapt: R → X is genericallyeither 1-to-1 or 2-to-1. In the former case, we are done. In the latter caseR = V andit is enough to show thatt−1(y) ⊂ Z consists of two points. Fixz in t−1(y). We candefine a regular functiona: R → C by Eqn. (1). Sincetrα has a pole aty, a must haveeither a pole or a zero at any point oft−1(y). Choose a sequence of representationsρj

converging toy. From Eqn. (1), we haveρj(α) =

(aj 1

0 1/aj

). Since generically a point

of X has two inverse images inR, for all but finitely manyj there are representations

ψj ∈ R which are conjugate to, but not equal to,ρj . Henceψj(α) =

(1/aj 1

0 aj

)because

as mentioned above, conjugates which agree on〈α, γ〉 are equal. Then theψj convergeto a pointz′ in t−1(y). But z is notz′ since ifa has a zero atz thena has a pole atz′ andvice versa. Sot−1(y) consists of two points and we are done.

If trβ is constant, fixλ ∈ C so thatλ + 1/λ = trβ. DefineV to be the subset ofRgiven by:

ρ(α) =

(a s0 1/a

), and ρ(β) =

(λ 01 1/λ

)

wherea, s ∈ C and proceed as before. ✷

2.3 Associated surfaces

There is a dual surface to any action ofπ1(M) on a treeT as follows. LetM be theuniversal cover ofM . Choose an equivariant mapf : M → T which is transverse to themidpoints of the edges ofT . If S is the inverse image underf of the midpoints of theedges ofT , thenS is an equivariant family of surfaces inM which descends to a surfaceS in M . In Section 1.3 of [CGLS] it is shown how any suchf can be modified so thatSbecomes essential. Ify is an ideal point, this construction gives an essential surface dualto the action onTy. This surface,S, is said to be associated toy. Note thatS need notbe connected or unique up to isotopy.

SupposeM has torus boundary, and∂M is essential inM . Supposey is an idealpoint where for some peripheral elementγ ∈ π1(∂M), trγ has a pole (there may be idealpoints where this does not happen). In this case, there is exactly one slope{α,−α} suchthattrα is finite aty (again, see [CGLS] for details). Thenα fixes a point ofTy. If S is anessential surface associated toTy, then some component ofS has non-empty boundarywith boundary classes{α,−α}. Let β be such that(α, β) is a basis forπ1(∂M). I willneed:

Proposition 2.3 (Culler and Shalen)The surface associated to an ideal pointy can bechosen so that the number of boundary components is equal to twice the order of pole oftrβ at y.


Proof This is essentially part of Section 5.6 of [CCGLS]. By Proposition 2.2, twicethe order of pole oftrβ at y is the same as the translation length,l(β), of β acting onTy. First, I claim |∂S| ≥ l(β), where|∂S| is the number of components of∂S. Letp: M → M be the covering map. Pick an arca in M which is a lift of a loop inMrepresentingβ and which intersectsp−1(S) in |∂S| points. Hence the image ofa in Tyintersects the midpoints of|∂S| edges ofTy. One of the endpoints of the image ofa inTy is sent to the other under the action ofβ. Therefore the translation length ofβ actingonTy is at most|∂S|.

Now I will produce a surface associated toy with at mostl(β) boundary components.Choose a connected component ofp−1(∂M), C, which we identify withR2 so thatαacts onC by unit translation in the first coordinate, andβ acts by unit translation inthe second coordinate. The abelian subgroupπ1(∂M) leaves invariant a unique lineLin Ty, and acts onL via translations. An elementγ = aα + bβ ∈ π1(∂M) acts onL by translation byb · l(β). Choose a mapf : C → L which is equivariant under theaction ofπ1(∂M) as follows. LetY be the second coordinate axis. First, project ontothe second coordinate to get a map fromC to Y . Then compose this with a linear mapfrom Y to L that expands by a factor ofl(β). There is a unique extension off to all ofp−1(∂M) which is equivariant underπ1(M). SinceTy is contractible, we can extendfto an equivariant map of all ofM to Ty. The dual surfaceS has|∂S| = l(β). Changingf so thatS becomes essential does not increase the number of boundary components,so there is a surface associated toy with |∂S| ≤ l(β). As we also have|∂S| ≥ l(β), Smust have exactlyl(β) boundary components. ✷

2.4 Associated plane curves

The authors of [CCGLS] introduced a plane curve associated to the character variety.This gives a nice set of coordinates for the computations in the proof of Theorem 4.1.Let M be a compact 3-manifold with torus boundary. Let∆ be the set of diagonalrepresentations ofπ1(∂M) into SL2C. For anyγ ∈ π1(∂M) there is a well definedeigenvalue functionξγ : ∆ → C∗ which takes a representationρ to the upper left handentry ofρ(γ). Fixing a basis(α, β) for π1(∂M), the pair of eigenvalue functions(ξα, ξβ)gives coordinates on∆, and allows us to identify it withC∗ ×C∗. Said another way, wecan identify a point(a, b) ∈ C∗ × C∗ with the representationρ such that:

ρ(α) =

(a 00 1/a

), and ρ(γ) =

(b 00 1/b

).

I will say that(a, b) = (ξα, ξβ) are the eigenvalue coordinates corresponding to the basis(α, β).

There is a natural mapt: ∆ → X(∂M) which is onto and generically 2-to-1. IfY is a one-dimensional subvariety ofX(∂M), we can take the closure oft−1(Y ) ⊂∆ ⊂ CP2 to get a plane curveD which is a double cover ofY . If X is an irreduciblecomponent ofX(M) such thati∗(X) ⊂ X(∂M) is one-dimensional, we can associate


the plane curveD(X) ≡ t−1(i∗(X)) to X. The union ofD(X) over all componentsXof X(M) with i∗(X) one-dimensional is called the associated plane curve and denotedDM . Note for a componentX0 of X(M) which contains the character of a discretefaithful representation,i∗(X0) is one-dimensional by Proposition 1.1.1 of [CGLS].

There is an alternate construction ofDM given in [CL] that we will need for the nextsection. LetRU (M) be the subset ofR(M) consisting of representations whose restric-tion to the subgroupπ1(∂M) is upper-triangular. Then there is a mapi∗: RU (M) → ∆which sendsρ to the pair consisting of the upper left hand entries ofρ(α) andρ(β). It isnot hard to see that the union of thei∗(R) over all componentsR ⊂ RU (M) with i∗(R)one-dimensional is exactlyDM .

2.5 Volume of a representation

Let N be a closed 3-manifold. The volume,vol(ρ), of a representationρ: π1(N) →PSL2C is defined as follows. Choose any smooth equivariant mapf : N → H3. Theform f∗(VolH3) on N descends to a form onN . The volumevol(ρ) is the absolutevalue of the integral of this form overN . Since any two such maps are equivariantlyhomotopic, the volume is independent off . If f andg are two such maps one can usethe straight line homotopyH defined byH(p, t) = tf(p)+(1−t)g(p), where this linearcombination is along the geodesic joiningf(p) to g(p).

The purpose of this section is to define the volume of a representation for a compact3-manifoldM whose boundary is a torus. The definition given in the closed case doesnot work here. As∂M is non-empty, the value of the integral used to define the volumedepends on the choice of equivariant map.

For convenience, I will assume throughout that the torus∂M is incompressible inM .Let M be the universal cover ofM with π: M → M the covering map. LetM denotethe quotient space ofM obtained by collapsing each component plane ofπ−1(∂M)to a point. The points ofM coming from the collapsed components ofπ−1(∂M) willbe denoted∂M ; the setM \ ∂M will be denotedint(M). The action ofπ1(M) on Minduces an action onM . This action is free onint(M), but each point of∂M is stabilizedby some peripheral subgroup ofπ1(M). Let H3 denote the union ofH3 with the sphereat infinityS2

∞. The idea for defining the volume of a representationρ: π1(M) → PSL2Cis to consider equivariant mapsf : M → H3 which send∂M into S2

∞ and sendint(M)intoH3, and then proceed as in the closed case.

Before I can define the precise types of maps to be considered, I need to define somenotation. LetT be a torus and fix a product structureT × [0,∞] on a neighborhood of∂M such that∂M = T ×{∞} (the reason for this choice of closed interval will becomeclear in a moment). This gives an equivariant product structure onπ−1(T × [0,∞]) inM . This product structure induces an equivariant cone structure on a neighborhood of∂M in M . If v is in ∂M , I will denote the component of this neighborhood containingvbyNv, and the cone structure onNv by asPv × [0,∞], wherePv is a plane that covers


T ×{0} andPv×{∞} is really just the cone pointv. The peripheral subgroup ofπ1(M)which fixesv will be denotedStab(v). Note thatStab(v) preservesNv.

Let ρ: π1(M) → PSL2C be a representation. Apseudo-developing mapfor ρ isa smooth equivariant mapf : M → H3 which sends∂M into S2

∞ andint(M) into H3

and which satisfies the following additional condition. For eachv in ∂M , I require thatfmaps each ray{p}×[0,∞] in the cone neighborhoodNv to a geodesic ray, and thatf pa-rameterizes this ray by arc length with respect to cone parameter in[0,∞] (by continuity,this ray has endpointf(v) in S2

∞). Note thatf(v) must be a fixed point ofρ(Stab(v)).As an example, ifρ is the holonomy representation for a complete hyperbolic structureon int(M), then a developing map for this structure extends to a pseudo-developing mapof M by sending eachv in ∂M to the unique point inS2

∞ fixed byρ(Stab(v)). Anotherexample is if we have a decomposition ofM into ideal tetrahedra; an equivariant mapwhich is piecewise straight with respect to this triangulation is a pseudo-developing map.Everyρ has a pseudo-developing map. To construct one, pick av in ∂M and definef(v)to be some point fixed byρ(Stab(v)). Extendf toNv by picking anyStab(v) equivari-ant map ofP ×{0} intoH3 and then extending geodesically along the cone structure ofNv. There is a unique equivariant extension off to the cone neighborhood of∂M . AsH3 is contractible,f extends to a pseudo-developing map forρ.

I will define the volume of a pseudo-developing mapf in the same way as in theclosed case. The pull backf∗(VolH3) descends to a form onM . The absolute value ofthe integral of this form overM is defined to bevol(f). To see that this integral is welldefined, pick av ∈ ∂M and choose a fundamental domainF = D×[0,∞] for the actionof Stab(v) onNv. We need that the integral off∗(VolH3) overF is defined and finite.Pick a horoball inH3 centered atf(v) which containsf(F ). Projectf(D × {0}) out tothe corresponding horosphereS along geodesic rays with limitf(v). Call this projectionθ. Let AreaS be the area form onS. Let A =

∫D×{0} |θ∗AreaS |, the total unoriented

area ofθ(D × {0}). As f restricted to{p} × [0,∞] is a geodesic parameterized byarc length, the same computation as computing the volume of a cusp shows that theintegral of |f∗(VolH3)| overF is bounded byA/2. Thus the integral of the absolutevalue off∗(Vol3H) overM is finite, and sovol(M) is well defined. Note that for the twoexamples of pseudo-developing maps given in the last paragraph, the volume of such amap is the volume that you would expect. It would also be possible to define the volumeof f without taking the absolute value of the given integral. All the lemmas in this sectionremain true with this altered definition; this fact will be needed in Section 2.6.

I would now like to say that the volume of a pseudo-developing map is independentof the choice of map using the same argument as in the closed case. However, thereis a problem. Supposef is a pseudo-developing map for a representationρ such thatρ(π1(∂M)) has exactly two distinct fixed points. Consider av ∈ ∂M . Let p and qin S2

∞ be the two fixed points ofρ(Stab(v)). Thenf(v) is eitherp or q, sayp. Wecan construct a pseudo-developing mapg for ρ with g(v) = q. Thenf andg are nothomotopic through equivariant maps fromM to H3 which send∂M intoS2

∞. I see no apriori way to compare the volumes of these two maps. However, we have:


Lemma 2.5.1If f and g are two pseudo-developing maps forρ which agree on∂M ,thenvol(f) = vol(g).

Proof Construct a homotopyH: M × [0, 1] → H3 betweenh andg through pseudo-developing maps as follows. Letv be in∂M . Consider the coneNv = Pv×[0,∞]. Beginto constructH by settingH(v, t) = f(v) = g(v) for all t. ExtendH overPv × {0}by anyStab(v) equivariant homotopy between the restrictions off andg to Pv × {0}.As before, cone along geodesic rays ending atf(v) to extendH overNv. We can nowextendH to the desired homotopy.

To see thatvol(f) = vol(g) considerMt = M \ T × (t,∞], whereT × [0,∞]is our collar on∂M . For t ∈ [0,∞], defineVt(f) =

∫Mtf∗(VolH3) and letVt(g) be

the corresponding quantity forg. Note|Vf (∞)| = vol(f) and similarly forg. The formVolH3 is closed, so by Stokes Theorem:

0 =

∫

Mt×[0,1]dH∗(VolH3) = Vf (t)− Vg(t) +

∫

(∂Mt)×[0,1]H∗(VolH3)

BecauseH was constructed by geodesically coning over the restriction ofH to Pv ×{0}, the integral

∫(∂Mt)×[0,1]H

∗(VolH3) goes to zero exponentially witht. Thereforevol(f) = vol(g). ✷

If ρ is a representation such that the volume of any two pseudo-developing mapsagree, I will say that the volume ofρ is defined and setvol(ρ) = vol(f) for anypseudo-developing mapf . If ρ is a representation whereρ(π1(∂M)) contains a non-trivial parabolic, thenρ(π1(∂M)) has a unique fixed point inS2

∞. By the lemma, anytwo pseudo-developing maps forρ have the same volume, and so the volume is definedfor suchρ. I will show:

Lemma 2.5.2If ρ is a representation whose character lies in an irreducible componentofX(M) which contains the character of a discrete faithful representation thenvol(ρ)is defined.

which follows immediately from:

Lemma 2.5.3Supposeρt, with t ∈ [0, 1] is a smooth one parameter family of represen-tations ofπ1(M) into PSL2C. If the volume ofρ0 is defined then so is the volume ofρ1.

Proof I will need the results of Section 4.5 of [CCGLS]. There, they construct a par-ticularly nice form of pseudo-developing map as follows. LetN denoteM with ∂Mcollapsed to a point. DecomposeN as a simplicial complex so that each simplex has atmost one vertex at the collapsed∂M . This induces an equivariant simplicial decomposi-tion of M . Pick a preferred vertexv∞ of ∂M and a fixed pointp of ρ(Stab(v∞)). For arepresentationρwe can construct a pseudo-developing mapf for ρ so thatf is piecewisestraight on the simplices ofM and which sendsv∞ to p. Simply pick an equivariant mapof the zero skeleton ofM sendingv∞ to p and extend linearly along the simplices. This


gives a pseudo-developing map because each simplex has at most one vertex in∂M andso the only points sent toS2

∞ are those in∂M .Pick two pseudo-developing mapsf1 andg1 for ρ1. We can build a smooth family

of mapsft, for t ∈ [0, 1] whereft is a pseudo-developing map forρt of the special typejust discussed. Letgt be a similar family of maps forg1. In Section 4.5 of [CCGLS] itis shown that work of Hodgson [Hod] implies that the derivative ofvol(ft) (or vol(gt))depends only on the restriction ofρt to π1(∂M). This implies that the derivatives ofvol(ft) and vol(gt) are the same. Asvol(f0) = vol(g0) = vol(ρ0), we must havevol(f1) = vol(g1). Sovol(ρ1) is well defined. ✷

For the proof of Theorem 3.1, I will need:

Lemma 2.5.4Supposeρ is a representation ofπ1(M) which factors through the fun-damental group of a Dehn fillingMγ of M . Then the volume ofρ with respect toM isdefined, and is equal to the volume ofρ with respect to the closed manifoldMγ .

Proof LetC be the solid torus added toM to makeMγ . LetMγ be the universal cover ofMγ , with φ the covering map. Pick an equivariant mapf from Mγ toH3. LetV = Mγ \φ−1(int(C)). LetW beM minus the open cone neighborhood of∂M . Adjusting collars,I will view V as a quotient ofW . The restriction off to V is aπ1(Mγ)-equivariant mapwhich induces aπ1(M)-equivariant mapF : W → H3. ExtendF over ∂M by anyπ1(M)-equivariant map. By coning along geodesics we can extendF over the coneneighborhood of∂M to a pseudo developing map forρ.

To compare volumes, choosef : Mγ → H3 so thatf(φ−1(C)) is one-dimensional(this is possible because there is map fromMγ to itself which is the identity outsidea neighborhood ofC and which collapsesC to a core curve inC). Thus the volumeof ρ with respect toMγ is the integral off∗(VolH3) overMγ \ C. The image of thecone neighborhood of∂M underF is at most two-dimensional, and sovol(f) is theintegral ofF ∗(VolH3) overM minus the collar on∂M . BecauseF is a lift of f thesetwo integrals have the same value. Thusvol(f) is equal to the volume ofρ with respecttoMγ .

Since the restriction ofF to ∂M was an arbitrary equivariant map, the volume ofρwith respect toM is defined and is equal to the volume ofρ with respect toMγ . ✷

2.6 Volume form

In this sectionM will be a hyperbolic 3-manifold with one cusp. In Section 2.5, I dis-cussed the volume of a representation ofπ1(M) into PSL2C. As volume is invariantunder conjugation, there is a mapvol from the irreducible characters ofX(M) to R+

wherevol(χ) is the volume of any representation with characterχ. The next proposi-tion is a reinterpretation of the results of Sections 4.4-4.5 of [CCGLS], and will be oneof the keys to the proof of Theorem 3.1. A normalization of a curveX is a smoothcurveY together with a regular birational mapf : Y → X. A normalization ofX can


be constructed by taking a smooth projective modelY ′ with birational isomorphismf : Y ′ → X and lettingY = f−1(X).

Theorem 2.6LetX0 be an irreducible component ofX(M) which contains the char-acter of a discrete faithful representation. LetY be a normalization ofi∗(X0) wherei∗: X(M) → X(∂M) is the map induced byi: ∂M → M . Then the mapvol on theirreducible characters ofX0 factors through a mapY → R+.

That is, there is a mapvol: Y → R+ such that the diagram

Yvol

uullllllllllllllllll

R+ X0vol

ooi∗

// i∗(X0)

f

OO

commutes. In fact,vol will be the absolute value of a generically smooth function fromY toR.

Proof I will use the notation of Section 2.4. In Section 4 of [CCGLS], the authors definea real-valued differential formη on∆ by

η = log |a| d arg(b)− log |b| d arg(a).

which measures the change in volume in the following sense. LetR0 denote the subsetof RU(M) which maps toX0. In the language of Section 2.5, we can define a mapV : R0 → R by settingV (ρ) to be the integral off∗(VolH3) overM wheref is anypseudo-developing map forρ. ThusV (ρ) = ±vol(ρ), and the results of Section 2.5show thatV is well defined. Work of Hodgson, see Section 4.5 of [CCGLS], showsthatV is smooth and thatdV is the pull-back of−1

2η alongi∗: R0 → DM . By eitherSection 4.4 or 4.5 of [CCGLS], the formη is exact onDM , in the sense that it is exact(where defined) on any smooth projective model ofDM . LetD′

M be subset ofDM onwhich η is defined. Therefore,V factors through a map from a normalization ofD′

M toR. More precisely, letDM be a normalization ofD′

M , andf : D′M → DM a birational

isomorphism. Then there is a smooth functionv: DM → R such thatv◦f ◦i∗ = V . Notethat the formη is invariant under the transformations which quotient∆ down toX(∂M)(these transformations are(a, b) 7→ (1/a, 1/b), (a, b) 7→ (−a, b), etc.). Thereforevdescends to a functionv′ on a normalization oft(D′

M ). As any character inX0 lifts toone in SL2C andη is defined on all of∆, the curvei∗(X0) is contained int(D′

M ). Sothe functionv′ is defined on a normalization ofi∗(X0). If χ ∈ X0 is the character of arepresentationρ ∈ R0, we have|v′(i∗(χ))| = |V (ρ)| = vol(ρ) = vol(χ). So|v′| is therequired function. ✷


2.7 Discrete faithful representations

It is important to remember that ifρ andρ′: π1(M) → PSL2C are holonomy repre-sentations of a one-cusped finite-volume hyperbolic 3-manifoldM , thenρ andρ′ neednot be conjugate in PSL2C. They are conjugate in O(3, 1) by Mostow rigidity, and it’snot difficult to see thatρ′ is conjugate in PSL2C to eitherρ or to the complex conjugateρ of ρ. By complex conjugate, I meanρ(γ) is the matrix whose entries are the com-plex conjugates of those ofρ(γ), for all γ ∈ π1(M). I will need the following littlelemma, which the reader may wish to skip until it is used in the proof of Theorem3.1.It is just an application of Lemma 6.1 of [CL], and describes the behavior of the mapi∗: X(M) → X(∂M) near the two discrete faithful characters. Letp be the point ofX(∂M) where the trace of any element ofπ1(∂M) is ±2. If χ is a discrete faithfulcharacter, theni∗(χ) = p. The lemma is:

Lemma 2.7Let M be a finite-volume hyperbolic 3-manifold with one cusp. Then thetwo discrete faithful characters have neighborhoods inX(M) whose images underi∗

are distinct branches ofi∗(X0) throughp.

Proof Let (α, β) be a basis ofπ1(∂M). If ρ0 is a discrete faithful representation inPSL2C we can conjugate it by an element of PSL2C so that

ρ0(α) = ±(1 10 1

)and ρ0(β) = ±

(1 a0 1

). (2)

The cusp shape ofM with respect to this basis isa (Eqn. (2) uniquely determinesa).The cusp shapea is always non-real, and, by changing the basis ofπ1(∂M) if necessary,we can assumea is not pure imaginary. Letχ0 ∈ X(M) be the character ofρ0. Asin Lemma 6.1 of [CL] it is not hard to show thatlimχ→χ0

fβ/fα = a2 (the key is tonote thatα andβ commute). From Eqn. (2) we see that the cusp shape ofρ0 is a. So,limχ→χ0

fβ/fα = a2 6= a2. Sincefα andfβ depend only on the image of a character inX(∂M), neighborhoods ofχ0 andχ0 must go to distinct branches ofi∗(X(M)) throughp. ✷

3 Degrees of maps of character curves

Theorem 3.1Let M be a finite-volume hyperbolic 3-manifold with one cusp. LetX0

be a component of thePSL2C character variety ofM which contains the character ofa discrete faithful representation. The inclusioni: ∂M → M induces a regular mapi∗: X0 → X(∂M). This map is a birational isomorphism onto its image.

Proof Fix a discrete faithful characterχdf in X0. By Proposition 1.1.1 of [CGLS],X0

has complex dimension 1, andi∗ is non-constant, soi∗(X0) also has dimension 1. AsX0

is irreducible, so isi∗(X0). The mapi∗: X0 → i∗(X0) is a regular map of irreduciblealgebraic curves, and so has a degree which is the number of points in(i∗)−1(p) for


genericp ∈ i∗(X0) (here, generic means except for a finite number of points ofi∗(X0)).A degree-1 map is always a birational isomorphism. Thus it suffices to show thatthereare infinitely many pointsp in i∗(X0) where(i∗)−1(p) consists of a single point.

We constructpj , j ∈ N, so that(i∗)−1(pj) consists of a single point as follows. ByThurston’s Hyperbolic Dehn Surgery Theorem, all but finitely many Dehn fillings ofMare hyperbolic (see [Thu] and [NZ], or [BP]). Choose an infinite sequence of distinctDehn fillings so that the resulting manifoldsMγ1 ,Mγ2 , . . . are all hyperbolic. Moreover,Thurston’s theorem says that we can choose theγj ’s so that the hyperbolic structuresof theMγj converge to the hyperbolic structure ofM . In particular, there are holonomycharactersχj of theMγj which converge toχdf. Additionally, we can assume for eachj that the core of the solid torus attached toM to formMγj is a geodesic inMγj . ByCorollary 3.28 in [Por],χdf is a smooth point ofX(M), and since theχj converge toχdf, infinitely manyχj lie in X0 (alternatively, for our purposes, we could just changeX0 if necessary).

By Proposition 2.6 the volume of a character depends only on its image inX(∂M),or more precisely in the smooth projective model ofi∗(X0). This is key to the proof.

Consider one of the holonomy charactersχj which comes from Dehn-fillingM alongthe curveγj in ∂M . From now on, consider the mapi∗: X0 → i∗(X0) as a map to asmooth projective model ofi∗(X0), though I will not change notation. Letpj = i∗(χj).Supposeχ is a point in(i∗)−1(pj) besidesχj . Now if ρj is a representation correspond-ing to χj thenρj(γj) = I. Let β be a curve in∂M which forms a basis withγj ofπ1(∂M). Thenβ is homotopic inMγj to the core of the solid torus attached toM toformMγj , and soβ is homotopic to a closed geodesic inMγj . Soρj(β) is a hyperbolicelement of PSL2C. In particular,tr(ρj(β)) 6= ±2. Now if ρ is a representation whosecharacter isχ thentr(ρ(γj)) = tr(ρj(γj)) = ±2 and tr(ρ(β)) = tr(ρj(β)) 6= ±2.Sinceρ(β) andρ(γj) commute, we must haveρ(γj) = I. Henceρ is also a representa-tion of π1(Mγj ).

Sinceχj andχ map to the same point ini∗(X0), they have the same volume. More-over by Lemma 2.5.4, for any representationψ: π1(M) → PSL2C which factors throughπ1(Mγj ), the volume ofψ does not depend on whether it is computed with respect toMγj orM . Now by Volume Rigidity forMγj , the representationρ: π1(Mγj ) → PSL2Cmust be discrete and faithful. Hence, as discussed in Section 2.7,ρ is conjugate toρj orρj . I claim that for largej, i∗(ρj) 6= i∗(ρj). We know that theχj converge toχdf andso theχj converge toχdf. By Lemma 2.7 we know neighborhoods ofχdf and χdf goto distinct branches ofi∗(X(M)). So for largej, i∗(χj) 6= i∗(χj). Hence for largej,(i∗)−1(pj) consists only ofχj . So the mapi∗: X0 → i∗(X0) has degree 1. ✷

It is easy to deduce the corresponding result for SL2C character varieties.

Corollary 3.2 LetM be a finite-volume hyperbolic 3-manifold with one cusp. LetX0

be a component of theSL2C character variety ofM which contains the character of adiscrete faithful representation. The inclusioni: ∂M → M induces a mapi∗: X0 →X(∂M). This map has degree onto its image at most|H1(M,Z2)|/2 where| · | denotes


number of elements. In particular, ifH1(M,Z2) = Z2 theni∗ is a birational isomor-phism onto its image.

Proof If a representationρ into PSL2C lifts to a representationρ into SL2C then thereare |H1(M,Z2)| distinct lifts which are constructed like this: Ifǫ ∈ H1(M,Z2) isthought of as a homomorphismǫ: π1(M) → Z2 = {I,−I} ⊂ SL2C then we canconstruct another lift ofρ by φ(g) = ǫ(g)ρ(g). By Poincare duality and the long ex-act sequence for the pair(M,∂M), the imageH1(∂M,Z2) → H1(M,Z2) is one-dimensional. So ifχ is a PSL2C character which lifts to an SL2C character, then the|H1(M,Z2)| distinct lifts map to precisely two points inX(∂M), unless the traces ofπ1(∂M) are all zero. Since all the traces are not zero generically onX0, the last theoremshows that the mapi∗: X0 → X(∂M) has degree at most|H1(M,Z2)|/2. ✷

Remark In the first example of [Dun] the mapi∗ has degree 4 onX0 and the first ho-mology isZ ⊕ Z4 ⊕ Z2. For components ofX(M) which do not contain a discretefaithful character, the mapi∗: X(M) → X(∂M) may not have degree 1. For instance,there are sometimes components ofX(M) which have dimension greater than 1 whoseimage underi∗ is one-dimensional (see Theorem 8.2 and 10.1 of [CL]). Even if weconsider an irreducible component ofX(M) of dimension 1 whose image underi∗ isone-dimensional,i∗ may still fail to have degree 1. This happens with the exterior of theknot74.

4 Cyclic Surgery Slopes and Boundary Slopes

LetK be a knot in a homotopy sphereΣ with exteriorM . Fix a meridianµ and longitudeλ in H1(∂M,Z). The slope ofγ ∈ H1(∂M,Z) with respect to the basis(µ, λ) will bedenotedrγ . In this section I will prove:

Theorem 4.1SupposeK is a small hyperbolic knot in a homotopy sphere. Suppose thereis aβ ∈ H1(∂M,Z) withπ1(Mβ) cyclic andβ 6= ±µ. Then there is an essential surfacein the exterior ofK whose boundary slope is non-integral and lies in(rβ − 1, rβ + 1).

By the Cyclic Surgery Theorem [CGLS],rβ is always an integer. Theorem 4.1 istrue more generally for a small hyperbolic knot in a manifoldΣ with cyclic fundamentalgroup whose exterior satisfiesH1(M,Z2) = Z2. In this setting, takeλ to be an arbitraryelement ofH1(∂M,Z) such that(µ, λ) is a basis (there is not always a natural choicewhenπ1(Σ) is non-trivial). Note the condition onH1(M,Z2) holds if π1(Σ) has oddorder. Theorem 4.1, including this more general case, follows easily from:

Theorem 4.2LetM be a one-cusped finite-volume hyperbolic 3-manifold withH1(M,Z2) =Z2. Suppose(µ, β) is a basis forH1(∂M,Z)whereπ1(Mβ) is cyclic. Forγ ∈ H1(∂M,Z),denote bysγ the slope with respect to the basis(µ, β). Then there is a boundary classγsuch that|sγ | < 1.


Note that hypotheses of Theorem 4.2 do not restrictπ1(Mµ), nor is it assumed thatM is small. I will now prove Theorem 4.1 assuming Theorem 4.2.

Proof (of Theorem 4.1)By the Cyclic Surgery Theorem, the meridianµ and the givenclassβ form a basis forH1(∂M,Z). Let γ be the boundary class ofM given by Theo-rem 4.2. Noterγ = sγ + rβ , and recall thatrβ is an integer. Since we are assumingMis small, Theorem 2.0.3 of [CGLS] shows thatβ is not a boundary class. Thereforerγ isa non-integral boundary slope in(rβ − 1, rβ + 1).

I will now prove Theorem 4.2:

Proof (of Theorem 4.2)I will use SL2C character varieties here, so fix a componentX0

of X(M) which contains the character of a discrete faithful representation. I will breakthe proof into three lemmas. The first is:

Lemma 4.3If K is a counterexample to the theorem, then the functionfµ/fβ is constanton X0.

So supposefµ/fβ is constant. From this, it is possible to determinei∗(X0) precisely.To state the answer, I will use the associated plane curve and the notation of Section 2.4.LetD0 be an irreducible component oft−1(i∗(X0)). Let (m, b) be the eigenvalue coor-dinates on∆ corresponding to the basis(µ, β). The second lemma is:

Lemma 4.4If fµ/fβ is constant onX0, there is a constantC 6= ±1 such thatD0 isexactly the set of zeros of the irreducible polynomial

P (m, b) ≡ bm2 − b− Cb2m+ Cm. (3)

The proof of the theorem is finished with:

Lemma 4.5The polynomialP of the last lemma can not defineD0 for a hyperbolicmanifoldM withH1(M,Z2) = Z2.

I will now prove these lemmas. The first follows immediately from the followingresult, which is more general in that it makes no assumption aboutH1(M,Z2):

Lemma 4.6LetM be a one-cusped finite-volume hyperbolic 3-manifold. Suppose(µ, β)is a basis forH1(∂M,Z)whereπ1(Mβ) is cyclic. Either there is a boundary classγ suchthat |sγ| < 1, or the functionfµ/fβ is constant onX0.

Proof (Lemma 4.6)Set g = fµ/fβ . Let Y be a smooth projective model ofX0. Ifg is constant onY , we are done. Otherwise, I will produce an essential surface withboundary classγ where|sγ| < 1. As sβ = 0, if β is a boundary class takeγ = β. Soassumeβ is not a boundary class. Lety ∈ Y be a pole ofg. Supposefβ has a zeroat y. For any rational functionh on Y , let Zy(h) denote the order of zero aty, whereZy(h) = 0 if h does not have a zero aty. As β is not a boundary class andπ1(Mβ) iscyclic, Proposition 1.1.3 of [CGLS] shows that the functionfµ also has a zero aty and


Zy(fµ) ≥ Zy(fβ). But theng does not have a pole aty, a contradiction. Sofβ must nothave a zero aty. Thusfµ must have a pole aty, and soy must be an ideal point wherethe associated surfaces have non-empty boundary (see Section 2.3). Letγ be a boundaryclass associated toy. By the proof of Lemma 1.4.1 of [CGLS], we have:

|sγ | =Πy(fβ)

Πy(fµ), (4)

whereΠy denotes the order of pole aty or is 0 if there is no pole. SinceΠy(g) =Πy(fµ)−Πy(fβ) > 0, we have|sγ | < 1. ✷

Now I will prove Lemma 4.4, which determines the equation definingD0, assumingthatfµ/fβ is constant.

Proof (Lemma 4.4)Let C ′ be the constant such thatfµ/fβ = C ′ on D0. With ourcoordinates(m, b) on∆, we have

fµ =

(m+

1

m

)2

− 4 =

(m− 1

m

)2

and fβ =

(b− 1

b

)2

.

So onD0 the equation (m− 1

m

)2

= C ′

(b− 1

b

)2

holds. SinceD0 is irreducible, for some square rootC of C ′ the equation(m− 1

m

)= C

(b− 1

b

)

holds onD0. Turning this into a polynomial condition, we have

P (m, b) ≡ bm2 − b− Cb2m+ Cm = 0 (5)

onD0.Next, I will showC 6= ±1. If C = ±1, P = (bm + C)(m − Cb). Then one of

the eigenvalue functionsξµ+β or ξµ−β would be constant onD0, which is impossible byProposition 2 of [CS2].

If C 6= ±1, it is an elementary exercise to check thatP is irreducible. ThusD0 mustbeexactlythe zero set ofP . ✷

Finally, I will prove Lemma 4.5, showing that the polynomialP of Lemma 4.4 cannot be the defining polynomial forD0 becauseH1(M,Z2) = Z2.

Proof (Lemma 4.5)As before, letY be a smooth projective model ofX0. SinceH1(M,Z2) =Z2, Corollary 3.2 shows that the mapi∗: X0 → i∗(X0) is a birational isomorphism. Sowe can also think ofY as a smooth projective model ofi∗(X0). Thus we have a rationalmapt: D0 → Y induced byt: D0 → i∗(X0). NoteP (0, 0) = 0, and so(0, 0) ∈ D0.


This is a smooth point ofD0 and the equation for the tangent line through(0, 0) isCm − b = 0. It follows that the eigenvalue functionsm = ξµ andb = ξβ restricted toD0 have simple zeros at(0, 0). So the functionstrµ = m+ 1/m and andtrβ = b+ 1/bonD0 have simple poles at(0, 0). Let y = t ((0, 0)). Theny ∈ Y is an ideal point ofX0 and bothtrµ andtrβ have simple poles aty. Looking again at the line tangent toD0

though(0, 0), we see thatµ − β is a boundary class associated toy (see Section 2.3).Also, the eigenvalue ofµ− β at (0, 0) isC. By the theorem in Section 5.7 of [CCGLS]the value ofξµ−β at y is a root of unity whose order divides the number of boundarycomponents of any surface associated toy. So sinceC 6= ±1, any surface associated toy has at least three boundary components. We now get our contradiction by showing thatthere is a surface associated toy with only two boundary components. LetTy be the treeassociated toy. By Proposition 2.2, the translation length ofµ onTy is twice the order ofpole oftrµ at y. From the dual role ofY as the smooth projective model of bothi∗(X0)

andX0, we have calculated that the order of pole oftrµ at y is 1, and so the translationlength ofµ is 2. But then Proposition 2.3 shows there is a surface associated toy withtwo boundary components, a contradiction. SoP can not be the equation definingD0.✷

Since we have proved Lemmas 4.3, 4.4, and 4.5, we have proven the theorem.✷

RemarkNote that the proof usedH1(M,Z2) = Z2 in a fundamental way. If the degreeof X0 → i∗(X0) wered, we could only have concluded that there was a surface associ-ated toy having as few as2d boundary components. The first example in [Dun] has anideal point whose image ini∗(X0) looks exactly like in the proof, in the sense thattrµhas a simple pole at the corresponding point ofD0. But there,H1(M,Z) isZ⊕Z4⊕Z2

and the associated surface has four boundary components. This illustrates the way thatCorollary 3.2 shows there is a surprising connection between the size ofH1(M,Z2) andthe character variety ofM .

5 The norm and the diameter of the set of boundary slopes

Lemma 4.6 has an interpretation in terms of the norm onH1(∂M,R) defined in [CGLS].Let V denoteH1(∂M,R) andL ⊂ V the latticeH1(∂M,Z). This norm onV has theproperty that forγ ∈ L, the number||γ|| is the degree offγ onX0. Letr be the minimumof ||γ|| over all non-zeroγ ∈ L. LetB denote the closed ball of radiusr about0 in V .The ballB is a finite-sided convex polygon which is invariant underv 7→ −v. Boundaryclasses associated to ideal points ofX0 correspond bijectively with vertices ofB, wherea boundary classγ corresponds to a vertexv wherev = aγ for a positive rational numbera (this is not quite explicit in [CGLS], but see Lemma 6.1 of [BZ1]). An elementγ ∈ Lsuch thatπ1(Mγ) is cyclic and which is not a boundary class has minimal norm, that is,||γ|| = r andγ ∈ B. In this language, a slight modification of the proof of Lemma 4.6gives:


Lemma 5.1Suppose(µ, β) is a basis ofL such thatπ1(Mµ) is cyclic,µ is not a bound-ary slope, andβ has minimal norm. Then either there is a boundary classγ, associatedto an ideal point ofX0, which satisfies|sγ| < 1 or the functionfµ/fβ is constant onX0.

Proof The needed modification to the proof of Lemma 4.6 is at the step that showsthat if fβ has a zero aty thenfµ also has a zero aty andZy(fµ) ≥ Zy(fβ). Here, byProposition 1.1.3 of [CGLS], we know iffµ has a zero aty thenZy(fµ) ≤ Zy(fβ). Thetotal number of zeros offµ is equal to the total number of zeros offβ since||µ|| = ||β||.Thus the sets of zeros (including multiplicity) offµ andfβ are the same, and we canapply the rest of the proof of Lemma 4.6 unchanged. ✷

I will now give an application to the following question. Consider a knot in a ho-motopy sphere with irreducible exteriorM . The set of boundary slopes is finite [Hat],and so has a well-defined diameterd as a subset ofQ ∪ {∞} (I will use the conventionthatd = ∞ if ∞ is a boundary slope). Hatcher and Thurston [HT] asked in the case ofknots inS3 whetherd is always greater than2. In [CS3], Culler and Shalen showed thatfor any knot, the diameterd ≥ 2. Another application of the proof of Theorem 4.1 is toshow:

Theorem 5.2If d = 2 for a hyperbolic knot in a homotopy sphere, the greatest and leastboundary slopes are not integers.

Proof Consider such a knot withd = 2. In [CS3, Proof of Main Theorem] it is shownthat for a suitable choice of meridian-longitude basis(µ, λ) for L,B is as in Fig. 1. Sincethis is not quite explicit there, let me elaborate. In the notation of [CS3], letv1 andv2 bethe vertices ofB which are the ends of the edge containingµ. Thenµ = tv1 +(1− t)v2for somet ∈ [0, 1], and they show that the diameter of the set of boundary slopes isbounded below by1/(2t(t − 1)) ≥ 2. If the diameter of the set of boundary slopesis 2, we must havet = 1/2. In this case, the area of the parallelogram with vertices{v1, v2,−v1,−v2} is 4 and soB is equal to this parallelogram. Sincet = 1/2, µ lies inthe middle of an edge ofB. Combined with the fact that the area ofB is 4, the sides ofB must be segments of vertical linesλ = ±1. Since0 is always a boundary slope, afterpossibly changing the signs ofµ andλ,B must be as in Fig. 1.

From the correspondence between boundary classes associated to ideal points ofX0

and vertices ofB, there are only two boundary slopes associated to ideal points ofX0,namely,−p/q and2− p/q where0 ≤ p ≤ q andgcd(p, q) = 1.

Showing that the greatest and least slopes are not integers is equivalent to showingq > 1. Supposeq = 1. There are two cases:p = 0 andp = 1. If p = 1, the two boundaryslopes are1 and−1 with respect to the basis(µ, λ). Sinceλ is inB it has minimal norm,and asµ is not a boundary class (sinced is finite), Lemma 5.1 shows thatfµ/fλ mustbe constant. We can now apply Lemmas 4.4 and 4.5 to get a contradiction. Ifp = 0, weapply the same argument withλ replaced byγ = µ+ β. ✷

It is possible to prove more:


N

l

m

q

p

(2q, 2q-p)

(q, 2q)

λ

γµ

Fig. 1. The fundamental polygon. The slopes ofthe dotted lines are2− p/q and−p/q.

Fig. 2.The Newton polygon ofP

Theorem 5.3Suppose the diameter of the set of boundary slopes of a hyperbolic knotin a homotopy sphere is 2. Let0 ≤ p ≤ q be such that the greatest and least slopes are2− p/q and−p/q respectively. Thenp is even,q odd, andq > 1.

The key to this version is a lemma which is a consequence of Theorem 3.1. Considerthe commutative diagram:

X0i∗ //

πM

��

i∗(X0) ⊂ X(∂M)

π∂M

��X0

i∗ // i∗(X0) ⊂ X(∂M)

(6)

The mapπ∂M from X(∂M) to X(∂M) has degree 4, so one might expect that therestriction ofπ∂M to i∗(X0) could also have degree 4. But I will show the degree ofthis restriction can be no more than 2. An involutionf of i∗(X0) which has the propertythat π∂M ◦ f = π∂M will be called asymmetryof i∗(X0). If the restriction ofπ∂Mto i∗(X0) has degree 4, the group of symmetries would have order 4. One possiblesymmetry,τ , is the restriction of the involution ofX(∂M) whose action in coordinatesis (trµ, trλ, trµ+λ) 7→ (−trµ, trλ,−trµ+λ). I will show:

Lemma 5.4For the exterior of a hyperbolic knot in a homotopy sphere, the only possiblesymmetry ofi∗(X0) is τ .

Proof The discussion in the proof of Corollary 3.2 shows thatπM has degree 1 or 2,depending on whether(πM )−1(X0) is irreducible. By Theorem 3.1 and Corollary 3.2,the horizontal maps in Eqn. (6) have degree 1. So the degrees ofπM andπ∂M are equal.If these degrees are 1, there are no symmetries. If the degrees are 2,i∗(X0) has onesymmetry, which I claim isτ . In this case, there is an involutionτ ′ of X0 for whichπM ◦τ ′ = πM , namely, multiplication of characters by the unique non-trivial homomorphismπ1(M) → Z2 = {I,−I} ⊂ SL2C. Thenτ ′ induces the symmetryτ of i∗(X). ✷


RemarkFor the exterior of a knot in a homotopy sphere, it is conceivable thati∗(X0)might not have the symmetryτ . Suppose(πM )−1(X0) consists of two irreducible com-ponents. Whilei∗

((πM )−1(X0)

)would be invariant underτ , the curvei∗(X0) might

not be. The first example of [Dun] is a one-cusped hyperbolic manifold where(πM )−1(X0)splits into multiple components. Applyingi∗ to one component yields a curve which isnot invariant under the analogue ofτ . However, this manifold is not exterior of a knot ina homotopy sphere.

From Lemma 5.4 we can prove Theorem 5.3:

Proof (Theorem 5.3)Let p andq be as in the statement. Theorem 5.3 shows thatq > 1, and I will assume

this throughout. Letγ = µ+ λ. Consider the function

g =fpλf

q−pγ

f qµon X0,

noting thatq − p > 0. The proof follows the same basic plan as that of Theorem 4.1.The analogue of Lemma 4.3 is:

Lemma 5.5If d = 2, theng is constant onX0.

Let (m, l) be the eigenvalue coordinates on∆ corresponding to(µ, λ). The analogueof Lemma 4.4 is:

Lemma 5.6If g is constant there is aC ∈ C so thatD0 is exactly the zeros of

P (m, l) ≡ mp(l2 − 1

)p (l2m2 − 1

)q−p − Clq(m2 − 1

)q. (7)

The final step, which is the one that differs most from the proof of Theorem 4.1, is touse Lemma 5.4 to show the following analogue of Lemma 4.5:

Lemma 5.7If the polynomialP of the last lemma definesD0 for the exterior of a knotin a homotopy sphere, thenp is even andq is odd.

I will begin with:

Proof (Lemma 5.5)This is a refinement of the proofs of Lemmas 4.6 and 5.1. Supposeg is not constant. Lety ∈ Y be a pole ofg, whereY is a smooth projective model ofX0. We know thatπ1(Mµ) is a cyclic and thatµ is not a boundary class. From Fig. 1 weknowµ, λ, andγ all have minimal norm. Therefore, as in the proof of Lemma 5.1, thesets of zeros (including multiplicity) offµ, fλ, andfγ are all the same. But theng wouldnot have a pole aty, a contradiction. Thus at least one offλ or fγ has a pole aty. Soy isan ideal point.

By Lemma 1.4.1 of [CGLS] there is a linear functionall onH1(∂M,Z) such that theorder of pole offα,Πyfα, is |l(α)| for all α in H1(∂M,Z). The slope associated toy iseither−p/q or 2 − p/q. In the first case,Πyf−pµ+qλ = 0 and so there is ad ∈ Z such


thatl(aµ+ bλ) = d(qa+pb). ThenΠyfµ = |d|q,Πyfλ = |d|p, andΠyfγ = |d|(p+q).So

Πyg = −qΠyfµ + pΠyfλ + (q − p)Πyfγ = 0.

If the slope associated toy is 2−p/q = (2q−p)/q thenl(aµ+bλ) = d(qa−(2q−p)b).ThenΠyfµ = |d|q,Πyfλ = |d|(2q − p), andΠyfγ = |d|(q − p); henceΠyg = 0.

Sog has no poles and must be constant. ✷

Now we deduce the equation definingD0.

Proof (Lemma 5.6)Just as in the proof of Lemma 4.4, there is a constantC so that the following equation

holds onD0: (l − 1

l

)p(ml − 1

ml

)q−p

= C

(m− 1

m

)q

, (8)

This is equivalent to the polynomialP given in the statement of Lemma 5.6 being zero.Now I will show thatD0 is exactly the zeros ofP . We need to take the point of

view of [CCGLS] in which information about ideal points ofX0 is deduced from theNewton polygon of the equation definingD0. TheNewton polygonof a polynomialQin variablesx andy is the convex hull inR2 of:

{(i, j) ∈ Z2 | the coefficient ofxiyj in Q is nonzero}.

The Newton polygon,N , of P is shown in Fig. 2.I claim that{P = 0} is exactlyD0. SupposeQ were a proper factor ofP which

definesD0. LetN ′ be the Newton polygon ofQ. In [CCGLS] it is shown that the slopesof the sides ofN ′ are precisely the boundary slopes of surfaces associated to ideal pointsof X0. Thus the slopes of the sides ofN ′ are−p/q and2 − p/q. So we know thatN ′

is a parallelogram whose sides are parallel to those ofN . The idea is thatN is thesmallest such parallelogram and thereforeN = N ′ andP = Q. Let diaml(N

′) denotethe diameter of the projection ofN ′ onto the axis corresponding to the exponent ofl.If we think of Q as the a polynomial in a single variablel overC[m], then the degreeof this polynomial,degl(Q), is equal todiaml(N

′). SinceQ is a factor ofP , we knowdeglQ ≤ degl P , and sodiaml(N

′) ≤ diaml(N) = 2q. Projecting onto the other axis,we also havediamm(N ′) ≤ diamm(N) = 2q. Moreover, if both diameters ofN ′ andN agree, we haveP = Q. Now consider a sideS of N ′ with slope−p/q. Because theendpoints ofS are inZ2, we must havediaml(S) ≥ q anddiamm(S) ≥ p. Similarly,a sideT of N ′ with slope2 − p/q must havediaml(T ) ≥ q anddiamm(T ) ≥ 2q − p.Thusdiaml(N

′) ≥ 2q = diaml(N) anddiamm(N ′) ≥ 2q = diamm(N). SoP = Qas desired. SoD0 must be exactly the zero set ofP . ✷

Now I will show that ifP definesD0 for the exterior of a knot in a homotopy sphere,thenp is even andq is odd. By Lemma 5.4 the only allowed symmetry is the one whoseaction onDM sends(m, l) to (−m, l). If q is even,p andq − p are odd. Note in this


case that Eqn. (8) and henceP are invariant under(m, l) 7→ (m,−l). But then thereis a symmetry ofi∗(X0) other thanτ , which is impossible. Thereforeq is odd, and wealready knowq 6= 1. If p andq are both odd, then Eqn. (8) is invariant under(m, l) 7→(−m,−l), so this case is ruled out as well. Sop is even andq is odd. ✷

RemarksLetQ denote the polynomial in Eqn. (7) withp = 1, q = 2, andC = 1. ThepolynomialQ actually occurs as the defining equation forD0 for N , the sister of theexterior of the figure-8 knot. In this case,Nµ has fundamental groupZ10. The results of[CS3], more generally, say thatd ≥ 2 for any knot in a manifold with cyclic fundamentalgroup whose exterior is irreducible and not cabled. It turns out that forN , the diameterof the set of all boundary slopes is exactly2, and this shows that the estimate of Cullerand Shalen is sharp in this more general context (see Example 1.4 of [CS3]). It wasthe observation that equation definingDN was{Q = 0} that led me to discover thisexample.

If one cares about the diameter of the set ofstrict boundary slopes, then the situationstays the same except thatλ is replaced by some random classν with integer slope. Inthis case, you can not rule outp andq both being odd sinceν may generateH1(M,Z2).This is also why you can not use the argument about symmetries to prove Theorem 4.1.

6 Proof of volume rigidity

This section provides a proof of:

Theorem 6.1 (Gromov-Thurston-Goldman)SupposeM is a compact hyperbolic 3-manifold. Ifρ1: π1(M) → PSL2C is a representation withvol(ρ1) = vol(M), thenρ1is discrete and faithful.

Proof The proof is essentially the same as that of Thurston’s strict version of Mostow’sTheorem [Thu, Theorem 6.4]. The only modification is that sinceH3/ρ1 may be nasty,rather than a compact manifold, it is necessary to do some things equivariantly. I willfollow [Thu], with some details coming from Toledo’s paper [Tol] using the same tech-nique. The same proof works in higher dimensions with the aid of [HM], but I stick tothe 3-dimensional case for simplicity. I assume some familiarity with Gromov’s proofof Mostow’s Theorem (for a nice account, see [Mun] or the very through [Rat, Chapter11]).

Let ρ0 be a discrete faithful representation forM . Pick a smooth equivariant mapffrom H3 acted on byρ0 to H3 acted on byρ1. If the integral

∫M f∗(Vol3

H) is negative,

choose an orientation reversing isometryr of H3 and replaceρ1 by r ◦ ρ1 ◦ r−1 andfby r ◦ f so that the integral is positive. Then the hypothesis onvol(ρ1) gives:

∫

MVolH3 = vol(M) = vol(ρ1) =

∫

Mf∗(VolH3). (9)

By a tetrahedron, I will mean a simplex inH3 with totally geodesic faces. I willdivide the proof into the following three claims:


Claim 1The mapf extends to an equivariant measurable mapf from S2∞ = ∂H3 to

itself.

Claim 2The mapf takes vertices of almost all regular ideal tetrahedra to vertices ofregular ideal tetrahedra.

Claim 3Because of Claim 2,f is essentially a Mobius transformation. That is, there isa Mobius transformationF so thatF = f almost everywhere.

The proof is then completed by noting that sinceF is equivariant, it conjugates theaction ofρ0 onS2

∞ to the action ofρ1 onS2∞. As ρ0 andρ1 are conjugate,ρ1 is discrete

and faithful.Let me start in on the proof of Claim 1. The key is that the regular ideal tetrahedron

is the unique tetrahedron of maximal volume [Rat, Theorem 10.4.7]. The idea behindClaim 1 is thatf must take the vertices of a non-ideal tetrahedron of near-maximalvolume to points which span a tetrahedron of near-maximal volume or elsef would bevolume shrinking and Eqn. (9) would be violated.

The proof will use measure homology, which I will briefly describe (for details seeSection 4 of [Mun] or Section 11.5 of [Rat]). LetC1(∆k,M) be the space ofC1 mapsof the standardk-simplex∆k intoM , supplied with theC1 topology. LetCk(M,R) bethe set of real-valued Borel measures of bounded total variation onC1(∆k,M). Thereis a natural boundary operator fromCk(M,R) to Ck−1(M,R), and the homology of theresulting complex is called themeasure homologyof M . There is an inclusion of theusualC1 singular chain complexC∗(M,R) into C∗(M,R) which sends a singular sim-plex (an element ofC1(∆k,M)) to the associated Dirac measure. This inclusion inducesan isomorphism on homology. The pairing betweenCk(M,R) and smooth differentialk-forms extends toCk(M,R).

LetG = PSL2C andΓ ⊂ G beπ1(M) acting viaρ0. The unit tangent bundle toMisX = Γ\G, and letµ be the Haar measure onG such thatµ(X) = vol(M). LetD bea polyhedron minus some faces which is a fundamental domain forM in H3 (each orbitof H3 underΓ has exactly one representative inD).

Letσ ⊂ H3 be a non-ideal tetrahedron. Letsmear σ be the measure cycle inC3(H3,R)

consisting of all translates ofσ with first vertex inD, uniformly weighted by1/vol(σ);that is, ifη is a differential 3-form:

〈η, smearσ〉 =∫

g∈X

(1

vol(σ)

∫

g·ση

)dµ.

Here, I abuse notation and denoteD × SO(3) ⊂ G, a fundamental domain for theunit tangent bundleX, by X. Now if σ− denotesσ with the opposite orientation, letzσ = 1/2(smearσ − smear σ−). If π : H3 → M is the covering map, we haveπ∗(zσ)is closed. Since measure homology is the same as standard homology,zσ representssome multiple of the fundamental class. As〈VolH3 , smear σ〉 = vol(M), the cyclezσ


represents the fundamental class. Nowvol(ρ1) = 〈f∗VolH3 , zσ〉 = 〈f∗VolH3 , smearσ〉,and so

vol(ρ1) =

∫

X

(1

vol(σ)

∫

g·σf∗VolH3

)dµ. (10)

There is a chain mapStr from C∗(H3,R) to itself which replaces a singular simplex by

a geodesic simplex with the same vertices (see Section 4 of [Mun] or Section 11.5 of[Rat] for a definition ofStr and its properties). I claim that we can replace

∫g·σ f

∗VolH3

in Eqn. (10) by∫Str(f(g·σ))VolH3 = vol(Str(f(g · σ))). The idea is this: Consider the

analogous question for singular simplicial homology, i.e. letz be a lift toC3(H3,Z) of a

cycle inC3(M,Z). The chainz is now a finite linear combination of singular simplices.While ∂z is not zero, you can choose elements ofΓ that pair up the simplices of∂z thatreflect the fact thatπ(∂z) = 0 in C2(M,Z). The difference betweenf∗(z) andStrf∗(z)depends onf∗(∂z). The way elements ofΓ pair up the simplices of∂z shows that wecan replace

∫g·σ f

∗VolH3 by∫Str(f(g·σ))VolH3 in (10).

Formally, there is a chain homotopy,H, from Str to the identity which is invariantunder isometries. Now

〈f∗VolH3 , zσ〉 = 〈VolH3 , f∗(zσ)〉 = 〈VolH3 , Str(f∗(zσ))−H(f∗(∂zσ))〉

and so to show the claim it is enough to show〈VolH3 , H(f∗(∂zσ))〉 = 0. Equivalently,define a cochainc by c(τ) = 〈VolH3 , H(f∗(τ))〉; we wantc(∂zσ) = 0. Sincef isequivariant andH commutes with isometries,c descends to a cochaincM on M . Asπ∗(∂zσ) = 0, cM (π∗(∂zσ)) = c(∂zσ) = 0, as desired. This proves the claim. Hence wehave from Eqn. (10):

vol(σ)vol(M) =

∫

Xvol(Str(f(g · σ))) dµ. (11)

where the volume ofStr(f(g · σ)) is signed volume. This is the formula which willguarantee thatf do not shrink large tetrahedra very much.

Fix a geodesic rayr with endpointb ∈ D. Let σi be a regular tetrahedron all ofwhose sides have lengthi with one vertexb and an edge lying onr. The idea is to usethe expanding sequences{g · σi} for g ∈ X to approximate what would happen to anideal tetrahedron. Letv3 denote the volume of a regular ideal tetrahedron, which is theunique tetrahedron of maximal volume. The next lemma is a quantitative version of thestatement “f does not shrink volume.”

Lemma 6.2Let ǫi = v3 − vol(σi), and letY = {g ∈ X : vol(Str(f(g · σi)) <vol(σi)− i2ǫi}. Thenµ(Y ) < µ(X)/i2.

Proof This is Lemma 2.3 of [Tol]. Sincevol(Str(f(g · σ))) < v3 = vol(σi) + ǫi,Eqn. (11) gives us

vol(σi)vol(M) =

∫

Xvol(Str(f(g · σi))) dµ


=

∫

Yvol(Str(f(g · σi))) dµ+

∫

X\Yvol(Str(f(g · σi))) dµ

< (vol(σi)− i2ǫi)µ(Y ) + v3µ(X \ Y )

Since

(vol(σi)− i2ǫi)µ(Y ) + v3µ(X \ Y ) = vol(σi)vol(M) + ǫi(µ(X \ Y )− i2µ(Y ))

we have

vol(σi)vol(M) < vol(σi)vol(M) + ǫi(µ(X \ Y )− i2µ(Y )).

Therefore0 < µ(X \ Y )− i2µ(Y ), and soµ(Y ) < µ(X)/i2. ✷

The next lemma shows that for largei, σi is a very good approximation of a regularideal tetrahedron.

Lemma 6.3Letσi be a regular tetrahedron inH3 all of whose sides have lengthi. Thenfor i large,ǫi = v3 − vol(σi) decreases exponentially withi.

Proof This is Lemma 6.4.1 of [Thu]. Letδ∞ be a fixed regular ideal tetrahedron. Letpbe the center of mass ofδ∞ and consider the four rays starting atp and ending at thevertices ofδ∞. The tetrahedron whose vertices are the points on these rays a distancetfrom p is regular, and will be denotedδi wherei ∈ R+ is the length of any of its edges(thusσi and δi are isometric). Any tetrahedron has a natural straight (or barycentric)

dvdt

p

v

q

θ

Max. normal velocity

v

q

d

tθ

p

Fig. 3.The tetrahedronδi Fig. 4.Finding the maximum normal velocity

parameterization coming from the affine structure of the hyperboloid model ofH3 (seeSection 11.4 of [Rat]). Parameterizingδi in this way, we get a one parameter family ofdiffeomorphismsϕi from the standard 3-simplex∆3 to δi. A point y = ϕi(x) on a faceof δi has a velocity which is defined as the derivative ofϕi(x) with respect toi. Thenormal velocity ofy is the component of velocity normal to the face ofδi containingy.


The derivativedvol(δi)dt is bounded by the area of∂δi times the maximum normal

velocity of∂δi. Let q be the center of mass of one of the faces ofδi. From Figs. 3 and 4,we see that the maximum normal velocity of∂δi is sin θ.

By the law of sines,sin θ = sinh dsinh t . Since any triangle in hyperbolic space has area

less thanπ, we have:dvol(δi)

dt< 4π

sinh d

sinh t

Sincesinh d is bounded,dvol(δi)dt decreases exponentially witht. Applying the law ofcosines to the triangle with two sides of lengtht and one side of lengthi shows thatasymptoticallyi is 2t plus a constant, and sodvol(δi)di decreases exponentially withi. ✷

By Lemma 6.2, if we fixi0 then the set ofg ∈ X for which vol(Str(f(g · σi))) <vol(σi)− i2ǫi for somei ≥ i0 has measure less thanµ(X)

∑∞i=i0

1/i2. Thus except ona small exceptional set, we have

vol(Str(f(g · σi))) ≥ vol(σi)− i2ǫi (12)

for all i ≥ i0.By Lemma 6.3,i2ǫi → 0 asi → ∞. So lettingi0 → ∞ we have that for almost all

g ∈ X, vol(Str(f(g · σi))) converges tov3. Let r(i) denote the point ofr at a distanceifrom b, the basepoint ofr.

Let τi be a tetrahedron with verticesb, r(i + 1), and the two vertices ofσi not onr(see Fig. 5). Note thatv3−vol(σi) > v3−vol(τi) and so the above arguments show that

r

b

a

c

σ

τi

i

r (i)r (i+1)

f(b)

f(r (i+1))

f (r (i))f(a)

f(c)

Fig. 5.The tetrahedraσi andτi Fig. 6.The tetrahedraStr(f(σi)) andStr(f(τi))

for almost allg ∈ X, vol(Str(f(g · τi))) converges tov3. Supposeg is such that bothvol(Str(f(g · σi))) andvol(Str(f(g · τi))) converge tov3. For notational convenience,takeg to be the identity. I claimf(r) converges to a point inS2

∞. Sincef is Lipschitz,it is enough to show that thef(r(i)) converge to a point inS2

∞. Regular ideal tetrahedraare the only tetrahedra of maximal volume, so sincevol(Str(f(σi))) goes tov3, we musthave the distance fromf(b) to f(r(i)) going to∞ asi goes to∞. Hence thef(r(i))


head out towardS2∞. To show they converge, as opposed to wandering about willy-nilly,

we need to show the visual angle off(r(i)) with respect tof(b) converges. The changein visual angle betweenf(r(i)) andf(r(i+1)) is the angle between the lines fromf(b)to f(r(i)) and fromf(b) andf(r(i + 1)). From Fig. 6 we see this change is less thanthe sum of the two indicated face angles ofStr(f(σi)) andStr(f(τi)).

The following lemma allows us to estimate these face angles:

Lemma 6.4There is a constantC > 0 such that ifσ is a tetrahedron withvol(σ) suffi-ciently close tov3, then for any face angleβ of σ:

v3 − vol(σ) > Cβ2

For largei, Eqn. (12) showsv3 − vol(Str(σi)) ≤ (v3 − vol(σi))+ i2ǫi = ǫi(1+ i2).From Lemma 6.4 we have that the change in visual angle for largei is less than

2

√ǫi(1 + i2)

C.

By Lemma 6.3 this is eventually exponentially decreasing withi, and so the visual anglesof thef(r(i)) converge. Hencef(g · r) converges for almost allg ∈ X. Therefore, foralmost all geodesic raysr in H3, f(r) converges to a point inS2

∞.Moreover, asf is Lipschitz any two rays which are asymptotic have images underf

which converge to the same point ofS2∞. Hence we have an extension off , f : S2

∞ →S2∞, and you can check thatf is measurable. Modulo the proof of Lemma 6.4, we have

proven Claim 1.Let us go back and prove the lemma.

Proof (Lemma 6.4)Let v be the vertex ofσ which is the endpoint of the angleβ. With-out changing a neighborhood ofv, push the other three vertices ofσ to S2

∞ (this onlydecreasesv3 − vol(σ)). Extend an edge throughv which is a side ofβ to S2

∞, as inFig. 7. Look at the partP added on by doing this. Nowvol(P ) ≤ v3 − vol(σ). ConsiderFig. 8 in the upper half space model where we are using Euclidean coordinates such thatdist(x, w) = dist(x, v) = 1.

We will estimate the volume ofP above the dotted line. By requiring thatvol(σ)be large, we can assume the dihedral angles ofσ are close toπ/3. The indicated crosssection is then about an equilateral triangle whose area is bounded below byC1d

2. Noted = 1− cosβ with respect to our Euclidean coordinate system, and so

vol(P ) ≥ Vol. above dotted line=∫ ∞

sinβ

C1(1− cosβ)2

z3dz

Evaluating the integral we get

vol(P ) ≥ C1(1− cos β)2

2 sin2 β≥ C2β

2

for some constantC2. Thusv3 − vol(σ) ≥ C2β2, as desired. ✷


Cross section

x

P

w

d

vβ

σ

β

β σv

P

Fig. 7.CreatingP Fig. 8.Estimating area ofP

Next, I will check Claim 2 thatf sends the vertices of almost every positively ori-ented regular ideal tetrahedron to the vertices of a positively oriented regular ideal tetra-hedron. Letδi denote a positively oriented regular tetrahedron with side lengthi, withcenter of massat a fixed pointb ∈ D, and whose vertices lie along four fixed geodesicraysr1, r2, r3, r4 emanating fromb. Arguing as above, we can show that for almost allg ∈ X, vol(Str(f(g · δi))) converges tov3. Moreover for almost allg this is true and,in addition,f(g · rj) converges to a pointpj ∈ S2

∞ for all j. Sincevol(Str(f(g · δi)))converges tov3, thepj must span a regular ideal tetrahedron. Since this is true for al-most allg ∈ X, the vertices of almost all regular ideal tetrahedra are sent to regular idealtetrahedra. This proof of Claim 2.

I will now prove Claim 3, thatf is essentially a Mobius transformation. The spaceT of regular oriented ideal tetrahedra with labeled vertices is a full measure subset ofS2∞ × S2

∞ × S2∞. Let TG be the subset ofT consisting of tetrahedra whichf takes

to regular oriented ideal tetrahedra. We have just shown thatTG has full measure. By

Fubini’s Theorem there is av0 ∈ S2∞ such that almost allT ∈ T with first vertexv0

are inTG. In fact, this is true for almost allv0, so we can assume thatf(v0) is defined(recall thatf is only defined by the process of looking at images of geodesic rays for afull measure subset ofS2

∞).

Without loss of generality, we can take bothv0 andf(v0) to be the point at infinity inthe upper half space model ofH3. Tetrahedra inT with first vertex at∞ are equivalentto oriented equilateral triangles inC with labeled vertices, which are parameterized by afull measure subset ofC×C. It will help the reader to think ofC as having finite measurewhen applying Fubini; we will only be concerned with which sets have measure zero,


a property which is invariant under diffeomorphism [Boo, Section VI.1]. For almostalllines l through0, almost all equilateral triangles with the edge between the first andsecond vertices parallel tol define tetrahedra which are inTG. Assume that one suchline is the real axis. LetS denote tetrahedra with first vertex at∞ and such that the edgebetween the second and third vertices (the first and second vertices of the correspondingtriangle) is parallel to the real axis.

We know thatSG ≡ S ∩ TG has full measure inS. Let ω be the 3

√−1 which has

positive imaginary part. Then{0, 1, ω} is an oriented equilateral triangle. LetL0 be allequilateral triangles in the tiling ofC by the triangle{0, 1, ω}. Let Lk be the same setof triangles scaled by2−k. Let L =

⋃k∈Z Lk be this nested family of equitriangular

lattices (See Fig. 9).

Fig. 9.Some of the triangles in the nested family of latticesL

I claim there is anr ∈ R such that for almost allz ∈ C, the entire countable set oftrianglesz+ rL are inTG. Consider the submersionπ: C×R×Z×Z×Z → S whichsends(z, r, k, n,m) to the equilateral triangle with vertices

(z + r2−k(n+mω), z + r2−k(n+ 1 +mω), z + r2−k(n+ (m+ 1)ω)

)

in z + rLk. We will think of Z as having a measure where the measure ofq is 1/q2.As π is a submersion,π−1(SG) has full measure. Thus by Fubini, for almost allr andz, we haveπ−1

(SG)∩ ({r} × {z} × Z× Z× Z) has full measure, that is equal to

{r} × {z} × Z × Z × Z, as desired. Without loss of generality assumer = 1 has thisproperty. So for almost allz ∈ C all triangles inz + L are inSG. This forcesf(z + L)to be family of nested equitriangular lattices (see Fig. 9). For eachz there is a complexnumberh(z) such that:

f(z + 2−k(n+mω)

)= f(z) + h(z)2−k(n+mω)

for all {n,m, k} ⊂ Z. I claim the functionh is invariant under the group of translationsof the formz 7→ z + 2−j(a+ bω), where{j, a, b} ⊂ Z. Let z′ = z + 2−j(a+ bω). We


have

f(z′) = f(z) + h(z)2−j(a+ bω) (13)

f(z′) + h(z′)2−j = f(z) + h(z)2−j(a+ 1 + bω) = f(z′ + 2−j) (14)

Subtracting Eqn. (13) from Eqn. (14) we geth(z′) = h(z). Our group of translationsis dense, and so acts ergodically. Thereforeh is constant almost everywhere. But thenf(z′) = f(z) + h2−j(a + bω) almost everywhere which implies thatf(z) − h · z isinvariant under our group of translations. So there is a constantc such thatf(z)−hz = calmost everywhere and thusf(z) = c+ hz almost everywhere.

Thusf is essentially a Mobius transformation. The corresponding Mobius transfor-mation conjugates the actions ofρ0 andρ1 on S2

∞ to one another. Thusρ0 andρ1 areconjugate, andρ1 is discrete and faithful. ✷

References

[BP] R. Benedetti and C. Petronio.Lectures on hyperbolic geometry. Springer, 1992.[Boo] W. M. Boothby. An introduction to differentiable manifolds and Riemannian geometry, volume

120 ofPure and Applied Mathematics. Academic Press Inc., Orlando, Fla., second edition, 1986.[BMZ] S. Boyer, T. Mattman, and X. Zhang. The fundamental polygons of twist knots and the(−2, 3, 7)

pretzel knot. In S. Suzuki, editor,Proceedings of Knots 96, pages 159 – 172. World Scientific,1997.

[BZ1] S. Boyer and X. Zhang. Finite Dehn surgery on knots.J. Amer. Math. Soc.9(1996), 1005–1050.[BZ2] S. Boyer and X. Zhang. Cyclic surgery and boundary slopes. In W. H. Kazez, editor,Geometric

Topology, volume 2(part 1) ofAMS/IP Studies in Advanced Mathematics, pages 62–79, 1997.[BZ3] S. Boyer and X. Zhang. On Culler-Shalen seminorm and Dehn filling. Preprint.[CCGLS] D. Cooper, M. Culler, H. Gillet, D. D. Long, and P. B. Shalen. Plane curves associated to

character varieties of 3-manifolds.Invent. Math.118(1994), 47–74.[CL] D. Cooper and D. D. Long. Remarks on theA-polynomial of a knot.J. Knot Theory Ramifications

5(1996), 609–628.[Cor1] K. Corlette. FlatG-bundles with canonical metrics.J. Differential Geom.28(1988), 361–382.[Cor2] K. Corlette. Archimedean superrigidity and hyperbolic geometry.Ann. of Math. (2)135(1992),

165–182.[Cul] M. Culler. Lifting representations to covering groups. Adv. in Math.59(1986), 64 – 70.[CGLS] M. Culler, C. M. Gordon, J. Luecke, and P. B. Shalen. Dehn surgery on knots.Ann. of Math. (2)

125(1987), 237–300.[CS1] M. Culler and P. B. Shalen. Varieties of group representations and splittings of 3-manifolds.Ann.

of Math. (2)117(1983), 109–146.[CS2] M. Culler and P. B. Shalen. Bounded, separating, incompressible surfaces in knot manifolds.

Invent. Math.75(1984), 537–545.[CS3] M. Culler and P. B. Shalen. Boundary slopes of knots. Preprint.[Dun] N. M. Dunfield. Examples of non-trivial roots of unity at ideal points of hyperbolic 3-manifolds.

Topology38(1999), 457–465. To appear, e-print math.GT/9801064.[FS] R. Fintushel and R. J. Stern. Constructing lens spaces by surgery on knots.Math. Z.175(1980),

33–51. Correction in178(1981), 143.[Gol] W. M. Goldman. Characteristic classes and representations of discrete subgroups of Lie groups.

Bull. Amer. Math. Soc. (N.S.)6(1982), 91–94.


[Gor] C. M. Gordon. Dehn filling: A survey. InKnot Theory, volume 43 ofBanach Center Publ.PolishAcad. Sci., Warsaw, 1998. To appear.

[GL] C. M. Gordon and J. Luecke. Knots are determined by theircomplements.Bull. Amer. Math.Soc. (N.S.)20(1989), 83–87.

[HM] U. Haagerup and H. J. Munkholm. Simplices of maximal volume in hyperbolicn-space.ActaMath.147(1981), 1–11.

[Hat] A. Hatcher. On the boundary curves of incompressible surfaces. Pacific J. Math.99(1982),373–377.

[HO] A. Hatcher and U. Oertel. Boundary slopes for Montesinos knots.Topology28(1989), 453–480.[HT] A. Hatcher and W. Thurston. Incompressible surfaces in2-bridge knot complements.Invent.

Math.79(1985), 225–246.[Hod] C. D. Hodgson.Degneration and regeneration of geometric structures on three-manifolds. PhD

thesis, Princeton, 1986.[Lue] J. Luecke. Dehn surgery on knots in the3-sphere. InProceedings of the International Congress

of Mathematicians, Vol. 1, 2 (Zurich, 1994), pages 585–594, Basel, 1995. Birkhauser.[MS] J. W. Morgan and P. B. Shalen. Valuations, trees, and degenerations of hyperbolic structures. I.

Ann. of Math. (2)120(1984), 401–476.[Mos] L. Moser. Elementary surgery along a torus knot.Pacific J. Math.38(1971), 737–745.[Mun] H. J. Munkholm. Simplices of maximal volume in hyperbolic space, Gromov’s norm, and Gro-

mov’s proof of Mostow’s rigidity theorem (following Thurston). InTopology Symposium, Siegen1979, number 788 in Lecture Notes in Math., pages 109–124. Springer, 1980.

[NZ] W. D. Neumann and D. Zagier. Volumes of hyperbolic three-manifolds.Topology24(1985), 307– 332.

[Por] J. Porti. Torsion de Reidemeister pour les varieteshyperboliques. Mem. Amer. Math. Soc.128(1997), x+139.

[Rat] J. G. Ratcliffe.Foundations of hyperbolic manifolds, volume 149 ofGraduate Texts in Mathe-matics. Springer-Verlag, New York, 1994.

[Sha] P. B. Shalen. Representations of 3-manifold groups. In Handbook of geometric topology. ElsevierPress.

[Thu] W. P. Thurston. The geometry and topology of 3-manifolds. Lecture notes. Revised version to bepublished by Princeton Univ. Press, 1978.

[Tol] D. Toledo. Representations of surface groups in complex hyperbolic space.J. Diff. Geom.29(1989), 125–133.

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arXiv:math/9802022v3 [math.GT] 25 Oct 1998