PSL (C)-character varieties and Seifert fibered cosmetic 2 surgeries Huygens C. Ravelomanana 7 1 0 2 Abstract n a J WestudysmallSeifertpossiblychiralcosmeticsurgeriesonnotnecessarilynull-homologousknot 5 inrationalhomologyspheres. UsingPSL2(C)-charactervarietytheorywegiveasharpboundonthe 2 number of slopes producing the same small Seifert manifold if the ambient manifold satisfies some representation theoretic conditions. ] T G Keyword: Dehn surgeries, cosmetic surgeries,character variety. . h t a 1 Introduction m [ 1 Let Y be a rational homology sphere and K be a knot in Y such that Y := Y \int(N(K)) is bound- K v ary irreducible and irreducible. Two Dehn surgeries Y (r) and Y (s) with distinct slopes are called 4 K K 8 “cosmetic”if they are homeomorphic. They are called“truly cosmetic”if the homeomorphism preserve 3 orientationand“chirallycosmetic”ifthehomeomorphismreverseorientation. Itisconjecturedthattruly 7 cosmetic surgery on such knot K does not exist [9]. In this paper we will focus on“general”cosmetic 0 . surgery and will not distinguish between chiral and truly cosmetic surgeries. 1 0 Let’s fix a slope s and define 7 1 : C(s)={slope r6=s| YK(r)∼=YK(s)}. v i X Here we allow the homeomorphism to reverse the orientation. If C(s) 6= ∅ then we have a cosmetic r surgery (possibly chiral). The following theorem then gives a bound on the number of element in C(s). a Theorem 1.1. Let K be a small knot in Y and YK(s) be small-Seifert. If Hom(π1(Y),PSL2(C)) contains only diagonalisable representations and ||s|| is not a multiple of s·λ. Then ♯C(s)≤1. Here , || || is a semi-norm on H1(∂YK;R) similar to the Culler-Shalen semi-norm, λ is the rational longitude of K and s·λ is the algebraic intersection. The knot K being small means that its exterior does not contain closed“incompressible surfaces”. The bound in the theorem is sharp since amphicheiral knot in S3 admits “chiral”cosmetic surgeries. Moreoverin [2], we canfind a constructionofa one-cuspedhyperbolic3-manifoldwitha pair ofdistinct slopeswhichgivesoppositelyorientedcopiesofthe lensspaceL(49/18). Aninfinitefamilyofhyperbolic manifolds which admit pairs {α,β} of reducible filling slopes, of which some pairs yield homeomorphic manifolds are presented in [10]. A straightforward consequence of the theorem which relates to the cosmetic surgery conjecture is the following. 1 Corollary 1.2. Under the hypothesis of the theorem, there are at most two distinct slopes which can produce the same oriented manifold after surgery. Result similar to this corollary has been proven for null homologous knot in 3-manifold with positive first Betti number [11], for knot in S3 [13, 12] and for knot in L-space integer homology sphere, [16]. The main ingredient for the proof of Theorem 1.1 is the theory of 3-manifold character variety started by Culler and Shalen [8] and which was essential in the proof of the“cyclic surgery theorem”[7], the “finite surgery theorem”[5] and is also an useful tool for studing topological properties of knot exterior. We willuse resultsfrom[1]togetherwithanormderivedfromPSL (C)-charactervarietysimilartothe 2 “Culler Shalen norm”. Organization. In section 2 we give some background on character varieties and the Culler-Shalen norm. The proof of Theorem 1.1 will be given in section 3. Acknowledgment. The author is gratefulto Steven Boyerfor advices, comments and helpful discus- sions. The author also thanks the University of Georgia for its support, the CIRGET group at UQA`M in Montr´eal. 2 Character varieties LetM denoteacompactorientable3-manifoldwhichisirreducible andboundaryirreducible withbound- ary consisting of a single torus. Typically M would be the knot exterior Y in the introduction. We K recall that irreducible means every embedded 2-sphere bounds a 3-ball and boundary irreducible means everysimpleclosedcurveon∂M whichboundsadiskinM boundsadiskin∂M. Aproperlyembedded surfaceinM willbe calledessential ifitisnotboundaryparallel,itisnota2-sphereandisπ -injective. 1 We say that M is small if it does not contain closed essential surfaces. A slope is the isotopy class of a simple closed curve on ∂M. Since M is boundary irreducible, π (∂M) ֒→ π (M) therefore using the 1 1 fact that H (M) ∼= π (∂M) we will think of a slope as both being an element of H (∂M), π (∂M) or 1 1 1 1 π (M). A slope will be called“boundary slope”if it correspondsto the boundary of anessentialsurface. 1 It will be called strict boundary slope”if it corresponds to the boundary of an essential surface which is not a (semi) fibre in any (semi) fibration of M over S1. The PSL (C)-representation variety of M is the set R(M) := hom(π (M),PSL (C)) equipped with 2 1 2 the compact-open topology. It consists of representations of π (M) to PSL (C). The space R(M) 1 2 has the structure of an affine complex algebraic set [8]. The group PSL (C) acts algebraically on 2 R(M) by conjugation. Two representations are called equivalent if they are conjugate to each other. If two representations are equivalent then they belong to the same irreducible component of R(M) [8]. The set of equivalence classes of representations corresponds to the quotient R(M)//PSL (C), where 2 the quotient is taken in the algebraic geometric category. In order to understand this set, Culler and ShalenintroducedthePSL (C)-character variety ofM usingthetracefunction. Foreachrepresentation 2 ρ∈R(M), the PSL2(C)-character of ρ is the map χρ defined by χρ :π1(M)→C, χρ(g) =trace(ρ(g))2. 2 ThesetofallcharactersX(M)= χ | ρ∈R(M) isalsoacomplexalgebraicsetinanaturalwaysuch ρ that the following map is regular, in the sense of algebraic geometry, (cid:8) (cid:9) t:R(M)−→X(M), t(ρ)=χ . ρ irr Moreoveritscorrespondstothe quotientR(M)//PSL (C)[8]. LetR (M)bethesubsetofirreducible 2 irr irr irr irr representations and let X (M) = t R (M) . The spaces R (M) and X (M) are Zariski open subsets of R(M) and X(M) respective(cid:16)ly [8]. (cid:17) For each γ ∈π (M) we consider the function defined by 1 fγ :X(M)−→C, fγ(χ)=trace(ρ(γ))2−4=χ(γ)−4. Thefunctionf isaregularfunctionandthezerosoff arethecharactersofrepresentationsρforwhich γ γ ρ(γ) is parabolic or ρ(γ) = [±Id]. We will use the same notation f for the restriction of f to a curve γ γ X ⊂X(M). 0 Let X ⊂ X(M) be a non-trivial irreducible curve component. Here non-trivial means that it contains 0 the character of an irreducible representation. Let X be the normalized projective completion of X . 0 0 There is an isomorphism between function fields b C(X )∼=C(X ), f 7−→f. 0 0 Wecanthendefinethedegreeoff tobethedegreeoff. Forx∈X wedenotebyZ (f )themultiplicity b b x γ ofxasazerooff . ByconventionZ (f )=∞iff ≡0. NowwedenoteΛ=π (∂M)seenasasubgroup γ x γ γ 1 b b b of π (M). We can also think of Λ as a lattice in H (∂M,R). An element γ ∈Λ satisfies [14], 1 1 b b b deg(f )= Z (f ). γ x γ xX∈Xb0 b b The degree is finite if f is non-constant on X . The key property of deg(f ) is that for each curve γ 0 γ X0 ⊂X(M) it defines a semi-norm ||.||X0 on H1(∂M,R) which for each γ ∈Λ satisfies [3, 14] b b b 0 if f |X is constant γ 0 ||γ|| = X0 (deg(fγ) if fγ|X0 6=0. This semi-norm is called the Culler-Shalen semi-norm associatedto the curve X . b 0 Note that if B is the ball of radius k centred at the origin, then B can be viewed as the unit ball for k k the norm 1||.|| thereforeB has the same propertiesas the unit ball. Some of these properties are[3, k X0 k Proposition 5.2, Proposition 5.3]: if it is a norm then the unit ball is a balanced convex polygon and if it is not a norm but is a non-zero semi-norm them the unit ball is an infinite band. LetX ,··· ,X beallthenon-trivialirreduciblecurvecomponentsinX(M). Wecandefinean“absolute” 1 k semi-norm ||.|| on H (∂M,R) by 1 ||.||=||.|| +···+||.|| . X1 Xk We will call this semi-norm the absolute semi-norm. There is a unique 4-dimensional subvariety R ⊂ R(M) for which t(R ) = X , see [3, Lemma 4.1]. If 0 0 0 ρ(α)=±I for some slope α, then we have an induced representationρ′ :π (M(α))−→PSL (C) and a 1 2 cohomology group H1(M(α);Adρ′) . 3 Let ν : Xν := X \{ideal points} −→ X be the map which corresponds to the affine normalization of 0 0 0 X . There is an affine normalization Rν −→ R , which we still denote by ν, such that the following 0 0 0 b diagram commutes. ν Rν R 0 0 ν t t Xν X 0 ν 0 The map tν and ν are all surjective, see [7]. Let N ⊂PSL (C) denote the subgroup 2 z 0 0 w ∗ ± ,± | z,w∈C ( 0 z−1 ! −w−1 0 ! ) For each γ ∈π (M) we are going to consider the following subset of X(M): 1 A(γ)= χ ∈X(M) | ρ(γ)=±I; ρ is non-abelian and conjugates into N . ρ (cid:8) (cid:9) B(γ)={χ ∈X(M) | ρ(γ)=±I; ρ is non-abelian and does not ρ conjugates into N }. Note that elements of A(γ) must be irreducible but not necessarily those of B(γ). The following Theorem and proposition relate Z (f ) and Z (f ) for two slopes α and r when x ∈ X x α x r 0 is a regular point. b b b Theorem 2.1. [1] Fix a slope α on ∂M and consider a non-trivial, irreducible curve X ⊂ X(M). 0 Suppose that x ∈ X is not an ideal point and corresponds to a character χ for some representation 0 ρ ρ∈R with non-abelian image and which satisfies ρ(α) ∈{±I}. Assume that H1(M(α);Ad )=0 and 0 ρ b ρ(π (∂M))*{±I}. 1 1. If β ∈π (∂M) and ρ(β)6=±I, then 1 Z (f )+1 if ρ conjugates into N, x β Z (f )≥ x α (Zx(fbβ)+2 otherwise. b b 2. If r ∈π (∂M) and Z (f )>Z (f ), then f |X 6=0, ρ(r)6=±I and 1 x α x β α 0 b b Zx(fβ)b+1 if ρ conjugates into N, Z (f )= x α (Zx(fbβ)+2 otherwise. b b The condition ρ(π (∂M))*{±I} may not be satisfied in general. For it to be true we will assume the 1 auxiliary assumption that the manifold Y has only diagonalisable PSL (C) representations. 2 4 Proposition 2.2. [7] Let α and β be non-zero elements of Λ. Suppose that x is a point of Xν such 0 that Z (f ) > Z (f ). Then for every ρ˜ ∈ Rν with tν(ρ˜) = x, the representation ρ = ν(ρ˜) satisfies x α x β 0 ρ(α)=±I. When we have zeros at ideal points we have the following property. Proposition 2.3. [7] Let x be an ideal point of X . Let α and β be non-zero elements of Λ. Suppose 0 that α is primitive and is not a boundary class, and that b Z (f )>Z (f ). x α x β Then there is a closed essential surface in M which is incompressible in M(α). 3 Proof of theorem 1.1 From now on we consider the case M =Y the knot exterior described in the introduction. K Lemma 3.1. Assume that rankZ(H1(YK)) = 1. Then for each ordinary point x ∈ X0 there is a representationρ∈R , with non-abelian image, such that χ =ν(x). 0 ρ b Proof. Let Z0 ⊂X(YK) (SL2(C)-charactervariety)be anirreducible curve componentof π−1(X0), and S a component of t−1(Z ). 0 0 Let X(Γ) be the SL (C)-character variety of a finitely generated group Γ. In [4, Proposition 2.8] it is 2 shown that if x is a reducible trivial character in a non-trivial curve inside X(Γ) then the first Betti numbersatisfiesb1(Γ)≥2. SincerankZ(H1(YK))=1byassumptionandb1(π1(YK))=rankZ(H1(YK)), any character in a non-trivial curve inside X(Y ) is non-trivial, in particular any element of Z is non- K 0 trivial. ThesameProposition2.8of[4]appliedtoπ (Y )impliesthatifacharacterz ∈Z isnon-trivial 1 K 0 then there is a representationρ∈S ∩t−1(z)with non-abelianimage. Since for eachx∈X , ν(x)∈X 0 0 0 we can take z ∈ π−1(ν(x)) to get such a representation ρ and then take the corresponding PSL (C) 2 b representationρ. Lemma 3.2. If Y is small then for each curve X ⊂X(Y ), ||.|| is not identically zero. K 0 K X0 Proof. By[3,Proposition5.5]if||.|| ≡0thenY containsaclosedessentialsurface,thisisnotpossible X0 K if Y is small. K Lemma 3.3. If s and r be two slopes on ∂YK such that π1(YK(s)) ∼= π1(YK(r)). Then there is a one to one correspondence between A(s) and A(r), and between B(s) and B(r), Proof. Let Ψ : π (Y (r)) → π (Y (s)) be an isomorphism, p : π (Y ) → π (Y (s)), and p : 1 K 1 K s 1 K 1 K r π (Y ) → π (Y (r)) be the obvious projections. Let χ ∈ A(s), we have a representation Φ (ρ) : 1 K 1 K ρ s π1(YK(s))→PSL2(C) obtained via the following factorisation of ρ 5 π (Y ) 1 K ρ p s π1(YK(s)) PSL2(C) Φ (ρ) s We also have an equivalent representation Φr(ρ) : YK(r) → PSL2(C) for the r-case. Let ρ′ be the composition ρ′ :=Φ(ρ)◦Ψ◦p r ρ′ π1(YK) PSL2(C) pr Φ(ρ) π (Y (r)) π (Y (s)) 1 K 1 K Ψ The maps p ,p and Ψ are all surjective so imρ = imρ′. In particular if ρ does not conjugate into N r s then neither does ρ′, and if ρ is irreducible then so is ρ′. The representation ρ′ satisfies ρ′(r) = ±I by construction. Next we need to check that if χρ1 =χρ2 then χρ′1 =χρ′2. We first assume that χ is an irreducible character. Therefore ρ =g ρ g−1 for some ρ1 2 1 g ∈SL (C). Then we deduce that 2 ′ ρ =Φ (ρ )◦Ψ◦p 2 s 2 r =Φ g ρ g−1 ◦Ψ◦p s 1 r = g Φ(cid:0)s(ρ1) g−(cid:1)1 ◦Ψ◦pr =(cid:0)g (Φs(ρ1)◦Ψ(cid:1)◦pr) g−1 =g ρ′g−1 1 which implies ρ′2 =g ρ′1 g−1. Therefore χρ′1 =χρ′2. If γ ∈π1(∂YK) we denote Xirr(γ) and Xred(γ) the sets Xirr(γ)= χ ∈X(Y ) | ρ(γ)=±I, and ρ is irreducible ρ K Xred(γ)=(cid:8)χ ∈X(Y ) | ρ(γ)=±I, and ρ is reducible (cid:9). ρ K (cid:8) (cid:9) We then have a well defined map F :Xirr(s)−→Xirr(r), ρ7−→ρ′. The map F sends A(s) to A(r), and B(s) to B(r). The next step is to extend F to the reducible representations. Now assume x = χ = χ is a reducible character. By analogy with [4] there exists a representation ρ1 ρ2 a:π1(YK)→C∗ such t−1(x)=Rxa∪Rxa−1 where Ra = AρA−1|A∈PSL (C), ρ∈Ua , x 2 x (cid:8) (cid:9) 6 Ra−1 = AρA−1|A∈PSL (C), ρ∈Ua−1 x 2 x n o a b Ua = representation ρ=± , x ( 0 a−1 !) Ua−1 = representation ρ=± a−1 b x ( 0 a !) If ρ1 and ρ2 are conjugate, by the same argument as for irreducible characters χρ′1 =χρ′2. Assume that ρ and ρ are not conjugate. Without loss of generality we can suppose that 1 2 a b a−1 b ρ =± and ρ =±A A−1, for some A∈PSL (C). 1 0 a−1 ! 2 0 a ! 2 Since Φ(ρ ) andρ , i=1,2 havethe same imagewe canuse the same matricesto representΦ(ρ ), i= i i i 1,2. Hence a◦Ψ◦p b◦Ψ◦p ′ r r ρ =Φ(ρ )◦Ψ◦p =± 1 1 r 0 a−1◦Ψ◦pr ! a−1◦Ψ◦p b◦Ψ◦p and ρ′ =Φ(ρ )◦Ψ◦p =±A r r A−1. 2 2 r 0 a◦Ψ◦pr ! Therefore trace(ρ′)=± a◦Ψ◦p +a−1◦Ψ◦p 1 r r trace(ρ′)=±(cid:2)a−1◦Ψ◦p +a◦Ψ◦p (cid:3) 2 r r It follows that trace(ρ′1)2 =trace(ρ′2)2, that is(cid:2)χρ′1 =χρ′2. Thus F is we(cid:3)ll defined on the set of reducible characters. Finally,we showthatF is bijective. If ρ′ ∈Xirr(r) (resp. Xred(r)) , wegetρ∈Xirr(s) (resp. Xred(s)) asfollow: wefirstdefineΦ (ρ)tobeΦ (ρ)=Φ (ρ′)◦Ψ−1 thenρ′ =Φ (ρ)◦p . Thisuniquelydetermine s s r s s ρ, therefore the map F is bijective. Lemma3.4. Letsbeaslopeon∂Y whichisnotaboundaryslope. AssumethatY (s)issmall-Seifert, K K YK is small, Y has only diagonalisablePSL2(C)-representationsand b1(YK(s))=0. If r is a slope such that YK(r)∼=YK(s), then ||r||=||s||. Proof. It is known from [6] that X(Y ) has no 0-dimensional component. Let X ,··· ,X be the curve K 1 k components of X(Y ). Since Y is small X(Y ) is the union of these components. If s ∈ Λ is not a K K K boundary slope, then Lemma 3.2 allows us to write the X -norm of s in terms of the zeros of f | for i s Xi each i∈{1,··· ,k} ||s||Xi = Zx(fs|Xci). xX∈Xci b 7 Let X be the abstract disjoint union of all the X , i ∈ {1,··· ,k}, then we have the following formula i for the absolute semi-norm b c ||s||= Z (f ) x s xX∈Xb b where f is understood to be the restriction to the appropriate component. Let x ∈ X, we define the s number m and m to be x 0 b b m =min Z (f ) | γ ∈Λ\{0} , and m = m . x x γ 0 x n o xX∈Xb b We can then deduce ||s||=m −m + Z (f )=m + Z (f )−m . 0 0 x s 0 x s x xX∈Xb xX∈Xb (cid:16) (cid:17) b b Let us suppose that x is an ideal point of X ⊂X. If Z (f )−m >0 then Z (f )> Z (f ) for some 0 x s x x s x γ γ ∈ Λ\{0}. Since s is primitive and is not a boundary class, Lemma 2.3 implies that there is a closed b b b b b surface in Y which is incompressible in Y (s). This situation does not occur if we assume Y (s) is K K K small-Seifert with b (M(s))=0. Therefore we always have Z (f )−m =0 at an ideal point. 1 x s x Let x∈X be an ordinary point, x is contained in some X0 andbby Lemma 3.1 there is a representation ρ∈R with non-abelian image, such that χ =ν(x). Let ρ˜=ν−1(ρ), we have the following equality 0 ρ b b ν(tν(ρ˜))=t(ρ)=ν(x). The normalization map ν : Xν → X is an“isomorphism”outside singular points, so if x is a smooth 0 0 point then tν(ρ˜) = x. This smoothness is provided by Theorem A of [4]. A direct consequence of this is that for an ordinary point x, ν(x) is contained in only one irreducible component. Therefore if we consider instead of X, the“natural”union X ∪···∪X , we can write the absolute semi-norm of s as 1 k b c c ||s||= Z (f ). x s x∈Xc1X∪···∪Xck b NowifweassumethatZ (f )>m thenbyLemma2.2therepresentationρ=ν(ρ˜)satisfiesρ(s)=±I. x s x Since Y (s) is Small seifert, H1(Y (s);Ad ) = 0. If we add the extra condition that Y have only K K ρ b AbelianPSL (C)-representationsthenρ(Λ)*{±I}andallthe hypothesisofTheorem2.1aresatisfied. 2 In particular if m =Z (f ) for some r ∈Λ\{0} then f |X 6=0, ρ(r)6=±I and x x r s 0 Zx(fbs)−mx =Zx(fs)−Zx(fr)=b1 if ρ conjugates into D, (Zx(fbs)−mx =Zx(fbs)−Zx(fbr)=2 otherwise. Since X(Y ) is the union of its 1-dimensional components we have K b b b ||s||=m + Z (f )−m =m +A(s)+2B(s). 0 x s x 0 xX∈Xb (cid:16) (cid:17) b Finally ♯A(s)=♯A(r) and ♯B(s)=♯B(r) from Lemma 3.3, since YK(s)∼=YK(r). Therefore ||r||=||s||. 8 Proof of Theorem 1.1. Let s be an exceptional slope on ∂Y and r ∈C(s). K We canassumethatb (Y (s))=0sincethereisonlyoneslopewhichproducesamanifoldwithpositive 1 K first Betti number. Let us suppose that s is a boundary slope. From [7, Theorem 2.0.3] we have the following possibilities: (i) Y (s) contains a closed essential surface of strictly positive genus. K (ii) Y (s) is the connected sum of two lens spaces. K (iii) There is a closed essential surface S ⊂ Y which compresses in Y (s) but which remains incom- K K pressible in Y (δ) as long as ∆(s,δ)>1. K (iv) YK(s)∼=S1×S2. Since YK(s) is small-Seifert with b1(YK(s))=0, only (iii) can occur. Then the fact that YK(s)∼=YK(r) implies thatS alsocompressesinY (r) so∆(s,r)≤1. The condition∆(s,r)≤1 implies thatthereare K at most three of such slopes. Assume r ∈C(s) is distinct from s, then ∆(s,r)=1 and we have either C(s)={r,s+r}, or C(s)={r}. Now by homological reasonthere is a constant c independent of s such that K |H1(YK(s);Z)|=cK∆(s,λ) andsimilarlyforr,see[15]fordetails. ThereforesinceH1(YK(s);Z)=H1(YK(r);Z)=H1(YK(s+r);Z), we must have ∆(s,λ)=∆(r,λ)=∆(s+r,λ). Thus allthreeelements s,r,s+r lie onthe sameline l inR2 whichis atfixeddistance fromthe rational longitude λ. This is impossible since s and r are linearly independent. It follows that the first case cannot occur and we have: C(s)={r}. Now if s is not a boundary slope we can apply Lemma 3.4 to obtain ||s||=||r||. Let k = ||s||, then every element of C(s) lies on the boundary of the ball B(0,k) of the absolute semi-norm. On the other hand if r ∈ C(s) then s and r must have the same µ component since they give homeomorphic manifolds. Hence they also lie on a line l parallel to λ. From [3, Proposition 5.2, Proposition5.3] B(0,k) is either a convex polygon or an infinite band. In each cases the line l intersect ∂B(0,r) twice (at s and r) unless k is a multiple of s·λ. 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