An equivariant Casson invariant of knots in homology ... - CiteSeerX

An equivariant Casson invariant of knots in homology spheres ∗Julien Marché †March 21, 2005AbstractFrom the construction of the Casson invariant of homology spheres using intersection onconfiguration spaces, we propose a construction of an equivariant Casson invariant for a knotK homologous to 0 in rational homology sphere M. Our construction is adapted from C.Lescop ([7]) and use the same ideas in an equivariant setting. We show that the invariantwe obtain is similar to the 2-loop part of the rational Kontsevich integral, and may be equal.We check that it codes the Casson invariants of the cyclic ramified coverings of M along Kin exactly the same way as found by S. Garoufalidis and A. Kricker for the 2-loop invariant([2]).1 IntroductionDealing with universal finite type invariants of links in 3-manifolds, two approaches are usedessentially, both linked with perturbative Chern-Simons theory. The first one uses integrationon configuration spaces of points lying in a prescribed way on the link and the 3-manifold. Theidea of the construction is rather elegant but the actual computation of the invariants are verytechnical (see for instance [5, 6, 7]). The second approach is based on the Kontsevich integral.Although computations are difficult in this setting too, the combinatorics of the Kontsevichintegral make it a more tractable invariant. By instance, formulas for invariants of lens spaces,some Seifert fibered spaces and torus knots can be achieved [1, 10]. We will focus on a propertyof finite type invariants of a knot in a 3-manifold called rationality which were shown using theKontsevich integral (see [3]).Let us explain this briefly: for any knot K homologous to 0 in a rational homology sphere M,the (unwheeled) Kontsevich integral is a series of uni-trivalent graphs with rational coefficientshaving the following form: Z # (M, K) = exp( ∑ i>0 z i). In this formula z i is a series of connectedgraphs having first Betti number (or loop number) equal to i. The first series z 1 is directlylinked to the Alexander polynomial ∆ of the pair (M, K) and hence is well understood. Therationality theorem states that each z i for i > 1 is coded by a finite sum of diagrams with loopnumber i with rational expressions of the form P (t)∆(t)attached to the edges (see [3]). We call thissum zirat (M, K). For a fixed integer i, this invariant is a kind of polynomial invariant, and codesa big part of the Kontsevich integral. This gives a good motivation for computing it, startingwith the 2-loop part z2 rat (M, K).∗ keywords: finite type invariants, configuration space, Kontsevich integral, rationality, Casson invariant.2000 Mathematics Subject Classification: 57M27.† Institut de Mathématiques de Jussieu, Équipe “Topologie et Géométries Algébriques” Case 7012, UniversitéParis VII, 75251 Paris CEDEX 05, France. e-mail: marche@math.jussieu.fr1

This happens to be a difficult but prometting task because this so-called 2-loop part of theKontsevich integral has some striking topological properties. Among them, we can mentionbeautiful formulas for cyclic branched coverings of M along K ([2]), surgery formulas alongclaspers([3, 9]), simple cabling formulas for degree 0 cablings or toric cablings ([11, 12]), and adegree inequality relating the degree of the 2-loop part to the genus of the knot when M is S 3([13]).A bad feature of this invariant is its intricate definition: to obtain the rational form of theKontsevich integral, one has to use a surgery presentation of the pair (M, K), and an equivariantversion of the Kontsevich integral together with an equivariant version of formal gaussianintegration.The attempt of this paper is to give a simpler, geometric definition of this invariant. This isjust an attempt as we will construct a new invariant and just conjecture that it coincide with the2-loop part of the Kontsevich integral, but we will give good reasons to believe this conjecture.The idea for the definition of this invariant is to go back to configuration space integrals.When constructing the first non-trivial invariant of a rational homology sphere M, i.e the Cassoninvariant, one can replace the integral of some differential form by an intersection of 3 suitable4-cycles in the configuration space of 2 points in M. This was explained to us by C. Lescop([7]). We can adapt this construction to the equivariant setting, considering cycles with twistedcoefficients.In order to give an understandable plan of the paper, let us present briefly this construction:we will give it in much more details in the article. Let K be an oriented knot homologous to0 in a rational homology sphere M. The complement of a tubular neighborhood of K in Mis a manifold N satisfying H 1 (N, Z) = Z. Let us fix a trivialization on N which is standardin some way on the boundary of N. Then, we consider the compactified configuration spaceC 2 (N) of 2 points in N. If these points are very close to each other or if one of them is in theboundary of N, we can define a vector in S 2 which is in some sense the direction from the firstpoint to the second. For almost any point in S 2 , the preimage of this point is a 3-cycle in theboundary of C 2 (N) with twisted coefficents which bounds a 4-cycle in C 2 (N). By intersectingthree such twisted 4-cycles we obtain an object which we identify with a 2-loop diagram. Wenote it Casson(M, K). More precisely, we obtain the following theorem.Let (M, K) be a pair with Alexander polynomial ∆. We note A = ⊗ 3Z[t ±1 ] Z[t±1 ,1∆(t) ] wherethe ring Z[t ±1 ] acts diagonally on the tensor product. The group S 3 acts on A by permutingthe factors and the group Z/2Z acts by replacing t by t −1 simultaneously in each term of thetensor product. Let Res r : A → Q be the map defined by the formula below for all integers rsuch that ∆ does not vanish on r-th roots of unity:Res r f ⊗ g ⊗ h = 1 r∑ω r =η r =δ r =ωηδ=1f(ω)g(η)h(δ).Theorem. For all pairs (M, K) there is a well-defined element of A obtained as before andnoted Casson(M, K). This element is invariant through the action of S 3 × Z/2 and contains theCasson invariant of the cyclic branched covers of M along K of order r noted Σ r (M, K) in thefollowing way.λ(Σ r (M, K)) = 1 8 σ r(M, K) + 1 6 Res r Casson(M, K).In this formula, σ r (M, K) is the sum of the equivariant signatures of (M, K) at r roots of unity.2

Remark 1.1. • Such a formula were proven in [2] with the invariant Casson(M, K) replacedby z2 rat (M, K). This is a good reason to believe that we have indeed the identityCasson(M, K) = z2 rat (M, K) but this is not a proof because there are non-trivial 2-loopdiagrams having zeros r-residues for all r, for instance t 2 ⊗ t ⊗ 1 − t ⊗ 1 ⊗ 1 ∈ A.• The motivation of this construction was to give a simple, geometric construction of the2-loop part of the Kontsevich integral: as a bonus, our construction has another advantage:the invariant Casson(M, K) has obviously “integral” coefficients, which is not clear for the2-loop part of the Kontsevich integral. For instance, if we can identify Casson(M, K) with(M, K), we will prove Rozansky’s integrality conjecture about 2-loop polynomial (see[14]).z rat2Aknowledgments: I would like to thank S. Garoufalidis, C. Lescop, G. Masbaum, T. Ohtsuki,and P. Vogel for inspiring many ideas of this article and for stimulating discussions.Contents1 Introduction 12 Construction of the equivariant Casson invariant 32.1 Configuration space of 2 points and its boundary . . . . . . . . . . . . . . . . . . 32.2 Definition of the cycle Σ a . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.3 The cycle Σ a bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.4 Construction of the invariant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.5 About trivializations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 Recovering Casson invariants 113.1 Forgetting coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.2 The lift map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.3 A formula for Casson(Σ r (M, K)) . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 Construction of the equivariant Casson invariant2.1 Configuration space of 2 points and its boundaryLet us fix once for all an oriented banded knot K homologous to 0 in a rational homology sphereM. We suppose that K has self-linking 0. By blowing up the knot in M, we obtain a 3-manifoldN whose boundary is identified with the unit normal bundle of K in M. Imitating the article[7], we want to identify a neigbourhood of the boundary of N with some subset of R 3 .We will describe this subset very explicitely: let S 1 be the unit circle of C embedded in thestandard way in C × R = R 3 , Z be the axis {0} × R and C be the circle Z ∪ {∞} in R 3 ∪ {∞}.Consider the map θ : R 3 \ Z → [0, 1) defined in the following way: any point M in R 3 \ Z hasa unique projection onto the circle S 1 which we call f(M). Then, we set θ(M) = ||M−f(M)||||M+f(M)||3

where || · || denotes the euclidian norm. This map extends continuously to C by setting θ(x) = 1for all x in C. Moreover the level surfaces of θ are tori for parameters in (0, 1), and we haveS 1 = θ −1 (0) and C = θ −1 (1).Let U be the solid torus θ −1 ([ 1 4 , 1]) in S3 and ϕ be an orientation preserving embedding of Uin M mapping C to K such that a parallel copy of C is sent to the prefered longitude of K.Then we haveM = U ∪ θ −1 ([ 1 4 , 3 4 ]) T M,Kwhere we have set T M,K = M \ ϕ(θ −1 (( 3 4 , 1])). The manifold T M,K is isomorphic to N and thechoice of ϕ is unique up to isotopy. We will say that ϕ is a “good” embedding. Our furtherconstructions are isotopy invariants, hence the precise choice of good embedding does not matter.Definition 2.1. Let us fix a pair (M, K) and a good embedding ϕ : U → M. A trivialization ofM \ K standard along the knot K is a trivialization of M \ K which coincide with the standardtrivialization of R 3 on ϕ(U \ C). In formulas, we have τ| ϕ(U\C) = ϕ ∗ τ R 3.In section 2.7, we will show that such a trivialization always exists.A direct computation shows that we have H 1 (M \K, Z) = Z. The generator of this cohomologygroup is the linking number with K and hence depends on an orientation of K. It is representedby a map f : M \ K → S 1 which is unique up to homotopy. If we choose a good embeddingϕ : U → M, we can suppose that the restriction of f ◦ ϕ to U \ C is exactly the previousprojection map f from R 3 \ Z to S 1 .As the pair (M, K) is fixed once for all, we simplify the notations and set T = T M,K . Let ∆ bethe diagonal in T × T . We call C 2 (T ) the manifold with corners obtained by blowing up ∆. Letus fix a trivialization τ standard along the knot K. We construct a new map p : ∂C 2 (T ) → S 2 .The set ∂T is identified with the level surface θ −1 ( 3 4 ). The boundary of C 2(T ) is decomposedinto three manifolds: the unit normal bundle of ∆ noted U∆, the spaces T × ∂T and ∂T × T .• For y ∈ ∂T and x ∈ T , we set x ′ = (1 − ε(θ(x)))f(x) + ε(θ(x))x where ε is a smooth mapfrom [0, 1] to itself which is equal to 0 on [0, 1 3 ] and to 1 on [ 2 3, 1]. In that way, the pointx ′ always belong to R 3 .Then, we set p(x, y) =y−x′||y−x ′ ||and p(y, x) =• For a point (x, v, w) in U∆, we setx′ −y||x ′ −y|| .p(x, v, w) = τ x(w − v)||τ x (w − v)|| .Lemma 2.2. The map p : ∂C 2 (T ) → S 2 is well-defined and continuous.Proof. The proof is clear as the 3 definitions of p coincide when the arguments are both closeto the boundary of T . The figure 1 may help visualizing the map p.2.2 Definition of the cycle Σ aIn this section, we use the map p to define twisted 3-cycles in ∂C 2 (T ).4

UZ∂TS 1CFigure 1: Identification of the boundary of the knotThe map f × f gives a map from C 2 (T ) to S 1 × S 1 . Hence, in this section, we will considererhomology with coefficients in a Z[t ±11 , t±1 2 ]-module. Let E be the Z[t±1 1 , t±1 2 ]-moduleZ[t ±11 , t±1 2 ]/(t 1t 2 − 1). It will be helpful to keep in mind the following interpretation: an elementof the covering space of C 2 (T ) associated to E is a couple points (x, y) in T with a homologyclass of path from f(x) to f(y) in S 1 .For all regular values a of p in S 2 , we have a codimension 2 submanifold Σ a of ∂C 2 (T ). We wantto show that for some values of a, the submanifold Σ a is a 3-cycle in ∂C 2 (T ) with coefficientsin E.Consider the covering π : S 1 ×R → S 1 ×S 1 defined by π(x, t) = (x, exp(it)x). The submanifoldΣ a is an element of H 3 (∂C 2 (T ), E) if and only if it lifts to this covering, in other words, thereis a map ˜f which makes the following diagram commute: f×fΣ a efS 1 × RπS 1 × S 1Lemma 2.3. Let n be the vector (0, 0, 1) ∈ S 2 ⊂ R 3 . There is a disc D around n in S 2 suchthat for any regular value a of p in D ∪ −D, the map f : Σ a → S 1 × S 1 lifts through π.Proof. Fix a regular value a close to n. Suppose we have some element (x, y) ∈ Σ a . If x and yare very close, then f(x) and f(y) are very close and there is no difficulty to lift the diagonalof S 1 through π. But when x and y are not close, it means that p is close to the standard mapy−x||y−x||(x, y) ↦→ . Geometrically, we see that if the vector p(x, y) is close to n, then f(x) and f(y)are close to each other. It means that we can globally lift Σ a through π. Of course, in order toidentify Σ a with an element of H 3 (∂C 2 (T ), E), we have to specify which lift we choose, but thereis a prefered choice: as f(x) and f(y) are close to each other on the circle, we choose the pathfrom f(x) to f(y) which stays in the interval [f(x), f(y)]. We see that the situation is exactlythe same close to −n, hence the lemma is proven.5

In that way, we obtain the following short exact sequence.0 H 2 (C 2 (T )) H 3 (∂C 2 (T )) H 3 (C 2 (T )) 0∼ sH 3 (U∆, ∂)H 3 (U∆)(2)Our aim is to show that the cycle Σ a maps to 0 in H 3 (C 2 (T )). We will use for that theintersection product 〈, 〉 : H 3 (∂C 2 (T )) ⊗ H 3 (∂C 2 (T )) → H 1 (∂C 2 (T ), F ⊗ Z F ) p 1∗→ H 1 (T, Z) ⊗ Λ.Thanks to the exact sequence (2) and the section s, the space H 3 (∂C 2 (T )) is naturally splittedin two parts noted H 2 (C 2 (T )) and H 3 (U∆). For dimensional reasons, the intersection formrestricted to these spaces is 0. Let us examine the pairing between H 2 (C 2 (T )) and H 3 (U∆).We have seen that the Z factor of the sequence (1) is represented by the product of two disjointSeifert surfaces. Geometrically, this class cannot intersect with H 3 (U∆). This proves that thepairing only depends on the image of the first factor in H 3 (U∆, ∂U∆) = Λ. This pairing isdefined by the formula 〈1, α〉 = α.Let us explain how Σ a appears in this decomposition: by definition of Σ a as a preimage ofa by p, it must have self-intersection 0. Suppose Σ a maps to f + x. Then, f is obtained byintersecting Σ a on a small sphere centered at x ∈ T , hence we have f = 1. The self-intersectionof Σ is then 2x which must be 0.This proves that Σ a bounds a 4-cycle in H 4 (C 2 (T ), ∂C 2 (T ); F ). This cycle is unique becauseof the vanishing of H 4 (C 2 (T )).2.4 Construction of the invariantLet D be the disc of lemma 2.3 and a, b, c be three regular values of p in D ∪ −D. Then, Σ a , Σ band Σ c are three disjoint 3-cycles in H 3 (∂C 2 (T ), F ). Each 3-cycle bounds a 4-cycle ∆ a , ∆ b , ∆ cin C 2 (T ), unique up to a 5-boundary. As they have disjoint boundaries, we can form theirintersection. We get an element of H 0 (C 2 (T ), ⊗ 3Z F ). We compute H 0(C 2 (T ), ⊗ 3Z F ) = ⊗ 3A Λ.In the last tensor product, the action of A is supposed to be diagonal.For convenience, let us give a geometric interpretation of this intersection. Let a, b, c be threeregular values of p in D, Σ i and ∆ i the corresponding cycles. Let f : T → S 1 be the usualrepresentant of the generator of H 1 (T, Z). A point in the interior of C 2 (T ) is represented bya 4-tuple (x, y, γ, g) where x and y are two distinct points of T , γ is a homotopy class of pathin S 1 from f(x) to f(y) and g is an element of Λ. We identify the 4-tuple (x, y, tγ, g) with the4-tuple (x, y, γ, tg) where tγ is the path γ with an extra loop around the circle.Then, at each triple intersection point of ∆ a , ∆ b and ∆ c , we get three 4-tuples such as(x, y, γ a , g a ), (x, y, γ b , g b ) and (x, y, γ c , g c ). By gluing these three paths in x and y we obtaina graph Θ with edges decorated by elements of Λ. More formally, we obtain an element ofΛ. We will call A this space.⊗ 3ALemma 2.5. The element ∆ a ∩∆ b ∩∆ c is independent of the choice of regular values in D∪−D.It only depends on the pair (M, K) and the homotopy class of trivialization τ and we note itCasson(M, K, τ). Let S be the automorphism of Λ sending t on t −1 . The group G = Z/2Z × S 3acts on A by the formula S k σ · x 1 ⊗ x 2 ⊗ x 3 = S k x σ(1) ⊗ S k x σ(2) ⊗ S k x σ(3) . The elementCasson(M, K, τ) is G-invariant.7

Proof. Suppose we have two triples of regular values of p in D ∪ −D, namely a, b, c and a ′ , b ′ , c ′ .Associated to these values, there are 6 3-cycles Σ a , Σ b , Σ c and Σ ′ a, Σ ′ b , Σ′ c. We note as usual∆ a , ∆ b , ∆ c the 4-cycles bounding the Σ i in C 2 (T ). The idea of the proof is to define cycles ∆ ′ iby adding cobordisms W i between ∆ i and ∆ ′ i in ∂C 2(T ) × [0, 1]. This is possible if and only ifthey represent the same homology class in H 3 (∂C 2 (T ), F ). From the lemma 2.4, we now thatall cycles Σ bound in C 2 (T ) and hence come from H 4 (C 2 (T ), ∂C 2 (T ); F ). In the exact sequence(1), we have seen that all cycles Σ in the middle term map to 1, hence, they all differ by amultiple of the generator η of the first space, that is a product of two disjoint Seifert surfaces.Such a multiple has to be 0 for the following reason: consider a parallel in the boundary of T ,and γ the surface made of pair of points lying on the parallel. We check immediately that ηintersect γ in precisely one point whereas the cycles Σ never intersect γ as they are made ofpairs of points in almost vertical position. This finally proves that all cycles Σ are homologousto each other.Let W i be a cobordism between the Σ i and the Σ ′ i . The difference between the intersectioncomputed from the Σ ′ i and from the Σ i is equal to the triple intersection of the W i . But we candecompose the intersection W a ∩ W b ∩ W c in the following sum: W a ∩ Σ b × [0, 1] ∩ Σ c × [0, 1] +Σ ′ a × [0, 1] ∩ W b ∩ Σ c × [0, 1] + Σ ′ a × [0, 1] ∩ Σ ′ b × [0, 1] ∩ W c. As all cycles Σ are disjoint, beingpreimages of distinct points of S 2 , we see that this triple intersection vanishes.We can now conclude the lemma: as the construction of the invariant Casson(M, K, τ) doesnot depend on the choice of regular values in D, it only depends on the homotopy class of τ,hence Casson(M, K, τ) is actually an invariant of the triple (M, K, τ).It is clear that the element Casson(M, K, τ) is S 3 invariant because the action of the permutationcorresponds to a permutation of the regular values a, b, c and the invariant does not dependof the choice of a triple.We use the same idea to prove the S-invariance of Casson(M, K, τ).Then, let us consider the automorphism S of C 2 (T ) sending (x, y) to (y, x). This automorphismchanges orientation. We check that p(S(x, y)) = p(y, x) = −p(x, y) on the boundary. Whenwe compute the Casson invariant by taking points close to −n instead of n, we change thepoints a, b, c to −a, −b, −c and then the cycles into SΣ a , SΣ b , SΣ c with opposite orientations.As we have seen above, the intersection remains the same, hence we have ∆ a ∩ ∆ b ∩ ∆ c =S∆ a ∩ S∆ b ∩ S∆ c . The map S changes the orientation of the cycles and of the ambient space,hence it does not change the sign of the intersection. Using the map S acting on A, we obtainS Casson(M, K, τ) = Casson(M, K, τ). We have finally proven that Casson(M, K, τ) is welldefinedand G-invariant.2.5 About trivializationsThe definition of Casson(M, K, τ) depends on the choice of the trivialization τ up to homotopy.In this part, we will adapt many considerations of [7].First, let us show that such a trivialization always exists. Any 3-manifold has a trivialization,hence let us fix an arbitrary one τ : T M → R 3 respecting orientations. We consider themonodromy map π 1 (M) → π 1 (SO(3)) = Z/2. This map is a group homomorphism with valuesin a commutative group, hence it factors through H 1 (M, Z). As the knot K is supposed to behomologous to 0, its monodromy has to be trivial.8

Let us examine this trivialization in the decomposition M = U ∪ T . It differs from thestandard trivialization of U \ ∞ by composition with a map G : U \ ∞ → SO(3). Moreover,both trivializations have trivial monodromies along a generator of π 1 (U \ ∞), hence the mapG induces a trivial map on fundamental groups. As π 2 (SO(3)) = 0 and U is homotopic to agluing of S 1 and S 2 in one point, the map G is contractible. This implies that we can deformthe trivialization τ on U \ ∞ such that it coincides with the standard trivialization of R 3 , whichis the definition of a trivialization of M \ K standard along the knot.Next, fix the pair (M, K) and the good diffeomorphism ϕ : U → M. Two trivializations ofT standard along the knot differ by a map G : (T, ∂T ) → (SO(3), 1). Let us consider the setH of all homotopy class of maps from (M, K) to (SO(3), 1). This is a group using the productstructure of SO(3).By adapting results of [7], lemma 2.32 p.28, we obtain the following lemma:Lemma 2.6. The group H is abelian and the map deg : H ⊗ Q → Q is an isomorphism.Proof. Let G be an element of H, and consider the map induced by G on fundamental groups.It gives an element of hom(π 1 (M, K), Z/2) = H 1 (M, K; Z/2). Moreover, the map H →H 1 (M, K; Z/2) is a group homomorphism. This map fits in the following exact sequence:deg0 Z iH H 1 (M, K; Z/2) 0The element ρ = i(1) is constructed in the following way: choose a point x inside T and fixa small ball B around x. Identifying B with the complement of 1 in S 3 , we deduce from thestandard covering map S 3 → SO(3) a map B → SO(3) sending the boundary of B on 1 andhaving degree 2. By extending this map on T by one, we obtain a element of H called ρ ofdegree 2. Hence, the map deg ◦i is the multiplication by 2. The fact that the above sequence isexact is a consequence of the cell decomposition of the pair (M, K).The subgroup [H, H] maps to 0 on H 1 (M, K; Z/2) and is a subgroup of Z of degree 0. Hence,it is 0 and H is a commutative group. This finally proves that the map deg2: H ⊗ Q → Q is anisomorphism.We will use the standard map p 1 from the set of trivializations of rational homology ballsto Z given in [7]. Here, we define p 1 (M, K, τ) as the p 1 invariant of the trivialization τ onB = T ∪ (U ∩ B(0, 2)), where B(0, 2) is a closed ball of radius 2 centered at 0.Briefly, it is defined in the following way: fix a 4-manifold W bounding B × {0} ∪ B(0, 2) ×{1} ∪ ∂B(0, 2) × [0, 1]. Using the trivialization τ of M together with the tangent vector of thecomponent [0, 1], we can construct a trivialization of the tangent space of W restricted to itsboundary. The obstruction of extending the complexified trivialization of T W ⊗ C to W is bydefinition p 1 (W, τ)[W, ∂W ]. We set p 1 (τ) = p 1 (W, τ) − 3σ(W ). This integer does not depend onW .We will use the following identity proved in [7]:Fix a triple (M, K, τ): for all maps G : (T, ∂T ) → (SO(3), 1), we have p 1 (M, K, Gτ) −p 1 (M, K, τ) = −2 deg(G).Our next task is to show that the same kind of formula appears for the invariant Casson(M, K, τ).Precisely, we will show the following lemma.9

Lemma 2.7. Let Θ be the element 1 ⊗ 1 ⊗ 1 ∈ A. Fix a triple (M, K, τ) and map G : (T, ∂T ) →(SO(3), 1), we have Casson(M, K, Gτ) − Casson(M, K, τ) = − 1 2deg GΘ.Proof. Fix a triple (M, K, τ). First, we show that the map from H to A defined by G ↦→Casson(M, K, Gτ) − Casson(M, K, τ) is a group homomorphism. Let G be an element of H, Σ aa cycle in H 3 (∂C 2 (T ), F ) associated to the trivialization τ and Σ ′ a the cycle associated to thetrivialization Gτ. We remark that these cycles are homologous. Indeed, as they differ only onthe submanifold U∆ of ∂C 2 (T ), their difference is in H 3 (U∆, F ). But they both map to 0 inH 3 (C 2 (T ), F ) which is isomorphic to H 3 (∂C 2 (T ), F ) through the inclusion. This finally impliesthat they are homologous.We can compute the invariant Casson(M, K, Gτ) in the following way. Consider the manifoldC 2 + (T ) obtained by thickening the component U∆ of the boundary of C 2(T ). Precisely,we have C 2 + (T ) = C 2(T ) ∪ U∆ × [0, 1] where we identify the manifold U∆ in the first factorwith U∆ × {0} and for all x ∈ ∂U∆ we identifiy {x} × [0, 1] with {x}. This new manifoldis obviously diffeomorphic to C 2 (T ). We call ∂ 1 C 2 + (T ) and ∂ 2C 2 + (T ) the submanifolds correspondingrespectively to U∆ × {0} and U∆ × {1}. We consider three cycles Σ a , Σ b , Σ c as lyingin ∂ 1 C 2 + (T ) and Σ′ a, Σ ′ b , Σ′ c as lying in ∂ 2 + C 2(T ). As Σ i is homologous to Σ ′ i , we can find threecobordisms Π i in U∆×[0, 1] from Σ i to Σ ′ i . Geometrically, this implies that we have the identityCasson(M, K, Gτ)−Casson(M, K, τ) = Π a ·Π b ·Π c . Let us take another element G ′ of the groupH. We find some 4-cycles Π ′ i which define cobordisms from Σ i to Σ G′i , the cycles defined with G ′ τ.to Σ GG′iBy applying the map G to the manifold U∆×[0, 1], we obtain a cobordism GΠ ′ i from ΣG iwhich has the same triple intersection. This proves that Π a ∪ GΠ ′ a · Π b ∪ GΠ ′ b · Π c ∪ GΠ ′ c is equalto Π a · Π b · Π c + Π ′ a · Π ′ b · Π′ c. Hence, the map sending G to Casson(M, K, Gτ) − Casson(M, K, τ)is a group homomorphism.The above map take its values in A which has no torsion, therefore it is completely determinedby its value on a generator of H ⊗ Q, for instance the element ρ.Let us construct the cobordisms Π i in such a situation. As ρ has support in a small ball, wecan suppose that the cobordisms Π i are trivial except in that ball. Let B be the standard ballin R 3 with radius 2π. The map ρ from B to SO(3) sends a point x at distance d from the originto the rotation of axis x and angle d. Call x, y, z be the coordinates of R 3 . We use the samenames for the corresponding directions in S 2 . We have the following situation: there are threecycles Σ x , Σ y and Σ z in B × S 2 which are respectively equal to B × {x}, B × {y} and B × {z}and three 3-cycles Σ ′ x, Σ ′ y and Σ ′ z obtained from the others by applying the map ρ. The cyclesΣ i and Σ ′ i coincide on ∂B × S2 , and then define a cycle in ∂B × S 2 × [0, 1]. We know thereare cobordisms Π i from Σ i to Σ ′ i and we need to compute their intersection. Let us add someintermediary steps:(Σ x , Σ y , Σ z ) → (Σ x , Σ y , Σ ′ z) → (Σ x , Σ ′ y, Σ ′ z) → (Σ ′ x, Σ ′ y, Σ ′ z)When two cycles are unchanged in some step, we use the trivial cobordism. As the cycles Σ iare disjoint, and so are the cycles Σ ′ i , only the middle term may give some contribution. Hencewe need to compute the following intersection Σ x × [0, 1] ∩ Π y ∩ Σ ′ z × [0, 1].From the defintion of ρ, we see that Σ x ∩ Σ ′ z is equal to C × {x} where C is a circle in Blying in the plane x = z. We also compute that Σ ′ y ∩ Σ x is equal to D × {x} where D is a circlelying in the plane x = y. Hence, generically, Σ x × [0, 1] ∩ Π y is a 2-cycle in B × {x} × [0, 1]bounding D × {x} × {1}. The triple intersection that we have computed is hence the linkingnumber between C and D which is -1.10

From the lemma 2.7, we obtain the following proposition:Proposition 2.8. Let (M, K) be a pair and τ be a trivialization standard along the knot. Thequantity Casson(M, K) = Casson(M, K, τ) − 1 4 p 1(τ)Θ ∈ A does not depend on the choice oftrivialization τ.3 Recovering Casson invariantsFrom now we have not shown anything about the non-triviality of the invariant Casson(M, K).This will come from the fact that our construction is a generalization of Lescop’s construction.3.1 Forgetting coefficientsWe recall that for a pair (M, K) with normalized Alexander polynomial ∆, we have constructedthe invariant Casson(M, K) as an element of A = ⊗ 3Z[t ±1 ] Z[t±1 , 1 ∆]. The evaluation at 1 fromZ[t ±1 , 1 ∆ ] to Z[ 1∆(1) ] induces a map Res 11 : A → Z[∆(1)] called evaluation.Proposition 3.1. For all pairs (M, K), we have Res 1 Casson(M, K) = 6 Casson(M).Proof. This is a consequence of Lescops articles [7, 8]. By surgery arguments, she showed thatthe first term of the invariant Z(M) she constructs is equal to 1 2λ(M) where λ is the Cassoninvariant. When we forget coefficients, our construction reduces strictly to her construction, upto the order of the automorphism group of the graph C which is 12. This proves the assertion.Hence, the invariant Casson(M, K) contains at least the Casson invariant of the underlyingmanifold. We will show in this section a formula relating the Casson invariant of a pair (M, K)and the Casson invariant of its cyclic branched coverings. It will prove that Casson(M, K) andz 2 (M, K) are both generating functions for the Casson invariant of cyclic branched coverings ofM along K. This gives a good motivation to believe that these invariants are actually equal.3.2 The lift mapLet us give an algebraic definition of the map Lift r given in [2].Consider a pair (M, K) with Alexander polynomial ∆. For any integer r, we can considerthe cyclic branched covering of M along K of order r. We note Σ r (M, K) the pair made ofthis manifold and the preimage of K. It may happen that the new manifold is not a rationalhomology sphere. This occurs exactly when the polynomial ∆ vanishes on some r-th root ofunity.From now, we suppose that this situation does not occur: we will say that the pair (M, K) isr-regular. We can compute the Alexander polynomial of the new pair Σ r (M, K) by remarkingthat its Alexander module is precisely H 1 (M \ K, A) where we replace the action of t in A bythe action of θ = t r .We have the following situation. The ring A r = Z[θ ±1 ] is included in A. By adding a primitiver-th root of unity ξ in A, we obtain a Galois extension A r ⊂ A[ξ] with Galois group Z/r. Thismap extends to A r [ 1 ∆ r] → A[ξ, 1 ∆ ] where we have set ∆ r(θ) = N(∆) = ∏ i ∆(ξi t).11

Hence, the Alexander polynomial of Σ r (M, K) is just the norm of ∆.We can define the Lift r map easily in this setting. Let A be the space ⊗ 3A A[ 1 ∆] where A actsdiagonally on the three copies of A[ 1 ∆]. This space has a natural algebra structure given by theformula a ⊗ b ⊗ c · a ′ ⊗ b ′ ⊗ c ′ = aa ′ ⊗ bb ′ ⊗ cc ′ .By defining A r = ⊗ 3A rA r [ 1 ∆ r] we obtain a new algebra with an inclusion map A r → A. Byadding the r-th root ξ to A, we obtain a Galois extension with Galois group (Z/r) 3 /(Z/r). Inthese notations, the map Lift r is just the map 1 r Tr : A → A r.Let us see how to compute it more concretely. Let x = P ∆ ⊗ Q ∆ ⊗ R ∆be an element of A. As thepolynomial ∆ r is the norm of ∆, there is a polynomial G ∈ A such that ∆ r = ∆G. We computex = GP∆ r⊗ GQ∆ r⊗ GR∆ r. The element ∆ r lies in A r and hence is not affected by the trace. Theremaining elements are combination of monomials like t a ⊗ t b ⊗ t c and moreover, we can supposethat c is 0. The trace of such elements is precisely Tr t a ⊗ t b ⊗ 1 = ∑ i,j (ξi t) a ⊗ (ξ j t) b ⊗ 1 =r 2 δ r|a δ r|b t a ⊗ t b ⊗ 1 = r 2 θ a/r ⊗ θ b/r ⊗ 1. This formula gives us a direct way for computing theLift r map.3.3 A formula for Casson(Σ r (M, K))We prove in this section a formula relating the Casson invariant of (M, K) and the Cassoninvariant of Σ r (M, K) which is completely analogous to the formula of [2].First, we recall that there is a well defined invariant σ(M, K) : S 1 → Z called signature whichis a locally constant map jumping at roots of the Alexander polynomial. For any integer r, wewrite σ r (M, K) = ∑ ω r =1σ(M, K)(ω).Proposition 3.2. For any integer r and any r-regular pair (M, K), we haveCasson(Σ r (M, K)) = 3 4 σ r(M, K)Θ + Lift r Casson(M, K).Proof. Fix an integer r and a r-regular pair (M, K). We also choose a trivialization τ of Mstandard along the knot K. Let T be the manifold obtained as usual from (M, K) by removinga regular neighbourhood of the knot and T r the corresponding manifold for Σ r (M, K). There isa covering map π r : Σ r (M, K) → (M, K). In the standard identifications of the boundaries of Tand T r with subsets of R 3 , we can suppose that π r is the map from C × R to itself defined by theformula π r (z, t) = (z r , t). The problem is that the trivialization π ∗ rτ is not standard along theknot of the pair Σ r (M, K). Precisely, the difference is described by a map from ∂T r to SO(3)with values in rotations around the z-axis. Let f be the map from T to S 1 used to construct themap p, by composing the trivialization π ∗ rτ with the rotation G(x) of angle −rf(x) around theaxis z, we obtain a trivialization of T r standard along the knot. We use it to define a map p r ,cycles Σ r i and ∆r i and then the invariant Casson(Σr (M, K), Gπ ∗ rτ). We would like to comparethis invariant with the following one:Let ∆ a ′, ∆ b ′, ∆ c ′ be the twisted 4-cycles in C 2 (T ) whose triple intersection give Casson(M, K, τ).We obtain corresponding cycles in C 2 (T r ) by taking the preimage by π r × π r of the cycles ∆ i ′and intersecting them. We remark that although there is no projection from C 2 (T r ) to C 2 (T ),there is a well defined preimage of a twisted cycle. Let us call π ∗ r(Σ i ′) the cycles of ∂C 2 (T r )obtained in this lifting procedure.As in lemma 2.5, to identify the intersections ∆ r a ∩∆ r b ∩∆r c with π ∗ r(∆ a ′)∩π ∗ r(∆ b ′)∩π ∗ r(∆ c ′), itis sufficient to show that the cycles Σ r a, Σ r b , Σr c, π ∗ rΣ a ′, π ∗ rΣ b ′, π ∗ rΣ c ′ are homologous and disjoint.12

It is clear that all these cycles are homologous. To ensure that they are disjoint, it is sufficientto take regular values a, b, c, a ′ , b ′ , c ′ having distinct scalar product with n, as we have the identityp(π r x, π r y) · n = p r (x, y) · n.We now compute Casson(Σ r (M, K), Gπ ∗ rτ) from the second definition.Let us look closely how the preimage map works on twisted 0-cycles of C 2 (T ): a 0-cycle is apair of distinct points x, y in T together with a path from f(x) to f(y) in S 1 and a coefficient1∆(t) , where ∆(t) is the Alexander polynomial. As before, we note ∆ r the norm of ∆, which is apolynomial in θ = t r satisfying ∆ r = ∆G. The coefficient 1 ∆ may be written G ∆ r, and 1 ∆ ris fixedby the lifting map, being a polynomial in θ. Once we have taken a preimage x r of x in T r , thepolynomial G gives a combination of paths from f(x r ) to a point f(y r ) for some y r ∈ T r . Whenwe divide it by ∆ r , and take the sum over all choices of x r , we obtain precisely the lift map fromC 0 (C 2 (T ), F ) → C 0 (C 2 (T r ), F r ). It is now easy to describe what happens when we take a tripleintersection of the preimages of ∆ a , ∆ b and ∆ c . Fix a triple intersection point of the ∆ i . Itcorresponds to two points x, y in C 2 (T ) with three paths from f(x) to f(y) with coefficients. Weneed to do the lift operation for the three paths and to take the remaining intersection points.It means that the point x r is the same for the three lifts (and we have r choices), and the targetpoints y r in T r need to coincide. This is precidely the definition of the map Lift r .This prove the formula Casson(Σ r (M, K), π ∗ rτ) = Lift r Casson(M, K, τ). When we normalizethe trivializations, we have the following correction factor:Casson(Σ r (M, K)) = − 1 4 (p 1(Gπ ∗ rτ) − rp 1 (τ)] + Lift r Casson(M, K)The equality p 1 (Gπ ∗ rτ) − rp 1 (τ) = 3σ r (M, K) is classical (see for instance [4]). Let us sketchthe proof briefly: fix a triple (M, K, τ) and a 4-manifold W bounding M. We suppose that Whas signature 0, such that p 1 (τ) is exactly the obstruction of extending some trivialization ofthe complexified tangent bundle of M restricted to its boundary. Take a Seifert surface S of Kin M, and push it into W . By taking the cycling covering of W over S of order r, we obtaina manifold W r bounding Σ r (M, K). The obstruction of extending the trivialization Gπ ∗ rτ ismultiplied by r, but the manifold W r has no longer signature 0 but σ r (M, K) (see [4]). Hence,from the definition of the p 1 invariant, one has p 1 (Gπ ∗ rτ) = rp 1 (τ) − 3σ r (M, K). Hence thecorrection factor in the formula above is precisely 3 4 σ r(M, K) as announced.References[1] D. Bar-Natan and R. Lawrence. A rational surgery formula for the LMO invariant.arXiv:math.GT/0007045, to appear in Israel J. Math.[2] S. Garoufalidis and A. Kricker. Finite type invariants of cyclic branched covers. Topology,43:1247–1283, 2004. arXiv:math.GT/0107220.[3] S. Garoufalidis and A. Kricker. A rational noncommutative invariant of boundary links.Geom. Topol., 8:115–204, 2004. arXiv:math.GT/0105028.[4] L. H. Kauffman. On knots, volume 115 of Annals of Mathematics Studies. PrincetonUniversity Press, Princeton, NJ, 1987.13

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