FLOER HOMOLOGY OF BRIESKORN HOMOLOGY SPHERES NIKOLAI SAVELIEV Abstract. Every Brieskorn homology sphere (p; q; r) is a double cover of the 3{sphere rami ed over a Montesinos knot k(p; q; r). We relate Floer homology of (p; q; r) to certain invariants of the knot k(p; q; r), among which are the knot
signature and the Jones polynomial. We also de ne an integer valued invariant of integral homology 3{spheres which agrees with the {invariant of W. Neumann and L. Siebenmann for Seifert bered homology spheres, and investigate its behavior with respect to homology 4{cobordism.
Let p; q, and r be pairwise coprime positive integers. A Brieskorn homology 3{ sphere (p; q; r) is the link of the singularity of f ?1 (0), where f : C 3 ! C is a map of the form f (x; y; z ) = xp + yq + z r . The complex conjugation in C 3 acts on (p; q; r) turning it into a double branched cover of S 3 branched over a Montesinos knot k(p; q; r). The Floer homology groups In (); 0 n 7, are abelian groups associated with an integral homology 3{sphere , see [12]. The Floer homology groups of (p; q; r) were studied by R. Fintushel and R. Stern [11] who showed in particular that the groups I ((p; q; r)) are free abelian, and if an denotes the rank of In ((p; q; r)) then an = 0 for odd n. Therefore, a0 + a2 + a4 + a6 = 2 ((p; q; r)) (1) where ((p; q; r)) is the Casson invariant, see [33]. We add to this knowledge the following result. Theorem 1. Let (p; q; r) be a Brieskorn homology sphere represented as a double branched cover of S 3 rami ed over a Montesinos knot k(p; q; r). Let an denote the rank of the free abelian group In ((p; q; r)), and sign k(p; q; r) the signature of the knot k(p; q; r). Then (2) ?a0 + a2 ? a4 + a6 = 41 sign k(p; q; r): Conjecturally, the groups In ((p; q; r)) are 4{periodic, that is, In ((p; q; r)) = In+4 ((p; q; r)) for all n. For instance, this is the case for all homology spheres (p; q; pqn 1) obtained by surgery on (p; q){torus knots. If this conjecture is true in general, the equations (1) and (2) will give a closed form formula for Floer homology of Brieskorn homology spheres, and will answer Question 3.5 of [23] (which is attributed to M. Atiyah). 1991 Mathematics Subject Classi cation. 57M25, 57R80. Key words and phrases. Floer homology, Seifert manifolds, Montesinos knots, Casson invariant, Jones polynomial, homology cobordism, knot signature. Research was supported in part by NSF Grant DMS-97-04204. 1
Theorem 1 establishes links between Floer homology and the Jones polynomial. The following result is obtained by combining equation (2) with D. Mullins formula for the Casson invariant of two{fold branched covers, see [21]. Theorem 2. Let (p; q; r) be a Brieskorn homology sphere represented as a double branched cover of S 3 with the branch set a Montesinos knot k(p; q; r). Let an denote the rank of the free abelian group In ((p; q; r)) and Vk the Jones polynomial of the knot k = k(p; q; r). Then d 1 a2 + a6 = 12 dt ln Vk (t): t=?1 For an integral homology 3{sphere , we de ned in [27] an invariant by the formula 7 X n n rankQ In (): (3) () = 21 (?1) n=0 For a Brieskorn homology sphere (p; q; r) with the orientation of an algebraic link, the invariant is of the form ((p; q; r)) = 21 (?a0 + a2 ? a4 + a6 ) where an = rank In((p; q; r)) as before. It should be also mentioned that the {invariant changes sign with the change of orientation. We conjectured in [27] that the { invariant equals the {invariant of W. Neumann [22] and L. Siebenmann [31] for all Seifert bered homology spheres. Now we can prove that this conjecture is true. Theorem 3. For any Brieskorn homology 3{sphere (p; q; r), the {invariant de ned by (3) and the {invariant coincide, ((p; q; r)) = ((p; q; r)). Theorem 3 is equivalent to Theorem 1 due to the observation that ((p; q; r)) = 1 sign k(p; q; r), see Section 7. The statement of the conjecture for a general Seifert 8 bered homology sphere (a1 ; : : : ; an ) now follows from Theorem 3 by induction with the help of the so called splicing additivity. Namely, it has been proved in [10], see also [27], that ((a1 ; : : : ; an )) = ((a1 ; : : : ; aj ; p)) + ((q; aj +1 ; : : : ; an )) where j is any integer between 2 and n ? 2, and the integers q = a1 aj and p = aj+1 an are the products of the rst j and the last (n ? j ) Seifert invariants, respectively. The same additivity holds for the invariant , see [27]. Corollary 4. For any Seifert bered homology 3{sphere (a1 ; : : : ; an ), the { and the {invariants coincide, ((a1 ; : : : ; an )) = ((a1 ; : : : ; an )). Thus, the invariant is an extension of the {invariant from Seifert bered homology spheres to arbitrary integral homology spheres. In fact, the invariant is de ned for a broader class of homology 3{spheres, including all integral graph homology spheres. The question whether equals on this bigger class remains open. W. Neumann conjectured in [22] that the {invariant is in fact a homology cobordism invariant. In [28] and [29] we proved this conjecture for certain classes of Seifert bered homology spheres. Due to the identi cation = , the corresponding results hold for the {invariant. ( +1)( +2) 2
2
Theorem 5. (1) Let (a1 ; : : : ; an) be a Seifert bered homology sphere homology cobordant to zero. Then ((a1 ; : : : ; an )) 0. (2) Let = (p; q; pqm 1) be a surgery on a (p; q){torus knot, and suppose that is homology cobordant to zero. Then () = 0. (3) For all Seifert bered homology 3{spheres which are known to be homology cobordant to zero, including the lists of Casson-Harer [7] and Stern [32], () = 0. It should be also mentioned that the homology cobordism invariance of has been checked as well on certain classes of integral homology 3{spheres which are not graph manifolds, see [30]. The idea of proof of Theorem 1 is shortly as follows. In [33], C. Taubes proved that, for any integral homology 3{sphere , () = 1=2 (I ()): (4) The Casson's {invariant on the left is de ned using a Heegaard splitting of and SU(2){representation spaces. The number on the right is the Euler characteristic of Floer homology I (); it can be de ned using SU(2) gauge theory as an in nite dimensional generalization of the classical Euler characteristic. Let now = (p; q; r) be endowed with an involution induced on C 3 by the complex conjugation. We work out {invariant versions of both invariants in (4) for = (p; q; r). We use a {invariant Heegaard splitting of and the corresponding equivariant SU(2){representation spaces to de ne an invariant () in a manner similar to that of A. Casson. On the other hand, an equivariant gauge theory produces an equivariant Euler characteristic which we denote (). Note that the latter can be de ned without actually working out any Floer homology. Then, a {invariant version of the Taubes result (4) is that ((p; q; r)) = 1=2 ((p; q; r)): (5) Our next step is to show that ((p; q; r)) = 18 sign k(p; q; r). We achieve this by pushing equivariant representations of the manifolds in the {invariant Heegaard splitting down to their quotients. We note that the push-down representations are only de ned on the complements of the rami cation sets, and they map all their meridians to trace-free matrices in SU(2). After that we identify ((p; q; r)) with one forth of the Casson{Lin invariant h(k(p; q; r)) of the knot k(p; q; r). The invariant h(k) was de ned in [17] for an arbitrary knot k by using trace{free representation spaces associated with the knot k complement. X.-S. Lin proved in [17] that the invariant h(k) equals one half of the knot k signature, which in our case immediately implies that ((p; q; r)) = 81 sign k(p; q; r). The crucial observation is that all the representations of 1 (p; q; r) are in fact equivariant, and the {invariant just counts them with signs di erent from those de ned by Casson for his {invariant. Finally, we identify 1=2 () with () by comparing the equivariant spectral
ow used to de ne () with the \regular" spectral ow of [33] with the help of the G{index theorem of [8]. It should be pointed out that many results and de nitions in the paper hold in a more general situation. A description of the relevant results will appear elsewhere. Recently, several manuscripts have appeared that discuss a similar circle of ideas. 3
Weiping Li in [16] constructed a symplectic Floer homology theory for knots in S 3 whose Euler characteristic equals the Casson{Lin invariant. Olivier Collin in his Oxford Thesis developed an instanton Floer homology for knots via orbifolds singular along a given knot. The Euler characteristic of such a Floer homology gives a Tristram{Levine equivariant signature of the knot, and for a particular choice, one gets the signature of the knot. An interesting question is the relation of this Floer homology for knots to the Floer homology of an associated cyclic branched cover. C. Herald and S. Cappell, R. Lee, and E. Miller have studied generalizations of the Casson{Lin invariant to representations of the knot groups with a xed trace, not necessarily zero. The paper is organized as follows. It begins with a short introduction to the relevant topology of Brieskorn homology spheres and Montesinos knots. In Section 2 we investigate equivariant SU(2){representation spaces associated with a { invariant Heegaard splitting of (p; q; r) and de ne the {invariant. The equality ((p; q; r)) = 81 sign k(p; q; r) is proven in Section 3. Section 4 is devoted to equivariant gauge theory and the de nition of the {invariant. In Section 5 we express ((p; q; r)) in terms of Floer homology as ((p; q; r)) = 2 ((p; q; r)). The equality of and 2 for Brieskorn homology spheres is proved in Section 6. One should mention that all the equalities of the invariants above are proved up to a universal (){sign. The sign is xed at the end of Section 6, which completes the proof of Theorem 1. Section 7 contains a detailed discussion on the {invariant of W. Neumann and L. Siebenmann. I thank Boris Apanasov, Josef Dodziuk, Igor Dolgachev, Ron Fintushel, Michael Gekhtman, Walter Neumann, Liviu Nicolaescu, Frank Raymond, Shuguang Wang, and Arthur Wasserman for helpful conversations and useful remarks. 1. Topology of Brieskorn homology spheres In this section we describe shortly topology of Brieskorn homology spheres and Montesinos knots. More detailed account will be given in Section 7. Let p; q; r be relatively prime integers greater than or equal to 2. The Brieskorn homology sphere (p; q; r) is de ned as the algebraic link (p; q; r) = f (x; y; z ) 2 C 3 j xp + yq + z r = 0 g \ S 5 ; where S 5 is a sphere in C 3 . This is a smooth naturally oriented 3{manifold with H ((p; q; r)) = H (S 3 ). Moreover, (p; q; r) is Seifert bered, see [24], with the Seifert invariants f b; (p; b1 ); (q; b2 ); (r; b3 ) g such that b1 qr + b2 pr + b3 pq = 1 + bpqr: (6) The complex conjugation on C 3 obviously acts on (p; q; r) as : (p; q; r) ! (p; q; r); (x; y; z) 7! (x; y; z): (7) The xed point set of this action is never empty. The quotient of (p; q; r) by the involution is S 3 , with the branching set the so called Montesinos knot k(p; q; r), see [6], [20], or [31]. The knot k(p; q; r) can be described by the following diagram 4
b half-twists
p,b1
q,b 2
r,b3
Figure 1 where a box with ; in it stands for a rational ( ; ){tangle, see [6], Fig. 12.9. The parameters b; (p; b1 ); (q; b2 ); (r; b3 ) are the Seifert invariants of the corresponding (p; q; r). According to [6], Theorem 12.28, these parameters together with (6) determine the knot k(p; q; r) uniquely up to isotopy. 2. The invariant In this section we rst shortly recall the de nition of Casson's {invariant following the exposition [1]. Our {invariant for (p; q; r) will be an equivariant version of . To de ne it, we introduce a {invariant Heegaard splitting of (p; q; r). Then we de ne the corresponding equivariant representation spaces and investigate closely the representation space of 1 (p; q; r). It turns out that all the representations in the latter space are {invariant. After computing the necessary dimensions and checking the transversality condition, we de ne the {invariant as an intersection number of equivariant representation spaces. 2.1. Casson invariant. Let M be an oriented homology 3{sphere with a Heegaard splitting M = M1 [ M2 where M1 and M2 are handlebodies of genus g 2 glued along their common boundary, a Riemann surface M0 . Let R(Mk ) = Hom(1 Mk ; SU(2))= SU(2); k = 0; 1; 2; ;; be the set of conjugacy classes of irreducible representations of 1 Mk is SU(2). Each R(Mk ); k = 0; 1; 2, is naturally an oriented manifold. The dimension of R(M1 ) and R(M2 ) is 3g ? 3, and that of R(M0 ) is 6g ? 6. The inclusions M0 Mk ; k = 1; 2, induce embeddings R(Mi ) R(M0 ). The points of intersection of R(M1 ) with R(M2 ) are in one-to-one correspondence with R(M ). If the intersection is transversal we de ne the Casson's {invariant as the integer X " (8) (M ) = 21 2R(M ) 5
where " = 1 is a sign obtained by comparing the orientations on T R(M1 ) T R(M2 ) and T R(M0 ). Note that the intersection is transversal if = (p; q; r), see R. Fintushel and R. Stern [11], Proposition 2.5. The formula (8) de nes the invariant up to a (1){sign, which only depends on the orientation of , and changes by (?1) if the orientation of changed. One can show that j((2; 3; 5))j = 1; the invariant is then normalized by the requirement that ((2; 3; 5)) = +1. An explicit formula for ((p; q; r)) and more generally, for ((a1 ; : : : ; an )), is given in [23], Lemma 1.5. 2.2. {invariant Heegaard splitting. We are going to construct a {invariant Heegaard splitting of (p; q; r). Let us rst x notations. By we denote a Brieskorn homology sphere (p; q; r) and by : ! the involution de ned by (7). The projection on the quotient space will be denoted by , so : ! = = S 3 . The projection maps the xed point set Fix() onto a Montesinos knot k = k(p; q; r) S 3 . The knot k S 3 can be represented as the closure of a braid on n strings. We think about k as consisting of the braid and n untangled arcs in S 3 forming its closure, see Figure 2. Let S S 3 be an embedded 2{sphere in S 3 splitting S 3 in two 3{balls, B1 and B2 , with common boundary S , and such that the entire braid belongs to int B1 . The intersection k \ S consists of 2n points, P1 ; : : : ; P2n . P1 k
S
P2g+1 β
B2
B1
P2g+2 P2
Figure 2 Now, we de ne a Heegaard splitting = M1 [M M2 as follows: M1 = ?1 (B1 ); M2 = ?1 (B2 ); M0 = ?1 (S ): (9) Obviously, M1 and M2 are handlebodies branched over the braid and the n untangled arcs, respectively. Their common boundary is M0 , which is a closed surface branched over the points P1 ; : : : ; P2n 2 S . The genus g of M0 can be gured out by comparing Euler characteristics, 2 (S ) = (M0 ) + ( Fix() \ S ); 0
6
where (S ) = 2; ( Fix() \ S ) = 2n, and (M0 ) = 2 ? 2g. Therefore, g = n ? 1. The Heegaard splitting we just de ned is {invariant in the sense that (Mk ) = Mk for i = 0; 1; 2. 2.3. Representations of Brieskorn homology spheres. Let (p; q; r) be a Brieskorn homology 3{sphere, and R((p; q; r)) = Hom (1(p; q; r); SU(2))= SU(2) (10) the space of the conjugacy classes of irreducible representations of its fundamental group in SU(2). In this subsection we are concerned with describing the space R((p; q; r)) and the involution induced on the representation space by . We will follow R. Fintushel and R. Stern [11] and rst identify the group SU(2) with the group S 3 of unit quaternions in the usual way, so that i 0 0 1 0 i i = 0 ?i ; j = ?1 0 ; k = i 0 : Under this identi cation, the trace tr(A) of an element A 2 SU(2) coincides with the number 2 Re A; A 2 S 3 , and the mapping r : S 3 ! [0; ]; r(A) = arccos(Re A), is a complete invariant of the conjugacy class of the element A. This conjugacy class is a copy of S 2 in S 3 unless A = 1, in which case the conjugacy class consists of just one point. The fundamental group 1 (p; q; r) has the following presentation, see [6] or [11], 1 (p; q; r) = hx; y; z; h j h central ; xp = h?b ; yq = h?b ; z r = h?b ; xyz = h?b i; where b1 ; b2 ; b3 , and b satisfy (6). Specifying an irreducible representation : 1 (p; q; r) ! SU(2) amounts to specifying a set f (h); (x); (y); (z )g of unit quaternions. In fact, we only need to specify the rst three quaternions in this set because (z ) will then be expressed in their terms as (z ) = ( (x) (y))?1 (h)?b . Since h is central and the representation is irreducible, (h) = 1. Let us denote "i = (h)bi = 1; i = 1; 2, and "3 = (h)b ?rb = 1. Then the relations xp = h?b ; yq = h?b and (xy)r = hb ?rb imply the following restrictions on (x) and (y): r( (x)) = `1 =p; r( (y)) = `2 =q; r( (x) (y)) = `3 =r; (11) where `i is even if "i = 1, `i is odd if "i = ?1, and 0 < `1 < p; 0 < `2 < q; 0 < `3 < r. After conjugation, we may assume that (x) = ei` =p . The quaternions (y) and (x) (y) should lie in their respective conjugacy classes, S2 = r?1 (`2 =q) and S 3 = r?1(`3 =r). On the other hand, (x) (y) lies in (x) S2 , therefore, in order for (x) and (y) to de ne a representation, the intersection (x) S2 \ S3 must be non-empty. Since (x) S2 is a 2-sphere centered at (x), the intersection (x) S2 \ S3 in S 3 (if non-empty) is a circle. This circle parametrizes a whole collection of representations 0 coming together with such that r( 0 (x)) = r( (x)); r( 0 (y)) = r( (y)), and r( 0 (x) 0 (y)) = r( (x) (y)). In fact, all these representations are conjugate to each other by simultaneous conjugation of (x) and (y) by the complex circle S 1 S 3 . This can be seen from the following elementary technical lemma. 1
2
3
1
2
3
1
7
3
Lemma 6. Let and be irreducible representations of the group 1(p; q; r) in
SU(2) such that (1) (h) = (h) and (x) = (x) 2 C , (2) r( (y)) = r( (y)), and (3) r( (x) (y)) = r( (x) (y)). Then the representations and are conjugate to each other, that is there exists a unit quaternion c such that (t) = c (t) c?1 for all t 2 1 (p; q; r): Moreover, the quaternion c may be chosen to be a complex number, c 2 C . Corollary 7. ( Compare [11] ). The representation space R((p; q; r)) is nite. The involution : (p; q; r) ! (p; q; r) induces the involution on the fundamental group, : 1 (p; q; r) ! 1 (p; q; r);
h 7! h?1 ; x 7! x?1 ; y 7! xy?1 x?1 ; z 7! xyz ?1 y?1 x?1 ;
see [6], Proposition 12.30, which in turn induces an involution on the corresponding representation space ( [ ] stands for conjugacy class), : R((p; q; r)) ! R((p; q; r)); [ ] 7! [ 0 ]; (12) 0 where (t) = ( (t)); t 2 1 (p; q; r): (13) Proposition 8. If 0 : 1(p; q; r) ! SU(2) is a representation de ned by the formula (13) then there exists an element 2 SU(2) such that 2 = ?1 and 0 (t) = (t) ?1 for all t 2 1 (p; q; r): The element is de ned uniquely up to multiplication by 1, and the elements corresponding to di erent representations are conjugate to each other. In particular, the action (12) on the space R((p; q; r)) of the conjugacy classes of irreducible representations of 1 (p; q; r) in SU(2) is trivial. Proof. After conjugation we may assume that (x) is a unit complex number. Then 0 (x) = ( (x)) = (x?1 ) = (x)?1 is a complex number as well. Let us consider another representation, , de ned as a conjugate of by the unit quaternion j , (t) = j ?1 0 (t) j for all t 2 1 (p; q; r): Next we want to verify that the representations and satisfy the conditions of Lemma 6. We have (x) = j ?1 0 (x) j = j ?1 (x)?1 j = j ?1 (x) j = j ?1 j (x); since (x) 2 C ; = (x): 8
Note that, for any unit quaternion a, we have r(a) = r(a) = r(a?1 ). Therefore, r( (y)) = r(j ?1 0 (y) j ) = r( 0 (y)) = r( (xy?1 x?1 )) = r( (x) (y)?1 (x)?1 ) = r( (y)?1 ) = r( (y)); and similarly, r( (x) (y)) = r(j ?1 0 (x) 0 (y) j ) = r( 0 (x) 0 (y)) = r( (x)?1 (x) (y)?1 (x)?1 ) = r(( (x) (y))?1 ) = r( (x) (y)): Since r( (h)) is equal to r( (h)) we can apply Lemma 6 to the representations and to nd a complex number c 2 C such that (t) = c (t) c?1 for all t 2 1 (p; q; r). Since (t) = j ?1 0 (t)j = c (t)c?1 , we get that 0 (t) = (t)?1 with = jc. If we apply twice to the representation , we will get again because is an involution. Therefore, is conjugate to itself by the element 2 . But the representation is irreducible, therefore, 2 = 1. The following easy computation shows that 2 is in fact ?1, 2 = jcjc = jj cc = ?jcj2 = ?1: The uniqueness of up to 1 follows from the irreducibility of . The fact that the elements de ned by di erent representations are conjugate follows from the fact that tr(jc) = 0 for any complex number c. 2.4. Heegaard splitting and representation spaces. Let = (p; q; r) be a Brieskorn homology sphere with a {invariant Heegaard splitting of genus g, = M1 [M M2 ; constructed in (9). According to Proposition 8, for any : 1 (p; q; r) ! SU(2) there exists an element 2 SU(2) such that = ?1 and 2 = ?1. Generally speaking, depends on but it follows from Proposition 8 that the elements corresponding to di erent 's are conjugate to each other. Therefore, the following de nition makes sense. For each manifold Mk ; k = 0; 1; 2, in the splitting (9), we de ne the equivariant representation space, R(Mk ) = f : 1Mk ! SU(2) j is irreducible and ( (t)) = (t)?1 ; t 2 1Mk g=S1 ; (14) as the space of all irreducible {invariant representations modulo the adjoint action of the 1{dimensional Lie group S1 SU(2) consisting of all g 2 SU(2) such that g 0= g. Note that whenever 0 = hh?1 ; h 2 SU(2), we have that R (Mk ) = R (Mk ), hence each of the spaces (14) only depends on the conjugacy class of and not on itself. 0
9
We rst describe the space R (M0 ). The classical result of Luroth [18] and Hurwitz [14], see also [4], Theorem 3.4, implies that a double branched cover of S 2 is uniquely determined by the cardinality of its branch set. Thus we are free to visualize the covering M0 ! S 2 by using the model shown in Figure 3, the action of being the rotation by 180 about the horizontal symmetry axis.
t
2g
t 2g-1
t3
t2
t1
Figure 3 One can easily see that
1 M0 = h t1 ; : : : ; t2g j t1 t2g = t2g t1 i where the generators t1 ; : : : ; t2g are represented by the respective curves in Figure 3. The induced involution : 1 M0 ! 1 M0 acts on the generators ti by inverting them, (ti ) = t?1 i . Let us introduce the surface M0 = M0 nf D; (D) g where D and (D) are a pair
of disjoint open discs as shown in Figure 4.
D t 2g+1
σ (D)
Figure 4 The group 1 M0 is a free group on 2g + 1 generators, and obviously R(M0 ) Hom(1 M0; SU(2))=S1 where Hom (1 M0 ; SU(2)) consists of all the representations : 1 M0 ! SU(2) such that = Ad . Let t2g+1 be the curve shown in Figure 4. Then (t2g+1 ) = 10
t?1 2g+1 , and the curves t1 ; : : : ; t2g+1 together give a basis of the free group 1 M0 . This
choice of generators allows us to make the following identi cation,
Hom (1 M0 ; SU(2)) = f (T1 ; : : : ; T2g+1 ) 2 SU(2)2g+1 j Ti?1 = Ti ?1 ; i = 1; : : : ; 2g + 1 g:
Lemma 9. Let 2 SU(2) be such that 2 = ?1. Then the subset S2 of SU(2) consisting of all a 2 SU(2) such that a?1 = a?1 is a two-dimensional sphere. Proof. Any element of SU(2) with 2 = ?1 has zero trace. Therefore, there exists x 2 SU(2) such that = xjx?1 (remember that we identify SU(2) with the group of unit quaternions), and then the map a 7! x?1 ax establishes a di eomorphism between Sj2 and S2 . Now, Sj2 consists of all b 2 SU(2) such that bj = jb and jbj2 = 1, hence b = u + vi + wk for some real u; v; w with u2 + v2 + w2 = 1. This of course determines a 2{sphere.
Therefore, Hom (1 M0 ; SU(2)) = (S 2 )2g+1 ; in particular, Hom (1 M0 ; SU(2)) gets a smooth manifold structure from (S 2 )2g+1 . A (2g + 1){tuple (T1 ; : : : ; T2g+1 ) 2 (S 2 )2g+1 de nes an equivariant representation of 1 M0 in SU(2) if T1 T2g+1 = 1. We call a point (T1 ; : : : ; T2g+1 ) 2 (S 2 )2g+1 reducible if there is a matrix A 2 SU(2) such that ATi A?1 are all diagonal matrices for i = 1; : : : ; 2g +1, and call it irreducible otherwise.
Lemma 10. Let S2 SU(2) be the 2{sphere of unit quaternions T such that T ?1 = T?1 , and the product map,
: (S2 )2g+1 ! SU(2); (T1 ; : : : ; T2g+1 ) = T1 T2g+1 : Then the di erential d is onto at any irreducible point (T1 ; : : : ; T2g+1 ) such that
T1 T2g+1 = 1.
Proof. Let us assume that = j and suppose that (T1 ; : : : ; T2g+1 ) is an irreducible point such that (T1 ; : : : ; T2g+1 ) = 1 2 SU(2). Without loss of generality, we may assume that i 0 e A iB T1 = 0 e?i and T2g+1 = iB A
with A a complex number and B a non{zero real number. The tangent space to Sj2 at X1 is the image under the left multiplication by T1 of the linear subspace of su(2) consisting of the matrices
i i e?i u1 = i e i ?i
11
with real and . For such a u1 , (d)(u1 ; 0; : : : ; 0) = T1 u1 T2 : : : T2g+1 = T1 u1 T1?1
i = i ei ?i i e ?i = i ? sin j + cos k; ; 2 R: (15) Thus, the image of d restricted to the tangent space to Sj2 at T1 is 2{dimensional. The tangent space to Sj2 at T2g+1 is the image under the left multiplication by T2g+1 of the linear subspace of su(2) consisting of the matrices i b u2g+1 = ?b ?i such that 2 R and Re(Ab + B ) = 0. For such a u2g+1 , (d)(0; : : : ; 0; u2g+1 ) = T1 T2g+1 u2g+1 = u2g+1 : Let us choose b = cos + i sin with as in (15), and 2 R. The equation Re(Ab + B ) = 0 can be solved for 2 R to get = Re(Ab)=B , and then the vector Re(Ab)=B i + cos j + sin k (16) will lie in the image of d. The vectors (15) and (16) span the entire Lie algebra su(2), therefore, d is onto. This lemma implies that R (M0 ) is a smooth (open) manifold of dimension 4g ? 2. A choice of basis t1 ; : : : ; tg in 1 M1 with the property that (ti ) = t?1 i identi es R(M1 ) with a smooth submanifold in (S2)g =S1 of dimension 2g ? 1. The construction for R (M2 ) is completely analogous.
Due to Seifert{Van Kampen Theorem, we have the following commutative diagrams of the fundamental groups
1(p; q; r) ??? 1 M1 x x ? ?
? ?
??? 1 M0
1 M2
and of the equivariant representation spaces
R?() ???! R(?M1 ) ? y
? y
R (M2 ) ???! R(M0 ) where
R() = f : 1 ! SU(2) j is irreducible; = ?1 g=S1 : 12
The maps in the latter diagram are injective, so we can think of R () as the intersection of R (M1 ) and R (M2 ) inside R (M0 ). Lemma 11. The space R() is a (regular) double cover of R((p; q; r)). The manifolds R (M1 ) and R (M2 ) intersect transversally inside the manifold R (M0 ) in a nite number of points. Proof. Let ; : 1 ! SU(2) be irreducible representations such that = ?1 and = ?1 . If they are conjugated, = h h?1 ; h 2 SU(2), then (h h?1 ) = h h?1 ?1 , therefore, = (h?1 h) (h?1 h)?1 = ?1 : Since is irreducible, = hh?1 , or h = h. Therefore, R((p; q; r)) = f : 1 ! SU(2) j is irreducible; = ?1 g=S1 ; where S1 is a subgroup of SU(2) consisting of all h such that h = h. The group S1 is a subgroup of S1 of index 2. This proves the rst part of the lemma. Each of the manifolds R (Mk ); k = 0; 1; 2, is the xed point submanifold of the involution 7! Ad on the corresponding manifold R(Mk ). The intersection of R(M1 ) with R(M2 ) in R(M0 ) is transversal, see e.g. [11], Proposition 2.5, and R(M1 ) \ R (M2 ) = R ((p; q; r)) is nite. 2.5. The de nition of . Let = (p; q; r) be a Brieskorn homology sphere. Choose a {invariant Heegaard splitting = M1 [ M2 along a Riemann surface M0 = M1 \ M2 as in (9). We de ne the {invariant of as one forth of the algebraic intersection number of R (M1 ) and R (M2 ) inside R (M0 ), so that
() = 1=4
X
2R ()
"
(17)
where " equals 1 depending on whether the orientations on the spaces T R (M1 ) T R(M2 ) and T R(M0 ) agree, see below. Due to (8) and the fact that R() double covers R(), the number () is an integer. The manifolds R (M0 ) and R (Mi ); i = 1; 2, are oriented as follows. First, we x an orientation on SU(2). There are two transversal submanifolds in SU(2), S1 = f g 2 SU(2) j g = g g and S2 = f g 2 SU(2) j g?1 = g?1 g: At 1 2 SU(2), we obviously have the decomposition (18) T1 SU(2) = T1 S1 T1 S2 of T1 SU(2) = su(2) into the (1){eigenspaces of the operator Ad : su(2) ! su(2). We orient S1 and S2 so that the orientations are consistent with the decomposition (18). Let us now x an orientation on M0 . This orientation orients the two boundary circles of M0 ; and ( ). As we have seen, a choice of basis in 1 M0 identi es Hom (1 M0 ; SU(2)) with (S2 )2g+1 , which allows to orient Hom (1 M0 ; SU(2)) as a product. This orientation is independent of the choice of a basis due to the fact that S2 is even{dimensional. The orientations on and on S1 2 SU(2) orient the 13
manifold R (M0 ). Note that this orientation is independent of the initial choice of orientations on SU(2) and S1 , as soon as the orientation of S2 is consistent with those via (18). The orientation of R (M0 ), however, depends on the orientation of M0 through the induced orientation of ; change the orientation of M0 , and the orientation of R (M0 ) changes by (?1). The manifolds R (M1 ) and R (M2 ) are oriented as submanifolds of (S2 )g =S1 . Their orientations depend on the choices of SU(2) and S1 orientations, but the orientation of T R (M1 ) T R (M2 ) at a point of intersection in R (M0 ) is insensitive to these choices. Switching the roles of M1 and M2 in the sum T R (M1 ) T R (M2 ) changes its orientation by (?1). Thus, the signs " in (17) are well{de ned given the orientation of M0 and the order of the handlebodies. Changing both choices leaves () invariant, so it is only the induced orientation of that must be speci ed to avoid an ambiguity. The sign of () changes with the change of orientation of . The invariant may a priori depend on the choice of Heegaard splitting. In fact, it does not, and this will follow from Corollary 16, which expresses in terms of the knot signature. 3. The invariant and knot signature The signature of a knot in S 3 has been interpreted by X.-S. Lin in [17] as an intersection number of trace{free SU(2){representation spaces associated with the knot complement. This number is usually referred to as Casson{Lin invariant. In this section we use this interpretation to show that, under proper normalizations, the invariant of a Brieskorn homology sphere (p; q; r) equals the signature of the Montesinos knot k(p; q; r). 3.1. De nition of Casson{Lin invariant. Let Bn be the braid group of rank n with the standard generators 1 ; : : : ; n?1 represented in a free group Fn on symbols x1 ; : : : ; xn as follows: i : xi 7! xixi+1 x?1 i
xi+1 7! xi xj 7! xj ; if j 6= i; i + 1: If 2 Bn then the automorphism of Fn representing maps each xi to a conjugate of some xj and preserves the product x1 xn . Let k S 3 be a knot represented as the closure of a braid 2 Bn . Let us x an embedding of k into S 3 as shown in Figure 2, the sphere S separating S 3 in two 3{balls, B1 and B2 , with the braid inside B1 and n untangled arcs inside B2 . The fundamental group 1 K of the knot k complement K = S 3 n k, has the presentation 1 K = h x1 ; : : : ; xn j xi = (xi ); i = 1; : : : ; n i; the generators x1 ; : : : ; xn being represented by the meridians of . The knot complement K can be now decomposed as K = M10 [M 0 M20 where M00 = S \ K and Mk0 = Bk \ K; k = 1; 2. The manifolds M10 and M20 are handlebodies of genus n = g + 1, and M00 is a 2{sphere with 2g + 2 punctures at the points 0
14
P1 ; : : : ; P2g+2 , see Figure 2. Due to Seifert{Van Kampen Theorem, we get the following commutative diagram of fundamental groups
??? 1xM10
1 K x ? ?
? ?
1 M20 ??? 1M00 where
1 M00 = h x1 ; : : : ; x2g+2 j x1 x2g+2 = 1 i; 1 M10 = h x1 ; : : : ; xg+1 j i; 1 M20 = h xg+2 ; : : : ; x2g+2 j i are isomorphic to free groups, and the generators x1 ; : : : ; x2g+2 of 1 M00 are represented by the loops in Figure 5. S
D
γ
Q x2g+2
P2g+2
x2
P2
x1
P1
Figure 5 In complete analogy with the de nition of Casson invariant, we are going to de ne SU(2){representation spaces of the groups 1 K and 1 Mk0 ; k = 0; 1; 2, and compute the corresponding intersection number. To make this program work, we impose the extra condition on the representations that all the meridians x1 ; : : : ; x2g+2 go to trace-free matrices in SU(2). Thus the representation spaces in question are
R0 (K ) = f : 1K ! SU(2) j is irreducible, tr (xi ) = 0 g= SU(2); R0 (Mk0 ) = f : 1Mk0 ! SU(2) j is irreducible, tr (xi ) = 0 g= SU(2); where k = 0; 1; 2. The commutative diagram of fundamental groups above induces the following commutative diagram of inclusions of representation spaces, 15
R0?(K ) ???! R0 (?M10 ) ? y
? y
R0 (M20 ) ???! R0 (M00 ) In particular, the irreducible trace-free representations of 1 K in SU(2) are in one-toone correspondence with the intersection points of R0 (M10 ) with R0 (M20 ) in R0 (M00 ). X.-S. Lin showed in [17] that R0 (Mk0 ); k = 0; 1; 2, are smooth (open) manifolds of dimensions dim R0 (M00 ) = 4g ? 2 and dim R0 (M10 ) = dim R0 (M20 ) = 2g ? 1, and that the intersection R0 (K ) = R0 (M10 ) \R0 (M20 ) in R0 (M00 ) is compact. As we will see later, this intersection is nite if the knot k(p; q; r) is a Montesinos knot. The Casson{Lin invariant for a knot k = k(p; q; r) is now de ned as
h(k) =
X
2R0 (K )
"0
(19)
where "0 = 1 is a sign obtained by comparing the orientations ( see below ) on T R0 (M10 ) T R0 (M20 ) and T R0 (M00 ). For a general knot, the number (19) is well de ned after some compact perturbations. The orientations of R(Mk0 ); k = 0; 1; 2, are de ned as follows. Let SU(2) be oriented by the standard basis i; j; k in its Lie algebra su(2). The submanifold of SU(2) consisting of the trace-free matrices is a 2{dimensional sphere, which we denote by S02 SU(2). It is naturally oriented as S02 = exp(S (0; 2 )) where S (0; 2 ) is the 2{sphere in su(2) centered at 0 and of radius 2 in the metric hu; vi = 21 tr(uvt ); u; v 2 su(2): Let us introduce the manifold M00 n D where D is an open disc in M00 away from the points P1 ; : : : ; P2g+2 , see Figure 5. An orientation of M00 orients the boundary circle, , of M00 n D. The choice of generators x1 ; : : : ; x2g+2 in 1 M00 identi es the representation space of M00 n D, f : 1 M00 ! SU(2) j tr (xi ) = 0 g, with the product (S02 )2g+2 . The manifold R0 (M00 ) is then identi ed with an SU(2){quotient of the open set of irreducible representations in ?1 (1) where : (S02 )2g+2 ! SU(2) is de ned by parallel transport around . The di erential d of the map is onto at all irreducible representations such that ( ) = 1, see [17], Lemma 1.5. This orients R0 (M00 ) so that the natural isomorphism TX S02 : : : TX g S02 = T1 SU(2) T(X ;:::;X g ) R0 (M00 ) T1 SU(2) (with the second T1 SU(2){factor corresponding to the SU(2){action) is orientation preserving. Note that the orientation of R0 (M00 ) changes by (?1) with the change of orientation on M00 (through the orientation of ). 1
2 +2
1
16
2 +2
The manifolds R0 (M10 ) and R0 (M20 ) are oriented as SU(2){quotients of open subsets in (S02 )g+1 , and their orientations are well de ned as soon as the orientations of SU(2) and S02 are xed as above. The switch of M1 and M2 changes the orientation of T R0 (M10 ) T R0 (M20 ). Thus the sign "0 in (19) is well de ned given the orientation of M00 and the speci cation of the handlebody to call M10 . Changing both choices does not change h(k), therefore, we only need to x an orientation of K to avoid an ambiguity. The sign of h(k) changes by (?1) if the orientation of K is changed. The following result follows from Corollary 2.10 of [17]. Proposition 12. For a knot k = k(p; q; r); h(k) = 2" sign k, where " = 1 is a universal constant independent of k. 3.2. Some equivariant gauge theory. Let = (p; q; r) be a Brieskorn homology sphere endowed with the involution (7) and a {invariant Heegaard splitting = M 1 [ M M2 . Let E ! be a (necessarily trivial) SU(2){vector bundle over with a xed trivialization. Let us x a {invariant Riemannian metric on and consider the space A of smooth connections on E . The choice of trivialization gives an isomorphism of A with 1 (; su(2)), the linear space of smooth di erential 1{forms on with su(2){coecients. We use the L21 {inner product on 1 (; su(2)) to de ne A as a smooth manifold modelled on a pre{Hilbert space, see [26]. The bundle E restricts to trivial SU(2){bundles Ek over Mk ; k = 0; 1; 2. The metric on induces {invariant Riemannian metrics on Mk , and we denote by Ak the corresponding space of smooth connections on Ek ; k = 0; 1; 2. The spaces Ak are de ned as smooth manifolds modelled on pre{Hilbert spaces 1 (Mk ; su(2)) with the L21 {inner product. The group G = C 1 (; SU(2)) is called the gauge group; it acts on A as (g; A) 7! g dg?1 + g A g?1 . Let B = A=G be the quotient space. Set B = A=G where A A is the subspace of irreducible connections A on which G acts with the stabilizer 1. It is the complement of a space of in nite codimension in A. Then B is an in nite dimensional manifold modelled on a pre{Hilbert space using the L21 {theory. The projection A ! B can be shown to be a principal G {bundle, so that 1 (B ) = 0 (G ) = 0 C 1(; SU(2)) = [ ; SU(2) ] = Z, the last isomorphism given by the mapping degree. For every k = 0; 1; 2, the gauge group Gk = C 1(; SU(2)) acts on Ak , and we de ne Bk as the Gk {quotient of the subspace Ak Ak of irreducible connections. Each of the spaces Bk can be endowed with the structure of an in nite dimensional simply{connected manifold. It is a classical result in di erential geometry that the holonomy map establishes a one{to{one correspondence between the gauge equivalence classes of irreducible
at connections in a bundle and the conjugacy classes of irreducible representations of the fundamental group of its base. 0
17
For every k = ;; 0; 1; 2, the involution can be lifted to a bundle endomorphism of Ek . Any endomorphism of Ek clearly induces an action on Ak by pull-back, and an action on Bk as well. Since any two liftings of di er by a gauge transformation, we have a well-de ned action : Bk ! Bk . Denote by Bk the manifold consisting of connections invariant with respect to . Let 2 SU(2) be such that 2 = ?1. The formula (x; ) 7! ((x); ) (20) de nes a lifting of : ! on Ek , which will again be denoted by : E ! E . Let Bk Bk be the manifold consisting of the gauge equivalence classes of irreducible connections A in Ek such that A = A. For instance, Proposition 8 implies that all irreducible at connections on belong to B . The following lemma can be easily checked, compare with Proposition 1 of [34]. Lemma 13. Let k = ;; 0; 1; 2. For any ; 0 2 SU(2) such that 2 = 0 2 = 0 ?1 ; Bk = Bk . Furthermore, Bk is bijective to Ak =Gk where Gk = Gk = 1 and Ak = f A 2 Ak j A = A g. With the inner L21 {product, Ak is a smooth manifold whose tangent space at any point A is identi ed with the pre{Hilbert space of smooth 1{forms ! 2 1 (; su(2)), respectively, ! 2 1 (Mk ; su(2)) if k = 0; 1; 2, such that !((x)) = Ad !(x) (= !(x)?1 ) for all x 2 , respectively, x 2 Mk . The gauge group Gk consists of all smooth maps g : ! SU(2) such that g((x)) = g(x), the maps g and ?g being identi ed. It is more convenient for us to work with the double covers Bk = Ak =Gk (21) of the manifolds Bk; k = ;; 0; 1; 2, where the gauge groups Gk consist of all the smooth maps g : ! SU(2), respectively, g : Mk ! SU(2), such that g = g. Again, Ak ! Bk; k = ;; 0; 1; 2, is a principal Gk {bundle, hence, 1 B = 0 G = [ ; SU(2) ]Z=2; 1 Bk = 0 Gk = [ Mk ; SU(2) ]Z=2; where [ ; ]Z=2 stands for the group of Z=2{equivariant homotopy class. The equivariant obstruction theory of [5] can be used to show that 1 B = Z Z, and the manifolds Bk are simply{connected for k = 0; 1; 2. 3.3. Representations of the knot k(p; q; r) complement and . Let = (p; q; r) be a Brieskorn homology sphere with a {invariant Heegaard splitting = M1 [M M2 where M1 = ?1 (B1 ); M2 = ?1 (B2 ); M0 = ?1 (S ) (22) as in (9). Let K be the knot k(p; q; r) complement in S 3 , then K = M10 [M 0 M20 where M10 = B1 \ K; M20 = B2 \ K; M00 = S \ K: (23) 0
0
18
Let us x orientations on ; K and M0 ; M00 so that the projection is orientation preserving. Let E ! be a trivialized SU(2){bundle over . The quotient of by the involution is S 3 . Since 6= 1, it is impossible to de ne the quotient bundle of E over S 3 . However, one can de ne it away from the knot k S 3 , that is, on the knot complement K . Given an irreducible equivariant at connection A 2 A in E , its push{down A0 is an irreducible SU(2){connection over K . In other words, A0 is an irreducible at SU(2){connection singular along k in the sense of P. Kronheimer and T. Mrowka [15]. The following result follows from Proposition 17 of [34]. Lemma 14. The at SU (2){connection A0 has holonomy 1=4 around k, i.e. the holonomy is trace free. Let Ek be the restriction of the bundle E to the submanifold Mk ; k = 0; 1; 2. Given an irreducible at connection in Ek , its push{down is an irreducible at connection over Mk0 whose holonomy around k \ Mk0 is trace{free. Proposition 15. The push{down of at connections induces orientation preserving di eomorphisms R (Mk ) ! R0 (Mk0 ); k = 0; 1; 2, and R () ! R0 (K ), which commute with the embeddings of representation spaces. Corollary 16. There exists a universal constant " = 1 such that for any Brieskorn homology sphere (p; q; r) and the corresponding Montesinos knot k(p; q; r), ((p; q; r)) = 41 h(k(p; q; r)) = 8" sign k(p; q; r): Proof of Proposition 15. The push{down of at connections de nes a map of representation spaces, 0 : R (M0 ) ! R0 (M00 ), as follows. Let : 1 M0 ! SU(2) be the holonomy representation of a at connection A over M0 , = holA . For every loop x in M00 , we de ne 0 ( )(x) = holA (~x) where x~ is a lift of the loop x to M0 . Note that the curve x~ is not necessarily closed. With respect to the generators x1 ; : : : ; x2g+2 and t1 ; : : : ; t2g+1 of the groups 1 M00 and 1 M0 , respectively, see Figure 3, 4 and 5, the map 0 can be given by the following formula (remember that we used the left multiplication by in (20) to identify the bers over the pairs of points P and (P )) 0 ( )(x1 ) = ; 0 ( )(x2 ) = (t1 )?1 ; 0 ( )(x3 ) = (t2 ); (24)
::::::
0 ( )(x2g+2 ) = (t2g+1 )?1 : If we think about R (M0 ) and R0 (M00 ) as the following smooth manifolds, R(M0 ) = f (T1 ; : : : ; T2g+1 ) j (T1 ; : : : ; T2g+1 ) is irreducible; Ti 2 S2 ; T1 T2g+1 = 1 g= S1 and 19
R0(M00 ) = f (X1 ; : : : ; X2g+2 ) j (X1 ; : : : ; X2g+2 ) is irreducible; Xi 2 S02 ; X1 X2g+2 = 1 g= SU(2);
the map 0 is given by the formula 0 (T1 ; : : : ; T2g+1 ) = (; T1 ?1 ; T2 ; : : : ; T2g+1 ?1 ): The map 0 is obviously smooth. To check its injectivity we suppose that there exists h 2 SU(2) such that = hh?1 ; T10 ?1 = hT1 ?1 h?1 ; T20 = hT2 h?1 ;
::::::
T20g+1 ?1 = hT2g+1 ?1 h?1 for some (T1 ; : : : ; T2g+1 ) and (T10 ; : : : ; T20g+1 ) in R (M0 ). Then the rst equation implies that h = h, and the remaining equations assure that Ti0 = hTi h?1 ; i = 1; : : : ; 2g + 1, with h 2 S1 . To show that 0 is surjective we take an arbitrary (2g + 2){tuple in R0 (M00 ), conjugate it to a (2g + 2){tuple of the form (; X2 ; : : : ; X2g+2 ), and de ne (T1 ; T2 ; : : : ; T2g+1 ) = (X2 ; ?1 X3 ; : : : ; X2g+2 ): Obviously, 0 (T1 ; : : : ; T2g+1 ) = (; X2 ; : : : ; X2g+2 ). Next we need to check that the map 0 is orientation preserving. We orient SU(2) and S02 as in the de nition of Casson{Lin invariant. As we have seen in Section 2.5, the orientation of R (M0 ) is independent of the orientation choices for S2 and SU(2), as soon as the orientation on S1 is consistent in the sense that the natural isomorphism
T1 SU(2) = T1 S1 T1 S2
(25) is orientation preserving. We use the already xed orientation on SU(2), and orient S1 in such a way that the left multiplication by ,
S2 ! S02 ; g 7! g;
is orientation preserving. Then the right multiplication by ?1 is also orientation preserving. In our construction of the map 0 we used the identi cation S02 = SU(2)=S1 where S2 is the orbit space of the adjoint SU(2){action on itself with S1 the stabilizer at . The natural isomorphism T SU(2) = T1 S1 T S02 (26) orients S1 , and this orientation is consistent with one obtained from (25) because the left multiplication by is orientation preserving on both SU(2) and S2 . The orientations of both R0 (M00 ) and R (M0 ) included a choice of orientations of manifolds M00 and M0 , respectively. The latter two are oriented consistently by the 20
requirement that the projection be orientation preserving. With all the choices made, the map 0 is an orientation preserving di eomorphism. The maps k : R (Mk ) ! R0 (Mk0 ); k = 0; 1; 2, are de ned after choosing suitable bases in the free groups 1 Mk0 and 1 Mk , by the formulas similar to (24). Both 1 and 2 are orientation preserving di eomorphisms with the above choices of the orientations. The maps k ; k = 0; 1; 2, commute with the embeddings of representation spaces due to the naturality of the construction. In particular, the sets of intersection points, R() = R(M1 ) \ R(M2 ) and R0(K ) = R0(M10 ) \ R0 (M20 ) are in one{to-one correspondence. With the order of handlebodies (M1 ; M2 ) and (M10 ; M20 ) xed by (22) and (23), the conclusions of both the proposition and the corollary follow. 4. The invariant We use gauge theory to de ne an invariant for Brieskorn homology spheres as an in nite dimensional equivariant Euler characteristic. In Section 5 we will express in terms of Floer homology, and in Section 6 will prove that, under proper normalizations, = 2 . 4.1. Equivariant gauge theory on (p; q; r). Let B = A =G be the manifold de ned in (21). Remember that is endowed with a {invariant Riemannian metric. The assignment to an su(2){valued smooth di erential p{form ! of the form ! = Ad ! de nes an involution on the space p (; su(2)). This involution splits
p(; su(2)) in the direct sum of its (1){eigenspaces. Denote the (+1){eigenspace by p (; su(2)) so that (27)
p (; su(2)) = f ! 2 p (; su(2)) j ! = !?1 g: If A 2 A is a {invariant at connection, the covariant di erentiation dA commutes with the splitting (27). Therefore, we have a splitting in de Rham cohomology. In particular, for any p, Hp(; dA ) H p(; dA) where Hp (; dA ) is the p{th equivariant cohomology groupp whose elements are canonically represented by {invariant harmonic forms ! 2 (; su(2)). A 1-form on A assigns to a connection A a homomorphism from TA A = 1 (; su(2)) into the real numbers. De ne such a homomorphism by the formula
a 2 1(; su(2))
Z
7! fA(a) = tr(a ^ FA)
2 where FA 2 (; su(2)) is the curvature of the connection A. Due to the Bianchi identity, fA annihilates the tangent space to the G {orbit through A, TA(G A) = f dA = d + [A; ] j 2 0 (; su(2)) g: (28) 21
Since f is G {equivariant it can be thought of as being the pull-back of a 1{form f on B . The zeroes of f on B are precisely the G {orbits of the {invariant at connections. Let denote the Hodge star operator associated with the chosen {invariant Riemannian metric on , and p : p(; su(2)) ! 3? (29) (; su(2)) its restriction to the {invariant forms. The restriction (29) is well{de ned due to the fact that is an orientation preserving isometric involution. The Riemannian metric on de nes an L2 {inner product on p (; su(2)) by the formula Z
ha; bi = ? tr(a ^ b):
With respect to this metric, the formal adjoint dA to the restriction of the operator dA on the {invariant forms is given by the usual formula, dA = ? dA :
p (; su(2)) ! p?1 (; su(2)). Thus the tangent space to B at an orbit [A], which is isomorphic to the orthogonal complement of TA (G A) in 1 (; su(2)), is simply the vector space TA = f ! 2 1(; su(2)) j dA a = ? dA = 0 g: If A 2 A is a at connection, the operator dA : TA ! TA is well{de ned. In general, we compose the operator dA with the orthogonal projection onto TA to get the operator rfA : TA ! TA; (30) a 7! dA a ? dAu(a); where u(a) is the unique solution in 1 (; su(2)) of the equation dA dAu(a) = (FA ^ a ? a ^ FA): (31) One easily checks that u(a) 2 0 (; su(2)), namely, if we plug the form Ad u(a) in the equation (31), we will get dA dA(Ad u(a)) = Ad (dA dA u(a)) = Ad ((FA ^ a ? a ^ FA )) = (FA ^ a ? a ^ FA ): Since the equation (31) has only one solution, we must have u(a) = Ad u(a). The operator rfA : TA ! TA is an elliptic operator on the closed manifold . It follows from the standard elliptic theory that the L2 {completion of rfA is a selfadjoint Fredholm operator whose domain is the L21 {Sobolev completion of TA. Its eigenvalues form a discrete subset of the real line which has no accumulation points, and which is unbounded in both directions. Each eigenvalue has nite multiplicity, compare Lemma 1.1 of [33]. A non-degenerate zero of the 1{form f on B is, by de nition, the orbit of a at connection in A for which ker rfA = 0. 22
Lemma 17. All zeroes of f on B are non{degenerate.
Proof. The statement follows from the observation that ker rfA = H1 (; ad A) H 1 (; ad A) for any at A 2 A and the fact that H 1 (; ad A) = 0 for = (p; q; r) and any irreducible at connection A, see [11], Proposition 2.5. 4.2. De nition of . The assignment to an orbit [A] 2 B of the operator rfA on the L2 {completion of TA de nes a smooth map from B into the Banach space of real, selfadjoint operators on a separable Hilbert space. Let [A0 ] and [A1 ] in B be ( non{degenerate ) zeroes of f. Consider a continuously di erentiable path of operators rfA(t) with [A(0)] = [A0 ] and [A(1)] = [A1 ]. The one{parameter family of spectra of operators rfA(t) can be thought of as a collection of spectral curves in the plane, see Figure 6, connecting the spectrum of rfA to the spectrum of rfA . These curves are continuously di erentiable functions of t, at least near zero. The number of eigenvalues which cross from \minus" to \plus" minus the number which cross from \plus" to \minus" is well{de ned and nite along a generic path. This number is called the spectral ow of the family along the path, see [3]. 0
1
spectrum
spectral flow = -1 t
0
1
Figure 6 The spectral ow only depends on the homotopy class rel f 0; 1 g of the path t ! [A(t)]. Therefore, it de nes a locally constant function on the space of continuous paths between [A0 ] and [A1 ]. We denote this function by sf ([A0 ]; [A1 ]). Lemma 18. The integer sf ([A0 ]; [A1 ]) is independent of the homotopy class of the path between [A0 ] and [A1 ] modulo 4. Proof. Let us consider a path A(t) in A connecting A0 to a gauge equivalent of itself. It can be thought of as a connection in an SU(2){bundle E (not necessarily trivial) on the 4{manifold S 1 , and then the spectral ow along the path A(t) will be equal to the index of the restricted self{duality operator, DA = dA + d?A : 1 ( S 1 ; su(2)) ! ( 0 2?) ( S 1 ; su(2)): The involution extends to S 1 as the product involution 1 with the quotient space S 3 S 1 and the xed point set B 0 S 3 S 1 an embedded torus with the 23
trivial normal bundle. According to [34], Theorem 18, index DA = 4c2 (E ) ? 32 ( ? )(S 3 S 1 ) + (B 0 ) + 12 B 0 : B 0 ; where stands for Euler characteristic, for signature ( the sign in ( ? ) is di erent from that in [34] because we work with self-dual rather than anti-self-dual connections ). Therefore, index DA = 0 mod 4. Now we are in position to de ne the invariant () up to a (){sign. Pick 2 R() and de ne
() = "
X
2R()
(?1)sf ( ; ) ; " = 1:
(32)
To de ne an overall sign of (), it is sucient to associate a sign to one particular
at connection orbit, 2 R(). In principle, we would like to take a path from the trivial connection to the connection and set " = sf (; ). The latter though is ill de ned because is reducible, hence, is a degenerate zero of f. In what follows we describe a way around this problem. To begin with, we replace the operator rfA : TA ! TA by the operator 0 d A : ( 0 1 ) (; su(2)) ! ( 0 1 ) (; su(2)): K = (33)
dA dA We are going to use KA to de ne the spectral ow sf (; ). The advantage of KA over rfA is that the operator KA and its spectrum depend continuously on A even if A is reducible. For any A, the operator KA is an elliptic operator on . Its L2 {completion is a self{adjoint Fredholm operator. It has pure point real spectrum A
without accumulation points, which is unbounded in both directions. The operators rfA and KA are related as follows. At any irreducible connection A 2 A , the principal G {bundle : A ! B induces the canonical exact sequence
d T A ! T B ! 0: 0 ! T1 G ?! A A The operator dA : TA A ! T1 G provides a G {equivariant splitting of this sequence, so that rfA : TA B ! TA B extends to a G {equivariant endomorphism of T1 G TAA = 0 (; su(2)) 1 (; su(2)) given by the formula 0 d A : ( 0 1 ) (; su(2)) ! ( 0 1 ) (; su(2)): K (f) =
dA rfA By construction, K (f)A has the same spectral ow as rfA . Note that K (f)A = KA if A is at, and that the di erence KA ? K (f)A is a relatively compact perturbation of KA. For a generic path t ! A(t), the spectral ow of KA (t) equals sf ( ; ) where [A(0)] = and [A(1)] = . Lemma 19. At the trivial connection , the operator K has a 1{dimensional kernel given by the d {constant elements in 0 (; su(2)). A
24
Proof. The kernel of K consists of the harmonic, hence constant, functions : ! su(2) which are {invariant. If we think of as an element of su(2), the latter condition means that Ad = . Since 2 = ?1; = c for some real constant c. Therefore, dim ker K = 1. From now on, we identify the kernel of K with ker K = H0 (; su(2)) = su (2) = f 2 su(2) j Ad = g: Let 1 (; su(2)) = im d ker d be the Hodge decomposition corresponding to the operator d : 0 (; su(2)) ! 1 (; su(2)): Note that the operator d : ker d ! ker d is invertible since is an integral homology 3{sphere. Choose a 2 ker d and de ne the following symmetric bilinear form on su (2),
a(; ) =
Z
tr ([; a] ^ [; (d )?1 a]):
(34)
This form plays an important part in the perturbation theory of K . Lemma 20. There exists " > 0 such that for any a 2 ker d and for any 0 s < ", the operator K+sa has exactly one eigenvalue in the interval (?"; "). To order s3 , the eigenvalue of K+sa in (?"; ") is s2 where is the eigenvalue of the form a . Proof. Let a 2 1 (; su(2)) be an su(2){valued 1{form on , then K+sa = K + s Ba where a 0 i B = a
ia ia
and ia : 0 (; su(2)) ! 1 (; su(2)) is given by ia (') = [a; ']. Suppose that for a small s 0, K+sa s = s s; (35) where 0 = 0 and s = ('s ; !s) 2 ( 0 1 ) (; su(2)). Up to the order s3 ; s = 1 s + 2 s2 + : : : and s = 0 + 1 s + 2 s2 + : : : so that the equation (35) is of the form K 0 = 0; (36) K 1 + Ba 0 = 1 0 ; (37) K 2 + Ba 1 = 2 0 + 1 1 : (38) Since K2 is the Laplace operator, equation (36) implies that 0 = ('0 ; 0) with '0 2 su(2) a constant. Equation (37) takes now the form ( d !1 = 1 '0 ; d '1 + d !1 = ?[a; '0 ]: We apply d and d respectively to the rst and second equations in the system. Since d d !1 = 1 d '0 = 0, we get that d !1 = 0 and 1 = 0. The second equation 25
implies d d '1 = ?d [a; '0 ] = ?[d a; '0 ] = 0. Therefore, 1 = ('1 ; !1 ) where '1 2 su(2) is a constant. Equation (38) is of the form (
d !2 + ia !1 = 2 '0 ; d '2 + d !2 = ?[a; '1 ] ? ia !1 ;
therefore, for any 2 su (2), hia !1; i = h2 '0 ; i ? hd !2; i = 2h'0 ; i; which implies that h!1 ; ia i = 2 h'0 ; i. On the other hand, d !1 = ?ia '0 , so that !1 = (d )?1 (ia '0 ), and 2 h'0 ; i = ?h(d )?1 (ia '0 ); ia i: In other words, the eigenvalues of K+sa up to the order s3 are of the form s2 where is the eigenvalue of the bilinear form ('0 ; ) 7! ?h(d )?1 (ia '0 ); ia i = which is isomorphic to (34).
Z
tr ([(d )?1 a; '0 ] ^ [a; ]);
This lemma justi es the following de nition. Choose a 2 1 (; su(2)) for which a is non{degenerate (the existence of such forms will follow from section 6.4). Fix = [A] 2 R(), and de ne " in (32) equal sign det( a ) times the spectral ow of the family t 7! (1 ? t) K+sa + t KA for s small enough. The invariant () is now well{de ned by the formula (32) due to the fact that for any a1 ; a2 2 1 (; su(2)) for which a and a are non{degenerate, the di erence in sign det( a ) and sign det( a ) is given by the spectral ow of KA between + s a1 and + s a2 , see Lemma 2.9 of [33]. In a less formal way, one can think of () as the sum 1
1
2
2
() =
X
2R()
(?1)sf (; )
where sf (; ) is de ned as follows. For a generic nearby connection 0 of , the operator K0 has one small non{zero eigenvalue. De ne p(0 ) = 1 if this eigenvalue is negative, and p(0) = 0 otherwise. Then sf (; ) = p(0 ) + sf (0; ) mod 4: Geometrically, to de ne sf (; ) we simply count the intersection points of the spectral curves with the straight line connecting the points (0; ?") and (1; ") where " > 0 is chosen much smaller than the absolute value of any non{zero eigenvalue of K and K , see Figure 7. 26
spectrum
spectral flow = -1 ε
−ε
t
Figure 7 We further note that the invariant () is metric independent. The de nition of a at connection requires no metric, so the metric can only a ect () through the spectral ow. In fact, it does not because the space of {invariant Riemannian metrics on is path connected. 5. The invariant and Floer homology In this section we express the invariant of Brieskorn homology spheres in terms of Floer homology. 5.1. Casson invariant via gauge theory. C. Taubes gave in [33] a gauge{theoretical interpretation of the Casson's invariant. We shortly recall his construction in the special case of a Brieskorn homology sphere = (p; q; r). Let E ! be a trivialized SU(2){bundle over . For any connection A in E de ne the following elliptic di erential operator 0 d A : ( 0 1 )(; su(2)) ! ( 0 1 )(; su(2)); K = (39) A
dA dA
where dA stands for the covariant derivative with respect to A, and is the Hodge operator associated with a Riemannian metric on . If the metric is {invariant and the connection A is {invariant in the sense of Section 3.2, then the restriction of the operator KA on the {invariant forms (; su(2)) coincides with the operator (33). In proper Sobolev completions of ( 0 1 )(; su(2)) the operator (39) is Fredholm. Therefore, for any pair ; 2 B of at connections in E one can de ne the spectral ow sf ( ; ) as follows. Let A(t); 0 t 1, be a path of connections in E such that [A(0)] = and [A(1)] = . Associated to A(t) is a path of Fredholm operators KA(t) . The spectral ow ( ; ) is the net number of the eigenvalues of KA(t) which change from under ?" to over " as A(t) varies from A(0) to A(1); here, " > 0 is chosen much smaller than the absolute value of any non{zero eigenvalue of KA(0) and KA(1) , compare to the de nition of sf ( ; ) in Section 4.2. Note that in our case of (p; q; r), all at connections except for the trivial one are irreducible and non{degenerate. The latter means that ker K = 0 unless 6= . The kernel of K is 3{dimensional, while the kernel of K is 1{dimensional. 27
The spectral ow sf ( ; ) is well{de ned modulo 8. C. Taubes proved in [33] that
() = 21
X
2R()
(?1)sf (; )
where at connections, as usual, are identi ed with SU(2){representations via holonomy. Furthermore, R. Fintushel and R. Stern showed in [11] that for any 2 R((p; q; r)), the spectral ow sf (; ) is even, thus simply counts the irreducible representations. For any oriented integral homology 3{sphere , A. Floer de ned in [12] a (Z=8){ graded instanton homology theory I () whose Euler characteristic equals 2 (). In the special case of = (p; q; r) this theory is of a particularly simple form. Namely, the group In ((p; q; r)) is trivial for odd j , and is a free abelian group generated by all irreducible 2 R((p; q; r)) such that sf (; ) = j mod 8 for even j. 5.2. Comparing the spectral ows. With any irreducible 2 R((p; q; r)) we have associated two numbers, sf (; ) mod 8 and sf (; ) mod 4. The aim of this section is to compare them. Let and be at (possibly trivial) connections in a trivial SU(2){bundle E over = (p; q; r). By pull-back we extend E to a trivial bundle (which is also denoted by E ) over the in nite cylinder R. Let us choose a {invariant metric on , and the corresponding product metric on R. Let further : E ! E be the lift of de ned in (20), and A(t) a path of {invariant connections in E forming a {invariant connection over R vanishing in the R{direction and equal to respectively and near the ends of R. Let us consider the self{duality operator (40) DA = dA d?A : 1( R; su (2)) ! ( 0 2?)( R; su (2)); and impose the global boundary condition (2.3) of Atiyah{Patodi{Singer [2]. This boundary value problem has well{de ned index, index DA ( ; ). The assignment to an su(2){valued di erential p{form ! on R of the form ! = Ad ! de nes an involution on the space p ( R; su (2)); p = 0; 1; 2; : : : This involution splits all the spaces p( R; su (2)), as well as the space 2?( R; su (2)), in the direct sum of the (1){eigenspaces of . According to this splitting, (41) DA = DA+ DA?: We denote the operator DA+ by DA , so that (42) DA : 1( R; su (2)) ! ( 0 2?) ( R; su (2)); compare to (27). The Atiyah{Patodi{Singer index of the operator (42) will be denoted by index DA ( ; ). 28
Lemma 21. Let A(t) be a 1{parameter family of {invariant connections connecting to , where is an irreducible at connection on (p; q; r). Then sf (; ) = ?3 ? index DA ( ; ) mod 8; sf (; ) = ?1 ? index DA ( ; ) mod 4: If A(t) connects two irreducible at connections, and , then sf ( ; ) = index DA( ; ) mod 8 and sf ( ; ) = index DA ( ; ) mod 4: Proof. Both statements concerning sf are proved in [11], Lemma 3.2; the other two follow by the same argument after one notes that sf (; ) = ?1. Let index( ; DA ) denote the di erence tr( j ker DA ) ? tr( j coker DA ). A standard argument proves the following result. Lemma 22. For any 1{parameter family A(t) of {invariant connections connecting an irreducible at connection to a connection , which is either trivial or at irreducible, index DA ( ; ) = 21 index DA ( ; ) + 12 index( ; DA )( ; ): Proof. According to (41), the operator DA splits in the direct sum DA = DA+ DA? with DA+ = DA . Therefore, index DA = tr(1 j ker DA ) ? tr(1 j coker DA ) = tr(1 j ker DA+ ) + tr(1 j ker DA? ) ? tr(1 j coker DA+) ? tr(1 j coker DA?); index( ; DA ) = tr( j ker DA ) ? tr( j coker DA ) = tr(1 j ker DA+ ) ? tr(1 j ker DA? ) ? tr(1 j coker DA+) + tr(1 j coker DA?): Adding these two formulas together completes the proof. The G{index theorem for manifolds with boundary proved by H. Donnelly in [8], gives the following formula for index( ; DA )( ; ), Z index( ; DA )( ; ) = 0 ? 12 (h + (0))() + 21 (?h + (0))() (43) N where N is the xed point set of the involution 1 : R ! R, the integrand 0 is a universal polynomial in characteristic forms, h is the trace of the map induced by on H 0 (; d ) H 1 (; d ), and (0) is the {invariant of the operator K , see (39). More precisely, the {invariant is de ned as follows. We de ne the {function of by the formula X (s) = sign tr( j W ) jj?s ; (44) 6=0 where W is the {eigenspace of the operator K , and tr( j W ) the trace of the operator restricted to W . Standard methods show that the series on the right converges for Re(s) large enough and has a meromorphic continuation to the entire complex s{plane such that (0) is nite. 29
The formula (43) will be further simpli ed by using the at cobordism techniques of R. Fintushel and R. Stern [11]. Any Brieskorn homology 3{sphere (p; q; r) with p; q; r 2, is the boundary of the complex surface V (p; q; r) = f (x; y; z) 2 C 3 j xp + yq + z r = 0 g which is non{singular except at the origin, and (p; q; r) = V (p; q; r) \ S 5 C 3 is a smooth manifold. It is Seifert bered over S 2 with the Seifert invariants fb; (p; b1 ); (q; b2 ); (r; b3 )g satisfying the equation (6). For simplicity, we will sometimes denote the Seifert invariants (p; b1 ); (q; b2 ); (r; b3 ) by respectively (a1 ; b1 ); (a2 ; b2 ); (a3 ; b3 ), and (p; q; r) = (a1 ; a2 ; a3 ) by . If we blow up the surface V (p; q; r) once at the origin, we get another surface, W , which is non{singular except at three points. The singularities of W are just cones on the lens spaces L(ai ; bi ); i = 1; 2; 3. Topologically, one can identify W with a 4{dimensional orbifold obtained as the mapping cylinder of the Seifert orbit map ! S 2 . The oriented boundary of W is . Let W0 denote W with open cones around the singularities removed. Then W0 is a smooth manifold with the oriented boundary @W0 = [?L(a1; b1 ) [?L(a2; b2 ) [?L(a3; b3 ) and the fundamental group 1 W0 = 1=hhi = hx; y; z j xp = yq = z r = xyz = 1i the (p; q; r){triangle group. Since is a homology sphere, there is a one{to{one correspondence between SU(2){representations of 1 and SO(3){representations of 1 W0 , which we again denote by . The involution (7) is the restriction on of the complex conjugation involution, which we still call , on W . The involution on W , in its turn, restricts to involutions on W0 and on each of the lens spaces L(ai ; bi ). The latter can be described as follows. Let (ai ; bi ) be a pair of relatively prime positive numbers, and S 3 C 2 be de ned by the equation jz j2 + jwj2 = 1. The group Z=ai acts on S 3 by the formula t(z; w) = (tz; tbi w); t = e2ij=ai 2 C ; 0 j ai ? 1: (45) This action is free, and its quotient is the lens space L(ai ; bi ) = S 3 =(Z=ai). The involution acting on L(ai ; bi ) is induced by the complex conjugation on S 3 ; (z; w) 7! (z ; w ). It turns L(ai ; bi ) into a double branched cover of S 3 with branch set a two bridge link of type (ai ; bi ), see [6], Chapter 12. Note that the manifold W0 itself is a double branched cover of W0 =, which is a 4{ball with three disjoint 4{balls in its interior removed. The branch set is a surface N0 in W0= providing a cobordism between the Montesinos knot k(p; q; r) on one side and three two bridge links of the types (ai ; bi ); i = 1; 2; 3, on the other. The surface N0 is homeomorphic to the real projective plane RP 2 with four disjoint discs removed. Let V denote the at SO(3){bundle over W0 determined by 2 R() with the rotation numbers `1 ; `2 ; `3 , see (11). When V is restricted over a boundary component L(ai ; bi ), it takes the form S 3 Z=ai su(2) ! L(ai ; bi ) 30
where Z=ai acts on S 3 by the formula (45), and on su(2) by sending 2 su(2) to Ad (t) ; t = exp(2i`i j=ai ); 0 j ai ? 1. An SO(3){bundle V is classi ed by its second Stiefel{Whitney class w2 2 H 2 (W0 ; Z=2) = Z=2 which is the obstruction to lifting it to an SU(2){bundle. This means that w2 (V ) = 0 i (h) = +1. The involution acting on W0 preserves the class w2; w2 = w2 . The lift of the involution acting on , see (20), de nes a lift Ad of acting on W0 , to the SO(3){bundle V so that all at SO(3){connections in V are Ad { invariant. Equivariant di erential forms, equivariant self{duality operators etc. are de ned over W0 and L(ai ; bi ); i = 1; 2; 3, in the same way as they were de ned over (p; q; r) R and (p; q; r). Lemma 23. Let : 1(p; q; r) ! SU(2) be an irreducible representation, and V the induced SO(3){bundle over W0 with at connection A induced by . Then index( ; DA )(W0 ) = 0. Proof. Let be an irreducible representation. Since index( ; DA )(W0 ) = tr( j ker DA ) ? tr( j coker DA ); the result follows from the fact that both ker DA and coker DA on W0 vanish, see Proposition 3.3 and the argument right after it in [11]. Let and be irreducible at connections. The Donnelly's theorem [8] can be applied again, this time to the manifolds W0 and ?W0 : Z 3 X 1 1 (46) 0 = 0 ? 2 (h + (0))() + 2 (?h + (0))(L(ai ; bi )); N i=1 0
0=
Z
?N0
0 + 12 (?h + (0))() ? 21
3
X
i=1
(h + (0))(L(ai ; bi ));
(47)
where ?N0 stands for the xed point set of the involution acting on ?W0 . Adding (43), (46), and (47) together and noting that h () = h () = 0 (because and are non{degenerate and irreducible), we get index( ; DA )( ; ) =
? 21
3
X
i=1
Z
N0 [N [?N0
0
3 X (h + (0))(L(ai ; bi )) + 12 (?h + (0))(L(ai ; bi )): (48) i=1
By [8], the right hand side of (48) is simply index( ; DB )(W 0 ) where B is the connection over W 0 = W0 [?W0 = W0 [ R [?W0 built from at connections A over W0 and A over ?W0 , respectively, and a 1{parameter family A of connections over connecting to . Thus, index( ; DA )( ; ) = index( ; DB )(W 0 ); and the latter is given by the right hand side of the formula (48) with the integral R R N [N [?N 0 replaced by N [?N 0 . 0
0
0
0
31
Proposition 24. Let and be irreducible at connection on (p; q; r) and B
the connection over W 0 built as above from at connections A and A and a 1{ parameter family A of {invariant connections over connecting to . Then index( ; DB )(W 0 ) = 0 mod 8. Proof. The formula (48) involves three kinds of terms, each of which a priori depends on the representation and/or the manifold (p; q; r). An easy calculation with the group cohomology H 0 (L(ai ; bi ); V ) = H 0 (Z=ai; V ) and H 1 (L(ai ; bi ); V ) = H 1 (Z=ai; V ) = 0 shows that h (L(ai ; bi )) = h (L(ai ; bi )) = ?1. Proposition 27 below assures that, for any lens space L(a; b) and any (trivial or irreducible) representation : 1 L(a; b) ! SU(2), the -invariant vanishes. It follows from Theorem 18 of [34] that, for any closed orientable manifold X with an orientation preserving involution , the integrand 0 is a polynomial in the Euler class of the xed point set B of and the Euler class ofRthe normal bundle of B in X . Since 0 is universal, we conclude in particular that N [?N 0 in (48) does not depend on the choice of ; , or B . Let us choose = , then index DA ( ; ) = sf ( ; ) = 0 mod 4 and index DA ( ; ) = sf ( ; ) = 0 mod 8, therefore, index( ; DA )( ; ) = 0 mod 8, see Lemma 22. On the other hand, the formula (48) implies that, modulo 8, 0
0 = index( ; DA )( ; ) =
Z
N0 [?N0
0
0 + 3;
so N [?N 0 = ?3 mod 8. Now, for any choice of at irreducible and , we have index( ; DA )( ; ) = ?3 + 3 = 0 mod 8. Corollary 25. There is a universal constant c mod 4 such that, for any irreducible
at connections ; 2 R((p; q; r)), sf (; ) = 21 sf (; ) + c mod 4 and sf ( ; ) = 21 sf ( ; ) mod 4; in particular, sf ( ; ) is always even. Proof. Since sf ( ; ) = index DA ( ; ) mod 4 and sf ( ; ) = index DA ( ; ) mod 8, the result for irreducible and follows from Proposition 24 and Lemmas 21 and 22. Since ?1 = sf (; )+ sf ( ; ) mod 4 and ?1 = sf (; )+ sf ( ; )+ sf ( ; ) mod 4 for any at irreducible connections and , we get that sf ( ; ) = sf (; ) ? sf (; ) mod 4; and similarly, sf ( ; ) = sf (; ) ? sf (; ) mod 8: Coupled with the fact that sf ( ; ) = 21 sf ( ; ) mod 4, these imply sf (; ) ? sf (; ) = 12 sf (; ) ? 12 sf (; ) mod 4; or, equivalently, sf (; ) ? 12 sf (; ) = sf (; ) ? 21 sf (; ) mod 4: R
0
0
32
Therefore, the di erence sf (; ) ? 12 sf (; ) mod 4 is the same for all at irreducible representations . We denote it by c. The exact value of c can be found by repeating the argument of Proposition 24 for the connection = . The answer will depend on index( ; DA )(W0 ), which is a rational homology invariant of W0 . This implies the universality of c in that it is independent of (p; q; r). Corollary 26. For any Brieskorn homology sphere, ((p; q; r)) = 2 ((p; q; r)) with a universal (){sign. 5.3. The {invariant for lens spaces. Let L(a; b) be a lens space, and : 1 L(a; b) ! SO(3) a representation of its fundamental group. In this section we compute the {function of the operator 0 d : ( 0 1 )(; su(2)) ! ( 0 1 )(; su(2)) K =
d d
de ned in (44) by the formula X (s) = sign tr( j W ) jj?s ; 6=0
(49)
where W is the {eigenspace of K . Proposition 27. For any lens space L(a; b) and any representation : 1 L(a; b) ! SO(3), the function (49) is identically zero, (s) = 0. First step in our proof is the following result, which is a version of Cancellation Lemma of [19]. Lemma 28. The {function (49) equals the {function of the operator d : ker d ! ker d ; in other words, X (s) = sign tr( j V ) jj?s ; (50) 6=0 where V ker d is the {eigenspace of the operator d . Proof. We can ignore the d {harmonic forms as we only consider non-zero eigenval-
ues in (49) and (50). By the Hodge decomposition, we can split the remaining forms into d {closed and d {coclosed, c and cc. Since H 1 (L(a; b); d ) = 0, we have the following direct sum decomposition :
1 (L(a; b); su(2)) = ker d ker d : Therefore, the spaces 1c and 1cc can be identi ed as 1c = ker d and 1cc = ker d . The space 0cc consists of those 0{forms 2 0 (L(a; b); su(2)) for which d 6= 0. A straightforward calculation shows that the decomposition ( 0cc 1c ) 1cc is stable with respect to the operator K , and that K restricted to 1cc is isomorphic to the operator d : ker d ! ker d . 33
Let us restrict our attention to the space 0cc 1c . There is a basis for 0cc consisting of eigenforms of the Laplace operator = d d . Let w be one of these. Then w = 2 w with some > 0. Consider the two elements of 0cc 1c de ned by (w; d w=) and (w; ?d w=). They are each eigenvectors of K restricted to
0cc 1c with eigenvalues of opposite sign :
d d d 0
w d d w= = w = w = d w= d w d w d w= :
Since the forms (w; d w=) form a basis for 0cc 1c we are done. Let V denote the space of all su(2){valued 1{forms on S 3 such that d = 0; d = . Since L(a; b) is a quotient of S 3 by a free action of the group Z=a, the {eigenspace of the operator d : ker d ! ker d is then a subspace in V consisting of su(2){valued 1{forms on S 3 satisfying the equation t = Ad (t) for all t 2 Z=a: The involution on S 3 C 2 induced by complex conjugation on C 2 de nes an involution on V by the formula 7! Ad : It descends to the involution : V ! V (51) which is used in the formula (49) de ning the {function. The fact that the involution (51) is well{de ned can be checked directly as follows. Assume for simplicity that = j , then For any 2 V we need to check that t = Ad (t) . Assume for simplicity that = j , then t = Adj t = Adj (t?1 ) ; = Adj Ad (t? ) ; since 2 V ; = Adj Adj Ad (t? ) Adj ; = Ad (t) Adj = Ad (t) : For any angle between 0 and 2 and any positive integer c we de ne an operator (c) on C 2 by the formula (c) : (z; w) 7! (ei z; eic w): (52) An elementary calculation shows that (c) = (c)?2 (c). The operator (c) induces by pull{back an operator (c) on su(2){valued 1{forms on S 3 (with trivial action on the coecients). Since (c) commutes with the (Z=a){action, it is well de ned as an operator on V for every . Moreover, (c) (c) : (c) = ?2 (53) 1
1
34
Let = , then the operators and (c) commute due to the fact that 2 (c) = 1 = 1. Moreover, ( (c))2 = 2 (c) = 1, hence (c) is an involution on V . Let V = V (+1; (c)) V (?1; (c)) be a decomposition of V in the direct sum of the (1){eigenspaces of (c). Then, of course, tr( j V ) = tr( j V (+1; (c))) + tr( j V (?1; (c))): We claim that tr( j V (?1; (c))) = 0 for any c. This follows from the fact that the following diagram
V (?1; (c)) V (?1; (c)) ???! ?
(c)? = y 2
?
(c)? = y 2
V (?1; (c)) V (?1; (c)) ???!
anticommutes, because, due to (53), (c) = (c) (c) = ? (c) on V (?1; (c)): = ? =2 2 =2 We now restrict our attention to computation of tr( j V (+1; )) where
V (+1; ) =
1
\
c=1
V (+1; (c))
Note that the number of di erent spaces on the right is nite. Since (c) = 1 on the (c) de nes an involution on it, which space V (+1; )) for all c, the operator = 2 commutes with , (c) = (c) (c) = (c) on V (+1; ): = ? =2 2 =2
(c), The space V (+1; )) can be split in the direct sum of (1){eigenspaces of = 2 so that (c))) + tr( j V (?1; (c))): tr( j V (+1; )) = tr( j V (+1; = =2 2
(c))) = 0 due to the existence of the following anticommuAgain, tr( j V (?1; = 2 tative diagram
(c)) (c)) ???! V (?1; = V (?1; = 2 2 ?
(c)? = y 4
?
(c)? = y 4
(c)) ???! (c)) V (?1; = V (?1; = 2 2
35
Its anticommutativity can be checked by using the formula (53), (c) = (c) (c) ; = 4 ?=2 =4 (c) (c) ; since (c) is an involution; = = 2 =4 =2 = ?=4 (c) : It is clear that next we can restrict the trace computation to the space )= V (+1; = 2
1
\
(c)); V (+1; = 2
c=1 then use =4 (c) as involutions, etc. Induction by the powers of 2 will nally show
that
tr( j V ) = tr( j V (+1))
where
V (+1) = Since the sums S 1 {invariant,
P
1
\
k) = V (+1; = 2
k=0 k ak =2 ; ak
1 1
\ \
k=0 c=1
k (c)): V (+1; = 2
2 Z, are dense in R, all 1{forms
2 V (+1) are
t =
for all t 2 S 1 and c 1; (54) where the action of t is given by the formula (52). On the other hand, t = Ad (t) for all t 2 Z=a (55) with the action (z; w) 7! (exp(2ij=a)z; exp(2ijb=a)w). The equations (54) and (55) together imply, in particular, that any 2 V (+1) must be a (+1){eigenvector of the operators Ad (t) . Suppose that is not trivial, then must be an S 1 {invariant in the sense of (54) real valued 1{form on S 3 such that d = 0 and d = . The trace of on such forms is obviously independent of and L(a; b). If = is a trivial representation, then V consists of su(2){valued 1{forms which are S 1 {invariant in the sense of (54) and belong to V . Again, tr( j V ) is independent of L(a; b). The 3{sphere S 3 admits an orientation reversing involution induced by complex conjugation on the second coordinate in C 2 . It induces an isomorphism of the spaces V (+1) and V? (+1) commuting with . Therefore, tr( j V (+1)) ? tr( j V? (+1)) = 0 for any and 6= 0, and the function (49) vanishes identically. This completes the proof of Proposition 27. 6. The invariant via gauge theory In this section, we complete the proof of our main theorem by proving the following result (and xing the signs). Proposition 29. Let = (p; q; r) be a Brieskorn homology sphere, and and the invariants de ned by (17) and (32). Then () = 1=2 (). 36
This result will be obtained by a modi cation of the Taubes argument of [33] in our {invariant setting. 6.1. Gauge theory and Heegaard splitting. Let = M1 [M M2 be a { invariant Heegaard splitting of of genus g de ned in (9). It is convenient to enlarge both M1 and M2 to overlap (still being {invariant) and think of the 3{manifold M0 = M1 \ M2 as a thickened Riemann surface, M0 = R0 [?1=2; 1=2]. Let E be a vector SU(2){bundle over with a xed trivialization, and Ek the trivial SU(2){bundle obtained by restricting E onto Mk ; k = 0; 1; 2. For each k, we introduced in (21) the space Ak of the {invariant smooth connections on Ek . These spaces are considered as pre{Hilbert manifolds with the L21 {Sobolev space structure. The inclusions i ;i M0 ?j??;j! M1 ; M2 ??! induce by pull-back the maps J A A (56) A ?!I A1 A2 ?! 0 0 where I = i1 i2 and J = j1 j2 . The sequence (56) is exact in that I is an embedding, J is a submersion, and im (I ) = J ?1 () where A0 A0 is the diagonal. Each of the arrows in (56) is equivariant with respect to the corresponding group action, Gk . Let A0; A and A0k; Ak be the subsets of connections which restrict to irreducible connections on R0 [ ?1=4; 1=4 ] M0 . In each case, A0; A ; A0k; Ak is open and dense and the complement of a set of in nite codimension. Introduce the quotients B ; Bk ; B0; , and Bk0; by the respective gauge groups. With the L21 {Sobolev structure, each of B0; ; Bk0; is naturally a pre{Hilbert manifold, and A0; ! B0;; A0k; ! Bk0; are principal bundles with the structure groups G and Gk, compare to [33], Proposition 4.1. The tangent space to Bk0; at an orbit [A] is the vector space TkA = f ! 2 1(Mk ; su(2)) j dA ! = 0 and i (!) = 0 g where i : @Mk ! Mk is the inclusion. Let us de ne another vector bundle, Lk ! Bk0;; k = 0; 1; 2, with the ber at an orbit [A] 2 Bk0; the vector space LkA = f ! 2 1(Mk ; su(2)) j dA! = 0 g: Give LkA the structure of pre{Hilbert space by using the L2 {inner product on
1 (Mk ; su(2)). Note that the bundle Lk is di erent from the tangent bundle Tk , and that the orbits of irreducible at connections on Ek are in fact zeroes of a smooth section fk of Lk . The section fk in question assigns [ A; FA ] 2 LkA to each [A] 2 Bk0; . The covariant derivative rfk of fk ; k = 0; 1; 2, is the linear map rfkA : TkA ! LkA; (57) a 7! (rfkA )(a) = dA a ? dAuk (a); 0
1
2
1
37
2
where uk (a) 2 0 (Mk ; su(2)) is the unique solution of the equation dA dA uk (a) = (FA ^ a ? a ^ FA ); i uk (a) = 0; compare to (31). Proposition 30. The operator rfkA is a bounded operator from TkA to LkA when these spaces are completed in the Sobolev L21 and L2 norms, correspondingly. This operator is Fredholm of index 2g ? 1 for k = 1; 2, and of index 4g ? 2 for k = 0. Proof. The assertion that rfkA is Fredholm is a standard elliptic theory result. The index of rfkA is independent of the connection A. We choose A to be the trivial connection , and reduce calculations to equivariant de Rham theory. The index of rfkA turns out to be equal to minus one half of the Euler characteristic of the cohomology H (@Mk ; su(2)) = H (Mk ; d ), compare with [33], Lemma A.2. The result now is a corollary of the following lemma. Lemma 31. The vector spaces H0(Mk ; d ); k = 0; 1; 2, are 1{dimensional. The vector spaces H1 (M1 ; d ); H1 (M2 ; d ) and H1 (M0 ; d ) have dimensions 2g; 2g and 4g, respectively. Proof. The rst assertion can be proved by the same argument as in Lemma 19. We prove the second statement of the lemma for H1 (M0 ; d ); for the other two spaces the proof is similar. Let us identify H1 (M0 ; d ) with the space of {invariant 1{cocycles on 1 M0 , i.e. the maps : 1 M0 ! su(2) such that (x y) = (x) + (y) and ( x) = Ad (x). Since the action of 1 M0 on su(2) is trivial, all the coboundaries vanish, and the cocycles factor through H1 (M0 ). Let us choose a basis t1 ; : : : ; t2g in 1 M0 such that : 1 M0 ! 1 M0 acts by reversing the generators ti ; (ti ) = t?1 i . The involution acts on the rst homology group H1 (M0 ) as minus identity, and the 1{cocycles are then simply the linear maps : H1 (M0 ) ! su(2) such that ? (t) = Ad (t). Since the (?1){ eigenspace of Ad acting on su(2) is 2{dimensional, and dim H 1 (M0 ) = 2g, we get that dim H1 (M0 ; d ) = 4g. We observe that R() B0; and R (Mk ) Bk0; , which means that irreducible
at connections A on E and Ek restrict to irreducible connections on R0 [?1=4; 1=4]. The latter follows from the fact that the inclusions R0 ! Mk ! induce epimorphisms on the fundamental groups, 1 R0 ! 1 Mk and 1 R0 ! 1 . Proposition 32. The representation space R(Mk ) is an embedded submanifold in Bk0;; k = 0; 1; 2. Proof. It is sucient to show that dim ker rfkA = index rfkA for [A] 2 R (Mk ). Any ! 2 coker rfkA would satisfy the conditions dA ! = 0; dA ! = 0; i ! = 0, and ! = Ad !. Thus, [!] 2 H1 (Mk ; @Mk ; dA ). Since H1 (Mk ; @Mk ; dA ) H 1 (Mk ; @Mk ; dA ) which vanishes by Lemma 4.10 of [33], we get [!] = 0. Thus, ! = dA for 2
0 (Mk ; su(2)) with dA dA = 0 and, as A is irreducible, = 0 on @Mk . This means that both and ! vanish on Mk and hence coker rfkA = 0. 38
The maps I and J in (56) commute with the actions of respective gauge groups. They induce the sequence J B 0; B 0; (58) B 0; ?!I B10; B20; ?! 0 0 which is exact in the following sense ( see [33], Lemma 4.2 ). Lemma 33. In the sequence (58), I is an embedding, J is a submersion, and im I = J ?1 () where B00; B00; is the diagonal. On the level of tangent spaces, (58) gives the following sequence of linear maps T 0 ! TA ?! (59) 1;A T2;A ?! T0;A ! 0 where (A1 ; A2 ) = I (A); (A0 ; A0 ) = J (A1 ; A2 ), and = I ; = (j1 ) ? (j2 ) . The map I has the following explicit description : at [A] 2 B 0; , I (a) = (i1 a ? dAk 1 (a); i2 a ? dAk 2 (a)); where k (a) 2 0 (Mk ; su(2)); = 1; 2, is the unique solution to the Neumann boundary problem dAk dAk k (a) = 0 and (ik a ? dAk k (a))j@Mk = 0: For = 1; 2, the map (jk ) at [Ak ] 2 Bk0; is given by the formula (jk ) (ak ) = jk ak ? dA k (ak ) where dA dA k (ak ) = 0 and (jk ak ? dA k (ak ))j@M = 0: The inclusion maps in (56) also induce the following sequence 1
2
0
0
0
0
0
0
L1;A L2;A ?! L0;A ! 0; (60) 0 ! LA ?! where (f ) = (i1 f; i2 f ) and (f1 ; f2 ) = j1 f1 ? j2 f2 . The following statement is an easy modi cation of Lemmas 4.3 and 4.4 of [33]. Lemma 34. Each of the sequences (59) and (60) is an exact Fredholm complex. 1
2
0
6.2. A spectral interpretation of the intersection number. Let [A]; [A0 ] 2 B 0; be two points in f?1(0), and j1 i1 [A] = [A0 ] and j1 i1 [A0 ] = [A00 ] be the corresponding points in R (M1 ) \ R (M2 ) R (M0 ). A calculation of the relative sign di erence between the intersection numbers of R (M1 ) \ R (M2 ) at the points [A0 ] and [A00 ] can be made as follows. Choose a path 0 : [0; 1] ! R (M0 ) of (the gauge equivalence classes of) { invariant at connections on M0 between [A0 ] and [A00 ]. Since R (M2 ) is path connected, we may assume that 0 extends to a path of {invariant at connections on M2 , that is, 0 = j2 2 with 2 : [0; 1] ! R (M2 ). We can extend it to a path of {invariant connections on the entire manifold . These connections cannot be made at on the entire since R () is discrete. In other words, we construct a path : [0; 1] ! B 0; of {invariant connections such that 2 = i2 , and (0) = [A]; (1) = [A0 ]. The path restricts on M1 to a path which we call 1 , so 1 = i1 . 39
The operators rf0 j de ne a path of bounded Fredholm operators of index 4g ? 2. The vector bundle ker(rf0 j ) can be identi ed with the restriction 0 T R (M0 ) of the tangent bundle of R (M0 ) to the path 0 . Similarly, the vector bundle ker(rf2 j ) is identi ed with 2 T R (M2 ). Over the path 1 : [0; 1] ! B10; , the operators rf1 j de ne a path of bounded Fredholm operators of index 2g ? 1. At the endpoints of 1 , the spaces ker(rf1 j (t) ), t = 0; 1, can be identi ed with the corresponding tangent spaces of R (M1 ). This ensures that coker(rf1 j (0) ) = 0 and coker(rf1 j (1) ) = 0. If the cokernel of some rf1j (t) is not trivial, one can perturb the path of operators relf 0,1 g to get a path of bounded Fredholm operators of index 2g ? 1 with trivial cokernels at each t, see Lemma 5.7 of [33]. Note that the orientation of the vector bundle ker(rf1 j ) at the endpoints t = 0 and t = 1 is consistent with the orientation on the corresponding tangent spaces to R (M1 ). It can be checked by choosing a homotopy relf 0,1 g of the path 1 to a path which lies in R (M1 ). Such a homotopy exists due to the fact that 1 B10; = 0 G1 = [ M1 ; SU(2) ]Z=2 = 1. Let V ! [0; 1] be the vector bundle V = ker(rf1j ) ker(rf2 j ); and h : V ! 0 T R (M0 ) the L2 {orthogonal projection in 0 T0 onto 0 T R (M0 ). Then h is a bundle homomorphism, and is an orientation preserving isomorphism V = 0 (T R (M1 ) T R (M2 )) at the endpoints. Let det h denote the corresponding section of the real line bundle 4g?2 (0 T R (M0 )) 4g?2 (V ): Note that (det h)?1 (0) is supported away from f 0,1 g. If det h is not transverse to the zero section, perturb h rel f 0,1 g to h0 with det h0 being transverse. The intersection numbers at [A0 ] and [A00 ] coincide if and only if the cardinality of (det h0 )?1 (0) is even. Our next step is to represent the cardinality of (det h0 )?1 (0) modulo 2 as the spectral ow of a family of Fredholm operators. Let [A] 2 B 0; ; [Ak ] = ik [A], and [A0 ] = jk [Ak ] for k = 1; 2. Introduce the pre{Hilbert spaces EA0 = T1;A T2;A L0;A ; EA1 = L1;A L2;A T0;A ; and the operator HA : EA0 ! EA1 ; (61) (a1 ; a2 ; u) 7! (rf1;A (a1 ); rf2;A (a2 ); Y (a1 ; a2 ) ? rf0;A (u)) where the adjoint operator, rf0;A : L0;A ! T0;A , is de ned via integration by parts, 0
0
2
1
1
1
1
1
1
1
1
2
2
0
1
1
2
2
0
0
0
0
0
Z
Z
tr(dA u ^ a); tr(u ^ a) ? hrf0;A (u); aiL = M @M and the identi cation T0;A = T0;A is provided by the metric h ; 0 iA = hdA ; dA 0iL + h ; 0 iL ; 2
0
0
0
0
0
0
0
40
0
2
2
(62)
see [33], Appendix, for details. The map Y is the map : T1;A T2;A ! T0;A of (59) followed by the map T0;A ! T0;A given by the L2 {pairing, and the map T0;A ! T0;A given by the pairing in the metric (62). Thus the assignment of Y (a1 ; a2 ) 2 T0;A to a pair (a1 ; a2 ) 2 T1;A T2;A de nes a compact operator. The operator HA is a bounded Fredholm operator of index 0 since it is a compact perturbation of the Fredholm, index 0, operator (a1 ; a2 ; u) 7! (rf1;A (a1 ); rf2;A (a2 ); ?rf0;A (u)); see Proposition 30. We now return to the section det h. A pair (a1 ; a2 ) 2 ker(rf1;A ) ker(rf2;A ) belongs to (det h)?1 (0) if and only if the L2 {orthogonal projection of (a1 ; a2 ) onto ker(rf0;A ) is zero, which is the same as to say that Y (a1 ; a2 ) ? rf0;A (u) = 0 for some u. The latter means precisely that the triple (a1 ; a2 ; u) belongs to the kernel of HA . The following proposition is now apparent. Proposition 35. Let [A]; [A0 ] 2 B 0; be two points in f?1(0). The intersection numbers at [A0 ] = j1 i1 [A] and [A00 ] = j1 i1 [A0 ] in R (M1 ) \ R (M2 ) R (M0 ) coincide if and only if the spectral ow of a path of operators HA(t) connecting HA to HA0 is even, sfH ([A]; [A0 ]) = 0 mod 2. 1
0
2
0
0
0
0
0
1
2
1
2
0
1
0
2
0
6.3. Deformation of the H {family. The goal of this subsection is to prove that, for any pair of at [A]; [A0 ] 2 B 0;, sfH ([A]; [A0 ]) = sf ([A]; [A0 ]) mod 2: (63) The left and right hand sides of (63) are, respectively, the spectral ows of the paths of operators HA(t) and rfA(t) , see (61) and (30). We will construct a homotopy of one path into the other keeping the spectral ow modulo 2 unchanged. To begin with, we compare the operators HA and rfA for A at. Lemma 36. Let [A] 2 B 0; be a point in f?1(0), then ker HA = ker(rfA) = 0. Proof. The kernel of HA consists of the triples (a1 ; a2 ; u) 2 ker(rf1;A ) ker(rf2;A ) coker(rf0;A ) such that the L2{projection of (a1; a2 ) onto ker(rf0;A ) vanishes. Suppose that f(A) = 0. Then (rfk;Ak )(ak ) = dAk ak for any k = 0; 1; 2; see (57). If (a1 ; a2 ) 2 ker(rf1;A ) ker(rf2;A ), then (a1 ; a2 ) lies in ker(rf0;A ), because (rf0;A )( (a1 ; a2 )) = dA (j1 a1 ? dA 1 ) ? dA (j2 a2 ? dA 2 ); see (59); = j1 (dA a1 ) ? j2 (dA a2 ) ? d2A 1 + d2A 2 ; = 0: Therefore, ker HA = f (a1 ; a2 ) 2 ker(rf1;A ) ker(rf2;A ) j (a1 ; a2 ) = 0 g: 1
0
1
0
2
0
0
2
0
0
0
0
2
1
1
41
0
2
0
Lemma 34 provides a unique a 2 TA such that (a1 ; a2 ) = (a) and a 2 ker(rfA). The latter can be seen as follows. Since (a) = (a1 ; a2 ), we have that ak = ik a ? dAk k for k = 1; 2, and 0 = dAk ak = dAk (ik a ? dAk k ) = dAk (ik a) = ik (dA a):
The form (rfA )(a) = dA a restricts as zero to both M1 and M2 , hence, dA a = 0. For each [A] 2 B0; we have the following diagram T T ???! T ???! 0 0 ???! TA ???! A A A ? ? y
? ? y
1
rA f
0 ???! LA ???!
? ? y
2
? ? y
0
?
? r 1;A1 r 2;A2 r A0 y LA1 LA2 ???! LA0 ???! 0 f
f
f
The maps ; and ; have been de ned, respectively, in (59) and (60). According to Lemma 34, the rows of this diagram are exact. The operators rfA and rfk;Ak ; k = 0; 1; 2, are Fredholm operators of indices index rfA = 0; index rfk;Ak = 2g ? 1; k = 1; 2, and index rf0;A = 4g ? 2, see Proposition 30. One can easily check right from the de nitions that the diagram commutes if A is a at connection. Unfortunately, this is not the case for a general A. However, the diagram can be made into commutative by compactly perturbing the operators rf1;A rf2;A and rf0;A . More precisely, one can construct operators Qk;Ak : T1;A T2;A ! L1;A ; k = 1; 2; and Q0;A : T0;A ! L0;A ; (64) 0
1
2
0
1
2
1
0
0
0
and the operator QA : EA0 ! EA1 , (65) QA (a1 ; a2 ; u) = (Q1;A (a1 ); Q2;A (a2 ); Y (a1 ; a2 ) ? Q0;A (u)) with the following signi cance. Lemma 37. For every [A] 2 B0;, there exist bounded operators (64) such that the operator QA given by (65) is a Fredholm operator of index 0. Moreover, the following hold: 1. The operators QA ?HA; (Q1;A Q2;A )?(rf1;A rf2;A ), and Q0;A ?rf0;A are relatively compact operators; they vanish whenever A is at. 2. The assignment of QA to A de nes a smooth section, Q, over B0; of the Banach space bundle Fred0 (E 0 ; E 1 ), whose ber over A consists of Fredholm operators EA0 ! EA1 of index 0. The sections Q and H are homotopic rel ?1 (0). f 3. The following diagram is a commutative diagram of Fredholm bundle maps over B0;, 1
2
1
2
42
0
1
2
0
0
T T ???! T ???! 0 0 ???! T ???! 1 2 0 ? ? y
? ? y
? ? y 1
? ? y 0
Q Q2
r
f
? ? y
Q
L1 L2 ???! L0 ???! 0 0 ???! L ???!
where the rows are exact, and the indices of both Q1 Q2 and Q0 are 4g ? 2. Proof. The entire construction of the operators Qk;Ak in [33], pages 574{577, works in the equivariant setting, so we omit the proof. Proposition 38. For any pair of at connections [A]; [A0 ] 2 B0;; sfH ([A]; [A0 ]) = sf ([A]; [A0 ]) mod 2. Proof. Let A; A0 2 A be any pair of at connections representing [A]; [A0 ] 2 B0;. Choose a smooth path [A(t)]; t 2 [0; 1]; from [A] to [A0 ]. Since the paths QA(t) and HA(t) are homotopic rel f 0; 1 g, their spectral ows agree, sfH ([A]; [A0 ]) = sfQ([A]; [A0 ]) mod 2: We wish to show now that sfQ([A]; [A0 ]) = sf ([A]; [A0 ]) mod 2. To this end, choose a splitting : T0 ! T1 T2 with the property that = 1. Likewise, choose a splitting : L0 ! L1 L2 with the property that = 1. Since the rows in the commutative diagram above are exact, we get the commutative diagram T T T ? T0 ???! 1 2 ?
? r Q0 yQ1 Q2 L0 ???! L1 L2 ? y
L
f
where and are isomorphisms. Since QA = HA for any at connection A; coker Q0 = 0 at the endpoints of the path A(t). We may assume that in fact coker Q0;A (t) = 0 for all t, see Lemma 5.7 of [33]. Due to Proposition 30, ker Q0;A (t) = R4g?2 , and we can consider the family of operators, rf Q0;A (t) : TA(t) T0;A (t) ! LA(t) L0;A (t) R4g?2 ; where : T0;A (t) ! R4g?2 is orthogonal projection to ker Q0;A (t) . Since ker(Q0;A (t) ) = 0, the spectral ows of rfA(t) and rf Q0;A (t) coincide. In its turn, the spectral ow of rf Q0;A (t) equals the spectral ow of the path of operators C (t) making the following diagram commutative, 0
0
0
0
0
0
0
0
0
T ? T0
???!
Q Q2
r Q0
? y
T1 ? T2 ? y 1
f
1 ! L1 L2 R4g?2 L L0 R4g?2 ?????? 43
0
Thus, for any pair (a1 ; a2 ) 2 T1;A (t) T2;A (t) , we have that C (t)(a1 ; a2 ) = (Q1;A (t) (a1 ; a2 ); Q2;A (t) (a1 ; a2 ); ( (a1 ; a2 ))): This path of operators has the same spectral ow as the path QA(t) de ned by (65), compare the proof of Proposition 35. 6.4. Comparing the signs. In this section we prove that the signs of () and () agree. Recall that our de nition of () made use of the symmetric bilinear form a on su (2) de ned in (34) by the formula 1
2
1
2
a (; ) =
Z
tr([; a] ^ [; (d )?1 a]):
Here, a 2 ker d is an su(2){valued 1{form on . Since d : ker d ! ker d is invertible (due to the fact that is an integral homology sphere), we may assume that a = d b; b 2 ker d . Moreover, both and in su (2) are proportional to , see Lemma 19, so it is sucient to compute a (; ) for = = . In this case,
a(; ) =
Z
tr([b; ] ^ [a; ]); a = d b:
(66)
Our immediate goal is to construct such a that a is non-degenerate. To do that, we use an equivariant Heegaard decomposition of , = M1 [ M2 ; M1 \ M2 = M0 = R0 [?1=2; 1=2] where M0 is a \thickened" Riemann surface. The Mayer{Vietoris isomorphism, H 1 (M1 ) H 1 (M2 ) = H 1 (M0 ) induces an isomorphism in {equivariant cohomology, (67) : H1 (M1 ; d ) H1 (M2 ; d ) ! H1 (M0 ; d ): If we think of H1 (Mk ; d ); k = 0; 1; 2, as the space of d {harmonic {invariant di erential 1{forms ! on Mk such that i@Mk (!) = 0, the isomorphism is given by the formula (v1 ; v2 ) = j1 v1 ? d 1 ? j2 v2 + d 2 where 1 ; 2 are {invariant harmonic su(2){valued functions on M0 such that i@M (jk vk ? d k ) = 0; k = 1; 2: The vector spaces H1 (M0 ; d ) and H1 (Mk ; d ); k = 1; 2, have dimensions 4g and 2g, respectively, see Lemma 31. A 1{form a making the form a non{degenerate will be constructed after choosing elements !1 2 (H1 (M1 ; d ) 0) and !2 2 (0 H1 (M2 ; d )). Write ?1 (!1 ? !2 ) = ( 1 ; 2 ); then 1 restricts to M0 as !1 + d 1 , and 2 restricts to M0 as !2 + d 2 , with 1 and 2 harmonic. Set a = d b where b = (1 ? ) 1 + 2 (68) and : [?1=2; 1=2] ! [0; 1] is a cut-o function such that = 0 on [?1=2; ?1=4] and = 1 on [1=4; 1=2]. Then a = d b = ? 0 (dt ^ ( 1 ? 2 )); 0
44
and
a (; ) = ? = =
Z
M0
Z
M0 Z M0
tr([; (1 ? ) 1 + 2 ] ^ 0 (dt ^ [; 1 ? 2 ]))
0 dt ^ tr([; 1 ] ^ [; 2 ]) dt ^ tr([; !1 ] ^ [; !2 ]):
The last expression is positive if !2 = ?!1 , where !1 is the dual of !1 with respect to the non{degenerate symplectic pairing on H1 (M0 ; d ),
hv; v0 i = ?
Z
M0
dt ^ tr(v ^ v0 );
and !1 2 H1 (M1 ; d ) is chosen in general position. Thus, to prove that the signs of () and () agree we need to check that the orientations of the vector spaces ker(rf1;+sa ) ker(rf2;+sa ) and ker(rf0;+sa) agree for a chosen as in (68) and all s > 0 suciently small. The vector spaces in question are oriented as follows. When a connection A 2 A restricts to one or more of Mk ; k = 0; 1; 2, so that [A] 2 R (Mk ), then ker(rfk;+sa) is naturally identi ed with T[A] R (Mk ). Then, following a generic path from A to + s a, we orient ker(rfk;+sa). This method for computing the orientations actually works because is a limit point of each of R (Mk ); k = 0; 1; 2, hence one can nd connections A of the form + s a0, for small s, which restrict to one of Mk so that [A] 2 R (Mk ). The kernels of rfk;+sa0 for small s can be controlled by identifying them with the kernels of certain homomorphisms independent of s. We rst consider the general case with a0 unrestricted. Lemma 39. Let a0 2 1(; su(2)) be xed. Introduce the homomorphisms Gk : H1 (Mk ; d ) ! su(2) , by the formula
Gk (v)() = ?
Z
Mk
tr([; a0 ] ^ v); k = 1; 2:
If Gk are surjective, then for all s > 0 suciently small, ker(rfk;+sa0 ) = ker Gk H1 (Mk ; d ); k = 1; 2: Proof. Let us x k = 1; the proof for k = 2 is completely analogous. Let us consider the linear Fredholm operator 0 d :E !E ; L = 0
d d
1
2
where E1 and E2 are the following Hilbert spaces : E1 = f ( 0 ; 1 ) 2 L21 ( 0 (M1 ; su(2))) L21 ( 1 (M1 ; su(2))) j i@M ( 0 ) = 0; i@M ( 1 ) = 0 g; E2 = L2 ( 0 (M1 ; su(2))) L2 ( 1 (M1 ; su(2))): 1
45
1
The kernel of rf1;+sa0 can be identi ed with the kernel of the perturbed operator,
L0 + s u, where u is a compact operator de ned from the equation 0 0 d (69) L0 + s u = d 0 d+sa 0 : +sa +sa The kernel of L0 + s u can be computed by using techniques from linear perturbation theory. Namely, let h[u] : ker L0 ! coker L0 be a linear map sending v 2 ker L0 to the orthogonal projection of u(v) to coker L0 . Suppose that this projection is surjective. Then, for all s suciently small, the kernel of L0 + s u is isomorphic to h[u]?1 (0) ker L0 . The latter isomorphism is obtained as follows : for each v 2 ker(L0 + s u), there exists a unique w 2 h[u]?1 (0) ker L0 such that v ? w is L2 {orthogonal in E1 to ker L0 . It follows from de Rham theory that ker L0 = H1 (M1 ; d ) and coker L0 = H0 (M1 ; d ) = su(2). According to (69), the map h[u] : H1(M1 ; d ) ! H0 (M1 ; d ) sends 1 2 H1 (M1 ; d ) to the orthogonal projection of ? [ a; 1 ] to H0 (M1 ; d ). Therefore, 1 2 h[u]?1 (0) if and only if ? [ a; 1 ] is orthogonal to ker d = su (2), i.e. for any 2 su (2), 0=? =?
Z
M1
Z
tr( [a; 1 ]) tr([; a] ^ 1 )
M1 G1 ( 1 )():
= Thus, if G1 is surjective, the perturbation theory identi es ker rf1;+sa0 with ker G1. A similar argument proves the following result. Lemma 40. Let a0 2 1(; su(2)) be xed. Introduce the homomorphisms G0; G00 : H1 (M0 ; su(2)) ! su(2) de ned by the formulas Z
Z
dt ^ tr([; a0 ] ^ v): M0 M0 Suppose that both G0 and G00 are surjective. Then for all s suciently small, (70) ker(rf0;+sa0 ) = ker G00 ker G0 H1 (M0 ; d ): Orient S2 and S1 so that when + s a0 restricts to M1 to lie in R (M1 ) then the orientation of ker(rf1;+sa0 ) = H1 (M1 ; d )= im G1 agrees with the orientation which is obtained by identifying ker(rf1;+sa0 ) with T[+sa0 ] R (M1 ). Then, whenever + s a0 restricts to M2 to lie in R (M2 ), the orientation of ker(rf2;+sa0 ) = H1 (M2 ; d )= im G2 agrees with the orientation of ker(rf2;+sa0 ) = T[+sa0] R(M2 ). Likewise, when + s a0 restricts to a at connection in R (M0 ), the orientations of H1 (M0 ; d )= im (G00 G0) and T[+sa0] R(M0 ) agree. Thus, as a0 varies in 1 (; su(2)) keeping G00 G0 and G1 ; G2 surjective, the orientations on ker(rfk;+sa0 ); k = 0; 1; 2, are determined by the maps G00 G0 46 G0 (v)() = ?
tr([; a0 ] ^ v) and
G00 (v)() = ?
and G1 ; G2. We need to prove now that these orientations agree if a0 = a as in (68) with !2 = ?!1 . With the orientations of S2 and S1 xed as above, the isomorphism : H 1 (M1 ; d ) H 1 (M2 ; d ) ! H 1 (M0 ; d ) is orientation preserving because = 1 where : H 1 (M1 ) H 1 (M2 ) ! H 1 (M0 ) is the usual Mayer{Vietoris isomorphism, and 1 : T1 S2 ! T1S2 is the identity homomorphism. We use the isomorphism to make the following identi cations,
H 1 (M1 ; d ) H 1 (M2 ; d ) = h1 h2 and H 1(M0 ; d ) = h1 h2 where h1 = (H 1 (M1 ; d ) 0) and h2 = (0 H 1 (M2 ; d )). The homomorphisms G1 ; G2 , and G00 G0 are then given by the following formulas (possibly after some generic deformations preserving the orientations, compare with [33], Lemma 7.3 and Lemma 7.5),
G1 G2 ; G00 G0 : h1 h2 ! su (2) su (2) (G1 G2 )(v1 ; v2 ) = (G1 (v1 ); G2 (v2 )); and 0 (G0 G0 )(v1 ; v2 ) = (G1 (v1 ) + G2 (v2 ); G1 (v1 ) ? G2 (v2 )): Therefore, ker(G00 G0 ) and ker(G1 G2 ) have opposite orientations if oriented by the requirement that the following isomorphisms be orientation preserving, (2) su (2) ker G1 ker G2 = h1 h2 ; 0 su (2) su (2) ker(G0 G0 ) = h1 h2 :
su
The orientation on ker G1 ker G2 was chosen, however to make the isomorphism (2)
ker G1 su(2) ker G2 = h1 h2 orientation preserving. Since dim ker G1 = 2g ? 1 is odd, it implies that : ker G1 ker G2 ! ker(G00 G0 ) is orientation preserving. su
6.5. Proof of main theorem. It follows from Corollary 16, Corollary 26 and Proposition 29 that (71) ((p; q; r)) = 18 sign k(p; q; r) with a universal (){sign. To x the sign, we compute the left and the right hand sides for (p; q; r) = (2; 3; 7). According to [11], the group In ((2; 3; 7)) is Z if n = 2; 6, and is trivial otherwise. Therefore, ((2; 3; 7)) = 1. On the other hand, standard methods show that sign k(2; 3; 7) = 8, see Figure 11. Thus the sign in (71) is plus. This completes the proof. 47
7. Application to the {invariant In this section we de ne the invariants of Neumann and Siebenmann for integral homology 3{spheres of plumbing type and prove that they coincide. Their common value is usually referred to as the {invariant. It follows right from Siebenmann's de nition of the {invariant and Theorem 1 that the () = () for all Brieskorn homology spheres . The question remains whether the same is true for arbitrary plumbed integral homology spheres. Note that our de nition of both the Neumann's and Siebenmann's invariants di ers from the original one by a factor of 1=8. In should be also mentioned that one can de ne both Neumann's and Siebenmann's invariants for arbitrary Z=2{homology 3{spheres of plumbing type, and prove that they coincide by exactly the same methods as ones used in this section. 7.1. De nition of Neumann's {invariant. The principal S 1 {bundles over S 2 are classi ed by the Euler class e 2 H 2 (S 2 ; Z) = Z. Let the Hopf bration of S 3 over S 2 be represented by e = ?1. This xes the orientations. Denote the associated D2 {bundles indexed by e 2 Z as Y (e). Let ? be a connected plumbing graph, that is a connected graph with no cycles, each of whose vertices vi carries an integer weight ei ; i = 1; : : : ; s. Associated to each vertex vi is the D2 {bundle Y (ei ). If the vertex vi has di edges connected to it in the graph ?, we choose di disjoint discs in the base of Y (ei ) and call the disc bundle over the j {th disc Bij , so that Bij = D2 D2 . When two vertices, vi and vk , are connected by an edge, we identify Bij with Bk` by exchanging the base and ber coordinates. This pasting operation is called plumbing, and the resulting smooth 4{manifold P (?) is said to be obtained by plumbing according to ?. The manifold P (?) is simply connected but, generally speaking, not parallelizable. Its intersection form in the natural basis associated to plumbing (namely, the basis represented by the zero-sections of the plumbed bundles) is given by the matrix A(?) = (aij )i;j=1;:::;s with the entries 8 > if i = j ; 1; if i is connected to j by an edge; : 0; otherwise. Disconnected graphs are also allowed. Namely, if ? = ?0 + ?1 is a disjoint union of ?0 and ?1 then P (?) is the boundary connected sum P (?0 )\P (?1 ). If ? is a plumbing graph as above, then (?) = @P (?) is an integral homology sphere if and only if det A(?) = 1, in which case it is called a homology sphere of plumbing type or a plumbed homology sphere. Any Seifert bered homology sphere is of plumbing type, with a star-shaped plumbing graph, see [24]. For any plumbed homology sphere (?), there exists a unique homology class w 2 H2 (P (?); Z) satisfying the following two conditions. First, w is characteristic, that is (dot represents intersection number) w:x = x:x mod 2 for all x 2 H2 (P (?); Z); (73) and second, all the coordinates of w are either 0 or 1 in the natural basis of H2 (P (?); Z). We call w the integral Wu class for P (?). It is proven in [22] that the 48
integer
() = 18 (sign P (?) ? w:w)
(74)
only depends on (?) and not on ?. We call it the Neumann's {invariant. One can easily see that condition (73) implies that for no two adjacent vertices in ? the corresponding coordinates of w are both equal to 1. Therefore, the class w can be chosen to be spherical, hence () = () mod 2 where () is the Rohlin invariant. Another description of the {invariant is as follows. In the natural basis of H2 (P (?); Z), the de ning property (73) translates to the linear system s
X
i=1
aij "j = aii mod 2; i = 1; : : : ; s:
(75)
Due to the fact that det A(?) = 1 mod 2, this system has a unique solution ("1 ; : : : ; "s ) over Z=2. The {invariant is then given by the formula !
X () = 18 sign P (?) ? aii : "i =1
The {invariant is easy to compute. The splicing additivity, see [10] and [27], reduces the problem to the computation of ((a1 ; : : : ; an )). There is an explicit formula for the latter, see [22] and [24], Theorem 6.2. Namely, in the following two cases, (1) all ai are odd; (2) exactly one of the ai is even, say a1 , the -invariant is given respectively by
((a1 ; : : : ; an ) =
n
X
i=1
(c(ai ; bi ) + sign bi ) ? 1
and ((a1 ; : : : ; an ) =
n
X
i=1
c(ai ? bi; ai ) ? 1
(76) (77)
where c(a; b) is the function described below and, in the second case, we are to choose the Seifert invariants so that (ai ? bi ) is odd for all i. The integers c(a; b) are de ned for coprime integer pairs (a; b) with a odd. They are uniquely determined by the recursion : c(a; 1) = 0 for any odd a; c(a 2b; b) = c(a; b); c(a; b + a) = c(a; b) + sign b(b + a); c(a; b) = ?c(?a; b) = ?c(a; ?b): 49
Several equivalent de nitions of c(a; b) are available, see [24]. For example,
(?1)i # f k j 0 < k < b ; i < ka b < i + 1g X ( + 1)( a + 1) = ? 1b a ? 1) ; ( ? 1)( b =?1 = b ? 1 ? 4 #f 1 i b ?2 1 b ?2 1 < ai < b mod b g; b odd: 7.2. Preferred involutions on plumbed manifolds. J. Montesinos observed in [20] that on any homology 3{sphere of plumbing type, there exists a preferred class I of orientation preserving involutions such that, for 2 I, the orbit space = is S 3 and the xed point set is a knot k in S 3 , which is called a knot of plumbing type. The goal of this section is to describe the involutions 2 I . Let D2 be the unit disc in C ; D2 = f z 2 C j jz j 1 g, and : D2 D2 ! D2 D2 the complex conjugation involution, (z; w) 7! (z ; w ). The boundary S 3 of D2 D2 can be represented by stereographic projection onto R3 [ 1 in such a way that induces on it the 180 -rotation about the y{axis. In this representation @D2 D2 is a regular neighborhood X of the unit circle in the xy{plane, the belt{sphere is the z{axis, and the belt{tube is Y = S 3 n int X . The xed point set of the involution on D2 D2 is the unit disc B = D1 D1 2 D D2 . The orbit space of is a 4{dimensional ball D4 , so that the projection p : D2 D2 ! D4 is a 2{fold branched cover of D4 with the branch set the 2{disc B 0 = p(B ). The restriction p @(D D ): @ (D2 D2 ) ! S 3 is the standard covering over the trivial knot @B 0 S 3 . The branch set B 0 D4 can be described as follows. Let S 3 = X [ Y as above. The restriction of p onto X turns X into a double branched cover of the 3{ball X 0 = X= with the branch set consisting of the two arcs shown in Figure 8.
c(a; b) =
X
2
p(-1, -1)
2
p(1, -1)
I
I’
p(-1, 1)
p(1, 1)
Figure 8
Figure 9
These two arcs represent the intersection B 0 \ X 0 S 3 . Similarly, Y 0 = Y= is a 3{ball with the branch set the arcs shown in Figure 9. The 3{balls X 0 and Y 0 are 50
glued together to form S 3 , the arcs in X 0 and Y 0 being glued along their endpoints into the trivial knot @B 0 S 3 . Now, let B 00 be a 2{disc in S 3 with @B 00 = @B 0 . Then the branch set B 0 D4 can be viewed as a 2{disc obtained by pushing the interior of B 00 radially into D4 so that B 0 \ S 3 = @B 0 = @B 00 . The complex conjugation involution on D2 D2 can be extended over the 2{disc bundles Y (e) over S 2 , which are the building blocks for the plumbed manifolds. The 4{manifold Y (e) can be obtained by pasting two copies of D2 D2 ,
Y (e) = D2 D2 [he D2 D2 ; along @D2 D2 where he : @D2 D2 ! @D2 D2 is given by the formula (ei ; ei ) 7! (e?i ; ei( ?e ) ); 0 ; < 2; 0 1: The complex conjugation involution on D2 D2 obviously commutes with the map he , so can be extended to an involution (also denoted by ) on Y (e). The quotient Y (e)= is glued from two 4{balls along a 3{ball, so that Y (e)= = D4 . The branch set in D4 can be described as follows. Let us represent Y (e)= = D4 as
D4 = D4 [D Cyl (h e ) [D D4 3
3
where Cyl (h e ) is the mapping cylinder of the map h e : D3 ! D3 induced by he on the quotient of @D2 D2 . The branch set in Y (e)= then consists of four bands glued together, see Figure 10 : two untwisted bands in the two copies of D4 , and two more bands in Cyl (h e ) generated by the arcs I 0 and I , see Figure 8. One can easily see that the former is untwisted while the latter one is twisted e half{twists in the right{hand direction (in case e is positive).
Figure 10 Let P (?) be a plumbed manifold obtained by plumbing D2 {bundles Y (ei ) according to a connected plumbing tree ?. The involutions on the manifolds Y (ei ) t together to form an involution on P (?) whose orbit space is D4 , and the branch set B (?) is obtained by plumbing twisted bands (like one in Figure 10) according to the tree ?, see Figure 11 and [31]. 51
2
0
0
-3
-7
2
-3
-7
Figure 11 In particular, this construction represents the boundary of a plumbed manifold
P (?) as a double branched cover of S 3 branched over the boundary of the surface B (?). It follows from the Smith theorem that the xed point set of an involution acting on a Z=2{homology sphere is itself a Z=2{homology sphere. Therefore, if is an integral homology sphere, the boundary of the surface B (?) is a knot, which we denote by k . In the special case of Seifert bered integral homology spheres (a1 ; : : : ; an ), one can do plumbing holomorphically to obtain P (?) as a complex manifold with holomorphic C {action, see [25], representing a resolution of the singularity of f ?1 (0) where f : C n ! C n?2 is a map of the form f (z1 ; : : : ; zn ) =
n
X
k=1
b1;k zkak ; : : : ;
n
X
k=1
bn?2;k zkak
!
(78)
with suciently general coecient matrix (bi;j ); bi;j 2 R, see [24]. The involution on P (?) will then be the complex conjugation involution. Therefore, any Seifert bered homology sphere (a1 ; : : : ; an ) is a double branched cover of S 3 with the branch set the knot k = k(a1 ; : : : ; an ), which is usually called a Montesinos knot, see [20]. A Brieskorn homology sphere (p; q; r) is a Seifert bered homology sphere with n = 3 bers, and it is a double branched cover of S 3 with branch set k(p; q; r). 7.3. De nition of Siebenmann's ~{invariant. Let be a Z=2{homology sphere of plumbing type, and an involution on from the preferred class I of involutions described in the previous section. Its orbit space = is S 3 , and the branch set is a knot k S 3 . L. Siebenmann observed in [31] that one can use Waldhausen's classi cation to show that if 0 is also in I , the knot k0 is related to k by a sequence of mutations. The knot signature is a mutation invariant, thus the integers sign k ; 2 I , all coincide. L. Siebenmann calls this common value the ~{invariant, 52
so that
~() = 18 sign k :
The fact that ~() = () mod 2 can be proven as follows. Fix 2 I and let F be an (orientable) Seifert surface of the knot k pushed radially into D4 . Let W be the double branched cover of D4 with the branch set F . This manifold is parallelizable, its boundary is , therefore, sign W = () mod 2. On the other hand, sign W = sign k , see e.g. [13]. Therefore, ~ is a lift of the Rohlin invariant into the integers de ned for all integral homology 3{spheres of plumbing type. 7.4. The invariants of Neumann and Siebenmann coincide. Now we have two lifts of the Rohlin invariant into the integers de ned for all plumbed integral homology spheres, one by W. Neumann and the other by L. Siebenmann. We prove in this section that these lifts in fact coincide. Proposition 41. For any integral homology sphere of plumbing type, () = ~(). As we have mentioned before, the invariants () and ~() are in fact de ned for all Z=2{homology spheres of plumbing type, and one can show that they are equal to each other for such spheres as well. Proof. Let be a plumbed homology sphere, = @P (?). The 4{manifold P (?) is a double branched cover of D4 whose branch set is the surface F = B (?) D4 such that @F \ S 3 = k . The surface F is orientable if and only if all the weights in the plumbing tree ? are even. Suppose that ? has only even weights. Then, since the intersection form of P (?) is even, the Wu class w vanishes, and ~() = 81 sign k = 81 sign P (?) = (): Now suppose that ? has at least one odd weight. According to C. Gordon and R. Litherland [13], Theorem 3 and Corollary 5, sign k = sign P (?) + 21 (F ): The number (F ) is de ned as minus the knot linking number, (F ) = ?lk(k ; kF ) where kF is a parallel copy of the knot k missing the surface F . Thus to complete the proof we only need to show that 1 lk(k ; kF ) = w:w 2 where w is the Wu class of the manifold P (?). Let us rst color in black the surface F , and orient its boundary @F = k . In a knot projection there are two types of double points, 53
Type 1
Type 2
η=1
η = -1
Figure 12 and one can easily see that 1=2 lk(k ; kF ) is the sum of the incidence numbers of double points of type 2. In a regular projection of k , like one seen in Figure 11, the double points only occur because of twisting the bands, and all the double points on the same band have the same type and the same incidence number. Let vi be a vertex in ? weighted by aii (remember that the aij 's are the entries of the matrix A(?), see (72)). Let "i = 1 if the double points on the band corresponding to vi are of type 2, and "i = 0 otherwise. Then s 1 lk(k ; kF ) = X 2 i=1 "i aii :
On the other hand, by comparing the orientations on the boundary of the i-th band with the orientations on the boundaries of bands adjacent to it one can easily check that s
X
j =1
"j aij = aii mod 2; i = 1; : : : ; s:
Therefore, the vector w = ("1 ; : : : ; "s ) is the Wu vector of the matrix A(?), and P w:w = "i aii. 54
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Department of Mathematics, University of Michigan, Ann Arbor, MI 48109
E-mail address :
[email protected]
56
