Non-integral toroidal surgery on hyperbolic knots in S^3

It is shown that a hyperbolic knot in S3 admits at most one nonintegral Dehn surgery producing a toroidal manifold. Let K be a knot in the 3-sphere S and M = MK the complement of an open regular neighborhood of K in S. As usual, the set of slopes on the torus ∂M (i.e. the set of isotopy classes of essential simple loops on ∂M) is parameterized by {m/n : m,n ∈ Z, n > 0, (m,n) = 1} ∪ {1/0}, so that 1/0 is the meridian slope and 0/1 is the longitude slope. A slope m/n is called non-integral if n ≥ 2. The manifold obtained by Dehn surgery on S along the knot K (equivalently, Dehn filling on M along the torus ∂M) with slope m/n, is denoted by M(m/n). Now suppose that K ⊂ S is a hyperbolic knot, i.e. the interior of M has a complete hyperbolic metric of finite volume. A basic question in Dehn surgery theory is: when can a surgery on K produce a non-hyperbolic 3-manifold? A special case of this question is: when can a surgery on K produce a toroidal 3-manifold, i.e. a 3-manifold which contains an (embedded) incompressible torus? In [GL1], Gordon and Luecke showed that if m/n is a non-integral slope and M(m/n) contains an incompressible torus, then n = 2. In this paper, we show that there is at most one such surgery slope. Theorem 1. For a hyperbolic knot in S, there is at most one surgery with nonintegral slope producing a manifold containing an incompressible torus. There are examples of hyperbolic knots in S which admit toroidal surgeries with non-integral slopes [EM]. The best known example is the (−2, 3, 7)-pretzel knot, on which the surgery with slope 37/2 gives a toroidal 3-manifold. It is a conjecture ([Go1], [K, Problem 1.77]) that for a hyperbolic knot in S, there is at most one surgery with non-integral slope producing a non-hyperbolic 3-manifold, and further, if there is such a surgery, it must be a toroidal surgery and the knot must belong to the collection of examples given in [EM]. Theorem 1 provides some supporting evidence for this conjecture. We now go on to the proof of Theorem 1. Our argument is based on applications of results and combinatorial techniques developed in [CGLS], [GL1], [Go2]. Received by the editors May 20, 1997 and, in revised form, August 3, 1998. 1991 Mathematics Subject Classification. Primary 57N10, 57M25. The first author was partially supported by NSF grant DMS 9626550. The first and second authors were supported in part by Research at MSRI NSF grant #DMS 9022140. c ©2000 American Mathematical Society 1869 1870 C. McA. GORDON, Y-Q. WU, AND X. ZHANG Recall that the distance between two slopes m1/n1 and m2/n2 is defined as the number ∆ = ∆(m1/n1,m2/n2) = |m1n2 − n1m2|, which is equal to the minimal geometric intersection number between simple loops representing the two slopes on ∂M . Suppose that there are two slopes m1/2 and m2/2 such that both M(m1/2) and M(m2/2) contain incompressible tori. It follows from [Go2, Theorem 1.1] that there are exactly four hyperbolic manifolds that admit two toroidal Dehn fillings at distance more than 5, but for homological reasons, only one of these is the complement of a knot in S, namely the figure 8 knot complement, and by [Th] every non-integral surgery on this manifold is hyperbolic. Hence we have ∆(m1/2,m2/2) = |2m1 − 2m2| = 2|m1 − m2| ≤ 5. Note that both m1 and m2 are odd integers. So |m1 − m2| is even and thus must be equal to 2. Hence the distance between the two slopes is exactly 4. Our task here is to show that this is impossible. Note, however, that 4 can be realized as the distance between integral toroidal surgery slopes for a hyperbolic knot in S. For instance the slopes 16 and 20 are both toroidal surgery slopes for the (−2, 3, 7)-pretzel knot. The reason that distance 4 is impossible in our situation is mainly due, as we will see, to the fact that the first homology of M(mi/2) with Z2 coefficients is trivial for i = 1, 2. By [GL1] and [GL2] (see [GL1, Theorem 1.2]), for i = 1, 2, there is an incompressible torus Ti in M(mi/2) such that M ∩ Ti = Ti is an incompressible, ∂-incompressible, twice punctured torus properly embedded in M with each component of ∂Ti having slope mi/2 in ∂M . Note that Ti separates M since M(mi/2) has finite first homology. By an isotopy of Ti, we may assume that T1 and T2 intersect transversely, and T1 ∩ T2 has the minimal number of components. So T1 ∩ T2 is a set of finitely many circle components and arc components properly embedded in Ti, i = 1, 2. Furthermore, no circle component of T1 ∩ T2 bounds a disk in Ti and no arc component of T1 ∩ T2 is boundary parallel in Ti, i = 1, 2, since Ti is incompressible and ∂-incompressible. We shall use the indices i and j to denote 1 or 2, with the convention that, when they are used together, {i, j} = {1, 2} as a set. Let Vi denote the solid torus that is attached to M in forming M(mi/2). The torus Ti intersects Vi in two disks Bi(1) and Bi(2), which cut Vi into two 2-handles, which we denote by Hi(1) and Hi(2). Correspondingly, ∂Ti = ∂Bi(1) ∪ ∂Bi(2) cuts ∂M into two annuli Ai(1) and Ai(2), where Ai(k) ⊂ ∂Hi(k). Each ∂Tj∩Ai(k) consists of exactly 8 essential arcs on Ai(k). The torus Ti separates M(mi/2) into two submanifolds which we denote by Xi(1) and Xi(2), with Hi(k) ⊂ Xi(k). Correspondingly Ti separates M into two pieces, denoted by Xi(1) and Xi(2). Thus Xi(k) = M ∩ Xi(k). Note that Xi(k) is obtained by attaching the 2-handle Hi(k) to Xi(k) along the annulus Ai(k), k = 1, 2. Also note that Fi(k) = ∂Xi(k) = Ti ∪Ai(k) is a closed surface of genus two, k = 1, 2. Lemma 2. For i = 1, 2, k = 1, 2, we have H1(Xi(k), Ti; Z2) = 0. Proof. Since H1(M(mi/2); Z2) = 0, and Ti is connected, we have 0 = H1(M(mi/2), Ti; Z2) = H1(Xi(1), Ti; Z2)⊕H1(Xi(2), Ti; Z2); hence each H1(Xi(k), Ti; Z2) = 0. Now, as in [CGLS, 2.5], and [Go2], we construct two graphs Γ1 and Γ2 in T1 and T2 respectively by taking the arc components of T1∩T2 as edges and Ti− int(Ti) = TOROIDAL SURGERY ON HYPERBOLIC KNOTS 1871 1 2 1 2 1 2 1 2 Bi(k)