THE CLASSIFICATION OF DEHN SURGERIES ON 2-BRIDGE KNOTS

Mark Brittenham and Ying-Qing Wu

Abstract. We will determine whether a given surgery on a 2-bridge knot is reducible, toroidal, Seifert fibered, or hyperbolic.

In [Th1] Thurston showed that if K is a hyperbolic knot, then all but finitely many surgeries on K are hyperbolic. In particular, for the Figure 8 knot, it was shown that exactly 9 nontrivial surgeries are non-hyperbolic. Let K_{p/q} be a 2-bridge knot associated to the rational number p/q. When p ≡ ±1 mod q, K is a torus knot, on which the surgeries are well understood. By [HT], all other 2-bridge knots are hyperbolic, admitting no reducible surgeries. Moreover, K_{p/q} admits a toroidal surgery if and only if p/q = [r_{1}, r_{2}] = 1/(r_{1} − 1/r_{2}) for some integers r_{1}, r_{2}. See Lemma 8 below for a complete list of all toroidal surgeries. The Geometrization Conjecture [Th2] asserts that if a closed orientable 3-manifold is irreducible and atoroidal, then it is either a hyperbolic manifold, or a Seifert fibered space whose orbifold is a 2-sphere with at most three cone points, called a small Seifert fibered space. The conjecture has been proved for two large classes of manifolds: the Haken manifolds [Th2], and those admitting an orientation preserving periodic map with nonempty fixed point set [Th3,Ho,KOS,Zh]. It can be shown that surgery on a 2-bridge knot yields a manifold which admits such a periodic map, so it has a geometric decomposition. Our main result will classify all surgeries on 2-bridge knots according to whether they are reducible, toroidal, Seifert fibered, or hyperbolic manifolds.

Typeset by AMS-TEX

1991 Mathematics Subject Classification. 57N10, 57M25, 57M50.

Key words and phrases. Dehn surgery, 2-bridge knots, Seifert fibered spaces, hyperbolic manifolds.

We use [b_{1}, …, b_{n}] to denote the partial fraction decomposition 1/(b_{1} −1/(b_{2} −…− 1/b_{n}) …). Recall that a 2-bridge knot K is a twist knot if it is equivalent to some K_{p/q} with p/q = [b, ±2] for some integer b. Since [b, ±2] = [b ∓ 1, ∓2], we may assume that b is even. Let K(γ) be the manifold obtained by γ surgery on K. We always assume that γ ≠ ∞, that is, the surgery is nontrivial. (4) and (5) in the following theorem are well known. They are included here for the sake of completeness.

Theorem 1. Let K be a 2-bridge knot.

(1) If K ≠ K_{[b_{1},b_{2}]} for any b_{1}, b_{2}, then K(γ) is hyperbolic for all γ.

(2) If K = K_{[b_{1},b_{2}]} with |b_{1}|, |b_{2}| > 2, then K(γ) is hyperbolic for all but one γ, which yields toroidal manifold. When both b_{1} and b_{2} are even, γ = 0. If b_{1} is odd and b_{2} is even, γ = 2b_{2}.

(3) If K = K_{[2n,±2]} and |n| > 1, K(γ) is hyperbolic for all but five γ: K(γ) is toroidal for γ = 0, ∓4, and is Seifert fibered for γ = ∓1, ∓2, ∓3.

(4) If K = K_{[2,−2]} is the Figure 8 knot, K(γ) is hyperbolic for all but nine γ: K(γ) is toroidal for γ = 0, 4, −4, and is Seifert fibered for γ = −1, −2, −3, 1, 2, 3.

(5) If K = K_{[b]} is a (2, b) torus knot, K(γ) is Seifert fibered unless γ = 2b. K(2b) is a reducible manifold.

The reader is referred to [GO] for definitions and basic properties concerning essential branched surfaces and essential laminations, which play a central role in the proof of the theorem. In [De1, De2] Delman constructed essential branched surfaces and laminations in 2-bridge knot complements, which are persistent in the sense that they remain essential after all nontrivial surgeries. Brittenham [Br] showed that if M is a small Seifert fibered space containing an essential branched surface F, then each component of M − IntN(F) is an I bundle over a compact surface G. This is a very useful criteria in determining which manifolds are small Seifert fibered spaces. Before proving the theorem, we need to review some results of [De1, De2].

Figure 1

For each rational number p/q, there is associated a diagram D(p/q), which is the minimal subdiagram of the Hatcher-Thurston diagram [HT, Figure 4] that contains all minimal paths from 1/0 to p/q. See [HT, Figure 5] and [De1]. D(p/q) can be constructed as follows. Let p/q = [a_{1}, …, a_{k}] be a partial fraction decomposition of p/q. To each a_{i} is associated a “fan” F_{a_{i}} consisting of a_{i} simplices, see Figure 1(a) and 1(b) for the fans F_{4} and F_{−4}. The edges labeled e_{1} are called initial edges, and the ones labeled e_{2} are called terminal edges. The diagram D(p/q) can be constructed by gluing the F_{a_{i}} together in such a way that the terminal edge of F_{a_{i}} is glued to the initial edge of F_{a_{i+1}}. Moreover, if a_{i}a_{i+1} < 0 then F_{a_{i}} and F_{a_{i+1}} have one edge in common, and if a_{i}a_{i+1} > 0 then they have a 2-simplex in common. See Figure 1(c) for the diagram of [2, −2, −4, 2].

To each vertex v_{i} of D(p/q) is associated a rational number r_{i}/s_{i}. It has one of the three possible parities: odd/odd, odd/even, or even/odd, denoted by o/o, o/e, and e/o, respectively. Note that the three vertices of any simplex in D(p/q) have mutually different parities.

We consider D(p/q) as a graph on a disk D, with all vertices on ∂D, containing ∂D as a subgraph. The boundary of D forms two paths from the vertex 1/0 to the vertex p/q. The one containing the vertex 0/1 is called the top path, and the one containing the vertex 1/1 is called the bottom path. Edges on the top path are called top edges. Similarly for bottom edges.

Let Δ_{1}, Δ_{2} be two simplices in D(p/q) with an edge in common. Assume that the

Figure 2

two vertices which are not on the common edge are of parity o/o. Then the arcs indicated in Figure 2(a) and (b) are called channels. A path α in D(p/q) is a union of arcs, each of which is either an edge of D(p/q) or a channel.

Let α be a path in D(p/q). Let v be a vertex on α. Let e_{1}, e_{2} be the edges of α incident to v. Then there is a fan F_{v} in D(p/q) with e_{1}, e_{2} as initial and terminal edges. The number of simplices in F_{v} is called the corner number of v in α, denoted by c(v; α) or simply c(v). A path α from 1/0 to p/q is an allowable path if it has at least one channel, and c(v) ≥ 2 for all v in α.

Now assume that K = K_{p/q} is a 2-bridge knot. Then q is an odd number. Recall that K_{p/q} = K_{p′/q} if p′ ≡ p^{±1} mod q, and K_{−p/q} is the mirror image of K_{p/q}. We may assume without loss of generality that p is even, and 1 < p < q. This is because K_{(q−p)/q} is equivalent to the mirror image of K_{p/q}, so the result of γ surgery on the first is the same as that of −γ surgery on the second. Note that q − p and p have different parity, since q is odd. The following result is due to Delman. See [De1] and [De2, Proposition 3.1].

Lemma 2. Given an allowable path α of D(p/q), there is an essential branched surface F in S^{3} − K which remains essential after all nontrivial surgeries on K. □

Lemma 3. If there is an allowable path α in D(p/q) such that c(v) > 2 for some vertex v in α, then K(γ) is not a small Seifert fibered space for any γ.

Proof. It was shown in [Br, Corollary] that if F is an essential branched surface in a small Seifert fibered space M, then each component of M − IntN(F) is an I-bundle over a compact surface G, such that the vertical surface ∂_{v}N(F) (also called cusps) is the I-bundle over ∂G. It has been shown in [De1] that for each vertex v of α there is a component W_{v} of S^{3} − IntN(F) such that W_{v} is a solid torus whose meridian disk intersects the cusps c(v) times. In particular, if c(v) > 2 then W_{v} is not an I bundle as above. Since F is an essential branched surface in K(γ), it follows that K(γ) is not a small Seifert fibered space. □

Lemma 4. Suppose p is even, q is odd, and 1 < p < q −1. If p/q does not have partial fraction decomposition of type [r_{1}, r_{2}], then D(p/q) has an allowable path α such that some vertex v on α has c(v) > 2.

Proof. Let [a_{1}, …, a_{n}] be the partial fraction decomposition of p/q such that all a_{i} are even. Then a_{1} ≥ 2. If a_{i} = 2 for all i, then p/q = (q − 1)/q, contradicting our assumption. Thus either some a_{i} < 0, or some a_{i} > 4. We separate the two cases.

CASE 1. Some a_{i} < 0.

Let a_{i} be the first negative number. Then a_{i−1} > 0, so there is a sign change. By [De2] there is a channel α_{0} in F_{a_{i−1}} ∪ F_{a_{i}} starting at a bottom edge and ending at a top edge, where F_{a_{i}} is the fan in D(p/q) corresponding to a_{i}. Let α_{1} be the part of the bottom path of D(p/q) from the vertex 1/0 to the initial point of α_{0}, and let α_{2} be the part of the top path from the end point of α_{0} to the vertex p/q. Then α = α_{1} ∪ α_{0} ∪ α_{2} is an allowable path in D(p/q). We need to show that if c(v) = 2 for all vertices v on this path, then p/q = [r_{1}, r_{2}] for some r_{1}, r_{2}.

Consider the vertices on α_{1}. Since c(v_{i}) = 2 for all v_{i}, each vertex v_{i} is incident to exactly one non boundary edge e_{i} of D(p/q), which must have the other end on a vertex v′_{i} in the top path. If some of these v′_{i} are different, then since all faces of D(p/q) are triangles, it is clear that some v_{j} on α_{1} would have at least two non boundary edges, which would be a contradiction. Similarly, each vertex on α_{2} has a unique non boundary edge, leading to a common vertex on the bottom path, so the

Figure 3

diagram D(p/q) looks exactly as in Figure 3(a). It is the union of two fans F_{r_{1}} and F_{r_{2}} with r_{1} > 0, and r_{2} < 0. Therefore, p/q = [r_{1}, r_{2}].

CASE 2. Some a_{i} ≥ 4.

In this case there is a channel α_{0} with both ends on the bottom path. Construct an allowable path α = α_{1} ∪ α_{0} ∪ α_{2} with α_{1}, α_{2} in the bottom path. Similar to Case 1, it can be shown that each vertex on α_{i} has a unique non boundary edge leading to a common vertex v′_{i} on the top path, so D(p/q) looks like that in Figure 3(b). In this case p/q = [r_{1}, r_{2}], with both r_{i} > 0. □

Lemma 5. Let K = K_{p/q} be a two bridge knot such that p/q = [r_{1}, r_{2}], and |r_{i}| ≥ 3 for i = 1, 2. Then no surgery on K is a small Seifert fibered space.

Proof. Since K is a knot, q is odd, so at least one of the r_{i} is an even number. We assume without loss of generality that r_{1} = 2n for some integer n, since K_{[r_{1},r_{2}]} is equivalent to K_{[r_{2},r_{1}]}, by turning the standard diagram for the first knot upside down.

Let L = k_{1} ∪ k_{2} be a 2-bridge link associated to the rational number p′/q′ = [2, r_{2}, −2]. See Figure 4, where r_{2} = −6. Notice that after −1/n surgery on k_{1}, the other component k_{2} becomes the knot K = K_{[2n,r_{2}]}. Therefore, doing γ surgery on K is the same as doing −1/n surgery on k_{1}, then doing some γ′ surgery on k_{2}.

By considering the mirror image of K if necessary we may assume that r_{2} < 0. If r_{2} is even, then [2, r_{2}, −2] is a partial fraction decomposition with even coefficient, and

Figure 5

r_{2} ≤ −4. There is an allowable path in D(p′/q′) with two channels, as shown in Figure 5(a), where r_{2} = −4. If r_{2} is odd, then p′/q′ = [2, r_{2} + 1, 2], in which case D(p′/q′) also has an allowable path with two channels. See Figure 5(b) for the case r_{2} = −3.

Let F be the essential branched surface in the link exterior associated to the above allowable path in D(p′/q′), as constructed in [De2]. There is one solid torus component V_{i} in S^{3}−IntN(F) for each k_{i}, containing k_{i} as a central curve. ¿From the construction of F one can see that each channel contributes two cusps, one on each ∂V_{i}. Actually from [De2, Figure 3.5] we see that the two cusps are around two points of L on a level sphere with same orientation. Since each k_{i} intersects the sphere at two points with different orientations, those two cusps must be around different components of L. See [Wu] for more details about surgery on 2-bridge links.

As the allowable path above has two channels, each V_{i} has two meridional cusps. Thus F remains an essential branched surface after surgery on L. Moreover, since the surgery on k_{1} has coefficient −1/n, which is non-integral, after surgery V_{1} becomes a solid torus whose meridional disk intersects the cusps at least four times. By [Br, Corollary], the surgered manifold is not a small Seifert fibered space. □ Lemma 6. Let L = k_{1} ∪k_{2} be a 2-bridge link associated to the rational number p′/q′ = [2, 2, −2]. Let L(γ_{1}, γ_{2}) be the manifold obtained by γ_{i} surgery on k_{i}. If γ_{1} = −1/n and γ_{2} = −1, −2 or −3, then L(γ_{1}, γ_{2}) is a small Seifert fibered space. Proof. By definition L(∞, γ_{2}) is the manifold obtained from S^{3} by γ_{2} surgery on k_{2}. After −1 surgery on k_{2}, the knot k_{1} becomes a trefoil knot in L(∞, −1) = S^{3}. Since the exterior of a torus knot is a Seifert fibered space with orbifold a disk with two cones, it is easy to see that all but one surgeries yield Seifert fibered spaces, each having an orbifold a disk with at most three cones. For this trefoil, the exceptional surgery has coefficient −6, yielding a reducible manifold. Thus L(−1/n, −1) is a small Seifert fibered space for any n.

After −2 surgery on k_{2}, k_{1} becomes a knot in ℝP^{3} = L(∞, −2). The link L is drawn in Figure 6(a), where the curve C is a curve on ∂N(k_{2}) of slope −2, so it bounds a disk in L(∞, −2). Thus a band sum of k_{1} and C forms a knot k′_{1} isotopic to k_{1} in L(∞, −2). The link L′ = k′_{1} ∪ k_{2} is shown in Figure 6(b). Using Kirby Calculus one can show that L(−1/n, −2) = L′(−2 − 1/n, −2). The exterior of k′_{1} in S^{3} is a Seifert fibered space with orbifold a disk with two cones, in which k_{2} is a singular fiber of index 3. Thus after −2 surgery on k_{2}, the manifold L(∞, −2) − IntN(k′_{1}) is still Seifert fibered, with orbifold a disk with two cones. The fiber slope on ∂N(k′_{1}) is 6. It follows that all but the 6 surgery on k′_{1} in L(∞, −2) yield small Seifert fibered manifolds. In particular, L(−1/n, −2) = L′(−2 − 1/n, −2) are small Seifert fibered manifolds for all n.

The proof for γ_{2} = −3 is similar. One can show that the band sum of k_{1} and the curve C of slope −3 on ∂N(k_{2}) is isotopic to the curve k′_{1} shown in Figure 6(c), which is a (3, −2) torus knot. By the same argument as above one can show that L(−1/n, −3) = L(−3 − 1/n, −3) are small Seifert fibered manifolds for all n. □

Corollary 7. Let K be a non-torus 2-bridge knot.

Figure 6

(1) K admits a small Seifert fibered surgery if and only if it is a twist knot;

(2) If K = K_{p/q} is a twist knot with p/q = [2n, ±2], then K(γ) is a small Seifert fibered space for γ = ∓1, ∓2 and ∓3.

(3) K(γ) is not Seifert fibered unless γ is an integer.

Proof. As noticed before, up to taking the mirror image we may assume that K = K_{p/q}, where 1 < p < q, p is even, and q is odd. Note that if p/q = (q − 1)/q then K is a (2, q) torus knot. Hence if K is a non-torus knot admitting some small Seifert fibered surgery, then by Lemma 3 and Lemma 4 we must have p/q = [b_{1}, b_{2}] for some integers b_{1}, b_{2}. Note that b_{i} ≠ ±1, otherwise K is a torus knot. Therefore by Lemma 5 one of the b_{i} must be ±2. In other words, K is a twist knot.

Consider the case p/q = [b, 2]. The proof for p/q = [b, −2] is similar. Since [b, 2] = [b − 1, −2], we may further assume that b = 2n is even. Let L = k_{1} ∪ k_{2} be a 2-bridge link associated to the rational number p′/q′ = [2, 2, −2]. Notice that after −1/n surgery on k_{1}, the knot k_{2} becomes the knot K = K_{[b,2]} in S^{3} = L(−1/n, ∞). Therefore by Lemma 6, K(γ) = L(−1/n, γ) are small Seifert fibered spaces for γ = −1, −2 and −3.

Part (3) follows from [Br]. □

Lemma 8. Let K be a non-torus 2-bridge knot. Suppose K(γ) is toroidal. Then (1) K = K_{[b_{1},b_{2}]} for some b_{1}, b_{2}.

(2) If |b_{i}| > 2 for i = 1, 2, there is exactly one such γ. When both b_{i} are even, γ = 0. When b_{1} is odd and b_{2} is even, γ = 2b_{2}.

(3) If K = K_{[2n,2]} and |n| > 1, K(γ) is toroidal if and only if γ = 0 or −4. For K = K_{[2n,−2]}, γ = 0 or 4.

(4) If K = K_{[2,−2]}, γ = 0, 4, or −4. Proof. We refer the reader to [HT] for notations. If K(γ) is toroidal, there is an essential punctured torus T in the knot exterior. By Theorem 1 of [HT], T is carried by some Σ[b_{1}, …, b_{k}], where [b_{1}, …, b_{k}] is an expansion of p/q. By the proof of Theorem 2 of [HT], we have 0 = 2 − 2g = n(2 − k). Therefore k = 2. This proves (1). The rest follows by determining all the possible expansions of type [b_{1}, b_{2}] for p/q. The boundary slopes of the surfaces can be calculated using Proposition 2 of [HT]. By the proof of [Pr, Corollary 2.1], an incompressible punctured torus T in the exterior of a 2-bridge knot will become an essential torus after surgery along the slope of ∂T. □ Lemma 9. If K is a nontorus 2-bridge knot, then K(γ) has a geometric decomposition, i.e. it is either toroidal, or Seifert fibered, or hyperbolic. Proof. A p/q 2-bridge knot can be obtained by taking two arcs of slope p/q on the “pillowcase”, then joining the ends with two trivial arcs. From this picture it is easy to see that K is a strongly invertible knot, i.e, there is an involution φ of S^{3} such that φ(K) = K, and the fixed point set of φ is a circle S intersecting K at two points. φ restricts to an involution of E(K) = S^{3}−IntN(K), which can be extended to an involution φ̂ of the surgered manifold. φ̂ has nonempty fixed point. Since K(γ) is irreducible [HT], the result follows from Thurston’s orbifold geometrization theorem [Th3], which says that if an irreducible, closed 3-manifold admits an orientable preserving periodic map with nonempty fixed point set, then M has a geometric decomposition. See the thesis of Q. Zhou [Zh] for a proof. □ Proof of Theorem 1. Part (5) is well known: If K is a (p, q) torus knot, then E(K) is a Seifert fibered space, so all but one Dehn filling are Seifert fibered. The exceptional one is the one with slope pq, producing a connected sum of two lens spaces. See [Mo].

Part (4) is also well known: If K is the Figure 8 knot, by [Th1] K(γ) is hyperbolic unless γ is an integer between −4 and 4. Since K = K_{[2,−2]} = K_{[−2,2]}, by Corollary 7 and Lemma 8 K(γ) is toroidal for γ = 0, −4, 4, and is Seifert fibered for γ = −3, −2, −1, 1, 2, 3.

If K is not a torus knot or twist knot, then by Corollary 7 and Lemma 8, K(γ) is not Seifert fibered, and it is toroidal if and only if K and γ are as described in (2). So (1) and (2) follows from Lemma 9.

It remains to prove (3). Consider the knot K = K_{[2n,2]}. Using Corollary 7, Lemmas 8 and 9, we need only show that K(γ) is hyperbolic if γ is an integer not between 0 and −4. Since K = K_{[2n,2]} is a hyperbolic knot [HT], and since K(0) in non hyperbolic, the 2π-theorem of Gromov-Thurston (see [BH]) and Lemma 9 imply that K(γ) is hyperbolic unless Δ(0, γ) = |γ| < 23.

Let L = k_{1}∪k_{2} be the 2-bridge link associated to the rational number p/q = [2, 2, −2]. We use L(−, γ_{2}) to denote the manifold obtained by removing k_{1} and performing γ_{2} surgery on k_{2}. As in the proof of Lemma 6, we have K_{[2n,2]}(γ) = L(−1/n, γ).

Neumann and Reid [NR] proved that L(−, γ) is hyperbolic unless γ is an integer between 0 and −4. Since L(∞, γ) is a lens space, hence non hyperbolic, the 2π-theorem and Lemma 9 imply that if γ is not between 0 and −4, then L(−1/n, γ) is hyperbolic unless Δ(∞, −1/n) = |n| < 23. Combining this with the above result, we see that K_{[2n,2]}(γ) is hyperbolic unless |n| < 23 and |γ| < 23, so there are only finitely many surgeries left to check. Finally, we use Jeff Weeks’ SnapPea program [We] to check this finite set of surgeries. It has been verified that if |n| > 1 then K_{[2n,2]}(γ) is hyperbolic unless γ = 0, −1, −2, −3, −4.

The knot K = K_{[2n,−2]} is the mirror image of K_{[−2n,2]}, so K_{[2n,−2]}(γ) is homeomorphic to K_{[−2n,2]}(γ). Therefore, if |n| > 1 then K(γ) is hyperbolic unless γ = 0, 1, 2, 3, 4. □

Acknowledgment. We are grateful to Alan Reid for some very useful comments to an earlier version of this paper.

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Department of Mathematics, University of Texas at Austin, Austin, TX 78712 E-mail address: britten@math.utexas.edu

Department of Mathematics, University of Iowa, Iowa City, IA 52242 E-mail address: wu@math.uiowa.edu