TOROIDAL AND ANNULAR DEHN FILLINGS

Cameron McA. Gordon^{1,2} and Ying-Qing Wu^{2}

Abstract. Suppose M is a hyperbolic 3-manifold which admits two Dehn fillings M(r_{1}) and M(r_{2}) such that M(r_{1}) contains an essential torus and M(r_{2}) contains an essential annulus. It is known that Δ = Δ(r_{1}, r_{2}) ≤ 5. We will show that if Δ = 5 then M is the Whitehead sister link exterior, and if Δ = 4 then M is the exterior of either the Whitehead link or the 2-bridge link associated to the rational number 3/10. There are infinitely many examples with Δ = 3.

§1. Introduction

Let M be a (compact, connected, orientable) 3-manifold with a torus boundary component T_{0}. If r is a slope (the isotopy class of an essential simple loop) on T_{0}, then as usual we denote by M(r) the 3-manifold obtained by r-Dehn filling on M, that is, attaching a solid torus J to M along T_{0} in such a way that r bounds a disk in J.

We shall say that a 3-manifold M is hyperbolic if M with its boundary tori removed has a complete hyperbolic structure of finite volume with totally geodesic boundary. If M has non-empty boundary, then Thurston’s Geometrization Theorem for Haken manifolds [Th1] [Th2] asserts that M is hyperbolic if and only if it contains no essential surface of non-negative euler characteristic, i.e., sphere, disk, torus or annulus.

There has been a considerable amount of investigation of the question of when a Dehn filling on a hyperbolic 3-manifold fails to be hyperbolic; if ∂M −T_{0} is non-empty then, as noted above, this is equivalent to the existence of an essential surface of nonnegative euler characteristic in the filled manifold. So suppose that M is hyperbolic, and that M(r_{1}) and M(r_{2}) contain essential surfaces F̂_{1} and F̂_{2} respectively, each of which is a sphere, disk, torus or annulus. Then for each of the ten possible pairs of surfaces, one can find upper bounds on Δ(r_{1}, r_{2}), the minimal geometric intersection number of r_{1} and r_{2}. In all cases, the best possible bounds are now known; see [GW] for a fuller discussion. It is also of interest to examine the examples (M; r_{1}, r_{2}) that realize values of Δ(r_{1}, r_{2}) at or near the maximal value; it seems to be the case that these are usually quite special.

Typeset by AMS-TEX

1991 Mathematics Subject Classification. Primary 57N10.

1 Partially supported by NSF grant #DMS 9626550.

2 Research at MSRI supported in part by NSF grant #DMS 9022140.

In the present paper we consider the case where F̂_{1} is an annulus and F̂_{2} is a torus. Here it is known that Δ(r_{1}, r_{2}) ≤ 5 [Go, Theorem 1.3]. To state our main result, let L_{1}, L_{2}, L_{3} be the links shown in Figure 7.1 (a), (b) and (c), respectively. These are the Whitehead link, the 2-bridge link associated to the rational number 3/10, and the Whitehead sister link. Let M_{i} be the exterior of L_{i}, i = 1, 2, 3. Also, let J_{α} denote the attached solid torus in M(r_{α}), α = 1, 2.

Theorem 1.1. Suppose M is a hyperbolic 3-manifold, such that M(r_{1}) contains an essential annulus, M(r_{2}) contains an essential torus, and Δ(r_{1}, r_{2}) ≥ 4. Then either

(1) Δ(r_{1}, r_{2}) = 4 and M is homeomorphic to M_{1} or M_{2}; or

(2) Δ(r_{1}, r_{2}) = 5 and M is homeomorphic to M_{3}.

Moreover, M_{i} is hyperbolic, i = 1, 2, 3, and admits two Dehn fillings M_{i}(r_{1}) and M(r_{2}), such that

(i) Δ(r_{1}, r_{2}) = 4 for i = 1, 2, and Δ(r_{1}, r_{2}) = 5 for i = 3;

(ii) M_{1}(r_{1}) contains an essential annulus and an essential torus, each intersecting J_{1} once. All the other M_{i}(r_{α}) contain an essential annulus and essential torus, each intersecting J_{α} twice.

This follows immediately from Theorems 6.10(1) and 7.5.

In particular, the upper bound of 5 on Δ(r_{1}, r_{2}) is realized by the exterior of the Whitehead sister link. The existence of annular and toroidal fillings on this manifold with Δ(r_{1}, r_{2}) = 5 has also been proved by Miyazaki and Motegi [MM].

We also show the following.

Theorem 7.8. There are infinitely many hyperbolic manifolds M_{s}, which admit two Dehn fillings M_{s}(r_{1}), M_{s}(r_{2}), each containing an essential annulus and an essential torus, with Δ(r_{1}, r_{2}) = 3.

We now sketch the proof and describe the organization of the paper. Throughout, α, β will denote the numbers 1 or 2, with the convention that if they both appear, then {α, β} = {1, 2}.

We may assume that the surface F̂_{α} meets the solid torus J_{α} in n_{α} meridional disks, and that n_{α} is minimal over all choices of F̂_{α}. The arcs of intersection of the corresponding punctured annulus and punctured torus in M define labeled graphs Γ_{α} ⊂ F̂_{α}, with n_{α} vertices, α = 1, 2. We assume that Δ(r_{1}, r_{2}) = 4 or 5. The proof consists of a detailed analysis of such graphs Γ_{α}, and the elimination of all but four possible pairs, which are then shown to correspond to the examples listed in Theorem 1.1.

In Section 2 we give some definitions, and establish some properties of the intersection graphs Γ_{α}, which will be used throughout the paper.

Section 3 deals with the case that one of the graphs has a single vertex. It is shown that if n_{α} = 1 then Δ(r_{1}, r_{2}) = 4, n_{β} = 2, and M is the exterior of the Whitehead link.

Sections 4 and 5 deal with the case that one of the graphs has two vertices. In Section 4 it is shown that if n_{2} = 2 then the graph Γ_{1} is 2-separable, i.e., there is an essential sub-annulus A′ of F̂_{1}, which contains two vertices of Γ_{1} and has boundary disjoint from Γ_{1}. Note that this is also (trivially) true if n_{1} = 2. In Section 5 it is shown that if Γ_{1} is 2-separable then M must be one of the three manifolds M_{i} in Theorem 1.1.

Section 6 deals with the generic case. It is shown that if both graphs have at least three vertices, then Δ(r_{1}, r_{2}) ≤ 3. This completes the proof of the first part of Theorem 1.1. We then prove in Section 7 that the manifolds M_{i} in Theorem 1.1 do admit two annular and toroidal Dehn fillings with Δ(r_{1}, r_{2}) = 4 or 5, and that there are infinitely many manifolds which admit two annular and toroidal Dehn fillings with Δ(r_{1}, r_{2}) = 3.

§2. Preliminary Lemmas

Throughout this paper, we will fix a hyperbolic 3-manifold M, with a torus T_{0} as a boundary component. Let r_{1}, r_{2} be slopes on T_{0}. Denote by Δ = Δ(r_{1}, r_{2}) the minimal geometric intersection number between r_{1} and r_{2}. Recall that α, β denote the numbers 1 or 2, with the convention that if they both appear, then {α, β} = {1, 2}.

Assumption. We will always assume that F̂_{1} is an essential annulus in M(r_{1}), and F̂_{2} is an essential torus in M(r_{2}). Moreover, we assume that M(r_{α}) contains no essential sphere, i.e. it is irreducible.

The last assumption is necessary in our proof, as some of the properties stated in Lemmas 2.3–2.5 are not true without it. However, it is not necessary in the hypothesis of Theorem 6.10, because if M(r_{1}) is reducible, then by [Wu1] or [Oh] we have Δ ≤ 3. Similarly, if M(r_{2}) is reducible it follows from [Wu3] that Δ ≤ 2.

Denote by F_{α} the punctured surface F̂_{α} ∩ M_{α}. Let n_{α}, α = 1, 2, be the number of boundary components of F_{α} on T_{0}. Choose F̂_{α} in M(r_{α}) so that n_{α} is minimal among all essential surfaces homeomorphic to F̂_{α}. Minimizing the number of components of F_{1}∩F_{2} by an isotopy, we may assume that F_{1}∩F_{2} consists of arcs and circles which are essential on both F_{α}. Denote by J_{α} the attached solid torus in M(r_{α}). Let u_{1}, …, u_{n_{1}} be the disks that are the components of F̂_{1} ∩ J_{1}, labeled successively when traveling along J_{1}. Similarly let v_{1}, …, v_{n_{2}} be the disks in F̂_{2} ∩ J_{2}. Let Γ_{α} be the graph on F̂_{α} with the u_{i}’s or v_{j}’s, respectively, as (fat) vertices, and the arc components of F_{1} ∩ F_{2} as edges. The minimality of the number of components in F_{1} ∩ F_{2} and the minimality of n_{α} imply that Γ_{α} has no trivial loops, and that each disk face of Γ_{α} in F̂_{α} has interior disjoint from F_{β}.

If e is an edge of Γ_{1} with an endpoint x on a fat vertex u_{i}, then x is labeled j if x is in u_{i} ∩ v_{j}. In this case e is called a j-edge at u_{i}, and in Γ_{2} it is an i-edge at v_{j}. The labels in Γ_{1} are considered as integers mod n_{2}. In particular, n_{2} + 1 = 1. When going around ∂u_{i}, the labels of the endpoints of edges appear as 1, 2, …, n_{2} repeated Δ times. Label the endpoints of edges in Γ_{2} similarly.

Each vertex of Γ_{α} is given a sign according to whether J_{α} passes F̂_{α} from the positive side or negative side at this vertex. Two vertices of Γ_{α} are parallel if they have the same sign, otherwise they are antiparallel. Note that if F̂_{α} is a separating surface, then n_{α} is even, and v_{i}, v_{j} are parallel if and only if i, j have the same parity.

We use val(v, G) to denote the valency of a vertex v in a graph G. If G is clear from the context, we denote it by val(v).

Definition 2.1. An edge of Γ_{α} is a positive edge if it connects parallel vertices. Otherwise it is a negative edge.

When considering each family of parallel edges of Γ_{α} as a single edge ê, we get the reduced graph Γ̂_{α} on F̂_{α}. It has the same vertices as Γ_{α}.

We use Γ_{α}^{+} (resp. Γ_{α}^{−}) to denote the subgraph of Γ_{α} whose vertices are the vertices of Γ_{α} and whose edges are the positive (resp. negative) edges of Γ_{α}. Similarly for Γ̂_{α}^{+} and Γ̂_{α}^{−}.

A cycle in Γ_{α} consisting of positive edges is a Scharlemann cycle if it bounds a disk with interior disjoint from the graph, and all the edges in the cycle have the same pair of labels {i, i+1} at their two endpoints, called the label pair of the Scharlemann cycle. A pair of edges {e_{1}, e_{2}} is an extended Scharlemann cycle if there is a Scharlemann cycle {e′_{1}, e′_{2}} such that e_{i} is parallel and adjacent to e′_{i}. There is an obvious extension of this definition to extended Scharlemann cycles of arbitrary length, but this is enough for our purposes.

A subgraph G of a graph Γ on a surface F is essential if it is not contained in a disk in F. The following lemma contains some common properties of the graphs Γ_{α}.

Lemma 2.2. (Properties of Γ_{α})

(1) (The Parity Rule) An edge e is a positive edge in Γ_{1} if and only if it is a negative edge in Γ_{2}.

(2) A pair of edges cannot be parallel on both Γ_{1} and Γ_{2}.

(3) If Γ_{α} has a set of n_{β} parallel negative edges, then on Γ_{β} they form mutually disjoint essential cycles of equal length.

(4) If Γ_{α} has a Scharlemann cycle, then F̂_{β} is separating, and n_{β} is even.

(5) If Γ_{α} has a Scharlemann cycle {e_{1}, …, e_{k}}, then on Γ_{β} the subgraph consisting of e_{1}, …, e_{k} and their end vertices, is essential.

(6) If n_{β} > 2, then Γ_{α} contains no extended Scharlemann cycle.

Proof. (1) can be found in [CGLS, Page 279].

(2) is [Go, Lemma 2.1]. It was shown that if a pair of edges is parallel on both graphs, then the manifold M contains a (1, 2) cable space, which is impossible because M is assumed hyperbolic. See also [GLi].

(3) is [Go, Lemma 2.3] and [GLi, Proposition 1.3]. A set of n_{β} parallel edges with ends v, v′ determine a permutation φ on the set of labels {1, …, n_{β}} such that an edge with label i at v has label φ(i) at v′. The edges a collection of mutually disjoint cycles of length L in Γ_{β}, where L is the length of the orbits of φ. It was shown that if some cycle is not essential then M is cabled, hence non-hyperbolic.

(4) and (5) follow from the proof of [CGLS, Lemma 2.5.2]. It was shown that using the disk bounded by the Scharlemann cycle one can find another surface F̂′_{β} in M(r_{β}) which has fewer intersections with J_{β}, and is cobordant to F̂_{β}, so if F̂_{β} were nonseparating then F̂′_{β} would still be essential, which would contradict the minimality of n_{β}. If the subgraph G consisting of e_{1}, …, e_{k} and their end vertices is inessential then one can show that M(r_{β}) would be reducible, which contradicts our assumption.

(6) If F̂_{β} is an annulus or torus, this is [Wu3, Lemma 5.4(3)]. If F̂_{β} is a torus, see [BZ, Lemma 2.9] or [GLu2, Theorem 3.2]. □

Let ê be a collection of at least n_{α} parallel negative edges on Γ_{β}, connecting v_{1} to v_{2}. Then ê defines a permutation φ : {1, …, n_{α}} → {1, …, n_{α}}, such that an edge e in ê has label k at v_{1} if and only if it has label φ(k) at v_{2}. Call φ the permutation associated to ê. If we interchange v_{1} and v_{2}, then the permutation is φ^{−1}; hence it is well defined up to inversion.

Lemma 2.3. (Properties of Γ_{1}) (1) If a family of parallel edges in Γ_{1} contains more than n_{2} edges, and if n_{2} > 2, then all the vertices of Γ_{2} are parallel, and the permutation associated to this family is transitive.

(2) If Γ_{1} contains two Scharlemann cycles with disjoint label pairs {i, i + 1} and {j, j + 1}, then i ≡ j mod 2.

(3) Suppose n_{2} > 2. Then a family of parallel positive edges in Γ_{1} contains at most n_{2}/2 + 2 edges, and if it does contain n_{2}/2 + 2 edges, then n_{2} ≡ 0 mod 4.

Proof. (1) If the parallel edges are positive, then they contain an extended Scharlemann cycle, which is impossible by Lemma 2.2(6). So assume that the edges are negative. It was shown in [Go, Lemma 4.2] that if the permutation is not transitive, then M contains a cable space, which is impossible in our case because M is hyperbolic. Hence the permutation is transitive. By the parity rule these edges connect parallel vertices in Γ_{2}, hence all the vertices of Γ_{2} are parallel.

(2) and (3) are the same as [Wu1, Lemmas 1.7, 1.4, and 1.8]. □

Lemma 2.4. Let ê be a family of parallel positive edges in Γ_{α}. If the label i appears twice among the labels of ê, then there is a pair of edges among this family which form a Scharlemann cycle with i as one of its labels. In particular, if ê has more than n_{β}/2 edges, then it contains a Scharlemann cycle.

Proof. Since the edges are positive, by the parity rule i cannot appear at both endpoints of a single edge in this family. Let e_{1}, e_{2}, …, e_{k} be consecutive edges of ê such that e_{1} and e_{k} have i as a label. If k ≥ 4, one can see that these edges contain an extended Scharlemann cycle in the middle, so by Lemma 2.2(6) we have n_{β} = 2, in which case e_{1}, e_{2} is a Scharlemann cycle with i as a label, as required. If k = 2 then e_{1}, e_{2} is a Scharlemann cycle and we are done. If k = 3 the parity rule implies that the endpoints of e_{1} and e_{3} on the same vertex have the same label, so again n_{β} = 2 and e_{1}, e_{2} is a Scharlemann cycle.

If ê has more than n_{β}/2 edges, then it has more than n_{β} endpoints, so some label must appear twice. □

Lemma 2.5. (Properties of Γ_{2}) (1) Any two Scharlemann cycles on Γ_{2} have the same label pair.

(2) Any family of parallel positive edges ê in Γ_{2} contains at most n_{1}/2 + 1 edges. Moreover, if it does contain n_{1}/2 + 1 edges, then two of them form a Scharlemann cycle.

(3) Any family of parallel edges ê in Γ_{2} contains at most n_{1} edges.

(4) No three i-edges of Γ_{2} are parallel for any i.

Proof. (1) This is [Wu3, Lemma 5.4(2)].

(2) If ê contains n_{1}/2 + 1 edges then they have n_{1} + 2 endpoints, so some label i appear twice. By Lemma 2.4 i is the label of a Scharlemann cycle in this family. This proves the second sentence of (2).

Now suppose ê contains n_{1}/2 + 2 positive edges. If n_{1} = 2 then there are only two nonparallel edges on the annulus F̂_{1} connecting the two vertices, so a family of n_{1}/2 + 2 = 3 edges in ê would contain two which are parallel on both graphs, contradicting Lemma 2.2(2). If n_{1} > 2, then 4 labels appear twice among the endpoints of the edges in ê, so by Lemma 2.4 they are all labels of Scharlemann cycles in ê, contradicting (1).

(3) If the edges in ê are positive, this follows from (2). So assume that ê consists of n_{1} + 1 parallel negative edges e_{1}, e_{2}, …, e_{n_{1}+1}. By Lemma 2.2(3) the first n_{1} edges form several essential cycles in Γ_{1}. Let e_{1} ∪ e_{j_{2}} … ∪ e_{j_{k}} be the one containing e_{1}. The edges e_{1} and e_{n_{1}+1} have the same label pair at their endpoints in Γ_{2}, hence have the same end vertices on Γ_{1}. By Lemma 2.2(2) they cannot be parallel on Γ_{1}, so after shrinking each vertex to a point, the path e_{n_{1}+1} is homotopic rel ∂ to the path e_{j_{2}} ∪ … ∪ e_{j_{k}}; in other words, the cycle e_{j_{2}} ∪ … ∪ e_{j_{k}} ∪ e_{n_{1}+1} is an inessential cycle in Γ_{1}, which contradicts Lemma 2.2(3) because it is a cycle corresponding to the n_{1} parallel edges e_{2}, …, e_{n_{1}+1} in Γ_{2}.

(4) If there are three i-edges which are parallel, then two of them would have the same label at one of their end vertices, so there would be at least n_{1} +1 parallel edges, contradicting (3). □

Let u be a vertex of Γ_{α}. Let P, Q be two edge-endpoints on ∂u. Let P_{0} = P, P_{1}, …, P_{k−1}, P_{k} = Q be the edge-endpoints encountered when traveling along ∂u in the direction induced by the orientation of u. Then the distance from P to Q is defined as ρ_{u}(P, Q) = k. To emphasize that the distance is measured on Γ_{α}, we may also use τ_{α}(P, Q) to denote ρ_{u}(P, Q). Notice that if the valency of u is m, then ρ_{u}(Q, P) = m − ρ_{u}(P, Q). The following lemma can be found in [Go].

Lemma 2.6. ([Go, Lemma 2.4]) Let a, b be components of ∂F_{α} ∩ T_{0}, and x, y components of ∂F_{β} ∩ T_{0}.

(i) Suppose P, Q ∈ a ∩ x and R, S ∈ b ∩ y. If τ_{α}(P, Q) = τ_{α}(R, S) then τ_{β}(P, Q) = τ_{β}(R, S).

(ii) Suppose that P ∈ a ∩ x, Q ∈ a ∩ y, R ∈ b ∩ x, and S ∈ b ∩ y. If τ_{α}(P, Q) = τ_{α}(R, S), then τ_{β}(P, R) = τ_{β}(Q, S). □

If e_{i} is an edge with exactly one endpoint on u, i = 1, 2, define ρ_{u}(e_{1}, e_{2}) = ρ_{u}(P_{1}, P_{2}), where P_{i} = e_{i} ∩ u.

A pair of edges e_{1}, e_{2} connecting two vertices u, v in Γ is an equidistant pair if ρ_{u}(e_{1}, e_{2}) = ρ_{v}(e_{2}, e_{1}). If u = v, then define e_{1}, e_{2} to be an equidistant pair if ρ_{u}(P_{1}, P_{2}) = ρ_{v}(Q_{2}, Q_{1}), where P_{i}, Q_{i} are the endpoints of e_{i}. One can check that this is independent of the choices of P_{i} and Q_{i}.

The following lemma follows immediately from the definition, and is convenient for applications. In particular, it applies to parallel loops based at a vertex.

Lemma 2.7. If e_{1}, e_{2} is a pair of parallel edges connecting a pair of parallel vertices in Γ, then e_{1}, e_{2} is an equidistant pair in Γ. □

Lemma 2.8. Let e_{1}, e_{2} be a pair of edges with ∂e_{1} = ∂e_{2} in both Γ_{1} and Γ_{2}. Then e_{1}, e_{2} is an equidistant pair in Γ_{1} if and only if it is an equidistant pair in Γ_{2}.

Proof. Suppose e_{i} joins vertices a and b in Γ_{1}, and vertices x and y in Γ_{2}, i = 1, 2. Denote the endpoints of e_{1}, e_{2} in Γ_{1} by P, S and Q, R, respectively, as in Figure 2.1(a) or (c). There are two cases, depending on whether or not P and Q have the same label. We give a proof in the case that P and Q have the same label, say x. The proof for the other case is similar, using Lemma 2.6(2) instead of 2.6(1).

In this case we have P, Q ∈ a ∩ x and R, S ∈ b ∩ y, so e_{1}, e_{2} being equidistant on Γ_{1} implies that

τ_{α}(P, Q) = ρ_{a}(e_{1}, e_{2}) = ρ_{b}(e_{2}, e_{1}) = τ_{α}(R, S).

By Lemma 2.6(1), we have τ_{β}(P, Q) = τ_{β}(R, S), which implies that ρ_{x}(e_{1}, e_{2}) = ρ_{y}(e_{2}, e_{1}), so e_{1}, e_{2} is also equidistant on Γ_{2}. Similarly, e_{1}, e_{2} being equidistant on Γ_{2} also implies that they are equidistant on Γ_{1}. □

Figure 2.1

Definition 2.9. Given two slopes r_{1}, r_{2} on T_{0}, choose a meridian-longitude pair m, l on the torus T_{0} so that r_{1} = m, and the slope r_{2} is represented by dm + Δl for some d with 1 ≤ d ≤ Δ/2. Then d = d(r_{1}, r_{2}) is called the jumping number of r_{1}, r_{2}. Notice that if Δ = 4, then d = 1, and if Δ = 5, then d = 1 or 2.

Lemma 2.10. Let P_{1}, …, P_{Δ} be the points of ∂u_{i} ∩ ∂v_{j}, labeled so that they appear successively on u_{i}. Let d = d(r_{1}, r_{2}) be the jumping number of r_{1}, r_{2}. Then on v_{j} these points appear in the order of P_{d}, P_{2d}, …, P_{Δd}, in some direction. In particular, if Δ = 4, then P_{1}, P_{2}, P_{3}, P_{4} also appear successively on v_{j}.

Proof. This follows by examining the order of the intersection points on the two curves r_{1} and r_{2}. If Δ = 4, we must have d = 1, hence the result follows. □

Lemma 2.11. ([HM, Proposition 5.1]) If C is a cycle in Γ_{α} consisting of positive i-edges, bounding a disk D which contains no vertices in its interior, then D contains a Scharlemann cycle. □

Lemma 2.12. (1) Suppose G is a reduced graph on an annulus A with V vertices. Then G has at most 3V − 2 edges, and has a vertex of valency at most 5.

(2) A reduced graph G on a torus has at most 3V edges.

Proof. (1) By embedding the annulus A in a sphere S, we can consider G as a graph on S. Let E, F be the number of edges and disk faces of G. Then V − E + F ≥ 2 (the inequality may be strict if there are some non-disk faces.) All but two faces of G have at least three edges, and each of the two exceptional ones has at least one edge. Hence we have

2 + 3(F − 2) ≤ 2E.

Solving those two inequalities, we get E ≤ 3V − 2.

If each vertex of G has valency at least 6, then 6V ≤ 2E, which is impossible because E ≤ 3V − 2.

(2) The calculation for G on torus is similar but simpler. We have V − E + F ≥ 0, and 3F ≤ 2E. The result follows by solving these inequalities. □

§3. Case 1: One of the graphs has a single vertex

From now on we assume that F̂_{1} is an essential annulus A in M(r_{1}), and F̂_{2} is an essential torus T in M(r_{2}). Recall that M(r_{α}) is assumed to be irreducible. By [Go, Theorem 1.3] we may, and will always, assume that Δ ≤ 5. In this section we will consider the case where one of the graphs has only one vertex.

The link in Figure 7.1(a) is called the Whitehead link.

Lemma 3.1. Suppose n_{1} = 1 and Δ ≥ 4. Then n_{2} = 2, Δ = 4, and the graphs Γ_{i} are as shown in Figure 3.1. The manifold M is the Whitehead link exterior.

Proof. In this case, there is only one edge in the reduced graph Γ̂_{1}, so all edges of Γ_{1} are parallel positive edges. By the parity rule, all edges of Γ_{2} are negative. In particular, n_{2} ≥ 2. Moreover, the Δn_{2}/2 parallel edges of Γ_{1} contains an extended Scharlemann cycle, so by Lemma 2.2(6) we must have n_{2} = 2. In this case there are exactly Δ edges in each graph. Since they are parallel on Γ_{1}, by Lemma 2.2(2) no two of them can be parallel on Γ_{2}. There are at most 4 nonparallel edges on the torus connecting two vertices. Hence Δ ≤ 4, and if Δ = 4, then up to homeomorphism of the torus T the graph is that shown in Figure 3.1. We will postpone until Lemma 7.2 the proof that if the intersection graphs are as shown in Figure 3.1 then the manifold M is the exterior of the Whitehead link. This will complete the proof of Lemma 3.1. □

Figure 3.1

Lemma 3.2. Suppose n_{2} = 1 and Δ ≥ 4. Then n_{1} = 2, Δ = 4, and the graphs Γ_{i} are as shown in Figure 3.3(c) and (d). The manifold M is the Whitehead link exterior.

Proof. The reduced graph Γ̂_{2} on the torus is a subgraph of that shown in Figure 3.2(a), which has only three edges. The parity rule implies that n_{1} > 1. The graph Γ_{2} contains Δn_{1}/2 ≥ 2n_{1} edges, and so has a family of more than n_{1}/2 parallel positive edges. By Lemmas 2.5(2) and 2.2(4) n_{1} must be even.

Figure 3.2

We want to show that n_{1} = 2. If Δ = 5 and n_{1} ≥ 4, then Γ_{2} contains a family of Δn_{1}/6 = \frac{1}{2}n_{1} + \frac{1}{3}n_{1} > \frac{1}{2}n_{1} + 1 parallel positive edges, which contradicts Lemma 2.5(2). Assume Δ = 4. Then the labels at a family of parallel arcs are as shown in Figure 3.2(b). In particular, the number of parallel edges is even by the parity rule. Thus if n_{1} ≥ 4, then there are at least 4 edges in one of the families. From the labeling in Figure 3.2 we see that this family contains an extended Scharlemann cycle, contradicting Lemma 2.2(6).

Now we assume that n_{1} = 2. In this case there are only two edges in Γ̂_{1}. Since no two edges can be parallel on both graphs, there are at most two edges in each family of parallel edges in Γ_{2}. Thus when Δ = 5, there is only one possible configuration, shown in Figure 3.3(a), and when Δ = 4, there are two possibilities, shown in Figure 3.3(b) – (c). The first two cases can be ruled out because there is an edge with the same label on both endpoints, which contradicts the parity rule. It follows that Figure 3.3(c) is the only possibility, which also uniquely determines the graph Γ_{1}, shown in Figure 3.3(d).

Figure 3.3

One can build the manifold M as in Lemma 7.2 to show that M is the exterior of the Whitehead link. However, there is a shortcut. Notice that the union of A × I with the Dehn filling solid torus V_{1} is a genus 3 handlebody. After attaching the 2-handles coming from the faces of Γ_{2} and then capping off the spherical boundary components, we get a manifold Y with boundary a torus. Since the original manifold M is assumed to be hyperbolic, this implies that Y must be homeomorphic to M(γ_{1}); in other words, there cannot be anything else attached to Y. Therefore M is uniquely determined by the intersection graphs. Combining this with the above result, we conclude that there is at most one manifold M such that M(γ_{1}) contains a torus intersecting the Dehn filling solid torus only once, M(γ_{2}) is annular, and Δ(γ_{1}, γ_{2}) > 3. On the other hand, the Whitehead link exterior does satisfy those conditions, because the −4 surgery is annular, and the Seifert surface of one component of the link extends, after 0 surgery, to a torus which intersects the Dehn filling solid torus just once. Therefore, this must be the manifold. □

Definition 3.3. The graph Γ_{1} is k-separable if there is an annulus A′ on A, such that (i) A′ contains exactly k vertices of Γ_{1}, (ii) ∂A′ consists of two essential curves in the interior of A, and (iii) ∂A′ is disjoint from Γ_{1}. Denote by Γ′_{1} the graph Γ_{1} ∩ A′ on A′, and by Γ′_{2} the subgraph of Γ_{2} consisting of edges which, when considered as edges in Γ_{1}, lie in A′. Notice that by definition Γ_{1} is always n_{1}-separable.

Lemma 3.4. Suppose Δ ≥ 4, and Γ_{1} is 1-separable. Then n_{1} = 1.

Proof. Let Γ′_{i} be the graphs on A′ and T defined in Definition 3.3. As in the proof of Lemma 3.1, the graphs Γ′_{i} must be the ones shown in Figure 3.1. Using the construction in the proof of Lemma 7.2, we see that the manifold W = (A′ × I) ∪ J_{1} ∪ (D × I) is the Whitehead link exterior, where D is a disk face of Γ′_{2}. Let T_{1} = ∂M − T_{0}. Then T_{1} contains the boundary of A, so it is nonempty. Let T′ be the component of ∂W in the interior of M. Note that T′ separates T_{0} from T_{1}, so T′ is incompressible in M because otherwise M would be reducible. Since M is hyperbolic, T′ must be parallel to a boundary component of M. Since W is not a product, we see that T′ must be parallel to T_{1}. In particular, T_{1} is a torus. Now the annulus A′ can be extended to an essential annulus in M(r_{1}) intersecting J_{1} just once. By the minimality of n_{1}, we must have n_{1} = 1. □

Proposition 3.5. Suppose Δ ≥ 4. If n_{α} = 1, then Δ = 4, n_{β} = 2, and the manifold M is the exterior of the Whitehead link shown in Figure 7.1.

Proof. This follows immediately from Lemmas 3.1 and 3.2. □

§4. The case that n_{2} = 2

So far we have seen that if Δ = 4 and one of the n_{α} = 1 then the manifold M is the exterior of the Whitehead link. The goal of this section is to show that if n_{2} = 2 then the graph Γ_{1} is 2-separable; see Proposition 4.6 below. In the next section it will be shown that if Γ_{1} is 2-separable then M must be the exterior of one of the links in Figure 7.1.

Lemma 4.1. Suppose that Δ ≥ 4. If all vertices of Γ_{2} are parallel, then n_{1} = n_{2} = 2, and hence Γ_{1} is 2-separable.

Proof. By Lemma 2.12 the reduced graph Γ̂_{2} has at most 3n_{1} edges. For each i there are Δn_{1} i-edges in Γ_{2}, so two of them are parallel. By Lemma 2.4 this implies that i is the index of some Scharlemann cycle in Γ_{2}. This is true for any index from 1 to n_{1}, but by Lemma 2.5(1) there are at most two labels that can appear as the labels of Scharlemann cycles in Γ_{2}, hence we have n_{1} = 2. (We cannot have n_{1} = 1 because of the parity rule.)

Now on Γ_{1} there are only two families of parallel edges. Consider the function

φ : {1, 2, … n_{2}} → {1, 2, … n_{2}}

defined so that if e is an edge in Γ_{1} with label i on u_{1} then it has label φ(i) on the vertex u_{2}. Clearly this is well defined. Note that some family contains at least 2n_{2} parallel edges, so by Lemma 2.3(1) either n_{2} = 2 and we are done, or the function φ is transitive, i.e. it has only one orbit. In particular, if n_{2} > 2 and φ(k) = l, then φ(l) ≠ k.

On Γ_{2} there are Δn_{1} > 3n_{1} edges, so there is a pair of adjacent parallel edges e_{1}, e_{2} connecting, say, the vertex v_{k} to v_{l}. Note that e_{1} and e_{2} have different labels at their endpoints on v_{k}, hence in Γ_{1} the edge e_{1} has label k, say, at the vertex u_{1}, and label l at the vertex u_{2}, while e_{2} has label l at u_{1} and label k at u_{2}. Therefore φ(k) = l and φ(l) = k, a contradiction. □

Lemma 4.2. If n_{2} = 2 then, up to homeomorphism of T, the reduced graph Γ̂_{2} is a subgraph of the graph shown in Figure 4.1.

Proof. If Γ̂_{2} contains two loops based at v_{1}, then they cut the torus into a disk, hence there would be no loops based at v_{2}, so any edge with an endpoint on v_{2} must have another endpoint on v_{1}; but since v_{1} and v_{2} have the same valency in Γ_{2}, this is impossible. Therefore Γ̂_{2} has at most one loop based at each v_{i}. It is now easy to see that Γ̂_{2} is isomorphic to a subgraph of that shown in Figure 4.1(a). □

Figure 4.1

We will use Figure 4.1(b) to indicate the reduced graph shown in Figure 4.1(a). The graph Γ_{2} is obtained by replacing the edges in Figure 4.1(b) by families of parallel edges. Denote by p_{i} the number of parallel edges in Γ_{2} represented by ê_{i} in the reduced graph Γ̂_{2}.

Lemma 4.3. If Δ ≥ 4 and n_{2} = 2, then the two vertices v_{1}, v_{2} of Γ_{2} are antiparallel.

Proof. Assume that v_{1}, v_{2} are parallel. Then by Lemma 4.2 we have n_{1} = 2. Since all edges of Γ̂_{2} are positive, by the parity rule there is no loop on Γ̂_{1}, so Γ̂_{1} consists of two families of parallel edges connecting u_{1} to u_{2}.

Notice that in this case either all edges of Γ̂_{1} have the same label on both endpoints, or they all have distinct labels on their two endpoints. In the first case Γ̂_{2} consists of two families of parallel loops, so some pair of edges would be parallel on both graphs, which is impossible by Lemma 2.2(2). In the second case there are no loops in Γ̂_{2}, so there are only 4 edges in Γ̂_{2}. It follows again from Lemma 2.2(2) that Δ = 4, and Γ̂_{2} consists of 4 pairs of parallel edges. See Figure 4.2.

Consider a pair of parallel edges e_{1}, e_{2} in Γ_{2}. Since e_{1}, e_{2} are a pair of parallel edges connecting parallel vertices, by Lemma 2.7 they form an equidistant pair in Γ_{2}, so by Lemma 2.8 they also form an equidistant pair in Γ_{1}, that is ρ_{u_{1}}(e_{1}, e_{2}) = ρ_{u_{2}}(e_{2}, e_{1}). Also, from Figure 4.2(b) we can see that ρ_{u_{2}}(e_{2}, e_{1}) = 8 − ρ_{u_{1}}(e_{1}, e_{2}), so we must have ρ_{u_{1}}(e_{1}, e_{2}) = ρ_{u_{2}}(e_{2}, e_{1}) = 4. On the other hand, observe that the endpoints of e_{1}, e_{2} on u_{1} have different labels, so ρ_{u_{1}}(e_{1}, e_{2}) must be an odd number, which is a contradiction. □

Figure 4.2

Lemma 4.4. Suppose Δ ≥ 4 and n_{2} = 2. If p_{2} + p_{3} ≥ p_{4} + p_{5}, then p_{2} = p_{3} = n_{1}.

Proof. By Lemma 2.5(3), p_{i} ≤ n_{1}, 1 ≤ i ≤ 6. Since v_{1} and v_{2} are antiparallel, we shall write v_{1} = v_{+}, and v_{2} = v_{−}. After relabeling if necessary, we may assume that the labels of the endpoints of the edges in ê_{1} are 1, 2, …, k at the bottom of v_{+}. If p_{2} + p_{3} = 2n_{1} − r, then since

Δn_{1} = 2k + (p_{2} + p_{3}) + (p_{4} + p_{5}) ≥ 2k + 2(2n − r),

we have r ≤ k, so the labels of the left r endpoints of edges in ê_{1} are 1, 2, …, r at the top of v_{+}; see Figure 4.3(a). If r > 2, then the left r edges of ê_{1} would contain an extended Scharlemann cycle, which would contradict Lemma 2.2(6) because n_{1} ≥ k ≥ r > 2. Therefore we have p_{2} + p_{3} ≥ 2n_{1} − 2. Also, the left edge of ê_{1} cannot have the same label at both endpoints because of the parity rule. Hence either p_{2} + p_{3} = 2n_{1}, which implies p_{2} = p_{3} = n_{1}, or p_{2} + p_{3} = 2n_{1} − 2. We need to rule out the second possibility.

Assume p_{2} + p_{3} = 2n_{1} − 2. Since it is assumed that p_{2} + p_{3} ≥ p_{4} + p_{5}, we have

p_{1} = p_{6} = \frac{1}{2}(Δn_{1} − (p_{2} + p_{3}) − (p_{4} + p_{5})) ≥ 2.

Examining the labels, we see that the two leftmost edges in ê_{1} form a Scharlemann cycle. Similarly for the two leftmost edges in ê_{6}. By Lemma 2.5(1), these two Scharlemann cycles must have the same label pairs, say {1, 2}; see Figure 4.3(b). The existence of Scharlemann cycles implies that n_{1} is even, and all vertices in Γ_{1} of the same parity are parallel.

By Lemma 2.5(2) ê_{1} contains at most n_{1}/2 + 1 edges, so p_{4} + p_{5} = Δn_{1} − (p_{2} + p_{3}) − 2p_{1} ≥ Δn_{1} − 3n_{1}. Since p_{4} + p_{5} ≤ p_{2} + p_{3}, we must have Δ = 4.

First assume that n_{1} ≥ 4. In this case, because of the orientation of v_{+} and v_{−}, the labels of the Scharlemann cycles must be as shown in Figure 4.3(b). It is easy to see that we cannot have p_{2} = p_{3} = n_{1} − 1, otherwise an edge e in the family ê_{2} would have labels of distinct parities on its two endpoints, which would contradict the parity rule because e would connect antiparallel vertices in both graphs. Hence one of p_{2}, p_{3} is n_{1} − 2 and the other is n_{1}. Without loss of generality we may assume that p_{2} = n_{1} − 2, and p_{3} = n_{1}. See Figure 4.3(b).

Let e_{1}, …, e_{n_{1}} be the parallel edges in the family ê_{3}, where e_{i} has label i at its endpoint on v_{+}. From Figure 4.3(b) we can see that the other endpoint of e_{i} is labeled i+2, hence these edges form two cycles e_{1}∪e_{3}∪e_{5}∪…∪e_{n_{1}−1} and e_{2}∪e_{4}∪e_{6}∪…∪e_{n_{1}} on Γ_{1}. Also, the edges of the Scharlemann cycle in the family ê_{1} form a cycle e′_{1} ∪ e′_{2} on Γ_{1} with vertices u_{1} and u_{2}. By Lemmas 2.2(3) and 2.2(5) all these three cycles are essential on the annulus A, which implies that the cycle e′_{1} ∪ e′_{2} must lie between the other two cycles. See Figure 4.3(c) for a possible picture of Γ̂_{1}, where n_{1} = 6.

There are Δ = 4 edge endpoints at v_{+} labeled 1, among which the two endpoints on e′_{1} and e′_{2} are non-adjacent. Since p_{1} ≤ n_{1}, the other two edges labeled 1 at v_{+} connect antiparallel vertices in Γ_{2}, hence connect parallel vertices in Γ_{1}. Since all the vertices with the same parity as u_{1} are on one side of the cycle e′_{1} ∪ e′_{2}, this implies that all the edges with label + at u_{1} are on one side of e′_{1} ∪ e′_{2}. It follows that the endpoints of e′_{1} and e′_{2} on u_{1} are adjacent among the endpoints of all such edges. Since we have seen that the corresponding endpoints of e′_{1} and e′_{2} are non-adjacent on v_{+} among all endpoints labeled 1, this contradicts Lemma 2.10.

Now assume n_{1} = 2. Since p_{1} ≤ n_{1}, and since p_{2} + p_{3} = 2n_{1} − 2 = 2 ≥ p_{4} + p_{5}, we must have p_{2} + p_{3} = p_{4} + p_{5} = p_{1} = 2. The graph Γ_{1} is shown in Figure 4.3(d). Now the same argument as above applies: Among the 4 points of ∂v_{+} ∩ ∂u_{1}, the two on e′_{1} and e′_{2} are non-adjacent on ∂v_{+} but adjacent on ∂u_{1}, which contradicts Lemma 2.10, completing the proof of the lemma. □

Figure 4.3

Lemma 4.5. Suppose Δ ≥ 4, n_{2} = 2, and n_{1} ≥ 2. Then p_{1} = p_{6} > 0.

Proof. Assume that p_{1} = p_{6} = 0. The reduced graph Γ̂_{2} has 4 edges ê_{2}, ê_{3}, ê_{4}, ê_{5}, each representing at most n_{1} edges of Γ_{2}. The valency of v_{+}, v_{−} in Γ_{2} is Δn_{1}, and Δ ≥ 4, so we must have Δ = 4, and each ê_{i} represents exactly n_{1} edges.

Consider the permutation φ : {1, …, n_{1}} → {1, …, n_{1}} determined by the parallel edges in ê_{i}. Since each family has exactly n_{1} edges, one can see that φ is independent of the choice of ê_{i}. Let L be the length of an orbit of φ. The edges in ê_{i} form cycles in Γ_{1}, each having length L. Thus the reduced graph Γ̂_{1} consists of several cycles of length L, and each edge of Γ̂_{1} represents exactly 4 edges of Γ_{1}.

CASE 1. L ≥ 3.

There are 4 edges in Γ_{1} from u_{1} to u_{φ(1)}, two of which have label + at u_{1}. On the other hand, there are 4 edges in Γ_{2} having label 1 at v_{+} and label φ(1) at v_{−}, which is a contradiction. See Figure 4.4(a)–(b) for the case that L = n_{1} = 3 and φ(1) = 2.

Figure 4.4

CASE 2. L = 2.

Let k = φ(1) = n_{1}/2. There is a pair of edges a, b in the family ê_{3}, such that a has label 1 at v_{+} and label k at v_{−}, while b has label k at v_{+} and label 1 at v_{−}. Let P, S, Q, R be the endpoints of a, b, as shown in Figure 4.4(c). Notice that the vertices u_{1} and u_{k} are parallel, so from Figure 4.4(d) we can see that ρ_{1}(a, b) = ρ_{k}(b, a). Hence a, b are an equidistant pair in Γ_{1}. By Lemma 2.8 a, b are also an equidistant pair in Γ_{2}. However, this is impossible because ρ_{+}(a, b) = k and ρ_{−}(b, a) = 7k.

CASE 3. L = 1.

In this case all edges of Γ_{1} are loops, hence the reduced graph Γ̂_{1} consists of n_{1} disjoint loops. Since n_{1} > 1, Γ_{1} is 1-separable, which contradicts Lemma 3.4. □

Proposition 4.6. Suppose Δ ≥ 4, n_{2} = 2, and n_{1} > 2. Then Γ_{1} is 2-separable.

Proof. By Lemmas 4.3–4.5 we may assume that v_{1} = v_{+} and v_{2} = v_{−} are antiparallel, p_{1} = p_{6} > 0, and p_{2} = p_{3} = n_{1}. We may assume without loss of generality that the labels of the edge endpoints are as shown in Figure 4.5(a).

The families of parallel edges in ê_{2} and ê_{3} determine the same permutation φ on the set of labels {1, …, n_{1}}. Let L be the length of an orbit of φ. Let k = φ(1). The argument splits into two cases: L ≥ 2, and L = 1. We will show that the first case leads to a contradiction, while the second implies that Γ_{1} is 2-separable.

CASE 1. L ≥ 2.

The parallel edges in ê_{2} form cycles in Γ_{1}. There is a loop (namely C in Figure 4.5(a)) in Γ_{2} at v_{+} with endpoints labeled 1 and n_{1}, so by the parity rule u_{1} and u_{n_{1}} are of opposite signs, hence 1 and n_{1} belong to distinct orbits of φ. Let θ be the cycle containing u_{1} and u_{k}, and let θ′ be the one containing u_{n_{1}} and u_{k−1}. By Lemma 2.2(3) all these cycles are essential in A. Since there is an edge C in Γ_{2} with endpoints labeled 1 and n_{1} (see Figure 4.5(a)), which connects u_{1} to u_{n_{1}} in Γ_{2}, there are no vertices inside the annulus between θ and θ′.

First assume that Δ = 4. Let A, B, C, D be the edges with an endpoint on v_{+} labeled 1, and let A, B, F, G be those with an endpoint on v_{−} labeled k. See Figure 4.5(a). On Γ_{2} these edges form a subgraph as shown in Figure 4.5(b). The edges C and G cut the annulus between θ and θ′ into two disks U_{1} and U_{2}, as shown in the figure.

The endpoints of the edges A, B, C, D appear successively on ∂v_{+} in Γ_{2}, so by Lemma 2.10 they also appear successively on ∂u_{1} in Γ_{1}. Similarly, the endpoints of A, B, F, G appear successively on ∂u_{k}. From Figure 4.5(b) one can see that exactly one of the edges D and F, say D, lies inside of D; in particular, we must have D ≠ F. On the other hand, in U_{1} the only vertices which are parallel to u_{1} are u_{k} and u_{1} itself, so D must have its other endpoint on u_{k}. But this is a contradiction, because in Γ_{2} the endpoint of D on v_{−} would then have label k, implying that D = F.

Now assume that Δ = 5. Let A, B, C, D, E be the edges with an endpoint on v_{+} labeled 1, and let A, B, F, H, G be those with an endpoint on v_{−} labeled k, as shown in Figure 4.5(a). By Lemma 2.10 there are two possible orders in which the endpoints of these edges appear on ∂u_{1} and ∂u_{k} respectively. (i) The jumping number d = d(r_{1}, r_{2}) = 1. In this case the endpoints appear as A, B, C, D, E on ∂u_{1}, and the proof is similar to that given above, noticing that D, E are in U_{1} if and only if F and H are outside of U_{1}. (ii) The jumping number d = 2, so the endpoints appear in the order A, D, B, E, C on ∂u_{1}, and A, H, B, G, F on ∂u_{k}. See Figure 4.5(c). One can see that the edges E and F lie in different disks U_{1}, U_{2}, while D and H lie between A and B; hence the argument above can be applied to the edges E and F give a similar contradiction.

Figure 4.5

CASE 2. L = 1.

In this case all the edges in ê_{2} and ê_{3} are loops in Γ_{1}. Orient all the loops from label + to label −. Let e_{i} and e′_{i} be the edges in ê_{2}, ê_{3} which form loops based at u_{i}. There are two possibilities, depending on whether they have the same orientation; see Figure 4.6. We claim that in either case all the non-loop edges with an endpoint at u_{i} are on one side of the loops.

First assume Δ = 4. If the two edges have different orientations, then since their + endpoints are adjacent among all endpoints on ∂v_{+} which are labeled i, by Lemma 2.10 these two endpoints are also adjacent on ∂u_{i} among all endpoints labeled +. Since the + and − labels alternate around ∂u_{i}, it follows that all the endpoints besides those of e_{i} and e′_{i} are on one side of the loops. If the two edges have the same orientation, then there is another edge between them. Since the + and − labels alternate, from Figure 4.6(b) we see that the number of labels on each side of the loops must be even. Therefore the remaining two endpoints must be on the same side of the loops.

Now assume Δ = 5. We have p_{4} + p_{5} + 2p_{1} = 3n_{1}. Since p_{4}, p_{5} ≤ n_{1}, and p_{1} ≤ n_{1}/2 + 1, we have only two possibilities: Either p_{4} = p_{5} = n_{1} and p_{1} = n_{1}/2, or p_{4} + p_{5} = 2n_{1} − 2 and p_{1} = n_{1}/2 + 1. In the first case there are four loops based at each vertex, so there are only two adjacent edge endpoints left, which must be on the same side of the loops. In the second case, the argument beginning at the second paragraph of the proof of Lemma 4.4, with p_{2}, p_{3} replaced by p_{4}, p_{5}, applies (although here we don’t have p_{4} + p_{5} ≥ p_{2} + p_{3}), to show that the reduced graph Γ_{1} contains a subgraph as shown in Figure 4.3(c), (which illustrates the case n_{1} = 6.) However, since p_{1} = n_{1}/2+1 ≥ 3, there is an edge in ê_{1} adjacent to the Scharlemann cycle, with endpoints labeled 3 and n_{1}. On Γ_{1} this edge connects u_{3} to u_{n_{1}}, which is impossible because u_{3} and u_{n_{1}} are on different sides of the loop e′_{1} ∪ e′_{2} in Figure 4.3(c). This completes the proof of the claim.

Consider the reduced graph Γ̂_{1}. We have shown that each vertex is the base of a loop. Let u′, u′′ be two vertices that are connected by an edge. The above claim says that there are no edges connecting u′, u′′ to any other vertices. Thus there is an annulus A′ containing only the vertices u′, u′′, and with boundary disjoint from Γ_{1}. By definition Γ_{1} is 2-separable. □

§5. The case that Γ_{1} is 2-separable

Assume that n_{1}, n_{2} ≥ 2, and Δ ≥ 4. In Section 4 we have shown that if n_{2} = 2 then Γ_{1} is 2-separable. Lemma 5.1 below shows that the converse is also true. The goal of this section is to show that if Γ_{1} is 2-separable, (in particular, if one of the n_{α} = 2), then n_{1} = n_{2} = 2, and the manifold M is the exterior of one of the links in Figure 7.1(b) or (c). See Proposition 5.5.

Throughout this section we will assume that Γ_{1} is 2-separable. We use A′, Γ′_{1}, Γ′_{2} to denote the annulus and graphs given in Definition 3.3. Let u_{1}, u_{2} be the vertices of Γ′_{1}. The labeling of u_{i} in Γ′_{1} may not be the same as that in Γ_{1}, however this does not affect the proof below. Notice also that a pair of edges is an equidistant pair in Γ′_{i} if and only if it is an equidistant pair in Γ_{i}, hence Lemma 2.8 still applies to Γ′_{i}. Also, since Γ′_{i} are subgraphs of Γ_{i}, most of the properties of Γ_{i} still hold for Γ′_{i}. For example, there is no extended Scharlemann cycle in Γ′_{1}.

Lemma 5.1. Suppose Δ ≥ 4. If Γ_{1} is 2-separable, then n_{2} = 2.

Proof. Let A′, Γ′_{i} be as in Definition 3.3. The reduced graph Γ̂′_{1} is a subgraph of the graph shown in Figure 5.1. If some edge represents more than n_{2} parallel edges of Γ′_{1}, then by Lemma 2.3(1) either n_{2} = 2 and we are done, or all vertices of Γ_{2} are parallel, which by Lemma 4.1 again implies that n_{2} = 2. We can thus assume that Δ = 4, and that each edge in Figure 5.1 represents exactly n_{2} parallel edges of Γ′_{1}. The n_{2} parallel loops based at u_{1} contains a Scharlemann cycle, so by Lemma 2.2(4) n_{2} is even. Also, notice that the labels of the 2n_{2} endpoints of these loops appear successively, so if n_{2} ≥ 4 then the 4 edges in the middle would form an extended Scharlemann cycle, which is impossible by Lemma 2.2(6). □

Lemma 5.2. Suppose Δ ≥ 4, and Γ_{1} is 2-separable. Then Γ′_{2} must be one of the graphs shown in Figure 5.2(d) or (e).

Proof. By Lemma 5.1 we have n_{2} = 2, by Lemma 4.3 the two vertices of Γ′_{2}, v_{+} and v_{−}, are antiparallel, and by Lemma 4.2 the reduced graph Γ̂′_{2} is a subgraph of the graph shown in Figure 4.1. Denote by p′_{i} the number of edges of Γ′_{2} in the family ê_{i} shown in Figure 4.1(b). Recall that p_{i} is the number of edges of Γ_{2} in ê_{i}. Without loss of generality we may assume that p_{2} + p_{3} ≥ p_{4} + p_{5}. Then by Lemma 2.5(3) we have p_{i} ≤ 2, and by Lemma 4.4 we have p_{2} = p_{3} = n_{1}. By considering the edges of Γ′_{2} in each family ê_{i}, it follows that p′_{i} ≤ 2, and p′_{2} = p′_{3} = 2. According as p′_{1} = 0, 1 or 2, and Δ = 4 or 5, we have the following possibilities.

(a) p′_{1} = 0, Δ = 4, p′_{4} + p′_{5} = 4;

(b) p′_{1} = 1, Δ = 4, p′_{4} + p′_{5} = 2;

(c) p′_{1} = 1, Δ = 5, p′_{4} + p′_{5} = 4;

(d) p′_{1} = 2, Δ = 4, p′_{4} + p′_{5} = 0;

(e) p′_{1} = 2, Δ = 5, p′_{4} + p′_{5} = 2;

Using the parity rule, each of the above uniquely determines Γ′_{2} up to a homeomorphism of the underlying torus T. See Figure 5.2(a)–(e). We want to show that cases (a), (b) and (c) cannot occur.

Consider the graph in Figure 5.2(a). It is easy to see that either all edges have the same label at both endpoints, or each edge has distinct labels at its two endpoints. In the first case Γ′_{1} consists of loops, so Γ_{1} is 1-separable. By Lemma 3.4 we must then have n_{1} = 1, which is a contradiction. In the second case, the two vertices of Γ′_{1} are parallel by the parity rule, the reduced graph Γ̂′_{1} consists of two edges from u_{1} to u_{2}, so one can see that any pair of edges is an equidistant pair on Γ′_{1}. On the other hand a pair of parallel edges {e_{1}, e_{2}} is not an equidistant pair on Γ′_{2} because ρ_{v_{+}}(e_{1}, e_{2}) = 1 and ρ_{v_{−}}(e_{2}, e_{1}) = 7. This contradicts Lemma 2.8, showing that Γ′_{2} cannot be the graph in Figure 5.2(a).

In all the remaining cases, p′_{1} ≥ 1, so there is a loop based at v_{+}. By the parity rule, the two vertices of Γ′_{1} are antiparallel. Thus any edge in Γ′_{2} from v_{+} to v_{−} must have the same label at its two endpoints, and is a loop in Γ′_{1}. The reduced graph of Γ′_{1} is a subgraph of that in Figure 5.1, so any two loops based at u_{1} are parallel. In particular, the edges e_{1}, e_{2} shown in Figure 5.2(a)–(b) are parallel loops in Γ′_{1}, so by Lemma 2.7 they form an equidistant pair in Γ′_{1}, and by Lemma 2.8 they must also be an equidistant pair in Γ′_{2}. However, in both cases we have ρ_{v_{+}}(e_{1}, e_{2}) = 2 and ρ_{v_{−}}(e_{2}, e_{1}) = 4. Hence cases (b) and (c) cannot happen. □

Figure 5.2

Lemma 5.3. Each of the graphs Γ′_{2} in Figure 5.2(d) and (e) uniquely determines the graph Γ′_{1} up to symmetry, as shown in Figure 5.3(a) and (b), respectively. The correspondence between the edges in Γ′_{1} and Γ′_{2} is also uniquely determined up to symmetry.

Proof. The reduced graph Γ̂′_{1} is a subgraph of the graph shown in Figure 5.1. The edges of Γ′_{2} come in parallel pairs, which are nonparallel in Γ′_{1} by Lemma 2.2(2). Using the fact that any loop in Γ′_{2} is a non-loop in Γ′_{1}, and vice versa, we see that the graphs Γ′_{1} corresponding to Γ′_{2} in Figure 5.2(d) and (e) are the ones in Figure 5.2(a) and (b), respectively. It remains to show that the correspondence between the edges of Γ′_{1} and Γ′_{2} is unique up to symmetry.

Label the edges of Γ′_{2} as in Figure 5.2(d). Up to symmetry we may assume that the edge f is the one shown in Figure 5.3(a). The points of ∂v_{+} ∩ ∂u_{1} appear in the order of e, f, a, c on ∂v_{+}, so by Lemma 2.10 they must also appears in this order on ∂u_{1}. This determines these edges in Γ′_{1}. Similarly, the points of ∂v_{+} ∩ ∂u_{2} appear in the order of f, e, b, d on ∂v_{+}, so they also appear in this order on ∂u_{2}. This determines the edges b, d. By looking at the points of ∂v_{−} ∩ ∂u_{1}, one can further determine the edges h, g. Hence in case (d) the edge correspondence is determined by Γ′_{2}.

The proof of case (e) is similar, except that in this case there is a jumping number d = d(r_{1}, r_{2}) given by Definition 2.9, so if P_{1}, P_{2}, …, P_{5} are the points of ∂v_{i} ∩ ∂u_{j}, numbered succesively on ∂v_{i}, (here i = + or −), then they appear as P_{d}, P_{2d}, …, P_{5d} on ∂u_{j}. Label the edges of Γ′_{2} as in Figure 5.2(e), and fix a choice of e on Γ′_{1}. Then there is only one more edge from u_{1} to u_{2} with endpoint labeled + on u_{1}, so that must be the edge f. The points of ∂v_{+} ∩ ∂u_{1} appear as e, j, f, a, c on ∂v_{+}, and as e, ∗, ∗, ∗, f on ∂u_{1}, so we must have d = 2, and the edges are in the order of e, a, j, c, f. Similarly one can determine the other edges of Γ′_{1}. See Figure 5.3(b). □

Figure 5.3

Let F′_{1} be the punctured annulus A′ − Int(u_{1} ∪ u_{2}). Let F′_{2} be the subsurface of F_{2} consisting of disk faces of Γ′_{2}. Consider the submanifold X of M which is a regular neighborhood of T_{0} ∪ F′_{1} ∪ F′_{2} with spherical boundary components capped off by 3balls. Since M is irreducible, each sphere on ∂N(T_{0} ∪ F′_{1} ∪ F′_{2} does bound a 3-ball in M, hence X is indeed a submanifold of M. The manifold X can be constructed as follows. Lut Y = (F′_{1} × I) ∪ (T_{0} × I). For each disk face D_{i} of Γ′_{2} attach a 2-handle H_{i} = D_{i} × I to Y according to the boundary curve of D_{i} on F′_{1} ∪ T_{0}, and finally cap off the spherical boundaries.

Lemma 5.4. The manifold X constructed above is uniquely determined by the graph Γ′_{2}.

Proof. By Lemma 5.3 the graph Γ′_{1} and the correspondence between the edges of Γ′_{1} and Γ′_{2} are uniquely determined by Γ′_{2}. Thus for each face D_{i} of Γ′_{2}, the boundary curve of D_{i} on F′_{1} ∪ T_{0} is determined. The lemma now follows from the above construction of X, because X is uniquely determined by the way the 2-handles H_{i} are attached to Y. □

Proposition 5.5. Suppose Δ ≥ 4, and Γ_{1} is 2-separable. Then n_{1} = n_{2} = 2, and there are exactly two such manifolds M. If Δ = 4, M is the exterior of the link in Figure 7.1(b), and if Δ = 5, M is the exterior of the link in Figure 7.1(c).

Proof. By Lemma 5.1 we have n_{2} = 2. Let F′_{1}, X, Y be as above. The surface F′_{1} separates Y into two sides, the black side Y_{1} and the white side Y_{2}. Let P_{i} be the punctured torus (∂Y −T_{0})∩Y_{i}. Each face of Γ′_{2} is colored according to the side it is on. Since F_{2} is transverse to F′_{1}, the colors on the faces of Γ′_{2} alternate. From Figure 6.2(d) and (e) we see that all the white disk faces have three edges, so they are nonseparating curves on the punctured torus P_{2}. Thus after attaching the 2-handles and capping off the spheres, the surface P_{2} becomes an annulus. On the black side, there is a Scharlemann cycle on Γ′_{2}, so after attaching the H_{i} and capping off the spheres, P_{1} become either an annulus or a pair of disks. The second case does not happen because otherwise the boundary of X contains a sphere separating the boundary components of M, which contradicts the fact that M is irreducible. It follows that X has exactly two boundary components, each being a torus. The annulus A′ is an essential annulus in X, with one boundary component on the outside torus T′, and two other boundary components on T_{0}, hence X is not a product T_{0} × I.

The torus T′ is incompressible in the global manifold M, because otherwise after compression it would become a sphere separating the boundary components of M, contradicting the irreducibility of M. Since M is atoroidal, and X is not a product, it follows that T′ must be parallel to another boundary components T_{1} of M. That is, M − IntX is a product T_{1} × I. In particular, by Lemma 5.4 M ≅ X is determined by Γ′_{2}. Since there is only one possible Γ′_{2} for each of Δ = 4, 5, given in Figure 5.2(d) and (e), we see that there is at most one such manifold M for each of Δ = 4, 5. The existence of such manifolds will be established by Theorem 7.5.

Since M − IntX ≅ T_{1} × I, the annulus A′ in X(r_{1}) can be extended to an essential annulus A′′ in M(r_{1}) intersecting J_{1} just twice. By the minimality of n_{1}, we must have n_{1} = 2. □

§6. The generic case

After the results in the previous sections, we may now assume that n_{α} ≥ 3 for α = 1, 2. The purpose of this section is to show that in this case we must have Δ ≤ 3.

Lemma 6.1. If n_{1} ≥ 3, and n_{2} = 3, then Δ ≤ 3.

Proof. By Lemma 4.2 this is true if all vertices of Γ_{2} are parallel, so we may assume that v_{1}, v_{2} are positive vertices, and v_{3} is a negative vertex of Γ_{2}. Assume Δ ≥ 4.

First assume that there is only one family of parallel loops based at v_{3}, containing k edges. By Lemma 2.5(3), k ≤ n_{1}. These loops contribute 2k endpoints at v_{3}, so there are Δn_{1} − 2k negative edges at v_{3}, hence the number of 3-edges in Γ_{1} is

E = (Δn_{1} − 2k) + k = Δn_{1} − k ≥ 3n_{1}.

By Lemma 2.2(2) no two negative 3-edges are parallel in Γ_{1} because they are parallel positive edges in Γ_{2}. Since n_{2} = 3, T is nonseparating, so by Lemmas 2.4 and 2.2(4) no two positive 3-edges are parallel in Γ_{1}. Thus all the 3-edges are mutually nonparallel, so the reduced graph Γ̂_{1} has at least 3n_{1} edges, which contradicts Lemma 2.12(1).

Now assume that there are at least two families of parallel loops based at v_{3}. These (together with v_{3}) cut T into disks, so there are no loops based at the other vertices, and there is at most one family of parallel edges from v_{1} to v_{2}. The argument in this case is now similar to that given above, considering the 1-edges of Γ_{1} instead of the 3-edges. □

Let G be a graph on a disk D. A vertex v of G is an interior vertex if there is no arc from v to ∂D with interior disjoint from G. It is a cut vertex if there is an arc α on D, cutting D into D_{1}, D_{2}, such that the graphs G_{i} = D_{i} ∩ G have the properties that G_{1} ∩ G_{2} = v, and neither G_{i} is a single vertex. In this case write G = G_{1} ∗_{v} G_{2}. A vertex v is a boundary vertex if it is neither an interior vertex nor a cut vertex. Define

a_{i} = a_{i}(G): the number of boundary vertices v with val(v, G) = i;

l = l(G): the number of trivial loops in G;

p = p(G): the number of pairs of adjacent parallel edges;

σ(G) = 3a_{1} + 2a_{2} + a_{3} + 4l + 2p;

τ(G) = a_{1} + a_{2} + l + p.

Lemma 6.2. Let G be a connected graph on a disk D, which is not a single vertex. Suppose each interior vertex v of G has val(v) ≥ 6. Then (i) σ(G) ≥ 6, and (ii) if G has no interior vertices then τ(G) ≥ 2.

Proof. The result is obvious if G has only one edge. Also, if G = G_{1} ∗_{v} G_{2} then the vertex v contributes at most 3 to σ(G_{i}) and at most 1 to τ(G_{i}), so

σ(G) ≥ (σ(G_{1}) − 3) + (σ(G_{2}) − 3) ≥ σ(G_{1}) + σ(G_{2}) − 6;
τ(G) ≥ (τ(G_{1}) − 1) + (τ(G_{2}) − 1) ≥ τ(G_{1}) + τ(G_{2}) − 2;

hence the result follows by induction. We can thus assume that G contains more than one edge, and has no cut vertices. By considering a sub-disk if necessary, we may assume that ∂D ⊂ G. Notice that in this case a_{1}(G) = 0.

(i) Taking the double of D, we get a graph G′ on a sphere S, with a_{0}(G′) = a_{1}(G′) = a_{3}(G′) = a_{5}(G′) = 0, a_{2}(G′) = a_{2}(G), and a_{4}(G′) = a_{3}(G). There are 2l faces with 1 edge, 2p faces with 2 edges, and all the others have at least 3 edges. Denote by V, E, F the number of vertices, edges, and faces of G′. We have the following inequalities:

2a_{2} + 4a_{3} + 6(V − a_{2} − a_{3}) ≤ 2E
2l + 2(2p) + 3(F − 2l − 2p) ≤ 2E
V − E + F = 2

Solving these inequalities gives σ(G) = 2a_{2} + a_{3} + 4l + 2p ≥ 6.

(ii) This is obvious if all edges of G are on the boundary of D. So assume that e is an edge which is not on the boundary of D. It cuts D into two disks D_{1}, D_{2}, and G into G_{1}, G_{2}. If l(G_{i}) + p(G_{i}) ≥ 1 for both i, then τ(G) ≥ 2 and we are done. Suppose l(G_{i}) = p(G_{i}) = 0 for some i. Then by induction we have τ(G_{i}) = a_{2}(G_{i}) ≥ 2, i.e. it has at least two boundary vertices of valency 2. If both endpoints of e have valency 2 in G_{i}, then shrinking e to a point gives a graph G′_{i}, which by induction must have another boundary vertex of valency 2. Hence G_{i} must have a boundary vertex of valency 2 which is not on ∂e. Similarly for the other graph G_{j}. Hence we have τ(G) ≥ 2. □

Lemma 6.3. Suppose n_{1} ≥ 3, n_{2} = 4, and Δ ≥ 4. Then Γ_{1} cannot have 4 length 2 Scharlemann cycles on mutually distinct label pairs.

Proof. There are only four possible label pairs {i, i + 1}, i = 1, …, 4, so if the lemma were false then they would all appear as label pairs of Scharlemann cycles in Γ_{1}. Denote by V_{i} the part of the Dehn filling solid torus between v_{i} and v_{i+1}. The existence of Scharlemann cycles in Γ_{1} implies that the torus T in M(r_{2}) is separating (Lemma 2.2(4)). Without loss of generality we may assume that V_{1}, V_{3} lie on the side containing some boundary components of M(r_{2}).

Let E^{i} = {e_{1}^{i}, e_{2}^{i}} be the Scharlemann cycle on Γ_{1} with label pair {i, i + 1}, and let D_{i} be the disk bounded by E^{i} on the punctured annulus F^{1}. Then e_{1}^{1} ∪e_{2}^{1} is a cycle on Γ_{2} with endpoints on v_{1}, v_{2}. Consider the manifold X = (T × I) ∪ V_{1} ∪ (D_{1} × I). It is easy to see that X is a cable space, whose boundary consists of two tori T ∪ T′, with T′∩J_{2} consisting of two disks. If T′ is compressible, then after compression it becomes a sphere containing T on one side and some boundary component of M on the other, which is therefore a reducing sphere in M(r_{2}); by [Wu3] in this case we would have Δ ≤ 2. So assume that T′ is incompressible. Then X′ = (T′ × I) ∪ V_{3} ∪ (D_{3} × I) is also a cable space, hence T′ is not parallel to a torus in ∂M(r_{2}). It follows that T′ is an essential torus in M(r_{2}) intersecting J_{2} in fewer than n_{2} disks, contradicting the choice of the essential torus T. □

A vertex u in Γ_{1} is a full vertex if its valency in Γ_{1}^{+} is Δn_{2}. In other words, u is full if all the edges incident to it are positive.

Lemma 6.4. Suppose n_{1} ≥ 3, n_{2} ≥ 4, and Δ ≥ 4. Then each full vertex u_{j} of Γ_{1} has valency at least 6 in Γ̂_{1}.

Proof. Assume that val(u_{j}, Γ̂_{1}) ≤ 5. There are a total of Δn_{2} edge endpoints at u_{j}, and each family of parallel edges contains at most n_{2}/2 + 2 edges (Lemma 2.3(3)), so we have

\frac{Δn_{2}}{5} ≤ \frac{n_{2}}{2} + 2.

This immediately gives n_{2} ≤ 6. If n_{2} = 6, there is a family of 5 parallel edges, with endpoints labeled 1, 2, 3, 4, 5 at u_{j}, say. Then one can see that this family contains two Scharlemann cycles on label pairs {1, 2} and {4, 5}, respectively, which contradicts Lemma 2.3(2). If n_{2} = 5 then T is non-separating, so by Lemmas 2.4 and 2.2(4) each family of parallel edges contains at most n_{2}/2 edges. Replacing the right hand side of the above inequality with n_{2}/2, we see that Δ < 3, contradicting the assumption. Hence we must have n_{2} = 4.

Consider the reduced negative graph Γ̂_{2}^{−}. Let E, F be the number of its edges and disk faces. There are four edges in Γ_{2}^{−} labeled j at each v_{i}, no two of which are parallel because otherwise there would be n_{1} + 1 parallel edges in Γ_{2}, which would contradict Lemma 2.5(3). Thus the valency of v_{i} in Γ̂_{2}^{−} is at least 4 for any i, so

4n_{2} ≤ 2E.

Each face of Γ̂_{2}^{−} must have at least 4 edges because two adjacent vertices on the boundary of the face have opposite signs; hence

4F ≤ 2E.

Calculating the Euler number we have

n_{2} − E + F ≥ 0.

These three inequalities hold if and only if they are all equalities. In particular, each vertex has valency 4 in Γ̂_{2}^{−}, and there are exactly 2n_{2} = 8 edges in Γ̂_{2}^{−}.

The four edges of Γ̂_{2}^{−} incident to v_{2} have the other endpoint on v_{1} or v_{3}. If there were three edges of Γ̂_{2}^{−} connecting v_{2} to v_{1}, then they would cut the torus T into a disk. Since val(v_{3}, Γ̂_{2}^{−}) = 4, some pair of edges of Γ̂_{2}^{−} incident to v_{3} would be parallel, which is absurd because Γ̂_{2}^{−} is a reduced graph. Therefore there are exactly two edges of Γ̂_{2}^{−} with endpoints on v_{1} and v_{2}. Similarly, for each i = 1, …, 4, there are exactly two edges in Γ̂_{2}^{−} connecting v_{i} to v_{i+1}.

The 8 edges of Γ̂_{2}^{−} contain the 16 j-edges, and by Lemma 2.5(4) each edge of Γ̂_{2}^{−} contains at most two j-edges, and hence exactly two j-edges. Combining this with the conclusion of the last paragraph, we see that there are 4 j-edges connecting v_{i} to v_{i+1} for any i. On Γ_{1} this means that among the 16 edges incident to u_{j}, exactly 4 of them have label pairs {i, i + 1}, for each i.

By Lemma 6.3, some pair {i, i + 1}, say {1, 2}, is not the label pair of a length 2 Scharlemann cycle on Γ_{1}. Thus by Lemma 2.4 the 4 edges with label pair {1, 2} must be mutually nonparallel. Notice that if an edge e is parallel to an edge with label pair {1, 2}, then it has label pair either {1, 2} or {3, 4}. Since there are only 5 families of parallel edges at u_{j}, 4 of which contain edges labeled {1, 2}, it follows that all the 8 edges labeled {2, 3} or {4, 1} are in the remaining family. However, this is impossible because by Lemma 2.3(3) each family contains at most n_{2}/2 + 2 = 4 edges. □

Lemma 6.5. Suppose n_{1} ≥ 3, n_{2} ≥ 4, Δ ≥ 4, and suppose Γ_{1}^{+} has a full vertex u_{k}. Then

(i) Γ̂_{1}^{+} has no vertices of valency at most 1;

(ii) if u_{i} is a boundary vertex of Γ̂_{1}^{+} with valency at most 3 in Γ̂_{1}^{+}, then i is a label of a Scharlemann cycle in Γ_{2}; and

(iii) each component of Γ̂_{1}^{+} contains at most one boundary vertex of valency at most 3.

Proof. Since u_{k} is a full vertex, all the k-edges in Γ_{2} are negative edges. By Lemma 2.5(4) no three such are parallel on Γ_{2}. Since there are Δn_{2} k-edges, we see that the reduced graph Γ̂_{2} contains at least Δn_{2}/2 ≥ 2n_{2} negative edges.

(i) By Lemma 2.12(2) there are at most 3n_{2} edges in the reduced graph Γ̂_{2}. We have shown that at least 2n_{2} of the edges are negative, so Γ_{2} has at most n_{2} positive edges. By Lemma 2.3(3) each edge of Γ̂_{1}^{+} represents at most n_{2}/2 + 2 ≤ n_{2} edges of Γ_{1}, so if some u_{i} has valency at most 1 in Γ̂_{1}^{+}, then it has more than 2n_{2} negative edges, so there are more than 2n_{2} positive i-edges in Γ_{2}, which then implies that 3 of those edges are parallel in Γ_{2}. This contradicts Lemma 2.5(4).

(ii) If u_{i} has more than n_{2} negative edges, then two of them are parallel positive edges on Γ_{2}, hence by Lemma 2.4 i is a label of some Scharlemann cycle in Γ_{2}, and we are done. Now assume that u_{i} has at most n_{2} negative edges, and hence at least (Δ − 1)n_{2} positive edges. By assumption it has at most three families of parallel positive edges, and by Lemma 2.3(3) each family contains at most n_{2}/2 + 2 edges, so we have (Δ − 1)n_{2} ≤ 3(n_{2}/2 + 2), which implies that Δ = n_{2} = 4, and each family contains exactly n_{2}/2 + 2 = 4 edges. Since u_{i} is a boundary vertex in Γ̂_{1}^{+}, these three families are adjacent, as shown in Figure 6.1.

Figure 6.1

By Lemma 2.2(6) Γ_{1} contains no extended Scharlemann cycle, so the labels at the other endpoints of the positive edges are as shown in Figure 6.1. From the figure we can see that there are three Scharlemann cycles with label pair {1, 2} and another three with label pair {3, 4}. On Γ_{2}, this implies that there are 6 i-edges connecting v_{1} to v_{2}, and another 6 connecting v_{3} to v_{4}. Since n_{2} = 4, there are no other vertices, so it is easy to see that there exist three parallel i-edges, contradicting Lemma 2.5(4).

(iii) If G is a component of Γ̂_{1}^{+}, and u_{i}, u_{j} are boundary vertices of valency at most 3 in G, then by (ii) both i, j are labels of Scharlemann cycles in Γ_{2}. Since u_{i}, u_{j} are parallel, no label pair of a Scharlemann cycle could contain both i and j; hence there must be two Scharlemann cycles in Γ_{2} with distinct label pairs, contradicting Lemma 2.5(1). □

Consider a graph Γ on a sphere S. A component G of Γ is called an extremal component if there is a disk D of S, such that D ∩ Γ = G.

Lemma 6.6. Suppose n_{1} ≥ 3, n_{2} ≥ 4, and Δ ≥ 4. Then Γ_{1} has no full vertices.

Proof. Suppose Γ_{1} has a full vertex u_{k}. Embed the annulus A in a sphere S, and consider Γ̂_{1} as a graph on S. If Γ̂_{1}^{+} is connected, then all vertices of Γ_{1} are full. By Lemma 2.12(1) there is a vertex of valency at most 5 in Γ̂_{1}^{+}, which would contradict Lemma 6.4. Therefore Γ̂_{1}^{+} is not connected, so it has at least two extremal components G_{1}, G_{2}. Let D_{i} be a disk on S such that D_{i} ∩ Γ̂_{1}^{+} = G_{i}. By Lemma 6.5(1) all vertices of G_{i} have valency at least 2. That is, G_{i} is not an isolated vertex, and a_{1}(G_{i}) = 0.

If some D_{i} contains no component of S − A, then G_{i} has no loops or parallel edges, so by Lemma 6.2 it has at least two vertices of valency at most 3, which contradicts Lemma 6.5(iii). It follows that each D_{i} contains exactly one component of S−A, hence l(G_{i}) + p(G_{i}) ≤ 1. This also shows that G_{1}, G_{2} are the only extremal components, otherwise choosing another extremal component as G_{1} would give a contradiction. Applying Lemma 6.2 to G_{i}, we have σ(G_{i}) = 3a_{1} + 2a_{2} + a_{3} + 4l + 2p ≥ 6. Since a_{1} = 0, and l + p ≤ 1, we have either a_{3} ≥ 2 or a_{2} ≥ 1. The first case is impossible by Lemma 6.5(iii). Hence a_{2} ≥ 1, that is, each G_{i} contains a boundary vertex u^{i} of valency 2. By Lemma 6.5(ii), the index of u^{i} is a label of all Scharlemann cycles in Γ_{2}. Without loss of generality, we may assume that the vertex u^{i} has label i, that is, u^{i} = u_{i}, i = 1, 2. Thus all the Scharlemann cycles in Γ_{2} have label pair {1, 2}.

CLAIM. There are exactly 4 Scharlemann cycles of length 2 in Γ_{2}, and there are no other positive edges in Γ_{2}.

Let k be the number of positive i-edges in Γ_{2}, i = 1, 2. These are the negative edges in Γ_{1} incident to u_{i}. Each edge of Γ̂_{1}^{+} incident to u_{i} represents at most n_{2}/2 + 2 parallel edges. Since u_{i} has valency 2 in Γ̂_{1}^{+}, we have k ≥ Δn_{2} −2(n_{2}/2+2) ≥ 3n_{2} −4. On the other hand, as shown in the proof of Lemma 6.5(i), the existence of a full vertex of Γ̂_{1}^{+} implies that there are at most n_{2} positive edges in the reduced graph Γ̂_{2}, and by Lemma 2.5(4) each such edge represents at most two i-edges. Therefore, we must have k ≤ 2n_{2}. These inequalities give n_{2} = Δ = 4, and k = 2n_{2} = 8. Moreover, each of the 4 positive edge in Γ̂_{2} must contain exactly two i-edges, hence the 8 i-edges form 4 pairs of Scharlemann cycles. As shown before the Claim, they all have the label pair {1, 2}.

By Lemma 2.2(5) the edges of a Scharlemann cycle form an essential loop in Γ_{1}, which separates the two extremal components G_{1}, G_{2} because u_{1}, u_{2} are boundary vertices and neither G_{i} is contained in a disk in the annulus A. Since all the 8 negative edges from u_{1} end up on u_{2}, there are no other negative edges in Γ_{1}. This completes the proof of the claim. In particular, all vertices of Γ_{1} other than u_{1}, u_{2} are full vertices.

Note that since there is an arc connecting the only two extremal components G_{1}, G_{2} of Γ̂_{1}^{+}, we must have Γ̂_{1}^{+} = G_{1} ∪ G_{2}. We have assumed that n_{1} ≥ 3, hence one of the G_{i}, say G_{1}, contains more than one vertex.

Embed A in a sphere S and consider G_{1} as a graph on S. Let V, E, F be the number of vertices, edges and faces of G_{1}. By Lemma 6.4 all the vertices but u_{1} have valency at least 6, so

2 + 6(V − 1) ≤ 2E.

There are two faces D_{1}, D_{2} of G_{1} which may contain some components of S − A, all others are incident to at least three edges. Since u_{1} has valency 2, and G_{1} is connected and contains at least two vertices, there is no loop based at u_{1}, so the disk face D_{i} containing u_{1} on its boundary has at least two edges. Hence we have

2 + 1 + 3(F − 2) ≤ 2E.

Solving these two inequalities, we have 3V −3E +3F ≤ 5, which contradicts the Euler number formula V − E + F ≤ 2. □

Lemma 6.7. Suppose n_{1} ≥ 3, n_{2} ≥ 4, and Δ ≥ 4. If Γ̂_{1}^{+} has no interior vertex, then n_{2} = Δ = 4. Moreover, each vertex of Γ_{1} has at most 8 negative edges incident to it, and there are 8 negative edges connecting a pair of vertices u_{1}, u_{2}, which form 4 Scharlemann cycles in Γ_{2}.

Proof. Let u_{i} be a vertex which has valency 2 in Γ̂_{1}^{+}, given by Lemma 6.2(i). By Lemma 2.3(3) at u_{i} there are at most 2(n_{2}/2 + 2) = n_{2} + 4 endpoints of positive edges, hence at least (Δ − 1)n_{2} − 4 negative edges. Choose a vertex u_{j} such that the number m of negative edges incident to it is maximal among all the vertices in Γ_{1}. The above shows that m ≥ (Δ − 1)n_{2} − 4. The parity rule says that there are m positive j-edges in Γ_{2}.

Let G be the subgraph of Γ_{2} consisting of the positive j-edges and all vertices of Γ_{2}. Then G has n_{2} vertices and m edges. Denote by F the number of disk faces of G. Calculating the Euler number, we have n_{2} − m + F ≥ 0, hence

F ≥ m − n_{2}.

By Lemma 2.11, each disk face of G contains a Scharlemann cycle, so there are at least t = 2F ≥ 2(m − n_{2}) edges of Scharlemann cycles. By Lemma 2.5(1), all these Scharlemann cycles have the same label pair, say {1, 2}. Let p ∈ {1, 2}. Then in Γ_{1} there are at least t negative edges incident to the vertex u_{p}. By the definition of m, u_{p} also has at least Δn_{2} − m endpoints of positive edges. Thus the valency of u_{p} is at least

t + (Δn_{2} − m) ≥ 2(m − n_{2}) + Δn_{2} − m = m − 2n_{2} + Δn_{2}
≥ (Δ − 1)n_{2} − 4 − 2n_{2} + Δn_{2}
= (Δ − 3)n_{2} − 4 + Δn_{2}
≥ Δn_{2}

Since the valency of u_{p} is Δn_{2}, all the above inequalities are equalities, so we must have (i) Δ = n_{2} = 4, (ii) m = (Δ − 1)n_{2} − 4 = 8, hence there are at most 8 negative edges at each vertex in Γ_{2}, and (iii) there are t = 2(m − n_{2}) = 8 edges of Scharlemann cycles with label pair {1, 2}, hence on Γ_{1} there are exactly 8 negative edges at u_{i}, i = 1, 2, all connecting u_{1} to u_{2}. □

Lemma 6.8. Suppose n_{1} ≥ 3, n_{2} ≥ 4, Δ ≥ 4, and suppose Γ̂_{1}^{+} has no interior vertex. If Γ_{1} has at least three vertices incident to 8 negative edges, then Γ_{1} is 2-separable.

Proof. We have seen from Lemma 6.7 that Δ = n_{2} = 4, and u_{1}, u_{2} have 8 negative edges, which form 4 Scharlemann cycles in Γ_{2} with label pair {1, 2}. Assume u_{i} is another vertex incident to 8 negative edges. By Lemma 2.5(1) i is not a label of a Scharlemann cycle, so by Lemma 2.4 no two of the 8 positive i-edges in Γ_{2} are parallel. Thus the reduced graph Γ̂_{2} has at least 8 positive edges. By Lemma 2.12 Γ̂_{2} has at most 3n_{2} = 12 edges, so it has at most 12 − 8 = 4 negative edges.

For each label j, by Lemma 2.5(4) each of the 4 negative edges in Γ̂_{2} contains at most two j-edges, so Γ_{2} has at most 8 negative j-edges. On Γ_{1} this implies that there are at most 8 positive edges incident to u_{j}. On the other hand, by Lemma 2.7 there are also at most 8 negative edges at u_{j}. Since there are a total of 16 edges at each vertex, this implies that there are exactly 8 positive and 8 negative edges at each vertex u_{j}. In particular, both Γ_{α} have the same number of positive edges and negative edges. There are a total of Δn_{1}n_{2}/2 = 8n_{1} edges, so Γ_{2} has 4n_{1} negative edges. By Lemma 2.5(3) each of the 4 negative edges in Γ̂_{2} represents exactly n_{1} negative edges.

Let E be a set of n_{1} parallel negative edges of Γ_{2}. By Lemma 2.2(3) they form several essential cycles of length L on Γ_{1}, which we will call the E-cycles. Choose E so that L is maximal. Let e_{1}, e_{2} be a Scharlemann cycle in Γ_{2}, with label pair {1, 2}. By Lemma 2.2(5) these edges form an essential cycle in Γ_{1}, hence cut the annulus A into two annuli A_{1}, A_{2}, each containing several E-cycles, and having u_{1}, u_{2} on its boundary. Let p (resp. n) be the number of E-cycles in A_{1} whose vertices are positive (resp. negative). Assume that u_{1} is a positive vertex. Then on A_{1} there are pL positive vertices, and nL + 1 negative vertices (including u_{2}). Recall that each vertex has 8 negative edges, which connect it to vertices of opposite sign. Moreover, all the negative edges at u_{1} go to u_{2}. Counting the number of negative edges, we have

8(pL) = 8(nL + 1),

which implies that L = 1.

When L = 1, all the positive edges in Γ_{1} are loops. Thus the loops at u_{1}, u_{2} and the edges connecting them form a component of Γ_{1}, contained in an annulus A′ with boundary disjoint from Γ_{1}. Therefore Γ_{1} is 2-separable. □

Lemma 6.9. Suppose n_{1} ≥ 3, n_{2} ≥ 4, and Δ ≥ 4. If Γ̂_{1}^{+} has no full vertices, then it is 2-separable.

Proof. By Lemma 6.7, n_{2} = Δ = 4, and there are four Scharlemann cycles on Γ_{2}, all having the same label pair {1, 2}, say. Let G_{i}, i = 1, 2, be the component of Γ̂_{1}^{+} containing u_{i}. If there is an extremal component G of Γ̂_{1}, G ≠ G_{1}, G_{2}, or if G_{i} is an extremal component but u_{i} is a cut vertex, then applying Lemma 6.2(ii) to G_{1}, G_{2} and G we see that there is a vertex u_{j} ≠ u_{1}, u_{2}, which has valency at most 2 in Γ̂_{1}^{+}. Such a vertex has at most 2(n_{2}/2+2) = 8 positive edges, hence at least Δn_{2} −2(n_{2}/2+2) = 8 negative edges, so the result will follow from Lemma 6.8. Therefore we assume that G_{1}, G_{2} are the only extremal components of Γ̂_{1}^{+}, and u_{1}, u_{2} are boundary vertices. Since there are edges connecting u_{1} to u_{2}, there are no other components, that is, Γ̂_{1}^{+} = G_{1} ∪ G_{2}.

The edges e_{1}, e_{2} of a Scharlemann cycle in Γ_{2} form an essential cycle on the annulus A, which cuts A into two annuli A_{1}, A_{2}. Since u_{i} is not a cut vertex, each G_{i} lies in one of the A_{j}. The two G_{i} must be on different A_{j}, otherwise one of the G_{i} would lie on a disk in A, so by Lemma 6.2(ii) it would have at least two vertices of valency at most 2 in Γ̂_{1}^{+}. Together with the vertices u_{1} or u_{2} on the other component G_{k}, this would give three vertices incident to at least 8 negative edges, contradicting Lemma 6.8.

Now the only edges connecting G_{1} to G_{2} must have at least one endpoint on u_{1} or u_{2}; but since all the 8 negative edges from u_{1} must go to u_{2} (Lemma 6.7), there are no other negative edges. It follows that all vertices other than u_{1}, u_{2} are full vertices. Since n_{1} ≥ 3, this contradicts the assumption. □

Theorem 6.10. Suppose M is a hyperbolic manifold, such that M(r_{1}) contains an essential annulus F̂_{1}, M(r_{2}) contains an essential torus F̂_{2}, and Δ(r_{1}, r_{2}) ≥ 4.

(1) If Δ = 4, M is the exterior of the link in Figure 7.1(a) or (b); otherwise, Δ = 5, and M is the exterior of the link in Figure 7.1(c).

(2) F_{α} can be chosen to intersect the Dehn filling solid torus at most twice.

(3) If some F_{α} intersects the Dehn filling solid torus only once, then M is the exterior of the Whitehead link in Figure 7.1(a).

Proof. If some n_{α} = 1, then by Proposition 3.5 the manifold M is the exterior of the Whitehead link in Figure 7.1(a), n_{β} = 2, and Δ = 4.

Assume n_{1}, n_{2} ≥ 2, and some n_{α} = 2. If n_{1} = 2, then by definition Γ_{1} is 2-separable. If n_{2} = 2 and n_{1} > 2, then by Proposition 4.6 Γ_{1} is also 2-separable. By Proposition 5.5, in this case we must have n_{1} = n_{2} = 2, and the manifold M is the exterior of the link in Figure 7.1(b) or (c) respectively, according to Δ = 4 or 5.

The remaining case is that n_{α} ≥ 3 for α = 1, 2. By Lemma 6.1, we cannot have n_{2} = 3. Assume n_{2} ≥ 4. By Lemma 6.6 the graph Γ̂_{1}^{+} has no full vertex, and by Lemma 6.9 this implies that Γ_{1} is 2-separable, which by Proposition 5.5 implies that n_{1} = 2, contradicting the assumption. Hence this case does not happen. □

§7. Some Examples

In this section we give some examples of manifolds admitting toroidal and annular surgeries. Consider the three links shown in Figure 7.1. The first one is the Whitehead link. It is the 2-bridge link associated to the rational number 3/8. The second is the 2-bridge link associated to 3/10. The third one is called the Whitehead sister link.

Figure 7.1

It is the (−2, 3, 8) pretzel link. The Whitehead link is well known to be hyperbolic; see for example [Th2]. We will show in Theorem 7.5 that the other two links are also hyperbolic. First consider the Whitehead link in Figure 7.1(a).

Lemma 7.1. Let M be the exterior of the Whitehead link L shown in Figure 7.1(a). Then there are two slopes r_{1}, r_{2} such that (i) M(r_{1}) contains an essential torus and an essential annulus, each intersecting the Dehn filling solid torus J_{1} once; (ii) M(r_{2}) contains an essential annulus and an essential torus, each intersecting J_{2} twice; and (iii) Δ(r_{1}, r_{2}) = 4.

Proof. The manifold M is shown in Figure 7.2(a), with T_{0} ⊂ ∂M the inside torus. Let φ be the π-rotation of (M, T_{0}) about the horizontal axis shown in Figure 7.2(a). Let α be the axis of φ in M. Then one can see that (M, α)/φ = (X, η), where X is a product S^{2} × I, and η consists of four arcs. See Figure 7.2(b). Thus M is the double cover of X branched along η. By an isotopy (X, η) is equivalent to the pair shown in Figure 7.2(c).

It is well known that, with respect to a certain framing on T_{0}, the manifold M(r) can be obtained by filling the inner component of X with a rational tangle to get a pair (B, η(r)), and then taking the double cover of B branched along η(r); see [M]. The tangle (B, η(1/0)) is shown in Figure 7.2(d). Consider the shaded disk D and Möbius band U shown in Figure 7.2(d). Cutting along D and U, we get a manifold X homeomorphic to T × I, on which D becomes an annulus A, and U becomes a torus T.

The double branched cover M(−1/2) can be obtained by taking two copies of X, gluing T to T, and A to A. Since T and A are incompressible in both sides, they become essential surfaces in M(−1/2). The core of the Dehn filling solid torus J branch covers the core of the rational tangle, shown as K in Figure 7.2(d). Since each of D and U intersects K in a single boundary point, the annulus A and the torus T also intersect J in a single disk. This proves (i).

Now consider the tangle (B, η(1/2)) shown in Figure 7.2(e). It is equivalent to that in Figure 7.2(f). There is a disk D cutting it into a (−1/2, −1/2)-Montesinos tangle and a (1/2)-rational tangle. The lift of D is then an essential annulus A in M(1/2), cutting M(1/2) into a Seifert fiber space Y whose orbifold is a disk with two cone points of order 2, and a solid torus on which A runs twice along the longitude. Thus A is an essential annulus, and the boundary of Y pushed into the interior of M(1/2) is an essential torus, which covers a Conway sphere S on the right hand side of the disk D in B. Notice that both D and S intersect the core of the rational tangle just once, hence A and T intersect the core of the Dehn filling solid torus exactly twice. Since Δ(−1/2, 1/2) = 4, this completes the proof of (ii) and (iii).

Notice that the framing we have chosen is not the standard meridian-longitude framing of the Whitehead link. By a more careful examination, one can show that M(−1/2) is the 0-surgery, and M(1/2) is the 4-surgery with respect to the standard framing. □

Figure 7.2

Lemma 7.2. Suppose M is a hyperbolic manifold such that M(r_{1}) is annular, M(r_{2}) is toroidal, and the intersection graphs Γ_{α} are as shown in Figure 3.1. Then M is the exterior of the Whitehead link.

Proof. One can prove this by using the argument in the proof of Lemma 3.2. Here we give an explicit construction of M. The method can be used to construct other manifolds from intersection graphs.

Let J_{1} be the Dehn filling solid torus in M(r_{1}). Taking the union of A × I with J_{1}, so that the fat vertex of Γ_{1} in A is identified with a meridian disk of J_{1}, we get a genus 2 handlebody X, as shown in Figure 7.3(a). The graph Γ_{2} in Figure 3.1(b) cuts the torus T into two disk faces, whose boundary curves are parallel on ∂X, each being isotopic to the curve C shown in Figure 7.3(a). Therefore the regular neighborhood of the union of the handlebody X and the punctured torus F_{2}, with the spherical boundary component capped off by a 3-ball, is the same as the manifold Y obtained by attaching a 2-handle H to X along the curve C. Let K_{1} be the central curve of the solid torus J_{1}. The manifold M is obtained by removing a regular neighborhood V of K_{1} from Y.

There is an involution φ of X, as shown in the figure, which carries C to itself, so φ extends to an involution of Y = X ∪ H. Let L be the axis of φ in Y. After the involution, the quotient of X is a 3-ball B = X/φ, and the image of L ∩ X consists of three arcs α_{1}, α_{2}, α_{3} in B, as shown in Figure 7.3(b). The quotient of C is an arc α on ∂B connecting the two endpoints of α_{3}. The image of the attached 2-handle H is a 3-ball B′, with L ∩ H projecting to an arc which is isotopic rel ∂ to α. It follows that there is a homeomorphism from Y/φ to B, under which the image of L is the union of α and the three arcs α_{i}, with α pushed into the interior of B. Denote by β the 1-manifold L/φ. The central curve K_{1} of J_{1} becomes an arc with endpoints on β, denoted by K. In Figure 7.3(b) K is the thick grey arc inside B. Consider β as a tangle in the 3-ball B. After an isotopy, the triple (B, β, K) is the same as that shown in Figure 7.2(b), where B is the 3-ball, and β is the tangle consisting of two arcs and a circle.

Let X = B − IntN(K), and η = β ∩ X. Then M is the double cover of X branched along η, which has been shown in the proof of Lemma 7.2 to be the exterior of the Whitehead link. □

Figure 7.3

We need a lemma to test the hyperbolicity of the exteriors of braids in solid tori.

A braid is trivial if it has only one strand.

Lemma 7.3. Let K be a knot in a solid torus V, which is a nontrivial braid, and let M be the exterior of K in V. Then M is hyperbolic unless (i) there is a solid torus V′ in V whose core is a nontrivial braid in V, and K is a nontrivial braid in V′; or (ii) K is a torus knot in V, i.e. it is isotopic to a curve on ∂V.

Proof. Clearly M is irreducible and ∂-irreducible. Thus by the Geometrization Theorem of Thurston for Haken manifolds [Th1], M is hyperbolic unless it is toroidal or Seifert fibered.

Suppose M is toroidal, containing an essential torus T. Let D be a meridian disk of V intersecting K in n disks, where n is the braid index of K. Let P be the punctured disk D−IntN(K). Isotope T so that it has minimal intersection with P. After cutting along P, M becomes a product P × I, and T becomes a set of essential annuli A with boundary on P × ∂I. Hence A is a product C × I, where C is a set of essential curves on P. It is now clear that after gluing P × 0 to P × 1, the annuli A are glued to give a torus bounding a solid torus V′ whose core is a braid in V, which is nontrivial because T is not boundary parallel. K is now a braid in V′, which must be nontrivial because T is not parallel to ∂N(K). Hence (i) holds in this case.

Now suppose M is Seifert fibered. After filling in N(K), the Seifert fibration extends to one on the solid torus V, whose orbifold is a disk with at most one cone point. Since K is nontrivial, it is not a singular fiber, hence it is isotopic to a regular fiber on ∂V, and (ii) follows. □

A braid b can be written as σ_{i_{1}}^{ϵ_{1}}σ_{i_{2}}^{ϵ_{2}} … σ_{i_{k}}^{ϵ_{k}}, where each σ_{i_{j}} is one of the standard generators of the braid group, and ϵ_{j} = ±1. The total exponent of b is defined as e(b) = ∑ ϵ_{j}.

Corollary 7.4. Suppose a knot K is an n-braid in a solid torus V, with total exponent e. If n is prime, and if e is not a nonzero multiple of n − 1, then the exterior of K is hyperbolic.

Proof. If K is an n_{1} braid in V′ whose core is an n_{2} braid in V, then K is an n_{1}n_{2} braid in V. If K is a (n, q) torus knot in V, then the total exponent is (n − 1)q. Hence the corollary follows from Lemma 7.3. □

Now let M_{i}, i = 1, 2, 3 be the exterior of the three links L_{i} shown in Figure 7.1. Denote by T_{0} the boundary torus corresponding to the right component of L_{i} in Figure 7.1.

Theorem 7.5. Each M_{i} is a hyperbolic manifold, and admits two Dehn fillings M_{i}(r_{1}) and M(r_{2}), such that

(i) Δ = 4 for i = 1, 2, and Δ = 5 for i = 3;

(ii) M_{1}(r_{1}) contains an essential annulus A and an essential torus T, each intersecting J_{1} once. All the other M_{i}(r_{k}) contain an essential annulus A and an essential torus T, each intersecting J_{k} twice.

Proof. If i = 1, this is Lemma 7.1. So assume i = 2 or 3.

Removing the left component of L_{i}, we get a solid torus V containing a knot K_{i}. For K_{2}, the braid index is 3, and the total exponent is 0. For K_{3}, the braid index is 5, and the total exponent is 6. It follows from Corollary 7.4 that the exteriors of both K_{i} are hyperbolic.

Figure 7.4

The proof of the rest part of the theorem is similar to that of Lemma 7.1, with Figure 7.2 replaced by Figures 7.4 and 7.5. The essential torus and essential annulus in M_{2}(r_{2}) are the double covers of the disk D_{1} in Figure 7.4(d) and the boundary sphere S_{1} on the right hand side of D, pushed slightly into the interior. Similarly for the others. In all cases, the tangle (B, η(r_{j})) is the sum of a (±1/2, ∓1/3) Montesinos tangle and a (1/2)-rational tangle, so M_{i}(r_{j}) is a graph manifold obtained by gluing a solid torus to a Seifert fiber space along an annulus which runs twice along the longitude of the solid torus. The disks D_{i} and the Conway spheres S_{i} intersect the core of the rational tangle at one interior point, so after double covering the annuli and the tori intersect the Dehn filling solid torus at two points. □

Figure 7.5

Consider the manifold M shown in Figure 7.6(a). It is the exterior of a three component link in S^{3}. Let L = l_{0} ∪ l_{1} be the link in the solid torus V, where l_{0} is the component with winding number 1 in V. Let T_{0} = ∂N(l_{0}), with the standard meridian-longitude framing.

Lemma 7.6. M is hyperbolic, and both M(0) and M(−3/2) are toroidal and annular.

Proof. L is a braid in V, so M is irreducible and ∂-irreducible. As in the proof of Lemma 7.3, it can be shown that if T is an essential torus in M, then T bounds a solid torus V′ in V such that the core of V′ is a braid in V, and L ∩ V′ is a braid in V′. If V′ contains the component l_{0} which has braid index 1 in V, then the core of V′ is a trivial braid in V. So if V′ also contains l_{1} then T = ∂V′ is parallel to ∂V, and if V′ does not contain l_{1} then T is parallel to ∂N(l_{0}), in either case contradicting the assumption that T is essential in M. Hence V′ must contain l_{1} but not l_{0}. The braid index of l_{1} in V′ is 2, otherwise T would be parallel to ∂N(l_{1}). Let K′ be the core of V′. Then the linking number lk(l_{0}, l_{1}) = 2 lk(l_{0}, K′) is even, which is absurd because from the picture we see that lk(l_{0}, l_{1}) = 1.

To prove that M is hyperbolic, it remains to show that M is not a Seifert fiber space. If M were Seifert fibered, then after the trivial filling on L, the Seifert fibration would extend to one on the solid torus V, so that each l_{i} is a fiber. Since a Seifert fibration of V has at most one singular fiber, and l_{0}, l_{1} are not isotopic, one can see that l_{0} must be a singular fiber while l_{1} is a regular fiber, so there would be an isotopy sending l_{0} to the core l′_{0} of V and l_{1} to a (2, 1) curve l′_{1} on ∂V. However, the braid index of l′_{0} ∪ l′_{1} is 3 but the braid index of l_{0} ∪ l_{1} is 1, so they cannot be isotopic. This completes the proof that M is hyperbolic.

As before, let φ be the π-rotation of M about the horizontal axis shown in Figure 7.6(a), and let α be the axis of φ in M. Then (M, α)/φ = (X, η), where X is a twice punctured 3-ball, and η consists of 6 arcs, as shown in Figure 7.6(b). M is the double cover of X branched along η. After an isotopy, the picture is shown in Figure 7.6(c), in which the image of T_{0} is the left sphere S_{0}. We have been careful to ensure that the standard meridian and longitude framings induce the ∞ and 0 slopes on ∂S_{0}. However, notice that the conventions about slopes on ∂S_{0} and T_{0} are opposite. Denote by (Y, η(r)) the pair obtained from (X, η) by filling S_{0} with a rational tangle of slope r. Then M(r) is the double cover of (Y, η(−r)).

Figure 7.6

It is easy to see that (Y, η(0)) is the tangle shown in Figure 7.6(d). A disk intersecting η(0) twice is shown in the figure. The double branched cover M(0) of (Y, η(0)) is the union of a trefoil knot exterior and a copy of T^{2} ×I, glued along an annulus which is meridional on the trefoil knot exterior. Hence M(0) is actually the exterior of the connected sum of a trefoil and a Hopf link. This manifold is toroidal and annular.

Figure 7.6(e) is the picture of (Y, η(3/2)). After an isotopy it is shown in Figure 7.6(f). In this case M(−3/2) is the union of a solid torus and a (2, 1) cable space, glued along an annulus which runs twice along the longitude of the solid torus, hence it is toroidal and annular. One can show that M(−3/2) is actually the exterior of a knot K in a solid torus V, where K is some (2, q) cable of a (2, 1) torus knot in V. □

We call the manifold M above the magic manifold. Table 7.7 lists some manifolds obtained by Dehn filling on T_{0}. We use X(p/q) to denote the exterior of the 2-bridge link associated to the rational number p/q, and X(p, q, r) the exterior of the (p, q, r) pretzel link. We have seen from Lemma 7.6 that M(0) and M(−3/2) are toroidal and annular. The first is the exterior of a link which is the connected sum of a trefoil and a Hopf link, and the second one is the exterior of some (2, q) cable of the (2, 1) torus knot in the solid torus, so it is also the exterior of a link in S^{3}.

The first five fillings are non-hyperbolic, and the last three are the exteriors of the three links in Figure 7.1, which by Theorem 6.10 are the only manifolds admitting annular and toroidal fillings of distance at least 4 apart. It is surprising that all the three exceptional manifolds in Theorem 6.10 arise from Dehn fillings on this manifold.

Table 7.7

There is an automorphism of M interchanging any pair of boundary components of M. Let T_{1} be the boundary of the component of L which wraps along V twice. Let (m_{i}, l_{i}) be the standard meridian-longitude pair of T_{i}. The homeomorphism sending T_{0} to T_{1} will send m_{0} to l_{1}, and l_{0} to m_{1}−l_{1}. Hence all the above Dehn fillings can also be realized by Dehn filling on T_{1}, and the slopes can be calculated accordingly. For example, the two toroidal and annular fillings on T_{1} have slopes −1 and 2, respectively.

We can apply Theorem 6.10 to prove the following result.

Proposition 7.7. If r ≠ 0, −1, −3/2, −2, ∞, then M(r) is hyperbolic.

Proof. Note that M has three boundary components, so it is not one of the three exceptional manifolds in Theorem 6.10. Let r_{1} = 0, and r_{2} = −3/2. By Lemma 7.6 each M(r_{i}) is toroidal and annular, so if r is a slope such that Δ(r, r_{i}) ≥ 4 for some i, then by Theorem 6.10 the manifold M(r) is atoroidal and anannular, and by [Wu1,Oh] and [GLu] it is also irreducible and ∂-irreducible, and hence M(r) is hyperbolic. Thus if M(r) is non-hyperbolic, then Δ(r, r_{i}) ≤ 3 for i = 1, 2.

Suppose r = a/b. Without loss of generality we may assume that a ≥ 0. Then we have

Δ(0, a/b) = a ≤ 3;
Δ(−3/2, a/b) = |2a + 3b| ≤ 3.

Solving these inequalities, we have a/b = 0, ∞, −1, −3/2, −2, or −3. From Table 7.7 we see that M(−3) is the exterior of the Whitehead sister link shown in Figure 7.1(c), which is hyperbolic by Theorem 7.5. □

Theorem 7.8. There are infinitely many hyperbolic manifolds M_{s}, which admit two Dehn fillings M_{s}(r_{1}), M_{s}(r_{2}), each containing an essential annulus and an essential torus, with Δ(r_{1}, r_{2}) = 3.

Proof. Let M be the manifold in Lemma 7.6. We are now considering Dehn fillings on both T_{0} and T_{1}, so denote by M(r, ∗) = M(r) the manifold obtained by r-filling on T_{0}, by M(∗, s) = M_{s} the manifold obtained by s-filling on T_{1}. Let r_{1} = 0, and r_{2} = −3/2.

With respect to certain framings on T_{1}, the manifold M(∗, s) can be obtained by filling the sphere S_{1} in Figure 7.6(c) with a rational tangle of slope s, then taking the double branched cover. From Figure 7.6(d) and 7.6(f) we can see that if s ∉ {0} ∪ {1/n | n ∈ ℤ}, then for i = 1, 2, the essential annulus and essential torus in M(r_{i}) = M(r_{i}, ∗) remain essential in M(r_{i})(s) = M_{s}(r_{i}). By Thurston [Th2], there is a finite set S such that if s ∉ S then M(∗, s) is hyperbolic. Actually using Proposition 7.7 and the symmetry of M we see that S contains exactly 5 slopes. Thus if s ∉ S ∪ {0} ∪ {1/n | n ∈ ℤ}, then M_{s} is hyperbolic, and M_{s}(r_{i}) is toroidal and annular, i = 1, 2. Clearly there are infinitely many such s. Since Δ(r_{1}, r_{2}) = Δ(0, −3/2) = 3, this proves the theorem.

Note that the framing on T_{1} above is not the standard meridian-longitude framing. □

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Department of Mathematics, University of Texas at Austin, Austin, TX 78712 and MSRI, 1000 Centennial Drive, Berkeley, CA 94720-5070

   E-mail address: gordon@math.uiowa.edu

Department of Mathematics, University of Iowa, Iowa City, IA 52242; and MSRI, 1000 Centennial Drive, Berkeley, CA 94720-5070

   E-mail address: wu@math.uiowa.edu