Klein Bottles and Dehn Filling on a Component of Two- component Link Exterior

pISSN 1225-6951 eISSN 0454-8124

⃝ Kyungpook Mathematical Journalc

Klein Bottles and Dehn Filling on a Component of Two- component Link Exterior

Nabil Sayari

D´epartement de Math´ematiques et de Statistique, Universit´e de Moncton, Moncton, Nouveau-Brunswick, Canada

e-mail : [email protected]

Abstract. Let M be the exterior of a hyperbolic link K ∪ L in a homology 3-sphere Y , such that the linking number lk(K, L) is non-zero. In this note we prove that if γ is a slope in ∂N (L) such that the manifold ML(γ) obtained by γ-Dehn filling along ∂N (L) contains a Klein bottle, then there is a bound on ∆(µ, γ), depending on the genus of K and on lk(K, L).

1. Introduction

Let M be a compact, connected, orientable and irreducible 3-manifold with a toral boundary component T . The unoriented isotopy class of a non-trivial simple closed curve on T is called its slope, and if γ1and γ2are slopes on T then ∆(γ1, γ2) will denote their minimal geometric intersection number.

Let γ be a slope on T , and let M (γ) be the 3-manifold obtained by γ-Dehn filling. Thus M (γ) = M∪ Jγ, where Jγ is a solid torus, glued to M along T in such a way that γ bounds a disk in Jγ.

Recall that a surface in a 3-manifold is called essential if it is properly embedded and either (i) incompressible, not parallel to a subsurface of the boundary of the 3-manifold, and not a 2-sphere, or (ii) a 2-sphere that does not bound a 3-ball.

In [6], Thurston has shown that if M is hyperbolic then M (γ) is also hyperbolic, for all but finitely many slopes γ. A slope γ is said to be exceptional if M (γ) is not hyperbolic. Then the exceptional slopes γ has been the subject of many important works in low dimensional topology. This paper is devoted to a special class of exceptional slopes, those which produce a Klein bottle in the resulting 3-manifold.

Note that a 3-manifold which contains a Klein bottle contains also an essential 2-torus or an essential 2-sphere, or is Seifert fibered, so is not hyperbolic.

In the case where M is the exterior of a hyperbolic knot in S³, Gordon and

Received November 20, 2019; revised July 17, 2020; accepted July 21, 2020.

2010 Mathematics Subject Classification: 57M25.

Key words and phrases: Dehn filling, essential surface, Klein bottle, Scharlemann cycle.

831

Luecke [2] showed that if M (γ) contains a Klein bottle, then ∆(µ, γ) = 1. Ichihara and Teragaito [Corollary 1.3, 4] also showed that ∆(λ, γ)≤ 4g, where g is the genus of the knot. It is then natural to ask if it is possible for an arbitrary hyperbolic link K∪ L, that some Dehn filling on L produces a 3-manifold containing a Klein bottle.

More precisely, let K∪ L be a hyperbolic link in a homology 3-sphere Y such that their linking number lk(K, L)̸= 0; we denote the genus of K in Y by g. Denote its exterior by M and let µ be the meridional slope on the boundary of a tubular neighborhood N (L) of L. Let M_L(γ) be the 3-manifold obtained by γ-Dehn filling along ∂N (L). Then our main result is stated as follows:

Theorem 1.1. If ML(γ) contains a Klein bottle, then

∆(µ, γ)≤ 4 + 6g− 3

|lk(K, L)|.

Remark 1.2. Theorem 1.1 can be generalized if we take a link with more compo- nents instead the knot K.

2. Notations and Definitions

Let M be a compact, connected, orientable and irreducible 3-manifold with a toral boundary component T . Let (Fi, ∂Fi) ⊂ (M, ∂M) (i = 1, 2) be an ori- entable surface, which is possibly compressible or ∂-compressible in M . Suppose that ∂Fi∩ T ̸= ∅ and that components of ∂F1∩ T and of ∂F2∩ T have distinct slopes γ₁ and γ₂ on T . We assume that F₁ and F₂intersect transversely and that

∂F₁ and ∂F₂intersect in the minimal number of points so that each component of

∂F₁ and each of ∂F₂ intersect just ∆(γ₁, γ₂) times in T . We obtain a surface bF_i when we cap off the components of ∂F_i∩ T with disks.

The intersections of F₁and F₂in M give rise to a pair of labeled graphs, G₁⊂ cF₁ and G₂⊂ cF₂. We obtain these graphs as follow: G₁is the graph obtained by taking as fat vertices the disks bF₁− IntF1 and as edges the arc components of F₁∩ F2 in Fb₁. Similarly, G₂ is the graph in bF₂ whose vertices are the disks bF₂− IntF2 and whose edges are the arc components of F₁∩ F2 in bF₂. Note that we neglect circle components of F1∩F2. We number the components of ∂F11, 2, ..., n1in the order in which they appear on T . Similarly, we number the components of ∂F2 1, 2, ..., n2. This gives a numbering of the vertices of G1 and G2. Furthermore, it induces a labelling of the endpoints of edges in G1and G2. On a vertex of G1 (resp. G2) one sees the labels 1 through n2repeated ∆(γ1, γ2) times (resp. 1 through n1 repeated

∆(γ1, γ2) times) appearing in order around the vertex. Two vertices on G1 are parallel if the ordering of the labels on each is clockwise or the ordering on each is anticlockwise, otherwise the vertices are called antiparallel. An edge of G1 is a positive edge if it connects parallel vertices. Otherwise it is a negative edge. The same applies to vertices of G₂. The orientability of F_i (i =1,2) and M gives us the

parity rule: An edge e in Gi is a positive edge if and only if the corresponding edge in G_j is a negative edge.

An edge e is an interior edge if it connects vertices, and a boundary edge oth- erwise. An edge with labels x and y at its endpoints is called an {x, y}-edge. If an edge has a label x at least one endpoint, it is called an x-edge. An x-cycle is a cycle σ of x-edges of G1 such that all the vertices of G1 in σ are parallel and such that the cycle can be oriented so that the tail of each edge has label x. A Scharlemann cycle is an x-cycle that bounds a disk face of G₁(called a Scharlemann disk ). The number of edges in a Scharlemann cycle, σ, is called the length of σ. A pair of adja- cent parallel positive edges{e1, e₂} in G1 is called an S-cycle if it is a Scharlemann cycle of length 2. A Scharlemann cycle is called a trivial loop if its length equals one, and a non-trivial Scharlemann cycle otherwise. Let t_ibe the number of trivial loops in the graph G_i, and b_i the number of boundary edges each of which has an end point in a fat vertex. Since b1= b2, we denote this number by b. Then we have the following inequality from Hayashi and Motegi [Theorem 2.1, 3].

Theorem 2.1. Suppose the graph G_i(i = 1, 2) contains no non-trivial Scharlemann cycles. Then we have:

∆(γ1, γ2)≤ 2 +(t1− χ( bF1))

n₁ +(t2− χ( bF2))

n₂ + b

n₁n₂.

3. Preparatory Lemmas

As in Section 1, let M be the exterior of a hyperbolic link K∪ L in a homology 3-sphere Y , such that the linking number lk(K, L)̸= 0. Let P = Σ ∩ M, where Σ is a Seifert surface of minimal genus g ≥ 1 for the knot K. Now, among all these minimal genus surfaces, we choose P such that p =|∂P ∩ ∂N(L)| is minimal. Then P is incompressible and boundary-incompressible in M . Note that p≥ |lk(K, L)|.

Let ML(γ) be the 3-manifold obtained by γ-Dehn filling along ∂N (L). Suppose that M_L(γ) contains a Klein bottle bS such that s =| bS∩ Jγ| is minimal and s > 1.

Let S = bS∩ M, then S is incompressible and boundary-incompressible in M.

Let bT = ∂N ( bS), where N ( bS) denotes a thin regular neighborhood of bS in M (γ).

Then bT is a 2-torus, intersecting Jγ in t = 2s points. Let T = bT∩ M.

As above, the intersection P ∩ T gives rise to a pair of labeled graphs, GP ⊂ Pb ⊂ ML(µ) and G_T ⊂ bT ⊂ ML(γ). The choice of bP , bS and N ( bS) guarantees that the intersection P ∩ T is essential: no circle component of P ∩ T bounds a disk in P or in T , and no arc component of P∩ T is boundary-parallel in either P or T . In particular neither GPnor GT contains trivial loops. Also, since T∩∂N(K) = ∅, then the graphs GP and GT have no boundary edges. Since the torus bT is separating, then bT divides ML(γ) in a black side, U , and a white side W . Therefore, the faces of GP are divided into black and white faces. Furthermore each black face of GP is a face of length 2, corresponding to the regular neighborhood of an arc component of P∩ S. See [2] (pages 626-628) for more details.

Lemma 3.1. s is odd.

Proof. If s is even, then we can obtain a closed non-orientable surface in M by at- taching suitable annuli in ∂M to S along ∂S, which impossible (since the homology sphere Y does not contain a closed non-orientable surface). 2 Lemma 3.2. GT does not contain a Scharlemann cycle.

Proof. Assume for contradiction that GT contains a Scharlemann cycle. Now by a standard argument of surgery (see for example [Lemma 2.5.2, 1]), we can construct a new surface P^′ of genus g such that ∂0P^′ = ∂0P and |∂P^′∩ ∂N(L)| = p − 2

contradicting the minimality of p. 2

Now let {σj}j=1,...,m denote the set of all Scharlemann cycles in GP and let {Dj}j=1,...,mbe the set of their corresponding Scharlemann disks in GP.

The following Lemma can be proved by using [Lemma 2.1, 5]. However for conve- nience of the readers we give a proof here.

Lemma 3.3. m≥ (∆(µ, γ) − 2)p − 2g + 1.

Proof. Let bP×[−ϵ, ϵ] be a small regular neighborhood of bP , and let bR be the closed surface of genus 2g obtained by taking two copies of bP glued together along a small regular neighborhood C of ∂₀P . bR = ( bP× {−ϵ}) ∪ C ∪ ( bP × {ϵ}). Without loss of generality we can assume that bR is contained in Y -∂N (K) (we push slightly bR in Y -∂N (K)). Let R = bR∩ M, then ∂R is a collection of 2p components, each having the meridional slope µ in ∂N (L). As above, we define the labelled graphs of intersection, GRin bR and GT in bT . We note that GRis made of two disjoint copies of GP.

By hypothesis the graph GP contains only m Scharlemann cycles, so GR con- tains exactly 2m Scharlemann cycles. Recall that for each Scharlemann cycle σj in GP there exists a disk face Dj such that GP∩ intDj =∅. For each j, let Bj a small disk contained in intDj. Now, we consider the annuli Hj in Bj× [−ϵ, ϵ], such that one boundary component of Hj lies in Bj× {−ϵ}, while the other lies in Bj× {ϵ}.

Let H = H₁∪ H2∪ · · · ∪ Hm. Let bF be the resulting surface obtained by deleting Bj’s and then attaching H to bR. It is easy to see that bF is a closed surface of genus 2g + m. As usual, let F = bF∩ M. Let (GF, G_T) be the pair of intersection graphs associated to the pair of surfaces (F, T ). Note that G_F has no Scharlemann cycle.

Now by Hayashi-Motegi’s inequality [Theorem 2.1, 3], applied to the two graphs obtained from F ∩ T we get:

∆(µ, γ)≤ 2 − χ( bF )

2p −χ( bT )

t = 2−2− 2(2g + m) 2p

m≥ (∆(µ, γ) − 2)p − 2g + 1.

Hence, the proof is completed. 2

Lemma 3.4. If σ, τ and κ are Scharlemann cycles in GP, then two of them must have the same pair of labels.

Proof. See [Theorem 6.5, 2]. 2

Lemma 3.5. If t ≥ 4 then GP does not contain a white Scharlemann disk. In particular every Scharlemann disk is a face of length 2.

Proof. See [Theorem 6.4, 2]. 2

4. Proof of Theorem 1.1

Proof of Theorem 1.1. Suppose that GP contains m Scharlemann cycles. By Lemma 3.5, any Scharlemann cycle in GP bounds a black face, hence any Scharlemann cycle σ in GP is a black S-cycle. Let{σj¹}j=1,...,m₁ denote the set of all black S-cycles in G_P having{x, x + 1} as labels, and let {σ²j}j=1,...,m2 be the set of all black S-cycles in G_P having{y, y + 1} as labels such that {x, x + 1} ̸= {y, y + 1}. Since the torus T is separating, then x and y must have the same parity.b

By Lemma 3.4, m1+ m2 = m. Recall that for each S-cycle σ_j¹ (resp. σ²_j) in GP there exists a black disk face D¹_j (resp. a black disk face D²_j) such that GP∩ intD¹_j =∅ (resp. GP∩ intDj²=∅).

Without loss of generality, we assume that m₁ ≥ m

2 and {x, x + 1} = {1, 2}.

Now let Γ be the subgraph of G_P consisting of the vertices of G_P and all the edges of all the black S-cycles in G_P having{1, 2} as labels. Denote by fk the faces of Γ.

Now an Euler characteristic count gives

χ( bP ) = 1− 2g ≤ p − 2m1+∑

f_k

χ(fk) = p− m1+ ∑

f_k̸=D¹j

χ(fk)

Now let F be the number of disk faces of Γ which are not a Scharlemann disk {Dj¹}j=1,...,m₁. Then

1− 2g ≤ p − m1+ ∑

fk̸=Dj¹

χ(fk)≤ p − m1+ F

Lemma 4.1. Γ contains exactly m1 disks faces.

Proof. We argue by contradiction. Suppose that Γ contains a cycle σ bounding a disk face E, such that E is not a black Scharlemann disk. By construction of Γ, all the vertices of σ are parallel and{1, 2} are the labels of σ.

Now, let e be an edge of GP ∩ E. e is called a diagonal edge if e is incident to a vertex v and has other end in a vertex w which is not adjacent to v on σ.

We divide in two cases depending upon whether a diagonal edge e exists or not.

Case 1: There is no diagonal edge in GP ∩ E.

Let e be an edge of GP ∩ E. Therefore, if k is the label at one endpoint of e then 2s + 3− k (mod 2s) is the label at the other endpoint.

In particular, if e is an (s + 1)-edge or an (s + 2)-edge, then e is an{s+1, s+2}- edge. Note that since s > 1, then{s + 1, s + 2} ̸= {1, 2}.

Now, let e and f be two adjacent{s + 1, s + 2}-edges of GP∩ E. If e and f are parallel, then{e, f} is an S-cycle with {s + 1, s + 2} as labels. Since s + 1 is even and{1, 2} are the labels of all the black S-cycles of Γ, then {e, f} is a white S-cycle, which contradicts Lemma 3.5. If not, G_P must contain a white Scharlemann cycle with{s + 1, s + 2} as labels and having the same length as σ, but this is impossible by Lemma 3.5.

Case 2: There is a diagonal edge e in GP∩ E.

Let e be a diagonal edge connecting two vertices v and w of σ. Let i be the label at the end of e at v and j be the label at the end of e at w. The diagonal edge e divides the disk E into two disks E1 and E2. Without loss of generality, we may assume that there is no other diagonal edge in GP∩ E1. Let σ^′ be the cycle bounding E1.

As in Case 1, if two adjacent{s + 1, s + 2}-edges of GP ∩ E1 are parallel, then GP must contain a white S-cycle, which contradicts Lemma 3.5. If not, let u be a vertex of σ^′ adjacent to the vertex v and u̸= w, then only one {s + 1, s + 2}-edge of G_P∩ E1connect u to v, hence i∈ {s + 1, s + 2}. Using the same argument we show that j ∈ {s + 1, s + 2}. Since v and w are parallel, then by the parity rule i must be different to j. Hence{i, j} = {s + 1, s + 2} and e must be an {s + 1, s + 2}-edge.

Then G_P must contain a white Scharlemann cycle with{s + 1, s + 2} as labels and having the same length as σ^′, but this also impossible by Lemma 3.5. 2 By the previous lemma, F = 0. Hence we obtain:

1− 2g ≤ p − m1

Since m₁≥ m

2, this yields

1− 2g ≤ p − m 2 m≤ 2(2g + p − 1) Now by Lemma 3.3, we get

(∆(µ, γ)− 2)p − 2g + 1 ≤ 2(2g + p − 1)

∆(µ, γ)≤ 4 + 6g− 3

p ≤ 4 + 6g− 3

|lk(K, L)|.

Hence, the proof of Theorem 1.1 is completed. 2

Acknowledgements. We would like to thank the editor and reviewers for careful reading, and constructive suggestions for our manuscript. We would like to thank Professor Rebhi Salem for helpful comments.

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