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Virtually fibred Montesinos links of type SL 2 Xiao Guo, Yu Zhang (cid:103) ∗ † Department of Mathematics, SUNY at Buffalo 9 0 0 2 Abstract n a J We find a larger class of virtually fibred classic Montesinos links of type 1 3 SL , extending a result of Agol, Boyer and Zhang. 2 ] T (cid:103) G 1 Introduction . h t a m A 3-manifold is called virtually fibred if it has a finite cover which is a surface bundle [ over the circle. A link in a connected 3-manifold is said to be virtually fibred if 2 its exterior is a virtually fibred 3-manifold. Thurston conjectured that all closed v 2 hyperbolic 3-manifolds and all hyperbolic links in closed 3-manifolds are virtually 3 4 fibred. This conjecture, which has been named as virtually fibred conjecture, is one of 2 . the most fundamental and difficult problems in 3-manifold topology. 4 0 8 Recall that a link K in S3 is called a generalized Montesinos link if the double 0 branched cover W of (S3,K) is a Seifert fibred 3-manifold. Such a link K is further : K v i said to be of type SL if the canonical geometric structure on W is from the SL - X 2 K 2 geometry. When every component of the branched set in W is not a fiber of the r K a (cid:103) (cid:103) Seifert fibration of W , K is called a classic Montesinos link. Recent work of Walsh K [Wa] and work of Agol-Boyer-Zhang [ABZ] combined together solved the virtually fibred conjecture for all generalized Montesinos links in S3 which are not classic Montesinos links of type SL . Agol-Boyer-Zhang [ABZ] also gave an infinite family 2 of virtually fibred classic Montesinos links of type SL . In this paper, we extend the 2 (cid:103) latter result of [ABZ] to a larger family of classic Montesinos links of type SL . Note 2 (cid:103) E-mail: xiaoguo@buffalo.edu ∗ (cid:103) E-mail: yz26@buffalo.edu † 1 q /p tangle q /p tangle q /p tangle 1 1 2 2 n n · · · Figure 1: Classic Montesinos link that every classic Montesinos link has a cyclic rational tangle decomposition of the form (q /p ,q /p ,...,q /p ) with all p 2 as shown in Figure 1. We prove 1 1 2 2 n n i ≥ Theorem 1.1 If a classic Montesinos link K has a cyclic rational tangle decompo- q q q 1 2 n sition of the form ( , , , ) with p 3 odd, then K is virtually fibred. p p ··· p ≥ This theorem is proved in [ABZ] when n is a multiple of p. Our approach follows closely to that of [ABZ]. At a number of places we need to deal with some new issues that arise. We shall describe our ways of dealing with these issues when we reach these places. Let K be a link as given in Theorem 1.1. The base orbifold of the Seifert fibred K B space W is a 2-sphere with n cone points each has order p. Let f : W be K K K → B the Seifert fibration which is invariant under the covering involution τ : W W . K K → The Euler number e(W ) of W and the orbifold Euler characteristic χ( ) of K K K K B B are given by the following formulas: n q n e(W ) = i=1 i, χ( ) = 2 n+ . K K − p B − p (cid:80) Note that W has the SL -geometry precisely when e(W ) = 0 and χ( ) < 0. K 2 K K (cid:54) B Thus K is of type SL precisely when 2 (cid:103) Case (1). n = 3, p 5, and q +q +q = 0, or (cid:103) 1 2 3 ≥ (cid:54) Case (2). n > 3 and q +...+q = 0. 1 n (cid:54) We shall split our proof of Theorem 1.1 into these two cases, given in Section 2 and Section 3 respectively. All main ingredients of the proof will occur already in Case (1); the proof of Case (2) will be a quick generalization. 2 In the remainder of this section, we shall give a few additional notes and no- tations which will be used throughout this paper. From now on we assume that K = K(q /p,...,q /p) is a link satisfying the conditions of Case (1) or (2) listed 1 n above. First note that K is a single component knot if q + ... + q is odd, and a two 1 n component link otherwise. The double branched cover W has the SL -geometry and K 2 its base orbifold is hyperbolic. Let K be the corresponding branched set in W K K B (cid:103) and let K∗ = f(K), where f : W is the Seifert quotient map. By the orbifold K K → B (cid:101) theorem we may assume that the covering involution τ on W is an isometry and K (cid:101) the branch set K = Fix(τ) is a geodesic which is orthogonal to the Seifert fibres by Lemma 2.1 of [ABZ]. Also K∗ is a geodesic in , and is an equator of containing K K B B (cid:101) all the n cone points which we denote by c ,c , ,c (indexed so that they appear 1 2 n ··· consecutively along K∗). The order of each cone point c is p. Since p is odd, the i restriction map of f to each component of K is a double covering onto K∗. (cid:101) 2 Proof of Theorem 1.1 in Case (1). In this section we prove Theorem 1.1 in the case n = 3, p 5 and q +q +q = 0. 1 2 3 ≥ (cid:54) Throughout this section, without further notice, i,j,k as well as numbers expressed by them will be considered as non-negative integers mod p, and i = k will also be (cid:54) assumed. 2.1 K is a knot. Here is an outline of the proof. Step 1 Take a specific p2-fold orbifold cover F of = S2(p,p,p) such that F is a K B smooth surface. There is a corresponding free cover Ψ : Y W of the same degree K → such that the base orbifold of the Seifert space Y is F and such that the following diagram commutes fˆ Y F −−−→ Ψ ψ  f  W  K K (cid:121) −−−→ B(cid:121) ˆ where f is the Seifert quotient map. Note that Y is a locally-trivial circle bundle over 3 F since F is a smooth surface. Let L = Ψ−1(K). Then L has exactly p2 components, which we denote by L : i,j { 1 (cid:54) i,j (cid:54) p . Let L∗ = fˆ(L ). Then L∗ : 1 (cid:54) j (cid:54) p are p mutually disjoint } i,j i,j { j,j } (cid:101) simple closed geodesics in F. p ◦ Step 2 Construct a surface semi-bundle structure on M = Y N(L ), where j,j − j∪=1 N(L ) is a small regular neighborhood of L in Y, 1 (cid:54) j (cid:54) p. j,j j,j Note that M is a graph manifold with non-empty boundary, with vertices Mj and 2 ◦ p ◦ Y , where Mj = fˆ−1(L∗ ) [ (cid:15),(cid:15)] N(L ), 1 (cid:54) j (cid:54) p. Y = M Mj. We 1 2 j,j × − − j,j 1 − j∪=1 2 construct a surface semi-bundle structure on M by using [WY] and following [ABZ]. Step 3 Isotope all L , i = k, such that they are transverse to the surface bundle in i,k (cid:54) Mj, denoted by j. This is one of the key parts of the proof. Let U = fˆ−1(L∗ ), the 2 F2 i,k i,k vertical torus over L∗ . By the construction in Step 2, there are exactly two singular i,k points in the induced foliation by j on every component of U Mj. We show that 2 i,k 2 F ∩ L can be oriented such that the arcs L Mj whose images intersect in F, all travel i,k 2 ∩ from one of fˆ−1(L )∗ (cid:15) and fˆ−1(L )∗ (cid:15) to the other. This property allows j,j j,j ×{− } ×{ } us to arrange these arcs in U Mj such that they always travel from the “ ” side i,k 2 ∩ − of the surface bundle in Mj to the “+” side. 2 ˘ ˘ Step 4 Construct a double cover of M, denoted by M, so that M has a surface ˘ bundle structure. Denote the corresponding double cover of Y as Y,the lift of Y as 1 ˘ ˘ ˘ ˘ Y Y , and the lift of L as L L . Perform certain Dehn twist operations 1,1 1,2 i,k i,k,1 i,k,2 ∪ ∪ ˘ ˘ ˘ on the surface bundles of Y Y so that the L are transverse to the new surface 1,1 1,2 i,k,s ∪ bundles, s = 1,2. This is another key part of the proof. We perform these Dehn twist operations ˘ ˘ along a union of vertical tori, Γ Y Y . Γ contains some boundary parallel tori 1,1 1,2 ⊂ ∪ ˘ ˘ of Y Y as in [ABZ], different form [ABZ], these boundary parallel tori intersect 1,1 1,2 ∪ ˘ ˘ some arcs of L Y two times with different direction or do not intersect at all, i,k,s 1,s ∩ ˘ ˘ s = 1,2. We call such arcs of L Y “bad” arcs. We construct four specific extra i,k,s 1,s ∩ tori as members of Γ to deal with the “bad” arcs, s = 1,2. ˘ By Steps 1-4, the exterior of the inverse image of K in Y has a surface bundle structure. It is a free cover of the exterior of K in W , which in turn is a free double K (cid:101) cover of the exterior of K in S3. Thus K is virtually fibred in S3. (cid:101) Now we fill in the details. 4 cˆ L∗1 L∗3 L∗5 1L∗2 L∗4 c c c 3,4 3,1 3,5 c3,2 c3,3 cˆ 2 Figure 2: Covering space F(cid:48) Step 1 Insteadoftakingthep-foldcycliccoverof asin[ABZ], weconstructthep2-fold K B cover F of = S2(p,p,p), by a composition of two p-fold cyclic covers. K B Let Γ = π (S2(p,p,p)) be the orbifold fundametal group of S2(p,p,p). It has a 1 1 presentation Γ =< x ,x ,x xp = xp = xp = x x x = 1 > . 1 1 2 3 1 2 3 1 2 3 | where x is represented by a small circular loop in S2(p,p,p) centered at c ,r = 1,2,3. r r Let ψ : F(cid:48) S2(p,p,p) be the p-fold cyclic orbifold cover of S2(p,p,p) corresponding 1 → to the homomorphism: h : Γ Z/p 1 1 → where h (x ) = 1,h (x ) = 1, and h (x ) = 0. F(cid:48) is as shown in Figure 2 (while 1 1 1 2 1 3 − p = 5). The order of h (x ) and h (x ) are both p, so ψ−1(c ) = cˆ and ψ−1(c ) = cˆ 1 1 1 2 1 1 1 1 2 2 are two points in F(cid:48) (not cone points). The order of h (x ) is 0, so ψ−1(c ) = 1 3 1 3 c ,c , ,c are the only cone points in F(cid:48), each has order p. We have F(cid:48) = 3,1 3,2 3,p { ··· } S2(p,p, ,p). ψ−1(K∗) is a set of p geodesics L∗,L∗, ,L∗, such that L∗ goes ··· 1 1 2 ··· p i p through the cone point c (cf. Figure 2). 3,i (cid:124) (cid:123)(cid:122) (cid:125) Let τ∗ be the deck transformation of ψ corresponding to ¯1 Z . Fix(τ∗) = 1 1 ∈ p 1 5 cˆ ,cˆ . We may assume that τ∗(L∗) = L∗ . Orient L∗ and give L∗ the induced { 1 2} 1 i i+1 1 i orientation, 1 < i (cid:54) p. F(cid:48) admits an orientation such that τ∗ is a counterclockwise 1 rotationby2π/pnearcˆ andaclockwiserotationby2π/pnearcˆ onF(cid:48),sinceh (x ) = 1 2 1 1 ¯1 and h (x ) = 1. Note that L∗ intersects L∗ at cˆ in an angle of 2π(i j)/p, 1 2 − i j 1 − also intersects L∗ at cˆ in an angle of 2π(j i)/p. (We always suppose that the j 2 − counterclockwise direction is the positive direction of angles.) Let Γ be the orbifold fundamental group of F(cid:48). It has the presentation 2 Γ = π (S2(p,p, ,p)) =< y ,y , ,y yp = yp = = yp = y y y = 1 >, 2 1 ··· 1 2 ··· p | 1 2 ··· p 1 2··· p p where y is re(cid:124)pre(cid:123)s(cid:122)ente(cid:125)d by a small circular loop in S2(p,p, ,p) centered at c . j 3,j ··· p Let ψ : F F(cid:48) be the p-fold cyclic orbifold cover of F(cid:48) corresponding to the 2 → (cid:124) (cid:123)(cid:122) (cid:125) homomorphism: h : Γ Z/p 2 2 → where h (y ) = 1, 1 (cid:54) j (cid:54) p. 2 j The order of h (y ) is p, so ψ−1(c ) = cˆ is a point, 1 (cid:54) j (cid:54) p. Hence F is a 2 j 2 3,j 3,j smooth closed orientable surface without cone points. For each r = 1,2, ψ−1(cˆ ) is a 2 r set of p points, which we denote by cˆ ,cˆ , ,cˆ . For each i = 1,...,p, ψ−1(L∗) is r,1 r,2 ··· r,p 2 i a set of p simple closed geodesics, which we denote by L∗ ,L∗ , ,L∗ . i,1 i,2 ··· i,p Denote the deck transformation of ψ corresponding to ¯1 Z/p by τ∗. Fix(τ∗) = 2 ∈ 2 2 cˆ ,cˆ , ,cˆ . We may assume that τ∗(L∗ ) = L∗ , τ∗(cˆ ) = cˆ , and { 3,1 3,2 ··· 3,p} 2 i,j i,j+1 2 1,j 1,j+1 τ∗(cˆ ) = cˆ . Then ψ−1(L∗) = L∗ : 1 (cid:54) j (cid:54) p . Now we fix an orientation for 2 2,j 2,j+1 2 i { i,j } each of L∗ ,L∗ , ,L∗ such that L∗ goes through cˆ , cˆ and cˆ in order, and 1,1 2,2 ··· p,p j,j 3,j 2,j 1,j give L∗ = (τ∗)j−i(L∗ ) the induced orientation. F admits an orientation such that τ∗ i,j 2 i,i 2 is a counterclockwise rotation by 2π/p near the fixed points cˆ , so L∗ ,L∗ , ,L∗ 3,i i,1 i,2 ··· i,p intersect at cˆ , and L∗ intersects L∗ at cˆ in an angle of 2π(j i)/p, 1 (cid:54) i,j (cid:54) p. 3,i i,j i,i 3,i − Since cˆ = ψ (cˆ ) is not a cone point on F(cid:48), a small regular neighborhood of cˆ r 2 r,j r,j on F is a copy of a small regular neighborhood of cˆ on F(cid:48), r = 1,2. Recall that L∗ r i intersects L∗ at cˆ in an angle of 2π(i j)/p, and at cˆ in an angle of 2π(j i)/p. j 1 − 2 − Then L∗ intersects L∗ at cˆ in an angle of 2π(i j)/p, and at cˆ in an angle of i,j j,j 1,j − 2,j 2π(j i)/p. See the schematic picture, Figure 3, for example p = 5. Here we didn’t − depict the genus of the surface, so some curves meet in the picture actually do not meet. For convenience, we only draw a part of L∗ in Figure 3, 1 (cid:54) j (cid:54) p. j,j Summarizing the above discussion, we have the following remark. 6 L L∗5,1 L∗2,1 L∗1,5 ∗1,1 L4∗,1 cˆ2,1 L∗3,1 L∗1,4 β21 cˆ1,1 L∗3,1 L∗4,1 cˆ3,1 L∗1,3 β11 L∗2,1 L∗5,1 L∗1,2 L∗2,2 L∗1,2 L∗5,2 cˆ2,2 L∗3,2 L∗4,2 L∗2,1 L∗2,5 β22 cˆ1,2 L∗4,2 L∗5,2 cˆ3,2 L∗2,4 β12 L∗3,2 L∗1,2 L∗2,3 L L∗2,3 L∗4,3 L∗3,2 ∗3,3 L∗1,3 cˆ2,3 L∗5,3 L∗3,1 β23 cˆ1,3 L∗5,3 L1∗,3 cˆ3,3 L∗3,5 β13 L∗4,3 L∗2,3 L∗3,4 L∗4,4 L∗3,4 L∗2,4 cˆ2,4 L∗5,4 L∗1,4 L∗4,3 L∗4,2 β24 cˆ1,4 L∗1,4 L∗2,4 cˆ3,4 L∗4,1 β14 L∗5,4 L∗3,4 L∗4,5 L L∗4,5 L∗1,5 L∗5,4 ∗5,5 L∗3,5 cˆ2,5 L∗2,5 L∗5,3 β25 cˆ1,5 L∗2,5 L∗3,5 cˆ3,5 L∗5,2 β15 L∗1,5 L∗4,5 L∗5,1 Figure 3: L∗ ,1 (cid:54) i,j (cid:54) p with p = 5 i,j 7 Remark 2.1 L∗ : 1 (cid:54) i,j (cid:54) p only intersects at the points cˆ ,r = 1,2,3,1 (cid:54) { i,j } { r,j j (cid:54) p on F, and L∗ goes through cˆ , cˆ and cˆ in order following the given } i,j 3,i 2,j 1,j orientation of L∗ , 1 (cid:54) i,j (cid:54) p. In particular L∗ goes through cˆ ,cˆ , and cˆ , so i,j j,j 3,j 2,j 1,j L∗ ,1 (cid:54) j (cid:54) p are mutually disjoint, and { j,j } cˆ in an angle of 2π(i j)/p, L∗ intersects L∗ at 1,j − i,j j,j  cˆ in an angle of 2π(i j)/p,  2,j − − L∗ intersects L∗ at cˆ in an angle of 2π(j k)/p, j,k j,j 3,j − − Note that i,j,k,i j,k j are considered as integers mod p. − − Let ψ = ψ ψ : F ψ2 S2(p,p, ,p) ψ1 S2(p,p,p), which is a p2-fold orbifold 1 2 ◦ −→ ··· −→ p covering. We have ψ−1(c ) = cˆ ,cˆ , ,cˆ , r = 1,2,3, and L∗ = ψ−1(K∗) = r r,1 r,2 r,p {(cid:124) (cid:123)(cid:122) ·(cid:125)·· } L∗ : 1 (cid:54) i,j (cid:54) p . { i,j } Let Ψ : Y(cid:48) W be the p-fold cover of W corresponding to ψ , Ψ : Y Y(cid:48) 1 K K 1 2 → → the p-fold cover of Y(cid:48) corresponding to ψ , and Ψ = Ψ Ψ . Then Ψ : Y W is 2 1 2 K ◦ → a p2-fold cover of W corresponding to ψ. Y has locally-trivial circle bundle Seifert K structure with base surface F. Set L = Ψ−1(K), which is a geodesic link in Y. We have the following commutative diagram: (cid:101) L L ∗ ˆ f Y F Ψ ψ f W K K B K K ∗ Note that L has exactl(cid:127)y p2 components. Let L = L : 1 (cid:54) i,j (cid:54) p , where i,j { } L = fˆ−1(L∗ ),1 (cid:54) i,j (cid:54) p. fˆ : L L∗ is a 2-fold cover. i,j i,j | i,j → i,j Step 2 By Remark 2.1, L∗ ,1 (cid:54) j (cid:54) p are mutually disjoint. Let Fj = L∗ [ (cid:15),(cid:15)], { j,j } 2 j,j × − for some small positive number (cid:15), be a neighborhood of L∗ in F, such that Fj’s are j,j 2 8 (L1 )∗ tail head (L3 )∗ tail head (L5 )∗ tail head i,k i,k i,k i k (cid:54) p−1 βk βi i k (cid:54) p−1 βk βk i k (cid:54) p−1 βi βk − 2 1 1 − 2 2 2 − 2 2 1 i k > p−1 βk βi i k > p−1 βk βk i k > p−1 βi βk − 2 2 2 − 2 1 1 − 2 1 2 Note: i k is considered as a nonnegative integer mod p − Table 1: The induced orientation on (L2l−1)∗,1 (cid:54) l (cid:54) 3. i,k p mutually disjoint. Let βj = L∗ (cid:15) , βj = L∗ (cid:15) , and F = F ( Fj). We 1 j,j ×{− } 2 j,j ×{ } 1 − j∪=1 2 may suppose that βj is on the left side of L∗ . By Remark 2.1, L∗ goes through cˆ , 1 j,j i,k 1.k cˆ and cˆ , so L∗ is separated into 2n = 6 arcs by F ,Fk, and Fi. We denote the six 2,k 3,i i,k 1 2 2 arcs of L∗ by (L1 )∗, (L2 )∗, , (L6 )∗, so that (L2 )∗ (L4 )∗ Fk, (L6 )∗ Fi, i,k i,k i,k ··· i,k i,k ∪ i,k ⊂ 2 i,k ⊂ 2 (L1 )∗ (L3 )∗ (L5 )∗ F , and cˆ (L2 )∗, cˆ (L4 )∗, and cˆ (L6 )∗. We i,k ∪ i,k ∪ i,k ⊂ 1 1,k ∈ i,k 2,k ∈ i,k 3,i ∈ i,k also require that as we travel along L∗ in its orientation, we will pass (L6 )∗, (L5 )∗, i,k i,k i,k , (L1 )∗ consecutively. From the construction, we have Remark 2.2 and Table 1. ··· i,k (L1 )∗ is the part of L∗ between cˆ and cˆ inside F ; i,k i,k 3,i 1,k 1 Remark 2.2 (L3 )∗ is the part of L∗ between cˆ and cˆ inside F ;  i,k i,k 1,k 2,k 1   (L5 )∗ is the part of L∗ between cˆ and cˆ inside F . i,k i,k 2,k 3,i 1    We need Table 1 in the following lemma and in Step 3. Lemma 2.3 F is connected. 1 Proof: It suffices to prove that the boundary components of F , which is the set 1 βi,βj,1 (cid:54) i,j (cid:54) p , can be mutually connected to each other by arcs in F . { 1 2 } 1 FromTable1,thearc(L5 )∗ connectsβj+1 andβj,andthearc(L1 )∗ connects j+1,j 2 1 j+1,j p 1 βj and βj+1, since j + 1 j = 1 (cid:54) − , where p 3. So βj+1 and βj+1 can be 1 1 − 2 ≥ 2 1 connected in F , 1 (cid:54) j +1 (cid:54) p. Also βj+1 and βj are connected by the arc (L5 )∗. 1 2 1 j+1,j Hence βi,βj,1 (cid:54) i,j (cid:54) p can be mutually connected to each other in F . (cid:3) { 1 2 } 1 Let Tj = fˆ−1(L∗ ) Y, which is a vertical torus over L∗ . Then L and L are j,j ⊂ j,j i,j j,i transverse to Tj for all 1 (cid:54) i (cid:54) p and i = j. The situation near L∗ in F is described (cid:54) j,j in Figure 4, while p = 5. 9 cˆ 1,j L∗j+3,j L∗j+1,j cˆ 3,j L∗j+4,j L∗j+2,j L∗j,j+3 L∗j,j L∗j,j+1 cˆ2,j L∗j,j+4 L∗j+2,j L∗j,j+2 L∗j+4,j L∗j+1,j L∗j,j L∗j+3,j L∗j,j×{−ǫ} L∗j,j×{ǫ} Figure 4: A neighborhood of L∗ in F. j,j Since L∗ is a geodesic, the torus Tj is a totally geodesic torus which inherits a j,j EuclideanstructurefromtheSL structureonY. Foranytwosimpleclosedgeodesics 2 a ,b Tj with H (Tj) =< a ,b >, Tj can be identified to S1 S1 where each j j 1 j j { } ⊂ × (cid:103) S1 is a geodesic isotopic to a and each S1 is a geodesic isotopic to j × {∗} {∗} × b ,1 (cid:54) j (cid:54) p. j Similar to Proposition 6.1 in [ABZ], we have the following proposition. Proposition 2.4 The exterior of L L L in Y is a surface semi- 1,1 2,2 p,p { ∪ ∪ ··· ∪ } bundle. Proof: To prove this, we apply [WY] and compare [ABZ]. Let Yj be the submanifold of Y lying over Fj = L∗ [ (cid:15),(cid:15)], and Y be the 2 2 j,j × − 1 submanifold of Y lying over F . Note that Y is connected since F is connected by 1 1 1 Lemma 2.3. Define Tj = fˆ−1(βj), Tj = fˆ−1(βj). 1 1 2 2 Let Y be the 3-manifold obtained by cutting Y open along Tj, and F the surface 0 2 0 obtained by cutting F open along all βj, 1 (cid:54) j (cid:54) p. The restriction of the Seifert 2 10