Taut foliations, left-orderability, and cyclic branched covers
TTAUT FOLIATIONS, LEFT-ORDERABILITY, AND CYCLIC BRANCHEDCOVERS
CAMERON GORDON AND TYE LIDMAN
Abstract.
We study the question of when cyclic branched covers of knots admit taut foliations,have left-orderable fundamental groups, and are not L-spaces.
Keywords: taut foliation, left-orderability, cyclic branched cover
MSC 2010 Classification: Introduction
The study of taut foliations has led to a number of important advances in knot theory and three-manifold topology. Consequently, any connection between taut foliations and other invariants inlow-dimensional topology can be quite useful. One example of this is that taut foliations guaranteethe non-triviality of certain analytically-defined invariants associated to three-manifolds, namelyvarious types of Floer homology. This non-triviality plays a key role in the proofs of Property P[KM04] and the Dehn surgery characterization of the unknot [KMOS07]. One such non-trivialitystatement is that if a rational homology sphere Y admits a co-orientable taut foliation, then Y isnot an L-space (i.e. rank (cid:100) HF ( Y ) > | H ( Y ; Z ) | , where (cid:100) HF denotes the hat-flavor of Heegaard Floerhomology) [OS04, Theorem 1.4]. Consequently, L-spaces form an interesting family of manifoldsthat have been intensely studied. Some useful examples of L-spaces to keep in mind for this paperare manifolds with finite fundamental group [OS05a, Proposition 2.3].Taut foliations are also related to the fundamental group of the underlying three-manifold. Forinstance, any loop transverse to the foliation must represent a non-trivial element in π . Morestructure can often be found. Thurston’s universal circle construction shows that for a co-orientabletaut foliation with hyperbolic leaves, there is an orientation-preserving action of the fundamentalgroup on a circle; for more details see [CD03]. Furthermore, in many cases, one can actually findan orientation-preserving action on the real line, such as when the manifold is an integer homologysphere [BB15, Lemma 0.4] or the foliation is R -covered [CD03, Corollary 7.11]. Since Homeo + ( R )is left-orderable (i.e. admits a left-invariant strict total order), [BRW05, Theorem 1.1] shows thatadmitting such an action implies that π ( Y ) is left-orderable.Thus, there is a strong relationship between left-orderability, taut foliations, and L-spaces. Infact, there are at present no counterexamples known to the possibility that, for an irreduciblerational homology sphere Y , the following three conditions are equivalent: π ( Y ) is left-orderable,(1.1) Y admits a co-orientable taut foliation,(1.2) Y is not an L-space.(1.3)The equivalence of (1.1) and (1.3) was explicitly conjectured in [BGW13]. The equivalence of(1.1), (1.2), and (1.3), has been established for many families of three-manifolds, including Seifertmanifolds [BGW13, BRW05, LS07], Sol manifolds [BGW13], and graph manifold homology spheres[BB15]. Also, it is shown in [BC17] that (1.1) and (1.2) are equivalent for graph manifolds. However,the only relation between (1.1), (1.2), and (1.3) that is unconditionally known is the implication a r X i v : . [ m a t h . G T ] M a y CAMERON GORDON AND TYE LIDMAN mentioned above: if Y is an L-space, then Y does not admit a co-orientable taut foliation [OS04,KR15, Bow16].One reason why the possible equivalence of (1.1), (1.2), and (1.3) is desirable is that they exhibitdifferent types of behavior. For example, left-orderability is well-understood under non-zero degreemaps, while Floer homology is well-understood under Dehn surgery. The equivalence of (1.1), (1.2),and (1.3) would allow one to use these properties in tandem. We mention that it would follow fromthe equivalence of (1.1) and (1.3) that a toroidal integer homology sphere can never be an L-space(see [CLW13, Section 4]). Definition 1.1.
A closed, connected, orientable three-manifold Y is excellent if Y admits a co-orientable taut foliation and π ( Y ) is left-orderable. Consequently, Y is prime and is not an L-space.On the other hand, Y is called a total L-space if all three conditions (1.1) , (1.2) , and (1.3) fail. Inother words, a total L-space is an L-space whose fundamental group is not left-orderable. We remark that if b ( Y ) > Y is prime, then Y always admits a co-orientable taut foliation[Gab83] and has left-orderable fundamental group [BRW05, Corollary 3.4], and thus is excellent.Many examples of total L-spaces have come from branched covers of knots, including all two-foldbranched covers of non-split alternating links [BGW13, OS05b] and some higher-order branchedcovers of two-bridge knots [DPT05, Pet09]. In this paper, we study when cyclic branched covers ofcertain other families of knots give excellent manifolds or total L-spaces. We will denote the n -foldcyclic branched cover of a knot K by Σ n ( K ) and will always assume n ≥
2. Note that by the Smithconjecture [MB84], Σ n ( K ) is simply-connected only if K is trivial.A key input throughout the paper is that, as mentioned above, the equivalence of (1.1), (1.2),and (1.3) is known for Seifert fibered manifolds. This enables us to give a complete analysis of thecase of torus knots. Throughout, we let T p,q denote the ( p, q )-torus link. Theorem 1.2.
Let n, p, q ≥ and suppose p, q are relatively prime. Then, Σ n ( T p,q ) is excellent ifand only if its fundamental group is infinite. Thus, Σ n ( T p,q ) is excellent except in the cases:(i) { p, q } = { , } , ≤ n ≤ ,(ii) { p, q } = { , } , ≤ n ≤ ,(iii) { p, q } = { , r } , r ≥ , n = 2 ,(iv) { p, q } = { , } , n = 2 ,(v) { p, q } = { , } , n = 2 ,in which case Σ n ( T p,q ) is a total L-space. Both left-orderability and taut foliations have been studied in the context of gluing manifoldswith toral boundary (see for instance [BC17, CLW13]). For this reason, we will also study certainfamilies of satellite knots. While bordered Floer homology is suitable machinery for studying theHeegaard Floer homology of manifolds glued along surfaces [LOT08], we will not focus on its usehere. Let K (cid:48) be a satellite knot with non-trivial companion K ; then the complement of K (cid:48) containsa copy of the exterior X of K . In Σ n ( K (cid:48) ), the preimage of X consists of copies of some (possiblytrivial) cyclic cover of X , and the proof of [GL84, Theorem 1] shows that if K (cid:48) is prime, then theboundary components of this preimage are incompressible in Σ n ( K (cid:48) ).Work of Li and Roberts [LR14] provides abundant examples of taut foliations on the exteriors ofnon-trivial knots in S (see Section 2 for more precise statements). We will utilize these foliationsheavily. In the case of cables, the remaining piece of Σ n ( K (cid:48) ) is Seifert fibered, and so we are ableto get a nearly complete result in this case. Let C p,q ( K ) denote the ( p, q )-cable of a knot K , where q denotes the longitudinal winding. Assume q > Theorem 1.3.
Let K be a non-trivial knot in S . Then, Σ n ( C p,q ( K )) is excellent, except possiblyif n = q = 2 . It turns out that in the case of n = q = 2, we are still able to partially extend Theorem 1.3. AUT FOLIATIONS, LEFT-ORDERABILITY, AND CYCLIC BRANCHED COVERS 3
Theorem 1.4.
For any non-trivial knot in S , we have that Σ ( C p, ( K )) is not an L-space. Fur-ther, if K is a torus knot or iterated torus knot, then Σ ( C p, ( K )) is excellent.Remark . In the published version of this article [GL14] and previous arXiv version, it wasclaimed that Σ ( C p, ( K )) was not excellent if K was the right-handed trefoil and p >
1. Theargument there was incorrect. See Remark 4.6 below for a more detailed explanation of the mistake.It is still unknown to the authors if Σ ( C p, ( K )) is excellent for arbitrary K . Any counterexamplewould also provide a counterexample to the conjectural equivalence of (1.1), (1.2), and (1.3).L-spaces seem to be rare among integer homology spheres: it is conjectured that the only ir-reducible integer homology sphere L-spaces are S and the Poincar´e homology sphere. This, andthe possibility that (1.3) implies (1.1), suggests that most integer homology spheres should haveleft-orderable fundamental group. Now, if a knot has trivial Alexander polynomial, then all itscyclic branched coverings are integer homology spheres. A well-known class of knots with trivialAlexander polynomial are the Whitehead doubles. For the cyclic branched coverings of these knots,we show the following. Theorem 1.6.
Let K be a non-trivial knot and let W h ( K ) denote the positive, untwisted Whiteheaddouble of K . Then, Σ ( W h ( K )) is an excellent manifold and π (Σ n ( W h ( K ))) is left-orderable forall even n . If π ( S α ( K )) is left-orderable for α = 1 , , or , then π (Σ n ( W h ( K ))) is left-orderablefor all n ≥ . In light of Theorems 1.3 and 1.6, we make the following conjecture.
Conjecture 1.7.
Let K be a prime, satellite knot. For all n (cid:29) , Σ n ( K ) is excellent. The condition that K be prime is necessary, as Σ n ( K ) is prime (in fact irreducible) if and onlyif K is prime; this follows from the equivariant sphere theorem [MSY82] and the Smith conjecture[MB84]. If K has a summand which is the figure-eight knot, then no cyclic branched cover of K will have left-orderable fundamental group [DPT05, Theorem 2(c)].As mentioned, there have been a number of results determining when cyclic branched covers oftwo-bridge knots have non-left-orderable fundamental group and/or are L-spaces [DPT05, Hu15,Pet09, Tra15]. Let K [ p ,...,p m ] denote the two-bridge knot of the type ab , where [ p , . . . , p m ] is thecontinued fraction expansion for ab . We follow the convention that ab = p + p + ... + 1 pm . We givethe first examples of cyclic branched covers of hyperbolic two-bridge knots which are excellentmanifolds. Theorem 1.8.
Suppose k, (cid:96) ≥ and n divides (2 k + 1) for n > . Then Σ n ( K [2(2 k +1) , (cid:96) +1] ) isexcellent. Our next result gives a family of cyclic branched covers of two-bridge knots which are total L-spaces. If a i ≥
1, 1 ≤ i ≤ m , then for any n ≥
2, the n -fold cyclic branched cover of the two-bridgeknot K [2 a , a ,..., a m ] is the two-fold branched cover of an alternating knot or link [MV01] and istherefore a total L-space; see [Ter14], also [DPT05, Pet09]. The simplest case where this positivitycondition does not hold is the family of two-bridge knots K [2 (cid:96), − k ] , (cid:96), k ≥
1. For these, it is knownthat Σ ( K [2 (cid:96), − k ] ) is an L-space [Pet09] and has non-left-orderable fundamental group [DPT05] andis therefore a total L-space. Recently, Teragaito has shown that Σ ( K [2 (cid:96), − k ] ) is an L-space [Ter14].We show that π (Σ ( K [2 (cid:96), − k ] )) is not left-orderable, so we have the following. Theorem 1.9.
For k, (cid:96) ≥ , Σ ( K [2 (cid:96), − k ] ) is a total L-space. Finally, Teragaito [Ter15] has also shown that the three-fold cyclic branched cover of a three-strand pretzel knot of the form P (2 k + 1 , (cid:96) + 1 , m + 1), where k, (cid:96), m ≥
1, is an L -space. We showthat its fundamental group is not left-orderable. We thus establish the following. CAMERON GORDON AND TYE LIDMAN
Theorem 1.10.
For k, (cid:96), m ≥ , Σ ( P (2 k + 1 , (cid:96) + 1 , m + 1)) is a total L-space. With this, we can improve [Ter15, Corollary 1.2] to the following.
Corollary 1.11.
Let K be a genus one, alternating knot. Then Σ ( K ) is a total L-space.Proof. We repeat the argument of [Ter15, Corollary 1.2], where it is shown that Σ ( K ) is an L-space. We are interested in showing π (Σ ( K )) is not left-orderable. If K is a genus one, alternatingknot, then K is either a genus one, two-bridge bridge knot or, up to mirroring, a pretzel knot P (2 k + 1 , (cid:96) + 1 , m + 1) with k, (cid:96), m ≥ (cid:3) Organization:
In Section 2, we review the relevant definitions and results that we will invokethroughout the paper. In Section 3 we prove Theorem 1.2. In Section 4 we prove Theorems 1.3and 1.4. In Sections 5, 6, 7, and 8 we prove Theorems 1.6, 1.8, 1.9, and 1.10 respectively. Finally,we briefly discuss two other families of cyclic branched covers in Section 9.
Acknowledgments:
We would like to thank Steve Boyer, Nathan Dunfield, Adam Levine, andRachel Roberts for helpful discussions. We are particularly grateful to Nathan Dunfield for provid-ing the information described in Remark 6.3(2). The first author was partially supported by NSFGrant DMS-1309021. The second author was partially supported by NSF Grant DMS-0636643.2.
Background
In this section, we will collect results from the literature and other technical lemmas which willbe used throughout the paper. All three-manifolds are assumed to be connected, compact, andorientable unless specified otherwise. For a knot K in S , we use the notation − K to denote themirror of K .2.1. Taut foliations and left-orders.
We begin with the relevant background on taut foliationsand left-orders. We will discuss these simultaneously so we can draw numerous parallels.A taut foliation F on a three-manifold is a foliation by surfaces such that for each leaf F ∈ F ,there exists a curve γ which is transverse to F and intersects the leaf F . Manifolds with tautfoliations are prime and have infinite fundamental group (see, for instance, [Cal07]).A group G is left-orderable if there exists a left-invariant, strict total order on G . We use theconvention that the trivial group is not left-orderable. Examples of left-orderable groups include Z , braid groups [Deh94], and Homeo + ( R ), while any group with torsion (e.g. a finite group) is notleft-orderable. One reason why the orderability of three-manifold groups is particularly well-suitedfor study is the following: Theorem 2.1 (Boyer-Rolfsen-Wiest, Theorem 1.1 in [BRW05]) . Let Y be a compact, connected,irreducible, P -irreducible three-manifold. Then, if there exists a non-zero homomorphism from π ( Y ) to a left-orderable group, then π ( Y ) is left-orderable. In particular, if there exists a non-zero degree map from Y to Y (cid:48) and π ( Y (cid:48) ) is left-orderable, then so is π ( Y ) . Note that for closed, orientable manifolds other than R P , irreducibility implies P -irreducibility.Since there are no non-trivial homomorphisms from π ( R P ) to a left-orderable group, this casewill not be a concern. Further, observe that π ( S × S ) is left-orderable, so we may further replaceirreducible with prime. Theorem 2.1 implies that a prime three-manifold with b > b > excellent if it admits a co-orientable taut foliation and has left-orderable fundamental group. Anyprime three-manifold with b > AUT FOLIATIONS, LEFT-ORDERABILITY, AND CYCLIC BRANCHED COVERS 5
Remark . As we will repeatedly invoke Theorem 2.1, we want to ensure that the cyclic branchedcovers we work with are prime. All knots that we will take cyclic branched covers of in thispaper (torus knots, cables, Whitehead doubles, two-bridge knots, and pretzel knots) will be prime.Therefore, all such cyclic branched covers will be prime; we will not mention this point again.We also would like notions of left-orders and taut foliations with appropriate boundary conditions.We will only consider rational slopes on boundary tori in this paper.
Definition 2.3.
Let M be a compact three-manifold with boundary a disjoint union of tori, T , . . . , T n ,for some n ≥ . Let α i be a slope on T i for each ≤ i ≤ n . The multislope α = ( α , . . . , α n ) is called a CTF multislope if M has a co-orientable taut foliation which meets T i transversely incircles of slope α i for ≤ i ≤ n . If M is the exterior of a knot K in S , we say that α is a CTFslope for K if α is a CTF slope for M . Observe that if α is a CTF multislope on M , the manifold M ( α ) admits a co-orientable tautfoliation. Recently, Li and Roberts have shown that sufficiently small slopes on knots in S arealways CTF slopes. Theorem 2.4 (Li-Roberts, Theorem 1.1 in [LR14]) . Let K be a non-trivial knot in S . Then,there exists an interval ( − a, b ) with a, b > such that if α ∈ ( − a, b ) then α is a CTF slope for K . In fact, they conjecture the existence of a universal interval for all knots.
Conjecture 2.5 (Li-Roberts, Conjecture 1.9 in [LR14]) . Let K be a non-trivial knot in S . If α ∈ ( − , , then α is a CTF slope for K . We point out that the Li-Roberts Conjecture is known for many families of knots, includinghyperbolic fibered knots [Rob01] and non-special alternating knots [Rob95].In analogy with CTF multislopes, we have the following definition for left-orderability, andconsequently excellence.
Definition 2.6.
We say that α is an LO multislope if π ( M ( α )) is left-orderable. We say that α is excellent if it is both an LO multislope and a CTF multislope. If M is the exterior of a knot K in S , we will say that α is an LO slope for K (respectively excellent slope for K ) if it is an LOslope (respectively excellent slope) for M . Observe that if α is an excellent multislope, the manifold M ( α ) is excellent.There is a strong relationship between taut foliations and left-orderability for homology spheres.Note that on an integer homology sphere, any foliation is automatically co-orientable. In [CD03],it is shown that an atoroidal homology sphere with a taut foliation has left-orderable fundamentalgroup. This was then extended to all integer homology spheres. Lemma 2.7 (Boileau-Boyer, Lemma 0.4 in [BB15]) . Suppose that Y is an integer homology sphereadmitting a taut foliation. Then, π ( Y ) is left-orderable. Therefore, if an integer homology sphere Y admits a taut foliation, Y is automatically excellent.Using this observation and Theorem 2.4, we can construct numerous excellent slopes on exteriorsof knots in S . Lemma 2.8.
Let K be a non-trivial knot in S . Then, there exists k such that if | k | > k , k isan excellent slope for K .Proof. By Theorem 2.4, there exists k such that if | k | > k , then k is a CTF slope for K .Since H ( S k ( K )) = 0, Lemma 2.7 implies that k is an LO slope for K . Thus, k is excellent for | k | > k . (cid:3) CAMERON GORDON AND TYE LIDMAN
Toroidal manifolds.
We are interested in the more general question of when we can gluepieces with toral boundary together to obtain manifolds with taut foliations and/or left-orderablefundamental group. Let M be a compact three-manifold with boundary a disjoint union of tori, S , . . . , S m , for m ≥
0. Further, let J be a disjoint union of incompressible, separating tori T , . . . , T n in M which are not boundary parallel. Let X , . . . , X n +1 be the components of M cutalong J . Choosing a slope on each component of ∂M ∪ J defines a multislope α i on the boundaryof X i for each 1 ≤ i ≤ n + 1. Lemma 2.9.
Let M be as above. If α i is a CTF multislope for each X i , then α | ∂M is a CTFmultislope for M . If in addition, α i is an LO multislope for each X i , then α | ∂M is an LO multislopefor M . Consequently, if α i is excellent for each X i , then α | ∂M is excellent for M . Lemma 2.9 can be rephrased in a simple way for the case that M is closed. In this case, it saysthat if one glues manifolds with toral boundary such that CTF multislopes (respectively excellentmultislopes) are identified, the resulting manifold still has a co-orientable taut foliation (respectivelyis excellent). The first claim in Lemma 2.9 is in fact trivial. One can simply glue the foliations oneach X i together. We are thus interested in showing that α | ∂M is an LO slope on M . This will bea generalization of the following theorem. Theorem 2.10 (Clay-Lidman-Watson, Theorem 2.7 in [CLW13]) . Let X and X be compact,oriented, connected three-manifolds with incompressible torus boundary. Suppose α and α are LOslopes on X and X respectively, and f : ∂X → ∂X is an orientation-reversing homeomorphismsuch that f ( α ) = α . If Y = X ∪ f X is irreducible, then Y has left-orderable fundamental group. The argument we use to prove Lemma 2.9 will essentially repeat that of Theorem 2.10. We referthe reader to the proof of that theorem for more details.
Proof of Lemma 2.9.
Let T , . . . , T n be the components of J and let S , . . . , S m be the componentsof ∂M , as above. We abuse notation and write M ( α ) for M ( α | ∂M ). We would like to show that π ( M ( α )) is left-orderable. We prove the result by induction on n . First, let n = 0. In this case,we have that ( α , . . . , α m ) is an LO slope by assumption. Now suppose that the result holds fora non-negative integer n . We will show the result holds for n + 1. Recall that each T i ∈ J isseparating. Let M cut along T have components X (cid:48) and X (cid:48)(cid:48) and let α (cid:48) and α (cid:48)(cid:48) be the inducedmultislopes on ∂X (cid:48) and ∂X (cid:48)(cid:48) respectively.Let α denote the element of α which is a slope on T . By induction, we have that α (cid:48) and α (cid:48)(cid:48) are LO slopes on X (cid:48) and X (cid:48)(cid:48) respectively. We study the normal closure of α (cid:48) , (cid:104)(cid:104) α (cid:48) (cid:105)(cid:105) , in π ( X (cid:48) ). Weconsider (cid:104)(cid:104) α (cid:48) (cid:105)(cid:105) ∩ π ( T ). This group is either (cid:104) α (cid:105) , π ( T ), or Z ⊕ k Z for some k ≥
2. Observe thatthe third possibility cannot occur, since in this case π ( X (cid:48) ( α (cid:48) )) would have torsion, contradictingthe assumption that π ( X (cid:48) ( α (cid:48) )) is left-orderable. A similar discussion applies for α (cid:48)(cid:48) on X (cid:48)(cid:48) .We first consider the case that (cid:104)(cid:104) α (cid:48) (cid:105)(cid:105) ∩ π ( T ) = π ( T ) in π ( X (cid:48) ). Observe that in this case thequotient maps π ( X (cid:48) ) → π ( X (cid:48) ) / (cid:104)(cid:104) α (cid:48) (cid:105)(cid:105) and π ( X (cid:48)(cid:48) ) → π ( T ). Since π ( M ) = π ( X (cid:48) ) ∗ π ( T ) π ( X (cid:48)(cid:48) ) , we obtain a quotient π ( M ) → π ( X (cid:48) ( α (cid:48) )). Since (cid:104)(cid:104) α | ∂M (cid:105)(cid:105) is in the kernel, we have an inducedquotient π ( M ( α )) → π ( X (cid:48) ( α (cid:48) )). Because M ( α ) admits a co-orientable taut foliation by assump-tion, we have that M ( α ) is prime. Since π ( X (cid:48) ( α (cid:48) )) is left-orderable by our induction hypothesis,Theorem 2.1 shows that π ( M ( α )) is left-orderable.The case that (cid:104)(cid:104) α (cid:48)(cid:48) (cid:105)(cid:105) ∩ π ( T ) = π ( T ) in π ( X (cid:48)(cid:48) ) is similar. Therefore, we assume that (cid:104)(cid:104) α (cid:48) (cid:105)(cid:105) ∩ π ( T ) = (cid:104) α (cid:105) in π ( X (cid:48) ) and (cid:104)(cid:104) α (cid:48)(cid:48) (cid:105)(cid:105) ∩ π ( T ) = (cid:104) α (cid:105) in π ( X (cid:48)(cid:48) ). Thus, π ( T ) has cyclic image in π ( X (cid:48) ( α (cid:48) )) and π ( X (cid:48)(cid:48) ( α (cid:48)(cid:48) )). We have an induced quotient π ( M ) → π ( X (cid:48) ( α (cid:48) )) ∗ Z π ( X (cid:48)(cid:48) ( α (cid:48)(cid:48) )) . AUT FOLIATIONS, LEFT-ORDERABILITY, AND CYCLIC BRANCHED COVERS 7
Observe that (cid:104)(cid:104) α | ∂M (cid:105)(cid:105) ⊂ π ( M ) is contained in the kernel of this quotient map. Therefore, weobtain a quotient π ( M ( α )) → π ( X (cid:48) ( α (cid:48) )) ∗ Z π ( X (cid:48)(cid:48) ( α (cid:48)(cid:48) )) . Each of π ( X (cid:48) ( α (cid:48) )) and π ( X (cid:48)(cid:48) ( α (cid:48)(cid:48) )) is left-orderable by the induction hypothesis. Therefore, thegroup π ( X (cid:48) ( α (cid:48) )) ∗ Z π ( X (cid:48)(cid:48) ( α (cid:48)(cid:48) )) is left-orderable by [BG09, Corollary 5.3]. By assumption M ( α )has a co-orientable taut foliation and therefore is prime. By Theorem 2.1 we have that π ( M ( α ))is left-orderable. This completes the proof. (cid:3) Branched covers.
Let K be a nullhomologous knot in a rational homology sphere Y withexterior X . Then the generator of H ( X, ∂X ) ∼ = H ( X ) ∼ = Z is represented by a properly embeddedorientable surface F with a single boundary component isotopic to K in a neighborhood of K . Let µ be a meridian of K . The map from Z to H ( X ) which sends a generator to µ has a unique splitting H ( X ) → Z defined by [ γ ] (cid:55)→ γ · F . The corresponding maps π ( X ) → Z and π ( X ) → Z /n definecanonical infinite and n -fold cyclic coverings X ∞ → X and X n → X respectively. Note that µ n lifts to a simple loop µ n ⊂ ∂X n . The n -fold cyclic branched covering of K is then defined to beΣ n ( K ) = X n ( µ n ). The obvious branched covering projection p n : Σ n ( K ) → Y factors through abranched cover p n,m : Σ n ( K ) → Σ m ( K ) for any m which divides n ; it is clear we have p n = p m ◦ p n,m .Observe that both p n and p n,m are non-zero degree maps. However, in the proof of Theorem 1.8,we will need something stronger. Lemma 2.11.
The induced map ( p n,m ) ∗ : π (Σ n ( K )) → π (Σ m ( K )) is surjective.Proof. We have inclusions π ( X ∞ ) ⊂ π ( X n ) ⊂ π ( X m ) ⊂ π ( X ) induced by the covering maps. If g ∈ π ( X ), then g = hµ k for some k ∈ Z and some h ∈ π ( X ∞ ). Let q m : π ( X m ) → π (Σ m ( K ))be induced by the quotient map X m → Σ m ( K ). Note that ker q m = (cid:104)(cid:104) µ m (cid:105)(cid:105) . Similarly, we have themap q n : π ( X n ) → π (Σ n ( K )).Let x ∈ π (Σ m ( K )). Then x = q m ( g ) for some g ∈ π ( X m ). Writing g = hµ k as above, g ∈ π ( X m ) implies k ≡ m ). Therefore, q m ( g ) = q m ( h ) ∈ π (Σ m ( K )). Since q n ( h ) ∈ π (Σ n ( K )), and ( p n,m ) ∗ ( q n ( h )) = q m ( h ), the result follows. (cid:3) Let X be the exterior of a knot K in S and let λ, µ be a longitude-meridian pair on ∂X . Underthe n -fold cyclic covering X n → X , we have that λ lifts to a simple closed curve (cid:101) λ and µ n lifts toa simple closed curve (cid:101) µ . We use (cid:101) λ, (cid:101) µ as a basis for slopes on ∂X n . Note that the inverse image ofthe slope µ + kλ is a single circle of slope (cid:101) µ + nk (cid:101) λ . Lemma 2.12.
Let K be a non-trivial knot in S and let M n denote the n -fold cyclic cover of theexterior of K . Then, there exists an integer k such that if | k | > k , then (cid:101) µ + nk (cid:101) λ is an excellentslope for M n .Proof. By Theorem 2.4, we have k such that for | k | > k , µ + kλ is a CTF slope for K . Let F besuch a taut foliation on the exterior of K . Then the preimage of F under the covering projectionfrom M n to M is a taut foliation on M n which intersects the boundary in simple closed curves ofslope (cid:101) µ + nk (cid:101) λ . Therefore, (cid:101) µ + nk (cid:101) λ is a CTF slope on M n .Observe that M n ( (cid:101) µ + nk (cid:101) λ ) is an n -fold cyclic branched cover of M ( µ + kλ ). Therefore, thereexists a non-zero degree map from M n ( (cid:101) µ + nk (cid:101) λ ) to M ( µ + kλ ). By Lemma 2.8, we have that π ( M ( µ + kλ )) is left-orderable for | k | > k . Since (cid:101) µ + nk (cid:101) λ is a CTF slope on M n , we must havethat M n ( (cid:101) µ + nk (cid:101) λ ) is irreducible. Theorem 2.1 implies that π ( M n ( (cid:101) µ + nk (cid:101) λ )) is left-orderable. Thus, (cid:101) µ + nk (cid:101) λ is excellent for | k | > k . (cid:3) Seifert manifolds.
A three-manifold Y is Seifert fibered if it admits a foliation by circles.Quotienting Y by the obvious S -action yields an orbifold, called the base orbifold . A rationalhomology sphere always has base orbifold S or R P and an integer homology sphere always has CAMERON GORDON AND TYE LIDMAN base orbifold S . Combining [BGW13, Proposition 5] and [BRW05, Theorem 1.3], we have thatfor rational homology spheres, if the base orbifold is R P , then Y is a total L-space.Whenever we work explicitly with the Seifert invariants of a manifold, we will restrict to thecase of base orbifold S . We will write our Seifert invariants as M ( β α , . . . , β m α m ). Here, we requirethat α i , β i are relatively prime, and if β i = 0 then α i = 1. We will often use normalized Seifertinvariants, as in [EHN81], which take the form M ( b, β α , . . . , β n α n ) where b ∈ Z ( b is the Euler number )and 0 < β i < α i for each i . We will usually point out explicitly when we are using normalizedSeifert invariants.We now make our conventions more explicit in terms of orientations. Suppose that r, s arerelatively prime positive integers. If 0 < r (cid:48) < r , 0 < s (cid:48) < s , and b are such that brs + r (cid:48) s + s (cid:48) r = − M ( b, r (cid:48) r , s (cid:48) s ) on S has regular fibers given by the positive ( r, s )-torusknots. On the other hand, the Seifert structure M ( − b, − r (cid:48) r , − s (cid:48) s ) on S has regular fibers given bythe ( − r, s )-torus knots. More generally, we have(2.1) − M (cid:18) b, β α , . . . , β n α n (cid:19) = M (cid:18) − b, − β α , . . . , − β n α n (cid:19) = M (cid:18) − n − b, α − β α , . . . , α n − β n α n (cid:19) . Note that if Y has normalized Seifert invariants M ( b, β α , . . . , β n α n ), then M ( − n − b, α − β α , . . . , α n − β n α n )give the normalized Seifert invariants of − Y .We will repeatedly use that (1.1), (1.2), and (1.3) are equivalent for Seifert manifolds. In fact,a stronger result holds. Recall that a horizontal foliation of a Seifert manifold is a codimensionone foliation which is transverse to the fibers. By definition, such foliations are taut. Further, fororientable Seifert fibered manifolds, a horizontal foliation is co-orientable if and only if the baseorbifold is orientable [BRW05, Lemma 5.5]. For the following, we specialize the result to the caseof closed, orientable manifolds. Theorem 2.13 ([BGW13],[BRW05],[LS07]) . Let Y be a closed, orientable Seifert fibered space.Then, the following are equivalent:(1) Y admits a co-orientable taut foliation,(2) either b ( Y ) = 0 , Y has base orbifold S , and admits a horizontal foliation, or b ( Y ) > ,(3) Y is not an L-space,(4) π ( Y ) is left-orderable. Note that if b ( Y ) = 0 and any of the conditions of Theorem 2.13 are satisfied, then Y hasbase orbifold S . A key part of the proof of Theorem 2.13 comes from the classification of Seifertmanifolds admitting horizontal foliations, due to Eisenbud-Hirsch-Neumann, Jankins-Neumann,and Naimi. We state only the case for closed, orientable manifolds with base orbifold S . Theorem 2.14 ([EHN81, JN85, Nai94]) . Let Y be a Seifert fibered space with base orbifold S and normalized Seifert invariants M ( b, β α , . . . , β n α n ) where n ≥ (and thus < β j < α j ). Then, Y admits a horizontal foliation if and only if one of the following holds:(1) − ( n − ≤ b ≤ − ,(2) b = − and there are integers < a < m such that after some permutation of the β j α j , wehave that β α < am , β α < m − am , and β j α j < m for ≤ j ≤ n ,(3) b = − ( n − and (2) holds for − Y = M ( − , α − β α , . . . , α n − β n α n ) .Remark . In [BRW05], the condition that a and m be relatively prime is also included. However,this condition is easily shown to be redundant.Given an explicit Seifert manifold, Theorem 2.14 provides a concrete means of determiningwhether it is excellent or if it is a total L-space. Note that the horizontal foliations guaranteed byTheorem 2.14 are necessarily co-orientable; further, in Theorem 2.14, there are no assumptions on AUT FOLIATIONS, LEFT-ORDERABILITY, AND CYCLIC BRANCHED COVERS 9 the first Betti number. Using this, we can easily understand the relationship between foliations onSeifert manifolds with torus boundary and foliations on their Dehn fillings. As discussed above,if α is an excellent multislope on M , then M ( α ) is an excellent manifold. The converse holds inmany cases for Seifert manifolds. Lemma 2.16.
Let M be an orientable Seifert fibered manifold with boundary tori T , . . . , T n andlet α be a multislope on M . If M ( α ) is Seifert fibered over S and admits a horizontal foliation,then α is an excellent multislope on M and M ( α ) is an excellent manifold.Proof. Our assumptions guarantee that M ( α ) is an excellent manifold by Theorem 2.13. By def-inition of excellence, α is an LO slope on M . It remains to show that α is a CTF slope for M .Consider a horizontal foliation F on M ( α ). Since the base orbifold is S , F is co-orientable. Be-cause the cores of the Dehn fillings are fibers in the Seifert fibration, they are necessarily transverseto the leaves of the horizontal foliation. The desired co-orientable foliation on M intersecting each T i in simple closed curves of slope α i is simply the restriction of F to M . (cid:3) Surgery on torus links.
Many of the cyclic branched covers that we will encounter later onwill contain the exterior of a torus link. Therefore, we are interested in which multislopes on toruslink exteriors are excellent.
Proposition 2.17.
Let k , . . . , k d be integers which are at least 2 and let r, s be relatively prime,positive integers. Here, we allow d = 1 only when r and s are both at least 2 and we allow r = s = 1 only when d ≥ . Then, − k = ( − k , . . . , − k d ) is an excellent multislope on the exterior of the toruslink T dr,ds .Proof. By Lemma 2.16, it suffices to show that S − k ( T dr,ds ) is a Seifert fibered space with baseorbifold S and admits a horizontal foliation. We do this by explicitly computing the Seifertinvariants of ( − k , . . . , − k d )-surgery on T dr,ds and applying Theorem 2.14. To do this, we firstwould like to find a Seifert structure on S such that T dr,ds consists of a collection of regular fibers,each isotopic to T r,s . We consider three cases. The first case is that r, s ≥ M (0 , β r , β s ), of S , where β s + β r = − β so that 0 < β < s . Then β (cid:48) = β + r satisfies 0 < β (cid:48) < r , and the normalized Seifertinvariants are now M ( − , β (cid:48) r , β s ). The torus link T dr,ds consists of d parallel regular fibers in theSeifert fibration, K , . . . , K d . Let µ i and λ i denote the meridian and longitude of K i respectively,and let ϕ i denote the fiber slope on ∂N ( K i ).We study the result of Dehn filling the boundary of the exterior of T dr,ds by the slopes γ i = a i µ i + b i ϕ i . By our orientation conventions, − µ i , ϕ i gives an oriented section-fiber basis for eachboundary torus of the link exterior. By construction, as long as a i (cid:54) = 0 for all i , the filled manifoldhas Seifert invariants M (cid:18) − , β (cid:48) r , β s , − b a , . . . , − b d a d (cid:19) . Recall that we are interested in ( − k , . . . , − k d )-surgery. This corresponds to filling the boundarytori by slopes − k i µ i + λ i . Note that the longitude λ i satisfies ϕ i = λ i + rsµ i . This can be seen, forinstance, by noting that the linking number of the ( r, s )-torus knot with a parallel regular fiber is rs . Therefore, − k i µ i + λ i = ( − k i − rs ) µ i + ϕ i . We conclude, if k , . . . , k d >
0, that the normalizedSeifert invariants for ( − k , . . . , − k d )-surgery are given by S − k ( T dr,ds ) = M (cid:18) − , β (cid:48) r , β s , k + rs , . . . , k d + rs (cid:19) . Let m = rs + 1 and a = β r + 1. Then β s < am , and β (cid:48) r = rs − − β rrs = m − a − m − < m − am . Also, if k i ≥
2, then k i + rs < m . Thus, Condition (2) of Theorem 2.14 holds, and we can concludethat S − k ( T dr,ds ) admits a horizontal foliation. This completes the proof for the case that both r and s are at least 2.The next case is when exactly one of r or s is 1. Without loss of generality, r = 1 and thus, byassumption, s ≥ d ≥
2. We consider the Seifert structure M ( − , s − s ) on S . In this case T d,ds is given by the union of d parallel regular fibers, K , . . . , K d (each fiber is an unknot). Byarguments similar to those in the previous case, we compute the normalized Seifert invariants for S − k ( T d,ds ) to be S − k ( T d,ds ) = M (cid:18) − , s − s , k + s , . . . , k d + s (cid:19) . Since d ≥ S − k ( T d,ds ) has at least three singular fibers. Because k , . . . , k d ≥
2, we see thatCondition (2) of Theorem 2.14 is satisfied by choosing a = 1 and m = s + 1.The final case to consider is r = s = 1 and d ≥
3. This case is similar to the previous one.We have that T d,d is a collection of d parallel regular fibers in the Seifert structure M ( −
1) on S .Therefore, we compute the Seifert invariants for S − k ( T d,d ) to be S − k ( T d,d ) = M (cid:18) − , k + 1 , . . . , k d + 1 (cid:19) . Since d ≥ S − k ( T d,d ) has at least three singular fibers. We again see that because k , . . . , k d ≥ a = 1 and m = 2. This completes theproof. (cid:3) Branched covers of torus knots
Since the cyclic branched cover of a torus knot is Seifert fibered, to determine if it is an excellentmanifold our general strategy will be to check the existence of horizontal foliations by Theorem 2.14and appeal to Theorem 2.13.Thus, in order to do this, we must compute the Seifert invariants of these manifolds. In fact,these have been explicitly calculated by N´u˜nez and Ram´ırez-Losada [NRL03, Theorem 1]. Due to itslength, we do not state it here. However, we will use their result throughout, so we refer the readerto [NRL03] for the precise statements. We will switch between the Seifert invariants of Σ n ( T p,q )and those of − Σ n ( T p,q ) by (2.1) to make use of both Conditions (2) and (3) of Theorem 2.14.We begin with a quick lemma that will help simplify our case analysis. Lemma 3.1.
Suppose that Σ r ( T p,q ) is an excellent manifold and let r divide n . Then Σ n ( T p,q ) isan excellent manifold.Proof. By assumption, π (Σ r ( T p,q )) is left-orderable. As discussed in Section 2.3, there exists anon-zero degree map from Σ n ( T p,q ) to Σ r ( T p,q ). Because Σ n ( T p,q ) is prime and π (Σ r ( T p,q )) isleft-orderable, Theorem 2.1 shows that π (Σ n ( T p,q )) is left-orderable. The result now follows fromTheorem 2.13. (cid:3) We now dispense with the easiest case of Theorem 1.2; we state this case separately since we willregularly appeal to it in the proof for the general case.
Proposition 3.2. If gcd( n, pq ) = 1 , then Σ n ( T p,q ) is excellent, unless { p, q, n } = { , , } .Proof. If gcd( n, pq ) = 1, then Σ n ( T p,q ) is a Seifert fibered integer homology sphere with base orbifold S and three singular fibers of multiplicities p , q , and n . It follows from [BRW05, Corollary 3.12]that π (Σ n ( T p,q )) is left-orderable unless { p, q, n } = { , , } . By Theorem 2.13, Σ n ( T p,q ) is excellentunless { p, q, n } = { , , } . (cid:3) With a view towards applying Lemma 3.1, we establish the following.
AUT FOLIATIONS, LEFT-ORDERABILITY, AND CYCLIC BRANCHED COVERS 11
Proposition 3.3.
Suppose that r divides either p or q . Then Σ r ( T p,q ) is excellent unless r = 2 and { p, q } = { , k } or r = 2 and { p, q } = { , } or r = 3 and { p, q } = { , } .Proof. Let r divide p or q . Without loss of generality, we may assume that r divides p . Let β , β be such that(3.1) β q + β p = − . By case (2) of [NRL03, Theorem 1], we have(3.2) Σ r ( T p,q ) = M (cid:18) β p/r , β q , . . . , β q (cid:19) , where there are r fibers of the form β q .We may choose β such that 0 < β < q . Then, by (3.1), we have(3.3) − ( p − ≤ β ≤ − . Therefore, rβ p satisfies − ( p − rp ≤ rβ p ≤ − rp . It follows that after normalizing the Seifert invariants, we have(3.4) − r ≤ b ≤ − . Case 1: r < p
In this case, the number of exceptional fibers of Σ r ( T p,q ) is ( r + 1). By (3.4), unless b = − b = − r , we have satisfied Condition (1) of Theorem 2.14, and therefore Σ r ( T p,q ) admits a horizontalfoliation. By Theorem 2.13, this manifold is excellent. Subcase (i): b = − < β < q and b = −
1, we must have − < rβ p <
0, giving β > − pr and rβ p = − p + rβ p ,where 0 < p + rβ < p . Thus, by (3.2), we have Σ r ( T p,q ) = M ( − , β + p/rp/r , β q , . . . , β q ). Let N bethe least positive integer such that β < − pr N . Clearly we have N ≥
2. We first claim that p + rβ p < r N − − r N − . If not, then we would have r N − ( p + rβ ) ≥ p ( r N − − , and thus r N β ≥ − p . This gives β ≥ − pr N , which is a contradiction.We next claim that β q < r N − . By (3.1), we see that β q < − β p . If the claim were false, wewould have − β p > r N − , and so β < − pr N − . This contradicts the choice of N .These two claims combine to show that Condition (2) of Theorem 2.14 is satisfied by taking m = r N − and a = 1. This shows that Σ r ( T p,q ) is excellent and completes the proof for Subcase(i). Subcase (ii): b = − r In this case, we will show that Condition (3) of Theorem 2.14 is satisfied for Σ r ( T p,q ), unless r = 2, p = 4, and q = 3. In other words, we consider Condition (2) of Theorem 2.14 for − Σ r ( T p,q ). Toobtain the normalized Seifert invariants for − Σ r ( T p,q ) = Σ r ( T − p,q ), we use the condition(3.5) p (cid:48) q + q (cid:48) p = 1 . Again, we choose q (cid:48) so that 0 < q (cid:48) < q . We have that − Σ r ( T p,q ) has (not yet normalized) Seifertinvariants M ( rp (cid:48) p , q (cid:48) q , . . . , q (cid:48) q ). Since Σ r ( T p,q ) has b = − r and r + 1 singular fibers, we have that for − Σ r ( T p,q ), b = −
1. Thecondition b = − − < rp (cid:48) p <
0, and thus it follows that the normalized Seifertinvariants for − Σ r ( T p,q ) are M ( − , p (cid:48) + p/rp/r , q (cid:48) q , . . . , q (cid:48) q ). We can satisfy Condition (2) of Theorem 2.14if we can show there exists m ≥ q (cid:48) q < m , p + rp (cid:48) p < m − m , or equivalently,(3.6) q (cid:48) q < m , rp (cid:48) p < − m . Subcase (a): p (cid:48) (cid:54) = − p (cid:48) ≤ −
2. Let m be the least positive integer such that the second inequality in (3.6)holds, i.e. rp (cid:48) p < − m . Then m ≥ rp (cid:48) p ≥ − m − . Also, since p (cid:48) < −
1, we have rp (cid:48) p < − p/r , and therefore(3.8) m ≤ pr , by the minimality of m . Recall that we are interested in showing that the first inequality in (3.6)holds for our choice of m . Because q (cid:48) q = 1 pq − p (cid:48) p ≤ pq + 1 r ( m − , by (3.7), it suffices to show that pq + r ( m − < m , or equivalently that mpq < − mr ( m − . Now, by (3.8), mpq ≤ p/rpq = 1 rq . Therefore, it is enough to show that rq < − mr ( m − , i.e., that q < r − mm − = ( r −
1) + m − . Since r ≥
2, this is clearly true.
Subcase (b): p (cid:48) = − − q + q (cid:48) p = 1, giving(3.9) q (cid:48) p = q + 1 . Let m = pr + 1. Then, rp (cid:48) p = − rp = − p/r < − m , so the first inequality in (3.6) holds. Claim: q (cid:48) q < m unless r = 2, p = 4, and q = 3. AUT FOLIATIONS, LEFT-ORDERABILITY, AND CYCLIC BRANCHED COVERS 13
Proof of Claim.
Since q (cid:48) q = q +1 qp , we must show that q + 1 qp < m = 1 p/r + 1 = 1 p (cid:32) r + p (cid:33) , or equivalently that qq +1 > r + p . Note that if q = 2, (3.9) gives that p = 3, contradicting the factthat r < p . Therefore q ≥ qq +1 ≥ . On the other hand,1 r + 1 p <
12 + 14 = 34 , unless r = 2 and p = 4. Finally, if r = 2 and p = 4, then r + p = and qq +1 > if q ≥
5. Thiscompletes the proof of the claim. (cid:3)
Therefore, we have shown that Condition (2) of Theorem 2.14 is satisfied by choosing m = pr + 1,excluding the case r = 2, p = 4, q = 3. This completes the proof of the theorem in Case 1. Case 2: r = p We will show that Condition (2) of Theorem 2.14 is satisfied for either Σ p ( T p,q ) or − Σ p ( T p,q ), exceptin the cases p = 2 , r = 2 or p = 3 , q = 2 , r = 3.We return to the setup before Case 1. From (3.2), r = p implies the normalized Seifert invariantsof Σ p ( T p,q ) are(3.10) Σ p ( T p,q ) = M (cid:18) β , β q , . . . , β q (cid:19) , i.e. β = b . Thus, the number of exceptional fibers of Σ p ( T p,q ) is p . If p = 2, then Σ p ( T p,q )is a lens space, and thus π (Σ p ( T p,q )) is finite. Therefore, suppose p ≥
3. By (3.3), we have − ( p − ≤ b ≤ −
1. Again, Condition (1) of Theorem 2.14 is satisfied unless b = − b = − ( p − Subcase (i): b = − p ( T p,q ) has normalized Seifert invariants M ( − , β q , . . . , β q ). In thiscase, since β = −
1, (3.1) implies that β = q − p . Therefore, β q = q − pq < p . Condition (2) ofTheorem 2.14 is now satisfied by taking m = p and a = 1. Subcase (ii): b = − ( p − β p = ( p − q −
1, and hence q − β q = (cid:16) q +1 q (cid:17) (cid:16) p (cid:17) . Thus, afterreversing orientation, we see from (2.1) and (3.10) that − Σ p ( T p,q ) has normalized Seifert invariants − Σ p ( T p,q ) = M (cid:18) − , (cid:18) q + 1 q (cid:19) p , . . . , (cid:18) q + 1 q (cid:19) p (cid:19) . Unless p = 3 and q = 2 (and consequently r = 3), Condition (2) of Theorem 2.14 is satisfied bytaking m = 2 , a = 1.We have now shown that Σ p ( T p,q ) is excellent when r divides p or q , except for the cases statedin the proposition. (cid:3) With Proposition 3.2 and Proposition 3.3, we are now ready to prove Theorem 1.2.
Proof of Theorem 1.2.
We are interested in the general case of when Σ n ( T p,q ) is excellent for ar-bitrary n . We first note that if π (Σ n ( T p,q )) is finite, then this group is not left-orderable. Inthis case, Σ n ( T p,q ) is a total L-space by Theorem 2.13. (This was originally established in [OS05a,Proposition 2.3]). By Proposition 3.2, we have reduced the problem to the case gcd( n, pq ) (cid:54) = 1. Without loss ofgenerality, we can assume that gcd( n, p ) (cid:54) = 1. By Proposition 3.3, we established that if n has afactor r which divides p , then Σ r ( T p,q ) is excellent unless p = 4 , q = 3 or p = 3 , q = 2, or p = 2.By Lemma 3.1, we thus have that Σ n ( T p,q ) is excellent unless p = 4 , q = 3 or p = 3 , q = 2 or p = 2. From now on, we assume that ( p, q ) is of one of these three forms. Write n = 2 k (cid:96) m w ,where gcd( w,
30) = 1. First, suppose w (cid:54) = 1. If gcd( q, w ) = 1, then Σ w ( T p,q ) is excellent fromProposition 3.2, and consequently so is Σ n ( T p,q ) by Lemma 3.1. Now suppose gcd( q, w ) (cid:54) = 1 andlet s be a prime factor of gcd( q, w ); note that in this case s ≥ q ≥
7. Thus, Σ s ( T p,q )is excellent by Proposition 3.3. By Lemma 3.1, Σ n ( T p,q ) is excellent. Thus, we now assume n = 2 k (cid:96) m . We will compile a list of the remaining cases we must check by hand.Let p = 4 , q = 3. Note that Σ ( T , ) is excellent by Proposition 3.2. Therefore, if m ≥
1, Σ n ( T , )is excellent by Lemma 3.1. Thus, assume m = 0. Note that since gcd( n, p ) (cid:54) = 1 by assumption, wemust have that k ≥
1. Further, we have that Σ ( T , ) is excellent by Proposition 3.3. Consequentlyif (cid:96) ≥
1, Lemma 3.1 shows that Σ n ( T , ) is excellent. Thus, for the case of p = 4 , q = 3, it remainsto consider n = 2 k for k ≥
2. Thus, by Lemma 3.1, it suffices to show that Σ ( T , ) is excellent,since π (Σ ( T , )) is finite. This case is handled by Proposition 3.3.Now, we consider the case p = 3 , q = 2 , n = 2 k (cid:96) m . Since by assumption gcd( n, p ) (cid:54) = 1, we have (cid:96) ≥
1. Proposition 3.2 shows that Σ ( T , ) is excellent, and thus Σ n ( T , ) is excellent if m ≥ n ( T , ) has finite fundamental group if n = 3. On the other hand, if we can showΣ n ( T , ) is excellent for n = 6, 9, and 15, this will complete the proof for the case of p = 3 , q = 2by again applying Lemma 3.1.Finally, suppose that p = 2 , n = 2 k (cid:96) m . By assumption, k ≥
1. The relevant triples withfinite fundamental group are ( n, p, q ) = (4 , ,
3) and (2 , , q ) for q ≥
3. First, consider the case q = 3. Arguments similar to those above show that it suffices to establish the excellence of Σ n ( T , )for n = 8 and 10. Now suppose that q ≥
5. If gcd( q,
5) = 1, then Σ ( T ,q ) is excellent byProposition 3.2. If 5 divides q , then Σ ( T ,q ) is excellent by applying Proposition 3.3. Thus, byLemma 3.1, if m ≥
1, then Σ n ( T ,q ) is excellent. Thus, assume n = 2 k (cid:96) with k ≥
1. If q ≥ ( T ,q ) is excellent by applying either Proposition 3.2 or Proposition 3.3 with the roles of p and q reversed (dependent on whether gcd(3 , q ) = 1). Therefore, if q ≥ (cid:96) ≥
1, Σ n ( T ,q ) is excellentby Lemma 3.1. Thus, for p = 2 , q ≥
7, it suffices to show that Σ ( T ,q ) is excellent. On the otherhand, if q = 5, then it suffices to show Σ ( T , ) and Σ ( T , ) are excellent.We summarize the above discussion with the list of remaining cases for which we need to establishexcellence to complete the proof of the theorem:(1) p = 2 , q ≥ , n = 4(2) p = 2 , q = 3 , n = 6(3) p = 2 , q = 3 , n = 8(4) p = 2 , q = 3 , n = 9(5) p = 2 , q = 3 , n = 10(6) p = 2 , q = 3 , n = 15(7) p = 2 , q = 5 , n = 6. Case 1: p = 2 , q ≥ , n = 4To compute the Seifert invariants of Σ ( T ,q ), we use case (3)(a) of [NRL03, Theorem 1]. Write q = 2 k − , k >
2. Let q ∗ = −
1; then, qq ∗ = 1 − k , and so k is as in [NRL03, Theorem 1]. Also, wehave β q + β p = −
1, and so we can take β = 1 , β = − k . From [NRL03, Theorem 1], we get thatΣ ( T ,q ) = M (cid:18) , k, − k q , − k q (cid:19) . AUT FOLIATIONS, LEFT-ORDERABILITY, AND CYCLIC BRANCHED COVERS 15
Let c = (cid:98) k q (cid:99) + 1. Then, b = k − c and Σ ( T ,q ) has normalized Seifert invariants(3.11) Σ ( T ,q ) = M (cid:18) k − c, , c − k q , c − k q (cid:19) . Note that c = (cid:22) k k − (cid:23) + 1 = (cid:26) k + 1 , k even k +12 , k odd . Subcase (i): k oddThen b = k − c = −
1. Also, in this case c − k q = k + 12 − k q = q ( k + 1) − k q = k − k − < . Therefore, Condition (2) of Theorem 2.14 is satisfied by taking m = 4 , a = 1. Subcase (ii): k evenHere b = k − ( k + 2) = −
2. Also, in this case1 − ( c − k q ) = (1 − c ) + k q = k q − k
2= 2 k − qk q = k k − . It follows from (2.1) and (3.11) that − Σ ( T ,q ) has normalized Seifert invariants M ( − , , k k − , k k − ).If q ≥
5, then k > k k − < . Condition (2) of Theorem 2.14 is therefore satisfied by taking m = 3 , a = 1. Case 2: p = 2 , q = 3 , n = 6By case (2) of [NRL03, Theorem 1], we have that H (Σ ( T , )) is infinite. Therefore, Σ ( T , ) isexcellent. Case 3: p = 2 , q = 3 , n = 8By case (3) of [NRL03, Theorem 1], letting β = − , β = 1 , k = 1, and 3 ∗ = − ( T , ) = M (cid:18) − , , , (cid:19) . Condition (2) of Theorem 2.14 holds with m = 2 , a = 1. Case 4: p = 2 , q = 3 , n = 9By case (3)(a) of [NRL03, Theorem 1], letting β = 1 , β = − , k = 1 , and 2 ∗ = − ( T , ) = M (cid:18) , , − , − , − (cid:19) = M (cid:18) − , , , , (cid:19) . Therefore, b = − (cid:3) Cables
Let K be a knot in S , with regular neighborhood N ( K ). Let C be a simple loop on ∂N ( C )with slope p/q , where q ≥
2. Then C is the ( p, q ) -cable of K , C p,q ( K ). If K is trivial then C p,q ( K )is the torus knot T p,q . Since C − p,q ( K ) = C p, − q ( K ) = − C p,q ( − K ), we will always assume that p ≥ q ≥ Theorem 1.3.
Let K be a non-trivial knot in S . Then, Σ n ( C p,q ( K )) is excellent, unless n = q = 2 . In the case n = q = 2, it turns out that Σ ( C p, ( K )) ∼ = X ∪ ∂ X , where X i is a copy of theexterior X of K , i = 1 ,
2. For p = 1 we get the following conditional result. Theorem 4.1. If is a CTF slope for K then Σ ( C , ( K )) is excellent. In particular, if the Li-Roberts Conjecture holds, then Σ ( C , ( K )) is excellent for all non-trivial K . The theorem will be proved at the end of the section.Let S = V ∪ T W be the standard genus 1 Heegaard splitting of S . Let C ⊂ int W be anisotopic copy of a ( p, q )-curve on T , where this curve intersects the meridional disk of W q times.Let J be a core of V . Let σ, τ ⊂ T be meridians of V, W , respectively. We will be more preciseabout our orientations of these curves shortly.In S = V ∪ W the curve C is the torus knot T p,q . In the corresponding Seifert fibration of S ,in which C is an ordinary fiber, J is the exceptional fiber of multiplicity p . Let π : Σ n ( T p,q ) → S be the branched covering projection. Then π − ( V ) has gcd( n, q ) components (cid:101) V i , 1 ≤ i ≤ gcd( n, q ),each an n gcd( n,q ) -fold covering of V . In the induced Seifert fibration on Σ n ( T p,q ) the core of each (cid:101) V i is an exceptional fiber of multiplicity p gcd( n,p ) . Let (cid:101) σ i be a meridian of (cid:101) V i , 1 ≤ i ≤ gcd( n, q ). Also, (cid:101) τ i = π − ( τ ) ∩ ∂ (cid:101) V i is connected for 1 ≤ i ≤ gcd( n, q ). Let s, ϕ ⊂ ∂W be an oriented section-fiberpair for the induced Seifert structure on W (i.e. s · ϕ = 1). We orient τ such that τ · ϕ = q in ∂W .Observe that ϕ is a ( p, q )-curve on ∂W . We then orient σ such that τ · σ = 1 on ∂W .Let X be the exterior of the knot K , and let λ, µ be a (0-framed) longitude-meridian pair on ∂X . If we remove int ( V ) from S and replace it by X , identifying ∂X with ∂W in such a way that λ, µ are identified with σ, τ , respectively, we get S , and the curve C becomes the ( p, q )-cable of K .Let (cid:102) W = π − ( W ) = Σ n ( T p,q ) − (cid:96) gcd( n,q ) i =1 int ( (cid:101) V i ), the n -fold cyclic branched cover of W branchedalong C . Let (cid:101) X be the n gcd( n,q ) -fold cyclic covering of X , and let (cid:101) λ, (cid:101) µ ⊂ ∂ (cid:101) X be as in the discussionin Section 2.3. Then Σ n ( C p,q ( K )) = (cid:102) W ∪ (cid:96) gcd( n,q ) i =1 (cid:101) X i , where (cid:101) X i is a copy of (cid:101) X and ∂ (cid:101) X i is gluedto ∂ (cid:101) V i in such a way that (cid:101) λ i , (cid:101) µ i are identified with (cid:101) σ i , (cid:101) τ i , respectively.4.1. The proof of Theorem 1.3.
The following proposition establishes Theorem 1.3 in the genericcase.
Proposition 4.2.
Let K be a knot in S . Suppose p (cid:54) = 1 and { n, p, q } (cid:54) = { , , r } , { , , } , { , , } or { , , } . Then Σ n ( C p,q ( K )) is excellent.Proof. Using the notation in the discussion before the statement of the proposition, (cid:102) W ( (cid:101) σ ) = (cid:102) W ( (cid:101) σ , . . . , (cid:101) σ gcd( n,q ) ) = Σ n ( T p,q ), which by Theorem 1.2 is excellent if { n, p, q } is not one of theexceptions listed in the proposition. Hence by Lemma 2.16, (cid:101) σ is excellent for (cid:102) W . By [Gab87], λ AUT FOLIATIONS, LEFT-ORDERABILITY, AND CYCLIC BRANCHED COVERS 17 is a CTF slope for X , and hence (cid:101) λ is a CTF slope for (cid:101) X . In particular, we have that (cid:101) X ( (cid:101) λ ) isprime. Since H ( X ( λ )) is infinite, so is H ( (cid:101) X ( (cid:101) λ )). Therefore, (cid:101) λ is an LO slope for (cid:101) X by [BRW05,Corollary 3.4]. Thus Σ n ( C p,q ( K )) is excellent by Lemma 2.9. (cid:3) To treat the cases where ( n, p, q ) is one of the exceptional triples in Theorem 1.2 we continue witha more detailed analysis of the description of Σ n ( C p,q ( K )) given just before the present subsection.Recall that ϕ ⊂ ∂W is a regular fiber in the Seifert fibration of S described above. Note that (cid:101) ϕ i = π − ( ϕ ) ∩ ∂ (cid:101) V i has gcd( n gcd( n,q ) , p ) = gcd( n, p ) components. Since we have oriented ϕ so that τ · ϕ = q on ∂W , we have (cid:101) τ i · (cid:101) ϕ i = nq gcd( n,q ) gcd( n,p ) on ∂ (cid:102) W . For notation, we let ω = nq gcd( n,q ) gcd( n,p ) .We have (cid:102) W ( (cid:101) σ , . . . , (cid:101) σ gcd( n,q ) ) = Σ n ( T p,q ). As in Section 3, we will use the description of the Seifertinvariants of cyclic branched covers of torus knots in [NRL03, Theorem 1]. We define (cid:101) γ i ⊂ ∂ (cid:101) V i as the section with respect to which the Seifert invariants of Σ n ( T p,q ) are described in [NRL03,Theorem 1]. Thus (cid:101) γ i · (cid:101) ϕ i = 1 in ∂ (cid:102) W , and if the core of (cid:101) V i has Seifert invariants rs in Σ n ( T p,q ), then (cid:101) σ i = ± ( s (cid:101) γ i + r (cid:101) ϕ i ). In some cases, the orientation of (cid:101) σ i will be easily determined, while often itwill not. In the latter situation, rather than repeat the constructions of [NRL03] in precise detailto obtain the exact orientation of (cid:101) σ i , it will be easier to simply treat both cases, even though onlyone can possibly arise.Observe that (cid:101) τ i = ω (cid:101) γ i + η (cid:101) ϕ i for some η ∈ Z . Recall that we must have that (cid:101) τ i · (cid:101) σ i = (cid:101) γ i · (cid:101) ϕ i = 1on ∂ (cid:102) W . This will determine the possible values of η .The strategy to complete the proof of Theorem 1.3 is as follows. We consider slopes α k = µ + kλ on ∂X , k ∈ Z . The corresponding slope on ∂ (cid:101) X i is (cid:101) α k = (cid:101) µ i + n gcd( n,q ) k (cid:101) λ i , which is identifiedwith (cid:101) τ i + n gcd( n,q ) k (cid:101) σ i on ∂ (cid:102) W . We will show that for k (cid:29) k (cid:28) (cid:102) W ( (cid:101) α k , . . . , (cid:101) α k ) is a Seifertfibered space with a horizontal foliation, and therefore ( (cid:101) α k , . . . , (cid:101) α k ) is an excellent multislope for (cid:102) W by Lemma 2.16. Since, for | k | sufficiently large, (cid:101) α k is an excellent slope for (cid:101) X by Lemma 2.12,Σ n ( C p,q ( K )) will be excellent by Lemma 2.9.The details are given in Propositions 4.3, 4.4, and 4.5 below.We first treat the case where gcd( n, q ) = 1 and p (cid:54) = 1. Then, π − ( V ) = (cid:101) V is connected, and wedrop the subscript i from (cid:101) V i , (cid:101) σ i , etc. Also, for the rest of this section, X ( n ) will denote the n -foldcyclic covering of X . Proposition 4.3.
Let K be a non-trivial knot in S . If { n, p, q } = { , , } or ( n, p, q ) =(2 , , , (3 , , , (4 , , , or (2 , , q ) , then Σ n ( C p,q ( K )) is excellent.Proof. Note that (cid:101) J = π − ( J ) is a fiber of multiplicity p gcd( n,p ) in Σ n ( T p,q ). Also, (cid:101) τ = ω (cid:101) γ + η (cid:101) ϕ where ω = nq gcd( n,p ) , since gcd( n, q ) = 1. Here, (cid:101) α k = (cid:101) µ + nk (cid:101) λ is identified with (cid:101) τ + nk (cid:101) σ on ∂ (cid:102) W . Thus,by the discussion preceding the statement of the proposition, it suffices to show that (cid:102) W ( (cid:101) τ + nk (cid:101) σ )admits a horizontal foliation for k (cid:29) k (cid:28) Case 1: ( n, p, q ) = (2 , , ( T , ) = M (1 , − , − , − ). Since (cid:101) J is the fiber of multiplicity 3, (cid:101) σ = ± ( − (cid:101) γ + (cid:101) ϕ ). We have ω = 10, and thus1 = (cid:101) τ · (cid:101) σ = ± (10 (cid:101) γ + η (cid:101) ϕ ) · ( − (cid:101) γ + (cid:101) ϕ ) = ± (3 η + 10) . Therefore, we have that η = −
3, and thus (cid:101) τ = 10 (cid:101) γ − (cid:101) ϕ and (cid:101) σ = − (cid:101) γ + (cid:101) ϕ .The slope (cid:101) α k = (cid:101) µ + 2 k (cid:101) λ on ∂X (2) is identified with (cid:101) τ + 2 k (cid:101) σ = (10 (cid:101) γ − (cid:101) ϕ ) + 2 k ( − (cid:101) γ + (cid:101) ϕ ) =(10 − k ) (cid:101) γ + ( − k ) (cid:101) ϕ on ∂ (cid:102) W . Hence (cid:102) W ( (cid:101) α k ) has Seifert invariants M (1 , − , − , k − − k ) = − M ( − , , , k − k − ). If k ≤
0, then 0 < k − k − < . By choosing m = 3 and a = 1, it follows fromCondition (3) of Theorem 2.14 that (cid:102) W ( (cid:101) α k ) has a horizontal foliation. This is what we wanted to show. Case 2: ( n, p, q ) = (2 , , (cid:101) σ = ± ( − (cid:101) γ + (cid:101) ϕ ), and (cid:101) τ = 6 (cid:101) γ + η (cid:101) ϕ . We have1 = (cid:101) τ · (cid:101) σ = ± (6 (cid:101) γ + η (cid:101) ϕ ) · ( − (cid:101) γ + (cid:101) ϕ ) = ± (6 + 5 η ) . Therefore, η = − (cid:101) τ = 6 (cid:101) γ − (cid:101) ϕ , (cid:101) σ = − (cid:101) γ + (cid:101) ϕ . Hence on ∂ (cid:102) W , (cid:101) α k = (cid:101) τ +2 k (cid:101) σ = (6 − k ) (cid:101) γ +(2 k − (cid:101) ϕ . Then (cid:102) W ( (cid:101) α k ) is the Seifert fibered space M (1 , − , − , k − − k ) = − M ( − , , , k − k − ).For k ≤
0, 0 < k − k − < . Letting m = 5 and a = 2, we see that (cid:102) W ( (cid:101) α k ) has a horizontal foliationby Condition (3) of Theorem 2.14. Case 3: ( n, p, q ) = (3 , , (cid:101) σ = ± ( − (cid:101) γ + (cid:101) ϕ ), and (cid:101) τ = 15 (cid:101) γ + η (cid:101) ϕ . In this case, we have1 = ± (15 (cid:101) γ + η (cid:101) ϕ ) · ( − (cid:101) γ + (cid:101) ϕ ) = ± (15 + 2 η ) . Thus, we have two possibilities. The first is that η = − (cid:101) τ = 15 (cid:101) γ − ϕ , and (cid:101) σ = − (cid:101) γ + (cid:101) ϕ . Thesecond case is that η = − (cid:101) τ = 15 (cid:101) γ − (cid:101) ϕ , and (cid:101) σ = 2 (cid:101) γ − (cid:101) ϕ .First, suppose that η = −
7. The slope (cid:101) α k = (cid:101) µ + 3 k (cid:101) λ on ∂X (3) is identified with (cid:101) τ + 3 k (cid:101) σ =(15 − k ) (cid:101) γ + (3 k − (cid:101) ϕ on ∂ (cid:102) W . Therefore (cid:102) W ( (cid:101) α k ) has Seifert invariants M (1 , − , − , k − − k ) = − M ( − , , , k − k − ). If k ≤
0, then 0 < − k − k < and (cid:102) W ( (cid:101) α k ) has a horizontal foliation byCondition (3) of Theorem 2.14.Next, suppose that η = −
8. The slope (cid:101) α k = (cid:101) µ + 3 k (cid:101) λ on ∂X (3) is identified with (cid:101) τ + 3 k (cid:101) σ =(15 + 6 k ) (cid:101) γ + ( − − k ) (cid:101) ϕ on ∂ (cid:102) W . Therefore (cid:102) W ( (cid:101) α k ) has Seifert invariants M (1 , − , − , − k − k +15 ) = − M ( − , , , k +86 k +15 ). If k (cid:28)
0, then 0 < k +86 k +15 < and (cid:102) W ( (cid:101) α k ) has a horizontal foliation byCondition (3) of Theorem 2.14.The cases ( n, p, q ) = (3 , , , (5 , , , ,
2) are completely analogous; we leave the detailsto the reader.
Case 4: ( n, p, q ) = (2 , , ( T , ) = M ( , − , − ). Since (cid:101) J has multiplicity 2, we have (cid:101) σ = ± (2 (cid:101) γ + (cid:101) ϕ ). Also, ω = nq gcd( n,p ) = 3. We again have two cases. The first case is η = 1, (cid:101) τ = 3 (cid:101) γ + (cid:101) ϕ , (cid:101) σ = 2 (cid:101) γ + (cid:101) ϕ . The slope (cid:101) α k = (cid:101) µ + 2 k (cid:101) λ on ∂X (2) is identified with (cid:101) τ + 2 k (cid:101) σ = (4 k + 3) (cid:101) γ + (2 k + 1) (cid:101) ϕ on ∂ (cid:102) W . So (cid:102) W ( (cid:101) α k ) = M (cid:18) − , − , k + 14 k + 3 (cid:19) = − M (cid:18) , , − k + 14 k + 3 (cid:19) = − M (cid:18) − , , , k + 24 k + 3 (cid:19) . Since k +14 k +3 < for k ≥ (cid:102) W ( (cid:101) α k ) has a horizontal foliation by Condition (3) of Theorem 2.14.The second case is η = 2, (cid:101) τ = 3 (cid:101) γ + 2 (cid:101) ϕ , (cid:101) σ = − (cid:101) γ − (cid:101) ϕ . The slope (cid:101) α k = (cid:101) µ + 2 k (cid:101) λ on ∂X (2) isidentified with (cid:101) τ + 2 k (cid:101) σ = (3 − k ) (cid:101) γ + (2 − k ) (cid:101) ϕ on ∂ (cid:102) W . So (cid:102) W ( (cid:101) α k ) = M (cid:18) − , − , k − k − (cid:19) = − M (cid:18) , , − k k − (cid:19) = − M (cid:18) − , , , k − k − (cid:19) . Since k − k − < for k ≤ (cid:102) W ( (cid:101) α k ) has a horizontal foliation by Condition (3) of Theorem 2.14. Case 5: ( n, p, q ) = (3 , , ( T , ) = M ( − , − , − , ). Since (cid:101) J has multiplicity 1, we have (cid:101) σ = ± ( (cid:101) γ + (cid:101) ϕ ). We have two cases. AUT FOLIATIONS, LEFT-ORDERABILITY, AND CYCLIC BRANCHED COVERS 19
First, (cid:101) τ = 2 (cid:101) γ + (cid:101) ϕ , (cid:101) σ = (cid:101) γ + (cid:101) ϕ . The slope (cid:101) α k = (cid:101) µ + 3 k (cid:101) λ on ∂X (3) is identified with (cid:101) τ + 3 k (cid:101) σ =(3 k + 2) (cid:101) γ + (3 k + 1) (cid:101) ϕ on ∂ (cid:102) W . Therefore, (cid:102) W ( (cid:101) α k ) has Seifert invariants M (cid:18) − , − , − , k + 13 k + 2 (cid:19) = M (cid:18) − , , , , − k + 2 (cid:19) . For k (cid:28)
0, the latter are normalized Seifert invariants, and thus (cid:102) W ( α k ) has a horizontal foliationby Condition (1) of Theorem 2.14.Second, we have (cid:101) τ = 2 (cid:101) γ + 3 (cid:101) ϕ , (cid:101) σ = − (cid:101) γ − (cid:101) ϕ . The slope (cid:101) α k = (cid:101) µ + 3 k (cid:101) λ on ∂X (3) is identified with (cid:101) τ + 3 k (cid:101) σ = (2 − k ) (cid:101) γ + (3 − k ) (cid:101) ϕ on ∂ (cid:102) W . Therefore, (cid:102) W ( (cid:101) α k ) has Seifert invariants M (cid:18) − , − , − , − k − k (cid:19) = M (cid:18) − , , , , − k (cid:19) . For k ≤
0, the latter invariants are normalized, and thus (cid:102) W ( α k ) has a horizontal foliation by Con-dition (1) of Theorem 2.14. Case 6: ( n, p, q ) = (4 , , ( T , ) = M ( − , , − , − ). Since (cid:101) J has multiplicity 1, we have (cid:101) σ = ± ( (cid:101) γ + (cid:101) ϕ ). Again, there are two cases.For the first case, we have (cid:101) τ = 6 (cid:101) γ + 5 (cid:101) ϕ , (cid:101) σ = (cid:101) γ + (cid:101) ϕ . The slope (cid:101) α k = (cid:101) µ + 4 k (cid:101) λ on ∂X (4) is identifiedwith (cid:101) τ + 4 k (cid:101) σ = (4 k + 6) (cid:101) γ + (4 k + 5) (cid:101) ϕ on ∂ (cid:102) W . Thus, (cid:102) W ( α k ) has Seifert invariants M (cid:18) − , − , − , k k (cid:19) = − M (cid:18) − , , , , k + 74 k + 6 (cid:19) . Hence for k ≤ − (cid:102) W ( (cid:101) α k ) has a horizontal foliation.For the second case, we have (cid:101) τ = 6 (cid:101) γ + 7 (cid:101) ϕ , (cid:101) σ = − (cid:101) γ − (cid:101) ϕ . In this case, (cid:102) W ( (cid:101) α k ) has Seifert invariants M (cid:18) − , − , − , − k − k (cid:19) = − M (cid:18) − , , , , k − k − (cid:19) . Hence for k ≤ (cid:102) W ( (cid:101) α k ) has a horizontal foliation. Case 7: ( n, p, q ) = (2 , , q ).Observe that Σ ( T ,q ) is a lens space. More specifically, from [NRL03, Theorem 1] we get thatΣ ( T ,q ) has Seifert invariants M ( − , β q , β q ) where β = q − . Since J has multiplicity 2, and (cid:101) J → J is a 2-fold connected covering, we have (cid:101) J is the fiber with Seifert invariant − . Hence (cid:101) σ = ± ( − (cid:101) γ + (cid:101) ϕ ). Again, there are two cases.First, we have (cid:101) τ = q (cid:101) γ + (1 − q ) (cid:101) ϕ and (cid:101) σ = − (cid:101) γ + (cid:101) ϕ . In Σ ( C ,q ( K )) ∼ = X (2) ∪ (cid:102) W , the slope of (cid:101) α k = (cid:101) µ + 2 k (cid:101) λ is identified with (cid:101) τ + 2 k (cid:101) σ = ( q − k ) (cid:101) γ + (2 k + 1 − q ) (cid:101) ϕ . Then, (cid:102) W ( α k ) has Seifertinvariants M (cid:18) − , ( q − / q , ( q − / q , q − k (cid:19) . Since ( q − / q < , and 0 < q − k < if k ≤ (cid:102) W ( (cid:101) α k ) admits a horizontal foliation by (2) ofTheorem 2.14.Second, we have (cid:101) τ = q (cid:101) γ + ( − − q ) (cid:101) ϕ and (cid:101) σ = (cid:101) γ − (cid:101) ϕ . Then, (cid:102) W ( α k ) has Seifert invariants M (cid:18) − , ( q − / q , ( q − / q , − k + q (cid:19) . Since ( q − / q < , and 0 < − k + q < if k (cid:28) (cid:102) W ( (cid:101) α k ) admits a horizontal foliation by (2) ofTheorem 2.14. (cid:3) We next consider the case gcd( n, q ) (cid:54) = 1, i.e. (cid:101) J is not connected. Proposition 4.4.
Let K be a non-trivial knot in S . Then the manifolds Σ ( C , ( K )) , Σ ( C , ( K )) ,and Σ ( C , ( K )) are excellent.Proof. We use the same arguments as in Proposition 4.3. As before, it suffices to show that (cid:102) W ( (cid:101) α k , . . . , (cid:101) α k ) admits a horizontal foliation. Recall that (cid:101) τ = ω (cid:101) γ + η (cid:101) ϕ , where ω = nq gcd( n,q ) gcd( n,p ) ,and we solve for the possible values of η using (cid:101) τ · (cid:101) σ = 1 . Case 1: ( n, p, q ) = (2 , , (cid:101) J has two components, each mapping homeomorphically to J . By [NRL03, Theo-rem 1], Σ ( T , ) = M ( , − , − ). Since J has multiplicity 3, each component of (cid:101) J is a fiber ofmultiplicity 3. Thus, (cid:101) σ i = ± ( − (cid:101) γ i + (cid:101) ϕ i ). In this case, there is only one choice for η , and we find (cid:101) σ i = − (cid:101) γ i + (cid:101) ϕ i and (cid:101) τ i = 4 (cid:101) γ i − (cid:101) ϕ i , for i = 0 , ( C , ( K )) ∼ = X ∪ X ∪ (cid:102) W , where X i is a copy of X , glued along the two boundary com-ponents of (cid:102) W . On ∂X i the slope α k = µ i + kλ i is identified with (cid:101) τ i + k (cid:101) σ i = (4 − k ) (cid:101) γ i + ( k − (cid:101) ϕ i .Hence (cid:102) W ( α k , α k ) is the Seifert fibered space M ( , k − − k , k − − k ) = − M ( − , , k − k − , k − k − ). Since0 < k − k − < if k ≤ (cid:102) W ( α k , α k ) has a horizontal foliation by Condition (3) of Theorem 2.14. Case 2: ( n, p, q ) = (4 , , ( T , ) = M ( − , − , − , ). The curve J has multiplicity 3, (cid:101) J hastwo components, each mapping to J by a covering map of degree 2, and each being a fiber ofmultiplicity 3. So, (cid:101) σ i = ± ( − (cid:101) γ i + (cid:101) ϕ i ). There is only one choice for η , and we see (cid:101) τ i = 4 (cid:101) γ i − (cid:101) ϕ i and (cid:101) σ i = − (cid:101) γ i + (cid:101) ϕ i .The manifold Σ ( C , ( K )) ∼ = X (2)0 ∪ X (2)1 ∪ (cid:102) W , where each X (2) i is a copy of X (2) . The slope (cid:101) α k = (cid:101) µ i + 2 k (cid:101) λ i on ∂X (2) i is identified with (cid:101) τ i + 2 k (cid:101) σ i = (4 − k ) (cid:101) γ i + (2 k − (cid:101) ϕ i . So (cid:102) W ( (cid:101) α k , (cid:101) α k ) is theSeifert manifold M (1 , − , k − − k , k − − k ) = − M ( − , , k − k − , k − k − ). For k ≤
0, we have k − k − < , sothe result follows from Condition (3) of Theorem 2.14. Case 3: ( n, p, q ) = (3 , , ( T , ) = M ( − , − , − , ). In this case, J has multiplicity 2, (cid:101) J hasthree components, and each component of (cid:101) J has multiplicity 2. Therefore, (cid:101) σ i = ± ( − (cid:101) γ i + (cid:101) ϕ i ).There are two cases.First, η = −
1. We have (cid:101) τ i = 3 (cid:101) γ i − (cid:101) ϕ i , (cid:101) σ i = − (cid:101) γ i + (cid:101) ϕ i . We have Σ ( C , ( K )) ∼ = X ∪ X ∪ X ∪ (cid:102) W ,where each X i is a copy of X . The gluing of the boundaries identifies α k = µ i + kλ i with (cid:101) τ i + k (cid:101) σ i =(3 − k ) (cid:101) γ i + ( k − (cid:101) ϕ i . Therefore (cid:102) W ( α k , α k , α k ) is the Seifert fibered space M (1 , k − − k , k − − k , k − − k ) = − M ( − , k − k − , k − k − , k − k − ). Since 0 < k − k − < if k ≤
0, Condition (3) of Theorem 2.14 impliesthat (cid:102) W has a horizontal foliation.Second, η = −
2. We have (cid:101) τ i = 3 (cid:101) γ i − (cid:101) ϕ i , (cid:101) σ i = 2 (cid:101) γ i − (cid:101) ϕ i . Therefore (cid:102) W ( α k , α k , α k ) is theSeifert fibered space − M ( − , k +22 k +3 , k +22 k +3 , k +22 k +3 ). Since 0 < k +22 k +3 < if k (cid:28)
0, Condition (2) ofTheorem 2.14 implies that (cid:102) W has a horizontal foliation. (cid:3) The following completes the proof of Theorem 1.3.
Proposition 4.5.
Let K be a non-trivial knot in S . Unless n = q = 2 , Σ n ( C ,q ( K )) is excellent. AUT FOLIATIONS, LEFT-ORDERABILITY, AND CYCLIC BRANCHED COVERS 21 C ‘ J ‘ C J
Figure 1.
The union of a regular fiber, C , and an exceptional fiber, J , of multi-plicity p = 5, in a Seifert fibration of S . C J
C J
Figure 2.
The union of a regular fiber, C , and an exceptional fiber, J , of multi-plicity p = 1 in a Seifert fibration of S . Proof.
Let d = gcd( n, q ) and let r = nd and s = qd . Note that gcd( r, s ) = 1. From the discussionimmediately preceding Proposition 4.2, we have that Σ n ( C ,q ( K )) = (cid:102) W ∪ (cid:96) di =1 X ( r ) i , where (cid:102) W = Σ n ( T ,q ) − d (cid:97) i =1 int ( (cid:101) V i ) = S − int ( N ( (cid:101) J )) . To identify (cid:101) J ⊂ S , it is convenient to note that there is an isotopy of S that interchanges thecomponents of C and J . With this picture in mind, it is then clear that (cid:101) J is the ( n, q )-torus link T n,q = T dr,ds . Let (cid:101) J , . . . , (cid:101) J d denote the components of (cid:101) J .As before, let (cid:101) σ i be the meridian of (cid:101) V i and (cid:101) τ i ⊂ ∂ (cid:101) V i , the lift of τ . Let (cid:101) ρ i ⊂ ∂ (cid:101) V i be the 0-framedlongitude of (cid:101) J i ⊂ Σ n ( T ,q ) ∼ = S . Since (cid:101) τ i · (cid:101) σ i = 1 on ∂ (cid:102) W , we have (cid:101) τ i = (cid:101) ρ i + c (cid:101) σ i for some c ∈ Z . Let α k be the slope µ + kλ on ∂X , and (cid:101) α k = (cid:101) µ i + rk (cid:101) λ i the corresponding slope on ∂X ( r ) i . Under the gluinghomeomorphism ∂X ( r ) i → ∂ (cid:101) V i ⊂ ∂ (cid:102) W , we have that (cid:101) α k is mapped to (cid:101) τ i + rk (cid:101) σ i = (cid:101) ρ i + ( c + rk ) (cid:101) σ i . ByProposition 2.17, for k sufficiently negative, the multislope ( (cid:101) α k , . . . , (cid:101) α k ) is excellent for (cid:102) W , unless n = q = 2. By Lemma 2.12, we have that for k (cid:28) (cid:101) α k is an excellent slope for for each X ( r ) i . Theresult again follows from Lemma 2.9. (cid:3) Cyclic branched covers of ( p, -cables. We now prove Theorems 4.1 and 1.4. Let p ≥ C ∪ J from the discussion just before Proposition 4.2 is illustrated inFigures 1 and 2, which show the cases p = 5 and p = 1 respectively. For the special case of p = 1,we relabel the components as C (cid:48) ∪ J (cid:48) , with associated meridian-longitude pair µ (cid:48) , λ (cid:48) .Now, for any p , the exterior of C ∪ J is in fact homeomorphic to C (cid:48) ∪ J (cid:48) . One way to see this is asfollows. The exterior of C ∪ J is the exterior of a regular fiber in the πp Seifert fibered solid torus.Therefore, applying r = p − negative meridional Dehn twists to this solid torus produces a π
122 CAMERON GORDON AND TYE LIDMAN CJ Figure 3.
The result of interchanging the components of C (cid:48) ∪ J (cid:48) . J (cid:48) C (cid:48) Figure 4.
The lift, (cid:101) J (cid:48) , of J (cid:48) in Σ ( T , ).Seifert fibered solid torus; this is the exterior of C (cid:48) ∪ J (cid:48) . In other words, there is a homeomorphism h : W − N ( C ) → W (cid:48) − N ( C (cid:48) ) such that h ( τ ) = τ (cid:48) and h ( σ ) = σ (cid:48) − rτ (cid:48) .We may interchange the components of C (cid:48) ∪ J (cid:48) to obtain the link in Figure 3. From this, we cansee that (cid:101) J (cid:48) ⊂ Σ ( T , ) ∼ = S is the Hopf link (see Figure 4).Let the components of (cid:101) J (cid:48) be J (cid:48) and J (cid:48) , and let the corresponding lifts of V (cid:48) be V (cid:48) and V (cid:48) .Then (cid:102) W (cid:48) ∼ = T × I , where ∂V (cid:48) i = T × { i } , for i = 0 ,
1. Consequently, we have the correspondingdecomposition Σ ( T p, ) = V ∪ V ∪ (cid:102) W . We would like to determine the gluing. If α is a slopeon ∂V (respectively ∂V (cid:48) ), let α i (respectively α (cid:48) i ) be the corresponding slope on ∂V i (respectively ∂V (cid:48) i ), for i = 0 ,
1. Note that the homeomorphism h lifts to a homeomorphism (cid:101) h : (cid:102) W → (cid:102) W (cid:48) suchthat (cid:101) h ( τ i ) = τ i and (cid:101) h ( σ i ) = σ (cid:48) i − rτ (cid:48) i , for i = 0 , β (cid:48) be the blackboard framed longitude of J (cid:48) in Figure 3. Then β (cid:48) = τ (cid:48) + σ (cid:48) , and β (cid:48) i = τ (cid:48) i + σ (cid:48) i is the 0-framed longitude of J (cid:48) i .Recall that Σ ( C p, ( K )) ∼ = X ∪ X ∪ (cid:102) W , where X i is the copy of the exterior X of K , for i = 0 ,
1, and the gluing homeomorphism ∂X i → ∂V i takes λ i to σ i and µ i to τ i , for i = 0 , (cid:101) h , Σ ( C p, ( K )) ∼ = X ∪ X ∪ (cid:102) W (cid:48) , where the gluing homeomorphism ∂X i → ∂V (cid:48) i = T × { i } now takes λ i to σ (cid:48) i − rτ (cid:48) i = σ (cid:48) i − r ( β (cid:48) i − σ (cid:48) i ) = ( r + 1) σ (cid:48) i − rβ (cid:48) i , and takes µ i to τ (cid:48) i = − σ (cid:48) i + β (cid:48) i , for i = 0 , λ i , µ i and σ (cid:48) i , β (cid:48) i , this gluing homeomorphism is given bythe matrix A = (cid:18) r + 1 − − r (cid:19) . Since (cid:102) W (cid:48) ∼ = T × I , with σ (cid:48) , β (cid:48) homotopic to β (cid:48) , σ (cid:48) , respectively, itfollows that Σ ( C p, ( K )) ∼ = X ∪ X , glued by the homeomorphism f : ∂X → ∂X that is givenwith respect to the ordered bases λ , µ and λ , µ , by the matrix A − BA , where B = (cid:18) (cid:19) , i.e.by (cid:18) r + 1 − (cid:19) = (cid:18) p − (cid:19) . In other words, we have(4.1) Σ ( C p, ( K )) = X ∪ f X , f ( aµ + bλ ) = ( pb − a ) µ + bλ . AUT FOLIATIONS, LEFT-ORDERABILITY, AND CYCLIC BRANCHED COVERS 23
J C
C J
Figure 5.
The Whitehead link C ∪ J .Observe that under this identification, a slope ab on ∂X is sent to the slope p − ab on ∂X . Proof of Theorem 4.1.
From (4.1), we have Σ ( C , ( K )) = X ∪ X , where the slope ab is identifiedwith 1 − ab . The slopes on ∂X and ∂X are therefore identified. If is a CTF slope for K , then is an excellent slope by Lemma 2.7. Thus, Σ ( C , ( K )) is excellent by Lemma 2.9. (cid:3) Proof of Theorem 1.4.
By (4.1), Σ ( C p, ( K )) ∼ = X ∪ X , where X and X are copies of the exterior X of K , glued along their boundaries so that the slope ab on ∂X is identified with the slope p − ab on ∂X . In particular, the meridians of X and X are identified.Let L ( X ) be the set of L-space slopes on ∂X , i.e. L ( X ) = { α | X ( α ) is an L-space } . By[KMOS07, OS11], L ( X ) = {∞} , [2 g − , ∞ ] , or [ −∞ , − g ] , where g is the genus of K . Here, the intervals are to be interpreted as being in Q ∪ {∞} . Hencethe meridian µ = ∞ of K is not in the interior L o ( X ). Therefore, by [HRW], Σ ( C p, ( K )) is notan L-space.If K is a torus knot or iterated torus knot then Σ ( C p, ( K )) is a graph manifold. By [BC17]and [HRRW15], for graph manifolds the properties of not being an L-space, having a co-orientabletaut foliation, and having left-orderable fundamental group, are equivalent. This proves the secondpart of the theorem. (cid:3) Remark . The mistake in the previous version of Theorem 1.4 occurs in the last three sentencesof the original argument. There it was incorrectly assumed that ∞ is not a foliation-detected slope;indeed, ∞ is a foliation-detected slope for the trefoil [BC17, Corollary A.7].5. Whitehead doubles
Let K be a knot in S and let W h ( K ) be the positive untwisted Whitehead double of K . Thiscan be described as follows. Consider the Whitehead link with a positive clasp, C ∪ J , as shown inFigure 5.Let N ( C ) be a tubular neighborhood of C , disjoint from J , with meridian m and 0-framedlongitude l . Let X be the exterior of K , with meridian µ and longitude λ . Remove intN ( C ) from S and replace it with X , identifying the torus boundaries in such a way that m is identified with λ and l is identified with µ . In the resulting S , the image of J is W h ( K ). Observe that if oneconsiders the negative untwisted double, where we instead use the Whitehead link with a negativeclasp, we obtain the mirror of the positive untwisted Whitehead double of the mirror of K . Forthis reason, we restrict our attention to positive untwisted Whitehead doubles.To prove Theorem 1.6 we first give an explicit description of Σ n ( W h ( K )). Since the componentsof the Whitehead link are interchangeable, we can redraw the link as in Figure 6.The n -fold cyclic branched cover of ( S , J ) is S , and the inverse image of C under the coveringprojection is the n -component chain link L n shown in Figures 7 and 8, which illustrate the cases J C
J C
Figure 6.
The Whitehead link after interchanging the components in Figure 5. C C C C Figure 7.
The 2-component chain link L . C C C C C C C C C C Figure 8.
The 5-component chain link L . n = 2 and n = 5 respectively. Since (cid:96)k ( J, C ) = 0, we have that π − ( C ) has n components, C i for i ∈ Z /n . Let m i , l i on ∂N ( C i ) be the lifts of m, l on ∂N ( C ). Let Y n = S − (cid:96) i ∈ Z /n intN ( C i ) bethe exterior of L n . Then, because (cid:96)k ( C, J ) = 0, we have that Σ n ( W h ( K )) is obtained by gluing to Y n , for each i , a copy X i of X along ∂N ( C i ) in such a way that m i is sent to λ i and l i is sent to µ i , where µ i , λ i are the corresponding copies of µ, λ on ∂X i .Let b be the blackboard framing of C corresponding to Figure 6; see Figure 9. Then, in H ( ∂N ( C )), we have b = 2 m + l . The lift b i of b in ∂N ( C i ) is the 0-framed longitude of C i (see Figure 10 for the case n = 2), and l i = − m i + b i . Proof of Theorem 1.6.
We first consider the case of two-fold branched covers. Note that the link L is the negative torus link T − , . Consider the slope µ i + n i λ i on X i . Note that this slopecorresponds to n i on K . For | n i | (cid:29) AUT FOLIATIONS, LEFT-ORDERABILITY, AND CYCLIC BRANCHED COVERS 25 Cb Figure 9.
The blackboard framing of C from Figure 6. C b C b Figure 10.
The lifts of the blackboard framing of C from Figure 6.above discussion, the slope µ i + n i λ i is identified with the slope ( n i − m i + b i . Since b i representsthe longitude on the i th component of T − , , we have that the slopes µ + n λ and µ + n λ on X and X respectively are identified with the multislope ( n − , n − ) on the exterior of T − , .By Proposition 2.17, we have that for n , n (cid:29)
0, ( − n , − n ) is an excellent multislope on theexterior of T , . Therefore, we have that ( n − , n − ) is an excellent multislope on the exterior of T − , for n , n (cid:29)
0. It now follows from Lemma 2.9 that Σ ( W h ( K )) is an excellent manifold.Since Σ n ( W h ( K )) is a branched cover of Σ ( W h ( K )), Theorem 2.1 shows that π (Σ n ( W h ( K ))is left-orderable for all n .We now study the higher-order branched covers. There is a degree one map from X to S × D which restricts to a homeomorphism ∂X → ∂ ( S × D ) mapping λ to a meridian of S × D and µ to a longitude of S × D . Applying this map to each X i except X defines a degree 1 map f : Σ n ( W h ( K )) → M , where M is obtained from Y n by filling in each N ( C i ) for i (cid:54) = 0, whileleaving X attached along ∂N ( C ) as before.Now, S − intN ( C ) is a solid torus whose meridian is b . Since b = 2 m + l is identified with2 λ + µ in ∂X , M is homeomorphic to S ( K ). Suppose first that π ( S ( K )) is left-orderable.Since f has degree one and Σ n ( W h ( K )) is irreducible, we have by Theorem 2.1 that π (Σ n ( W h ( K )))is left-orderable.Now assume that n ≥ X i except X and X . This gives a degree 1 map from Σ n ( W h ( K )) to Q , the manifold obtained by attaching X and X to E = S − ( int ( N ( C )) ∪ int ( N ( C ))), the exterior of the Hopf link, which is homeomorphicto T × I . In other words, Q ∼ = X ∪ h X , for some gluing homeomorphism h : ∂X → ∂X . To determine h ∗ : H ( ∂X ) → H ( ∂X ), we note that the homeomorphism ∂N ( C ) → ∂N ( C ) givenby the product structure on E sends m to b and b to m , since the b i are longitudes for C i .Then, we have that h ∗ : µ (cid:55)→ l = − m + b (cid:55)→ − b + m (cid:55)→ − µ + 2 λ ) + λ = − µ − λ and λ (cid:55)→ m (cid:55)→ b (cid:55)→ µ + 2 λ . Thus, with respect to the ordered bases µ , λ and µ , λ , h ∗ is given by the matrix (cid:18) − − (cid:19) .Let α be the slope k = µ + kλ on ∂X . Then, h ∗ ( α ) = ( k − µ + (2 k − λ on ∂X .This will be of the form k (cid:48) if and only if k = 1 or k = 3, in which cases we have h ∗ ( ) = and h ∗ ( ) = .Observe that Q is irreducible. It follows from Theorem 2.10 that if π ( S ( K )) or π ( S ( K )) isleft-orderable, then so is π ( Q ). Therefore, since we have a degree one map from Σ n ( W h ( K )) onto Q , we may again apply Theorem 2.1 to conclude that π (Σ n ( W h ( K )) is left-orderable. (cid:3) We have the following obvious corollary.
Corollary 5.1.
Let K be a knot in S that satisfies the Li-Roberts Conjecture. Then π (Σ n ( W h ( K )) is left-orderable for all n ≥ . Some excellent cyclic branched covers of two-bridge knots
Consider the two-bridge knots corresponding to rational numbers of the form2(2 k + 1)(2 (cid:96) + 1) + ε (2 (cid:96) + 1) = [2(2 k + 1) , ε (2 (cid:96) + 1)] , where k, (cid:96) ≥ ε = ±
1. By [Min82, Corollary 11.2(b)], Σ k +1 ( K [2(2 k +1) ,ε (2 (cid:96) +1)] ) is an integerhomology sphere. By Lemma 2.11, Σ n ( K [2(2 k +1) ,ε (2 (cid:96) +1)] ) is also an integer homology sphere if n divides (2 k + 1). We are interested in the case of ε = +1 for Theorem 1.8.Before proving Theorem 1.8, we give a corollary which easily follows from the theorem. Corollary 6.1. If gcd( n, k + 1) > then π (Σ n ( K [2(2 k +1) , (cid:96) +1] )) is left-orderable.Proof. Let m = gcd( n, k + 1). Then, we have a non-zero degree map from Σ n ( K [2(2 k +1) , (cid:96) +1] ) toΣ m ( K [2(2 k +1) , (cid:96) +1] ). Theorem 1.8 applies to Σ m ( K [2(2 k +1) , (cid:96) +1] ) and the result now follows fromTheorem 2.1. (cid:3) Let B ( n, k, (cid:96), sign( ε )) denote the manifold Σ n ( K [2(2 k +1) ,ε (2 (cid:96) +1)] ). Theorem 1.8 will be proved bystudying the following surgery description of B ( n, k, (cid:96), ± ) when n divides (2 k + 1). Lemma 6.2.
Suppose n divides k + 1 , n > , and let d = (2 k + 1) /n . Let P (2 (cid:96) + 1 , . . . , (cid:96) + 1) be the n -stranded pretzel knot with (2 (cid:96) + 1) right-handed half-twists in each strand. Then (1) B ( n, k, (cid:96), +) ∼ = − S d ( P (2 (cid:96) + 1 , . . . , (cid:96) + 1)) , (2) B ( n, k, (cid:96), − ) ∼ = S − d ( P (2 (cid:96) + 1 , . . . , (cid:96) + 1)) .Proof. (1) Let K = − K [2(2 k +1) , (cid:96) +1] ; see Figure 11, which illustrates the case k = 1, (cid:96) = 2. Considerthe 2-component link J ∪ C shown in Figure 12. Performing k +1 -surgery on C transforms ( S , J ) to( S , K ). Since J is unknotted, the n -fold cyclic branched covering of ( S , J ) is S ; let (cid:101) C = π − ( C )be the inverse image of C under the branched covering projection π . Since (cid:96)k ( C, J ) = ±
2, and n is odd, (cid:101) C is connected. In fact, since the components of J ∪ C are interchangeable, we can seefrom Figure 12 that (cid:101) C is the pretzel knot P (2 (cid:96) + 1 , . . . , (cid:96) + 1); see Figure 13, which shows the case k = 1, (cid:96) = 2, n = 3. Let µ, λ be a meridian and 0-framed longitude of C respectively. We have that AUT FOLIATIONS, LEFT-ORDERABILITY, AND CYCLIC BRANCHED COVERS 27
Figure 11.
The knot − K [6 , . CJ J C
Figure 12.
The 2-component link J ∪ C . We can obtain K from J by doing k +1 -surgery on C . Figure 13.
The pretzel knot P (5 , , (cid:101) λ is connected, while since (cid:96)k ( µ, J ) = 0, we have π − ( µ ) consists of n copies of (cid:101) µ , the meridian of (cid:101) C . Also, C bounds a disk D that meets J in two points (see Figure 12 with the roles of J and C reversed). Then π − ( D ) is an orientable surface with boundary (cid:101) C . It follows that π − ( λ ) = (cid:101) λ isthe 0-framed longitude of (cid:101) C .Let α = µ + (2 k + 1) λ be the slope k +1 on the boundary of a neighborhood of C . Since n divides2 k + 1, π − ( α ) consists of n copies of the slope (cid:101) α = (cid:101) µ + d (cid:101) λ on the boundary of a neighborhood of (cid:101) C . It follows that the n -fold cyclic branched covering of K is obtained by d -surgery on (cid:101) C , whichproves (1).The proof of (2) is completely analogous. (cid:3) Proof of Theorem 1.8.
Since P = P (2 (cid:96) + 1 , . . . , (cid:96) + 1) is alternating and negative, it follows from[Rob95, Theorem 0.2] that S r ( P ) has a taut foliation for all r ∈ Q , r ≥
0. (Note that theconvention for the signs of crossings in [Rob95] is the opposite of the usual one.) In particular B ( n, k, (cid:96), +) ∼ = − S d ( P ) has a taut foliation. Since B ( n, k, (cid:96), +) is an integer homology sphere, it isexcellent by Lemma 2.7. (cid:3) Remark . (1) The manifolds B ( n, k, (cid:96), ± ) in Lemma 6.2 are not L-spaces by [OS05a, Theo-rem 1.5].(2) In the case ε = −
1, [Rob95, Theorem 0.2] does not apply to S − d ( P (2 (cid:96) + 1 , . . . , (cid:96) + 1)); wedo not know if B ( n, k, (cid:96), − ) has a taut foliation, for any k, (cid:96) ≥ k = (cid:96) = 1 and k = 1, (cid:96) = 2, π ( B (2 k +1 , k, (cid:96), − )) has a non-trivial representation into P SL (2 , R ). Since B (2 k + 1 , k, (cid:96), − ) is aninteger homology sphere, the obstruction to lifting to (cid:94) SL ( R ) vanishes. Since (cid:94) SL ( R ) is aleft-orderable group (it is a subgroup of Homeo + ( R )), we may apply Theorem 2.1 to thelift. Therefore, π ( B (2 k + 1 , k, (cid:96), − )) is left-orderable. Dunfield has also shown that, in thecase (cid:15) = +1, π ( B (2 k + 1 , k, (cid:96), +) has a non-trivial P SL (2 , R )-representation for k = 1,1 ≤ (cid:96) ≤ ε = +1, we have 2(2 k + 1)(2 (cid:96) + 1) + ε ≡ π (Σ n ( K [2(2 k +1) , (cid:96) +1] )) is left-orderable for all sufficiently large n . We do not know if thecorresponding manifolds have co-orientable taut foliations or whether or not they are L-spaces.Arguments similar to those that appear in [Hu15] are used in [Tra15] to show (in partic-ular) that π (Σ n ( K [2(2 k +1) , − (2 (cid:96) +1)] ) is also left-orderable for n sufficiently large. Moreover,in [Tra15] explicit lower bounds for n are given, both for ε = +1 and ε = −
1. However,these bounds go to infinity as (cid:96) increases. Again we do not know if these manifolds haveco-orientable taut foliations, or whether or not they are L-spaces.7.
Total L-spaces arising from two-bridge knots
Proof of Theorem 1.9.
That these manifolds are L-spaces is proved in [Ter14].It thus remains to show that π (Σ n ( K [2 (cid:96), − k ] )) is not left-orderable. We first note that [2 (cid:96), − k ] = k(cid:96) − k = [2 (cid:96) − , , k − π (Σ n ( K [2 (cid:96) − , , k − )). Replacing k and (cid:96) in the presentation given there by k − (cid:96) −
1, foreach i ∈ Z /n , the relators r i from this presentation can be written as( x − ki x ki +1 ) (cid:96) ( x − ki +2 x ki +1 ) (cid:96) − ( x − ki +2 x k − i +1 ) . Since there is an automorphism of π (Σ n ( K [2 (cid:96) − , , k − )) given by sending x i to x i +1 for all i ∈ Z /n ,no x i is trivial. Also, we see that in r i , the occurrences of a given generator either all have positiveexponent or all have negative exponent. AUT FOLIATIONS, LEFT-ORDERABILITY, AND CYCLIC BRANCHED COVERS 29
We take n = 4 and write a = x , b = x , c = x , d = x . Then, π (Σ ( K [2 (cid:96), − k ] )) has apresentation with generators a, b, c, d and relations abcd = 1 , (7.1) ( a − k b k ) (cid:96) ( c − k b k ) (cid:96) − ( c − k b k − ) = 1 , (7.2) ( b − k c k ) (cid:96) ( d − k c k ) (cid:96) − ( d − k c k − ) = 1 , (7.3) ( c − k d k ) (cid:96) ( a − k d k ) (cid:96) − ( a − k d k − ) = 1 , (7.4) ( d − k a k ) (cid:96) ( b − k a k ) (cid:96) − ( b − k a k − ) = 1 . (7.5)The signs of the exponents of the occurences of the generators in the relators (7.2), (7.3), (7.4),and (7.5), respectively, are( − , + , − , ◦ ) , ( ◦ , − , + , − ) , ( − , ◦ , − , +) , and (+ , − , ◦ , − ) , where the coordinates correspond to ( a, b, c, d ), and ◦ indicates that the corresponding generatordoes not appear. Also, the relator (7.1) has exponent signs (+ , + , + , +).Now suppose for contradiction that π (Σ ( K [2 (cid:96), − k ] )) is left-orderable. The above observationsprovide a number of restrictions on the purported order. For instance, we cannot have a < b > c <
1, since this implies the left hand side of (7.2) is positive, and consequently not 1. Byanalyzing all of these conditions, we may assume that after possibly applying a cyclic automorphismto ( a, b, c, d ), we have a, b > c, d < b > c − >
1, relation (7.2) gives(7.6) a − k b k < . We rewrite (7.5) as d − k ( a k d − k ) (cid:96) − ( a k b − k ) (cid:96) a k − = 1 . Since a > d − >
1, we get(7.7) a k b − k < . We now claim that(7.8) da > . If not, then da <
1, and hence, since d <
1, we have a < d − ≤ d − k . Therefore, a − d − k >
1. Again,we rewrite (7.5) as(7.9) a k ( d − k a k ) l − ( b − k a k ) (cid:96) ( a − d − k ) = 1 . Note that a > d − >
1, and further, by (7.6), b − k a k >
1. Therefore, the product of the termsabove on the left hand side of (7.9) is positive. This gives a contradiction. Therefore, we have that da > da >
1, (7.1) gives(7.10) bc < . We rewrite (7.2) as ( b k a − k ) (cid:96) − b k ( c − k b k ) (cid:96) − c − k b k − a − k = 1 . By (7.7), we have that a − k > b − k , and so we get1 > ( b k a − k ) (cid:96) − b k ( c − k b k ) (cid:96) − c − ( k − ( c − b − ) . However, since b > , c − > b k a − k > c − b − > (cid:3) If K is a two-bridge knot corresponding to a rational number of the form [2 a , a , . . . , a k ] where a i >
0, for 1 ≤ i ≤ k , then Σ n ( K ) is the branched double cover of an alternating link [MV01].Therefore, for any n ≥
2, Σ n ( K ) is an L-space by [OS05b, Proposition 3.3] and π (Σ n ( K )) is notleft-orderable [BGW13, Theorem 4] (see also [Gre11, Ito13, LL12]). The results of [Hu15, Tra15]say that for certain two-bridge knots K , we have that π (Σ n ( K )) is left-orderable for all sufficientlylarge n . The situation for cyclic branched covers of torus knots is described in Theorem 1.2. Theseresults all suggest the following question (compare to Conjecture 1.7). Question 7.1.
Let K be a knot in S . If π (Σ m ( K )) is left-orderable, is π (Σ n ( K )) left-orderablefor n ≥ m ? Recall that the answer to Question 7.1 is yes if K is prime and m divides n . An interestingexample from this point of view is 5 = K [4 , − . It is shown in [Hu15] (see also [Tra15]) that π (Σ n (5 )) is left-orderable for all n ≥
9. On the other hand, π (Σ n (5 )) is not left-orderable for n = 2 (since Σ (5 ) is a lens space), n = 3 [DPT05], and n = 4, by Theorem 1.9. Question 7.2. Is π (Σ n (5 )) left-orderable for ≤ n ≤ ? Masakazu Teragaito has informed us that Mitsunori Hori has shown that Σ (5 ) is an L-space,and thus the conjectural equivalence of (1.1) and (1.3) would suggest that π (Σ (5 )) is not left-orderable. 8. Three-strand pretzel knots
Proof of Theorem 1.10.
Let K = P (2 k + 1 , (cid:96) + 1 , m + 1) where k, (cid:96), m ≥
1. That Σ ( K ) is anL-space is proved in [Ter15]. We will construct an explicit presentation of π (Σ ( K )). Let M bethe exterior of K . It is shown in [Tro63] that π ( M ) has a presentation with generators x, y, z andrelators(8.1) ( xy − ) m x ( xy − ) − m ( yz − ) k +1 z − ( yz − ) − ( k +1) (8.2) ( yz − ) k y ( yz − ) − k ( zx − ) (cid:96) +1 x − ( zx − ) − ( (cid:96) +1) (8.3) ( zx − ) (cid:96) z ( zx − ) − (cid:96) ( xy − ) m +1 y − ( xy − ) − ( m +1) where x, y, z are meridians of K .It is straightforward to verify that the product of the relators (8.1), (8.2) and (8.3) is the identityin the free group on x, y and z , and so relator (8.3) may be eliminated. Let X be the 2-complexcorresponding to the resulting presentation; thus X has a single 0-cell c , three 1-cells e x , e y , e z corresponding to x, y, z , respectively, and two 2-cells D , D corresponding to the relators (8.1)and (8.2). Let p : X → X be the 3-fold cyclic cover. Then p − ( c ) = { c , c , c } , say, and p − ( e x ) = e (0) x ∪ e (1) x ∪ e (2) x , where e ( i ) x is a path in X from c i to c i +1 , i ∈ Z /
3; similarly for p − ( e y )and p − ( e z ). Each 2-cell D j , j ∈ { , } , lifts to three 2-cells D ( i ) j , i ∈ Z / π ( X ) we choose a maximal tree in the 1-skeleton X (1)3 ; we take thisto be e (0) z ∪ e (1) z . The path e ( i ) x now represents an element x i ∈ π ( X ), and similarly for e ( i ) y , e ( i ) z .Note that z = z = 1 ∈ π ( X ). The 2-cells D ( i ) j give the following relations, where i ∈ Z / x i y − i ) m x i ( x i +1 y − i +1 ) − m ( y i +1 z − i +1 ) k +1 z − i ( y i z − i ) − ( k +1) = 1(8.5) ( y i z − i ) k y i ( y i +1 z − i +1 ) − k ( z i +1 x − i +1 ) (cid:96) +1 x − i ( z i x − i ) − ( (cid:96) +1) = 1 . Note that π ( X ) ∼ = π ( M ), where M is the three-fold cyclic cover of M . Let µ be a meridian of K and π : M → M the covering map. Then, recall that Σ ( K ) is given by M ( π − ( µ )). Thus, to geta presentation of π (Σ ( K )) we must adjoin the branching relations x x x = y y y = z z z = 1. AUT FOLIATIONS, LEFT-ORDERABILITY, AND CYCLIC BRANCHED COVERS 31 i x x x y y y − ◦ − + ◦ ◦ + − ◦ − +2 − ◦ + + ◦ − Table 1.
Signs of the exponents of the generators as they appear in the relation(8.6) for i ∈ Z /
3. The notation ◦ indicates that the generator does not appear inthe corresponding relation. i x x x y y y − ◦ + − ◦ ◦ + − ◦ + − − ◦ + − ◦ + Table 2.
Signs of the exponents of the generators as they appear in the relation(8.7) for i ∈ Z /
3. The notation ◦ indicates that the generator does not appear inthe corresponding relation.Since we have that z = z = 1, we must have that z = 1 as well. Eliminating z , z , and z from(8.4) and (8.5), we obtain(8.6) ( x i y − i ) m x i ( x i +1 y − i +1 ) − m y k +1 i +1 y − ( k +1) i = 1(8.7) y k +1 i y − ki +1 x − ( (cid:96) +1) i +1 x (cid:96)i = 1 . As in the proof of Theorem 1.9, the relations (8.6) and (8.7) have the property that for eachgenerator, the exponents of all the occurrences of that generator in the relator have the same sign.These signs are given in Tables 1 and 2, for relations (8.6) and (8.7) respectively.Suppose that there exists a left-invariant order, < , on π (Σ ( K )). Since there is an automorphismof π (Σ ( K )) sending x i to x i +1 , if some x i = 1 then x = x = x = 1. Relation (8.7) then gives y k +1 i = y ki +1 , i ∈ Z /
3, which implies that each y i has finite order. Since we are assuming that π (Σ ( K )) is left-orderable, y = y = y = 1 and so π (Σ ( K )) = 1, a contradiction. Similarly, itfollows that no y i is trivial in π (Σ ( K )).We say that an element g ∈ π (Σ ( K )) is positive (respectively negative ) if g > g < i ∈ Z /
3, if x i and x i +1 have oppositesign then y i and y i +1 have the same sign. On the other hand, the relations x x x = 1, y y y = 1show that the x i cannot all have the same sign, and the y i cannot all have the same sign. This isclearly a contradiction. (cid:3) More examples
We discuss a few more families of cyclic branched covers of knots.9.1.
Knot epimorphisms.
Other examples of knots K with π (Σ n ( K )) left-orderable arise fromthe following easy lemma. Lemma 9.1.
Let
K, K (cid:48) be prime knots in S . Suppose there exists an epimorphism ϕ : π ( S − K ) → π ( S − K (cid:48) ) such that ϕ ( µ ) = µ (cid:48) for meridians µ, µ (cid:48) of K, K (cid:48) respectively. If π (Σ n ( K (cid:48) )) isleft-orderable, then so is π (Σ n ( K )) .Proof. Since there is an exact sequence1 → π (Σ n ( K )) → π ( S − K ) / (cid:104)(cid:104) µ n (cid:105)(cid:105) → Z /n → , and similarly for K (cid:48) , we have that ϕ induces an epimorphism from π (Σ n ( K )) to π (Σ n ( K (cid:48) )). Theresult now follows from Theorem 2.1. (cid:3) Examples to which Lemma 9.1 applies are given in [KS05, KS08], which list all pairs of distinctprime knots
K, K (cid:48) with at most 10 crossings such that there exists an epimorphism π ( S − K ) → π ( S − K (cid:48) ) (written K ≥ K (cid:48) in the notation of [KS05, KS08]). The authors note in [KS08] thatin all cases, there is actually a meridian-preserving epimorphism. Also, the knot K (cid:48) is either 3 ,4 , or 5 . It follows from Theorem 1.2 that for the knots with K ≥ , we have π (Σ n ( K )) isleft-orderable for n ≥
6. It also follows from [Hu15] (see also [Tra15]) that for the knots K with K ≥ , we have π (Σ n ( K )) is left-orderable for n ≥
9. Since all cyclic branched covers of 4 havenon-left-orderable fundamental group [DPT05], we are not able to say anything about π (Σ n ( K ))for K ≥ .9.2. Unknotting number one, determinant one knots.Proposition 9.2.
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