Hyperbolic Knots And Cyclic Branched Covers

Publ. Mat. 49 (2005), 257–284 HYPERBOLIC KNOTS AND CYCLIC BRANCHED COVERS Luisa Paoluzzi Abstract We collect several results on the determination of hyperbolic knots by means of their cyclic branched covers. We construct examples of knots having two common cyclic branched covers. Finally, we briefly discuss the problem of determination of hyperbolic links. 1. Introduction To each knot Kin S3and each integer n≥2 one can associate, in a natural way, a 3-manifold M(K, n) which is the total space of the n-fold cyclic cover of S3branched along Kand which is called —by abuse of language— the n-fold cyclic branched cover of K. There are two standard ways to construct M(K, n): •Let U(K) denote a small open tubular neighbourhood of K. Take the complement S3\ U(K) of Kin S3and let Mnbe its unique n-fold cyclic cover. The boundary torus of S3\ U(K) lifts to the boundary torus of Mn. The closed manifold M(K, n) is obtained by gluing a solid torus along the boundary of Mnin such a way that the meridian of the solid torus is identified with the lift of the meridian of K. Note that the action on Mnof the cyclic group of covering transformations extends to an action on M(K, n), such that each non trivial element of the group acts on the added solid torus as a rotation about its core, in particular it fixes pointwise the core. Remark, moreover, that M(K, n) admits a projection onto S3—the space of orbits of the action— whose restriction to Mncoincides with the covering and such that the core of the solid torus projects precisely to Kin S3. 2000 Mathematics Subject Classification. Primary: 57M25; Secondary: 57M12, 57M50. Key words. Hyperbolic knots and links, cyclic branched covers, hyperbolic and Seifert structures, Jaco-Shalen-Johannson decomposition. 258 L. Paoluzzi •Consider the manifold Σ obtained by cutting S3open along a Seifert surface Ffor K. The boundary of Σ consists of two copies F+and F−of F. The manifold M(K, n) is obtained by gluing cyclically together ncopies of Σ in such a way that F+in the i-th copy is glued to F−in the (i+ 1)-st copy, where indices are taken mod n(see [Ro, Chapter 6]). The cyclic action, in this case, is just the cyclic permutation of the ncopies of Σ. Note that the common boundary of the F+’s and F−’s is pointwise fixed. The manifolds M(K, n) are easily seen to form an infinite family of topological invariants for K, depending on one integer parameter n≥2. In the very special case where Kis the unknot, the manifolds M(K, n) are readily seen to be all homeomorphic to S3. In this case, Smith’s conjecture assures that, for each fixed n≥2, the unknot is the unique knot Ksuch that M(K, n) is the 3-sphere. A more precise statement is the following: Theorem (Smith’s conjecture).Let φ:S3−→ S3be a finite order orientation preserving diffeomorphism which is not the identity. If its fixedpoint set is not empty then it is the unknot. The proof of this result was established by Waldhausen [W] for diffeomorphisms of order 2 and was finally achieved in [MB] thanks to the contributions of several mathematicians. In general, however, these topological invariants are not strong enough to distinguish a knot Kup to equivalence, i.e. there exist knots Kand K0 whose n-fold cyclic branched covers are homeomorphic for all neven if the pairs (S3, K) and (S3, K0) are not homeomorphic. This is, for instance, the case, already known to Viro [V], of certain composite knots. Indeed, take a non invertible oriented knot K; the two knots K=K]K and K0=K](−K), where −K denotes Kendowed with the opposite orientation, have homeomorphic n-fold cyclic branched covers for all n (indeed, M(K]K, n)∼ =M(K, n)]M(K, n)∼ =M(K](−K), n), because the orientation of the knot cannot be detected by the cover) without being equivalent. Even for prime knots, cyclic branched covers can be very weak invariants. This fact is well-illustrated by the behaviour of double covers. Examples of non equivalent prime knots with the same 2-fold cyclic branched covers have long been known: the mutants of Conway [C] and Kinoshita-Terasaka [KT], and the Montesinos knots [Mo] are possibly the most famous examples. These examples are based on the same principle, i.e. performing an isotopy along an incompressible torus of M(K, 2) Hyperbolic Knots and Cyclic Branched Covers 259 does not change M(K, 2) but results in a modification —called Conway mutation— along a Conway sphere of the quotient knot which may change the knot. Recall that a Conway sphere Sis a sphere which meets the knot Kin four points and such that S\ U(K) is incompressible and ∂-incompressible in S3\ U(K). Note that a Conway sphere for K can also be seen as the quotient of an incompressible torus of M(K, 2) which meets the preimage of Kin four points. A Conway mutation consists in the following operation: cut the pair (S3, K) along a Conway sphere to obtain two tangles inside two balls; reglue the two balls after rotating one of them of an angle π. Note that there are three possible Conway mutations along a Conway sphere. In Figure 1, the Conway mutation transforming the knot of Conway into that of Kinoshita-Terasaka is shown. The three possible Conway mutations along a Conway sphere are also illustrated. possible mutations Kinoshita-Teresaka Conway π π π π Figure 1 In the case of Montesinos knots, the double cover is a Seifert fibred manifold with base S2and orientable fibration. Assume that all the singular fibres are positioned along the equator of S2. It is easy to see that such a manifold admits an involution with non-empty fixed-point 260 L. Paoluzzi set which acts on the base as a reflection with respect to the equator and which reverses a given orientation of the fibres. In particular it fixes setwise each fibre that projects to a point on the equator and acts on it by fixing precisely two points. By removing all singular fibres (and at least one fibre) from the equator, one obtains a fibred solid torus with some fibres removed. It is easy to see that the quotient of such a drilled solid torus by the involution is a 3-sphere from which some balls have been removed. The image of the fixed-point set consists of pairs of arcs joining adjacent boundary components. Filling back in the solid tori containing the singular fibres corresponds to gluing the removed balls. These added balls contain rational tangles connecting the arcs. The double cover is a particularly bad invariant for Montesinos knots. Indeed, there are many different ways to position the singular fibres of a given manifold along the equator. Note that, if the manifold has at least four singular fibres, one can exchange the position of two of them by performing an isotopy along an incompressible torus which separates the two fibres from the remaining ones. This results in a permutation of two rational tangles, which may give rise to different quotient knots. Indeed, one can construct arbitrary many non-equivalent Montesinos knots with the same 2-fold cyclic branched cover. Tk T1 T2 Tj aistands for aicrossings a1 a2 a3 an a2 a1 a3 Figure 2 Figure 2 shows schematically a Montesinos knot and a rational tangle. More generally, Montesinos and Whitten [MW] showed that constructing different knots with the same 2-fold cyclic branched cover is, in principle, extremely easy. Their idea is to paste together, along their boundaries, link complements admitting different symmetries of order 2. If the gluing is performed in an equivariant way, the symmetries induce Hyperbolic Knots and Cyclic Branched Covers 261 involutions of the resulting closed manifold. Under appropriate conditions, the quotient of the manifold by the action of the involutions with non empty fixed-point set is the 3-sphere and different knots are recovered as the images of the fixed-point sets of different involutions. Covers of larger orders seem to be better invariants for prime knots. For examples, Thurston’s hyperbolic Dehn surgery theorem (see, for instance, [BeP] for this and other basic results in hyperbolic geometry) implies that prime knots which are also hyperbolic (i.e. their complement admits a complete hyperbolic structure of finite volume) are determined by (any of) their cyclic branched covers of sufficiently large orders. More precisely, the above statement is a consequence of the following corollary of Thurston’s hyperbolic Dehn surgery theorem whose proof can be found in [Ko, Lemma 4]: Proposition (Shortest geodesic).If Kis a hyperbolic knot, then its cyclic branched covers are hyperbolic manifolds provided that nis sufficiently large. Moreover, the preimage of Kin its n-fold cyclic branched cover Mis the unique shortest geodesic of M, provided again that nis sufficiently large. Indeed, assume now that Mis the n-fold cyclic branched cover of two knots. Both knots lift to the unique shortest geodesic for the unique (according to Mostow’s rigidity theorem) hyperbolic metric for M. Using Thurston’s orbifold geometrization theorem (see [BoP] and [CHK] for a proof), both groups of deck transformations can be chosen to act by isometries for the hyperbolic metric. In particular they both act as groups of rotations of order nabout the shortest geodesic and thus coincide. Kojima [Ko] proved, more generally, that the above property is true for arbitrarily prime knots, showing that cyclic branched covers of large orders are indeed good invariants for prime knots: Theorem (Kojima).For each prime knot Kthere exists a constant nK such that two prime knots Kand K0are equivalent if their n-fold cyclic branched covers are homeomorphic for some n > max(nK, nK0). Kojima’s result is based on an induction argument on the length of the Jaco-Shalen-Johannson decomposition [JS], [J] of the complement of the knot. When the decomposition is trivial the result follows from the above proposition in the case where the complement admits a hyperbolic structure, and from a combinatorial analysis in the case where it admits a Seifert fibred structure (torus knot). 262 L. Paoluzzi However, there are drawbacks in considering covers of large orders. Covers of large orders are clearly more complex (to construct and understand) than those of small orders. Moreover, Sakuma [Sa1] and Nakanishi [N] constructed, for each fixed n, pairs of non equivalent prime knots with the same n-fold cyclic branched cover, so there is no universal bound on the constant nKin the above theorem. Besides, using a result of Zimmermann’s [Z2] that we shall discuss in more detail in Section 3, one can construct, for each fixed n≥2, pairs of non-equivalent hyperbolic knots with the same n-fold cyclic branched cover, so that there is no universal bound even in the case of hyperbolic knots. As a consequence, Kojima’s result cannot be exploited in practice. Thus the natural question, put originally by Boileau and Flapan [BF], is to understand whether a finite set of cyclic branched covers (preferably of small orders) and of fixed cardinality is sufficient to determine prime knots. In their paper, Boileau and Flapan answered this question for the class of π-hyperbolic knots, i.e. hyperbolic knots whose double cover admits a hyperbolic structure. They showed that these knots are determined by their 2-fold and 4-fold cyclic branched cover if they are not strong invertible (this condition was eventually proved to be superfluous by Zimmermann in [Z2]). For the most basic class of prime knots, that is, for hyperbolic knots, the analysis of their cyclic branched covers not only leads to a positive answer to the aforementioned question, but also to a complete understanding of their behaviour. The following result [P3], [Z1], [P2] answers Boileau and Flapan’s question: Theorem 1. Three cyclic branched covers suffice to determine hyperbolic knots. Moreover, two cyclic branched covers are not sufficient in general. In this paper we shall collect the results concerning the determination of hyperbolic knots by means of their cyclic branched covers, obtained mainly by Zimmermann, Mecchia and Reni, and the author. We shall mainly discuss the case of covers of “small” orders: the interested reader is referred to [RZ2] for a survey concerning covers of “large” orders (i.e. admitting a hyperbolic structure). More specifically, in [RZ2] Reni and Zimmermann discuss the following problem: given a hyperbolic manifold M, find all the knots Kfor which M=M(K, n) for some n≥2, and describe how they are related. They show that, if M is a cyclic branched cover of large order of more than one knot, then the quotient knots are symmetric (compare Section 3). More precisely, a central result of [RZ2, Theorem 6] is the following: Hyperbolic Knots and Cyclic Branched Covers 263 Theorem (Reni-Zimmermann).Assume that M=M(K, n), where n > 3. If M=M(K0, n0), then either the symmetry group of Kis non trivial or K=K0and n=n0. In the following two sections, we shall start by pointing out the difference in behaviour between the 2-fold and n-fold, n > 2, cyclic branched covers, this difference reflects the fact that the “geometric” structure of the double cover of a hyperbolic knot can be extremely complex, while (almost) all covers of larger orders are hyperbolic. Section 2 is devoted to the case of double covers while Section 3 deals with covers of larger orders. In Section 4 various sufficient conditions for a knot to be determined by one or two of its cyclic branched covers will be given. In Section 5, we shall construct new examples of hyperbolic knots with two common cyclic branched covers. The last part of the paper (Section 6) is substantially new and concerns the determination of hyperbolic links using their cyclic branched covers. The author is indebted to M. Boileau for pointing out this possible application. 2. Flexibility of double covers: a sample of possible situations It was observed in the introduction that the 2-fold cyclic branched covers of knots are the simplest invariants to construct but also the least useful in general. Their behaviour depends on their geometric structure, which is somehow reflected in the the knot by the presence (or absence) of Conway spheres. Recall that a knot is called Conway irreducible if it does not admit any Conway sphere and Conway reducible otherwise. We summarise here the different situations that can occur. Kis Conway irreducible. Because of Thurston’s orbifold geometrization theorem, the 2-fold branched cover M(K, 2) of Kis a geometric manifold and there are three possible cases: (i) M(K, 2) is a lens space and thus admits a spherical structure. In this case Kis a 2-bridge knot. It was proved by Hodgson and Rubinstein [HR, 4.9 Theorem] that 2-bridge knots are determined by their double covers so Kis the unique knot admitting M(K, 2) as 2-fold cyclic branched cover. 264 L. Paoluzzi (ii) M(K, 2) is not a lens space but admits a Seifert fibration. In this case Kis a Montesinos knot with at most three rational tangles. M(K, 2) can be the 2-fold cyclic branched cover of at most one knot K0 non equivalent to Kand K0is a torus knot. (iii) M(K, 2) admits a hyperbolic structure, i.e. Kis π-hyperbolic. It was proved by Reni [Re2, Corollary 1] that M(K, 2) can be the 2-fold cyclic branched cover of at most nine non equivalent knots which are all hyperbolic. In this case, if M(K, 2) does not determine K,Kadmits at least one symmetry of order 2 with non empty fixed-point set. Given K, Mecchia and Reni [MR, Theorem 2] showed that it is possible to recover the remaining knots which admit M(K, 2) as 2-fold cyclic branched cover by lifting in all possible ways the iterated quotients of Kvia symmetries of order 2. Note that there are examples of three [Z1, Chapter 5] respectively four [RZ1, Chapter 4] pairwise non equivalent π-hyperbolic knots with the same double cover. The three non-equivalent knots of [Z1, Chapter 5] are all strongly invertible. The image of each knot together with the axis of the strong inversion, in the quotient by the action of the strong inversion, is a theta-curve, the same theta-curve for all three knots. In other words, each knot projects to a different edge of the thetacurve and each pair of edges forms a trivial knot which is the image of an axis. It is not known yet whether the upper bound of nine non equivalent hyperbolic knots is attained, however Mecchia and Zimmermann [MZ] constructed examples of nine non equivalent π-hyperbolic knots with the same double cover, contained in a Z-homology sphere. Remark. We call symmetry of a knot Ka finite order diffeomorphism of the pair (S3, K) which preserves the orientation of the 3-sphere. A symmetry is n-periodic if its order is nand its fixed-point set is non empty and disjoint from K. A symmetry is a strong inversion if its order is 2 and its fixed-point set is non empty and intersects Kin exactly two points. A strong inversion reverses the orientation of K. Kis Conway reducible. Under this hypothesis, M(K, 2) contains an incompressible torus. (iv) M(K, 2) is geometric and thus admits a Seifert fibration. In this case Kis a Montesinos knot with at least four tangles. Two Montesinos knots have the same 2-fold cyclic branched cover if their rational tangles are the same up to permutation. As described in the Introduction, any permutation of tangles is a Conway mutation of a special type, moreover on the double cover, any Conway mutation corresponds to an Hyperbolic Knots and Cyclic Branched Covers 265 isotopy along an incompressible torus. It is well-known that the number of pairwise non equivalent Montesinos knots grows exponentially with the number of (pairwise distinct) rational tangles. (v) M(K, 2) has a non trivial Jaco-Shalen-Johannson decomposition. As for Montesinos knots, M(K, 2) can be the double cover of arbitrarily many hyperbolic knots. To recover them it is necessary to localise, on the geometric pieces of the decomposition for M(K, 2), the situations seen for Montesinos and π-hyperbolic knots [P1]. Note that, using the techniques of Montesinos and Whitten [MW], it is not difficult to construct hyperbolic knots which share the same 2-fold cyclic branched cover of a satellite knot. Explicit examples can be found in [RZ2, Chapter 5] and [P1, Chapter 3]. Remark that sometimes it is not possible to detect the existence of satellite knots double covered by M(K, 2), just by looking at the hyperbolic knot K[P1, Theorem 3.2]. ∆8∆6 ∆4 ∆8∆4∆6 Figure 3 To illustrate the different behaviours we show different sets of hyperbolic knots with the same 2-fold cyclic branched covers. Figure 3 shows three π-hyperbolic knots with a common theta-curve quotient (∆2denotes the full twist on three strands). Figure 4 shows two Montesinos knots differing by a permutation of their rational tangles. Figure 5 shows different Conway reducible hyperbolic knots which are not Montesinos knots. In this last case the knots can differ by a Conway mutation 272 L. Paoluzzi The construction described above will be used again in Section 5 to obtain hyperbolic knots with two common cyclic branched covers and certain additional properties. 4. Pairs of covers which determine a hyperbolic knot In the previous section, we have seen under which conditions two cyclic branched covers of orders strictly larger than 2 suffice to determine hyperbolic knots. In this section we shall consider the case when one of the two cyclic branched covers is the double one. The following result gives a variety of sufficient conditions: Theorem 3. Let n > 2and Kbe a hyperbolic knot. Kis determined by its 2-fold and n-fold cyclic branched covers (together) if one of the following conditions holds: (i) Kis a 2-bridge knot; (ii) Kis a Montesinos or, more generally, an arborescent knot; (iii) Kis π-hyperbolic and nis even; (iv) Kis π-hyperbolic and admits a unique incompressible Seifert surface up to isotopy, for instance Kis fibred. Note that in case (i) the result of Hodgson and Rubinstein [HR] says that the double cover alone suffices to determine K. For case (ii) note that if Kis a Montesinos knot with at most 3 rational tangles and which is not determined by M(K, 2) then it shares its double cover with a torus knot K0. However, the hyperbolic manifold M(K, n) cannot be the n-fold cyclic branched cover of K0. If Kis a Montesinos knot with at least four rational tangles or an arborescent knot, the result is proved in [P2, Corollary 2]. Case (iii) is similar to that of higher order covers and was proved by Zimmermann [Z2, Theorem 3]. The last case (iv) is proved in [P4, Theorem 2]. The conditions of Theorem 3 are in some sense best possible because of the following examples: Theorem 4. (a) For each fixed n > 2, it is possible to construct pairs of non equivalent Conway reducible hyperbolic knots with the same 2-fold and n-fold cyclic branched covers [P2, Theorem 1]. (b) For each fixed odd n > 2, it is possible to construct pairs of non equivalent π-hyperbolic knots with the same 2-fold and n-fold cyclic branched covers [P4]. Hyperbolic Knots and Cyclic Branched Covers 273 Point (ii) of Theorem 3 implies that the 2-fold cyclic branched cover of a Conway reducible hyperbolic knot which is not determined by its 2-fold and n-fold cyclic branched covers together must have a non trivial Jaco-Shalen-Johannson decomposition containing at least one hyperbolic piece. Indeed, there is a “special” hyperbolic piece of the decomposition (see [P2, Claim 5] for a more detailed explanation). Such a piece contains the fixed-point set of the lift of the n-periodic symmetry of the knot, whose existence is given by Theorem 2. The covering involutions of the knots constructed in [P2] —those of point (a)— coincide (up to conjugation) on the special hyperbolic piece. A different situation can arise, namely there are pairs of non equivalent Conway reducible hyperbolic knots with the same 2-fold and n-fold cyclic branched covers (nodd) whose covering involutions do not coincide on the special hyperbolic piece. These examples will be discussed in the next section. 5. Hyperbolic knots with common cyclic branched covers: an example Let Kbe a Conway reducible hyperbolic knot which is not determined by its 2-fold and n-fold cyclic branched covers, i.e. there exists a (necessarily) hyperbolic Conway reducible hyperbolic knot K0which has the same 2-fold and n-fold cyclic branched covers as K. It was proved in [P2] that the 2-fold cyclic branched cover Mof Kadmits a non trivial Jaco-Shalen-Johannson decomposition containing a hyperbolic piece N in which the fixed-point sets of the lifts to Mof the n-periodic symmetries of Kand K0lie. Let Gbe the group of isometries of Nwhich are induced by diffeomorphisms of Mwhich fix Nsetwise. Gcontains two involutions τand τ0induced by the covering involutions for Kand K0 respectively and two isometries of order nwith non empty fixed-point sets, hand h0, induced by the lifts of the n-periodic symmetries of K and K0. Two situations can arise: Proposition 2. Up to conjugation, we can assume that h=h0and either τ=τ0, or τand τ0commute and nis odd. Proof: Let us start by showing that, up to conjugation, h=h0. If nis odd, this was proved in [P3,§3.3.1], so we can assume that nis even. In this case the linking number of Kand the axis of its n-periodic symmetry is odd so that the fixed-point sets of hand h0are connected. Reasoning by contradiction as in [P3,§3.3.1], we can find a cyclic subgroup Hof G which commutes with hhiand intersects it only in the identity. Since τcommutes with h, we see that the normaliser of hin Gcontains a 274 L. Paoluzzi subgroup of the form Z2⊕Zn⊕Zn, but this is impossible for even n > 2, since a group of isometries which leaves invariant a geodesic must be a finite subgroup of Z2n(Q/Z⊕Q/Z) where the element of order 2 acts on the direct sum by sending each element to its inverse. Up to conjugation (of h0and τ0), we can thus assume that h=h0. Note that the same argument as in [P3,§3.4] assures that the lifts of the two n-periodic symmetries coincide on the whole of M. The group generated by h,τand τ0in Gmust be of the form Zn⊕Dt, where Dt denotes the dihedral group of order 2t. Since this group preserves the fixed-point set of h, if n > 2 is even, and using the fact that Fix(h) is connected, we see, arguing as above, that tmust be 1, i.e. τ=τ0. Assume now that nis odd. If tis odd, τand τ0are conjugate, so, up to a conjugation inside Dtwhich preserves h, we have again that τ=τ0. To end the proof of the proposition we only need to show that, if tis even, then t= 2, and τand τ0commute. In this case, we can perform equivariant hyperbolic Dehn surgery along the boundary of Nin such a way that the quotient of the resulting manifold by the action of h (τ, respectively τ0) is the 3-sphere, the fixed-point sets of τand τ0are connected, and the group of isometries of the resulting manifold is a subgroup of G, containing h,τand τ0. We are thus in the situation described in [P4, Proposition 2] and we deduce that t= 2. This finishes the proof of Proposition 2. Examples of non-equivalent pairs of hyperbolic knots satisfying the first condition were constructed in [P2]. Here we shall construct knots realising the second situation. We start by observing that the structure of these knots is rather special. Proposition 3. The quotient of Nby the action of τlooks schematically as in Figure 8(a) (where the case n= 3 is pictured), i.e. it consists of n boundary spheres cyclically connected by two subarcs of K. Proof: Since τand τ0commute on N,τ0induces an involution τ0of N/hτi with non empty fixed-point set which fixes setwise each boundary component. If γis a subarc of Kcontained in N/hτiwhich joins two distinct boundary components then τ0(γ) is a different subarc of Kjoining again the same boundary components. Since τ0commutes with h,τ0cannot act as a strong inversion (i.e. by fixing some point) on K∩N/hτi, so that if γis a subarc of Kcontained in N/hτiwhose endpoints are on the same component then τ0(γ) is a different subarc of Kwhose endpoints are again on the given component. This situation is however impossible. In fact the n-periodic symmetry of Kmust freely permute the Hyperbolic Knots and Cyclic Branched Covers 275 boundary components of N/hτiwhose number must then be a multiple of nand cannot be 1. Consider now the quotient of Kby the action of its n-periodic symmetry; this quotient is schematically illustrated in Figure 8(b). Since it is a trivial knot, the number of Conway spheres cannot exceed 1 and the conclusion follows. (a) (b) Figure 8 In the proof of Proposition 3 we have seen that the lifts of the two n-periodic symmetries coincide on M. In the example that we are about to construct we assume also that the two covering involutions commute on the entire manifold M. This allows us to assume that we are in a situation similar to that described in Proposition 1, i.e. Kand K0 are obtained as lifts of a three component link with special properties. Notice that Proposition 3 above implies that, in such a link, the Conway spheres intersect two distinct components. To obtain the desired examples it is then sufficient to construct a hyperbolic link satisfying the condition described in [Z1, Property 5] (see also part two of Proposition 1). As in [Z1] (see also Section 3), we start by considering the three component link 63 1(see Figure 9). In this case, however, we shall replace its three rational tangles (instead of six trivial tangles) by a π-hyperbolic tangle constructed as follows. 276 L. Paoluzzi 63 1 Figure 9 Consider the prime knot K= 10155. The following claims describe its properties. Claim 1. Kadmits two strong inversions whose product is a free involution. The knot and its 2 inversions are drawn in Figure 10: the axis of the first strong inversion lies in the plane of the figure, while the axis of the second one is orthogonal to the plane of the figure and meets the knot in the two marked points. Note that the two axes form a Hopf link. The existence of a free symmetry was established in [Ha]. Claim 2. Kis hyperbolic. It suffices to show that Kis not a satellite link, for it is prime and cannot be a torus knot (this last fact is easily seen by considering the Alexander polynomial). The following fact is due to Schubert [Sc] and was pointed out to the author by M. Boileau and D. Lines: the bridge number of a companion of a satellite knot is strictly less than the bridge number of the knot. It is easily seen that the bridge number of Kis at most 3 (in fact it is equal to 3), so Kcan only be the companion of a 2-bridge knot. Moreover, the bridge number of the knot is greater or equal to the bridge number of the companion times the wrapping number, i.e. the minimal number of times the knot intersects a meridian disc of the companion. Since Kis prime the wrapping number cannot be 1 and the claim follows. Hyperbolic Knots and Cyclic Branched Covers 277 Figure 10 Claim 3. The two strong inversions of Kare not conjugate in the group of symmetries of K, moreover the only symmetry which normalizes them is the free involution. Since Kis hyperbolic, its group of symmetries, which contains a subgroup of the form Z2⊕Z2according to Claim 1, must be dihedral. The order of its maximal cyclic group (of free or periodic symmetries) is even and the free involution ρis the central element of the group. If the two strong inversions ηand ξwere conjugate in the group of symmetries, the element gconjugating them would exchange their fixed-point sets and have even order. Indeed, if gηg−1=ξthen gξg−1=gηρg−1=ξρ =η. Thus, under the above assumption, the order of the symmetry group would be of the form 8t; in particular it would admit an element of order 4. Since the centre of the group acts freely, the maximal cyclic subgroup of order a power of 2 must act freely as well. However, the only possible free periods for Khave order 2 (see [Ha]). This contradiction shows that the two strong involutions of Kcannot be conjugate. The second part of the claim follows from the fact that if the group of symmetries of a knot is dihedral of order at least 6, then the strong inversions cannot lie in its maximal cyclic subgroup. Note that hyperbolicity of Kand the structure of its symmetry group can be checked using J. Weeks’ SnapPea. Consider the quotient of S3\ U(K) by the action of the Z2⊕Z2group generated by the strong involutions of K: by construction it is a π-hyperbolic 2-tangle Tadmitting no non trivial symmetries, which we show in Figure 11(a). Replace the rational tangles of 63 1with copies of Tin such a way to preserve the symmetry of 63 1of order 3 which cyclically 278 L. Paoluzzi exchanges its components. We get a new link L(compare Figure 11(b)). An easy check, following the lines of [Z1], shows that the conditions of [Z1, Property 5] are satisfied. Note that the knots Kand K0we have just constructed possess the minimum number of Conway spheres necessary for a Conway reducible hyperbolic knot not to be determined by its n-fold cyclic branched cover, if n≥3. T L (a) (b) Figure 11 It is worth to stress two things which are not straightforward. Since each component meets a Conway sphere for Lin at most 2 points, when we take the 2-fold cyclic cover of S3ramified along one component, Conway spheres lift to Conway spheres so that the lift of the remaining two components is again a hyperbolic link. This is not the case for the examples constructed by Zimmermann in [Z1]. For the same reason, if we use this link to repeat Zimmermann’s construction for two coprime integers n, m > 2 the hyperbolic knots (which have common n-fold and m-fold cyclic branched covers) we obtain are Conway irreducible (as those of [P4]). 6. Determination of hyperbolic links Let Lbe an oriented link in S3with more than one component. It is well-known that Ladmits many different n-fold cyclic branched covers if n > 2. This depends on the fact that the manifold S3\ U(L) admits Hyperbolic Knots and Cyclic Branched Covers 279 different cyclic covers, each corresponding to a surjection π1(S3\ U(L)) Zn. Such a surjection factors through homology H1(S3\ U(L)) Zn so it is determined by the images of (the homological classes of) the oriented meridians of L. To obtain the branched cover, it suffices to Dehn fill the boundary components in such a way that meridinal discs are glued to the lifts of meridians of L. An n-fold cyclic branched cover is called strong if all meridians are sent to generators of Zn. We shall consider here only a very special type of strong n-fold cyclic branched cover, namely the one obtained by sending all meridians to the same generator of Zn. Such cover —that we shall call pure— is clearly unique. Let Lbe a hyperbolic link and consider its pure n-cyclic branched cover M(L, n). If M(L, n) is a hyperbolic manifold (this is indeed always the case if n≥4) then the deck transformation is an isometry of M(L, n) and acts as a rotation of the same angle about each component of the preimage of Lin M(L, n). We want to show that if M(L, n) does not determine Land n≥3, then the situation is similar to the one seen for knots. Lemma 1. Let n > 2. If the pure n-fold cyclic branched cover of a hyperbolic link Lis hyperbolic and does not determine L, then Ladmits a symmetry of order nwith non empty fixed-point set and trivial quotient, i.e. the quotient of Lby the action of the symmetry is the trivial knot. In particular, the number of components of Lmust divide n. Proof: Assume that M(L, n) is the pure n-cyclic branched cover of another, necessarily hyperbolic, link L0and let hrespectively h0the rotations of angle 2π/n about the preimages in M(L, n) of Land L0respectively, which are generators of the groups of deck transformations. Since we are assuming that Land L0are not equivalent, we can deduce that the groups generated by hand h0are not conjugate. Note that no non trivial powers of hand h0can be conjugate for otherwise, the element conjugating the power would map Fix(h) to Fix(h0) and conjugate a rotation of angle 2π/n about Fix(h) to a rotation of the same angle about Fix(h0). Let qbe a maximal prime power dividing nand different from 2 and let g=hn/q. Since hgiand hh0n/qiare not conjugate, using [Su, Chapter 2, 280 L. Paoluzzi 1.5] we can find an element ˆg, conjugate to either gor h0n/q via an element f, which normalizes gand such that hgi∩hg0i={1}(for similar reasoning see, for instance, [P3, Chapter 4]). Since ˆgnormalizes hgi, we see that ˆg(Fix(h)) = Fix(h) so that ˆgnormalizes hhi. Indeed, since his a rotation of the same angle about each component of its fixed-point set we see that hand ˆg2commute (note that ˆgconjugates heither to itself or to its inverse according to whether it preserves the orientation of Fix(h)). In particular h(Fix(ˆg)) = Fix(ˆg) = Fix(ˆg2) for g2is not the identity, so that hnormalizes ˆgand, more generally, any rotation about Fix(ˆg). Let ˆ hbe fhf−1if ˆgis conjugate to gand fh0f−1if ˆgis conjugate to h0n/q. Since ˆ his a rotation of the same angle around each component of its fixed-point set, as above we deduce that h2and ˆ hcommute. In particular ˆ h(Fix(h)) = Fix(h) = Fix(h2) so that ˆ hnormalizes the group generated by h. Reasoning as in [P3, Chapter 4] one sees that the groups generated by hand ˆ hintersect only in the identity and, since they normalize each other, they must commute. The conclusion now follows as in the proof of [Z2, Corollary 1] and Lemma 1 is proved. A result of Reni [Re1, Theorem 1] says that a hyperbolic link is determined by a strongly cyclic branched cover provided that its order n is at least 3, and not a multiple of the number of its components and that the cover is hyperbolic. Note that the hypothesis that the order is at least 3 is indeed necessary although it is omitted in [Re1]. The above Lemma 1 and the result of Reni allow us to give some sufficient conditions on the determination of a hyperbolic link: Theorem 5. Let n > m > r ≥2be three integers. A hyperbolic link L is determined by its n-fold, m-fold and r-fold pure cyclic branched covers (together) if either r > 2or nand mare coprime. Proof: Assume that r= 2. In this case, for homological reasons [Sa2], any other link doubly covered by M(L, 2) has the same number of components as L. If Lis a knot the conclusion follows from Theorem 1, else the conclusion follows from the fact that at least one between nand m cannot be a multiple of the number of components of L. Note that to apply Reni’s result we must check whether the cover is hyperbolic. Because of Thurston’s orbifold geometrization theorem, this is always the case if the order of the cover is at least 4, and the cover of order 3 is not hyperbolic only for a finite number of links, in fact, only for the figure-eight knot (see [D]). If r>2, Lemma 1 assures the existence of symmetries for Lwith trivial quotients, non empty fixed-point sets and orders n,mand r > 2. Hyperbolic Knots and Cyclic Branched Covers 281 If these symmetries commute, then the proof follows the same lines of [P3, Proposition]. Observe that commutativity of the symmetries is not necessary and it suffices to require that one of the three symmetries is normalized by the other two. To understand in which cases they do not commute and none of them is normalized by the remaining two, it suffices to analyse the structure of the symmetry group G of L. Since the symmetries have non empty fixed-point sets, Gis linear, i.e. a finite subgroup of SO(4). Any finite subgroup of SO(4) is of the form H1×Z2H2, where Hiis a finite subgroup of S3, the group of unitary quaternions, hence cyclic, dihedral, alternating on 4 or 5 elements, symmetric on 4 elements, quaternionic, binary tetrahedral, binary octahedral or binary dodecahedral. In particular, for the symmetries not to commute, they must belong to the same Hiand their orders are necessarily 3, 4, 5, 6, 8 or 10. If n,mand rare coprime, Reni’s result applies and we are done; else (n, m, r) must be equal to either (10,6,4) and the elements are contained in the binary dodecahedral group or to (8,6,4) and the elements are contained in the binary octahedral group. Note that, in this case Lcannot be a knot, for the group of symmetries of a hyperbolic knot is either cyclic or dihedral; moreover, according to Reni’s result Lmust have precisely two components. This implies that either the group A5or the group S4leaves setwise invariant each component of L. This is again impossible, since the group of symmetries which fixes setwise each component of a hyperbolic link must be a finite subgroup of Z2n(Q/Z⊕Q/Z), where the involution conjugates each element of the product to its inverse, thus cannot be A5or S4. This finishes the proof of Theorem 5. As was the case for knots, the 2-fold cyclic branched cover of a hyperbolic link is much more rich and complex, thus we shall not try to discuss whether three pure cyclic branched covers always suffice to determine a hyperbolic link. Note that “determination” here means among all links. References [BeP] R. Benedetti and C. Petronio,“Lectures on hyperbolic geometry”, Universitext, Springer-Verlag, Berlin, 1992. [BF] M. Boileau and E. Flapan, On π-hyperbolic knots which are determined by their 2-fold and 4-fold cyclic branched coverings, Topology Appl. 61(3) (1995), 229–240. [BoP] M. Boileau and J. Porti, Geometrization of 3-orbifolds of cyclic type, Appendix A by Michael Heusener and Joan Porti, Ast´erisque 272 (2001), 208 pp.