LERF, tameness and Simon s conjecture.

LERF, tameness and Simon s conjecture. D. D. Long & A. W. Reid November 18, 2003 Abstract This paper discusses applications of LERF to tame covers of certain hyperbolic 3-manifolds. For example, if M = H 3 /Γ is an orientable noncompact finite volume hyperbolic 3-manifold and Γ is LERF, then the cover of M corresponding to any subgroup of Γ generated by two parabolic elements is tame. 2000 Mathematics Subject Classification 57M50, 20H10 1 Introduction A non-compact 3-manifold M is called tame if M is homeomorphic to the interior of a compact 3-manifold. One of the main unresolved questions in the theory of infinite volume hyperbolic 3- manifolds is the question of whether M = H 3 /Γ is tame, where Γ is a finitely generated, torsion-free Kleinian group. There has been considerable progress on this recently; see [5] for example. However tameness is still not yet established for manifolds arising from Kleinian groups which are free on two elements. Our interest here are those Kleinian groups that arise as subgroups of finite co-volume Kleinian groups. This forms part of a broader conjecture of Simon, who conjectured (see [22]) that if M is a compact 3-manifold, and H is any finitely generated subgroup of π 1 (M), then the cover corresponding to H is tame. Simon proved this in many cases (see 4 for a discussion of this). One of the aims of this paper is to connect tameness, Simon s conjecture and the group theoretic property of LERF. Recall that if G is a group, and H a subgroup of G, then G is called H-separable if for every g G \ H, there is a subgroup K of finite index in G such that H K but g / K. G is called LERF or subgroup separable if G is H-separable for all finitely generated subgroups H. As is well-known LERF is a powerful property in the setting of low-dimensional topology (see [21]), but a priori there would appear to be no connection between LERF and tameness. However, it is known, using work of Canary [7], that an affirmative solution to the tameness conjecture, and the weaker condition of GFERF for Kleinian groups implies LERF. Recall that a Kleinian group Γ is called GFERF if Γ is H-separable for every geometrically finite subgroup H. The aforementioned implication follows, since an affirmative solution to the tameness conjecture would imply that every geometrically infinite subgroup of a finite covolume Kleinian group is virtually a fiber in a fibration over the circle (see [7] Corollary 8.3). Such groups are known to be separable by work of Bonahon and Thurston (see [4] and [24]). The aim of this paper is to provide some partial converses to this; that is, that LERF can be used to tame covers. One of our main results is. This work was partially supported by the N. S. F. This work was partially supported by the N. S. F. and a grant from the Texas Advanced Research Program. 1

Theorem 1.1 Let M = H 3 /Γ be a non-compact orientable finite volume hyperbolic 3-manifold and H a subgroup Γ generated by two parabolic elements. If Γ is LERF then the cover corresponding to H is tame. Indeed, once the covering spaces associated to the groups H as in Theorem 1.1 are known to be tame, they will be geometrically finite (by [7]). In [18], tameness (indeed geometrical finiteness) for H as in Theorem 1.1 is established when H is of the 2nd kind. On the other hand, [18] also gives an example of a group H generated by two parabolics which is of the first kind. In light of Theorem 1.1 such a group cannot be a subgroup of a finite co-volume LERF Kleinian group. More generally our methods also show that modulo a certain property of subgroups of Kleinian groups Γ as in Theorem 1.1 that Simon s conjecture holds in this setting; see Theorem 3.6 for more details. This and the proof of Theorem 1.1 depend on the following useful topological consequence of LERF in this setting. Proposition 1.2 Let M = H 3 /Γ be a non-compact orientable finite volume hyperbolic 3-manifold and assume that Γ is LERF. Let H and K be finitely generated subgroups of Γ with H < K < Γ and where both [Γ : K] and [K : H] are infinite. Then there is a subgroup of finite index K 0 in K, a finite group G with a surjective homomorphism θ : K 0 G such that: θ admits an extension ˆθ : Γ 0 G with ˆθ(H) = 1. If the rank of H is k, then H 3 /ker ˆθ has at least k + 1 cusps. Agol [2] has proved Theorem 1.1 without the LERF assumption for M the complement of a 2- bridge link. The authors (unpublished) have also shown that Theorem 1.1 holds without the LERF assumption for arithmetic hyperbolic 3-manifolds. The rest of the paper is organized as follows. We gather together some preliminary material in 2 that is needed in proving in Theorem 1.1 in 3. This includes a proof of the above proposition (Proposition 2.4) which, as we discuss, is a useful tool for taming covers in conjunction with work of Canary, [7]. In 4 we prove some results concerning Simon s Conjecture both in the non-hyperbolic and hyperbolic settings (assuming LERF and some additional control in the hyperbolic setting). Acknowledgement: The second author thanks Jeff Brock for helpful discussions regarding tameness. 2 Preliminaries Here we record some preliminary material that will be needed. All manifolds are assumed orientable unless otherwise stated. 2.1 Given a compact manifold M with incompressible boundary whose interior admits a complete hyperbolic structure, we will often abuse notation and use M for the non-compact hyperbolic 3-manifold arising as a quotient of hyperbolic space. If M is non-compact and finite volume, then given a cusp C of M, a torus cusp cross-section T C of C, and an essential closed curve α on T C we define α-filling on M to be the result of possibly orbifold Dehn filling along the curve α. The following lemma is from [9] and [19]. We include a proof for convenience. Lemma 2.1 Let M 1 = H 3 /Γ 1 be 1-cusped finite volume hyperbolic 3-manifold, M 2 = H 3 /Γ 1 a 1-cusped finite volume hyperbolic 3-orbifold with a torus cusp, and Γ 1 < Γ 2. If Γ 1 is generated by 2

parabolic elements, then the cover M 1 M 2 is regular. In the case when M 2 is a manifold, the cover is cyclic. Proof. Let P be a maximal peripheral subgroup of Γ 2. Since M 2 has a torus cusp, then P is abelian. Since both M 1 and M 2 have a single cusp, Γ 2 = P.Γ 1 ; that is every element in Γ 2 can be expressed as a product p γ for some p P and γ Γ 1. Let g = p γ be such an element, and x P Γ 1. Then since P is abelian, notice that g 1 xg = γ 1 p 1 xpγ = γ 1 xγ. This latter element lies in Γ 1. Hence the normal closure of P Γ 1 in Γ 2 is contained in Γ 1. Now Γ 1 is generated by parabolic elements, and so normally generated by P Γ 1. It therefore follows that normal closure of P Γ 1 in Γ 2 coincides with Γ 1. Hence Γ 1 is a normal subgroup of Γ 2 as required. In the case when M 2 is a manifold, it can be shown that H 1 (M 2 ; Z) = Z, and that the covering group is isomorphic to P/(P Γ 1 ) which is therefore cyclic. The following is well-known. Lemma 2.2 Let M be a compact orientable 3-manifold with incompressible boundary consisting of a disjoint union of m tori. Then the first Betti number is at least m. 2.2 The following powerful consequence of LERF will be useful for us and seems independently interesting. Theorem 2.3 Suppose that Γ is LERF, and H a finitely generated subgroup. Suppose θ : H G is a homomorphism onto a finite group. Then there is subgroup of finite index Γ 0 of Γ containing H and a homomorphism ˆθ : Γ 0 G with ˆθ H = θ. Proof. We can assume that H has infinite index in Γ. Let K = ker θ, a finite index subgroup of H, and hence finitely generated. Since Γ is LERF, there is a finite index subgroup Γ < Γ such that Γ H = K. Define = {gγ g 1 : g H}. Note that since K is normal in H, K < and moreover, since Γ has finite index in Γ, this is a finite intersection. Hence has finite index in Γ (and also Γ). Let Γ 0 denote the group generated by H and. It is easy to check that, by construction, is a normal subgroup of Γ 0, so that Γ 0 = H.. The canonical projection ˆθ : Γ 0 Γ 0 / defines the extension, since by the first isomorphism theorem Γ 0 / = H/K = G. Furthermore, note that, by definition ˆθ H = θ as required. Definition: A homomorphism ˆθ as in Theorem 2.3 is called an extension of θ. This theorem will be useful in the following setting. Recall that if G is a finitely generated group, the rank of G is the cardinality of a minimal generating set for G. By a compression body group we will mean a free product A B where A is free of some rank n 0 and B is a free product of surface groups (i.e. the fundamental groups of closed surfaces). Note this is somewhat non-standard, in that we allow non-orientable closed surfaces. However this will be convenient in what follows. Proposition 2.4 Let M = H 3 /Γ be a non-compact orientable finite volume hyperbolic 3-manifold and assume that Γ is LERF. Let H and K be finitely generated subgroups of Γ with H < K < Γ and where both [Γ : K] and [K : H] are infinite. Then there is a subgroup of finite index K 0 in K, a finite group G with a surjective homomorphism θ : K 0 G such that: 3

θ admits an extension ˆθ : Γ 0 G with ˆθ(H) = 1. If the rank of H is k, then H 3 /ker ˆθ has at least k + 1 cusps. Proof. The group K has the structure of a (possibly trivial) free product A B where A is a compression body group and B = π 1 (W 1 )... π 1 (W s ), with each W i an orientable 3-manifold with π 1 (W i ) freely indecomposable. Since M is orientable, Γ contains no Klein bottle group. Hence the free product decomposition of K cannot contain any Klein bottle group. We now break the proof up into a discussion of various cases depending on the above free product decomposition. We begin with the simplest cases. K is Z or Z Z: Recalling that H is required to have infinite index, we see that the former case implies H is trivial, the latter that H is trivial or rank one. The result holds easily for H the trivial group (rank 0), since we can choose the trivial homomorphism of K which extends over the whole group Γ. M (and all its coverings) have at least one cusp and this completes the proof in this case. If K = Z Z, and H is infinite cyclic, then the group K is conjugate into a maximal peripheral subgroup of Γ. By peripheral separability [17], there is a finite index subgroup Γ 0 of Γ containing K, so that H 3 /Γ 0 has at least two cusps. In this case, we may again take the homomorphism of K on to the trivial group and the trivial extension to Γ 0 to prove the proposition. K is free of rank at least two: Hence H is a free subgroup of infinite index. Since free groups are LERF, we can find infinitely many finite index subgroups of K containing H. Furthermore, we can arrange that H is a free factor (cf. [21]). Hence we can arrange that H is a free factor of a subgroup K 0 = H Q of finite index in K, with Q free of rank N (chosen very much larger than 2). Since Q surjects any group which is N generator, applying Theorem 2.3, we have, for any finite group G which is N-generator, a homomorphism θ : K 0 G with θ(h) = 1 which may be extended to a finite index subgroup of Γ. This proves the first part of the proposition in this case. To complete the proof, we now specify a group G so that the second part of Proposition 2.4 will also hold. Fix p > k a prime and choose G = (Z/pZ) N. As above, we have an extension of θ to a homomorphism ˆθ on a finite index subgroup Γ 0 of Γ with ˆθ(H) = 1. We claim that the cover corresponding to ker ˆθ does have at least k + 1 cusps. To see this consider the preimage of a cusp C in the cover corresponding to ker ˆθ. Let P be the maximal peripheral subgroup of Γ 0 associated to C. Now ˆθ(P ) is at worst Z/pZ Z/pZ, which, by choice of N (being very much larger than 2) has index at least p in G. Now covering space theory shows that the number of preimages of the cusp in the cover corresponding to ker ˆθ is G / ˆθ(P ), and so this is at least p. By choice of p this is at least k + 1 and so the result follows in this case. K is a (non-free) compression body group: Thus K = A B where A is a free group and B is a free product of non-trivial closed surface groups, all different from those of RP 2 and the Klein bottle. The above considerations also show that we may suppose that K does not consist of one copy of Z Z. Now since K is a subgroup of a LERF group, K is also LERF (in fact since free products preserve LERF any compression body group is LERF). Since H is assumed to have infinite index in K, we can find subgroups of arbitrarily large finite index in K containing H. By the Kurosh subgroup theorem, these groups will also be compression body groups, and so every finite index subgroup of K is of the form F n Z where F n is a free group of some rank n, and Z is a free product of a finite number of surface groups. To complete the proof in this case we use the following. Lemma 2.5 Given an infinite tower of finite index subgroups of F n Z as above, then the first Betti number goes to infinity in this tower. 4

Proof. If the rank of the free part goes to infinity then this is clear. Thus assume the rank of the free part remains bounded and let Z = Σ g1... Σ gt where Σ gj is a closed surface group of genus g j (orientable or non-orientable). By the Kurosh subgroup theorem, a subgroup of finite index in F n Z splits as a free product of a free group and conjugates of subgroups of the surface subgroups. Since we are assuming the rank of the free part is bounded, we can therefore assume that the subgroups arising from conjugates of subgroups of Z must be finite index subgroups of the surface groups, and so themselves are surface groups. Now in the cases under consideration, χ(f n Z) is nonzero and since Euler characteristic is multiplicative in finite covers, a subgroup of index N has Euler characteristic Nχ(F n Z) and one sees easily that if the rank of the free part remains bounded while N, then either the number of surface components or the genus (orientable or non-orientable) of one of these components must increase without bound. Either of these case proves the lemma. With this lemma in hand, the proof of the proposition in this case is completed in a manner similar to the previous case. As before, let p > k be a prime, choose N very large, say with N k > 2, and let G = (Z/pZ) N. Now from, Lemma 2.5 we can find a finite index subgroup K 0 of K which contains H and a homomorphism θ : K 0 G. Note in this case θ(h) need not be trivial, but θ(h) is at worst (Z/pZ) k. Hence G/θ(H) = (Z/pZ) N k 1 since N k > 2. Indeed, by construction, (Z/pZ) N k contains a copy of (Z/pZ) 3. Relabeling we can define a homomorphism θ : K 0 (Z/pZ) N k in which θ(h) = 1. By Theorem 2.3 we can find a finite index subgroup Γ 0, an extension of θ to Γ 0, and as in the previous case we consider the preimage of a cusp C in the cover corresponding to ker ˆθ. The arguement is finished exactly as before. K = π 1 (W ) is freely indecomposable: Since K < Γ, then K is LERF, and again, H is contained in subgroups of K of arbitrarily large finite index. The result will follow in this case on showing that the first betti number in such a tower also goes to infinity. Note that W and cannot consist only of tori. For if so, then the interior of W will admit a complete hyperbolic structure of finite volume by Thurston s uniformization theorem for Haken manifolds. This implies [Γ : K] < which contradicts our assumption on K (note also π 1 (W ) = Z or Z Z has been dealt with by our previous considerations). Hence W has some component of genus at least 2. Now a generalization of Lemma 2.2 shows that the first Betti number of W is at least 1 2dim( W ). We claim that this must go to infinity in a tower of finite covers. This is clear if the number of boundary components goes to infinity (by the above remark). If the number of boundary components stays bounded, for some component of the boundary the genus must go to infinity (recall there is a component of genus at least 2). Hence this proves the claim. The argument is now completed exactly as in the previous cases. The proof for the general case, that is, when K = A B where A is a compression body group and B = π 1 (W 1 )... π 1 (W s ), with each W i an orientable 3-manifold with π 1 (W i ) freely indecomposable can be handled using a combination of the above methods. 2.3 Recall that if a group Γ is LERF and H < Γ is a finitely generated subgroup, then H = {K : [Γ : K] <, H < K}. Lemma 2.6 Let M = H 3 /Γ be a finite volume hyperbolic 3-manifold. Assume that Γ is LERF and that H is a finitely generated subgroup of infinite index in Γ with the property that every subgroup 5

K of finite index in Γ with H < K is normal in Γ. Then H is the fundamental group of a compact 2-manifold S and M is virtually fibered over the circle with fiber S. Proof. Since Γ is LERF, we have from above that H = K where K ranges over all subgroups of finite index containing H. Since each K is normal in Γ it follows that H is normal in Γ, and by hypothesis Γ/H is infinite. It follows from [14] Theorem 11.1 that H is the fundamental group of a compact 2-manifold. The virtual fibration statement also follows from Theorem 11.1 of [14]. A useful corollary of this in our setting is the following. Corollary 2.7 Let M and H be as above, and assume further that H is a free subgroup generated by two parabolic elements. Then there is a non-normal subgroup of finite index in Γ containing H. Proof. If not, then by Lemma 2.6, H is the fundamental group of a compact 2-manifold S and M is virtually fibered over the circle with fiber S. However, this is a contradiction. For since H is generated by two parabolics, it follows that the compact 2-manifold S is a twice-punctured disk. This has a unique hyperbolic structure, in particular it is aways totally geodesic. Therefore the group of such a manifold cannot arise as a virtual fiber group. 3 Proof of Theorem 1.1 The proof of Theorem 1.1 relies heavily on the following result of Canary (see [7], Proposition 8.4). For convenience we shall henceforth call a Kleinian group Γ tame if H 3 /Γ is tame. Theorem 3.1 Let N = H 3 /Γ be a finite volume hyperbolic 3-manifold, and H a finitely generated subgroup of Γ. If there exists an epimorphism φ : Γ Z such that H lies entirely in the kernel of φ, then H is tame. Immediately from this we observe that in the context of Theorem 1.1, we may as well assume that M has first Betti number at most 2. In particular, Lemma 2.2 implies that M has at most 2 cusps. Furthermore, we can assume that H is free; for any 2-generator subgroup will be finite index, peripheral or free by [16] Theorem VI.4.1. The first is obviously tame, and the second is tame by [22] and [15]. Henceforth, let H =< x, y > be free, with x and y non-commuting parabolic elements of Γ. Notation: Let Γ be a finite covolume Kleinian group and t Γ a parabolic element. Denote by P t the maximal peripheral subgroup of Γ containing t. Proof of Theorem 1.1 As remarked above we can assume that M has at most 2 cusps. Let µ Γ \ H be a parabolic element. Note that such elements are easily seen to exist, for since H is free, for every parabolic element t Γ, there exists at least an infinite cyclic subgroup of P t which is not contained in H. In particular there are (primitive) parabolic elements of Γ no power of which lie in H. Denote by Γ µ the subgroup of Γ generated by {x, y, µ}. The first lemma deals with the case that Γ µ has infinite index for some µ as above. Lemma 3.2 Let Γ and H be as in Theorem 1.1 and µ a parabolic element not in H. If Γ µ has infinite index in Γ, then H is tame. Proof. Recall first that a Kleinian G group satisfies Property (B) of [4] if for any non-trivial free product decomposition A B of G, there exists a parabolic element of G that is not conjugate into either A or B. If Γ µ is freely indecomposable or satisfies Property (B) of [4], then [4] implies that Γ µ 6

is tame. Thus assume neither of these cases pertains, and Γ µ = A B is a non- trivial free product decomposition. We will first show that [Γ µ : H] is infinite. Note that the rank of Γ µ is at most 3. Hence the rank of A (say) is at most 2 and the rank of B is 1. Suppose first that Γ µ has rank 2, so that A and B are both infinite cyclic. If H had finite index in Γ µ, then since both are free of rank 2, the index must be 1, that is Γ µ = H. However, µ is not an element of H, a contradiction. So H has infinite index in Γ µ in this case. If now A has rank 2, then (as observed earlier), since A has infinite index, either A is itself free or it is isomorphic to Z Z. In the first case Γ µ is a free group of rank 3 and so H obviously has infinite index. If A = Z Z, then since H is free it must be infinite index in Γ µ, for otherwise it would contain a Z Z. Hence in any of these cases, [Γ µ : H] is infinite and we may apply Proposition 2.4 to produce a subgroup Γ 0 containing H having at least 3 cusps. Hence Theorem 3.1 applies to show H is tame. The proof of Theorem 1.1 is completed by our next lemma. Lemma 3.3 Let Γ and H be as in Theorem 1.1. Assume that for every parabolic element µ / H, the subgroup Γ µ has finite index in Γ. Then H is tame. The remainder of this section is dedicated to proving Lemma 3.3. Proof of Lemma 3.3 We are assuming that Γ µ has finite index in Γ. Let N µ = H 3 /Γ µ denote the cover of M corresponding to Γ µ. Recall that we are supposing that M has at most two cusps. We will first reduce the 1-cusped case to the 2-cusped case. To this end, suppose that M has a single cusp. Claim 1: There is a 2-cusped cover N of M such that H < π 1 (N). Proof of Claim 1: Suppose that every cover N of M with H < π 1 (N) has 1 cusp. Since Γ is LERF there are subgroups of arbitrarily large finite index containing H, and furthermore, that every subgroup containing H contains a subgroup of the form Γ µ for some µ in Γ. Also, notice that, since each Γ µ is generated by parabolics, Lemma 2.1 shows that each of the covers N µ M is a finite cyclic cover. Since every subgroup of finite index containing H contains some Γ µ, it follows that all subgroups of finite index in Γ containing H are normal subgroups of Γ. However, this contradicts Corollary 2.7. Hence there exists a 2-cusped cover as required. Thus we may as well now assume that M is 2-cusped and establish the lemma. We can also assume that the groups Γ µ are all 2-cusped, or again we are done. Since Γ is LERF we can pass to a subgroup Γ 0 of arbitrarily large finite index (index > 8 will suffice for our purposes) which contains H. Denote by M 0 = H 3 /Γ 0, and let µ be a primitive parabolic element in Γ 0 \ H. Note that we can arrange that µ has the following properties: (a) µ has length at least 2π in the Euclidean structure on the associated cusp torus, denoted by T µ. Call such a parabolic long. (b) µ-filling on T µ produces an irreducible 3-manifold with incompressible torus boundary. To see this, consider a maximal peripheral subgroup P =< α, λ > Γ 0. If P H = 1, then it is clear we can find infinitely many long parabolics. Furthermore, by Theorem 1.3 of [11] (see also 7

Theorem 3.1 of [12]), the number of slopes that produce manifolds that are reducible or in which the boundary compresses is at most 8. This allows us to choose the required µ in this case. If P H is non-trivial, it is infinite cyclic, say P H =< t >. Now t = α p λ q for integers p and q (not necessarily coprime). If either of p or q is zero, then it is clear how to find such an element. For example, if q = 0, then any element α p λ k for (k, p) = 1 will be primitive and eventually long. Such an element cannot lie in H. If neither p or q are zero, we can find infinitely many integers k such that (k + p, q) = 1 and so α k t is primitive and will be long for large k. As above, such an element cannot lie in H. Note this provides infinitely many long µ. Moreover, as above, an application of [11] allows us to choose a k which produces an irreducible manifold and in which the boundary will remain incompressible. This completes the proof of (b). We can also assume that P µ, and P x (say) are not Γ 0 -conjugate. For since there are two Γ 0 -conjugacy classes of maximal peripheral subgroup, if P x and P y are Γ 0 -conjugate we choose some µ from the other conjugacy class. If P x and P y are not Γ 0 -conjugate we choose µ in the conjugacy class of P y and not in P x. Claim 2: P x and P y cannot be Γ µ -conjugate. Proof of Claim 2: Assume to the contrary that P x and P y are Γ µ -conjugate. By choice of µ, since there are two cusps, P µ cannot be Γ µ -conjugate to P x. Perform µ-filling on the associated cusp torus of P µ. By (b) above, this results in an irreducible 3-manifold with a single torus boundary component, which is incompressible. By (a), since µ is long, the resultant manifold will carry a metric of negative curvature ([13]). Let the manifold resulting from µ-filling on N µ be denoted by N µ. It follows that N µ is a finite volume hyperbolic 3-manifold with a single cusp. Moreover, x and y will remain parabolic after filling, and so since µ has been killed, the fundamental group of N µ is generated by two parabolic elements. Appealing to a result of Adams [1] implies that N µ is homeomorphic to the complement of a hyperbolic 2-bridge knot in S 3. Now we can perform a µ orbifold filling on M, resulting in a 1-cusped orbifold M with a torus cusp. We obtain an orbifold covering N µ M. By Lemma 2.1, this covering is regular. In particular the covering group is a subgroup of the group of isometries of N µ, a hyperbolic 2-bridge knot complement. However this has order at most 8 (see [10] for example). But the choice of covering degree was made very large and hence we have a contradiction. Thus we can now assume that P x and P y are not Γ µ -conjugate. To complete the proof in this case we argue as follows. By assumption, P µ cannot be Γ µ -conjugate to P x, it is therefore Γ µ -conjugate to P y. Hence the cusps are represented by the peripheral subgroups P x and P y. An application of LERF, together with another application of the observation above that every subgroup of finite index contains some Γ ν allows us to pass to a finite cover and to arrange that x is a primitive parabolic in some Γ µ1 < Γ µ. Now perform x-filling on N µ1. If the result is hyperbolic, we can argue as above to obtain a contradiction for the covering degree of N µ1 over the result of x-filling on M is still large. Thus assume that the result of x-filling on N µ1, denoted by N µ 1, is non-hyperbolic. In this case, y and µ 1 remain peripheral. If the boundary torus is incompressible, then [3] Corollary 6 implies that N µ 1 is a 2-bridge torus knot. We will deal with this case below and assume first that the boundary torus of N µ 1 is compressible. Claim 3: π 1 (N µ 1 ) is infinite cyclic. 8

Proof of Claim 3: N µ 1 has a single torus boundary component which, by hypothesis, is compressible. Performing this compression results in a 2-sphere, if this 2-sphere bounds a homotopy ball we are done. If not, the 2-sphere splits the manifold into a connect sum V #W where V is a solid torus and W a compact 3-manifold. This yields a free product decomposition π 1 (N µ 1 ) = Z B for some compact 3-manifold group B. Now π 1 (N µ 1 ) is generated by (the images of) y and µ 1, and so has rank at most 2. Hence we are done unless B is a nontrivial cyclic group. Observe that the homology of N µ 1 is noncyclic when computed with appropriate finite field coefficients. In addition, since y and µ 1 are peripheral in N µ1, they are peripheral in the surgered manifold, so that by a standard Poincaré duality argument, the subgroup they generate in the homology with finite field coefficients is at most a cyclic group. However, y and µ 1 generate the whole fundamental group of N µ 1, a contradiction. Thus N µ 1 is a homotopy solid torus as required by the claim. We now complete the proof in the case that the boundary torus of N µ 1 is compressible. By Corollary 2.7, Γ µ1 has a non-normal subgroup Λ of finite index containing H. Let ν be a parabolic element in Λ which commutes with µ 1. In particular P ν is not Λ-equivalent to P x. Now Λ contains the group Γ ν with finite index. Claim 4: N ν = H 3 /Γ ν has 3 cusps, and so H is tame. Proof of Claim 4: We can assume that N ν has 2 cusps. Performing x-filling on N ν determines a cover of N µ 1. Since π 1 (N µ 1 ) is infinite cyclic, it follows that both have infinite cyclic fundamental group. We therefore deduce that N ν N µ1 is a finite cyclic cover. Hence N ν H 3 /Λ N µ1 is a cyclic cover, so that Λ is a normal subgroup of finite index. This contradicts our choice of Λ. Since H < Γ ν, we again obtain a homomorphism onto Z for which H maps trivially, and so H is tame by Theorem 3.1. Claim 5: In the case when N µ 1 is a 2-bridge torus knot, H is tame. To do this we make use of following standard facts about the groups PSL(2, F p ), where p is prime and F p denotes the field with p elements.(see [23] Chapter 3, 6). (i) For p odd, PSL(2, F p ) = p(p2 1) 2. (ii) For all p every abelian subgroup A of PSL(2, F p ) of odd order is cyclic. If A is of even order then either A is cyclic or isomorphic to Z/2Z Z/2Z. (iii) For p 3, the largest order of an element of finite order in PSL(2, F p ) is p. Proof of Claim 5: A 2-bridge torus knot is a Seifert fibered space over the disk with 2 exceptional fibers of order 2 and q. Denoting the free product of the cyclic group of order 2 with a cyclic group of order q by F q, we have epimorphisms: Γ µ1 π 1 (N µ 1 ) F q. Note that since µ 1 and y are peripheral, under the latter homomorphism the images remain peripheral; that is conjugate into the subgroup generated by the homotopy class of the boundary of the disk. From [20], we can find infinitely primes p and surjections f p : F q PSL(2, F p ) with f p (µ 1 ) and f p (y) being elements of order p. Composing homomorphisms, we obtain for these p a homomorphism Φ p : Γ µ1 PSL(2, F p ) with Φ p (x) = 1, Φ p (y) and Φ p (µ 1 ) of order p. Hence K p = ker Φ p H is a normal subgroup of index p (with cyclic quotient). Standard facts about free groups imply that K p 9

is a free subgroup of rank p + 1. Consider the covering of N µ1 corresponding to ker Φ p. We claim that this cover has > p + 1 cusps. If this is the case, then we obtain a homomorphism ker Φ p Z for which K p maps trivially. Hence K p is tame by Theorem 3.1. Since tameness is preserved by passage to super and subgroups of finite index, we conclude that H is tame. To see that the cover corresponding to ker Φ p has enough cusps, we use the standard facts about the groups PSL(2, F p ) recorded above. Assuming p odd, with these facts, note that the number of components in the preimage of the cusp torus associated to P y is (p 2 1)/2. For since the image of P y is abelian, and y has order p in the quotient, it follows that the image of P y is cyclic of order p. In the preimage of the cusp torus associated to P x, since x maps trivially, either P x is cyclic of order at most p, or isomorphic to the Klein 4-group. Then choosing p large, we can arrange that there are at least (p 2 1)/2 cusps in this case. From this we see that there are at least p 2 1 cusps in the cover corresponding to ker Φ p. For p > 2, p 2 1 > p + 1. This completes the proof of the claim. The proof of Lemma 3.3 is now complete. Note that the argument of Lemma 3.2 proves that if M = H 3 /Γ is a non-compact orientable finite volume hyperbolic 3-manifold, Γ is LERF, and H any 2-generator subgroup of Γ with the property that there exists an element µ Γ \ H such that [Γ : Γ µ ] is infinite, then H is tame. For as before, since µ / H, then Γ µ H, and the argument is completed as before using Proposition 2.4; parabolicity plays no role in that argument. It appears more difficult to generalize Lemma 3.3 to arbitrary 2-generator subgroups. 3.1 The method of proof of Theorem 1.1 can be used in other situations (again assuming LERF). One such instance is the following. We continue to use the notation Γ µ =< H, µ >. Theorem 3.4 Let M = H 3 /Γ be a non-compact orientable finite volume hyperbolic 3-manifold, and assume that Γ is LERF. If H is a free subgroup of Γ with the property that there exists an element µ Γ \ H such that Γ µ = H < µ >, then H is tame. Proof. Since µ / H, Γ µ H. Since Γ µ is free of rank greater than that of H, it follows that H has infinite index in Γ µ. We can apply Proposition 2.4 and Theorem 3.1 to show H is tame. Note that if H is a virtual fiber subgroup of a group Γ as above, then no such element exists. However, such subgroups are tame. There are several other problems in trying to generalize Theorem 1.1. One is that it could be the case that for every element µ Γ \ H, there exists an integer n(µ) such that µ n(µ) H. Conjecturally this should force H to have finite index in Γ (cf. [8]). Another problem arises in that part of the proof of Theorem 1.1 when Γ µ has finite index in Γ for all µ. A natural conjecture arising from these considerations is. Conjecture 3.5 Let M = H 3 /Γ be a non-compact orientable finite volume hyperbolic 3-manifold. Then if H is a finitely generated subgroup of infinite index in Γ, then either there exists an µ Γ\H such that both the indices [Γ : Γ µ ] and [Γ µ : H] are infinite or H is a virtual fiber group. Theorem 3.6 Let M = H 3 /Γ be a non-compact orientable finite volume hyperbolic 3-manifold. Let H be a finitely generated subgroup of Γ. Assuming Conjecture 3.5 and that Γ is LERF implies that H is tame. 10

Proof. We can clearly assume that H is a finitely generated subgroup of infinite index in Γ, and furthermore, that H is not a virtual fiber group. Suppose that H has rank k, then an application of Proposition 2.4 yields a finite index subgroup Γ 0 of Γ which contains H and has at least k + 1 cusps. Now H is tame by Theorem 3.1. 4 Simon s conjecture in the non-hyperbolic setting Let M be a compact orientable 3-manifold. Following the notation of [22], say that M has almostcompact coverings if for any finitely generated subgroup H of π 1 (M), the covering corresponding to H is tame. Simon s Conjecture: If M is a compact orientable irreducible 3-manifold, then M has almostcompact coverings. Theorem 3.6 immediately implies: Corollary 4.1 Let M = H 3 /Γ be a non-compact orientable finite volume hyperbolic 3-manifold, and H be a finitely generated subgroup of Γ. Assuming Conjecture 3.5 and that Γ is LERF implies Simon s Conjecture for M. When M is non-hyperbolic, more can be said using [22]. The following does not appear to be well-known, although it is a straightforward consequence of the work in [22]. Theorem 4.2 Let M be a compact orientable 3-manifold that admits a geometry other than the hyperbolic geometry. Then Simon s Conjecture holds for M. Proof. We may clearly assume that π 1 (M) is infinite. In which case since M is geometric, it is either a Seifert fiber space or a SOLV manifold. If M has a finite cover that is of the form F S 1 where F is a compact surface (possibly with boundary), then Corollary 3.3 of [22] applies to prove almost-compact coverings for F S 1 and hence M. In particular this proves the theorem for M Euclidean or covered by S 2 R. If M is covered by a torus bundle over the circle, then π 1 (M) has the finitely generated intersection property and so [22] Theorem 3.7 establishes almost-compact coverings in this case (this establishes the theorem for NIL and SOLV). Thus we can assume that M is a Seifert fiber space which is not virtually F S 1. Now M contains an immersed torus, and since π 1 (M) is LERF for example [21], this immersed torus can be promoted to an embedded non-separating torus in a finite cover. We can therefore write M as a union of T 2 I and M 0, where M 0 is a compact Seifert fiber space with non-empty boundary. These latter are virtually of the form F S 1, and so they, and T 2 I have almost-compact coverings ([22] establishes this for T 2 I). Now we can apply Corollary 3.2 of [22] to complete the proof. Recall that a graph manifold is a compact 3-manifold with no non-seifert fibered pieces in its JSJ decomposition. From Theorem 4.2 and Corollary 3.2 of [22] we can deduce. Corollary 4.3 Let M be a graph manifold. Then M has almost-compact coverings. Unlike the hyperbolic case where no example is known not be LERF, there are examples of graph manifolds M for π 1 (M) is not LERF [6]. 11

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