FREE QUOTIENTS AND THE FIRST BETTI NUMBER OF SOME HYPERBOLIC MANIFOLDS

ALEXANDER LUBOTZKY ^{∗}

Institute of Mathematics Hebrew University Jerusalem 91904, Israel alexlub@math.huji.ac.il

Abstract. In this note we present a very simple method of proving that some hyperbolic manifolds M have finite sheeted covers with positive first Betti number. The method applies to the standard arithmetic subgroups of SO(n, 1) (a case which was proved previously by Millson [Mi]), to the non-arithmetic lattices in SO(n, 1) constructed by Gromov and Piatetski-Shapiro [GPS] and to groups generated by reflections. In all these cases we actually show that Γ = π_{1}(M) has a finite index subgroup which is mapped onto a nonabelian free group.

1. Introduction

Let M^{n} be a finite-volume hyperbolic manifold. A well known conjecture of Thurston (cf. [Bo, p. 88] and [T, p. 380]) asserts that M^{n} has a finite sheeted covering N with a nonzero first Betti number, β_{1}(N) > 0. Various authors have applied several different techniques to prove this conjecture for many cases where π_{1}(M^{n}) is an arithmetic lattice in P O(n, 1). The conjecture is known to be true for all arithmetic lattices in P O(n, 1), provided n ≠ 3 and 7 (see [Lu3], [Ra] and the references therein). While π_{1}(M^{n}) is always a lattice in P O(n, 1), it is not necessarily an arithmetic one. In this note we present a very simple method which proves the conjecture for the known examples of nonarithmetic lattices in P O(n, 1), at least when n ≠ 3. When it can be applied, the method actually gives more; it shows that Γ = π_{1}(M^{n}) is mapped onto a virtually-(nonabelian)-free group. This implies that Γ has a finite index subgroup Γ_{0} which is mapped onto a nonabelian free group. In particular, M^{n} has finite sheeted covers with arbitrary large β_{1}. Our main result says:

Theorem 3.5. Let M be an oriented n-dimensional finite-volume hyperbolic manifold. Assume M has a codimension one totally geodesic submanifold F. Then Γ = π_{1}(M) has a virtually (nonabelian) free quotient. In particular, M has finite sheeted covers with arbitrarily large first Betti numbers.

∗ Received December 27, 1995; Accepted February 4, 1996

Our method works for the “standard” arithmetic lattices in P O(n, 1) (see Section 3.1 below; they include all the nonuniform arithmetic lattices for every n and all arithmetic lattices if n is even). But, it does not seem to work for the other arithmetic hyperbolic manifolds which lack totally geodesic hypersurfaces.

If Γ is mapped onto a virtually-free group then its subgroup growth is the same as that of a free group, i.e., super exponential. Thus, the results of this paper partially confirm a conjecture in [Lu2] which asserts that the subgroup growth of every lattice in P O(n, 1) is super exponential.

When Γ is arithmetic and mapped onto a virtually-free group, it has a negative solution to the congruence subgroup problem. Moreover, its congruence kernel contains a free pro-finite group of countable rank (see [Lu1]). This holds, therefore, for the standard arithmetic lattices in P O(n, 1).

The paper is organized as follows: In Section 2, we establish two simple group theoretic lemmas which play the key role in our method. In Section 3, we apply it first to the standard arithmetic lattices; here we simplify the proof of Millson [Mi] for this case. His proof uses in an essential way a symmetry between the two parts of the manifold (an idea which is also reproduced in [H1]). Our proof does not use this symmetry. We can therefore adapt it to the nonarithmetic lattices constructed by Gromov and Piatetski-Shapiro ([GPS]) where there is no such symmetry. This plus a common generalization is shown in Section 3. In Section 4, we bring in few more applications to Haken 3-manifolds.

Acknowledgment. The work was supported by the Bi-National Science Foundation U.S.A. – Israel. Most of the work was done while the author visited the University of Chicago, whose warm hospitality is gratefully acknowledged. I am also grateful to Shmuel Weinberger, Shahar Mozes, Peter Shalen, and Zlil Sela for their help and insight.

2. Some simple group theoretic lemmas

Let’s start with a definition:

Definition 2.1. (i) A group Λ will be called virtually-free if it contains a subgroup of finite index which is isomorphic to a nonabelian free group. (ii) A group Γ has a virtually positive first Betti number (Γ has vβ_{1} > 0, for short) if Γ has a finite index subgroup Λ which is mapped onto ℤ – the infinite cyclic group. (iii) A group Γ has a virtually-free quotient (Γ has vF Q, for short) if Γ is mapped onto a virtually free group. Note that vF Q implies vβ_{1} > 0; moreover, it implies that for every n, Γ has finite index subgroups which are mapped onto ℤ^{n}, i.e., Γ has finite index subgroups with arbitrary large first Betti numbers.

Lemma 2.2. Let Γ = A_{1} ∗_{C} A_{2} be a free product with amalgam. Assume Γ has a finite quotient π: Γ → S, such that C̄ = π(C) <_{≠} π(A_{i}) = Ā_{i}, i = 1, 2.

Then:

(i) Γ has vβ_{1} > 0.

(ii) If ([Ā_{1}: C̄] − 1)([Ā_{2}: C̄] − 1) > 1 then Γ has vF Q.

(iii) K = Ker(π) has a quotient which is a free group of rank r = [S: C̄]−

1 − ∑_{i=1}^{2}([S: Ā_{i}] − 1).

Proof. By the universal property of a free product with amalgam, Γ is mapped onto Λ = Ā_{1} ∗_{C̄} Ā_{2}. This is a free product with amalgam of finite

groups, hence it is virtually free (cf. [Se2, Prop. 11, p. 120]). Moreover, a simple calculation using [Se2, Exercise 3, p. 123] shows that H = Ker(Λ → S) is a free group of rank r as in (iii). This proves (iii). Now, (i) follows, since by our assumptions [S: C] ≥ 2[S: A_{i}] for i = 1, 2 and hence r ≥ 1. The assumption of (ii) implies further that r ≥ 2 and (ii) is also proved.

It is interesting to observe that while the assumptions of the lemma are fairly simple, they are crucial. Not every residually finite free product with amalgam has a virtually free quotient:

Example 2.3. Let p be a prime and Γ = SL_{2}(ℤ[\frac{1}{p}]). It is well known

that Γ = A_{1} ∗_{C} A_{2} where A_{1} = SL_{2}(ℤ), A_{2} = \{( \begin{matrix}a & pb \\ p^{−1}c & d \end{matrix} ) | ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈

SL_{2}(ℤ)\} ≃ SL_{2}(ℤ) and C = A_{1} ∩ A_{2} = \{( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ SL_{2}(ℤ)|c ≡ 0(mod p)\}

(see [Se2, Corollary 2, p. 80]). It is also known that every normal subgroup of Γ is either central or of finite index ([Ma, IV 4.9]). In particular, Γ has no infinite virtually free quotient. One can see that the assumptions of Lemma 2.2 do not hold here: Indeed, Γ is known to have the congruence subgroup property([Se1]), so its finite quotients are all obtained via congruence subgroups, Γ/Γ(m) where Γ(m) = {γ ∈ Γ|γ ≡ 1(mod m)} and m is prime to p. For these quotients, Γ/Γ(m) ≃ SL_{2}(ℤ/mℤ) and the images of A_{1}, A_{2} and C are all onto and in particular are all the same!

An analogue to Lemma 2.2 holds for HNN- constructions. Let A be a group with a subgroup B and a monomorphism φ : B → A. The HNNconstruction Γ = A_{∗_{B}} is the group with presentation Γ = ⟨A, t|t^{−1}bt = φ(b), for b ∈ B⟩.

Lemma 2.4. Let Γ = A_{∗_{B}} be an HNN-construction as above. Assume Γ has a finite quotient π: Γ → S with B̄ = π(B) <_{≠} π(A) = Ā. Then:

(i) Γ has vF Q.

(ii) H = Ker(π) is mapped onto a free group of rank r = [S: B̄] − [S: Ā] + 1.

Proof. By the universal property of the HNN-construction, Γ is mapped onto Λ = Ā_{∗B̄}. This is a fundamental group of a finite graph of finite groups and hence virtually free ([Se2, Prop. 11, p. 120]). Again [Se2, Exercise 3, p. 123] gives that H = Ker(Λ → S) is a free group of rank r.

It is convenient to reformulate the above lemmas in terms of the pro-finite topology of Γ. Recall the following definitions:

Definition 2.5. Let Γ be a group.

(i) The pro-finite topology of Γ is the topology for which the finite index subgroups of Γ serve as a fundamental system of neighborhoods of the identity. (ii) If H is a subgroup of Γ then H^{∗} denotes the closure of H with respect to the pro-finite topology of Γ. It is equal to the intersection of all the finite index subgroups of Γ containing H.

In this language, Lemmas 2.2 and 2.4 give:

Lemma 2.6. (i) Let Γ = A_{1} ∗_{C} A_{2} be a free product with amalgam. If C^{∗} <_{≠}

A_{1}^{∗}, A_{2}^{∗} then Γ has vβ_{1} > 0. If, in addition, ([A_{1}^{∗}: C^{∗}] − 1)([A_{2}^{∗}: C^{∗}] − 1) > 1, then Γ has vF Q. (ii) Let Γ = A_{∗_{B}} be a HNN-construction. If B^{∗} <_{≠} A^{∗} then Γ has vF Q.

3. Virtually free quotients of hyperbolic lattices

In this section we apply the group theoretic results of the previous section to prove nonvanishing of the first Betti number for some hyperbolic groups.

3.1 The standard arithmetic lattices We start with reproving the result of Millson [Mi] concerning the standard arithmetic lattices in SO(n, 1).

Let us first recall the construction (see [Mi] for more details).

Let K be a totally real number field of degree m over the rational numbers, 𝔒 its ring of integers and σ_{1}, …, σ_{m} the embeddings of K into ℝ. Let f(X_{1}, X_{2}, …, X_{n+1}) = a_{1}X_{1}^{2} + … + a_{n}X_{n}^{2} − a_{n+1}X_{n+1}^{2} be a diagonal quadratic form with a_{i} ∈ 𝔒. Assume f^{σ_{1}} has signature (n, 1) and f^{σ_{i}} are positive definite for i = 2, 3, …, m. The subgroup Γ of GL_{n+1}(𝔒) preserving f is a lattice in Π_{i=1}^{m}O_{ℝ}(f^{σ_{i}}) and so is its projection to O_{ℝ}(f^{σ_{1}}) ≃ O(n, 1). For an ideal P of 𝔒, the congruence subgroup Γ(P) = {A ∈ Γ|A ≡ I(mod P)} is a normal finite index subgroup of Γ and for almost all P, Γ(P) is torsion free. Fix such a P whose norm is different than 2 and let Γ_{n} = Γ(P). Γ_{n} is in SO_{0}(n, 1) — the connected component of SO(n, 1) (see [Mi, p. 239]). As explained there the reflection τ through the plane X_{1} = 0 normalizes Γ_{n}; its centralizer in Γ_{n} will be denoted Γ_{n−1}. Γ_{n−1} is the congruence subgroup mod P of the subgroup of SO(n−1, 1) preserving the form f_{0} which is equal to f restricted to the hyperplane X_{1} = 0. It is a lattice in SO_{0}(n − 1, 1). Moreover, Y_{n−1} = Γ_{n−1} \ SO_{0}(n − 1, 1)/SO(n − 1) is an embedded totally geodesic hyperplane in the manifold Y_{n} = Γ_{n} \ SO_{0}(n, 1)/SO(n) (see [Mi, p. 240] or [GPS, 2.8. A]). Y_{n−1} might separate Y_{n} or not. (In fact, most of the work in [Mi] is to ensure that, if P is deep enough, then Y_{n−1} does not separate which gives the desired nonzero first Betti number). We can work with both cases: In the first case Γ_{n} = A ∗_{Γ_{n−1}} B where A and B are the

fundamental groups of the two “halves” 𝔞 and 𝔟, respectively, of Y_{n}. In the second case Γ_{n} = A_{∗Γ_{n−1}}, an HNN-construction of A over Γ_{n−1}. In either case Γ_{n−1} is a proper subgroup of infinite index in A (and in B in the first case). Actually, A (and B) is Zariski dense in SO(n, 1) ([GPS, 1.7]).

We can now prove:

Theorem 3.1. Γ_{n} has vF Q.

Proof. As before either Γ_{n} = A ∗_{Γ_{n−1}} B or Γ = A_{∗_{Γ_{n−1}}} . We only have to

check that we can apply Lemma 2.2 and 2.4 respectively.

We have that Γ_{n} ⊆ GL_{n+1}(𝔒) with Γ_{n−1} ⊆ GL_{n}(𝔒) and Γ_{n−1} = Γ_{n} ∩ GL_{n}(𝔇). Thus, Γ_{n−1} is a proper closed subgroup of Γ_{n} with its congruence topology and Γ_{n−1} is of infinite index in A and B. It follows that for some ideal 𝔮 of 𝔒, [π(A) : π(Γ_{n−1})] ≥ 3 (and [π(B) : π(Γ_{n−1})] ≥ 3 in the first case) where π is the natural projection π : GL_{n+1}(𝔒) → GL_{n+1}(𝔒/𝔮). We thus have all the assumptions of Lemmas 2.2 and 2.4.

3.2 The hybrid manifolds In [GPS], Gromov and Piatetski-Shapiro used a method of “interbreeding” two arithmetic groups Γ_{n} and Γ′_{n} in G = SO(n, 1) to get a new lattice Γ in G. Under suitable assumptions (essentially when Γ_{n} and Γ′_{n} come from different ℚ-forms of G) the resulting lattice Γ is a nonarithmetic lattice of G.

The arithmetic lattices which are suitable for this interbreeding process are the standard ones – those we discussed in Subsection 3.1. We will also use the notations of Subsection 3.1. Let f and f′ be two quadratic forms defined over the same number field K and take P to be the same ideal. We assume that the restrictions f_{0} and f′_{0} of f and f′, respectively, to the hyperplanes X_{1} = 0 are equal to each other. Let Y_{n} and Y′_{n} be two manifolds as in (3.1). By our assumptions the hypersurfaces Y_{n−1} and Y′_{n−1} are isometric and the groups Γ_{n−1} and Γ′_{n−1} are isomorphic and actually can be identified with each other.

The hypersurface Y_{n−1} (resp. Y′_{n−1}) can either separate Y_{n} (resp. Y′_{n}) or not. Assume either both separate or both do not.

In the separating case the “hybrid manifold” is obtained by gluing one “half” of Y_{n} with a “half” of Y′_{n} along Y_{n−1} ≃ Y′_{n−1}. As explained in [GPS], the resulting manifold V is an oriented hyperbolic manifold whose fundamental group Γ is a lattice in SO(n, 1). (It can be either cocompact or not). In group theoretical terms one sees that Γ = A ∗_{Γ_{n−1}} A′ where A

(resp. A′) is the fundamental group of the half taken from Y_{n} (resp. Y′_{n}).

We can now argue in a way similar to the proof of Theorem 3.1: A and A′ are subgroups of GL_{n+1}(𝔒) which contain Γ_{n−1} as a proper subgroup of infinite index. Moreover, Γ_{n−1} is closed in the congruence topology and so, for most congruence quotients of GL_{n+1}(𝔒), the image of Γ_{n−1} is different (and of index > 2) from the images of A and A′. By Lemma 2.2, Γ is mapped onto a virtually free group.

Let us consider now the nonseparating case: In this case Y_{n} (resp. Y′_{n}) is cut along Y_{n−1} (resp. Y′_{n−1}) and the resulting manifold V_{n} (resp. V′_{n}) has two boundary components, each isometric to Y_{n−1} (resp. Y′_{n−1}). As Y_{n−1} ≃ Y′_{n−1} we can glue V_{n} to V′_{n} to get a complete hyperbolic manifold V (see [GPS]) whose fundamental group we denote by Γ. Here is the group theoretic description of Γ: as explained in (3.1), Γ_{n} = π_{1}(Y_{n}) = A_{∗_{Γ_{n−1}}} = ⟨A, t⟩ and Γ′_{n} = π_{1}(Y′_{n}) = A′_{∗_{Γ{′ \atop n−1}}} = ⟨A′, t′⟩. By gluing one boundary component of Y_{n} with one boundary component of Y′_{n} we get a manifold U = Y_{n} ∪ Y′_{n} with π_{1}(U) = A ∗_{Γ_{n−1}=Γ′_{n−1}} A′. Then we close U by gluing its two boundary

components one to the other to get the resulting manifold V with

π_{1}(V) = ⟨π_{1}(U), t′′| t^{′′−1}(t^{−1}Γ_{n−1}t)t′′ = t^{′−1}Γ_{n−1}t′⟩.

Note that π_{1}(U) is still a subgroup of GL_{n+1}(𝔒). This is not clearly the case for π_{1}(V). Still, t′′ ∈ π_{1}(V) which conjugates t^{−1}Γ_{n−1}t (≤ A ≤ GL_{n}(𝔒)) to t^{−1}Γ_{n−1}t′ (≤ A′ ≤ GL_{n}(𝔒)) preserves the congruence structure. Thus if we take all this mod 𝔮 for some ideal 𝔮 prime to P, we get a map from π_{1}(V) onto the HNN-construction of π_{1}(U)(mod 𝔮) over the image of t^{−1}Γ_{n−1}t^{−1} (mod q). The latter is a proper subgroup of the first (for most q). Thus Γ is mapped onto a virtually free group (by [Se2, Prop. 11 p. 120 and Exercise 3 p. 123]).

To summarize we have:

Theorem 3.2. Let M be a hyperbolic manifold obtained as the hybrid manifold of two arithmetic manifolds as in [GPS]. Then π_{1}(M) is mapped onto a virtually-(nonabelian) free group.

Corollary 3.3. M has finite sheeted covers with arbitrary large first Betti number.

Remarks 3.4. (a) In case v_{n} = Y_{n} \ v′_{n} = Y_{n−1} and Y′_{n} \ Y′_{n−1} are connected (the nonseparating case) it is clear that the resulting hybrid manifold has a nonzero first Betti number. It is less clear (but proved by our method) that by passing to finite covers this number can be made arbitrary large. For arithmetic lattices there is a general argument which ensures that once β_{1} ≠ 0, then β_{1} is arbitrary large for some congruence subgroups (see [Bo, 2.8 and 4.2]). For nonarithmetic lattices we do not know any such general principle (and it is clearly not true for a general manifold).

(b) One can easily generalize the construction of hybrid manifolds: it is possible to cut few manifolds, each one in few disjoint hypersurfaces and then to glue them together in various different ways à la [GPS] (see also [VS, p. 228–231]). Theorem 3.3 would hold for such constructions as well.

3.3 Manifolds with totally geodesic hypersurfaces A common generalization of Theorems 3.1 and 3.2 is:

Theorem 3.5. Let M be an oriented n-dimensional finite-volume hyperbolic manifold. Assume M has a codimension one totally geodesic submanifold F. Then Γ = π_{1}(M) has vF Q and, in particular, M has finite sheeted covers with arbitrarily large first Betti number.

Proof. Γ is a lattice in P O(n, 1) and after conjugation we can assume that π_{1}(F) is mapped into P O(n − 1, 1). In fact it is a lattice there. The hypersurface F may cut M into two part 𝔞 and 𝔟 or be a nonseparating one. In the first case Γ = A ∗_{π_{1}(F)} B where A (resp. B) is the fundamental group

of 𝔞 (resp. 𝔟). In the second case Γ = A_{∗_{π_{1}(F)}} where this time A is the fundamental group of the open manifold obtained by cutting M along F. In either case π_{1}(F) is of infinite index in A (and B). By the Borel density theorem π_{1}(F) is Zariski dense in P O(n−1, 1) and as P O(n−1, 1) is a maximal algebraic subgroup of P O(n, 1), we deduce that A (and B) is Zariski dense in P O(n, 1). This implies that with respect to the pro-finite topology of Γ, π_{1}(F)^{∗} is of infinite index in A^{∗} (and in B^{∗}). Indeed, if π_{1}(F)^{∗} would be of finite index in A^{∗}, then by [MS, Proposition 3] the Zariski closure of π_{1}(F) would be of finite index in the Zariski closure of A. But SO(n − 1, 1) is of infinite index in SO(n, 1). We can now apply Lemma 2.6 to deduce that Γ has vF Q.

3.4 Groups generated by reflections An easy corollary of Theorem 3.5 is:

Corollary 3.6. Let Γ be a torsion-free lattice in P O(n, 1). Assume there exists a reflection τ ∈ P O(n, 1) normalizing Γ. Then Γ has vF Q.

Proof. The reflection τ acts on the manifold Γ \ ℍ^{n} and its fixed point set is a codimension one totally geodesic submanifold. Theorem 3.5 now applies for Γ = π_{1}(Γ \ ℍ^{n}).

This in particular says that if Γ is generated by reflections then a finite index subgroup of Γ has vF Q. For n ≥ 4, all known examples of nonarithmetic groups are either “hybrid manifolds” (3.2) or are commensurable to ones generated by reflections (cf. [VS, pp. 227-228]). So, all of them satisfy Thurston’s conjecture, i.e., have finite covers with positive Betti number. For n = 3, there is another kind of nonarithmetic lattices: those obtained by (closing) knots’ complements à la Thurston. Most of them are nonarithmetic and for those Thurston’s conjecture is open. Note however that all torsion-free nonuniform lattices in P O(3, 1) have positive Betti number, so Thurston’s conjecture is open only for the cocompact ones. We do not know, however, whether an arbitrary nonuniform lattice in P O(3, 1) has a finite index subgroup which is mapped onto a non-abelian free group. This is the case for the arithmetic ones as we now show.

3.5 The Bianchi groups The group P O(3, 1) is locally isomorphic to SL_{2}(ℂ) and every arithmetic nonuniform lattice in it is commensurable (up to conjugacy) to SL_{2}(O) where O is the ring of integers in Q(\sqrt{−d}) for some 0 < d ∈ ℤ. The fact that SL_{2}(O) has a finite index subgroup with a nonabelian free quotient was proved by Grunewald and Schwermer [GS] using number theoretical results. It also follows from our Theorem 3.1 since SL_{2}(O) are commensurable to the standard arithmetic lattices discussed there. We however show here that the proof is “constructive” allowing us to present explicit subgroups of finite index with free quotients of explicitly calculated ranks. (In fact Theorems 3.1 and 3.2 are special cases of 3.5, we have proved them separately as the direct proof is effective).

Theorem 3.7. Let O = O_{d} be the ring of integers in Q(\sqrt{−d}), 0 < d ∈ ℤ and Γ_{d} a finite index torsion free subgroup of SL_{2}(O). Then Γ_{d} has vF Q. Moreover if p ∈ ℤ is a prime which does not split in O, then if p is large enough, Γ_{d}(p) = ker(SL_{2}(O) → SL_{2}(O/pO)) is mapped onto a free group of rank p^{3} + p − 1.

Proof. Let Δ_{d} = Γ_{d} ∩ SL_{2}(ℝ) = Γ_{d} ∩ SL_{2}(ℤ). Then Δ_{d} \ SL_{2}(ℝ)/SO(2) is a totally geodesic hypersurface in Γ_{d} \ SL_{2}(ℂ)/SU(2). Thus either Γ_{d} = A ∗_{Δ_{d}} B for some subgroups A and B of Γ_{d} containing Δ_{d} as a subgroup

of infinite index (see (3.1)) or Γ_{d} = A_{∗_{Δ_{d}}}, an HNN-construction for some Δ_{d} < A < Γ_{d}, where (A : Δ_{d}) is infinite. As Δ_{d} is closed in the congruence topology of Γ_{d}, Lemma 2.6 implies that Γ_{d} has vF Q.

Moreover, let π be the projection of Γ_{d} to SL_{2}(O/pO) = D_{p} (which is onto for almost all p). If p is large enough π(A) (and π(B)) is equal to D_{p} while π(Δ_{p}) is equal to SL_{2}(ℤ/pℤ) ≨ D_{p}. Indeed, for large enough p, π(A) contains π(Δ_{p}) properly and SL_{2}(p) is a maximal subgroup of D_{p} = SL_{2}(p^{2}). We can apply Lemma 2.2 (iii) and Lemma 2.4 (ii) to ker(π) = Γ_{d}(p), to deduce that Γ_{d}(p) is mapped onto a free group of rank r = \frac{|SL_{2}(p^{2})|}{|SL_{2}(p)|} − 1 = p^{3} + p − 1.

4. Haken 3-manifolds

Throughout this section, M is a compact orientable irreducible 3-manifold with an infinite fundamental group. M is called Haken if it contains a 2-sided incompressible surface. A well known conjecture of Waldhausen asserts that M is always virtually Haken, i.e., has a finite cover which is Haken. This is a weaker conjecture than the one mentioned in the introduction asserting that M has a finite cover with positive β_{1}, since β_{1}(M) > 0 implies M is Haken.

In this section we give few results asserting that under suitable conditions, if M is Haken it has a finite cover with positive Betti number.

4.1 Haken nonhyperbolic 3-manifolds

Proposition 4.1. If M is Haken and nonhyperbolic then π_{1}(M) is either virtually-solvable or has vF Q. In any case it has a finite cover with a positive first Betti number.

Proof. The Jaco-Shalen-Johanson decomposition Theorem gives a splitting of M by incompressible tori into pieces which are either Seifert fibered, solv or, by the work of Thurston, have hyperbolic structure (see [H2] and [T]). If this set of tori is non empty then π_{1}(M) is decomposed as a free product with amalgam or HNN- construction where the edge group is abelian (isomorphic to ℤ ⊕ ℤ – the fundamental group of a torus). The edge group cannot be dense in the vertex groups (with the pro finite topology) as the vertex groups are non abelian and π_{1}(M) is residually finite ([H]). Thus Lemma 2.6 implies that Γ has vF Q. If this set of tori is empty then, as M is not hyperbolic, M is Seifert fibered or a solv manifold. In case it is solv, its fundamental group is solvable. In case it is Seifert fibered, it is either nil or Euclidean in which case its fundamental group is virtually nilpotent, otherwise π_{1}(M) is mapped onto a nonabelian free group.

The last proposition is a slight improvement of the main result of [Ko] (see also [LN, Corollary on page 212]).

Anyway, the conjecture Haken ⇒ vβ_{1} > 0 has to be proven only for hyperbolic manifolds.

4.2 Haken manifold where π_{1}(M) is LERF

A group Γ is called LERF (or: subgroup separable) if for every finitely generated subgroup H and every g ∈ Γ\H, there exists a homomorphism φ from Γ to a finite group such that φ(g) ∉ φ(H). It is widely believed that fundamental groups of hyperbolic (3-)manifolds are LERF. It is therefore of interest to observe:

Proposition 4.2. If M is Haken and π_{1}(M) is LERF but not virtually solvable, then Γ = π_{1}(M) has vF Q.

Proof. LERF means that for every finitely generated subgroup H of Γ, H^{∗} = H. Thus, if M is Haken Γ = A ∗_{π_{1}(F)}B or Γ = A_{∗_{π_{1}(F)}} where π_{1}(F) is a proper

subgroup of A (and B). By LERF-ness, π_{1}(F)^{∗} <_{≠} A^{∗} (and π_{1}(F)^{∗} <_{≠} B^{∗})

and hence by Lemma 2.6, Γ has vF Q. (It is impossible that (A^{∗}: π_{1}(F)^{∗}) = (B^{∗}: π_{1}(F)^{∗}) = 2 as in such a case π_{1}(F) would be normal in Γ which is impossible).

4.3 Some remarks on Haken ⇒ vβ_{1} > 0

Assume M is a closed hyperbolic Haken 3-manifold. So there exists a cocompact torsion free lattice Γ is P SL_{2}(ℂ) s.t. M = Γ\ℍ^{3}. M has an incompressible surface F. If F does not separate M, then β_{1}(M) > 0. So we assume F separates M into two parts 𝔄_{1} and 𝔄_{2}. Denote A_{i} = π_{i}(𝔄_{i}), i = 1, 2 and C = π_{1}(F). Then C is a subgroup (of infinite index) in A_{i}(i = 1, 2) and Γ = A_{i} ∗_{C} A_{2}. To prove that Γ has vβ_{1} > 0, it suffices to show that in the pro-finite topology of Γ, the closure of C is a proper subgroup of the closure of A_{i}(i = 1, 2), (see Lemma 2.6 (i)). An elegant argument of Long and Niblo [LN] shows that C is closed in the pro-finite topology of A_{i}. Indeed, let B_{i} be the double of 𝔄_{i} along its boundary F. Then B_{i} is a hyperbolic manifold. Hence Γ_{i} = A_{i} ∗_{C} A_{i} is a residually finite group. There is an involution τ_{i} of Γ_{i} sending one copy of A_{i} to the other and fixing C. Moreover, C is exactly the set of fixed points of τ_{i}. It is easy to see that such a set is closed in the pro-finite topology of Γ_{i} and hence in that of A_{i} (for i = 1 and 2).

This argument plus our methods gives:

Proposition 4.3. If 𝔄 is an irreducible, orientable 3-manifold with an incompressible boundary F, and M is a double of 𝔄 along F. Then π_{1}(M) has vF Q.

But, by no means the above argument shows that in our original M, C is closed in the pro-finite topology of Γ. Look for example at Example 2.3 above. There, C is closed (in fact of finite index) in A_{i} for i = 1, 2, but, C is dense in Γ = A_{1} ∗_{C} A_{2}. The topology induces from Γ to A_{i} may be weaker than the pro finite topology of A_{i}! An interesting test case is the following: Assume M is a double of a manifold 𝔄 with a boundary F such that the identification of the two halves is done after twisting the F by an automorphism φ. Algebraically, this means Γ = A ∗_{C=φ̃(C)} A where φ̃ is the

automorphism of C = π_{1}(F) induced by φ. Even in this case we do not know how to prove that vβ_{1} > 0.

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