L^{2}-Homology and asphericity

Jonathan A. Hillman

School of Mathematics and Statistics, The University of Sydney, Sydney, NSW 2006, AUSTRALIA

ABSTRACT

We use L^{2} methods to show that if a group with a presentation of deficiency one is an extension of Z by a finitely generated normal subgroup then the 2-complex corresponding to any presentation of optimal deficiency is aspherical and to prove a converse of the Cheeger-Gromov-Gottlieb theorem relating Euler characteristic and asphericity. These results are applied to several open problems on knot groups and on the homotopy types of certain 4-manifolds.

AMS (MOS) Subject Classification (1991): Primary 57M05, Secondary 57N13, 57Q45 Keywords and phrases: aspherical, deficiency, geometric dimension, L^{2}-homology, knot

§0. Introduction

One of the applications of L^{2}-cohomology in [CG] was to show that if X is a finite aspherical complex such that π_{1}(X) has an infinite amenable normal subgroup A then χ(X) = 0. (This generalised a theorem of Gottlieb, who assumed that A was a central subgroup [Gt]). When A is an elementary amenable group this has also been proven by a localization argument, and there is then a converse: if X is a [π, m]_{f}-complex (a finite m-dimensional complex with π_{1}(X) ≅ π and with (m − 1)-connected universal cover X̃) and π has a nontrivial torsion free elementary amenable group then X is aspherical if and only if χ(X) = 0 ([Hi3] - see also [Hi1, Hi2, Li and Ro]). Here we shall show that the converse also holds when A is merely infinite and amenable, as another easy application of L^{2} methods. We shall first give a similar but easier argument for groups π which are extensions of Z by a finitely generated normal subgroup which relates the conditions “π has deficiency 1” and “there is an aspherical [π, 2]_{f}-complex”. The result restricts further possible counterexamples to the Whitehead problem on subcomplexes of aspherical 2complexes and has interesting consequences for knot theory. In the final section we shall apply the earlier work to the study of abelian subgroups of 2-knot groups and to the homotopy characterization of certain closed 4-manifolds.

If π is a group then ζπ and π′ shall denote the centre and the commutator subgroup of π, respectively. If π is finitely presentable def(π) shall denote its deficiency. A P D_{n}^{(+)}- group is an (orientable) Poincaré duality group of dimension n. A [π, m]_{f}-complex X is aspherical if and only if π_{m}(X) = 0. In that case we shall say that π has geometric dimension at most m, written g.d.π ≤ m.

§1. Extensions of Z by finitely generated normal subgroups

The L^{2}-Betti numbers β_{i}^{(2)}(X) of a finite complex X are defined in [At]. (See also [CG],

[Ec2] and [Lü]). They are multiplicative in finite covers, and for i = 0 or 1 depend only on π_{1}(X). In [CG] a limiting process is used to define L^{2}-Betti numbers β_{i}^{(2)}(Y; π) for general actions of countable group π on a space Y, and it is shown that β_{i}^{(2)}(π) = β_{i}^{(2)}(K_{π}; π), where K_{π} is (any) contractible complex on which π acts freely. If X is a finite Poincaré duality complex then these Betti numbers satisfy Poincaré duality. The alternating sum of the L^{2}-Betti numbers is the Euler characteristic χ(X) [At]. The usual Betti numbers of a space or group shall be denoted by β_{i}(X; Q) = dim_{Q}H_{i}(X; Q).

Theorem 1. Let π be a finitely presentable group such that β_{i}^{(2)}(π) = 0 for i < m, and let X be a [π, m]_{f}-complex. If χ(X) = 0 then X is aspherical. Proof. Since X is (m − 1)-connected β_{i}^{(2)}(X) = β_{i}^{(2)}(π) = 0 for i < m and so χ(X) = (−1)^{m}β_{m}^{(2)}(X). Hence β_{m}^{(2)}(X) = 0 also, and so the L^{2}-homology of X is trivial. Since X is m-dimensional π_{m}(X) = H_{m}(X̃; Z) is a subgroup of the m^{th} L^{2}-homology group of X. Therefore π_{m}(X) = 0 and so X is aspherical. //

If X = S^{1} ∨ S^{1} then X is an aspherical [F (2), 1]_{f}-complex and β_{0}^{(2)}(F (2)) = 0, but χ(X) = −1 ≠ 0. Thus the implication in the statement of this theorem cannot be reversed, in general.

Theorem 2. Let π be an infinite finitely presentable group. Then def(π) ≤ 1 + β_{1}^{(2)}(π), with equality only if g.d.π ≤ 2. Proof. Let X be a finite 2-complex correponding to a presentation P for π. Then def(P) = 1 − χ(X) = 1 + β_{1}^{(2)}(π) − β_{2}^{(2)}(X) ≤ 1. If def(P) = 1 + β_{1}^{(2)}(π) then β_{2}^{(2)}(X) = 0, so π_{2}(X) = H_{2}(X̃; Z) = 0 and X is aspherical. Hence g.d.π ≤ 2. //

Let G = F (2) × F (2). Then β_{1}^{(2)}(G) = 0, by Proposition 2.7 of [CG]. Moreover def(G) = 0 and g.d.G = 2; in fact < u, v, x, y|ux = xu, uy = yu, vx = xv, vy = yv > is an optimal presentation. Thus the implication in the final sentence of the statement of this theorem cannot be reversed, in general. (Compare the Corollary of Theorem 3 below).

The following lemma is part of Theorem 2.1 of [Lü] (proven there under slightly stronger hypotheses).

Lemma [Lück]. If a finitely presentable group π is an extension of Z by a finitely generated normal subgroup N then β_{1}^{(2)}(π) = 0. Proof. Suppose that N is generated by g elements and let N_{n} be the preimage in π of the subgroup nZ ≤ Z. Then [π : N_{n}] = n, so β_{1}^{(2)}(N_{n}) = nβ_{1}^{(2)}(π). But each N_{n} is also finitely presentable and is generated by g + 1 elements. Hence β_{1}^{(2)}(N_{n}) ≤ g + 1, and so β_{1}^{(2)}(π) = 0. //

If the Whitehead conjecture is false then either there is a finite nonaspherical 2complex X such that X ∪_{f} D^{2} is contractible for some f : S^{1} → X or there is an infinite ascending chain of nonaspherical 2-complexes whose union is contractible [Ho]. In the finite case χ(X) = 0 and so π = π_{1}(X) has deficiency 1; moreover, π has weight 1 since it is normally generated by the conjugay class represented by f. Such groups are 2-knot groups. Conversely, the exterior of a ribbon n-knot or of a ribbon concordance between classical knots is homotopy equivalent to such a 2-complex. (The asphericity of such ribbon exteriors has been raised in Question 2 of [Co] and Question 6.5 of [Go]). If π′ is finitely generated then β_{1}^{(2)}(π) = 0, by Lück’s Lemma, and so X is aspherical, by Theorem 1.

A group is called knot-like if it has abelianization Z and deficiency 1 [Ra]. Rapaport asked whether the commutator subgroup of a knot-like group must be free if it is finitely generated, and established this in the 2-generator, 1-relator case [Ra]. Our next corollary provides a substantial partial answer to this question.

Corollary. Let π be a finitely presentable group which is an extension of Z by a finitely generated normal subgoup N, and suppose that β_{2}(π; Q) = β_{1}(π; Q)−1. Then def(π) = 1 if and only if g.d.π ≤ 2. If def(π) = 1 and N is almost finitely presentable then N is free. Proof. If def(π) = 1 then g.d.π ≤ 2 by Lück’s Lemma and Theorem 2. Conversely, if X is a finite aspherical 2-complex with π_{1}(X) ≅ π then χ(X) = 1 − β_{1}(π; Q) + β_{2}(π; Q) = 0. After collapsing a maximal tree in X we may assume it has a single 0-cell, and then the presentation read off the 1- and 2-cells has deficiency 1. The final assertion then follows from Corollary 8.6 of [Bi]. //

In particular, if the group of a fibred 2-knot has a presentation of deficiency 1 then its commutator subgroup must be free. This settles Problem 14 of [Hi2]. Any 2-knot with such a group is s-concordant to a fibred homotopy ribbon knot, by Theorem VIII.7 of [Hi2]. Must it in fact be a ribbon knot? (This is Problem 13 of [Hi2]).

The kernel of the homomorphism from F (2) × F (2) = | < u, v, x, y|ux = xu, uy = yu, vx = xv, vy = yv > | to Z which sends u and y to 0 and v and x to a generator is generated by u, vx^{−1} and y, but is not free, as u and y generate a rank two abelian subgroup. (Thus this kernel is finitely generated but not almost finitely presentable. See page 119 of [Bi]). Silver has given examples of high-dimensional knot groups whose commutators subgroups are finitely generated but not finitely presentable [Si1]. He has also suggested that every knot-like group should have a finitely presentable HNN base. If this were true the Corollary would settle Rapaport’s question completely, for if the commutator subgroup is finitely generated then it is the unique HNN base [Si2]. (See also Problem 22 of [Hi2]).

§2. Infinite amenable normal subgroups

The following result is stated without proof on page 226 of [Gr]. I am grateful to Peter Linnell for explaining how it follows from the argument of §3 of [CG].

Theorem [Gromov]. Let π be a group with an infinite amenable normal subgroup A. Then β_{i}^{(2)}(π) = 0 for all i. Proof. Since K_{π/A} × K_{π} (with the diagonal π-action) is π-freely homotopy equivalent to K_{π} we have β_{i}^{(2)}(π) = β_{i}^{(2)}(K_{π/A} × K_{π}; π), for all i. This is in turn equal to β_{i}^{(2)}(K_{π/A}; π), by Proposition 2.2 of [CG]. Now the cell-stabilizers of the action of π on K_{π/A} are all A, and by Theorem 0.2 of [CG], β_{i}^{(2)}(A) = 0, for all i. Since Σβ_{i}^{(2)}(K_{π/A}; π) ≤ Σβ_{i}^{(2)}(A), by Theorem 0.1 of [CG], it follows that β_{i}^{(2)}(π) = β_{i}^{(2)}(K_{π/A}; π) = 0, for all i. //

Since the amenability of A is used only to ensure that β_{i}^{(2)}(A) = 0 for all i, it is sufficient to assume that A be subnormal in π and β_{i}^{(2)}(A) = 0 for all i. (Note however that if a group has an infinite amenable subnormal subgroup then it has an infinite amenable normal subgroup).

The next result gives a converse to the Cheeger-Gromov extension of Gottlieb’s Theorem.

Theorem 3. Let X be a [π, m]_{f}-complex and suppose that π has an infinite amenable normal subgroup. Then X is aspherical if and only if χ(X) = 0. Proof. Since χ(X) = (−1)^{m}β_{m}^{(2)}(X) this follows immediately from Gromov’s Theorem and Theorem 1. //

Corollary. Let π be a finitely presentable group with an infinite amenable normal subgroup A. Then def(π) = 1 if and only if g.d.π ≤ 2. Proof. If X is an aspherical [π, 2]_{f}-complex then χ(X) = 0 by Gromov’s Theorem, so def(π) = 1 − χ(X) = 1. Conversely if X is the finite 2-complex corresponding to a presentation of deficiency 1 then χ(X) = 0 and so X is aspherical by Theorem 3.//

In [Hi3] it is shown that if def(π) = 1 and the subgroup A is elementary amenable then either A ≅ Z or π is metabelian. Is this true in general? (If the Tits alternative holds for groups of finite cohomological dimension this would be so).

§3. Applications to 4-manifolds and 2-knots.

If L : μS^{2} → S^{4} is a 2-link M(L) shall denote the closed orientable 4-manifold obtained by surgery on the components of L. The link group is then πL = π_{1}(M(L)). If K is a 2-knot (μ = 1) M(K)′ shall denote the infinite cyclic covering space, with π_{1}(M(K)′) = πK′.

The arguments of §1 enable us to show that no 2-knot group π with deficiency 1 can be a P D_{4}-group (thereby removing the hypothesis π′ ≠ π′′ from Theorem III.9 of [Hi2]). For if X is a finite 2-complex corresponding to a deficiency 1 presentation of π then β_{2}^{(2)}(π) − β_{1}^{(2)}(π) ≤ β_{2}^{(2)}(X) − β_{1}^{(2)}(π) = χ(X) = 0. If π were a P D_{4}-group then we would also have 0 = χ(π) = β_{2}^{(2)}(π) − 2β_{1}^{(2)}(π). Hence all these L^{2}-Betti numbers are 0 and so X is aspherical, and so c.d.π = 2 ≠ 4.

The following theorem is implicit in the addendum to [Ec].

Theorem [Eckmann]. Let M be a finite P D_{4}-complex with χ(M) = 0 and let π = π_{1}(M). If β_{1}^{(2)}(π) = 0 then the natural map from H^{2}(π; Z[π]) to H^{2}(M; Z[π]) is an isomorphism. In particular, if moreover H^{s}(π; Z[π]) = 0 for s ≤ 2 then M is aspherical. Proof. Since M is a P D_{4}-complex χ(M) = 2β_{0}^{(2)}(π) − 2β_{1}^{(2)}(π) + β_{2}^{(2)}(M). Since π is infinite β_{0}^{(2)}(π) = 0, and β_{1}^{(2)}(π) = 0 by hypothesis. Hence β_{2}^{(2)}(M) = χ(M) = 0 also. It now follows from Proposition 1.2 of [Ec2] (the natural map from unreduced L^{2}-cohomology to ordinary cohomology factors through reduced L^{2}-cohomology) and diagram (4) on page 504 of [Ec2] that the natural map from H^{2}(M; Z[π]) to H^{2}(M̃; Z) is 0 (where M̃ is the universal cover of M) and so the map from H^{2}(π; Z[π]) to H^{2}(M; Z[π]) is an isomorphism. The final assertion follows by equivariant Poincaré duality in the universal covering space. //

We shall use the theorems of Eckmann and Gromov in conjunction to obtain strong restrictions on the possible abelian normal subgroups of 2-knot groups.

Theorem 4. Let L : μS^{2} → S^{4} be a 2-link with group π = πL. If μ ≥ 2 then ζπ is finite. If μ = 1 and ζπ is infinite, then either π is a P D_{4}-group (and so ζπ is torsion free of rank 1 or 2) or H^{2}(π; Z[π]) ≠ 0 and ζπ is finitely generated of rank 1 or is a torsion group. Proof. We have β_{2}^{(2)}(π) − 2β_{1}^{(2)}(π) = χ(M(L)) = 2(1 − μ). Hence if ζπ is infinite, so β_{1}^{(2)}(π) = 0 by Gromov’s Theorem, we must have μ = 1, and so L is a 2-knot. Moreover π has one end, i.e., H^{s}(π; Z[π]) = 0 for s ≤ 1. Hence if also H^{2}(π; Z[π]) = 0 then M(L) is aspherical, by Eckmann’s Theorem. (In this case the centre is torsion free, and is of rank at most 2: see Chapter VI of [Hi2]). If ζπ has an infinite cyclic subgroup A such that ζπ/A is infinite then this cohomological condition holds, for then ζ(π/A) is infinite, so π/A has one end, and so π is simply connected at ∞, by Theorem 1 of [Mi]. Thus if H^{2}(π; Z[π]) ≠ 0 then ζπ is finitely generated of rank 1 or is a torsion group. //

This provides a substantial partial answer to Problem 16 of [Hi2], which asked whether the centre of a 2-knot group must be finite cyclic, Z, Z ⊕ (Z/2Z) or Z^{2} and whether the centre of the group of a 2-link with more than one component must be trivial. Most of the results in the case μ = 1 follow also via the localization arguments of [Hi2], on observing that π/A cannot have two ends (since a knot group cannot be virtually Z^{2}) and has finite centre if it has infinitely many ends; however Gromov’s Theorem is needed to consider the possibility that ζπ may be an infinite torsion group. No examples of the latter type are known. Note also that the argument used in the rank 1 case settles Problem 19 of [Hi2], for if the group π of a 2-knot K has a rank one abelian normal subgroup A such that π/A has one end then β_{1}^{(2)}(π) = 0 and H^{s}(π; Z[π]) = 0 for s ≤ 2, and so M(K) is aspherical.

The theorems of Gromov and Eckmann together settle Problem 11 of [Hi5]. We may similarly simplify the hypotheses of some theorems in Chapter VI of [Hi5]. In the next result h shall denote the Hirsch length, a natural measure of the size of an elementary amenable group. (See [Hi3,5] for details).

Theorem 5. Let M be a closed 4-manifold with χ(M) = 0. Suppose that π = π_{1}(M) has an elementary amenable normal subgroup ρ with h(ρ) ≥ 2 and H^{2}(π; Z[π]) = 0. Then M is aspherical. If h(ρ) = 2 then ρ is virtually abelian, while if h(ρ) ≥ 3 then M is homeomorphic to an infrasolvmanifold. Proof. Since β_{i}^{(2)}(π) = 0 for all i, by the theorems of Eckmann and Gromov, M is aspherical. Hence ρ must be torsion free and the theorem follows from Theorems VI.2 and VI.11 of [Hi5]. //

If ρ is torsion free, of infinite index and h(ρ) = 2 then the hypothesis H^{2}(π; Z[π]) = 0 follows from [Mi]. (See Theorem VI.11 of [Hi5]). The hypothesis that the index be infinite is necessary; every group with a presentation of the form < a, t|tat^{−1} = a^{n} > is torsion free and solvable of Hirsch length 2, and is the fundamental group of some closed orientable 4manifold with Euler characteristic 0. If M is a closed orientable 4-manifold with χ(M) = 0 and such that π = π_{1}(M) is amenable, has one end and H^{2}(π; Z[π]) ≠ 0 must π be one of these groups? (Corollary 4.2 of [Ec2] implies that Problem 17 of [Hi2] may be reduced to a special case of this question). If h(ρ) ≥ 3 can the hypothesis on H^{2}(π; Z[π]) be dropped completely? Can the hypotheses of this theorem be rephrased in terms of amenable normal subgroups, using homological dimension over Q rather than Hirsch length as a measure of the size of such groups? (This is Problem 12 of [Hi5], and would follow from a Tits alternative for groups of finite cohomological dimension).

Corollary. Let K be a 2-knot whose group π = πK has an abelian normal subgroup A of rank > 1. Then M(K) is aspherical and so A is torsion free. Proof. An easy LHS spectral sequence argument shows that H^{s}(π; Z[π]) = 0 for s ≤ 2. Hence M(K) is aspherical, by Theorem 5, and so π is a P D_{4}^{+}-group and A is torsion free.//

In fact either π′ is a P D_{3}^{+}-group with centre A∩π′ of rank 1 but not finitely generated (which seems unlikely) or A is free abelian, of rank 2, 3 or 4 [Hi2].

If π′ is almost finitely presentable (F P_{2}) we can do better. The strategy of the next result is adapted from that of [Hi4].

Theorem 6. Let M be a closed 4-manifold with χ(M) = 0 and such that π = π_{1}(M) is an extension of Z by an almost finitely presentable infinite normal subgroup N with a nontrivial finite normal subgroup F. Then M is homotopy equivalent to the mapping torus of a self homeomorphism of RP^{2} × S^{1}. Proof. Let M̃ be the universal covering space of M. Since N is infinite and finitely generated π has one end, and so H_{i}(M̃; Z) = 0 for i ≠ 0 or 2. Let Π = π_{2}(M) = H_{2}(M̃; Z). We wish to show that Π ≅ Z, and that w = w_{1}(M) maps F isomorphically onto Z^{×} = {±1}. Since β_{1}^{(2)}(π) = 0 by Lück’s Lemma, Poincaré duality and Eckmann’s Theorem together give an isomorphism of left Z[π]-modules Π ≅ \overline{H^{2}(π; Z[π])}. An application of the LHS spectral sequence for π as an extension of Z by N then gives Π ≅ \overline{H^{1}(N; Z[N])}, which is a free abelian group.

Let f be a nontrivial element of F. Since F is normal in π the centralizer C_{π}(f) of f has finite index in π, and we may assume without loss of generality that F is generated by f and is central in π. It follows from the spectral sequence for the projection of M̃ onto M̃/F that there are isomorphisms H_{s+3}(F; Z) ≅ H_{s}(F; Π) for all s ≥ 4, since M̃/F is a 4-dimensional complex. Here F acts trivially on Z, but we must determine its action on Π.

Now central elements n of N act trivially on H^{1}(N; Z[N]) and hence via w(n) on Π. (See [Hi4]). Thus if w(f) = 1 the sequence 0 → Z/|f|Z → Π → Π → 0 is exact, where the right hand homomorphism is multiplication by |f|. As Π is torsion free this contradicts f ≠ 1. Therefore if f is nontrivial it has order 2 and w(f) = −1. Hence w : F → Z^{×} is an isomorphism and there is an exact sequence 0 → Π → Π → Z/2Z → 0, where the left hand homomorphism is multiplication by 2. Since Π is a free abelian group it must be infinite cyclic, and so M̃ ≃ S^{2}. The theorem now follows from Theorems VII.4 and VII.7 of [Hi5]. //

Corollary. Let K be a 2-knot whose group π = πK has a nontrivial abelian normal subgroup A, and suppose that π′ is finitely presentable. Then either π′ is finite or A∩π′ = 1 and A ≅ Z or M(K) is aspherical and A is torsion free. In all cases M(K)′ is an orientable P D_{3}-complex. Proof. If A ∩ π′ is finite then either A ∩ π′ = 1 or π′ is finite, by Theorem 6. Hence we may assume that A ∩ π′ is infinite. But then π′ has one end and so H^{s}(π; Z[π]) = 0 for s ≤ 2, by an LHS spectral sequence argument. Hence M(K) is aspherical, by Eckmann’s Theorem. The final assertion follows from [Hi6]. //

If we assume only that π′ is finitely generated then this Corollary and Theorem IV.5 of [Hi2] together imply that either π′ is finite or A ∩ π′ = 1 or M(K) is aspherical or A is a torsion group. There are no known examples of 2-knot groups π with π′ finitely generated but not finitely presentable. (See [Si1] for higher dimensional examples).

L^{2}-Cohomology is used in [Hi6] to give a simple characterization of P D_{4}-complexes which are homotopy equivalent to mapping tori, which largely settles Problem 8 of [Hi2] and reduces Problem 7 of [Hi5] to that of characterizing P D_{3}-complexes which are homotopy equivalent to 3-manifolds. In particular, a closed 4-manifold M is homotopy equivalent to the total space of a surface bundle over a torus if and only if χ(M) = 0 and π = π_{1}(M) is an extension of Z^{2} by a finitely presentable normal subgroup. The next result is an alternative characterization of 4-manifolds covered by such bundle spaces, and settles Problem 14 of [Hi5].

Theorem 7. Let M be a closed 4-manifold with χ(M) = 0 and such that π = π_{1}(M) has a normal subgroup G of infinite index which is a P D_{2}-group. Then M is aspherical. If ζG = 1 then M is finitely covered by a manifold which is simple homotopy equivalent to the total space of a surface bundle over the torus. Proof. It follows easily from the LHS spectral sequence for π as an extension of H = π/G by G that H^{s}(π; Z[π]) = 0 for s ≤ 2. After passing to a finite covering space if necessary, we may assume that the image of H in Out(G) is torsion free, since Out(G) is virtually of finite cohomological dimension. The image of H in Out(G) is isomorphic to π/G.C_{π}(G) and H is an extension of this group by G.C_{π}(G)/G ≅ C_{π}(G)/ζG.

If ζG ≠ 1 then it is an infinite (elementary) amenable normal subgroup of π and so M is aspherical, by the theorems of Eckmann and Gromov. (The asphericity also follows from Theorem II.6 of [Hi5], since Z[π] has a safe extension). If ζG = 1 and H has an element of infinite order then β_{1}^{(2)}(π) = 0 by Theorem 3.1 of [Lü] and so M is aspherical by Eckmann’s Lemma. Moreover v.c.d.H ≤ v.c.d.K +c.d.C_{π}(G) < ∞ and so H is virtually a P D_{2}-group, by Theorem 9.11 of [Bi]. In fact H must be virtually Z^{2}, since we then have 0 = χ(π) = χ(G)χ_{virt}(H). If H is torsion then its image in Out(G) is finite, hence trivial, so π = G.C_{π}(G) and H ≅ C_{π}(G)/ζ(G) ≅ C_{π}(G). Since G ∩ C_{π}(G) = ζG is trivial π ≅ G × C_{π}(G) ≅ G × H. But then β_{1}^{(2)}(G × H) = 0, by Proposition 2.7 of [CG], and so G×H is a P D_{4}-group, hence torsion free, contrary to the assumption that H is an infinite torsion group. This completes the argument. //

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