WHITEHEAD S ASPHERICITY QUESTION AND ITS RELATION TO OTHER OPEN PROBLEMS

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1 WHITEHEAD S ASPHERICITY QUESTION AND ITS RELATION TO OTHER OPEN PROBLEMS A. J. BERRICK AND J. A. HILLMAN A. This note explores J. H. C. Whitehead s 1941 question as to whether a subcomplex of an aspherical 2-complex need also have vanishing higher homotopy groups. Methods from L 2 - cohomology are brought to bear on the question, and relate it to other open problems on low-dimensional complexes some introduced here as well as open problems on group theory, such as the Kervaire-Laudenbach conjecture, and on group algebras, like the Bass trace conjecture. In [43], J. H. C. Whitehead asked whether any subcomplex of an aspherical 2-dimensional complex must be aspherical. That this open question has an affi rmative answer is widely referred to as the Whitehead Conjecture. It is one of a number of longstanding open problems on low-dimensional complexes. For example, it is well known [8] that the Whitehead Conjecture and the Eilenberg-Ganea Conjecture (that every group G with cdg = 2 has a 2-dimensional K(G, 1)- complex) cannot both be true. In this work, we employ techniques of L 2 -cohomology to explore relationships of the conjecture to several open problems, exposing some new linkages. The first exploration concerns a reduction by Howie [28] of the Whitehead Conjecture to two possible cases. One of these, which we call the knot-like case, corresponds to a finitely presented group of weight and deficiency both equal to 1 having cohomological dimension at most 2. This relates to several open problems, including the Kervaire- Laudenbach Conjecture, that if a free product G Z has weight 1 then G = 1. Examples testing the sharpness of various implications here are discussed in the paper s second section. In the third section we exhibit some suffi cient conditions for asphericity of a 2-complex in terms of its L 2 -Betti numbers and those of its fundamental group, and observe a network of implications in the case of groups of equal weight and deficiency. Date: October 1, Key words and phrases. acyclic cover, Cockcroft condition, cohomological dimension, L 2 -Betti number, subaspherical complex, Whitehead Conjecture. 1

2 2 AJB & JAH OCTOBER 1, 2018 The next section focuses on a subcomplex of an aspherical 2-complex, where the Cockcroft condition comes into play. This leads to a characterization of aspherical 2-complexes that includes the weak Bass Conjecture as one of its components. This is perhaps surprising in that the conjecture was originally formulated algebraically, as the equality of the Kaplansky and augmentation traces of finitely generated projective modules over group rings [2]. The characterization gives rise to a reformulation of the Whitehead Conjecture as the conjecture that the fundamental group of a finite subcomplex of a contractible 2-complex has rational cohomological dimension at most 2. Suppose that a subcomplex X of an aspherical 2-complex is a counterexample to the Whitehead Conjecture. Then, after Adams [1], Brandenburg, M. Dyer and Strebel [12], J.M. Cohen [13] and Howie [26], it is known that nontrivial perfect subgroups of π 1 (X) (finitely generated, in Howie s case) and associated acyclic covering spaces of X (in the other cases) must enter the picture. This raises the question of whether an acyclic cover can be associated to the case of a finitely generated perfect group. With the help of our L 2 -Betti number results, we are to a large extent able to answer that question in the following two sections, with the assistance, intriguingly, of the generalized Kervaire-Laudenbach Conjecture. 1. T - W C Let π be a finitely presentable group, with a presentation P = x i, i g w j, j r φ. The deficiency of P and π are def(p) = g r and def(π) = max{def(p) P presents π}, respectively. Let X = X(P) be the corresponding 2-complex, with one 0-cell, g 1-cells and r 2-cells. Then π 1 (X) =π and χ(x) = 1 def(p). The weight wt(π) of a group π is the minimum cardinality of a subset whose normal closure is π. It is easy to see that if π is finitely presentable then (with π as the commutator subgroup of π) (1.1) wt(π) wt(π/π ) rank(π/π ) def(π). The last inequality is because if X is the 2-complex associated to a presentation of π of maximum deficiency, then rank(π/π ) = β 1 (X) 1 χ(x) = def(π). In particular, if def(π) = wt(π) = w then π/π = Z w. We recall that Howie [28] showed that if the Whitehead Conjecture is false then there is a counterexample of one of the following two types: (a) there is a finite 2-complex X with π 2 (X) 0 and a map f : S 1 X such that Y = X f e 2 is contractible; or

3 WHITEHEAD ASPHERICITY QUESTION October 1, (b) there is an infinite ascending chain K n K n+1 of finite 2- complexes with π 2 (K n ) 0 for all n, and such that each inclusion is nullhomotopic. (Thus, n 1 K n is contractible.) Subsequently, Luft [36] proved that if there is a counterexample of type (a) then there is also one of type (b). Here, we concentrate on (a) above, from the perspective of L 2 -Betti numbers. The negation of Howie s case (a) is statement (i) of the following result. Proposition 1.1. The following statements are equivalent. (i) Every finite 2-complex X such that for some map f : S 1 X the complex X f e 2 is contractible, has X itself aspherical. (ii) Every finitely presentable group π of weight 1 and deficiency 1 has cd Q π 2. Proof. (i) (ii). Let π be a group of weight 1 that has a presentation P of deficiency 1, X(P) be its associated 2-complex with π 1 (X(P)) = π, and f : S 1 X(P) represent a conjugacy class whose closure is π. Then Y (P ) = X(P) f e 2 is a simply-connected 2-complex with χ(y (P)) = 1 + χ(x(p)) = 2 def(π) = 1; this makes Y (P) contractible. Therefore, by (i), the 2-complex X(P) serves as a K(π, 1), so that in fact gd(π) 2. (ii) (i). If X is a finite 2-complex such that Y = X f e 2 is contractible, then χ(x) = χ(y ) 1 = 0 and π = π 1 (X) is the closure of the conjugacy class represented by the attaching map f of the final 2- cell of Y. Therefore π is finitely presentable and def(π) 1 χ(x) = 1. Since π has weight 1 we see that def(π) = 1 and π/π = Z. Hence H 2 (X; Z) = H 2 (X; Q) = 0, and so Q Z[π] π 2 (X) H 2 (X; Q) = 0. It follows from (ii) that cd Q π 2. As in Theorem 2.8 of [25] the latter two properties together imply that π 2 (X) = 0 and so X is aspherical. (See also the proof of Theorem 4.4 below.) We refer to these equivalent assertions as the knot-like case of the Whitehead Conjecture because, for each n 2, every group π satisfying the hypotheses of statement (ii) above is an n-knot group (that is, π = πk = π 1 (S n+2 K), where K is a locally flat n-sphere in S n+2 ) [32]. (Note that the term knot-like was used more broadly in [39], for groups with deficiency 1 and infinite cyclic abelianization.) Lemma 1.2. Let π be a finitely presentable group of weight 1 and deficiency 1. If every nontrivial finitely generated perfect subgroup of π is infinite and subnormal, then cd Q π 2. Proof. As deduced from the inequalities (1.1) above, we have π/π = Z and so any perfect subgroup of π, being a subgroup of π, must have infinite index in π. If cd Q π > 2, then by definition the X(P) associated to π fails to be aspherical. According to [26], π then has a nontrivial

4 4 AJB & JAH OCTOBER 1, 2018 finitely generated perfect subgroup H. Since H is by hypothesis infinite and subnormal, and of infinite index, by [6] Theorem 3.2 we have cd Q π 2. Note that the class of groups π satisfying (a) above properly includes the class of groups having no nontrivial finitely generated perfect subgroups, which in turn properly includes the class of locally indicable groups (for which every nontrivial finitely generated subgroup has infinite abelianization). As noted in [7], an interesting question is whether this knot-like case of the Whitehead Conjecture is equivalent to, or strictly weaker than, the finite case of the Whitehead Conjecture, namely the assertion that any subcomplex of a finite aspherical 2-complex is also aspherical. We now introduce L 2 -Betti numbers i [35] into this situation. In Theorem 2.4 of [25] it is shown that if X is a finite 2-complex with χ(x) = 0 and 1 (π 1 (X)) = 0 then X is aspherical. This result shows that the knot-like case is implied by the following hypothesis, first posed as the central question of [7]. Hypothesis 1.3. If π is a finitely presentable group of weight 1 and deficiency 1, then 1 (π) = 0. If πk is the group of a tame classical knot K S 3 then 1 (πk) = 0 (see 4.3 of [35]), and so the 2-complex associated to any deficiency 1 presentation of a classical knot group (not necessarily the standard Wirtinger presentation) is aspherical. Hypothesis 1.3 implies that this situation holds for any n-knot group of deficiency 1. Hypothesis 1.3 also implies the Kervaire-Laudenbach Conjecture for groups of deficiency 1: if def(π) = wt(π) = 1 then π is indecomposable. For, since by the inequalities (1.1) above π/π is Z, if π is a proper free product, then one component must be perfect while the other must have abelianization Z. Then 1 (π) > 0, by Theorem 1.35 of [35]. These observations may be summarized in the following network of implications. Whitehead Conjecture finite Whitehead Conjecture Hypothesis 1.3 = knot-like Whitehead Conjecture deficiency 1 Kervaire- Laudenbach Conjecture The natural class of groups to consider in the higher-weight cases are the groups with weight = deficiency. Such groups are groups of w-component n-links (for all n 2), and whether they all have cohomological dimension at most 2 is a question of long standing. (See

5 WHITEHEAD ASPHERICITY QUESTION October 1, Question 3 of [12].) In this context, 1 (π) w 1, by Corollary 3.4 below, with equality for the free group of rank w. Hence the natural extension of Hypothesis 1.3 to groups of higher weight is as follows. Hypothesis 1.4. If π is a finitely presentable group of weight w and deficiency w, then 1 (π) = w 1. If this is so, then every subcomplex of a finite contractible 2-complex is aspherical, and every group π with wt(π) = def(π) has finite geometric dimension at most 2. (See Lemma 3.5.) What might be a good strategy for attacking Hypothesis 1.3? If X is a finite 2-complex with π 1 (X) = π and f : V X is the inclusion of a subcomplex V S 1 such that the closure of the conjugacy class of [f] is the whole of π, then V is a retract of X (since π/π = Z) and X/V is contractible. One is tempted to compare the L 2 -chain complex of the universal cover X with that of the induced cover of the subcomplex V. We may also reformulate Hypothesis 1.3 in more topological terms and extend it as follows. Hypothesis 1.5. Suppose that X is a finite 2-complex, with retract a subcomplex V such that X/V is contractible. Then 1 (π 1 (X)) 1 (π 1 (V )). This is certainly the case if the kernel of the epimorphism from π 1 (X) to π 1 (V ) induced by the retraction is finitely generated. For, when the kernel is finite, then π 1 (V ) is isomorphic to a subgroup of finite index in π 1 (X), and so 1 (π 1 (X)) is just 1 (π 1 (V )) divided by that index. When, on the other hand, the finitely generated kernel is infinite, then 1 (π 1 (X)) = 0 by [35] Theorem E Dicks and Linnell [14] have recently shown that if π is a torsion-free one-relator group on g generators then 1 (π) = g 2 and 2 (π) = 0. (Their computations extend to a larger class of groups, and they show that 2 (π) = 0 for every one-relator group π [14].) In particular, if π is a one-relator group and def(π) = wt(π) = 1 then π is torsion-free and g = 2, and so 1 (π) = 0. Let P be a g-generator, r-relator presentation for π, with corresponding 2-complex X = X(P) having one 0-cell, g 1-cells and r 2-cells. We recall that the cellular chain complex of the universal covering space X is a finite complex C = C ( X) of left Z[π]-modules, with C 0 = Z[π], C 1 = Z[π] g, C 2 = Z[π] r and C q = 0 otherwise. The nonzero differentials are given by 1 (e i ) = x i 1 for i g and 2 (f j ) = ΣJ(P) kj e k for j r, where J(P) = [φ( w j x k )] is the r g Jacobian matrix of free derivatives corresponding to this presentation.the work [14] suggests that we should attempt to show that the Jacobian matrix J(P) representing the second differential in the chain

6 6 AJB & JAH OCTOBER 1, 2018 complex C ( X(P)) has an r r submatrix that is invertible when coefficients are extended to the algebra of operators U(π) affi liated to the von Neumann algebra N (π). One might expect this to be generically true for presentations of deficiency 1. However some qualification is necessary, as the following examples show. Our primary interest (Hypothesis 1.3) is whether weight 1 (w) and deficiency 1 (d) together imply that 1 = 0 (b). It is immediate that the first two conditions imply that the abelianization is infinite cyclic (a). We shall give a number of examples illustrating the independence (or otherwise) of each of these four properties from conjunctions of the others. There are four maximal combinations: (1) bdw a. This is obvious, since dw a, as just observed. (2) abw d. The semidirect product Z/3Z 1 Z is solvable and has abelianization Z. Hence it has weight 1 and 1 = 0. However it cannot have deficiency 1, since it has nontrivial torsion. (See Corollary 3.2 below.) (3) abd w. Let p, q, r be distinct primes, and let π be the group with presentation x, y, z x p = y q = z r. Then π/π = Z and def(π) = 1. Since π has an infinite central subgroup (generated by the image of x p ), 1 (π) = 0. However π does not have weight 1, since the free product Z/pZ Z/qZ Z/rZ obtained by factoring out the centre has weight 2 [29]. (The authors thank Jack Button for directing us to this example.) (4) (a)dw b? This is Hypothesis 1.3. If we consider these conditions in pairs, then there are another four combinations not already covered. (1) bd a. The group Z 2 is a counterexample. (2) bw a. The group Z/2Z is a counterexample. (3) ad b. Let Hig be Higman s infinite superperfect group, with presentation a, b, c, d ab = ba 2, bc = cb 2, cd = dc 2, da = ad 2. Then Z Hig has abelianization Z and deficiency 1. However 1 (Z Hig) 1, by Theorem 1.35 of [35]. (4) aw b. The group π with presentation a, b, t a 3 = b 7 = 1, aba 1 = b 2, tat 1 = a 2 is the normal closure of the image of t, and so has weight 1. It is an HNN extension with base a, b of order 21 and associated subgroups of order 3, and so is virtually free. Hence 1 (π) = χ virt (π). It is not hard to show that χ virt (π) = = 2 7,

7 WHITEHEAD ASPHERICITY QUESTION October 1, using the additivity of the Euler characteristic over amalgamated free products and HNN extensions. Thus 1 (π) = 2. 7 The group just given is in fact an n-knot group, for all n 2, and has infinitely many ends [21]). Can the condition deficiency = 1 in Hypothesis 1.3 be replaced by π/π = Z and π has one end? If so, this would give a natural criterion for the asphericity of the closed 4-manifold M(K) obtained by surgery on a locally flat 2-knot K. Such a manifold M(K) is orientable, πk = π 1 (M(K)) is the knot group, and χ(m(k)) = 0 (see Chapter 14 of [25]). If 1 (πk) = 0 and H s (πk; Z[πK]) = 0 for s 2, then M(K) is aspherical and πk is a PD 4 -group, by Corollary of [25]. Since these cohomological conditions are clearly necessary, Hypothesis 1.3 would strengthen this result to give a very natural criterion: M(K) is aspherical if and only if H s (πk; Z[πK]) = 0 for s 2. (Note that H 0 (πk; Z[πK]) = 0 since πk is infinite; H 1 (πk; Z[πK]) = 0 if and only if πk has one end; if K is a nontrivial classical knot then H 2 (πk; Z[πK]) 0.) The group π = Z/2Z Z/3Z with presentation a, b a 2 = b 3 = 1 is the normal closure of the element represented by ab, and so has weight 1. (It is also the quotient of the trefoil knot group by its centre.) Since the abelianization of π is finite, by the inequalities (1.1) def(π) 1, and so this group is not a counterexample to the Kervaire-Laudenbach Conjecture for groups of deficiency 1. One situation in which our strategy works is when π = Fr r Z is a semidirect product, with a finitely generated free normal subgroup and infinite cyclic quotient. It follows very easily from the multiplicativity of the L 2 -Betti numbers under passage to subgroups of finite index that 1 (π) = 0 [35]. We would like to derive this instead from properties of the Jacobian matrix. Such groups have presentations of the form P = t, x i, i r tx j t 1 = w j, j r φ. The r r block of J(P) corresponding to differentiation with respect to the variables x i has the form ti r M, where the entries of M lie in the subring Z[Fr r ] = Z x i. Every element of C 2 can be written as a finite sum γ = Σ n mt k v k with v k a vector with entries in Z[Fr r ], where we may assume that v n 0. Since (ti r M)(γ) = t n+1 v n + o(n), where the terms of o(n) have degree at most n in t, it follows that ti r M is injective. Hence J is injective and X is aspherical. Can we extend this argument to other groups of deficiency and weight 1? It would clearly suffi ce that J has an r r block of the form t n A + o(n 1), where A is injective and the entries of the summand o(n 1) all have degree < n in t. In so far as the Whitehead conjecture implies that groups π with weight 1 and deficiency 1 have finite 2-dimensional K(π, 1) complexes (fgd 2), it is of some interest to consider also this condition. For

8 8 AJB & JAH OCTOBER 1, 2018 each combination of the conditions a, b, d and w not including both d and w there are examples which have nontrivial torsion, and so cannot have finite cohomological dimension. We note just one explicitly: since the finite group I = SL(2, 5) is perfect and has deficiency 0 the free product π = Z I has abelianization Z and deficiency 1. (It has weight 2 and 1 (π) = χ virt (π) = 1 1 = 119.) I W, L 2 -B The word complex below always refers to a connected CW-complex. A complex is said to be homotopy finite if it is homotopy equivalent to a finite complex (not necessarily of the same dimension) and finitely dominated if it is a homotopy retract of a finite complex. The following theorem and its corollary strengthen Theorem 2.4 and Corollary of [25]. Theorem 3.1. Let X be a 2-complex, with π = π 1 (X). Then 2 (X) 2 (π). If also X is finitely dominated, then X is aspherical if and only if 2 (X) = 2 (π). Proof. The inequality follows from the fact that a K(π, 1) may be obtained from X by adjoining cells of dimension at least 3. Since the cofibre has dimension at least 3, the homology exact sequence gives a (weak) epimorphism θ : H(2) (2) 2 (X) H 2 (π). This gives the first assertion. In the case where X is finitely dominated, these modules are finite. When they have equal dimension, by [35] (1.13) θ is also injective. Moreover, because X has dimension 2, there are no two-dimensional integral or L 2 -boundaries in the chain complexes for X, giving an injection of H 2 ( X; (2) Z) in H 2 (X). Then the commuting diagram π 2 (X) = H 2 ( X; Z) π 2 (K(π, 1)) = 0 (2) H = θ 2 (X) H (2) 2 (π) forces π 2 (X) to vanish, making X aspherical. Finally, the only if direction is just the definition of 2 (π). Corollary 3.2. Let X be a homotopy finite 2-complex, with π = π 1 (X). If 1 (π) + χ(x) 0, then X is aspherical. Proof. Because X is homotopy equivalent to a finite complex, X say, the Euler characteristics of X and X agree, as do their L 2 -Betti numbers by [35] Theorem Moreover, since X is a finite complex

9 WHITEHEAD ASPHERICITY QUESTION October 1, we have that χ(x ) = χ (2) (X ). Therefore χ(x) = χ(x ) = χ (2) (X ) = 0 (X ) 1 (X ) + 2 (X ) 3 (X ) = 0 (π) 1 (π) + 2 (X), because i (X) = i (X ) and for i = 0, 1 i (X) = i (π), while, since X is a 2-complex, 3 (X) = 0. Combining the nonnegativity of 0 (π) and 2 (X) with the hypothesis that 1 (π) + χ(x) 0 we deduce that 2 (X) = 0. The result now follows from the theorem. Remark 3.3. When π satisfies the weak Bass conjecture (for example, if cd Q (π) 2 [16]), a weaker hypothesis on X suffi ces. For, if X is finitely dominated, then by [5] we still have χ(x) = χ (2) (X), so that again 0 (π)+ 2 (X) = 1 (π)+χ(x) 0 forces 2 (X) = 0. Hence, again, X is aspherical. Here is a purely group-theoretic version. Corollary 3.4. Let π be a nontrivial finitely presentable group. Then defπ (π) 2 (π), with equality if and only if gdπ 2. Proof. First, if π is finite, then the left-hand side is nonpositive, while the right is equal to 1 since π is amenable, whence its higher L 2 -Betti numbers vanish. In this situation π has infinite geometric dimension. Let X be the 2-complex associated to a presentation of an infinite group π of maximum deficiency. Then by the preceding theorem 1 (π) 2 (π) 1 (π) 2 (X), and X is aspherical if and only if equality holds. Since 0 (X) = 0 (π) = 0 and 1 (X) = 1 (π), and because X is a finite 2-complex, the right-hand side is equal to χ (2) (X) = χ(x) = defπ 1; whence the result. The next lemma is in part asserted in the discussion of Question 3 of [12]. Lemma 3.5. Let X be a finite 2-complex with fundamental group π, and let T be a maximal tree in the 1-skeleton. Then the following are equivalent. (a) A set E of w 2-cells may be adjoined to X to form a contractible 2-complex; (b) π has weight w and deficiency w and X/T is the 2-complex associated to a maximal deficiency presentation; (c) π has weight w, π/π = Z w and H 2 (X; Z) = 0.

10 10 AJB & JAH OCTOBER 1, 2018 Proof. (a) (b). If X E is contractible, then χ(x) = 1 w and π has weight at most w. Since 1 χ(x) def(π) and from (1.1) w wt(π) def(π), we see that w = def(π). (b) (c). If w = defπ, then χ(x) = 1 w and wt(π/π ) = rank(π/π ), so π/π = Z w. Therefore, β 2 (X) = χ(x) 1 w = 0. Since X is a 2-complex, H 2 (X; Z) is torsion-free, and so H 2 (X; Z) = 0. Hence H 2 (π; Z) = 0 also. (c) (a). If π/π = Z w and H 2 (X; Z) = 0 then χ(x) = 1 w. Hence if π also has weight w attaching w 2-cells along representatives of a normally generating set gives a 1-connected 2-complex Y = X E with χ(y ) = 1, which is therefore contractible. The next result combines the previous results of this section. Corollary 3.6. Consider the following assertions. (a) (i) Hypothesis 1.4: For all w 1, if π is a finitely presentable group of weight w and deficiency w, then 1 (π) = w 1. (ii) For all w 1, if π is a finitely presentable group of weight w and deficiency w, then gd(π) 2. (iii) For all w 1, if π is a finitely presentable group of weight w and deficiency w, then cd Q (π) 2. (b) (i) Every subcomplex Y of a finite contractible 2-complex has 2 (Y ) = 0. (ii) Every subcomplex Y of a finite contractible 2-complex is aspherical. Among these assertions, the following implications hold. (a)(i) = (a)(ii) (a)(iii) (b)(i) = (b)(ii) Proof. (a)(i) (a)(ii). This follows from (3.4), because 2 (π) (, 0] [0, ] = {0}, making π of geometric dimension 2. (a)(ii) (a)(iii). Of course, always gd(π) 2 implies cd Q (π) 2. To prove results about a subcomplex Y of a finite contractible 2- complex W, we note that, by cellular approximation, W may be formed by first attaching 1-cells, giving a finite complex X containing the 1- skeleton of W, and then passing to W by attaching 2-cells. From the homology exact sequence of the pair (X, Y ), we observe that 2 (Y ) = 2 (X). Now, by (3.5) π := π 1 (X) has its weight and deficiency equal, to w say. (a)(iii) (b)(ii). Continuing with this notation, as in (3.5), we have H 2 (X; Z) = H 2 (X; Q) = 0, and so Q Z[π] π 2 (X) H 2 (X; Q) = 0.

11 WHITEHEAD ASPHERICITY QUESTION October 1, As in Theorem 2.8 of [25] this combines with cd Q (π) 2 to imply that π 2 (X) = 0 and so X is aspherical. Since X has the homotopy type of the wedge of Y with a bouquet of circles, Y is aspherical too. (b)(i) (b)(ii). This is immediate from (3.1), since 2 (π) 0. (b)(ii) (a)(ii). Let π be a finitely presentable group of weight w and deficiency w, and choose X as the 2-complex associated to a maximal deficiency presentation. Then by (3.5) X is a subcomplex of a finite contractible 2-complex, and so, by (b)(ii), aspherical. Hence, gd(π) 2. (b)(i) (a)(i). If, in the previous argument, we also know that 2 (X) = 0, then from (3.1) it follows that 2 (π) = 0. Hence, by (3.4), we obtain that 1 (π) = def(π) 1. (a)(i) (b)(i). Given (a)(i), let Y be a subcomplex of a finite contractible 2-complex W. Then form X from Y as indicated above. By (3.5) π := π 1 (X) has its weight and deficiency equal, to w say, and so by (a)(i) equal to (π). Since gd(π) 2, π is torsion-free and therefore 0 (X) = 0 (π) = 0. Hence, 2 (Y ) = 2 (X) = χ (2) (X) + 1 (π) = χ(x) + def(π) 1 = T C This section explores the relationships between the following notions. Definitions 4.1. A 2-complex S is called subaspherical if it is a subcomplex of an aspherical 2-complex, and subcontractible if it is a subcomplex of a contractible 2-complex. For a subgroup H of G = π 1 (S), write S H for a 2-complex covering S and having fundamental group H (a regular covering space if and only if H is normal in G). Let R be a commutative ring with 1. When R = Z, it is often omitted from the terminology that follows. A space X is called R- acyclic if H (X; R) = 0, and a group N is R-superperfect if H 1 (N; R) = H 2 (N; R) = 0. From e.g. [30], we recall that a 2-complex S is called almost acyclic if H 1 (S) is torsion-free and H 2 (S) = 0, and that this is equivalent to the condition that H 2 (S; F p ) = 0 for every prime p, or equally that H 2 (S; R) = 0 for any coeffi cient ring R.

12 12 AJB & JAH OCTOBER 1, 2018 A 2-complex X with fundamental group N is R-Cockcroft if, in the commuting diagram 0 R Z π 2 (X) N R Z π 2 (X) = H 0 (N; H 2 ( X; R)) H 2 (X; R) H 2 (N; R) 0 with exact rows (the lower row coming from the Serre spectral sequence of the universal covering space fibration), the right vertical Hurewicz map (equivalently, the first lower horizontal map) is zero. For a group H, H D(R) (H is R-conservative) if for any RHhomomorphism α : P Q of RH-projectives, id α : R RH P R RH Q monic implies that α is monic. After [41], we call E(R) the class of groups G such that R regarded as a G-trivial module has an RG-projective resolution (A i, i ) for which the induced map id R RG 2 : R RG A 2 R RG A 1 is injective. Lemma 4.2. Let S be a 2-complex. (a) If S is subaspherical, then S is (Z)-Cockcroft; that is, the Hurewicz homomorphism π 2 (S) H 2 (S; Z) is trivial. (b) If the cover of S associated to a subgroup of π 1 (S) is Cockcroft, then so is S. (c) The universal cover of S is Cockcroft if and only if S is aspherical. (d) If S is a subcomplex of an aspherical 2-complex T with π 1 (S) π 1 (T ) trivial, then S is subcontractible. (e) If S is almost acyclic, then so too is any subcomplex of S. In particular, any subcontractible 2-complex is almost acyclic. Proof. Assertion (a) is due to Cockcroft. (b) holds because the Hurewicz homomorphism for S factors through that of any of its covers. For the universal cover, the Hurewicz map is an isomorphism, so its triviality makes the universal cover contractible; this yields (c). Assertion (d) follows from the factorization of S T through the contractible universal cover T of T. The claim (e), observed by Howie [26], is a consequence of the homology exact sequence of a pair of 2- complexes. In certain situations, the apparent generalization from Cockcroft to R-Cockcroft is bogus, as we now see. Lemma 4.3. For a ring R, the following are equivalent. (a) For 2-complexes, R-Cockcroft coincides with Cockcroft. (b) Z is a subring of R. Proof. (b) (a). This follows from the composition π 2 (X) π 2 (X) R H 2 (X) H 2 (X) R H 2 (X; R),

13 WHITEHEAD ASPHERICITY QUESTION October 1, since H 2 (X), as a submodule of C 2 (X), is a free Z-module. (a) (b). Let R be a ring such that the image of Z in R is the finite ring Z/m. Consider the 2-complex X associated to the usual presentation of the group Z Z/m. Thus, X has one 0-cell, and two cells each in dimensions one and two; this gives χ(x) = 1. Since H 1 (X) has torsion-free rank 1, it follows that H 2 (X) = Z. Meanwhile, H 2 (π 1 (X)) = Z/m. A chase of the diagram (4.2) π 2 (X) H 2 (X) H 2 (π 1 (X)) H 2 (X) R H 2 (X; R) (with exact row) then reveals X to be R-Cockcroft but not Cockcroft. Recall that, in contrast to the Cockcroft pattern for associated covers, the cohomological dimension of subgroups L of π 1 (X) behaves in the opposite way: if cdπ 1 (X) 2 then cdl 2. Similar facts hold with rational coeffi cients. Theorem 4.4. A 2-complex X with fundamental group G is aspherical if and only if all of the following hold: (i) cd Q G 3; (ii) H 3 (G; Q) = 0; (iii) the weak Bass conjecture holds for QG; (iv) X is Cockcroft; and (v) π 2 (X) Q is a finitely generated QG-module. Proof. Necessity of the conditions follows because X aspherical implies that G must have geometric dimension at most 2, whence by [16] the Bass conjecture holds for QG. Therefore, we prove suffi ciency. First, in [30] Theorem 3.5 it is shown that (i), (ii) and (iv) imply that Q QG π Q = 0, where π Q = π 2 (X) Q. Here is an alternative proof. From the cellular chain complex C = C ( X; Q) of X, 0 π Q C C1 C0 Q 0, we have from (i) that π Q =H 2 ( X; Q) is also projective over QG. Now, the Hurewicz homomorphism takes the form π 2 (X) = H 2 ( X) H 2 (X) with the second map induced by the augmentation map, so that its rational version factors as π Q Q QG π Q H 2 (X; Q). From the Serre homology spectral sequence for the fibration X X K(G, 1), the kernel of the latter map is the group H 3 (G; Q), which vanishes by (ii). Thus the Cockcroft hypothesis (iv) implies that Q QG π Q = 0.

14 14 AJB & JAH OCTOBER 1, 2018 In other words, the augmentation of π Q is zero. Because G satisfies the weak Bass trace conjecture, it follows that the Kaplansky trace of the projective QG-module π Q (here assumed finitely generated) equals its augmentation trace, and so vanishes. By a result of Kaplansky [2] (8.9), this forces the module π Q = H 2 ( X; Q) to be zero. Since the 2-complex X has π 2 (X) = H 2 ( X; Z) a free Z-module, it follows that π 2 (X) is also zero, whence X is aspherical. Remarks 4.5. (a) If X is finitely dominated, then a Schanuel lemma argument shows that (v) is equivalent to G being of type FP 3 over Q. (b) If X is finitely dominated and cd Q G 2, then (v) holds and G is finitely generated. (c) For G finitely generated, if hd Q G 2 (which always implies that H 3 (G; Q) = 0), then cd Q G 3, by [9]. (d) If cd Q G 2, then (i), (ii) and (by [16]) (iii) all hold. (e) If X is finite and G has type FF over Q, then, as in [13] (but arguing rationally), π 2 (X) Q is stably free, and appeal to the Bass conjecture may be avoided. By (b), (c) and (d) of the remarks above, we have the following special case, which generalizes a theorem of [22]. Corollary 4.6. A finitely dominated 2-complex X is aspherical if and only if it is Cockcroft and cd Q π 1 (X) 2. Remarks 4.7. (a) For finitely dominated 2-complexes Corollary 4.6 strengthens Corollary 3.6 of [30] which, as well as considering integral coeffi cients, further requires the existence of a transfinite subnormal series with locally F pα -indicable factors. (b) Corollary 4.6 strengthens [31] Corollary 4.6, which for finite subaspherical (hence Cockcroft) X imposes the condition that gdπ 1 (X) 2. Interestingly, [31] relates the Whitehead Conjecture to Wall s D(2) Problem (concerning whether a finite 3-complex X with cdx 2 is homotopy equivalent to a 2-complex), and to the Wiegold Conjecture that a finitely generated perfect group has a single normal generator. See also [37] for further linkage to the D(2) Problem. Both of these papers use plus-construction arguments, as we do below. (c) Homotopy finiteness conditions are needed only because there is no complete homological finiteness criterion for 2-complexes. If n 2 and X is a homotopy finite n-dimensional complex, then X is homotopy equivalent to a finite n-dimensional complex, by [42] Theorem F. The Eilenberg-Ganea Conjecture posits that the restriction n 2 is superfluous when X is aspherical. (Note that the Eilenberg-Ganea Conjecture and the Whitehead Conjecture cannot both hold [8].) We readily obtain the following further consequence of Corollary 4.6.

15 WHITEHEAD ASPHERICITY QUESTION October 1, Corollary 4.8. The Whitehead conjecture is equivalent to the conjecture that the fundamental group of a finite subcomplex of a contractible 2-complex has rational cohomological dimension at most 2. Proof. If the conjecture holds, then because the subcomplex is aspherical its fundamental group has geometric dimension at most 2. In the other, more surprising direction, suppose that the conjecture is false. Then, by the reduction of Luft [36] mentioned above, there is a finite subcomplex of a contractible 2-complex that fails to be aspherical, yet from (4.2)(a) must be Cockcroft. By Corollary 4.6, the rational cohomological dimension of its fundamental group must exceed 2. For finitely dominated complexes Corollary 4.6 answers in the affirmative Question 2 of [12], which asks whether two Cockcroft 2- complexes with isomorphic fundamental groups must be both or neither aspherical. Previously, this had been shown for finite complexes [11]. In fact, consideration of resolutions reveals the following. Lemma 4.9. If Q[G 1 ] = Q[G 2 ], then cd Q G 1 = cd Q G 2. Given a map f : X 1 X 2 with X 2 path-connected, let f : X1 X 2 be its pullback over the universal cover X 2 of X 2. We say that f is rationally simply-connected if the homotopy fibre F f has π 1 (F f) Q = 0. (Passage to the universal cover is to avoid possible problems of connectivity of the fibre.) Proposition Let f : X 1 X 2 be a rationally simply-connected map between 2-complexes. If X 1 is aspherical, then so too is X 2. Proof. First observe that when k = 1 both (i) cd Q π 1 (X k ) 2, and (ii) π 2 (X k ) Q = 0. We show that (i) and (ii) hold also when k = 2. The condition on the homotopy fibre of f implies that π 1 (f) induces an isomorphism of rational group rings. By the lemma, this yields (i) for k = 2. Concerning (ii), we have an epimorphism π 2 (X 1 ) Q π 2(f) Q π 2 (X 2 ) Q. However, its domain is zero. Thus (i) and (ii) indeed hold for X 2. Moreover, the triviality of π 2 (X 2 ) Q H 2 (X 2 ; Q) makes X 2 Cockcroft. Then Theorem 4.4 (see Remark 4.5) finishes the proof. 5. A In this section we explore how asphericity may be deduced from the combination of an acyclic cover and certain finiteness conditions. The following is well-known (e.g. [40]).

16 16 AJB & JAH OCTOBER 1, 2018 Lemma 5.1. Let S N be a regular covering space of a 2-complex S, associated to a nontrivial normal subgroup N of G = π 1 (S). Then the 2-complex S N is R-acyclic if and only if both (i) N is R-superperfect, and (ii) S N (hence S) is R-Cockcroft. When these conditions hold, cd R (G/N) 2; and if R = Z then gd(g/n) 3. Proof. The first assertion follows readily from the diagram 4.2 above. When the conditions hold, R-acyclicity of the homotopy fibre S N of the map S K(G/N, 1) implies that K(G/N, 1) has the same cohomological dimension over R as the 2-complex S. In the case R = Z, because the fibre sequence S N S K(G/N, 1) has acyclic fibre, the map S K(G/N, 1) corresponds to the plusconstruction applied to S with respect to N, and so may be obtained by attaching cells of dimension at most 3. Hence, K(G/N, 1) has the homotopy type of a 3-complex. Proposition 5.2. If an acyclic regular proper cover S N of a 2-complex S is finitely dominated, then S is aspherical. If also S has countable fundamental group, then S N is its universal cover. Proof. Suppose that S is not aspherical. Then N 1 (since if S N is both acyclic and simply-connected then it is contractible); and so, by Lemma 5.1, cd(g/n) = 1 or 2, where G = π 1 (S). Thus, G/N is nontrivial and torsion-free, so that we may choose an infinite cyclic subgroup Z, giving a group extension and associated fibration N Ḡ Z S N S Ḡ S 1. Recall M. Mather s theorem [38] that a space X is finitely dominated iff X S 1 is homotopy finite. Now, the fibration S N S 1 S Ḡ S 1 S 1 has both base and fibre homotopy finite. Hence S Ḡ S 1 is also homotopy finite (see [33] Lemma 7.2), making the 2-complex S Ḡ finitely dominated as well. Because S N is acyclic, it follows that the Euler characteristic χ(s Ḡ ) = χ(s 1 ) = 0. Also, because S N is finitely dominated, the group N must be finitely presented (by Lemma 1.3 of [42]). Then by [18] 0 (Ḡ) = 1 (Ḡ) = 0. Thus, by Remark 3.3, SḠ is aspherical. Hence S is aspherical too.

17 WHITEHEAD ASPHERICITY QUESTION October 1, Thus, G has gdg 2. Assume that G is also countable. Then, since N is by assumption finitely presented and perfect, by [6] Theorem 3.3 N must be trivial or of finite index in G. However, as noted above G/N is nontrivial and torsion-free. Hence, N is trivial, making S N the universal cover. Remarks 5.3. The two ingredients for this argument are: (a) S Ḡ is finitely dominated, and (b) 1 (Ḡ) = 0. Concerning (a): If every finite subcomplex of a complex X is aspherical, then X is aspherical. For, the image of any map from a sphere is a compact subspace. Concerning (b): Let S be a 2-complex with an acyclic regular proper cover S N. If G = π 1 (S) contains a locally finite, infinite, ascendant subgroup L, then (b) holds and (a) fails. For, by [27] every element of finite order in G lies in N, hence L N Ḡ. Since L is amenable, (b) follows from [34] Theorem 3.3. If also (a) were to hold, then because S would be forced to be aspherical by the proof of the proposition, G would be of finite geometric dimension, hence torsion-free; this contradiction thwarts (a). (In fact, [3] even disallows S subaspherical when G itself is locally finite.) Here is a strengthening of the previous proposition. Proposition 5.4. Let S N be an acyclic regular proper cover of a 2- complex S. If N is finitely generated and 1 (S N ), 2 (S N ) are finite, then S is aspherical. In particular, if S N is finitely dominated then S is aspherical. Proof. fibration Proceed as in the proof of Proposition 5.2, to obtain the S N S S 1, where S has the same homotopy type as the 2-complex S Ḡ. Now, from the hypotheses on N, S N, by [35] Exercise 6.15 we have χ (2) (S Ḡ ) = χ (2) (S ) = χ (2) (S N ) χ(s 1 ) = 0 Since as before for i = 0, 1 i (S Ḡ ) = i (Ḡ) = 0, we obtain that also 2 (S Ḡ ) = 0. Again, because S Ḡ is a 2-complex, 2 (S Ḡ ) = 0 forces it to be aspherical. Hence S is aspherical too. When S N is finitely dominated, by [42] Theorem A N is finitely generated. Finite-dimensionality of the homology module H i ( S N ; l 2 (N)) (i = 1, 2) also follows from domination by a finite complex, for which the homology modules are necessarily finite-dimensional.

18 18 AJB & JAH OCTOBER 1, 2018 By [36], if there are counterexamples to the Whitehead problem, then there are such with S finite and subcontractible. Then by Lemma 5.1 a nontrivial normal subgroup N of G = π 1 (S) is superperfect if S N is acyclic; in that event, cd(g/n) 2. Proposition 5.5. Suppose that S is finitely dominated. Let 1 N G = π 1 (S) with S N Q-acyclic. Suppose also that β 2 (G/N) = β 1 (G/N) 1 and that G contains an infinite subgroup that is either (i) amenable and ascendant, or (ii) finitely generated, subnormal and of infinite index in G. Then S is aspherical. Proof. Since by Lemma 5.1 cd Q (G/N) 2, the hypothesis on G/N implies that χ(s) = χ(k(g/n, 1)) = 0, while the hypotheses on the subgroup each imply that 0 (G) = 1 (G) = 0, as in [6] Theorem 3.2. Hence, by Remark 3.3, S is aspherical. Note that we are not assuming that G/N is finitely presentable. The hypothesis on G/N is satisfied if, for instance, G/N has an infinite elementary amenable normal subgroup Ē, by Theorem 2.7 of [25]. (In this theorem it is observed that, because it has finite cohomological dimension, Ē is soluble; the almost coherence of Ē required for this theorem then follows from [20].) In particular, suppose that S N is (Z)- acyclic. Then cd(g/n) 2 implies that G/N is torsion-free. So, an Ē exists if for some ordinal α we have G (α+1) N G (α) G (α). (For then, G (α) N/N = G (α) /N G (α) is nontrivial abelian normal in G/N.) This condition fails to hold if N PG or if N = PG = G (β) (β minimal) for β either zero or a limit ordinal. We can arrange to have an infinite elementary amenable normal subgroup of G/N as follows. Corollary 5.6. With S finitely dominated, and S N an acyclic regular cover of S, suppose that G = π 1 (S) contains an amenable normal subgroup E that does not lie in N. Then S is aspherical, provided that either (i) E is elementary amenable, or (ii) N is normally finitely generated in G. Proof. The hypotheses imply that EN/N, because of its isomorphism to E/(E N), is a nontrivial, hence infinite (because cd(g/n) 2), amenable (elementary amenable in case (i)) normal subgroup of G/N. Since E is therefore infinite, (i) of the proposition is satisfied by E. It remains to force χ(g/n) = 1 β 1 (G/N) + β 2 (G/N) = 0. (i). As noted above, by [25] Theorem 2.7 we have β 2 (G/N) = β 1 (G/N) 1, and the proposition applies. (ii). In this situation, we exploit the fact that the fibrations S N S K(G/N, 1)

19 WHITEHEAD ASPHERICITY QUESTION October 1, and S N S S 1 K(G/N, 1) S 1 correspond, because the 2-complex S N is acyclic, to plus-construction fibrations. Now, by (ii) the plus-construction applies to S S 1 by adjoining only a finite number of 2- and 3-cells to a finite complex homotopy equivalent to S S 1, resulting in a finite complex homotopy equivalent to K(G/N, 1) S 1. Thus, K(G/N, 1) is finitely dominated. Since by [16] Theorem 3.3 and [17] Theorem 2 χ(g/n) = χ (2) (G/N). Meanwhile, as in [6] Theorem 3.2 we have for all i 0 that i (G/N) = 0. Hence, we obtain the remaining condition in the proposition that χ(g/n) = 0. Observe that this strengthens Theorem 3.1 of [15], which proves (a special case of) the corollary with the condition that E be elementary amenable replaced by torsion-free abelian. In the corollary, can one omit the condition that E not lie in N; for example, what results if N is elementary amenable? 6. E Having seen the utility of acyclic covers in discussing asphericity, we now discuss the matter of their existence. Examples. 1. After [1], when S is a nonaspherical subcomplex of an aspherical 2-complex, there is a normal subgroup r(g) (called the locally indicable residual in [30]) such that S r(g) is acyclic. In [1], r(g) is the intersection of all characteristic subgroups K of G such that G/K D(Z); however, one can also take r(g) as the intersection of all normal subgroups N of G such that G/N D(Z), since this intersection is characteristic. Incidentally, by [30] Theorems 1 and 3(i), D(Z) = D(Q). 2. When S is a nonaspherical subcomplex of an aspherical 2-complex, by the main theorem (3.6) of [12] the perfect radical PG of G has S PG acyclic. 3. According to [26], any nonaspherical subcomplex of an aspherical 2-complex admits a cover whose fundamental group is nontrivial, finitely generated and perfect. Comparison of the third example with the first two examples above prompts the question: Can a regular acyclic cover (of a nonaspherical subaspherical complex) have finitely generated fundamental group? This question is addressed in (6.6) below. Lemma 6.1. Let X be a subcomplex of a 2-complex T with H 2 (T ; R) = 0. Then H 2 (X; R) = 0, and π 1 (X) E(R). Proof. The vanishing of the homology group is immediate from the fact that T is a 2-complex. For the second claim, consider the chain complex of the universal cover X of X. Again from 2-dimensionality,

20 20 AJB & JAH OCTOBER 1, 2018 this is the initial segment of a free resolution of R over π 1 (X) such that id R RG 2 is injective. Our interest in almost acyclic complexes comes from (3.5). For such complexes we can establish the existence of acyclic covers, and indeed minimal such. Proposition 6.2. Let S be an almost acyclic 2-complex. Then π 1 (S) has a normal subgroup N whose associated regular cover S N of S is acyclic, and has a minimal such N. Moreover, for minimal such N we have N = r(n). Proof. Arguing as in [23], we obtain the existence of a subgroup N minimal among normal subgroups M (the argument of [23] does not assume normality) such that for any coeffi cient ring R we have H 2 (S M ; R) = 0. From the lemma it follows that N E(R) for all R. We wish to deduce that also H 2 (S PN ; R) = 0. From the Serre spectral sequence of the cover S PN S N K(N/PN, 1), there is an exact sequence (coeffi cients in R) 0 H 3 (N/PN) R R[N/PN] H 2 (S PN ) H 2 (S N ) Combining the fact that N E(R) for all R with [41] (2.7) yields that cd R (N/PN) 2, so H 3 (N/PN) = 0. By construction of N, H 2 (S N ) = 0. Thus, R R[N/PN] H 2 (S PN ) = 0. However, because S PN is a 2-complex and PN is perfect, there is an exact sequence of R[N/PN]-modules 0 H 2 (S PN ) C 2 (S PN ) C 1 (S PN ) C 0 (S PN ) R 0. Then, since cd R (N/PN) 2, H 2 (S PN ) is a projective R[N/PN]- module. Meanwhile, as in [41] p.317 (see also [9] Theorem 8.15), N/PN D(R). Therefore, H 2 (S PN ; R) = 0, for any coeffi cient ring R. Since PN is normal in π 1 (S), by minimality we must have PN = N, so that H 1 (S N ; Z) = π 1 (S N ) ab = N ab = 0. Likewise, as in [30] (3.1), H 2 (S N ; R) = 0 forces H 2 (S r(n) ; R) = 0 too; so by minimality N = r(n). Example. Let S = X(P) be the 2-complex associated to a balanced presentation P of a non-perfect finite group G. Then H 2 (S) = 0, but H 1 (S) is not torsion-free. According to [19] Theorem 4.1, there is no proper subgroup N of S for which H 2 (S N ) = 0. Thus, in the result above, the condition that H 1 (S) is torsion-free cannot be dropped. Howie [26] has shown that subcontractible S must be aspherical if π 1 (S) is locally indicable. The proposition above allows an apparent weakening of this condition. Corollary 6.3. An almost acyclic 2-complex is aspherical if and only if it has an almost acyclic cover whose fundamental group is locally indicable.

21 WHITEHEAD ASPHERICITY QUESTION October 1, Proof. If the complex S is aspherical then its universal cover is contractible and so satisfies the conditions. In the other direction, according to the proposition, the fundamental group M of the cover contains a subgroup N whose associated cover S N is acyclic, and r(n) = N. Since M is locally indicable, N must be too, forcing r(n) = 1. Therefore, S N is both simply-connected and acyclic; that is, the universal cover of S is contractible. Lemma 6.4. Suppose that S is a subcomplex of a finite contractible 2-complex Y, where S and Y have a single, common, 0-cell and Y is formed from S by the adjunction of r 1-cells and s 2-cells. Then and r s. Proof. The geometry yields χ(s) = 1 + r s, χ(s) + s r = χ(y ) = 1, while the vanishing of π 1 (Y ) forces r s. This is because, since S has a single 0-cell, the adjunction of r 1-cells gives a complex S ( r S 1 ) with fundamental group G Fr r. If G is nontrivial, then for the adjunction of s 2-cells to kill this fundamental group, the group G Fr r must have weight at most s. Since the abelianization of G Fr r contains a free abelian summand of rank r, this requires r s. As a consequence, we have the following generalization of [10], which deals with the case where S is acyclic. Proposition 6.5. If S is a subcomplex of a finite contractible 2-complex such that χ(s) = 1, and the generalized Kervaire-Laudenbach conjecture holds for G = π 1 (S), then S is contractible. Proof. By the lemma, we are in the situation where r = s and the group G Fr r has weight r. Now, the generalized Kervaire-Laudenbach conjecture asserts that the only group with wt(g Fr r ) = r is the trivial group. Therefore, in applying (6.2) we have N = G = 1 and S = S N is thus both acyclic and simply connected, hence contractible. Theorem 6.6. Let S be a subcomplex of a finite contractible 2-complex, such that some acyclic regular proper cover S N of S has finitely generated fundamental group N. Then S is aspherical, or else π 1 (S) is a counterexample to the generalized Kervaire-Laudenbach conjecture. Proof. Writing G = π 1 (S), we first dispose of the case where N is finite. Here, N is Q-acyclic. So, by (5.1), S is Cockcroft, and cd Q (G/N) 2, while cd Q (N) = 0. It follows from standard Lyndon-Hochschild-Serre spectral sequence arguments that cd Q (G) = cd Q (G/N). Thus, by (4.6), S is aspherical. (This shows that gd(g) 2 and so G is torsion-free, making N trivial.)

22 22 AJB & JAH OCTOBER 1, 2018 We next suppose N to be infinite and χ(s) 0. Since S N is acyclic, by (5.1) S is Cockcroft, and cd(g/n) 2; so that N must have infinite index in G. Then, since N is finitely generated, by [35] Theorem 3.3, 0 (G) = 1 (G) = 0. Therefore, χ(s) = χ (2) (S) = 2 (G) 0. Hence χ(s) = 0, and by (3.2) (or [24] Theorem 1) S is aspherical. Now the remaining case is where N is infinite and, by the lemma, χ(s) = 1. According to the proposition immediately above, G must then be a counterexample to the generalized Kervaire-Laudenbach conjecture. R [1] J. F. Adams: A new proof of a theorem of W. H. Cockcroft, J. London Math. Soc. 30 (1955), [2] H. Bass: Euler characteristics and characters of discrete groups, Invent. Math. 35 (1976), [3] W. H. Beckmann: Completely Aspherical 2-Complexes, PhD dissertation, Cornell Univ. (1980). [4] A. J. Berrick: An Approach to Algebraic K-Theory, Pitman Research Notes in Math 56 (London, 1982). [5] A. J. Berrick, I. Chatterji & G. Mislin: Homotopy idempotents on manifolds and Bass conjectures, Proc. Conf. in Honor of G. Nishida, Kinosaki 2003, Geometry & Topology Monographs 10 (2007), [6] A. J. Berrick & J. A. Hillman: Perfect and acyclic subgroups of finitely presentable groups, J. London Math. Soc. 68 (2003), [7] A. J. Berrick & J. A. Hillman: The Whitehead Conjecture and L 2 -Betti numbers, L Enseignement Math. (2) 54 (2008), [8] M. Bestvina & N. Brady: Morse theory and finiteness properties of groups, Invent. Math. 129 (1997), [9] R. Bieri: Homological Dimension of Discrete Groups, Queen Mary College Math Notes (London, 1976). [10] N. Brady, I. J. Leary and B. E. A. Nucinkis: On algebraic and geometric dimensions for groups with torsion, J. London Math. Soc. (2) 64 (2001), [11] J. Brandenburg & M. Dyer: On J.H.C. Whitehead s aspherical question I, Comment. Math. Helv. 56 (1981), [12] J. Brandenburg, M. Dyer & R. Strebel: On J.H.C. Whitehead s aspherical question II, Low-dimensional Topology, Contemp. Math. 20, Amer. Math. Soc. (Providence RI, 1983), [13] J. M. Cohen: Aspherical 2-complexes, J. Pure Appl. Algebra 12 (1978), [14] W. Dicks and P. Linnell: L 2 -Betti numbers of one-relator groups, Math. Ann. 337 (2007), [15] M. N. Dyer: Localization of group rings and applications to 2-complexes, Comment. Math. Helv. 62 (1987), [16] B. Eckmann: Cyclic homology of groups and the Bass conjecture, Comment. Math. Helv. 61 (1986),