The strong Bass conjecture for group elements of finite order and for residually finite groups

Thomas Schick^{∗}

Last compiled October 5, 2000; last edited Oct 5, 2000 or later

Abstract

Let k be an integral domain and ring and G a discrete group. Then the Hattori-Stallings trace τ_{P} of a finitely generated projective kG-module P is a function on the conjugacy classes of elements of G with values in k.

The strong Bass conjecture asserts that τ_{P} (g) is zero if the order of g is not invertible in k (with the convention that ∞ is never invertible).

In this note, we prove the strong Bass conjecture for arbitrary integral domains k and for group elements of finite order. This is a generalization of the corresponding result of Linnell for k = ℤ and (like Linnell’s proof) is based on a calculation of Cliff.

The establish the strong Bass conjecture (for all group elements) for domains in which no prime is invertible, if G is residually finite. If G is residually a finite p-group for every prime p, then we prove the strong Bass conjecture for kG for every integral domain k.

MSC:

Throughout this note, k is a commutative ring with 1 and G is a discrete group. If g, h ∈ G we write g ∼ h if and only if g is conjugate to h in G. The order of g (i.e. the smallest positive integer n with g^{n} = 1, or infinity if no such integer exists) is denoted with o(g).

1. Definition. For any ring R, we define the trace tr: M(n, R) → R as the sum of diagonal entries of the matrix (note that in general this does not satisfy the trace property tr(AB) = tr(BA)).

For g ∈ G define evaluation the map

v_{g}: kG → G: ∑_{h∈G} λ_{h}h ↦ λ_{g}.

Let e ∈ M(n, kG) be an idempotent, i.e. e^{2} = e. The Hattori-Stallings trace τ_{e} is the class function defined by

τ_{e}(g) = ∑_{h∼g} v_{h}(tr(e)).

∗e-mail: thomas.schick@math.uni-muenster.de www: http://wwwmath.uni-muenster.de/u/lueck/schick.html Fax: ++49 -251/83 38370

If P is a finitely generated projective kG-module, choose a “complement” Q such that P ⊕ Q ≅ (kG)^{n} for some n and let e ∈ M(n, kG) be an idempotent with image P. Set τ_{P} := τ_{e}.

It is well known that τ_{P} is well defined, and actually only depends of the class which P represents in K_{0}(kG).

2. Conjecture. The statement of the strong Bass conjecture is the following: Assume k is an integral domain with 1 and G is a group. Let e ∈ M(n, kG) be an idempotent. If g ∈ G and either o(g) = ∞ or o(g) is not a unit in k, then τ_{e}(g) = 0.

There are weaker versions of this conjecture (still called the strong Bass conjecture) where k is assumed to be a subring of ℂ with k ∩ ℚ = ℤ.

Peter Linnell proved the strong Bass conjecture for k = ℤ and group elements of finite order [11, Lemma 4.1] (the group being arbitrary). This was generalized by Schafer [12] to k the ring of integers in an algebraic number field. Schafer gives a detailed account of Linnell’s proof and also observes the corollary [12, Corollary 3] that the strong Bass conjecture for kG holds if k is the ring of integers in a number field and G is residually finite, for all elements of G.

Our goal is to give generalizations of these results to arbitrary domains k.

3. Lemma. Let k → R be a ring homomorphism. We get a commutative diagram

\begin{matrix}K_{0}(kG) & ⟶_{τ} & Cl(G, k) \\ ↓ & & ↓ \\ K_{0}(RG) & ⟶_{τ} & Cl(G, k). \end{matrix}  (4)

where Cl(G, k) denotes the class functions on G with values in k.

Proof. This follows immediately from the definition of τ.

5. Remark. Observe that it is by no means true that a group homomorphism ϕ: G → G′ induces a commutative diagram similar to (4). The reason is that two non-conjugate elements of G may very well be mapped to conjugate elements of G′. Instead, if e ∈ M(n, kG) is an idempotent, and ϕ(e) ∈ m(n, kG′) its image, then

τ_{ϕ(e)}(g) = ∑_{h∼g} τ_{e}(h),

where the sum is over the G′-conjugacy classes of g, but we pick only one representative for each G-conjugacy class.

Note, however, that this formula implies that the strong Bass conjecture for kG is true if and only if it is true for all finitely generated subgroups of G (since M(n, kG) = ⋃_{G′≤G} M(n, kG′), where the union is over all finitely generated subgroups G′ of G).

Assume e ∈ M(n, kG) is an idempotent. Then there is a finitely generated subring k′ of k such that e is contained in the subring M(n, k′G). Because of (4) the strong Bass conjecture is true for e considered as element of M(n, kG) if and only if it is true for e considered as element of M(n, k′G). Therefore, we will study finitely generated rings first. This idea is inspired by the treatment of the strong Bass conjecture by Farrell and Linnell in [6].

6. Lemma. Let k be a commutative ring in which every prime number is invertible, i.e. which contains ℚ. Then k is not finitely generated.

Proof. Let m be a maximal ideal of k. If ℚ ⊂ k, then ℚ ⊂ k/m, which is a field. By [1, Ex. 6, p. 84] k/m is not finitely generated. Therefore the same is true for k.

7. Lemma. Let k be a commutative ring of characteristic zero, i.e. which contains ℤ. If all but finitely many prime numbers are invertible in k, then k is not finitely generated.

Proof. Let p_{1}, …, p_{n} be the primes which are not invertible in k. Set S := {p_{1}^{n_{1}} ⋯ p_{n}^{n_{n}} | n_{j} ≥ 0}. This is a multiplicative subset of k. Set k′ := S^{−1}k. Then k′ contains ℚ and is therefore not finitely generated. However, k′ is a finitely generated k-algebra (generated by 1/p_{1}, …, 1/p_{n}). It follows that k is not finitely generated, either.

8. Lemma. Let k be a finitely generated domain. Assume the prime p is not invertible in k. Given 0 ≠ x ∈ k, there is a surjective ring homomorphism ϕ : k ↠ R such that ϕ(x) ≠ 0 and the characteristic of R is a power of p.

Proof. Using [1, Corollary 1.5], choose a maximal ideal 𝔪 of k which contains p. Since k is finitely generated, it is Noetherian [1, Corollary 7.7]. By [1, Corollary 10.18] ⋂_{n∈ℕ} 𝔪^{n} = {0}. Choose n such that x ∉ 𝔪^{n}. Set R := k/𝔪^{n} with the canonical projection ϕ: k → R. Then ϕ(x) ≠ 0. Since p ∈ 𝔪, p^{n} ∈ 𝔪^{n}, and therefore Char(R) | p^{n}, i.e. is a power of p.

The following lemma is implicit in Cliff [4, Proof of Lemma 4], compare also [12, Lemma 1].

9. Lemma. Let p be a prime and R a ring with Char(R) = p^{n}. Assume e ∈ M(n, R[G]) is an idempotent. Then there are r_{1}, …, r_{N} ∈ R and g_{1}, …, g_{N} ∈ G such that for every k ≥ 0

τ_{e} = ∑_{j=1}^{N} r_{j}^{p^{k}}[g_{j}^{p^{k}}].

The notation means that τ_{e} has the value ∑_{g_{j}^{p^{k}} ∼x} r_{j}^{p^{k}} on the conjugacy class of

x ∈ G.

Simplifying the arguments of Linnell [11, Proof of Lemma 4.1] we conclude from this the following corollary.

10. Corollary. Let p be a prime and R a ring with Char(R) = p^{n}. Assume e ∈ M(n, RG) is an idempotent. If g ∈ G with τ_{e}(g) ≠ 0, then there is j > 0 such that g ∼ g^{p^{j}}.

Proof. Inspecting the formulas of Lemma 9 we see that for every j ≥ 0 there is i_{j} ∈ {1, …, N} such that g ∼ g_{i_{j}}^{p^{j}}, since these are the only conjugacy classes which can contribute to a non-zero value of τ_{e}(g).

Since there are only finitely many possibilities for i_{j}, we find j_{1} < j_{2} such that i_{j_{1}} = i_{j_{2}}. Renumbering if necessary, assume i_{j_{1}} = i_{j_{2}} = 1. Then g ∼ g_{1}^{p^{j_{1}}} and g ∼ g_{2}^{p^{j_{2}}} , therefore

g^{p^{j_{2}−j_{1}}} ∼ (g_{1}^{p^{j_{1}}} )^{p^{j_{2}−j_{1}}} = g_{1}^{p^{j_{1}} p^{j_{2}−j_{1}}} = g_{1}^{p^{j_{2}}} ∼ g.

Since j_{2} − j_{1} > 0, this concludes the proof.

In view of this corollary the following well known lemma (compare Schafer [12, Lemma 4]) is of interest.

11. Lemma. Assume that g ∈ G and g ∼ g^{p^{j}} for some j > 0. Then either o(g) = ∞ or p does not divide o(g).

12. Theorem. Let k be a domain with 1. Assume that e ∈ M(n, kG) is an idempotent and g ∈ G with τ_{e}(g) ≠ 0. Let P be the set of all primes which is not invertible in k. If p ∈ P then there is n_{p} > 0 such that g ∼ g^{p^{np}}.

If Char(k) = 0, there are infinitely many different primes p and numbers n_{p} > 0 such that g ∼ g^{p^{np}} (even if |P| < ∞).

In particular, if o(g) < ∞, then o(g) is invertible in k.

Proof. As above, let R be a finitely generated subring of k such that e is contained in M(n, RG). Because of Lemma 3 we have τ_{e}(g) ≠ 0 where we consider e to be an element of M(n, RG).

If p ∈ P then p is not invertible in R. However, if Char(k) = 0 and therefore Char(R) = 0 then by Lemma 7 there are infinitely many primes which are not invertible in R.

For each prime p which is not invertible in R, we find by Lemma 8 a projection ϕ_{p}: R → R_{p} such that ϕ_{p}(τ_{e}(g)) = τ_{ϕ_{p}(e)}(g) ≠ 0, where the characteristic of the ring R_{p} is a power of p. Observe that ϕ_{p}(e) ∈ M(n, R_{p}) and ϕ_{p}(τ_{e}(g)) = τ_{ϕ_{p}(e)}(g) by Lemma 3.

The statement now follows from Corollary 10 and Lemma 11.

13. Remark. Using the construction of Linnell [11, Proof of Lemma 4.1], if g ∈ G is of infinite order and τ_{e}(g) ≠ 0 for an idempotent e ∈ M(n, kG), then g is contained in a subgroup A of G which is isomorphic to the ring ℤ with all elements of P inverted, where P is defined as in Theorem 12. If Char(k) = 0 and P is finite, we can enlarge P to some infinite set of primes (depending on g). Moreover, A is contained in a finitely generated subgroup of G. Note that we can not prove that the elements of A are contained in the union of only finitely many conjugacy classes of G, because we don’t know that τ_{e}(a) ≠ 0 for every a ∈ A.

We use the following obvious lemmas.

14. Lemma. Assume p ∈ ℕ is a prime, j > 0, and g ∈ G with g ∼ g^{p^{j}}. If H is a p-group of bounded exponent, e.g. a finite p-group, and ϕ: G → H is a group homomorphism, then ϕ(g) = 1.

Proof. Observe that (by iteration) g ∼ g^{p^{nj}} for every n ≥ 0. Consequently for each n ≥ 0 there is g_{n} ∈ G with g_{n}^{p^{nj}} = g. Choose N > 0 such that h^{p^{N}} = 1 for every h ∈ H (N exists by the definition of a p-group of bounded exponent). If nj ≥ N then ϕ(g) = ϕ(g_{n})^{p^{nj}} = ((ϕ(g_{n}))^{p^{N}})^{p^{nj−N}} = 1.

15. Lemma. Assume P is a set of prime numbers, and we are given integers j_{p} > 0 for each p ∈ P. Let g ∈ G be a group element with g ∼ g^{p^{jp}} for each p ∈ P. Assume H is a torsion group of bounded exponent N (e.g. a finite group). If ϕ: G → H is a group homomorphism and p is a prime such that p | o(ϕ(g)), then p ∉ P.

Proof. As in the proof of Lemma 14, for each p ∈ P and n ≥ 0 there are g_{p,n} ∈ G such that g_{p,n}^{p^{njp}} = g. Write N = p_{1}^{n_{1}} ⋯ p_{l}^{n_{l}} with distinct primes p_{1}, …, p_{l}. For 1 ≤ i ≤ l, if p_{i} ∈ P, choose m_{i} such that m_{i}j_{i} ≥ n_{i}. Then

ϕ(g)^{N/p_{i}^{n_{i}}} = (ϕ(g_{p_{i},m_{i}})^{p_{i}^{m_{i}j_{i}}} )^{N/p_{i}^{n_{i}}} = (ϕ(g_{p_{i},m_{i}})^{p_{i}^{m_{i}j_{i}−n_{i}}} )^{N} = 1.

Since p_{i} is not a factor of N/p_{i}^{n_{i}}, p_{i} is not a factor of o(ϕ(g)).

16. Corollary. Let k be a domain with 1. Let P be the set of all primes which is not invertible in k.

The strong Bass conjecture for kG is true for all elements of finite order in G.

If G is a group such that x ∼ x^{k} for k > 1 implies that x ∈ G has finite order, then the strong Bass conjecture is true for kG.

If P contains every prime number, e.g. if k ⊂ ℂ and k ∩ ℚ = ℤ, and if G is residually finite then the strong Bass conjecture is true for kG.

More generally, assume that for every 1 ≠ g ∈ G there is a finite quotient ϕ: G → H_{g} such that ϕ(g) ≠ 1 and every prime factor of |H_{g}| ∈ P. (This is the case e.g. Char(k) = p^{k} and G is residually a finite p-group.) Then the strong Bass conjecture is true for kG.

Let G be a group which is residually a finite p-group for every prime p. Then RG satisfies the strong Bass conjecture for every commutative ring R with 1.

Proof. This follows immediately from Theorem 12 and from Lemma 15.

17. Example. If G is the fundamental group of a connected orientable threemanifold, then the strong Bass conjecture is true for kG. This follows from Corollary 16 and [13, Theorem 1] which says that x ∼ x^{p^{k}} only if x ∈ G is of finite order.

The paper [7] gives many groups which are residually a finite p-group for every prime p, e.g. free groups, free nilpotent groups, free solvable groups, free polynilpotent groups, finitely generated residually torsion-free nilpotent groups. (Observe that because of Remark 5 we can restrict our attention to finitely generated groups.)Let F be a free group generated by b_{1}, …, b_{n} and c ∈ F an element which generates its own centralizer. The following groups are finitely generated torsion-free residually nilpotent.

(1) ⟨r, b_{1}, …, b_{n} | c = r^{n}⟩ (by [3, Theorem 1]).

(2) F ∗_{c=r} A, where A is a finitely generated free abelian group and r is one of its generators (by [2, Theorem 8]).

(3) ⟨a_{1}, …, a_{k} | a_{1}^{n_{1}} ⋯ a_{l}^{n_{l}} = 1⟩, with l > 3 and n_{i} ∈ ℤ (by [2, p. 414]).

(4) Artin’s pure braid groups P_{n}, and more generally the fundamental groups of (complements of) fiber-type arrangements like the generalized pure braid groups associated to the Coxeter groups A_{l}, C_{l}, G_{2}, I_{2}(p) (by [5, Theorem 12.6]).

(5) Let G be a primitive link group (compare [10, Definition 4.21]) or a primitive one-relator group, i.e. a one relator group whose relation is primitive (compare [10, Definition 4.18] and [8]). Set G^{ω} := ⋂_{n∈ℕ} γ_{n}(G), where γ_{n}(G) are the lower central series subgroups of G. Then clearly G/G^{ω} is residually nilpotent. By [8] and [9, Theorems 1 and 2], G/G^{ω} is residually torsion-free nilpotent.

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