Journal of Lie Theory Volume 8 (1998) 219–228 ◯^{C}1998 Heldermann Verlag

Idempotents in complex group rings: theorems of Zalesskii and Bass revisited

Marc Burger and Alain Valette

Communicated by K. H. Hofmann

Abstract. Let Γ be a group, and let ℂΓ be the group ring of Γ over ℂ. We first give a simplified and self-contained proof of Zalesskii’s theorem [23] that the canonical trace on ℂΓ takes rational values on idempotents. Next, we contribute to the conjecture of idempotents by proving the following result: for a group Γ, denote by P_{Γ} the set of primes p such that Γ embeds in a finite extension of a pro-p-group; if Γ is torsion-free and P_{Γ} is infinite, then the only idempotents in ℂΓ are 0 and 1. This implies Bass’ theorem [2] asserting that the conjecture of idempotents holds for torsion-free subgroups of GL_{n}(ℂ).

1. Introduction

For a group Γ and a field F, we denote by F Γ the group ring over F; evaluation at the identity 1 ∈ Γ defines the canonical trace on F Γ:

τ_{Γ} : F Γ → F : a ↦ a(1)

(a ∈ F Γ; most often we shall write τ for τ_{Γ}). In this paper we shall deal mainly, but not exclusively, with the case F = ℂ, the field of complex numbers. In that case, we shall also consider the reduced C*-algebra C_{r}^{∗}Γ of Γ, i.e. the norm closure of ℂΓ acting by left convolution on the Hilbert space ℓ^{2}(Γ). The canonical trace on ℂΓ extends to C_{r}^{∗}Γ by the formula

τ(T) = ⟨T (δ_{1})|δ_{1}⟩  (1)

(T ∈ C_{r}^{∗}Γ; here δ_{1} denotes the characteristic function of {1}). For a unital algebra A over a field F, denote by K_{0}(A) the Grothendieck group of projective, finite type modules over A; if A is endowed with a trace Tr : A → F, then Tr defines a homomorphism Tr_{∗} : K_{0}(A) → F. The starting point of this paper was the following conjecture, due to Baum and Connes [3].

ISSN 0949–5932 / $2.50 ◯^{C} Heldermann Verlag Conjecture 1.1. For any group Γ, the range of τ_{∗} : K_{0}(C_{r}^{∗}Γ) → ℂ is the subgroup of ℚ generated by the \frac{1}{|H|} ’s, where H runs over finite subgroups of Γ.

Since τ_{∗}(K_{0}(ℂΓ)) clearly contains this subgroup of ℚ, we see that conjecturally τ_{∗}(K_{0}(CΓ)) should coincide with this subgroup. The main evidence for this conjecture is:

Corollary 1.2. τ_{∗}(K_{0}(ℂΓ)) is a subgroup of ℚ, for any group Γ.

This is an easy consequence of the following nice result of Zalesskii [23] (see also [17], Theorem 3.5 in Chapter 2), for which we present a simplified and self-contained proof in section 3.

Theorem 1.3. If e ∈ ℂΓ is an idempotent, then τ(e) is a rational number.

Note that conjecture 1.1 implies the following conjecture of Farkas ([6], # 17): if e ∈ ℂΓ is an idempotent, and if some prime number p divides the denominator of τ(e) but not its numerator, then Γ should contain an element of order p.

Assume now that Γ is a torsion-free group. Then Conjecture 1.1 says that τ_{∗}(K_{0}(C_{r}^{∗}Γ)) = ℤ. By a standard argument involving positivity and faithfulness of τ on C_{r}^{∗}Γ, which for completeness we recall in section 2, this implies the Kaplansky-Kadison conjecture on idempotents (see [21] for a survey):

Conjecture 1.4. If Γ is a torsion-free group, then C_{r}^{∗}Γ has no idempotent except 0 and 1.

In particular, there should not be any nontrivial idempotent in ℂΓ when Γ is torsion-free. Denote by BΓ the classifying space of Γ, and by RK_{0}(BΓ) its even K-homology with compact support. In [3], Baum and Connes define an index map (or analytical assembly map)

μ_{0}^{Γ} : RK_{0}(BΓ) → K_{0}(C_{r}^{∗}Γ)

which they conjecture to be an isomorphism when Γ is torsion-free. In this case, Conjectures 1 and 2 are known to follow from the surjectivity (^{1}) of μ_{0}^{Γ}. At this juncture, we mention that this conjecture of Baum and Connes was recently proved by Higson and Kasparov ([10]; see also [20]) for torsion-free amenable groups; in particular, for such an amenable torsion-free group Γ, the group ring ℂΓ has no non-trivial idempotent: there is no algebraic proof of this result.

Our contribution to the conjecture of idempotents is the following:

Theorem 1.5. For a group Γ, denote by P_{Γ} the set of prime numbers p such that Γ embeds in a finite extension of a pro-p-group. If Γ is torsion-free and P_{Γ} is infinite, then there is no non-trivial idempotent in ℂΓ.

We shall see that Theorem 1.5 implies the following result of Bass ([2], Corollary 9.3 and Theorem 9.6):

1On the other hand, the injectivity of μ_{0}^{Γ} implies deep results in topology, e.g. the Novikov conjecture on homotopy invariance of higher signatures for manifolds with fundamental group Γ.

Corollary 1.6. If Γ is torsion-free and linear in characteristic 0, then ℂΓ has no non-trivial idempotent.

Actually Bass proves this for torsion-free linear groups in any characteristic, but our proof only works in characteristic 0.

2. Kaplansky’s theorem

Kaplansky’s theorem (see [12]) is the ancestor of all results on values of the trace on idempotents in group algebras. Existing proofs involve embedding ℂΓ in a suitable completion (see e.g. [16]). For completeness, we shall give a proof, by embedding ℂΓ in the von Neumann algebra vN(Γ), i.e. the commutant of the right regular representation of Γ on ℓ^{2}(Γ) (^{2}).

Theorem 2.1. 1. Let e be an idempotent in vN(Γ). Then 0 ≤ τ(e) ≤ 1, with equality if and only if e is a trivial idempotent.

2. If e is an idempotent in ℂΓ, then τ(e) belongs to the field \overline{ℚ} of algebraic numbers.

Proof. 1. The trace τ on vN(Γ) enjoys the following properties:

• positivity: τ(T^{∗}T) ≥ 0 for T ∈ C_{r}^{∗}Γ;

• faithfulness: τ(T^{∗}T) = 0 if and only if T = 0.

Fix an idempotent e ∈ vN(Γ). Then the element z = 1 + (e^{∗} − e)^{∗}(e^{∗} − e) is self-adjoint and invertible in vN(Γ). Set f = ee^{∗}z^{−1}. Using the fact that z commutes with e, one sees that f = f^{∗}. From ee^{∗}z = (ee^{∗})^{2}, one deduces f = f^{2}; from ez = ee^{∗}e, one deduces fe = e; clearly ef = f. So f is a self-adjoint idempotent and τ(f) = τ(e). Since τ(f) = τ(f^{∗}f) and τ(1 − f) = τ((1 − f)^{∗}(1 − f)), it follows from 1 = τ(f) + τ(1 − f) and positivity of τ that 0 ≤ τ(e) ≤ 1. If τ(e) = 0, then by faithfulness f = 0, hence e = 0; replacing e by 1 − e, one gets the other case of equality.

2. The group of all automorphisms of ℂ acts on ℂΓ. If e = e^{2} ∈ ℂΓ, then τ(σ(e)) = σ(τ(e)) for every σ ∈ Aut ℂ, so that 0 ≤ σ(τ(e)) ≤ 1 by the first part of the theorem. Since Aut ℂ acts transitively on transcendental numbers, this implies τ(e) ∈ \overline{ℚ}.

Remark 2.2. In the beginning of the proof of Theorem 2.1, the argument (taken from [7], 3.2.1) really shows that, in a unital C*-algebra A, any idempotent is equivalent to a self-adjoint idempotent. What is needed is the fact that every element of A of the form 1 + x^{∗}x is invertible in A.

2The double commutant theorem shows that vN(Γ) is the weak closure of ℂΓ acting in the left regular representation; the canonical trace extends to vN(Γ) by formula (1).

Remark 2.3. The theorems of Kaplansky and Zalesskii are trivial for finite groups. Indeed, if Γ is a finite group of order n, denote by Tr the standard trace on M_{n}(ℂ), and by λ : ℂΓ → M_{n}(ℂ) the left regular representation. Then

τ(a) = \frac{Tr λ(a)}{n} (a ∈ ℂΓ).

In particular, if e is an idempotent in ℂΓ, we get

τ(e) = \frac{Rank λ(e)}{n},

a rational number between 0 and 1. A similar argument appears in lemma 1.2 of Chapter 2 of [17].

Remark 2.4. Say that a group is locally residually finite if every finitely generated subgroup is residually finite. For example, abelian groups are locally residually finite, and so are linear groups (in any characteristic!), by a theorem of Mal’cev [14] (see [1] for a recent proof). We observe that the theorems of Kaplansky and Zalesskii are basically obvious for a locally residually finite group Γ. Indeed, let e ∈ ℂΓ be a non-zero idempotent, and denote by H the subgroup generated by supp e. Since H is residually finite, we can find in H a normal subgroup N of finite index, such that N ∩ (supp e) = 1. Let π : ℂH → ℂ(H/N) be the homomorphism induced by the quotient map H → H/N. Denote by τ_{H/N} the canonical trace on ℂ(H/N), so that

τ_{H/N}(π(a)) = ∑_{n∈N} a(n) (a ∈ ℂH).

Because of the assumption on N, we have

τ(e) = τ_{H/N}(π(e));

by the case of finite groups, we deduce that τ(e) is a rational number in [0, 1].

3. Zalesskii’s theorem

We follow Zalesskii’s original strategy, i.e. we first prove a result in positive characteristic, and then lift it to characteristic 0. Thus we shall prove the following extension of Theorem 1.3:

Theorem 3.1. Let F be a field. Let e ∈ F Γ be an idempotent. Then τ(e) belongs to the prime field of F.

Proof. char F = p. This part of the proof is basically Zalesskii’s beautiful argument. Start with the remark that, if A is an algebra over F endowed with a trace Tr : A → F, then one enjoys “Frobenius under the trace”: for every x, y ∈ A:

Tr((x + y)^{p}) = Tr(x^{p}) + Tr(y^{p}).  (2)

To see it, expand (x + y)^{p} in 2^{p} monomials, and let the cyclic group of order p act by cyclic permutations on this set of monomials. The trace Tr is constant along orbits, so the traces along orbits with p elements sum up to 0; therefore only the two monomials x^{p} and y^{p} contribute to Tr((x + y)^{p}).

Write now |γ| for the order of an element γ in Γ. Define a family of traces on F Γ by

Tr_{k}(a) = ∑_{|γ|=p^{k}} a(γ) (k ∈ ℕ; a ∈ F Γ);

notice that Tr_{0} = τ. Write e = ∑_{γ∈Γ} e(γ).γ; since e = e^{p}, formula (2) yields

Tr_{k}(e) = ∑_{|γ|=p^{k}} e(γ)^{p} Tr_{k}(γ^{p}).  (3)

But, for k ≥ 1:

Tr_{k}(γ^{p}) = \{ \begin{matrix}1 & if |γ| = p^{k+1} \\ 0 & otherwise; \end{matrix}

while, for k = 0:

τ(γ^{p}) = \{ \begin{matrix}1 & if either γ = 1 or |γ| = p \\ 0 & otherwise. \end{matrix}

For k ≥ 1, we get from (3):

Tr_{k}(e) = ∑_{|γ|=p^{k+1}} e(γ)^{p} = (Tr_{k+1}(e))^{p}.

Since e has finite support, we clearly have Tr_{k}(e) = 0 for k large enough. Going backwards, we get:

Tr_{1}(e) = Tr_{2}(e) = … = 0.

For k = 0, we get from (3):

τ(e) = e(1)^{p} + ∑_{|γ|=p} e(γ)^{p} = (τ(e))^{p} + (Tr_{1}(e))^{p} = (τ(e))^{p},

so that τ(e) lies in the prime field of F.

This concludes the proof of Theorem 3.1 in positive characteristic.

We now want to lift this proof to characteristic 0.

Lemma 3.2. If e is an idempotent in ℂΓ, there exists an idempotent f in \overline{ℚ}Γ such that supp e ⊃ supp f and τ(e) = τ(f).

Proof. Set S = {st : s, t ∈ supp e} and consider the affine algebraic variety in ℂ^{S} defined by the following set of equations:

x_{γ} = ∑_{s,t∈supp e: st=γ} x_{s}x_{t}, γ ∈ S  (4)
x_{γ} = 0, γ ∈ S − supp e  (5)
x_{1} = τ(e).  (6)

This variety has to be understood as follows: suppose that x ∈ ℂΓ is defined by this set of equations inside S, and by 0 outside S. Then (4) says that x is an idempotent, (5) prescribes the support, and (6) prescribes the trace. By Kaplansky’s theorem, this variety is defined over \overline{ℚ}, and it has a point over ℂ (namely e); by the Nullstellensatz, it has points over \overline{ℚ}.

We shall need a particular case of the Frobenius density theorem [9]; see [19] for interesting historical comments on this not so well-known result.

Lemma 3.3. Let f ∈ ℤ[X] be an irreducible, monic polynomial; denote by Gal(f/ℚ) the Galois group of f over ℚ. The set of prime numbers p such that f is a product of linear factors over 𝔽_{p}, has an analytical density of \frac{1}{| Gal(f/ℚ)|}.

Proof. Let K be the splitting field of f over ℚ, denote by

ζ_{K}(s) = ∏_{℘}(1 − \frac{1}{N(℘)^{s}})^{−1} (s > 1)

the Dedekind ζ -function of K, where the product is over prime ideals ℘ in the ring of integers ℜ of K. We shall use the fact that

lim_{s→1^{+}} \frac{ln ζ_{K}(s)}{ln \frac{1}{s−1}} = 1,

which follows easily from the fact that ζ_{K}(s) has a simple pole at s = 1 (see 1(2) and 1(4) in Chapter V of [4]; note that we do not need the exact value of the residue at s = 1). But

ln ζ_{K}(s) = ∑_{℘} ∑_{k=1}^{∞} \frac{N(℘)^{−ks}}{k} = ∑_{℘} N(℘)^{−s} + ψ(s),

where ψ is a continuous function on [1, ∞[. For an ordinary prime p, denote by ℘_{1}, …, ℘_{g_{p}} the prime ideals in ℜ lying above p, so that

pℜ = (℘_{1} … ℘_{g_{p}})^{e_{p}},

all ℘_{i}’s have the same norm N(℘_{i}) = p^{f_{p}} (1 ≤ i ≤ g_{p}), and

e_{p}f_{p}g_{p} = [K : ℚ] = | Gal(f/ℚ)|

(see [18], Proposition 1 in Chapter VI). Then

∑_{℘} N(℘)^{−s} = ∑_{p} g_{p}.p^{−f_{p}s}

= | Gal(f/ℚ)| ∑_{p:f_{p}=1,e_{p}=1} p^{−s} + ∑_{p:f_{p}=1,e_{p}>1} g_{p}.p^{−s} + ∑_{p:f_{p}>1} g_{p}.p^{−f_{p}s}.

The first sum is exactly over primes p such that f is a product of linear factors over 𝔽_{p}; the second sum is over some primes which are ramified in K, so that it is a finite sum (see [4], 5(4) of Chapter III); the third sum converges at s = 1. Finally, recalling that

lim_{s→1^{+}} \begin{matrix}∑ & p & \frac{1}{p^{s}} \\ ln & \frac{1}{s−1} \end{matrix} = 1,

we get

1 = lim_{s→1^{+}} \frac{∑_{℘} N(℘)^{−s}}{∑_{p} p^{−s}} = | Gal(f/ℚ)| lim_{s→1^{+}} \frac{∑_{p:f_{p}=1,e_{p}=1} p^{−s}}{∑_{p} p^{−s}};

this concludes the proof.

Note that this proof shows that the primes p for which f is a product of linear factors over 𝔽_{p}, are responsible for the pole of ζ_{K} at s = 1. Proof of Theorem 3.1: char F = 0. Clearly we may assume that F is a subfield of ℂ. By lemma 3.2, we may assume that F is a finite algebraic extension of ℚ. Enlarging F if necessary, we may assume this extension to be Galois. Let ℜ be the ring of integers of F. For a prime ideal ℘ of ℜ not dividing denominators of coefficients of e, we may reduce modulo ℘ and get an idempotent \overline{e} ∈ (ℜ/℘)Γ. By the first part of the proof, τ(\overline{e}) is an element of the prime field of ℜ/℘; the same holds with e replaced by σ(e), for every σ ∈ Gal(F/ℚ). Write τ(e) = \frac{α}{d}, where α ∈ ℜ and d ∈ ℕ, and let f ∈ ℤ[X] be the minimal polynomial of α over ℚ. The preceding argument shows that, for all primes p but a finite number, the polynomial f splits completely into linear factors over 𝔽_{p}. By lemma 3.3, this means that f has degree 1, so that α ∈ ℤ, and τ(e) ∈ ℚ.

Remark 3.4. Compared with the original proof of Zalesskii [23], the main simplification in the above proof lies in lemma 3.2, which allows us to assume immediately, when the characteristic of F is 0, that F is a number field (a similar argument also based on the Nullstellensatz appears in [2], Corollary 8.3). In this way one bypasses the results in commutative algebra saying that the Jacobson radical of finitely generated, commutative domain is zero, and that the quotient of such a domain by a maximal ideal is a finite field. Also, lemma 3.3 makes clear that only a very modest part of the Frobenius density theorem is needed in the lifting argument from characteristic p to characteristic 0 (for cyclotomic extensions, lemma 3.3 was probably known to Dirichlet). Proof of Corollary 1.2: Let e be an idempotent in ℂΓ ⊗ M_{n}(ℂ); we have to show that (τ_{Γ} ⊗ Tr_{n})(e) is rational. Let H be a finite group which has an irreducible representation of degree n; we view ℂΓ ⊗ M_{n}(ℂ) as a subalgebra of ℂ(Γ × H). Then (τ_{Γ} ⊗ Tr_{n})(e) = \frac{|H|}{n}.τ_{Γ×H}(e) is a rational number, since τ_{Γ×H}(e) is.

4. On the conjecture of idempotents

For a group Γ, we define a set N_{Γ} of positive integers as follows:

N_{Γ} = {n ∈ ℕ − {0, 1} : there exists x ∈ Γ − {1} which is conjugate to x^{n}}.

The method of proof of the next lemma is due to Formanek [8].

Lemma 4.1. Let F be a field of positive characteristic p. Assume that Γ has no p-torsion and, for every k ≥ 1 : p^{k} ∉ N_{Γ}. Let e be an idempotent in F Γ. Then τ(e) = 0 or 1.

Proof. For x ∈ Γ − {1}, denote by C_{x} the conjugacy class of x, and define a trace Tr_{x} on F Γ by

Tr_{x}(a) = ∑_{γ∈C_{x}} a(γ) (a ∈ F Γ).

Write e = ∑_{γ∈Γ} e(γ).γ; since augmentation F Γ → F is a character, we have ∑_{γ∈Γ} e(γ) ∈ {0, 1}. Now

∑_{γ∈Γ} e(γ) = τ(e) + ∑_{[x]} Tr_{x}(e),

where the last sum is over a set of representatives for non-trivial conjugacy classes. So it is enough to show

Tr_{x}(e) = 0.

By formula (2), we have for all k ≥ 1:

Tr_{x}(e) = Tr_{x}(e^{p^{k}}) = ∑_{γ∈Γ} e(γ)^{p^{k}} Tr_{x}(γ^{p^{k}}) = ∑_{γ∈supp e;γ^{p^{k}} ∈C_{x}} e(γ)^{p^{k}}.

We notice that, for a fixed γ ∈ Γ, there is at most one k ≥ 1 such that γ^{p^{k}} ∈ C_{x}. Indeed, suppose by contradiction that γ^{p^{j}} and γ^{p^{k}} belong to C_{x}, for j < k. Then γ^{p^{j}} is conjugate to (γ^{p^{j}})^{p^{k−j}}, and since p^{k−j} ∉ N_{Γ} this implies γ^{p^{j}} = 1; since Γ has no p-torsion this means that γ = 1, which contradicts x ≠ 1.

This remark shows, by taking k large enough, that Tr_{x}(e) = 0, which concludes the proof of the lemma.

At this point we re-obtain a result of Formanek ([8], Theorem 9; see also [17], Theorem 3.9 in Chapter 2).

Proposition 4.2. Suppose that, for infinitely many primes p, one has p^{k} ∉ N_{Γ} for every k ≥ 1. Then ℂΓ has no non-trivial idempotent.

Proof. We first notice that the assumption implies that Γ is torsion-free. Indeed, if Γ admits an element x of order N ≥ 2, then for every prime p not dividing N and every integer k ≥ 1 such that p^{k} ≡ 1 (mod N), we have p^{k} ∈ N_{Γ} since x^{p^{k}} = x.

Let now e be an idempotent in ℂΓ; in view of Kaplansky’s theorem, it is enough to show that τ(e) = 0 or 1. By lemma 3.2, we may assume that e ∈ F Γ, where F is a finite algebraic extension of ℚ. Denote by ℜ the ring of integers of F. Let p be a prime as in the assumption, not dividing denominators of coefficients of e, and let ℘ be a maximal ideal of ℜ lying above p; reducing modulo ℘, we obtain an idempotent \overline{e} ∈ (ℜ/℘)Γ to which lemma 4.1 applies. So, for infinitely many ℘’s, we have τ(e) ≡ 0 or 1 (mod ℘); hence the result.

Remark 4.3. Let Γ be a torsion-free group which is hyperbolic in the sense of Gromov; it is then known that N_{Γ} is empty, so that ℂΓ has no non-trivial idempotent. Note that more is true in this case; indeed, Ji [11] showed that the Banach algebra ℓ^{1}(Γ) has no non-trivial idempotent; and Delzant [5] proved that ℂΓ has no zero divisor for many torsion-free hyperbolic groups.

Recall that, for an arbitrary group Γ, we defined a set P_{Γ} of primes by

P_{Γ} = {p : Γ embeds in a finite extension of a pro-p-group}.

Lemma 4.4. Let Γ be a non-trivial torsion-free group. If p ∈ P_{Γ} and n ∈ N_{Γ}, then p does not divide n.

Proof. Since p ∈ P_{Γ}, there exists a decreasing sequence (Γ^{(k)})_{k≥0} of finite index normal subgroups of Γ, with Γ^{(0)} = Γ, ⋂_{k=0}^{∞} Γ^{(k)} = {1} and Γ^{(1)}/Γ^{(k)} a finite pgroup. Set a_{p} = [Γ : Γ^{(1)}] and p^{b_{k}} = [Γ^{(1)} : Γ^{(k)}]. Let x ∈ Γ − {1} be conjugate to x^{n}; denote by |x|_{k} the order of the image of x in the quotient-group Γ/Γ^{(k)}. Since Γ is torsion-free, one has

lim_{k→+∞} |x|_{k} = +∞.

On the other hand, |x|_{k} divides a_{p}.p^{b_{k}}, meaning that, for k large enough, p divides |x|_{k}. Now |x|_{k} = |x^{n}|_{k}, so that n and |x|_{k} are relatively prime; in particular p does not divide n.

Proof of Theorem 1.5: Lemma 4.4 ensures that, if p ∈ P_{Γ} and k ≥ 1, then p^{k} ∉ N_{Γ}. The desired result then follows from Proposition 4.2.

Proof of Corollary 1.6: If Γ is a finitely generated subgroup of GL_{n}(ℂ), then all but a finite number of primes belong to P_{Γ}, by a result of Merzljakov [15]; see also [22], Theorem 4.7; [13], lemma 3.

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Departement Mathematik ETH Zűrich Ra̋mistrasse 101 CH-8092 Zűrich, Switzerland burger(a)math.ethz.ch Institut de Mathématiques Uni. Neuchâtel Rue Emile Argand 11 CH-2007 Neuchâtel, Switzerland alain.valette(a)maths.unine.ch

Received October 7, 1996 and in final form January 26, 1998