The spectral measure of certain elements of the complex group ring of a wreath product

Warren Dicks^{∗} Departament de Matemàtiques

Universitat Autònoma de Barcelona

08193 Bellaterra (Barcelona)

Spain Thomas Schick^{†} FB Mathematik Universität Münster Einsteinstr. 62 48159 Münster Germany

Last edited: March 22, 2001. Last compiled: March 23, 2001.

Abstract

We use elementary methods to compute the L^{2}-dimension of the eigenspaces of the Markov operator on the lamplighter group and of generalizations of this operator on other groups. In particular, we give a transparent explanation of the spectral measure of the Markov operator on the lamplighter group found by Grigorchuk-Zuk [4]. The latter result was used by Grigorchuk-Linnell-Schick-Zuk [3] to produce a counterexample to a strong version of the Atiyah conjecture about the range of L^{2}-Betti numbers.

We use our results to construct manifolds with certain L^{2}-Betti numbers (given as convergent infinite sums of rational numbers) which are not obviously rational, but we have been unable to determine whether any of them are irrational.

1 Notation and statement of main result

In this section we introduce notation that will be fixed throughout and will be used in the statement of the main result.

Let U denote a discrete group with torsion. Let e be a nontrivial projection (so e = e^{∗} = e^{2}, e ≠ 0, 1) in ℂ[U]. For

example, U could be finite and nontrivial, and e could be the ‘average’ of the elements of U,

avg(U) := \frac{1}{|U|} ∑_{u∈U} u.

∗e-mail: dicks@mat.uab.es www: http://mat.uab.es/dicks/ Fax: ++34 -93/581 2790 Research funded by the DGI (Spain) through Grant BFM2000-0354.

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

This will be the example we shall make the most use of.

Let W = W (U, e) denote the inverse of the coefficient of 1 in the expression of e as a ℂ-linear combination of elements of U. By results of Kaplansky and Zaleskii, W is a rational number greater than 1. For example, if U is finite and nontrivial, and e = avg(U), then W = |U|.

For integers m, n, with 1 ≤ m ≤ n − 1, let λ_{m,n} := 2 cos(\frac{m}{n}π).

For any integer n ≥ 2, let M_{n} := {λ_{m,n} | 1 ≤ m ≤ n − 1, m coprime to n}. We write

U ≀ ℤ := (⊕_{i∈ℤ}U) ⋊ C_{∞},

where C_{∞} denotes an infinite cyclic group with generator t = t_{U} which acts on ⊕_{i∈ℤ}U by the shift, i.e. t^{−1}((g_{n})_{n∈ℤ})t = (g_{n−1})_{n∈ℤ}. For each u ∈ U, let a_{u} denote (…, 1, u, 1, …) ∈ ⊕_{i∈ℤ}U where u occurs with index 0. Throughout, we identify u with a_{u}. Thus U is a subgroup of U ≀ℤ. Notice that U ≀ℤ is generated by t and U.

Set

T = T (U, e) := (et + t^{−1}e) ∈ ℂ[U ≀ ℤ].

If U is finite and nontrivial, and e = avg(U), then T is two times the Markov operator of U ≀ ℤ with respect to the symmetric set of generators {ut, (ut)^{−1} | u ∈ U}.

Let N (U ≀ℤ) denote the (von Neumann) algebra of bounded linear operators on the Hilbert space l^{2}(U ≀ ℤ) which commute with right multiplication by elements of U ≀ ℤ. We identify each element x of ℂ[U ≀ ℤ] with an element of l^{2}(U ≀ ℤ) in the natural way, and also with the element of N (U ≀ ℤ) given by left multiplication by x. Thus ℂ[U ≀ ℤ] is viewed as a subset of l^{2}(U ≀ ℤ) and as a subalgebra of N (U ≀ ℤ). For a ∈ N (U ≀ ℤ) the (regularized) trace of a is defined as

tr_{U≀ℤ}(a) := ⟨a(1), 1⟩_{l^{2}(U≀ℤ)}.

Similar notation applies for any group.

Note that, if a ∈ N (U ≀ ℤ) leaves invariant l^{2}(G) for a subgroup G, then we can consider a to be an element of N (G), and here tr_{G}(a) and tr_{U≀ℤ}(a) coincide.

Note also that, if a lies in ℂ[U ≀ ℤ], then tr_{U≀ℤ}(a) is the coefficient of 1 in the expression of a as a ℂ-linear combination of elements of U ≀ ℤ.

The element (left multiplication by) T of N (U ≀ ℤ) is self-adjoint. For each μ ∈ ℝ, let pr_{μ}: l^{2}(U ≀ ℤ) → l^{2}(U ≀ ℤ) denote the orthogonal projection onto ker(T − μ), so pr_{μ} ∈ N (U ≀ ℤ). The number

dim_{U≀ℤ} ker(T − μ) := ⟨pr_{μ}(1), 1⟩_{l^{2}(U≀ℤ)} = tr_{U≀ℤ}(pr_{μ})

is called the L^{2}-multiplicity of μ as an eigenvalue of T.

Our main result is the following.

1.1 Theorem. With all the above notation, for any μ ∈ ℝ,

dim_{U≀ℤ} ker(T − μ) = \begin{cases}\frac{(W −1)^{2}}{W^{n}−1} & if n ≥ 2 and μ ∈ M_{n}, \\ 0 & if μ ∉ ⋃_{n≥2} M_{n}. \end{cases}

Moreover, l^{2}(U ≀ ℤ) is the Hilbert sum of the eigenspaces of T, i.e. the spectral measure of T off its eigenspaces is zero.

In [4, Corollary 3], Grigorchuk-Zuk proved the case of this result in which U is (cyclic) of order two and e = avg(U), so W = 2. This was used in [3] to give a counterexample to a strong version of the Atiyah conjecture about the range of L^{2}-Betti numbers. The argument in [4] is based on automata and actions on binary trees, while our proof is based on calculating traces of projections in the group ring ℂ[U ≀ ℤ].

2 Preliminary matrix calculations

In this section, we introduce more notation which will be used throughout, and verify some identities which will be used in the proof.

For positive integers i, j, let

α_{i,j} := δ_{|i−j|,1} = \begin{cases}1 & if i − j = ±1, \\ 0 & otherwise. \end{cases}

For each integer n ≥ 2, let A_{n} denote the n − 1 × n − 1 matrix

A_{n} = (α_{i,j})_{1≤i,j≤n−1} = (\begin{matrix}0 & 1 & 0 & . . . . . . . . . . . \\ 1 & 0 & 1 & 0 & . . . . . . . . \\ 0 & 1 & 0 & 1 & 0 & …. \\ . & . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \\ . . . . . . . . & 0 & 1 & 0 & 1 \\ . . . . . . . . . . . & 0 & 1 & 0 \end{matrix}) .

Recall that λ_{m,n} denotes 2 cos(\frac{m}{n}π).

2.1 Lemma. For each n ≥ 2, the family of eigenvalues of A_{n}, with multiplicities, is {λ_{m,n} | 1 ≤ m ≤ n − 1}.

Proof. For a complex number μ different from 0, 1, −1, one checks immediately by induction on n, and determinant expansion of the first row, that

det(A_{n} + (μ + μ^{−1})I_{n−1}) = \frac{μ^{n} − μ^{−n}}{μ − μ^{−1}} .

Now, for 1 ≤ m ≤ n − 1, taking μ = −e^{\frac{m}{2n} 2πi} shows that λ_{m,n} is an eigenvalue of A_{n}. Since we have n−1 distinct eigenvalues for A_{n}, they all have multiplicity one.

For n ≥ 2, A_{n} is a real symmetric matrix, so there exists a real orthogonal matrix B_{n} = (β_{i,j}^{(n)})_{1≤i,j≤n−1} such that B_{n}A_{n}B_{n}^{∗} is a diagonal matrix D_{n}; here the diagonal entries are λ_{m,n}, 1 ≤ m ≤ n − 1, and we may assume the entries occur in this order, so D_{n} = (δ_{i,j}λ_{j,n})_{1≤i,j≤n−1}. Since B_{n}B_{n}^{∗} = I_{n−1} and B_{n}A_{n} = D_{n}B_{n} we have the identities

∑_{j=1}^{n−1} β_{i,j}^{(n)}β_{k,j}^{(n)} = δ_{i,k}, 1 ≤ i, k ≤ n − 1,  (2.2)

∑_{j=1}^{n−1} β_{i,j}^{(n)}α_{j,k} = λ_{i,n}β_{i,k}^{(n)}, 1 ≤ i, k ≤ n − 1.  (2.3)

3 Proof of the main result

We shall frequently use the following, which is well known and easy to prove.

3.1 Lemma. Let G and H be discrete groups, and let p ∈ N (G) and q ∈ N (H). Embed G and H in the canonical way into G × H, so p and q become elements of N (G × H). Then

tr_{G×H}(pq) = tr_{G}(p) · tr_{H}(q).

We need even more notation.

For each i ∈ ℤ, we define, in ℂ[U ≀ ℤ], e_{i} := t^{−i}et^{i} and f_{i} := 1 − e_{i}.

It is easy to see that all the e_{i}, f_{j} are projections which commute with each other; moreover,

tr_{U≀ℤ}(e_{i}) = tr_{U≀ℤ}(e) = \frac{1}{W} and tr_{U≀ℤ}(f_{i}) = 1 − \frac{1}{W} .  (3.2)

For n ≥ 2, let q_{n} := f_{1}e_{2}e_{3} ⋯ e_{n−2}e_{n−1}f_{n}. It is clear that q_{n} is a projection. Moreover, the factors lie in ℂ[t^{−i}Ut^{i}], 1 ≤ i ≤ n, so, by Lemma 3.1,

tr_{U≀ℤ}(q_{n}) = tr_{U≀ℤ}(f_{1}) tr_{U≀ℤ}(e_{2}) ⋯ tr_{U≀ℤ}(e_{n−1}) tr_{U≀ℤ}(f_{n}).

By (3.2),

tr_{U≀ℤ}(q_{n}) = (1 − \frac{1}{W} )^{2}(\frac{1}{W} )^{n−2} = \frac{(W − 1)^{2}}{W^{n}} .  (3.3)

3.4 Lemma. If 1 ≤ m < n and 1 ≤ m′ < n′ then

q_{n′}t^{−m′}t^{m}q_{n} = δ_{n,n′}δ_{m,m′}q_{n}.

Proof. Note that t^{m}q_{n}t^{−m} = f_{1−m}e_{2−m} ⋯ e_{n−m−1}f_{n−m}, and this is a projection. Thus

(t^{m}q_{n}t^{−m} | n ≥ 2, 1 ≤ m < n) = (f_{−i}e_{−i+1} ⋯ e_{j−1}f_{j} | −i ≤ 0, 1 ≤ j).

This is a family of pairwise orthogonal projections, since, if −i, −i′ ≤ 0, 1 ≤ j, j′, then either (i, j) = (i′, j′), or the product of f_{−i}e_{−i+1} ⋯ e_{j−1}f_{j} and f_{−i′}e_{−i′+1} ⋯ e_{j′−1}f_{j′} is zero since it contains a factor e_{α}f_{α} = 0 for at least one α ∈ {−i, −i′, j, j′}. Since t is invertible, the result follows.

Notice that, for 1 ≤ m < n,

T (t^{m}q_{n}) = ett^{m}q_{n} + t^{−1}et^{m}q_{n}
= t^{m+1}e_{m+1}q_{n} + t^{m−1}e_{m}q_{n}
= t^{m+1}(1 − δ_{m,n−1})q_{n} + t^{m−1}(1 − δ_{m,1})q_{n}.

Hence

T (t^{m}q_{n}) = ∑_{i=1}^{n−1} α_{m,i}t^{i}q_{n}.  (3.5)

For 1 ≤ m ≤ n − 1, define r_{m,n} := ∑_{i=1}^{n−1} β_{m,i}^{(n)}t^{i}q_{n} and p_{m,n} := r_{m,n}r_{m,n}^{∗}. Observe that, if we identify the ith standard basis vector with t^{i}q_{n}, 1 ≤ i ≤ n−1, then r_{m,n} is an eigenvector of A_{n} with eigenvalue λ_{m,n}. Moreover, we have just checked that T acts like A_{n} on the span of the t^{m}q_{n}. This partially explains why the r_{m,n} give rise to pairwise orthogonal projections with image contained in the eigenspace of T for the eigenvalue λ_{m,n}, which is essentially the statement of the following lemma.

3.6 Lemma. (p_{m,n} | n ≥ 2, 1 ≤ m ≤ n − 1) is a family of pairwise orthogonal projections in ℂ[U ≀ ℤ] which is complete, that is, ∑_{n≥2} ∑_{m=1}^{n−1} tr_{U≀ℤ}(p_{m,n}) = 1. Moreover, if 1 ≤ m ≤ n − 1, then T (p_{m,n}) = λ_{m,n}p_{m,n}.

Proof. Let 1 ≤ m ≤ n − 1 and 1 ≤ m′ ≤ n′ − 1.

Here

r_{m,n}^{∗} = q_{n}^{∗} ∑_{i=1}^{n−1}(t^{i})^{∗}β_{m,i}^{(n)∗} = q_{n} ∑_{i=1}^{n−1} t^{−i}β_{m,i}^{(n)}.

Thus

r_{m′,n′}^{∗}r_{m,n} = q_{n′} ∑_{j=1}^{n′−1} t^{−j}β_{m′,j}^{(n′)} ∑_{i=1}^{n−1} β_{m,i}^{(n)}t^{i}q_{n}

= δ_{n,n′}q_{n} ∑_{i=1}^{n−1} β_{m′,i}^{(n)}β_{m,i}^{(n)} by Lemma 3.4

= δ_{n,n′}q_{n}δ_{m,m′} by (2.2).

It follows that the p_{m,n} are pairwise orthogonal.

Moreover,

tr_{U≀ℤ}(p_{m,n}) = tr_{U≀ℤ}(r_{m,n}r_{m,n}^{∗}) = tr_{U≀ℤ}(r_{m,n}^{∗}r_{m,n})

= tr_{U≀ℤ}(q_{n}) = \frac{(W − 1)^{2}}{W^{n}} by (3.3).

Now,

\begin{aligned}∑_{n≥2} ∑_{m=1}^{n−1} tr_{U≀ℤ}(p_{m,n}) & = ∑_{n≥2} ∑_{m=1}^{n−1} \frac{(W − 1)^{2}}{W^{n}} = ∑_{n≥2}(n − 1)\frac{(W − 1)^{2}}{W^{n}} \\ & = ∑_{n≥1} n\frac{(W − 1)^{2}}{W^{n+1}} = (1 − \frac{1}{W} )^{2} ∑_{n≥1} n(\frac{1}{W} )^{n−1} = 1, \end{aligned}

since, for |x| < 1, ∑_{n≥1} nx^{n−1} = (∑_{n≥0} x^{n})′ = (\frac{1}{1−x})′ = \frac{1}{(1−x)^{2}} .

Also,

T (r_{m,n}) = T (∑_{j=1}^{n−1} β_{m,j}^{(n)}t^{j}q_{n}) = ∑_{j=1}^{n−1} β_{m,j}^{(n)}T (t^{j}q_{n})

= ∑_{j=1}^{n−1} β_{m,j}^{(n)} ∑_{k=1}^{n−1} α_{j,k}t^{k}q_{n} by (3.5)

= ∑_{k=1}^{n−1}(∑_{j=1}^{n−1} β_{m,j}^{(n)}α_{j,k})t^{k}q_{n}

= ∑_{k=1}^{n−1} λ_{m,n}β_{m,k}^{(n)}t^{k}q_{n} by (2.3)

= λ_{m,n}r_{m,n}.

Thus T (r_{m,n}) = λ_{m,n}r_{m,n}, and, on right multiplying by r_{m,n}^{∗}, we see T (p_{m,n}) = λ_{m,n}p_{m,n}.

We have now ‘diagonalized’ T in the sense that we have decomposed l^{2}(U ≀ℤ) into the Hilbert sum of subspaces of the form p_{m,n}(l^{2}(U ≀ ℤ)) on which T acts as multiplication by the scalar λ_{m,n}.

Hence, for each μ ∈ ℝ, ker(T − μ) is the Hilbert sum of those p_{m,n}(l^{2}(U ≀ ℤ)) such that λ_{m,n} = μ. Thus either ker(T − μ) = 0 or μ = λ_{m_{0},n_{0}} for some m_{0}, n_{0} with 1 ≤ m_{0} ≤ n_{0} − 1.

We now consider the latter case. Here, for all (m, n), λ_{m,n} = μ if and only if \frac{m}{n} = \frac{m_{0}}{n_{0}} . We may assume that m_{0} and n_{0} are coprime, so μ ∈ M_{n_{0}}. Also, λ_{m,n} = μ if and only if (m, n) = (im_{0}, in_{0}) for some i ≥ 1. Thus ker(T − μ) is the Hilbert sum of the p_{im_{0},in_{0}}(l^{2}(U ≀ ℤ)) with i ≥ 1; hence

dim_{U≀ℤ}(ker(T − λ_{m_{0},n_{0}})) = ∑_{i≥1} dim_{U≀ℤ}(p_{im_{0},in_{0}}(l^{2}(U ≀ ℤ)))

= ∑_{i≥1} tr_{U≀ℤ}(p_{im_{0},in_{0}}) = ∑_{i≥1} \frac{(W − 1)^{2}}{W^{in_{0}}} = \frac{(W − 1)^{2}}{W^{n_{0}} − 1} .

Theorem 1.1 now follows.

3.7 Remarks. The hypothesis in Theorem 1.1 that U has torsion could be weakened to the assumption that ℂ[U] has a nontrivial projection; however, if U is torsion-free, it is conjectured, and known in many cases, that ℂ[U] does not contain any nontrivial projections.

It easy to show that the hypothesis in Theorem 1.1 that e is a nontrivial projection in ℂ[U] can be weakened to the assumption that e is a nontrivial projection in N (U); here, the hypothesis that U has torsion should be weakened to the assumption that U is nontrivial.

4 Direct products of wreath products

We now produce even more unusual examples by taking direct products of the groups studied so far.

4.1 Theorem. Let U and V be groups with torsion, and G = (U ≀ ℤ) × (V ≀ ℤ). Let e be a nontrivial projection in ℂ[U] and f a nontrivial projection in ℂ[V]. Let X = (tr_{U}(e))^{−1} and Y = (tr_{V} (f))^{−1}, so X > 1, Y > 1. Let T = T (U, e) ∈ ℂ[U ≀ ℤ] ⊂ ℂ[G], and S = T (V, f) ∈ ℂ[V ≀ ℤ] ⊂ ℂ[G]. Then

dim_{G}(ker(T − S))

= (X − 1)^{2}(Y − 1)^{2}( ∑_{m≥1} ∑_{n≥1} \frac{gcd(m, n)}{X^{m}Y^{n}}) − (X − 1)(Y − 1). (4.2)

Proof. By Lemma 3.6, there is a complete family (p_{m,n} | n ≥ 2, 1 ≤ m < n) of pairwise orthogonal projections in ℂ[U ≀ ℤ], such that, if 1 ≤ m < n, then T (p_{m,n}) = λ_{m,n}p_{m,n}, and, by (3.3), tr_{U≀ℤ}(p_{m,n}) = \frac{(X−1)^{2}}{X^{n}}.

Similarly, there is a complete family (q_{m,n} | n ≥ 2, 1 ≤ m < n) of pairwise orthogonal projections in ℂ[V ≀ ℤ] such that, if 1 ≤ m < n, then S(q_{m,n}) = λ_{m,n}q_{m,n}, and tr_{V ≀ℤ}(q_{m,n}) = \frac{(Y −1)^{2}}{Y^{n}}.

By Lemma 3.1, there is a complete family

(p_{m,n}q_{m′,n′} | n, n′ ≥ 2, 1 ≤ m < n, 1 ≤ m′ < n′)

of pairwise orthogonal projections in ℂ[G], such that, if 1 ≤ m < n and 1 ≤ m′ < n′ then

T (p_{m,n}q_{m′,n′}) = λ_{m,n}p_{m,n}q_{m′,n′} and S(p_{m,n}q_{m′,n′}) = λ_{m′,n′}p_{m,n}q_{m′,n′},

and

tr_{G}(p_{m,n}q_{m′,n′}) = \frac{(X − 1)^{2}}{X^{n}} \frac{(Y − 1)^{2}}{Y^{n′}} .

Thus l^{2}(G) is the Hilbert sum of the subspaces of the form p_{m,n}q_{m′,n′}(l^{2}(G)) where T − S acts as multiplication by the scalar λ_{m,n} − λ_{m′,n′}.

Hence ker(T − S) is the Hilbert sum of the p_{m,n}q_{m′,n′}(l^{2}(G)) such that λ_{m,n} = λ_{m′,n′}.

Therefore,

dim_{G}(ker(T − S)) = ∑_{n≥1} ∑_{n′≥1} b(n, n′)\frac{(X − 1)^{2}}{X^{n}} \frac{(Y − 1)^{2}}{Y^{n′}}

where b(n, n′) is the number of pairs (m, m′) such that 1 ≤ m < n, 1 ≤ m′ < n′, and \frac{m}{n} = \frac{m′}{n′}. But such pairs correspond bijectively to the fractions of the form \frac{m_{0}}{gcd(n,n′)}, 1 ≤ m_{0} < gcd(n, n′). Thus b(n, n′) = gcd(n, n′) − 1. Hence

dim_{G}(ker(T − S)) = ∑_{n≥1} ∑_{n′≥1} \frac{(gcd(n, n′) − 1)(X − 1)^{2}(Y − 1)^{2}}{X^{n}Y^{n′}}

= ∑_{n≥1} ∑_{n′≥1} \frac{gcd(n, n′)(X − 1)^{2}(Y − 1)^{2}}{X^{n}Y^{n′}} − ∑_{n≥1} ∑_{n′≥1} \frac{(X − 1)^{2}(Y − 1)^{2}}{X^{n}Y^{n′}} .

Since ∑_{n≥1} \frac{1}{X^{n}} = X^{−1} \frac{1}{1−X^{−1}} = \frac{1}{X−1}, the result follows.

4.3 Remarks. Recall that, for any positive integer n, ϕ(n) denotes the number of primitive nth roots of unity, so |M_{n}| = ϕ(n).

For X > 1, Y > 1, the double infinite sum occurring in (4.2) has an expession as a single infinite sum,

∑_{m≥1} ∑_{n≥1} \frac{gcd(m, n)}{X^{m}Y^{n}} = ∑_{k≥1} \frac{ϕ(k)}{(X^{k} − 1)(Y^{k} − 1)},

since

∑_{k≥1} \frac{ϕ(k)}{(X^{k} − 1)(Y^{k} − 1)} = ∑_{k≥1} ϕ(k) ∑_{i≥1} X^{−ik} ∑_{j≥1} Y^{−jk} = ∑_{m≥1} ∑_{n≥1} \frac{a(m, n)}{X^{m}Y^{n}}

where

a(m, n) = ∑_{{k≥1:k|m,k|n}} ϕ(k) = ∑_{k| gcd(m,n)} ϕ(k) = gcd(m, n).

It follows that

dim_{G}(ker(T − S)) = (X − 1)^{2}(Y − 1)^{2} ∑_{k≥2} \frac{ϕ(k)}{(X^{k} − 1)(Y^{k} − 1)}.

5 L^{2}-Betti numbers

We previously observed that, by results of Kaplansky and Zaleskii, the traces of projections in complex, or rational, group algebras are rational numbers in the interval [0, 1]. In order to maximize the scope of Theorem 4.1 for producing examples of L^{2}-Betti numbers, we need the following result which shows that the traces of projections in rational group algebras are precisely the rational numbers in the interval [0, 1]. We write C_{n} for a cyclic group of order n, written multiplicatively, with generator t = t_{n}.

5.1 Lemma. Let q be a rational number in the interval [0, 1]. Then there is an expression q = \frac{m}{n} where the denominator has the form n = 2^{r}s with s odd and 2^{r} ≥ s − 1, and, for any such expression, ℚ[C_{n}] contains some projection e with trace q, and ne ∈ ℤ[C_{n}].

Proof. By multiplying the numerator and denominator of q by a sufficiently high power of 2, we see that q has an expression of the desired type. Now consider any expression q = \frac{m}{n} where n = 2^{r}s with s odd and 2^{r} ≥ s − 1.

We first show, by induction on r, that, if 0 ≤ c ≤ 2^{r}, then ℚ[C_{2^{r}}] = ℚ[t | t^{2^{r}} = 1] has an ideal whose dimension over ℚ is c. Since the orthogonal complement is then an ideal of dimension 2^{r} − c over the rationals, it amounts to the same if we consider only c ≤ 2^{r−1}. For r = 0, we can take the zero ideal; thus, we may assume that r ≥ 1 and the result holds for smaller r. Now ℚ[C_{2^{r}}]

has a projection e = \frac{1+t^{2^{r−1}}}{2} ; this is avg(U) for the subgroup U of order 2 in C_{2^{r}}. As rings

eℚ[C_{2^{r}}] ≃ ℚ[C_{2^{r}}]/(1 − e) ≃ ℚ[C_{2^{r−1}}].

By the induction hypothesis, the latter has an ideal of dimension c over ℚ, and viewed in eℚ[C_{2^{r}}] this is an ideal of ℚ[C_{2^{r}}]. This completes the proof by induction. Hence, if 0 ≤ c ≤ 2^{r}, then ℚ[C_{2^{r}}] has a projection e(c) with tr_{C_{2^{r}}} (e(c)) = \frac{c}{2^{r}}.

Let f = avg(C_{s}) ∈ ℚ[C_{s}], so tr_{C_{s}}(f) = \frac{1}{s}, and tr_{C_{s}}(1 − f) = \frac{s−1}{s}.

By identifying

ℚ[C_{n}] = ℚ[C_{n}^{s} × C_{n}^{2^{r}}] = ℚ[C_{2^{r}} × C_{s}],

we see that, for 0 ≤ c ≤ 2^{r}, we have projections e(c)f and e(c)(1 − f) in ℚ[C_{n}], with traces \frac{c}{2^{r}} \frac{1}{s} = \frac{c}{n} and \frac{c}{2^{r}} \frac{s−1}{s} = \frac{c(s−1)}{n}, respectively, by Lemma 3.1.

We claim there exist integers a, b with 0 ≤ a, b ≤ 2^{r} such that a + (s − 1)b = m. We know that 0 ≤ m ≤ n = 2^{r}s. If m ≥ 2^{r}(s − 1), then m ∈ [2^{r}(s − 1), 2^{r}s], and we can take b = 2^{r} and a = m − (s − 1)b = m − 2^{r}(s − 1) ∈ [0, 2^{r}]. If m < 2^{r}(s−1), then, by the division algorithm, m = (s−1)b+a with 0 ≤ b < 2^{r}, and 0 ≤ a ≤ s − 2 < 2^{r}. This proves the claim.

Now let e = e(a)f + e(b)(1 − f), a sum of orthogonal projections. Thus, e is a projection and

tr_{C_{n}}(e) = tr_{C_{n}}(e(a)f) + tr_{C_{n}}(e(b)(1 − f)) = \frac{a}{n} + \frac{b(s − 1)}{n} = \frac{a + b(s − 1)}{n} = \frac{m}{n},

as desired.

It remains to show that e lies in \frac{1}{n}ℤ[C_{n}], but it is well known that this holds for all the idempotents of ℚ[C_{n}]. Alternatively, it is straightforward to check that all the projections involved in the foregoing proof have the right denominators.

We now obtain the following special case of Theorem 4.1.

5.2 Corollary. Let p and q be rational numbers with 0 < p, q < 1. There exist positive integers m and n, and projections

e = e^{∗} = e^{2} ∈ ℚ[C_{m}], f = f^{∗} = f^{2} ∈ ℚ[C_{n}]

with tr_{U}(e) = p, tr_{V} (f) = q. Let

G(p, q) := (C_{m} ≀ ℤ) × (C_{n} ≀ ℤ),
T := T (U, e) ∈ ℂ[U ≀ ℤ] ⊂ ℂ[G], and S := T (V, f) ∈ ℂ[V ≀ ℤ] ⊂ ℂ[G].

Let Z = Z(p, q) := mn(T − S), and let

\begin{aligned}κ & = κ(p, q) := (p^{−1} − 1)^{2}(q^{−1} − 1)^{2} ∑_{k≥2} \frac{ϕ(k)}{(p^{−k} − 1)(q^{−k} − 1)} \\ & = (p^{−1} − 1)^{2}(q^{−1} − 1)^{2} (∑_{i≥1} ∑_{j≥1} gcd(i, j)p^{i}q^{j}) − (p^{−1} − 1)(q^{−1} − 1). \end{aligned}

Then Z ∈ ℤ[G] and dim_{G}(ker Z) = κ.

5.3 Remarks. Let 0 < p, q < 1 be rational numbers. Let G = G(p, q), Z = Z(p, q) and κ = κ(p, q) as in Corollary 5.2.

By the Higman Embedding Theorem, any recursively presented group can be embedded in a finitely presented group, so G can be embedded in a finitely presented group H. (Here it is easy to find an explicit suitable finitely presented group; see, for example, [2] or [3, Lemma 3]. This explicit supergroup has the additional nice property of being metabelian, that is, 2-step solvable. Moreover, one can precisely describe its finite subgroups.)

By Corollary 5.2, Z ∈ ℤ[G] ⊆ ℤ[H] and dim_{H}(ker Z) = dim_{G}(ker Z) = κ. It is then well known how to construct a finite CW-complex or a closed

manifold M with π_{1}(M) ≃ H and with third L^{2}-Betti number κ; see, for example, [3].

Thus κ(p, q) is an L^{2}-Betti number of a closed manifold. It is conceivable that this is a counterexample to Atiyah’s conjecture [1] that L^{2}-Betti numbers of closed manifolds are rational, but we have not been able to decide whether κ(p, q) is rational or not.

5.4 Example. Consider κ(\frac{1}{2}, \frac{1}{2}) = ∑_{k≥2} \frac{ϕ(k)}{(2^{k}−1)^{2}} = 0.1659457149 …. If we sum the first 400 terms, then elementary methods show that the remaining tail is less than 10^{−201}. This allows us to calculate the first 199 terms of the continued fraction expansion of κ(\frac{1}{2}, \frac{1}{2}). One consequence we find is that if κ(\frac{1}{2}, \frac{1}{2}) is rational then both the numerator and the denominator exceed 10^{100}. It seems reasonable to assert that κ(\frac{1}{2}, \frac{1}{2}) is not obviously rational.

6 Power series

Throughout this section, let ℂ((x, y)) denote the field of (formal) Laurent series in two variables (with complex coefficients).

The expression

Φ(x, y) := ∑_{m≥1} ∑_{n≥1} gcd(m, n)x^{m}y^{n}

arising from (4.2) can be viewed as an element of ℂ((x, y)). By Remarks 5.3, if there exist rational numbers p, q in the interval (0, 1) such that (the limit of) Φ(p, q) is irrational, then there exists a counterexample to the Atiyah conjecture; so it is of interest to know whether Φ(p, q) is always rational for such rational numbers p, q. One (traditionally successful) way to show that such an expression is rational would be to show that Φ(x, y) itself is rational, that is, lies in the subfield ℚ(x, y) of rational Laurent series over the rationals. In this section, we will eliminate this possibility by showing that Φ(x, y) is transcendental over ℂ(x, y). In fact, we will show the stronger result that the specialization Φ(x, x) is transcendental over ℂ(x).

The following result is well known, but we have not found a reference. The proof is left to the reader.

6.1 Lemma. Suppose that f ∈ ℂ((x)) is algebraic over ℂ(x) of degree d. Then the subfield ℂ(x, f) is closed under the usual derivation operation, F ↦ F′ = \frac{dF}{dx}, on ℂ((x)). Moreover, ℂ(x, f) is a d-dimensional vector space over ℂ(x), so the d + 1 higher-order derivatives f^{(i)} := (\frac{d}{dx})^{i}(f), 0 ≤ i ≤ d, are ℂ(x)-linearly dependent. Hence f satisfies some non-trivial order d differential equation over ℂ(x).

We can now apply this lemma to get a transcendentality criterion.

6.2 Proposition. Suppose that a: ℕ → ℂ, n ↦ a(n), has the property that, for each N ∈ ℕ, there exist infinitely many m ∈ ℕ such that, whenever j ∈ ℤ satisfies 1 ≤ |j| ≤ N,

|a(m)| > N |a(m + j)| .

Then the power series ∑_{n≥0} a(n)x^{n} ∈ ℂ((x)) does not satisfy any non-trivial

differential equation over ℂ(x), so is transcendental over ℂ(x).

Proof. Let f := ∑_{n≥0} a(n)x^{n} ∈ ℂ((x)), and suppose that f satisfies a nontrivial differential equation over ℂ(x),

∑_{i=0}^{d} q_{i}f^{(i)} = 0  (6.3)

where q_{i} ∈ ℂ(x), not all zero. By multiplying through by a common denominator, we may assume that all the q_{i} lie in ℂ[x]. (Notice it is natural not to have a “constant term” on the right-hand side of (6.3) since it could be eliminated by iterated derivation of the equation.)

Viewing (6.3) as a collection of equations, one for each power x^{n}, we see that there exists some N ∈ ℕ, and polynomials p_{k}(t) ∈ ℂ[t] such that

∑_{k=0}^{N} p_{k}(n)a(n + k) = 0 for all n ∈ ℕ.  (6.4)

Choose k_{0}, with 0 ≤ k_{0} ≤ N, and n_{0} ∈ ℕ such that |p_{k_{0}}(n)| ≥ |p_{k}(n)| for all n ≥ n_{0}, and all k with 0 ≤ k ≤ N. In other words, p_{k_{0}} eventually dominates all the p_{k}, 0 ≤ k ≤ N.

It follows from the hypothesis on the a(n) that there exists m ∈ ℕ such that m ≥ n_{0} + k_{0}, and |a(m)| > N |a(m + j)| for all j ∈ ℤ with 1 ≤ |j| ≤ N. Now take n = m − k_{0}. Then n ≥ n_{0}, and

|a(n + k_{0})| > ∑_{k=0}^{k_{0}−1} |a(n + k)| + ∑_{k=k_{0}+1}^{N} |a(n + k)| .

Thus

|p_{k_{0}}(n)a(n + k_{0})| > ∑_{k=0}^{k_{0}−1} |p_{k}(n)a(n + k)| + ∑_{k=k_{0}+1}^{N} |p_{k}(n)a(n + k)|

≥ |(∑_{k=0}^{N} p_{k}(n)a(n + k)) − p_{k_{0}}(n)a(n + k_{0})|

= |0 − p_{k_{0}}(n)a(n + k_{0})| by (6.4).

This contradiction shows that f does not satisfy any non-trivial differential equation over ℂ(x), so, by Lemma 6.1, f is not algebraic over ℂ(x).

We now record some important results from number theory that we shall require.

6.5 Lemma. For each positive integer i, let p_{i} denote the ith prime number. There exists an integer Q_{0} such that, for all Q ≥ Q_{0}, the following hold.

(1) Q! ≤ (\frac{Q}{2})^{Q}.

(2) \frac{3}{4} ≤ \frac{p_{Q}}{Q log Q} ≤ \frac{5}{4}.

(3) ∏_{i=1}^{Q}(1 − \frac{1}{p_{i}}) ≥ \frac{1}{Q}.

Proof. In the following, f(Q) = o(g(Q)) means lim_{Q→∞} f(Q)/g(Q) = 0, and f(Q) ∼ g(Q) means lim_{Q→∞} f(Q)/g(Q) = 1.

(1) By Stirling’s formula, Q! ∼ \sqrt{2π}Q^{Q+ \frac{1}{2}} e^{−Q}, and the latter is o((\frac{Q}{2})^{Q}), since e > 2. One can argue directly that ∑_{i=1}^{Q} log i ≤ ∫_{1}^{Q+1} log x dx, so

log Q! ≤ (Q + 1) log(Q + 1) − Q,

so

Q! ≤ (Q + 1)^{Q+1}e^{−Q} = Q^{Q}(1 + \frac{1}{Q})^{Q}(Q + 1)e^{−Q} = o((\frac{Q}{2})^{Q}),

since e > 2.

(2) By the Prime Number Theorem, p_{Q} ∼ Q log Q; see [5, Theorem 8, pages 10, 367].

(3) By Mertens’ Theorem, ∏_{i=1}^{Q}(1 − \frac{1}{p_{i}}) ∼ \frac{e^{−γ}}{log p_{Q}}, where γ is Euler’s constant;

see [5, Theorem 429, page 351]. By the Prime Number Theorem, log p_{Q} ∼ log Q,

so ∏_{i=1}^{Q}(1 − \frac{1}{p_{i}}) ∼ \frac{e^{−γ}}{log Q}. Since \frac{1}{Q} = o(\frac{1}{log Q}), we see that \frac{1}{Q} = o( ∏_{i=1}^{Q}(1 − \frac{1}{p_{i}})).

The result now follows.

6.6 Theorem. Φ(x, x) = ∑_{m≥1} ∑_{n≥1} gcd(m, n)x^{m+n} and ∑_{n≥1} ∑_{d|n} \frac{ϕ(d)}{d}nx^{n} are

transcendental over ℂ(x).

Proof. For each positive integer n, let a(n) := n ∑_{d|n} \frac{ϕ(d)}{d}. Thus

a(n) = n ∑_{d|n} \frac{ϕ(d)}{d} = ∑_{d|n} \frac{n}{d}ϕ(d) = ∑_{d|n} ∑_{\{i:1≤i≤n,d|i\}} ϕ(d) = ∑_{i=1}^{n} ∑_{\{d:d|i,d|n\}} ϕ(d)

= ∑_{i=1}^{n} gcd(i, n) = ∑_{i=1}^{n} gcd(i, n − i) = ∑_{i=1}^{n−1} gcd(i, n − i) + n.

Now

Φ(x, x) = ∑_{i≥1} ∑_{j≥1} gcd(i, j)x^{i+j} = ∑_{n≥1} ∑_{i=1}^{n−1} gcd(i, n − i)x^{n},

so

(∑_{n≥1} a(n)x^{n}) − Φ(x, x) = ∑_{n≥1} nx^{n} = x(∑_{n≥0} x^{n})′ = \frac{x}{(1 − x)^{2}} .

Thus ∑_{n≥1} a(n)x^{n} and Φ(x, x) differ by an element of ℚ(x), so it suffices to show that ∑_{n≥1} a(n)x^{n} is transcendental over ℂ(x).

By Proposition 6.2, it suffices to show that, for each N ∈ ℕ, there exist infinitely many m ∈ ℕ such that, whenever j ∈ ℤ satisfies 1 ≤ |j| ≤ N,

|a(m)| > N |a(m + j)| .

We may suppose that N is fixed.

Remember the p_{i} is the ith prime number. For each Q ∈ ℕ, let

m_{Q} := ∏_{i=1}^{Q} p_{i} ∏_{i=1}^{N} p_{i}^{N}.

We may now suppose that j is fixed with 1 ≤ |j| ≤ N, and it suffices to show that

lim_{Q→∞} \frac{a(m_{Q} + j)}{a(m_{Q})} = 0.

We use the notation of Lemma 6.5, concerning Q_{0}. Let

C_{1} = ∏_{i=1}^{Q_{0}} p_{i} ∏_{i=1}^{N} p_{i}^{N}.

Now suppose that Q is an integer with Q ≥ max{Q_{0}, N}, let m = m_{Q} and let m′ = ∏_{i=1}^{Q} p_{i}.

We wish to bound a(m) = m ∑_{d|m} \frac{ϕ(d)}{d} from below. Recall that, for any

positive integer n, \frac{ϕ(n)}{n} = ∏(1− \frac{1}{p}), where the product is over the distinct prime

divisors p of n. Thus a(m) ≥ m ∑_{d|m} \frac{ϕ(m)}{m} = m d(m)\frac{ϕ(m)}{m}, where d(m) denotes

the number of divisors d of m. Also, \frac{ϕ(m)}{m} = ∏_{i=1}^{Q}(1− \frac{1}{p_{i}}), which, by Lemma 6.5(3),

is at least \frac{1}{Q}. Thus a(m) ≥ m d(m)\frac{1}{Q}. Notice that d(m) ≥ d(m′), since m′ divides m. From the definition of m′, we see that d(m′) = 2^{Q}. Thus

a(m) ≥ m2^{Q} \frac{1}{Q}.

We next wish to bound a(m + j) from above. Let Ω(m + j) be the number, counting multiplicity, of prime factors of m, and let

m + j = p_{i_{1}}p_{i_{2}} ⋯ p_{i_{Ω(m+j)}}

be the factorization of m + j into prime factors. Then d(m + j) ≤ 2^{Ω(m+j)}, and

\begin{aligned}a(m + j) & = (m + j) ∑_{d|(m+j)} \frac{ϕ(d)}{d} ≤ (m + j) ∑_{d|(m+j)} 1 = (m + j) d(m + j) \\ & ≤ (m + j)2^{Ω(m+j)} ≤ (m + N)2^{Ω(m+j)} ≤ 2m2^{Ω(m+j)}. \end{aligned}

Consider 1 ≤ l ≤ Ω(m + j). If i_{l} ≤ Q, then p_{i_{l}} divides m so p_{i_{l}} divides j. But 1 ≤ |j| ≤ N, so p_{i_{l}} ≤ N, so i_{l} ≤ N. Hence p_{i_{l}}^{N} divides m, but p_{i_{l}}^{N} ≥ 2^{N} > N ≥ |j|, so p_{i_{l}}^{N} cannot divide j, so cannot divide m + j. Thus, the number of i_{l} which are less than Q is at most N^{N}. Let z = z(Q, j) denote the number of l such that i_{l} ≥ Q, so Ω(m + j) ≤ z + N^{N}, and

a(m + j) ≤ 2m2^{Ω(m+j)} ≤ 2m2^{z+N^{N}} .

Thus

\frac{a(m + j)}{a(m)} ≤ \frac{2m2^{z+N^{N}}}{m2^{Q} \frac{1}{Q}} = Q2^{z−Q}2^{N^{N} +1}.

Hence it remains to show that lim_{Q→∞} Q2^{z−Q} = 0, or equivalently,

lim_{Q→∞} Q − z − log_{2} Q = ∞.

Since z is the number, counting multiplicity, of prime factors p_{i_{l}} of m + j with p_{i_{l}} ≥ p_{Q},

p_{Q}^{z} ≤ m + j ≤ m + N ≤ 2m.

We can write

m = ∏_{i=1}^{Q} p_{i} ∏_{i=1}^{N} p_{i}^{N} ≤ ∏_{i=1}^{Q_{0}} p_{i} ∏_{i=2}^{Q}(\frac{5}{4} i log i) ∏_{i=1}^{N} p_{i}^{N} = C_{1} ∏_{i=2}^{Q}(\frac{5}{4} i log i)

≤ C_{1}(\frac{5}{4})^{Q}Q!(log Q)^{Q} ≤ C_{1}(\frac{5}{4})^{Q}(\frac{Q}{2})^{Q}(log Q)^{Q},

by Lemma 6.5(1). Thus

(\frac{3}{4} Q log Q)^{z} ≤ p_{Q}^{z} ≤ 2m ≤ 2C_{1}(\frac{5}{4})^{Q}(\frac{Q}{2})^{Q}(log Q)^{Q}.

Hence

(\frac{3}{4}Q log Q)^{z−Q} ≤ 2C_{1}(\frac{4}{3})^{Q}(\frac{1}{2})^{Q}(\frac{5}{4})^{Q} = 2C_{1}(\frac{5}{6})^{Q},

so (z − Q)(log \frac{3}{4} + log Q + log log Q) ≤ log 2C_{1} − Q log(\frac{6}{5}), and

−(Q − z) ≤ \frac{log 2C_{1} − Q log(\frac{6}{5})}{log \frac{3}{4} + log Q + log log Q} ∼ − log(\frac{6}{5})\frac{Q}{log Q}.

It follows that

lim_{Q→∞} Q − z − log_{2} Q ≥ lim_{Q→∞} log(\frac{6}{5})\frac{Q}{log Q} − log_{2} Q = ∞,

as desired.

References

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[2] G. Baumslag. A finitely presented metabelian group with a free abelian derived group of infinite rank. Proc. Amer. Math. Soc., 35:61–62, 1972.

[3] R. Grigorchuk, P. Linnell, T. Schick, and A. Zuk. On a question of Atiyah. C. R. Acad. Sci., Ser I, Math., 331:663–668, 2000. Earlier version available via http://arxiv.org/abs/math.GT/0009175.

[4] R. Grigorchuk and A. Zuk. The lamplighter group as a group generated by a 2-state automaton, and its spectrum. to appear in Geometriae Dedicata, 2001.

[5] G. H. Hardy and E. M. Wright. An introduction to the theory of numbers, fifth edition. Oxford University Press, New York, 1979.