VON NEUMANN ALGEBRAS AND LINEAR INDEPENDENCE OF TRANSLATES

PETER A. LINNELL

Abstract. For x, y ∈ ℝ and f ∈ L^{2}(ℝ), define (x, y)f(t) = e^{2πiyt}f(t + x) and if Λ ⊆ ℝ^{2}, define S(f, Λ) = {(x, y)f | (x, y) ∈ Λ}. It has been conjectured that if f ≠ 0, then S(f, Λ) is linearly independent over ℂ; one motivation for this problem comes from Gabor analysis. We shall prove that S(f, Λ) is linearly independent if f ≠ 0 and Λ is contained in a discrete subgroup of ℝ^{2}, and as a byproduct we shall obtain some results on the group von Neumann algebra generated by the operators {(x, y) | (x, y) ∈ Λ}. Also we shall prove these results for the obvious generalization to ℝ^{n}.

1. Introduction

Let n be a positive integer, let G_{n} be the abelian group {(x, y) | x, y ∈ ℝ^{n}} with the operation addition (so G_{n} ≅ ℝ^{2n}), and for x, y ∈ ℝ^{n}, let x · y denote the dot product x_{1}y_{1} + ⋯ + x_{n}y_{n}. Let ℂ ∗ G_{n} denote the twisted group ring (a twisted group ring is a particular kind of crossed product) which has ℂ-basis \{ḡ | g ∈ G_{n}\}, and multiplication satisfying \overline{(a, b)} \overline{(x, y)} = e^{2πia·y}\overline{(a + x, b + y)}. For g ∈ G_{n}, we shall often write g instead of ḡ if there is no danger of confusion, and then g^{−1} will mean ḡ^{−1} rather than \overline{g^{−1}}. Let L^{2}(ℝ^{n}) denote the Hilbert space of square integrable functions {f: ℝ^{n} → ℂ | ∫_{ℝ^{n}} |f(t)|^{2} dt < ∞} with two functions f_{1}, f_{2} ∈ L^{2}(ℝ^{n}) being equal if and only if f_{1}(t) = f_{2}(t) almost everywhere, and let B(L^{2}(ℝ^{n})) denote the set of bounded linear operators on L^{2}(ℝ^{n}). Then ℂ ∗ G_{n} acts on the left of L^{2}(ℝ^{n}) according to the rule (x, y) f(t) = e^{2πiy·t}f(t + x), and extending to the whole of ℂ ∗ G_{n} by ℂ-linearity. To check that this indeed defines an action, we need only verify that (a, b)((x, y)f(t)) = ((a, b)(x, y))f(t), which is indeed true because both sides equal e^{2πi(a·y+b·t+y·t)}f(t + a + x). Thus we obtain a homomorphism from ℂ ∗ G_{n} into B(L^{2}(ℝ^{n})). Since ℂ ∗ G_{n} is a simple ring by Lemma 2.1, this homomorphism must be a monomorphism and so we may view ℂ ∗ G_{n} as a ℂ-subalgebra of B(L^{2}(ℝ^{n})). We shall consider the following conjecture.

Conjecture 1.1. Let 0 ≠ θ ∈ ℂ ∗ G_{n} and 0 ≠ f ∈ L^{2}(ℝ^{n}). Then θf ≠ 0.

Motivation for studying this problem comes from Gabor analysis and in particular the conjecture on page 2790 of [4]. If G ⩽ G_{n}, then ℂ ∗ G will denote the ℂ-subalgebra of ℂ ∗ G_{n} which has ℂ-basis {ḡ | g ∈ G}. Of course when talking about discrete subsets of G_{n}, we are giving G_{n} the usual topology from ℝ^{2n}. We shall prove

Theorem 1.2. Let G be a discrete subgroup of G_{n}. If 0 ≠ θ ∈ ℂ ∗ G and 0 ≠ f ∈ L^{2}(ℝ^{n}), then θf ≠ 0.

1991 Mathematics Subject Classification. Primary: 46L10; Secondary 42C99.

Key words and phrases. group von Neumann algebra, Gabor analysis, Heisenberg group.

Of course it follows immediately that if G is a discrete subgroup of G_{n}, g ∈ G_{n}, 0 ≠ θ ∈ gℂ ∗ G and 0 ≠ f ∈ L^{2}(ℝ^{n}), then θf ≠ 0. This means we can rephrase the above result in terminology closer to that of [4] as follows. For x, y ∈ ℝ^{n} and f ∈ L^{2}(ℝ^{n}), define (x, y)f(t) = e^{2πiy·t}f(t + x) and if Λ ⊆ ℝ^{2n}, define S(f, Λ) = {(x, y)f | (x, y) ∈ Λ}. Then Theorem 1.2 yields

Proposition 1.3. Let n be a positive integer, let Λ be a subset of ℝ^{2n} of the form g +G where G is a discrete subgroup of ℝ^{2n}, and let 0 ≠ f ∈ L^{2}(ℝ^{n}). Then S(f, Λ) is linearly independent.

As a byproduct, we shall obtain results on the von Neumann algebra generated by ℂ ∗ G, which we shall denote by W ∗ G. Thus W ∗ G is the weak closure of ℂ ∗ G in B(L^{2}(ℝ^{n})) and is rather similar to the group von Neumann algebra of G. For f, g ∈ L^{2}(ℝ^{n}), let ⟨f, g⟩ denote the inner product ∫_{ℝ^{n}} f(t)ḡ(t) dt, where ¯ denotes complex conjugation, and let U(L^{2}(ℝ^{n})) denote the set of closed densely defined linear operators [5, §2.7] acting on L^{2}(ℝ^{n}). Then the adjoint α^{∗} of α ∈ U(L^{2}(ℝ^{n})) satisfies ⟨αf, g⟩ = ⟨f, α^{∗}g⟩ whenever f, g ∈ L^{2}(ℝ^{n}) and αf, α^{∗}g are defined. Of course ^{∗} restricts to an involution on both B(L^{2}(ℝ^{n})) and W ∗ G. If G is a discrete subgroup of G_{n}, then W ∗ G is a finite von Neumann algebra by Lemma 3.2; also in many cases this can be deduced from Rieffel’s paper [7]. In this situation, we let U ∗ G indicate the operators of U(L^{2}(ℝ^{n})) which are affiliated to W ∗ G [3, p. 150]. The results of [3] (especially theorem 1 and the proof of theorem 10) now show that (U ∗ G)^{∗} = U ∗ G, U ∗ G is a ∗-regular ring containing W ∗ G, and every element of U ∗ G can be written in the form γδ^{−1} where γ, δ ∈ W ∗ G. In particular every nonzero divisor in W ∗ G is invertible in U ∗ G. We shall prove

Theorem 1.4. Let G be a discrete subgroup of G_{n}. Then W ∗ G is a finite von Neumann algebra, every nonzero element of ℂ ∗ G is invertible in U ∗ G, and the set {γδ^{−1} | γ ∈ ℂ ∗ G, 0 ≠ δ ∈ ℂ ∗ G} is a division subring of U ∗ G.

Let L be a locally compact group, let G be a torsion free subgroup of L, and let L^{2}(L) denote the Hilbert space of square integrable functions on L with respect to the left Haar measure on L. Then G acts on the left of L^{2}(L) according to the rule gf(l) = f(g^{−1}l) for g ∈ G, f ∈ L^{2}(L), l ∈ L. For f ∈ L^{2}(L) \ 0, a closely related problem to Conjecture 1.1 is to determine whether the set {gf | g ∈ G} is linearly independent over ℂ. If the von Neumann algebra W ∗ G generated by G is a finite von Neumann algebra, then by using the techniques of this paper, it is possible in many cases to show that the set {gf | g ∈ G} is linearly independent. On the other hand if W ∗ G is not a finite von Neumann algebra, then the techniques of this paper cannot be applied. It will usually be the case that W ∗ G is not finite if G is not discrete and has no abelian subgroup of finite index. A specific example would be to let L be the Heisenberg group consisting of upper unitriangular 3 by 3 matrices with entries in ℝ, in other words matrices of the form

(\begin{matrix}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{matrix})

where a, b, c ∈ ℝ, and to let G = L. Then it is not known in this case whether for f ∈ L^{2}(L) \ 0, the set {gf | g ∈ G} is linearly independent.

I am very grateful to Chris Heil and Jonathan Rosenblatt for bringing the problem studied in this paper to my attention, and for some helpful email correspondence.

2. Notation, Terminology and Assumed Results

The identity of a group will be denoted by either 0 or 1. If n is a positive integer and R is a ring, then M_{n}(R) will denote the n by n matrices over R, and we shall let δ_{ij} indicate the Kronecker delta, so δ_{ij} = 0 if i ≠ j and δ_{ij} = 1 if i = j. The identity matrix of M_{n}(R) will be denoted by I_{n}, and the zero matrix of M_{n}(R) will be denoted by 0_{n}. We shall view vectors in ℝ^{n} as column vectors rather than row vectors. A lattice in ℝ^{n} will mean a discrete subgroup of ℤ-rank n; in other words a discrete subgroup of finite covolume (note that this is a different definition of lattice from that of [4, p. 2791]). If α = ∑_{g∈G_{n}} λ_{g}g ∈ ℂ ∗ G_{n} where λ_{g} ∈ ℂ for all g ∈ G_{n}, then the support of α, denoted supp α, is the set {g ∈ G_{n} | λ_{g} ≠ 0}. We shall use the notation ∥f∥_{2} for the norm \sqrt{⟨f, f⟩} of an element f ∈ L^{2}(ℝ^{n}), and \overline{X} for the closure of a subset X in L^{2}(ℝ^{n}). The commutant of a subset A of B(L^{2}(ℝ^{n})) is A′ = {x ∈ B(L^{2}(ℝ^{n})) | ax = xa for all a ∈ A}. If A = A^{∗}, then A′ is a von Neumann algebra and by von Neumann’s double commutant theorem [1, theorem 1.2.1], A is dense in A′′ in the weak operator topology. Thus another description of W ∗ G is the double commutant of ℂ ∗ G in B(L^{2}(ℝ^{n})). In the case W ∗ G is a finite von Neumann algebra, we can now describe U ∗ G as those unbounded operators in U(L^{2}(ℝ^{n})) which commute with every element of (W ∗G)′.

Lemma 2.1. ℂ ∗ G_{n} is a simple ring.

Proof. Suppose 0 ≠ I ⊲ ℂ ∗ G_{n} with I ≠ ℂ ∗ G_{n}, and choose 0 ≠ α ∈ I with minimal support. If g ∈ supp α, then 1 ∈ supp ḡ^{−1}α and ḡ^{−1}α ∈ I, so we may assume that 1 ∈ supp α. Since I ≠ ℂ ∗ G_{n}, we may choose a ∈ G_{n} such that 1 ≠ a ∈ supp α. Then there exists g ∈ G_{n} such that ḡāḡ^{−1} ≠ ā, and now we have 0 ≠ ḡαḡ^{−1} −α ∈ I. This contradicts the minimality of supp α because | supp(ḡαḡ^{−1} − α)| < | supp α|, and the result follows.

If R is a ring and σ is an automorphism of R, then R_{σ}[X] will denote the twisted polynomial ring over R in the indeterminate X, so multiplication is defined by ∑ a_{i}X^{i} ∑ b_{j}X^{j} = ∑_{n}(∑_{i+j=n} a_{i}σ^{i}b_{j})X^{n}. We say that R is an Ore domain if it is contained in a division ring D, called the division ring of fractions of R, such that every element of D can be written in the form rs^{−1} and also in the form s^{−1}r, with r, s ∈ R and s ≠ 0. Of course the division ring D containing R is unique up to R-isomorphism. Also if R is contained in a ring D′ such that every nonzero element of R is invertible, then the set {rs^{−1} | r, s ∈ R and s ≠ 0} is the division ring of fractions containing R. The following two elementary results are well known.

Lemma 2.2. Let R be an Ore domain with division ring of fractions D, and let σ be an automorphism of R. Then σ extends uniquely to an automorphism of D, which we shall also call σ, and if α, β ∈ D_{σ}[X], then there exists r ∈ R \ 0 such that rα, rβ ∈ R_{σ}[X].

Lemma 2.3. Let G be a subgroup of G_{n}. Then ℂ∗G is an Ore domain, and if I, J are nonzero left ideals of ℂ ∗ G, then I ∩ J ≠ 0.

Finally we require the following:

Lemma 2.4. Let G be a discrete subgroup of G_{n}, let H ⊲ G such that G/H is infinite cyclic, and let x ∈ G such that Hx is a generator for G/H. If ζ ∈ ℂ and |ζ| = 1, then there exists y ∈ G_{n} such that ȳh̄ȳ^{−1} = h̄ for all h ∈ H and ȳx̄ȳ^{−1} = ζx̄ in ℂ ∗ G_{n}.

Proof. Since G is discrete, we may choose m ∈ ℤ and a subset {h_{1}, …, h_{m}} which generates H and is linearly independent over ℝ. Note that {h_{1}, …, h_{m}, x} is also linearly independent over ℝ. Choose t ∈ ℝ such that e^{2πit} = ζ, and define a bilinear form β: G_{n} → ℝ by β((a, b), (c, d)) = a · d − b · c, where a, b, c, d ∈ ℝ^{n}. Note that in ℂ ∗ G_{n}, we have

(a, b)(c, d)(a, b)^{−1} = e^{2πi(a·d−b·c)}(c, d).

It is easily checked that β is nondegenerate, so there exists y ∈ G_{n} such that β(y, h_{i}) = 0 for all i and β(y, x) = t. This completes the proof.

3. Faithful Traces

In this section, we show that W ∗G has a faithful weakly continuous tracial state, which in particular will establish that W ∗ G is a finite von Neumann algebra. Throughout this section, n will be a positive integer. The purpose of the next lemma is to reduce to the case when G is a lattice in ℝ^{2n} containing 1 × ℤ^{n}; its proof is modelled on [4, §2, p. 2790].

We shall think of ℝ^{2n} as ℝ^{n} ⊕ℝ^{n}, so we can view ℝ^{n} as a subgroup of ℝ^{2n} in the usual way via the map x ↦ (x, 0). We then have a monomorphism ψ: G_{n} → G_{2n} and this induces a monomorphism ℂ ∗ G_{n} → ℂ ∗ G_{2n}, which we shall also call ψ.

Given f, g ∈ L^{2}(ℝ^{n}), we can form the element f ⊗ g ∈ L^{2}(ℝ^{2n}) defined by (f ⊗ g)(x, y) = f(x)g(y) for x, y ∈ L^{2}(ℝ^{n}), and then the functions of the form ∑_{i=1}^{m} f_{i} ⊗ g_{i} are dense in L^{2}(ℝ^{2n}). If θ ∈ B(L^{2}(ℝ^{n})), then we have a well defined operator θ⊗1 ∈ B(L^{2}(ℝ^{2n})) satisfying (θ⊗1)(f ⊗g) = (θf)⊗g for all f, g ∈ L^{2}(ℝ^{n}), and this yields a weakly continuous ∗-monomorphism θ ↦ θ ⊗ 1: B(ℝ^{n}) → B(ℝ^{2n}).

Note that when we view ℂ ∗ G_{n} and ℂ ∗ G_{2n} as subalgebras of B(L^{2}(ℝ^{n})) and B(L^{2}(ℝ^{2n})) respectively, then ψ(θ) = θ ⊗ 1 for all θ ∈ ℂ ∗ G_{n}. Furthermore if G ≤ G_{n}, then ψ induces isomorphisms W ∗ G → W ∗ ψG and (assuming W ∗ G is a finite von Neumann algebra) U ∗G → U ∗ψG, which means we may identify G with the subgroup ψG of G_{2n}; we shall do this without further comment and without using ψ in the future.

Let {e_{1}, …, e_{2n}} denote the standard basis for ℝ^{2n}, so e_{i} has a 1 in the ith position and zeros elsewhere, and e_{i} · e_{j} = δ_{ij}. If G ⩽ G_{n}, then we define {ℂG} = {λg | λ ∈ ℂ and g ∈ G}, a subset of ℂ ∗ G.

Lemma 3.1. Let G be a discrete subgroup of G_{n}. Then there exists a lattice H in G_{2n} containing 1 × ℤ^{2n} and a unitary operator u ∈ B(L^{2}(ℝ^{2n})), such that u{ℂG}u^{−1} ⊆ {ℂH}.

Proof. Choose an ℝ-basis {g_{1}, …, g_{2n}} for G_{n} such that {g_{1}, …, g_{r}} is a ℤ-basis for G, where r is the rank of G. Let E = {(e_{1}, 0), …, (e_{2n}, 0), (0, e_{1}), …, (0, e_{2n})}, let

F = \{(e_{1}, e_{n+1})/\sqrt{2}, (e_{2}, e_{n+2})/\sqrt{2}, …, (e_{n}, e_{2n})/\sqrt{2}, (e_{n+1}, e_{1})/\sqrt{2},

(e_{n+2}, e_{2})/\sqrt{2}, …, (e_{2n}, e_{n})/\sqrt{2}, (−e_{n+1}, e_{1})/\sqrt{2}, (−e_{n+2}, e_{2})/\sqrt{2}, …,
(−e_{2n}, e_{n})/\sqrt{2}, (−e_{1}, e_{n+1})/\sqrt{2}, (−e_{2}, e_{n+2})/\sqrt{2}, …, (−e_{n}, e_{2n})/\sqrt{2}\},

and let

K = \{g_{1}, …, g_{2n}, (−e_{n+1}, e_{1})/\sqrt{2}, (−e_{n+2}, e_{2})/\sqrt{2}, …, (−e_{2n}, e_{n})/\sqrt{2},

(−e_{1}, e_{n+1})/\sqrt{2}, (−e_{2}, e_{n+2})/\sqrt{2}, …, (−e_{n}, e_{2n})/\sqrt{2}\}, so E, F and K are ℝ-bases of G_{2n}. For i = 1, …, 4n, we shall let ê_{i}, f_{i}, k_{i} denote the ith basis elements of E, F, K respectively, and we shall let K be the lattice in G_{2n} which has ℤ-basis K. Let A_{i} denote the coordinates of k_{i} with respect to the basis F, and let a_{ji} denote the jth coordinate of A_{i}. Then for 2n + 1 ≤ i ≤ 4n, a_{ji} = 1 if j = i and a_{ji} = 0 if j ≠ i. Now define h_{i} = ∑_{j=1}^{4n} a_{ji}ê_{j} ∈ G_{2n}, and let H be the subgroup of G_{2n} generated by the h_{i}. Then H is a lattice in G_{2n} which contains 1 × ℤ^{2n}.

Let T be the transition matrix from E to F, and let J = J_{2n} = (\begin{matrix}0_{n} & I_{n} \\ I_{n} & 0_{n} \end{matrix}) ∈

M_{2n}(ℝ). Thus if T has entries t_{ij}, then f_{j} = ∑_{i=1}^{4n} t_{ij}ê_{i}, and if we think of the A_{i} as column vectors, then the coordinates of k_{i} with respect to E are T A_{i}. Also

T = (\begin{matrix}I_{2n}/\sqrt{2} & −J_{2n}/\sqrt{2} \\ J_{2n}/\sqrt{2} & I_{2n}/\sqrt{2} \end{matrix})

= (\begin{matrix}I_{2n} & 0_{2n} \\ −J_{2n} & I_{2n} \end{matrix}) (\begin{matrix}J_{2n}/\sqrt{2} & 0_{2n} \\ 0_{2n} & J_{2n}\sqrt{2} \end{matrix}) (\begin{matrix}0_{2n} & −I_{2n} \\ I_{2n} & 0_{2n} \end{matrix}) (\begin{matrix}I_{2n} & 0_{2n} \\ −J_{2n} & I_{2n} \end{matrix}) .

Let τ, α, β, γ: G_{2n} → G_{2n} be the linear mappings determined by the matrices

T, (\begin{matrix}J_{2n}/\sqrt{2} & 0_{2n} \\ 0_{2n} & J_{2n}\sqrt{2} \end{matrix}) , (\begin{matrix}I_{2n} & 0_{2n} \\ −J_{2n} & I_{2n} \end{matrix}) , (\begin{matrix}0_{2n} & −I_{2n} \\ I_{2n} & 0_{2n} \end{matrix})

respectively with respect to the basis E, so τê_{i} = f_{i} for all i. Then τH = K ⊇ G, so it will be sufficient to show that there exists a unitary operator u ∈ B(L^{2}(ℝ^{2n})) such that u^{−1}ℂgu = ℂτg for all g ∈ G_{2n}. Since τ = βαγβ, it will be sufficient to do this with α, β, γ in place of τ. We now use metaplectic transformations [8, p. 578]. Write g = (x, y) where x, y ∈ ℝ^{2n}, and then we have three cases to consider.

1. The matrix α = (\begin{matrix}J_{2n}/\sqrt{2} & 0_{2n} \\ 0_{2n} & \sqrt{2}J_{2n} \end{matrix}). For f ∈ L^{2}(ℝ^{2n}) and t ∈ ℝ^{2n}, we define

uf(t) = 2^{−n/2}f(Jt/\sqrt{2}) (we are considering t as a column vector in ℝ^{2n} here). Then u is ℂ-linear and ∥uf∥_{2} = ∥f∥_{2} for all f ∈ L^{2}(ℝ^{2n}), hence u is a unitary operator. Also u^{−1}f(t) = 2^{n/2}f(\sqrt{2}Jt) because J_{2n}^{2} = I_{2n}, consequently

u^{−1}guf(t) = u^{−1}g2^{−n/2}f(Jt/\sqrt{2}) = u^{−1}e^{2πiy·t}2^{−n/2}f(J(t + x)/\sqrt{2})
= e^{2πi\sqrt{2}Jy·t}f(t + Jx/\sqrt{2}) because J_{2n} is symmetric
= (αg)f(t)

for all t ∈ ℝ^{2n} and for all f ∈ L^{2}(ℝ^{2n}). Thus u^{−1}gu = αg as required.

2. The matrix β = (\begin{matrix}I_{2n} & 0_{2n} \\ −J_{2n} & I_{2n} \end{matrix}). Here we define uf(t) = e^{−πiJt·t}f(t). Then

u is ℂ-linear and ∥uf∥_{2} = ∥f∥_{2}, so u is a unitary operator. Since u^{−1}f(t) = e^{πiJt·t}f(t),

u^{−1}guf(t) = u^{−1}ge^{−πiJt·t}f(t) = u^{−1}e^{2πiy·t}e^{−πiJ(t+x)·(t+x)}f(t + x)
= e^{πiJt·t}e^{2πiy·t}e^{−πiJ(t+x)·(t+x)}f(t + x)
= e^{−πiJx·x}e^{2πit·(y−Jx)}f(t + x) because J_{2n} is symmetric
= e^{−πiJx·x}(βg)f(t)

for all t ∈ ℝ^{2n}, and we have shown that u^{−1}gu ∈ ℂ(βg).

3. The matrix γ = (\begin{matrix}0_{2n} & −I_{2n} \\ I_{2n} & 0_{2n} \end{matrix}). Here we use the Fourier transform; specifically

uf(t) = ∫_{ℝ^{2n}} e^{2πit·s}f(s) ds, and then u^{−1}f(t) = ∫_{ℝ^{2n}} e^{−2πit·s}f(s) ds. Observe that (x, 0)uf(t) = u(0, x)f(t) and (0, y)uf(t) = u(−y, 0)f(t), consequently

u^{−1}(x, y)u = u^{−1}(0, y)(x, 0)u = (−y, 0)(0, x) = e^{−2πix·y}(−y, x) = e^{−2πix·y}γ(x, y)

and we deduce that u^{−1}ℂgu = ℂγg.

This completes the proof of Lemma 3.1

Lemma 3.2. Let G be a discrete subgroup of G_{n}, and define τ: ℂ ∗ G → ℂ by τg = 0 when 1 ≠ g ∈ G, and τ1 = 1. Then

(i) τ extends to a weakly continuous ℂ-linear map W ∗ G → ℂ. (ii) If α, β ∈ W ∗ G, then τ(αβ) = τ(βα). (iii) If α ∈ W ∗ G and x ∈ G_{n}, then τ(x̄αx̄^{−1}) = τ(α). (iv) If e is a nonzero projection in W ∗ G, then 0 < τe ≤ 1.

(v) Let e, f be projections in W ∗ G. If eL^{2}(ℝ^{n}) ⊆ fL^{2}(ℝ^{n}), then τe ≤ τf. (vi) Let e, f be projections in W ∗ G, and let h be the projection of L^{2}(ℝ^{n}) onto

\overline{eL^{2}(ℝ^{n}) + fL^{2}(ℝ^{n})}. Then h ∈ W ∗ G and if eL^{2}(ℝ^{n}) ∩ fL^{2}(ℝ^{n}) = 0, then τe + τf = τh.

Proof. Since G is a discrete subgroup of G_{n}, there is by Lemma 3.1 a lattice H in G_{2n} containing 1 × ℤ^{2n} and a unitary operator u ∈ B(L^{2}(ℝ^{2n})) such that uℂ ∗ Gu^{−1} ⊆ ℂ ∗ H. If we can find a weakly continuous ℂ-linear map τ: ℂ ∗ H → ℂ with the required properties, then the weakly continuous ℂ-linear map α ↦ τ(uαu^{−1}) for α ∈ ℂ∗G will suffice. Therefore we may assume that G is a lattice in G_{n} containing 1 × ℤ^{n}. If a is a positive number, we shall let C(a) denote the standard unit cube in ℝ^{n} with side of length a; thus C(a) = {(a_{1}, …, a_{n}) | 0 ≤ a_{i} ≤ a for all i}.

(i) Choose a positive integer b such that hC(1/b) ∩ C(1/b) = ∅ whenever (h, k) ∈ G \ (1 × ℤ^{n}), which is possible because G is a lattice in G_{n} containing 1 × ℤ^{n}, and set C = C(1/b). Let c = b^{n}, let C_{1}, …, C_{c} denote the c translates of C which are contained in the unit cube C(1), and for each i, let χ_{i} denote the characteristic function of C_{i}. For θ ∈ W ∗ G, define

τθ = ∑_{i=1}^{c}⟨θχ_{i}, χ_{i}⟩ = ∑_{i=1}^{c} ∫_{C_{i}} θχ_{i}(t) dt.

Let g ∈ G and write g = (h, k) where h, k ∈ ℝ^{n}. Then gχ_{i}(t) = e^{2πik·t}χ_{i}(t + h), so if h ≠ 0 we have gχ_{i}(t) = 0 for all t ∈ C_{i} and hence τg = 0. On the other hand if h = 0 and k ≠ 0, then k ∈ ℤ^{n} \ 0 because G is a lattice containing 1 × ℤ^{n}, consequently

τg = ∑_{i=1}^{c} ∫_{C_{i}} e^{2πik·t} dt = ∫_{C} e^{2πik·t} dt = 0.

Finally τ1 = 1 and (i) is proven.

(ii) If x, y ∈ G, then xy = 1 if and only if yx = 1. Therefore τx̄ȳ = τȳx̄ = 0 if xy ≠ 1 and x̄ȳ = ȳx̄ if xy = 1, hence τx̄ȳ = τȳx̄ for all x, y ∈ G and we deduce that ταβ = τβα for all α, β ∈ ℂ ∗ G. Since τ is weakly continuous and W ∗ G is the weak closure of ℂ ∗ G, we see that ταβ = τβα for all α, β ∈ W ∗ G which proves

(iii) Define σ: W ∗ G → ℂ by σ(α) = τ(xαx^{−1}). Observe that σ1 = 1 and if 1 ≠ g ∈ G, then σg = τ(xgx^{−1}) = 0 because xgx^{−1} = ζg for some ζ ∈ ℂ with |ζ| = 1. Thus σg = τg for all g ∈ G and since σ is a weakly continuous ℂ-linear map, we deduce that σ(α) = τ(α) for all α ∈ W ∗ G, which is the required result.

(iv) Note that if e is a projection in W ∗ G, then e^{∗}e = e and hence

τe = ∑_{i=1}^{c}⟨e^{∗}eχ_{i}, χ_{i}⟩ = ∑_{i=1}^{c}⟨eχ_{i}, eχ_{i}⟩ ≥ 0.

Let K = ℤ^{n} × ℤ^{n} ⩽ G_{n}. If k ∈ K, then τe = τ(k^{−1}ek) by (iii) and we deduce that

τe = τ(k^{−1}ek) = ∑_{i=1}^{c}⟨k^{−1}e^{∗}ekχ_{i}, χ_{i}⟩ = ∑_{i=1}^{c}⟨ekχ_{i}, ekχ_{i}⟩.

Let χ denote the characteristic function of C(1), and suppose τe = 0. Then ekχ_{i} = 0 for all i, hence ekχ = 0 for all k ∈ K. Now the set {kχ | k ∈ K} forms a Hilbert basis for L^{2}(ℝ^{n}) so if τe = 0, we see that ef = 0 for all f ∈ L^{2}(ℝ^{n}) and we deduce that e = 0. Also 1 − e is a projection if e is a projection, so applying the above to 1 − e we obtain 0 ≤ τ(1 − e), hence τe ≤ 1 and (iv) follows.

(v) Let h be the projection of L^{2}(ℝ^{n}) onto the orthogonal complement of eL^{2}(ℝ^{n}) in fL^{2}(ℝ^{n}). Then e + h = f, hence τe + τh = τf. Thus h ∈ W ∗ G and the result follows from (iv).

(vi) Let u be a unitary operator in (ℂ ∗ G)′. Then ueu^{−1} = e and ufu^{−1} = f. Since h is the projection of L^{2}(ℝ^{n}) onto \overline{eL^{2}(ℝ^{n}) + fL^{2}(ℝ^{n})}, we see that uhu^{−1} is the projection of L^{2}(ℝ^{n}) onto \overline{ueu^{−1}L^{2}(ℝ^{n}) + ufu^{−1}L^{2}(ℝ^{n})} = \overline{eL^{2}(ℝ^{n}) + fL^{2}(ℝ^{n})} and we deduce that uhu^{−1} = h. Therefore uh = hu. Now (ℂ∗G)′ is a von Neumann algebra, so any element of (ℂ ∗ G)′ is a ℂ-linear sum of unitary elements, hence xh = hx for all x ∈ (ℂ ∗ G)′ and we conclude that h ∈ W ∗ G.

We now claim that h = e ∪ f [2, p. 4]. Since eL^{2}(ℝ^{n}) ⊆ hL^{2}(ℝ^{n}), we see that e = he and hence e ≤ h. Similarly f ≤ h and so e ∪ f ≤ h. Now let g = e ∪ f. Then gL^{2}(ℝ^{n}) ⊇ eL^{2}(ℝ^{n}), fL^{2}(ℝ^{n}) and hence gL^{2}(ℝ^{n}) ⊇ hL^{2}(ℝ^{n}). We deduce that g ≥ h, consequently g = h and the claim is established.

If eL^{2}(ℝ^{n}) ∩ fL^{2}(ℝ^{n}) = 0, then (e ∩ f)L^{2}(ℝ^{n}) ⊆ eL^{2}(ℝ^{n}) ∩ fL^{2}(ℝ^{n}) = 0. Since W ∗ G is a von Neumann algebra, we may apply the parallelogram law to deduce that e ∼ e∪f −f [2, §1,§13]. Thus there is an element w ∈ W ∗G such that w^{∗}w = e and ww^{∗} = e ∪ f − f. Since τ(w^{∗}w) = τ(ww^{∗}), we deduce that τe = τ(e ∪ f − f) and hence τe + τf = τ(e ∪ f) = τh. This completes the proof.

4. Proofs

Theorems 1.2 and 1.4 are now immediate consequences of the following result.

Lemma 4.1. Let G be a discrete subgroup of ℝ^{n} and let θ ∈ ℂ ∗ G \ 0. Then (i) If 0 ≠ f ∈ L^{2}(ℝ^{n}), then θf ≠ 0. (ii) θ is invertible in U ∗ G. (iii) The set {γδ^{−1} | γ ∈ ℂ ∗ G, 0 ≠ δ ∈ ℂ ∗ G} is a division subring of U ∗ G, and

is equal to {δ^{−1}γ | γ ∈ ℂ ∗ G, 0 ≠ δ ∈ ℂ ∗ G}.

Proof. Since G is a discrete subgroup of G_{n}, it is a free abelian group of rank at most 2n. We shall prove the result by induction on the rank of G, the result being trivially true if the rank of G is zero, because then G = 1. Thus we may assume that the rank of G is strictly positive, and then there exists H ⊲ G such that G/H ≅ ℤ.

Since H has strictly smaller rank than G, we may assume that the result is true for H. Let τ: W ∗ G → ℂ be the weakly continuous tracial state obtained from Lemma 3.2.

(i) For α ∈ ℂ ∗ G, let ker α = {f ∈ L^{2}(ℝ^{n}) | αf = 0}, and let N (α) be the projection from L^{2}(ℝ^{n}) onto ker α. Suppose u is a unitary element in (ℂ ∗ G)′. Then u^{−1}N (α)u = N (u^{−1}αu) = N (α). Since (ℂ ∗ G)′ is a von Neumann algebra, every element of (ℂ ∗ G)′ is a linear combination of unitary elements of (ℂ ∗ G)′ and we deduce that N (α) commutes with every element of (ℂ ∗ G)′. Therefore N (α) ∈ W ∗ G.

Let ν = sup{τ(N (α)) | 0 ≠ α ∈ ℂ ∗ G}. If ν = 0 then N (θ) = 0 by Lemma 3.2(iv), hence ker θ = 0 and the result follows, so we may assume that 0 < ν ≤ 1. Therefore we may choose α ∈ ℂ ∗ G such that τN (α) > ν/2. Since G/H is infinite cyclic, there exists x ∈ G such that Hx generates G/H, and then we may write α = ∑_{i=−∞}^{i=∞} α_{i}x^{i} where α_{i} ∈ ℂ ∗ H and α_{i} = 0 for all but finitely many i. By replacing α with x^{m}α for some integer m, we may assume that α_{1} ≠ 0 and α_{i} = 0 for all i < 0.

By induction, there is a division subring D of U ∗H containing ℂ∗H which is the division ring of fractions of ℂ∗H. Let σ be the automorphism β ↦ xβx^{−1} : ℂ∗H → ℂ ∗ H. By Lemma 2.2 we may extend σ to an automorphism of D, which we shall also call σ. We now have a natural ring homomorphism θ: D_{σ}[X] → U ∗ G, defined by θX = x and θd = d for all d ∈ D, which maps (ℂ ∗ H)_{σ}[X] into ℂ ∗ G. By [6, lemma 16], there exists ζ ∈ ℂ with |ζ| = 1 and β′, γ′ ∈ D_{σ}[X] such that

β′ ∑_{i} α_{0}^{−1}α_{i}X^{i} + γ′ ∑_{i} α_{0}^{−1}α_{i}ζ^{i}X^{i} = 1.

By Lemma 2.2, there exists 0 ≠ r ∈ ℂ ∗ H such that rβ′α_{0}^{−1}, rγ′α_{0}^{−1} ∈ (ℂ ∗ H)_{σ}[X], so setting β = rβ′α_{0}^{−1} and γ = rγ′α_{0}^{−1}, we have β, γ ∈ (ℂ ∗ H)_{σ}[X] and

β ∑_{i} α_{i}X^{i} + γ ∑_{i} α_{i}ζ^{i}X^{i} = r.

Set α′ = ∑_{i} α_{i}ζ^{i}x^{i}. Applying the homomorphism θ, we now have βα+γα′ = r. By Lemma 2.4 there exists y ∈ G_{n} such that yhy^{−1} = h for all h ∈ H and yxy^{−1} = ζx, and then we have yαy^{−1} = α′. Thus ker α′ = y(ker α)y^{−1}, consequently N (α′) = yN (α)y^{−1} and using Lemma 3.2(iii), we deduce that τN (α′) = τN(α) > ν/2.

Suppose f ∈ ker α ∩ ker α′. Then αf = α′f = 0, hence rf = 0 because r = βα + γα′, and we can invoke our inductive hypothesis to deduce that f = 0. Therefore ker α ∩ ker α′ = 0. If π is the projection onto \overline{ker α + ker α′}, we now see from Lemma 3.2(vi) that

τπ = τN (α) + τN (α′) > ν/2 + ν/2 = ν.

Using Lemma 2.3 we may choose δ so that 0 ≠ δ ∈ ℂ ∗ Gα ∩ ℂ ∗ Gα′, and then ker δ ⊇ ker α + ker α′, hence τN (δ) ≥ τπ > ν by Lemma 3.2(v). This contradicts the definition of ν and (i) is proven.

(ii) This follows from (i) and the remarks immediately preceding Theorem 1.4. (iii) This follows from (ii), Lemma 2.3 and the comments immediately preceding

Lemma 2.2.

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Math, VPI, Blacksburg, VA 24061–0123, USA

   E-mail address: linnell@math.vt.edu
   URL: http://www.math.vt.edu/people/linnell/