Kadison–Singer algebras, II: General case

Abstract

A new class of operator algebras, Kadison–Singer (KS-) algebras, is introduced. These highly noncommutative, non self-adjoint algebras generalize triangular matrix algebras. They are determined by certain minimally generating lattices of projections in the von Neumann algebras corresponding to the commutant of the diagonals of the KS-algebras. It is shown that these lattices and their reduced forms are often homeomorphic to classical manifolds such as the sphere.

KS-algebras were previously introduced (1). Examples of KS-algebras with hyperfinite diagonals were given and studied. In this article, we shall continue our study of KS-algebras and mostly deal with the case when the diagonal is a finite von Neumann algebra. We shall use the notation and definitions previously introduced (1).

Suppose is a Hilbert space and is the algebra of all bounded operators on . Recall that a KS-algebra is a maximal reflexive subalgebra of with respect to a given von Neumann algebra as its diagonal algebra. Previously (1), we constructed KS-algebras with hyperfinite factors as their diagonal algebras and provided many previously undescribed reflexive lattices. Our main result in this paper is to prove that the reflexive lattice generated by a double triangle (a special lattice with only three nontrivial projections) is, in general, homeomorphic to the two-dimensional sphere S² (plus two distinct points corresponding to zero and I), and the corresponding reflexive algebra is a KS-algebra. In particular, we show that the algebra that leaves three free projections invariant is a KS-algebra. This shows that many factors are (minimally) generated by a reflexive lattice of projections that is topologically homeomorphic to S². Noncommutative algebraic structures on S², determined by the projections, give rise to non-isomorphic factors and KS-factors.

The paper contains five sections. In section two, maximal triangularity is discussed in different aspects. In section three, we describe the reflexive lattice generated by three free projections and show that it is homeomorphic to S². In section four, we show that this lattice is a KS-lattice and thus the corresponding algebra is a KS-algebra. Certain generalizations of the result is also discussed. In section five, we introduce a notion of connectedness of projections in a lattice of projections in a finite von Neumann algebra and show that all connected components form another lattice, called a “reduced lattice.” Reduced lattices of most of our examples were computed.

Maximality conditions

In the definition of KS-algebras, we require that the algebra be maximal in the class of reflexive algebras with the same diagonal. Our examples of KS-algebras given previously (1) are “maximal triangular” in the class of all algebras with the same diagonal, that is, an algebraic maximality without reflexiveness or closedness assumptions. In general, the algebraic maximality assumption is a much stronger requirement. We call a subalgebra of “maximal triangular with respect to its diagonal C*- (or von Neumann) algebra ” if, for any subalgebra of , contains and has the same diagonal as , then is equal to . This may lead to unique, interesting classes of non self-adjoint algebras. Many similar questions as those introduce previously (2), that is, the closedness of , can be asked accordingly.

In the following, we give a canonical method to construct a maximal triangular algebra in the class of weak-operator, closed algebras with respect to a given von Neumann algebra as its diagonal.

Suppose is a von Neumann algebra acting on a Hilbert space and H₁,…,H_n are positive elements in such that generate as a von Neumann algebra. From previous work (3), we know that many von Neumann algebras can be generated by such positive elements, especially all type III and properly infinite von Neumann algebras. Let be the direct sum of n + 1 copies of . Then . With this identification, we shall view both and as subalgebras of . Let E_ij and i,j = 1,…,n + 1 be a matrix unit system for . We shall write elements in in a matrix form with respect to this unit system (with entries from ).

Theorem One.

Define

where * denotes all possible elements in . Then is maximal upper triangular with respect to the diagonal .

Proof:

It is easy to check that . Similar techniques we used in the proof of theorem three (1) will show that is maximal upper triangular with the given diagonal. We omit the details here.

With given in theorem one, suppose P is a projection in with and . It is easy to see that P must be diagonal and (I - P_ii)TP_jj = 0 for all i < j and any T in . Thus, for some k. We know that such a P lies in . This shows that . Clearly, this implies that is not reflexive. It is interesting to know if contains a subalgebra that is a KS-algebra with the same diagonal.

The semidirect product of a von Neumann algebra with a semigroup S (embedded in the automorphism group of ) will give us another construction of triangular algebras. In general, such construction will not give us a maximal triangular algebra. Whether there is a KS-algebra containing such a triangular algebra for certain ergodic actions is another interesting question.

On the other hand, if we start with a “minimal” lattice of projections in a von Neumann algebra so that the lattice generate the von Neumann algebra, then is a KS-algebra? In general, . Thus, may not be a KS-lattice. But we shall see that is often a KS-algebra in the next section.

Reflexive lattices generated by three projections

Lattices generated by two projections are always reflexive (4). But lattices generated by three projections are complicated. Most of factors acting on a separable Hilbert space are known to be generated by three projections or a projection and a positive operator (3).

Example One:

Suppose is a factor acting on and write , where is a subfactor of . We assume that is generated by a projection P and a positive element H with 0 ≤ H ≤ I and supp(H) = supp(I - H) = I. Here, we view as a subalgebra of and as the relative commutant of in . Let E₁₁, E₁₂, E₂₁, and E₂₂ be the standard matrix unit system for . Define P₁ = E₁₁, , P₃ = E₁₁ + E₂₂P, and P₄ = P₂∧P₃. Assume that P₁∧P₂ = 0 and P₁∨P₂ = I. Then , generated by {P₁,P₂,P₃}, is a distributive lattice and thus reflexive (5). From our construction, we know that is generated by as a von Neumann algebra. One also easily checks that any proper sublattice of does not generate . Thus, is a KS-lattice and is a KS-factor. From this construction, we can realize most of the factors as diagonals of KS-algebras. For example, one may choose as a factor of type II₁ generated by a projection P of trace and a positive operator H such that P and H are free, and H has the same distribution (with respect to the trace on ) as the function on [0,1] (with respect to Lebesgue measure). Let τ be the trace on . In this case, , , and . Then is a KS-factor of type II₁.

It is hard to determine when a lattice is reflexive even for a finite lattice. Finite distributive lattices are reflexive (5). But most of the lattice are not distributive. The simplest, non-distributive lattice is a “double triangle” where it contains zero; I; and three projections P₁, P₂, and P₃ so that P_i∨P_j = I and P_i∧P_j = 0 for any i ≠ j and i,j = 1,2,3. Any lattice that contains a double triangle sublattice is not distributive. Three free projections with trace in a factor of type II₁ (together with zero, I) form a double triangle lattice. In the following, we first describe factors generated by free projections. For basic theory on freeness and distributions, we refer to ref. 6.

Let G_n be the free product of with itself n times, for n≥2 or = ∞. When n≥3, G_n is an i.c.c. group so its associated group von Neumann algebra is a factor of type II₁ acting on l²(G_n) (7). If U₁,…,U_n are canonical generators for corresponding to the generators of G_n with , then , j = 1,…,n, are projections of trace . Let be the lattice consisting of these n free projections, zero, and I.

Clearly, is a minimal lattice that generates as a von Neumann algebra. Is and n≥3, a KS-algebra? What is ? When n = 2, Halmos (4) showed that is reflexive and, thus, is maximal and, hence, a KS-algebra. We shall answer the above questions for the case when n = 3 and show that is a KS-algebra and is homeomorphic to S², the two-dimensional sphere.

We shall realize as the von Neumann algebra generated by and its relative commutant in and write projection generators of in terms of 2 × 2 matrices (with respect to the standard matrix units in ) given by the following equations:

The freeness among P₁, P₂, and P₃ require that H₁, H₂, and V be free, H₁ and H₂ have the same distribution as on [0,1] with respect to Lebesgue measure and V a Haar unitary element. Then the subalgebra of is the von Neumann algebra generated by H₁, H₂, and V. Now and .

When is a subalgebra of , we may also view where is the commutant of in . Thus, all operators can be written as 2 × 2 matrices with entries from . In fact, when for some Hilbert space , then .

Because , any operator T belonging to must be upper triangular. The following lemma follows from the invariance of P₂ and P₃ under T. The computation is straight forward.

Lemma One.

With notation given above, , where if, and only if,

By using unbounded operators affiliated with , one can construct many finite rank operators in . Unbounded operators affiliated with a finite von Neumann form an algebra (7). Any finitely, many unbounded operators have a common dense domain. Let ξ and η be vectors in the common domain of , , and the adjoint . We shall use x⊗y to denote the rank one operator defined by x⊗y(z) = 〈z,x〉y, for any with x and y arbitrarily given. Now let , , and (determined by lemma one ). Then is a finite rank operator (at most rank four). This shows that contains many finite rank (and thus compact) operators. In fact, lemma three below will show that contains “almost” a copy of .

The following is a technical result that will be used frequently. The result might be well known. We only sketch a proof here.

Lemma Two.

Suppose U is a Haar unitary element in a factor of type II₁, and A is an element in (or an unbounded operator affiliated with) such that A and U are free with each other. Then any nonzero scalar λ can not be a point spectrum of AU.

Proof:

Suppose is a nonzero point spectrum of AU. By symmetry and freeness of A and ωU, we know that λ must be a point spectrum for A(ωU) for any with |ω| = 1. This implies that ω^-1λ is a point spectrum of AU. Suppose P_β is the spectral projection of AU supported at . Then P_λ is equivalent to P_ωλ for any |ω| = 1. A direct computation shows that if λ ≠ λ_j for j = 1,2,…,n then P_λ∧(P_λ1∨⋯∨P_λn) = 0. From finiteness of , it is easy to conclude that P_λ = 0 when λ ≠ 0.

In the following, we shall describe all elements in . Unbounded operators will be used in our computation. All unbounded operators affiliated with the factor form an algebra. From function calculus, many unbounded operators we encounter in this paper can be viewed as (positive) functions defined on (0,1), with respect to Lebesgue measure. When H (= H₁ or H₂) is identified with then I - H, , , , et cetera, can be viewed as trigonometric functions and they are all determined by any one of them. Lemma two also tells that many linear combinations of free (non-self-adjoint) operators such as are invertible (with unbounded inverses).

When S, T are unbounded operators affiliated with or a finite von Neumann algebra and , then SXT is an unbounded operator that can be viewed as the weak-operator limit of bounded operators of the form SE_ϵXF_ϵT for projections E_ϵ and F_ϵ in (or the finite von Neumann algebra) so that SE_ϵ and F_ϵT are bounded and E_ϵ and F_ϵ have a strong operator limit I (as ϵ → 0). Thus, for any operator X in a weak-operator dense subalgebra of , the operator SXT is a bounded operator.

By using unbounded operators, we may restate the above lemma in the following form.

Lemma Three.

With given above, let be an unbounded operator affiliated with . If then there is an such that

Conversely, if such that and S^-1AS are bounded operators, then the above T belongs to .

The above lemma shows that is quite large, in particular, is infinite dimensional. The following result follows easily from the above lemma and shows that all nontrivial projections in have trace .

Corollary Two.

For any , we have that Q∧P₁ = 0, Q∨P₁ = I, and .

Proof:

For any , . Thus, Q∧P₁ is invariant under all T given in lemma three and, thus, the A in (1,1) entry of T. This implies that Q∧P₁ = P₁ or zero. Similarly, we can show that Q∨P₁ = I or P₁.

This corollary actually shows that, for any distinct projections, Q₁, Q₂ in , Q₁∧Q₂ = 0, and Q₁∨Q₂ = I.

For any or an unbounded operator X affiliated with a (finite) von Neumann algebra, we shall use supp(X) to denote the support of X, that is, the range projection of X^∗X. When supp(X) = supp(X^∗) = I, X has an (unbounded) inverse.

Theorem Three.

For any projection Q in , there are K and U in such that

where (or K) and U are determined by the polar decomposition of for some . Moreover, for any given a in , the polar decomposition determines U and K uniquely, which give rise to a projection Q (in the above form) in .

Proof:

Suppose Q is given in the theorem. From corollary two, we know that supp(I - K) = I. From lemma three, for any (the commutant of in , for some Hilbert space ) such that S^-1AS and are bounded, then

here . Thus, (I - Q)TQ = 0. This implies that

Because supp(I - K) = I, I - K is invertible. We have

This gives us

The above equation holds for all A in a weak-operator dense subalgebra of (). Thus, there is an such that

This implies that

Conversely, when K and U are given by this equation, all above equations hold. From lemma three, one checks easily that Q, given in the theorem, lies in .

is a KS-Algebra

In this section, we shall prove that is a KS-lattice, which implies that is a KS-algebra.

Lemma Four.

For any two distinct projections, Q₁ and Q₂ in , we have that .

Proof:

By theorem three, we may assume that, for i = 1,2,

[1]

where and a₁ ≠ a₂. Then we have . Replacing S by (a₁ - a₂)S in lemma three, we know that .

Lemma Five.

For any three distinct projections Q₁, Q₂ and Q₃ in , we always have that P₁∈Lat(Alg({Q₁,Q₂,Q₃})).

Proof:

We may assume that and Q_i is given [1] in the proof of lemma four, for i = 1,2,3 and a₁, a₂, and a₃ are distinct scalars.

To prove this lemma, we only need to show that if

then A₂₁ = 0. Now assume that the above A belongs to Alg({Q₁,Q₂,Q₃}). Then (I - Q_i)AQ_i = 0, for i = 1,2,3. The (1,2) entries of the equation AQ_i = Q_iAQ_i give us that

This implies that

Thus, for i,j = 1,2,3 and from , we have

From this, we conclude that

By using the relation again, we easily get

which implies that

This gives us that A₂₁ = 0 and the lemma follows.

Now the follow theorem follows easily from our lemmas and theorem three.

Theorem Four.

With the above notation, we have that is a KS-algebra and is determined by the following: and P ≠ 0,I,P₁ if, and only if,

where K and U are uniquely determined by the following polar decomposition with any : . As a consequence, we have and as a tends to ∞, the projection P converges strongly to P₁. Thus, is homeomorphic to S².

Proof:

We only need to show the minimality of . Clearly, for any sublattice containing only two projections Q₁ and Q₂ in , can not generate the type II₁ factor . Lemma four shows that any reflexive sublattice of containing more than two nontrivial projections must agree with . Thus, is a KS-lattice.

Although the above theorem is stated for , similar results hold when P₁, P₂, and P₃ are not assumed to be free. There are many other factors that can be generated by three projections. Let be a factor of type II₁ with trace τ and and , a subfactor of with generators H₁, H₂, and V, where 0 ≤ H₁, H₂ ≤ I, and V is a unitary operator. Suppose . Let be a projection in . Furthermore, we assume that

are projections in such that P₂∨P₃ = I, P₂∧P₃ = 0, , , and are bounded invertible operators. All above conditions are satisfied when H₁ and H₂ are free positive elements with disjoint spectra in (0,1) (V = I), and and have disjoint spectra.

Theorem Five.

With the above notation and assumptions, we have that Alg({P₁,P₂,P₃}) is a KS-algebra and Lat(Alg({P₁,P₂,P₃})) is determined by the following: P∈Lat(Alg({P₁,P₂,P₃})) and P ≠ 0,I,P₁ if, and only if,

where L and U are uniquely determined by the following polar decomposition with any : . As a consequence, we have .

The proof of this theorem will be the same as that of theorem four. Moreover, as a → ∞, the projection P → P₁ in strong-operator topology. Thus Lat(Alg({P₁,P₂,P₃}))∖{0,I} is homeomorphic to the one point compactification of , that is, homeomorphic to S².

When a lattice contains four or more projections in a von Neumann algebra, the situation is not clear. Even for (the lattice generated by four free projections), we know from our above result that contains several copies of S². But we do not have a complete characterization of this lattice. The following theorem shows that contains no nonzero compact operators. We believe that , when n≥4, does not contain any compact operators.

Theorem Six.

Let be the lattice generated by countably, infinitely many free projections of trace in , whereas G_∞ is the free product of countably infinitely many copies of then does not contain any nonzero compact operators.

Proof:

Let U_j and j = 1,2,… be the standard generators of and the free projections in . Suppose is a compact operator. From (I - P_j)TP_j = 0, we know that (I + U_j)T(I - U_j) = 0 for j = 1,2,…. For any x and y in l²(G_∞), we have

Because U_j has weak-operator limit zero as j → ∞ and T compact, we have that 〈U_jx,y〉 → 0, 〈Tx,U_jy〉 → 0, and TU_jx → 0 (as j → ∞). This implies that 〈Tx,y〉 = 0 and thus T = 0.

Reduced lattices

Suppose is a lattice of projections in a finite von Neumann algebra with a faithful normal trace τ. Two projections P and Q in are said to be “connect” if, for any ϵ > 0, there are elements P₁,P₂,…,P_n in such that P₁ = P, P_n = Q, |τ(P_j - P_j+1)| < ϵ, and either P_j ≤ P_j+1 or P_j≥P_j+1, for j = 1,…,n - 1. Define the connected component O(P) of P to be the set of all projections in that are connected with P. Let be the set of all connected components in . We shall see that carries an induced lattice structure from and we call the reduced lattice of . It is clear that if is a continuous nest, then contains only one point. A basic fact on connected components is given in the following.

Proposition Seven.

Suppose is a lattice of projections in a finite von Neumann algebra , . If O(P) ≠ O(Q), then O(P)∩O(Q) = ∅.

The proof of this proposition follows easily from the definition. If O(P) and O(Q) are two elements in , then for any Q₁∈O(Q) it is easy to see that O(P∨Q) = O(P∨Q₁). Thus, O(P∨Q) depends on the components O(P) and O(Q), not the choices of P and Q in the components. We define O(P)∨O(Q) = O(P∨Q). Similarly, we define that O(P)∧O(Q) = O(P∧Q). It is easy to show that is a lattice. The following theorem is immediate.

Theorem Eight.

Suppose is a lattice of projections in a finite von Neumann algebra with a (faithful normal) trace τ. If contains only finitely many trace values, then is the same as (i.e., every connected component contains only one element in ).

From this theorem, we know that . Suppose is a KS-lattice and generates a finite von Neumann algebra. If contains only one point, then we call “contractible.” Thus, continuous nests are contractible. In a forthcoming paper, we shall construct other contractible KS-lattices and also show that manifolds of higher dimensions can appear as the reduced lattices of KS-lattices.

Reduced lattices can not contain continuous nests. The following theorem shows that all possible trace values can appear in a reduced lattice.

Theorem Nine.

Suppose is the lattice given in section four (1), then .

The proof is a direct computation. We omit the details here.