Taken from C*-algebras and Operator Theory by Gerald Murphy.
A useful way of thinking of the theory of C*-algebras is as “non-commutative topology”. This is justified by the correspondence between abelian C*-algebras and locally compact Hausdorff spaces given by the Gelfand representation. The algebras studied in this chapter, von Neumann algebras, are a class of C*-algebras whose study can be thought of as “non-commutative measure theory”. The reason for the analogy in this case is that the abelian von Neumann algebras are (up to isomorphism) of the form , where is a measure space.
The theory of von Neumann algebras is a vast and very well-developed area of the theory of operator algebras. We shall be able only to cover some of the basics. The main results of this chapter are the von Neumann double commutant theorem and the Kaplansky density theorem.
4.1 The Double Commutant Theorem
There are a number of topologies on ( a Hilbert space), apart from the norm topology, that play a crucial role, and each has valuable properties that the others lack. The two most important are the strong (operator) topology and the weak (operator) topology. This section is concerned with the former (we shall introduce the weak topology in the next section). One of the reasons for the usefulness of the strong topology is the “order completeness” property asserted in Vigier’s theorem which is analogous to the order completeness property of the order completeness property of the real numbers .
Henceforth, we shall be using results concerning locally convex spaces.
Let be a Hilbert space, and . Then the function
,
is a semi-norm on . The locally convex topology on generated by the separating family is called the strong topology on . Thus, a net converges strongly to an operator on iff for all . It follows that the strong topology is weaker than the norm topology on (This is a theorem in topology. Suppose that are topologies on a set . Suppose further that whenever is a net convergent with respect to the topology , then is also convergent with respect to the topology . Then — is weaker than .).
With respect to the strong topology, is a topological vector space, so the operations of addition and scalar multiplication are strongly continuous. This is not the case in general for the multiplication and involution operations.
Example 4.1.1
Let be an infinite-dimensional Hilbert space with an orthonormal basis . Set . If , then , so . Thus, the sequence is strongly convergent to to zero in . Now , so . Therefore for , so the sequence does not converge strongly to zero. This shows that the operation on is not strongly continuous, and therefore the strong and norm topologies on do not coincide.
Observe also that , so the sequence is not convergent to zero, and therefore the norm is not strongly continuous.
The operation of multiplication , is not strongly continuous either (Cf. Exercise 4.3).
The preceding example shows that the strong topology behaves badly in some respects, but it also has some very good qualities, as we shall prove in the next theorem.
Let be an arbitrary Hilbert space and suppose that is an increasing net in that converges strongly to (so also belongs to ). Then and (for ). The corresponding statement for decreasing nets in is also true. Both of these observations follow from the fact that if a net converges strongly to an operator , then ().
Theorem 4.1.1 (Vigier)
Let be a net of hermitian operators on a Hilbert space . Then is strongly convergent if it is increasing and bounded above, or if it is decreasing and bounded below.
Proof
We prove only the case where is increasing, since the decreasing case can be got from multiplying by minus 1.
Suppose then that is increasing and bounded above. By truncating the net (that is, by choosing an index and considering the truncated net ) we may suppose that is also bounded below, by say. We may further assume that all are positive (by considering the net if necessary). There is a positive number such that . It follows that the increasing net is bounded above by , so this net is convergent (by monotone convergence). Using the Polarisation Identity
we see that is a convergent net for all . Letting denote its limit, it is easy to check that the function
,
is a sesquilinear form on . Moreover,
,
so is bounded. Hence, there is an operator on such that for all (Theorem 2.3.6). Clearly (same theorem), is hermitian, and for all . Also
,
(the term comes from a triangle inequality) and so . Thus, converges strongly to
Remark 4.1.1
If is a net of projections on a Hilbert space strongly convergent to an operator , then is a projection. For is self-adjoint and , so .
Theorem 4.1.2
Suppose that is a net of projections on a Hilbert space .
- If is increasing, then it is strongly convergent to the projection of onto the closed vector subspace .
- If is decreasing, then it is strongly convergent to the projection of onto .
Proof
Left as an exercise
Just as for normed vector spaces, we say a family of elements of a locally convex space is summable to a point is the net (where runs over all non-empty finite subsets of ) is convergent to , and in this case we write .
Theorem 4.1.3
Let be a family of projections on a Hilbert space that are pairwise orthogonal (that is if .) Then is summable in the strong topology on to a projection , such that
.
If , then the map
, ,
is a unitary.
Proof
If is a finite non-empty subset of $latex\Lambda$, then is a projection. Therefore, is an increasing net of projections, hence strongly convergent to a projection ; that is, the family is strongly summable to . Moreover
.
The observation concerning the case when is clear
If is a subset of an algebra , we define its commutant to be the set of all elements of that commute with all the elements of . Observe that is a subalgebra of (straightforward). The double commutant of is . Similarly, . Always and (let . If then . That is . Now let so that for all , we have . Now for any , so commutes with everything in . That is . The text claims that in fact . Now let . Now , for all . However so therefore for all . That is also.). If is a normed algebra, then is closed (with more structure we could show this as follows. Define by . Now is closed as is .). If is a *-algebra and is self-adjoint, then is a *-subalgebra of (straightforward).
Lemma 4.1.4
Let be a Hilbert space and a *-subalgebra of containing . Then is strongly dense in .
Proof
Let , , and . Then is a closed vector subspace of which is invariant, and therefore reducing ( and invariant), for all , since is self-adjoint. Thus, if is the projection of onto , then , so . Hence , and therefore there is a sequence such that .
For each positive integer the map
, ,
is a unital *-homomorphism, so is a subalgebra of containing . Moreover, , for if and then . Hence , so . Therefore, . Suppose now that . Then by the first paragraph of this proof there is a sequence in such that . Hence, for .
We show that this implies that is in the strong closure of . If is a strong neighbourhood of , we must show that is non-empty, and to do this we may suppose that is a basic neighbourhood of . Therefore, there are elements and a positive number such that.
.
Hence, there is a sequence in such that
.
Consequently, for some the operator , so
Let be a Hilbert space. If is a strongly closed *-subalgebra of , we call a von Neumann algebra on . Since the strong topology is weaker than the norm topology, a strongly closed set is also norm closed. Hence, a von Neumann algebra is a C*-algebra.
Obviously, is a von Neumann algebra on , as is . If is a family of Hilbert spaces and is a von Neumann algebra on for each index , then it is an easy exercise to show that the direct sum is a von Neumann algebra on .
If is a *-algebra on a Hilbert space , then its commutant is a von Neumann algebra on (Let be a net in , strongly convergent to , then . This is a consequence of the fact that is strongly closed.)
If is a von Neumann algebra on and is a projection in , then is a von Neumann algebra on (. We want to show that this set is strongly closed. Let be a net in strongly convergent to . We want to show that for some in . Strong operator convergence implies that for any we have that . We know that , as is strongly closed. The rest follows through as multiplication is continuous). Also, is a von Neumann algebra on .
If is infinite-dimensional, then is not a von Neumann algebra. To see this, let be an orthonormal basis for , and for each finite non-empty subset of let . Then is a finite rank projection and the net converges strongly to on . If were a von Neumann algebra, this would imply that it contains , and so , contradicting our assumption on .
A fundamental result concerning von Neumann algebras is the following, known as the Double Commutant Theorem.
Theorem 4.1.5 (von Neumann’s Double Commutant Theorem)
Let be a *-algebra on a Hilbert space and suppose that . Then is a von Neumann algebra on iff .
Proof
Immediate from Lemma 4.1.4
The intersection of a family of von Neumann algebras on a Hilbert space is also a von Neumann algebra. Thus, for any set there is a smallest von Neumann algebra containing . We call the von Neumann algebra generated by . If is self-adjoint and contains , then . If in addition consists of commuting elements, then is abelian (for in this case ). This implies that there are non-trivial examples of abelian von Neumann algebras. We give an explicit example.
Example 4.1.2
Let be a compact Hausdorff space, and suppose that is a finite positive regular Borel measure on . We saw in Example 2.5.1 (https://jpmccarthymaths.wordpress.com/2011/01/19/c-algebras-and-operator-theory-2-5-the-spectral-theorem/) that the map
, ,
is an isometric *-homomorphism. It’s range is a C*-subalgebra of . Denote by the C*-subalgebra of all multiplication operators on with continuous symbol. The commutant of is . Hence, is a von Neumann algebra on the Hilbert space . Since (because is abelian) and , therefore . Consequently, , so is strongly dense in be Lemma 4.1.4.
Let be a closed vector subspace of a Hilbert space and let be the projection of onto . If , let be compression (restriction) of to . It is easy to verify that the map
,
is a *-isomorphism ().
If is a *-algebra on and is in , set .
Lemma 4.1.6
Let be a *-algebra on a Hilbert space , and a projection in . Then and are *-algebras on and , respectively, and the map
, ,
is a *-isomorphism. Moreover, if also , then .
Proof
We only show that — the rest is left to the exercises.
Suppose that and . Then there exist and and , respectively, such that and . Now we have
, and
,
so that which implies that . With respect to this same basis
,
and we have that ; equivalently
,
which implies that so that . Therefore , so .
Conversely, suppose now that , and write for some . Let
.
Then if ,
.
So we have
, and
.
But , so
commutes with .
This implies that so that . Consequently, , and therefore . This shows that the inclusion holds
The reader should be aware that some authors define a von Neumann algebra on a Hilbert space to be a *-algebra on such that . This automatically ensures that . However, proofs appear to run more smoothly if von Neumann algebras are defined as we have done. Moreover, we can frequently reduce to the case where , by the trick explained in Remark 4.1.2. Using our definition von Neumann algebras are still unitial, but the unit may not be the identity of the underlying Hilbert space.
Theorem 4.1.7
If is a non-zero von Neumann algebra, then it is unital.
Proof
Suppose that acts on a Hilbert space , and let be an approximate unit for . By Theorem 4.1.1, converges strongly to a self-adjoint operator, say, and obviously , since is strongly closed. If and , then , so . Hence, is a unit for
Remark 4.1.2
Let be a von Neumann algebra on a Hilbert space and let be the unit of . Of course, is a projection in . The map
, ,
is a *-isomorphism (by Lemma 4.1.6), and is a von Neumann algebra on containing , so . This device will be frequently used to reduce to the case where the von Neumann algebra is it’s own double commutant (we don’t necessarily have if ).
If is an operator on a Hilbert space , then its range projection is the projection of on . We have , since ( First equality. Suppose . That is for all we have . The rest by the polar decomposition of ).
Theorem 4.1.9
If is a von Neumann algebra, then it contains the range projections of all of its elements.
Proof
Let act on and let . Since , to show that the range projection of is in , we may suppose that . Obviously we may also suppose that . Let for . Then is an increasing sequence of positive elements in the closed unit ball of , so by Theorem 4.1.1, is strongly convergent to a positive operator, say. If , then
Therefore, converges to strongly. But for all so .
The sequence is in , so . The continuous functions
, ,
form an increasing sequence and converge pointwise to the identity function , so by Dini’s Theorem, they converge uniformly. Therefore, by the functional calculus, ; that is . Hence, , so . Therefore
Theorem 4.1.10
Let be a von Neumann algebra on a Hilbert space and and element of with polar decomposition . Then .
Proof
Let be a unitary in the unital C*-algebra . Then is a partial isometry on such that and (The projection onto the initial space of is
But the initial space of is , and this projection lies in , so
that is the initial space of is the initial space of . Therefore ). It follows, therefore, from the uniqueness of the polar decomposition that , so and commute (). But is the linear span of its isometries (Suppose that a C*-algebra is unital and let . Then is a set of unitaries that spans that space.), so must commute with all elements of , and therefore . By Lemma 4.1.4 there is a net and a net in such that the net converges strongly to on . If , then by Theorem 4.1.9 . Since ( by Theorem 2.3.4), we have ; that is, . Therefore, is the strong limit of the net which lies in , so
Theorem 4.1.11
Suppose that is a von Neumann algebra on a Hilbert space .
- is the closed linear span of its projections.
- If and is a normal element of , then for every Borel set of , where is the spectral resolution of the identity for .
- If and , then if and only if commutes with all the projections in .
Proof
We may suppose in all cases that (certainly in conclusions 2 and 3. I imagine we will be adapting Remark 4.1.2 to conclusion 1. if . ). We prove condition 2. first. Let be a normal element with spectral resolution of the identity for denoted by . If , then and (as is self-adjoint — take adjoints to ), so for every (go on the Stone-Weierstrass!). In particular, for every Borel set of . Therefore, .
Condition 1. follows directly from condition 2., using the fact the closed linear span of the characteristic functions is for each normal element of .
Condition 3. follows immediately from condition 1., since is a von Neumann algebra, and therefore it is the closed linear span of its projections
We give an immediate and important application of this result in the next theorem. First we make some observations.
Remark 4.1.3
If is a Hilbert space, then . For it is obvious that , and since is a von Neumann algebra containing , Theorem 4.1.5 implies that , so .
Remark 4.1.4
If is a C*-algebra acting on a Hilbert space and , consider the set .
We say acts non-degerately on if . Equivalently, for each non-zero there exists such that .
If acts non-degenerately on and is an approximate unit for , then converges to strongly. To show this we must prove that , for all . This is clear if for some and (because — we can fill in the gaps by continuity). Now
If acts irreducibly on and , then it acts non-degenerately, since is a non-zero vector subspace of invariant for , and therefore equals ( acting irreducibly on means that the only closed vector subspaces of are and ).
Theorem 4.1.12
Let be a non-zero C*-algebra acting on a Hilbert space . The following conditions are equivalent:
- acts irreducibly on .
- .
- is strongly dense in .
Proof
If is a projection in , then if and only if the closed vector subspace of is invariant for . Since is a von Neumann algebra, it is the closed linear span of its projections by Theorem 4.1.11, so if acts irreducibly , then has no projections except the trivial ones, and therefore . Therefore 1. implies 2. The reverse implication is clear.
If we suppose now that , then (), and therefore . By Lemma 4.1.4 the C*-algebra is strongly dense in . However, since acts irreducibly on , it acts non-degenerately, and therefore if is an approximate unit for , it is strongly convergent to . Therefore, , and , where the closure is strong closure (). Hence 2. implies 3.
Finally, if is strongly dense in , then , so 3. implies 2.
If are projections in a C*-algebra , we say they are Murray-von Neumann equivalent, and we write , if there exists such that and . It is indeed straightforward to show that this is indeed an equivalence relation on the set of projections of ( so is reflexive. Clearly is symmetric. Now suppose and ; and and . I don’t think I need to full result actually…). The relation is of fundamental importance in the classification theory of von Neumann algebras (see the Addenda), and in -theory for C*algebras. However, for the moment we only need one small result concerning .
Remark 4.1.5
If are infinite-rank projections on a separable Hilbert space , then (I can see how this goes back to the talk on .). To see this, choose orthonormal basis and for and . Let be the unitary such that for all . Define by setting on and on . It is easily verified that and .
Theorem 4.1.15
If is a separable infinite-dimensional Hilbert space, then is the unique non-trivial closed ideal of .
Proof
Let be a non-zero closed ideal of . By Theorems 2.4.5 and 2.4.7, . Now if , then by Corollary 4.1.14 (not covered as it uses hereditary C*-algebras), does not contain all of the projections of . Hence, has an infinite-rank projection, say. If is any other infinite-rank projection on , then as we say in Remark 4.1.5 there is an element such that and . Therefore, , so also belongs to . Hence, contains all the projections of , whether their rank is finite or infinite, and therefore . Thus, we have shown that the only closed ideals of are , and
Remark 4.1.6
Let be a separable infinite-dimensional Hilbert space. If is a dense sequence on , it is easy to check that is the closed linear span of the operators () using the fact that is the closed linear span of the rank-one projections (Theorems 2.4.5 and 2.4.6). Hence is separable. However, is non-separable (and therefore is non-separable). To see this, choose an orthonormal basis for . For each set of positive integers, let be the projection onto . Clearly, if (). Hence, the family of operators cannot be in the closure of the range of any sequence in .
Theorem 4.1.16
If is an infinite-dimensional separable Hilbert space, then the Calkin algebra is a simple C*-algebra.
Proof
Let denote the Calkin algebra and the quotient map. To see that is simple, let be a closed ideal in . Then is a closed ideal in (as the quotient map is a homomorphism), and therefore by Theorem 4.1.15, or . Hence, or
Remark 4.1.7
Let be operators on a Hilbert space such that . Then there is an operator such that . To see this, observe that
.
Hence, we get a well-defined norm-decreasing linear map by setting . Clearly, we can extend to a bounded linear on . Setting , we get the required result
3 comments
Comments feed for this article
July 6, 2011 at 3:33 pm
Von Neumann Algebras: The Weak and Ultraweak Topologies « J.P. McCarthy: Math Page
[…] the conjugate map which is continuous (is this O.K.?) to get .). We showed in Example 4.1.1 (http://irishjip.wordpress.com/2011/06/24/von-neumann-algebras-the-double-commutant-theorem/) that the involution is not strongly continuous in general, so the weak and strong topologies do […]
July 11, 2011 at 2:54 pm
Von Neumann Algebras: The Kaplansky Density Theorem « J.P. McCarthy: Math Page
[…] first that is a C*-algebra on . By applying Remark 4.1.2, we may suppose that […]
October 5, 2011 at 2:13 pm
Representations of C*-Algebras: Irreducible Representations and Pure States « J.P. McCarthy: Math Page
[…] Proof : Condition 1. is immediate from Theorem 4.1.12. […]