### Introduction to von Neumann algebras, Lecture 7 (von Neumann algebras as dual spaces, various topologies)

#### by Orr Shalit

Until this point in the course, we concentrated on constructions of von Neumann algebras, examples, and properties of von Neumann algebras *as algebras*. In this lecture we turn to study subtler topological and Banach-space theoretic aspects of von Neumann algebras. We begin by showing that every von Neumann algebra is the Banach-space dual of a Banach space. For this to have any hope of being true, it must be true for the von Neumann algebra ; we therefore look there first.

(The reference for this lecture is mostly Takesaki, Vol. I, Chapters 2 and 3).

#### 1. Trace class operators and as the double dual of the compacts

Fix a Hilbert space (no need to worry about dimension; on the other hand, even finite dimensional spaces are interesting). In this lecture we will write to denote the space of all bounded functionals on a normed space . Even though our normed spaces will be linear subspaces of , we will not have an opportunity to be confused regarding whether the denotes the adjoint operator.

For every we define

- A linear functional given by .
- A linear operator given by .

It customary to also write , for reasons which I hope are obvious (or by “physicists”). Both and are sometimes denotes by .

Let denote the norm closed two sided ideal of compact operators in . Every vector functional can also be considered as an element of . We shall see below that in a natural way (where “in a natural way” is meant in a loose way). Let us begin by taking a closer look at compact operators.

**Lemma 1:** *For every , there exist two orthonormal sequences and a sequence of positive numbers that is either finite or convergent to zero, such that is given as the norm convergent sum *

.

*If , then has the form *

.

**Proof: **For self-adjoint operators, this is simply the spectral theorem for compact self-adjoint operators (see here for the formulation I am using). The result for positive operators follows at once.

For a general compact operator , let be the polar decomposition of . Then is compact, and . Then

,

so putting (and recalling that is a partial isometry with ), we are done.

Now let us consider a functional . Since for every , the rank one operator is compact, we can apply to . This gives rise to a sesqui-linear form

for .

(For example, . )

Now, , so is bounded. It follows from a familiar consequence of the Riesz representation theorem (see Exercise A below) that there is some such that

for all ,

or, in other words,

for all .

If you never encountered the “familiar consequence of Riesz theorem” that we used above, then prove it (now, please!).

**Exercise A:** Let be a Hilbert space, and let be a bounded sesqui-linear form, meaning that there is some such that for all and all ,

- ,
- ,
- .

Prove that there exists with , such that , for all .

Note that , so that the map is a bounded linear map. Moreover, it is not hard to see that this map is injective (as a map from to ).

**Lemma 2:** . Consequently, if , are orthonormal sequences, and , then ; moreover, is the unique element in mapped to .

**Proof:** We check

,

while, on the other hand,

.

It follows that . The remainder follows from the above remarks, that is a bounded, linear and injective map.

Now suppose that is finite dimensional. Then

.

Therefore (by linearity), for all . In other words, every linear functional on is given by , for some . Our discussion below will show that this is also true, in the appropriate sense, when has infinite dimension; however, not every will give rise to a bounded operator. The ones that do are said to be the * trace class operators*, and corresponds precisely to .

**Exercise B: **If , and , then define . Prive that .

**Lemma 3:** *For all , the operator is compact. Moreover, for every orthonormal sequence , *

* . *

*In fact, . *

**Proof:** For all , let us write for some complex number of absolute value one. Then for every finite set of indices ,

.

This shows that the series converges.

To show that is compact, we assume that it is a positive operator, and leave it to the reader to reduce the general case to this one.

Now, suppose that (note that in this case, the first part of the proof shows that can be supported on at most countably many basis vectors; we may therefore assume that is separable. This is not crucial.)

By the spectral theorem (Lecture 1), we may assume that is a multiplication operator for . To prove that is compact, it suffices to prove that the spectral measure of satisfies that (which is ) has finite rank for all (recall Exercise M in Lecture 1). But on the range of , so for all in the range; if the range was infinite dimensional this would contradict the summability of the series established above.

**Exercise C**: Complete the above proof, by showing that is compact in the general case (use Exercise A).

Now we can compute . For every , it is clear that , and in particular it restricts to a bounded functional on . Moreover, since , it certainly holds that for every sequence and every two orthonormal sequences , the series

converges to a bounded functional in and in . Now we shall see that all bounded functionals of have this form.

**Theorem 4: ***Let . Then there is a **sequence and two orthonormal sequences , such that *

*.*

**Proof:** From what we have gathered until now, we know that there is an operator such that

for all . By Lemma 3, . By Lemma 1, for a series of positive numbers that is either finitely supported or converges to zero, and two orthonormal sequences. For every finite , we can define . Then is a compact operator, and

.

But . We see that .

**Definition 5: **Let denote the space of all bounded operators for which there is some such that . Equivalently, is the space of all such that

for any (or every) orthonormal sequence . The space is called the space of * trace class operators*. For every we define the

*to be the norm of the linear functional for which .*

**trace norm****Remark 6: **By Theorem 4, every extends naturally in a unique way to a bounded functional on (the extension is unique only among “natural” extensions). For every , if satisfies , we write , and more generally, for any we write . Recall that in functional analysis, one writes the “pairing” between a space and its dual as

for .

In our setting, we identify with , and if , we write

.

**Exercise D:** Prove that .

**Exercise E: **Every is WOT continuous on , and hence extends to a unique such functional on .

Every (where and are orthonormal sequences) extends to be a bounded linear functional on , defined by . Therefore, every determines a bounded linear functional on by

.

The map is called * the canonical map*.

**Theorem 7: ***The canonical map is an isometric isomorphism of onto . *

**Proof: **For every pair of unit vectors , the functional . This can be used to show that the canonical map is an injective map, an in fact it is norm-nondecreasing:

which approximates .

On the other hand, if , then for any net of finite rank projections increasing to , so , therefore the canonical map is isometric.

Now if , then we define a bounded sesqui-linear form on :

.

One finds that there is some bounded operator , such that

for all .

Since the linear span of the functionals of the form is dense in , we see that is the image of under the canonical map.

We henceforth identify with , and with . It is also common to denote , and to refer to it as * the predual* of (an alert student should worry about the word “the”; until we show that the predual of a von Neumann algebra is unique, we can refer to it more precisely as

*or, to be completely on the safe side,*

**the canonical predual**

**the****).**

*standard predual***Exercise F: (Tying all loose ends) **Prove that for every positive ,

,

where is an orthonormal basis for . Show that the right hand sum is independent of the particular orthonormal basis. (Recall that we defined to be the norm of the corresponding functional in ). For a not-necessarily positive , prove that , and that

Converges absolutely, and to the same value, for every choice of orthonormal basis . Prove that is an ideal in . Prove that for every and ,

.

#### 2. The -weak operator, the -strong operator topologies, and more

Since , we can consider the weak-* topology on it.

**Definition 8:** The * -weak operator topology* on (or just

*, or*

**-weak***) is the weak-* topology induced on by its predual .*

**ultraweak topology**Thus, in the -weak operator topology, if and only if

for all .

Equivalently, the -weak topology is determined by the seminorms

where satisfy .

(Recall that a topology is said to be * generated by a family of seminorms* if convergence of a net is determined by convergence for all . Thus, the strong (operator) topology is the topology generated by the family of seminorms , .)

**Definition 9:** The * -strong operator topology *(or simply the -strong topology) is the topology generated by the seminorms

,

where .

**Definition 10:** The strong * topology is the topology generated by the seminorms , . The -strong * topology is defined similarly.

**Exercise G:** The *whatever*-strong topology is strictly stronger than the *whatever*-weak topology. The –*whatever* topology is strictly stronger than the *whatever* topology, but they coincide on the unit ball . The *whatever* * topology is strictly stronger than the *whatever* topology. All are strictly weaker than the norm topology.

****

**Theorem 11:** *Let be a von Neumann algebra. Then is closed in all of the above topologies. Consequently, is a dual Banach space. To be precise, if we let denote the subspace of consisting of all -weakly continuous functionals, then can be isometrically isomorphically identified with . *

**Proof:** Since is WOT closed, it is closed in the weakest, and hence in all, of the topologies. To see that is a dual space, we consider it as a weak-* closed subspace of . Let

for all .

Then, being weak-* closed,

for all .

(see Proposition 13 in this old lecture). But by standard results on dual spaces (see Theorem 7 in this old lecture),

by a natural map (where is the quotient map). Thus

,

and it remains to observe that this isomorphism respects the -weak functionals. For this, note that the restriction map , given by , induces an isomorphism of with its image – the weak-* functionals on .

#### 3. The second dual and the universal enveloping von Neumann algebra of a C*-algebra

We saw above that . In this section, we will see that the double dual of any C*-algebra is a von Neumann algebra. In contrast with everything we have done until this point, our C*-algebras will be just abstract C*-algebras: Banach *-algebras that satisfy the C*-identity: . We will use basic results of the theory (with due apologies) when we need them.

**Lemma 12: ***Let be a C*-algebra, let be a *-representation, and define . Let be the canonical embedding. Then there exists a unique linear map , which is surjective, continuous with respect to the and -weak topologies, which extends (in the sense that ). Moreover, maps the unit ball of onto the unit ball of . *

**Proof: **Consider as a map between the Banach spaces . Let be the adjoint map. Let be the restriction of to the space of -weakly continuous functionals. So . Now define . This satisfies for all , that

.

This shows that . The continuity is a “general nonsense” fact which always holds for adjoints: indeed, if in , then for all .

Finally, to show the map is surjective, it suffices to show that it takes the closed unit ball onto the closed unit ball . Since is continuous, it takes onto a -weakly compact set. But this compact set contains the open unit ball of (because that’s what *-homomorphisms do, being the composition of a quotient and an injective (hence isometric) *-homomorphism). But by Kaplansky (and Exercise F) the open unit ball of is -weakly dense in .

Let be a unital C*-algebra. Now we recall *the universal representation*

,

where (here, given a state , is the GNS representation of ). Since is an isometric isomorphism (by the Gelfand-Naimark theorem), we can identity with and consider as a C*-subalgebra of .

**Definition 13:** The algebra is called **the universal enveloping von Neumann algebra of . **

**Theorem 14:** *The map (given by Lemma 12) is isometric, hence . Every bounded functional on extends to a -weakly continuous continuous functional on . The universal enveloping von Neumann algebra has the following universal property: if is a *-representation, then there exists a unique -weakly continuous of onto such that . *

**Proof: **For simplicity of notation, put , , and we shall use the notation of Lemma 12 and its proof. By construction, every state on extends to a vector state on . By Exercise H below, every bounded functional is the linear combination of four states, and from this it is easy to show every functional extends to a vector functional, and therefore extends -weakly.

We will now show that is surjective, this will show that the map is injective, and since it maps an open unit ball onto an open unit ball it must be isometric and surjective. To see that is surjective, we just recall that it is defined to be . But is the restriction map given by (being the conjugate of an inclusion ), and since every extends to a map in (by the previous paragraph), this shows that is surjective.

Finally, given , let be as in Lemma 12. Then is a surjective and -weakly continuous linear map. Since it is a *-homomorphism on the -dense subspace , it is a representation. We get , and restricting to we get the final assertion.

**Exercise H: **Prove that every is the linear combination of four states. Conclude that every can be written as , for some representation and .

**Exercise I:** Prove that if is a *-representation, , and if have the same universal property as , then is -weakly continuously *-isomorphic to by a map that fixes .