Introduction to von Neumann algebras, Lecture 2 (Definitions, the double commutant theorem, etc.)
by Orr Shalit
In this second lecture we start a systematic study of von Neumann algebras.
1. Some topologies on
On we have the topology generated by the operator norm. There are other natural topologies that may be used.
Definition 1: The weak operator topology (WOT, sometimes also called the weak topology) is the topology on generated by the sets
where , and (recall that this means that a basis for this topology is given by finite intersections of the above sets). In fact, it suffices to consider sets of the form (why?) Equivalently, a net converges to in the weak operator topology if and only if for all .
Remark: It is common to call the weak operator topology “the weak topology”, for short (and also because that’s how von Neumann called it). Since is a Banach space, the term “weak topology” can also mean the topology generated on by the space of bounded functionals on ; however, that weak topology is rarely used, so this abuse of terminology hardly ever causes confusion .
Definition 2: The strong operator topology (SOT, sometimes also called the strong topology) is the topology on generated by the sets
where , and . As above, it suffices to consider sets of the form . Equivalently, a net converges to in the strong operator topology if and only if (in the norm) for all .
Remark: Once again, the terminology “strong topology” is used in Banach space theory to refer to the norm topology, in contrast to the weak topology. However, in our setting, since “weak topology” does not mean the topology on induced by the Banach space dual , “strong topology” is usually not used for the norm topology, but for the strong operator topology.
A beginner might be tempted to think that, since the norm topology comes from a norm and in particular is metric space topology, it should be more convenient to work with than the above two topologies (it turns out that the weak and strong topologies are not metrizable). Why would one want to work with a weaker topology? One reason is that a closed set has better chances of being compact in a weaker topology . Another reason is that closures are bigger, hence richer. For example, we have already seen that the spectral projections of a selfadjoint operator , are not in the norm closed algebra generated by , but rather in the weak/strong closed algebra generated by (these turn out to be the same, see below).
There are other natural topologies, we shall encounter them later.
It is easy to see that the weak and strong topologies are Hausdorff, and that addition and scalar multiplication are jointly continuous. If , then the maps and are WOT and SOT continuous. The adjoint operation is WOT continuous. Indeed, if wot, then
However, the adjoint operation is not SOT continuous. For example, if denotes the unilateral shift on (that is, ), then converges SOT to as , but does not.
Exercise A: Prove that the strong topology is strictly stronger than the weak topology.
Multiplication is not jointly continuous, WOT nor SOT, but it is SOT continuous when restricted to bounded sets.
Exercise B: Show that multiplication is not jointly SOT continuous (this requires finding a tricky example). However, show that the map given by is SOT continuous (here, denotes the closed unit ball of ). What about joint WOT continuity of multiplication?
Exercise C: Explain why the previous exercise implies that the strong topology is not metrizable. What about the weak topology – is it metrizable?
Theorem 3: The closed unit ball of is WOT compact.
Proof: The proof is similar to the proof of Alaoglu’s theorem. For every , define
Let , equipped with the product topology. By Tychonoff’s theorem, is compact. We consider as a subspace of the space of all functions , equipped with the topology of pointwise convergence, which is the same thing as the product topology.
Now let be defined by
is injective, and by definition, is a homeomorphism between with the weak operator topology and with the topology of pointwise convergence. To show that is compact, it suffices to show that is closed.
Now, if , then it is not hard to show that must be a sesqui-linear form on and it is bounded in the sense that . It follows from a familiar consequence of the Riesz representation theorem that there is some , such that for all , therefore , as required.
Exercise D: Show that with the strong topology is not compact.
Exercise E: If is separable, then is metrizable in both the weak and the strong operator topologies. The metric for with the weak topology is
where is a dense sequence in the unit ball of . Fill in the rest of the details.
2. The double commutant theorem
We recall some definitions, and give a couple of new ones.
Definition 4: A -subalgebra of that is SOT closed and contains is called a von Neumann algebra.
Definition 5: For a set , the commutant of is the set defined
for all .
Exercise F: Prove that and that .
Definition 6: The null space of a subset is the set
Definition 7: A subset is said to be non-degenerate if (where denote the closed linear span of a set ).
Exercise G: A -algebra is non-degenerate if and only if .
Theorem 8 (von Neumann’s double commutant theorem): Let be a -subalgebra with a trivial null space. Then
In particular, a -algebra is a von Neumann algebra if and only if .
Before the proof, we will prove a basic lemma, which will also emphasize an important role that the commutant plays.
Lemma: Let be a -subalgebra, let be an orthogonal projection, and set . Then is invariant for if and only if .
Proof: For , if and only if for all , which happens if and only if . Therefore, if and only if , or . Therefore, if is invariant for , then for every , the adjoint too, so , therefore . This argument works backwards as well.
Proof of the double commutant theorem: As is WOT closed, it holds that
Now let . To show that , we have to find, for every , an operator such that for all .
Case 1: . Let be the orthogonal projection onto the subspace . Then by the lemma, . Now, , so the assumption implies that , or .
Now since and , we have
This means that there is some such that .
Case 2: . Define ( times), and for every operator , define in (the last equality should be clear, if it is not STOP NOW!).
Put . Then is a subalgebra of . A calculation shows that , and that .
Now put , where . Applying Case 1 to the algebra and the vector , we find such that , and this implies for all .
Corollary 9: A unital -subalgebra is a von Neumann algebra if and only if .
Exercise H: Prove that every von Neumann algebra is the commutant of the image of a unitary representation.
Corollary 10: Let , and suppose that is its polar decomposition. Then both and are in (the von Neumann algebra generated by ).
Proof: We have already seen that . To show that , we will show that commutes with every (here we are using the double commutant theorem). Given such , we have
It follows that for every . To finish the proof we need to show that for every . By the definition of , for , so . On the other hand, if , then , so . Thus for , and we are done.
Example: If , then , and . More generally, if , then , and .
The first equality follows from the fact that has no invariant subspaces, thus the only projections in are and . Since a von Neumann algebra is generated by its projections, .
Similarly, if (the algebra of compact operators), then and (that the SOT/WOT closure of is is also easy to prove directly, but note how it just falls out of the air here).
Example: Let be a probability space, and consider the algebra , where we are using the notation introduced in the previous lecture:
If you check carefully the notes of the previous lecture, you will see that we never really proved that this algebra is SOT closed. We shall now see that , and it will follow that is a von Neumann algebra. It is common to abuse notation and write .
Of course, because is commutative. Let . If was in , then and we would be able to recover as . Here, denotes the constant function with value everywhere on , which is in because is a probability measure. We therefore define , and hope to be able to show that – originally only known to be in – is bounded, and that . But this is easy: for every function ,
therefore . It follows (for example, by considering characteristic functions of sets of finite measure) that , and the equation above shows that agrees with on the space , which is dense in . Since both operators are bounded, they are equal.
Surely one can prove directly that is a strongly closed algebra, but the above simple computation shows the beauty and power of the double commutant theorem: an analytical problem (involving, perhaps, the limits of convergent unbounded nets) is reduced to a rather simple minded algebraic problem: computing the commutant.
Example: Suppose that is a locally compact Hausdorff space, and that is a regular Borel measure on . Identify the algebra of continuous, compactly supported functions on , with the algebra of operators . Using an argument similar to the one above, one can show that .
Exercise I: Prove that the von Neumann algebra generated by a selfadjoint operator that has a cyclic vector, is unitarily equivalent to , where is a probability measure on . Give an example of an operator (necessarily, non-cyclic) such that the von Neumann algebra that it generates is not unitarily equivalent to (for any probability space ). Your example is still related to – how so?
3. Existence of limits of monotone nets
Theorem 11: Let be an increasing net of selfadjoints in a von Neumann algebra . Suppose that this net is bounded above, in the sense that there is some some operator such that for all . Then is SOT convergent to some , and is the least upper bound for in . In case where every is a projection, then is the projection onto the closure of the union of the ranges of .
Proof: For every ,
and the left hand side is a bounded increasing net of real numbers; it therefore has a limit , which we write as
Now if we define
Then , so it is a bounded sesquilinear form. It follows from a familiar consequence of Riesz theorem that there exists a such that for all . It easily follows that , and that for all . Moreover, if for all , then , for all , so . We therefore have that .
SOT convergence is trickier. Without loss of generality, let us assume that for all (if not, then just consider the net for some fixed index ).
Now , so .
where the inequality is justified by the following claim.
Claim: For every positive operator , it holds that , in the sense that for all .
Proof of claim: This follows immediately from the functional calculus, because for a bounded nonnegative function , it clearly holds that almost everywhere.
That concludes the proof of the claim, so we have that in the SOT.
The final assertion (regarding projections) is left as an exercise.
Exercise J: Prove that if is a bounded increasing net of projections, then is the projection onto the union of the ranges of the .
Remark: Here are another two ways to see that . First, recall that if and if is any operator, then . Indeed:
Now, we simply apply this to , and :
But, by our characterization of the norm of a selfadjoint operator, together with the spectral mapping theorem
4. Kaplansky’s density theorem(s)
By the double commutant theorem, if , then there is a net such that .
Technical problem: Convergent nets may be unbounded.
Technical solution: Kaplansky’s density theorem below shows that one may assume that the net is bounded by .
Exercise K: Use Kaplansky’s density theorem (Theorem 14 below) to prove directly that is SOT closed (“directly” means without making use of the double commutant theorem).
Proposition 12: Let be a convex set. Then, the WOT and SOT closures of coincide.
Remark: Recall that in a normed space, a convex set is weakly closed if and only if it is closed in the norm, hence the weak and norm closures of a convex set are the same (short explanation: every weakly closed set is strongly closed, and conversely, if a strongly closed set is convex, then, by the Hahn-Banach theorem, is equal to the intersection of closed half spaces, hence it is weakly closed as the intersection of weakly closed sets).
Proof of Proposition 12: Clearly, . Let . To show that , we fix , and find such that for all – this will show that there is a point for every open set of the form , whence .
Assume that , otherwise repeat the trick from the proof of the double commutant theorem. Write . Now, , so it follows that
where the last equality follows from the remark preceding the proof. Thus, by definition, there is some such that , as required.
From general functional analysis (in a locally convex topological vector space, a linear functional is continuous if and only if its kernel is closed) we obtain:
Corollary 13: The SOT and the WOT have the same continuous linear functionals.
Theorem 14 (Kaplansky’s density theorem): Let be a non-degenerate *-subalgebra of . Then
- (i.e., the closed unit ball of is SOT dense in the closed unit ball of ).
- If is a unital C*-algebra, then .
Proof: The heart of the matter is to prove item 2: that is, to prove . This done, the rest of the assertions follow by various tricks or by adapting the arguments. For example, suppose that we know that holds for every . To prove that every is the strong limit of a net of elements , we define
Then we know that there is some bounded net of selfadjoints
converging to , and it follows that and that .
Thus we turn to the heart of the matter, which is proving that . We may assume that is closed in the norm. Using Proposition 12, it is easy to prove that (Hey there! If you don’t see why Proposition 12 is needed, then make a note to return to this point at the end of the proof) . We therefore assume that and , and we will find a net of selfadjoint contractions in that converges to .
By the double commutant theorem, there is a net such that in the SOT, and therefore WOT. Since the adjoint is WOT continuous, in the WOT, and therefore weakly. The elements of this sequence are selfadjoint, thus,
where the last equality follows from Proposition 12. It follows that is the SOT limit of a net in , say . The net might still be missing the crucial property . To fix this, we need the following technical lemma.
Lemma: For every , and every net , if , then .
The lemma will be proved below; in the meanwhile, let be defined by if is in the interval , and otherwise . Then by the functional calculus, are selfadjoint contractions, and . By the lemma, on the other hand, . This shows that , and the proof of assertion 2 is complete.
Exercise L: Complete the proof of Theorem 14.
Proof of the lemma: For the span of this proof, let us agree that all functions are real valued. Let us say that a function is strongly continuous if implies for every net of selfadjoint operators . Let us define
is strongly continuous .
If and , then , because multiplication is SOT continuous on . In Exercise K below, you will be asked to prove that if is a sequence of strongly continuous functions, and if converge uniformly to , then is also strongly continuous. It follows that is a norm closed subalgebra of . Thus, to prove that , all that we have to do is to find enough functions in to separate points, and then apply the (locally compact version of the) Stone-Weierstrass theorem.
We will finish the proof by showing that and that are in . Since , and using our observation above on the product of strongly continuous functions, we just have to deal with . If , then
Now let converge SOT to . Then for every ,
and both summands on the RHS tend to zero. For example, to see that
we set , and then we obtain that because SOT. But then , so the whole expression converges to .
Exercise M: Prove that if is a sequence of strongly continuous functions, and if converge uniformly to , then is also strongly continuous.
Corollary 15: A non-degenerate *-algebra is a von Neumann algebra if and only if the unit ball of is WOT compact.
5. Additional exercises
Definitions: A family of operators is said to be topologically irreducible when and are the only closed subspaces invariant under the action of all . The family is said to be algebraically irreducible if for every and every , there exists such that such that .
Exercise N: Let . Prove that is topologically irreducible if and only if (equivalently, if and only if ).
Exercise O: Prove that a C*-algebra is topologically irreducible if and only if it is algebraically irreducible.
Thanks to the result of the above exercise, neither the terminology “topologically irreducible” nor “algebraically irreducible” is used; a C*-algebra satisfying either requirement is simply said to be irreducible.
Exercise P: Generalize the result in Exercise O in the following way: if a C*-algebra is irreducible, then for any and there exists such that for all . Moreover if there exists a selfadjoint (respectively, unitary) operator such that for all , then there is a selfadjoint (respectively, unitary) operator such that for all .
Exercise O: Prove Corollary 15.