5b. Von Neumann algebras
Instead of further building on the above results, which are already quite non-trivial, let us return to our modest status of apprentice operator algebraists, and declare ourselves rather unsatisfied with Definition 5.1, on the following intuitive grounds: \begin{thought} Our assumption that [math]A\subset B(H)[/math] is norm closed is not satisfying, because we would like [math]A[/math] to be stable under polar decomposition, under taking spectral projections, and more generally, under measurable functional calculus. \end{thought} Here all these “defects” are best visible in the context of Theorem 5.3, with the algebra [math]A=C(X)[/math] found there, with [math]X=\sigma(T)[/math], being obviously too small. In fact, Theorem 5.3 teaches us that, when looking for a fix, we should look for a weaker topology on [math]B(H)[/math], as for the algebra [math]A= \lt T \gt [/math] generated by a normal operator to be [math]A=L^\infty(X)[/math].
So, let us get now into this, topologies on [math]B(H)[/math], and fine-tunings of Definition 5.1, based on them. The result that we will need, which is elementary, is as follows:
For a subalgebra [math]A\subset B(H)[/math], the following are equivalent:
- [math]A[/math] is closed under the weak operator topology, making each of the linear maps [math]T\to \lt Tx,y \gt [/math] continuous.
- [math]A[/math] is closed under the strong operator topology, making each of the linear maps [math]T\to Tx[/math] continuous.
In the case where these conditions are satisfied, [math]A[/math] is closed under the norm topology.
There are several statements here, the proof being as follows:
(1) It is clear that the norm topology is stronger than the strong operator topology, which is in turn stronger than the weak operator topology. At the level of the subsets [math]S\subset B(H)[/math] which are closed things get reversed, in the sense that weakly closed implies strongly closed, which in turn implies norm closed. Thus, we are left with proving that for any algebra [math]A\subset B(H)[/math], strongly closed implies weakly closed.
(2) Consider the Hilbert space obtained by summing [math]n[/math] times [math]H[/math] with itself:
The operators over [math]K[/math] can be regarded as being square matrices with entries in [math]B(H)[/math], and in particular, we have a representation [math]\pi:B(H)\to B(K)[/math], as follows:
Assume now that we are given an operator [math]T\in\bar{A}[/math], with the bar denoting the weak closure. We have then, by using the Hahn-Banach theorem, for any [math]x\in K[/math]:
Now observe that the last formula tells us that for any [math]x=(x_1,\ldots,x_n)[/math], and any [math]\varepsilon \gt 0[/math], we can find [math]S\in A[/math] such that the following holds, for any [math]i[/math]:
Thus [math]T[/math] belongs to the strong operator closure of [math]A[/math], as desired.
Observe that in the above the terminology is a bit confusing, because the norm topology is stronger than the strong operator topology. As a solution, we agree to call the norm topology “strong”, and the weak and strong operator topologies “weak”, whenever these two topologies coincide. With this convention made, the algebras [math]A\subset B(H)[/math] in Proposition 5.7 are those which are weakly closed. Thus, we can now formulate:
A von Neumann algebra is an operator algebra
These algebras will be our main objects of study, in what follows. As basic examples, we have the algebra [math]B(H)[/math] itself, then the singly generated algebras, [math]A= \lt T \gt [/math] with [math]T\in B(H)[/math], and then the multiply generated algebras, [math]A= \lt T_i \gt [/math] with [math]T_i\in B(H)[/math]. But for the moment, let us keep things simple, and build directly on Definition 5.8, by using basic functional analysis methods. We will need the following key result:
For an operator algebra [math]A\subset B(H)[/math], we have
We can prove this by double inclusion, as follows:
“[math]\supset[/math]” Since any operator commutes with the operators that it commutes with, we have a trivial inclusion [math]S\subset S''[/math], valid for any set [math]S\subset B(H)[/math]. In particular, we have:
Our claim now is that the algebra [math]A''[/math] is closed, with respect to the strong operator topology. Indeed, assuming that we have [math]T_i\to T[/math] in this topology, we have:
Thus our claim is proved, and together with Proposition 5.7, which allows us to pass from the strong to the weak operator topology, this gives [math]\bar{A}\subset A''[/math], as desired.
“[math]\subset[/math]” Here we must prove that we have the following implication, valid for any [math]T\in B(H)[/math], with the bar denoting as usual the weak operator closure:
For this purpose, we use the same amplification trick as in the proof of Proposition 5.7. Consider the Hilbert space obtained by summing [math]n[/math] times [math]H[/math] with itself:
The operators over [math]K[/math] can be regarded as being square matrices with entries in [math]B(H)[/math], and in particular, we have a representation [math]\pi:B(H)\to B(K)[/math], as follows:
The idea will be that of doing the computations in this representation. First, in this representation, the image of our algebra [math]A\subset B(H)[/math] is given by:
We can compute the commutant of this image, exactly as in the usual scalar matrix case, and we obtain the following formula:
We conclude from this that, given an operator [math]T\in A''[/math] as above, we have:
In other words, the conclusion of all this is that we have:
Now given a vector [math]x\in K[/math], consider the orthogonal projection [math]P\in B(K)[/math] on the norm closure of the vector space [math]\pi(A)x\subset K[/math]. Since the subspace [math]\pi(A)x\subset K[/math] is invariant under the action of [math]\pi(A)[/math], so is its norm closure inside [math]K[/math], and we obtain from this:
By combining this with what we found above, we conclude that we have:
Since this holds for any [math]x\in K[/math], we conclude that any operator [math]T\in A''[/math] belongs to the strong operator closure of [math]A[/math]. By using now Proposition 5.7, which allows us to pass from the strong to the weak operator closure, we conclude that we have:
Thus, we have the desired reverse inclusion, and this finishes the proof.
Now by getting back to the von Neumann algebras, from Definition 5.8, we have the following result, which is a reformulation of Theorem 5.9, by using this notion:
For an operator algebra [math]A\subset B(H)[/math], the following are equivalent:
- [math]A[/math] is weakly closed, so it is a von Neumann algebra.
- [math]A[/math] equals its algebraic bicommutant [math]A''[/math], taken inside [math]B(H)[/math].
This follows from the formula [math]A''=\bar{A}[/math] from Theorem 5.9, along with the trivial fact that the commutants are automatically weakly closed.
The above statement, called bicommutant theorem, and due to von Neumann [1], is quite interesting, philosophically speaking. Among others, it shows that the von Neumann algebras are exactly the commutants of the self-adjoint sets of operators:
Given a subset [math]S\subset B(H)[/math] which is closed under [math]*[/math], the commutant
We have two assertions here, the idea being as follows:
(1) Given [math]S\subset B(H)[/math] satisfying [math]S=S^*[/math], the commutant [math]A=S'[/math] satisfies [math]A=A^*[/math], and is also weakly closed. Thus, [math]A[/math] is a von Neumann algebra. Note that this follows as well from the following “tricommutant formula”, which follows from Theorem 5.10:
(2) Given a von Neumann algebra [math]A\subset B(H)[/math], we can take [math]S=A'[/math]. Then [math]S[/math] is closed under the involution, and we have [math]S'=A[/math], as desired.
Observe that Proposition 5.11 can be regarded as yet another alternative definition for the von Neumann algebras, and with this definition being probably the best one when talking about quantum mechanics, where the self-adjoint operators [math]T:H\to H[/math] can be though of as being “observables” of the system, and with the commutants [math]A=S'[/math] of the sets of such observables [math]S=\{T_i\}[/math] being the algebras [math]A\subset B(H)[/math] that we are interested in. And with all this actually needing some discussion about self-adjointness, and about boundedness too, but let us not get into this here, and stay mathematical, as before.
As another interesting consequence of Theorem 5.10, we have:
Given a von Neumann algebra [math]A\subset B(H)[/math], its center
This follows from the fact that the commutants are weakly closed, that we know from the above, which shows that [math]A'\subset B(H)[/math] is a von Neumann algebra. Thus, the intersection [math]Z(A)=A\cap A'[/math] must be a von Neumann algebra too, as claimed.
In order to develop some general theory, let us start by investigating the finite dimensional case. Here the ambient algebra is [math]B(H)=M_N(\mathbb C)[/math], any linear subspace [math]A\subset B(H)[/math] is automatically closed, for all 3 topologies in Proposition 5.7, and we have:
The [math]*[/math]-algebras [math]A\subset M_N(\mathbb C)[/math] are exactly the algebras of the form
We have two assertions to be proved, the idea being as follows:
(1) Given numbers [math]n_1,\ldots,n_k\in\mathbb N[/math] satisfying [math]n_1+\ldots+n_k=N[/math], we have indeed an obvious embedding of [math]*[/math]-algebras, via matrix blocks, as follows:
In addition, we can twist this embedding by a unitary [math]U\in U_N[/math], as follows:
(2) In the other sense now, consider a [math]*[/math]-algebra [math]A\subset M_N(\mathbb C)[/math]. It is elementary to prove that the center [math]Z(A)=A\cap A'[/math], as an algebra, is of the following form:
Consider now the standard basis [math]e_1,\ldots,e_k\in\mathbb C^k[/math], and let [math]p_1,\ldots,p_k\in Z(A)[/math] be the images of these vectors via the above identification. In other words, these elements [math]p_1,\ldots,p_k\in A[/math] are central minimal projections, summing up to 1:
The idea is then that this partition of the unity will eventually lead to the block decomposition of [math]A[/math], as in the statement. We prove this in 4 steps, as follows:
\underline{Step 1}. We first construct the matrix blocks, our claim here being that each of the following linear subspaces of [math]A[/math] are non-unital [math]*[/math]-subalgebras of [math]A[/math]:
But this is clear, with the fact that each [math]A_i[/math] is closed under the various non-unital [math]*[/math]-subalgebra operations coming from the projection equations [math]p_i^2=p_i^*=p_i[/math].
\underline{Step 2}. We prove now that the above algebras [math]A_i\subset A[/math] are in a direct sum position, in the sense that we have a non-unital [math]*[/math]-algebra sum decomposition, as follows:
As with any direct sum question, we have two things to be proved here. First, by using the formula [math]p_1+\ldots+p_k=1[/math] and the projection equations [math]p_i^2=p_i^*=p_i[/math], we conclude that we have the needed generation property, namely:
As for the fact that the sum is indeed direct, this follows as well from the formula [math]p_1+\ldots+p_k=1[/math], and from the projection equations [math]p_i^2=p_i^*=p_i[/math].
\underline{Step 3}. Our claim now, which will finish the proof, is that each of the [math]*[/math]-subalgebras [math]A_i=p_iAp_i[/math] constructed above is a full matrix algebra. To be more precise here, with [math]n_i=rank(p_i)[/math], our claim is that we have isomorphisms, as follows:
In order to prove this claim, recall that the projections [math]p_i\in A[/math] were chosen central and minimal. Thus, the center of each of the algebras [math]A_i[/math] reduces to the scalars:
But this shows, either via a direct computation, or via the bicommutant theorem, that the each of the algebras [math]A_i[/math] is a full matrix algebra, as claimed.
\underline{Step 4}. We can now obtain the result, by putting together what we have. Indeed, by using the results from Step 2 and Step 3, we obtain an isomorphism as follows:
Moreover, a more careful look at the isomorphisms established in Step 3 shows that at the global level, that of the algebra [math]A[/math] itself, the above isomorphism simply comes by twisting the following standard multimatrix embedding, discussed in the beginning of the proof, (1) above, by a certain unitary matrix [math]U\in U_N[/math]:
Now by putting everything together, we obtain the result.
In relation with the bicommutant theorem, we have the following result, which fully clarifies the situation, with a very explicit proof, in finite dimensions:
Consider a [math]*[/math]-algebra [math]A\subset M_N(\mathbb C)[/math], written as above:
Let us decompose indeed our algebra [math]A[/math] as in Theorem 5.13:
The center of each matrix algebra being reduced to the scalars, the commutant of this algebra is then as follows, with each copy of [math]\mathbb C[/math] corresponding to a matrix block:
By taking once again the commutant we obtain [math]A[/math] itself, and we are done.
As another interesting application of Theorem 5.13, clarifying this time the relation with operator theory, in finite dimensions, we have the following result:
Given an operator [math]T\in B(H)[/math] in finite dimensions, [math]H=\mathbb C^N[/math], the von Neumann algebra [math]A= \lt T \gt [/math] that it generates inside [math]B(H)=M_N(\mathbb C)[/math] is
This is something which is routine, by using the linear algebra and spectral theory developed in chapter 1, for the matrices [math]M\in M_N(\mathbb C)[/math]. To be more precise:
(1) The fact that [math]A= \lt T \gt [/math] decomposes into a direct sum of matrix algebras is something that we already know, coming from Theorem 5.13.
(2) By using standard linear algebra, we can compute the block sizes [math]n_1,\ldots,n_k\in\mathbb N[/math], from the knowledge of the spectral theory of the associated matrix [math]M\in M_N(\mathbb C)[/math].
(3) In the normal case, [math]TT^*=T^*T[/math], we can simply invoke the spectral theorem, and by suitably changing the basis, we are led to the conclusion in the statement.
Let us get now to infinite dimensions, with Theorem 5.15 as our main source of inspiration. The same argument applies, provided that we are in the normal case, and we have the following result, summarizing our basic knowledge here:
Given a bounded operator [math]T\in B(H)[/math] which is normal, [math]TT^*=T^*T[/math], the von Neumann algebra [math]A= \lt T \gt [/math] that it generates inside [math]B(H)[/math] is
The measurable functional calculus theorem for the normal operators tells us that we have a weakly continuous morphism of [math]*[/math]-algebras, as follows:
Moreover, by the general properties of the measurable calculus, also established in chapter 3, this morphism is injective, and its image is the weakly closed algebra [math] \lt T \gt [/math] generated by [math]T,T^*[/math]. Thus, we obtain the isomorphism in the statement.
More generally now, along the same lines, we have the following result:
Given operators [math]T_i\in B(H)[/math] which are normal, and which commute, the von Neumann algebra [math]A= \lt T_i \gt [/math] that these operators generates inside [math]B(H)[/math] is
This is once again routine, by using the spectral theory for the families of commuting normal operators [math]T_i\in B(H)[/math] developed in chapter 3.
As a fundamental consequence now of the above results, we have:
The commutative von Neumann algebras are the algebras
We have two assertions to be proved, the idea being as follows:
(1) In one sense, we must prove that given a measured space [math]X[/math], we can realize the [math]A=L^\infty(X)[/math] as a von Neumann algebra, on a certain Hilbert space [math]H[/math]. But this is something that we know since chapter 2, the representation being as follows:
(2) In the other sense, given a commutative von Neumann algebra [math]A\subset B(H)[/math], we must construct a certain measured space [math]X[/math], and an identification [math]A=L^\infty(X)[/math]. But this follows from Theorem 5.17, because we can write our algebra as follows:
To be more precise, [math]A[/math] being commutative, any element [math]T\in A[/math] is normal, so we can pick a basis [math]\{T_i\}\subset A[/math], and then we have [math]A= \lt T_i \gt [/math] as above, with [math]T_i\in B(H)[/math] being commuting normal operators. Thus Theorem 5.17 applies, and gives the result.
(3) Alternatively, and more explicitly, we can deduce this from Theorem 5.16, applied with [math]T=T^*[/math]. Indeed, by using [math]T=Re(T)+iIm(T)[/math], we conclude that any von Neumann algebra [math]A\subset B(H)[/math] is generated by its self-adjoint elements [math]T\in A[/math]. Moreover, by using measurable functional calculus, we conclude that [math]A[/math] is linearly generated by its projections. But then, assuming [math]A=\overline{span}\{p_i\}[/math], with [math]p_i[/math] being projections, we can set:
Then [math]T=T^*[/math], and by functional calculus we have [math]p_0\in \lt T \gt [/math], then [math]p_1\in \lt T \gt [/math], and so on. Thus [math]A= \lt T \gt [/math], and [math]A=L^\infty(X)[/math] comes now via Theorem 5.16, as claimed.
The above result is the foundation for all the advanced von Neumann algebra theory, that we will discuss in the remainder of this book, and there are many things that can be said about it. To start with, in relation with the general theory of the normed closed algebras, that we developed in the beginning of this chapter, we have: \begin{warning} Although the von Neumann algebras are norm closed, the theory of norm closed algebras does not always apply well to them. For instance for [math]A=L^\infty(X)[/math] Gelfand gives [math]A=C(\widehat{X})[/math], with [math]\widehat{X}[/math] being a certain technical compactification of [math]X[/math]. \end{warning} In short, this would be my advice, do not mess up the two theories that we will be developing in this book, try finding different rooms for them, in your brain. At least at this stage of things, because later, do not worry, we will be playing with both.
Now forgetting about Gelfand, and taking Theorem 5.18 as such, tentative foundation for the theory that we want to develop, as a first consequence of this, we have:
Given a von Neumann algebra [math]A\subset B(H)[/math], we have
We know from Proposition 5.12 that the center [math]Z(A)\subset B(H)[/math] is a von Neumann algebra. Thus Theorem 5.18 applies, and gives the result.
It is possible to further build on this, with a powerful decomposition result as follows, over the measured space [math]X[/math] constructed in Theorem 5.20:
But more on this later, after developing the appropriate tools for this program, which is something non-trivial. Among others, before getting into such things, we will have to study the von Neumann algebras [math]A[/math] having trivial center, [math]Z(A)=\mathbb C[/math], called factors, which include the fibers [math]A_x[/math] in the above decomposition result. More on this later.
General references
Banica, Teo (2024). "Principles of operator algebras". arXiv:2208.03600 [math.OA].