12b. Von Neumann algebras

[math] \newcommand{\mathds}{\mathbb}[/math]

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In order to discuss all this, we will need some basic von Neumann algebra theory, coming as a complement to the [math]C^*[/math]-algebra theory from chapter 1. Let us start with a key result in functional analysis, as follows:

Proposition

For an operator algebra [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.

Show Proof

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) But this latter fact is something standard, which can be proved via an amplification trick. Consider the Hilbert space obtained by summing [math]n[/math] times [math]H[/math] with itself:

[[math]] K=H\oplus\ldots\oplus H [[/math]]

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:

[[math]] \pi(T)=\begin{pmatrix} T\\ &\ddots\\ &&T \end{pmatrix} [[/math]]

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]:

[[math]] \begin{eqnarray*} T\in\bar{A} &\implies&\pi(T)\in\overline{\pi(A)}\\ &\implies&\pi(T)x\in\overline{\pi(A)x}\\ &\implies&\pi(T)x\in\overline{\pi(A)x}^{\,||.||} \end{eqnarray*} [[/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]:

[[math]] ||Sx_i-Tx_i|| \lt \varepsilon [[/math]]

Thus [math]T[/math] belongs to the strong operator closure of [math]A[/math], as desired.

■

In the above the terminology, while standard, is a bit confusing, because the norm topology is stronger than the strong operator topology. As a solution, we agree in what follows to call the norm topology “strong”, and the weak and strong operator topologies “weak”, whenever these two topologies coincide. With this convention, the algebras from Proposition 12.6 are those which are weakly closed, and we can formulate:

Definition

A von Neumann algebra is a [math]*[/math]-algebra of operators

[[math]] A\subset B(H) [[/math]]

which is closed under the weak topology.

As basic examples, we have the algebra [math]B(H)[/math] itself, then the singly generated von Neumann algebras, [math]A= \lt T \gt [/math], with [math]T\in B(H)[/math], and then the multiply generated von Neumann algebras, namely [math]A= \lt T_i \gt [/math], with [math]T_i\in B(H)[/math]. At the level of the general results, we first have the bicommutant theorem of von Neumann, as follows:

Theorem

For a [math]*[/math]-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].

Show Proof

Since the commutants are automatically weakly closed, it is enough to show that weakly closed implies [math]A=A''[/math]. For this purpose, we will prove something a bit more general, stating that given a [math]*[/math]-algebra of operators [math]A\subset B(H)[/math], the following holds, with [math]A''[/math] being the bicommutant inside [math]B(H)[/math], and with [math]\bar{A}[/math] being the weak closure:

[[math]] A''=\bar{A} [[/math]]

We prove this equality 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:

[[math]] A\subset A'' [[/math]]

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:

[[math]] \begin{eqnarray*} T_i\in A'' &\implies&ST_i=T_iS,\ \forall S\in A'\\ &\implies&ST=TS,\ \forall S\in A'\\ &\implies&T\in A \end{eqnarray*} [[/math]]

Thus our claim is proved, and together with Proposition 12.6, which allows us to pass from the strong to the weak operator topology, this gives the desired inclusion:

[[math]] \bar{A}\subset A'' [[/math]]

“[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:

[[math]] T\in A''\implies T\in\bar{A} [[/math]]

For this purpose, we use the same amplification trick as in the proof of Proposition 12.5. Consider the Hilbert space obtained by summing [math]n[/math] times [math]H[/math] with itself:

[[math]] K=H\oplus\ldots\oplus H [[/math]]

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:

[[math]] \pi(T)=\begin{pmatrix} T\\ &\ddots\\ &&T \end{pmatrix} [[/math]]

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:

[[math]] \pi(A)=\left\{\begin{pmatrix} T\\ &\ddots\\ &&T \end{pmatrix}\Big|T\in A\right\} [[/math]]

We can compute the commutant of this image, exactly as in the usual scalar matrix case, and we obtain the following formula:

[[math]] \pi(A)'=\left\{\begin{pmatrix} S_{11}&\ldots&S_{1n}\\ \vdots&&\vdots\\ S_{n1}&\ldots&S_{nn} \end{pmatrix}\Big|S_{ij}\in A'\right\} [[/math]]

We conclude from this that, given an operator [math]T\in A''[/math] as above, we have:

[[math]] \begin{pmatrix} T\\ &\ddots\\ &&T \end{pmatrix}\in\pi(A)'' [[/math]]

In other words, the conclusion of all this is that we have:

[[math]] T\in A''\implies \pi(T)\in\pi(A)'' [[/math]]

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:

[[math]] P\in\pi(A)' [[/math]]

By combining this with what we found above, we conclude that we have:

[[math]] T\in A''\implies \pi(T)P=P\pi(T) [[/math]]

Now since this holds for any [math]x\in K[/math], we conclude that any [math]T\in A''[/math] belongs to the strong operator closure of [math]A[/math]. By using now Proposition 12.5, which allows us to pass from the strong to the weak operator closure, we conclude that we have [math]A''\subset\bar{A}[/math], as desired.

■

In order to develop now some general theory, let us start by investigating the finite dimensional case. Here the ambient operator algebra is [math]B(H)=M_N(\mathbb C)[/math], and any subspace [math]A\subset B(H)[/math] is automatically closed, for all 3 topologies from Proposition 12.6. Thus, we are left with the question of investigating the [math]*[/math]-algebras of usual matrices [math]A\subset M_N(\mathbb C)[/math]. But this is a purely algebraic question, whose answer is as follows:

Theorem

The [math]*[/math]-algebras [math]A\subset M_N(\mathbb C)[/math] are exactly the algebras of the form

[[math]] A=M_{r_1}(\mathbb C)\oplus\ldots\oplus M_{r_k}(\mathbb C) [[/math]]

depending on parameters [math]k\in\mathbb N[/math] and [math]r_1,\ldots,r_k\in\mathbb N[/math] satisfying

[[math]] r_1+\ldots+r_k=N [[/math]]

embedded into [math]M_N(\mathbb C)[/math] via the obvious block embedding, twisted by a unitary [math]U\in U_N[/math].

Show Proof

We have two assertions to be proved, the idea being as follows:

(1) Given numbers [math]r_1,\ldots,r_k\in\mathbb N[/math] satisfying [math]r_1+\ldots+r_k=N[/math], we have an obvious embedding of [math]*[/math]-algebras, via matrix blocks, as follows:

[[math]] M_{r_1}(\mathbb C)\oplus\ldots\oplus M_{r_k}(\mathbb C)\subset M_N(\mathbb C) [[/math]]

In addition, we can twist this embedding by a unitary [math]U\in U_N[/math], as follows:

[[math]] M\to UMU^* [[/math]]

(2) In the other sense now, consider an arbitrary [math]*[/math]-algebra of the [math]N\times N[/math] matrices, [math]A\subset M_N(\mathbb C)[/math]. Let us first look at the center of this algebra, which given by:

[[math]] Z(A)=A\cap A' [[/math]]

It is elementary to prove that this center, as an algebra, is of the following form:

[[math]] Z(A)\simeq\mathbb C^k [[/math]]

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:

[[math]] p_1+\ldots+p_k=1 [[/math]]

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]:

[[math]] A_i=p_iAp_i [[/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:

[[math]] A=A_1\oplus\ldots\oplus A_k [[/math]]

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:

[[math]] A_1+\ldots+ A_k=A [[/math]]

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]r_i=rank(p_i)[/math], our claim is that we have isomorphisms, as follows:

[[math]] A_i\simeq M_{r_i}(\mathbb C) [[/math]]

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:

[[math]] Z(A_i)=\mathbb C [[/math]]

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:

[[math]] A =A_1\oplus\ldots\oplus A_k \simeq M_{r_1}(\mathbb C)\oplus\ldots\oplus M_{r_k}(\mathbb C) [[/math]]

Moreover, a careful look at the isomorphisms established in Step 3 shows that at the global level, of the algebra [math]A[/math] itself, the above isomorphism comes by twisting the standard multimatrix embedding [math]M_{r_1}(\mathbb C)\oplus\ldots\oplus M_{r_k}(\mathbb C)\subset M_N(\mathbb C)[/math], discussed in the beginning of the proof, (1) above, by a certain unitary [math]U\in U_N[/math]. Thus, we obtain the result.

■

As an application of Theorem 12.9, clarifying the relation with linear algebra, or operator theory in finite dimensions, we have the following result:

Proposition

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

[[math]] A=M_{r_1}(\mathbb C)\oplus\ldots\oplus M_{r_k}(\mathbb C) [[/math]]

with the sizes of the blocks [math]r_1,\ldots,r_k\in\mathbb N[/math] coming from the spectral theory of the associated matrix [math]M\in M_N(\mathbb C)[/math]. In the normal case [math]TT^*=T^*T[/math], this decomposition comes from

[[math]] T=UDU^* [[/math]]

with [math]D\in M_N(\mathbb C)[/math] diagonal, and with [math]U\in U_N[/math] unitary.

Show Proof

This is something standard, by using the basic linear algebra theory and spectral theory for the usual matrices [math]M\in M_N(\mathbb C)[/math].

■

Let us get now to infinite dimensions, with Proposition 12.10 as our main source of inspiration. We have here the following result:

Theorem

Given an 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

[[math]] \lt T \gt =L^\infty(\sigma(T)) [[/math]]

with [math]\sigma(T)[/math] being its spectrum, formed of numbers [math]\lambda\in\mathbb C[/math] such that [math]T-\lambda[/math] is not invertible.

Show Proof

This is something standard as well, by using the spectral theory for the normal operators [math]T\in B(H)[/math], coming from chapter 1.

■

More generally, along the same lines, we have the following result, dealing this time with commuting families of normal operators:

Theorem

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

[[math]] \lt T_i \gt =L^\infty(X) [[/math]]

with [math]X[/math] being a certain measured space, associated to the family [math]\{T_i\}[/math].

Show Proof

This is again routine, by using this time the spectral theory for the families of commuting normal operators [math]T_i\in B(H)[/math]. See for instance Blackadar .

■

As an interesting abstract consequence of this, we have:

Theorem

The commutative von Neumann algebras are the algebras of type

[[math]] A=L^\infty(X) [[/math]]

with [math]X[/math] being a measured space.

Show Proof

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 commutative algebra [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 already know, coming from the multiplicity operators [math]T_f(g)=fg[/math] from the proof of the GNS theorem, the representation being as follows:

[[math]] L^\infty(X)\subset B(L^2(X)) [[/math]]

(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 12.12, because we can write our algebra as follows:

[[math]] A= \lt T_i \gt [[/math]]

To be more precise, [math]A[/math] being commutative, any element [math]T\in A[/math] is normal. Thus, 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 12.12 applies, and gives the result.

■

Moving ahead now, we can combine Proposition 12.8 with Theorem 12.13, and by building along the lines of Theorem 12.9, but this time in infinite dimensions, we are led to the following statement, due to Murray-von Neumann and Connes:

Theorem

Given a von Neumann algebra [math]A\subset B(H)[/math], if we write its center as

[[math]] Z(A)=L^\infty(X) [[/math]]

then we have a decomposition as follows, with the fibers [math]A_x[/math] having trivial center:

[[math]] A=\int_XA_x\,dx [[/math]]

Moreover, the factors, [math]Z(A)=\mathbb C[/math], can be basically classified in terms of the [math]{\rm II}_1[/math] factors, which are those satisfying [math]\dim A=\infty[/math], and having a faithful trace [math]tr:A\to\mathbb C[/math].

Show Proof

This is something that we know to hold in finite dimensions, as a consequence of Theorem 12.9. In general, this is something heavy, the idea being as follows:

(1) This is von Neumann's reduction theory main result, whose statement is already quite hard to understand, and whose proof uses advanced functional analysis.

(2) This is heavy, due to Murray-von Neumann and Connes, the idea being that the other factors can be basically obtained via crossed product constructions.

■

All this is certainly quite advanced, taking substantial time to be fully understood. For general reading on von Neumann algebras we recommend the book of Blackadar , but be aware tough that, while being at the same time well-written, condensed and reasonably thick, that book is only an introduction to Theorem 12.14. So, if we want to learn the full theory, with the complete proof of Theorem 12.14, you will have to go, every now and then, through the original papers of Murray-von Neumann and Connes.

By the way, talking von Neumann and Connes, we can only warmly recommend their books, and , as a complement to anything that you might want to learn on operator algebras, from Blackadar or from somewhere else. The point indeed is that von Neumann algebras come from quantum mechanics, meaning they are designed to help with quantum mechanics, and also happen to have 0 serious applications to pure mathematics, which by the way is not really science anyway, and if there are 2 people who understood all this, what von Neumann algebras are potentially good for, these are von Neumann himself, and Connes. So, read their books, and .

For the discussion to be complete, yet another 2 people who understood what von Neumann algebras are good for, in more modern times, are Jones and Voiculescu. So, have their main writings, and , available nearby, ready for some reading. And finally, talking quantum mechanics, always a pleasure to recommend, as usual, the books of Feynman , Griffiths and Weinberg . And with a preference for Griffiths , that's got all the quantum mechanics that you need to know, clearly explained, and comes with a cat on the cover too. But Feynman and Weinberg are excellent too, and as a general rule, as long as you stay away from quantum mechanics books which claim to be “rigorous”, “axiomatic”, “mathematical”, and so on, things fine.

Always remember here, as per Feynman saying, that “no one understands quantum mechanics”. With this being of course a euphemism for something of type “quantum mechanics as we know it is wrong, sorry for that, and we're working on the fix.

Now back to work, and our noncommutative geometry questions, as a first application of the above, we can extend our noncommutative space setting, as follows:

Theorem

Consider the category of “noncommutative measure spaces”, having as objects the pairs [math](A,tr)[/math] consisting of a von Neumann algebra with a faithful trace, and with the arrows reversed, which amounts in writing [math]A=L^\infty(X)[/math] and [math]tr=\int_X[/math].

• The category of usual measured spaces embeds into this category, and we obtain in this way the objects whose associated von Neumann algebra is commutative.
• Each [math]C^*[/math]-algebra given with a trace produces as well a noncommutative measure space, by performing the GNS construction, and taking the weak closure.
• In what regards the finitely generated group duals, or more generally the compact matrix quantum groups, the corresponding identification is injective.
• Even more generally, for noncommutative algebraic manifolds having an integratiuon functional, like the spheres, the identification is injective.

Show Proof

This is clear indeed from the basic properties of the GNS construction, from Theorem 12.2, and from the general theory from Theorem 12.14.

■

Before getting back to matrix models, we would like to formulate the following result, in relation with our axiomatization questions from chapters 1-4:

Theorem

In the context of noncommutative geometries coming from quadruplets [math](S,T,U,K)[/math], we have von Neumann algebras, with traces, as follows,

[[math]] \xymatrix@R=50pt@C=50pt{ L^\infty(S)\ar[r]\ar[d]\ar[dr]&L^\infty(T)\ar[l]\ar[d]\ar[dl]\\ L^\infty(U)\ar[u]\ar[ur]\ar[r]&L^\infty(K)\ar[l]\ar[ul]\ar[u] } [[/math]]

with [math]L^\infty(S)\subset L^\infty(U)[/math] being obtained by taking the first row algebra.

Show Proof

This follows indeed from the results that we already have, from chapters 1-4 above, by using the general formalism from Theorem 12.15.

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This statement, which is quite interesting, philosophically speaking, raises the question of axiomatizing, or rather re-axiomatizing, the quadruplets [math](S,T,U,K)[/math] that we are interested in directly in terms of the associated von Neumann algebras, as above. Indeed, in view of our general quantum mechanics motivations, we are after all mostly interested in integrating over our quantum manifolds, and so with this is mind, the von Neumann algebra formalism seems to be the one which is best adapted to our questions.

However, this is wrong. The above result is something theoretical, because it assumes the existence of Haar measures on our spaces [math]S,T,U,K[/math], which itself is something coming as a theorem. Thus, while all this is nice, the good way of doing things is with [math]C^*[/math]-algebras, as we did in chapters 1-4. And the von Neumann algebras from Theorem 12.16 remain something more advanced and specialized, coming afterwards.

As a side comment here, and for ending with some physics, the question “does the algebra or the Hilbert space come first” is a well-known one in quantum mechanics, basically leading to 2 different schools of thought. We obviously adhere here to the “algebra comes first” school. But let us not get here into this, perhaps enough controversies discussed, so far in this book. For more on this, get to know about the Bohr vs Einstein debate, which is the mother of all debates, in quantum mechanics.

And then, once this learned, as an instructive exercise: what do you think, from your reading so far of this book, do we rather side with Bohr, or with Einstein?

General references

Banica, Teo (2024). "Affine noncommutative geometry". arXiv:2012.10973 [math.QA].