Functional analysis

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9a. Kaplansky density

Welcome to this second half of the present book. We will get back here to a more normal pace, at least for most of the text to follow, our goal being to discuss the basics of the von Neumann algebra theory, due to Murray, von Neumann and Connes , , , , , , , , or at least the “basics of the basics”, the whole theory being quite complex, and then the most beautiful advanced theory which can be built on this, which is the subfactor theory of Jones , , , , , , .

The material here will be in direct continuation of what we learned in chapter 5, namely bicommutant theorem, commutative case, finite dimensions, and a handful of other things. The idea will be that of building directly on that material, and using the same basic techniques, namely functional analysis and operator theory.

As an important point, all this is related, but in a subtle way, to what we learned in chapters 6-8 too. To be more precise, what we will be doing in chapters 9-12 here will be more or less orthogonal to what we did in chapters 6-8. However, and here comes our point, the continuation of all this, chapters 13-16 below following Jones, will stand as a direct continuation of what we did in chapters 6-8, with Jones' subfactors being something more general than the random matrices and quantum groups from there.

Getting started, as a first objective we would like to have a better understanding of the precise difference between the norm closed [math]*[/math]-algebras, or [math]C^*[/math]-algebras, [math]A\subset B(H)[/math], and the weakly closed such algebras, which are the von Neumann algebras, from a functional analytic viewpoint. Let us begin with some generalities. We first have:

Proposition

The weak operator topology on [math]B(H)[/math] is the topology having the following equivalent properties:

• It makes [math]T\to \lt Tx,y \gt [/math] continuous, for any [math]x,y\in H[/math].
• It makes [math]T_n\to T[/math] when [math] \lt T_nx,y \gt \to \lt Tx,y \gt [/math], for any [math]x,y\in H[/math].
• Has as subbase the sets [math]U_T(x,y,\varepsilon)=\{S:| \lt (S-T)x,y \gt | \lt \varepsilon\}[/math].
• Has as base [math]U_T(x_1,\ldots,x_n,y_1,\ldots,y_n,\varepsilon)=\{S:| \lt (S-T)x_i,y_i \gt | \lt \varepsilon,\forall i\}[/math].

Show Proof

The equivalences [math](1)\iff(2)\iff(3)\iff(4)[/math] all follow from definitions, with of course (1,2) referring to the coarsest topology making that things happen.

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Similarly, in what regards the strong operator topology, we have:

Proposition

The strong operator topology on [math]B(H)[/math] is the topology having the following equivalent properties:

• It makes [math]T\to Tx[/math] continuous, for any [math]x\in H[/math].
• It makes [math]T_n\to T[/math] when [math]T_nx\to Tx[/math], for any [math]x\in H[/math].
• Has as subbase the sets [math]V_T(x,\varepsilon)=\{S:||(S-T)x|| \lt \varepsilon\}[/math].
• Has as base the sets [math]V_T(x_1,\ldots,x_n,\varepsilon)=\{S:||(S-T)x_i|| \lt \varepsilon,\forall i\}[/math].

Show Proof

Again, the equivalences [math](1)\iff(2)\iff(3)\iff(4)[/math] are all clear, and with (1,2) referring to the coarsest topology making that things happen.

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We know from chapter 5 that an operator algebra [math]A\subset B(H)[/math] is weakly closed if and only if it is strongly closed. Here is a useful generalization of this fact:

Theorem

Given a convex set [math]C\subset B(H)[/math], its weak operator closure and strong operator closure coincide.

Show Proof

Since the weak operator topology on [math]B(H)[/math] is weaker by definition than the strong operator topology on [math]B(H)[/math], we have, for any subset [math]C\subset B(H)[/math]:

[[math]] \overline{C}^{\,strong}\subset\overline{C}^{\,weak} [[/math]]

Now by assuming that [math]C\subset B(H)[/math] is convex, we must prove that:

[[math]] T\in\overline{C}^{\,weak}\implies T\in\overline{C}^{\,strong} [[/math]]

In order to do so, let us pick vectors [math]x_1,\ldots,x_n\in H[/math] and [math]\varepsilon \gt 0[/math]. We let [math]K=H^{\oplus n}[/math], and we consider the standard embedding [math]i:B(H)\subset B(K)[/math], given by:

[[math]] i T(y_1,\ldots,y_n)=(Ty_1,\ldots,Ty_n) [[/math]]

We have then the following implications, which are all trivial:

[[math]] T\in\overline{C}^{\,weak}\implies iT\in\overline{iC}^{\,weak}\implies iT(x)\in\overline{iC(x)}^{\,weak} [[/math]]

Now since the set [math]C\subset B(H)[/math] was assumed to be convex, the set [math]iC(x)\subset K[/math] is convex too, and by the Hahn-Banach theorem, for compact sets, it follows that we have:

[[math]] iT(x)\in\overline{iC(x)}^{\,||.||} [[/math]]

Thus, there exists an operator [math]S\in C[/math] such that we have, for any [math]i[/math]:

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

But this shows that we have [math]S\in V_T(x_1,\ldots,x_n,\varepsilon)[/math], and since [math]x_1,\ldots,x_n\in H[/math] and [math]\varepsilon \gt 0[/math] were arbitrary, by Proposition 9.2 it follows that we have [math]T\in \overline{C}^{\,strong}[/math], as desired.

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We will need as well the following standard result:

Proposition

Given a vector space [math]E\subset B(H)[/math], and a linear form [math]f:E\to\mathbb C[/math], the following conditions are equivalent:

• [math]f[/math] is weakly continuous.
• [math]f[/math] is strongly continuous.
• [math]f(T)=\sum_{i=1}^n \lt Tx_i,y_i \gt [/math], for certain vectors [math]x_i,y_i\in H[/math].

Show Proof

This is something standard, using the same tools at those already used in chapter 5, namely basic functional analysis, and amplification tricks:

[math](1)\implies(2)[/math] Since the weak operator topology on [math]B(H)[/math] is weaker than the strong operator topology on [math]B(H)[/math], weakly continuous implies strongly continuous. To be more precise, assume [math]T_n\to T[/math] strongly. Then [math]T_n\to T[/math] weakly, and since [math]f[/math] was assumed to be weakly continuous, we have [math]f(T_n)\to f(T)[/math]. Thus [math]f[/math] is strongly continuous, as desired.

[math](2)\implies(3)[/math] Assume indeed that our linear form [math]f:E\to\mathbb C[/math] is strongly continuous. In particular [math]f[/math] is strongly continuous at 0, and Proposition 9.2 provides us with vectors [math]x_1,\ldots,x_n\in H[/math] and a number [math]\varepsilon \gt 0[/math] such that, with the notations there:

[[math]] f(V_0(x_1,\ldots,x_n,\varepsilon))\subset D_0(1) [[/math]]

That is, we can find vectors [math]x_1,\ldots,x_n\in H[/math] and a number [math]\varepsilon \gt 0[/math] such that:

[[math]] ||Tx_i|| \lt \varepsilon,\forall i\implies|f(T)| \lt 1 [[/math]]

But this shows that we have the following estimate:

[[math]] \sum_{i=1}^n||Tx_i||^2 \lt \varepsilon^2\implies |f(T)| \lt 1 [[/math]]

By linearity, it follows from this that we have the following estimate:

[[math]] |f(T)| \lt \frac{1}{\varepsilon}\sqrt{\sum_{i=1}^n||Tx_i||^2} [[/math]]

Consider now the direct sum [math]H^{\oplus n}[/math], and inside it, the following vector:

[[math]] x=(x_1,\ldots,x_n)\in H^{\oplus n} [[/math]]

Consider also the following linear space, written in tensor product notation:

[[math]] K=\overline{(E\otimes 1)x}\subset H^{\oplus n} [[/math]]

We can define a linear form [math]f':K\to\mathbb C[/math] by the following formula, and continuity:

[[math]] f'(Tx_1,\ldots,Tx_n)=f(T) [[/math]]

We conclude that there exists a vector [math]y\in K[/math] such that the following happens:

[[math]] f'\big((T\otimes1)y\big)= \lt (T\otimes 1)x,y \gt [[/math]]

But in terms of the original linear form [math]f:E\to\mathbb C[/math], this means that we have:

[[math]] f(T)=\sum_{i=1}^n \lt Tx_i,y_i \gt [[/math]]

[math](3)\implies(1)[/math] This is clear, because we have, with respect to the weak topology:

[[math]] \begin{eqnarray*} T_n\to T &\implies& \lt T_nx_i,y_i \gt \to \lt Tx_i,y_i \gt ,\forall i\\ &\implies&\sum_{i=1}^n \lt T_nx_i,y_i \gt \to\sum_{i=1}^n \lt Tx_i,y_i \gt \\ &\implies&f(T_n)\to f(T) \end{eqnarray*} [[/math]]

Thus, our linear form [math]f[/math] is weakly continuous, as desired.

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Here is one more well-known result, that we will need as well:

Theorem

The unit ball of [math]B(H)[/math] is weakly compact.

Show Proof

If we denote by [math]B_1\subset B(H)[/math] the unit ball, and by [math]D_1\subset\mathbb C[/math] the unit disk, we have a morphism as follows, which is continuous with respect to the weak topology on [math]B_1[/math], and with respect to the product topology on the set on the right:

[[math]] B_1\subset\prod_{||x||,||y||\leq1}D_1\quad,\quad T\to( \lt Tx,y \gt )_{x,y} [[/math]]

Since the set on the right is compact, by Tychonoff, it is enough to show that the image of [math]B_1[/math] is closed. So, let [math](c_{xy})\in\bar{B}_1[/math]. We can then find [math]T_i\in B_1[/math] such that:

[[math]] \lt T_ix,y \gt \to c_{xy}\quad,\quad\forall x,y [[/math]]

But this shows that the following map is a bounded sesquilinear form:

[[math]] H\times H\to\mathbb C\quad,\quad (x,y)\to c_{xy} [[/math]]

Thus, we can find an operator [math]T\in B(H)[/math], and so [math]T\in B_1[/math], such that [math] \lt Tx,y \gt =c_{xy}[/math] for any [math]x,y\in H[/math], and this shows that we have [math](c_{xy})\in B_1[/math], as desired.

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Getting back to operator algebras, we have the following result, due to Kaplansky, which is something very useful, and of independent interest as well:

Theorem

Given an operator algebra [math]A\subset B(H)[/math], the following happen:

• The unit ball of [math]A[/math] is strongly dense in the unit ball of [math]A''[/math].
• The same happens for the self-adjoint parts of the above unit balls.

Show Proof

This is something quite tricky, the idea being as follows:

(1) Consider the self-adjoint part [math]A_{sa}\subset A[/math]. By taking real parts of operators, and using the fact that [math]T\to T^*[/math] is weakly continuous, we have then:

[[math]] \overline{A_{sa}}^{\,w}\subset\left(\overline{A}^{\,w}\right)_{sa} [[/math]]

Now since the set [math]A_{sa}[/math] is convex, and by Theorem 9.3 all weak operator topologies coincide on the convex sets, we conclude that we have in fact equality:

[[math]] \overline{A_{sa}}^{\,w}=\left(\overline{A}^{\,w}\right)_{sa} [[/math]]

(2) With this result in hand, let us prove now the second assertion of the theorem. For this purpose, consider an element [math]T\in\overline{A}^{\,w}[/math], satisfying [math]T=T^*[/math] and [math]||T||\leq1[/math]. Consider as well the following function, going from the interval [math][-1,1][/math] to itself:

[[math]] f(t)=\frac{2t}{1+t^2} [[/math]]

By functional calculus we can find an element [math]S\in\left(\overline{A}^{\,w}\right)_{sa}[/math] such that:

[[math]] f(S)=T [[/math]]

In other words, we can find an element [math]S\in\left(\overline{A}^{\,w}\right)_{sa}[/math] such that:

[[math]] T=\frac{2S}{1+S^2} [[/math]]

Now given arbitrary vectors [math]x_1,\ldots,x_n\in H[/math] and an arbitrary number [math]\varepsilon \gt 0[/math], let us pick an element [math]R\in A_{sa}[/math], subject to the following two inequalities:

[[math]] ||RTx_i-STx_i||\leq\varepsilon\quad,\quad \left|\left|\frac{R}{1+S^2}x_i-\frac{S}{1+S^2}x_i\right|\right|\leq\varepsilon [[/math]]

Finally, consider the following element, which has norm [math]\leq1[/math]:

[[math]] L=\frac{2R}{1+R^2} [[/math]]

We have then the following computation, using the above formulae:

[[math]] \begin{eqnarray*} L-T &=&\frac{2R}{1+R^2}-\frac{2S}{1+S^2}\\ &=&2\left(\frac{1}{1+R^2}\big(R(1+S^2)-(1+S^2)R\big)\frac{1}{1+S^2}\right)\\ &=&2\left(\frac{1}{1+R^2}(R-S)\frac{1}{1+S^2}+\frac{R}{1+R^2}(S-R)\frac{S}{1+S^2}\right)\\ &=&\frac{2}{1+R^2}(R-S)\frac{1}{1+S^2}+\frac{L}{2}(S-R)T \end{eqnarray*} [[/math]]

Thus, we have the following estimate, for any [math]i\in\{1,\ldots,n\}[/math]:

[[math]] ||(L-T)x_i||\leq\varepsilon [[/math]]

But this gives the second assertion of the theorem, as desired.

(3) Let us prove now the first assertion of the theorem. Given an arbitrary element [math]T\in\overline{A}^{\,w}[/math], satisfying [math]||T||\leq1[/math], let us look at the following element:

[[math]] T'=\begin{pmatrix}0&T\\ T^*&0\end{pmatrix}\in M_2(\overline{A}^{\,w}) [[/math]]

This element is then self-adjoint, and we can use what we proved in the above, and we are led in this way to the first assertion in the statement, as desired.

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We can go back now to our original question, from the beginning of the present chapter, namely that of abstractly characterizing the von Neumann algebras, and we have:

Theorem

An operator algebra [math]A\subset B(H)[/math] is a von Neumann algebra precisely when its unit ball is weakly compact.

Show Proof

This is something which is now clear, coming from the Kaplansky density results established in Theorem 9.6. To be more precise:

(1) In one sense, assuming that [math]A\subset B(H)[/math] is a von Neumann algebra, this algebra is weakly closed. But since the unit ball of [math]B(H)[/math] is weakly compact, we are led to the conclusion that the unit ball of [math]A[/math] is weakly compact too.

(2) Conversely, assume that an operator algebra [math]A\subset B(H)[/math] is such that its unit ball is weakly compact. In particular, the unit ball of [math]A[/math] is weakly closed. Now if [math]T[/math] satisfying [math]||T||\leq1[/math] belongs to the weak closure of [math]A[/math], by Kaplansky density we conclude that we have [math]T\in A[/math]. Thus our algebra [math]A[/math] must be a von Neumann algebra, as claimed.

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There are several other abstract characterizations of the von Neumann algebras, inside the class of [math]C^*[/math]-algebras, and we will be back to this, on several occasions, and notably at the end of the present chapter, with such a characterization involving the predual.

9b. Projections, order

In order to further investigate the von Neumann algebras, the key idea, coming from the analysis of the finite dimensional algebras from chapter 5, will be that of looking at the projections. Let us start with some generalities. In analogy with what happens in finite dimensions, we have the following notions, over an arbitrary Hilbert space [math]H[/math]:

Definition

Associated to any two projections [math]P,Q\in B(H)[/math] are:

• The projection [math]P\wedge Q[/math], projecting on the common range.
• The projection [math]P\vee Q[/math], projecting on the span of the ranges.

Abstractly speaking, these two operations can be thought of as being inf and sup type operations, and all the known algebraic formulae for inf and sup hold in this setting. For the moment we will not need all this, and we will be back to it later. Let us record however the following basic formula, which is something very useful:

Proposition

We have the following formula,

[[math]] P+Q=P\wedge Q+P\vee Q [[/math]]

valid for any two projections [math]P,Q\in B(H)[/math].

Show Proof

This is clear from definitions, because when computing [math]P+Q[/math] we obtain the projection [math]P\vee Q[/math] on the span on the ranges, modulo the fact that the vectors in the common range are obtained twice, which amounts in saying that we must add [math]P\wedge Q[/math].

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With the above notions in hand, we have the following result:

Theorem

Consider two projections [math]P,Q\in B(H)[/math].

• In finite dimensions, over [math]H=\mathbb C^N[/math], we have, in norm:
  [[math]] (PQ)^n\to P\wedge Q [[/math]]
• In infinite dimensions, we have the following convergence, for any [math]x\in H[/math],
  [[math]] (PQ)^nx\to (P\wedge Q)x [[/math]]
  but the operators [math](PQ)^n[/math] do not necessarily converge in norm.

Show Proof

We have several assertions here, the proof being as follows:

(1) Assume that we are in the case [math]P,Q\in M_N(\mathbb C)[/math]. By substracting [math]P\wedge Q[/math] from both [math]P,Q[/math], we can assume [math]P\wedge Q=0[/math], and we must prove that we have:

[[math]] P\wedge Q=0\implies(PQ)^n\to 0 [[/math]]

Our claim is that we have [math]||PQ|| \lt 1[/math]. Indeed, we know that we have:

[[math]] ||PQ||\leq||P||\cdot||Q||=1 [[/math]]

Assuming now by contradiction that we have [math]||PQ||=1[/math], since we are in finite dimensions, we must have, for a certain norm one vector, [math]||x||=1[/math]:

[[math]] ||PQx||=1 [[/math]]

Thus, we must have equalities in the following estimate:

[[math]] ||PQx||\leq||Qx||\leq||x|| [[/math]]

But the second equality tells us that we must have [math]x\in Im(Q)[/math], and with this in hand, the first equality tells us that we must have [math]x\in Im(P)[/math]. But this contradicts [math]P\wedge Q=0[/math], so we have proved our claim, and the convergence [math](PQ)^n\to 0[/math] follows.

(2) In infinite dimensions now, as before by substracting [math]P\wedge Q[/math] from both [math]P,Q[/math], we can assume [math]P\wedge Q=0[/math], and we must prove that we have, for any [math]x\in H[/math]:

[[math]] P\wedge Q=0\implies(PQ)^nx\to 0 [[/math]]

For this purpose, we use a trick. Consider the following operator:

[[math]] R=PQP [[/math]]

This operator is positive, because we have [math]R=(PQ)(PQ)^*[/math], and we have:

[[math]] ||R||\leq||P||\cdot||Q||\cdot||P||=1 [[/math]]

Our claim, which will finish the proof, is that for any [math]x\in H[/math] we have:

[[math]] R^nx\to0 [[/math]]

In order to prove this claim, let us diagonalize [math]R[/math], by using the spectral theorem for self-adjoint operators, from chapter 3. If all the eigenvalues are [math] \lt 1[/math] then we are done. If not, this means that we can find a nonzero vector [math]x\in H[/math] such that:

[[math]] ||Rx||=||x|| [[/math]]

But this condition means that we must have equalities in the following estimate:

[[math]] ||PQPx||\leq||QPx||\leq||Px||\leq||x|| [[/math]]

The point now is that this is impossible, due to our assumption [math]P\wedge Q=0[/math]. Indeed, the last equality tells us that we must have [math]x\in Im(P)[/math], and with this in hand, the middle equality tells us that we must have [math]x\in Im(Q)[/math]. But this contradicts [math]P\wedge Q=0[/math], so we have proved our claim, and the convergence [math](PQ)^nx\to 0[/math] follows.

(3) Finally, for a counterexample to [math](PQ)^n\to0[/math], in infinite dimensions, we can take [math]H=l^2(\mathbb N)[/math], and then find projections [math]P,Q[/math] such that [math](PQ)^ne_k\to 0[/math] for any [math]k[/math], but with the convergence arbitrarily slowing down with [math]k\to\infty[/math]. Thus, [math](PQ)^n\not\to0[/math].

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As a consequence, in connection with the von Neumann algebras, we have:

Theorem

Given two projections [math]P,Q\in B(H)[/math], the projections

[[math]] P\wedge Q\quad,\quad P\vee Q [[/math]]

both belong to the von Neumann algebra generated by [math]P,Q[/math].

Show Proof

This comes from the above. Indeed, in what regards [math]P\wedge Q[/math], this is something that follows from Theorem 9.10. As for [math]P\vee Q[/math], here the result follows from the result for [math]P\wedge Q[/math], and from the formula [math]P+Q=P\wedge Q+P\vee Q[/math], from Proposition 9.9.

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The idea now will be that of studying the von Neumann algebras [math]A\subset B(H)[/math] by using their projections, [math]p\in A[/math]. Let us start with the following result:

Theorem

Any von Neumann algebra is generated by its projections.

Show Proof

This is something that we know from chapter 5, coming from the measurable functional calculus, which can cut any normal operator into projections.

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There are many other things that can be said about projections, in the general setting. In what follows we will just discuss the most important and useful such results. A first such result, providing us with some geometric intuition on projections, is as follows:

Theorem

Given a von Neumann algebra [math]A\subset B(H)[/math], and a projection [math]p\in A[/math], we have the following equalities, between von Neumann algebras on [math]pH[/math]:

• [math]pAp=(A'p)'[/math].
• [math](pAp)'=A'p[/math].

Show Proof

This is not exactly obvious, but can be proved as follows:

(1) As a first observation, the von Neumann algebras [math]pAp[/math] and [math]A'p[/math] commute on [math]pH[/math]. Thus, we must prove that we have the following implication:

[[math]] x\in(A'p)'\implies x\in pAp [[/math]]

For this purpose, consider the element [math]y=xp[/math]. Then for any [math]z\in A'[/math] we have:

[[math]] \begin{eqnarray*} zy &=&zxp\\ &=&zpxp\\ &=&xpzp\\ &=&xpz\\ &=&yz \end{eqnarray*} [[/math]]

But this shows that we have [math]y\in A[/math], and so we obtain, as desired:

[[math]] x=pyp\in pAp [[/math]]

(2) As before, one of the inclusions being clear, we must prove that we have:

[[math]] x\in(pAp)'\implies x\in A'p [[/math]]

By using the standard fact that any bounded operator appears as a linear combination of 4 unitaries, that we know from the beginning of chapter 4, it is enough to prove this for a unitary element, [math]x=u[/math]. So, assume that we have a unitary as follows:

[[math]] u\in(pAp)' [[/math]]

In order to prove our claim, consider the following vector space:

[[math]] K=\overline{ApH} [[/math]]

This space being invariant under both the algebras [math]A,A'[/math], we conclude that the projection [math]q=Proj(K)[/math] onto it belongs to the center of our von Neumann algebra:

[[math]] q\in Z(A) [[/math]]

Our claim now, which will quickly lead to the result that we want to prove, is that we can extend the above unitary [math]u\in(pAp)'[/math] to the space [math]K=\overline{ApH}[/math] via the following formula, valid for any elements [math]x_i\in A[/math], and any vectors [math]\xi_i\in pH[/math]:

[[math]] v\left(\sum_ix_i\xi_i\right)=\sum_ix_iu\xi_i [[/math]]

In order to prove this latter claim, we can use the following computation:

[[math]] \begin{eqnarray*} \left|\left|v\left(\sum_ix_i\xi_i\right)\right|\right|^2 &=&\sum_{ij} \lt x_iu\xi_i,x_ju\xi_j \gt \\ &=&\sum_{ij} \lt x_j^*x_iu\xi_i,u\xi_j \gt \\ &=&\sum_{ij} \lt px_j^*x_ipu\xi_i,u\xi_j \gt \\ &=&\sum_{ij} \lt upx_j^*x_ip\xi_i,u\xi_j \gt \\ &=&\sum_{ij} \lt px_j^*x_ip\xi_i,\xi_j \gt \\ &=&\sum_{ij} \lt x_j^*x_i\xi_i,\xi_j \gt \\ &=&\sum_{ij} \lt x_i\xi_i,x_j\xi_j \gt \\ &=&\left|\left|\sum_ix_i\xi_i\right|\right|^2 \end{eqnarray*} [[/math]]

Thus [math]v[/math] is well-defined by the above formula, and is an isometry of [math]K[/math]. Now observe that this element [math]v[/math] commutes with the algebra [math]A[/math] on the space [math]ApH[/math], and so on [math]K[/math]. Thus [math]vq\in A'[/math], and so [math]u=vqp[/math], which proves that we have [math]u\in A'p[/math], as desired.

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As a second result now, once again in the general setting, we have:

Proposition

Given a von Neumann algebra [math]A\subset B(H)[/math], the formula

[[math]] p\simeq q\iff\exists u,\ \begin{cases} uu^*=p\\ u^*u=q \end{cases} [[/math]]

defines an equivalence relation for the projections [math]p\in A[/math].

Show Proof

This is something elementary, which follows from definitions, with the transitivity coming by composing the corresponding partial isometries.

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As a third result, once again in the general setting, which once again provides us with some intuition, but this time of somewhat abstract type, we have:

Theorem

Given a von Neumann algebra [math]A\subset B(H)[/math], we have a partial order on the projections [math]p\in A[/math], constructed as follows, with [math]u[/math] being a partial isometry,

[[math]] p\preceq q\iff\exists u,\ \begin{cases} uu^*=p\\ u^*u\leq q \end{cases} [[/math]]

which is related to the equivalence relation [math]\simeq[/math] constructed above by:

[[math]] p\simeq q\iff p\preceq q,\ q\preceq p [[/math]]

Thus, [math]\preceq[/math] is a partial order on the equivalence classes of projections [math]p\in A[/math].

Show Proof

We have several assertions here, the idea being as follows:

(1) The fact that we have indeed a partial order is clear, with the transitivity coming, as before, by composing the corresponding partial isometries.

(2) Regarding now the relation with [math]\simeq[/math], via the equivalence in the statement, the implication [math]\implies[/math] is clear. Thus, we are left with proving [math]\Longleftarrow[/math], which reads:

[[math]] p\preceq q,\ q\preceq p\implies p\simeq q [[/math]]

Our assumption is that we have partial isometries [math]u,v[/math] such that:

[[math]] uu^*=p\quad,\quad u^*u\leq q [[/math]]

[[math]] v^*v\leq p\quad,\quad vv^*=q [[/math]]

We can construct then two sequences of decreasing projections, as follows:

[[math]] p_0=p\quad,\quad p_{n+1}=v^*q_nv [[/math]]

[[math]] q_0=q\quad,\quad q_{n+1}=u^*p_nu [[/math]]

Consider now the limits of these two sequences of projections, namely:

[[math]] p_\infty=\bigwedge_ip_i\quad,\quad q_\infty=\bigwedge_iq_i [[/math]]

In terms of all these projections that we constructed, we have the following decomposition formulae for the original projections [math]p,q[/math]:

[[math]] p=(p-p_1)+(p_1-p_2)+\ldots+p_\infty [[/math]]

[[math]] q=(q-q_1)+(q_1-q_2)+\ldots+q_\infty [[/math]]

Now observe that the summands are equivalent, with this being clear from the definition of [math]p_n,q_n[/math] at the finite indices [math]n \lt \infty[/math], and with [math]p_\infty\simeq q_\infty[/math] coming from:

[[math]] v^*q_\infty v=p_\infty\quad,\quad q_\infty vv^*q_\infty=q_\infty [[/math]]

Thus we obtain that we have [math]p\simeq q[/math], as desired, by summing.

(3) Finally, the fact that the order relation [math]\preceq[/math] factorizes indeed to the equivalence classes under [math]\simeq[/math] follows from the equivalence established in (2).

■

Summarizing, in view of Theorem 9.12, and of Theorem 9.15, we can formulate: \begin{conclusion} We can think of a von Neumann algebra [math]A\subset B(H)[/math] as being a kind of object belonging to “mathematical logic”, consisting of equivalence classes of projections [math]p\in A[/math], ordered via the relation [math]\preceq[/math], and producing [math]A[/math] itself via transport by partial isometries, and then linear combinations, and weak limits. \end{conclusion} Which is something quite remarkable, who on Earth could have guessed, say when we were back in chapter 5, struggling with the basics of the von Neumann algebra theory, or even at the beginning of the present chapter 9, again struggling with some sort of basics, of the more advanced theory, that we will end up with something that luminous.

Well, that person on Earth who found this was von Neumann himself, back in the 1930s. And his Conclusion 9.16, called “von Neumann vision” of the operator algebras, has been extremely useful ever since, and is still largely used nowadays.

Very nice all this, first class mathematics, but in what concerns us, however, we will rather stick to our [math]A=L^\infty(X)[/math] viewpoint, with [math]X[/math] being a quantum measured space, and the most often being a “quantum manifold”. This is more of a “continuous” philosophy, and in order to keep it intact, and powerful, we will have to take sometimes distances with the von Neumann philosophy, especially in what concerns the terminology.

In short, we will be definitely users of the von Neumann projection technology, which is extremely powerful, and is quite often the only available tool, but keeping in mind however that we are dealing with continuous objects [math]X[/math], and choosing the terminology and notations accordingly, inspired from continuous geometry.

9c. States, isomorphism

Getting back now to general questions concerning the von Neumann algebras, one question that we met on several occasions, and that we would like to clarify now, is the relation between abstract isomorphism and spatial isomorphism.

To be more precise, we would like to understand when two von Neumann algebras [math]A\subset B(H)[/math] and [math]B\subset B(K)[/math] are isomorphic, in an algebraic and topological sense, but without reference to the ambient Hilbert spaces [math]H,K[/math]. With the idea in mind that, once this understood, we will be able to talk about the von Neumann algebras [math]A[/math] as being abstract objects, a bit as were the [math]C^*[/math]-algebras, discussed in chapter 7.

In order to discuss this, let us start with some technical preliminaries. Here is a definition that I have been postponing for long, but which is now unavoidable:

Definition

We call ultraweak and ultrastrong topologies on [math]B(H)[/math] the topologies defined exactly as the weak and strong operator topologies, but by using infinite families of vectors [math](x_i)_{i\in\mathbb N}\subset H[/math] instead of finite families [math](x_i)_{i=1,\ldots,N}\subset H[/math].

And up to you to tell me if you love such things or not, and I will be here listening, like Sigmund Freud. Anyway. With this convention, we have the following result:

Proposition

Given a vector space [math]E\subset B(H)[/math], and a linear form [math]f:E\to\mathbb C[/math], the following conditions are equivalent:

• [math]f[/math] is ultraweakly continuous.
• [math]f[/math] is ultrastrongly continuous.
• [math]f(T)=\sum_{i=1}^\infty \lt Tx_i,y_i \gt [/math], for certain vectors [math]x_i,y_i\in H[/math].

Show Proof

This is similar to the proof of Proposition 9.4, as follows:

[math](1)\implies(2)[/math] Since the ultraweak operator topology is weaker than the ultrastrong operator topology, ultraweakly continuous implies ultrastrongly continuous.

[math](2)\implies(3)[/math] Assume that [math]f:E\to\mathbb C[/math] is ultrastrongly continuous. By continuity we can find vectors [math]x_i\in H[/math] and a number [math]\varepsilon \gt 0[/math] such that:

[[math]] \sum_{i=1}^\infty||Tx_i||^2 \lt \varepsilon^2\implies |f(T)| \lt 1 [[/math]]

It follows from this that we have the following estimate:

[[math]] |f(T)| \lt \frac{1}{\varepsilon}\sqrt{\sum_{i=1}^\infty||Tx_i||^2} [[/math]]

Consider now the direct sum [math]H^{\oplus\infty}[/math], and inside it, the following vector:

[[math]] x=(x_i)\in H^{\oplus\infty} [[/math]]

Consider also the following linear space, written in tensor product notation:

[[math]] K=\overline{(E\otimes 1)x}\subset H^{\oplus\infty} [[/math]]

We can define a linear form [math]f':K\to\mathbb C[/math] by the following formula, and continuity:

[[math]] f'((Tx_i)_i)=f(T) [[/math]]

We conclude that there exists a vector [math]y\in K[/math] such that:

[[math]] f'\big((T\otimes1)y\big)= \lt (T\otimes 1)x,y \gt [[/math]]

But in terms of the original linear form [math]f:E\to\mathbb C[/math], this means that we have:

[[math]] f(T)=\sum_{i=1}^\infty \lt Tx_i,y_i \gt [[/math]]

[math](3)\implies(1)[/math] This is indeed clear from definitions.

■

As a consequence of the above result, we have:

Theorem

Given a von Neumann algebra [math]A\subset B(H)[/math], and a positive linear form [math]f:A\to\mathbb C[/math], the following are equivalent:

• [math]f[/math] is normal, in the sense that [math]f\left(\sup_ix_i\right)=\sup_i f(x_i)[/math], for any increasing sequence of positive elements [math]x_i\in A[/math].
• [math]f[/math] is completely additive, in the sense that [math]f\left(\bigvee_ip_i\right)=\sum_if(p_i)[/math], for any family of pairwise orthogonal projections [math]p_i\in A[/math].
• [math]f[/math] is ultraweakly continuous, or equivalently, [math]f[/math] is a vector state, [math]f= \lt Tx,x \gt [/math], when suitably extending it to the space [math]H\otimes l^2(\mathbb N)[/math].

Show Proof

This is something very standard, as follows:

[math](1)\implies(2)[/math] Given a family of pairwise orthogonal projections [math]\{p_i\}[/math], we can consider the following increasing sequence of positive elements:

[[math]] x_n=\sum_{i=1}^np_i [[/math]]

By using now the formula in (1) for these elements we obtain, as desired:

[[math]] \begin{eqnarray*} f\left(\bigvee_ip_i\right) &=&f\left(\sup_nx_n\right)\\ &=&\sup_nf(x_n)\\ &=&\sup_n\sum_{i=1}^nf(p_i)\\ &=&\sum_if(p_i) \end{eqnarray*} [[/math]]

[math](2)\implies(3)[/math] This is something more technical, that we will prove in several steps. Let us fix a projection [math]q\in A[/math], and consider a vector [math]\xi\in Im(q)[/math] such that:

[[math]] \lt q\xi,\xi \gt \gt 1 [[/math]]

Our claim is that there exists a projection [math]p\leq q[/math] such that, for any [math]x\in A[/math]:

[[math]] f(pxp)\leq \lt pxp\xi,\xi \gt [[/math]]

Indeed, in order to prove this, let us pick, by using the Zorn lemma, a maximal family of pairwise orthogonal projections [math]\{p_i\}\subset A[/math] such that, for any [math]i[/math], we have:

[[math]] f(p_i)\geq \lt p_i\xi,\xi \gt [[/math]]

By using our complete additivity assumption, we have then:

[[math]] \begin{eqnarray*} f\left(\bigvee_ip_i\right) &=&\sum_if(p_i)\\ &\geq&\sum_i \lt p_i\xi,\xi \gt \\ &=&\left \lt \left(\bigvee_ip_i\right)\xi,\xi\right \gt \end{eqnarray*} [[/math]]

Now consider the following projection, which is nonzero:

[[math]] p=q-\bigvee_ip_i [[/math]]

By maximality of the family [math]\{p_i\}[/math], for any nonzero projection [math]r\leq p[/math], we have:

[[math]] f(r) \lt \lt r\xi,\xi \gt [[/math]]

We therefore obtain the following estimate, valid for any [math]x\in A_+[/math], as desired:

[[math]] f(pxp)\leq \lt pxp\xi,\xi \gt [[/math]]

Now by Cauchy-Schwarz we obtain that for any [math]x\in A[/math], [math]||x||\leq1[/math], we have:

[[math]] \begin{eqnarray*} |f(xp)|^2 &\leq&f(px^*xp)f(1)\\ &\leq& \lt px^*xp\xi,\xi \gt \\ &=&||xp\xi||^2 \end{eqnarray*} [[/math]]

Thus the following linear form is strongly continuous on the unit ball of [math]A[/math]:

[[math]] x\to f(px) [[/math]]

In order to finish now, once again by using the Zorn lemma, let us pick a maximal family of pairwise orthogonal projections [math]\{p_i\}\subset A[/math] such that [math]x\to f(p_ix)[/math] is strongly continuous on the unit ball of [math]A[/math], for any [math]i[/math]. By maximality we have then:

[[math]] \sum_if(p_i) =f\left(\sum_ip_i\right) =f(1) =1 [[/math]]

Now given [math]\varepsilon \gt 0[/math], let us choose a finite subset of our index set, [math]F\subset I[/math], such that for all the finite subsets [math]F\subset J\subset I[/math], we have an inequality as follows:

[[math]] 1-f\left(\sum_{j\in J}p_j\right)\leq\varepsilon [[/math]]

By Cauchy-Schwarz we have then, for any [math]x\in A[/math], [math]||x||=1[/math], the following estimate:

[[math]] \begin{eqnarray*} \left|f\left(x\left(1-\sum_{j\in J}p_j\right)\right)\right|^2 &\leq&f\left(1-\sum_{j\in J}p_j\right)f(xx^*)\\ &\leq&\varepsilon \end{eqnarray*} [[/math]]

We conclude from this that we have the following estimate:

[[math]] \left|\left|f-f\left(.\left(1-\sum_{j\in J}p_j\right)\right)\right|\right|\leq\sqrt{\varepsilon} [[/math]]

Thus we obtain [math]f\in A_*[/math], as desired.

[math](3)\implies(1)[/math] This is something trivial, coming from definitions.

■

We can now go back to our original question, and we have:

Theorem

Given two von Neumann algebras [math]A\subset B(H)[/math] and [math]B\subset B(K)[/math], acting on possibly different Hilbert spaces [math]H,K[/math], any algebraic isomorphism

[[math]] \Phi:A\simeq B [[/math]]

is spatial up to amplification, in the sense that we have a formula as follows,

[[math]] \Phi(T)\otimes1=U(T\otimes1)U^* [[/math]]

for a certain Hilbert space [math]L[/math], and a certain unitary [math]U:H\otimes L\to K\otimes L[/math].

Show Proof

This is something standard, coming from Theorem 9.19, as follows:

(1) As a first observation, assuming that a positive unital linear form [math]f:A\to\mathbb C[/math] is a vector state, given by a certain vector [math]x\in H[/math], then by Theorem 9.19 the linear form [math]f\Phi^{-1}[/math] is also a vector state, say given by a vector [math]y\in K[/math].

(2) We conclude from this that we have a unitary as follows, intertwining the corresponding actions of the von Neumann algebras [math]A[/math] and [math]B[/math]:

[[math]] U_x:\overline{Ax}\to\overline{By} [[/math]]

Now by making the above vector [math]x\in H[/math] vary, and performing a direct sum, we obtain with [math]L=l^2(\mathbb N)[/math] an isometry as in the statement, namely:

[[math]] U:H\otimes L\to K\otimes L [[/math]]

Our construction shows that [math]U[/math] intertwines indeed the actions of the von Neumann algebras [math]A[/math] and [math]B[/math], and what is left to do is to study the unitarity of [math]U[/math].

(3) We will prove now that, up to a suitable replacement, the above operator [math]U[/math] can be taken to be unitary, still intertwining the actions of the von Neumann algebras [math]A[/math] and [math]B[/math]. For this purpose, consider the action of von Neumann algebra [math]A[/math] on the direct sum Hilbert space [math](H\otimes L)\oplus(K\otimes L)[/math] given by the following matrices:

[[math]] x'=\begin{pmatrix} x\otimes1&0\\ 0&\Phi(x)\otimes1 \end{pmatrix} [[/math]]

Since [math]U[/math] intertwines the actions of the von Neumann algebras [math]A[/math] and [math]B[/math], in terms of [math]2\times 2[/math] matrices, we are led to the following conclusion:

[[math]] \begin{pmatrix} 0&0\\ U&0 \end{pmatrix}\in A' [[/math]]

Thus, the following happens inside the von Neumann algebra [math]A'[/math]:

[[math]] \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}\preceq \begin{pmatrix} 0&0\\ 0&1 \end{pmatrix} [[/math]]

On the other hand, the same reasoning applied to the isomorphism [math]\Phi^{-1}[/math] shows that we have as well, once again inside the von Neumann algebra [math]A'[/math]:

[[math]] \begin{pmatrix} 0&0\\ 0&1 \end{pmatrix}\preceq \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix} [[/math]]

(4) We are now in position to finish. By combining the above two conclusions, we obtain an equivalence of projections inside [math]A'[/math], as follows:

[[math]] \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}\simeq \begin{pmatrix} 0&0\\ 0&1 \end{pmatrix} [[/math]]

Now pick a partial isometry implementing this equivalence. This partial isometry must be of the following form, with [math]U'[/math] being now a unitary:

[[math]] V=\begin{pmatrix} 0&0\\ U'&0 \end{pmatrix} [[/math]]

Thus, we have a unitary as follows, which intertwines the actions of [math]A[/math] and [math]B[/math]:

[[math]] U':H\otimes L\to K\otimes L [[/math]]

But this is the unitary we were looking for, and we are done.

■

The above result is something quite fundamental, allowing us to talk about von Neumann algebras [math]A[/math] as abstract objects, without reference to the exact Hilbert space [math]H[/math] where the elements [math]a\in A[/math] live as operators [math]a\in B(H)[/math], and with this being of course possible modulo some functional analysis knowledge. We will heavily use this point of view in chapter 10 below, and then in chapters 13-16, when talking about [math]{\rm II}_1[/math] factors.

9d. Predual theory

We have seen so far, as a consequence of the Kaplansky density theorem, that an operator algebra [math]A\subset B(H)[/math] is a von Neumann algebra precisely when its unit ball is weakly compact. This is certainly useful, but there are many other possible characterizations of the von Neumann algebras, as operator algebras, which are useful as well.

To be more precise, going ahead now with more abstract functional analysis, that we will be using in what follows, on several occasions, let us formulate:

Definition

Given a von Neumann algebra [math]A\subset B(H)[/math], we set

[[math]] A_*=\Big\{f:A\to\mathbb C,\ {\rm weakly\ continuous}\Big\} [[/math]]

regarded as a linear subspace, [math]A_*\subset A^*[/math], of the usual dual, given by

[[math]] A^*=\Big\{f:A\to\mathbb C,\ {\rm norm\ continuous}\Big\} [[/math]]

and we call this space [math]A_*[/math] predual of our von Neumann algebra [math]A[/math].

Our first goal will be that of proving that we have the following duality formula, between the linear space [math]A_*[/math] constructed above, and the algebra [math]A[/math] itself:

[[math]] A=(A_*)^* [[/math]]

In order to do so, let us first discuss the case of the full operator algebra [math]A=B(H)[/math] itself. This is actually the key case, with the extension to the arbitrary von Neumann algebras [math]A\subset B(H)[/math] being something coming afterwards, quite straightforward.

We will need some standard operator theory, developed in chapter 4. First, we have the following result, regarding the trace class operators, established there:

Theorem

The space of trace class operators, which appears as an intermediate space between the finite rank operators and the compact operators,

[[math]] F(H)\subset B_1(H)\subset K(H) [[/math]]

is a two-sided [math]*[/math]-ideal of [math]K(H)[/math]. The following is a Banach space norm on [math]B_1(H)[/math],

[[math]] ||T||_1=Tr|T| [[/math]]

satisfying [math]||T||\leq||T||_1[/math], and for [math]T\in B_1(H)[/math] and [math]S\in B(H)[/math] we have:

[[math]] ||ST||_1\leq||S||\cdot||T||_1 [[/math]]

Also, the subspace [math]F(H)[/math] is dense inside [math]B_1(H)[/math], with respect to this norm.

Show Proof

This is indeed something standard, explained in chapter 4.

■

We will need as well the following result, regarding this time the Hilbert-Schmidt operators, which is also from chapter 4:

Theorem

The space of Hilbert-Schmidt operators, which appears as an intermediate space between the trace class operators and the compact operators,

[[math]] F(H)\subset B_1(H)\subset B_2(H)\subset K(H) [[/math]]

is a two-sided [math]*[/math]-ideal of [math]K(H)[/math]. In terms of the singular values [math](\lambda_n)[/math], the Hilbert-Schmidt operators are characterized by the following formula:

[[math]] \sum_n\lambda_n^2 \lt \infty [[/math]]

Also, the following formula, taking as input two Hilbert-Schmidt operators,

[[math]] \lt S,T \gt =Tr(ST^*) [[/math]]

defines a scalar product of [math]B_2(H)[/math], making it a Hilbert space.

Show Proof

As before, this is something standard, explained in chapter 4.

■

We will need as well the following technical result, also from chapter 4:

Theorem

We have the following commutation formula,

[[math]] Tr(ST)=Tr(TS) [[/math]]

valied for any two Hilbert-Schmidt operators [math]S,T\in B_2(H)[/math].

Show Proof

As before, this is something standard, explained in chapter 4.

■

With the above ingredients in hand, and getting back now to von Neumann algebras, and to our predual questions raised before, we first have the following result:

Theorem

The linear space [math]B(H)_*\subset B(H)^*[/math] consisting of the linear forms [math]f:B(H)\to\mathbb C[/math] which are weakly continuous is given by

[[math]] B(H)_*=\left\{T\to Tr(ST)\Big| S\in B_1(H)\right\} [[/math]]

and we have the following duality formula

[[math]] B(H)=(B(H)_*)^* [[/math]]

as a duality in the usual Banach space sense.

Show Proof

There are several things to be proved, the idea being as follows:

(1) First of all, any linear form of type [math]T\to Tr(ST)[/math], with [math]S[/math] being trace class, is weakly continuous. Thus, if we denote by [math]B(H)_\circ[/math] the subspace of [math]B(H)[/math] in the statement, consisting of such linear forms, we have an inclusion as follows:

[[math]] B(H)_\circ\subset B(H)_* [[/math]]

(2) In order to prove now the reverse inclusion, consider an arbitrary weakly continuous linear form [math]f\in B(H)_*[/math]. We can then find vectors [math](x_i)[/math] and [math](y_i)[/math] such that:

[[math]] f(T)=\sum_i \lt Tx_i,y_i \gt [[/math]]

Let us consider now the following operators, going by definition from the Hilbert space [math]l^2(\mathbb N)[/math] to our Hilbert space [math]H[/math], and which are both Hilbert-Schmidt:

[[math]] Q:e_i\to x_i\quad,\quad R:e_i\to y_i [[/math]]

In terms of these operators, our linear form can be written as follows:

[[math]] f(T)=Tr(R^*TQ) [[/math]]

On the other hand, by using the formula in Theorem 9.24 we obtain:

[[math]] Tr(R^*TQ)=Tr(TQR^*) [[/math]]

Thus, with [math]S=QR^*[/math], which is trace class, we have the following formula:

[[math]] f(T)=Tr(TS) [[/math]]

Thus, we have proved that we have an inclusion as follows:

[[math]] B(H)_*\subset B(H)_\circ [[/math]]

(3) Summing up, from (1) and (2) we conclude that we have an equality as follows, which proves the first assertion in the statement:

[[math]] B(H)_*=B(H)_\circ [[/math]]

(4) It remains to prove that [math]B(H)[/math] is indeed the dual of [math]B(H)_*[/math]. For this purpose, we use the above identification, which ultimately identifies [math]B(H)_*[/math] with the space of trace class operators [math]B_1(H)[/math]. So, assume that we have a linear form, as follows:

[[math]] f:B_1(H)\to\mathbb C [[/math]]

It is then routine to show that [math]f[/math] must come from evaluation on a certain operator [math]T\in B(H)[/math], and this leads to the conclusion that [math]B(H)[/math] is indeed the dual of [math]B(H)_*[/math].

■

More generally now, for the arbitrary von Neumann algebras [math]A\subset B(H)[/math], we have:

Theorem

Given a von Neumann algebra [math]A\subset B(H)[/math], if we set

[[math]] A_*=\Big\{f:A\to\mathbb C,\ {\rm weakly\ continuous}\Big\} [[/math]]

regarded as a linear subspace, [math]A_*\subset A^*[/math], of the usual dual, given by:

[[math]] A^*=\Big\{f:A\to\mathbb C,\ {\rm norm\ continuous}\Big\} [[/math]]

then we have the duality formula [math]A=(A_*)^*[/math], in the usual Banach space sense.

Show Proof

This can be proved in several steps, as follows:

(1) First of all, we know from the above that the result holds for the von Neumann algebra [math]A=B(H)[/math] itself, in the sense that we have:

[[math]] B(H)=(B(H)_*)^* [[/math]]

(2) The point now is that for any von Neumann subalgebra [math]A\subset B(H)[/math], or more generally for any weakly closed linear subspace [math]A\subset B(H)[/math], we have an equality as follows, coming as a consequence of the Hahn-Banach theorem:

[[math]] A=A^{\perp\perp} [[/math]]

(3) Thus, modulo some standard algebra, and some standard identifications for quotient spaces and their duals, we are led to the conclusion in the statement.

■

In fact, we have the following result, due to Sakai:

Theorem

The von Neumann algebras are exactly the [math]C^*[/math]-algebras which have a predual, in the above sense.

Show Proof

This is a variation of the above, which caps the above series of results, and closes any further discussions, and for details here, we refer to Sakai's book .

■

There are many other things that can be said, of purely abstract nature, on the von Neumann algebras. We will be back to this, from time to time, in what follows.

General references

Banica, Teo (2024). "Principles of operator algebras". arXiv:2208.03600 [math.OA].