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

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

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

#### 1. Trace class operators and $B(H)$ as the double dual of the compacts

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

For every $f,g \in H$ we define

1. A linear functional $w_{f,g} \in B(H)^*$ given by $w_{f,g}(T) = \langle Tf, g \rangle$.
2. A linear operator $t_{f,g} \in B(H)$ given by $t_{f,g}(h) = \langle h, g \rangle f$.

It customary to also write $t_{f,g} = f g^*$, for reasons which I hope are obvious (or $t_{f,g} = |f\rangle \langle g|$ by “physicists”). Both $w_{f,g}$ and $t_{f,g}$ are sometimes denotes by $f \otimes g^*$.

Let $K(H)$ denote the norm closed two sided ideal of compact operators in $B(H)$. Every vector functional $w_{f,g}$ can also be considered as an element of $K(H)^*$. We shall see below that $K(H)^{**} = B(H)$ in a natural way (where “in a natural way” is meant in a loose way). Let us begin by taking a closer look at compact operators.

Lemma 1: For every $T \in K(H)$, there exist two orthonormal sequences $\{f_n\}, \{g_n\} \subset H$ and a sequence of positive numbers $\{c_n\}$ that is either finite or convergent to zero, such that $T$ is given as the norm convergent sum

$T = \sum c_n t_{f_n, g_n}$.

If $T \geq 0$, then $T$ has the form

$T = \sum c_n t_{g_n, g_n}$.

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

For a general compact operator $T$, let $T = U|T|$ be the polar decomposition of $T$. Then $|T|$ is compact, and $|T| = \sum c_n t_{g_n,g_n}$. Then

$T = U \sum c_n t_{g_n,g_n} = \sum c_n t_{Ug_n,g_n}$,

so putting $f_n = U g_n$ (and recalling that $U$ is a partial isometry with $ker ^\perp = Im |T|$), we are done.

Now let us consider a functional $w \in K(H)^*$. Since for every $f,g \in H$, the rank one operator $t_{f,g}$ is compact, we can apply $w$ to $t_{f,g}$. This gives rise to a sesqui-linear form

$[f,g] = w(t_{f,g})$    for    $f,g \in H$.

(For example, $[f_1+f_2, g] = w(t_{f_1+f_2,g}) = w(t_{f_1,g}) + w(t_{f_2,g}) = [f_1,g] + [f_2,g]$. )

Now, $|[f,g]| = |w(t_{f,g})| \leq \|w\| \|t_{f,g}\| = \|w\|\|f\|\|g\|$, so $[\cdot, \cdot]$ is bounded. It follows from a familiar consequence of the Riesz representation theorem (see Exercise A below) that there is some $t(w) \in B(H)$ such that

$[f,g] = \langle t(w)f, g\rangle$     for all     $f,g \in H$,

or, in other words,

$w(t_{f,g}) = \langle t(w) f, g \rangle$    for all   $f,g \in H$.

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

Exercise A: Let $H$ be a Hilbert space, and let $[\cdot, \cdot ] : H \times H \to \mathbb{C}$ be a bounded sesqui-linear form, meaning that there is some $C$ such that for all $f, f_1, f_2, g, \in H$ and all $c \in \mathbb{C}$,

1. $[c f_1 + f_2, g] = c[f_1, g ] + [f_2, g]$,
2. $[f,g] = \overline{[g,f]}$,
3. $|[f,g]|\leq C\|f\| \|g\|$.

Prove that there exists $T \in B(H)$ with $\|T\|\leq C$, such that $[f,g] = \langle Tf, g \rangle$, for all $f,g \in H$.

Note that $\|t(w)\|\leq \|w\|$, so that the map $w \mapsto t(w)$ is a bounded linear map. Moreover, it is not hard to see that this map is injective (as a map from $K(H)^*$ to $B(H)$).

Lemma 2: $t(w_{f,g}) = t_{f,g}$. Consequently, if $(c_n) \in \ell^1$, $\{f_n\}, \{g_n\}$ are orthonormal sequences, and $w = \sum c_n w_{f_n,g_n}$, then $t(w) = \sum c_n t_{f_n,g_n}$; moreover, $w$ is the unique element in $K(H)^*$ mapped to $\sum c_n t_{f_n,g_n}$.

Proof: We check

$\langle t(w_{f,g}) f', g' \rangle = w_{f,g}(t_{f',g'}) = \langle t_{f',g'}(f), g \rangle = \langle \langle f, g' \rangle f' , g \rangle = \langle f, g' \rangle \langle f' , g \rangle$,

while, on the other hand,

$\langle t_{f,g}f', g' \rangle = \langle f', g \rangle \langle f, g' \rangle$.

It follows that $t(w_{f,g}) = t_{f,g}$. The remainder follows from the above remarks, that $w \mapsto t(w)$ is a bounded, linear and injective map.

Now suppose that $H$ is finite dimensional. Then

$w(t_{f,g}) = \langle t(w) f, g \rangle = g^* t(w) f = Trace(g^*t(w) f) = Trace(t(w) fg^*) = Trace(t(w) t_{f,g})$.

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

Exercise B: If $w \in K(H)^*$, and $X \in B(H)$, then define $w_X(A) = w(XA)$. Prive that $t(w_X) = t(w) X$.

Lemma 3: For all $w \in K(H)^*$, the operator $t(w)$ is compact. Moreover, for every orthonormal sequence $\{e_n\}_{n \in J} \subseteq H$

$\sum_n |\langle t(w) e_n , e_n \rangle|< \infty$

In fact, $\sum_n |\langle t(w) e_n , e_n \rangle| \leq \|w\|$

Proof: For all $n$, let us write $|\langle t(w) e_n , e_n \rangle| = c_n \langle t(w) e_n, e_n \rangle$ for some complex number $c_n$ of absolute value one. Then for every finite set of indices $F \subseteq J$,

$\sum_{n \in F} |\langle t(w) e_n , e_n \rangle| = \sum_{n \in F} c_n \langle t(w) e_n, e_n \rangle =$

$= \sum_{n \in F} c_n w(t_{e_n, e_n}) \leq \|w\| \|\sum_{n \in F} c_n t_{e_n,e_n}\| \leq \|w\|$.

This shows that the series converges.

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

Now, suppose that $t(w) \geq 0$ (note that in this case, the first part of the proof shows that $t(w)$ can be supported on at most countably many basis vectors; we may therefore assume that $H$ is separable. This is not crucial.)

By the spectral theorem (Lecture 1), we may assume that $t(w) = M_f$ is a multiplication operator for $f \geq 0$. To prove that $t(w) = M_f$ is compact, it suffices to prove that the spectral measure $E$ of $t(w)$ satisfies that $E(\{z : |z| > \epsilon\})$ (which is $\equiv \chi_{\{f>\epsilon\}}$) has finite rank for all $\epsilon >0$ (recall Exercise M in Lecture 1). But $f>\epsilon$ on the range of $E(\{z : |z| > \epsilon\})$, so $\langle M_f g, g \rangle \geq \epsilon \|g\|^2$ for all $g$ in the range; if the range was infinite dimensional this would contradict the summability of the series established above.

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

Now we can compute $K(H)^*$. For every $f,g \in H$, it is clear that $w_{f,g} \in B(H)^*$, and in particular it restricts to a bounded functional on $K(H)$. Moreover, since $\|w_{f,g}\| = \|f\|\|g\|$, it certainly holds that for every sequence $(c_n) \in \ell^1$ and every two orthonormal sequences $\{f_n\}, \{g_n\}$, the series

$\sum_n c_n w_{f_n, g_n}$

converges to a bounded functional in $K(H)^*$ and in $B(H)^*$. Now we shall see that all bounded functionals of $K(H)^*$ have this form.

Theorem 4: Let $w \in K(H)^*$. Then there is a sequence $(c_n) \in \ell^1$ and two orthonormal sequences $\{f_n\}, \{g_n\}$, such that

$w = \sum_n c_n w_{f_n, g_n}$.

Proof: From what we have gathered until now, we know that there is an operator $t(w) \in B(H)$ such that

$w(t_{f,g}) = \langle t(w) f, g \rangle$

for all $f,g \in H$. By Lemma 3, $t(w) \in K(H)$. By Lemma 1, $t(w) = \sum_n c_n t_{f_n,g_n}$ for a series $(c_n)$ of positive numbers that is either finitely supported or converges to zero, and $\{f_n\}_{n \in J}, \{g_n\}_{n \in J}$ two orthonormal sequences. For every finite $F \subseteq J$, we can define $A = \sum_{n \in F} t_{g_n,f_n}$. Then $A$ is a compact operator, and

$w(A) = \sum_{n \in F} w(t_{g_m,f_m}) = \sum_{n \in F} \langle t(w)g_m, f_m \rangle = \sum_{n \in F} c_n$.

But $\|A\| \leq 1$. We see that $\sum_{n\in J} c_n \leq \|w\|$.

Definition 5: Let $L^1(H)$ denote the space of all bounded operators $T \in B(H)$ for which there is some $w \in K(H)^*$ such that $T = t(w)$. Equivalently, $L^1(H)$ is the space of all $T \in B(H)$ such that

$Trace(|T|) := \sum_n \langle |T|e_n, e_n \rangle < \infty$

for any (or every) orthonormal sequence $\{e_n\}$. The space $L^1(H)$ is called the space of trace class operators. For every $T \in L^1(H)$ we define the trace norm $\|T\|_1$ to be the norm $\|w\|$ of the linear functional $w$ for which $T = t(w)$.

Remark 6: By Theorem 4, every $w \in K(H)^*$ extends naturally in a unique way to a bounded functional on $B(H)$ (the extension is unique only among “natural” extensions). For every $T \in L^1(H)$, if $w \in K(H)^*$  satisfies $T = t(w)$, we write $Trace(T) = w(1)$, and more generally, for any $A \in B(H)$ we write $w(A) = Trace(TA)$. Recall that in functional analysis, one writes the “pairing” between a space $X$ and its dual $X^*$ as

$\langle x, f \rangle := f(x)$    for     $x \in X, f \in X^*$.

In our setting, we identify $L^1(H)$ with $K(H)^*$, and if $T = t(w)$, we write

$w(A) = \langle A, w \rangle = \langle A, T \rangle = Trace(AT)$.

Exercise D: Prove that $B(H)^* \supsetneq L^1(H)$.

Exercise E: Every $w \in K^*(H)$ is WOT continuous on $B(H)_1$, and hence extends to a unique such functional on $B(H)$.

Every $w = \sum c_n w_{f_n,g_n} \in K(H)^*$ (where $(c_n) \in \ell^1$ and $\{f_n\}, \{g_n\}$ are orthonormal sequences) extends to be a bounded linear functional on $B(H)$, defined by $A \mapsto w(A) = \sum c_n \langle Af_n, g_n \rangle$. Therefore, every $A \in B(H)$ determines a bounded linear functional on $K(H)^*$ by

$\langle w, A \rangle = \langle A, w \rangle = w(A)$.

The map $A \mapsto \langle \cdot , A \rangle \in (K(H)^*)^*$ is called the canonical map.

Theorem 7: The canonical map is an isometric isomorphism of $B(H)$ onto $K(H)^{**} \cong L^1(H)^*$

Proof: For every pair of unit vectors $f,g \in H$, the functional $w_{f,g} \in K(H)^*$. This can be used to show that the canonical map is an injective map, an in fact it is norm-nondecreasing:

$\|\langle \cdot, A \rangle \| \geq |\langle w_{f,g}, A \rangle| = |\langle Af, g \rangle|$ which approximates $\|A\|$.

On the other hand, if $w = \sum c_n w_{f_n,g_n} \in K(H)^*$, then $w(A) = \lim_n w(P_n A P_n)$ for any net of finite rank projections $P_n$ increasing to $I_H$, so $\| \langle \cdot , A \rangle\| \leq \|A\|$, therefore the canonical map is isometric.

Now if $\phi \in K(H)^{**}$, then we define a bounded sesqui-linear form on $H$:

$\langle \langle f, g \rangle \rangle = \phi(w_{f,g})$.

One finds that there is some bounded operator $T_\phi \in B(H)$, such that

$\langle \langle f, g \rangle \rangle = \langle T_\phi f , g \rangle$ for all $f,g \in H$.

Since the linear span of the functionals of the form $w_{f,g}$ is dense in $K(H)^*$, we see that $\phi$ is the image of $T_\phi$ under the canonical map.

We henceforth identify $K(H)^*$ with $L^1(H)$, and $B(H)$ with $L^1(H)^*$. It is also common to denote $L^1(H) = B(H)_*$, and to refer to it as the predual of $B(H)$ (an alert student should worry about the word “the”; until we show that the predual of a von Neumann algebra is unique, we can refer to it more precisely as the canonical predual or, to be completely on the safe side, the standard predual).

Exercise F: (Tying all loose ends) Prove that for every positive $T \in L^1(H)$,

$\|T\|_1 = Trace(T) := \sum_n \langle T e_n , e_n \rangle$,

where $\{e_n\}$ is an orthonormal basis for $H$. Show that the right hand sum is independent of the particular orthonormal basis. (Recall that we defined $\|T\|_1$ to be the norm of the corresponding functional in $K(H)^*$). For a not-necessarily positive $T \in L^1(H)$, prove that $\|T\|_1 = \||T|\|_1 = Trace(|T|)$, and that

$Trace(T) = \sum \langle T e_n , e_n \rangle$

Converges absolutely, and to the same value, for every choice of orthonormal basis $\{e_n\}$. Prove that $L^1(H)$ is an ideal in $B(H)$. Prove that for every $A \in B(H)$ and $T \in L^1(H)$,

$\langle T, A \rangle = Trace(TA)$.

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

Since $B(H) = L^1(H)^*$, we can consider the weak-* topology on it.

Definition 8: The $\sigma$-weak operator topology on $B(H)$ (or just $\sigma$-weak, or ultraweak topology) is the weak-* topology $\sigma(B(H), L^1(H))$ induced on $B(H) = L^1(H)^*$ by its predual $B(H)_* = L^1(H)$.

Thus, $T_n \to T$ in the $\sigma$-weak operator topology, if and only if

$\langle T_n, X \rangle \to \langle T, X\rangle$ for all $X \in L^1(H)$.

Equivalently, the $\sigma$-weak topology is determined by the seminorms

$p(T) =| \sum_n \langle T x_n, y_n \rangle|$

where $\{x_n\}, \{y_n\} \subseteq H$ satisfy $\sum \|x_n\|^2, \sum_n \|y_n\|^2 < \infty$.

(Recall that a topology is said to be generated by a family $\{p_\alpha\}$ of seminorms if convergence of a net $x_n \to x$ is determined by convergence $p_\alpha(x-x_n) \to 0$ for all $\alpha$. Thus, the strong (operator) topology is the topology generated by the family of seminorms $p(T) = \|Tx\|$, $x \in H$.)

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

$p(T) = \left( \sum_n \|T x_n\|^2\right)^{1/2}$,

where $\sum \|x_n\|^2 < \infty$.

Definition 10: The strong * topology is the topology generated by the seminorms $p(T) = \|Tx\| + \|T^*x\|$, $x \in H$. The $\sigma$-strong * topology is defined similarly.

Exercise G: The whatever-strong topology is strictly stronger than the whatever-weak topology. The $\sigma$whatever topology is strictly stronger than the whatever topology, but they coincide on the unit ball $B(H)_1$. The whatever * topology is strictly stronger than the whatever topology. All are strictly weaker than the norm topology.

****

Theorem 11: Let $A \subseteq B(H)$ be a von Neumann algebra. Then $A$ is closed in all of the above topologies. Consequently, $A$ is a dual Banach space. To be precise, if we let $A_*$ denote the subspace of $A^*$ consisting of all $\sigma$-weakly continuous functionals, then $A$ can be isometrically isomorphically identified with $(A_*)^*$

Proof: Since $A$ is WOT closed, it is closed in the weakest, and hence in all, of the topologies. To see that $A$ is a dual space, we consider it as a weak-* closed subspace of $B(H) = L^1(H)^*$. Let

$A_\perp = \{t \in L^1(H) : \langle a, t \rangle = 0$  for all  $a \in A \}$.

Then, being weak-* closed,

$A = (A_\perp)^\perp = \{b \in B(H) : \langle b, t \rangle = 0$ for all $t \in A_\perp\}$.

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

$(L^1(H) / A_\perp)^* \cong (A_\perp)^\perp = A \subseteq B(H) = L^1(H)^*$

by a natural map $\phi \mapsto \phi \circ \pi$ (where $\pi : L^1(H) \to L^1(H)/A_\perp$ is the quotient map). Thus

$A \cong (L^1(H)/A_\perp)^*$,

and it remains to observe that this isomorphism respects the $\sigma$-weak functionals. For this, note that the restriction map $L^1(H) \to A^*$, given by $\phi \mapsto \phi\big|_A$, induces an isomorphism of $L^1(H)/A_\perp$ with its image – the weak-* functionals on $A$.

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

We saw above that $K(H)^{**} = B(H)$. In this section, we will see that the double dual of any C*-algebra is a von Neumann algebra. In contrast with everything we have done until this point, our C*-algebras will be just abstract C*-algebras: Banach *-algebras that satisfy the C*-identity: $\|a^*a\| = \|a\|^2$. We will use basic results of the theory (with due apologies) when we need them.

Lemma 12: Let $A$ be a C*-algebra, let $\pi : A \to B(H)$ be a *-representation, and define $M = \pi(A)''$. Let $i : A \to A^{**}$ be the canonical embedding. Then there exists a unique linear map $\tilde{\pi} : A^{**} \to M$, which is surjective, continuous with respect to the $\sigma(A^{**},A^*)$ and $\sigma$-weak topologies, which extends $\pi$ (in the sense that $\tilde{\pi} \circ i = \pi$). Moreover, $\tilde{\pi}$ maps the unit ball of $A^{**}$ onto the unit ball of $M$

Proof: Consider $\pi$ as a map between the Banach spaces $A \to M$. Let $\pi^* : M^* \to A^*$ be the adjoint map. Let $\pi_* = (\pi^*)\big|_{M_*}$ be the restriction of $\pi^*$ to the space of $\sigma$-weakly continuous functionals. So $\pi_* : M_* \to A^*$. Now define $\tilde{\pi} = (\pi_*)^* : A^{**} \to (M_*)^* = M$. This satisfies for all $x \in M_*$, that

$\langle x, \tilde{\pi} \circ i(a) \rangle = \langle \pi_*(x), i(a) \rangle = \langle a, \pi^*(x) \rangle = \langle \pi(a), x \rangle$.

This shows that $\tilde{\pi} = \pi\circ i$. The continuity is a “general nonsense” fact which always holds for adjoints: indeed, if $u_n \to u$ in $\sigma(A^{**},A^*)$, then $\langle \tilde{\pi} u_n, x \rangle = \langle u_n, \pi^*(x) \rangle \to \langle u, \pi^*(x) \rangle = \langle \tilde{\pi}(u), x \rangle$ for all $x \in M_*$.

Finally, to show the map is surjective, it suffices to show that it takes the closed unit ball $(A^{**})_1$ onto the closed unit ball $M_1$. Since $\tilde{\pi}$ is continuous, it takes $(A^{**})_1$ onto a $\sigma$-weakly compact set. But this compact set contains the open unit ball of $\pi(A)$ (because that’s what *-homomorphisms do, being the composition of a quotient and an injective (hence isometric) *-homomorphism). But by Kaplansky (and Exercise F) the open unit ball of $\pi(A)$ is $\sigma$-weakly dense in $M_1$.

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

$\pi_u = \bigoplus_{\rho \in S(A)}\pi_\rho : A \to B(H_u)$,

where $H_u = \oplus_{\rho \in S(A)} H_\rho$ (here, given a state $\rho \in S(A)$, $(\pi_\rho, H_\rho, \xi_\rho)$ is the GNS representation of $\rho$). Since $\pi_u$ is an isometric isomorphism (by the Gelfand-Naimark theorem), we can identity $A$ with $\pi_u(A)$ and consider $A$ as a C*-subalgebra of $\pi_u(A)''$.

Definition 13: The algebra $M_u:=\pi_u(A)''$ is called the universal enveloping von Neumann algebra of $A$

Theorem 14: The map $\tilde{\pi}_u : A^{**} \to M_u$ (given by Lemma 12) is isometric, hence $M_u \cong A^{**}$. Every bounded functional on $A$ extends to a $\sigma$-weakly continuous continuous functional on $M_u$. The universal enveloping von Neumann algebra $M_u$ has the following universal property: if $\phi : A \to B(H)$ is a *-representation, then there exists a unique $\sigma$-weakly continuous $\overline{\phi}$ of $M_u$ onto $\phi(A)''$ such that $\phi = \overline{\phi} \circ \pi_u$

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

We will now show that $\pi_*$ is surjective, this will show that the map $\tilde{\pi} = (\pi_*)^*$ is injective, and since it maps an open unit ball onto an open unit ball it must be isometric and surjective. To see that $\pi_*$ is surjective, we just recall that it is defined to be $(\pi^*)\big|_{M_*}$. But $\pi^*$ is the restriction map $M^* \to A^*$ given by $f \mapsto f\big|_{A}$ (being the conjugate of an inclusion $A \to M$), and since every $f \in A^*$ extends to a map in $M_*$ (by the previous paragraph), this shows that $\pi_*$ is surjective.

Finally, given $\phi : A \to B(H)$, let $\tilde{\phi} : A^{**} \to \phi(A)''$ be as in Lemma 12. Then $\overline{\phi} := \tilde{\phi} \circ \tilde{\pi}^{-1}$ is a surjective and $\sigma$-weakly continuous linear map. Since it is a *-homomorphism on the $\sigma$-dense subspace $A$, it is a representation. We get $\tilde{\phi} = \overline{\phi} \circ \tilde{\pi}$, and restricting to $A$ we get the final assertion.

Exercise H: Prove that every $f \in A^*$ is the linear combination of four states. Conclude that every $f \in A$ can be written as $f(a) = \langle \pi(a)x,y \rangle$, for some representation $\pi : A \to B(H)$ and $x,y \in H$.

Exercise I: Prove that if $\psi : A \to B(H)$ is a *-representation, $N = \psi(A)''$, and if $(\psi, N)$ have the same universal property as $(M_u, \pi_u)$, then $N$ is $\sigma$-weakly continuously *-isomorphic to $M_u$ by a map that fixes $A$.