43.4 Normal Maps

Proof. Assume without loss of generality that there exist $\xi, \eta \in H$ such that $\dpn{x, \omega}{M}= \dpn{x\xi, \eta}{H}$ for all $x \in M$. In which case, by polarisation,

for all $x \in M$.$\square$

Proof. By the Cauchy-Schwarz inequality, for each $x, y \in M$,

so the mapping $(x, y) \mapsto \omega(y^{*}x)$ is a bounded sesquilinear form on $M\xi$. By the Riesz Representation Theorem, there exists $z \in B(\ol{M\xi})$ such that $\omega(y^{*}x) = \dpn{zx\xi, y\xi}{H}$ for all $x, y \in M$. In particular, $0 \le \omega(x^{*}x) = \dpn{zx\xi, x\xi}{H}$ for all $x \in M$, so $z \ge 0$. Moreover, for any $x, y, t \in M$,

so $z \in M'_{+}$. Finally, let $p: H \to \ol{M\xi}$ be the orthogonal projection onto $\ol{M\xi}$, and $x' = (zp)^{1/2}\in M'$, then for each $x \in M$,

$\square$

Proof, [Lemma 2.5.6., AP]. Let $\eta \in H$ with $\varphi(1_{M}) < \dpn{\eta, \eta}{H}$. Let $\mathcal{P}\subset M$ be a maximal family of mutually orthogonal projections with $\varphi(p) \ge \dpn{p\eta, \eta}{H}$ for all $p \in \mathcal{P}$. By assumption,

Let $p := 1_{M} - \sum_{q \in \mathcal{P}}q > 0$, then by maximality of $\mathcal{P}$, $\varphi(q) < \dpn{q\eta, \eta}{H}$ for all subprojections $q \le p$.

Let $x \in (pMp)_{+}$ with $0 \le x \le 1$. By Theorem 43.7.2, there exist projections $\seq{q_n}\subset pMp$ such that $x = \sum_{n = 1}^{\infty} \frac{q_{n}}{2^{n}}$. In which case^[1],

Therefore $\dpn{x\eta, \eta}{H}\ge \varphi(x)$ for all $x \in (pMp)_{+}$. By the Radon-Nikodym theorem for positive functionals, there exists $\xi \in pH$ with $\varphi(x) = \dpn{x\xi, \xi}{H}$ for all $x \in pMp$.$\square$

Proof, [Theorem 2.5.5, AP]. (2) $\Rightarrow$ (3): Let $\mathcal{P}\subset M$ be a maximal family of mutually orthogonal projections such that for each $p \in \mathcal{P}$, there exists $\xi_{p} \in pH$ such that $\omega(x) = \dpn{x\xi_p, \xi_p}{H}$ for all $x \in pMp$. Let $q = \sum_{p \in \mathcal{P}}p$, then by maximality of $\mathcal{P}$ and Lemma 43.4.3 applied to $(1 - q)M(1 - q)$, $\sum_{p \in \mathcal{P}}p = 1_{M}$.

For each $x \in M$ and $p \in \mathcal{P}$, by the Cauchy-Schwarz inequality,

so the mapping $y \xi_{p} \mapsto \omega(yp)$ extends into a bounded linear functional on $\ol{M\xi_p}$. By the Riesz Representation Theorem, there exists $\eta_{p} \in \ol{M\xi_p}$ such that $\omega(xp) = \dpn{x\xi_p, \eta_p}{H}$ for all $x \in M$.

Let $\mathcal{F}\subset 2^{\mathcal{P}}$ be the collection of all finite subsets of $\mathcal{P}$, directed by inclusion. For each $F \in \mathcal{F}$, let $q_{F} = \sum_{p \in F}p$ and $\omega_{F} \in M^{*}$ be defined by $\omega_{F}(x) = \omega(q_{F}xq_{F})$, then for each $x \in M$,

so $\omega_{F}$ is weak-operator continuous. Finally, for any $x \in M$,

By (2), $\lim_{F \in \mathcal{F}}\omega(1 - q_{F})^{1/2}= 0$. Therefore $\omega = u\text{-}\lim_{F \in \mathcal{F}}\omega_{F}$.$\square$

Proof. By (2) of Definition 43.4.4.$\square$

Proof, [Proposition 2.5.8, AP]. (2) $\Rightarrow$ (3): Let $\omega \in N^{*}$ be a weak-operator continuous functional. By Lemma 43.4.1, there exist positive weak-operator continuous functionals $\bracsn{\omega_k}_{0}^{3}$ such that $\omega = \sum_{k = 0}^{3} i^{k} \omega_{k}$. By assumption, $\omega_{k} \circ \phi$ is weak-operator continuous on bounded subsets of $M$. In which case, $\omega \circ \phi = \sum_{k = 0}^{3} i^{k} \omega_{k} \circ \phi$ is weak-operator continuous on bounded subsets of $M$. Therefore $\phi$ is weak-operator continuous on bounded subsets of $M$.

(3) $\Rightarrow$ (1), (4) $\Rightarrow$ (1): By (1) of Theorem 43.3.1.

(3) $\Rightarrow$ (4)^[2]: Suppose that $M$ acts on the Hilbert space $H$, and $N$ acts on the Hilbert space $K$. Let $\angles{x_\alpha}_{\alpha \in A}\subset M$ be a bounded net converging to $x \in M$ in the strong operator topology, then for each $\xi \in H$,

By polarisation, $\wotlim_{\alpha \in A}(x - x_{\alpha})^{*}(x - x_{\alpha}) = 0$. As $\angles{x_\alpha}_{\alpha \in A}$ is bounded,

as well. In particular, for any $\xi \in K$,

Therefore $\phi$ restricted to bounded subsets of $M$ is strong-operator continuous.$\square$

Proof. As $\phi$ is an isomorphism, it is an order isomorphism. By (1) of Definition 43.4.6, $\phi$ is normal. By (3) and (4) of Definition 43.4.6, its restriction to bounded subsets of $M$ is both weak-operator and strong-operator continuous.$\square$

Proof, [Proposition 2.5.11, AP]. Let $\varphi = \psi \circ \phi \le \omega$, then $\varphi$ is normal by Lemma 43.4.5. For any bounded, upward-directed family $A \subset M$,

As $\psi$ is faithful, $\sup(\phi(A)) = \phi(\sup(A))$. Therefore $\phi$ is normal by (1) of Definition 43.4.6.$\square$

Proof. Let $N = \pi(M)''$ be the von Neumann algebra generated by $\pi(M)$. By the Kaplansky Density Theorem, $\ol{B_N(0, 1)}^{B(H)}= \ol{B_{\pi(M)}(0, 1)}^{\text{\small WOT}}$. By Proposition 5.16.3 and Proposition 43.1.3, $\ol{B_M(0, 1)}^{M}$ is weak-operator compact, so $\pi(\ol{B_M(0, 1)}^{M})$ is weak-operator compact and contains $\pi(B_{M}(0, 1))$. Thus

and $\ol{B_{\pi(M)}(0, 1)}^{B(H)}= \ol{B_{N}(0, 1)}^{B(H)}$, so $\pi(M) = \ol{\pi(M)}^{B(H)}= N$ by Theorem 39.13.3.$\square$

Proof. (1): Suppose that $\varphi$ is normal. Since $\pi(M)\xi$ is dense in $H$, the Arzelà-Ascoli Theorem implies that the weak operator topology on $\ol{B_{\pi(M)}(0, 1)}$ coincides with the topology of pointwise convergence on $\bracsn{\eta \otimes \theta|\eta, \theta \in \pi(M)\xi}$. For each $x, y, z \in M$,

By separate continuity of composition in the weak operator topology, the functional $x \mapsto \dpn{\pi(x)\pi(y)\xi, \pi(z)\xi}{H}$ is weak-operator continuous. As such, $\pi|_{\ol{B_M(0, 1)}}: \ol{B_M(0, 1)}\to \ol{B_{\pi(M)}(0, 1)}$ is weak-operator continuous.

(2): By Proposition 43.4.9.$\square$