Housekeeping: moved von Neumann algebras to its own part.
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Bokuan Li
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\section{Normal Maps}
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\label{section:ultraweak}
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\begin{lemma}[Jordan Decomposition]
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\label{lemma:weak-operator-jordan}
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Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $\omega: M \to \complex$ be a weak-operator continuous linear functional, then there exist positive weak-operator continuous functionals $\bracsn{\omega_k}_0^3$ such that $\omega = \sum_{k = 0}^3 i^k \omega_k$.
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\end{lemma}
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\begin{proof}
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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 \hyperref[polarisation]{proposition:polarisation-complex},
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\[
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\dpn{x, \omega}{M} = \dpn{x\xi, \eta}{H} = \frac{1}{4}\sum_{k = 0}^3 i^k \dpn{x(\xi + i^k\eta), \xi + i^k\eta}{H}
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\]
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for all $x \in M$.
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\end{proof}
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\begin{lemma}[Radon-Nikodym]
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\label{lemma:vna-radon-nikodym}
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Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, $\omega: M \to \complex$ be a positive linear functional, and $\xi \in H$ such that $\omega(x) \le \dpn{x\xi, \xi}{H}$ for all $x \in M_+$, then there exists $x' \in M'_+$ such that $\omega(x) = \dpn{x(x'\xi), x'\xi}{H}$ for all $x \in M$.
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\end{lemma}
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\begin{proof}
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By the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, for each $x, y \in M$,
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\[
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|\omega(y^*x)|^2 \le \omega(x^*x)\omega(y^*y) \le \norm{x\xi}_H^2 \norm{y\xi}_H^2
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\]
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so the mapping $(x, y) \mapsto \omega(y^*x)$ is a bounded sesquilinear form on $M\xi$. By the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}, 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 $x, y, t \in M$,
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\[
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\dpn{ztx\xi, y\xi}{H} = \omega(y^*tx) = \omega((t^*y)^*x) = \dpn{zx\xi, t^*y\xi}{H} = \dpn{tzx\xi, y\xi}{H}
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\]
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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 $x \in M$,
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\[
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\omega(x) = \dpn{zx\xi, \xi}{H} = \dpn{zpx\xi, \xi}{H} = \dpn{x(zp)^{1/2}\xi, (zp)^{1/2}\xi}{H}
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\]
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\end{proof}
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\begin{lemma}
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\label{lemma:completely-additive-projection-choose}
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Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $\varphi: M \to \complex$ be a positive linear functional. If for any mutually orthogonal projections $\mathcal{P} \subset M$, $\varphi\paren{\sum_{p \in \mathcal{P}}p} = \sum_{p \in \mathcal{P}}\varphi(p)$, then there exist a non-zero projection $p \in M$ and $\xi \in pH$ such that $\varphi(x) = \dpn{x\xi, \xi}{H}$ for all $x \in pMp$.
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\end{lemma}
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\begin{proof}[Proof, {{\cite[Lemma 2.5.6.]{AnantharamanPopa}}}. ]
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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,
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\begin{align*}
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\varphi\paren{\sum_{p \in \mathcal{P}}p} &\ge \angles{\sum_{p \in \mathcal{P}}p\eta, \eta}_H \\
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\varphi\paren{1_M - \sum_{p \in \mathcal{P}}p}&= \varphi(1_M) - \varphi\paren{\sum_{p \in \mathcal{P}}p} \\
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&< \dpn{\eta, \eta}{H} - \angles{\sum_{p \in \mathcal{P}}p\eta, \eta}_H \ge 0
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\end{align*}
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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 $q \le p$.
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Let $x \in (pMp)_+$ with $0 \le x \le 1$. By \autoref{theorem:vn-projection-norm-dense}, there exist projections $\seq{q_n} \subset pMp$ such that $x = \sum_{n = 1}^\infty \frac{q_n}{2^n}$. In which case\footnote{Here, only continuity of $\varphi$ is used.},
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\[
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\dpn{x\eta, \eta}{H} = \sum_{n = 1}^\infty 2^{-n}\dpn{q_n\eta, \eta}{H} \ge \sum_{n = 1}^\infty \frac{\varphi(q_n)}{2^n} = \varphi(x)
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\]
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Therefore $\dpn{x\eta, \eta}{H} \ge \varphi(x)$ for all $x \in (pMp)_+$. By the \hyperref[Radon-Nikodym theorem for positive functionals]{lemma:vna-radon-nikodym}, there exists $\xi \in pH$ with $\varphi(x) = \dpn{x\xi, \xi}{H}$ for all $x \in pMp$.
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\end{proof}
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\begin{definition}[Normal Linear Functional]
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\label{definition:vna-normal-functional}
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Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $\omega: M \to \complex$ be a positive linear functional, then the following are equivalent:
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\begin{enumerate}
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\item For any bounded, upward-directed family $A \subset M$, $\omega(\sup(A)) = \sup(\omega(A))$.
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\item For any mutually orthogonal projections $\mathcal{P} \subset M$, $\omega\paren{\sum_{p \in \mathcal{P}}p} = \sum_{p \in \mathcal{P}}\omega(p)$.
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\item $\omega$ is the norm limit of a sequence of weak-operator continuous positive linear functionals.
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\item $\omega$ is weak-operator continuous on bounded subsets of $M$.
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\item $\omega$ is strong-operator continuous on bounded subsets of $M$.
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\end{enumerate}
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\end{definition}
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\begin{proof}[Proof, {{\cite[Theorem 2.5.5]{AnantharamanPopa}}}. ]
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(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 \autoref{lemma:completely-additive-projection-choose} applied to $(1 - q)M(1 - q)$, $\sum_{p \in \mathcal{P}}p = 1_M$.
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For each $x \in M$ and $p \in \mathcal{P}$, by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz},
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\[
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|\omega(xp)|^2 \le \omega(1) \cdot \omega(px^*xp) = \omega(1) \cdot \norm{x\xi_p}_H^2
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\]
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so the mapping $y \xi_p \mapsto \omega(yp)$ extends into a bounded linear functional on $\ol{M\xi_p}$. By the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}, there exists $\eta_p \in \ol{M\xi_p}$ such that $\omega(xp) = \dpn{x\xi_p, \eta_p}{H}$ for all $x \in M$.
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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 $\omega_F(x) = \omega(q_Fxq_F)$, then for each $x \in M$,
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\[
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\omega_F(x) = \sum_{p \in F}\omega(q_Fxp) = \sum_{p \in F}\dpn{q_Fx\xi_p, \eta_p}{H}
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\]
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so $\omega_F$ is weak-operator continuous. Finally, for any $x \in M$,
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\begin{align*}
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|\omega(x) - \omega_F(x)| &\le |\omega(x) - \omega(xq_F)| + |\omega(xq_F) - \omega(q_Fxq_F)| \\
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&= |\omega(x(1_M - q_F))| + |\omega((1 - q_F)xq_F)| \\
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&\le 2\norm{x}_M\omega(1)^{1/2} \cdot \omega(1 - q_F)^{1/2}
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\end{align*}
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By (2), $\lim_{F \in \mathcal{F}}\omega(1 - q_F)^{1/2} = 0$. Therefore $\omega = u\text{-}\lim_{F \in \mathcal{F}}\omega_F$.
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\end{proof}
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\begin{lemma}
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\label{lemma:normal-functional-bound}
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Let $M$ be a von Neumann algebra, $\varphi, \omega \in M^*$ be positive linear functionals. If $\omega$ is normal and $\varphi \le \omega$, then $\varphi$ is also normal.
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\end{lemma}
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\begin{proof}
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By (2) of \autoref{definition:vna-normal-functional}.
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\end{proof}
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\begin{definition}[Normal Positive Map]
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\label{definition:normal-linear-map}
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Let $M, N$ be von Neumann algebras and $\phi: M \to N$ be a positive linear map, then the following are equivalent:
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\begin{enumerate}
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\item For any bounded, upward-directed family $A \subset M$, $\phi(\sup(A)) = \sup(\phi(A))$.
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\item For any normal, positive linear functional $\omega \in N^*$, $\omega \circ \phi$ is a normal, positive linear functional.
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\item $\phi$ restricted to bounded subsets of $M$ is continuous with respect to the weak operator topologies.
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\end{enumerate}
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If $\phi$ is in addition a *-homomorphism, then the above are also equivalent to
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\begin{enumerate}[start=3]
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\item $\phi$ restricted to bounded subsets of $M$ is continuous with respect to the strong operator topologies.
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\end{enumerate}
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If the above holds, then $\phi$ is \textbf{normal}.
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\end{definition}
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\begin{proof}[Proof, {{\cite[Proposition 2.5.8]{AnantharamanPopa}}}. ]
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(2) $\Rightarrow$ (3): Let $\omega \in N^*$ be a weak-operator continuous functional. By \autoref{lemma:weak-operator-jordan}, 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 $M$.
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(3) $\Rightarrow$ (1), (4) $\Rightarrow$ (1): By (1) of \autoref{theorem:existence-of-projections-vna}.
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(3) $\Rightarrow$ (4)\footnote{Central here is the observation that $x_\alpha \to 0$ in the strong operator topology i.f.f. $x_\alpha^*x_\alpha \to 0$ in the weak operator topology. }: 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 $\xi \in H$,
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\[
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\lim_{\alpha \in A}\dpn{(x - x_\alpha)^*(x - x_\alpha)\xi, \xi}{H} = \lim_{\alpha \in A}\normn{(x - x_\alpha)\xi}_H^2 = 0
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\]
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By \hyperref[polarisation]{proposition:polarisation-complex}, $\wotlim_{\alpha \in A}(x - x_\alpha)^*(x - x_\alpha) = 0$. As $\angles{x_\alpha}_{\alpha \in A}$ is bounded,
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\[
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\wotlim_{\alpha \in A}\phi((x - x_\alpha)^*(x - x_\alpha)) = 0
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\]
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as well. In particular, for any $\xi \in K$,
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\[
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\lim_{\alpha \in A}\norm{\phi(x - x_\alpha)\xi}_K^2 = \lim_{\alpha \in A}\dpn{\phi(x - x_\alpha)^*\phi(x - x_\alpha)\xi, \xi}{K} = 0
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\]
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Therefore $\phi$ restricted to bounded subsets of $M$ is strong-operator continuous.
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\end{proof}
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\begin{proposition}
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\label{proposition:vna-isomorphism-normal}
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Let $M, N$ be von Neumann algebras, then every isomorphism $\phi: M \to N$ is normal. In particular, the weak and strong operator topologies on $M$, restricted to bounded subsets, are independent of its representation.
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\end{proposition}
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\begin{proof}
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As $\phi$ is an isomorphism, it is an order isomorphism. By (1) of \autoref{definition:normal-linear-map}, $\phi$ is normal. By (3) and (4) of \autoref{definition:normal-linear-map}, its restriction to bounded subsets of $M$ is both weak-operator and strong-operator continuous.
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\end{proof}
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\begin{proposition}
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\label{proposition:vna-normal-bound}
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Let $M, N$ be von Neumann algebras, $\omega \in M^*$ and $\psi \in N^*$ be normal positive linear functionals with $\psi$ being faithful, and $\phi: M \to N$ be a positive linear map with $\psi \circ \phi \le \omega$, then $\phi$ is normal.
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\end{proposition}
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\begin{proof}[Proof, {{\cite[Proposition 2.5.11]{AnantharamanPopa}}}. ]
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Let $\varphi = \psi \circ \phi \le \omega$, then $\varphi$ is normal by \autoref{lemma:normal-functional-bound}. For any bounded, upward-directed family $A \subset M$,
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\[
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\psi(\sup(\phi(A))) =\sup(\psi(\phi(A))) = \sup(\varphi(A)) = \varphi(\sup(A)) = \psi(\phi(\sup(A)))
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\]
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As $\psi$ is faithful, $\sup(\phi(A)) = \phi(\sup(A))$. Therefore $\phi$ is normal by (1) of \autoref{definition:normal-linear-map}.
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\end{proof}
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\begin{proposition}
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\label{proposition:normal-image-vna}
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Let $M$ be a von Neumann algebra, $H$ be a complex Hilbert space, and $\pi: M \to B(H)$ be a normal unital *-homomorphism, then $\pi(M)$ is a von Neumann algebra.
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\end{proposition}
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\begin{proof}
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Let $N = \pi(M)''$ be the von Neumann algebra generated by $\pi(M)$. By the \hyperref[Kaplansky Density Theorem]{theorem:kaplansky-density}, $\ol{B_N(0, 1)}^{B(H)} = \ol{B_{\pi(M)}(0, 1)}^{\text{\small WOT}}$. By \autoref{proposition:compact-extensions} and \autoref{proposition:bh-ultraweak-bounded}, $\ol{B_M(0, 1)}^M$ is weak-operator compact, so $\pi(\ol{B_M(0, 1)}^M)$ is weak-operator compact and contains $\pi(B_M(0, 1))$. Thus
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\[
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\ol{B_{\pi(M)}(0, 1)}^{B(H)} \supset \pi(\ol{B_M(0, 1)}^M) \supset \ol{B_{\pi(M)}(0, 1)}^{\text{\small WOT}} \supset \ol{B_N(0, 1)}^{B(H)}
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\]
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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 \autoref{theorem:cstar-quotient}.
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\end{proof}
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\begin{lemma}
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\label{lemma:tracial-von-neumann-gns}
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Let $M$ be a von Neumann algebra, $\varphi \in S(M)$ be a state, and $(H, \pi, \xi)$ be the \hyperref[GNS triple]{definition:gns-triple} associated with $\varphi$, then $\pi$ is normal if and only if $\varphi$ is normal. In which case, $\pi(M) \subset B(H)$ is a von Neumann algebra.
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\end{lemma}
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\begin{proof}
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(1): Suppose that $\varphi$ is normal. Since $\pi(M)\xi$ is dense in $H$, the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli} 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 $x, y, z \in M$,
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\[
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\dpn{\pi(x)\pi(y)\xi, \pi(z)\xi}{H} = \dpn{\pi(z^*)\pi(x)\pi(y)\xi, \xi}{H} = \varphi(z^*xy)
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\]
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By \hyperref[separate continuity of composition]{proposition:bh-operator-topologies-facts} 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.
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(2): By \autoref{proposition:normal-image-vna}.
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\end{proof}
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\begin{definition}[Separable]
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\label{definition:separable-vna}
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Let $M$ be a von Neumann algebra, then $M$ is \textbf{separable} if there exists a separable Hilbert space $H$ and a faithful normal representation $\pi: M \to B(H)$.
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\end{definition}
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