Exercises on von Neumann Algebra

$\textbf{4.4}$ Let $A$ be a von Neumann algebra on a Hilbert space $H$, and suppose that $\tau$ is a bounded linear functional on $A$. We say that $\tau$ is normal if, whenever an increasing net $(u_\lambda)_{\lambda\in \Lambda}$ in $A_{sa}$ converges strongly to an operator $u\in A_{sa}$, we have $\lim_{\lambda}\tau(u_\lambda)=\tau(u)$. Show that every $\sigma$-weakly continuous functional $\tau\in A^*$ is normal. $\newcommand{\vN}{\text{von Neumann algebra}}\renewcommand{\l}[1]{\lVert #1\rVert}$ $\textbf{Lemma}.$ Suppose $u,v$ are two bounded linear operators on some Hilbert space, then

\begin{equation*} \lvert uv\rvert\leq \lVert u\rVert\lvert v\rvert. \end{equation*}

Indded,Let $v=w\lvert u\rvert$ and $uv=w'\lvert uv\rvert$ be the Polar Decomposition, then $\lvert uv\rvert^2=w'^*uw\lvert v\rvert$, thus \begin{align*} \lvert uv\rvert^2 & =\lvert v\rvert w^* u^* w' w'^*uw\lvert v\rvert\\ &\leq \lVert w'^*uw\rVert^2\lvert v\rvert^2\leq\l{u}^2\lvert v\rvert^2, \end{align*} Hence $ \lvert uv\rvert\leq \lVert u\rVert\lvert v\rvert.$

$\textbf{Theorem}.$ Let $A$ be a $\vN$ on Hilbert space containing $\operatorname{id}_H$, then for any $\sigma$-continuous linear $\tau$ on $A$, there exists $u\in L^1(H)$ such that $$\tau(v)=\operatorname{tr}(uv)$$ for all $v\in A$.

Proof. Suppose $\{e_\alpha\}_{\alpha\in\Gamma}$ is a basis for $H$, $u\in L^1(H)$ such taht $\tau(v)=\operatorname{tr}(uv)$ for all $v\in A$. and a bounded net $(v_\lambda)_{\lambda\in \Lambda}$ in $A_{sa}$ converges strongly to an operator $v\in A_{sa}$. For every $\varepsilon>0$, there exists a finite subset $\Gamma_0$ of $\Gamma$ such that $$\sum_{\alpha\notin \Gamma_0}\langle \lvert u\rvert e_\alpha,e_\alpha\rangle\leq \varepsilon,$$ and there exists $\lambda_0\in \Lambda$ such that $\forall\lambda\geq\lambda_0$ $$\l{(v_\lambda-v) u e_\alpha}\leq \varepsilon, \alpha\in\Gamma_0.$$ Therefore, \begin{align*}\l{(v_\lambda-v) u}_1 &=\sum_{\alpha\in \Gamma}\langle \lvert (v_\lambda-v) u\rvert e_\alpha,e_\alpha\rangle\\ & \leq \sum_{\alpha\notin \Gamma_0}\langle \lvert u\rvert e_\alpha,e_\alpha\rangle\l{v_\lambda-v} + \sum_{\alpha\in \Gamma_0}\l{(v_\lambda-v) u e_\alpha}\\ &\leq M\varepsilon \end{align*} for some constant $M$. Hence $\lvert \tau(v_\lambda)-\tau(v)\rvert\leq\l{(v_\lambda-v) u}_1\to 0$. $\newcommand{\vphi}{\varphi} \newcommand{\rank}{\operatorname{rank}} \def\ni{^{-1}}$ $\textbf{4.7}$ Let $A$ be a $C^*$-algebra. (1)Show that if $p, q$ are equivalent projections in $A$, and $r$ is a projection orthogonal to both (that is, $rp = rq = 0$), then the projections $r + p$ and $r + q$ are equivalent. (2)If $H$ is a separable Hilbert space and $p$ is a projection not of finite rank, set $\operatorname{rank}(p) =\infty$. If $p$ has finite rank, set $\operatorname{rank}(p) = \operatorname{dim}p(H)$. Show that $r \sim q$ in $B(H)$ if and only if $ \operatorname{rank}(p) = \operatorname{rank}(q).$ Thus, the equivalence class of a projection in a C*-algebra can be thought of as its "generalised rank." (3)We say a projection $p$ in a $C^*$-algebra $A$ is finite if for any projection $q$ such that $q\sim p$ and $q \leq p$ we necessarily have $q = p.$ Otherwise, the projection is said to be infinite. Show that if $p, q$ are projections such that $q\leq p$ and $p$ is finite, then $q$ is finite. (4)A projection $ p$ in a von Neumann algebra $A$ is abelian if the algebra $pAp$ is abelian. Show that abelian projections are finite. (5)A von Neumann algebra is said to be finite or infinite according as its unit is a finite or infinite projection. If $H$ is a Hilbert space, show that the von Neumann algebra $B(H)$ is finite or infinite according as $H$ is finite- or infinite- dimensional. Proof. (1) Suppose $p=u^*u $ and $q=uu^*$, then $ru^*u r=rpr=0$ and $ ruu^*r=rqr=0$, so $ur=ru^*=ru=u^*r=0$.Hence \begin{align*} & r+p=r^2+u^*u+u^*r+ru=(r+u^*)(r+u);\\ & r+q=r^2+uu^*+ur+ru^*=(r+u)(r+u^*), \end{align*} that is, $r+p\sim r+q.$ (2)If $\operatorname{ran}(p)$ and $\operatorname{ran}(q)$ have a same dimension less than $\aleph_0$, choose orthonormal bases $\{e_n\}_{n=1}^N$ and $\{f_n\}_{n=1}^\infty$ for $ p(H)$ and $q(H)$, respectively($N$ may be infty). Let $$ u:\operatorname{ran}(p)\to \operatorname{ran}(q), u(e_n)=f_n, n=1,2,\cdots,N.$$ and let $$v(x)=\begin{cases} u(x), x\in \operatorname{ran}(p)=(\ker(p))^\perp;\\ 0, x\in\ker(p), \end{cases} $$ then $v^* v=p$(because they coincide on $\ker(p)$ and $(\ker(p))^\perp$) and $v v^*=q$. Conversely, if $p\sim q$, then $p=u^*u $ and $q=uu^*$ for some $u\in B(H)$, so \begin{align*} \rank(p)=\rank(u^*qu )\leq\rank(q);\\ \rank(q)=\rank(upu^*)\leq\rank(p). \end{align*} (3)Since $(p-q)r=0, (p-q)q=0$, $r+(p-q)\sim q+(p-q)=p$ by (1). Because $r+(p-q)\leq p$, $r+(p-q)=p$, i.e. $r=q.$ (4)Since $pAp$ is an abelian $C^*$-algebra with the identity $p$, there is a $*$-isomorphism $\vphi: pAp\to C(\Omega)$, where $\Omega$ is the compact set of all non-zero algebra homomorphism from $pAp$ to $\mathbb{C}$. Suppose $q\leq p$, $p=u^*u$ and $q=uu^*$, then \begin{align*} u & =qu=pqu\\ &=pqup\in pAp, \end{align*} thus \begin{align*} p=\vphi\ni(\vphi(p))=\vphi\ni(\vphi(u)^*\vphi(u))=\vphi\ni(\lvert \vphi(u)\rvert^2);\\ q=\vphi\ni(\vphi(q))=\vphi\ni(\vphi(u)\vphi(u)^*)=\vphi\ni(\lvert \vphi(u)\rvert^2),\\ \end{align*} so $p=q$. (5)If $H$ is finite-demensional, then $\rank(1)<\infty$, so $1$ is finite, i.e., $B(H)$ is finite. If $H$ is infinite-demensional, suppose $\{e_n\}_{n=0}^\infty \sqcup \{e_\lambda\}_{\lambda\in\Lambda}$ is an orthonormal basis for $H$, and set $$u(e_n)=e_{n+1}(n=0,1,\cdots), u(e_\lambda)=e_\lambda(\lambda\in\Lambda).$$ Then $$u^*(e_n)=e_{n-1}(n=1,2,\cdots),$$ $$ u^*(e_0)=0, u^*(e_\lambda)=e_\lambda.$$ Thus $$u^*u(e_n)=e_n(n=0,1,\cdots),$$ $$ uu^*(e_n)=e_n(n=1,2,\cdots), uu^*(e_0)=0.$$ Let $p=uu^*$, then $p\sim u^*u=1,$ and $p\leq 1$, but $p\neq 1$, that is $1$ is infinite, so is the Hilbert space. 1. https://math.stackexchange.com/questions/2355169/non-trivial-commutant-implies-proper-projection/2358469#2358469 2.https://math.stackexchange.com/questions/2362496/what-generates-ell-infty

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