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

Solutions to Some Exercises in Murphy’s book (Chapter 3)

$\newcommand{\zhang}[1]{{\Large{\text{#1}}}} \def \A{\mathscr A} \def \F{\mathbb{F}} \def \N{\mathbb{N}} \def \R{\mathbb{R}} \def \C{\mathbb{C}}\def \H{\mathscr{H}}\def \d{\mathrm{d}} \def \Inv{\mathrm{Inv}} \def \ni{^{-1}} \def \jx{\lim_{n\to\infty}} \def \yb{\frac{1}{2}} \newcommand{\x}{\times} \newcommand{\wt}{\textbf{Question.}} \newcommand{\bh}[1]{\textbf{#1.}} \newcommand{\xl}[2]{#1_1, #1_2, \cdots, #1_{#2}} \newcommand{\xl}[2]{#1_1, #1_2, \cdots, #1_{#2}} \newcommand{\tm}[1]{\textbf{ #1.}} \renewcommand{\l}[1]{\lVert #1\rVert}\newcommand{\zm}{Proof.} \newcommand{\zb}{}$
$\bh{3.5}$An element a of $A_+$ is strictly positive if the hereditary $C^*$-subalgebra of $A$ generated by $a$ is $A$ itself, that is, if $(aAa)^- = A$.

(a) Show that if $A$ is unital, then $a \in A_+$ is strictly positive if and only if $a$ is invertible.

(b) If $H$ is a Hilbert space, show that a positive compac operator on $H$ is strictly positive in $K(H)$ if and only if it has dense range.

(c) Show that if $a$ is strictly positive in $A$, then $\tau( a) > 0$ for all non-zero positive linear functionals $\tau$ on $A$.

$\zm$ (a) If $a$ is invertible, then $aAa=A$, thus $a$ is strictly positive.
Conversely, if $a$ is strictly positive, then $(aAa)^-=A$.
Since the set $G$ of all invertible elements is open, $aAa\cap G\neq\emptyset$. Hence there exists some $b\in A$ such that $aba$ is invertible, thus $a$ is invertible.

(b)If $u$ is strictly positive in $K(H)$, that is $(uK(H)u)^-=K(H)$, then $\forall x\in H,$ there exists a sequence $\{v_n\}$ in $B(H)$ such that $$x\otimes x=\lim_{n\to \infty} uv_nu.$$
Hence $$\l{x}^2x=x\otimes x(x)=\lim_{n\to\infty}u(v_nu(x))\in (uH)^-.$$
Therefore, $x\in (uH)^-$ and $H=(uH)^-$.
Conversely, if $(uH)^-H$, then for all $x\in H,$ then there exists a sequence $\{h_n\}$ in $H$ such that $uh_n\to x$, hence $$u(h_n\otimes h_n )u= (uh_n)\otimes(uh_n)\to x\otimes x.$$
Therefore, every rank-one projection belongs to $(uK(H)u)^-$.
Since all the rank-one projections is the total subset of $K(H)$, $K(H)=(uK(H)u)^-$.

(c)Suppose $\tau$ is a positive functional such that $\tau(a)=0$,
then $$\tau((a^{\frac{1}{2}})^*a^\yb)=0,$$ so $$\tau(a^* a)=\tau((a^\frac{1}{2})^3 a^\yb)=0.$$
Hence for all $b\in A$, $$\tau(aba)=0.$$ Therefore, $\tau=0$ for $A=(aAa)^-$.

$\bh{3.3}$If $\varphi:A\to B$ is a positive linear map between $C^*$ algebra, show that $\varphi$ is bounded.

$\zm$ For all positive linear functional $\tau$ on $B$, $\tau\circ\varphi$ is a positive linear functional on $A$ and thus bounded, i.e.,$$\lVert\tau\circ\varphi(x)\rVert\leq M_\tau ~~ (\forall x\in A \text{ with } \l{x}\leq 1)$$ for some positive constant $M_\tau$.

Every bounded linear functional on $B$ is a linear combination of 4 positive linear functional on $B$ by Jordan Decomposition, thus for all $b^*\in B^*,$ $$\l{b^*(\varphi(x))}\leq M_{b^*} ~~( \forall x\in A \text{ with } \l{x}\leq 1)$$ for some positive constant $M_{b^*}$. Hence $\l{\varphi(x)}\leq M$ for some positive constant $M$ by the Principle of Uniform Boundedness, i.e. $\varphi$ is bounded.