A factor von Neumann algebra is a prime algebra.

Question

Let $\mathcal{H}$ be a complex Hilbert space and $\mathcal{B}(\mathcal{H})$ denotes the algebra of all bounded linear operators on $\mathcal{H}.$ Recall that a von Neumann algebra $\mathcal{U}\subseteq\mathcal{B}(\mathcal{H} ) $ is said to be a factor if its center is $\mathbb{C}I,$ where $I$ is the identity operator.

My attmept: I want to show that for any $A, B\in\mathcal{U},$ $A\mathcal{U}B=\{0\}$ implies that either $A=0$ or $B=0.$ Since $I\in\mathcal{U},$ we have $AB=0.$ Assume on contrary that $A\neq0$ and $B\neq0.$ Then, there exists some $x_0, y_0\in\mathcal{H}$ such that $ A(x_0)\neq 0$ and $ B(y_0)\neq 0.$ Therefore, we have $A(B(y_0))=0.$ From here we must get a contradiction. But I got stuck here and I don't know how to use the facts that $\mathcal{U}$ is a factor or Hahn-Banach Theorem.

Answer 1

This is a partial answer under the stronger assumption that the commutant of the von Neumann algebra $\mathcal{U}$ is trivial, i.e. $\mathcal{U}=B(\mathcal{H}).$ The corrections have been made due to two comments by @MaoWao.

If $\mathcal{U}$ is satisfies $\mathcal{U}'=\mathbb{C}I$ then for every $v\neq 0$ the set $V:=\{Tv\,:\, T\in \mathcal{U}\}$ is linearly dense in $\mathcal{H}.$ Indeed, assume that $P$ is the orthogonal projection onto the closure of the linear span of $V.$ Clearly $V$ is invariant for $S\in \mathcal{U}.$ Hence the ${\rm span}\,V$ is invariant as well as its closure. Therefore $SP=PSP$ for any $S\in \mathcal{U}.$ If $S=S^*$ we get $SP=PS.$ Since every element in $\mathcal{U}$ can be represented as $S_1+iS_2$ where $S_1,\, S_2\in \mathcal{U}$ are selfadjoint thus $SP=PS$ for any $S\in\mathcal{U}.$ Thus $P\in \mathcal{U}'$ wich implies $P=I.$ Therefore the set $V$ is linearly dense.

Assume $A\mathcal{U}B=0.$ If $B\neq 0$ there exists $x\in \mathcal{H}$ such that $v:=Bx\neq 0.$ Thus $ATv=ATBx=0$ for any $T\in \mathcal{U}.$ Hence $A$ vanishes on $V=\{Tv\,:\, T\in \mathcal{U}\}$ and consequently on its linear span. As the span of $V$ is dense we conclude that $A=0.$

Answer 2

If $A\mathcal{U}B=\{0\}$, then either $A=0$ or $B=0$.

Pf: If $A=0$ we're done. If $A\neq 0$, then let $V = \overline{\textrm{span}\{TB(H): T\in\mathcal{U}\}}$. Since $TB(H)\subseteq\ker{A}$ for each $T\in\mathcal{U}$, we have $V\subseteq\ker{A}\neq H$. Also, $$T(V)\subseteq \overline{\textrm{span}\{TSB(H): S\in\mathcal{U}\}} \subseteq \overline{\textrm{span}\{SB(H): S\in\mathcal{U}\}}= V $$ for each $T\in\mathcal{U}$. Thus, $V$ is an invariant subspace of $\mathcal{U}$. An equivalent way of expressing this is $\big(Tz = zT\hspace{3mm} \forall T\in\mathcal{U}\big)$, where $z$ is the $\perp$-projection onto $V$. It is up to the reader to show that $z$ is in the SOT-closure of $\mathcal{U}$. Consequently, $z$ is a central projection.

Since $\mathcal{U}$ is a factor, either $z=0$ or $z=1$. Since $V\neq H$, we must have $z=0$, and so $V=\{0\}$.

At this point, we get $T(B(H))=\{0\}$ for all $T\in\mathcal{U}$. For the identity $1\in\mathcal{U}$, we get $B(H)=\{0\}$. Hence, $B=0$.

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Another way of saying the same thing would be as below.

Suppose $A\neq 0$ and $B\neq 0$. Let $p$ be the $\perp$-projection onto $\overline{\textrm{imag}(B)}$ and $q$ be the $\perp$-projection onto $\ker(A)^\perp$. By standard arguments, since the source and range projections of any element in a von Neumann algebra fall in the algebra, then $p,q\in\mathcal{U}$. Clearly, $q\mathcal{U}p=\{0\}$. Let $$1-z:=\bigvee\{r\in\mathcal{U}: rAp=\{0\}, r\hspace{2mm}\textrm{is a projection}\} $$ then $z$ is a central projection. In fact, $z$ is the central cover of $p$. Since $\mathcal{U}$ is a factor, either and $z=0$ or $z=1$. Since $0\neq p\leq z$ and $0\neq q\leq 1-z$, we end up with a contradiction.

Answer 3

The more general result is as follows. Given $a\in M\subset B(H)$ we denote by $c(a)$ the central carrier of $a$. This is the orthogonal projection $$ c(a)=[\overline{MaH}] $$ where we used the square brackets to denote "orthogonal projection onto this subspace". It is a well-known fact that $c(a)$ is the smallest projection in the centre of $M$ satisfying $a\, c(a)=a$ (proof of all this at the end).

Then we have

Theorem. Let $M$ be a von Neumann algebra, $a,b\in M$. The following statements are equivalent:

• $aMb=0$;
• $c(a)c(b)=0$.

Proof. If $aMb=0$, we have $MbH\subset\ker a$, which implies that $c(b)\leq p_a$, where $p_a$ is the orthogonal projection onto $\ker a$. Then $c(b)a=ac(b)=0$. It follows that $c(b)\leq 1-c(a)$, and so $c(a)c(b)=0$. The steps are reversible, which proves the other implication. $\ \ \ \square$

When $M$ is a factor and $a,b$ are nonzero we have that $c(a)=c(b)=1$, and so $aMb=0$ implies that at least one of $a,b$ is zero.

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Facts about the central carrier. Given $a\in M$, let $c(a)$ be the orthogonal projection onto $\overline{MaH}$. We have

1. $c(a)\in Z(M)$. Indeed, given $T\in M'$, selfadjoint, we have $$TMaH=MaTH\subset MaH.$$ This implies that $c(a)Tc(a)=Tc(a)$. Taking adjoints we get that $c(a)T=Tc(a)$. As the selfadjoint elements span all of $M'$, it follows that $c(a)\in (M')'=M''=M$. If now $b\in M$ is selfadjoint, then $bMaH\subset MaH$, so $c(a)bc(a)=bc(a)$, and as before we get that $c(a)\in M'$. So $c(a)\in M\cap M'=Z(M)$.

2. $c(a)=\min\{q\in Z(M):\ qa=a\}$. Indeed, since $aH\subset MaH$ we get that $c(a)a=a$. As $c(a)\in Z(M)$ we also get $ac(a)=a$. Now let $q\in Z(M)$ with $qa=a$. Then $qMaH=MqaH=MaH$ and so $qc(a)=c(a)$. Thus $c(a)\leq q$.