Proposition 39.7.5.label Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$, then the following are equivalent:
- (1)
$Z(P)Z(Q) \ne 0$.
- (2)
$PAQ \ne \bracsn{0}$.
- (3)
There exists non-zero projections $P_{0} \le P$ and $Q_{0} \le Q$ such that $P_{0} \sim Q_{0}$.
Proof, [Proposition 24.7, Zhu93]. (1) $\Rightarrow$ (2): For each $T \in A$, let $R(T)$ be the orthogonal projection onto $\ol{T(H)}$. By Proposition 39.7.4,
Given that $Z(P)Z(Q) \ne 0$, $Z(P)(H) \not\perp Z(Q)(H)$. Since $Z(P)(H) = \ol{\bigcup_{S \in A}SP(H)}$ and $Z(Q)(H) = \ol{\bigcup_{T \in A}TQ(H)}$, there exists $S, T \in A$ such that $SP(H) \not\perp TQ(H)$. As such, there exists $x, y \in H$ with
so $PAQ \ne \bracsn{0}$.
(2) $\Rightarrow$ (3): Let $T \in A$ with $PTQ \ne 0$. Let $P_{0} = R(PTQ)$ and $Q_{0} = R(QT^{*}P)$, then $0 \ne P_{0} \le P$, $0 \ne Q_{0} \le Q$, and $P_{0} \sim Q_{0}$ by Lemma 39.7.11.
(3) $\Rightarrow$ (1): By Lemma 39.7.6, $Z(P_{0}) = Z(Q_{0})$, so
$\square$
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