Proposition 37.7.10.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. (1)

    $Z(P)Z(Q) \ne 0$.

  2. (2)

    $PAQ \ne \bracsn{0}$.

  3. (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 37.7.3,

\[Z(P) = \sup_{T \in A}R(TP) \quad Z(Q) = \sup_{T \in A}R(TQ)\]

so $Z(P)Z(Q) = \sup_{S, T \in A}R(SP)R(TQ) \ne 0$. Thus there exists $S, T \in A$ such that $R(SP)R(TQ) \ne 0$. As such, there exists $x, y \in H$ with

\[0 \ne \dpn{SPx, TQy}{H}= \dpn{QT^*SPx, y}{H}\]

so $PAQ \ne 0$.

(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 37.7.7.

(3) $\Rightarrow$ (1): By Lemma 37.7.9, $Z(P_{0}) = Z(Q_{0})$, so

\[Z(P)Z(Q) = Z(P) \wedge Z(Q) \ge Z(P_{0}) \vee Z(Q_{0}) \ne 0\]

$\square$

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