37.7.1 Central Support of Projections

Definition 37.7.2 (Central Support).label Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P \in \text{Proj}(A)$, then $Z(P) = \inf_{Q \in \text{Proj}(Z(A)), Q \ge P}Q$ is the central support of $P$.

Definition 37.7.3 (Centrally Orthogonal).label Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $\seqi{P}\subset \text{Proj}(A)$, then $\seqi{P}$ is centrally orthogonal if $\bracsn{Z(P_i)}_{i \in I}$ is mutually orthogonal.

Proposition 37.7.4.label Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P \in \text{Proj}(A)$. For each $T \in A$, let $R(TP)$ be the orthogonal projection onto $\ol{TP(H)}$, then

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

Proof, [Proposition 24.6, Zhu93]. Since $Z(P) \in Z(A)$, $Z(P)(H)$ is a reducing subspace of every operator in $A$. As $P \le Z(P)$, $TP(H) \subset T(Z(P)(H)) \subset Z(P)(H)$, so $Z(P) \ge R(TP)$ for all $T \in A$, and $Z(P) \ge \sup_{T \in A}R(TP)$.

On the other hand, for each $S, T \in A$, $S(TP(H)) \subset \bigcup_{R \in A}RP(H)$. As $A$ is a von Neumann algebra, the range of $\sup_{T \in A}R(TP)$ is a reducing subspace for every operator in $A$. Therefore $\sup_{T \in A}R(TP) \in Z(A)$, and $Z(P) \le \sup_{T \in A}R(TP)$.$\square$

Proposition 37.7.5.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. (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 37.7.4,

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

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

\[0 \ne \dpn{TQx, SPy}{H}= \dpn{PS^*TQx, y}{H}\]

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 37.7.11.

(3) $\Rightarrow$ (1): By Lemma 37.7.6, $Z(P_{0}) = Z(Q_{0})$, so

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

$\square$

Lemma 37.7.6.label Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$ with $P \sim Q$, then:

  1. (1)

    $Z(P) = Z(Q)$.

  2. (2)

    For any central projection $R \in \text{Proj}(A)$, $PR \sim QR$.

Proof. Let $V \in A$ with $P = V^{*}V$ and $Q = VV^{*}$, then $V$ is a partial isometry with initial space $P(H)$ and final space $Q(H)$.

(1): Since $Z(P) \ge P$ and $Z(P) \in Z(A)$,

\[Z(P)Q = Z(P)VV^{*} = VZ(P)V^{*} = VV^{*} = Q\]

and $Z(P) \ge Q$, and $Z(P) \ge Z(Q)$. By symmetry, $Z(Q) \ge Z(P)$, so $Z(P) = Z(Q)$.

(2): Let $R$ be a central projection, then

\[PR = V^{*}VR = V^{*}RV = (VR)^{*}(VR) \sim (VR)(VR)^{*} = VRV^{*} = VV^{*}R = QR\]

$\square$

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