Lemma 37.8.5.label Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, $\seqi{P}\subset \text{Proj}(A)$ be centrally orthogonal, and $P = \sum_{i \in I}P_{i}$, then
- (1)
For each $T \in A$, $PTP = \sum_{i \in I}P_{i}TP_{i}$.
- (2)
If $\seqi{P}$ are abelian, then $P$ is also abelian.
- (3)
If $\seqi{P}$ are finite, then $P$ is also finite.
Proof, [Lemma 26.2, Zhu93]. (1): For each $i \in I$, $Z(P_{i}) \ge P_{i}$, so $Z(P_{i})P_{i} = P_{i}$. For any $i, j \in I$ with $i \ne j$, $Z(P_{i})$ and $Z(P_{j})$ are orthogonal, so $Z(P_{i})P_{j} = Z(P_{i})Z(P_{j})P_{j} = 0$.
Let $T \in A$, then by Proposition 37.7.4, $Z(P_{i})(H) \supset TP_{i}(H)$ for all $i \in I$. Therefore
(2): Let $S, T \in A$, then by (1),
(3): Let $R \in \text{Proj}(A)$ with $P \sim R \le P$, and $V \in A$ with $R = V^{*}V$ and $P = VV^{*}$, then for each $i \in I$, $Z(P_{i})R \sim Z(P_{i})P = P_{i}$, and $Z(P_{i})R \le Z(P_{i})P = P_{i}$. As $\seqi{P}$ are finite, $Z(P_{i})R = P_{i}$ for all $i \in I$. Therefore
$\square$
Post a Comment