Lemma 39.8.4.label Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$.
- (1)
If $P$ is minimal, then $P$ is abelian.
- (2)
If $P$ is abelian, then $P$ is finite.
- (3)
If $P$ is finite and $P \sim Q$, then $Q$ is finite.
- (4)
If $P$ is finite and $Q \le P$, then $Q$ is finite.
- (5)
If $P$ is minimal and $P \sim Q$, then $Q$ is minimal.
- (6)
If $P, Q$ are minimal with $P \sim Q$, then for any $U, V \in A$ with $P = U^{*}U = V^{*}V$ and $Q = UU^{*} = VV^{*}$, there exists $\lambda \in \partial B_{\complex}(0, 1)$ such that $V = \lambda U$.
Proof, [Section 26.1, Zhu93]. (1): $PAP = \complex P$ is abelian.
(2): Let $R \in \text{Proj}(A)$ with $P \sim R \le P$, then there exists $V \in A$ such that $R = V^{*}V$ and $P = VV^{*}$. Since $V$ has initial space $R(H) \subset P(H)$ and final space $P(H)$, $V = PVP$ and $V^{*} = PV^{*}P$. As $PAP$ is abelian,
(3): Let $R \in \text{Proj}(A)$ with $Q \sim R \le Q$. Let $V \in A$ with $Q = V^{*}V$ and $P = VV^{*}$, then $V$ is a partial isometry with initial space $Q(H)$ and final space $P(H)$. In which case, $P = VQV^{*}$, and $VRV^{*} \le VQV^{*} = P$. Let $U = (VRV^{*})V$, then
and as $Q = V^{*}PV$,
so $VRV^{*} \sim R \sim Q \sim P$. Given that $P$ is finite, $VRV^{*} = P$. Therefore
(4): Let $R \in \text{Proj}(A)$ with $Q \sim R \le Q \le P$, then $P \sim (P - Q) + R \le P$, so $P - Q + R = P$, and $Q = R$.
(5): Since $P \sim Q$, there exists $V \in A$ with $P = V^{*}V$ and $Q = VV^{*}$. Let $R \in \text{Proj}(A)$ with $0 < R \le Q$, then $0 \le V^{*}RV \le V^{*}QV = P$. By minimality of $P$, $V^{*}RV = P$, so
(6): Let $R = U^{*}V$, then since $U$ and $V$ are partial isometries with initial space $P(H)$ and final space $Q(H)$,
so $PRP \in PAP = \complex P$. Thus there exists $\lambda \in \complex$ such that $R = \lambda P$. In which case,
and
so $|\lambda| = 1$.$\square$
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