37.7 The Projection Lattice

Proof. (1): For each $T \in A'$ and $Q \in S$, $TQ = QT$, so $Q(H)$ is a reducing subspace for $T$. As this holds for all $Q \in S$, the closed subspace generated by $\bigcup_{Q \in S}Q(H)$ is a reducing subspace for $T$. Therefore $PT = TP$, and $P \in A$ by the Bicommutant Theorem.

(2): For each $T \in A'$ and $Q \in S$, $TQ = QT$, so $Q(H)$ is a reducing subspace for $T$. As this holds for all $Q \in S$, $\bigcap_{Q \in S}Q(H)$ is a reducing subspace for $T$. Therefore $PT = TP$, and $P \in A$ by the Bicommutant Theorem.$\square$

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 $Z(P) \le \sup_{T \in A}R(TP)$.$\square$

Proof. (1): Let $x \in \ker(PQ)$, then $Q(x) \in \ker(P)$, so $x = Q(x) + (1 - Q)(x) \in \ker(Q) + \ker(P) \cap Q(H)$.

(2): Since $PQ = QP$, $(PQ)^{2} = P^{2}Q^{2} = PQ$ and $(PQ)^{*} = Q^{*}P^{*} = QP = PQ$, $PQ$ is a projection. As $PQ(H) = P(Q(H)) \subset P(H)$ and $PQ(H) = Q(P(H)) \subset Q(H)$, $PQ(H) \subset P(H) \cap Q(H)$. On the other hand, $PQ$ is the identity on $P(H) \cap Q(H)$, so $PQ(H) = P(H) \cap Q(H)$.$\square$

Proof. (1) $\Rightarrow$ (2): Let $V \in A$ with $P = V^{*}V$ and $Q = VV^{*}$. By Proposition 38.2.3, $V$ is a partial isometry with initial space $\ker(P)^{\perp}$, and $V^{*}$ is a partial isometry with initial space $\ker(Q)^{\perp}$. Therefore $V$ is a partial isometry from $P(H)$ to $Q(H)$.

(2) $\Rightarrow$ (1): By Proposition 38.2.3, $P = V^{*}V$ is a projection onto $\ker(V)^{\perp}$, and $Q = VV^{*}$ is a projection onto $\ker(V^{*})^{\perp} = V(H)$.$\square$

Proof. Let $T = VQ$ be the polar decomposition of $T$, then $V$ is a partial isometry from $\ol{T^*(H)}$ to $\ol{T(H)}$. Since $V \in A$, $P \sim Q$.$\square$

Proof. Using Theorem 37.3.1, assume without loss of generality that $I \in A$. In which case, by (1) of Lemma 37.7.4,

and since $P \vee Q$ and $Q$ commute,

by (2) of Lemma 37.7.4. Similarly,

so

Therefore

by Lemma 37.7.7.$\square$

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 $Q(H)$. In which case, since $Z(P) \ge P$ and $Z(P) \in Z(A)$,

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

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,

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

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

$\square$

Proof, [Lemma 25.1, Zhu93]. Let $U, V \in A$ be partial isometries such that $P = U^{*}U$, $UU^{*} \le Q$, $Q = V^{*}V$, and $VV^{*} \le P$. Denote $Q_{0} = Q$ and $P_{0} = P$. For each $n \in \natz$, inductively define $P_{n+1}= VQ_{n}V^{*}$ and $Q_{n+1}= UP_{n}U^{*}$, then:

1. (i)

   For each $n \in \natz$, $P_{n}, Q_{n} \in \text{Proj}(A)$.

2. (ii)

   For each $n \in \natz$, $P_{n} \le P$ and $Q_{n} \le Q$.

3. (iii)

   For each $n \in \natz$, $P_{n+1}\le P_{n}$ and $Q_{n+1}\le Q_{n}$.

As $\seq{P_n}, \seq{Q_n}\subset \text{Proj}(A)$ are non-increasing sequences, by Theorem 37.3.1, there exists $P_{\infty}, Q_{\infty} \in \text{Proj}(A)$ such that $P_{n} \to P_{\infty}$ and $Q_{n} \to Q_{\infty}$ in the strong operator topology as $n \to \infty$.

For each $n \in \natz$, $U(P_{n} - P_{n+1})U^{*} = Q_{n+1}- Q_{n+2}$, so

and $P_{n} - P_{n+1}\sim Q_{n+1}- Q_{n+2}$. Similarly, $Q_{n} - Q_{n+1}\sim P_{n+1}- P_{n+2}$. As $P_{n+1}= VQ_{n}V^{*}$ for all $n \in \natz$, $P_{\infty} \sim Q_{\infty}$ after passing through a strong-operator limit.

For each $N \in \natz$, $\sum_{n = 0}^{N} (P_{n} - P_{n+1}) = P - P_{N+1}$, so $P = P_{\infty} + \sum_{n = 0}^{\infty} (P_{n} - P_{n+1})$. Similarly, $Q = Q_{\infty} + \sum_{n = 0}^{\infty} (Q_{n} - Q_{n+1})$. Therefore

because $\sim$ is preserved through direct sums.$\square$