37.7.2 Murray-von Neumann Equivalence

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. For each $i \in I$, let $V_{i} \in A$ such that $P_{i} = V_{i}^{*}V_{i}$ and $Q_{i} = V_{i}V_{i}^{*}$, then $V_{i}$ is a partial isometry with initial space $P_{i}(H)$ and final space $Q_{i}(H)$. As $\seqi{P}$ is mutually orthogonal and $\seqi{Q}$ is mutually orthogonal, the sum $\sum_{i \in I}V_{i}$ converges in strong operator topology to an operator $V$, where

and

$\square$

Proof. Let $T = V|T|$ 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 37.7.7,

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

by (2) of Lemma 37.7.7. Similarly,

so

Therefore

by Lemma 37.7.11.$\square$

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

by Lemma 37.7.10.$\square$

Proof, [Theorem 25.4, Zhu93]. By Zorn’s lemma, there exists maximal families $\seqi{P}, \seqi{Q}\subset \text{Proj}(A)$ such that:

1. (i)

   $\seqi{P}$ is mutually orthogonal.

2. (ii)

   $\seqi{Q}$ is mutually orthogonal.

3. (iii)

   For each $i \in I$, $P_{i} \sim Q_{i}$.

4. (iv)

   For each $i \in I$, $P_{i} \le P$ and $Q_{i} \le Q$.

Let $P_{0} = \sum_{i \in I}P_{i}$ and $Q_{0} = \sum_{i \in I}Q_{i}$, then $P_{0} \sim Q_{0}$ by Lemma 37.7.10. By maximality, there exists no non-zero $P', Q' \in \text{Proj}(A)$ such that $P' \le P - P_{0}$, $Q' \le Q - Q_{0}$, and $P' \sim Q'$. By Proposition 37.7.5, $Z(P - P_{0}) Z(Q - Q_{0}) = 0$.

Let $R = Z(Q - Q_{0})$, then $Q - Q_{0} \le R$ and $P - P_{0} \le (I - R)$, so $(P - P_{0})R = 0$ and $(Q - Q_{0})R = Q - Q_{0}$. By Lemma 37.7.6,

and

$\square$

Proof. By the comparability theorem, there exists a central projection $R$ such that $PR \preceq QR$ and $Q(I - R) \preceq P(I - R)$. As $A$ is a factor, either $R = 0$ or $R = I$. In which case, $P \preceq Q$ or $Q \preceq P$.$\square$