37.8 Type Decomposition

Proof. (1) $\Rightarrow$ (2): Let $Q \in \text{Proj}(A)$ with $Q \le P$, then $Q = PQP$. As $PAP = \complex P$, either $PQP = 0$ or $PQP = P$.

(2) $\Rightarrow$ (1): Given that there exists no projections strictly between $0$ and $P$, the only non-zero projection in $PAP$ is $P$ itself. By Theorem 37.6.2, the linear span of projections in $PAP$ is norm-dense in $PAP$. Therefore $PAP = \complex P$.$\square$

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 $Q = R$.$\square$

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

$\square$

Proof, [Theorem 26.3, Zhu93]. ($\vnI$): By Zorn’s lemma, there exists a maximal family $\seqi{P}\subset \text{Proj}(A)$ of centrally orthogonal abelian projections. Let $P = \sum_{i \in I}P_{i}$, then $P$ is abelian by (2) of Lemma 37.8.5.

Let $P_{\vnI}= Z(P)$, then $A_{\vnI}:= P_{\vnI}AP_{\vnI}$ is a von Neumann algebra with identity $P_{\vnI}$. Let $R \in \text{Proj}(Z(A_{\vnI})) \setminus \bracs{0}$, then since $0 < R \le P_{\vnI}$, $RP \le R$ is a non-zero abelian projection. Therefore $A_{\vnI}$ is of type $\vnI$.

($\vnII$): Assume without loss of generality that $I \in A$. Since $\seqi{P}$ is maximal and $(I - P_{\vnI}) \in Z(A)$, $(I - P_{\vnI})A(I - P_{\vnI})$ has no non-zero abelian projections.

By Zorn’s lemma, there exists a maximal family $\seqj{Q}\subset \text{Proj}((I - P_{\vnI})A(I - P_{\vnI}))$ of centrally orthogonal finite projections. Let $Q = \sum_{j \in J}Q_{j}$, then $Q$ is finite by (3) of Lemma 37.8.5.

Let $P_{\vnII}= Z(Q)$ and $A_{\vnII}= P_{\vnII}AP_{\vnII}$, then $A_{\vnII}$ is a von Neumann algebra with identity $P_{\vnII}$. Let $R \in \text{Proj}(Z(A_{\vnII})) \setminus \bracs{0}$, then since $0 < R \le P_{\vnII}$, $RQ \le R$ is a non-zero finite projection by (4) of Lemma 37.8.4.

($\vnIII$): Let $P_{\vnIII}= I - P_{\vnI}- P_{\vnII}$ and $A_{\vnIII}= P_{\vnIII}AP_{\vnIII}$. Since $P_{\vnIII}\in Z(A)$ and $\seqi{P}$, $\seqj{Q}$ are maximal, $A_{\vnIII}$ has no non-zero finite projections. Therefore $A_{\vnIII}$ is of type $\vnIII$, and $A = A_{\vnI}\oplus A_{\vnII}\oplus A_{\vnIII}$.

($\vnII_{1}$): By Zorn’s lemma, there exists a maximal family $\bracsn{R_k}_{k \in K}\subset \text{Proj}(A_{\vnII})$ of orthogonal central finite projections. Let $P_{\vnII_1}= \sum_{k \in K}R_{k}$, then $P_{\vnII_1}$ is a central finite projection by (3) of Lemma 37.8.5. Hence $A_{\vnII_1}= P_{\vnII_1}AP_{\vnII_1}$ is of type $\vnII_{1}$.

($\vnII_{\infty}$): Let $P_{\vnII_\infty}= P_{\vnII}- P_{\vnII_1}$ and $A_{\vnII_\infty}= P_{\vnII_\infty}AP_{\vnII_\infty}$, then $P_{\vnII_\infty}$ is a central projection. By maximality of $\bracsn{R_k}_{k \in K}$, $A_{\vnII_\infty}$ admits no non-zero finite central projections. Therefore $A_{\vnII_\infty}$ is of type $\vnII_{\infty}$, $A_{\vnII}= A_{\vnII_1}\oplus A_{\vnII_\infty}$, and

(Uniqueness): Let $A = A_{\vnI}' \oplus A_{\vnII_1}' \oplus A_{\vnII_{\infty}}' \oplus A_{\vnIII}'$ be a decomposition of $A$ into von Neumann algebras of type $\vnI$, $\vnII_{1}$, $\vnII_{\infty}$, and $\vnIII$, respectively.

Let $P_{\vnI}'$, $P_{\vnII_1}'$, $P_{\vnII_\infty}'$, and $P_{\vnIII}'$ be the identity elements of $A_{\vnI}'$, $A_{\vnII_1}'$, $A_{\vnII_{\infty}}'$, and $A_{\vnIII}'$, respectively, then

is an orthogonal direct sum, and

1. ($\vnI$)

   Let $P_{1} = P'_{\vnI}(I - P_{\vnI})$, then by construction of $P_{\vnI}$, there exists no non-zero abelian projection $R \in \text{Proj}(A)$ with $R \le P_{1}$. As both $P_{\vnI}'$ and $(I - P_{\vnI})$ are central, $P_{1} \in A_{\vnI}'$, so $P_{1} = 0$ because $A_{\vnI}'$ is of type $\vnI$. Thus $P_{\vnI}' \le P_{\vnI}$. By symmetry, $P_{\vnI}= P_{\vnI}'$ and $A_{\vnI}= A_{\vnI}'$.

2. ($\vnII$, $\vnIII$)

   Let $P'_{\vnII}= P_{\vnII_1}' \oplus P_{\vnII_\infty}'$, $A_{\vnII}' = A_{\vnII_1}' \oplus A_{\vnII_\infty}'$, and $P_{2} = P'_{\vnII}(I - P_{\vnI}- P_{\vnII})$. By construction of $P_{\vnII}$, there exists no non-zero finite projection $R \in \text{Proj}(A)$ with $R \le P_{2}$. Since $P_{2} \in A_{\vnII}'$ and $A_{\vnII}'$ is of type $\vnII$, $P_{2} = 0$ and $P_{\vnII}' \le P_{\vnII}$. By symmetry, $P_{\vnII}= P_{\vnII}'$. Thus $P_{\vnIII}= P_{\vnIII}'$, $A_{\vnII}= A_{\vnII}'$, and $A_{\vnIII}= A_{\vnIII}'$.

3. ($\vnII_{1}$, $\vnII_{\infty}$)

   Let $Q_{2} = P'_{\vnII_1}(P_{\vnII}- P_{\vnII_1})$, then there exists no non-zero finite central projection $R \in \text{Proj}(A)$ with $R \le Q_{2}$. However, since $A_{\vnII_1}'$ is of type $\vnII_{1}$, $P'_{\vnII_1}$ is itself a finite projection, and every subprojection of $P'_{\vnII_1}$ is finite by (4) of Lemma 37.8.4. Thus $Q_{2} = 0$ and $P_{\vnII_1}' \le P_{\vnII_1}$. By symmetry, $P_{\vnII_1}' = P_{\vnII_1}$. Therefore $P_{\vnII_\infty}' = P_{\vnII_\infty}$, $A_{\vnII_1}' = A_{\vnII_1}$, and $A_{\vnII_\infty}' = A_{\vnII_\infty}$.

$\square$