Typo fixes in the type decomposition section.

src/op/vn/type-decomp.tex

@@ -9,7 +9,7 @@
 
 \begin{definition}[Abelian Projection]
 \label{definition:abelian-projection}
-    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P \in \text{Proj}(A)$, then $P$ is \textbf{minimal} if $PAP$ is abelian. 
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P \in \text{Proj}(A)$, then $P$ is \textbf{abelian} if $PAP$ is abelian. 
 \end{definition}
 
 \begin{definition}[Minimal Projection]
@@ -88,7 +88,7 @@
 
     (2): Let $S, T \in A$, then by (1),
     \begin{align*}
-        PSP \cdot PTP &= \sum_{i, j \in I}P_iSP_i \cdot P_jSP_j = \sum_{i \in I}P_iSP_i \cdot P_iTP_i \\
+        PSP \cdot PTP &= \sum_{i, j \in I}P_iSP_i \cdot P_jTP_j = \sum_{i \in I}P_iSP_i \cdot P_iTP_i \\
         &= \sum_{i \in I}P_iTP_i \cdot P_iSP_i = PTP \cdot PSP
     \end{align*}
 
@@ -157,16 +157,16 @@
     
     (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 $P_{\vnIII_\infty}'$ be the identity elements of $A_{\vnI}$, $A_{\vnII_1}$, $A_{\vnII_{\infty}}$, and $A_{\vnIII}$, respectively, then
+    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
     \[
-    I = P_{\vnI}' \oplus P_{\vnII_1}' \oplus P_{\vnII_\infty}' \oplus P_{\vnIII_\infty}'
+    I = P_{\vnI}' \oplus P_{\vnII_1}' \oplus P_{\vnII_\infty}' \oplus P_{\vnIII}'
     \]
 
     is an orthogonal direct sum, and
     \begin{enumerate}
-        \item[($\vnI$)] 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 $P_{\vnI}' \ge P_{\vnI}$. By symmetry, $P_{\vnI} = P_{\vnI}'$ and $A_{\vnI} = A_{\vnI}'$.
-        \item[($\vnII$, $\vnIII$)] Let $P'_{\vnII} = P_{\vnII_1} \oplus P_{\vnII_\infty}$ 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 $P_{\vnII}' \ge P_{\vnII}$. By symmetry, $P_{\vnII} = P_{\vnII}'$. Thus $P_{\vnIII} = P_{\vnIII}'$, $A_{\vnII} = A_{\vnII}'$, and $A_{\vnIII} = A_{\vnIII}'$.
-        \item[($\vnII_1$, $\vnII_\infty$)] 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 $P'_{\vnII_1}$ is itself a central projection, every subprojection of $P'_{\vnII_1}$ is finite by (4) of \autoref{lemma:projection-types-gymnastics}, so $Q_2 = 0$, and $P_{\vnII_1}' \ge P_{\vnII_1}$. By symmetry, $P_{\vnII_1}' = P_{\vnII_1}$. Therefore $P_{\vnII_\infty} = P_{\vnII_\infty}$, $A_{\vnII_1}' = A_{\vnII_1}$, and $A_{\vnII_\infty}' = A_{\vnII_\infty}$.
+        \item[($\vnI$)] 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 $A_{\vnI} = A_{\vnI}'$.
+        \item[($\vnII$, $\vnIII$)] Let $P'_{\vnII} = P_{\vnII_1}' \oplus P_{\vnII_\infty}'$ 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 $A_{\vnIII} = A_{\vnIII}'$.
+        \item[($\vnII_1$, $\vnII_\infty$)] 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 $P'_{\vnII_1}$ is itself a central projection, every subprojection of $P'_{\vnII_1}$ is finite by (4) of \autoref{lemma:projection-types-gymnastics}, so $Q_2 = 0$, 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 $A_{\vnII_\infty}' = A_{\vnII_\infty}$.
     \end{enumerate}
     
 \end{proof}