Oopsies daisies.

src/vna/ii1/fg.tex

@@ -3,19 +3,19 @@
 
 \begin{definition}[Induced von Neumann Algebra]
 \label{definition:induced-vna-commutant}
-    Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $q \in M'$, then $Mq$ is a von Neumann algebra on $qH$, known as the \textbf{induced von Neumann algebra} of $M$ with respect to $q$. 
+    Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $q \in \text{Proj}(M')$, then $Mq$ is a von Neumann algebra on $qH$, known as the \textbf{induced von Neumann algebra} of $M$ with respect to $q$. 
 \end{definition}
 
 \begin{proposition}
 \label{proposition:induced-vna-commutant-gymnastics}
-    Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $p \in M$, then:
+    Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $p \in \text{Proj}(M)$, then:
     \begin{enumerate}
         \item For any central projection $e \in Z(pMp)$, $e = Z(e)p$. 
         \item $Z(pMp) = Z(M)p$.
     \end{enumerate}
 \end{proposition}
 \begin{proof}[Proof, {{\cite[Proposition 4.2.1]{AnantharamanPopa}}}. ]
-    (1): Since $(p - e)Me = (p - e)pMpe = 0$, $(p - e)yey^* = 0$ for all $y \in U(M)$, so $(p - e)z(e) = 0$ by \autoref{proposition:central-support-vna}, and $e = z(e)p$. 
+    (1): Since $(p - e)Me = (p - e)pMpe = 0$, $(p - e)yey^* = 0$ for all $y \in U(M)$, so $(p - e)z(e) = 0$ by \autoref{proposition:central-support-vna}, and $e = Z(e)p$. 
 
     (2): By \autoref{theorem:vn-projection-norm-dense}.
 \end{proof}
@@ -37,7 +37,7 @@
     \tilde x y = pxp \cdot y = pxp \cdot yp = yp \cdot pxp = y \tilde x
     \]
 
-    so $\tilde x \in M'$, $x = p\tilde x p \in pMp$, and $pMp \subset (M'p)'$.
+    By the \hyperref[Bicommutant Theorem]{theorem:bicommutant}, $\tilde x \in M$, $x = p\tilde x p \in pMp$, and $pMp \supset (M'p)'$.
 \end{proof}
 
 
@@ -64,7 +64,7 @@
 \begin{proof}[Proof, {{\cite[Lemma 4.2.3]{AnantharamanPopa}}}. ]
     Let $m \in \natp$ and $p \in \text{Proj}(M_m(M))$ with $(\text{Tr}_m \otimes \tau)(p) = s$, so that $M^s = pM_m(M)p$, and $\tau_s = s^{-1}(\text{Tr}_m \otimes \tau)|_{M^s}$ denotes the normalised tracial state on $M^s$. 
     
-    Let $n \in \natp$ and $q \in \text{Proj}(M_n(M^s))$ with $(\text{Tr}_n \otimes \tau_s)(p) = t$, so that $(M^s)^t = qM_n(M^s)q$. Identify $M_n(M^s) \subset M_n(M_m(M))$, so that $q \in M_n(M_m(M))$, then 
+    Let $n \in \natp$ and $q \in \text{Proj}(M_n(M^s))$ with $(\text{Tr}_n \otimes \tau_s)(q) = t$, so that $(M^s)^t = qM_n(M^s)q$. Identify $M_n(M^s) \subset M_n(M_m(M))$, so that $q \in M_n(M_m(M))$, then 
     \begin{align*}
     (M^s)^t &= qM_n(M^s)q = qM_n(pM_m(M)p)q \\
     &= q(1_{M_n(\complex)} \otimes p)M_n(M_m(M))(1_{M_n(\complex)} \otimes p)q