Added the fundamental group.

src/op/ii1/fg.tex

@@ -20,5 +20,80 @@
     (2): By \autoref{theorem:vn-projection-norm-dense}.
 \end{proof}
 
+\begin{proposition}
+\label{proposition:vna-induced-commutants}
+    Let $H$ be a complex Hilbert space and $M \subset B(H)$ be a von Neumann algebra, then
+    \begin{enumerate}
+        \item For any $p \in \text{Proj}(M)$, the commutant $(pMp)'$ in $B(pH)$ is $pM'$. 
+        \item For any $q \in \text{Proj}(M')$, the commutant $(Mq)'$ in $B(qH)$ is $qM'q$. 
+    \end{enumerate}
+    
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 4.2.2]{AnantharamanPopa}}}. ]
+    (1): For any $x \in M$ and $y \in M'$, $pxp \cdot yp = ypxp = yp \cdot pxp$, so $pMp \subset (M'p)'$. 
+
+    For any $x \in (M'p)' \subset B(pH)$, let $\tilde x = xp = pxp \in B(H)$, then for any $y \in M'$, 
+    \[
+    \tilde x y = pxp \cdot y = pxp \cdot yp = yp \cdot pxp = y \tilde x
+    \]
+
+    so $\tilde x \in M'$, $x = p\tilde x p \in pMp$, and $pMp \subset (M'p)'$.
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:amplified-ii1-factor}
+    Let $(M, \tau)$ be a $\vnII_1$ factor, then $M_n(M)$ is a $\vnII_1$ factor, with normalised trace $\frac{1}{n}\text{Tr}_n \otimes \tau$. 
+\end{proposition}
+
+
+\begin{definition}[Amplification]
+\label{definition:ii1-amplification}
+    Let $(M, \tau)$ be a $\vnII_1$ factor, $t > 0$, $n \in \natp$ with $n \ge t$, and $p \in \text{Proj}(M_n(M))$ with $(\text{Tr}_n \otimes \tau)p = t$, then the von Neumann algebra $M^t := pM_n(M)p \subset M_n(M)$ is an \textbf{amplification} of $M$. 
+    
+    The spatial isomorphism class of $M^t$ is independent of the choice of $n \ge t$ and $p \in \text{Proj}(M_n(M))$ with $(\text{Tr}_n \otimes \tau)p = t$.
+\end{definition}
+\begin{proof}
+    By \autoref{proposition:ii1-trace-projection-equivalence-classes}.
+\end{proof}
+
+\begin{lemma}
+\label{lemma:ii1-amplification-homomorphism}
+    Let $(M, \tau)$ be a $\vnII_1$ factor and $s, t > 0$, then $(M^s)^t \iso M^{st}$. 
+\end{lemma}
+\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 
+    \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
+    \end{align*}
+     
+
+    Since $q \in M_n(M^s)$, $p$ is the identity on entries of $q$, so
+    \[
+    q(1_{M_n(\complex)} \otimes p) = q = (1_{M_n(\complex)} \otimes p)q
+    \]
+
+    and $(M^s)^t = qM_n(M_m(M))q = qM_{mn}(M)q$. Finally, 
+    \[
+        (\text{Tr}_{mn} \otimes \tau)(q) = (\text{Tr}_n \otimes \text{Tr}_{m} \otimes \tau)(q) = s(\text{Tr}_n \otimes \tau_s)(q) = st
+    \]
+
+    Therefore $(M^s)^t = M^{st}$. 
+\end{proof}
+
+
+\begin{definition}[Fundamental Group]
+\label{definition:ii1-fundamental-group}
+    Let $M$ be a $\vnII_1$ factor, then the \textbf{fundamental group} of $M$ is the family
+    \[
+    \fF(M) = \bracsn{t > 0| M^t \iso M} \subset \real_{>0}^\times
+    \]
+\end{definition}
+\begin{proof}
+    By \autoref{lemma:ii1-amplification-homomorphism}, $\fF(M)$ is a group. 
+\end{proof}
 
 

src/op/ii1/index.tex

@@ -2,4 +2,4 @@
 \label{chap:ii1-factors}
 
 \input{./trace.tex}
-
+\input{./fg.tex}