From d3a719833c57a7694d2da708a8747e1b976cfb1c Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Fri, 6 Nov 2026 17:31:54 -0500 Subject: [PATCH] Added the fundamental group. --- src/op/ii1/fg.tex | 75 ++++++++++++++++++++++++++++++++++++++++++++ src/op/ii1/index.tex | 2 +- 2 files changed, 76 insertions(+), 1 deletion(-) diff --git a/src/op/ii1/fg.tex b/src/op/ii1/fg.tex index 45ef9cd..b4f25ba 100644 --- a/src/op/ii1/fg.tex +++ b/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} diff --git a/src/op/ii1/index.tex b/src/op/ii1/index.tex index 063d82e..62dbbdb 100644 --- a/src/op/ii1/index.tex +++ b/src/op/ii1/index.tex @@ -2,4 +2,4 @@ \label{chap:ii1-factors} \input{./trace.tex} - +\input{./fg.tex}