From 400382be1dc6efa768accbd5de098afc345790c9 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Fri, 6 Nov 2026 17:50:54 -0500 Subject: [PATCH] Oopsies daisies. --- src/vna/ii1/fg.tex | 10 +++++----- 1 file changed, 5 insertions(+), 5 deletions(-) diff --git a/src/vna/ii1/fg.tex b/src/vna/ii1/fg.tex index b4f25ba..cd5122c 100644 --- a/src/vna/ii1/fg.tex +++ b/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