Revisions.

src/op/example/hii1.tex

@@ -26,10 +26,10 @@
         \item $\braks{\bigcup_{i \in I}\phi^i(M_i)}'' = M$, where the bicommutant is taken in $M$.
     \end{enumerate}
 
-    The pair $((M, \tau), \bracsn{\phi^i}_{i \in I}) = \lim_{\longleftarrow}(M_i, \tau_i)$ is the \textbf{inductive limit} of $(\bracsn{(M_i, \tau_i)}_{i \in I}, \bracsn{\phi^i_j|i, j \in I, i \le j})$. 
+    The pair $((M, \tau), \bracsn{\phi^i}_{i \in I}) = \lim_{\longrightarrow}(M_i, \tau_i)$ is the \textbf{inductive limit} of $(\bracsn{(M_i, \tau_i)}_{i \in I}, \bracsn{\phi^i_j|i, j \in I, i \le j})$. 
 \end{definition}
 \begin{proof}
-    Let $(M_0, \bracsn{\phi^i|i \in I})$ be the inductive limit of $(\bracsn{M_i}_{i \in i}, \bracsn{\phi^i_j|i \le j})$ as vector spaces. Since $\bracsn{\phi^i_j|i, j \in I, i \le j}$ are trace-preserving and $\bracsn{\tau_i}_{i \in I}$ are faithful, $\bracsn{\phi^i_j|i, j \in I, i \le j}$ are also unital isometric embeddings. As such, $\bracsn{M_i}_{i \in I}$ may be identified as subspaces of $M_0$, the norms on $\bracsn{M_i}_{i \in I}$ extend to $M_0$, and $\phi^i: M_i \to M_0$ are isometric embeddings for all $i \in I$. 
+    Let $(M_0, \bracsn{\phi^i|i \in I})$ be the inductive limit of $(\bracsn{M_i}_{i \in I}, \bracsn{\phi^i_j|i \le j})$ as vector spaces. Since $\bracsn{\phi^i_j|i, j \in I, i \le j}$ are trace-preserving and $\bracsn{\tau_i}_{i \in I}$ are faithful, $\bracsn{\phi^i_j|i, j \in I, i \le j}$ are also unital isometric embeddings. As such, $\bracsn{M_i}_{i \in I}$ may be identified as subspaces of $M_0$, the norms on $\bracsn{M_i}_{i \in I}$ extend to $M_0$, and $\phi^i: M_i \to M_0$ are isometric embeddings for all $i \in I$. 
     
     For each $i, j \in I$ with $i \le j$, $\phi^i_j: M_i \to M_j$ is trace-preserving. By universal property of the inductive limit, there exists a linear functional $\tau: M_0 \to \complex$ such that $\tau \circ \phi^i = \tau_i$ for all $i \in I$. Similarly, $\bracsn{\phi^i_j|i, j \in I, i \le j}$ are *-homomorphisms, the products and involutions from $\bracsn{M_i}_{i \in I}$ extend to $M_0$, and $\phi^i: M_i \to M_0$ is a unital trace-preserving *-homomorphism for all $i \in I$. In particular, $1_{M_0} = \phi^i(1_{M_i})$ for all $i \in I$. 
     
@@ -62,7 +62,7 @@
     \[
     \xymatrix{
     M_i \ar@{->}[d]_{\phi^i} \ar@{->}[rd]^{\psi^i} &  \\
-    M_0 \ar@{->}[r]_{\psi_0} & N
+    M_0 \ar@{->}[r]_{\phi_0} & N
     }
     \]
 
@@ -71,15 +71,15 @@
     U: H \to K \quad \dpn{UL_x1_{M_0}, \pi(y)\xi}{K} = \sigma(y^*\phi_0(x))
     \]
 
-    then for any $x \in M$, $\norm{UL_x1_{M_0}}_K^2 = \sigma(\phi_0(x^*x)) = \tau(x^*x) = \norm{L_x1_{M_0}}_H^2$. Thus $U$ extends into an isometry from $H$ to $K$. Moreover, for each $x \in M_0$ and $y, z \in N$, 
+    then for any $x \in M_0$, $\norm{UL_x1_{M_0}}_K^2 = \sigma(\phi_0(x^*x)) = \tau(x^*x) = \norm{L_x1_{M_0}}_H^2$. Thus $U$ extends into an isometry from $H$ to $K$. Moreover, for each $x \in M_0$ and $y, z \in N$, 
     \begin{align*}
-        \dpn{(\pi \circ \psi_0)(x)\pi(y)\xi, \pi(z)\xi}{K} &= \sigma(z^*\phi_0(x)y) = \sigma(yz^*\phi_0(x)) = \sigma((zy^*)^*\phi_0(x)) \\
+        \dpn{(\pi \circ \phi_0)(x)\pi(y)\xi, \pi(z)\xi}{K} &= \sigma(z^*\phi_0(x)y) = \sigma(yz^*\phi_0(x)) = \sigma((zy^*)^*\phi_0(x)) \\
         &= \dpn{UL_x1_{M_0}, \pi(zy^*)\xi}{K} = \dpn{L_x1_{M_0}, U^*\pi(zy^*)\xi}{H}
     \end{align*}
 
     so $x \mapsto \dpn{\psi_0(x)\pi(y)\xi, \pi(z)\xi}{K}$ is a weak-operator continuous functional on $M_0$. Since $\pi(N)\xi$ is dense in $K$, by the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, the weak operator topology on $\ol{B_{\pi(N)}(0, 1)}$ is induced by maps of the form $x \mapsto \dpn{\pi(x)\pi(y)\xi, \pi(z)\xi}{K}$ with $y, z \in N$. As such, $\psi_0: M_0 \to N$ is weak-operator continuous when restricted to bounded sets. 
 
-    By the \hyperref[Kaplanski Density Theorem]{theorem:kaplansky-density} and \autoref{theorem:uniform-continuous-extension}, $\psi_0$ extends into a *-homomorphism $\phi: M \to N$. Since $(N, \sigma)$ is a tracial von Neumann algebra, $\psi$ is also trace-preserving. Finally, if $\Phi: M \to N$ is another such map, then it is also weak-operator continuous on bounded sets by \autoref{lemma:tracial-von-neumann-gns}. Therefore $\Phi = \phi$ by uniqueness of continuous extensions. 
+    By the \hyperref[Kaplansky Density Theorem]{theorem:kaplansky-density} and \autoref{theorem:uniform-continuous-extension}, $\psi_0$ extends into a *-homomorphism $\phi: M \to N$. Since $(N, \sigma)$ is a tracial von Neumann algebra, $\psi$ is also trace-preserving. Finally, if $\Phi: M \to N$ is another such map, then it is also weak-operator continuous on bounded sets by \autoref{lemma:tracial-von-neumann-gns}. Therefore $\Phi = \phi$ by uniqueness of continuous extensions. 
 \end{proof}
 
 \begin{definition}[Hyperfinite $\vnII_1$ Factor]
@@ -121,8 +121,8 @@
 
     By the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, 
     \begin{align*}
-        \normn{(x - \lambda_\alpha)\xi}_H^2 &= \int_{U(M_{2^n}(\complex))} \dpn{u(x - \lambda_\alpha)u^*\xi, (x - \lambda_\alpha)\xi}{H} \mu(du) \\
-        &\le \norm{(x - \lambda_\alpha)\xi}_H\braks{\int_{U(M_{2^n}(\complex))}\normn{u(x - x_\alpha) u^*\xi}_H^2 \mu(du)} \\
+        \normn{(x - \lambda_\alpha)\xi}_H^2 &= \int_{U(M_{2^n}(\complex))} \dpn{u(x - x_\alpha)u^*\xi, (x - \lambda_\alpha)\xi}{H} \mu(du) \\
+        &\le \norm{(x - \lambda_\alpha)\xi}_H\braks{\int_{U(M_{2^n}(\complex))}\normn{u(x - x_\alpha) u^*\xi}_H^2 \mu(du)}^{1/2} \\
         &= \normn{(x - \lambda_\alpha)\xi}_H  \norm{(x - x_\alpha)\xi}_H
     \end{align*}