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Bokuan Li
2026-11-01 23:38:40 -05:00
parent bfa79c45ce
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src/op/example/hii1.tex
+8 -8
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@@ -26,10 +26,10 @@
\item $\braks{\bigcup_{i \in I}\phi^i(M_i)}'' = M$, where the bicommutant is taken in $M$. \item $\braks{\bigcup_{i \in I}\phi^i(M_i)}'' = M$, where the bicommutant is taken in $M$.
\end{enumerate} \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} \end{definition}
\begin{proof} \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 $i \in I$. 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 @@
\[ \[
\xymatrix{ \xymatrix{
M_i \ar@{->}[d]_{\phi^i} \ar@{->}[rd]^{\psi^i} & \\ M_i \ar@{->}[d]_{\phi^i} \ar@{->}[rd]^{\psi^i} & \\
M_0 \ar@{->}[r]_{\psi_0} & N M_0 \ar@{->}[r]_{\phi_0} & N
} }
\] \]
@@ -71,15 +71,15 @@
U: H \to K \quad \dpn{UL_x1_{M_0}, \pi(y)\xi}{K} = \sigma(y^*\phi_0(x)) 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*} \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} &= \dpn{UL_x1_{M_0}, \pi(zy^*)\xi}{K} = \dpn{L_x1_{M_0}, U^*\pi(zy^*)\xi}{H}
\end{align*} \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. 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} \end{proof}
\begin{definition}[Hyperfinite $\vnII_1$ Factor] \begin{definition}[Hyperfinite $\vnII_1$ Factor]
@@ -121,8 +121,8 @@
By the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, By the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz},
\begin{align*} \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) \\ \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)} \\ &\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 &= \normn{(x - \lambda_\alpha)\xi}_H \norm{(x - x_\alpha)\xi}_H
\end{align*} \end{align*}
src/op/example/lg.tex
+13 -13
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@@ -39,7 +39,7 @@
\begin{lemma} \begin{lemma}
\label{lemma:left-convolver-cc} \label{lemma:left-convolver-cc}
Let $G$ be a locally compact group and $f \in C_c(G; \complex)$, $f$ is a lefy convolver. Let $G$ be a locally compact group and $f \in C_c(G; \complex)$, $f$ is a left convolver.
\end{lemma} \end{lemma}
\begin{proof} \begin{proof}
By \hyperref[Young's Inequality]{theorem:young-fourier}, for each $g \in L^2(G; \complex)$, $\norm{f * g}_{L^2(G; \complex)} \le \norm{f}_{L^1(G; \complex)}\norm{g}_{L^2(G; \complex)}$. By \hyperref[Young's Inequality]{theorem:young-fourier}, for each $g \in L^2(G; \complex)$, $\norm{f * g}_{L^2(G; \complex)} \le \norm{f}_{L^1(G; \complex)}\norm{g}_{L^2(G; \complex)}$.
@@ -48,7 +48,7 @@
\begin{lemma} \begin{lemma}
\label{lemma:left-convolver} \label{lemma:left-convolver}
Let $G$ be a locally compacy group and $f: G \to \complex$ be a left convolver, then: Let $G$ be a locally compact group and $f: G \to \complex$ be a left convolver, then:
\begin{enumerate} \begin{enumerate}
\item $L_f \in B(L^2(G; \complex))$. \item $L_f \in B(L^2(G; \complex))$.
\item $f \in L_{\text{loc}}^1(G; \complex)$. \item $f \in L_{\text{loc}}^1(G; \complex)$.
@@ -64,8 +64,8 @@
In which case, for almost every $x \in G$, In which case, for almost every $x \in G$,
\begin{align*} \begin{align*}
\limv{N} f * (g_N - g)(x) &= \int_G f(y)(g_N - g)(y^{-1}x)dy \\ f * (g_N - g)(x) &= \int_G f(y)(g_N - g)(y^{-1}x)dy \\
&\le \int_G|f(y)|\sum_{n \ge N}|(g_n - g)(y^{-1}x)| dy |f * (g_N - g)(x)| &\le \int_G|f(y)|\sum_{n \ge N}|(g_n - g)(y^{-1}x)| dy
\end{align*} \end{align*}
where $\int_G|f(y)|\sum_{n \in \natp}|(g_n - g)(y^{-1}x)| dy < \infty$ by (ii). As such, by the \hyperref[Dominated Convergence Theorem]{theorem:dct}, $f * g_N \to f * g$ almost everywhere. By (i), $f * g = h$ almost everywhere. Therefore the \hyperref[Closed Graph Theorem]{theorem:closed-graph} implies that $L_f$ is continuous. where $\int_G|f(y)|\sum_{n \in \natp}|(g_n - g)(y^{-1}x)| dy < \infty$ by (ii). As such, by the \hyperref[Dominated Convergence Theorem]{theorem:dct}, $f * g_N \to f * g$ almost everywhere. By (i), $f * g = h$ almost everywhere. Therefore the \hyperref[Closed Graph Theorem]{theorem:closed-graph} implies that $L_f$ is continuous.
@@ -87,7 +87,7 @@
\begin{definition}[Space of Left Convolvers] \begin{definition}[Space of Left Convolvers]
\label{definition:left-convolver-algebra} \label{definition:left-convolver-algebra}
Let $G$ be a group, $LC(G) := \bracsn{L_f| G \to \complex\text{ is a left convolver}} \subset B(l^2(G; \complex))$, and $f, g: G \to \complex$ be left convolvers, then: Let $G$ be a group, $LC(G) := \bracsn{L_f| f: G \to \complex\text{ is a left convolver}} \subset B(l^2(G; \complex))$, and $f, g: G \to \complex$ be left convolvers, then:
\begin{enumerate} \begin{enumerate}
\item $f \in l^2(G; \complex)$. \item $f \in l^2(G; \complex)$.
\item $L_f^* = L_{f^*}$, where $f^*(x) = \ol{f(x^{-1})}$ for all $x \in G$. \item $L_f^* = L_{f^*}$, where $f^*(x) = \ol{f(x^{-1})}$ for all $x \in G$.
@@ -97,13 +97,13 @@
The set $LC(G)$ is the \textbf{algebra of left convolvers} on $G$. The set $LC(G)$ is the \textbf{algebra of left convolvers} on $G$.
\end{definition} \end{definition}
\begin{proof}[Proof, {{\cite[Propositon 1.3.5]{AnantharamanPopa}}}. ] \begin{proof}[Proof, {{\cite[Proposition 1.3.5]{AnantharamanPopa}}}. ]
(1): $f * \delta_e = f \in l^2(G; \complex)$. (1): $f * \delta_e = f \in l^2(G; \complex)$.
(2): Let $x, y \in G$, then (2): Let $x, y \in G$, then
\begin{align*} \begin{align*}
\dpn{L_f\delta_x, \delta_y}{l^2(G; \complex)} &= (f * \delta_x)(y) = f(x^{-1}y) \\ \dpn{L_f\delta_x, \delta_y}{l^2(G; \complex)} &= (f * \delta_x)(y) = f(yx^{-1}) \\
\dpn{\delta_x, L_{f^*}\delta_y}{l^2(G;\complex)} &= \ol{(f^* * \delta_y)(x)} = f^*(y^{-1}x) \dpn{\delta_x, L_{f^*}\delta_y}{l^2(G;\complex)} &= \ol{(f^* * \delta_y)(x)} = \ol{f^*(y^{-1}x)} = f(yx^{-1})
\end{align*} \end{align*}
@@ -114,7 +114,7 @@
(4): Let $T \in \ol{LC(G)}^{\text{\small SOT}}$ and $\angles{f_\alpha}_{\alpha \in A}$ be left convolvers such that $T = \sotlim_{\alpha \in A}L_{f_\alpha}$. Let $f = T\delta_e \in l^2(G; \complex)$, then (4): Let $T \in \ol{LC(G)}^{\text{\small SOT}}$ and $\angles{f_\alpha}_{\alpha \in A}$ be left convolvers such that $T = \sotlim_{\alpha \in A}L_{f_\alpha}$. Let $f = T\delta_e \in l^2(G; \complex)$, then
\[ \[
f = T\delta_e = \paren{\sotlim_{\alpha \in A}L_{f_\alpha}}\delta_e = l^2\text{-}\lim_{\alpha \in A}f_\alpha * e f = T\delta_e = \paren{\sotlim_{\alpha \in A}L_{f_\alpha}}\delta_e = l^2\text{-}\lim_{\alpha \in A}f_\alpha * \delta_e
\] \]
For any $g \in l^2(G; \complex)$, For any $g \in l^2(G; \complex)$,
@@ -132,7 +132,7 @@
\begin{proof}[Proof, {{\cite[Theorem 1.3.6]{AnantharamanPopa}}}. ] \begin{proof}[Proof, {{\cite[Theorem 1.3.6]{AnantharamanPopa}}}. ]
($\subset$): For any $x, y, z \in G$ and $f \in l^2(G; \complex)$, ($\subset$): For any $x, y, z \in G$ and $f \in l^2(G; \complex)$,
\[ \[
(\lambda(x)\rho(y)f)(z) = (\rho(h)f)(x^{-1}z) = f(x^{-1}zy) = (\lambda(x)f)(zy) = (\rho(y)\lambda(x)f)(z) (\lambda(x)\rho(y)f)(z) = (\rho(y)f)(x^{-1}z) = f(x^{-1}zy) = (\lambda(x)f)(zy) = (\rho(y)\lambda(x)f)(z)
\] \]
so $\lambda(G) \subset R(G)'$. By \autoref{definition:left-convolver-algebra}, $L(G) \subset LC(G) \subset R(G)'$. so $\lambda(G) \subset R(G)'$. By \autoref{definition:left-convolver-algebra}, $L(G) \subset LC(G) \subset R(G)'$.
@@ -147,7 +147,7 @@
\begin{definition} \begin{definition}
\label{definition:group-algebra-fourier-series} \label{definition:group-algebra-fourier-series}
Let $G$ be a group. For each $x \in G$, let $u_x = \lambda(x)$ and identify it with $\delta_x$, then $\lambda(G) = \bracsn{u_x|x \in G}$ is the \textbf{canonical unitaries} of $L(G)$. Let $G$ be a group. For each $x \in G$, let $u_x = \lambda(x)$ and identify it with $\delta_x$, then $\lambda(G) = \bracsn{u_x|x \in G}$ are the \textbf{canonical unitaries} of $L(G)$.
For each $f \in l^2(G)$, $f = \sum_{x \in G}f(x)u_x$. In particular, for each $T \in L(G) = LC(G) \subset l^2(G; \complex)$, there exists $\bracsn{T_x}_{x \in G} \subset \complex$ such that $T$ may be written as $T = \sum_{x \in G}T_x u_x$\footnote{This sum does not converge in the strong-operator or the weak-operator topology. }. In which case, the scalars $\bracsn{T_x}_{x \in G}$ are the \textbf{Fourier coefficients} of $T$. For each $f \in l^2(G)$, $f = \sum_{x \in G}f(x)u_x$. In particular, for each $T \in L(G) = LC(G) \subset l^2(G; \complex)$, there exists $\bracsn{T_x}_{x \in G} \subset \complex$ such that $T$ may be written as $T = \sum_{x \in G}T_x u_x$\footnote{This sum does not converge in the strong-operator or the weak-operator topology. }. In which case, the scalars $\bracsn{T_x}_{x \in G}$ are the \textbf{Fourier coefficients} of $T$.
\end{definition} \end{definition}
@@ -187,7 +187,7 @@
\label{proposition:wreath-product-icc} \label{proposition:wreath-product-icc}
Let $H$ be a non-trivial group, $\Gamma$ be a group acting on a set $I$, and $\Gamma_f \subset \Gamma$ be the subgroup of elements whose conjugacy class is finite. If Let $H$ be a non-trivial group, $\Gamma$ be a group acting on a set $I$, and $\Gamma_f \subset \Gamma$ be the subgroup of elements whose conjugacy class is finite. If
\begin{enumerate}[label=(\alph*)] \begin{enumerate}[label=(\alph*)]
\item The orbits of $\Gamma \curvearrowleft I$ are finite. \item The orbits of $\Gamma \curvearrowleft I$ are infinite.
\item The restricted action $\Gamma_f \curvearrowright I$ is faithful. \item The restricted action $\Gamma_f \curvearrowright I$ is faithful.
\end{enumerate} \end{enumerate}
@@ -198,7 +198,7 @@
Let $g = (\xi, \gamma) \in H \wr_I \Gamma$. First suppose that $\xi \ne e_{H^{(I)}}$, then the support of $\xi$ is non-empty. By (a), the support of $\xi$ has infinite orbit under $\Gamma$, so $g^G \supset g^\Gamma$ is infinite. Let $g = (\xi, \gamma) \in H \wr_I \Gamma$. First suppose that $\xi \ne e_{H^{(I)}}$, then the support of $\xi$ is non-empty. By (a), the support of $\xi$ has infinite orbit under $\Gamma$, so $g^G \supset g^\Gamma$ is infinite.
Now suppose that $\xi = e_{H^{(I)}}$ and $\gamma \ne e_{\Gamma}$. If $\gamma \in \Gamma \setminus \Gamma_f$, then the conjugacy class of $\gamma$ is infinite, and the conjugacy class of $g$ is also infinite. Now suppose that $\gamma \in \Gamma_f$. By (b), there exists $i \in I$ with $\gamma i_0 \ne i_0$. As $H$ is non-trivial, there exists $\xi_0 \in H^{(I)} \setminus \bracsn{e_{H^{(I)}}}$ with support $\bracsn{i_0}$. In which case, if $g_0 = (\xi_0, e_{\Gamma})$, then Now suppose that $\xi = e_{H^{(I)}}$ and $\gamma \ne e_{\Gamma}$. If $\gamma \in \Gamma \setminus \Gamma_f$, then the conjugacy class of $\gamma$ is infinite, and the conjugacy class of $g$ is also infinite. Now suppose that $\gamma \in \Gamma_f$. By (b), there exists $i_0 \in I$ with $\gamma i_0 \ne i_0$. As $H$ is non-trivial, there exists $\xi_0 \in H^{(I)} \setminus \bracsn{e_{H^{(I)}}}$ with support $\bracsn{i_0}$. In which case, if $g_0 = (\xi_0, e_{\Gamma})$, then
\[ \[
g_0^{-1}gg_0 = (\xi_0^{-1}\gamma\xi_0, \gamma) g_0^{-1}gg_0 = (\xi_0^{-1}\gamma\xi_0, \gamma)
\] \]