From c5b9539850c68f1dfcac62d523794edfc0613748 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Fri, 6 Nov 2026 15:52:00 -0500 Subject: [PATCH] Course related house keeping. --- src/cat/cat/cat-func.tex | 21 +++++ src/dyn/meas/definitions.tex | 74 ++++++++++++++++ src/fa/lc/compact.tex | 25 ++++-- src/fa/norm/normed.tex | 42 +++++++++ src/op/system/aos.tex | 136 ++++++++++++++++++++++++++++++ src/op/system/inj.tex | 24 ++++++ src/topology/uniform/complete.tex | 8 +- 7 files changed, 319 insertions(+), 11 deletions(-) create mode 100644 src/dyn/meas/definitions.tex create mode 100644 src/op/system/inj.tex diff --git a/src/cat/cat/cat-func.tex b/src/cat/cat/cat-func.tex index 67b3b1e..9965f9e 100644 --- a/src/cat/cat/cat-func.tex +++ b/src/cat/cat/cat-func.tex @@ -39,3 +39,24 @@ \item[(FN2')] For any $A, B, C \in \obj{\mathfrak{A}}$, $f \in \mor{A, B}$, and $g \in \mor{B, C}$, $\lambda(g \circ f) = \lambda(f) \circ \lambda(g)$. \end{enumerate} \end{definition} + + +\begin{definition}[Injective] +\label{definition:injective-cat} + Let $\catc$ be a category with a notion of embedding and $I \in \obj(\catc)$, then $I$ is \textbf{injective} if for any objects $X, Y \in \obj(\catc)$, embedding $f: X \to Y$, and morphism $\phi: X \to I$, there exists an extension $\Phi: Y \to I$ such that the following diagram commutes: + \[ + \xymatrix{ + & I \\ + X \ar@{->}[ru]^{\phi} \ar@{^{(}->}[r]_{f} & Y \ar@{->}[u]_{\Phi} + } + \] +\end{definition} + +\begin{definition}[Sufficiently Many Injectives] +\label{definition:suff-many-injective} + Let $\catc$ be a category with a notion of embedding, then $\catc$ has \textbf{sufficiently many injectives} if every object embeds into an injective object. +\end{definition} + + + + diff --git a/src/dyn/meas/definitions.tex b/src/dyn/meas/definitions.tex new file mode 100644 index 0000000..0ef22e5 --- /dev/null +++ b/src/dyn/meas/definitions.tex @@ -0,0 +1,74 @@ +\section{Measurable Dynamical Systems} +\label{section:measurable-dynamics-intro} + +\begin{definition}[Measure Preserving] +\label{definition:measure-preserving} + Let $(X, \cm, \mu)$ be a measure space and $T: X \to X$ be measurable, then $T$ is \textbf{measure preserving} if $\mu(T^{-1}(E)) = \mu(E)$ for all $E \in \cm$. +\end{definition} + +\begin{definition}[Automorphism of Measure Space] +\label{definition:measure-preserving-automorphism} + Let $(X, \cm, \mu)$ be a measure space and $T: X \to X$ be measurable, then $T$ is an \textbf{automorphism} if $T$ is bijective and measure preserving. The set $\text{Aut}(X, \cm, \mu)$ is the \textbf{automorphism group} of $(X, \cm, \mu)$. +\end{definition} + +\begin{definition}[Measure Preserving Action] +\label{definition:measure-preserving-action} + Let $\Gamma$ be a group and $(X, \cm, \mu)$ be a measure space, then a \textbf{measure preserving action} of $\Gamma$ on $(X, \cm, \mu)$ is a group homomorphism $\alpha: \Gamma \to \text{Aut}(X, \cm, \mu)$. The pair $(\Gamma, (X, \cm, \mu)) = \Gamma \curvearrowright (X, \cm, \mu)$ is a \textbf{measurable dynamical system}. +\end{definition} + +\begin{definition}[Invariant] +\label{definition:measurable-invariant-set} + Let $\Gamma$ be a group acting on a measure space $(X, \cm, \mu)$ and $E \in \cm$, then $E$ is \textbf{$\Gamma$-invariant} if $gE = E$ for all $g \in \Gamma$. For any separable metric space $Y$ and measurable function $f: X \to Y$, $f$ is \textbf{$\Gamma$-invariant} if $f \circ g = f$ a.e. for all $g \in \Gamma$. +\end{definition} + +\begin{definition}[Ergodic] +\label{definition:ergodic} + Let $\Gamma$ be a group acting on a measure space $(X, \cm, \mu)$, then the action is \textbf{ergodic} if every $\Gamma$-invariant set is either null or conull. +\end{definition} + +\begin{proposition} +\label{proposition:measurable-group-composition} + Let $\Gamma$ be a group acting on a measure space $(X, \cm, \mu)$ and $Y$ be a separable metric space, then $\Gamma$ induces an action\footnote{W.r.t. the topology of convergence in measure.} + \[ + \Gamma \to \text{Aut}(L^0(X; Y)) \quad g \cdot f = f \circ g + \] + + For any Banach space $E$\footnote{Separability not required since $L^p$ here refers to strongly measurable functions.}, this induces a representation + \[ + \Gamma \to L^p(X; E) \quad g \cdot f = f \circ g + \] + + for all $p \in [1, \infty]$. Moreover, the following are equivalent: + \begin{enumerate} + \item The action of $\Gamma$ on $(X, \cm, \mu)$ is ergodic. + \item For every $f \in L^0(X; \real)$, $f$ is $\Gamma$-invariant if and only if $f$ is constant almost everywhere. + \end{enumerate} +\end{proposition} +\begin{proof} + (1) $\Rightarrow$ (2): For each $t \in \real$, let $A_t = \bracsn{f \le t}$, then since the action of $\Gamma$ is ergodic, $A_t$ is either null or conull. Let + \[ + \alpha = \sup\bracsn{t \in \real| A_t \text{ null}} \in \real + \] + + then $A_\alpha \supset \bigcup_{n \in \natp}A_{\alpha + 1/n}$ is conull. Therefore $\bracsn{f = \alpha} = A_\alpha \setminus \bigcup_{n \in \natp} A_{\alpha - 1/n}$ is conull. +\end{proof} + + +\begin{example} + \label{example:ergodic-fourier} + Let $\mathbb{T} = \mathbb{R}/\mathbb{Z} = \partial B_\complex(0, 1)$, equipped with the normalised Lebesgue measure. For each $\alpha \in \real$, let $R_\alpha: \mathbb{T} \to \mathbb{T}$ be defined by $x \mapsto x + \alpha \mod 1$, then $R_\alpha$ induces an $\integer$-action on $\mathbb{T}$, defined by $n \cdot x = x + n\alpha$, which is ergodic if and only if $\alpha \in \real \setminus \mathbb Q$. +\end{example} +\begin{proof} + Let $f \in L^2(\mathbb{T}; \complex)$, then + \[ + f(x) = \sum_{n \in \integer} \hat f(n) e^{2\pi i n x} \quad R_\alpha f(x) = \sum_{n \in \integer} \hat f(n) e^{2\pi i n (x + \alpha)} + \] + + where the series converges in $L^2(\mathbb{T}; \complex)$. If $f$ is $R_\alpha$-invariant, then + \[ + f(x) = \sum_{n \in \integer} \hat f(n) e^{2\pi i n (x + \alpha)} = \sum_{n \in \integer} \hat f(n) e^{2\pi i n \alpha} e^{2\pi i n x} + \] + + As such, $\hat f(n)e^{2\pi i n \alpha} = \hat f(n)$ for all $n \in \integer$. As $\alpha \not\in \mathbb Q$, $e^{2\pi i n \alpha} \ne 1$ for all $n \in \integer \setminus \bracsn{0}$, so $\hat f(n) = 0$ for all $n \in \integer \setminus \bracsn{0}$. Therefore $f = \hat f(0)$ almost everywhere. By \autoref{proposition:measurable-group-composition}, $R_\alpha$ is ergodic. + +\end{proof} diff --git a/src/fa/lc/compact.tex b/src/fa/lc/compact.tex index 5741621..df2a3ea 100644 --- a/src/fa/lc/compact.tex +++ b/src/fa/lc/compact.tex @@ -29,7 +29,7 @@ \begin{definition}[Extreme Subset] \label{definition:extreme-subset} - Let $E$ be a vector space over $\real$, $K \subset E$ be convex, and $A \subset K$, then $A$ is \textbf{extreme set} if for any $x \in A$ and $y, z \in K$ such that $x \in (y, z)$, $y, z \in A$ as well. + Let $E$ be a vector space over $\real$, $K \subset E$ be convex, and $A \subset K$ be convex, then $A$ is \textbf{extreme} if for any $x \in A$ and $y, z \in K$ such that $x \in (y, z)$, $y, z \in A$ as well. \end{definition} \begin{lemma} @@ -40,6 +40,14 @@ Let $y, z \in K$ such that $x \in (y, z)$. Since $A$ is extreme, $y, z \in A$. As $x$ is extremal in $A$, $x = y = z$, so $x$ is extremal in $K$. \end{proof} +\begin{lemma} +\label{lemma:extremal-face-map} + Let $E, F$ be locally convex spaces over $\real$, $K \subset E$ be convex, $T: K \to F$ be affine, and $A \subset T(K)$ be extreme, then $T^{-1}(A)$ is also extreme. +\end{lemma} +\begin{proof} + Let $x \in T^{-1}(A)$ and $y, z \in K$ with $x \in (y, z)$, then $Tx \in (Ty, Tz)$. As $A$ is extreme, $Ty, Tz \in A$, so $y, z \in T^{-1}(A)$. +\end{proof} + \begin{lemma} \label{lemma:extremal-face} Let $E$ be a locally convex space over $\real$, $K \subset E$ be non-empty and compact, and $\phi \in E^*$. Let $\alpha = \sup\bracs{\dpn{x, \phi}{E}|x \in K}$, then $A = \bracs{\phi = \alpha} \cap K$ is a non-empty extreme subset of $K$. @@ -50,18 +58,19 @@ \begin{theorem}[Krein-Milman] \label{theorem:krein-milman} - Let $E$ be a separated locally convex space over $\real$ and $K \subset E$ be a compact convex set, then $K$ is the closed convex hull of its extreme points. + Let $E$ be a separated locally convex space over $\real$ and $\emptyset \ne + K \subset E$ be a compact convex set, then + \begin{enumerate} + \item Every non-empty, closed, extreme subset of $K$ contains an extreme point of $K$. + \item $K$ is the closed convex hull of its extreme points. + \end{enumerate} \end{theorem} \begin{proof}[Proof, {{\cite[Theorem 1.12.5]{Bogachev}}}. ] - Assume without loss of generality that $K \ne \emptyset$. - - Let $K_0 \subset K$ be a closed, extreme subset of $K$, and $\mathcal{E}(K_0) \subset 2^K$ be the collection of all non-empty closed extreme subsets of $K$ contained in $K_0$. Since $K_0 \in \mathcal{E}(K_0)$, $\mathcal{E}(K_0) \ne \emptyset$. Since $K$ is compact, for any chain $\mathcal{C} \subset \mathcal{E}$, $\bigcap_{A \in \mathcal{C}}A$ is also non-empty, closed, and extreme. + (1): Let $K_0 \subset K$ be a non-empty, closed, extreme subset of $K$, and $\mathcal{E}(K_0) \subset 2^K$ be the collection of all non-empty closed extreme subsets of $K$ contained in $K_0$. Since $K_0 \in \mathcal{E}(K_0)$, $\mathcal{E}(K_0) \ne \emptyset$. Since $K$ is compact, for any chain $\mathcal{C} \subset \mathcal{E}$, $\bigcap_{A \in \mathcal{C}}A$ is also non-empty, closed, and extreme. By Zorn's lemma, there exists a minimal element $A$ of $\mathcal{E}(K_0)$. Let $x, y \in A$, $\phi \in E^*$, and $\alpha = \sup_{z \in A}\dpn{z, \phi}{E}$, then $\bracs{\phi = \alpha} \cap K$ is a non-empty, closed, and extreme subset of $K$ by \autoref{lemma:extremal-face}, so $A \cap \bracs{\phi = \alpha}$ is also extreme. By minimality of $A$, $A \subset \bracs{\phi = \alpha}$. Thus by the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, $A$ consists of exactly one point. In which case, for any $y, z \in E$ with $A \subset (y, z) \subset K$, $y = z \in A$. Therefore there exists an extreme point of $K$ in $K_0$. - Since $K$ itself is an extreme subset, a minimal element of $\mathcal{E}(K)$ represents an extreme point, so $K$ admits at least one extreme point. - - Now, let $C$ be the collection of all extreme points in $K$. Assume for contradiction that $\ol{\conv}(C) \subsetneq K$, then by the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, there exists $x \in K$ and $\phi \in E^*$ such that $\sup_{y \in \ol{\conv}(C)}\dpn{y, \phi}{E} < \dpn{x, \phi}{E}$. Let $\alpha = \sup_{z \in K}\dpn{z, \phi}{E}$, then by \autoref{lemma:extremal-face}, $K \cap \bracs{\phi = \alpha}$ is a non-empty, closed, and extreme subset of $K$. By the preceding discussion, there exists an extreme point of $K$ in $K \cap \bracs{\phi = \alpha}$, which contradicts the fact that $\sup_{y \in \ol{\conv}(C)}\dpn{y, \phi}{E} < \alpha$. + (2): Let $C$ be the collection of all extreme points in $K$. Assume for contradiction that $\ol{\conv}(C) \subsetneq K$, then by the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, there exists $x \in K$ and $\phi \in E^*$ such that $\sup_{y \in \ol{\conv}(C)}\dpn{y, \phi}{E} < \dpn{x, \phi}{E}$. Let $\alpha = \sup_{z \in K}\dpn{z, \phi}{E}$, then by \autoref{lemma:extremal-face}, $K \cap \bracs{\phi = \alpha}$ is a non-empty, closed, and extreme subset of $K$. By (1), there exists an extreme point of $K$ in $K \cap \bracs{\phi = \alpha}$, which contradicts the fact that $\sup_{y \in \ol{\conv}(C)}\dpn{y, \phi}{E} < \alpha$. \end{proof} \begin{theorem}[Markov-Kakutani Fixed Point Theorem] diff --git a/src/fa/norm/normed.tex b/src/fa/norm/normed.tex index 62a06f5..0b5e2b3 100644 --- a/src/fa/norm/normed.tex +++ b/src/fa/norm/normed.tex @@ -161,3 +161,45 @@ then the identity $(K^d, \norm{\cdot}_1) \to (K^d, [\cdot])$ is continuous. Let $S = \bracsn{x \in K^d|\ \norm{x}_1 = 1}$, then $S$ is compact with respect to $\norm{\cdot}_1$ and hence compact with respect to $[\cdot]$. By continuity of $[\cdot]$, $\min_{x \in S}[x]$ exists and is non-zero. \end{proof} + +\begin{theorem} +\label{theorem:reflexive-banach-spaces} + Let $E$ be a Banach space over $K \in \RC$, then the following are equivalent: + \begin{enumerate} + \item $E$ is reflexive. That is, $E = E^{**}$. + \item $\ol{B_E(0, 1)}^E$ is weakly compact. + \item The weak and weak* topologies on $E^*$ coincide. + \item $E^*$ is reflexive. + \end{enumerate} +\end{theorem} +\begin{proof} + (1) $\Rightarrow$ (2): By the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}. + + (2) $\Rightarrow$ (1): By \hyperref[Goldstine's Theorem]{theorem:goldstine-weak} and \autoref{proposition:compact-extensions}, + \[ + \ol{B_{E^{**}}(0, 1)}^{E^{**}} = \ol{B_E(0, 1)}^{\sigma(E^{**}, E^*)} = \ol{B_E(0, 1)}^E + \] + + (1) $\Leftrightarrow$ (3): By \autoref{lemma:duality-dual}. + + (4) $\Rightarrow$ (1): By (1) $\Rightarrow$ (4), $E^{**}$ is reflexive. Since $\ol{B_{E}(0, 1)}^{E}$ is norm closed in $E^{**}$, it is weakly closed in $E^{**}$, and hence weak*-closed in $E^{**}$. By \hyperref[Goldstine's Theorem]{theorem:goldstine-weak}, $\ol{B_E(0, 1)}^E = \ol{B_{E^{**}}(0, 1)}^{E^{**}}$. +\end{proof} + +% This should be a three space property. +\begin{corollary} +\label{corollary:reflexive-banach-three-spaces} + Let $E$ be a Banach space over $K \in \RC$ and $F \subset E$ be a closed subspace, then $F$ and $E/F$ are reflexive if and only if $E$ is reflexive. +\end{corollary} +\begin{proof} + ($F$): Assume that $E$ is reflexive. As $F$ is closed, it is weakly closed. Thus $\ol{B_F(0, 1)} = \ol{B_E(0, 1)} \cap F$ is weakly compact. By (2) of \autoref{theorem:reflexive-banach-spaces}, $E$ is reflexive. + + ($E/F$): Assume that $E$ is reflexive. Let $\pi: E \to E/F$ be the canonical map, then $\pi$ is also weakly continuous. Hence $\pi(\ol{B_E(0, 1)})$ is weakly compact in $E/F$. By (2) of \autoref{theorem:reflexive-banach-spaces}, $E/F$ is reflexive. + + ($E$): Assume that $F$ and $E/F$ are reflexive. Let $x_0 \in E^{**}$, and $T: F^\perp \to (E/F)^*$ be the canonical isomorphism, then $x_0 \circ T^{-1} \in (E/F)^{**}$. Given that $E/F$ is reflexive, there exists $x \in E$ such that $x_0 \circ T^{-1} = x + F$. In particular, for any $\phi \in F^\perp$, + \[ + \dpn{\phi, x_0}{E^*} = \dpn{T\phi, x_0 \circ T^{-1}}{(E/F)^*} = \dpn{x + F, T\phi}{E/F} = \dpn{x, \phi}{E} = \dpn{\phi, x}{E^{**}} + \] + + so $x_0 - x \in F^{\perp\perp}$. By the \hyperref[Bipolar Theorem]{theorem:bipolar}, $F^{\perp\perp} = F^{**} = F$. Therefore $x_0 - x \in F$, and $x_0 \in E$. +\end{proof} + diff --git a/src/op/system/aos.tex b/src/op/system/aos.tex index d3bdceb..e7d55f0 100644 --- a/src/op/system/aos.tex +++ b/src/op/system/aos.tex @@ -70,7 +70,143 @@ \end{align*} \end{proof} +\begin{proposition} +\label{proposition:abstract-operator-system-norm} + Let $S$ be an abstract operator system with order unit $e$. For each $s \in M_n(S)$, let\footnote{Infimum exists because $I_{M_{2n}(S)} \otimes e$ is the order unit in $M_{2n}(S)$.} + \[ + \norm{s}_{M_n(S)} = \inf\bracs{r > 0 \bigg | \begin{bmatrix} + rI_{M_n(S)} \otimes e &s \\ + s^* & rI_{M_n(S)} \otimes e + \end{bmatrix} \ge 0} + \] + then + \begin{enumerate} + \item $\norm{\cdot}_{M_n(S)}$ is a norm. + \item For each $s \in M_n(S)$, $\norm{s}_{M_n(S)} = \normn{s^*}_{M_n(S)}$. + \item $\bracsn{s \in M_n(S)|s \ge 0}$ is closed with respect to $\norm{\cdot}_{M_n(S)}$. + \end{enumerate} +\end{proposition} +\begin{proof} + (1): By replacing $S$ with $M_n(S)$, assume without loss of generality that $n = 1$. Let $s \in S$ with $\norm{s}_{M_n(S)} = 0$, then for each $r > 0$, + \[ + \begin{bmatrix} + re &s \\ + s^* & re + \end{bmatrix} \ge 0 \quad \forall r > 0 + \] + As such, for each $\lambda \in \complex$, + \[ + 0 \le \begin{bmatrix} + 1 & \ol \lambda + \end{bmatrix}\begin{bmatrix} + re &s \\ + s^* & re + \end{bmatrix}\begin{bmatrix} + 1 \\ + \lambda + \end{bmatrix} = r(1 + |\lambda|^2)e + \lambda s + (\lambda s)^* + \] + As the above holds for all $r > 0$, $\lambda s + (\lambda s)^* \ge 0$. In particular, $0 \le s + s^* \le 0$ and $0 \le is + (is)^* \le 0$, so $\text{Re}(s) = \text{Im}(s) = 0$, and $s = 0$. + + (2): For any $s \in S$ and $r > 0$, + \[ + \begin{bmatrix} + 0 & 1 \\ + 1 & 0 + \end{bmatrix}\begin{bmatrix} + re &s \\ + s^* & re + \end{bmatrix}\begin{bmatrix} + 0 & 1 \\ + 1 & 0 + \end{bmatrix} = \begin{bmatrix} + re &s^* \\ + s & re + \end{bmatrix} + \] + + so + \[ + \begin{bmatrix} + re &s \\ + s^* & re + \end{bmatrix} \ge 0 \quad \Leftrightarrow \quad \begin{bmatrix} + re &s^* \\ + s & re + \end{bmatrix} \ge 0 + \] + + (3): Let $s \in S^+$, where the closure is taken with respect to $\normn{\cdot}_{M_1(S)}$. By (2), $\ol{S^+} \subset \ol S_{sa} = S$. Let $r > 0$, then there exists $t \in S^+$ and $\normn{s - t}_{S} < r$. As such, + \begin{align*} + + 0 &\le \begin{bmatrix} + re &s - t \\ + s - t & re + \end{bmatrix} \\ + 0 &\le \begin{bmatrix} + 1 & 1 + \end{bmatrix} \begin{bmatrix} + re &s - t \\ + s - t & re + \end{bmatrix}\begin{bmatrix} + 1 \\ 1 + \end{bmatrix} = 2re + 2s - 2t + \end{align*} + + so $re + s \ge 0$. By the Archimedean property, $s \ge 0$. + +\end{proof} + +\begin{theorem}[Choi-Effros] +\label{theorem:choi-effros} + Let $S$ be an abstract operator system with order unit $e$, then there exists a complex Hilbert space $H$, an operator system $S_1 \subset B(H)$, and a complete order isomorphism $\Phi: S \to S_1$ with $\phi(e) = I_H$. +\end{theorem} +\begin{proof} + For each $n \in \natp$, let + \[ + K_n = \bracsn{\phi: S \to M_n(\complex)| \phi \text{ is unital, completely positive}} + \] + + and $K = \bigsqcup_{n \in \natp}K_n$\footnote{The matrix state space.}. Define\footnote{The summand on each $\phi \in K_n$ is $M_n(\complex)$.} + \[ + \Phi: S \to [l^\infty(K); M_n(\complex)] \quad \Phi(s)_\phi = \phi(s) + \] + + then $[l^\infty(K); M_n(\complex)] \subset B([l^2(K); \complex^n])$, and $\Phi$ is unital and completely positive. + + Let $n \in \natp$ and $s \in M_n(S)$ with $\Phi_n(s) \ge 0$. Suppose for contradiction that $s < 0$, then by the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, there exists $\tilde \phi \in M_n(S)^*$ such that $\tilde \phi (s) \not\ge 0$ but $\tilde \phi(t) \ge 0$ for all $t \in M_n(S)$ with $t \ge 0$. As such, $\tilde \phi$ is a positive linear functional on $M_n(S)^*$. By \autoref{proposition:matrix-order-positive-characterisation}, the associated map $\phi: S \to M_n(\complex)$, defined by $\phi(t)_{ij} = \tilde \phi(E_{ij} \otimes t)$, is completely positive. However, + \[ + \sum_{i, j = 1}^n \dpn{\phi(s_{ij})e_j, e_i}{\complex^n} = \tilde \phi(s) < 0 + \] + + so $\phi_n(s) \not\ge 0$ as well. After composing with the range projection of $\phi(e)$, assume without loss of generality that $\phi(e)$ is invertible. In which case, $\phi' = \phi(e)^{-1/2}\phi \phi(e)^{-1/2}$ is unital and completely positive with $\phi_n'(s) \not\ge 0$, contradiction. +\end{proof} + +\begin{definition}[Maximal Operator System Structure] +\label{definition:maximal-operator-system} + Let $S$ be an operator system with order unit $e$. For each $n \in \natp$, $C_n = \ol{M_n(\complex)_+ \otimes S_+} \subset M_n(S)$, then $\seq{C_n}$ defines a matrix order on $S$, and $(S, \seq{C_n}, e)$ is an operator system. The triple $\text{OMAX}(S) = (S, \seq{C_n}, e)$ is the \textbf{maximal} operator system structure on $S$. +\end{definition} + +\begin{example} + \label{example:matrix-os-embedding-not-homomorphism} + + Let + \[ + \phi: M_n(\complex) \to M_{n+1}(\complex) \quad a \mapsto \begin{bmatrix} + a & 0 \\ + 0 & \frac{\tau(a)}{n} + \end{bmatrix} + \] + + then $\phi$ is an embedding of operator systems, but $\phi(M_n(\complex))$ is not a $C^*$-subalgebra. +\end{example} + +\begin{example} + \label{example:arveson-nc-function} + + Let $S = \text{span}\bracsn{1, z, \ol z} \subset C(\mathbb{T})$, then $C^*(S) = C(\mathbb{T})$, but $S$ also embeds into and generates $C(\mathbb D)$. In this particular case, $C(\mathbb T) = C^*_{\min}(S)$. +\end{example} diff --git a/src/op/system/inj.tex b/src/op/system/inj.tex new file mode 100644 index 0000000..ec5efc3 --- /dev/null +++ b/src/op/system/inj.tex @@ -0,0 +1,24 @@ +\section{Injective Operator Systems} +\label{section:injective-ops} + + +By \hyperref[Arveson's Extension Theorem]{theorem:arveson-extension}, $B(H)$ is injective for any Hilbert space $H$. Further, the \hyperref[Choi-Effros Theorem]{choi-effros} implies that every operator system embeds into $B(H)$, for a suitable Hilbert space $H$. Therefore the category of operator systems, with unital complete order isomorphisms, has sufficiently many injectives. + +\begin{theorem} +\label{theorem:opsys-predual} + Let $S, T$ be operator systems. If $T$ admits a predual, then the space + \[ + UCP(S; T) = \bracsn{\phi: S \to T| \phi \text{ unital, completely positive}} + \] + + is compact in the point-weak* topology. +\end{theorem} + + +\begin{definition}[Injective] +\label{definition:injective-operator-system} + Let +\end{definition} + + + diff --git a/src/topology/uniform/complete.tex b/src/topology/uniform/complete.tex index f30e79e..bb5ea96 100644 --- a/src/topology/uniform/complete.tex +++ b/src/topology/uniform/complete.tex @@ -16,17 +16,19 @@ \begin{proposition} \label{proposition:complete-closed} - Let $X$ be a uniform space and $A \subset X$ be closed, then $A$ is complete. + Let $X$ be a Hausdorff complete uniform space and $A \subset X$, then $A$ is closed if and only if it is complete. \end{proposition} \begin{proof} - Let $\fF$ be a Cauchy filter on $A$, then there exists $x \in X$ such that $\fF$ converges to $x$. Since $A$ is closed, $x \in X$ by (4) of \autoref{definition:closure}. + ($\Rightarrow$): Suppose that $A$ is closed. Let $\fF$ be a Cauchy filter on $A$, then there exists $x \in X$ such that $\fF$ converges to $x$. Since $A$ is closed, $x \in X$ by (4) of \autoref{definition:closure}. + + ($\Leftarrow$): Suppose that $A$ is complete. Let $x \in \ol A$, then there exists a filter $\fF \subset 2^A$ with $\fF \to x$. By the \hyperref[Cauchy criterion]{proposition:cauchycriterion}, $\fF$ is Cauchy, so there exists $y \in A$ such that $\fF \to y$. Since $X$ is Hausdorff, $x = y \in A$, so $A = \ol A$. \end{proof} \begin{lemma} \label{lemma:complete-dense} - Let $(X, \fU)$ be a uniform space and $A \subset X$ be a dense subset, then $X$ is \textbf{complete} if and only if every Cauchy filter on $A$ converges to at least one point in $X$. + Let $(X, \fU)$ be a uniform space and $A \subset X$ be a dense subset, then $X$ is complete if and only if every Cauchy filter on $A$ converges to at least one point in $X$. \end{lemma} \begin{proof} Let $\fF \subset X$ be a Cauchy filter. Using \autoref{proposition:minimalcauchyexistence}, assume without loss of generality that $\fF$ is minimal. By \autoref{proposition:cauchyinterior}, $\fF$ admits a base consisting of open sets. Hence $E \cap A \ne \emptyset$ for all $E \in \fF$. Let $\fF_A = \bracs{E \cap A| E \in \fF}$, then $\fF_A$ is a filter base on $A$.