From c5589509e6df59c7447228763d6cf5ccb87d7658 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Mon, 19 Oct 2026 14:27:52 -0400 Subject: [PATCH] Moved Stinespring. --- src/op/system/dilation.tex | 216 +++++++++++++++++++------------------ 1 file changed, 109 insertions(+), 107 deletions(-) diff --git a/src/op/system/dilation.tex b/src/op/system/dilation.tex index e78dc33..ecbab47 100644 --- a/src/op/system/dilation.tex +++ b/src/op/system/dilation.tex @@ -47,113 +47,6 @@ Similarly, if $(K, S)$ is a power dilation of $T$ and $(L, T)$ is a power dilation of $S$, then $(L, T)$ is a power dilation of $R$. \end{lemma} -\begin{theorem}[Stinespring's Dilation Theorem] -\label{theorem:stinespring-dilation} - Let $A$ be a unital $C^*$-algebra, $H$ be a complex Hilbert space, and $\phi: A \to B(H)$ be a completely positive map, then there exists a triple $(K, V, \pi)$, where: - \begin{enumerate}[label=(\alph*)] - \item $K$ is a complex Hilbert space. - \item $V: H \to K$ is a continuous linear map. - \item $\pi: A \to B(K)$ is a unital *-homomorphism. - \end{enumerate} - - such that: - \begin{enumerate} - \item $\norm{\phi(1_A)}_{B(H)} = \norm{V}_{L(H; K)}^2$. - \item For all $a \in A$, $\phi(a) = V^*\phi(a)V$. - \end{enumerate} - - Moreover, if $\phi$ is unital, then: - \begin{enumerate}[start=2] - \item $V$ is an isometry. - \item Under the identification that $H = V(H) \subset K$, $(K, \pi(a))$ is a dilation of $\phi(a)$ for all $a \in A$. - \end{enumerate} - - The triple $(K, V, \pi)$ is a \textbf{Stinespring representation} of $\phi$. If $\bracsn{\pi(a)VH|a \in A}$ is dense in $K$, then $(K, V, \pi)$ is \textbf{minimal}. -\end{theorem} -\begin{proof}[Proof, {{\cite[Theorem 4.1]{Paulsen}}}. ] - (a): Let $A \otimes H$ and define the Hermitian form - \[ - \dpn{\cdot, \cdot}{\phi}: (A \otimes H) \times (A \otimes H)\to \complex \quad \dpn{a \otimes \xi, b \otimes \eta}{\phi} = \dpn{\phi(b^*a)\xi, \eta}{H} - \] - - For each $\sum_{j = 1}^n a_j \otimes \xi_j \in A \otimes H$, let $\xi = (\xi_1, \cdots, \xi_n) \in H^n$ and - \[ - a = \sum_{i, j = 1}^n E_{ij} \otimes a_{i}^* a_j = \braks{\sum_{i = 1}^n E_{1i} \otimes a_{i}}^*\braks{\sum_{j = 1}^n E_{1j} \otimes a_j} \ge 0 - \] - - then since $\phi$ is completely positive, - \[ - \angles{\sum_{j = 1}^n a_j \otimes \xi_j, \sum_{j = 1}^n a_j \otimes \xi_j}_\phi = \sum_{i, j = 1}^n \dpn{\phi(a_i^*a_j)\xi_j, \xi_i}{H} = \dpn{\phi_n(a)\xi, \xi}{H^n} \ge 0 - \] - - and $\dpn{\cdot, \cdot}{\phi}$ is a pseudo-inner product. - - Let $N = \bracsn{u \in A \otimes H| \dpn{u, u}{\phi} = 0}$. For any $u \in N$ and $v \in A \otimes H$, by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, - \[ - \dpn{u, v}{\phi}| \le \dpn{u, u}{\phi}^{1/2} \dpn{v, v}{\phi}^{1/2} = 0 - \] - - so $N = \bracsn{u \in A \otimes H|\dpn{u, v}{\phi} = 0 \forall v \in K_0}$. - - Let $K_0 = (A \otimes H)/N$ and $K$ be the completion of $K_0$ with respect to $\dpn{\cdot, \cdot}{\phi}$, then $K$ is a complex Hilbert space. - - (b): Let $V: H \to K_0$ be defined by $\xi \mapsto 1_A \otimes \xi + N$, then for each $\xi \in H$, - \[ - \norm{V\xi}_{K_0}^2 = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H} \le \norm{\phi(1_A)}_{B(H)}\norm{\xi}_H^2 - \] - - so $\norm{V}_{L(H; K)}^2 \le \norm{\phi(1_A)}_{B(H)}$. - - (c): For each $a \in A$, let - \[ - \pi(a): K_0 \to K_0 \quad \pi(a)(b \otimes \xi + N) = (ab) \otimes \xi + N - \] - - then for any $u = \sum_{j = 1}^n b_j \otimes \xi_j \in A \otimes H$, - \begin{align*} - \norm{\pi(a)(u+N)}_{K_0}^2 &= \sum_{i, j = 1}^n \dpn{(ab_j) \otimes \xi_j, b_i \otimes \xi_i}{\phi} = \sum_{i, j = 1}^n \dpn{\phi(b_i^*a^*ab_j)\xi_j, \xi_i}{H} \\ - &\le \sum_{i, j = 1}^n \norm{a^*a}_A \dpn{\phi(b_i^*b_j)\xi_j, \xi_i}{H} = \norm{a}_A^2 \cdot \norm{u + N}_{K_0}^2 - \end{align*} - - so $\norm{\pi(a)}_{B(K_0)} \le \norm{a}_A$. By the \hyperref[Linear Extension Theorem]{theorem:linear-extension-theorem-normed}, $\pi(a)$ extends uniquely to an operator in $B(H)$, and $\pi: A \to B(H)$ is a bounded linear map. - - By definition, $\pi$ is a unital homomorphism of Banach algebras. Finally, for any $a \in A$ and $b \otimes \xi, c \otimes \eta \in A \otimes H$, - \begin{align*} - \dpn{\pi(a^*)(b \otimes \xi + N), c \otimes \eta + N}{K_0} &= \dpn{(a^*b) \otimes \xi, c \otimes \eta}{\phi}\\ - &= \dpn{\phi(c^*a^*b)\xi, \eta}{H} = \dpn{b \otimes \xi, (ac) \otimes \eta}{\phi} \\ - &= \dpn{b \otimes \xi + N, \pi(a)(c \otimes \eta + N)}{K_0} - \end{align*} - - By linearity, $\dpn{\pi(a^*)u, v}{K_0} = \dpn{u, \pi(a)v}{K_0}$ for all $u, v \in K_0$. As $K_0$ is dense in $K$, $\pi(a^*) = \pi(a)^*$, so $\pi$ is a unital *-homomorphism. - - (1): For each $\xi \in H$, - \[ - \norm{V\xi}_{K_0}^2 = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H} - \] - - so - \[ - \norm{V}_{K_0}^2 = \sup_{\substack{\xi \in H \\ \norm{\xi}_H \le 1}}\norm{V\xi}_{K_0}^2 = \sup_{\substack{\xi \in H \\ \norm{\xi}_H \le 1}}\dpn{\phi(1_A)\xi, \xi}{H} = w(\phi(1_A)) = \norm{\phi(1_A)}_{B(H)} - \] - - by \autoref{proposition:bh-numerical-radius-gymnastics}. - - (2): For each $a \in A$ and $\xi, \eta \in H$, - \[ - \dpn{V^*\pi(a)V\xi, \eta}{H} = \dpn{\pi(a)V\xi, V\eta}{K} = \dpn{a \otimes \xi, 1_A \otimes \eta}{\phi} = \dpn{\phi(a)\xi, \eta}{H} - \] - - so $V^*\pi(a)V = \phi(a)$. - - (3): If $\phi$ is unital, then for each $\xi \in H$, - \[ - \dpn{V\xi, V\xi}{H} = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H} = \dpn{\xi, \xi}{H} - \] - - so $V$ is an isometry. - -\end{proof} - \begin{theorem}[Sz.-Nagy's Dilation Theorem] \label{theorem:sznagy-dilation} @@ -393,3 +286,112 @@ by (3). \end{proof} + + +\begin{theorem}[Stinespring's Dilation Theorem] +\label{theorem:stinespring-dilation} + Let $A$ be a unital $C^*$-algebra, $H$ be a complex Hilbert space, and $\phi: A \to B(H)$ be a completely positive map, then there exists a triple $(K, V, \pi)$, where: + \begin{enumerate}[label=(\alph*)] + \item $K$ is a complex Hilbert space. + \item $V: H \to K$ is a continuous linear map. + \item $\pi: A \to B(K)$ is a unital *-homomorphism. + \end{enumerate} + + such that: + \begin{enumerate} + \item $\norm{\phi(1_A)}_{B(H)} = \norm{V}_{L(H; K)}^2$. + \item For all $a \in A$, $\phi(a) = V^*\phi(a)V$. + \end{enumerate} + + Moreover, if $\phi$ is unital, then: + \begin{enumerate}[start=2] + \item $V$ is an isometry. + \item Under the identification that $H = V(H) \subset K$, $(K, \pi(a))$ is a dilation of $\phi(a)$ for all $a \in A$. + \end{enumerate} + + The triple $(K, V, \pi)$ is a \textbf{Stinespring representation} of $\phi$. If $\bracsn{\pi(a)VH|a \in A}$ is dense in $K$, then $(K, V, \pi)$ is \textbf{minimal}. +\end{theorem} +\begin{proof}[Proof, {{\cite[Theorem 4.1]{Paulsen}}}. ] + (a): Let $A \otimes H$ and define the Hermitian form + \[ + \dpn{\cdot, \cdot}{\phi}: (A \otimes H) \times (A \otimes H)\to \complex \quad \dpn{a \otimes \xi, b \otimes \eta}{\phi} = \dpn{\phi(b^*a)\xi, \eta}{H} + \] + + For each $\sum_{j = 1}^n a_j \otimes \xi_j \in A \otimes H$, let $\xi = (\xi_1, \cdots, \xi_n) \in H^n$ and + \[ + a = \sum_{i, j = 1}^n E_{ij} \otimes a_{i}^* a_j = \braks{\sum_{i = 1}^n E_{1i} \otimes a_{i}}^*\braks{\sum_{j = 1}^n E_{1j} \otimes a_j} \ge 0 + \] + + then since $\phi$ is completely positive, + \[ + \angles{\sum_{j = 1}^n a_j \otimes \xi_j, \sum_{j = 1}^n a_j \otimes \xi_j}_\phi = \sum_{i, j = 1}^n \dpn{\phi(a_i^*a_j)\xi_j, \xi_i}{H} = \dpn{\phi_n(a)\xi, \xi}{H^n} \ge 0 + \] + + and $\dpn{\cdot, \cdot}{\phi}$ is a pseudo-inner product. + + Let $N = \bracsn{u \in A \otimes H| \dpn{u, u}{\phi} = 0}$. For any $u \in N$ and $v \in A \otimes H$, by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, + \[ + \dpn{u, v}{\phi}| \le \dpn{u, u}{\phi}^{1/2} \dpn{v, v}{\phi}^{1/2} = 0 + \] + + so $N = \bracsn{u \in A \otimes H|\dpn{u, v}{\phi} = 0 \forall v \in K_0}$. + + Let $K_0 = (A \otimes H)/N$ and $K$ be the completion of $K_0$ with respect to $\dpn{\cdot, \cdot}{\phi}$, then $K$ is a complex Hilbert space. + + (b): Let $V: H \to K_0$ be defined by $\xi \mapsto 1_A \otimes \xi + N$, then for each $\xi \in H$, + \[ + \norm{V\xi}_{K_0}^2 = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H} \le \norm{\phi(1_A)}_{B(H)}\norm{\xi}_H^2 + \] + + so $\norm{V}_{L(H; K)}^2 \le \norm{\phi(1_A)}_{B(H)}$. + + (c): For each $a \in A$, let + \[ + \pi(a): K_0 \to K_0 \quad \pi(a)(b \otimes \xi + N) = (ab) \otimes \xi + N + \] + + then for any $u = \sum_{j = 1}^n b_j \otimes \xi_j \in A \otimes H$, + \begin{align*} + \norm{\pi(a)(u+N)}_{K_0}^2 &= \sum_{i, j = 1}^n \dpn{(ab_j) \otimes \xi_j, b_i \otimes \xi_i}{\phi} = \sum_{i, j = 1}^n \dpn{\phi(b_i^*a^*ab_j)\xi_j, \xi_i}{H} \\ + &\le \sum_{i, j = 1}^n \norm{a^*a}_A \dpn{\phi(b_i^*b_j)\xi_j, \xi_i}{H} = \norm{a}_A^2 \cdot \norm{u + N}_{K_0}^2 + \end{align*} + + so $\norm{\pi(a)}_{B(K_0)} \le \norm{a}_A$. By the \hyperref[Linear Extension Theorem]{theorem:linear-extension-theorem-normed}, $\pi(a)$ extends uniquely to an operator in $B(H)$, and $\pi: A \to B(H)$ is a bounded linear map. + + By definition, $\pi$ is a unital homomorphism of Banach algebras. Finally, for any $a \in A$ and $b \otimes \xi, c \otimes \eta \in A \otimes H$, + \begin{align*} + \dpn{\pi(a^*)(b \otimes \xi + N), c \otimes \eta + N}{K_0} &= \dpn{(a^*b) \otimes \xi, c \otimes \eta}{\phi}\\ + &= \dpn{\phi(c^*a^*b)\xi, \eta}{H} = \dpn{b \otimes \xi, (ac) \otimes \eta}{\phi} \\ + &= \dpn{b \otimes \xi + N, \pi(a)(c \otimes \eta + N)}{K_0} + \end{align*} + + By linearity, $\dpn{\pi(a^*)u, v}{K_0} = \dpn{u, \pi(a)v}{K_0}$ for all $u, v \in K_0$. As $K_0$ is dense in $K$, $\pi(a^*) = \pi(a)^*$, so $\pi$ is a unital *-homomorphism. + + (1): For each $\xi \in H$, + \[ + \norm{V\xi}_{K_0}^2 = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H} + \] + + so + \[ + \norm{V}_{K_0}^2 = \sup_{\substack{\xi \in H \\ \norm{\xi}_H \le 1}}\norm{V\xi}_{K_0}^2 = \sup_{\substack{\xi \in H \\ \norm{\xi}_H \le 1}}\dpn{\phi(1_A)\xi, \xi}{H} = w(\phi(1_A)) = \norm{\phi(1_A)}_{B(H)} + \] + + by \autoref{proposition:bh-numerical-radius-gymnastics}. + + (2): For each $a \in A$ and $\xi, \eta \in H$, + \[ + \dpn{V^*\pi(a)V\xi, \eta}{H} = \dpn{\pi(a)V\xi, V\eta}{K} = \dpn{a \otimes \xi, 1_A \otimes \eta}{\phi} = \dpn{\phi(a)\xi, \eta}{H} + \] + + so $V^*\pi(a)V = \phi(a)$. + + (3): If $\phi$ is unital, then for each $\xi \in H$, + \[ + \dpn{V\xi, V\xi}{H} = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H} = \dpn{\xi, \xi}{H} + \] + + so $V$ is an isometry. + +\end{proof} +