Reworked Stinespring's Dilation Theorem.
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Bokuan Li
+64
-42
@@ -49,86 +49,108 @@
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\begin{theorem}[Stinespring's Dilation Theorem]
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\label{theorem:stinespring-dilation}
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Let $A$ be a unital $C^*$-algebra, $H$ be a complex Hilbert space, $\phi: A \to B(H)$ be a completely positive map, then there exists a complex Hilbert space $K$, a unital *-homomorphism $\pi: A \to B(K)$, and $V \in L(H; K)$ such that
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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:
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\begin{enumerate}[label=(\alph*)]
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\item $K$ is a complex Hilbert space.
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\item $V: H \to K$ is a continuous linear map.
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\item $\pi: A \to B(K)$ is a unital *-homomorphism.
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\end{enumerate}
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such that:
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\begin{enumerate}
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\item $\normn{\phi(1_A)}_{B(H)} = \norm{V}_{L(H; K)}^2$.
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\item For all $a \in A$, $\phi(a) = V^*\pi(a)V$.
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\item $\norm{\phi(1_A)}_{B(H)} = \norm{V}_{L(H; K)}^2$.
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\item For all $a \in A$, $\phi(a) = V^*\phi(a)V$.
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\end{enumerate}
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Moreover, if $\phi$ is unital, then $V$ is an isometry, and
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Moreover, if $\phi$ is unital, then:
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\begin{enumerate}[start=2]
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\item Under the identification of $H = V(H) \subset K$, $(K, \pi(a))$ is a dilation of $\phi(a)$ for all $a \in A$.
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\item $V$ is an isometry.
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\item Under the identification that $H = V(H) \subset K$, $(K, \pi(a))$ is a dilation of $\phi(a)$ for all $a \in A$.
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\end{enumerate}
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If $\phi$ is a state, then $(K, \pi)$ is the GNS representation of $A$ associated with $\phi$.
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The triple $(K, V, \pi)$ is a \textbf{Stinespring representation} for $\phi$. If $\bracsn{\pi(a)VH|a \in A}$ is dense in $K$, then $(K, V, \pi)$ is \textbf{minimal}.
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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}.
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\end{theorem}
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\begin{proof}[Proof, {{\cite[Theorem 4.1]{Paulsen}}}. ]
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($K$): For each $(a \otimes \xi), (b \otimes \eta) \in K_0$, let
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(a): Let $A \otimes H$ and define the Hermitian form
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\[
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\dpn{a \otimes \xi, b \otimes \eta}{\phi} = \dpn{\phi(b^*a)\xi, \eta}{H}
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\dpn{\cdot, \cdot}{\phi}: (A \otimes H) \times (A \otimes H)\to \complex \quad \dpn{a \otimes \xi, b \otimes \eta}{\phi} = \dpn{\phi(b^*a)\xi, \eta}{H}
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\]
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and extend by linearity. For each $\seqf{(a_j \otimes \xi_j)} \subset A \otimes H$, let
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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
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\[
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T = \sum_{i, j = 1}^n E_{ij} \otimes a_i^*a_j = \braks{\sum_{i = 1}^n E_{1i} \otimes a_{i}}^*\braks{\sum_{i = 1}^n E_{1i} \otimes a_{i}} \in M_n(A)
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a = \sum_{i, j = 1}^n E_{ij} \otimes a_{i}^* a_j = \braks{\sum_{i = 1}^n E_{1i} \otimes a_{i}}^*\braks{\sum_{j = 1}^n E_{1j} \otimes a_j} \ge 0
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\]
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Then $T \ge 0$. As $\phi$ is completely positive,
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then since $\phi$ is completely positive,
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\[
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\angles{\sum_{j = 1}^n a_j \otimes \xi_j, \sum_{j = 1}^n a_j \otimes \xi_j}_{\phi} = \angles{\phi_n(T)\begin{bmatrix}
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\xi_1 \\
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\vdots \\
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\xi_n
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\end{bmatrix}, \begin{bmatrix}
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\xi_1 \\
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\vdots \\
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\xi_n
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\end{bmatrix}}_{H^n} \ge 0
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\angles{\sum_{j = 1}^n a_j \otimes \xi_j, \sum_{j = 1}^n a_j \otimes \xi_j}_\phi = \sum_{i, j = 1}^n \dpn{\phi(a_i^*a_j)\xi_j, \xi_i}{H} = \dpn{\phi_n(a)\xi, \xi}{H^n} \ge 0
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\]
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so $\dpn{u, u}{\phi}$ is a pseudo inner product. Let
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and $\dpn{\cdot, \cdot}{\phi}$ is a pseudo-inner product.
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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},
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\[
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N = \bracsn{u \in A \otimes H| \dpn{u, u}{\phi} = 0}
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\dpn{u, v}{\phi}| \le \dpn{u, u}{\phi}^{1/2} \dpn{v, v}{\phi}^{1/2} = 0
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\]
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then by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz},
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so $N = \bracsn{u \in A \otimes H|\dpn{u, v}{\phi} = 0 \forall v \in K_0}$.
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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.
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(b): Let $V: H \to K_0$ be defined by $\xi \mapsto 1_A \otimes \xi + N$, then for each $\xi \in H$,
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\[
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N = \bracsn{u \in A \otimes H| \dpn{u, v}{\phi} = 0 \forall v \in A \otimes H}
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\norm{V\xi}_{K_0}^2 = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H} \le \norm{\phi(1_A)}_{B(H)}\norm{\xi}_H^2
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\]
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Let $K_0 = (A \otimes H)/N$ and $K$ be the completion of $K_0$, then $K$ is a Hilbert space.
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so $\norm{V}_{L(H; K)}^2 \le \norm{\phi(1_A)}_{B(H)}$.
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($\pi$): For each $a \in A$, let
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(c): For each $a \in A$, let
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\[
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\pi_0(a): K_0 \to K_0 \quad \pi_0(a)(b \otimes \xi + N) = (ab) \otimes \xi + N
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\pi(a): K_0 \to K_0 \quad \pi(a)(b \otimes \xi + N) = (ab) \otimes \xi + N
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\]
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For any $b \in A$, $b^*a^*ab \le b^*\norm{a}_A^2b = \norm{a}_A^2b^*b$. As such, for any $u = \sum_{j = 1}^n b_j \otimes \xi_j \in A \otimes H$,
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then for any $u = \sum_{j = 1}^n b_j \otimes \xi_j \in A \otimes H$,
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\begin{align*}
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\angles{\pi(a)u, \pi(a)u}_{\phi} &= \sum_{i, j = 1}^n \dpn{\pi(a)(b_i \otimes \xi_i), \pi(a)(b_j \otimes \xi_j)}{\phi} \\
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&= \sum_{i, j = 1}^n \dpn{(ab_i) \otimes \xi_i, (ab_j) \otimes \xi_j}{\phi} = \sum_{i, j = 1}^n \dpn{\phi(b_j^*a^*ab_i) \otimes \xi_i,\xi_j}{H} \\
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&\le \sum_{i, j = 1}^n \norm{a^*a}_{A}\dpn{\phi(b_j^*b_i)\xi_i, \xi_j}{H} = \norm{a}_A^2 \dpn{u, u}{\phi}
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\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} \\
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&\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
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\end{align*}
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Thus $\pi_0(a)$ is well-defined with $\norm{\pi_0(a)}_{B(K_0)} \le \norm{a}_A$, and extends into a continuous linear operator $\pi(a) \in B(K)$. The map $\pi: A \to B(K)$ is then a *-homomorphism.
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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.
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($V$): Let $V: H \to K$ be defined by $\xi \mapsto 1_A \otimes \xi + N$.
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By definition, $\pi$ is a unital homomorphism of Banach algebras. Finally, for any $a \in A$ and $b \otimes \xi, c \otimes \eta \in A \otimes H$,
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\begin{align*}
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\dpn{\pi(a^*)(b \otimes \xi + N), c \otimes \eta + N}{K_0} &= \dpn{(a^*b) \otimes \xi, c \otimes \eta}{\phi}\\
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&= \dpn{\phi(c^*a^*b)\xi, \eta}{H} = \dpn{b \otimes \xi, (ac) \otimes \eta}{\phi} \\
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&= \dpn{b \otimes \xi + N, \pi(a)(c \otimes \eta + N)}{K_0}
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\end{align*}
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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.
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(1): For each $\xi \in H$,
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\[
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\dpn{V\xi, V\xi}{K} = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H}
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\norm{V\xi}_{K_0}^2 = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H}
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\]
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so $\normn{\phi(1_A)}_{B(H)} = \norm{V}_{L(H; K)}^2$. In particular, if $\phi$ is unital, then $V$ is an isometry by \hyperref[polarisation]{proposition:polarisation-complex}.
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so
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\[
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\norm{V}_{K_0}^2 = \sup_{\substack{\xi \in H \\ \norm{\xi}_H \le 1}}\norm{V\xi}_{K_0}^2 = \sup_{\substack{\xi \in H \\ \norm{\xi}_H \le 1}}\dpn{\phi(1_A)\xi, \xi}{H} = w(\phi(1_A)) = \norm{\phi(1_A)}_{B(H)}
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\]
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(2): Finally, for any $\xi, \eta \in H$,
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\begin{align*}
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\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} \\
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&= \dpn{\phi(a)\xi, \eta}{H}
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\end{align*}
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by \autoref{proposition:bh-numerical-radius-gymnastics}.
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(2): For each $a \in A$ and $\xi, \eta \in H$,
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\[
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\dpn{V^*\pi(a)V\xi, \eta}{H} = \dpn{\pi(a)V\xi, V\eta}{K} = \dpn{a \otimes \xi, 1_A \otimes \eta}{\phi} = \dpn{\phi(a)\xi, \eta}{H}
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\]
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so $V^*\pi(a)V = \phi(a)$.
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(3): If $\phi$ is unital, then for each $\xi \in H$,
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\[
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\dpn{V\xi, V\xi}{H} = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H} = \dpn{\xi, \xi}{H}
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\]
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so $V$ is an isometry.
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\end{proof}
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