This commit is contained in:
Bokuan Li
@@ -242,3 +242,25 @@ The following is a slight generalisation of one form of the above example, but s
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which is not positive. By \hyperref[Choi's Theorem]{theorem:choi-matrix-positive}, $\phi$ is not completely positive.
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\end{proof}
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\begin{proposition}[Choi-Kraus]
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\label{proposition:choi-kraus}
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Let $\phi: M_n(\complex) \to M_m(\complex)$ be a completely positive map, then there exists $\bracsn{V_j}_1^{mn} \subset L(\complex^m; \complex^n)$ such that
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\[
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\phi(a) = \sum_{j = 1}^{nk}V_j^* a V_j
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\]
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for all $a \in M_n(\complex)$.
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\end{proposition}
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\begin{proof}
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Let $(\pi, V, K)$ be a minimal Stinespring representation for $\phi$. Since $\phi(M_n(\complex))V\complex^m$ is dense in $K$, $K = \phi(M_n(\complex))V\complex^m$, and $\dim(K) \le n^2m$.
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As $M_n(\complex)$ is simple, the only (non-zero) irreducible representation of $M_n(\complex)$ is the identity. Since $(K, \pi)$ is a direct sum of irreducible representations, after appending additional factors, assume without loss of generality that $K = \bigoplus_{k = 1}^{nm}\complex^n$ and that $\pi(a) = I_{M_{mn}(\complex)} \otimes a$ for all $a \in M_n(\complex)$.
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For each $1 \le k \le mn$, let $P_k \in B(K)$ be the orthogonal projection onto the subspace corresponding to the $k$-th summand, and $Q_k \in L(K; \complex^n)$ be the orthogonal projection onto the $k$-th summand itself, then
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\begin{align*}
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\phi(a) &= V^*\pi(a)V = \sum_{k = 1}^{mn}V^* (P_j\pi(a)P_j)V = \sum_{k = 1}^{mn}V^*Q_j^*aQ_jV \\
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&= \sum_{k = 1}^{mn} (Q_jV)^* a (Q_jV)
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\end{align*}
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\end{proof}
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+185
-138
@@ -48,12 +48,175 @@
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\end{lemma}
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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, and $\phi: A \to B(H)$ be a completely positive map, then there exists a triple $(\pi, V, K)$, 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 $\norm{\phi(1_A)}_{B(H)} = \norm{V}_{L(H; K)}^2$.
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\item For all $a \in A$, $\phi(a) = V^*\pi(a)V$.
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\end{enumerate}
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Moreover, if $\phi$ is unital, then:
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\begin{enumerate}[start=2]
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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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The triple $(\pi, V, K)$ is a \textbf{Stinespring representation} of $\phi$. If $\bracsn{\pi(a)VH|a \in A}$ is dense in $K$, then $(\pi, V, K)$ is \textbf{minimal}\footnote{$K$ can always be trimmed so that $(\pi, V, K)$ is minimal.}.
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\end{theorem}
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\begin{proof}[Proof, {{\cite[Theorem 4.1]{Paulsen}}}. ]
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(a): Let $A \otimes H$ be the algebraic tensor product and define the Hermitian form
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\[
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\dpn{\cdot, \cdot}{\phi}: (A \otimes H) \times (A \otimes H)\to \complex \quad \dpn{a \otimes \xi, b \otimes \eta}{\phi} = \dpn{\phi(b^*a)\xi, \eta}{H}
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\]
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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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\[
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a = \sum_{i, j = 1}^n E_{ij} \otimes a_{i}^* a_j = \braks{\sum_{i = 1}^n E_{1i} \otimes a_{i}}^*\braks{\sum_{j = 1}^n E_{1j} \otimes a_j} \ge 0
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\]
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then since $\phi$ is completely positive,
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\[
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\angles{\sum_{j = 1}^n a_j \otimes \xi_j, \sum_{j = 1}^n a_j \otimes \xi_j}_\phi = \sum_{i, j = 1}^n \dpn{\phi(a_i^*a_j)\xi_j, \xi_i}{H} = \dpn{\phi_n(a)\xi, \xi}{H^n} \ge 0
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\]
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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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\[
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|\dpn{u, v}{\phi}| \le \dpn{u, u}{\phi}^{1/2} \dpn{v, v}{\phi}^{1/2} = 0
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\]
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so $N = \bracsn{u \in A \otimes H|\dpn{u, v}{\phi} = 0 \forall v \in A \otimes H}$.
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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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\[
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\norm{V\xi}_{K_0}^2 = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H} \le \norm{\phi(1_A)}_{B(H)}\norm{\xi}_H^2
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\]
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so $\norm{V}_{L(H; K)}^2 \le \norm{\phi(1_A)}_{B(H)}$.
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(c): For each $a \in A$, let
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\[
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\pi(a): K_0 \to K_0 \quad \pi(a)(b \otimes \xi + N) = (ab) \otimes \xi + N
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\]
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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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\norm{\pi(a)(u+N)}_{K_0}^2 &= \sum_{i, j = 1}^n \dpn{(ab_j) \otimes \xi_j, (ab_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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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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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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\[
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\norm{V\xi}_{K_0}^2 = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H}
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\]
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so
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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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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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(4): Suppose that $\phi$ is unital, then $V$ is an isometry by (3). Under the identification that $H = V(H) \subset K$, let $P \in B(K)$ be the orthogonal projection onto $H$, then for each $a \in A$ and $\xi, \eta \in H$,
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\[
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\dpn{\phi(a)\xi, \eta}{H} = \dpn{\pi(a)V\xi, V\eta}{K} = \dpn{\pi(a)P\xi, P\eta}{K} = \dpn{P\pi(a)\xi, \eta}{K}
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\]
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Therefore $\phi(a) = P\pi(a)|_H$, and $(K, \pi(a))$ is a dilation of $\phi(a)$.
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\end{proof}
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\begin{proposition}[Uniqueness of Minimal Stinespring Representations]
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\label{proposition:stinespring-minimal-uniqueness}
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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, and $(\pi_1, V_1, K_1), (\pi_2, V_2, K_2)$ be minimal Stinespring representations for $\phi$, then there exists a unitary equivalence $U: K_1 \to K_2$ such that $UV_1 = V_2$.
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\end{proposition}
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\begin{proof}[Proof, {{\cite[Proposition 4.2]{Paulsen}}}. ]
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Suppose there exists a unitary equivalence $U: K_1 \to K_2$ such that $UV_1 = V_2$, then for each $\seqf{a_j} \subset A$ and $\seqf{\xi_j} \subset H$,
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\[
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U\braks{\sum_{j = 1}^n \pi_1(a_j)V_1\xi_j} = \sum_{j = 1}^n \pi_2(a_j)UV_1\xi_j = \sum_{j = 1}^n \pi_2(a_j)V_2\xi_j
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\]
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By minimality, $\pi_1(A)V_1H$ is dense in $K_1$, and $\pi_2(A)V_2H$ is dense in $K_2$. As such, it is sufficient to show that the mapping $\sum_{j = 1}^n \pi_1(a_j)V_1\xi_j \mapsto \sum_{j = 1}^n \pi_2(a_j)V_2\xi_j$ is a well-defined isometry. To this end, note that
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\begin{align*}
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\norm{\sum_{j = 1}^n \pi_1(a_j)V_1\xi_j}_{K_1}^2 &= \sum_{i, j = 1}^n \dpn{\pi_1(a_j)V_1\xi_j, \pi_1(a_i)V_1\xi_j}{K_1} \\
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&= \sum_{i, j = 1}^n \dpn{V_1^*\pi_1(a_i^*a_j)V_1\xi_j, \xi_j}{H} = \sum_{i, j = 1}^n \dpn{\phi(a_i^*a_j)\xi_j, \xi_j}{H} \\
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&= \sum_{i, j = 1}^n \dpn{V_2^*\pi_2(a_i^*a_j)V_2\xi_j, \xi_j}{H} \\
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&= \sum_{i, j = 1}^n \dpn{\pi_2(a_j)V_2\xi_j, \pi_2(a_i)V_2\xi_j}{K_2} = \norm{\sum_{j = 1}^n \pi_2(a_j)V_2\xi_j}_{K_2}^2
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\end{align*}
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\end{proof}
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\begin{proposition}[Multiplicative Domain of Stinespring Representations]
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\label{proposition:stinespring-multiplicative-domain}
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Let $A$ be a unital $C^*$-algebra, $H$ be a complex Hilbert space, $\phi: A \to B(H)$ be a unital completely positive map, and $(\pi, V, K)$ be a minimal Stinespring representation of $\phi$, then:
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\begin{enumerate}
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\item For any $a \in A$, $\phi(a)^*\phi(a) = \phi(a^*a)$ if and only if $V(H)$ is invariant for $\pi(a)$\footnote{This also shows that multiplicative domains of $\phi$ are subalgebras of $A$. }. In other words, $a$ is in the right multiplicative domain of $\phi$ if and only if $V(H)$ is invariant for $\pi(a)$.
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\item For any $a \in A$, $a$ is in the multiplicative domain of $\phi$ if and only if $V(H)$ is reducing for $\pi(a)$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(1): Since $\phi$ is unital, $V$ is an isometry. Identify $H = V(H) \subset K$ and let $P = VV^* \in B(K)$ be the orthogonal projection onto $H$.
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Let $a \in A$ and $\xi, \eta \in H$, then
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\begin{align*}
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\dpn{\phi(a^*)\phi(a)\xi, \eta}{H} &= \dpn{\pi(a^*)VV^*\pi(a)V\xi, V\eta}{K} = \dpn{\pi(a)^*P\pi(a)P\xi, P\eta}{K} \\
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\dpn{\phi(a^*a)\xi, \eta}{H} &= \dpn{\pi(a^*a)V\xi, V\eta}{K} = \dpn{\pi(a)^*\pi(a)P\xi, P\eta}{K}
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\end{align*}
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By \autoref{lemma:invariant-projection-test}, $\pi(a)^*P\pi(a)P = \pi(a)^*\pi(a)P$ if and only if $P(K) = V(H)$ is an invariant subspace for $\pi(a)$. Therefore $\phi(a)^*\phi(a) = \phi(a^*a)$ if and only if $V(H)$ is invariant for $\pi(a)$.
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\end{proof}
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\begin{theorem}[Sz.-Nagy's Dilation Theorem]
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\label{theorem:sznagy-dilation}
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Let $H$ be a complex Hilbert space and $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, then there exists a unitary power dilation of $T$.
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Let $H$ be a complex Hilbert space and $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, then there exists a unitary power dilation $(K, U)$ of $T$ such that:
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\begin{enumerate}
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\item $K$ is the smallest closed reducing subspace for $U$ containing $H$ such that $T^n = P_HU^n|_H$\footnote{Where $P_H$ is the orthogonal projection onto $H$.} for all $n \in \natz$.
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\item[(U)] For any unitary power dilation $(K', U')$ satisfying (1), there exists a unitary $V: K \to K'$ such that $V|_{H} = I_H$ and $VUV^* = U'$.
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\end{enumerate}
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\end{theorem}
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\begin{proof}[Proof, {{\cite[Theorem 1.1]{Paulsen}}}. ]
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Using \autoref{lemma:bh-dilation-transitive} it is sufficient to show the following statements:
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\begin{proof}[Constuctive proof without uniqueness, {{\cite[Theorem 1.1]{Paulsen}}}. ]
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Using \autoref{lemma:bh-dilation-transitive}, it is sufficient to show the following statements:
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\begin{enumerate}[label=(\roman*)]
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\item For any $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, there exists an isometric power dilation of $T$.
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\item For any isometry $V \in B(H)$, there exists a unitary power dilation of $V$.
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@@ -130,6 +293,25 @@
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Under the identification that $H = \bracsn{(x, 0)|x \in H} \subset H^2$, let $Q \in B(H^2)$ be the projection onto the first summand, then $QU^n|_{H} = V^n$ for all $n \in \natp$. Therefore $U$ is a unitary power dilation of $V$.
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\end{proof}
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\begin{proof}[Proof, using Stinespring's Dilation Theorem, {{\cite[Theorem 4.3]{Paulsen}}}. ]
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(1): Let $\mathbb{T} = \partial B_\complex(0, 1)$. By \autoref{theorem:polynomial-map-positive}, the mapping
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\[
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\phi: \complex[z, \ol z] \to B(H) \quad \phi(p + \ol q) = p(T) + q(T)^*
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\]
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extends into a unital positive map $C(\mathbb{T}; \complex) \to B(H)$, which is completely positive by \hyperref[Stinespring's Theorem]{theorem:stinespring-commutative-completely-positive}.
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Let $(\pi, W, K)$ be a minimal \hyperref[Stinespring representation]{theorem:stinespring-dilation} for $\phi$ and $U := \pi(z) \in U(B(K))$, then for each $n \in \natz$,
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\[
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T^n = \phi(z^n) = P_H\pi(z^n)|_H = P_HU^n|_H
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\]
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so $(K, U)$ is a unitary power dilation of $T$.
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(U): By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, $(\pi, V, K)$ is minimal if and only if $\bigcup_{n \in \integer}U^n H$ is dense in $K$, if and only if $U$ admits no non-trivial reducing subspace.
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By \hyperref[uniqueness of minimal Stinespring representations]{proposition:stinespring-minimal-uniqueness}, such a dilation is unique.
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\end{proof}
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\begin{theorem}[Von Neumann's Inequality]
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\label{theorem:von-neumann-inequality-polynomial}
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@@ -284,138 +466,3 @@
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\end{proof}
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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, and $\phi: A \to B(H)$ be a completely positive map, then there exists a triple $(\pi, V, K)$, 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 $\norm{\phi(1_A)}_{B(H)} = \norm{V}_{L(H; K)}^2$.
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\item For all $a \in A$, $\phi(a) = V^*\pi(a)V$.
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\end{enumerate}
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Moreover, if $\phi$ is unital, then:
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\begin{enumerate}[start=2]
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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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The triple $(\pi, V, K)$ is a \textbf{Stinespring representation} of $\phi$. If $\bracsn{\pi(a)VH|a \in A}$ is dense in $K$, then $(\pi, V, K)$ 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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(a): Let $A \otimes H$ be the algebraic tensor product and define the Hermitian form
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\[
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\dpn{\cdot, \cdot}{\phi}: (A \otimes H) \times (A \otimes H)\to \complex \quad \dpn{a \otimes \xi, b \otimes \eta}{\phi} = \dpn{\phi(b^*a)\xi, \eta}{H}
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\]
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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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\[
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a = \sum_{i, j = 1}^n E_{ij} \otimes a_{i}^* a_j = \braks{\sum_{i = 1}^n E_{1i} \otimes a_{i}}^*\braks{\sum_{j = 1}^n E_{1j} \otimes a_j} \ge 0
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\]
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then since $\phi$ is completely positive,
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\[
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\angles{\sum_{j = 1}^n a_j \otimes \xi_j, \sum_{j = 1}^n a_j \otimes \xi_j}_\phi = \sum_{i, j = 1}^n \dpn{\phi(a_i^*a_j)\xi_j, \xi_i}{H} = \dpn{\phi_n(a)\xi, \xi}{H^n} \ge 0
|
||||
\]
|
||||
|
||||
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 $N = \bracsn{u \in A \otimes H|\dpn{u, v}{\phi} = 0 \forall v \in A \otimes H}$.
|
||||
|
||||
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, (ab_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.
|
||||
|
||||
(4): Suppose that $\phi$ is unital, then $V$ is an isometry by (3). Under the identification that $H = V(H) \subset K$, let $P \in B(K)$ be the orthogonal projection onto $H$, then for each $a \in A$ and $\xi, \eta \in H$,
|
||||
\[
|
||||
\dpn{\phi(a)\xi, \eta}{H} = \dpn{\pi(a)V\xi, V\eta}{K} = \dpn{\pi(a)P\xi, P\eta}{K} = \dpn{P\pi(a)\xi, \eta}{K}
|
||||
\]
|
||||
|
||||
Therefore $\phi(a) = P\pi(a)|_H$, and $(K, \pi(a))$ is a dilation of $\phi(a)$.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\begin{proposition}[Multiplicative Domain of Stinespring Representations]
|
||||
\label{proposition:stinespring-multiplicative-domain}
|
||||
Let $A$ be a unital $C^*$-algebra, $H$ be a complex Hilbert space, $\phi: A \to B(H)$ be a unital completely positive map, and $(\pi, V, K)$ be a minimal Stinespring representation of $\phi$, then:
|
||||
\begin{enumerate}
|
||||
\item For any $a \in A$, $\phi(a)^*\phi(a) = \phi(a^*a)$ if and only if $V(H)$ is invariant for $\pi(a)$\footnote{This also shows that multiplicative domains of $\phi$ are subalgebras of $A$. }. In other words, $a$ is in the right multiplicative domain of $\phi$ if and only if $V(H)$ is invariant for $\pi(a)$.
|
||||
\item For any $a \in A$, $a$ is in the multiplicative domain of $\phi$ if and only if $V(H)$ is reducing for $\pi(a)$.
|
||||
\end{enumerate}
|
||||
\end{proposition}
|
||||
\begin{proof}
|
||||
(1): Since $\phi$ is unital, $V$ is an isometry. Identify $H = V(H) \subset K$ and let $P = VV^* \in B(K)$ be the orthogonal projection onto $H$.
|
||||
|
||||
Let $a \in A$ and $\xi, \eta \in H$, then
|
||||
\begin{align*}
|
||||
\dpn{\phi(a^*)\phi(a)\xi, \eta}{H} &= \dpn{\pi(a^*)VV^*\pi(a)V\xi, V\eta}{K} = \dpn{\pi(a)^*P\pi(a)P\xi, P\eta}{K} \\
|
||||
\dpn{\phi(a^*a)\xi, \eta}{H} &= \dpn{\pi(a^*a)V\xi, V\eta}{K} = \dpn{\pi(a)^*\pi(a)P\xi, P\eta}{K}
|
||||
\end{align*}
|
||||
|
||||
By \autoref{lemma:invariant-projection-test}, $\pi(a)^*P\pi(a)P = \pi(a)^*\pi(a)P$ if and only if $P(K) = V(H)$ is an invariant subspace for $\pi(a)$. Therefore $\phi(a)^*\phi(a) = \phi(a^*a)$ if and only if $V(H)$ is invariant for $\pi(a)$.
|
||||
\end{proof}
|
||||
|
||||
|
||||
@@ -74,3 +74,50 @@
|
||||
0 &\le \phi\braks{\sum_{j, k = 1}^n E_{jk} \otimes z^{j-k}} = \sum_{j, k = 1}^n E_{jk} \otimes A_{j - k}
|
||||
\end{align*}
|
||||
\end{proof}
|
||||
|
||||
\begin{theorem}
|
||||
\label{theorem:polynomial-map-positive}
|
||||
Let $H$ be a complex Hilbert space, $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, and $\mathbb{T} = \partial B_\complex(0, 1)$, then the mapping
|
||||
\[
|
||||
\phi: \complex[z, \ol z] \to B(H) \quad p \mapsto p(T)
|
||||
\]
|
||||
|
||||
extends into a completely positive map on $C(\mathbb{T}; \complex)$.
|
||||
\end{theorem}
|
||||
\begin{proof}[Proof, {{\cite[Theorem 2.6]{Paulsen}}}. ]
|
||||
Assume without loss of generality that $\norm{T}_{B(H)} < 1$. Using \autoref{proposition:trig-moment-sequence-operator}, it is sufficient to show that the Toeplitz matrix
|
||||
\[
|
||||
A = \sum_{i, j = 1}^n E_{ij} \otimes A_{i - j} = \begin{bmatrix}
|
||||
I & T^* &\cdots & (T^*)^n \\
|
||||
T &\ddots &\ddots &\vdots \\
|
||||
\vdots &\ddots &\ddots & T^* \\
|
||||
T^n &\cdots &T &I
|
||||
\end{bmatrix}
|
||||
\]
|
||||
|
||||
is positive for all $n \in \natp$. To this end, let
|
||||
\[
|
||||
R = \begin{bmatrix}
|
||||
0 &\cdots &\cdots &\cdots &0 \\
|
||||
T & \ddots & & &\vdots \\
|
||||
0 & \ddots &\ddots & &\vdots \\
|
||||
\vdots &\ddots &\ddots &\ddots &\vdots \\
|
||||
0 &\cdots &0 & T & 0
|
||||
\end{bmatrix} = \sum_{j = 1}^{n - 1}E_{(j+1)j} \otimes T
|
||||
\]
|
||||
|
||||
then for each $k \in \natz$, $R^k = \sum_{j = 1}^{n-k}E_{(j+k)j} \otimes T^n$. In particular, $R^{n+1} = 0$ and
|
||||
\[
|
||||
A = I + \sum_{k = 1}^n R^k + \sum_{k = 1}^n (R^*)^k = \sum_{k = 0}^\infty R^k + \sum_{k = 0}^\infty (R^*)^k - I
|
||||
\]
|
||||
|
||||
|
||||
As $\norm{R}_{M_n(B(H))} = \norm{T}_{B(H)} < 1$, by recognising the \hyperref[Neumann series]{lemma:neumann-series} and using the resolvent identity,
|
||||
\begin{align*}
|
||||
A &= (I - R)^{-1} + (I - R^*)^{-1} - I \\
|
||||
&= (I - R)^{-1}(I - R^* + I - R + (I - R)(I - R)^*)(1 - R^*)^{-1} \\
|
||||
&= (I - R)^{-1}(I - RR^*)(1 - R^*)^{-1} \ge 0
|
||||
\end{align*}
|
||||
|
||||
\end{proof}
|
||||
|
||||
|
||||
Reference in New Issue
Block a user