diff --git a/src/op/c-star/involution.tex b/src/op/c-star/involution.tex index 9643649..dfd5276 100644 --- a/src/op/c-star/involution.tex +++ b/src/op/c-star/involution.tex @@ -42,19 +42,7 @@ \end{proof} -\begin{definition}[$M_n(A)$] -\label{definition:matrix-cstar} - Let $A$ be a $C^*$-algebra, $M_n(A) = M_n(\complex) \otimes A$ be the space of $n \times n$ matrices over $A$, equipped with the product - \[ - M_n(A) \times M_n(A) \quad (M \otimes a)(N \otimes b) = MN \otimes ab - \] - and the involution - \[ - *: M_n(A) \to M_n(A) \quad M \otimes a \mapsto M^* \otimes a^* - \] - Let $H$ be a complex Hilbert space and $\pi: A \to B(H)$ be a faithful representation of $A$, then $\pi$ induces a faithful representation $M_n(A) \otimes A \to B(H^n)$. Under the identification that $M_n(A) \subset B(H^n)$, the inherited norm\footnote{Which is equal to the projective norm because $M_n(\complex)$ is finite-dimensional and hence nuclear.} makes $M_n(A)$ a $C^*$-algebra. -\end{definition} \begin{definition}[Completely] \label{definition:cstar-completely} diff --git a/src/op/example/bounded.tex b/src/op/example/bounded.tex index 8ea3824..126f1e9 100644 --- a/src/op/example/bounded.tex +++ b/src/op/example/bounded.tex @@ -94,138 +94,4 @@ (6): Let $A$ be the von Neumann algebra generated by $T$. By \autoref{theorem:existence-of-projections-vna}, $A$ is a unital $C^*$-algebra, so $P = \sqrt{T^*T} \in A$. To see that $V \in A$, it is sufficient to apply the \hyperref[Bicommutant Theorem]{theorem:bicommutant}. To this end, let $S \in A'$, then $TS = ST = SVP$ and $VSP = VPS = TS$, so $SV$ and $VS$ agree on $\ol{P(H)}$. Since $\ker(V) = \ker(P) = \ol{P(H)}^\perp$, $SV|_{\ker(P)} = 0$. On the other hand, as $SP = PS$, $S(\ker(P)) \subset \ker(P) = \ker(V)$, so $VS|_{\ker(P)} = 0$ as well. Therefore $V \in A'' = A$. -\end{proof} - - -\begin{definition}[Unitary Dilation] -\label{definition:unitary-dilation} - Let $H$ be a complex Hilbert space, $V \in B(H)$ be isometric, $P = I_H - VV^*$ be the projection onto $V(H)^\perp$, and - \[ - U = \begin{bmatrix} - V & P \\ - 0 & V^* - \end{bmatrix} - \] - - then: - \begin{enumerate} - \item $U^*U = UU^* = I_{H^2}$. In other words, $U$ is an isometry on $H^2$. - \item Under the identification that $H = H \times \bracsn{0} \subset H^2$, let $Q \in B(H^2)$ be the projection onto $H$, then $V^n = QU^n|_H$ for all $n \ge 0$. - \end{enumerate} - - The operator $U \in B(H^2)$ is a \textbf{unitary dilation} of $V$. -\end{definition} -\begin{proof} - (1): By direct computation, - \begin{align*} - UU^* &= \begin{bmatrix} - V & P \\ - 0 & V^* - \end{bmatrix} \cdot - \begin{bmatrix} - V^* & 0 \\ - P & V - \end{bmatrix} = \begin{bmatrix} - VV^* + P & PV \\ - V^*P & V^*V - \end{bmatrix} = I_{H^2} \\ - U^*U &= \begin{bmatrix} - V^* & 0 \\ - P & V - \end{bmatrix} \cdot - \begin{bmatrix} - V & P \\ - 0 & V^* - \end{bmatrix} - = \begin{bmatrix} - V^*V & V^*P \\ - PV & VV^* + P - \end{bmatrix} = I_{H^2} \\ - \end{align*} - - - (2): For each $n \in \natp$, - \[ - U^n = \begin{bmatrix} - V & P \\ - 0 & V^* - \end{bmatrix}^n = \begin{bmatrix} - V^n & \cdots \\ - 0 & (V^*)^n - \end{bmatrix} - \] - - so $QU^n|_{H} = V^n$. -\end{proof} - -\begin{definition}[Compression] -\label{definition:bh-compression} - Let $H$ be a complex Hilbert space, $K \subset H$ be a closed subspace, $P \in B(H)$ be the orthogonal projection onto $K$, $S \in B(K)$, and $T \in B(H)$, then $S$ is a \textbf{compression} of $T$ if $S = PT|_K$. -\end{definition} - -\begin{definition}[Dilation] -\label{definition:bh-dilation} - Let $H$ be a complex Hilbert space and $T \in B(H)$, then a \textbf{dilation} of $T$ is a Hilbert space $K \supset H$ and an operator $S \in B(K)$ such that $T = PS|_H$, where $P \in B(K)$ is the orthogonal projection onto $H$. If $T^n = PS^n|_H$ for all $n \in \natz$, then $S$ is a \textbf{power dilation} of $T$. -\end{definition} - - - -\begin{definition}[Isometric Dilation] -\label{definition:isometric-dilation} - Let $H$ be a complex Hilbert space, $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, $D_T = (1 - T^*T)^{1/2}$, and - \[ - V: l^2(\natp; H) \to l^2(\natp; H) \quad (Vx)_n = \begin{cases} - Tx_1 &n = 1 \\ - D_Tx_1 &n = 2 \\ - x_{n-1} &n \ge 3 - \end{cases} - \] - - then: - \begin{enumerate} - \item $V$ is isometric on $l^2(\natp; H)$. - \item Under the identification that $H = H \oplus 0 \oplus \cdots$, let $Q$ be the orthogonal projection onto $H$, then $T^n = QV^n|_H$ for all $n \ge 0$. - \end{enumerate} - - The operator $V$ is an \textbf{isometric dilation} of $T$. -\end{definition} -\begin{proof} - (1): Since $\norm{T}_H \le 1$, $1 - T^*T \ge 0$, and $(1 - T^*T)^{1/2}$ is well-defined. For each $x \in H$, - \begin{align*} - \norm{Tx}_H^2 + \norm{D_Tx}_H^2 &= \dpn{T^*Tx, x}{H} + \dpn{(1 - T^*T)x, x}{H} \\ - &= \dpn{x, x}{H} = \norm{x}_H^2 - \end{align*} - - Thus for each $x = \seq{x_n} \in l^2(H)$, - \begin{align*} - \norm{Vx}_{l^2(\natp; H)}^2 &= \norm{Tx_1}_H^2 + \norm{D_Tx_1}_H^2 + \sum_{n = 3}^\infty \norm{x_{n-1}}_H^2 \\ - &= \norm{x}_H^2 + \sum_{n = 2}^\infty \norm{x_{n}}_H^2 = \norm{x}_{l^2(\natp; H)}^2 - \end{align*} -\end{proof} - -\begin{theorem}[Sz.-Nagy's Dilation Theorem] -\label{theorem:sznagy-dilation} - 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 $T$. -\end{theorem} -\begin{proof}[Proof, {{\cite[Theorem 1.1]{Paulsen}}}. ] - Let $K = l^2(\natp; H)^2$ and identify $H = (H \oplus \bracsn{0} \oplus \cdot) \times \bracsn{0}$. Let $V \in B(l^2(\natp; H))$ be the isometric dilation of $T$, and $U \in B(l^2(\natp; H)^2)$ be the unitary dilation of $V$. Identify $l^2(\natp; H)$ as a subspace of $K$ as $l^2(\natp; H) \times \bracsn{0}$, and let $P_{l^2(\natp; H)}$ be the orthogonal projection onto $l^2(\natp; H)$, then for each $n \ge 0$, - \begin{align*} - V^n &= P_{l^2(\natp; H)}U^n|_{l^2(\natp; H)} \\ - T^n = P_{H}V^n|_{H} &= P_HU^n|_{H} - \end{align*} -\end{proof} - -\begin{corollary}[Von Neumann's Inequality] -\label{corollary:von-neumann-inequality-polynomial} - Let $H$ be a complex Hilbert space and $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, then for any $p \in \complex[x]$, - \[ - \norm{p(T)}_{B(H)} \le \sup_{z \in B_\complex(0, 1)}|p(z)| - \] -\end{corollary} -\begin{proof}[Proof, {{\cite[Corollary 1.2]{Paulsen}}}. ] - By \hyperref[Sz.-Nagy's Dilation Theorem]{theorem:sznagy-dilation}, there exists a Hilbert space $K \supset H$ and a unitary operator $U \in B(K)$ such that $T^n = P_HU^n|_H$ for all $n \in \natp$, where $P_H$ is the orthogonal projection onto $H$. - - Thus for any $p \in \complex[x]$, $p(T) = P_Hp(U)|_H$. As $U$ is unitary, the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus} shows that $\norm{p(U)}_{B(K)} \le \sup_{z \in B_\complex(0, 1)}|p(z)|$. Therefore $\norm{p(T)}_{B(H)} \le \sup_{z \in B_\complex(0, 1)}|p(z)|$. -\end{proof} - - +\end{proof} \ No newline at end of file diff --git a/src/op/index.tex b/src/op/index.tex index a96fcf0..a69429f 100644 --- a/src/op/index.tex +++ b/src/op/index.tex @@ -5,4 +5,5 @@ \input{./c-star/index.tex} \input{./vn/index.tex} \input{./example/index.tex} +\input{./system/index.tex} \input{./notation.tex} \ No newline at end of file diff --git a/src/op/system/completely.tex b/src/op/system/completely.tex new file mode 100644 index 0000000..d3081b3 --- /dev/null +++ b/src/op/system/completely.tex @@ -0,0 +1,514 @@ +\section{Completely Positive and Completely Bounded Maps} +\label{section:completely-positive-bounded} + +\begin{proposition} +\label{proposition:matrix-cstar} + Let $A$ be a $C^*$-algebra, $M_n(A) = M_n(\complex) \otimes A$ be the space of $n \times n$ matrices over $A$, + \[ + M_n(A) \times M_n(A) \to M_n(A) \quad (M \otimes a)(N \otimes b) = MN \otimes ab + \] + and the involution + + \[ + *: M_n(A) \to M_n(A) \quad M \otimes a \mapsto M^* \otimes a^* + \] + + then there exists a unique norm on $M_n(A)$ such that $M_n(A)$ is a $C^*$-algebra. +\end{proposition} +\begin{proof} + After applying a \hyperref[unitisation]{proposition:c-star-unitisation}, assume without loss of generality that $A$ is unital. + + By the \hyperref[Gelfand-Naimark Theorem]{theorem:gns}, there exists a faithful representation $(H, \pi)$ of $A$. Under the identification that $H^n = \complex^n \otimes H$, $\pi$ induces a faithful unital *-representation + \[ + \pi_n: M_n(A) \to B(H^n) \quad \pi_n(M \otimes a)(\lambda \otimes \xi) = (M\lambda) \otimes (\pi(a)\xi) + \] + + whose image is a uniformly closed subalgebra of $B(H^n)$. As such, $B(H^n)$ induces a norm on $M_n(A)$, making it a $C^*$-algebra. + + By \autoref{corollary:c-star-unique-norm}, this norm is unique. +\end{proof} + +\begin{definition}[Extended Map] +\label{definition:matrix-extended-map} + Let $A$ be a unital $C^*$-algebra, $B$ be a $C^*$-algebra, $S \subset A$ be an operator system, $\phi: S \to B$ be a linear map, and $n \in \natp$, then the mapping + \[ + \phi_n = \text{Id}_{M_n(\complex)} \otimes \phi: M_n(S) \to M_n(B) + \] + + is the \textbf{extended map}/\textbf{amplification} of $\phi$ to $M_n(S)$. +\end{definition} + + + +\begin{definition}[Completely Positive Map] +\label{definition:completely-positive-map} + Let $A, B$ be $C^*$-algebras with $A$ being unital, $S \subset A$ be an operator system, $\phi: S \to B$ be a linear map, and $n \in \natp$, then $\phi$ is \textbf{$n$-positive} if $\phi_n: M_n(S) \to M_n(B)$ is positive. The map $\phi$ is \textbf{completely positive} if $\phi$ is $n$-positive for all $n \in \natp$. +\end{definition} + +\begin{definition}[Completely Bounded Map] +\label{definition:completely-bounded-map} + Let $A, B$ be $C^*$-algebras with $A$ being unital, $S \subset A$ be an operator system, and $\phi: S \to B$ be a linear map, then + \[ + \norm{\phi}_{\text{cb}} = \sup_{n \in \natp}\norm{\phi_n}_{L(M_n(S); M_n(B))} + \] + + is the \textbf{completely bounded norm} of $\phi$. + + The map $\phi$ is \textbf{completely bounded} if $\norm{\phi}_{\text{cb}} < \infty$, and \textbf{completely contractive} if $\norm{\phi}_{\text{cb}} \le 1$. +\end{definition} + +\begin{lemma} +\label{lemma:completely-mental-gymnastics} + Let $A, B$ be $C^*$-algebras with $A$ being unital, $S \subset A$ be an operator system, $\phi: S \to B$ be a linear map, and $n \in \natp$. + \begin{enumerate} + \item If $\phi$ is $n$-positive, then $\phi$ is $k$-positive for all $1 \le k \le n$. + \item For each $1 \le k \le n$, $\norm{\phi_k}_{L(M_k(S); M_k(B))} \le \norm{\phi_n}_{L(M_n(S); M_n(B))}$. + \end{enumerate} +\end{lemma} +\begin{proof} + For any $n \in \natp$ and $1 \le k < n$, there is a natural *-homomorphism + \[ + e: M_k(\complex) \to M_n(\complex) \quad M \mapsto \begin{bmatrix} + M & 0 \\ + 0 & 0 + \end{bmatrix} + \] + + such that the following diagram commutes: + \[ + \xymatrix{ + M_k(S) \ar@{->}[d]_{e \otimes \text{Id}_S} \ar@{->}[r]^{\phi_k} & M_k(B) \ar@{->}[d]^{e \otimes \text{Id}_B} \\ + M_n(S) \ar@{->}[r]_{\phi_n} & M_n(B) + } + \] + + (1): Since $e \otimes \text{Id}_S$ is the restriction of a *-homomorphism, it is positive. On the other hand, for any $x \in M_k(B)$, $x \ge 0$ if and only if $(e \otimes \text{Id}_B)(x) \ge 0$. Given that $\phi_n$ is positive, $\phi_k$ must also be positive. + + (2): The maps $e \otimes \text{Id}_S$ and $e \otimes \text{Id}_B$ are both isometric. As such, the norm of $\phi_k$ is bounded above by the norm of $\phi_n$. +\end{proof} + + +% Omitted + +\begin{lemma} +\label{lemma:norm-matrix-test} + Let $A$ be a unital $C^*$-algebra, then for any $x \in A$ and $y \in A_{sa}$, $x^*x \le y$ if and only if + \[ + \begin{bmatrix} + 1 &x \\ + x^* &y + \end{bmatrix} \ge 0 + \] + + in $M_2(A)$. In particular, $\norm{x}_A \le 1$ if and only if + \[ + \begin{bmatrix} + 1 &x \\ + x^* &1 + \end{bmatrix} \ge 0 + \] + + in $M_2(A)$. +\end{lemma} +\begin{proof}[Proof, {{\cite[Lemma 3.1]{Paulsen}}}. ] + Using the \hyperref[Gelfand-Naimark Theorem]{theorem:gns}, assume without loss of generality that there exists a complex Hilbert space $H$ such that $A \subset B(H)$. For any $\xi, \eta \in H$, + \[ + \angles{ \begin{bmatrix} + 1 &x \\ + x^* &y + \end{bmatrix}\begin{bmatrix} + \xi \\ + \eta + \end{bmatrix}, \begin{bmatrix} + \xi \\ + \eta + \end{bmatrix}}_{H^2} = \norm{\xi}_H^2 + \dpn{x\eta, \xi}{H} + \dpn{\xi, x\eta}{H} + \dpn{y\eta, \eta}{H} + \] + + ($\Rightarrow$): If $y \ge x^*x$, then by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, + \begin{align*} + \angles{ \begin{bmatrix} + 1 &x \\ + x^* &y + \end{bmatrix}\begin{bmatrix} + \xi \\ + \eta + \end{bmatrix}, \begin{bmatrix} + \xi \\ + \eta + \end{bmatrix}}_{H^2} &\ge \norm{\xi}_H^2 + \dpn{x\eta, \xi}{H} + \dpn{\xi, x\eta}{H} + \dpn{x\eta, x\eta}{H} \\ + &\ge \norm{\xi}_H^2 - 2\norm{\xi}_H\norm{x\eta}_H + \norm{x\eta}_H^2 \ge 0 + \end{align*} + + ($\Leftarrow$): If the matrix is positive, then for each $\eta \in H$, + \begin{align*} + 0 &\le \angles{ \begin{bmatrix} + 1 &x \\ + x^* &y + \end{bmatrix}\begin{bmatrix} + -x\eta \\ + \eta + \end{bmatrix}, \begin{bmatrix} + -x\eta \\ + \eta + \end{bmatrix}}_{H^2} = \norm{x\eta}_H^2 -2\dpn{x\eta, x\eta}{H} + \dpn{y\eta, \eta}{H} \\ + &\le \dpn{y\eta, \eta}{H} - \dpn{x\eta, x\eta}{H} = \dpn{(y - x^*x)\eta, \eta}{H} + \end{align*} + + so $y \ge x^*x$. +\end{proof} + +\begin{proposition} +\label{proposition:unital-2-positive-contractive} + Let $A, B$ be unital $C^*$-algebras, $S \subset A$ be an operator system, $\phi: S \to B$ be a unital $2$-positive map, then $\norm{\phi}_{L(S; B)} \le 1$. +\end{proposition} +\begin{proof}[Proof, {{\cite[Proposition 3.2]{Paulsen}}}. ] + Let $x \in S$ with $\norm{x}_A \le 1$, then + \[ + \phi_2\begin{bmatrix} + 1 &x \\ + x^* &1 + \end{bmatrix} = \begin{bmatrix} + 1 &\phi(x) \\ + \phi(x)^* &1 + \end{bmatrix} \ge 0 + \] + + By \autoref{lemma:norm-matrix-test}, $\norm{\phi(x)}_B \le 1$. +\end{proof} + +\begin{proposition}[Kadison-Schwarz Inequality] +\label{proposition:cauchy-schwarz-2-positive} + Let $A, B$ be unital $C^*$-algebras, $\phi: A \to B$ be a unital $2$-positive map, then for each $x \in A$, + \[ + \phi(x)^*\phi(x) \le \phi(x^*x) + \] +\end{proposition} +\begin{proof} + Since + \[ + \begin{bmatrix} + 1 &x \\ + x^* &x^*x + \end{bmatrix} = \begin{bmatrix} + 1 &x \\ + 0 &0 + \end{bmatrix}^*\begin{bmatrix} + 1 &x \\ + 0 &0 + \end{bmatrix} \ge 0 + \] + + and $\phi$ is unital and $2$-positive, + \[ + \begin{bmatrix} + 1 &\phi(x) \\ + \phi(x)^* &\phi(x^*x) + \end{bmatrix} = \phi_2\begin{bmatrix} + 1 &x \\ + x^* &x^*x + \end{bmatrix} \ge 0 + \] + + By \autoref{lemma:norm-matrix-test}, $\phi(x)^*\phi(x) \le \phi(x^*x)$. +\end{proof} + +\begin{lemma}[Schur Complement Condition for Positivity] +\label{lemma:schur-complement-positive} + Let $H$ be a complex Hilbert space, $x, y \in B(H)$, and $z \in G(B(H))$, then + \[ + \begin{bmatrix} + x &y \\ + y^* &z + \end{bmatrix} \ge 0 + \] + if and only if $z \ge 0$ and $x - yz^{-1}y^* \ge 0$. +\end{lemma} +\begin{proof} + ($\Leftarrow$): Suppose that $z \ge 0$ and $x - yz^{-1}y^* \ge 0$, then since $z$ is invertible, + \begin{align*} + \begin{bmatrix} + x &y \\ + y^* &z + \end{bmatrix} &= \begin{bmatrix} + I &yz^{-1} \\ + 0 &I + \end{bmatrix}\begin{bmatrix} + x - yz^{-1}y^* &0 \\ + 0 &z + \end{bmatrix}\begin{bmatrix} + I &0 \\ + z^{-1}y^* &I + \end{bmatrix} \\ + &= \begin{bmatrix} + I &0 \\ + z^{-1}y^* &I + \end{bmatrix}^*\begin{bmatrix} + x - yz^{-1}y^* &0 \\ + 0 &z + \end{bmatrix}\begin{bmatrix} + I &0 \\ + z^{-1}y^* &I + \end{bmatrix} + \end{align*} + + Since the middle factor is positive, the given matrix is positive as well. + + ($\Rightarrow$): Now suppose that the given matrix is positive, then for any $\xi \in H$, + \[ + \dpn{z\xi, \xi}{H} = \angles{\begin{bmatrix} + x &y \\ + y^* &z + \end{bmatrix}\begin{bmatrix} + 0 \\ + \xi + \end{bmatrix}, \begin{bmatrix} + 0 \\ + \xi + \end{bmatrix}}_{H^2} \ge 0 + \] + + so $z \ge 0$. As such, + \begin{align*} + \begin{bmatrix} + x - yz^{-1}y^* &0 \\ + 0 &z + \end{bmatrix} &= \begin{bmatrix} + I &-yz^{-1} \\ + 0 &I + \end{bmatrix}\begin{bmatrix} + x &y \\ + y^* &z + \end{bmatrix}\begin{bmatrix} + I &0 \\ + -z^{-1}y^* &I + \end{bmatrix} \\ + &= \begin{bmatrix} + I &0 \\ + -z^{-1}y^* &I + \end{bmatrix}^*\begin{bmatrix} + x &y \\ + y^* &z + \end{bmatrix}\begin{bmatrix} + I &0 \\ + -z^{-1}y^* &I + \end{bmatrix} \\ + \end{align*} + + Since the middle factor is positive, the left hand side is also positive. Therefore $x - yz^{-1}y^* \ge 0$ as well. +\end{proof} + +\begin{proposition} +\label{proposition:completely-positive-norm} + Let $A, B$ be $C^*$-algebras with $A$ being unital, $S \subset A$ be an operator system, and $\phi: S \to B$ be a completely positive map, then + \[ + \norm{\phi}_{\text{cb}} = \norm{\phi}_{L(S; B)} = \norm{\phi(1_A)}_B + \] +\end{proposition} +\begin{proof} + After taking a \hyperref[unitisation]{proposition:c-star-unitisation}, assume without loss of generality that $B$ is unital. + + It is sufficient to show that $\norm{\phi}_{\text{cb}} \le \norm{\phi(1_A)}_B$. To this end, let $T \in M_n(S)$ with $\norm{T}_{M_n(S)} \le 1$, then + \[ + \begin{bmatrix} + 1_{M_n(A)} &T \\ + T^* &1_{M_n(A)} + \end{bmatrix} \ge 0 + \] + + in $M_{2}(M_n(S)) = M_{2n}(S)$ by \autoref{lemma:norm-matrix-test}. Since $\phi$ is completely positive, + \[ + \phi_{2n}\begin{bmatrix} + 1_{M_n(A)} &T \\ + T^* &1_{M_n(A)} + \end{bmatrix} = \begin{bmatrix} + \phi_n(1_{M_n(A)}) &\phi_n(T) \\ + \phi_n(T)^* & \phi_n(1_{M_n(A)}) + \end{bmatrix} \ge 0 + \] + + First suppose that $\phi_n(1_{M_n(A)})$ is invertible, then + + \begin{align*} + \phi_n(1_{M_n(A)}) &\le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)} \cdot 1_{M_n(B)} \\ + 1_{M_n(B)} &\le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)} \cdot \phi_n(1_{M_n(A)})^{-1} \\ + \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^{-1} &\le \phi_n(1_{M_n(A)})^{-1} + \end{align*} + + By the \hyperref[Schur complement condition]{lemma:schur-complement-positive}, + \begin{align*} + \phi_n(1_{M_n(A)}) &\ge \phi_n(T)\phi_n(1_{M_n(A)})^{-1}\phi_n(T)^* \\ + &\ge \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^{-1}\phi_n(T)\phi_n(T)^* \\ + \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^2 &\ge \normn{\phi_n(T)\phi_n(T)^*}_{M_n(B)} = \normn{\phi_n(T)}_{M_n(B)}^2 + \end{align*} + + and $\normn{\phi_n(T)}_{M_n(B)} \le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}$. Therefore $\norm{\phi}_{\text{cb}} \le \normn{\phi(1_A)}_B$. + + Now suppose that $\phi$ is an arbitrary completely positive map. Let $\psi \in S(A)$ be a state and $\eps > 0$, then $\phi + \eps 1_B\psi$ is also a completely positive map. In which case, + \[ + \norm{\phi}_{\text{cb}} \le \inf_{\eps > 0}\norm{\phi + \eps1_B\psi}_{\text{cb}} \le \inf_{\eps > 0}\norm{\phi(1_A) + \eps 1_B}_B = \norm{\phi(1_A)}_B + \] +\end{proof} + +\begin{example} + \label{example:transpose-not-completely-positive} + Let $T: M_2(\complex) \to M_2(\complex)$ be the transpose map, then $T$ is positive, but not $2$-positive. +\end{example} +\begin{proof} + Let + \[ + C = \begin{bmatrix} + 1&0&0&1 \\ + 0&0&0&0 \\ + 0&0&0&0 \\ + 1&0&0&1 + \end{bmatrix} = (1, 0, 0, 1) \otimes (1, 0, 0, 1) \ge 0 + \] + + then + \[ + T_2(C) = \begin{bmatrix} + 1&0&0&0 \\ + 0&0&1&0 \\ + 0&1&0&0 \\ + 0&0&0&1 + \end{bmatrix} + \] + + is invertible, with negative determinant, so $T_2(C)$ is not positive. +\end{proof} + +\begin{example} + \label{example:homomorphism-completely-positive} + Let $A, B$ be $C^*$-algebras and $\phi: A \to B$ be a *-homomorphism, then $\phi$ is completely positive. +\end{example} +\begin{proof} + For each $n \in \natp$, the induced map $\phi_n: M_n(A) \to M_n(B)$ is a *-homomorphism. +\end{proof} + +\begin{example} + \label{example:multiply-completely-bounded} + + Let $A$ be a $C^*$-algebra, $y, z \in A$, $\phi: A \to A$ be defined by $x \mapsto yxz$, then + \begin{enumerate} + \item $\phi$ is completely bounded. + \item If $y = z^*$, then $\phi$ is completely positive. + \end{enumerate} +\end{example} +\begin{proof} + (1): For each $n \in \natp$ and $T \in M_n(A)$, + \[ + \phi_n(T) = (1_{M_n(\complex)} \otimes y)T(1_{M_n(\complex)} \otimes z) + \] +\end{proof} + +\begin{proposition} +\label{proposition:commutative-mn-facts} + Let $X$ be a compact Hausdorff space, then: + \begin{enumerate} + \item $C(X; M_n(\complex))$ equipped with pointwise product, pointwise involution, and the norm + \[ + \norm{f}_{C(X; M_n(\complex))} = \sup_{x \in X}\norm{f(x)}_{M_n(\complex)} + \] + + is a $C^*$-algebra. + \item The mapping + \[ + M_n(C(X; \complex)) \to C(X; M_n(\complex)) \quad M \otimes f \mapsto M \cdot f + \] + + is a *-isomorphism. + \item For any $f \in C(X; M_n(\complex))$, $f \ge 0$ if and only if $f(x) \ge 0$ in $M_n(\complex)$ for all $x \in X$. + \end{enumerate} +\end{proposition} +\begin{proof} + (1): For any $f \in C(X; M_n(\complex))$, + \[ + \sup_{x \in X}\norm{f(x)}_{M_n(\complex)}^2 = \sup_{x \in X}\norm{f(x)^*f(x)}_{M_n(\complex)} = \norm{f^*f}_{C(X; M_n(\complex))} + \] + + (3): If $f \ge 0$, then there exists $g \in C(X; M_n(\complex))$ such that $f(x) = g^*(x)g(x)$ for all $x \in X$, so $f(x) \ge 0$ in $M_n(\complex)$ for all $x \in X$. + + On the other hand, if $f(x) \ge 0$ in $M_n(\complex)$ for all $x \in X$, then $g(x) = f(x)^{1/2}$ is defined for all $x \in X$. By the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $g$ is continuous from $X$ to $M_n(\complex)$. Therefore there exists $g \in C(X; M_n(\complex))$ such that $f = g^*g$. +\end{proof} + + +\begin{theorem} +\label{theorem:bounded-functional-completely-bounded} + Let $A$ be a unital $C^*$-algebra, $S \subset A$ be a subspace, $X$ be a compact Hausdorff space, and $\phi \in L(S; C(X; \complex))$, then: + \begin{enumerate} + \item $\norm{\phi}_{\text{cb}} = \norm{\phi}_{L(S; C(X; \complex))}$. + \item If $S$ is an operator system and $\phi$ is positive, then $\phi$ is completely positive. + \end{enumerate} +\end{theorem} +\begin{proof}[Proof, {{\cite[Proposition 3.8]{Paulsen}}}. ] + By \autoref{proposition:commutative-mn-facts}, the norm and order relation on $C(X; M_n(\complex)) \iso M_n(C(X; \complex))$ are pointwise. As such, assume without loss of generality that $X$ is a single point and $C(X; \complex) = \complex$. + + Let $a = (a_{ij}) \in M_n(S)$. For each $\xi, \eta \in \complex^n$, + \begin{align*} + \dpn{\phi_n(a)\xi, \eta}{\complex^n} &= \sum_{i, j = 1}^n \xi_j \ol{\eta_i} \dpn{a_{ij}, \phi}{S} = \angles{\sum_{i, j = 1}^n a_{ij}\xi_j\ol{\eta_i}, \phi}_S + \end{align*} + + where + \begin{align*} + \sum_{i, j = 1}^n a_{ij}\xi_j\ol{\eta_i} &= (\one \otimes \ol{\eta}) \cdot a \cdot (\xi \otimes \one) + \end{align*} + + (1): + \begin{align*} + |\dpn{\phi_n(a)\xi, \eta}{\complex^n}| &\le \norm{\phi}_{S^*} \cdot \norm{\sum_{i, j = 1}^n a_{ij}\xi_j\ol{\eta_i}}_{S} \\ + \norm{\sum_{i, j = 1}^n a_{ij}\xi_j\ol{\eta_i}}_{S} &\le \norm{\xi}_{\complex^n}\norm{\eta}_{\complex^n}\norm{a}_{M_n(S)} \\ + \norm{\phi_n(a)}_{M_n(\complex)} &\le \norm{\phi}_{S^*}\norm{a}_{M_n(S)} \\ + \norm{\phi}_{\text{cb}} &\le \norm{\phi}_{S^*} + \end{align*} + + (2): If $\xi = \eta$, then + \[ + \sum_{i, j = 1}^n a_{ij}\eta_j\ol{\eta_i} = (\xi \otimes \one)^* \cdot a \cdot (\xi \otimes \one) \ge 0 + \] + + so + \[ + \dpn{\phi_n(a)\xi, \eta}{\complex^n} = \dpn{(\xi \otimes \one)^* \cdot a \cdot (\xi \otimes \one), \phi}{S} \ge 0 + \] +\end{proof} + + +\begin{theorem}[Stinespring] +\label{theorem:stinespring-commutative-completely-positive} + Let $X$ be a compact Hausdorff space, $B$ be a $C^*$-algebra, $\phi: C(X; \complex) \to B$ be a positive map, then $\phi$ is completely positive. +\end{theorem} +\begin{proof} + Let $P \in C(X; M_n(\complex))$ with $P \ge 0$ and $\eps > 0$. By \autoref{proposition:uniform-continuous-compact} and \autoref{theorem:lch-paracompact}, there exists a partition of unity $\seqf{\psi_j} \subset C(X; [0, 1])$ and positive matrices $\seqf{p_j} \subset M_n(\complex)$ such that + \[ + \norm{P(x) - \sum_{j = 1}^n p_j\psi_j(x)}_{M_n(\complex)} < \eps + \] + + for all $x \in X$. For each $1 \le j \le n$, $\phi_n(p_j\psi_j) = p_j\phi(\psi_j) \ge 0$. As $\eps > 0$ is arbitrary, $\phi_n(P) \ge 0$. +\end{proof} + +\begin{theorem}[Choi] +\label{theorem:choi-matrix-positive} + Let $n \in \natp$, $B$ be a $C^*$-algebra, and $\phi: M_n(\complex) \to B$ be a linear map. For each $1 \le i, j \le n$, let $E_{ij}$ be the standard matrix units for $M_n(\complex)$, then $C_\phi = (\phi(E_{ij})) \in M_n(B)$ is the \textbf{Chi matrix} of $\phi$. The mapping $\phi \mapsto C_\phi$ is an isomorphism from $L(M_n(\complex); B)$ to $M_n(B)$, and the following are equivalent: + \begin{enumerate} + \item $\phi$ is completely positive. + \item $\phi$ is $n$-positive. + \item $C_\phi$ is positive in $M_n(B)$. + \end{enumerate} +\end{theorem} +\begin{proof} + (2) $\Rightarrow$ (3): The matrix $E = (E_{ij})$ is positive in $M_n(\complex)$. + + (3) $\Rightarrow$ (1): + + + + $\phi, \psi \in S(G)$, $K \subset G$ be compact, and $V \in \cn_G(1_G)$ be relatively compact, then + \begin{align*} + + \end{align*} + +\end{proof} + + + diff --git a/src/op/system/dilation.tex b/src/op/system/dilation.tex new file mode 100644 index 0000000..673fd62 --- /dev/null +++ b/src/op/system/dilation.tex @@ -0,0 +1,122 @@ +\section{Dilations} +\label{section:bh-dilation} + +\begin{definition}[Compression] +\label{definition:bh-compression} + Let $H$ be a complex Hilbert space, $K \subset H$ be a closed subspace, $P \in B(H)$ be the orthogonal projection onto $K$, $S \in B(K)$, and $T \in B(H)$, then $S$ is a \textbf{compression} of $T$ if $S = PT|_K$. +\end{definition} + +\begin{definition}[Dilation] +\label{definition:bh-dilation} + Let $H$ be a complex Hilbert space and $T \in B(H)$, then a \textbf{dilation} of $T$ is a pair $(K, S)$ where: + \begin{enumerate} + \item $K$ is a Hilbert space containing $H$. + \item $S \in B(K)$. + \item If $P \in B(K)$ is the orthogonal projection onto $H$, then $T = PS|_H$. + \end{enumerate} + + If $T^n = PS^n|_H$ for all $n \in \natz$, then $S$ is a \textbf{power dilation} of $T$. +\end{definition} + +\begin{lemma} +\label{lemma:bh-dilation-transitive} + Let $H$ be a complex Hilbert space, $R \in B(H)$, $(K, S)$ be a dilation of $R$, and $(L, T)$ be a dilation of $S$, then $(L, T)$ is a dilation of $R$. + + 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}[Sz.-Nagy's Dilation Theorem] +\label{theorem:sznagy-dilation} + 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 $T$. +\end{theorem} +\begin{proof}[Proof, {{\cite[Theorem 1.1]{Paulsen}}}. ] + Using \autoref{lemma:bh-dilation-transitive} it is sufficient to show the following statements: + \begin{enumerate}[label=(\roman*)] + \item For any $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, there exists an isometric power dilation of $T$. + \item For any isometry $V \in B(H)$, there exists a unitary power dilation of $V$. + \end{enumerate} + + (i): Since $\norm{T}_{B(H)} \le 1$, $1 - T^*T \ge 0$, and $(1 - T^*T)^{1/2}$ is well-defined. Let $D_T = (1 - T^*T)^{1/2}$, and + \[ + V: l^2(\natp; H) \to l^2(\natp; H) \quad (Vx)_n = \begin{cases} + Tx_1 &n = 1 \\ + D_Tx_1 &n = 2 \\ + x_{n-1} &n \ge 3 + \end{cases} + \] + + then for each $x \in H$, + \begin{align*} + \norm{Tx}_H^2 + \norm{D_Tx}_H^2 &= \dpn{T^*Tx, x}{H} + \dpn{(1 - T^*T)x, x}{H} \\ + &= \dpn{x, x}{H} = \norm{x}_H^2 + \end{align*} + + Thus for each $x = \seq{x_n} \in l^2(H)$, + \begin{align*} + \norm{Vx}_{l^2(\natp; H)}^2 &= \norm{Tx_1}_H^2 + \norm{D_Tx_1}_H^2 + \sum_{n = 3}^\infty \norm{x_{n-1}}_H^2 \\ + &= \norm{x_1}_H^2 + \sum_{n = 2}^\infty \norm{x_{n}}_H^2 = \norm{x}_{l^2(\natp; H)}^2 + \end{align*} + + So $V$ is an isometry. Moreover, $[V^n(x, 0, \cdots)]_1 = T^nx$ for all $n \in \natz$. Therefore under the identification that $H = \bracsn{(x, 0, \cdots)|x \in H} \subset l^2(\natp; H)$, $V$ is an isometric power dilation of $T$. + + (ii): Let $P = I_H - VV^*$ be the projection onto $V(H)^\perp$, and $U \in B(H^2)$ be defined by + \[ + U = \begin{bmatrix} + V & P \\ + 0 & V^* + \end{bmatrix} + \] + + then + \begin{align*} + UU^* &= \begin{bmatrix} + V & P \\ + 0 & V^* + \end{bmatrix} \cdot + \begin{bmatrix} + V^* & 0 \\ + P & V + \end{bmatrix} = \begin{bmatrix} + VV^* + P & PV \\ + V^*P & V^*V + \end{bmatrix} = I_{H^2} \\ + U^*U &= \begin{bmatrix} + V^* & 0 \\ + P & V + \end{bmatrix} \cdot + \begin{bmatrix} + V & P \\ + 0 & V^* + \end{bmatrix} + = \begin{bmatrix} + V^*V & V^*P \\ + PV & VV^* + P + \end{bmatrix} = I_{H^2} \\ + \end{align*} + + so $U$ is unitary. Moreover, for each $n \in \natp$, + \[ + U^n = \begin{bmatrix} + V & P \\ + 0 & V^* + \end{bmatrix}^n = \begin{bmatrix} + V^n & \cdots \\ + 0 & (V^*)^n + \end{bmatrix} + \] + + Under the identification that $H = \bracsn{(x, 0)|x \in H} \subset H^2$, $QU^n|_{H} = V^n$ for all $n \in \natp$. Therefore $U$ is a unitary power dilation of $V$. +\end{proof} + +\begin{theorem}[Von Neumann's Inequality] +\label{theorem:von-neumann-inequality-polynomial} + Let $H$ be a complex Hilbert space and $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, then for any $p \in \complex[x]$, + \[ + \norm{p(T)}_{B(H)} \le \sup_{z \in B_\complex(0, 1)}|p(z)| + \] +\end{theorem} +\begin{proof}[Proof, {{\cite[Corollary 1.2]{Paulsen}}}. ] + By \hyperref[Sz.-Nagy's Dilation Theorem]{theorem:sznagy-dilation}, there exists a Hilbert space $K \supset H$ and a unitary operator $U \in B(K)$ such that $T^n = P_HU^n|_H$ for all $n \in \natp$, where $P_H$ is the orthogonal projection onto $H$. + + Thus for any $p \in \complex[x]$, $p(T) = P_Hp(U)|_H$. As $U$ is unitary, the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus} shows that $\norm{p(U)}_{B(K)} \le \sup_{z \in B_\complex(0, 1)}|p(z)|$. Therefore $\norm{p(T)}_{B(H)} \le \sup_{z \in B_\complex(0, 1)}|p(z)|$. +\end{proof} diff --git a/src/op/system/index.tex b/src/op/system/index.tex index c36bfb5..ab42cd4 100644 --- a/src/op/system/index.tex +++ b/src/op/system/index.tex @@ -1,4 +1,6 @@ \chapter{Operator Systems} \label{chap:opertaor-systems} -\input{./definitions.tex} +\input{./dilation.tex} +\input{./os.tex} +\input{./completely.tex} diff --git a/src/op/system/definitions.tex b/src/op/system/os.tex similarity index 99% rename from src/op/system/definitions.tex rename to src/op/system/os.tex index c904064..8a960a1 100644 --- a/src/op/system/definitions.tex +++ b/src/op/system/os.tex @@ -39,7 +39,7 @@ \label{proposition:positive-automatic-continuity} Let $A, B$ be $C^*$-algebra, $S \subset A$ and $T \subset B$ be operator systems, and $\phi: S \to T$ be a positive map, then \begin{enumerate} - \item For any $x \in S_{sa}$, $\norm{\phi(x)}_{B} \le \norm{\phi(1_A)}_B \cdot \norom{x}_A$. + \item For any $x \in S_{sa}$, $\norm{\phi(x)}_{B} \le \norm{\phi(1_A)}_B \cdot \norm{x}_A$. \item $\norm{\phi}_{L(S; T)} \le 2\norm{\phi(1_A)}_B$. \end{enumerate} \end{proposition}