From 2fa30181ed093d7cd6161888be906887365692e3 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Tue, 6 Oct 2026 09:58:28 -0400 Subject: [PATCH] Cleaned up multiplicative domains. --- src/op/system/completely.tex | 60 -------------------- src/op/system/module.tex | 106 +++++++++++++++++++++++++++++++++++ 2 files changed, 106 insertions(+), 60 deletions(-) create mode 100644 src/op/system/module.tex diff --git a/src/op/system/completely.tex b/src/op/system/completely.tex index d35c3eb..6f26c58 100644 --- a/src/op/system/completely.tex +++ b/src/op/system/completely.tex @@ -326,63 +326,3 @@ is invertible, with negative determinant, so $T_2(C)$ is not positive. \end{proof} - -\begin{definition}[Module Map] -\label{definition:module-map} - Let $A, B, C$ be $C^*$-algebras with $C \subset A, B$ and $1_A = 1_B = 1_C$, then $A$ and $B$ are $C$-bimodules with respect to multiplication. - - Let $\phi: A \to B$ be a linear map, then $\phi$ is a \textbf{$C$-bimodule map} if $\phi(c_1ac_2) = c_1\phi(a)c_2$ for all $a \in A$ and $c_1, c_2 \in C$. -\end{definition} - -\begin{theorem} -\label{theorem:completely-positive-bimodule} - Let $A, B$ be unital $C^*$-algebras, $\phi: A \to B$ be a unital completely positive map, then: - \begin{enumerate} - \item For any $a \in A$, $\phi(a)^*\phi(a) = \phi(a^*a)$ if and only if $\phi(ba) = \phi(b)\phi(a)$ for all $b \in A$. The space $R = \bracsn{a \in A|\phi(a)^*\phi(a) = \phi(a^*a)}$ is a subalgebra of $A$, and $\phi|_R$ is a homomorphism. - \item For any $a \in A$, $\phi(a)\phi(a)^* = \phi(aa^*)$ if and only if $\phi(ab) = \phi(a)\phi(b)$ for all $b \in A$. The space $L = \bracsn{a \in A| \phi(a)\phi(a)^* = \phi(aa^*)}$ is a subalgebra of $A$, and $\phi|_L$ is a homomorphism. - \item $L \cap R$ is a $C^*$-subalgebra of $A$, and $\phi|_{L \cap R}$ is a *-homomorphism. - \end{enumerate} -\end{theorem} -\begin{proof} - (1, $\Leftarrow$): Suppose that $\phi(ba) = \phi(b)\phi(a)$ for all $b \in A$, then $\phi(a^*) = \phi(a)^*\phi(a)$ after plugging in $b = a^*$. - - (1, $\Rightarrow$): Suppose that $\phi(a^*a) = \phi(a)^*\phi(a)$. By the \hyperref[Kadison-Schwarz inequality]{proposition:cauchy-schwarz-2-positive}, - \begin{align*} - \phi_2\begin{bmatrix} - a & b^* \\ - 0 & 0 - \end{bmatrix}^*\phi_2\begin{bmatrix} - a & b^* \\ - 0 & 0 - \end{bmatrix} \le \phi_2\begin{bmatrix} - a^*a & a^*b^* \\ - ba & bb^* - \end{bmatrix} - \end{align*} - - so - \[ - \begin{bmatrix} - \phi(a^*)\phi(a) & \phi(a^*)\phi(b) \\ - \phi(b)\phi(a) & \phi(b)\phi(b^*) - \end{bmatrix} \le - \begin{bmatrix} - \phi(a^*a) & \phi(a^*b) \\ - \phi(ba) & \phi(bb^*) - \end{bmatrix} - \] - - and - \[ - \begin{bmatrix} - 0 & \phi(a^*b^*) - \phi(a^*)\phi(b) \\ - \phi(b)\phi(a) - \phi(ba) & \phi(b)\phi(b^*) - \phi(bb^*) - \end{bmatrix} \ge 0 - \] - - Since the top left entry is zero, $\phi(b)\phi(a) - \phi(ba) = 0$. - -\end{proof} - - - diff --git a/src/op/system/module.tex b/src/op/system/module.tex new file mode 100644 index 0000000..911007e --- /dev/null +++ b/src/op/system/module.tex @@ -0,0 +1,106 @@ +\section{Multiplicative Domains} +\label{section:completely-positive-multiplicative-domain} + + +\begin{definition}[Module Map] +\label{definition:module-map} + Let $A, B, C$ be $C^*$-algebras with $C \subset A, B$ and $1_A = 1_B = 1_C$, then $A$ and $B$ are $C$-bimodules with respect to multiplication. + + Let $\phi: A \to B$ be a linear map, then $\phi$ is a \textbf{$C$-bimodule map} if $\phi(c_1ac_2) = c_1\phi(a)c_2$ for all $a \in A$ and $c_1, c_2 \in C$. +\end{definition} + +\begin{lemma} +\label{lemma:matrix-positive-topleft-corner-zero} + Let $A$ be a $C^*$-algebra and $x, y \in A$, then + \[ + \begin{bmatrix} + 0 & x \\ + x^* & y + \end{bmatrix} \ge 0 + \] + + if and only if $x = 0$ and $y \ge 0$. +\end{lemma} +\begin{proof} + ($\Rightarrow$): Using the \hyperref[Gelfand-Naimark Theorem]{theorem:gns}, assume without loss of generality that there exists a complex Hilbert space $H$ such that $A \subset B(H)$ is a $C^*$-subalgebra. In which case, for any $\xi, \eta \in H$, + \begin{align*} + \angles{\begin{bmatrix} + 0 & x \\ + x^* & y + \end{bmatrix}\begin{bmatrix} + \xi \\ \eta + \end{bmatrix}, \begin{bmatrix} + \xi \\ \eta + \end{bmatrix}}_{H^2} &= \angles{\begin{bmatrix} + x\eta \\ x^*\xi + y\eta + \end{bmatrix}, \begin{bmatrix} + \xi \\ \eta + \end{bmatrix}}_{H^2} \\ + &= \dpn{x^*\xi, \eta}{H} + \dpn{x\eta, \xi}{H} + \dpn{y\eta, \eta}{H} \\ + &= 2\text{Re}(\dpn{x^*\eta, \xi}{H}) + \dpn{y\eta, \eta}{H} + \end{align*} + + If $x \ne 0$, then there exists $\xi_0, \eta_0 \in H$ such that $\dpn{x^*\eta_0, \xi_0}{H}\ne 0$. Therefore + \[ + \inf_{\xi \in H}\text{Re}(\dpn{x^*\eta_0, \xi}{H}) = -\infty + \] + + and the given matrix is not positive. +\end{proof} + +\begin{definition}[Multiplicative Domains] +\label{definition:positive-multiplicative-domains} + Let $A, B$ be unital $C^*$-algebras and $\phi: A \to B$ be a unital\footnote{For without this assumption the domains will not be populated.} $4$-positive map, then + \begin{enumerate} + \item For any $a \in A$, $\phi(a)^*\phi(a) = \phi(a^*a)$ if and only if $\phi(ba) = \phi(b)\phi(a)$ for all $b \in A$. The space + \[ + R = \bracsn{a \in A|\phi(a)^*\phi(a) = \phi(a^*a)} + \] + + is a subalgebra of $A$, and $\phi|_R$ is a homomorphism. + \item For any $a \in A$, $\phi(a)\phi(a)^* = \phi(aa^*)$ if and only if $\phi(ab) = \phi(a)\phi(b)$ for all $b \in A$. The space + \[ + L = \bracsn{a \in A| \phi(a)\phi(a)^* = \phi(aa^*)} + \] + + is a subalgebra of $A$, and $\phi|_L$ is a homomorphism. + \item $C = L \cap R$ is a $C^*$-subalgebra of $A$ with $1_C = 1_A$, and $\phi|_{C}$ is a unital *-homomorphism. + \end{enumerate} + + The spaces $R$ and $L$ are the \textbf{right} and \textbf{left} \textbf{multiplicative domains} of $\phi$, respectively. The $C^*$-subalgebra $C$ is the \textbf{multiplicative domain} of $\phi$. +\end{definition} +\begin{proof}[Proof, {{\cite[Theorem 3.18]{Paulsen}}}. ] + (1, $\Leftarrow$): Suppose that $\phi(ba) = \phi(b)\phi(a)$ for all $b \in A$, then $\phi(a^*) = \phi(a)^*\phi(a)$ after plugging in $b = a^*$. + + (1, $\Rightarrow$): Suppose that $\phi(a^*a) = \phi(a)^*\phi(a)$. By the \hyperref[Kadison-Schwarz inequality]{proposition:cauchy-schwarz-2-positive}, + \begin{align*} + \phi_2\begin{bmatrix} + a & b^* \\ + 0 & 0 + \end{bmatrix}^*\phi_2\begin{bmatrix} + a & b^* \\ + 0 & 0 + \end{bmatrix} &\le \phi_2\begin{bmatrix} + a^*a & a^*b^* \\ + ba & bb^* + \end{bmatrix} \\ + \begin{bmatrix} + \phi(a^*)\phi(a) & \phi(a^*)\phi(b) \\ + \phi(b)\phi(a) & \phi(b)\phi(b^*) + \end{bmatrix} &\le + \begin{bmatrix} + \phi(a^*a) & \phi(a^*b) \\ + \phi(ba) & \phi(bb^*) + \end{bmatrix} + \end{align*} + + so + \[ + \begin{bmatrix} + 0 & \phi(a^*b^*) - \phi(a^*)\phi(b) \\ + \phi(b)\phi(a) - \phi(ba) & \phi(b)\phi(b^*) - \phi(bb^*) + \end{bmatrix} \ge 0 + \] + + By \autoref{lemma:matrix-positive-topleft-corner-zero}, $\phi(b)\phi(a) - \phi(ba) = 0$. +\end{proof}