Cleaned up multiplicative domains.

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}
-
-
-

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}