Cleaned up multiplicative domains.

2026-10-06 09:58:28 -04:00
parent b29575d59a
commit 2fa30181ed
2 changed files with 106 additions and 60 deletions
@@ -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}
-
-
-