Added multiplicative domains.

src/op/system/dilation.tex

@@ -392,11 +392,34 @@
     \]
 
     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 $()
+    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 $P = V^* \in B(K)$ 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}, $\pi(a)^*P\pi(a)P = \pi(a)^*P\pi(a)P$ 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}