From fcd5b1a92f3b449120d70e479ee7fbb60960dcb6 Mon Sep 17 00:00:00 2001
From: Bokuan Li <bokuan.li@mail.mcgill.ca>
Date: Mon, 19 Oct 2026 15:04:19 -0400
Subject: [PATCH] Added multiplicative domains.

---
 src/op/system/dilation.tex | 25 ++++++++++++++++++++++++-
 src/op/vn/vn.tex           | 12 +++++++++++-
 2 files changed, 35 insertions(+), 2 deletions(-)

diff --git a/src/op/system/dilation.tex b/src/op/system/dilation.tex
index 3b58bdf..a00fa91 100644
--- a/src/op/system/dilation.tex
+++ b/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}
 
diff --git a/src/op/vn/vn.tex b/src/op/vn/vn.tex
index 7841e1c..e24a5db 100644
--- a/src/op/vn/vn.tex
+++ b/src/op/vn/vn.tex
@@ -100,9 +100,19 @@
     \begin{enumerate}
         \item $T(M) \subset M$. 
         \item $PTP = TP$. 
+        \item $T^*PTP = T^*TP$. 
     \end{enumerate}
 \end{lemma}
-% Proof omitted due to obviousness
+\begin{proof}
+    (3) $\Leftarrow$ (1): Let $\xi \in H$, then
+    \begin{align*}
+        \dpn{T^*TP\xi, P\xi}{H} &= \dpn{TP\xi, TP\xi}{H} = \norm{TP\xi}_H^2 \\
+        \dpn{T^*PTP\xi, P\xi}{H} &= \dpn{PTP\xi, TP\xi}{H} = \dpn{PTP\xi, PTP\xi}{H} = \norm{PTP\xi}_H^2
+    \end{align*}
+
+    As $\norm{TP\xi}_H = \norm{PTP\xi}_H$, $TP\xi \in H$, so $T(M) \subset M$. 
+    
+\end{proof}
 
 
 \begin{definition}[Reducing Subspace]