Adjustments.

src/op/c-star/gns.tex

@@ -61,9 +61,9 @@
 
     For each $x \in A$, let $\pi_\phi(x)$ be the continuous extension of $\pi_\phi^0(x)$ to an element of $B(H_\phi)$, then:
     \begin{enumerate}
-        \item $(H_\phi, \dpn{\cdot, \cdot}{\phi})$ is a Hibert space.
+        \item $H_\phi$ equipped with the unique continuous extension of $\dpn{\cdot, \cdot}{\phi}$ is a Hibert space.
         \item $(H_\phi, \pi_\phi)$ is a well-defined representation of $A$. 
-        \item $\xi_\phi = 1_A + N_\phi$ is a unit vector in $H_\phi$, and $\bracsn{\pi_\phi(x)\xi_\phi| x \in A}$ is dense in $H_\phi$. Moreover, for each $x, y \in A$,
+        \item $\xi_\phi = 1_A + N_\phi$ is a unit cyclic vector in $H_\phi$ such that for each $x, y \in A$,
         \[
         \dpn{x, y}{\phi} = \dpn{\pi_\phi(x)\xi_\phi, \pi_\phi(y)\xi_\phi}{H_\phi}
         \]
@@ -108,7 +108,7 @@
 
 \begin{proposition}
 \label{proposition:gns-irreducible}
-    Let $A$ be a unital $C^*$ algebra, $\phi \in S(A)$ be a state, and $(H, \pi, \xi)$ be its GNS triple, then $\phi$ is a pure state if and only if $(H, \pi)$ is irreducible. 
+    Let $A$ be a unital $C^*$-algebra, $\phi \in S(A)$ be a state, and $(H, \pi, \xi)$ be its GNS triple, then $\phi$ is a pure state if and only if $(H, \pi)$ is irreducible. 
 \end{proposition}
 \begin{proof}
     ($\Rightarrow$): Suppose that $(H, \pi)$ is reducible, then there exists a non-trivial invariant subspace $M \subset H$. Let $P \in B(H)$ be the orthogonal projection onto $M$, $\xi_1 = P\xi/\norm{P\xi}_H$, and $\xi_2 = (\xi - P\xi)/\norm{\xi - P\xi}_H$. For each $x \in A$, let
@@ -118,14 +118,21 @@
 
     then $\phi_1$ and $\phi_2$ are vector states on $A$. Since $M$ and $M^\perp$ are invariant for $\pi$, 
     \begin{align*}
-        \dpn{x, \phi}{A} &= \dpn{\pi(x)\xi, \xi}{A} = \dpn{\pi(x)P\xi, P\xi}{A} + \dpn{\pi(x)(I - P)\xi, (I - P)\xi}{A} \\
+        \dpn{x, \phi}{A} &= \dpn{\pi(x)\xi, \xi}{H} = \dpn{\pi(x)P\xi, P\xi}{H} + \dpn{\pi(x)(I - P)\xi, (I - P)\xi}{H} \\
         &= \norm{P\xi}_H^2\dpn{\pi(x)\xi_1, \xi_1}{H} + \norm{(1 - P)\xi}_H^2 \dpn{\pi(x)\xi_2, \xi_2}{H} \\
-        &= \norm{P\xi}_H^2 \dpn{x, \phi_1}{A} + \norm{(1 - P)\xi}_H^2 \dpn{x, \phi_2}{A}
+        &= \norm{P\xi}_H^2 \dpn{x, \phi_1}{A} + \norm{(I - P)\xi}_H^2 \dpn{x, \phi_2}{A}
     \end{align*}
 
-    so $\phi$ is a strict convex combination of two states, and is not pure.
+    so $\phi$ is a strict convex combination of two states. If $\phi_1 = \phi$, then for any $x, y \in A$, 
+    \begin{align*}
+    \dpn{\pi(x)\xi, \pi(y)\xi}{H} &= \norm{P\xi}_H^{-2}\dpn{\pi(x)P\xi, \pi(y)\xi}{H} \\
+    \dpn{[\pi(x) - \norm{P\xi}_H^{-2}P\pi(x)]\xi, \pi(y)\xi}{H} &= 0
+    \end{align*}
 
-    ($\Leftarrow$): Suppose that $(H, \pi)$ is irreducible. Let $t \in (0, 1)$ and $\phi_1, \phi_2 \in S(A)$ with $\phi = t\phi_1 + (1 - t)\phi_2$. Let $T: \pi(A)H \to H$ be defined by
+    so $I - \norm{P\xi}_H^{-2}P = 0$, which contradicts the fact that $P$ is a non-trivial projection. 
+    
+
+    ($\Leftarrow$): Suppose that $(H, \pi)$ is irreducible. Let $t \in (0, 1)$ and $\phi_1, \phi_2 \in S(A)$ with $\phi = t\phi_1 + (1 - t)\phi_2$. Let $T: \pi(A)\xi \to H$ be defined by
     \[
     \dpn{T\pi(x)\xi, \pi(y)\xi}{H} = \dpn{x, y}{\phi_1} = \dpn{y^*x, \phi_1}{A}
     \]
@@ -133,7 +140,7 @@
     then since $\phi_1 \le t^{-1}\phi$, $T$ extends continuously into a positive operator in $B(H)$. Moreover, for any $x, y, z \in A$, 
     \begin{align*}
     \dpn{\pi(z)T\pi(x)\xi, \pi(y)\xi}{H} &= \dpn{T\pi(x)\xi, \pi(z^*y)\xi}{H} \\
-    &= \dpn{y^*zx, \phi_1}{A} = \dpn{T\pi(z)\pi(x)\xi, \xi}{H}
+    &= \dpn{y^*zx, \phi_1}{A} = \dpn{T\pi(z)\pi(x)\xi, \pi(y)\xi}{H}
     \end{align*}
     
     so $T \in \pi(A)'$. By \hyperref[Schur's Lemma]{lemma:cstar-schur}, there exists $\lambda \in \complex$ such that $T = \lambda I$. However, since $\dpn{T\xi, \xi}{H} = \dpn{1_A, \phi_1}{A} = 1$, $\lambda = 1$, $T = I$, and $\phi_1 = \phi$. By symmetry, $\phi_2 = \phi$ as well.