Final revision.

src/harmonic/rep/l1.tex

@@ -56,27 +56,27 @@
         &= \ol{\dpn{\pi(f)\eta, \xi}{H}} = \dpn{\xi, \pi(f)\eta}{H}
     \end{align*}
     
-    (1, Non-Degenerate): For any $\xi \in H \setminus \bracsn{0}$, since $\pi_0$ is strong-operator continuous, there exists $V \in \cn_G(1)$ such that $\norm{\pi(x)\xi - \xi}_H < \norm{\xi}_H$ for all $x \in V$. In which case, if $f = |V|^{-1}\one_V$\footnote{$|V|$ is the value of the Haar measure at $V$.}, then for any $\eta \in H$, 
+    (1, Non-Degenerate): For any $\xi \in H \setminus \bracsn{0}$, since $\pi_0$ is strong-operator continuous, there exists $V \in \cn_G(1)$ such that $\norm{\pi_0(x)\xi - \xi}_H < \norm{\xi}_H$ for all $x \in V$. In which case, if $f = |V|^{-1}\one_V$\footnote{$|V|$ is the value of the Haar measure at $V$.}, then for any $\eta \in H$, 
     \begin{align*}
-        \dpn{\pi(f)\xi - \xi, \eta}{H} &= \frac{1}{|V|}\int_V \dpn{\pi(x)\xi - \xi, \eta}{H} dx \\
+        \dpn{\pi(f)\xi - \xi, \eta}{H} &= \frac{1}{|V|}\int_V \dpn{\pi_0(x)\xi - \xi, \eta}{H} dx \\
-        |\dpn{\pi(f)\xi - \xi, \eta}{H}| &\le \frac{1}{|V|}\int_V \norm{\pi(x)\xi - \xi}_H\norm{\eta}_Hdx < \norm{\xi}_H\norm{\eta}_H
+        |\dpn{\pi(f)\xi - \xi, \eta}{H}| &\le \frac{1}{|V|}\int_V \norm{\pi_0(x)\xi - \xi}_H\norm{\eta}_Hdx < \norm{\xi}_H\norm{\eta}_H
     \end{align*}
 
     so $\norm{\pi(f)\xi - \xi}_H < \norm{\xi}_H$, and $\pi(f)\xi \ne 0$. 
 
-    (3): Using \autoref{proposition:group-algebra-approximate-identity}, let $\angles{\phi_\alpha}_{\alpha \in A}$ be an approximate identity in $L^1(G; \complex)$. For each $x \in G$ and $f \in L^1(G; \complex)$, $L_x\phi_\alpha * f \to L_x f$ in $L^1(G; \complex)$ implies that $\angles{\pi(\phi_\alpha)}_{\alpha \in A}$ converges pointwise on
+    (3): Using \autoref{proposition:group-algebra-approximate-identity}, let $\angles{\phi_\alpha}_{\alpha \in A}$ be an approximate identity in $L^1(G; \complex)$. For each $x \in G$ and $f \in L^1(G; \complex)$, $L_x\phi_\alpha * f \to L_x f$ in $L^1(G; \complex)$ implies that $\angles{\pi(L_x\phi_\alpha)}_{\alpha \in A}$ converges pointwise on
     \[
     K := \text{span}\bracsn{\pi(g)\xi|\xi \in H, g \in L^1(G; \complex)}
     \]
 
-    Given that $\pi$ is non-degenerate, $K$ is dense in $H$. As $\bracsn{L_x\phi_\alpha|\alpha \in A}$ is equicontinuous and $H$ is complete, the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli} implies that there exists $\pi_0(x) \in \ol{B_{B(H)}(0, 1)}$ such that $L_x\phi_\alpha \to \pi_0(x)$ in the strong operator topology. In particular, $\pi_0(1) = \text{Id}$. 
+    Given that $\pi$ is non-degenerate, $K$ is dense in $H$. As $\bracsn{\pi(L_x\phi_\alpha)|\alpha \in A}$ is equicontinuous and $H$ is complete, the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli} implies that there exists $\pi_0(x) \in \ol{B_{B(H)}(0, 1)}$ such that $\pi(L_x\phi_\alpha) \to \pi_0(x)$ in the strong operator topology. In particular, $\pi_0(1) = \text{Id}$. 
     
     For each $f \in L^1(G; \complex)$, $\xi \in H$, and $x, y \in G$, 
     \[
     \pi_0(xy)\pi(f)\xi = \pi(L_{xy}f)\xi = \pi(L_xL_yf)\xi = \pi_0(x)\pi_0(y)\pi(f)\xi
     \]
 
-    By density of $K$ in $H$, $\pi_0(xy) = \pi_0(x)\pi_0(y)$, so $\pi$ is a homomorphism. For any $x \in G$, $\text{Id} = \pi_0(1) = \pi_0(x)\pi_0(x^{-1})$, so $\pi_0(x)$ is isometric and invertible, and hence unitary. 
+    By density of $K$ in $H$, $\pi_0(xy) = \pi_0(x)\pi_0(y)$, so $\pi_0$ is a homomorphism. For any $x \in G$, $\text{Id} = \pi_0(1) = \pi_0(x)\pi_0(x^{-1})$, so $\pi_0(x)$ is isometric and invertible, and hence unitary. 
 
     Finally, for any $f \in L^1(G; \complex)$ and $\xi, \eta \in H$, by the \hyperref[Dominated Convergence Theorem]{theorem:dct}, 
     \begin{align*}
@@ -91,9 +91,9 @@
     Let $G$ be a locally compact group, $(H, \pi_0)$ be a unitary representation of $G$, and $\pi: L^1(G; \complex) \to B(H)$ be the *-representation determined by $\pi_0$, then the von Neumann algebras generated by $\pi_0(G)$ and $\pi(L^1(G; \complex))$ coincide. 
 \end{theorem}
 \begin{proof}
-    By (3) of \autoref{definition:representation-l1-group-algebra}, $\pi_0(G)$ lies in the strong-operator closure of $L^1(G; \complex)$. 
+    By (3) of \autoref{definition:representation-l1-group-algebra}, $\pi_0(G)$ lies in the strong-operator closure of $\pi(L^1(G; \complex))$. 
 
-    On the other hand, let $K \subset G$ be compact, $\eps > 0$, and $\seqf{(\xi_j, \eta_j)} \subset H^2$, then there exist non-empty sets $\bracsn{V_k}_1^m \subset 2^G$ such that $K = \bigsqcup_{j = 1}^n V_j$ and $|\dpn{[\pi_0(x) - \pi_0(y)]\xi_j, \eta_j}{H}| \le \eps$ for all $x, y \in V_k$, $1 \le j \le n$, and $1 \le k \le m$. 
+    On the other hand, let $K \subset G$ be compact, $\eps > 0$, and $\seqf{(\xi_j, \eta_j)} \subset H^2$, then there exist non-empty sets $\bracsn{V_k}_1^m \subset 2^G$ such that $K = \bigsqcup_{k = 1}^m V_k$ and $|\dpn{[\pi_0(x) - \pi_0(y)]\xi_j, \eta_j}{H}| \le \eps$ for all $x, y \in V_k$, $1 \le j \le n$, and $1 \le k \le m$. 
     
     Let $\seqf{x_k} \subset G$ with $x_k \in V_k$ for all $1 \le k \le m$, then 
     \begin{align*}

src/harmonic/rep/unit.tex

@@ -31,7 +31,7 @@
 
 \begin{lemma}
 \label{lemma:lcg-intertwining}
-    Let $G$ be a locally compact group and $(H, \pi)$ be a unitary representation of $G$, then $\mathcal{C}(\pi, \pi)$ is the commutant of $\pi(G)'$, and hence a von Neumann algebra. 
+    Let $G$ be a locally compact group and $(H, \pi)$ be a unitary representation of $G$, then $\mathcal{C}(\pi, \pi) = \pi(G)'$, and hence a von Neumann algebra. 
 \end{lemma}
 
 \begin{definition}[Invariant Subspace]