More corrections.

src/harmonic/index.tex

@@ -2,4 +2,5 @@
 \label{part:harmonic}
 
 \input{./lcg/index.tex}
+\input{./rep/index.tex}
 \input{./notation.tex}

src/harmonic/rep/index.tex

@@ -0,0 +1,6 @@
+\chapter{Representation of Locally Compact Groups}
+\label{chap:lcg-rep}
+
+\input{./unit.tex}
+\input{./l1.tex}
+

src/harmonic/rep/l1.tex

@@ -0,0 +1,108 @@
+\section{Representations of $L^1(G)$}
+\label{section:rep-l1-group-algebra}
+
+
+\begin{definition}[Representation of $L^1(G; \complex)$]
+\label{definition:representation-l1-group-algebra}
+    Let $G$ be a locally compact group and $(H, \pi_0)$ be a unitary representation of $G$, then
+    \begin{enumerate}
+        \item The mapping
+        \[
+        \pi: L^1(G; \complex) \to B(H) \quad \dpn{\pi(f)\xi, \eta}{H} = \int_G f(x)\dpn{\pi_0(x)\xi, \eta}{H}dx
+        \]
+        
+        defines a non-degenerate *-representation of $L^1(G; \complex)$.
+        \item For any $x \in G$ and $f \in L^1(G; \complex)$, 
+        \[
+        \pi_0(x)\pi(f) = \pi(L_xf) \quad \pi(f)\pi_0(x) = \Delta_G(x^{-1})\pi(R_{x^{-1}}f)
+        \]
+    \end{enumerate}
+
+    Conversely, if $\pi: L^1(G; \complex) \to B(H)$ is a non-degenerate *-representation of $L^1(G; \complex)$, then:
+    \begin{enumerate}[start=2]
+        \item There exists a unique unitary representation $\pi_0: G \to B(H)$ such that
+        \[
+        \dpn{\pi(f)\xi, \eta}{H} = \int_G f(x)\dpn{\pi_0(x)\xi, \eta}{H}dx
+        \]
+
+        for all $f \in L^1(G; \complex)$ and $\xi, \eta \in H$. Moreover, for any approximate identity $\angles{\phi_\alpha}_{\alpha \in A} \subset L^1(G; \complex)$ and $x \in G$, $\pi_0(x) = \sotlim_{\alpha \in A}\pi(L_x\phi_\alpha)$. 
+    \end{enumerate}
+
+    The mapping $\pi: L^1(G; \complex) \to B(H)$ is the \textbf{representation of $L^1(G; \complex)$ determined by $\pi_0$}. 
+\end{definition}
+\begin{proof}[Proof, {{\cite[Theorem 3.9, Theorem 3.11]{FollandHarmonic}}}. ]
+    (1, Well-Defined): For each $x \in G$, $\pi_0(x)$ is unitary, so for any $f \in L^1(G; \complex)$ and $\xi, \eta \in H$, 
+    \[
+        \abs{\dpn{\pi(f)\xi, \eta}{H}} \le \int_G \abs{f(x)\dpn{\pi_0(x)\xi, \eta}{H}}dx \le \norm{\xi}_H\norm{\eta}_H\norm{f}_{L^1(G; \complex)}
+    \]
+    
+    and the representation is well-defined. 
+    
+    (1, Homomorphism): For any $f, g \in L^1(G; \complex)$ and $\xi, \eta \in H$, since $\pi_0$ is a unitary representation, 
+    \begin{align*}
+        \dpn{\pi(f * g)\xi, \eta}{H} &= \int_G (f * g)(x)\dpn{\pi_0(x)\xi, \eta}{H}dx \\
+        &= \iint_{G \times G}f(y)g(y^{-1}x)\dpn{\pi_0(x)\xi, \eta}{H}dydx \\
+        &= \int_{G}f(y)\int_Gg(x)\dpn{\pi_0(y)\pi_0(x)\xi, \eta}{H}dxdy\\
+        &= \int_{G}f(y)\int_Gg(x)\dpn{\pi_0(x)\xi, \pi_0(y^{-1})\eta}{H}dxdy \\
+        &= \int_G f(y) \dpn{\pi(g)\xi, \pi_0(y^{-1})\eta}{H} dy\\
+        &= \int_G f(y) \dpn{\pi_0(y)\pi(g)\xi,\eta}{H} dy = \dpn{\pi(f)\pi(g)\xi, \eta}{H}
+    \end{align*}
+    
+    (1, *-Homomorphism): For any $f \in L^1(G; \complex)$ and $\xi, \eta \in H$, since $\pi_0$ is unitary,
+    \begin{align*}
+        \dpn{\pi(f^*)\xi, \eta}{H} &= \int_G \Delta_G(x^{-1})\ol{f(x^{-1})}\dpn{\pi_0(x)\xi, \eta}{H} dx \\
+        &= \int_G \ol{f(x)} \dpn{\pi_0(x^{-1})\xi, \eta}{H}dx = \int_G \ol{f(x)} \dpn{\xi, \pi_0(x)\eta}{H}dx \\
+        &= \int_G \ol{f(x)} \cdot \ol{\dpn{\pi_0(x)\eta, \xi}{H}} dx = \overline{\int_G f(x)\dpn{\pi_0(x)\eta, \xi}{H} dx} \\
+        &= \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 $\norm{\pi_0(x)\xi - \xi}_H < \norm{\xi}_H$ 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_0(x)\xi - \xi, \eta}{H} dx \\
+        |\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 $\angles{\pi(L_x\phi_\alpha)}_{\alpha \in A}$ 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 $\bracsn{\pi(L_x\phi_\alpha)|\alpha \in A}$ 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 $\pi_0$ 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*}
+        \dpn{\pi(f)\pi(\phi_\alpha)\xi, \eta}{H} &= \dpn{\pi(f * \phi_\alpha)\xi, \eta}{H} = \angles{\pi\paren{\int_G f(y)L_y\phi_\alpha dy}\xi, \eta}_{H} \\
+        &= \int_G f(y) \dpn{\pi(L_y\phi_\alpha)\xi, \eta}{H}dy \\
+        \dpn{\pi(f)\xi, \eta}{H} &= \int_G f(y)\dpn{\pi_0(y)\xi, \eta}{H}dy
+    \end{align*}
+\end{proof}
+
+\begin{theorem}
+\label{theorem:l1-group-von-neumann}
+    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 $\pi(L^1(G; \complex))$. 
+
+    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 $K = \bigsqcup_{k = 1}^m V_k$ 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*}
+        \abs{\angles{\braks{\pi(\one_K) - \sum_{k = 1}^m |V_k|\pi_0(x_k)}\xi_j, \eta_j}_{H}} &\le \sum_{k = 1}^m \int_{V_k}|\dpn{[\pi_0(y) - \pi_0(x_k)]\xi_j, \eta_j}{H}|dy \\
+        &\le \sum_{k = 1}^m \eps |V_k| = \eps |K|
+    \end{align*}
+
+    so $\pi(\one_K)$ is in the von Neumann algebra generated by $\pi_0(G)$. By inner regularity of the Haar measure, $\pi(\one_A)$ is in the von Neumann algebra generated by $\pi_0(G)$ for all $A \in \cb_G$ with $\mu(A) < \infty$. By \hyperref[density of simple functions]{proposition:lp-simple-dense}, $\pi(L^1(G; \complex))$ is in the von Neumann algebra generated by $\pi_0(G)$. 
+    
+\end{proof}
+
+

src/harmonic/rep/unit.tex

@@ -0,0 +1,122 @@
+\section{Unitary Representations of Locally Compact Groups}
+\label{section:lcg-unitary}
+
+\begin{definition}[Unitary Representation]
+\label{definition:lcg-unitary}
+    Let $G$ be a locally compact group, then a \textbf{unitary representation} of $G$ is a pair $(H, \pi)$, where $H$ is a non-zero complex Hilbert space and $\pi: G \to U(B(H))$ is a strong-operator continuous homomorphism. 
+\end{definition}
+
+\begin{definition}[Left Regular Representation]
+\label{definition:lcg-left-regular}
+    Let $G$ be a locally compact group, then the mapping
+    \[
+    \pi_L: G \to U(B(L^2(G))) \quad x \mapsto L_x
+    \]
+
+    is the \textbf{left regular representation} of $G$. 
+\end{definition}
+
+\begin{definition}[Intertwining Operator]
+\label{definition:lcg-intertwining}
+    Let $G$ be a locally compact group, $(H_1, \pi_1)$ and $(H_2, \pi_2)$ be unitary representations of $G$, and $T \in L(H_1; H_2)$, then $T$ is an \textbf{intertwining operator} for $\pi_1$ and $\pi_2$ if the following diagram commutes 
+    \[
+    \xymatrix{
+    H_1 \ar@{->}[d]_{\pi_1(x)} \ar@{->}[r]^{T} & H_2 \ar@{->}[d]^{\pi_2(x)} \\
+    H_1 \ar@{->}[r]_{T} & H_2
+    }
+    \]
+
+    for all $x \in G$. The set $\mathcal{C}(\pi_1, \pi_2)$ is the \textbf{space of intertwining operators} for $\pi_1$ and $\pi_2$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:lcg-intertwining}
+    Let $G$ be a locally compact group and $(H, \pi)$ be a unitary representation of $G$, then $\mathcal{C}(\pi, \pi) = \pi(G)'$, and hence a von Neumann algebra. 
+\end{lemma}
+
+\begin{definition}[Invariant Subspace]
+\label{definition:lcg-invariant-subspace}
+    Let $G$ be a locally compact group, $(H, \pi)$ be a unitary representation of $G$, $M \subset H$ be a closed subspace, and $P \in B(H)$ be the orthogonal projection onto $M$, then the following are equivalent:
+    \begin{enumerate}
+        \item For each $x \in G$, $\pi(x)(M) \subset M$. 
+        \item $P \in \pi(G)'$. 
+    \end{enumerate}
+
+    If the above holds, then $M$ is a \textbf{invariant subspace} for $\pi$. 
+\end{definition}
+\begin{proof}
+    Since $\pi$ is unitary, $\pi(x)^* = \pi(x^{-1})$ for all $x \in G$. As such, $\pi(G)$ is a self-adjoint subset of $G$, so $M$ is invariant for $\pi(G)$ if and only if it is a reducing subspace for $\pi(G)$, if and only if $\pi(G)$ commutes with $P$. 
+\end{proof}
+
+\begin{definition}[Subrepresentation]
+\label{definition:lcg-subrepresentation}
+    Let $G$ be a locally compact group, $(H, \pi)$ be a unitary representation of $G$, and $M \subset H$ be an invariant subspace, then the mapping
+    \[
+    \pi_M: G \to U(B(M)) \quad x \mapsto \pi(x)|_M
+    \]
+
+    is a \textbf{unitary subrepresentation} of $G$. 
+
+    If $\pi$ admits a non-trivial invariant subspace, then $\pi$ is \textbf{reducible}. Otherwise, it is \textbf{irreducible}. 
+\end{definition}
+
+\begin{definition}[Direct Sum]
+\label{definition:lcg-rep-direct-sum}
+    Let $G$ be a locally compact group, $\bracsn{(H_i, \pi_i)}_{i \in I}$ be unitary representations of $G$, then the unitary representation
+    \[
+    \pi: G \to B([l^2(I); H_i]) \quad \pi(x)(\xi)_i = \pi_i(x)(\xi_i)
+    \]
+
+    is the \textbf{direct sum} of $\bracsn{(H_i, \pi_i)}_{i \in I}$, denoted $\bigoplus_{i \in I}\pi_i$. 
+\end{definition}
+
+\begin{definition}[Cyclic]
+\label{definition:lcg-rep-cyclic}
+    Let $G$ be a locally compact group, $(H, \pi)$ be a unitary representation of $G$, and $\xi \in H$, then $\xi$ is a \textbf{cyclic vector} if $\text{span}\bracsn{\pi(x)\xi|x \in G}$ is dense in $H$. The representation $(H, \pi)$ is \textbf{cyclic} if it admits a cyclic vector. 
+\end{definition}
+
+
+
+\begin{proposition}
+\label{proposition:lcg-rep-invariant-gymnastics}
+    Let $G$ be a locally compact group, $(H, \pi)$ be a unitary representation of $G$, and $M \subset H$ be an invariant subspace, then:
+    \begin{enumerate}
+        \item $M^\perp$ is also an invariant subspace. 
+        \item $(H, \pi)$ is the direct sum of $(M, \pi_M)$ and $(M^\perp, \pi_{M^\perp})$. 
+    \end{enumerate}
+
+    Moreover, 
+    \begin{enumerate}[start=3]
+        \item $(H, \pi)$ is a direct sum of cyclic representations. 
+    \end{enumerate}
+\end{proposition}
+% Proof omitted.
+
+\begin{lemma}[Schur]
+\label{lemma:lcg-schur}
+    Let $G$ be a locally compact group and $(H, \pi), (K, \tau)$ be unitary representations of $G$, then
+    \begin{enumerate}
+        \item $(H, \pi)$ is irreducible if and only if the commutant $\pi(G)' = \complex I$. 
+        \item If $(H, \pi)$ and $(K, \tau)$ are irreducible, then $(H, \pi)$ and $(K, \tau)$ are unitarily equivalent if and only if $\mathcal{C}(\pi, \tau)$ is one-dimensional. Otherwise, $\mathcal{C}(\pi, \tau) = \bracsn{0}$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Theorem 3.5]{FollandHarmonic}}}. ]
+    (1): $(H, \pi)$ is irreducible if and only if $\pi(G)'$ contains no non-trivial projections. Since $\pi(G)'$ is a von Neumann algebra, \autoref{theorem:vn-projection-norm-dense} implies that the span of projections in $\pi(G)'$ is dense in $\pi(G)'$. Therefore $(H, \pi)$ is irreducible if and only if $\pi(G)' = \complex I$. 
+
+    (2): For any $T \in \mathcal{C}(\pi, \tau)$, $T^* \in \mathcal{C}(\tau, \pi)$, so $T^*T \in \mathcal{C}(\pi, \pi) = \pi(G)'$, and $TT^* \in \mathcal{C}(\tau, \tau) = \tau(G)'$. Given that $(H, \pi)$ and $(K, \tau)$ are irreducible, there exist $\lambda, \mu \in \complex$ such that $T^*T = \lambda I_H$ and $TT^* = \mu I_K$. As such, either $T = 0$ or $\lambda^{-1/2}T$\footnote{Pick any branch.} is unitary, so $\mathcal{C}(\pi, \tau)$ is non-zero if and only if $(H, \pi)$ and $(K, \tau)$ are unitarily equivalent. 
+
+    If $\mathcal{C}(\pi, \tau) \ne \bracsn{0}$, then $\mathcal{C}(\pi, \tau)$ consists of scalar multiples of unitary operators. In which case, for any unitary operators $S, T \in \mathcal{C}(\pi, \tau)$, $S^{-1}T = S^*T \in \mathcal{C}(\pi, \pi)$. Thus there exists $\lambda \in \complex$ such that $S^{-1}T = \lambda I_H$, so $T = \lambda S$, and $\mathcal{C}(\pi, \tau)$ is one-dimensional.
+\end{proof}
+
+\begin{corollary}
+\label{corollary:lcg-schur-abelian-irrep}
+    Let $G$ be a locally compact group. If $G$ is abelian, then every irreducible representation of $G$ is one-dimensional. 
+\end{corollary}
+\begin{proof}[Proof, {{\cite[Corollary 3.6]{FollandHarmonic}}}. ]
+    Let $(H, \pi)$ be an irreducible representation of $G$. Since $G$ is abelian, $\pi(G) \subset \pi(G)'$. By \hyperref[Schur's Lemma]{lemma:lcg-schur}, $\pi(G)' = \complex I$. As every one-dimensional subspace of $H$ is invariant, $H$ itself must be one-dimensional. 
+\end{proof}
+
+
+
+
+

src/op/c-star/gns.tex

@@ -106,7 +106,41 @@
     For each $x, y \in A$, $\dpn{x, y}{\phi} = \dpn{\pi_\phi(x)(1_A + N_\phi), \pi_\phi(y)(1_A + N_\phi)}{H_\phi}$ by well-definedness of the inner product on $H_\phi$.
 \end{proof}
 
-\begin{theorem}[Gelfand-Naimark-Segal]
+\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. 
+\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
+    \[
+    \dpn{x, \phi_1}{A} = \dpn{\pi(x)\xi_1, \xi_1}{H} \quad \dpn{x, \phi_2}{A} = \dpn{\pi(x)\xi_2, \xi_2}{H}
+    \]
+
+    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} \\
+        &= \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}
+    \end{align*}
+
+    so $\phi$ is a strict convex combination of two states, and is not pure.
+
+    ($\Leftarrow$): 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
+    \[
+    \dpn{T\pi(x)\xi, \pi(y)\xi}{H} = \dpn{x, y}{\phi_1} = \dpn{y^*x, \phi_1}{A}
+    \]
+    
+    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}
+    \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. 
+\end{proof}
+
+
+\begin{theorem}[Gelfand-Naimark]
 \label{theorem:gns}
     Let $A$ be a unital $C^*$-algebra, then:
     \begin{enumerate}

src/op/c-star/homomorphism.tex

@@ -17,6 +17,39 @@
 \end{definition}
 
 
+
+\begin{definition}[*-Representation]
+\label{definition:star-rep}
+    Let $A$ be an involutive Banach algebra, then a \textbf{*-representation} of $A$ is a pair $(H, \pi)$, where $H$ is a complex Hilbert space, and $\pi: A \to B(H)$ is a *-homomorphism. 
+\end{definition}
+
+
+\begin{definition}[Non-Degenerate]
+\label{definition:star-rep-nondegen}
+    Let $A$ be an involutive Banach algebra and $(H, \pi)$ be a *-representation of $A$, then the following are equivalent:
+    \begin{enumerate}
+        \item $\text{span}\bracsn{\pi(x)\xi|x \in A, \xi \in H}$ is dense in $H$. 
+        \item There exists no non-zero $\xi \in H$ such that $\pi(x)\xi = 0$ for all $x \in A$.
+    \end{enumerate}
+    
+    If the above holds, then $(H, \pi)$ is \textbf{non-degenerate}.
+\end{definition}
+\begin{proof}
+    $\neg (1) \Rightarrow \neg (2)$: Let $\xi \in \bracsn{\pi(x)\eta|x \in A, \eta \in H}^\perp$, then for any $x \in A$ and $\eta \in H$, 
+    \[
+    \dpn{\pi(x)\xi, \eta}{H} = \dpn{\xi, \pi(x^*)\eta}{H} = 0
+    \]
+
+    so $\pi(x)\xi = 0$ for all $x \in A$. 
+    
+    $\neg (2) \Rightarrow \neg (1)$: Let $\xi \in H \setminus \bracsn{0}$ with $\pi(x)\xi = 0$ for all $x \in A$, then for any $x \in A$ and $\eta \in H$, 
+    \[
+    \dpn{\pi(x)\eta, \xi}{H} = \dpn{\eta, \pi(x^*)\xi}{H} = 0
+    \]
+
+    so $\bracsn{\pi(x)\eta|x \in A, \eta \in H} \subset \bracsn{\xi}^\perp$, and cannot be dense in $H$.
+\end{proof}
+
 \begin{proposition}
 \label{proposition:star-homomorphism-contractive}
     Let $A, B$ be unital $C^*$-algebras and $\phi: A \to B$ be a unital *-homomorphism, then for each $x \in A$, 
@@ -31,7 +64,7 @@
     (2): By (1) and \autoref{corollary:c-star-unique-norm}, 
     \begin{align*}
     \norm{\phi(x)}_B^2 &= \sup\bracsn{|\lambda|\ | \lambda \in \sigma_B(\phi(x^*x))} \\
-    &\ge \sup\bracsn{|\lambda|\ | \lambda \in \sigma_A(x^*x)} = \norm{x}_A^2
+    &\le \sup\bracsn{|\lambda|\ | \lambda \in \sigma_A(x^*x)} = \norm{x}_A^2
     \end{align*}
     
 \end{proof}
@@ -44,14 +77,14 @@
     Let $y \in \ol{\Phi(A)} \cap B_{sa}$, then there exists $x \in A_{sa}$ such that $\norm{y - \Phi(x)}_B \le \norm{y}_B/2$. Let
     \[
     f: \complex \to \complex \quad z \mapsto \begin{cases}
-        z & |z| \le 2\norm{y}_F \\
+        z & |z| \le 2\norm{y}_B \\
-        2\norm{y}_F \cdot \sgn z = 2\norm{y}_F \cdot \frac{z}{|z|} & |z| \ge 2\norm{y}_F
+        2\norm{y}_B \cdot \sgn z = 2\norm{y}_B \cdot \frac{z}{|z|} & |z| \ge 2\norm{y}_B
     \end{cases}
     \]
 
-    then $f \in C(\complex; \complex)$. Since $\norm{\Phi(x)}_B \le \norm{y}_B + \norm{y - \Phi(x)}_B \le 2\norm{y}_B$, $\sigma_B(\Phi(x)) \subset \ol{B_\complex(0, 2\norm{y}_B)}$, and $f|_{\sigma_B(\Phi(x))}$ is the identity. Thus by the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $\Phi(x) = f(\Phi(x)) = \Phi(f(x))$. By the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-continuous}, $\sigma_A(f(x)) = f(\sigma_A(x))$. By \autoref{theorem:c-star-normal-spectral-radius}, $\norm{f(x)}_A = [f(x)]_{sp} \le \norm{f}_u = 2\norm{y}_F$. 
+    then $f \in C(\complex; \complex)$. Since $\norm{\Phi(x)}_B \le \norm{y}_B + \norm{y - \Phi(x)}_B \le 2\norm{y}_B$, $\sigma_B(\Phi(x)) \subset \ol{B_\complex(0, 2\norm{y}_B)}$, and $f|_{\sigma_B(\Phi(x))}$ is the identity. Thus by the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $\Phi(x) = f(\Phi(x)) = \Phi(f(x))$. By the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-continuous}, $\sigma_A(f(x)) = f(\sigma_A(x))$. By \autoref{theorem:c-star-normal-spectral-radius}, $\norm{f(x)}_A = [f(x)]_{sp} \le \norm{f}_u = 2\norm{y}_B$. 
 
-    The above setup implies that for every $y \in \ol{\Phi(A)} \cap B_{sa}$, there exists $z \in A_{sa}$ such that $\norm{y - \Phi(z)}_{B} \le \norm{y}_B/2$, and $\norm{z}_A \le 2\norm{y}_B$. By the \hyperref[method of successive approximations]{theorem:successive-approximation}, $\phi(A_{sa}) = \ol{\Phi(A)} \cap B_{sa}$. Therefore $\Phi(A) = \ol{\Phi(A)}$. 
+    The above setup implies that for every $y \in \ol{\Phi(A)} \cap B_{sa}$, there exists $z \in A_{sa}$ such that $\norm{y - \Phi(z)}_{B} \le \norm{y}_B/2$, and $\norm{z}_A \le 2\norm{y}_B$. By the \hyperref[method of successive approximations]{theorem:successive-approximation}, $\Phi(A_{sa}) = \ol{\Phi(A)} \cap B_{sa}$. Therefore $\Phi(A) = \ol{\Phi(A)}$. 
 \end{proof}
 
 \begin{definition}[Representation of $C^*$-Algebra]
@@ -61,7 +94,7 @@
 
 \begin{definition}[Unitary Equivalence]
 \label{definition:representation-unitary-equivalent}
-    Let $A$ be a $C^*$-algebra and $(H_1, \pi_1), (H_2, \pi_2)$ be representations of $A$, then $(H_1, \pi_1)$ and $(H_2, \pi_2)$ are \textbf{unitarily equivalent} if there exists an isometry $U \in L(H_1; H_2)$ such that the following diagram commutes
+    Let $A$ be a $C^*$-algebra and $(H_1, \pi_1), (H_2, \pi_2)$ be representations of $A$, then $(H_1, \pi_1)$ and $(H_2, \pi_2)$ are \textbf{unitarily equivalent} if there exists a unitary map $U \in L(H_1; H_2)$ such that the following diagram commutes
     \[
     \xymatrix{
     H_1 \ar@{->}[r]^{U} \ar@{->}[d]_{\pi_1(x)} & H_2 \ar@{->}[d]^{\pi_2(x)} \\
@@ -74,4 +107,75 @@
 
 
 
+\begin{definition}[Invariant Subspace]
+\label{definition:cstar-invariant-subspace}
+    Let $A$ be a $C^*$-algebra, $(H, \pi)$ be a representation of $A$, $M \subset H$ be a closed subspace, and $P \in B(H)$ be the orthogonal projection onto $M$, then the following are equivalent:
+    \begin{enumerate}
+        \item For each $x \in A$, $\pi(x)(M) \subset M$. 
+        \item $P \in \pi(A)'$. 
+    \end{enumerate}
+
+    If the above holds, then $M$ is an \textbf{invariant subspace} for $\pi$. 
+\end{definition}
+\begin{proof}
+    Since $\pi(A)$ is a self-adjoint subspace of $B(H)$, $M$ is invariant for $\pi(A)$ if and only if it is a reducing subspace for $\pi(A)$, if and only if $\pi(A)$ commutes with $P$. 
+\end{proof}
+
+\begin{definition}[Intertwining Operator]
+\label{definition:cstar-intertwining}
+    Let $A$ be a $C^*$-algebra, $(H_1, \pi_1)$ and $(H_2, \pi_2)$ be representations of $A$, and $T \in L(H_1; H_2)$, then $T$ is an \textbf{intertwining operator} for $\pi_1$ and $\pi_2$ if the following diagram commutes 
+    \[
+    \xymatrix{
+    H_1 \ar@{->}[d]_{\pi_1(x)} \ar@{->}[r]^{T} & H_2 \ar@{->}[d]^{\pi_2(x)} \\
+    H_1 \ar@{->}[r]_{T} & H_2
+    }
+    \]
+
+    for all $x \in A$. The set $\mathcal{C}(\pi_1, \pi_2)$ is the \textbf{space of intertwining operators} for $\pi_1$ and $\pi_2$. 
+
+    In particular, $\mathcal{C}(\pi_1, \pi_1) = \pi(A)'$, and hence is a von Neumann algebra.
+\end{definition}
+
+\begin{definition}[Direct Sum]
+\label{definition:cstar-rep-direct-sum}
+    Let $A$ be a $C^*$-algebra and $\bracsn{(H_i, \pi_i)}_{i \in I}$ be representations of $A$, then the representation
+    \[
+    \pi: A \to B([l^2(I); H_i]) \quad \pi(x)(\xi)_i = \pi_i(x)(\xi_i)
+    \]
+
+    is the \textbf{direct sum} of $\bracsn{(H_i, \pi_i)}_{i \in I}$, denoted $\bigoplus_{i \in I}\pi_i$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:cstar-rep-invariant-gymnastics}
+    Let $A$ be a $C^*$-algebra, $(H, \pi)$ be a unitary representation of $G$, and $M \subset H$ be an invariant subspace, then:
+    \begin{enumerate}
+        \item $M^\perp$ is also an invariant subspace. 
+        \item $(H, \pi)$ is the direct sum of $(M, \pi_M)$ and $(M^\perp, \pi_{M^\perp})$. 
+    \end{enumerate}
+
+    Moreover, 
+    \begin{enumerate}[start=3]
+        \item $(H, \pi)$ is a direct sum of cyclic representations. 
+    \end{enumerate}
+\end{proposition}
+% Proof omitted.
+
+\begin{lemma}[Schur]
+\label{lemma:cstar-schur}
+    Let $A$ be a $C^*$-algebra and $(H, \pi), (K, \tau)$ be unitary representations of $G$, then
+    \begin{enumerate}
+        \item $(H, \pi)$ is irreducible if and only if $\pi(A)' = \complex I$. 
+        \item If $(H, \pi)$ and $(K, \tau)$ are irreducible, then $(H, \pi)$ and $(K, \tau)$ are unitarily equivalent if and only if $\mathcal{C}(\pi, \tau)$ is one-dimensional. Otherwise, $\mathcal{C}(\pi, \tau) = \bracsn{0}$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Theorem 3.5]{FollandHarmonic}}}. ]
+    (1): $(H, \pi)$ is irreducible if and only if $\pi(A)'$ contains no non-trivial projections. Since $\pi(A)'$ is a von Neumann algebra, \autoref{theorem:vn-projection-norm-dense} implies that the span of projections in $\pi(A)'$ is dense in $\pi(A)'$. Therefore $(H, \pi)$ is irreducible if and only if $\pi(A)' = \complex I$. 
+
+    (2): For any $T \in \mathcal{C}(\pi, \tau)$, $T^* \in \mathcal{C}(\tau, \pi)$, so $T^*T \in \mathcal{C}(\pi, \pi) = \pi(A)'$, and $TT^* \in \mathcal{C}(\tau, \tau) = \tau(A)'$. Given that $(H, \pi)$ and $(K, \tau)$ are irreducible, there exist $\lambda, \mu \in \complex$ such that $T^*T = \lambda I_H$ and $TT^* = \mu I_K$. As such, either $T = 0$ or $\lambda^{-1/2}T$\footnote{Pick any branch.} is unitary, so $\mathcal{C}(\pi, \tau)$ is non-zero if and only if $(H, \pi)$ and $(K, \tau)$ are unitarily equivalent. 
+
+    If $\mathcal{C}(\pi, \tau) \ne \bracsn{0}$, then $\mathcal{C}(\pi, \tau)$ consists of scalar multiples of unitary operators. In which case, for any unitary operators $S, T \in \mathcal{C}(\pi, \tau)$, $S^{-1}T = S^*T \in \mathcal{C}(\pi, \pi)$. Thus there exists $\lambda \in \complex$ such that $S^{-1}T = \lambda I_H$, so $T = \lambda S$, and $\mathcal{C}(\pi, \tau)$ is one-dimensional.
+\end{proof}
+
+