\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}