First draft of L1 representations.
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Bokuan Li
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\section{Unitary Representations of Locally Compact Groups}
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\label{section:lcg-unitary}
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\begin{definition}[Unitary Representation]
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\label{definition:lcg-unitary}
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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.
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\end{definition}
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\begin{definition}[Left Regular Representation]
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\label{definition:lcg-left-regular}
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Let $G$ be a locally compact group, then the mapping
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\[
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\pi_L: G \to U(B(L^2(G))) \quad x \mapsto L_x
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\]
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is the \textbf{left regular representation} of $G$.
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\end{definition}
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\begin{definition}[Intertwining Operator]
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\label{definition:lcg-intertwining}
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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
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\[
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\xymatrix{
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H_1 \ar@{->}[d]_{\pi_1(x)} \ar@{->}[r]^{T} & H_2 \ar@{->}[d]^{\pi_1(x)} \\
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H_1 \ar@{->}[r]_{T} & H_2
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}
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\]
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for all $x \in G$. The set $\mathcal{C}(\pi_1, \pi_2)$ is the \textbf{space of intertwining operators} for $\pi_1$ and $\pi_2$.
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\end{definition}
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\begin{lemma}
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\label{lemma:lcg-intertwining}
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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.
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\end{lemma}
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\begin{definition}[Invariant Subspace]
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\label{definition:lcg-invariant-subspace}
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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:
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\begin{enumerate}
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\item For each $x \in G$, $\pi(x)(M) \subset M$.
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\item $P \in \pi(G)'$.
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\end{enumerate}
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If the above holds, then $M$ is a \textbf{invariant subspace} for $\pi$.
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\end{definition}
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\begin{proof}
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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 $P$.
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\end{proof}
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\begin{definition}[Subrepresentation]
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\label{definition:lcg-subrepresentation}
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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
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\[
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\pi_M: G \to U(B(M)) \quad x \mapsto \pi(x)|_M
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\]
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is a \textbf{unitary subrepresentation} of $G$.
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If $\pi$ admits a non-trivial invariant subspace, then $\pi$ is \textbf{reducible}. Otherwise, it is \textbf{irreducible}.
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\end{definition}
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\begin{definition}[Direct Sum]
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\label{definition:lcg-rep-direct-sum}
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Let $G$ be a locally compact group, $\bracsn{(H_i, \pi_i)}_{i \in I}$ be unitary representations of $G$, then the unitary representation
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\[
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\pi: G \to B([l^2(I); H_i]) \quad \pi(x)(\xi)_i = \pi_i(x)(\xi_i)
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\]
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is the \textbf{direct sum} of $\bracsn{(H_i, \pi_i)}_{i \in I}$, denoted $\bigoplus_{i \in I}\pi_i$.
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\end{definition}
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\begin{definition}[Cyclic]
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\label{definition:lcg-rep-cyclic}
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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.
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\end{definition}
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\begin{proposition}
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\label{proposition:lcg-rep-invariant-gymnastics}
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Let $G$ be a locally compact group, $(H, \pi)$ be a unitary representation of $G$, and $M \subset H$ be an invariant subspace, then:
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\begin{enumerate}
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\item $M^\perp$ is also an invariant subspace.
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\item $(H, \pi)$ is the direct sum of $(M, \pi_M)$ and $(M^\perp, \pi_{M^\perp})$.
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\end{enumerate}
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Moreover,
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\begin{enumerate}[start=3]
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\item $(H, \pi)$ is a direct sum of cyclic representations.
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\end{enumerate}
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\end{proposition}
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% Proof omitted.
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\begin{lemma}[Schur]
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\label{lemma:lcg-schur}
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Let $G$ be a locally compact group and $(H, \pi), (K, \tau)$ be unitary representations of $G$, then
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\begin{enumerate}
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\item $(H, \pi)$ is irreducible if and only if the commutant $\pi(G)' = \complex I$.
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\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, $\mathcal{C}(\pi, \tau) = \bracsn{0}$.
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\end{enumerate}
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\end{lemma}
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\begin{proof}[Proof, {{\cite[Theorem 3.5]{FollandHarmonic}}}. ]
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(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 $\pi(G)' = \complex I$.
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(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 exists $\lambda, \mu \in \complex$ 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.
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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 $S, T \in \mathcal{C}(\pi, \tau)$, $S^{-1}T = S^*T \in \mathcal{C}(\pi)$. 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.
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\end{proof}
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\begin{corollary}
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\label{corollary:lcg-schur-abelian-irrep}
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Let $G$ be a locally compact group. If $G$ is abelian, then every irreducible representation of $G$ is one-dimensional.
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\end{corollary}
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\begin{proof}[Proof, {{\cite[Corollary 3.6]{FollandHarmonic}}}. ]
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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.
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\end{proof}
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