32.1 Unitary Representations of Locally Compact Groups

Definition 32.1.1 (Unitary Representation).label Let $G$ be a locally compact group, then a 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.

Definition 32.1.2 (Left Regular Representation).label 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 left regular representation of $G$.

Definition 32.1.3 (Intertwining Operator).label 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 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 space of intertwining operators for $\pi_{1}$ and $\pi_{2}$.

Lemma 32.1.4.label 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.

Definition 32.1.5 (Invariant Subspace).label 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:

  1. (1)

    For each $x \in G$, $\pi(x)(M) \subset M$.

  2. (2)

    $P \in \pi(G)'$.

If the above holds, then $M$ is a invariant subspace for $\pi$.

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 $P$.$\square$

Definition 32.1.6 (Subrepresentation).label 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 unitary subrepresentation of $G$.

If $\pi$ admits a non-trivial invariant subspace, then $\pi$ is reducible. Otherwise, it is irreducible.

Definition 32.1.7 (Direct Sum).label 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 direct sum of $\bracsn{(H_i, \pi_i)}_{i \in I}$, denoted $\bigoplus_{i \in I}\pi_{i}$.

Definition 32.1.8 (Cyclic).label Let $G$ be a locally compact group, $(H, \pi)$ be a unitary representation of $G$, and $\xi \in H$, then $\xi$ is a cyclic vector if $\text{span}\bracsn{\pi(x)\xi|x \in G}$ is dense in $H$. The representation $(H, \pi)$ is cyclic if it admits a cyclic vector.

Proposition 32.1.9.label Let $G$ be a locally compact group, $(H, \pi)$ be a unitary representation of $G$, and $M \subset H$ be an invariant subspace, then:

  1. (1)

    $M^{\perp}$ is also an invariant subspace.

  2. (2)

    $(H, \pi)$ is the direct sum of $(M, \pi_{M})$ and $(M^{\perp}, \pi_{M^\perp})$.

Moreover,

  1. (4)

    $(H, \pi)$ is a direct sum of cyclic representations.

Lemma 32.1.10 (Schur).label Let $G$ be a locally compact group and $(H, \pi), (K, \tau)$ be unitary representations of $G$, then

  1. (1)

    $(H, \pi)$ is irreducible if and only if the commutant $\pi(G)' = \complex I$.

  2. (2)

    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, $\mathcal{C}(\pi, \tau) = \bracsn{0}$.

Proof, [Theorem 3.5, Fol16]. (1): $(H, \pi)$ is irreducible if and only if $\pi(G)'$ contains no non-trivial projections. Since $\pi(G)'$ is a von Neumann algebra, Theorem 39.6.2 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 $\lambda^{-1/2}T$[1] 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.$\square$

Corollary 32.1.11.label Let $G$ be a locally compact group. If $G$ is abelian, then every irreducible representation of $G$ is one-dimensional.

Proof, [Corollary 3.6, Fol16]. Let $(H, \pi)$ be an irreducible representation of $G$. Since $G$ is abelian, $\pi(G) \subset \pi(G)'$. By Schur’s Lemma, $\pi(G)' = \complex I$. As every one-dimensional subspace of $H$ is invariant, $H$ itself must be one-dimensional.$\square$

  1. Pick any branch.keyboard_return

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