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