Proposition 7

Let G be a topological group and let $\varrho$ be a unitary representation of G in a complex Hilbert space E. The representation $\varrho$ is semisimple if and only if every non-zero subrepresentation of $\varrho$ contains an irreducible subrepresentation.

Suppose that $\varrho$ is semisimple. Let $(F_i)_{i\in I}$ be a family of irreducible subrepresentations of $\varrho$ such that E is the Hilbert sum of the subspaces $F_i$. Let F be a nonzero subrepresentation of E. There exists $i\in I$ such that the restriction to F of the orthogonal projector with image $F_i$ is nonzero. This orthogonal projector then defines by passing to subspaces a nonzero G-morphism $p_i$ of F into $F_i$ (Prop. 4 of V, p. 383). By Schur’s Lemma, the G-morphism $p_i$ is surjective (Cor. 5 of V, p. 388). The orthogonal complement in F of the kernel of $p_i$ is a subrepresentation of F isomorphic to $F_i$ (Prop. 3 of V, p. 383), hence irreducible.

Let us prove the converse assertion. Let $\mathscr{F}$ be the set of closed subspaces F of E stable under G such that the subrepresentation of $\varrho$ in F is irreducible. Let $\mathscr{O}$ be the set of subsets $\mathscr{G}$ of $\mathscr{F}$ such that the subspaces belonging to $\mathscr{G}$ are pairwise orthogonal.

The set $\mathscr{O}$ is of finite character (E, III, p. 34, Def. 2). Let then $\mathscr{G}$ be a maximal element of $\mathscr{O}$ (E, III, p. 35, Th. 1). Let $E_1$ be the subspace of E which is the Hilbert sum of the irreducible subrepresentations F of $\mathscr{G}$. If one had $E_1\not = E$, the orthogonal complement of $E_1$ would not be zero, and the subrepresentation of G in $E^{\circ}_1$ would contain by hypothesis an irreducible subrepresentation. The space F of the latter would be orthogonal to the elements of $\mathscr{G}$, so that $\mathscr{G}\cup \{F\} \in \mathscr{O}$; this contradicts the maximality of $\mathscr{G}$, therefore $E = E_1$, which completes the proof.