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Invariant orthogonal complements in unitary representations

Statement

Let $G$ be a topological group, let $π : G \to U (H)$ be a strongly continuous unitary representation of $G$ on a complex Hilbert space $H$ (Strongly continuous unitary representations, invariant linear subspaces and intertwiners, Hilbert space), and let $M \subseteq H$ be a closed invariant subspace. Then the orthogonal complement $M^{⊥}$ (Orthogonality and the orthogonal complement) is a closed invariant subspace of $H$ . This assertion is choice free.

Facts & Assumptions

Given: a topological group $G$ , a strongly continuous unitary representation $π$ on a complex Hilbert space $H$ , and a closed invariant subspace $M \subseteq H$ .

[F1]

Each $π (g)$ is a bijective isometry with $π (g)^{- 1} = π (g)^{*}$ , so $⟨ π (g) x, y ⟩ = ⟨ x, π (g)^{- 1} y ⟩$ for all $x, y \in H$ ; and $M$ is invariant, meaning $π (g) M = M$ for every $g \in G$ . (Strongly continuous unitary representations, invariant linear subspaces and intertwiners)

[F2]

For a subset $S$ of an inner-product space, $S^{⊥} = {v : ⟨ v, s ⟩ = 0$ for every $s \in S}$ is a linear subspace, and orthogonality is symmetric. (Orthogonality and the orthogonal complement, Real and complex inner-product spaces and their induced length)

[F3]

The inner product is jointly continuous, so for each fixed $y$ the map $x \mapsto ⟨ x, y ⟩$ is continuous. (The inner product is jointly continuous)

[F4]

In a metric space open balls are open, arbitrary unions of open sets are open, and a set is closed exactly when its complement is open; consequently ${0}$ is closed in $C$ , since its complement is the union of the open balls $B (z, ∣ z ∣)$ over $z \neq 0$ . (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement)

[F5]

A closed set is the complement of an open set, and arbitrary intersections of closed sets are closed because arbitrary unions of open sets are open. (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison)

Proof

technique · direct

1.1F1F2

Let $x \in M^{⊥}$ , $y \in M$ and $g \in G$ . Then $⟨ π (g) x, y ⟩ = ⟨ x, π (g)^{- 1} y ⟩$ , and $π (g)^{- 1} y \in M$ because $π (g)^{- 1} M = M$ ; hence $⟨ π (g) x, y ⟩ = 0$ and $π (g) x \in M^{⊥}$ , so $π (g) M^{⊥} \subseteq M^{⊥}$ . Replacing $g$ by $g^{- 1}$ gives $M^{⊥} \subseteq π (g) M^{⊥}$ as well, so $π (g) M^{⊥} = M^{⊥}$ : the complement is invariant.

1.2F3F4F5

The complement of $M^{⊥}$ in $H$ is the union over $y \in M$ of the sets ${x : ⟨ x, y ⟩ \neq 0}$ , each of which is the preimage under the continuous map $x \mapsto ⟨ x, y ⟩$ of the open set $C ∖ {0}$ ; a union of open sets is open, so the complement of $M^{⊥}$ is open and $M^{⊥}$ is closed.

2.1F2step 1.1step 1.2∎

Together with the fact that $M^{⊥}$ is a linear subspace, steps 1.1 and 1.2 show that $M^{⊥}$ is a closed invariant subspace of $H$ , with no use of any choice principle.

Depends on

Used by

Dependency tree · two levels

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Sources

Emmanuel Kowalski, An Introduction to the Representation Theory of Groups, §§5.2–5.6 (standard reference, not scraped)
Vera Serganova, Representation Theory, Chapter III §§1.6–2.1 (standard reference, not scraped)