DefinitionDefinition: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30

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Matrix coefficient of a unitary representation

Definition

Let $π : G \to U (H)$ be a group homomorphism and let $ξ, η \in H$ . Its matrix coefficient associated with $(ξ, η)$ is $c_{ξ, η} : G \to C, c_{ξ, η} (g) = ⟨ π (g) ξ, η ⟩ .$ We use the convention that the Hilbert pairing is linear in its first argument and conjugate-linear in its second. Thus $c_{ξ, η}$ is linear in $ξ$ and conjugate-linear in $η$ .

When $G$ is a topological group and $π$ is strongly continuous, every such coefficient is continuous:

Facts & Assumptions

[A1]

If $G$ is a topological group and $π$ is strongly continuous, then for every $v \in H$ the orbit map $g \mapsto π (g) v$ is norm-continuous (Strongly continuous unitary representations, invariant linear subspaces and intertwiners).

[A2]

For vectors $x, y$ in a complex inner-product space, $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

[A3]

Continuity of a map into $C$ is measured with the metric $d_{C} (z, w) = ∣ z - w ∣$ (The Euclidean metric, convergence, Cauchy sequences, and continuity on the complex plane).

Proof

Given: A group homomorphism $π : G \to U (H)$ and vectors $ξ, η \in H$ ; for the continuity assertion, assume that $G$ is a topological group and $π$ is strongly continuous.

Proof technique: direct.

1.1A1

Fix $g_{0} \in G$ . By strong continuity, the orbit map for $ξ$ is continuous at $g_{0}$ , so $∥ π (g) ξ - π (g_{0}) ξ ∥ \to 0$ as $g \to g_{0}$ .

2.1A2A3step 1.1∎

Linearity in the first argument and Cauchy–Schwarz give $∣ c_{ξ, η} (g) - c_{ξ, η} (g_{0}) ∣ = ∣ ⟨ π (g) ξ - π (g_{0}) ξ, η ⟩ ∣ \leq ∥ π (g) ξ - π (g_{0}) ξ ∥ ∥ η ∥ \to 0$ by step 1.1. By [A3] this is continuity of the coefficient at the arbitrary point $g_{0}$ , hence on $G$ . This also holds when $η = 0$ .

Matrix coefficient of a unitary representation

Definition

Facts & Assumptions

Proof

Depends on

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