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Finite-rank conjugation orbits are operator-norm continuous

Statement

Assume the Axiom of Choice. Let $G$ be a topological group and let $π : G \to U (H)$ be a strongly continuous unitary representation on a complex Hilbert space $H$ (Strongly continuous unitary representations, invariant linear subspaces and intertwiners, Hilbert space). Let $T \in B (H)$ be a bounded linear operator (A bounded linear operator between normed spaces) whose range $T (H)$ is finite dimensional (that is, $T$ is a finite-rank operator). Then the conjugation orbit

$g ⟼ π (g) T π (g)^{- 1}$

is continuous from $G$ to $B (H)$ for the operator norm (The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

Facts & Assumptions

Given: a topological group $G$ , a strongly continuous unitary representation $π$ on a complex Hilbert space $H$ , and a bounded finite-rank operator $T$ .

[A1]

The Axiom of Choice is assumed. (The Axiom of Choice)

[F1]

For every $v \in H$ the orbit map $g \mapsto π (g) v$ is norm continuous, and each $π (g)$ is a bijective isometry, so $π (g)^{- 1} = π (g)^{*}$ and $∥ π (g) v ∥ = ∥ v ∥$ . (Strongly continuous unitary representations, invariant linear subspaces and intertwiners)

[F2]

Countable Choice holds under AC, and it is the hypothesis of the Riesz representation theorem. (AC supplies the countable and dependent choices used in Banach integration, Riesz representation for Hilbert spaces)

[F3]

A finite-dimensional inner product space has an orthonormal basis, and for an orthonormal basis $e_{0}, \dots, e_{r - 1}$ of a finite-dimensional subspace every vector $v$ of that subspace satisfies $v = \sum_{i < r} ⟨ v, e_{i} ⟩ e_{i}$ . (Every finite-dimensional real or complex inner product space has an orthonormal basis, Bessel's inequality for a finite orthonormal list and Parseval's identity for an orthonormal basis)

[F4]

Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ . (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs)

[F5]

The operator norm is a bound and a least bound: $∥ S x ∥ \leq ∥ S ∥ ∥ x ∥$ , and $∥ S ∥$ is the least such constant. (The operator norm as the least bound and as the unit-sphere or unit-ball supremum, A bounded linear operator between normed spaces)

Proof

technique · direct

1.1A1F2

The assumed Axiom of Choice supplies Countable Choice, so the Riesz representation theorem is available for bounded linear functionals on $H$ .

1.2F3

Since $T (H)$ is finite dimensional, it has an orthonormal basis $a_{0}, \dots, a_{r - 1}$ , and for every $x \in H$ the vector $T x$ lies in $T (H)$ , so $T x = \sum_{i < r} ⟨ T x, a_{i} ⟩ a_{i}$ .

1.3F1F4F5

For fixed $a, b \in H$ the rank-one operator $R_{a, b} x : = ⟨ x, b ⟩ a$ is bounded with $∥ R_{a, b} x ∥ \leq ∥ a ∥ ∥ b ∥ ∥ x ∥$ by Cauchy–Schwarz, so $∥ R_{a, b} ∥ \leq ∥ a ∥ ∥ b ∥$ ; and for $g \in G$ one has $π (g) R_{a, b} π (g)^{- 1} = R_{π (g) a, π (g) b}$ , because $⟨ π (g)^{- 1} x, b ⟩ = ⟨ x, π (g) b ⟩$ and $π (g)$ is linear.

2.1F2F4step 1.1step 1.2

Each functional $x \mapsto ⟨ T x, a_{i} ⟩$ is bounded, since $∣ ⟨ T x, a_{i} ⟩ ∣ \leq ∥ T ∥ ∥ x ∥$ ; by Riesz representation there is for each $i$ a unique vector $b_{i} \in H$ with $⟨ T x, a_{i} ⟩ = ⟨ x, b_{i} ⟩$ for all $x$ , so $T = \sum_{i < r} ⟨ \cdot, b_{i} ⟩ a_{i} = \sum_{i < r} R_{a_{i}, b_{i}}$ .

3.1step 1.3step 2.1

Conjugating the finite sum of step 2.1 and using step 1.3 termwise gives $π (g) T π (g)^{- 1} = \sum_{i < r} π (g) R_{a_{i}, b_{i}} π (g)^{- 1} = \sum_{i < r} R_{π (g) a_{i}, π (g) b_{i}}$ for every $g \in G$ .

4.1F1F4F5step 3.1

For $g, h \in G$ and vectors $a, b, a^{'}, b^{'}$ , $R_{a, b} - R_{a^{'}, b^{'}} = R_{a - a^{'}, b} + R_{a^{'}, b - b^{'}}$ , so $∥ R_{a, b} - R_{a^{'}, b^{'}} ∥ \leq ∥ a - a^{'} ∥ ∥ b ∥ + ∥ a^{'} ∥ ∥ b - b^{'} ∥$ ; applying this with $a = π (g) a_{i}$ , $b = π (g) b_{i}$ , $a^{'} = π (h) a_{i}$ , $b^{'} = π (h) b_{i}$ and using unitarity bounds $∥ π (g) T π (g)^{- 1} - π (h) T π (h)^{- 1} ∥$ by $\sum_{i < r} (∥ b_{i} ∥ ∥ π (g) a_{i} - π (h) a_{i} ∥ + ∥ a_{i} ∥ ∥ π (g) b_{i} - π (h) b_{i} ∥)$ .

5.1F1step 4.1∎

The finitely many orbit maps $g \mapsto π (g) a_{i}$ and $g \mapsto π (g) b_{i}$ are norm continuous at $h$ by strong continuity, so the bound of step 4.1 tends to zero as $g \to h$ ; hence $g \mapsto π (g) T π (g)^{- 1}$ is continuous in operator norm at every $h \in G$ .

Depends on

Used by

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Sources

Vera Serganova, Representation Theory, Chapter III §§1.6–2.1 (standard reference, not scraped)