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Haar averaging projects contractively onto the bounded intertwiners

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let K be a compact Hausdorff group with normalized Haar probability measure μ (Normalized Haar probability on a compact group), and let π:K→U(H) and σ:K→U(J) be strongly continuous unitary representations of K on complex Hilbert spaces H and J (Strongly continuous unitary representations, invariant linear subspaces and intertwiners, Hilbert space). Write Hom⁡K(H,J):={T∈B(H,J):Tπ(g)=σ(g)T for every g∈K} for the space of bounded intertwiners and let A:B(H,J)→B(H,J) be the Haar averaging operator of Haar averaging of bounded operators as a weak operator integral, so that ⟨A(T)v,w⟩=∫K⟨σ(k)Tπ(k)−1v,w⟩ dμ(k) for all T∈B(H,J), v∈H, w∈J. Then:

  1. A is idempotent, A∘A=A, and its range is exactly Hom⁡K(H,J): A(T) intertwines for every bounded T, and A(T)=T for every bounded intertwiner T;
  2. A is a contraction, ∥A(T)∥≤∥T∥ for every T∈B(H,J), hence ∥A∥≤1;
  3. ∥A∥=1 if Hom⁡K(H,J)≠{0}, and ∥A∥=0 if Hom⁡K(H,J)={0}.

Facts & Assumptions

Given: AC, a compact Hausdorff group K with normalized Haar probability μ, strongly continuous unitary representations π on H and σ on J, the averaging map A of the definition item, and a bounded linear operator T∈B(H,J).

[F1]

The operator A(T) is well defined by the weak operator integral: for all v∈H and w∈J the displayed pairing formula holds; the integrand is continuous on K, hence bounded and integrable against μ; A is linear in T; and ∥A(T)∥≤∥T∥ (Haar averaging of bounded operators as a weak operator integral, A bounded linear operator between normed spaces). The definition uses Countable Choice, which follows from AC, through the Riesz representation theorem for J (Riesz representation for Hilbert spaces, Real and complex inner-product spaces and their induced length, AC supplies the countable and dependent choices used in Banach integration).

[F2]

The normalized Haar probability is left invariant and μ(K)=1; for h∈K the left translation Th(k):=hk is a measurable self-map with μ(Th−1E)=μ(h−1E)=μ(E), hence measure preserving, so ∫Kf(hk) dμ(k)=∫Kf(k) dμ(k) for every integrable f (Normalized Haar probability on a compact group, Measure-preserving transformations and systems, Integral invariance under measure-preserving maps, Measure spaces).

[F3]

π and σ are group homomorphisms into the unitary groups, so π(e)=idH, π(h−1k)−1=π(k)−1π(h), σ(hk)=σ(h)σ(k), and σ(h)−1=σ(h−1); each σ(h) is unitary with ⟨σ(h)x,y⟩=⟨x,σ(h)−1y⟩. A bounded operator T:H→J is an intertwiner, written T∈Hom⁡K(H,J), exactly when Tπ(g)=σ(g)T for every g∈K (Strongly continuous unitary representations, invariant linear subspaces and intertwiners).

[F4]

For bounded operators ∥Bv∥≤∥B∥ ∥v∥ for every vector v, the operator norm is the supremum of ∥Bv∥ over the closed unit ball (also when the domain is zero), and ∥B∥=0 exactly when B=0 (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
1.1F1F2F3

Let T∈B(H,J), h∈K, v∈H and w∈J. Using unitarity of σ(h) and the pairing formula of [F1], ⟨σ(h)A(T)v,w⟩=⟨A(T)v,σ(h)−1w⟩=∫K⟨σ(k)Tπ(k)−1v,σ(h)−1w⟩ dμ(k)=∫K⟨σ(h)σ(k)Tπ(k)−1v,w⟩ dμ(k)=∫K⟨σ(hk)Tπ(k)−1v,w⟩ dμ(k) by the homomorphism property of [F3]. Substituting k=h−1k′ and using left invariance of μ as recorded in [F2], this equals ∫K⟨σ(k′)Tπ(h−1k′)−1v,w⟩ dμ(k′)=∫K⟨σ(k′)Tπ(k′)−1π(h)v,w⟩ dμ(k′)=⟨A(T)π(h)v,w⟩, by [F3] and the pairing formula of [F1] again. As v,w are arbitrary, σ(h)A(T)=A(T)π(h), and as h is arbitrary, A(T) is an intertwiner.

1.2F1F3

Let T∈Hom⁡K(H,J) and k∈K. Then σ(k)Tπ(k)−1=σ(k)σ(k)−1T=T by the intertwining relation and the homomorphism properties of [F3]. Hence the integrand of the pairing formula is the constant k↦⟨Tv,w⟩, whose integral against the probability measure μ is again ⟨Tv,w⟩; therefore ⟨A(T)v,w⟩=⟨Tv,w⟩ for all v,w and A(T)=T.

1.3F1F4

For every T∈B(H,J) the definition of A gives ∥A(T)∥≤∥T∥ by [F1], so the operator norm of the linear map A satisfies ∥A∥≤1 by the supremum description in [F4].

2.1F1step 1.1step 1.2

Let T∈B(H,J). By step 1.1 the operator A(T) lies in Hom⁡K(H,J), and step 1.2 applied to that intertwiner gives A(A(T))=A(T). Hence A∘A=A, so A is idempotent, and range⁡(A)⊆Hom⁡K(H,J); conversely every S∈Hom⁡K(H,J) satisfies S=A(S) by step 1.2, so Hom⁡K(H,J)⊆range⁡(A). Thus the range of A is exactly Hom⁡K(H,J).

3.1F4step 1.2step 1.3step 2.1∎

If Hom⁡K(H,J)≠{0}, choose a nonzero intertwiner S. Then S=A(S) by step 1.2, so ∥S∥=∥A(S)∥≤∥A∥ ∥S∥ by [F4], and since ∥S∥≠0 this gives ∥A∥≥1; with step 1.3, ∥A∥=1. If instead Hom⁡K(H,J)={0}, then range⁡(A)={0} by step 2.1, so A(T)=0 for every T and ∥A∥=0 by [F4]. Together with the contraction bound this proves the precise norm statement.

Depends on

Used by

Cited to discharge well-definedness by Haar averaging of bounded operators as a weak operator integral.

Dependency tree · two levels

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