TheoremStatement: Literature-sourcedProof: Literature-sourcedPipeline-generatedprecheck passaudited 2026-10-02

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Schur orthogonality for general compact groups

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $K$ be a compact Hausdorff topological group (Topological group: multiplication and inversion are continuous, Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not) with normalized Haar probability measure $μ$ (Normalized Haar probability on a compact group), and let $π : K \to U (V)$ and $ρ : K \to U (W)$ be irreducible strongly continuous unitary representations of $K$ on nonzero complex Hilbert spaces $V$ and $W$ (Strongly continuous unitary representations, invariant linear subspaces and intertwiners, Hilbert space). Let $d : = \dim_{C} V$ be the degree of $π$ , so that $d < \infty$ and $d \geq 1$ . Matrix coefficients are written $k \mapsto ⟨ π (k) v, w ⟩$ and $k \mapsto ⟨ ρ (k) v^{'}, w^{'} ⟩$ with the pairing linear in the first variable (Matrix coefficient of a unitary representation, Real and complex inner-product spaces and their induced length). Then:

(Inequivalent pair.) If $π$ and $ρ$ are not unitarily equivalent (Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces), then $\int_{K} ⟨ π (k) v, w ⟩ \overline{⟨ ρ (k) v^{'}, w^{'} ⟩} d μ (k) = 0$ for all $v, w \in V$ and all $v^{'}, w^{'} \in W$ : the matrix coefficients of inequivalent irreducibles are orthogonal in $L^{2} (K)$ .
(A single irreducible.) For all $v, w, v^{'}, w^{'} \in V$ , $\int_{K} ⟨ π (k) v, w ⟩ \overline{⟨ π (k) v^{'}, w^{'} ⟩} d μ (k) = \frac{1}{d} ⟨ v, v^{'} ⟩ \overline{⟨ w, w^{'} ⟩} .$ The case $d = 1$ is included.
(Equivalent models.) If $U : V \to W$ is a unitary intertwiner, that is $U π (k) = ρ (k) U$ for every $k \in K$ , then $\int_{K} ⟨ π (k) v, w ⟩ \overline{⟨ ρ (k) U v^{'}, U w^{'} ⟩} d μ (k) = \frac{1}{d} ⟨ v, v^{'} ⟩ \overline{⟨ w, w^{'} ⟩}$ for all $v, w, v^{'}, w^{'} \in V$ .

Facts & Assumptions

Given: AC; a compact Hausdorff group $K$ with normalized Haar probability $μ$ ; irreducible strongly continuous unitary representations $π$ on the nonzero complex Hilbert space $V$ and $ρ$ on the nonzero complex Hilbert space $W$ ; vectors $v, w \in V$ and $v^{'}, w^{'} \in W$ .

[F1]

Finite dimensionality: the representation spaces of irreducible strongly continuous unitary representations of $K$ are finite dimensional, so $d : = \dim V < \infty$ and $\dim W < \infty$ , and $d \geq 1$ because $V \neq {0}$ (Irreducible unitary representations of compact groups are finite dimensional, Strongly continuous unitary representations, invariant linear subspaces and intertwiners).

[F2]

Schur's lemma: every bounded self-intertwiner of an irreducible strongly continuous unitary representation is a scalar multiple of the identity, and a nonzero bounded intertwiner between two such representations forces them to be unitarily equivalent; hence inequivalent irreducibles admit no nonzero bounded intertwiner (Schur lemma for complex unitary representations).

[F3]

Haar averaging: the weak operator integral $⟨ A (T) x, z ⟩ = \int_{K} ⟨ σ (k) T π_{0} (k)^{- 1} x, z ⟩ d μ (k)$ defines a linear contraction $A$ on the bounded operators between the carrier spaces of strongly continuous unitary representations $π_{0}$ and $σ$ , and its range is exactly the space of bounded intertwiners (Haar averaging of bounded operators as a weak operator integral, Haar averaging projects contractively onto the bounded intertwiners); here $A$ is linear and $∥ A (T) ∥ \leq ∥ T ∥$ (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[F4]

Trace: for a finite-dimensional complex vector space with an orthonormal basis $e_{1}, \dots, e_{d}$ , the trace of an endomorphism $T$ satisfies $tr (T) = \sum_{i = 1}^{d} ⟨ T e_{i}, e_{i} ⟩$ and $tr (c I) = c d$ ; similar endomorphisms have equal trace, so $tr (π (k) T π (k)^{- 1}) = tr (T)$ for every $k$ (The basis-independent trace of an endomorphism of a finite-dimensional vector space, Similar matrices have the same trace, Every finite-dimensional real or complex inner product space has an orthonormal basis).

[F5]

Orthonormal expansions: for an orthonormal basis $e_{1}, \dots, e_{d}$ of a finite-dimensional inner product space and any vector $y$ , one has $y = \sum_{i = 1}^{d} ⟨ y, e_{i} ⟩ e_{i}$ ; the pairing is linear in the first variable and conjugate-linear in the second, so a finite sum pulls out of the first variable, $⟨ \sum_{i = 1}^{d} c_{i} e_{i}, z ⟩ = \sum_{i = 1}^{d} c_{i} ⟨ e_{i}, z ⟩$ (Bessel's inequality for a finite orthonormal list and Parseval's identity for an orthonormal basis, Real and complex inner-product spaces and their induced length).

[F6]

Scalar integral: for a probability measure, $\int_{K} 1 d μ = 1$ , finite sums and scalar multiples of integrable functions integrate termwise (The Lebesgue integral is linear on $L^{1} (μ)$ , Integrable real and complex functions, and their integrals).

[F7]

A rank-one operator $T x = ⟨ x, v^{'} ⟩ v$ is linear (Linear map between vector spaces over the same field) and bounded with $∥ T x ∥ \leq ∥ v^{'} ∥ ∥ x ∥ ∥ v ∥$ by the Cauchy--Schwarz inequality (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

[F8]

A unitary intertwiner $U : V \to W$ is a bijective linear isometry satisfying $U π (k) = ρ (k) U$ ; hence $⟨ ρ (k) U v^{'}, U w^{'} ⟩ = ⟨ U π (k) v^{'}, U w^{'} ⟩ = ⟨ π (k) v^{'}, w^{'} ⟩$ (Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces, Real and complex inner-product spaces and their induced length).

Proof

technique · direct

1.1F1

Fix $v, w \in V$ and $v^{'}, w^{'} \in W$ . By [F1] the dimensions $d = \dim V$ and $\dim W$ are finite and $d \geq 1$ .

1.2F2F3F7

Inequivalent case. Assume first that $π$ and $ρ$ are not unitarily equivalent, and define $T : W \to V$ by $T x : = ⟨ x, v^{'} ⟩ v$ . By [F7] the operator $T$ is bounded of rank one. Applying the Haar averaging operator $A$ of [F3] to the pair $(ρ, π)$ of representations, the operator $A (T) : W \to V$ is a bounded intertwiner, so $A (T) ρ (k) = π (k) A (T)$ for every $k$ . Since $π$ and $ρ$ are inequivalent irreducible representations, Schur's lemma [F2] forces $A (T) = 0$ .

1.3F2F3F7

Same-representation case. Now take $ρ = π$ on $W = V$ and $T : V \to V$ , $T x = ⟨ x, v^{'} ⟩ v$ as before, a bounded finite-rank operator. The average $A (T) : V \to V$ is a bounded self-intertwiner of the irreducible representation $π$ , so Schur's lemma [F2] provides a scalar $c$ with $A (T) = c I$ .

1.4F4F5

The rank-one trace is $tr (T) = ⟨ v, v^{'} ⟩$ : in the orthonormal basis $e_{1}, \dots, e_{d}$ , the matrix of $T$ has entries $⟨ T e_{j}, e_{i} ⟩ = ⟨ e_{j}, v^{'} ⟩ ⟨ v, e_{i} ⟩$ , so $tr (T) = \sum_{i = 1}^{d} ⟨ e_{i}, v^{'} ⟩ ⟨ v, e_{i} ⟩ = \sum_{i = 1}^{d} ⟨ v, e_{i} ⟩ ⟨ e_{i}, v^{'} ⟩ = ⟨ \sum_{i = 1}^{d} ⟨ v, e_{i} ⟩ e_{i}, v^{'} ⟩ = ⟨ v, v^{'} ⟩$ , pulling the finite sum out of the first variable by [F5] and using the orthonormal expansion $v = \sum_{i = 1}^{d} ⟨ v, e_{i} ⟩ e_{i}$ of [F5].

2.1F3F7step 1.2

Evaluating at $w^{'}$ and $w$ , the weak pairing formula of [F3] for the pair $(ρ, π)$ and the definition of $T$ give $⟨ A (T) w^{'}, w ⟩ = \int_{K} ⟨ π (k) T ρ (k)^{- 1} w^{'}, w ⟩ d μ (k) = \int_{K} ⟨ ρ (k)^{- 1} w^{'}, v^{'} ⟩ ⟨ π (k) v, w ⟩ d μ (k)$ . Unitarity of $ρ (k)$ gives $⟨ ρ (k)^{- 1} w^{'}, v^{'} ⟩ = ⟨ w^{'}, ρ (k) v^{'} ⟩ = \overline{⟨ ρ (k) v^{'}, w^{'} ⟩}$ , so the integral equals $\int_{K} ⟨ π (k) v, w ⟩ \overline{⟨ ρ (k) v^{'}, w^{'} ⟩} d μ (k)$ . Since $A (T) = 0$ by step 1.2, this integral is $0$ .

2.2F3F4F6step 1.3

The trace of $A (T)$ computes $c$ and the trace of $T$ : by [F4] applied to an orthonormal basis $e_{1}, \dots, e_{d}$ of $V$ , $tr (A (T)) = \sum_{i = 1}^{d} ⟨ A (T) e_{i}, e_{i} ⟩ = \sum_{i = 1}^{d} \int_{K} ⟨ π (k) T π (k)^{- 1} e_{i}, e_{i} ⟩ d μ (k) = \int_{K} tr (π (k) T π (k)^{- 1}) d μ (k) = \int_{K} tr (T) d μ (k) = tr (T)$ , using the weak pairing formula of [F3] for $⟨ A (T) e_{i}, e_{i} ⟩$ , termwise integration by [F6], the trace of the conjugated endomorphism in [F4], and $μ (K) = 1$ in [F6]. On the other hand $tr (A (T)) = tr (c I) = c d$ by [F4].

3.1F3step 2.2step 1.4step 2.1algebra

Combining steps 1.3, 2.2 and 1.4 gives $c d = tr (A (T)) = tr (T) = ⟨ v, v^{'} ⟩$ , hence $c = d^{- 1} ⟨ v, v^{'} ⟩$ . Evaluating the weak pairing formula of [F3] at $w^{'}$ and $w$ exactly as in step 2.1 then gives $\int_{K} ⟨ π (k) v, w ⟩ \overline{⟨ π (k) v^{'}, w^{'} ⟩} d μ (k) = ⟨ A (T) w^{'}, w ⟩ = c ⟨ w^{'}, w ⟩ = \frac{1}{d} ⟨ v, v^{'} ⟩ \overline{⟨ w, w^{'} ⟩}$ , where the last equality uses $⟨ w^{'}, w ⟩ = \overline{⟨ w, w^{'} ⟩}$ .

4.1F8step 2.1step 3.1∎

Equivalent models. If $U : V \to W$ is a unitary intertwiner, then by [F8] the coefficient $⟨ ρ (k) U v^{'}, U w^{'} ⟩$ equals $⟨ π (k) v^{'}, w^{'} ⟩$ , so the integral in claim 3 reduces to the integral of claim 2 and equals $d^{- 1} ⟨ v, v^{'} ⟩ \overline{⟨ w, w^{'} ⟩}$ . Together with steps 2.1 and 3.1 this proves the orthogonality of matrix coefficients for inequivalent irreducibles, the normalized $d^{- 1}$ formula for a single irreducible including $d = 1$ , and the equivalent-model form.

Depends on

Used by

Isotypic projections are mutually orthogonal equivariant projections Theorem

Dependency tree · two levels

95 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Vera Serganova, Representation Theory, Chapter III §§1.6–2.1 (standard reference, not scraped)
David Vogan, Review of Harmonic Analysis on Compact Groups, §§2.1–2.16 (standard reference, not scraped)
Emmanuel Kowalski, An Introduction to the Representation Theory of Groups, §§5.2–5.6 (standard reference, not scraped)