LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02

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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.

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Well-defined induced inner product

Statement

Assume AC. For $F_{1}, F_{2} \in C_{c} (G, H; V)$ , $q = x H \mapsto ⟨ F_{1} (x), F_{2} (x) ⟩$ is independent of $x$ , continuous, and compactly supported. Its $μ_{ρ}$ integral is a positive-definite inner product; the norm vanishes only when $F = 0$ .

Facts & Assumptions

Given: Strongly continuous unitary $σ$ , rho-derived measure $μ_{ρ}$ , and $F_{1}, F_{2} \in C_{c} (G, H; V)$ .

[A1]

AC is assumed as stated (The Axiom of Choice).

[F1]

Covariant sections satisfy $F (x h) = σ (h)^{- 1} F (x)$ (Continuous covariant model and measurable completion).

[F2]

The representation $σ$ in the induced model is unitary on $V$ (Continuous covariant model and measurable completion).

[F3]

The quotient map is open and $G / H$ is LCH (Compact lifts and averaging onto C_c(G/H)).

[F4]

The rho-derived Radon measure has full support (Weil formula with a rho-function).

Proof

technique · direct

1.1F1F2F3A1

For $h \in H$ , covariance and unitarity give $⟨ F_{1} (x h), F_{2} (x h) ⟩ = ⟨ σ (h)^{- 1} F_{1} (x), σ (h)^{- 1} F_{2} (x) ⟩ = ⟨ F_{1} (x), F_{2} (x) ⟩ .$ Thus the scalar is independent of the representative. Its continuous lift to $G$ descends continuously because the quotient map is open; its support lies in the intersection of the compact quotient supports.

2.1A1F4step 1.1algebra

Radon finiteness on that compact support makes the integral finite. For $F_{1} = F_{2} = F$ , the integral is nonnegative. If it were zero but $F$ were nonzero at some $x$ , continuity would make $∥ F ∥^{2}$ positive on a nonempty open subset of $G / H$ , which has positive measure by full support [F4], a contradiction. Hence the norm is positive definite; integrating the pointwise sesquilinear form gives the asserted inner product. ∎

Sources

Bekka–de la Harpe–Valette, Kazhdan’s Property (T), Appendix E §E.1, Definition E.1.6, PDF pp. 412–413. Full relevant text was inspected.

Depends on

Used by

Density of averaged covariant generators Lemma
Unitary cocycle-corrected left action Lemma
Unitary induction from a closed subgroup Theorem

Dependency tree · two levels

15 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

Bekka–de la Harpe–Valette, Kazhdan’s Property (T), Appendices B and E (standard reference, not scraped)
Bruhat, Lectures on Lie Groups and Representations of Locally Compact Groups, Chapters 1 and 7 (standard reference, not scraped)
David Vogan, Unitary Representations of Locally Compact Groups and Induced Representations (standard reference, not scraped)