Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-08
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.

A Kazhdan pair for a compact group via Haar averaging

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let K be a compact Hausdorff topological group (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, Topological group: multiplication and inversion are continuous) with normalized Haar probability measure μ (Normalized Haar probability on a compact group). For every 0<ε≤1, (K,ε) is a Kazhdan pair (Kazhdan pairs, Kazhdan sets and Kazhdan constants); in particular, K is a Kazhdan set and has property (T) (Kazhdan's property (T)).

Explicitly, if (π,H) is a strongly continuous unitary representation (Strongly continuous unitary representations, invariant linear subspaces and intertwiners) on a Hilbert space H (Hilbert space) and ξ∈H is a unit vector with sup⁡x∈K∥π(x)ξ−ξ∥<ε≤1, then its Haar average η=∫Kπ(x)ξ dμ(x), as in Compact groups have property (T) by Haar averaging, is nonzero and K-invariant, with ∥η−ξ∥≤sup⁡x∈K∥π(x)ξ−ξ∥. Thus the whole group K is a Kazhdan set with tolerance 1; the averaging bound is not claimed to be optimal.

Facts & Assumptions

Given: AC, a compact Hausdorff topological group K with normalized Haar probability μ, and a strongly continuous unitary representation (π,H).

[F1]

For every 0<δ≤1, the preceding compact-group theorem says (K,δ) is a Kazhdan pair. It also gives the Haar average η=∫Kπ(x)ξ dμ(x) as a nonzero invariant vector with ∥η−ξ∥≤sup⁡K∥π(x)ξ−ξ∥ whenever ξ is a unit vector and that supremum is below 1 (Compact groups have property (T) by Haar averaging, Normalized Haar probability on a compact group, The Axiom of Choice).

[F2]

A pair (K,δ) is Kazhdan when every strongly continuous unitary representation with a (K,δ)-invariant unit vector has a nonzero invariant vector; a compact Kazhdan set with one positive tolerance gives property (T). Almost invariance supplies such unit vectors for every compact test and every positive tolerance. (Kazhdan pairs, Kazhdan sets and Kazhdan constants, Kazhdan's property (T), Almost invariant vectors for a unitary representation).

Proof

Proof technique: apply the earlier Haar-average theorem and specialize its uniform estimate to the compact test set K.

1.1F1givenalgebra

If ξ is a unit vector and M:=sup⁡x∈K∥π(x)ξ−ξ∥<ε≤1, then M<1. By [F1], the Haar average η=∫Kπ(x)ξ dμ(x) is nonzero and K-invariant, and ∥η−ξ∥≤M.

2.1F1F2∎

By [F1], (K,ε) is a Kazhdan pair for every 0<ε≤1. Taking ε=1 makes K a Kazhdan set; its pair property applied to almost invariant unit vectors gives property (T) by [F2]. A representation on the zero Hilbert space has no unit vector, so the pair implication is vacuous there.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

60 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