Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)
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.

Haar regularization of transitive unitary cocycles

Statement

Assume AC. Let G,H,μ,s be as in the Haar-lift lemma, K separable, and c:G×G/H→U(K) Borel with c(g1g2,x)=c(g1,g2x)c(g2,x) for each pair and almost every x. Suppose g↦c(g,⋅) is continuous in local convergence in measure in the strong topology of U(K). Then there exist a strongly continuous unitary representation σ:H→U(K) and a Borel B:G/H→U(K) with c(g,x)=B(gx)σ(s(gx)−1gs(x))B(x)−1 for every g and almost every x. This formula gives a strict Borel cocycle on all of G×G/H. The representation σ is unique up to unitary equivalence under Borel changes of fibre gauge.

Facts & Assumptions

Given: AC, a second-countable LCH group G, a closed subgroup H, the quotient G/H with a Borel section s, a nonzero quasi-invariant measure class (a representative μ), a separable Hilbert space K, and a Borel cocycle c.

[F1]

The Haar-lift lemma supplies: q−1(E) is Haar null iff μ(E)=0; the coordinate map Θ(x,h)=s(x)h is a Borel isomorphism; and every Borel F:G/H×H→U(K) satisfying F(x,hk)=F(x,h) for all k and a.e. (x,h) equals B(x) a.e. for a Borel B:G/H→U(K) (Haar null classes and Borel descent on a homogeneous space, Borel cross-sections for closed subgroups of second-countable locally compact Hausdorff groups).

[F2]

Steinhaus–Pettis: for a separable K, U(K) in the strong topology is a second-countable topological group and every Borel homomorphism H→U(K) is strongly continuous (Steinhaus and Pettis: Borel homomorphisms of second-countable locally compact groups are continuous, Strongly continuous unitary representations, invariant linear subspaces and intertwiners).

[F3]

Completed-product Tonelli/Fubini for σ-finite measures; left translations preserve the Haar measure and right translations scale it by the modular function; Haar null sets of the completed product are preserved by the coordinate changes used below (Tonelli and Fubini for the completed product, with only almost-everywhere section measurability, Strong continuity of left and modular right translations on L1 and L2, Monotone convergence for the integral).

[F4]

For a Borel U(K)-valued function u on G and ξ∈K, the translates g↦u(gt)ξ are continuous in local measure: on a fixed finite-Haar-measure set C one has ∫C∥u(gt)ξ−u(g0t)ξ∥2 dt→0. This follows by approximating t↦u(t)ej on relatively compact sets by continuous compactly supported K-valued functions (using a countable orthonormal basis and the density of Cc in L2) and then applying the L2 translation continuity, uniformly over g in a compact neighbourhood, where the modular factor ΔG(t) is bounded (Strong continuity of left and modular right translations on L1 and L2, Completeness of the complex Haar L1 and L2 spaces and density of Cc, A Hilbert space with a dense sequence has a finite or countable orthonormal basis, LCH Urysohn cutoff).

[F5]

Proof

technique · direct

Given: AC, the cocycle c and the continuity hypothesis.

1.1F1F3

Lift to G: put C(u,v)=c(uv−1,q(v)), a Borel U(K)-valued function on G×G. For the cocycle law applied at g1=uv−1, g2=vw−1 in the second variable q(w), the equivariance q(gx)=gq(x) of the quotient map gives (vw−1)q(w)=q(vw−1w)=q(v), so C(u,v)C(v,w)=c(uv−1,(vw−1)q(w))c(vw−1,q(w))=c(uw−1,q(w))=C(u,w) wherever the a.e. cocycle identity of c holds at that triple. The set of triples (u,v,w) for which it may fail is the preimage of the cocycle law's null set under the homeomorphism (u,v,w)↦(uv−1,vw−1,w) of G3, whose Jacobian is a positive modular factor; by [F3] that preimage is null. Hence C(u,v)C(v,w)=C(u,w) for Haar-a.e. (u,v,w).

2.1F3step 1.1

Choose w0 by Fubini so that C(u,v)C(v,w0)=C(u,w0) for Haar-a.e. (u,v), and set b(u)=C(u,w0)=c(uw0−1,q(w0)), a Borel U(K)-valued function. Then C(u,v)=C(u,w0)C(v,w0)−1=b(u)b(v)−1 for Haar-a.e. (u,v), i.e. c(g,q(t))=b(gt)b(t)−1 for Haar-a.e. (g,t).

3.1F4F5step 2.1

Upgrade to every fixed g: let G0 be the conull set of g for which the identity of [step 2.1] holds for a.e. t; it is dense because Haar measure is positive on nonempty open sets, so every g0 is a limit of a net (gi) in G0. Along that net the left-hand classes t↦c(gi,q(t)) converge in local measure to t↦c(g0,q(t)) by the continuity hypothesis: for a finite-Haar-measure set C⊆G, the finite measure q∗(1C dt) is absolutely continuous with respect to μ by [F1]; truncating its Radon–Nikodym density and exhausting the σ-finite base shows that local convergence in μ-measure implies convergence for this finite measure. Thus pullback is continuous in local Haar measure. The right-hand classes t↦b(git)b(t)−1 converge in local measure to t↦b(g0t)b(t)−1 by [F4]; multiplication by the fixed field b(t)−1 preserves this convergence, as follows by approximating b(t)−1ξ on each finite-measure set by finite-valued vectors and using the uniform norm bound on unitaries. Since the two sides agree at each gi, uniqueness of local-measure limits gives c(g0,q(t))=b(g0t)b(t)−1 for a.e. t. As g0 was arbitrary, the identity holds for every fixed g and Haar-a.e. t.

4.1F3step 3.1

Stabilizer constants: for h∈H put Ah(t)=b(t)−1b(th). For every g and Haar-a.e. t, [step 3.1] applied to gt and to t, together with q(th)=q(t) and the cocycle law, gives Ah(gt)=Ah(t); Tonelli and the measure-preserving change (g,t)↦(gt,t) show that Ah(u)=Ah(t) for Haar-a.e. (u,t), so Ah is Haar-a.e. constant, equal to some σ(h)∈U(K); hence b(th)=b(t)σ(h) for Haar-a.e. t.

5.1F2step 4.1

σ is a homomorphism: applying [step 4.1] twice, b(t)σ(hk)=b(thk)=b(t)σ(h)σ(k) a.e., so σ(hk)=σ(h)σ(k). It is Borel: integrating the matrix coefficients of the Borel U(K)-valued function Ah against a fixed positive probability density on G returns the matrix coefficients of σ(h) (because Ah=σ(h) a.e.) and is Borel in h by Tonelli; by [F2], applied to the second-countable group H and the target U(K), σ is strongly continuous.

6.1F1step 4.1step 5.1

Descent: define F(x,r)=b(s(x)r)σ(r)−1. For h∈H, F(x,rh)=b(s(x)rh)σ(rh)−1=b(s(x)r)σ(h)σ(h)−1σ(r)−1=F(x,r) up to the a.e. statements of [step 5.1]; hence by [F1] there is a Borel B:G/H→U(K) with b(t)=B(q(t))σ(s(q(t))−1t) for Haar-a.e. t.

7.1F1step 3.1step 6.1

Substituting [step 6.1] into [step 3.1] at the points t and gt gives, for every fixed g, c(g,q(t))=B(q(gt))σ(s(q(gt))−1gt) σ(s(q(t))−1t)−1B(q(t))−1 for a.e. t; using q(gt)=gq(t) and the exact section identity gs(x)=s(gx)h(g,x) this is the displayed formula for a.e. x; the strict section identity then makes the displayed expression an exact Borel cocycle on all of G×G/H.

7.2step 4.1step 6.1

Uniqueness of σ: if (Bi,σi) both factorize c, set bi(t)=Bi(q(t))σi(s(q(t))−1t); then b2(gt)−1b1(gt)=b2(t)−1b1(t) for every g and a.e. t, so [step 4.1] makes b2−1b1 a constant unitary T, and right-H covariance gives σ2(h)T=Tσ1(h) for every h. Since a change of gauge multiplies the lifted factorizations on the left, the class of σ is unchanged.

8.1step 5.1step 6.1step 7.1step 7.2F5∎

Steps 5.1, 6.1, 7.1 and 7.2 give a strongly continuous σ, a Borel B with the displayed factorization, its exact cocycle form, and the uniqueness up to gauge, as claimed.

Remarks

The continuity hypothesis is used only in the upgrade step [3.1]; the a.e. cocycle law and the left-invariance arguments are pure Haar-Tonelli computations. No value of an a.e. class is ever evaluated at a prescribed null coset.

Depends on

Used by

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