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Criterion for an invariant quotient measure

Statement

Assume AC. The quotient G/H has a nonzero G-invariant Radon measure if and only if ΔG∣H=ΔH. When they agree, ρ=1 in the Weil formula supplies such a measure.

Facts & Assumptions

Given: LCH G, closed H, fixed left Haar measures, and AC.

[F1]

The averaging map TH:Cc(G)→Cc(G/H) is onto (Compact lifts and averaging onto C_c(G/H)).

[F2]

Positive functionals on Cc have Radon representing measures, and Radon measures are determined by their Cc integrals (Positive functionals on C_c(X) are integration against a Radon measure, Assuming Dependent Choice, uniqueness of the RMK representing measure among Radon measures).

[F3]

Any two left Haar measures are positive scalar multiples (Uniqueness of left Haar measure up to scale).

[F4]

Right translation by h scales a left Haar integral by Δ(h)−1 (Right translation scales left Haar measure).

[F5]

ρ=1 is a rho-function exactly when ΔG∣H=ΔH; its Weil measure satisfies the quotient formula (Rho-function for a closed subgroup, Weil formula with a rho-function).

[F6]

AC implies DC as required by the cited measure results (AC implies DC implies countable choice).

[A1]

AC is assumed (The Axiom of Choice).

Proof

technique · direct
1.1F1F2A1

Assume ν is a nonzero invariant Radon measure on X=G/H. Define L(f)=∫XTHf dν for real f∈Cc(G). This functional is positive. If it were zero, surjectivity [F1] would make every Cc(X) integral against ν zero, and [F2] would force ν=0. Thus L is nonzero.

1.2F1F5

Conversely suppose the modular functions agree on H. Then ρ=1 satisfies the covariance in [F5]. Let μ1 be the Weil measure. For a∈G and ϕ=THf, its translate satisfies ∫Xϕ(a−1q)dμ1(q)=∫Gf(a−1x)dx=∫Gf(x)dx=∫Xϕ(q)dμ1(q), by left invariance.

2.1F2step 1.1

For a∈G, let Laf(x)=f(a−1x). Then TH(Laf)(xH)=THf(a−1xH), so invariance of ν gives L(Laf)=L(f). By [F2], L is represented by a Radon measure λ on G; it is left invariant and nonzero, hence a left Haar measure.

3.1F3F4step 2.1choose

By [F3], λ=c dx for c>0. For h∈H, right translation gives TH(Rhf)=ΔH(h)−1THf by [F4] applied in H. Hence L(Rhf)=ΔH(h)−1L(f). Since λ=c dx, [F4] applied in G also gives L(Rhf)=ΔG(h)−1L(f). Choose f with L(f)>0; equality forces ΔG(h)=ΔH(h).

4.1A1F1F2F3F4F5F6step 1.1step 2.1step 3.1step 1.2choose

Surjectivity [F1] gives this equality for every Cc(X) test function. The Radon uniqueness in [F2] shows a∗μ1=μ1; the Weil measure is nonzero. This proves sufficiency and the equivalence. ∎

Sources

Bekka–de la Harpe–Valette, Kazhdan’s Property (T), Appendix B §B.1, Corollary B.1.7, PDF pp. 355–356; Bruhat, Lectures on Lie Groups and Representations of Locally Compact Groups, Chapter 7 §3.3, Proposition 3, PDF pp. 74–75. Full relevant text was inspected.

Depends on

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