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Let $G$ be a locally compact Hausdorff group with a left invariant Haar measure $\mu$ and a closed subgroup $H$. It is well-known (and not hard to prove) that $G/H$ possesses an invariant measure if and only if the modular functions of $G$ and $H$ agree on $H$.

I am looking for the exact reason for which the following naive approach to defining such a measure on $G/H$ fails.

Let $f\in C_c(G/H)$ and define

$$\int_{G/H} f\, d\sigma:=\int_G \chi f\circ\pi \, d\mu$$ where $\pi\colon G\to G/H$ is the canonical projection map and $0\le \chi\in C_c(G)$ has the property that $\int_G \chi d\mu=1$ (or perhaps with some similar property.)

Assuming that $G/H$ does have an invariant measure, can it be expressed in the above form?

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What is your proof that such a formula is an invariant measure? –  KConrad Feb 23 at 20:33
    
Keith, I think the problem is that this "definition" doesn't necessarily assign a (finite) number to every $f$. But if it does, then its invariance is clear from the invariance of the Haar measure on $G$. Such a definition exists for groups diffemorphic to $\mathbb{R}^n$ by Fubini. –  user47358 Feb 24 at 0:40
    
For general quotient, you will have to work with quasi-invariant measures. –  Marc Palm Feb 24 at 12:51
    
Surely you'll need some additional conditions on $\chi$ to ensure invariance. As it stands, $\chi$ could concentrate on only "half" of the cosets of $H$. –  Andreas Blass Apr 25 at 14:13
    
you need to use of half forms see G. Heckman and H. Schlichtkrull: Harmonic Analysis and Special Functions on Symmetric Spaces. Academic Press 1994. –  Hassan Jolany May 25 at 14:10

2 Answers 2

I think the impulse to "lift" a function on $G/H$ to $G$, and define the integral on $G/H$ as the integral on $G$ of that lift, is entirely reasonable... but doesn't quite work out, no matter how one struggles. A prescription for seeing the problem in various (futile) attempts resides in the key point for the successful characterization: letting $\alpha$ be the averaging map $(\alpha f)(x)=\int_H f(hx)\,dh$, with the sided-ness of the measure easily determined... the requirement is exactly as suggested in the question: $$ \int_G f(g)\,dg \;=\; \int_{G/H} (\alpha f)(\dot{g})\,d\dot{g} $$ where the $\dot{g}$ is the coordinate on the quotient. To make this be an unequivocal characterization of the integral/measure on the quotient, one must check that the averaging map is a surjection of $C^o_c(G)$ to $C^o_c(G/H)$, which is indeed so, in general.

Then, to return to the question of making a "section" of the averaging map... if the condition on "modular functions" is met, and if continuity is not necessary, then this is certainly possible. But this might be a less than optimal approach, since we might want to characterize measures/integrals via Riesz' theorem about functionals on $C^o_c$.

NB, as @KConrad suggests, the modular-function condition is necessary, as one fills out the alleged proof that such a formula gives an invariant measure. The popular cases, such as $\mathbb R \to \mathbb R/\mathbb Z$ with section given by multiplying by characteristic function of an interval, so subliminally assume the modular-function condition that we may not have realized.

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I understand that the invariance of the quotient measure depends on the definition of $\chi$. In the case of modular groups, I think that such a $\chi$ exists (by taking a "section" of the averaging map), eg for compact groups $\chi=1$ seems to work. –  user47358 Feb 24 at 5:21
    
@user47349, if you register an account we can merge this unregistered user into it, and then you'll be able to accept the answer. Email community@stackexchange.com and tell them the account numbers you'd like merged. –  Scott Morrison Feb 24 at 8:13

For general quotient, you will have to work with quasi-invariant measures. There is a multiplier, which is constant if the measure is invariant. For it to be constant, it is necessary and sufficient that $\Delta_G|_{H} = \Delta_H$.

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