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Suppose you have a locally compact group G with a discrete subgroup H. Of course G has a unique (up to scalar) Haar measure, but it seems that G/H has and induced Haar measure as well.

How does one induce a measure on G/H from the Haar measure on G? Any sources that describe this?

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closed as off-topic by Yemon Choi, David Roberts, Andrey Rekalo, Willie Wong, Neil Strickland Jun 27 '13 at 17:57

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See here ... mathoverflow.net/questions/14278/haar-measure-on-a-quotient .. –  Anweshi Apr 18 '10 at 21:42
    
Hadn't seen that question. I note that no one there seems to mention that there might not be a G-invariant measure on G/H ... –  Yemon Choi Apr 19 '10 at 3:23
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voting to close - intended reason would be "no longer relevant" –  Yemon Choi Jun 26 '13 at 20:31

4 Answers 4

Given a locally compact group $G$ and a closed subgroup $H$, there is a $G$-invariant measure on $G/H$ if, and only if, $\Delta_G|_H=\Delta_H$ (where $\Delta_G$ is the so-called modular function of $G$, which measures the difference between the left and right Haar measures). Every discrete group $H$ is unimodular (i.e. $\Delta_H$=1), so, according to the above, you'd have to check that $\Delta_G|_H$ is trivial. in fact, when $H$ is a discrete subgroup of $G$, there's a $G$-invariant measure on $G/H$ if, and only if, This will occur if $G$ itself is unimodular (e.g when $G$ is abelian, or compact, or a reductive group), but may occur more generally, for example with $G$ being the upper-triangular matrices in $\text{SL}(2,\mathbf{R})$ and $H$ the unipotent upper-triangular matrices in $\text{SL}(2,\mathbf{Z})$. In most cases, to induce a measure on $X=G/H$ you can simply pick a fundamental domain $F\subseteq G$, and integrate over it (according to this, this can be done at least when $G$ is $\sigma$-compact). More generally, you can normalize things so that for any integrable function $f$ on $G$ $$\int_X\left(\int_H f(gh)d\mu_H(h)\right)d\mu_{G/H}(gH)=\int_Gf(g)d\mu_G(g).$$

ADDED: I've come across a reference that could help: It's in the Encyclopaedia of Mathematical Science series, specifically "Lie group and Lie algebras II". Chapter 1 is "Discrete subgroups of locally compact groups". Here's a link to google books.

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It may be worth mentioning that for any closed subgroup H in a locally compact group G, although the quotient space G/H might not have a G-invariant measure, it does always have a G-quasi-invariant measure $\mu$; that is, $g\cdot\mu$ is absolutely continuous with respect to $\mu$, for any $g\in G$. This can be found in, for instance, Ch.8 of "Classical harmonic analysis and locally compact groups" by Reiter and Stegemann; but there should be other sources too. –  Yemon Choi Apr 18 '10 at 1:30
    
See also en.wikipedia.org/wiki/Quasi-invariant_measure –  Yemon Choi Apr 18 '10 at 1:30
    
I was always a bit confused by this. Because $H$ is discrete (hence closed, assuming $G$ is Hausdorff), $G/H$ is a locally compact group, and so possesses a Haar measure. Now, am I correct in my understanding that a $G$-invariant measure on $G/H$ is a Borel measure $\mu$ on $G/H$ such that $\mu(xE)=\mu(E)$ for each $x\in G$ and $E\subseteq G/H$ a Borel set? But $x$ acts on $gH$ by $(x,gH)\mapsto xgH$, so $xE$ is the same as $(xH)E$, right? So Haar measure on $G/H$ is $G$-invariant in this sense, isn't it? What am I not getting right? –  Keenan Kidwell Apr 18 '10 at 1:32
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@Keenan: G/H is only a group if H is a normal subgroup... –  Yemon Choi Apr 18 '10 at 2:34
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@Rob H.: it is not true that there is an invariant measure on G/H iff G is unimodular. Take for instance H to be the trivial group. If however, there exists an invariant Radon measure on G/H which is finite, then one can show that G must be unimodular. –  doug Apr 18 '10 at 11:59

Chapter 14 in Royden's Real Analysis has a discussion of group-invariant measures on homogeneous spaces.

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The following works in the simpler case when $G$ is not just locally compact, but compact. The idea is to view a measure as a linear functional on the space of continuous functions, and embed continuous functions on the quotient $G/H$ into continuous functions on $G$.

More precisely, if $q: G \to G/H$ is the quotient map, then there is an induced map $$ q^* : C(G/H) \to C(G),$$ which just embeds functions on $G/H$ as functions on $G$ which are invariant under right-translation by $H$. Denote Haar measure on $G$ by $f \mapsto \int_G f d\mu$ for $f \in C(G)$. Then define a measure on $G/H$ by $$ g \mapsto \int_G q^*(g) d\mu.$$ This should give the Haar measure on $G/H$ with total mass the same as the total mass of the Haar measure on $G$ that you started with.

The problem with this when $G$ is not compact is that the Haar measure is a linear functional on $C_c(G)$, the continuous functions with compact support, and the map $q^*$ doesn't take $C_c(G/H)$ to $C_c(G)$, in general.

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Could you give some text references that expose your answer ? –  user10198 Oct 20 '10 at 12:24
    
Have a look at Echterhoff and Deitmar "Principles of Harmonic Analysis", it should be in the first chapter. –  plusepsilon.de Jun 25 '11 at 11:47

I couldn't figure out which other answer to make this be a comment for... so:

Already mentioned in other answers, but really working better than might have been acknowledged, is that the averaging map $\alpha:C^o_c(G)\rightarrow C^o_c(H\backslash G)$ by $\alpha f(x)=\int_H f(hx) dh$ is (readily provably) a surjection, for a closed subgroup $H$ of $G$. Thus, to attempt to define a right $G$-invariant integral/measure on $C^o_c(H\backslash G)$, define $\int_{H\backslash G} \alpha f dx=\int_G f$. The immediate issue, of course, is well-definedness, which holds only when the modular function of $G$ restricted to $H$ is equal to the modular function of $H$.

This is one of those quasi-standard riffs that gets submerged... perhaps I saw it in Weil's book on integration on topological groups.

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