Haar Measure on a Quotient - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T20:12:45Z http://mathoverflow.net/feeds/question/21704 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21704/haar-measure-on-a-quotient Haar Measure on a Quotient Krystal 2010-04-18T00:16:43Z 2011-06-26T21:52:58Z <p>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. </p> <p>How does one induce a measure on G/H from the Haar measure on G? Any sources that describe this? </p> http://mathoverflow.net/questions/21704/haar-measure-on-a-quotient/21710#21710 Answer by Rob Harron for Haar Measure on a Quotient Rob Harron 2010-04-18T01:14:22Z 2010-04-18T18:12:30Z <p>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. <s>in fact, when $H$ is a discrete subgroup of $G$, there's a $G$-invariant measure on $G/H$ if, and only if,</s> 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 <a href="http://books.google.com/books?id=V9z2cjHOhVoC&amp;lpg=PA33&amp;dq=%2522discrete%2520subgroup%2522%2520unimodular%2520haar&amp;pg=PA35#v=onepage&amp;q&amp;f=false" rel="nofollow">this</a>, 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 $G$ $$\int_X\left(\int_H f(gh)d\mu_H(h)\right)d\mu_{G/H}(gH)=\int_Gf(g)d\mu_G(g).$$</p> <p>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 <a href="http://books.google.com/books?id=hYYpI2ETqqwC&amp;lpg=PA8&amp;dq=invariant%2520measure%2520%2522discrete%2520subgroup%2522&amp;pg=PP1#v=onepage&amp;q&amp;f=false" rel="nofollow">google books</a>.</p> http://mathoverflow.net/questions/21704/haar-measure-on-a-quotient/21711#21711 Answer by KConrad for Haar Measure on a Quotient KConrad 2010-04-18T01:18:16Z 2010-04-18T01:18:16Z <p>Chapter 14 in Royden's Real Analysis has a discussion of group-invariant measures on homogeneous spaces.</p> http://mathoverflow.net/questions/21704/haar-measure-on-a-quotient/21806#21806 Answer by MTS for Haar Measure on a Quotient MTS 2010-04-19T02:08:25Z 2010-04-19T02:08:25Z <p>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 $G$.</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/21704/haar-measure-on-a-quotient/42892#42892 Answer by unknown (google) for Haar Measure on a Quotient unknown (google) 2010-10-20T12:24:33Z 2010-10-20T12:24:33Z <p>MTS</p> <p>Could you give some text references that expose your answer ?</p> <p>THX</p> http://mathoverflow.net/questions/21704/haar-measure-on-a-quotient/68881#68881 Answer by paul garrett for Haar Measure on a Quotient paul garrett 2011-06-26T21:52:58Z 2011-06-26T21:52:58Z <p>I couldn't figure out which other answer to make this be a comment for... so: </p> <p>Already mentioned in other answers, but really working better than might have been acknowledged, is that the <em>averaging</em> 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 $H$. </p> <p>This is one of those quasi-standard riffs that gets submerged... perhaps I saw it in Weil's book on integration on topological groups.</p>