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I am a physicist, and I have the following problem. Consider a locally compact group G acting over a measure space $(X, {\cal B}, \mu)$, and assume that $\mu$ is G-invariant. My problem is how to "quotient" the measure $\mu$ for obtaining a measure $\mu/G$ on the quotient space $X/G$, i.e., the space wose elements are the orbits of G. The simple answer consisiting of defining $\mu/G(\Delta):=\mu(\Delta)$ for $\Delta \in X/G$ is not appropriate, as one can see from the following example.

Let $X:=\mathbb{R}^2$ be the configuration space of two one-dimensional particles, let $\mu$ be the Lebesgue measure, and let $G:=\mathbb{R}$ be the group of translations: $a(x, y)=(x + a, y + a)$ for $a \in G$ and $(x, y) \in X$. The orbit of a point $(x, y)$ is the line at $45^°$ passing for $(x, y)$. It is easy to see that the Lebesgue measure of any set of orbits is either $0$ or $\infty$ .

My tentative answer is the following. Let S be a section of the partition of the orbits, i.e., a subset of X composed by an element for every orbit. Let the map $h_S:G \times X/G \to X$ be defined as follows: $h_S(g, \xi)=g s_\xi$, where $s_\xi$ is the element of the section $S$ belonging to the orbit $\xi$. It is easy to see that $h_S$ is a bijection between $G \times X/G$ and $X$. It is also easy to see that a measure $\nu_S$ on $X/G$ exists such that $\mu=h_S(\alpha \times \nu_S)$, where $\alpha$ is the Haar measure on G. The measure $\nu_S$ is the measure I am looking for. The problem is to prove rigorously that it does not depend on the chosen section S.

I guess that this problem has already been addressed. Can anybody give me some reference?

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I don't quite understand what you're trying to accomplish. In what sense is the measure which you defined at first is "inappropriate"? What properties do you want it to have? –  Mark Sep 28 '10 at 21:56
    
Another problem with your tentative proposal is that sections need not globally exist. A terrible quotient is given by the action of $\mathbb R$ on the torus by moving along an irrational slope. But there are less bad quotients that you still might not capture. If everything is compact and all measures are invariant, you probably also want $Vol(X/G) = Vol(X)/Vol(G)$, so you'll need to be careful to define the volume of "half a point". –  Theo Johnson-Freyd Sep 28 '10 at 22:10
    
Do you know whether the action of $G$ is proper? I imagine this would simplify things a bit. –  David Roberts Sep 28 '10 at 23:28
    
Also with the help of the comment of Huntsman I found a reference very good for me: Wijsman, Invariant Measures on group and their use in statistics, §7.3 (available at projecteuclid.org/… The group I am interested in is Euclidean group acting on configuration space $X=R^{3N}$. The group is unimodular, and therefore a measure on X/G satysfying equation 7.3.6 exists. This property corresponds to what I was intuitively looking for. –  Bruno Galvan Sep 29 '10 at 22:55
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2 Answers

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There is a fair amount of work on this. Since measures are roughly the same as cohomology, the standard approach in quantum field theory (one situation where such integrals are needed) roughly boils down to computing equivariant cohomology. The physics buzzword for this is "BRST".

For the case of finite-dimensional manifolds, I think the best mathematical analysis is:

  • A. Weinstein, 2009. The volume of a differentiable stack. Lett. Math. Phys. 90(1-3) pp.353--371.

In particular, he explains what are the correct type of smooth measures when a Lie group acts on a manifold with compact stabilizers (the group itself need not be compact).

It is important to keep in mind that the search for invariant measures is simply the wrong way to go about defining measures on the quotient in general. In particular, if you consider a group $G$ acting on a space $M$, then if $G$ is unimodular (left Haar measure equals right Haar measure), then the smooth measures on the stacky quotient $M/G$ are in bijection with the invariant measures on $M$ (a choice of bijection corresponds to a choice of Haar measure), but if $G$ is not unimodular (e.g. the two-dimensional nonabelian Lie group) then the correct thing to do is pick a measure on $M$ that fails to be invariant in precisely a way that cancels the failure of unimodularity.

More generally, the Weinstein point of view is that you should not think about a space and a group separately — that's too much data — but just think about the groupoid that encodes the action. Then there is a well-developed theory of "equivalence" of groupoids, and Weinstein measures push and pull well across equivalences.

A particular example of an equivalence is given by a "full open neighborhood": let $U \subseteq M$ be any open set that intersects every orbit of $M/G$, and restrict your groupoid to it. (Note: $G$ does not act on $U$, but we define a "partial action" in which two points in $U$ are equivalent iff they're equivalent in $M$, and with the same stabilizers.) Then the restricted groupoid is equivalent to the original one.

So, let's take $M = \mathbb R^2$ and $G = \mathbb R$ as in your example. Then we can restrict to the full subset $U = \{(x,y) \text{ s.t. } |x+y|<1$.

Now it is clear that we can pick an invariant Lebesgue measure with lots of finite-sized orbits. We should divide the volume of any region — this is part of the definition, and in general see [ibid.] — by the length of the orbits: each orbit pulls back to some open subset of $G = \mathbb R$, and measure its length there against Haar measure. You do get well-defined groupoid measure doing this. At the end of the day, it's what you want it to be: the measure of a cylinder over a subset of the line $\{x+y = 0\}$ is just the length of the subset (up to your choice of normalizations).

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The references given in the comments above and in the answer below are, in my opinion, somewhat misleading because all of them discuss the case of a transitive action (but maybe this is what you meant). In the general situation the problem is precisely that the space of orbits may be quite nasty. Let us look at the irrational circle rotation, i.e., the action of $\mathbb Z$ defined as $$ T^n x = x + n\alpha \;\; (\text{mod}\; 1)\;, $$ where $\alpha$ is an irrational number. Then the set obtained by taking one point from each orbit is a classical example of a non-measurable set, and there is no way one can define a non-trivial measure on it. Another example like this can be obtained by a little modification of your action: this is the action of $\mathbb R$ on $\mathbb R^2/\mathbb Z^2$ defined by projecting the action $$ T^t(x,y) = (x+t,y+\alpha t) \;, $$ onto the torus (here $\alpha$ is again an irrational number).

The reason why the orbit spaces are "bad" in these examples is that there are no non-trivial measurable sets (non-trivial in the sense that both the set and its complement are non-negligible with respect to the considered measure) which would be unions of orbits. Actions with this property are called $ergodic$. For an ergodic action the only reasonable quotient space is the one-point space.

For actions which are not ergodic there is a reasonable non-trivial quotient measure space, which is called the $space\; of\; ergodic\; components$. Its construction, roughly speaking, consists in first taking the $\sigma$-algebra of all measurable sets which are (up to subsets of measure 0) unions of orbits, and then using the fact that in all reasonable probability spaces any such $\sigma$-algebra can be realized as the preimage $\sigma$-algebra of a uniquely defined measurable quotient map.

You can find details, for instance, in the article

MR1784210 (2001i:28021) Greschonig, Gernot(A-WIEN); Schmidt, Klaus(A-WIEN) Ergodic decomposition of quasi-invariant probability measures. (English summary) Dedicated to the memory of Anzelm Iwanik. Colloq. Math. 84/85 (2000), , part 2, 495--514.

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Of course there is an appropriate replacement for the quotient measure space even (or especially) in the ergodic case. Look at the cross-product von Neumann Algebra. Though this is perhaps a little sophisticated for this question –  Owen Sizemore Sep 28 '10 at 23:38
    
@Owen: I've never been happy with the cross-product, because it is a lossy operation from the point of view of geometry. For example, consider the group of order $2$ acting on a point. The cross-product vN algebra is just the algebra of functions on 2 points, even though the stacks are different (e.g. the "counting measure" for one gives total volume $1/2$ and for the other total volume $2$). You can distinguish these if in addition to the algebra you remember the (symmetric!) tensor structure on the category of representations ("sheaves"), or equivalently the "sesquialgebra" structure. –  Theo Johnson-Freyd Oct 1 '10 at 4:26
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