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What does it mean and how is it defined if the action of a group is meant to be ergodic? Thank you for your replies!

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Actually, the question could be more refined. The action of the group may be ergodic with respect to symplectic measure on every leaf. –  Hamurabi Jun 15 '13 at 15:34
    
Maybe you should ask a new and more focused question: you seem to have a quite specific context in mind. Where does the question come from and what exactly are you trying to understand? –  Martin Jun 15 '13 at 15:38

1 Answer 1

up vote 8 down vote accepted

Here is a context of group action where ergodicity arises naturally. As pointed out by Martin, the definition given in the grey box applies to more general situations where the transformations are not required to be measure-preserving, and the group is not countable discrete.

Take a probability measure space $(X,\mathcal{A},\mu)$ and let a countable discrete group $G$ act on it by measure-preserving transformations.

First, this means that each $g\in G$ induces a measurable map $g:X\longrightarrow X$ such that $\mu(g^{-1}(A))=\mu (A)$ for every measurable set $A\in\mathcal A$. It is standard to write $g\cdot x$ for $g(x)$.

Second, we have a group action: the identity of $G$ induces the identity on $X$ and $g\cdot (g'\cdot x)=(gg')\cdot x$ for every $g,g'\in G$ and $x\in X$.

The action $G\curvearrowright(X,\mathcal A,\mu)$ is called ergodic if $$ g\cdot A=A\quad \forall g\in G\quad\Rightarrow\quad \mu(A)=0\mbox{ or } 1. $$ That is, up to measure $0$ sets, the only invariant measurable sets under the action of $G$ are $\emptyset$ and $X$.

Remarks: 1) As pointed out by Stéphane Laurent, the particular case $G=\mathbb{Z}$ corresponds to the action of a single measure-preserving transformation $T$. Then the action is ergodic if and only if $T$ is ergodic in the usual way. 2) By Koopman representation $u_g(f):=f\circ g$, we get a unitary representation $g\longmapsto u_g$ of $G$ on $L^2(X)$. In particular, $u_g(1_A)=1_A\circ g=1_{g^{-1}(A)}$ for every $A$ measurable. So ergodicity reads: there are no notrivial projections $p\in L^\infty(X)$ such that $u_g( p)=p$ for every $g\in G$. Identifying $f\in L^\infty(X)$ with multiplication $m_f$ by $f$ on $L^2(X)$, we get in $B(L^2(X))$: $u_gm_fu_g^*=m_{f\circ g}$. And now ergodicity reads: there are no nontrivial projections in $L^\infty(X)\subseteq B(L^2(X))$ which commute with $G\subseteq B(L^2(X))$.

Example: take a standard Borel space $(X,\mathcal{B},\mu)$ equipped with a $\sigma$-finite probability measure, i.e. a standard measure space. Then every essentially free ergodic group action $G\curvearrowright(X,\mathcal B,\mu)$ by an infinite countable discrete group gives rise to a $\rm{II}_1$ factor von Neumann algebra $L^\infty(X)\rtimes G$ by the so-called group-measure space construction of von Neumann. Whenever $G$ is amenable and $X\simeq [0,1]$, we get this way, up to isomorphism, the unique (by Connes' deep work) injective $\rm{II}_1$ factor $\mathcal{R}$.

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Since the question was tagged Lie groups by the OP: many actions of Lie groups that are of geometric interest are not really measure-preserving, but they preserve the measure class (that is, null sets are preserved). The same definition of ergodicity applies. –  Martin Jun 15 '13 at 15:02
    
It's also nice to mention that the classical ergodicity for a measure-preserving transformation $T$ is the case when $G={\{T^n\}}_{n \in \mathbb{Z}}$ . –  Stéphane Laurent Jun 15 '13 at 16:05
    
@StéphaneLaurent Right. Even if it is obvious, I should have mentioned that. Thanks for the note. –  1015 Jun 15 '13 at 16:10
    
@julien do you have a reference for a bunch of constructions of type III factors via group-measure space construction? –  no identity Jun 15 '13 at 21:32
    
@Norbert The only reference I can think of is Connes' book on Krieger factors (p.475 et seq), which generalize the Powers and Araki-Woods factors by ergodic means. –  1015 Jun 15 '13 at 22:15

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