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\title{\huge Relative ergodicity results for product groups}
%\author{Greg Hjorth}        % Enter your name between curly braces
\date{\today}          % Enter your date or \today between curly braces
\maketitle

\tableofcontents


\section{Introduction}
\label{introduction}

These notes present various relative ergodicity results, to the effect that 
actions of
Bernoulli shifts of a comparatively larger number of
copies of the free group must be relatively ergodic to free actions of
smaller numbers of copies of the free group. Ultimately this is used to give
a ``quote-unquote" elementary proof of a theorem previously established by
Adams and Kechris, that there are $\leq_B$-incomparable countable
Borel equivalence relations.


\section{Definitions}
\label{definitions}

\begin{definition}
We let $E_0$ be the equivalence relation of eventual agreement on infinite
binary sequences; thus for $f, g\in \{0, 1\}^\N$ we set
\[f E_0 g\]
if there is some $N$ such that for all $n>N$
\[f(n)=g(n).\]
\end{definition}

\begin{definition}
Given groups standard Borel groups
$G$, $H$, with $G$ acting in a Borel manner on standard Borel $X$, we say that
\[\alpha: X\times G\rightarrow H\]
is a {\it cocycle} if it is Borel and for all $x\in X$, $g_1, g_2\in G$ 
\[\alpha(x, g_1g_2)=\alpha(g_2\cdot x, g_1)\alpha(g_1, x).\]
Two cocycles
\[\alpha,\beta: X\times G\rightarrow H\]
are {\it cohomologous} if there is a Borel
\[\psi: X\rightarrow H\]
such that for all $x\in X, g\in G$
\[\beta(x, g)=\psi(g\cdot x)\alpha(x, g)\psi(x)^{-1}.\]
Frequently we will have a Borel probability measure $\mu$ on $X$ and we
can talk about a $\mu$-measurable cocycle, which will be a cocycle defined on
some conull subset of $(X, \mu)$, and two cocycles being
$\mu$-measurably cohomologous if they are outright cohomologous on a
conull set.
When there is no ambiguity about the measure, we will simply write
{\it measurable} in place of the more lengthy $\mu$-measurable.
\end{definition}

In the present paper we will only really be considering the case of
$G$ and $H$ both countable. Admittedly many of the results extend
to the wider context of locally compact.



\begin{definition} Let $E$ and $F$ be equivalence relations on standard
Borel $X$ and $Y$. A Borel function
\[\rho:X\rightarrow Y\]
is a {\it homomorphism} of $E$ to $F$ if for all $x_1, x_2\in X$
\[x_1 E x_2\Rightarrow \rho(x_1) F \rho(x_2);\]
in other words, $\rho\times \rho[E]\subset F$.
An equivalence relation $E$ on a standard Borel probability space $(X, \mu)$
is said to be {\it $F$-ergodic} if whenever $\rho$ is a homomorphism from
$E$ to $F$ then there is a single $y_0$ such that for $\mu$-a.e. $x\in X$ we
have $\rho(x) Fy_0$; in other words, $[\rho(x)]_F$ is constant on a conull 
set.

We write $E\leq_B F$, $E$ {\it is Borel reducible to $F$,}
if there is a homomorphism $\rho$ from $E$ to $F$ which has
the stronger property that $x_1 E x_2$ if {\it and only if}
$\rho(x_1) F\rho(x_2)$; in other words, $\rho$ can be thought of as pushing
down to an injection between the quotient spaces,
\[X/E \hookrightarrow Y/F.\]
It is known that an equivalence relation with all classes countable is
hyperfinite, that is to say writeable as an increasing union of Borel
equivalence relations with finite classes, if and only if it is Borel
reducible to $E_0$.

At some point it will be convenient to consider classes of reductions somewhat
broader than Borel. For instance, we say that $E$ is {\it universally 
measurably
reducible to $F$} if there is a universally measurable
\[\rho: X\rightarrow Y\]
such that for all $x_1,x_2\in X$
\[x_1 E x_2\Leftrightarrow \rho(x_1) F \rho(x_2).\]

Among the {\it countable}
(that is to say, all classes are countable) Borel equivalence relations
there are maximal ones with respect to the ordering of
Borel reducibility. We arbitrarily choose one of these, and denote
it by $E_\infty$ and assume it lives on the standard Borel
space $X_\infty$. While it is immediate from the definitions that any
countable Borel equivalence relations $E$ must be Borel reducible to
$E_\infty$, it is not true that $E\leq_B E_\infty$ implies $E$ is countable;
in this case we say that $E$ is {\it essentially} countable.



  Proofs of these and related facts can be found
in \cite{dokelo} and \cite{jakelo}.


A homomorphism $\rho$ from $E$ on $X$ to
$F$ on $Y$ is said to {\it factor through} a third equivalence relation 
$F'$ on
$Y'$ if there are homomorphisms
\[\rho_1: X\rightarrow Y',\]
\[\rho_2: Y'\rightarrow Y,\]
from $E$ to $F'$ and from $F'$ to $F$, with
\[\rho=\rho_2\circ \rho_1.\]
\end{definition}


An important example of a cocycle arises from a homomorphism of one
orbit equivalence relation into another. Suppose $G$ acts in a Borel
manner on $X$ and $H$ acts {\it freely} and in a Borel manner on $Y$
and
\[\rho: X\rightarrow Y\]
is a homomorphism of $E_G$ to $E_H$. Then we obtain an {\it induced cocyle}
\[\alpha: X\times G\rightarrow H\]
by letting
\[\alpha(x, g)=h\]
if $h\in H$ is the unique group element with
\[h\cdot \rho(x)=\rho(g\cdot x).\]

The characterstic identity of the definition of a cocyle is easily checked. 
We do
need to argue that $\alpha(\cdot, \cdot)$ is indeed Borel, but this follows by
the classical
uniformization theorem for Borel sets in the plane with countable sections
(see for instance the relevant passages of \cite{kechrisclassical}).



\section{Relative ergodicity result}
\label{theorem1}

\begin{lemma}
\label{trivialization}
Let $G$ and $H$ be countable groups.
Suppose $G$ acts by measure preserving transformations on
standard Borel
$(X, \mu)$ and $H$ acts freely by Borel automorphisms on standard Borel
$Y$. Suppose that the induced cocycle
\[\alpha: X\times G\rightarrow H\]
is cohomologous to a cocycle into an abelian subgroup of $H$ and suppose
further that $E_G$ is $E_0$-ergodic.

Then there is a single $y_0\in Y$ such that for $\mu$-a.e. $x\in X$ we have
\[\rho(x) E_H y_0.\]
\end{lemma}

\begin{proof}
Restating the assumptions of the lemma gives us some abelian $H_0 < H$
and Borel
\[\psi: X\rightarrow H\]
such that for all $x\in X$, $g\in G$
\[\psi(g\cdot x)\alpha(x, g)\psi(x)^{-1}\in H_0.\]
Therefore if we define
\[\rho_0: X\rightarrow Y\]
\[x\mapsto \psi(x)\cdot \rho(x),\]
then for any $x, g$ and
\[h_0=\psi(g\cdot x)\alpha(x, g)\psi(x)^{-1}\]
we must have
\[h_0\cdot \rho_0(x)=\psi(g\cdot x)\alpha(x, g)\psi (x)^{-1}\rho_0(x)\]
\[=\psi(g\cdot x)\alpha(x, g)\psi (x)^{-1}\psi(x)\rho(x)=
\psi(g\cdot x)\alpha(x, g)\rho(x)\]
\[=\psi(g\cdot x)\rho(g\cdot x)\]
by definition of $\alpha(\cdot, \cdot)$, which in turn equals
\[\rho_0(g\cdot x).\]
Thus the induced cocycle for the {\it new} homomorphism $\rho_0$ of
$E_G$ to $E_H$ always takes its values inside $H_0$.

In other words, $\rho_0$ is in fact a homomorphism from $E_G$ to
$E_{H_0}$, from the orbit equivalence relation induced by the action of
$G$ on $X$ to the equivalence relation induced by the action of $H_0$ on
$Y$.

We can accordingly define a measure
\[\nu=(\rho_0)_*(\mu)\]
on $Y$ by
\[\nu(B)=\mu(\rho_0^{-1}[B]).\]
As remarked in \cite{jakelo}, it follows from \cite{cofewe} that
any countable abelian group of Borel automorphisms gives rise to an
equivalence relation which is hyperfinite, and thus Borel reducible to
$E_0$. In particular, $E_{H_0}\leq_B E_0$.

The lemma now follows by $E_0$-ergodicity of $E_G$.
\end{proof}

\begin{lemma}
\label{homomorphismlemma}
Suppose $G$ and $H$ are countable groups and $K\triangleleft G$ is
normal. Suppose that $G$ acts by measure preserving transformations
on probability space $(X, \mu)$ and the action of $K$ is by itself already
ergodic.

Suppose that
\[\alpha: X\times G\rightarrow H\]
is a cocycle with
\[\alpha(x, k)=1\]
for all $k\in K$ and $\mu$-a.e. $x\in X$.

Then there is a homomorphism
\[\pi: G\rightarrow H\]
such that for all $g\in G$ and $\mu$-a.e. $x\in X$
\[\alpha(x, g)=\pi(g).\]
\end{lemma}

\begin{proof}

For a.e. $x$ and any $g\in G,$ $k\in K$ we have
\[\alpha( k\cdot x, g)=\alpha(k\cdot x, g)\alpha(x, k)\]
\[=\alpha(x,gk)=\alpha(x,\bar{k}g),\]
for $\bar{k}=gkg^{-1}\in K$, which in turn on a conull set
equals
\[\alpha(g\cdot x, \bar{k})\alpha(x,g)=\alpha(x,g).\]
Thus the function
\[x\mapsto \alpha(x, g)\]
is $K$-invariant, and so by the ergodicity of $K$ constant almost
everywhere.

If we define $\pi(g)$ to be the almost everywhere constant value of
$\alpha(x, g)$ then we obtain for any $g_1, g_2\in G$ that for a
conull set of $x\in X$
\[\pi(g_1g_2)=\alpha(x, g_1g_2)\]
\[=\alpha(g_2\cdot x, g_1)\alpha(x, g_2)=\pi(g_1)\pi(g_2).\]
\end{proof}

As a word towards notation, if $G$ is a discrete group and $S$ a discrete 
set, we let
$\{0, 1\}^{G\times S}$ be the collection of functions
\[f: G\times S\rightarrow \{0, 1\}\]
equipped with the product topology. We allow $G$ to act on this space of 
functions
by
\[(\sigma\cdot f)(\tau, s)=f(\sigma^{-1}\tau, s).\]

\begin{theorem}
\label{theoremI}

Let $S$ be a countable set.
Let $G=(\F_2)^n\times \Z$ act freely on
\[X=_{\rm df} \{0, 1\}^{G\times S}=\prod_{G\times S} \{0,1\}\]
by shift; view $X$ as coming equipped with the usual
$G$-invariant product measure.

Let $m<n$ and suppose that $H=(\F_2)^m\times \Z$ acts freely
by Borel automorphisms on standard Borel $Y$.

Then $E_G$ is $E_H$-ergodic.
\end{theorem}

\begin{proof}
We prove this by induction on $n$, with the base case $n=1$ following
by \ref{trivialization} and appendix \ref{E_0}.
Assuming now that $n>1$ let us suppose the theorem is true for
all $n'<n$.

At each $i\leq n, j\leq m$ let
\[G_i=\{1\}\times\{1\}\times...{\F_2}^{({\rm i^{th} place})}\times 
\{1\}...\times\{0\}\triangleleft G,\]
the $i^{\rm th}$ copy of the free group in $G$,
\[H_j=\{1\}\times\{1\}\times...{\F_2}^{({\rm j^{th} place})}\times 
\{1\}...\times\{0\}\triangleleft H,\]
the $j^{\rm th}$ copy of the free group in $H$,
and let
\[K=\{1\}\times\{1\}...\times \{1\}\times \Z\triangleleft G.\]

We take
\[\beta: X\times G\rightarrow H\]
to be the induced cocyle, and
at each $i\leq n$ we let
\[\beta_i:X\times G_i\rightarrow H\]
be the restriction $\beta|_{X\times G_i}$.
Defining
at each $j\leq m$
\[\pi_j:H\twoheadrightarrow H_j\]
to be the canonical surjection, we let
\[\beta_{i,j}: X\times G_i\rightarrow H_j\]
to be the cocycle given by
\[\beta_{i, j}=\pi_j\circ \alpha_i.\]
We similarly let
\[\hat{\beta}: X\times K\rightarrow H\]
be the cocycle induced for $K$ and let $\hat{\beta}_j: X\times K\rightarrow 
H_j$
be defined by $\hat{\beta}_j=\pi_j\circ \hat{\beta}$.


\medskip

For now on let us fix some $i\leq n$.

\smallskip

\noindent {\bf Case 1} At each $j\leq m$ we have that $\beta_{i, j}$
is measurably
cohomologous to a cocycle into an abelian subgroup of $H_j$.

\smallskip

Then at each $j$ fix measurable
\[\psi_j: X\rightarrow H_j\]
and abelian $H_j^0 < H_j$ such that
\[\psi_j(g\cdot x)\alpha_{i, j}(x, g)\psi_j(x)^{-1}\in H_j^0\]
for all $g\in G_i$ and almost every $x\in X$.
Thus if we define
\[\vec \psi: X\rightarrow H\]
\[x\mapsto \psi_1(x)\rho_2(x)\cdots \psi_m(x)\]
then we have that
\[\vec \psi(g\cdot x) \beta_i(x, g)
\vec \psi(x)^{-1} \in H_1^0\times H_2^0\times\cdots H_m^0\times Z.\]
In particular, $\beta_i(\cdot, \cdot)$ is (almost everywhere) cohomologous 
to a
cocycle into an abelian subgroup of $H$.

With this established, the lemma follows by \ref{trivialization}
and appendix \ref{E_0}.

\medskip

\noindent {\bf Case 2} There is some $j\leq m$ with
$\beta_{i, j}$ not measurably cohomologous to a cocycle into an
abelian subgroup of $H_j$.

\smallskip

 From now until the finish we fix this $j$.
The next argument parallels a similar point in
\cite{kechrisclassification}.

\smallskip

\noindent{\bf Claim:} We can find a measurable assignment
\[x\mapsto \mu_x\]
\[X\rightarrow M^+_{1, 3}(H_j)\]
with
\[\hat{\beta}_j(x, k)\cdot \mu_x=\mu_{k\cdot x}\]
for all $k\in K$ and a.e. $x\in X$.

\smallskip

\noindent{\bf Proof of claim:} Following appendix \ref{skew}
we can certainly find a measurable assignment
\[x\mapsto \mu_x\]
\[X\rightarrow M^+_{1}(\partial H_j)\]
with
\[\hat{\beta}_j(x, k)\cdot \mu_x=\mu_{k\cdot x}\]
for all $k\in K$ and a.e. $x\in X$.
Let us assume for a contradiction that we cannot find such an assignment with
the measures not just concentrating on at most 2 ends almost everywhere. 
Appealing
to the ergodicity of the action of $K$ on $X$ we have that either almost 
everywhere
$\mu_x$ concentrates on two ends or almost everywhere the measure concentrates
on exactly one measure. Which ever case holds, we may assume that there is no
other assignment concentrating on a greater number of ends.

For organizational simplicity, we will simply assume that $\mu_x$ almost 
everywhere
concentrates on two ends, $\{e_x, e_x'\}$; without any loss we can further
simplify the situation by assuming 
$\mu_x(\{e_x\})=\mu_x(\{e_x'\})=\frac{1}{2}$.
We then have that
\[\{\hat{\beta}_j(x, k)\cdot e_x, \hat{\beta}_j(x, k)\cdot 
e_x'\}=\{e_{k\cdot x}, e_{k\cdot x}'\}\]
all $k$, a.e. $x$, and that for any other such assignment
\[X\rightarrow [\partial H_j]^2 \]
\[x\mapsto \{f_x, f_x'\}\]
with $\{\hat{\beta}_j(x, k)\cdot f_x, \hat{\beta}_j(x, k)\cdot f_x'\}
=\{f_{k\cdot x}, f_{k\cdot x}'\}$
all $k$, a.e. $x$ we have
\[\{e_x, e_x'\}=\{f_x, f_x'\}\]
almost everywhere.
Now note the following:

  \leftskip 0.8in

\noindent For any $g\in G_i$, if we let
\[\{f_x, f_x'\}=\{\beta_{i,j}(x, g)^{-1}e_{g\cdot x}, \beta_{i,j}(x, 
g)^{-1}e_{g\cdot x}'\}\]
then for all $k\in K$ and a.e. $x\in X$
\[\{\hat{\beta}_j(x, k)\cdot f_x, \hat{\beta}_j(x, k)\cdot f_x'\}
=\{f_{k\cdot x}, f_{k\cdot x}'\}.\]

\leftskip 0in

To see this we use that $gk=kg$ and appeal to the defining identity of
cocycles to obtain
\[\beta(x, gk)=\beta(x, kg)\]
\[\therefore \beta(k\cdot x, g)\beta(x, k)=\beta(g\cdot x, k)\beta(x, g)\]
\[\therefore \beta(k\cdot x, g)\beta(x, k)\beta(x, g)^{-1}=\beta(g\cdot x, k)\]
\[\therefore \beta(x, k)\beta(x, g)^{-1}=\beta(k\cdot x, 
g)^{-1}\beta(g\cdot x, k),\]
which in particular gives
\[\hat{\beta}_j(x, k)\beta_{i,j}(x, g)^{-1}=
\beta_{i, j}(k\cdot x, g)^{-1}\hat{\beta}_j(g\cdot x, k),\]
and hence
\[\{\hat{\beta}_j(x, k)\cdot f_x, \hat{\beta}_j(x, k)\cdot f_x'\}=_{\rm df}
\{\hat{\beta}_j(x, k)\beta_{i,j}(x, g)^{-1}e_{g\cdot x},
\hat{\beta}_j(x, k)\beta_{i,j}(x, g)^{-1}e_{g\cdot x}'\}\]
\[=\{\beta(k\cdot x, g)^{-1}\beta(g\cdot x, k)\cdot e_{g\cdot x} ,
\beta(k\cdot x, g)^{-1}\beta(g\cdot x, k)\cdot e_{g\cdot x}'\}\]
\[= \{\beta(k\cdot x, g)^{-1}\cdot e_{kg\cdot x} ,
\beta(k\cdot x, g)^{-1}\cdot e_{kg\cdot x}'\}\]
\[= \{\beta(k\cdot x, g)^{-1}\cdot e_{g\cdot(k\cdot x)} ,
\beta(k\cdot x, g)^{-1}\cdot e_{g\cdot(k\cdot x)}'\}=_{\rm df}
\{f_{k\cdot x}, f_{k\cdot x}'\}.\]

That point then granted, we have by our assumptions on $\{e_x, e_x'\}$
that
\[\{e_x, e_x'\}=\{f_x, f_x'\}\]
almost everywhere. Since our choice of $g\in G_i$ was arbitrary, we have
discovered that for every $g\in G_i$ that
\[\mu_x=\beta_{i, j}(x, g)^{-1}\mu_{g\cdot x};\]
in other words,
\[\mu_{g\cdot x}=\beta_{i, j}(x, g)\mu_x.\]
In particular the map
\[x\mapsto \mu_x\]
\[X\rightarrow M^+_1(\partial \F_2)\]
is a homomorphism from $E^X_{G_i}$ to the equivalence relation
$E_{H_j}^{M^+_1(\partial H_j)}$. From \ref{A6} the latter equivalence
relation is hyperfinite, and then
from \ref{E_0} we have therefore that there is a single
$\hat{\mu}\in M^+_1(\partial H_j)$ such that for almost all $x\in X$
\[\mu_x F_{H_j}^{M^+_1(\partial H_j)} \hat{\mu}.\]
Thus we may in a Borel manner assign a group element $h_x$ to each
$x$ such that almost everywhere
\[h_x\cdot \mu_x=\hat{\mu}.\]
Thus if we replace $\beta_{i, j}(\cdot, \cdot)$ by the
cohomologous
\[\bar{\beta}: X\times G_i\rightarrow H_j\]
\[(x, g)\mapsto h_{g\cdot x} \beta_{i, j}(x, g) h_x^{-1}\]
then for all $g$ and a.e. $x$
\[\bar{\beta}(x, g)\cdot \hat{\mu}=
h_{g\cdot x}\beta_{i, j}(x, g)h_x^{-1}\cdot\hat{\mu}\]
\[=h_{g\cdot x}\beta_{i, j}(x, g)\cdot\mu_x=h_{g\cdot x}\cdot\mu_{g\cdot x}
=\hat{\mu}.\]
Thus almost everywhere
$\bar{\beta}$ maps into the stabilizer of $\hat{\mu}$, which from
\ref{A6} must be abelian.
With a contradiction to the assumption on $i$ and $j$.
\hfill ($\square$Claim)

\medskip

We can now quote \ref{A1} and find a
Borel
\[s: M^+_{1, 3}(\partial H_j)\rightarrow H_j\]
such that $s(h\cdot \mu)=hs(\mu)$, $h\in H_j$,
$\mu\in M^+_{1, 3}(\partial H_j)$, and then appealing to
the claim we define measurable
\[\hat{s}:X\rightarrow H_j\]
\[x\mapsto s(\mu_x).\]
We have for all $k\in K$ and a.e. $x\in X$
\[\hat{s}(k\cdot x)=s(\mu_{k\cdot x})=s(\hat{\beta}_j(x, 2)\cdot \mu_x)\]
\[=\hat{\beta}_j(x, k)\cdot s(\mu_x)=\hat{\beta}_j(x, k)\cdot \hat{s}(x)\]
\[\therefore \hat{s}(k\cdot x)^{-1}\hat{\beta}_j(x, k)\hat{s}(x)=1.\]

Thus let us replace the homomorphism
\[\rho: X\rightarrow Y\]
by
\[\hat{\rho}: X\rightarrow Y\]
\[x\mapsto \hat{s}^{-1}\cdot \rho(x);\]
then we have a new homomorphism from $E_G^X$ to $E_H^Y$ which has the
property that $\hat{\rho}(x)=\rho(x)$ almost everywhere. Consequently it
suffices to show to prove the theorem for this new homomorphism $\hat{\rho}$.
By analogy with before we let
\[\gamma: X\times G\rightarrow H\]
to be the induced cocyle, and
let
\[\gamma_j: X\times G\rightarrow H_j\]
be defined by $\gamma_j=\pi_j\circ \gamma$.

\medskip

\noindent{\bf Claim:} $\gamma_j|_{X\times K}$ is trivial almost everywhere.

\medskip

Note then for almost all $x$, all $k\in K$, if we choose
$\vec h=\gamma(x, k)$ then
\[\vec h\cdot \hat{\rho}(x)=\hat{\rho}(k\cdot x),\]
\[\therefore \vec h\cdot \hat{s}(x)^{-1}\cdot \rho(x)=
\hat{s}(k\cdot x)^{-1}\rho(k\cdot x)\]
\[\therefore \hat{s}(k\cdot x)\vec h\cdot \hat{s}(x)^{-1}\cdot \rho(x)=
\rho(k\cdot x)\]
which implies that
\[\hat{s}(k\cdot x)\vec h\cdot \hat{s}(x)^{-1}=\hat{\beta}(x, k)\]
\[\therefore \pi_j(\vec h)=\pi_j(\hat{s}(k\cdot x)^{-1}\hat{\beta}(x, 
k)\hat{s}(x))\]
\[=\hat{s}(k\cdot x)^{-1}\hat{\beta}_j(x, k)\hat{s}(x)=1.\]
\hfill ($\square$Claim)

\medskip

Thus by \ref{homomorphismlemma} we obtain an actual
homomorphism
\[\hat{\pi}: G\rightarrow H_j\]
such that almost every where
\[\gamma_j(x, g)=\hat{\pi}(g).\]
We want to show that $\hat{\pi}[G_{i'}]=\{1\}$ for $i'\neq i$. For this
purpose it suffices to prove the following claim:

\medskip

\noindent{\bf Claim:} If $\sigma_1, \sigma_2, \sigma_3\in \F_2$, all 
non-trivial,
and $\sigma_1$ and $\sigma_2$ commute as do $\sigma_1$ and $\sigma_3$, then
$\sigma_2$ and $\sigma_3$ commute as well.

\medskip

\noindent{\bf Proof of claim:}
Since subgroups of free groups are free, we have $\langle \sigma_1, 
\sigma_2\rangle
\cong\Z$, and thus we can find some $w\tau w^{-1}$ such that
\[\langle w\tau w^{-1}\rangle\supset \langle \sigma_1, \sigma_2\rangle.\]
We may assume by taking $\tau$ as short as possible that $\tau\neq 
\sigma^k$ for
any $k>1$, $\sigma\in\F_2$; and then we may assume that
$\tau\neq v\sigma v^{-1}$ any $\sigma\in\F_2$ and $v\neq 1$.
Thus there will be $\ell_1, \ell _2$ with
\[\sigma_1=w\tau^{\ell_1}w^{-1},\]
\[\sigma_2=w\tau^{\ell_2}w^{-1}.\]
Applying the same analysis to $\sigma_1, \sigma_3$ we
may find $u, \delta$ such that
\[\langle u\delta u^{-1}\rangle\supset \langle \sigma_1, \sigma_3\rangle,\]
but $\delta\neq v\sigma v^{-1}$, any $v\neq 1$, $\delta\neq \sigma^k $ any
$k>1$.
Then we have
\[\sigma_1=u\delta^{k_1}u^{-1},\]
\[\sigma_3=u\delta^{k_3}u^{-1},\]
some $k_1, k_3$. Thus in particular
\[\tau^{\ell_1}w^{-1}=u\delta^{k_1}u^{-1},\]
and then the minimality assumptions on $\tau$ and $\delta$ give
\[w=u,\]
\[\tau=\delta,\]
and we are happy.
\hfill ($\square$Claim)

\medskip

Thus we have established that the cocycle $\gamma_j$, derived from 
$\hat{\rho}$ and
projection to the $j^{\rm th}$ copy of the free group in $H$, is almost 
everywhere
trivial for elements of $G$ which have the identity on the $i^{\rm th}$ 
coordinate.
In other words, if we let
\[\hat{G}={\F_2}\times {\F_2 }\times...\{1\}^{({\rm i^{th} place})}
\times {\F_2}...\times  {\F_2}\triangleleft G,\]
and
\[\hat{H}={\F_2}\times {\F_2 }\times...\{1\}^{({\rm j^{th} place})}
\times {\F_2}...\times {\F_2 }\triangleleft H,\]
then for all $\vec g\in \hat{G}$ and almost all $x\in X$ there will exist 
some $\vec h\in \hat{H}$
with
\[\vec h\cdot \hat{\rho}(x)=\hat{\rho}(\vec g\cdot x).\]

In otherwords, $\hat{\rho}$ is a measurable
homomorphism of $E_{\hat{G}}^X$ to $E_{\hat{Y}}^Y$, and we
can appeal to the inductive assumption.

\end{proof}



Standing back somewhat from the specifics of the proof one can abstract
the following:

\begin{theorem}
\label{theoremII}
Let $G=G_1\times G_2\times...G_{n+1}$ and $H=H_1\times...H_m$ be countable
groups where $m\leq n$ and each $H_i$ is either the free group or abelian and
$G_{n+1}$ is amenable.

Suppose $G$ acts by mpts on standard Borel probability space $(X, \mu)$,
where the action of $G_{n+1}$ is ergodic and the action of each $G_i$, 
$i\leq n$
is $E_0$-ergodic. Suppose $H$ acts freely by Borel transformations on a
standard Borel space $Y$.

Then $E^X_G$ is $E^Y_H$-ergodic.
\end{theorem}

\begin{corollary} For $m\leq n+1$ and $S$ countable,
the shift action of $(\F_2)^{n+1}$ on $\{0, 1\}^{(\F_2)^{n+1}\times S)}$
is relatively ergodic with respect to any $E_Y^{(\F_2)^m}$, where $(\F_2)^m$
acts freely by Borel transformations on standard Borel automorphisms.
\end{corollary}

\begin{corollary} For $m, n, Y$ as above, and
\[\pi: (\F_2)^{n+1}\twoheadrightarrow (\F_2)^{m+1}\]
the projection onto the first $m+1$ coordinates, if we
let $(\F_2)^{n+1}$ act on
\[X=\{0, 1\}^{(\F_2)^{n+1}}\times \{0, 1\}^{(\F_2)^{m+1}}\]
by
\[g\cdot (x_1, x_2)=(g\cdot x_1, \pi(g)\cdot x_2),\]
then $E_{(\F_2)^{n+1}}^X$ is $E^Y_{(\F_2)^m}$-ergodic.
\end{corollary}

\begin{proof} Write $(\F_2)^{n+1}$ as $(\F_2)^{m+1}\times S_0$, where
$S_0=(\F_2)^{m-n}$. We take one more step in this direction and
let $S=S_0\cup\{*\}$, where $*$ is any point not in $S_0$, and observe that,
under its natural inclusion into $(\F_2)^{n+1}$,
the action of $(F_2)^{m+1}$ on $X$ is isomorphic in a measure preserving
fashion to the shift action of $(\F_2)^{m+1}$ on
\[\{0, 1\}^{(\F_2)^{m+1}\times S}.\]
\end{proof}






Thus we obtain a funny kind of ``invariant" which can be used to
distinguish between different kinds of equivalence relations induced
by the free measure preserving actions of some $(\F_2)^\ell$ on a
standard Borel probability space. We can consider the largest $m<\ell$
such that we obtain a relative ergodicity result with respect to free
actions of $(\F_2)^m$.

The example of the last corollary shows that this largest $m$ can be any 
natural
number less than $\ell$.
















\section{Amenable classes of structures}

In what follows we will adopt the convention that whenever we consider
a class of structures $\c$ we have in mind that they are all on some fixed
countable set. Most of the time that set could be taken to be $\N$, but
it is notationally convenient to allow domains such as $\N^n$.

\begin{definition}
We say that a countable Borel equivalence relation
$E$ on $X$ is {\it amenable} if there is an $E$-invariant
assignment
\[x\mapsto \varphi_x\]
assigning a mean to each equivalence
class $[x]_E$ which has the property that whenever $\mu$ is a
Borel probability
measure on $X$ and
\[f: E\rightarrow [-1, 1]\]
is Borel then
\[x\mapsto \varphi_{x}(f_x)\]
is $\mu$-measurable, where $f_x: [x]_E\rightarrow [-1,1]$
is defined by
\[y\mapsto f(x, y).\]
\end{definition}


\begin{definition}
We say that a standard Borel class of structures ${\cal C}$ is {\it 
essentially
countable} if
\[\cong|_{\cal C}\leq_B E_\infty;\]
that is to say, if its isomorphism relation is essentially countable.
\end{definition}

\begin{definition}
We say that a class of countable structures ${\cal C}$, each having underlying
set $\N^n$, some $n\in\N$, is
{\it amenable} if there is an assignment
\[{\cal M}\mapsto \varphi_{\cal M}\] of
means on $\N^n$ to structures in ${\cal C}$
such that whenever $\pi: \m\cong\n$
is an isomorphism, then
\[\varphi_\n\circ \pi=\varphi_\m,\]
and moreover whenever $\mu$ is a standard
Borel probability measure on $X$ and $f: {\cal C}\times \N^n\rightarrow 
[-1,1]$
is Borel then
\[\m\mapsto\varphi_\m(f_\m)\]
is $\mu$-measurable.

\end{definition}

\begin{example} (Compare \cite{jakelo}.)
Assume the continuum hypothesis, and let $\c$ be the structures on
$\N$ which are isomorphic to
\[(\Z, <, U),\]
the integers under the usual ordering, equipped with some unary predicate
$U\subset \Z$ which will depend on the particular $\m\in\c$. For each 
$\m\in \c$ we
have may in a Borel manner choose an isomorphism
\[\psi:\m\cong\Z,\]
and then at each $n$ let
\[\varphi_{\m, n}:[-1,1]^\m\rightarrow [-1, 1]\]
\[f\mapsto \frac{1}{2n+1}\sum_{i=-n}^{i=n} f(i).\]
Using Modobodzki's limit medial we may find a universally measurable 
ultrafilter
${\cal U}$ on $\N$ and for each $\m$ let
\[\varphi_m=\int_{\cal U}\varphi_{\m, n}: :[-1,1]^\m\rightarrow [-1, 1]\]
\[f\mapsto \int_{\cal U}\frac{1}{2n+1}\sum_{i=-n}^{i=n} f(i).\]

\end{example}






We have to face up to the fact that ${\cal C}$ may be amenable
without being essentially countable: For instance consider the models of 
\cite{makkai}, consisting of
expansions of $(\N\cdot \Z, <^{\rm lex})$; as discussed in 
\cite{hjorthproduct},
this isomorphism relation is far from essentially countable, but if we
assume CH then we can, in parallel to the example above, use Modobodzki to 
obtain an invariant
and suitably measurable assignment of means. Worse: The countable class of 
structures
may be amenable and essentially countable without  ${\cong}|_{\cal C}$ 
being Borel reducible
or universally measurably reducible to an amenable countable Borel equivalence
relation. If we take any class of structures ${\cal B}$ whatsoever and
replace it by ${\cal C}=\{\m\dot{\cup}(\Z, <): \m\in{\cal B}\}$ then under 
CH we may
assign suitably invariant means by concentrating on the $(\Z, <)$, while 
maintaining
$\cong|_{\cal B}\leq_B \cong|_{\cal C}$.

This difficulty is the subject of some rather tortured lemmas below. Under
suitable assumptions amenability, to the effect that the amenability is 
suitably ubiquitous,
we do indeed obtain that essentially countability will entail reduction to an
amenable countable Borel equivalence relation.

The second problem is that without CH there is no known way to obtain any
examples of measure amenable structures.

This will ultimately be bypassed by a standard metamathematical trick. The
statements we will be proving are all sufficiently absolute, and it turns 
out that
we may simply impose the assumption that the continuum hypothesis holds.

Perhaps there is a more elegant solution, and a better way to formulate
a relevant concept of amenability.

The next lemma is implicit in \cite{hjorthkechris}.

\begin{lemma}
\label{D1}
  Let ${\cal C}$ be a class of structures with
\[\theta: {\cal C}\rightarrow X_\infty\]
witnessing $\cong|_{\cal C}\leq_B E_\infty$. Then we may find a countable
Borel equivalence relation $F$ on standard Borel $Y$ with $\gamma: 
\c\rightarrow Y$
witnessing $\cong|_\c\leq_B F$
and we may in a
Borel manner assign to each $\m\in{\cal C}$ a pair $(A_\m, \rho_m)$
such that

\leftskip 0.4in

\noindent (a) $A_\m\subset \bigcup_n \m^n$;

\noindent (b) $\rho_\m: A_\m\rightarrow Y$;

\noindent (c) $\m\mapsto (A_\m, \rho_\m)$ is $\cong$-invariant, in the
sense that if $\pi:\m\cong\n$ then
\[\pi[A_\m]=A_\n,\]
\[\rho_\m=\rho_\n\circ \pi;\]

\noindent (d) $\rho_\m(\vec a) F (\m)$ all $\vec a\in A_\m$;

\noindent (e) $Y$ equals the range of $\gamma$;

\noindent (f) there is Borel $\psi: Y\rightarrow \c$ such that 
$\psi(\gamma(\m))\cong\m$ all
$\m\in\c$.

\leftskip 0in

\end{lemma}

\begin{proof} Following \cite{hjorthkechris} we may find a countable fragment
$F\subset{\cal L}_{\omega_1, \omega}$ such that for each $\m\in\c$ there 
will be
some $\vec a\in \m^<\N$ such that $\langle \m; \vec a\rangle$ is $F$-atomic.
The following are all routine consequences of the definition of atomicity:

\leftskip 0.4in

(i) the set $\{\langle\m, \vec a\rangle: \langle\m, \vec a\rangle$ is 
$F$-atomic $\}$
is Borel;

(ii) in a given fragment $F_n=F(c_1, ..., c_n)$ (where $c_1,..., c_n$ are 
fresh
constant symbols), the collection of $T\subset F_n$ which are complete and 
admit an
atomic model is Borel;

(iii) for each $T$ as in (ii) we may in a Borel way choose some $\langle\m, 
\vec a\rangle$
with $\langle\m, \vec a\rangle\models T$.

\leftskip 0in

Given these facts, we let
\[A_\m=\{\vec a: \langle\m, \vec a\rangle {\rm \: is } F{\rm -atomic}\},\]
\[\rho_\m(\vec a)={\rm Th}_F(\langle\m, \vec a\rangle),\]
\[Y=\{{\rm Th}_F(\langle\m, \vec a\rangle): \vec a\in A_\m\},\]
and for each $T\in Y$ we choose $\m\in \c$ as (iii).
\end{proof}

\def\c{{\cal C}}

\begin{definition}
For $\c$ a class of structures, we let $\c^{[n]}=\{\m^n:\m\in\c\}$, the 
class of all structures
arising as $n$-fold products of a structure from $\c$.
\end{definition}


\begin{lemma}
\label{D2}
If $\c$ is an amenable class of structures on $\N$ then
$\c^{[n]}$ is an amenable class of structures on $\N^n$.
\end{lemma}

\begin{proof}
Let $\m\mapsto \varphi_\m$ be the invariant and universally measurable 
assignment
of means. We then define $\varphi_\m^\ell$ recursively on $\ell$, beginning 
with
\[\varphi_\m^1=\varphi_\m\]
and continuing with
\[\varphi_\m^{\ell +1}(f)=\varphi_\m(a\mapsto \varphi_\m^\ell(f_a)),\]
where $f_a: \m^\ell\rightarrow [-1, 1]$ is defined by
$f_a(b_1, b_2, ..., b_\ell) =f(a, b_1, b_2, ..., b_\ell)$.
The invariance property for the various $\varphi_\m^\ell$'s is easily checked.
The main concern is measurability.

Given any measure $\mu$ on $\c^{[n]}$, for any $a_1,..., a_{n-1}\in \N$ we
have
\[\m\mapsto \varphi_\m(a_n\mapsto f(a_1, a_2, ..., a_n))\]
$\mu$-measurable, and hence we can find some $\mu$-conull set $A_1\subset\c$
such these functions are all Borel on this set.
Then the functions
\[A_1\rightarrow [-1,1]\]
\[\m\mapsto \varphi_\m(a_{n-1}\mapsto \varphi_\m(a_n\mapsto 
\varphi_\m(f(a_1, a_2,...a_{n-1}, a_n)))\]
are $\mu$-measurable, and we again can go to a conull $A_2\subset A_1$ on 
which they
are all Borel, and continue in this fashion until we reach $A_n$.
\end{proof}

\begin{lemma}
\label{D3}
Let $\c$ be an essentially countable class of amenable structures on
$\N$ with the futher property that
given any Borel and $\cong$-invariant assignment
\[\c \rightarrow {\cal P}(\N^n)\]
\[\m\mapsto A_\m\]
we can choose an assignment of means
\[\m\mapsto \varphi_\m\]
which not only witnesses amenability but has the further property of always
concentrating on $A_\m$ -- that is to say, $\varphi_\m(\chi_{A_\m})$ is always
1.

Then $\cong|_\c$ is Borel reducible to an amenable
countable Borel equivalence relation.
\end{lemma}

\begin{proof}
Following \ref{D1} we may choose $\m\mapsto (A_\m, \rho_\m), \gamma, \psi, 
Y, F$ all as indicated
there. At each $n$ let $Y_n=\{y\in Y: {\rm \: for\: }\langle\m, \vec 
a\rangle=\gamma(y),
\vec a\in \m^n\}$; it suffices to show that each $F|_{Y_n}$ is amenable.

But for this purpose we can choose an assignment of means
\[\m\mapsto \varphi_\m\]
such that each $\varphi_\m$ concentrates on $\m^n\cap A_\m$. (In the case
that $\m^n\cap A_\m$ is empty, then $\m$ plays no part in $Y_n$ and we could
for instance ask that $\varphi_\m$ simply concentrate on $\m^n$.)

Then for $y\in Y_n$ we let
\[\varphi_y: [-1, 1]^{[y]_F}\rightarrow [-1, 1]\]
\[f\mapsto \varphi_{\psi}(\vec a\mapsto f(\rho_\m(\vec a))).\]


\end{proof}


\begin{corollary}
\label{D4}
Let $\c$ be a class of amenable structures and suppose that for all $\m\in 
\c$ and
$a\in \m$
\[\m\models \forall b\bigvee_{t{\rm \: a \: term}} t(a)=b;\]
that is to say, the algebraic closure of any $a\in \m$ is the whole structure.

Then $\cong|_\c$ is Borel reducible to an amenable countable Borel 
equivalence relation.
\end{corollary}

\begin{proof} It is easily seen, and observed in the course of 
\cite{hjorthkechris}, that
since structures in $\c$ are finitely generated we must have $\cong|_{\c}$
essentially countable. Let $(t_i)_{i\in\n}$ be some fixed enumeration of 
terms.

Given any invariant Borel assignment $\m\mapsto A_\m\subset \m^n$ and 
invariant measurable
\[\m\mapsto \varphi_\m\]
we can define for each $\m\in \c$
\[\sigma_\m: \m\mapsto A_\m\]
by
\[\sigma_\m(a) =(t_{i(1)}(a), t_{i(2)}(a),..., t_{i(n)}(a))\]
where the sequence $i(1), i(2), ..., i(n)$ is chosen lexicographically 
least to
ensure
\[(t_{i(1)}(a), t_{i(2)}(a),..., t_{i(n)}(a))\in A_\m.\]
\end{proof}


\section{Non-reduction}

Throughout this section we assume the continuum hypothesis. We do this without
entertaining any philosophical bias either for or against CH, but simply
because the statements we are attempting to prove are all $\Ubf{\Pi}^1_1$ and
easily absolute between the universe and some extension where we have forced a
wellorder of $2^{\aleph_0}$ of ordertype $\aleph_1$.


We first need a technical result. This result  can be most easily proved
by collecting together the observations of the last section.

\begin{lemma} Let $G$ and $H=H_1\times H_2\times...H_N$ be countable
groups acting by Borel transformations on standard Borel $X$ and $Y$, with
the $H$ acting freely, and let
\[\rho: X\rightarrow Y\]
be a homomorphism of $E^X_G$ to $E^Y_H$
with
\[\beta: X\times G\rightarrow H\]
the induced cocycle, and
\[\beta_j=\pi_j\circ \beta: X\times G\rightarrow H_j\]
the induced cocycle for the various $H_j$.

Suppose that at each $j\leq N$ we have either

\leftskip 0.4in

\noindent (a) $\beta_j$ maps into an abelian subgroup of $H_j$, or

\noindent (b) there is a hyperfinite equivalence relation $F_j$ on some 
standard
Borel $Z_j$ and a homomorphism $\hat{\rho}: X\rightarrow Z_j$ from $E_G^X$ to
$F_j$ with the property that for all $x\in X, g\in G$
\[\hat{\rho}_j(g\cdot x)=\hat{\rho}_j(x)\Leftrightarrow \beta_j(x, g)=1.\]

\leftskip 0in

Then there is an amenable countable Borel equivalence relation $\hat{F}$ on 
$\hat{Z}$
such that $\rho$ factors through $\hat{F}$.
\end{lemma}

\begin{proof} We may as well assume that $N=2$, $H_1$ is abelian, and that 
there is
homomorphism
\[\hat{\rho}: X\rightarrow Z\]
from $E^X_G$ to some hyperfinite $F$ with the additional property
for all $x\in X, g\in G$
\[\hat{\rho}(g\cdot x)=\hat{\rho}(x)\Leftrightarrow \beta_2(x, g)=1.\]
We make the further harmless assumption that each $\hat{\rho}[[x]_G]$ is 
infinite;
the case for those which are finite can be dealt with separately using an 
argument
which is similar to and somewhat easier than the one below.

Since hyperfinite equivalence relations are induced by $\Z$-actions (see 
\cite{dokelo},
\cite{jakelo}), we may in a Borel fashion assign
\[z\mapsto <_z,\]
where each $<_z$ is a linear order of ordertype $\Z$ on $[z]_F$ with the
the invariance property that
\[z_1 F z_2 \Rightarrow <_{z_1}=<_{z_2}.\]

We then write
\[x R_\ell x'\]
if $\hat{\rho}(x) F \hat{\rho}(x')$ and $\hat{\rho}(x)$, $\hat{\rho}(x')$ 
are exactly
$\ell$ many places apart in the linear order 
$<_{\hat{\rho}(x)}=<_{\hat{\rho}(x')}$ restricted
to $\hat{\rho}[[x]_G]$. In other
words, if there are $x_0, x_1, ..., x_\ell\in [x]_G$ with
\[x_0=x,\]
\[x_\ell =x',\]
\[\hat{\rho}(x_0)=\hat{\rho}(x),\]
\[\hat{\rho}(x_\ell)=\hat{\rho}(x'),\]
each
\[\hat{\rho}(x_i)<_{\hat{\rho}(x)} \hat{\rho}(x_{i+1})\]
and for all $\hat{x}\in [x]_G$ with
\[\hat{\rho}(x_0)<_{\hat{\rho}(x)} \hat{\rho}(\hat{x})<_{\hat{\rho}(x)} 
\hat{\rho}(x_\ell)\]
there will be some $i$ with
\[\hat{\rho}(\hat{x})=\hat{\rho}(x_i).\]
This definition makes sense for $\ell\geq 0$, and we can extend it in the 
natural
way to$\ell<0$ by setting
\[x' R_\ell x\]
if $x R_{-\ell} x'$.

\def\d{{\cal D}}

We will build a class of amenable structures $\c$ and then a
class of expansions $\d$. We will show that $\rho$ factors through
$\cong|_\d$ and then appeal to \ref{D4} to argue that $\cong|_\d$ is
bi-borel reducible to an amenable countable Borel equivalence relation.

The language for $\c$ will have unary functions $(F_\ell)_{\ell\in \Z}$, 
$(\hat{F}_h)_{h\in H_1}$,
and an equivalence relation $E^*$. The structures in $\c$ are those 
structures on
$\N$ that satisfy:

\leftskip 0.4in

\noindent (i) $\forall a ([a]_{E^*}=\{\hat{F}_h(a): h\in H_1\})$;
for all $h_1, h_2\in H_1$
\[\hat{F}_{h_1}\circ \hat{F}_{h_2}=\hat{F}_{h_1 + h_2},\]
\[\forall b (\hat{F}_0(b)=b);\]

\noindent (ii) $\forall a, b$
\[\bigvee_{\ell \in \Z}\bigvee_{h\in H_1}(\hat{F}_h\circ F_\ell(a)=b);\]

\noindent (iii) for all $\ell\in \Z, h_1, h_2, h_3, h_4\in H_1$
\[\forall a, b\bigvee_{\bar{h}\in H_1} (\hat{F}_{\bar{h}}\circ 
\hat{F}_{h_1}\circ F_\ell
\circ \hat{F}_{h_2}(a) = \hat{F}_{h_3}\circ F_\ell \circ \hat{F}_{h_4}(a));\]

\noindent (iv)
\[\forall a\bigvee_{h\in H_1}(\hat{F}_h(a)=F_0(a));\]

\noindent (v) for all $\ell_1, \ell_2\in \Z$
\[\forall a\bigvee_{h\in H_1}(\hat{F}_h\circ F_{\ell_1 + \ell_2}(a) = 
F_{\ell_1} \circ F_{\ell_2} (a)).\]

\leftskip 0in

In essence (i) states that the $\{\hat{F}_h: h\in H_1\}$ act transitively 
on each $E^*$-equivalence
class, and that the action
respects the group structure suggested by $H_1$.
(ii) states in particular that the algebraic closure of any point is the 
whole structure. (iii)-(v)
state that in essence that $\Z$, via the assignment $\ell\mapsto F_\ell$, 
acts on the structure's
collection of
all $E^*$ equivalence classes.

We will in fact that this class of structures is not only amenable but
``2-amenable" in the sense of \cite{jakelo}.

We begin with a Folner sequence $(D_n)_{n\in\N}$ for $H_1$, that is to say 
a collection
with the property that
\[D_n\subset D_{n+1},\]
\[H_1=\bigcup_n D_n,\]
and for all $h\in H_1$
\[\frac{|D_n h\Delta D_n|}{|D_n|}\rightarrow 0.\]
For $\n \in \N$, $\m\in \c$, $a\in \m$, we first let
\[f^{a, \m}_n=\frac{1}{|D_n|}\chi_{\{F_h(a): h\in D_n\}};\]
that is to say, $f^{a, \m}_n(b)$ equals $1/D_n$ if there exists
$h\in D_n$ with $\m\models \hat{F}_h(a)=b$, and equals zero otherwise.
We iterate and define for each $n, m$
\[f^{a, \m}_{n, m}:\m\rightarrow [0,1]\]
\[f^{a, \m}_{n, m}=\frac{1}{2n+1}(\sum_{\ell\in [-m, m]} f_n^{F_\ell(a), 
m}).\]

At each $\m\in \c$ and $a, b\in \m$ with $a E^* b$, it follows from (i) and 
the properties of the
Folner sequence that
\[{\rm lim}_{n\rightarrow \infty} |\! | f^{a, \m}_n-f^{b, \m}_n |\! |_{\ell_1}
\rightarrow 0,\]
and then by (ii) we have that for {\it all} $a, b\in \m$
\[{\rm lim}_{m\rightarrow \infty}({\rm limsup}_{n\rightarrow \infty}
  |\! | f^{a, \m}_{n, m}-f^{b, \m}_{n, m}  |\! |_{\ell_1}\rightarrow 0.\]
We massage this into the form required by $\S$2.5 \cite{jakelo} by 
recalling that
each $\m\in \c$ has $\N=\{1, 2, 3, ...\}$ as its underlying set, and letting
\[f^\m_{n, m}=f^{1, \m}_{n, m},\]
and thereby obtain that $\c$ is 2-amenable.

As discussed in \cite{kechrisamenable},
it follows from CH and Modobodki's theorem that $\c$ is amenable in our 
sense above.

The class $\d$ arises by expanding the structures in $\c$ through
the addition of fresh unary predicates, $(U_n)_{n\in\N}$. For this purpose
let us fix countable Borel separating family $(B_n)_{n\in\N}$ for $Y$.
To each $\m\in \c$ and and function
\[\sigma: \m\rightarrow Y\]
we associate the structure $\m^\sigma$ which includes the language of $\c$,
on which it behaves just like $\m$, but interprets the fresh unary 
predicates by
obeying the dictate that
\[(\n^\sigma\models U_n(a)) \Leftrightarrow \sigma(a)\in B_n.\]
We let $\d$ be the collection
$\{\m^\sigma | \m \in \c, \sigma: \m\rightarrow Y\}$.
These satisfy the assumptions of \ref{D4} and so it suffices to show that
$\rho$ factors through $\cong|_\d$.

For each $x$ we define  a pair $n_x, \sigma_x: \n_x\rightarrow Y$. We will 
not literally
have that $\n_x$ is in $\c$ because its underlying set, although countable, 
will
not equal $\N$. Nonetheless we can simply choose an isomorphic copy 
$\n_x^*\in \c$
and argue that
\[\m_x=(\n_x^*)^{\sigma_x^*}\in \d,\]
for some natural choice of $\sigma^*_x$,
is as required.

The underlying set of $\n_x$ will be the set of pairs $(\ell, y)$ such for 
some
$\hat{x}\in [x]_G$ we have

\leftskip 0.4in

\noindent (a) $x R_\ell \hat{x}$;

\noindent (b) $y E_{H_1} \rho(\hat{x})$.

\leftskip 0in

\def\bh{\bar{h}}
\def\hr{\hat{\rho}}

We set
\[(\ell, y) E^* (\bar{\ell}, \bar{y})\]
if and only if $\ell=\bar{\ell}$. We naturally enough define the
$\hat{F}_h$ functions as suggested by the action of $H_1$, so that
\[\hat{F}_h(\ell, y)=(\ell, h\cdot y).\]
The $F_\ell$ functions require more care.

First fix an enumeration
$(\bh_i)_{i\in\N}$ of $H_1\times H_2$. Then let
\[F_\ell((y, \bar{\ell}))=(\bar{h}_{i_0}\cdot y, \ell+\bar{\ell})\]
where $i_0$ is least such that for some $ \hat{x}\in [x]_G$ we have
\[x R_{\ell +\bar{\ell}} \hat{x},\]
\[ (\bh_{i_0}\cdot y) E_{H_1}\rho(\hat{x}).\]



We then, in the most obvious way possible, define
\[\sigma_x: \n_x\rightarrow Y\]
\[(\ell, y)\mapsto y.\]
In a Borel manner it is possible to choice for each $x$ a bijection
\[\pi_x: \N\cong\{(\ell, y): \ell\in \Z, \exists \hat{x} \in [x]_G (y 
E_{H_1} \rho{\hat{x}},
x R_\ell \hat{x})\},\]
and we finish the definion by letting $\n_x^*$ be the copy on $\N$ of 
$\n_x$ provided
by $\pi_x$ and then setting
\[\m_x=(\n_x^*)^{\sigma_x\circ\pi_x}.\]



\medskip

\noindent{\bf Claim(I):} If $\hat{x}_1, \hat{x}_2\in
[x]_G$ with
\[x R_{\ell } \hat{x}_1,\]
\[x R_{\ell } \hat{x}_2,\]
then
\[[\rho(\hat{x}_1)]_{H_1}=[\rho(\hat{x}_2)]_{H_1}.\]

\smallskip

\noindent{\bf Proof of claim:}  Fix $g\in G$ with $g\cdot \hat{x}_1=
\hat{x}_2$.

It follows from the definition of
$R_{\ell }$ that $\hr(\hat{x}_1) =\hr(\hat{x}_2)$, and hence
\[\beta_2(\hat{x}_1, g)=1\]
\[\therefore \beta(\hat{x}_1, g)\in H_1\times\{1\}\]
\[\therefore [\rho(\hat{x}_1)]_{H_1}=[\rho(\hat{x}_2)]_{H_2}.\]
\hfill ($\square$Claim)

\medskip

\noindent{\bf Claim(II):} For any $x\in X$ we have $\n_x^*$ in $\c$.

\smallskip

\noindent{\bf Proof of claim:} Appealing to claim(I) we have
\[\{y: (\ell, y)\in \n_x\}\]
is an $H_1$-equivalence class for any $\ell \in Z$. This quickly implies 
(i) in
the definition of $\c$. (ii) follows from the careful
choice of the $F_\ell$ functions.
For (iii)-(v) we observe that the $E^*$-equivalence classes are arranged 
into a
$\Z$-chain by $\{x R_\ell(\cdot): \ell\in \Z\}$ and that the action of the
$F_\ell$ functions respects that ordering.
\hfill ($\square$Claim)

\medskip

\noindent{\bf Claim(III):} $\m_{x_1}\cong\m_{x_2}\Rightarrow \rho(x_1) E_H 
\rho(x_2)$.

\smallskip

\noindent{\bf Proof of claim:} Since the functions
$\sigma_{x_1}\circ\pi_{x_1}$ and $\sigma_{x_2}\circ\pi_{x_2}$
must have a point in common.
\hfill($\square$Claim)

\medskip

\noindent{\bf Claim(IV):} $\rho(x_1) E_H \rho(x_2)\Rightarrow \m_{x_1}\cong 
\m_{x_2}$.

\smallskip

\noindent{\bf Proof of claim:} If $x_1 E_G x_2$ then we can find some 
$\hat{\ell}$ with
\[x_1 R_{\hat{\ell}} x_2 ,\] and hence for all $\ell$
\[x_1 R_{\ell +\hat{\ell}} (\cdot) = x_2 R_{\ell}(\cdot).\]
We simply define an isomorphism
\[\psi: \n_{x_1}\cong \n_{x_2} \]
by
\[\psi(\ell, y) =(\ell-\bar{\ell}, y).\]
This isomorphism clealy intertwines $\sigma_{x_1}$ and $\sigma_{x_2}$.
\hfill ($\square$Claim)


Thus $\rho$ factors through $\cong|_\d$ and we finish by an appeal to
\ref{D4}.





\end{proof}

\begin{theorem}
\label{theoremIII}
Let $G=G_1\times G_2\times...G_N\times K$ and
$H=(\F_2)^M\times...\times\bar{K}$ be countable groups,
$K\cong \Z$, $\bar{K}$ abelian, $M<N$, $G$ acting by mpts on
standard Borel $(X, \mu)$ and $H$ acting freely by measure preserving
transformations on standard Borel $Y$, with the action of $K$ on
$(X, \mu)$ being ergodic.\footnote{Here and beyond we identify $K$ with
$\{1\}\times...\times\{1\}\times K\triangleleft G$, we identify $G_i$ with
$\{1\}\times...\times G_i \times \{1\}\times...\{1\}\times 
\{0\}\triangleleft G$,
we identify $\bar{K}$ with
$\{1\}\times...\times\{1\}\times \bar{K}\triangleleft H$, and in keeping 
with the
notation of \ref{theoremI} we let
$H_j=\{1\}\times\{1\}\times...
{\F_2}^{({\rm j^{th} place})}\times \{1\}...\times\{0\}\triangleleft H$.}
Let $\rho: X\rightarrow Y$ be a homomorphism of $E_G^X$ to $E_H^Y$.

Then at some $i\leq N$, $\rho$ viewed as a homomorphism from $E^X_{G_i}$ to
$E^Y_H$ factors through a hyperfinite equivalence relation $\mu$-a.e.
\end{theorem}

\begin{proof}
We use $\rho$ to define induced cocyles
\[\beta_j: X\times G\rightarrow H_j\]
\[\beta_{i, j}: X\times G_i\rightarrow H_j\]
\[\beta_{N+1, j}: X\times K\rightarrow H_j\]
for $i\leq N$, $j\leq M$ in the usual way.


\medskip

\noindent{\bf Definition:} We say that a coordinate $j\leq M$ is {\it passive}
for $G_i$ is either:

\leftskip 0.4in

\noindent (a) $\beta_{i, j}$ is cohomologous to a cocycle into an abelian 
subgroup of
$H_j$; or

\noindent (b) there is a standard Borel $Z_j$ and hyperfinite countable 
$F_j$ on
$Z_j$ and a homomorphism
\[\hat{\rho}_j: X\rightarrow Z_j\]
from $E_{G_i}$ to $F_j$ such for all $g\in G_i$ and $\mu$-a.e. $x\in X$
\[\hat{\rho}_j(g\cdot x) =\hat{\rho}_j(x) \Leftrightarrow \beta_{i, j}(x, 
g)=1.\]

\leftskip 0in

\medskip

Naturally we want to advance this terminology and say that a coordinate is
{\it active} if it is not passive.

\medskip

\noindent{\bf Claim(I)} If $j\leq M$ is active for $G_i$ then there is a Borel
assignment of measures
\[X\mapsto M^+_{1, 3}(\partial H_j)\]
\[x\mapsto \mu_x\]
such that for all $k\in K$ and $\mu$-a.e. $x\in X$
\[\mu_{k\cdot x} = \beta_{N+1, j}(x, k) \cdot \mu_x.\]

\smallskip

\noindent{\bf Proof of claim:} Following appendix \ref{B} we may certainly
find a
Borel
assignment of measures
\[X\mapsto M^+_{1}(\partial H_j)\]
\[x\mapsto \mu_x\]
such that for all $k\in K$ and $\mu$-a.e. $x\in X$
$\mu_{k\cdot x} = \beta_{N+1, j}(x, k) \cdot \mu_x.$
The task is to do this so that the measures do not simply concentrate on 
two or fewer ends.

Suppose then that we can find no such suitably invariant assignment
with $\mu_x\in M^+_{1, 3}(\partial H_j)$ almost everywhere. Following the 
proof of a parallel
step in \ref{theoremI} we may conclude that
\[\mu_{g\cdot x} =\beta_{i, j}(x, g)\cdot \mu_x\]
all $g\in G_i$ and $\mu$-a.e. $x\in X$.

Following \ref{A6} and the notation discussed there,
we obtain a split in cases:

\medskip

\noindent{\bf Case(a)} Almost everywhere $\mu_x\in B_2$, the set of $\nu\in
M^+_{1}(\partial H_j)$ with non-trivial stabilizer under the action of $H_j$.

\medskip

Then from \ref{A6} we have $E^{B_2}_{H_j}$ smooth, and so by appealing to
the ergodicity of the $K$ action we can find a single $\nu_0$ such that
\[\mu_x E^{B_2}_{H_j} \nu_0\]
almost everywhere. Appealing to the selection theorem for Borel sets with
countable sections, we may find a Borel function
\[\gamma: X\rightarrow H_j\]
such that
\[\gamma(x)\cdot \mu_x=\nu_0\]
almost everywhere.

Now consider the new cocycle
\[\alpha_{i, j}:X\times G_i\rightarrow H_j\]
\[(x, g)\mapsto \gamma(g\cdot x) \beta_{i, j}(x, g) \gamma(x)^{-1}.\]
This is certainly cohomologous to $\beta_{i, j}$ and we have that almost
everywhere
\[\alpha_{i, j}(x, g)\cdot \nu_0= \gamma(g\cdot x)
\beta_{i, j}(x, g) \gamma(x)^{-1}\cdot\nu_0\]
\[=\gamma(g\cdot x) \beta_{i, j}(x, g)\cdot\mu_x=\gamma(g\cdot x)\cdot 
\mu_{g\cdot x}=\nu_0\]
and hence
\[\alpha_{i, j}(x, g)\in (H_j)_{\nu_0},\]
the stabilizer of $\nu_0$.

 From \ref{A6} this stabilizer is abelian, and $H_j$ is passive for $G_i$.

\medskip


\noindent{\bf Case(b)} On a positive measure subset of $X$ the 
associated  $\mu_x$ is
not in $B_2$.

\medskip

Then appealing to the ergodicity of the action of $K$ we can obtain that 
almost
everywhere $\mu_x$ has trivial stabilizer. Thus almost everywhere we have
\[\mu_{g\cdot x}=\mu_x\]
if and only if
\[\beta_{i, j}(x, g)\cdot \mu_x=\mu_x\]
if and only if
\[\beta_{i, j}(x, g)=1.\]

\def\hr{\hat{\rho}}

Thus if we let $Z_j=B_0\cup B_1$, and define
\[\hr_j: X\rightarrow Z_j\]
\[x\mapsto \mu_x\]
almost everywhere, then by \ref{A6} we obtain after all that
$H_j$ is passive for $G_i$.
\hfill ($\square$Claim)

\medskip

\noindent{\bf Claim(II):} If $j\leq M$ is active for some $G_i$ then it is 
passive for
every other $G_{i'}$.

\smallskip

\noindent{\bf Proof of claim:} This parallels a similar point from
\ref{theoremI}.

Appealing to the last claim and \ref{A1} we may find a Borel
\[\hat{s}: X\rightarrow H_j\]
such that for any $k\in K$ we have almost everywhere
\[\hat{s}(k\cdot x) =\beta_{N+1, j}(x, k)\cdot \hat{s}(x),\]
and thus if we replace
\[\beta_j: X\times G\rightarrow H_j\]
by the cohomologous cocycle
\[\alpha_j: X\times G\rightarrow H_j\]
\[(x, g)\mapsto \hat{s}(g\cdot x)^{-1}\beta_j(x, g)\hat{s}(x)\]
then $\alpha_j$ is trivial for $K$. Appealing to
\ref{homomorphismlemma} we have a homomorphism
\[\pi: G\rightarrow H_j\]
with $\pi(g) =\alpha_j(x, g)$ almost everywhere, and we finish as
in the parallel part of the proof of \ref{theoremI}.
\hfill ($\square$Claim)

Thus by the pigeonhole principle we obtain some
$i\leq N$ such that no $j\leq M$ is active for $G_i$.
 From this it follows that the cocycle
\[\beta|_{X\times G_i}: X\times G_i\rightarrow H\]
is cohomologous to some
\[\beta^*: X\times G_i\rightarrow H\]
such that each of the projections
\[\beta^*_j: X\times G_i\rightarrow H_j\]
satisfy either (a) or (b) from the preceding lemma.

Fix $\gamma^*: X\rightarrow H$ such that
\[\gamma(g\cdot x)\beta(x, g)\gamma(x)^{-1}=\beta^*(x, g)\]
almost everywhere.
If we define
\[\rho^*: X\rightarrow Y\]
\[x\mapsto \gamma(x)\cdot \rho(x)\]
then $\rho^*$ is a homomorphism from
$E_{G_i}^X$ to $E^Y_H$ and $\beta^*$ is the induced cocycle.

Thus by the lemma we may find an amenable countable Borel
equivalence $F'$ on standard Borel $Z'$ and homomorphisms
\[\rho^*_0: X\rightarrow Z'\]
\[\rho^*_1: Z'\rightarrow Y\]
of $E^X_{G_i}$ to $F$ and of $F$ to $E^Y_H$ such that
\[\rho^*_1\circ \rho^*_0=\rho^*.\]
As in the proof of \ref{trivialization} we may find $\hat{Z}\subset Z'$ 
such that
$Z'|_{\hat{Z}}$ is hyperfinite and almost everywhere $\rho^*_0(x)\in \hat{Z}$.

We then take one further step and let $Z=\hat{Z}\times H$,
\[\rho_0: x\mapsto (\rho^*_0(x), \gamma(x)),\]
\[\rho_1: (z, h)\mapsto h^{-1}\cdot \rho^*_1(z)\] and
set $(z, h) F (z', h')$ if and only if $z F' z'$. It is easily
checked that $F$ is hyperfinite, and for almost every $x$ we have
\[\rho_1\circ \rho_0(x)= \rho_1(\rho^*_0(x), \gamma(x))\]
\[= \gamma(x)^{-1}\cdot \rho^*_1(\rho^*_0(x)))=\gamma(x)^{-1}\cdot 
\rho^*(x)=\rho(x),\]
as required.


\end{proof}



\begin{corollary} (After Gaboriau?)
Let $G=(\F_2)^N\times \Z$ and
$H=(\F_2)^M\times...\times\Z$, $M<N$, $G$ acting freely by mpts on
standard Borel probability space
$(X, \mu)$ and $H$ acting freely by measure preserving
transformations on standard Borel $Y$, with the action of $\Z$ on
$(X, \mu)$ being ergodic.
Then
\[E^X_G\not\leq_B E^Y_H.\]
\end{corollary}



\begin{proof} In light of the theorem and \ref{B2}, it suffices to show 
that if
$F$ is a countable Borel equivalence relation on a standard
Borel space $Z$, and
\[\rho: Z\rightarrow \{0, 1\}^\N\]
is a homomorphism of $F$ to $E_0$, and $\rho^{-1}(w)$ is countable all
$w\in 2^\N$, then $F$ is hyperfinite.

But this follows from the fact that
among the countable Borel equivalence relations
the class of hyperfinite equivalence relations are closed under downward under
$\leq_B$-reducibility and inclusion; see $\S$1 \cite{jakelo}.
\end{proof}

\begin{corollary} (After Adams-Kechris)
There are $\leq_B$-incomparable countable Borel equivalence relations.
\end{corollary}

\begin{proof} On the one side we consider  the Bernoulli shift
of $G=(\F_2)^3\times\Z$ on $\{0, 1\}^G$ equipped with the usual product
measure and let $X$ be the free part.

On the other side we first being with
\[H_1=(\F_2)^2\times \Z\]
\[H_2=\Z\times (\F_2)^2\]
along with their corresponding Bernoulli shifts on
$\{0, 1\}^{H_1}$ and $\{0, 1\}^{H_2}$.
We let $H=(\F_2)^2\times \Z\times (\F_2)^2$.
We let $Y_1, Y_2$ be the respectively free parts of these Bernoulli shifts and
take $Y=Y_1\times Y_2$.
We let $H$ act on $Y$ by
\[(h_1, \ell, h_2)\cdot (y_1, y_2)=((h_1, \ell)\cdot y_1, (h_2, \ell)\cdot 
y_2).\]

Since the product of mixing actions of $\Z$ are mixing, we obtain that the 
induced
action of $\Z$ on $Y$ is mixing and therefore certainly ergodic, and hence 
by the
last corollary
\[E^Y_H\not\leq E^X_G.\]

For the converse, suppose
\[\rho: X\rightarrow Y\]
is a homomorphism of $E_G^X$ to $E_H^Y$. We may write $\rho=(\rho_1, \rho_2)$
where
\[\rho_1: X\rightarrow Y_1\]
\[\rho_2: X\rightarrow Y_2\]
are homomorphisms of $E^X_G$ ot $E_H^{Y_1}$ and $E_H^{Y_2}$ respectively.

Appealing to \ref{theoremI} we obtain $X_1\subset X$ of measure one such that
$\rho_1|_{X_1}$ has countable image, and then making the same appeal once more
we have $X_2$ of measure one with $\rho_2|_{X_2}$ also having countable image.
Thus $\rho[X_2]$ is a countable subset of $Y$ and we are done.


\end{proof}





\appendix

\section{The boundary of the free group}

We adopt the notational convention that the generators of $\F_2$
be $a$ and $b$.

\begin{definition} We let $\partial \F_2$ be the {\it boundary of the free 
group},
consisting of all infinite {\it reduced} words in $a$ and $b$ -- that is to 
say, those
words with no consecutive appearances of $a$ and $a^{-1}$ or $b$ and $b^{-1}$.
We let $\F_2$ act in the obvious way on this collection, by concatenation 
and then
cancellation to obtain an infinite reduced word. Note then that $\partial\F_2$
is a compact space on which the free group acts by homeomorphisms.

$[\partial \F_2]^2$ be the collection of all unordered pairs or singletons 
from
$\F_2$. This is given a Borel structure in the following way: First we 
identify
$\partial \F_2$ with $\{\{e\}: e\in\partial \F_\}\subset [\partial 
\F_2]^2$. We then
fix a Borel linear ordering, $<$,  of $\partial \F_2\times \partial F_2$ 
and identify
\[ [\partial \F_2]^2\setminus \F_2\] with
\[\{(e_1, e_2)\in \partial \F_2\times \partial \F_2: e_1 < e_2\}.\]
We extend the action of $\F_2$ on $\partial \F_2$ by letting
\[\sigma\cdot \{e, e'\}=\{\sigma\cdot e, \sigma\cdot e'\}.\]

\end{definition}

The elements of $\partial \F_2$ are also sometimes called the {\it ends} of 
the
free group.

\begin{definition} Let $M^+_1(\partial\F_2)$ be the space of probability 
measures on
$\partial \F_2$. We let $M^+_{1,3}(\partial\F_2)$ be those probability 
measures which
do not simply concentrate on two ends -- that is to say, for all $e', 
e''\in \partial\F_2$ we
have $\mu(\{e', e''\})<1$.
\end{definition}

Note that if we are given three distinct
ends $e_1, e_2, e_3\in \partial \F_2$ then we may
select a common central branch point in a way which is firstly invariant 
under permutations
of the sequence $(e_1, e_2, e_3)$ and which secondly, and more importantly,
commutes with the action of $\F_2$. For instance, if we are presented with the
ends
\[e_1=(aba^{-1}bba...)\]
\[e_2=(aba^{-1}a^{-1}ba...)\]
\[e_3=(aba^{-1}a^{-1}bb...)\]
then we would be led to choose $s_0(e_1, e_2, e_3)=aba^{-2}b$ as their 
central point.
Then if we were to act on each of them with the group element
$\sigma=ba^2 b^{-1}a^{-1}$ we obtain
\[f_1=\sigma\cdot e_1=(babba...)\]
\[f_2=\sigma\cdot e_2=(bba..)\]
\[f_3=\sigma\cdot e_3=(bbb...)\]
and are then confronted with $s_0(f_1, f_2, f_3)=b^2$ as their branching 
point,
which is indeed equal to $\sigma s_0(e_1, e_2, e_3)$.

For measures in $M^+_{1, 3}(\partial \F_2)$ one can
use integration to extend this technique to find a
{\it center of the measure}:

\begin{theorem} (Lyons)
\label{A1}
There is a Borel function
\[s: M^+_{1, 3} (\partial \F_2)\rightarrow \F_2\]
such that for all $\mu\in M^+_{1, 3}(\partial\F_2)$, $\sigma\in F_2$
\[s(\sigma\cdot \mu)=\sigma s(\mu).\]
\end{theorem}

\begin{proof} See \cite{lyons} or \cite{kechrisclassification}.
\end{proof}

\begin{lemma}
\label{A2}
The orbit equivalence relation $E^{\partial \F_2}_{\F_2}$ given
by the action of $\F_2$ on $\partial \F_2$ is hyperfinite.
\end{lemma}

\begin{proof}
First consider the tail equivalence relation $F$ on $\partial \F_2$. So 
given two
ends $e, e'$ we set
\[eFe'\]
if there is some single end $f$ and two words $w_1, w_2$ with
\[e=w_1^\smallfrown f,\]
\[e'=w_2^\smallfrown f.\]
It is known (see \cite{dokelo}, \cite{jakelo})
that these kinds of tail equivalence relations are always hyperfinite.

Then since $E^{\partial \F_2}_{\F_2}\subset F$ we obtain that it must be
hyperfinite as well.
\end{proof}

\begin{lemma}
\label{A3}
The orbit equivalence relation $E_{\F_2}^{[\partial \F_2]^2}$
induced by the left concatenation of
$\F_2$ on $[\partial \F_2]^2$ is hyperfinite.
\end{lemma}

\begin{proof} First view $[\partial \F_2]^2$ as a subset of
$\partial\F_2\times\partial\F_2$ in the way indicated at the start of
this section and consider the orbit equivalence
relation $E_{\F_2}^{\partial\F_2\times\partial\F_2}$ given by the diagonal
action of $\F_2$ on this product space.         If we think of 
$\partial\F_2\times\partial\F_2$
as being a subset of infinite strings taken from $\F_2\times \F_2$
then the argument from \ref{A2} indeed shows
$E_{\F_2}^{\partial\F_2\times\partial\F_2}$ to be hyperfinite.

We clearly therefore have that the restricted equivalence relation
$E_{\F_2}^{\partial\F_2\times\partial\F_2}|_{[\partial \F_2]^2}$ is
hyperfinite. Now since $E_{\F_2}^{[\partial \F_2]^2}$ has finite index
over the hyperfinite equivalence relation
$E_{\F_2}^{\partial\F_2\times\partial\F_2}|_{[\partial \F_2]^2}$
we finish by \cite{jakelo}.
\end{proof}

\begin{lemma}
\label{A4} For any $e\in\partial \F_2$, the stabilizer of $e$ in
$\partial\F_2$, $(\F_2)_e$, is abelian.
\end{lemma}

\begin{proof} Note that if $e$ has non-trivial stabilizer then it must be
eventually periodic. That is to say, viewed as infinite word it will have the
form
\[e=w^\smallfrown \sigma^\smallfrown\sigma^\smallfrown \sigma...\]
for some $\sigma, w\in \F_2$. Let us assume we have represented $e$ in
this way and taken $\sigma$ to be as short a word as possible -- which is
basically to say that $\sigma$ does not have the form 
$\tau^\smallfrown\tau^\smallfrown...
\tau$ for any other $\tau\in\F_2$.

After possibly replacing $e$ by $w^{-1}\cdot e$ and conjugating
$(\F_2)_e$ by $w$, we may as well assume $w$ is the empty word and that
$e$ actually has the form 
$\sigma^\smallfrown\sigma^\smallfrown\sigma^\smallfrown...$.
Then it is easily seen that $(\F_2)_e=\{\sigma^\ell: \ell\in \Z\}$.
\end{proof}

\begin{lemma}
\label{A5} For any $\{e, e'\}\in [\partial\F_2]^2$, the stabilizer,
$(\F_2)_{\{e, e'\}}$, of $\{e, e'\}$
in $\F_2$ is abelian.
\end{lemma}

\begin{proof} Following the proof of \ref{A4} we may assume
\[e=\sigma^\smallfrown\sigma^\smallfrown\sigma...\]
where $\sigma\neq u^k$ for any $u\in \F_2$, $k>1$. Observe in particular
that any string of the form 
$\sigma^\smallfrown\sigma^\smallfrown\sigma...\sigma$
is already a reduced word and in particular $\sigma$ does not have the form
\[\sigma=v\sigma_0 v^{-1}\]
for any non-trival $v\in\F_2$.

For any $\tau\in (\F_2)_{\{e, e'\}}$ we must either have $\tau\cdot e=e'$ and
$\tau\cdot e'=e$ or $\tau\cdot e=e$ and $\tau\cdot e'=e'$. In either case we
end up discovering that $\tau^2\cdot e=e$.

Thus by the proof of the last lemma we in fact have to have 
$\tau^2=\sigma^{m}$
some $m$. It then follows by the observations regarding $\sigma$ observed
above that $\tau^2$ and hence $\tau$ cannot have the form $v\tau_0 v^{-1}$
for any non-trivial $v\in\F_2$.

Thus $\tau^\smallfrown\tau$ is a reduced word and equal to $\tau^2$. From 
this it
follows that $m$ is even and $\tau=\sigma^{\frac{m}{2}}.$
\end{proof}

The somewhat convoluted formulation of the next lemma is intentional. The 
statement
given helps organize the proof.

\begin{lemma}
\label{A6} We may divide $M^+_1(\partial\F_2)$ into three invariant pieces, 
$B_0, B_1, B_2$,
such that:

\leftskip 0.4in

\noindent (a) the orbit equivalence relation of $\F_2$ on $B_0$ is
smooth (i.e. $\leq_B$ id$(\R)$) and the action is free (i.e. each $\mu$ has 
trivial stabilizer);


\noindent (b) the orbit equivalence relation of $\F_2$ on $B_1$ is 
hyperfinite and
the action is free;

\noindent (c) the orbit equivalence relation of $\F_2$ on $B_2$ is smooth 
and each $\mu\in B_2$
has abelian stabilizer.


\leftskip 0in

\end{lemma}

\begin{proof}
We begin with $B_0=M^+_{1, 3}(\partial\F_2)$. \ref{A1} certainly implies 
that the action on
$B_0$ is free, and we see that the equivalence relation is smooth by taking
\[A_0=\{\mu\in M^+_{1, 3}(\partial\F_2): s(\mu)=1\}\]
as our selector.

We move on and let $C$ be the set of linear combinations of the form
\[\mu=\alpha_1 \delta_{e_1}+ \alpha_2 \delta_{e_2}\]
where $e_1, e_2\in \partial\F_2$, $\delta_{e_i}$ is the
Dirac point mass measure on $e_i$, $\alpha_1+\alpha_2=1$,
and
$\alpha_1, \alpha_2\geq 0$.We subdivide $C$ into $(C^r)_{r\in [1/2, 1]}$,
where $\alpha_1 \delta_{e_1}+ \alpha_2 \delta_{e_2}\in C^r$ if 
max$\{\alpha_1, \alpha_2\}=r$.

We then let $B_1^r=\{\alpha_1 \delta_{e_1}+ \alpha_2 \delta_{e_2}\in C^r$: 
one of $e_1, e_2$ has non-trivial
stabilizer in the action of $\F_2$ on $\partial\F_2\}$. We let $B_1$ be the 
union of
these $B_1^r$'s.
It follows from
the argument of \ref{A5} that the action of $\F_2$ on each $B_1^r$ is free.
We are left with trying to show that the orbit equivalence relation on such a
$B_1^r$ is hyperfinite.

In the case that $r=\frac{1}{2}$ this directly reduces to \ref{A3}; in the 
case
that $r>\frac{1}{2}$ we have an invariant way to select from the pair 
$\{e_1, e_2\}$
by chosing the end with highest weight, and it reduces to \ref{A2}.

Finally we take $B_2^r\{\alpha_1 \delta_{e_1}+ \alpha_2 \delta_{e_2}\in 
C^r$: both $e_1, e_2$ have non-trivial
stabilizer in the action of $\F_2$ on $\partial\F_2\}$; we are then forced to
let $B_2$ be the union of
these $B_2^r$'s.
Note that the analysis of \ref{A4} shows that there are only countably many 
possibilities for
$e_1$ and $e_2$; and so each $B_2^r$ is the union of countably many orbits, 
and hence
almost trivially smooth.
\end{proof}

In fact we can sharpen \ref{A6} by observing that the stabilizers will be 
isomorphic to
$\Z$ when non-trivial.





\section{Skew invariant measures}

\label{B}

\label{skew}

\begin{lemma}
Let $T$ be a measure preserving transformation of a standard
Borel probability space $(X, \mu)$ and let $\Z$ act on $X$ by
\[k\cdot x=T^k (x).\]
Let $\alpha: X\times \Z\rightarrow \F_2$ be a cocycle.

Then we may find a measurable assignment
\[X\rightarrow M_1^+(\partial\F_2)\]
\[x\mapsto \mu_x\]
such that for all $k\in \Z$ and almost all $x\in X$
\[\mu_{k\cdot x}=\alpha(x, k)\cdot \mu_x.\]

\end{lemma}

\begin{proof} (Sketch only; chapter 4 of \cite{zimmer} contains
much more general results.)

Let $\ell_1^\infty(X, M^+_1(\partial\F_2))$ be the space of all measurable
assignments
\[\gamma: X\rightarrow M^+_1(\partial\F_2).\]
We equip this space with the topology provided by the semi-norms of the
form
\[\gamma\mapsto \int_X(x\mapsto \int_{\partial \F_2} f_x d\gamma(x))d\mu\]
where $x\mapsto f_x$ is some measurable function from $X$ to the
functions in $C(\partial \F_2)$ of norm $\leq 1$.

In this topology $\ell_1^\infty(X, M^+_1(\partial \F_2))$ becomes
a compact convex set.

We let $\Z$ act on $\ell_1^\infty(X, M^+_1(\partial \F_2))$ by
\[(k\cdot \gamma)(x)=\alpha(-k\cdot x, k)\cdot \gamma(-k\cdot x).\]
If we  let
\[T_N:\ell_1^\infty(X, M^+_1(\partial \F_2))\rightarrow \ell_1^\infty(X, 
M^+_1(\partial \F_2))\]
\[  \gamma\mapsto \frac{1}{N+1}\sum_{k\leq N}k\cdot \gamma\]
and take any point in the intersection of the closures of the images of the 
operators
$T_N$ then we have a fixed point as required.
\end{proof}

This lemma can be used to give a neat proof that free actions
of $\F_2$ on probability spaces are never hyperfinite, and that neat
proof can in turn be used to show that the above lemma cannot be proved in
the Borel, or even Baire measurable, category.

\begin{corollary}
\label{B2}
If $\F_2$ acts freely by mpts on standard Borel probability
space $(X, \mu)$, then $E^X_{\F_2}$ is not hyperfinite.
\end{corollary}

\begin{proof}
Appealing to the ergodic decomposition theorem, we may assume without loss of
generality that $\F_2$ acts ergodically. Write $\F_2=\langle a, b\rangle$,
so that $a$ and $b$ are the generators.
We suppose for a contradiction that there is a Borel action of $\Z$ on $X$ 
with
$E_\Z^X=E_{\F_2}^X$ off of a null set.

We have a Borel cocycle
\[\alpha: X\times \F_2\rightarrow \Z\]
given by
\[\alpha(x, \sigma)=k\]
if $k\cdot x=\sigma\cdot x$.
Appealing the last lemma we may find a measure one set $X_0\subset X$ and a
Borel assignment
\[x\mapsto \mu_x\]
\[X_0\rightarrow M^+_1(\partial \F_2)\]
such that
\[\mu_{\ell \cdot x}=\alpha(x, \ell)\cdot \mu_x\]
all $x\in X_0$, $\ell\in \Z$. Observing that for each $x$ there will be
$\ell_a^x, \ell_b^x$ with
\[\alpha(x, \ell_a^x)=a,\]
\[\alpha(x, \ell_b^x)=b,\] we
obtain
\[\mu_{a\cdot x}=a\cdot \mu_x,\]
\[\mu_{b\cdot x}=b\cdot \mu_x,\]
all $x\in X_0$.

\medskip

\noindent {\bf Claim:} $\mu_x\notin \partial M_{1, 3}^+(\partial \F_2)$ 
a.e. $x$.

\medskip

\noindent {\bf Proof of claim:} Otherwise ergodicity gives almost every
$\mu_x\in M_{1,3}^+(\partial \F_2)$ and appealing to \ref{A1} we
may find a Borel function
\[s^*: X_0\rightarrow \F_2\]
such that for all $\sigma\in \F_2$ and
a.e. $x$
\[s^*(\sigma\cdot x)=\sigma\cdot s^*(x).\]
%Then we can appeal to ergodicity once more to find some single 
$\sigma_0\in\F_2$ with
%\[\forall^\mu x\in X(\exists x_0\in F_2\cdot x (s^*(x_0)=\sigma_0).\]
Letting $A_\sigma=\{x\in X: s^*(x)=\sigma\}$ we have
\[(A_\sigma)_{\sigma\in \F_2}\]
partitioning a conull subset of $X$ into $\aleph_0$ many disjoint sets of
equal measure, with a contradiction to $\mu(X)=1$. \hfill ($\square$Proof 
of claim)

\medskip

So we actually may assume there is a Borel function
\[X_0\rightarrow [\partial\F_2]^2\]
\[x\mapsto \{e_x, e_x'\}\]
such that
\[a\cdot \{e_x, e_x'\}=\{e_{a\cdot x}, e_{a\cdot x}'\},\]
\[b\cdot \{e_x, e_x'\}=\{e_{b\cdot x}, e_{b\cdot x}'\},\]
almost every $x$.

Let $A_0$ be the set of $x\in X_0$ such that neither $e_x$ nor
$e_x'$ begin with $a$ or $a^{-1}$. For each $x\in X_0$ we may find
some $n\in \N$ with
\[b^n\cdot x\in A_0.\]
Thus $\mu(A_0)=0$.

But considering that for $n_1\neq n_2\in \N$
\[a^{n_1}\cdot A_0\cap a^{n_2}\cdot A_0=\emptyset\]
we have our contradiction.
\end{proof}

\begin{counterexample}
There is a Borel action of $\Z$ on a Polish space $X$ and a
Borel cocycle $\alpha: X\times \Z\rightarrow \F_2$ such that
there is {\rm no} comeager $X_0$ and Borel assignment
\[X_0\rightarrow M_1^+(\partial\F_2)\]
\[x\mapsto \mu_x\]
with for all $x\in X_0$, $n\in \Z$
\[\mu_{n\cdot x}=\alpha(x, n)\cdot \mu_x.\]
\end{counterexample}
\begin{proof}
One constructs the example in this way. Again letting $a$ and $b$ be the
generators of $\F_2$ we start with the free part of the shift action of
$\F_2$ on $\{0, 1\}^{\F_2}$; notice that this
action is {\it generically ergodic}, in the sense that
every invariant Borel set is either meager or comeager.
Following
\cite{suwewr} we may find a comeager subset $X$ of this free part
on which

\leftskip 0.4in

\noindent (a) $E_{\F_2}|_X$ is hyperfinite;

\leftskip 0in

after some possible further contraction of $X$ we may additionally suppose

\leftskip 0.4in

\noindent (b) $X$ is $\F_2$-invariant;

\noindent (c) $\{a^\ell\cdot x: \ell \in Z\}$ is dense in $X$ for all $x\in 
X$;

\noindent (d) $\{b^\ell\cdot x: \ell \in Z\}$ is dense in $X$ for all $x\in 
X$.

\leftskip 0in

As indicated by say $\S 3.1$\cite{hjorthclassification}, it follows easily 
from
(c) and (d) that the orbit equivalence relations $E_{\langle a\rangle}^X$
and $E_{\langle b\rangle}^X$, induced by the respective cyclic subgroups
$\langle a\rangle$ and $\langle b \rangle$, are {\it properly generically 
ergodic},
and hence non-smooth, and in fact
non-smooth even when restricted to any comeager subset of $X$.
 From (a) we get a Borel action of $\Z$ with
\[E_\Z^X=E_{\F_2}^X.\]

Now for a contradiction suppose that on a comeager subset $X_1\subset X$ we 
can
find an assignment
\[x\mapsto \mu_x\] which is suitably skew invariant under the cocyle
from $\Z$ to $\F_2$. As in the corollary above we would obtain
\[\mu_{a\cdot x}=a\cdot\mu_x,\]
\[\mu_{b\cdot x}=b\cdot \mu_x\]
through out $X_1$. We may further assume that $X_1$ is invariant under the
action of $\F_2$.

\medskip


\noindent {\bf Claim:} $\mu_x\notin \partial M_{1, 3}^+(\partial \F_2)$ for 
a comeager
set of $x\in X_1$.

\medskip

\noindent {\bf Proof of claim:} Otherwise
generic ergodicity gives almost every
$\mu_x\in M_{1,3}^+(\partial \F_2)$ and, as before, we
may find a Borel function
\[s^*: X_1\rightarrow \F_2\]
such that for all $\sigma\in \F_2$ and
a comeager set of $X$
\[s^*(\sigma\cdot x)=\sigma\cdot s^*(x).\]
Letting $A_1=\{x\in X: s^*(x)=1\}$ we have that $A_1$ meets all the orbits
on a comeager subset $X_2$ of $X_1$ in exactly one point.
Then we may find appeal to the uniformization theorem for Borel sets with
countable sections to see that the function
\[f: X_2\rightarrow A_1\]
which assigns to each $x\in X_2$ the unique $f(x) E_{\F_2} x$ with
$f(x)\in A_1$ will be $\F_2$ invariant, and hence constant on a comeager 
subset
by generic ergodicity, with a contradiction to every orbit being meager.
\hfill ($\square$Proof of claim)

So as before may on a comeager subset $X'$ of $X_1$ assign a pair of ends
$\{e_x, e'_x\}$ with
\[\{e_{\sigma \cdot x}, e_{\sigma\cdot x}'\}=\sigma\cdot \{e_x, e_x'\}\]
all $\sigma\in \F_2$, $x\in X'$. We can again let $A_0$ be the set of
$x$ for which neither $e_x$ nor $e_x'$ begins with an $a$ or an $a^{-1}$.
By considering the
action of $\langle b\rangle$ we analogously obtain that it will be non-meager.
Thus by generic ergodicity of the action of $\langle a\rangle$ we obtain
that for a comeager collection of $x$ there will be some $\ell\in \Z$ with
$a^\ell\cdot x\in A_0$. Considering the definition of $A_0$ and the invariance
assumption on the $\{e_x, e'_x\}_{x\in X'}$ we obtain that there can be at 
most
one $\ell\in Z$ with $a^\ell \cdot x \in A_0$, and thus $A_0$ is a Borel 
selector
for a comeager subset of $X$, contradiction.
\end{proof}











\section{$E_0$-ergodicity}
\label{E_0}


\begin{lemma}
\label{folk}
  (Folklore; see \cite{jakelo}.) A countable equivalence
relation $E$ is hyperfinite if and only if $E\leq_B E_0$.
\end{lemma}

\begin{theorem} (Jones, Schmidt; see \cite{jonesschmidt})
\label{jonesschmidt}
Let $G$ be a countable group acting ergodically by mpts on a standard Borel
probability space $(X, \mu)$. Then $E^X_G$ is $E_0$ ergodic
if and only if there is no sequence of sets $(A_n)_{n\in\N}$,
with each
\[\mu(A_k)=\frac{1}{2},\]
and for all $g\in G$
\[\mu(g\cdot A_k \Delta A_k)\rightarrow 0\]
as $k\rightarrow \infty$.
\end{theorem}

Without going into the depths of the proof, we may first tackle
the {\it if} direction by supposing $\theta: X\rightarrow \{0, 1\}^\N$
is a homomorphism of $E^X_G$ to $E_0$. If $[\theta(x)]_{E_0}$ is not almost 
everywhere
constant, then ergodicity implies that
$\mu(\theta^{-1}[\{p\}])$, the measure of the pullback of the point $p$,
is null for each $p\in \{0, 1\}^\N$. Then it is not hard to show that for
each $k$ there exists a set $B_k\subset \{0, 1\}$ such that 
$\mu(\theta^{-1}[B_k])$
always equals a half and that for each $f, g\in \{0, 1\}^\N$ with $f(n)=g(n)$
all $n>k$ we have
\[f\in B_k\Leftrightarrow g\in B_k.\]
Then $A_k=\theta^{-1}[B_k]$ gives us the required sequence.

For the converse, suppose that $(A_k)$ is our asympotically invariant
sequence of sets of measure one half. After possibly refining the sequence 
we may
assume that for all $g\in G$ there is a measure one set of $x$ for which
\[\exists N\forall k>N(x\in A_k\Leftrightarrow g\cdot x\in A_k).\]
Then we devise a measurable homomorphism by assigning to $x\in X$
\[f_x: \N\rightarrow\{0, 1\}\]
with
\[f_x(k)=1\Leftrightarrow x\in A_k.\]

\begin{theorem} (Schmidt; see \cite{schmidt})
Let $S$ be a countable set and equip $\{0, 1\}^{\F_2\times S}$
with the product measure $\mu$. Then the shift action of $\F_2$ is
$E_0$ ergodic.
\end{theorem}

This result, which actually extends to any non-amenable countable
group in the place of $\F_2$, can be proved in various ways.
It suffices to show that we can write the induced representation
of $\F_2$ on $L^2(\{0, 1\}^{\F_2\times S}, \mu)$ as a
product of the trivial representation and $\aleph_0$ many copies of
the regular left representation,
\[L^2(\{0, 1\}^{\F_2\times S}, \mu)\cong 1\otimes \bigotimes_{\aleph_0}
\ell^2(\F_2).\]
For this it suffices to show that the square integrable
functions orthogonal to the
constant function, that is to say those $f\in L^2(\{0, 1\}^{\F_2\times S}, 
\mu)$
with
\[\int f \mu=0,\]
admit an orthonormal basis which is set wise invariant under the action
of $\F_2$.

Towards this goal we define for each finite $F\subset \F_2\times S$ the
function
\[A_F: \{0, 1\}^{\F_2\times S}\rightarrow \{1, -1\}\]
by
\[A_F(x)=-1\]
if the cardinality
of $\{(\sigma, s)\in F: x(\sigma, s)=1\}$ is odd, and
\[A_F(x)=1\]
if the cardinality is even.

Obviously this collection is set wise invariant. The Stone-Weierstrass theorem
immediately implies that they span the functions orthogonal to the constant
functions. We omit the routine calculation that they are orthogonal.

\medskip

Things would certainly be nicer if mixing actions of $\F_2$ were necessarily
$E_0$-ergodic.
However it is known from say \cite{valette} that this is untrue. We give an
elementary counterexample.

\begin{pcounterexample}
We first
consider the shift action of $\F_2$ on $\{0, 1\}^{\F_2}$ and define a sequence
of measures, $(\mu_n)_{n\in\N}$. We will make sure that these measures are
$\F_2$-invariant and that they are becoming increasingly ``sticky", with sets
of measure a half which are nearly invariant for ever larger subset of
$\F_2$. We finish with the diagonal action of $\F_2$ on
\[\prod_\N (\{0, 1\}^{\F_2\times S}, \mu_n).\]

It suffices to define the measure $\mu_n$ on sets of the form
\[C_s=\{x\in \{0, 1\}^{\F_2}: x|_T=s\}\]
for $s: T\rightarrow \{0, 1\}$ and $T\subset \F_2$ a finite set which
forms a tree under the usual Cayley graph for $\F_2$.

For such an $s$ and $T$ we let $(e_i)_{i=1}^{i=|T|-1}$ enumerate
the edges in $T$. We let $K_0$
be the number of edges for which $s$ takes the same
value at both vertices at either end and we let $K_1$ be the number of
edges for which $s$ assumes differing values at the two ends. We set
\[\mu_n(C_s)=\frac{1}{2}(\frac{n-1}{n})^{K_0}(\frac{1}{n})^{K_1}.\]

We omit the routine verification that this will indeed define a measure,
  and
simply observe that for each $g\in\F_2$
\[\mu_n(\{x\in \{0, 1\}^{\F_2}\}\Delta g\cdot\{x\in \{0, 1\}^{\F_2}\})\]
gravitates towards zero as $n$ marches towards infinity. Thus by 
\ref{jonesschmidt}
the diagonal action of $\F_2$ on the product space
$\prod_\N (\{0, 1\}^{\F_2\times S}, \mu_n)$ fails to be $E_0$-ergodic.
Since each individual action of $\F_2$ on some individual
$(\{0, 1\}^{\F_2\times S}, \mu_n)$ is mixing, so too the overall diagonal 
action
of $\F_2$ on the infinite product will be mixing.
\end{pcounterexample}























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\end{thebibliography}

\leftskip 0.5in







greg@math.ucla.edu

\bigskip



MSB 6363

UCLA

CA 90095-1555

\leftskip 0in

\end{document}




