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\title{Non-treeability for product group actions}         % Enter your title between curly braces
\author{Greg Hjorth\footnote{Partially supported by NSF grant DMS  0140503}}        % Enter your name between curly braces
\date{\today}          % Enter your date or \today between curly braces
\maketitle
\abstract{A product of two lcsc non-compact groups  $G_1\times G_2$ acting freely by measure preserving 
transformation on a standard Borel probability space gives rise to a non-treeable 
equivalence relation unless both groups are amenable.} 

\bigskip
\bigskip


The result of the present paper can be placed as one part in a series. 

\begin{theorem} \label{adams} 
(Adams, \cite{adams}) Let $E_1, E_2$ be countably infinite measure preserving equivalence relations 
on standard Borel probability spaces $X_1, X_2$ with $E_1\times E_2$ non-amenable. 
Then $E_1\times E_2$ is non-treeable. 
\end{theorem} 

\begin{theorem} 
(Gaboriau, \cite{gaboriau}) \label{gaboriau} 
Let $G_1, G_2$ be countably infinite groups 
with $G_1\times G_2$ non-amenable. 
Then any measure preserving action of $G_1\times G_2$ 
on a standard Borel probability space is non-treeable. 
\end{theorem} 

\begin{theorem} \label{kechris}
(Kechris, \cite{kechrisproduct}) 
Let $G_1, G_2$ be locally compact non-compact second countable groups 
one of which 
contains a non-compact closed amenable subgroup and with $G_1\times G_2$ non-amenable. 
Then any measure preserving action of $G_1\times G_2$ on a standard Borel 
probability space is non-treeable. 
\end{theorem} 

\begin{theorem} \label{pepe}
(Pemantle, Peres, \cite{pepe}) 
Let $X$ and $Y$ be infinite, locally finite graphs. Suppose that 
$G\subset {\rm Aut}(X)$ is a closed nonamenable subgroup of 
${\rm Aut}(X)$, and $H\subset {\rm Aut}(Y)$ has an infinite orbit. 
Then there is no $G\times H$-invariant probability measure on the set 
of spanning trees of the direct product graph $G\times H$. 
\end{theorem} 

While here we show: 

\begin{theorem} \label{me}
Let $G_1, G_2$ be locally compact second countable groups with $G_1\times G_2$ acting in a measure preserving,  measurable manner on a standard Borel probability space with the orbit equivalence relation 
$E_{G_1}$ non-amenable and $G_2$ not acting essentially transitively on any of its ergodic componenents. 
Then the orbit equivalence relation induced by the product group action, $E_{G_1\times G_2}$, is not 
treeable.  
\end{theorem} 

\begin{corollary}
Let $G_1, G_2$ be locally compact non-compact second countable groups with $G_1\times G_2$ acting in a free, measure preserving,  measurable manner on a standard Borel probability space. Then if $G_1\times G_2$ is 
non-amenable then $E_{G_1\times G_2}$ is not treeable.
\end{corollary} 






Our proof is a refinement of Kechris', which in turn had points in common with Adams', and 
as thus traces back to ideas from Zimmer, \cite{zi1}, and from there to earlier work 
of Margulis and Mostow.  
The original proof of Gaboriau's result, however, used a radically different 
idea to these three results, and the proof of Pemantle and Peres was different again. 
It is clear that \ref{me} implies \ref{kechris} and \ref{gaboriau}; in 
the last section we use some observations from 
\cite{kechrismiller} to show that \ref{pepe} follows as well, but this is less of an 
achievement since their proof is by far the simplest and most elegant of the sequence. 
Again as pointed out by \cite{kechrismiller}, \ref{pepe} can be used to derive 
\ref{gaboriau}, though it should also be said that 
\ref{gaboriau} arises as one small piece of a much broader body of work on 
the concept of {\it cost}. 
\ref{me} also implies \ref{adams}, since any countable measure preserving equivalence relation is induced 
by the measure preserving action of a countable group (\cite{feldmanmoore}). 


\bigskip 

{\bf Acknowledgments:} I am grateful to Nicholas Monod for a number of 
helpful conversations around the topic of the Poisson boundary, and indeed 
for suggesting it might have some applications to the theory of Borel reducibility. 
I am also grateful to Alexander Kechris for pointing out the relevance of 
$\S$29, 30 from \cite{kechrismiller}. I am grateful to Scot Adams for a 
number of discussions about these techiques in general, as well as some 
specific comments about this specific paper. And 
finally I am grateful to the referee for an 
unusually insightful reading; indeed the final statement of \ref{me} is sharper than the 
original version and borrows directly from a suggestion of the referee. 





\section{Borel reducibility}       % Enter section title between curly braces


In this paper we only consider actions by groups which are locally compact, second countable, and 
Hausdorff -- {\it lcsc} groups for short. Any such group is completely metrizable (see for example
\cite{morris}, chapter 3, exercise 3). 
Thus in particular it is Polish, and there is a robust structure theorem 
for its Borel sets and Borel actions (see for instance 
\cite{beckerkechris}, \cite{classical}). 

We will find it convenient to organize the later arguments around the notion 
of Borel reducibility. 

\begin{definition} Let $E, F$ be Borel equivalence relations on 
standard Borel spaces $X, Y$. We say that 
{\it $E$ is Borel reducible to $F$}, written $E\leq_B F$ 
if there is a Borel 
function 
\[\theta: X\rightarrow Y\] 
such that for all $x_1, x_2\in X$ 
\[x_1 E x_2 \Leftrightarrow \theta(x_1) F \theta(x_2).\] 
We say that $E$ is {\it smooth} if it is Borel reducible to id$(\R)$, the identity 
relation on $\R$. 

\end{definition} 

\begin{definition} An equivalence relation $E$ on standard Borel $X$ is 
{\it treeable} if there is a Borel subset $\g\subset X\times X$ such that: 

\leftskip 0.4in 

\no (i) $\g$ is symmetric; 

\no (ii) $\g$ is irreflexive, in the sense that 
$(x, x)$ is never in $\g$; 

\no (iii) $\g$ is acyclic (in other words, if $x_1, x_2, ..., x_n\in \g$ with 
each $(x_i, x_{i+1})\in \g$ and $x_i\neq x_j$ for $i\neq j$, 
then $(x_n, x_1)\notin \g$); 

\no (iv) the connected components of $\g$ form the equivalence classes of $E$. 

\leftskip 0in 

\no An equivalence relation is {\it countable} if every equivalence relation is 
countable. 




\end{definition} 

It has been customary to only consider treeability in the context of countable 
Borel equivalence relations, but it makes sense to consider treeable equivalence 
relations with uncountable classes and certain key aspects of the 
existing theory simply adapts. 


\begin{lemma} (essentially \cite{jakelo}) 
Let $E$ on $X$ be a treeable Borel equivalence relation and let 
$A\subset X$ be Borel set meeting each orbit in at most a countable set. 
Then $E|_A$, the restriction of $E$ to $A$, is again treeable. 
\end{lemma} 

\begin{proof} 
Let $B$ be the set of points which are either in $A$ or lie between two distinct points in 
$A$. By Novikov-Lusin (see \cite{classical}), this set is Borel, and it is clear that 
$E|_B$ is treeable. Now \cite{jakelo} shows that the restriction of a countable treeable 
equivalence relation to a Borel subset is again treeable. 
\end{proof} 

\begin{corollary} \label{f2} 
Let $G$ be lcsc group acting in a Borel manner on a 
standard Borel space $X$. Then $E^X_G$ is treeable if and only if $E^X_G$ is 
Borel reducible to an orbit equivalence relation induced by a free, 
Borel action of $\F_2$, the free group on two generators. 
\end{corollary} 

\begin{proof} 
Following \cite{sections}, we can fix Borel set $A\subset X$ meeting each 
orbit in a countable non-empty set. 
By the uniformization theorem for 
Borel sets in the plane with countable sections\footnote{Again, \cite{classical} 
is an ideal place to find an account of classical facts such as this one.} we can 
in a Borel manner select to each $x\in X$ some $x'=\phi(x)\in [x]_G\cap A$, and hence 
obtain $E^X_G\leq_B E^X_G|_A$. 

($\Rightarrow$:) 
It suffices to show that $E^X_G|_A$ is Borel 
reducible to a suitable action of $\F_2$. 
But by the last lemma $E^X_G|_A$ is again treeable, and then since it is 
countable we have that it is Borel reducible to a free, Borel action of 
$\F_2$ by \cite{jakelo}. 

($\Leftarrow$:) Assume $\theta: X\rightarrow Y$ gives a reduction of 
$E^X_G$ to $E^Y_{\F_2}$, where $Y$ is a free, Borel $\F_2$ space. 
Then in particular $\theta|_A$ reduces $E^X_G|_A$ to a treeable equivalence 
relation, and hence, by \cite{jakelo} again, $E^X_G|_A$ admits some 
treeing, $\g_A$. We then define a treeing by $x\g x'$ if and only if 
either: 

\leftskip 0.4in 

\no (i) $x, x'\in A$, $x\g_A x'$; or 

\no (ii) $x\in A$, $x'\notin A$, $\phi(x')=x$; or 

\no (iii) $x\in A$, $x'\notin A$, $\phi(x)=x'$. 

\leftskip 0in 

\end{proof} 


The treeable relations are most naturally contrasted with the 
hyperfinite. 

\begin{definition} An equivalence relation $E$ is 
{\it hyperfinite} if there is an increasing union of Borel 
equivalence relations, $(F_n)$, with finite classes, such that 
\[E=\bigcup_{n\in\N} F_n.\] 
\end{definition} 

Hyperfinite equivalence relations can be alternately characterized 
as the equivalence relations induced by a Borel action of $\Z$ or the 
Borel equivalence relations for which every infinite equivalence class 
can be assigned, in a uniformly Borel fashion, the structure of a 
$\Z$-chain.\footnote{See \cite{miller} for an extended analysis of 
some of the subtle distinctions which are possible here.} This last 
characterization makes it clear that hyperfiniteness is a special 
kind of treeability. 

However whereas most of the structural properties of hyperfinitess 
are understood, at least up to discounting null sets, the structural 
properties of the treeable equivalence relations 
have not been adequately documented.  

\begin{question} 
Is a countable Borel equivalence relation with finite index over treeable 
again treeable? 

See here \cite{jakelo} for a discussion of this problem and terminology. 
In that 
paper they do show that finite index over {\it hyperfinite} is again 
hyperfinite. 
\end{question} 

\begin{question} 
Which lcsc groups have all their free Borel, measure preserving actions 
on standard Borel probability spaces treeable? 
\end{question}

In the case of hyperfinite this is largely understood: If every 
such action of lcsc $G$ is Borel reducible to a hyperfinite equivalence 
relation 
then $G$ is amenable (see e.g. 
\cite{zi1}), and 
conversely if $G$ is amenable then after possibly discounting a null 
set any such action will be Borel reducible to a hyperfinite equivalence 
relation (see \cite{zi1}, \cite{orwe}). 
(It remains notoriously open, however, even for discrete groups, whether 
any such action will be hyperfinite {\it everywhere}, and not just 
{\it almost everywhere}.) 


On the other hand, there is at least one structural property known 
true for hyperfinite and known false for treeable: Increasing unions 
of hyperfinite equivalence relations are hyperfinite almost everywhere by 
\cite{cofewe}; 
an increasing union of countable treeable equivalence relations need not be 
treeable on any conull set (see \cite{jakelo} or \cite{gaboriau}). 



An especially important example of a hyperfinite equivalence relation 
is given by $E_0$. 

\begin{notation} We let $E_0$ be the equivalence relation of eventual 
agreement on infinite binary sequences. So for $\vec x, \vec y\in 
\{0, 1\}^\N (=_{\rm df} 2^\N)$ we let 
\[\vec x E_0 \vec y\] 
if $\exists N\forall m>N (x_m=y_m)$. 
\end{notation} 

$E_0$ is clearly hyperfinite; conversely \cite{dojake} shows 
that every 
hyperfinite equivalence relation is Borel reducible to $E_0$. 

\begin{notation} We let $E_1$ be the equivalence relation of eventual 
agreement on infinite sequences of reals: For $\vec x, \vec y\in \R^\N$ 
we let 
\[\vec x E_1 \vec y\] 
if $\exists N\forall m>N (x_m=y_m)$. 
\end{notation} 

$E_1$ is an example of a treeable equivalence relation which is not 
Borel reducible to any countable Borel equivalence relation. In fact 
by \cite{kechrislouveau}, not Borel reducible to any Borel action of a 
Polish group. 

\begin{notation} We let $E_{\infty\t}$ arise from the orbit equivalence relation 
of $\F_2$ on $F(2^{\F_2})$, the part of $2^{\F_2}(=_{\rm df} \{0, 1\}^{\F_2})$ 
on which every non-trivial element of $\F_2$ has no fixed points under the 
shift action. 
\end{notation} 

For our purposes the exact definition of $E_{\infty\t}$ is unimportant. We 
only need the fact established in \cite{jakelo} that it is induced by a free 
action of $\F_2$ and that any countable, Borel, treeable 
equivalence relation is Borel reducible to $E_{\infty\t}$. With this notation 
granted, I can finish this brief survey with one last, and, for the author, most 
central question: 

\begin{question} Up to $\leq_B$-reducibility, how many Borel equivalence relations 
$E$ are there with 
\[E_0<_B E < E_{\infty\t}{\rm ?}\] 
\end{question} 



\section{Mixing properties} 

\begin{definition} A locally compact group $G$ equipped with a Haar measure $\nu$ is 
said to act {\it measurably} on a standard Borel probability space $(X, \mu)$ if for 
any Borel $A\subset X$ the set $\{(g, x): g\cdot x\in A\}$ is measurable in 
$(G\times X, \nu\times\mu)$. 
Let $G$ be a lcsc group acting in a properly ergodic, 
measurable, non-singular manner 
on a standard Borel probability space $(X, \mu)$. The action of $G$ on 
$(X, \mu)$ is {\it mildly mixing} if whenever $G$ acts in a properly ergodic, 
measurable, non-singular, manner on a standard Borel 
probability space $(X', \mu')$ the diagonal action of $G$ on $(X\times X', \mu\times \mu')$ 
is again ergodic. (\cite{scwa}) 

For $\m$ a class of separable Banach spaces, the action of $G$ on 
$(X, \mu)$ is {\it $\m$-ergodic} if whenever $E\in \m$ and we have a continuous 
homomorphism 
\[\pi: G\rightarrow {\rm Iso}(E),\]
\[g\mapsto \pi_g,\]
from $G$ to the isometry group of $E$, and we have a measurable 
\[\varphi: X\rightarrow E\] 
which is equivariant in the sense that 
\[\varphi(g\cdot x)=\pi_g(\varphi(x))\] 
all $x\in X, g\in G$, then $\varphi$ is constant almost everywhere. 
\end{definition} 


A couple of remarks about these definitions. 

There is essentially only one standard Borel structure on a separable 
Banach space, and that is the Borel structure arising from the Polish 
topology induced by the Banach norm. This is the sense in which we require 
$\varphi: X\rightarrow E$ to be measurable. 

Since the isometry group of separable Banach space is Polish, it follows 
from Pettis' lemma, \cite{pettis}, that any measurable 
homomorphism $G\rightarrow {\rm Isom}(E)$ is 
necessarily continuous. 

Future discussions of $\m$-ergodicity will probably be 
clarified by noting that a standard argument shows that an a.e. 
equivariant map from $X$ to $E$ can be replaced by an everywhere 
equivariant map. 

\begin{lemma} Let $G$ be a locally compact group equipped with 
right invariant Haar measure acting in a non-singular 
measurable manner on standard Borel probability space $(X, \mu)$. 
Let $\pi: G\rightarrow {\rm Isom}(E)$ be a continuous homomorphism and 
$\hat{\varphi}: X\rightarrow E$ a measurable map with a.e. $x\in X$, $g\in G$ 
\[\hat{\varphi}(g\cdot x)=\pi_g(\hat{\varphi}(x)).\] 
Then we can find conull, invariant 
$X_0\subset X$ and $\varphi: X_0\rightarrow E$ agreeing a.e. with 
$\hat{\varphi}$ and such that {\it for all} $x\in X_0$, $g\in G$ 
\[{\varphi}(g\cdot x)=\pi_g({\varphi}(x)).\] 
\end{lemma} 

\begin{proof} 
Let $A\subset G\times X$ be the set of $(g, x)$ with 
$\hat{\varphi}(g\cdot x)=\pi_g(\hat{\varphi}(x)).$ We have assumed that 
$A$ is conull, and then in particular we have that 
for any $h\in G$ the set of $(g, x)$ with $(g, h\cdot x)\in A$ is again 
conull. 
Then by Fubini the set 
\[X_0=\{x\in X: {\rm a.e.}\: g, h\in G((g, h\cdot x)\in A)\}\] 
is again conull; by right invariance $X_0$ is $G$-invariant. 

Given any $x\in X_0$ we choose some $h\in G$ such that 
for a.e. $g\in G$ we have $(g, h\cdot x)\in A$ and let 
$\varphi(x)=\pi_h^{-1}(\hat{\varphi}(h\cdot x))$; this 
definition does not depend on the choice of $h$, since given 
a competing choice $h'$ we may find $g_1, g_2$ such that 
$g_1h=g_2h'$ and $(g_1, h\cdot x), (g_2, h'\cdot x)\in A$, 
and note that 
\[ \pi_h^{-1}(\hat{\varphi}(h\cdot x)) 
=\pi_h^{-1}\pi_{g_1}^{-1}(\hat{\varphi}(g_1h\cdot x)) \]
\[=\pi_{h'}^{-1}\pi_{g_2}^{-1}(\hat{\varphi}(g_2h'\cdot x))
=\pi_{h'}^{-1}(\hat{\varphi}(h'\cdot x)) . \] 
A similar calculation gives that $\varphi$ is equivariant on 
all of $X_0$. 
\end{proof} 

A parallel and well known argument gives that a measurable action of a 
lcsc group $G$ on a standard Borel probability space $(X, \mu)$ agrees a.e. 
with a Borel action of $G$ on $X$ (see \cite{zi1}). We will generally only 
assume our actions are measurable, but in light of this well known argument it 
is reasonable to blur the distinction between Borel actions and merely measurable 
actions. 

\begin{lemma} Mild mixing implies $\m$-ergodicity for $\m$ the class of 
all separable Banach spaces. 
\end{lemma} 

\begin{proof} Given $\pi: G\rightarrow {\rm Isom}(E)$ and equivariant 
$f: X\rightarrow E$, we obtain a $G$-invariant function 
\[X\times X\rightarrow \R\] 
\[(x, y)\mapsto \nh f(x)-f(y)\nh.\] 
If the action of $G$ on $X\times X$ is ergodic, which is the case assuming the 
original action was mild mixing, then this function must be constant a.e., which 
would imply it is zero a.e. 
\end{proof} 

The converse to this lemma fails. There is an action of $\F_2$ which is 
$\m$-ergodic for $\m$ the class of all separable Banach spaces but is not 
mildly mixing. This counterexample uses extraneous ideas, and so we put it off 
until the end of $\S $3. 

Nevertheless, $\m$-ergodicity does imply the analog of mild mixing if we 
restrict to taking products with {\it measure preserving actions}. 
Here we provide a definition. 

\begin{definition} 
A 
properly ergodic, 
measurable, non-singular action of a lcsc group $G$ on
$(X, \mu)$ is {\it moderately mixing} 
if whenever $G$ acts in a properly ergodic,
measurable, measure preserving manner on a standard Borel
probability space $(X', \mu')$ the diagonal action
of $G$ on $(X\times X', \mu\times \mu')$
is again ergodic. 
\end{definition} 

\begin{lemma} \label{lemmaproduct} 
Let $G$ be a lcsc group 
acting in a measurable, non-singular, ergodic manner on a standard 
Borel probability space $(X,\mu)$. 
The action of $G$ on $(X, \mu)$ is moderately mixing if and only if 
it is $\{\ell^2\}$-ergodic. 
\end{lemma} 

\begin{proof} $(\Leftarrow):$ 
Suppose $G$ acts by measure preserving transformations on a 
standard Borel probability space $(Y, \nu)$. 
For a contradiction assume 
$A\subset X\times Y$ 
is measurable, invariant, and neither 
null nor conull. At each $x\in X$ we let $A_x=\{y\in Y: (x, y)\in A\}$. Define 
\[\pi: X\rightarrow L^2(Y)\]
\[x\mapsto \chi_{A_x},\] 
the function which assigns to 
each $x$ the characteristic function of the slice 
$A_x$. 
This is a.e. $G$-equivariant, 
and so the assumptions on $X$ imply it is constant 
a.e. {\it as an element in $L^2(Y)$}. 
Thus we have some $A\subset Y$ such that $A_x=A$ a.e. $x$, and in particular 
for every $g\in G$ the measure of $A\Delta g\cdot A$ is zero. Replacing 
$A$ by $\{y\in Y: {\rm a.e.}\: g\in G(g\cdot y \in A)\}$, 
gives a truly invariant subset of $Y$, 
with a contradiction to ergodicity. 

$(\Rightarrow)$: Assume 
\[\pi: G\rightarrow {\rm LinIsom}(\ell^2)\]
\[g\mapsto \pi_g\] 
is 
a continuous homomorphism and we have $\pi$-equivariant 
measurable $\phi: X\rightarrow \ell^2$ with $\phi(g\cdot x) 
=\pi_g\cdot \phi(x)$ a.e. $x, g$. We assume 
$\pi$ is not constant and try to construct a 
counterexample to 
moderate mixing. 

\smallskip 

{\bf Claim:} There is an {\it irreducible} unitary representation 
\[\rho: G\rightarrow {\rm LinIsom}(\h)\]
\[g\mapsto \rho_g\] 
and non-trivial measurable $\psi: X\rightarrow \h$ with $\psi(g\cdot x)
=\rho_g\cdot\phi(x)$ a.e. $x, g$. 

\smallskip 


{\bf Proof of Claim:} Following say \cite{mautner} we 
find a standard measure space $(Z, m)$ and write $\ell^2$ as a 
direct integral of irreducible representations of $G$: 
\[\ell^2=\int_Z \h_z dz\]
\[\phi=\int_Z \phi^z dz.\]   
Then for each $z\in Z$ we have a map 
\[p_z: \ell^2\rightarrow \h_z\] 
which coordinate so that for any $v\in \ell^2$ we 
have 
\[v=\int_Z p_z(v) dz\] 
and for any $g\in G$ 
\[g\cdot v=\int_Z \phi^z_g\cdot p_z(v)dz.\] 
Thus for a.e. $g, x$ we have at $m$-a.e. $z$ 
\[p_z(\phi_g\cdot \pi(x))=\phi^z_g\cdot p_z(\pi(x));\] 
we may assume that $p_z\circ \pi$ is non-trivial at 
every $z$.  
Thus we may appeal to Fubini to find a single $z$ such that 
for a.e. $g, x$ $p_z(\phi_g\cdot \pi(x))=\phi^z_g\cdot p_z(\pi(x))$ 
and finish with $\h=\h_z$, $\rho=\phi^z$, $\psi=p_z(\pi)$. 
\hfill (Claim$\square$) 

\smallskip 


Now there is a split in cases: 

\smallskip 

{\bf Case 1:} The representation is finite dimensional. 

\smallskip 

Clearly there is a $U(\h)$-invariant invariant 
probability measure $\nu'$ on the unit ball in $\h$. 
For each $x\in X$ we let $(Y_x, \nu_x)$ be the ergodic 
component of $\psi(x)$ in $((\h)_1, \nu')$ under the action of 
$G$. By ergodicity of the $G$ action on $X$, we obtain that this 
ergodic component is constant $\mu$-a.e; let $(Y, \mu)$ be this 
ergodic component. 

Now we obtain a non-trivial $G$-invariant function 
\[f: X\times Y\rightarrow \R\]
\[(x, v)\mapsto \nh \psi(x)-v\nh.\] 

\smallskip 

{\bf Case 2:} The representation is infinite dimensional. 

\smallskip 

Following \cite{zi1} 5.2.13 we may find a standard 
Borel probability space $(Y, \nu)$ on which $G$ acts measurably, 
measure preservingly, and ergodically, and for which the 
representation $\rho$ is isomorphic to a direct summand of the usual 
representation of $G$ on $L^2(Y, \nu)$; at that point we 
may very well assume $\h$ is a direct summand of $L^2(Y, \nu)$, 
and hence every $f\in \h$ is a function in $L^2(Y, \nu)$. 

We obtain a non-trivial function 
with 
\[f: X\times Y\rightarrow \C\]
\[(x, y)\mapsto (\psi(x))(y).\] 
Unwinding the definitions we see that a.e.  
\[f(g\cdot x, g \cdot y) = 
(\psi(g\cdot x))(g\cdot y)\]
\[=g\cdot (\psi(x))(g\cdot y)= \psi(x) (g^{-1}g\cdot y)= \psi(x)(y),\] 
since for any function $F\in L^2(Y, \nu)$ we have by definition 
that $g\cdot F(z)=F(g^{-1}z)$ any $z\in Y$. 
\end{proof} 



Schmidt and Walters have obtained a purely combinatorial characterization of 
mild mixing: 

\begin{theorem} (\cite{scwa}) Let $G$ act in a measurable, non-singular, properly ergodic 
manner on $(X, \mu)$. The action is mild mixing if and only if for all $A$ measurable 
and neither null nor conull 
\[{\rm liminf}_{g\rightarrow \infty} \mu(A\Delta (g\cdot A))>0.\] 
\end{theorem} 

It would be interesting if there is a similar combinatorial characterization 
of when a properly ergodic non-singular action preserves ergodicity with respect to 
products with measure preserving, ergodic actions on standard Borel probability spaces. 

We work towards a general result, due to Kaimanovich, and after Burger and Monod. 
In this section we see any locally compact group has an $\m$-ergodic action, for 
$\m$ the class of separable Banach spaces. 
In the next section we review the amenability properties of this action. 


\begin{definition} Let $\mu$ be a probability measure on a lcsc group $G$ that 
has the same measure class as a left invariant Haar measure. Consider the 
one sided product, 
\[\prod_{n\in \N} (G, \mu),\] 
equipped with the measure $\lambda$ arising from the infinite product of $\mu$. 
We equip this space with the shift action 
\[T: \prod_{n\in \N} G\rightarrow  \prod_{n\in \N} G\]
\[(g_1, g_2, g_3, ...)\mapsto (g_2, g_3, g_4,...).\] 
\end{definition} 

\begin{lemma} \label{label1} 
(Kaimanovich; \cite{kaimanovich}) 
If $E$ is a separable Banach space and 
\[F: \prod_\N G\rightarrow E\] 
\[\pi : G\rightarrow {\rm Iso} (E)\] 
are measurable with 
\[\pi_{g_1} F((g_2, g_3, ...))= F((g_1, g_2, ...)), \] 
then $F$ is constant a.e. 
\end{lemma} 

Note that we do not yet assume $\pi$ is a homomorphism, though in 
actual fact it will be in the cases of interest to us. 
The measurability assumption on $F$ is that pullbacks of sets in $E$ 
which are open in the Banach space metric are $\lambda$ measurable. 

We need to define another semigroup action along with another action 
of the group $G$. 

\begin{definition} Let 
\[S: \prod_{n\in \N} G\rightarrow  \prod_{n\in \N} G\]
\[(g_1, g_2, g_3, ...)\mapsto (g_1g_2, g_3, g_4,...).\] 
Note that the assumptions on $\mu$ guarantee that this 
semigroup action is non-singular with respect to the 
product measure $\lambda$. 

We then let $G$ act on $\prod_\N G$ by 
\[\gamma\cdot (g_1, g_2, g_3,...)=(\gamma g_1, g_2, g_3,...).\] 
This $G$ action clearly commutes with $S$ and is again non-singular by 
choice of $\mu$. 
\end{definition} 

\begin{definition} 
Let $\Gamma(G, \mu)$ be the space of ergodic components of $S$; that is to 
say, we write $\prod_\N G$ as a  measurable union of disjoint pieces, 
\[\prod_\N G = \dot{\bigcup_{x\in \Gamma(G, \mu)}} Y_x,\] 
where each $Y_x$ is $S$-invariant and $S$ acts ergodically on $Y_x$. (See 
\cite{halmos}). We then let 
\[b: \prod G\twoheadrightarrow \Gamma(G, \mu)\] 
be the natural surjection, with $b(\vec g)=x$ if $\vec g\in Y_x$. 
We let $\nu=b^*[\lambda]$ be the push forward of $\lambda$, defined 
by $\nu[A]=\lambda[b^{-1}[A]]$. 

Since the action of $G$ commutes with $S$ we obtain a well defined action 
of $G$ on $\Gamma(G, \mu)$ with 
\[\gamma\cdot b(\vec g)=b(\gamma\cdot \vec g).\] 
Again this action is non-singular, since the pre-existing action on 
$(\prod_N G, \lambda)$ is non-singular. 
\end{definition} 

Therefore we have a daunting total of four non-singular actions: 
The semigroup action provided by $T$; the semigroup action provided by 
$S$; the action on the original $\prod_\N G$ by $G$; the induced action of 
$G$ on the quotient object $\Gamma(G, \mu)$. Ultimately we will only be 
interested in this final action of $G$, but it is the remarkable discovery of 
\cite{kaimanovich} that the other three actions cooperate to prove: 


\begin{theorem} \label{theorem1} 
(Kaimanovich, \cite{kaimanovich}) 
Let $\m$ be the class of separable Banach spaces. Then 
the action of $G$ on $\Gamma(G, \mu)$ is $\m$-ergodic. 
\end{theorem} 

\begin{proof} (In any reasonable mathematical sense, this proof 
has been previously given in \cite{kaimanovich}; however since our 
set up and notation is rather different, we recall it here for the reader's 
convenience.) 
Fix $E\in \m$ and $\pi: G\rightarrow {\rm Iso}(E)$ a continuous homomorphism. 
Let $f: \Gamma(G, \mu)\rightarrow E$ be $G$-equivariant and measurable in the 
usual sense. We then define $F=f\circ b$ 
\[F: \prod_\N G\rightarrow E,\] 
\[\vec g \mapsto f(b(\vec g)).\] 
$F$ is clearly measurable; on the other hand we obtain a.e. that 
\[\pi_{g_1}F((g_2, g_3, ...))=\pi_{g_1}(f(b((g_2, g_3, ...))) \]
\[=f(g_1\cdot b((g_2, g_3,...))),\] 
by the $G$-equivariance of $f$, 
\[=f(b(g_1\cdot (g_2, g_3,...))),\] 
by definition of the $G$-action on $\Gamma(G, \mu)$, 
\[=f(b(g_1g_2, g_3, ...))=f(b(g_1, g_2, g_3,...))= F((g_1, g_2, g_3,...)),\] 
by invariance of $b$ under $S$; hence by \ref{label1} $f$ is constant a.e. 
\end{proof} 

By suitably composing any $\vec g\in \prod_\N G$ we can view such a sequence 
as providing a random walk through $G$. The shift action $S$ then corresponds to 
discounting the choice of the first element in this walk. The space of 
ergodic components $\Gamma (G, \mu)$ is something like the space of random 
walks considered up to ``direction" or ``eventual behavior". We have failed 
to emphasize this point of view above; it comes through much more strongly in 
the original treatment of \cite{kaimanovich}. 

Kaimanovich's paper also proves a better result: The $G$-action on 
$\Gamma(G, \mu)$ is doubly ergodic. That is to say, any measurable, 
$G$-equivariant 
function $f: \Gamma(G, \mu)\times \Gamma(G, \mu)\rightarrow E$ is 
constant a.e. The proof of this stronger result, for which we have no 
need, is similar and follows from a strengthening of \ref{label1}. 

\section{Amenability} 

\begin{definition} (See \cite{zi1}; but note that, unlike Zimmer, 
we write our actions on the left.) 
Let $G$ be a lcsc group acting measurably on $(X, \mu)$ and let 
$H$ be a Polish group. 
A measurable function 
\[\alpha: X\times G\rightarrow H\] 
is a {\it cocycle} if for a.e. $x\in X$ and all $g, h\in G$ 
\[\alpha(x, gh)=\alpha(hx, g)\alpha(x, h).\] 
In the case that $H={\rm Iso}(E)$, the isometry group of 
some separable Banach space, we obtain the {\it dual cocycle} 
\[\alpha^*: X\times G\rightarrow {\rm Iso}(E^*)\] 
by taking the adjoint in the usual way. 
\end{definition} 


\begin{definition} 
We say that the action of $G$ on $(X,\mu)$ is {\it amenable} if 
whenever we have such a cocycle and a measurable assignment 
\[x\mapsto A_x,\] 
\[ X\rightarrow K_c(E^*_1),\] 
assigning to each point a weak$^*$ compact, convex subset of the 
unit ball $E^*$ with the invariance property 
\[\alpha^*(x, g)A_x=A_{gx}\] 
a.e. $x$, all $g$, then we can find a measurable assignment 
\[X\rightarrow E^*\]
\[x\mapsto a_x,\] 
with each $a_x\in A_x$ and $\alpha^*(x, g)a_x=a_{g\cdot x}$ a.e. $x$, all $g$. 
\end{definition} 

We are thinking of $E$ as having the weak$^*$ topology, and so in particular 
measurability of the assignment $x\mapsto A_x$ is the requirement that it be 
measurable with respect to the induced Effros Borel structure: That is to say, 
if $U\subset E^*$ is weak$^*$ open, then the set of $x$ with $A_x\cap U\neq \emptyset$ 
must be $\mu$-measurable. 

There is also a parallel definition we can give for amenability of equivalence relations. We 
do not give this definition here, but simply quote an alternative characterization from 
\cite{cofewe}. 

\begin{theorem} (Connes-Feldman-Weiss; \cite{cofewe}) 
\label{cofewe} An equivalence relation $E$ on a standard Borel probabibility space is 
{\it  amenable} if and only if there is a conull subset $A$ with $E|_A\leq_B E_0$. 
\end{theorem} 


\begin{lemma} \label{zi1} (Zimmer; \cite{zi1}) 
If $S$ is an amenable 
$G$-space and $X$ is any standard Borel probability 
space on which $G$ acts measurably and non-singularly, then $X\times S$ is an 
amenable $G$-space in the usual diagonal action. 
\end{lemma} 

\begin{theorem} \label{zi2} (Zimmer; \cite{zi2}) 
If $G$ is lcsc, then $\Gamma(G, \mu)$, as defined in 
$\S$2, is an amenable $G$-space. 
\end{theorem} 

Putting together \ref{zi1} and \ref{zi2} along with 
\ref{theorem1} we get: 

\begin{corollary} \label{big} 
An lcsc group $G$ admits a non-singular, measurable action on 
a standard Borel probability space $(S, m)$ such that for any other 
standard Borel probability space $(X, \nu)$ on which $G$ acts measurably 
and in a measure preserving, properly ergodic fashion, we have: 

\leftskip 0.4in 

\no (i) the diagonal action of $G$ on $(S\times X, m\times \nu)$ is properly 
ergodic; 


\no (ii) the diagonal action of $G$ on $(S\times X, m\times \nu)$ is 
amenable. 

\end{corollary} 

Without placing additional restrictions on $G$, this corollary would 
seem to be the strongest we can hope for in this direction. In particular, 
we cannot have a mildly mixing action of $\F_2$ which is amenable. 

\begin{proposition} \label{folklore} (Folklore) 
Let the free group $\F_2$ on two generators act amenably on a 
standard Borel probability space $(S, m)$ by non-singular transformations. 
Then the action of $\F_2$ on $(S^3, m^3)$ is not properly ergodic. 
\end{proposition} 

\begin{proof} 
Note that $\F_2$ acts continuously on its boundary, $\partial \F_2$, and hence we have 
an induced homomorphism $\rho: \F_2\rightarrow {\rm Iso} C(\partial \F_2)$,  from 
$\F_2$ to the isometry group of the continuous functions on 
the boundary. 
We then obtain an induced cocycle 
\[S\times \F_2 \rightarrow {\rm Iso} (C(\partial \F_2))\]
\[(s, \sigma)\mapsto \rho(\sigma).\] 
The action is amenable, and hence for $M(\partial \F_2)$ the space of 
probability measures on $\partial \F_2$ (and the dual of $C(\partial \F_2)$) 
we obtain a measurable equivariant assignment 
\[S\rightarrow M(\partial \F_2),\] 
\[s\mapsto \nu_s.\] 
At the level of $S^3$ we obtain a corresponding equivariant 
\[S^3\rightarrow M(\partial \F_2),\] 
\[(s_1, s_2, s_3)\mapsto \frac{1}{3}(\nu_{s_1}+\nu_{s_2} + \nu_{s_3}).\] 
On an invariant non-null set, $A\subset S^3$, the resulting measure 
$\nu_{s_1}+\nu_{s_2} + \nu_{s_3}$ does not concentrate on just two points. 
Following say $\S$C.4 \cite{hjorthkechris2} (or see \cite{adams2} and 
\cite{adamslyons} for the primary references) we may find a measurable 
equivariant map from the measures on $M(\partial\F_2)$ concentrating on 
more than two points, and hence there is a measurable equivariant map 
\[\varphi: A\rightarrow \F_2.\] 
This gives us a selector $A_0\subset A$ defined by 
\[A_0=\{(s_1, s_2, s_3)\in A: \varphi(s_1, s_2, s_3)=e\},\] 
where $e$ is the identity in $\F_2$;  the existence of such a selector 
contradicts 
proper ergodicity of $\F_2$ on $A$ and hence on $S^3$. 
\end{proof} 







There is one last fact in this direction we need. 











\begin{lemma} \label{E0} If lcsc $G$ acts on standard Borel probability space 
$(X, \mu)$ by measure preserving transformations with every point 
having an amenable stabilizer, then if $E^X_G\leq_B E_0$, then the 
group $G$ is amenable. 
\end{lemma} 

\begin{proof} From say 2.15 \cite{jakelo} we have that the 
equivalence relation is amenable; and then it is known from 
\cite{adelgo} 
that an action with amenable stabilizers and amenable equivalence 
relation is amenable, and thus $G$ is amenable by 4.3.3\cite{zi1}. 
\end{proof} 










\section{The result} 

\begin{theorem} \label{4.1} 
Let $G_1, G_2$ be lcsc groups acting on a standard Borel probability space 
$(X, \mu)$ with: 

\leftskip 0.4in 

\no (0) the actions of $G_1$ and $G_2$ commute; 

\no (i) the action of $G_1\times G_2$ is measurable (as a function from $G_1\times G_2\times X$ to $X$); 

\no (ii) the action of $G_1\times G_2$ is measure preserving; 

\no (iii) the orbit equivalence relation $E_{G_1}$ is not amenable; 

\no (iv) the action of $G_2$ is properly ergodic on all its ergodic components. 


\leftskip 0in 

\no Then the resulting orbit equivalence relation $E_{G_1\times G_2}^X$ is non-treeable. 

\end{theorem} 

\begin{proof} By going to ergodic components we may actually assume that the 
action of $G_1\times G_2$ is ergodic. 
Assume for a contradiction that the equivalence relation is treeable, 
and then appealing to \ref{f2} we obtain that some Borel 
$\theta: X\rightarrow F(2^{\F_2})$ witnessing 
$E^X_{G_1\times G_2}$ Borel reducible to the free Borel  action of $\F_2$ on the 
standard Borel probability space $F(2^{\F_2})$. 
We in particular obtain a corresponding cocycle 
\[\alpha: X\times G_1\times G_2\rightarrow \F_2.\] 
Following the ergodic decomposition theorem of \cite{halmos}, we 
write 
\[X={\dot{\bigcup_{z\in Z}}}X_z\] 
as a decomposition of $X$ into $G_2$-invariant ergodic pieces\footnote{ Here 
$Z$ is the standard Borel space arising as equivalence classes of the smooth equivalence 
relation $x_1 E x_2$ if and only if $\forall B\in \b_0({\rm a.e.} \: g\in G(g\cdot x_1\in B)\Leftrightarrow 
{\rm a.e.} \: g\in G(g\cdot x_2\in B))$, for some countable Boolean algebra $\b_0$ which 
generates the Borel sets on $X$}; applying the measure 
disintegration theorem (see for instance \cite{classical}) we obtain a measure 
$\hat{\mu}$ on $Z$ along with $G_2$-invariant 
measures $(\mu_z)_{z\in Z}$ such that 
$\mu=\int_Z \mu_z \hat{\mu}(z)$. The assumptions of the theorem imply that the 
action of $G_2$ on each $X_z$ is properly ergodic. This is a critical observation for the 
proof -- if $G_2$ were to act essentially transitively on its ergodic components, then 
it would make no real contribution to the complexity of the structure of the equivalence 
relation, the argument in the first claim below would fail, and in fact the theorem without 
this assumption would admit a counterexample. 

Now appealing to \ref{big} we let $(S,m)$ be a standard Borel probability space on 
which $G_2$ acts ergodically with the diagonal action on $X\times S$ amenable and such 
that at each fiber $X_z$ we have the induced action of $G_2$ on $X_z\times S$ still properly 
ergodic. Let 
\[\hat{\alpha}: (X\times S)\times G_2\rightarrow \F_2\] 
be derived from $\alpha$ in the obvious way: 
\[\hat{\alpha}(x, s, g)=\alpha(x, e_{G_1}, g),\] 
where $e_{G_1}$ is the identity in $G_1$. By the amenability of the action 
of $G_2$ on $X\times S$ we may find a measurable assignment of probability measures 
\[X\times S\rightarrow M(\partial \F_2)\]
\[(x, s)\mapsto \nu_{x, s}\] 
such that $\hat{\alpha}(x, s, g)\cdot \nu_{x, s}=\nu_{g\cdot x, g\cdot s}$ 
a.e. $x\in X, s\in S$ all $g\in G_2$. 

\def\mtwo{M_{\leq 2}(\partial \F_2)} 
\def\mthree{M_{\geq 3}(\partial \F_2)} 

We let $\mtwo$ be the probability measures concentrating on at most 
two points; we let $\mthree=M(\partial \F_2)\setminus\mtwo$ be the measures 
whose support is greater than two points. 

The next argument through the next couple of 
claim parallels \cite{adams} and is similar to ideas of Zimmer, and from 
there Margulis and eventually Mostow. 

\smallskip 

\no {\bf Claim:} For a.e. $x, s$, $\nu_{x, s}\in \mtwo$. 

\smallskip 

\no {\bf Proof of claim:} Otherwise we can obtain 
some ergodic component $X_z$ such that for a.e. $x\in X_z$ and 
a.e. $s\in S$ we have $\nu_{x, s}\in \mthree$. Appealing to the existence 
of an equivariant Borel map from $\mthree$ to $\F_2$, as given discussed in the 
course of \ref{folklore} (again see for instance theorem C4 of 
\cite{hjorthkechris2}, or 
\cite{adams2}, \cite{adamslyons}) we may find a measurable 
$\hat{\alpha}$-equivariant $\varphi: X_z\times S\rightarrow \F_2$. 

This will provide a contradiction to {\it proper} ergodicity of the action. We let 
$(\sigma_i)_{i\in\N}$ enumerate $\F_2$ and at each $(x, s)\in X_z\times S$ we 
let $i(x, s)$ be the least $i$ for which there exists a non-null collection of 
$g\in G_2$ with $\varphi(g\cdot x, g\cdot s)=\sigma_i$; since the measure 
quantifier preserves Borel sets (see for instance \cite{classical}), this 
function is Borel, and moreover the set 
\[A_1=\{(x, s): \varphi(x, s)=\sigma_{i(x, s)}\}\] 
will provide a Borel selector for $E^{X_z\times S}_{G_2}$; since 
$\hat{\alpha}(x, s, g)\cdot \theta(x)= \theta(g\cdot x)$ by the definition of the 
cocycle, we have 
\[\theta(x_1)=\theta(x_2)\] 
for all $(x_1, s_1), (x_2, s_2)\in A_1$ with  
$(x_1, s_1)E_{G_2}^{X_z\times S} (x_2, s_2)$; then using 
say Jankov, von Neumann uniformization (see for instance 
\cite{classical}) we obtain a measurable function $f: X_z\times S\rightarrow F(2^{\F_2})$ 
assigning to each 
$(x, s)$ the unique value $\theta(x')$ any $(x', s')\in A_1$ that is $E_{G_2}$ equivalent to 
$(x, s)$. Our 
action is ergodic, and $f$ is $G_2$-invariant, 
so $f$ is constant a.e. But that means in particular 
that the equivalence class $[\theta(x)]_{\F_2}$ is constant almost everywhere, and 
hence, by assumption on $\theta$, the equivalence class $[x]_{G_2}$ is constant 
almost everywhere on $X_z$. This 
%gives a routine contradiction 
%preserving and having compact stabilizers, since it would imply that for 
%some compact subgroup $H< G_2$ the existence of a left invariant probability 
%measure on 
%$H\setminus G_2$. 
contradicts the proper ergodicity of the action of $G_2$ on $X_z\times S$. 
\hfill (Claim$\square$) 

\smallskip 

By a straightforward exhaustion argument (compare \cite{adams}, \cite{kechrisproduct}, or 
even lemma 6.6 \cite{hjorthkechris2}) we may find a measurable, equivariant assignment 
\[X\times S\rightarrow M(\partial \F_2)\]
\[(x, s)\mapsto \nu_{x, s}\] 
which is maximal almost everywhere. In other words, if 
\[(x, s)\mapsto \nu_{x, s}'\] 
is any other measurable, equivariant assignment with 
\[{\rm supp}(\nu_{x, s})\subset {\rm supp}(\nu_{x, s}')\] 
(where here supp$(\mu)$ denotes the support of measure $\mu$)  
and $\hat{\alpha}(x, s, g)\cdot \nu_{x, s}'=\nu_{g\cdot x, g\cdot s}'$ a.e. $x$ a.e. $g$, 
then 
\[\nu_{x, s}'=\nu_{x, s}\] 
almost everywhere. 
Since $\nu_{x, s}\in \mtwo$ a.e, 
we may further more assume that 
these measures assign all points in their support equal mass -- 
which is to say, a.e. $\nu_{x, s}$ equals either $\delta_a$ or 
$\frac{1}{2}(\delta_{a_1}+ \delta_{a_2})$. 

\smallskip 

\no {\bf Claim:} For $g\in G_1$, the assignment 
\[(x, s)\mapsto \alpha(g\cdot x, g, e_{G_2})^{-1}\cdot \nu_{g\cdot x, s}\] 
is again $\hat{\alpha}$-equivariant. 

\smallskip 

\no {\bf Proof of claim:} Since for any $h\in G_2$ we have 
\[\hat{\alpha}(x, s, h)\cdot (\alpha(g\cdot x, g, e_{G_2})^{-1}\cdot \nu_{g\cdot x, s})\] 
\[=\alpha(x, h, e_{G_1})\cdot (\alpha(g\cdot x, g, e_{G_2})^{-1}\cdot \nu_{g\cdot x, s}),\] 
which by the defining identity of the cocycle and the fact that the 
actions of $G_1$ and $G_2$ commute equals
\[\alpha(hg\cdot x, g, e_{G_2})^{-1}\cdot (\alpha(g\cdot x, h, e_{G_1})\cdot \nu_{g\cdot x, s})\] 
\[=\alpha(hg\cdot x, g, e_{G_2})^{-1}\cdot \nu_{hg\cdot x, h\cdot s},\] 
by $G_2$-equivariance of the cocycle, which in turn by the commutativity of the 
$G_1, G_2$ actions equals 
\[=\alpha(gh\cdot x, g, e_{G_2})^{-1}\cdot \nu_{gh\cdot x, h\cdot s},\]
as required. \hfill (Claim$\square$) 

\smallskip 

Thus by maximality of the assignment 
\[(x, s)\mapsto \nu_{x, s}\] 
we have 
\[\alpha(g\cdot x, g, e_{G_2})^{-1}\cdot \nu_{g\cdot x, s}=\nu_{x, s}\] 
almost everywhere, and hence that $(x, s)\mapsto \nu_{x, s}$ must 
be equivariant under the entire action of $G_1\times G_2$. In particular 
we obtain that for almost every $(x, s)\in X\times S$ and almost 
every $g\in G_1$ we have some $\alpha(x, g, e_{G_2})\cdot \nu_{x,s}=
\nu_{g\cdot x, s}$. Using Fubini we may find 
some specific $s_0\in S$ such that 
for a.e. $g\in G_1$, $x\in X$ 
\[\alpha(x, g, e_{G_2})\cdot \nu_{x,s_0}=
\nu_{g\cdot x, s_0}\]
\[\therefore \nu_{g\cdot x, s_0} E^{M(\partial \F_2)}_{\F_2} \nu_{x, s_0}.\] 
Since the equivalence relation on $M(\partial \F_2)$ induced by the 
action of $\F_2$ is hyperfinite  (see for instance the appendix of 
\cite{hjorthkechris2}), we may write 
\[E^{M(\partial \F_2)}_{\F_2}=\bigcup_{n\in \N} F_n,\] 
where $(F_n)_{n\in \N}$ is an increasing union of Borel equivalence relation with 
finite classes. 

Following \cite{sections} we can fix a Borel $C\subset X$ meeting each $E^X_{G_1}$ equivalence class 
in a countable non-empty set and a Borel function 
\[f:X\rightarrow C\] 
with $xE_{G_1}^X f(x)$ all $x\in X$. In particular this gives $E^X_{G_1}\leq_B E^X_{G_1}|_C$. 
We then use $f$ to transfer the measure $\mu$ to a measure 
$\hat{\mu}$ on $C$ with 
\[\hat{\mu}(A)=\mu(f^{-1}[A]).\] 
Note that $E^X_{G_1}|_C$ will be non-amenable with respect to 
$\hat{\mu}$ since $E^X_{G_1}$ is non-amenable with respect to 
$\mu$. 



We then set 
\[x_1 \hat{F_n} x_2\] 
if $x_1 E^X_{G_1} x_2$ and $\nu_{x_1, s_0} F_n \nu_{x_2, s_0}$. Since 
$x_1 E^X_{G_1} x_2$ implies $\nu_{x_1, s_0} E^{M(\partial \F_2)}_{\F_2} \nu_{x_2, s_0}$ we 
have 
\[\bigcup_n \hat{F_n} = E^X_{G_1}.\] 

Now there are two cases, both leading to the conclusion that 
$E_{G_1}^{X}$ is Borel reducible to $E_0$ on a co-null set, and hence a 
contradiction to its non-amenability by \ref{E0}. 

\smallskip 


\no {\bf Case(1):} There is an $n$ and a $\hat{\mu}$ non-null set on which 
$\hat{F}_n$ is non-smooth. 

\smallskip 

\no {\bf Claim:} There is a $\hat{\mu}$ non-null set of $x\in C$ for which 
\[\theta[[x]_{\hat{F}_n}],\] 
the pointwise image of the set $[x]_{\hat{F}_n}$ under $\theta$, is infinite. 

\smallskip 

\no {\bf Proof of claim:} Otherwise apply Lusin-Novikov (as say found 
in \cite{classical}) to obtain a Borel function 
\[\rho: \theta[C]\rightarrow C\] 
such that for every $y$ we have 
\[\theta(\rho(y))=y.\] 
Then $\rho\circ \theta[C]$ meets each 
$\hat{F}_n$ equivalence class in a finite non-empty set. Then since 
Borel equivalence relations with finite classes are smooth and since 
$\hat{F}_n$ is Borel reducible to $\hat{F}_n|_{\rho\circ\theta[C]}$ we 
obtain that $\hat{F}_n$ is smooth, with a contradiction to case 
assumption. 
\hfill ($\square$Claim) 

\smallskip 

\no {\bf Claim:} There is a $\hat{\mu}$ non-null set of $x$'s for which the 
corresponding measure $\nu_{x, s_0}$ has infinite stabilizer. 

\smallskip 

\no {\bf Proof of claim:} Suppose first that $\nu_{x, s_0}$ has finite stabilizer. Then for 
any other $\nu$ with $\nu E_{\F_2} \nu_{x, s_0}$ there are only finitely many 
$\sigma\in \F_2$ with $\sigma\cdot \nu_{x, s_0}=\nu$. Thus since each $F_n$ is finite we obtain 
$\{\sigma\in \F_2: \sigma\cdot \nu_{x, s_0} F_n \nu_{x, s_0}\}$ finite, and 
hence $\theta[[x]_{\hat{F}_n}]$ is finite. 

In conclusion we have thus obtained that $x$ is outside the $\hat{\mu}$ positive set of the last 
claim. 
\hfill ($\square$Claim) 

\smallskip 

As noted for instance in the appendix of 
\cite{hjorthkechris2}, 
there are 
only countably many measures in $\mtwo$ with infinite stabilizer, and 
hence by ergodicity 
we may assume there is some fixed $\nu$ such that a.e. 
\[\nu_{x, s_0} E^{\mtwo}_{\F_2} \nu.\] 
Then we 
may find a measurable assignment 
\[X\rightarrow \F_2\]
\[x\mapsto \sigma_x\] 
such that $\sigma_x\cdot \nu_{x, s_0}=\nu$ a.e. Thus if we replace 
the reduction $\theta$ by 
\[\hat{\theta}: X\rightarrow F(2^{\F_2}),\]
\[x\mapsto \sigma_x\cdot \theta(x)\] 
then we obtain a reduction of $E_{G_1}^X$ a.e. into an orbit equivalence relation 
the induced by the stabilizer 
of $\nu$, which will necessarily be cyclic abelian, and hence 
hyperfinite by \cite{jakelo}, with a contradiction to 
non-amenability of $G_1$ by \ref{E0}. 

\smallskip 

\no {\bf Case(2):} At every $n$ there is a conull set on which $\hat{F}_n$ is smooth. 

\smallskip 

Hence $E^X_{G_1}$ is, on a conull set, hypersmooth -- that is to say, the increasing union 
of smooth equivalence relations. Hence by the Kechris-Louveau dichotomy theorem of 
\cite{kechrislouveau} we have that it is either $\geq_B E_1$ or $\leq_B E_0$. The former 
is impossible, since \cite{kechrislouveau} shows that $E_1$ is not Borel reducible to any 
Polish group action, whilst the latter gets us into the same contradiction to the 
non-amenability of $E_{G_1}$. 
\end{proof} 

\section{Spanning trees for product groups} 

\begin{definition} For $X$ a set, $T\subset X\times X$ is said to be a 
{\it spanning tree} if it is symmetric, irreflexive, acyclic, and moreover 
any two distinct points in $X$ are connected by some path in $T$. 
\end{definition} 

We will only be interested in the case that 
$X$ is countable. In this event, the collection of such spanning trees is a Borel 
subset of $\{0, 1\}^{X\times X}$ via the identification of a subset 
of $X\times X$ with its characteristic function. The group of all 
permutations of $X$ acts on the standard Borel space of 
spanning trees by $g\cdot T=\{(g\cdot x_1, g\cdot x_2): 
(x_1, x_2)\in T\}$. 

The following is a continuous analog of a lemma  presented in the 
discrete case with credit to Russ Lyons by the manuscript \cite{kechrismiller}. 




\begin{lemma} Let $\Omega$ be a countable set, let Sym$(\Omega)$ be the group of 
all permutations of $\Omega$, and let $G<{\rm Sym}(\Omega)$ be a subgroup. Suppose: 

\leftskip 0.4in 

\no (a) $G$ is closed in the topology of pointwise convergence; 

\no (b) for any $a\in \Omega$ the subgroup 
$G_a (=_{\rm df}\{g\in G: g\cdot a =a\})$ 
is compact; 

\no (c) there is a  $G$-invariant Borel probability measure on the 
spanning trees on $\Omega$. 

\leftskip 0in 

\no Then there is an action of $G$ on some standard Borel probability space $Y$ with: 

\leftskip 0.4in 

\no (i) the action  measurable (as a function from $G\times Y$ to $Y$); 

\no (ii) the action measure preserving; 

\no (iii) for any $y\in Y$ the stabilizer of $y$ in $G$, $G_y$, is compact; 

\no (iv) the equivalence relation $E^Y_G$ is treeable.  

\leftskip 0in 


\end{lemma} 

\begin{proof} 
Let $\Omega_0\subset\Omega$ be an orbit under $G$. I claim that we can replace $\Omega$ by 
$\Omega_0$ by obtaining an invariant probability measure on the spanning trees on 
$\Omega_0$. To do this, we only need to find a Borel and invariant assignment of a 
spanning tree on $\Omega_0$ to spanning trees on $\Omega$, and this follows by the 
kind of argument one sees in \cite{jakelo}: 
Given a spanning tree $T$ on $\Omega$, we assign to each $y\in \Omega$ 
the point $p_T(y)$ arising as the place of first entrance into $\Omega_0$ under a $T$-path 
from $y$ to $\Omega_0$; we can do this so if the path from 
$y$ to $p_T(y)$ passes through $y'$ then $p_T(y)=p_T(y')$; 
we then define $\hat{T}$ on $\Omega_0$ by 
$x_1 \hat{T} x_2$ if and only if $x_1\neq x_2$ and 
there are $y_1, y_2$ with $p_T(y_1)=x_1$, $p_T(y_2)=x_2$, 
and $y_1 T y_2$. 

Thus from now on we assume that the action of $G$ acts transitively on $\Omega$ and 
we fix some specific $a\in \Omega$ and let $G_a$ be its stabilizer in $G$. 
We use $\to$ to denote the 
space 
of spanning trees on $\Omega$. 
We let $\mu$ be the $G$-invariant measure $\to$. 

Following \cite{veech} we let $(X, \nu)$ be a standard Borel probability space on 
which $G$ acts freely. We need only show that the orbit equivalence relation 
arising from the action of $G$ on $y=\to\times X$ is treeable. 


Consider then the equivalence relation $E_{G_a}$ on $\to\times X$ arising from 
the group $G_a$. The responsible 
group is compact and hence the equivalence relation is smooth, and 
thus we may find a Borel selector $A\subset \to\times X$ meeting each 
$E_{G_a}$-equivalence class in exactly one point. 

We need only define the treeing on $A$. 
For $(T_1, x_1), (T_2, x_2)\in A$ set $(T_1, x_2)\g (T_2, x_2)$ if and only if 
there is $g\in G$ with 

\leftskip 0.4in 

\no $g\cdot T_1 =T_2$, 

\no $g\cdot x_1=x_2$, 

\no $(a, g(a))\in T_2$, or, equivalently, $(a, g^{-1}(a))\in T_1$. 

\leftskip 0in 

This is a Borel graphing, since the set of $g$ will be of size at most 
1, and we can appeal to the uniformization theorem for 
Borel sets in the plane with small sections. To check it is a treeing we need to 
show existence of unique paths between $((T, x), (T', x'))\in E^Y_G\cap A\times A$. 

We will construct a path and then go on to verify that any other $\g$-path must 
arise by the same process and in fact be the same. 

                    
                    
We choose the 
unique $h\in G$ with $h\cdot x=x'$; the assumption of $E_G$ equivalence 
entails $h\cdot T=T'$. We then form a non-self intersecting path 
$a_0=a, a_1, a_2,..., a_n=h(a)$ from $a$ to $h(a)$ inside the tree $T$. 
We consider the set of 
all $g_1$ with $g_1^{-1}(a_0)=a_1$; this exactly 
determines $g_1$ up to a left coset 
of $G_a$, so we can choose some such $g_1$ with 
$g_1\cdot (T, x)=(T_1, x_1)\in A$, and observe that 
$g_1$ witnesses $(T, x)\g (T_1, x_1)$. 
Applying the same recipe we may choose some $g_2$ with 
$g_2^{-1}(a_0)=g_1(a_2)$ and $(T_2, x_2)=g_2\cdot (T_1, x_1)\in A$; 
since $g_1^{-1}(a_0)=a_1$ 
\[g_1(a_1)=a_0,\] 
and from and $(a_1, a_2)\in T$ we obtain 
\[(g_1(a_1), g_1(a_2))\in T_1=g_1\cdot T\]
\[\therefore (a_0, g_1(a_2))=(a_0, g_2^{-1}(a_0))\in T_1\]
\[\therefore (T_1, x_1)\g (T_2, x_2).\] 

We continue successively, obtaining at each $i<n$ 
the pair $(T_i, x_i)$ 
along with $g_1, g_2, ...g_{i-1}$ 
\[(T_i, x_i)=g_{i}g_{i-2}...g_1\cdot (T, x),\] 
\[g_{i}g_{i-2}...g_1(a_{i})=a_{0},\] 
and hence 
\[(a_0, g_{i}g_{i-1}...g_1(a_{i+1}))=
(g_{i}g_{i-1}...g_1(a_{i}), g_{i}g_{i-1}...g_1(a_{i+1})) 
\in g_{i-1}g_{i-2}...g_1\cdot T = T_i.\] 
We then choose $g_{i+1}$ subject to the requirement that 
\[g_{i+1}^{-1}(a_0)=g_{i}g_{i-1}...g_1(a_{i+1})\] 
and $(T_{i+1}, x_{i+1})=_{\rm df} g_{i+1}\cdot (T_i, x_i)\in A$. 
This keeps the happy game playing for another round, since 
$g_{i+1}g_{i}...g_1(a_{i+1})=a_0$ and 
$(T_{i+1}, x_{i+1})\g (T_i, x_i)$. Eventually we end up with 
$g=g_ng_{n-1}...g_1$ with the properties that $g\cdot (T, x)=(T_n, x_n)\in A$ and 
$g^{-1}\cdot a_0=a_n$, which in particular entails 
$G_ag=G_ah$; then the assumption that $A$ selects exactly one point from 
each coset will entail $(T_n, x_n)=(T', x')$ as required. 

Alternatively, given any other $\g$ path, 
$(S_0, y_0)=(T, x), (S_1, y_1), ..., (S_m, y_m)=(T', x')$ we reverse the engineering 
above and find $h_1, h_2, ...h_m$ with each $h_i\cdot y_{i-1}=y_i$ and then 
let $b_i=h_1^{-1}h_2^{-1}...h_i^{-1}(a_0)$. 
The assumption of $\g$-adjacency implies each 
\[(h^{-1}_{i+1}(a_0), a_0)\in S_i=h_ih_{i-1}...h_1\cdot T\] 
\[\therefore  (h_1^{-1}h_2^{-1}...h_i^{-1}h_{i+1}^{-1}(a_0),h_1^{-1}h_2^{-1}...h_i^{-1}(a_0)\in T) \] 
\[\therefore (b_{i+1}, b_i)\in T.\]
Note however we must also end up with $h_mh_{m-1}...h_1=h$, and hence $b_m=a_n$, 
and in particular $a, b_1, b_2, ..., b_m=a_n$ traces out a path in $T$ from $a$ to $a_n$. 
It then follows from $T$ being a tree and the assumptions on our set $A$ that either 
that either $m=n$ and each $a_i=b_i$ or there is a repeated vertex -- $(S_i, y_i)=(S_j, y_j)$ 
some $i\neq j\leq n$ -- along the way. 
\end{proof} 

This lemma granted, \ref{pepe} follows from 
\ref{4.1}. In the situation they describe we have a product of lcsc 
automorphism groups, the first of which is assumed to be 
non-amenable and the second of which is implicity assumed to be non-compact. 
Their free action on any probability space must be non-treeable by 
\ref{4.1}, and hence by the lemma above there can be no invariant 
probability measure on the spanning trees of $X\times Y$ in the 
statement of \ref{pepe}. 













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