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\title{\huge Borel equivalence relations which are highly unfree\footnote{{\it Key words and phrases:} 
Ergodic theory, Borel equivalence relations, free group actions, Borel reducibility. 
{\it AMS Subject Classification:} 03E15, 28D15, 37A15.} }         
\author{Greg Hjorth}        % Enter your name between curly braces
\date{\empty}         % Enter your date or \today between curly braces
\maketitle


\maketitle 

\abstract{There is an ergodic, measure preserving, countable 
Borel equivalence relation $E$ on a standard Borel probability 
space $(X, \mu)$ such that 
$E|_C$ is not essentially free on any conull $C\subset X$. } 


\section{introduction} 

The results in this note are a direct response to developments from \cite{popa} and 
\cite{thomas}. Following these papers we now understand much better the extent to which 
Borel equivalence relations may be {\it reduced} to Borel equivalence relations arising from a free 
group action. 

\begin{definition} If $E, F$ are Borel equivalence relations on standard Borel spaces $X, Y$ (that is to 
say, they are Borel as subsets of $X\times X$ and $Y\times Y$), then we write 
\[E\leq_B F,\] 
$E$ {\it Borel reducible} to $F$, 
if there is a Borel $\varphi: X\rightarrow Y$ with 
\[x_1 E x_2 \Leftrightarrow \varphi(x_1) F \varphi(x_2).\] 

A Borel equivalence relation is {\it countable} if every equivalence is countable. (For a discussion 
of the general theory of countable Borel equivalence relations the reader should consult 
\cite{jakelo}.) An equivalence relation $E$ on a standard Borel probability space $(X, \mu)$ is 
{\it measure preserving} if every Borel partial bijection $f: A\rightarrow B$ included in the 
graph of $E$ is measure preserving. The equivalence relation is {\it ergodic} if every 
$E$-invariant Borel set is either null or conull. 
\end{definition} 

\begin{theorem} (Feldman-Moore; \cite{feldmanmoore}) Let $E$ be a countable Borel 
equivalence relation on a standard Borel space $X$. Then there is a countable group 
acting by Borel automorphisms whose orbit equivalence relation equals $E$. 
\end{theorem} 

The authors of \cite{feldmanmoore} go on to ask whether every such 
equivalence relation can be obtain be a {\it free} action of a countable group. Adams in 
\cite{adams} presented a negative solution to their question, 
but the proof worked, in some sense, by placing two 
mutually orthogonal measures on the space in question with respect to which the equivalence 
relation displayed very different measure theoretic properties. 

One way to describe the example as 
is follows. Take a non-amenable group such as $\F_2$, the free group on two generators, and 
let it act freely and by measure preserving transformations on a standard Borel probability space 
$(X_0 \mu_0)$ with induced equivalence relation $F_1$. Then let $\Z$ act freely and by 
measure preserving transformations 
on a standard Borel probability space $(X_1, \nu_1)$ with induced equivalence relation $F_2$.  
Let $X$ be the disjoint 
union of $X_0$ and $X_1$ and let $E$ be the disjoint union of $F_0$ and $F_1$. The action of $\F_2$  on $X_0$ is {\it non-amenable} in the sense of $\S 4.3$ \cite{zimmer} simply because 
$\F_2$ is non-amenable, and hence the orbit equivalence relation 
can only be induced by a measurable action of a 
non-amenable group in light of 4.3.3 \cite{zimmer}; on the 
other hand, the action of $\Z$ is amenable, and any group acting freely to 
give rise to the same orbit equivalence relation must itself be amenable, again by 4.3.3 \cite{zimmer}. 




Adams was to led to 
recast a measure theoretic version of the question from \cite{feldmanmoore} -- considering only measure preserving, ergodic equivalence relations and then only asking  
for a free action after discarding a null set. A negative answer was obtained in \cite{furman}: 

\begin{theorem} (Furman; \cite{furman}) There is an ergodic, measure preserving, countable 
Borel equivalence relation $E$ on a standard Borel probability space $(X, \mu)$ such that on 
any conull $C\subset X$ we have $E|_C$ not induced by the free, Borel action of a countable 
group. 
\end{theorem} 

\begin{definition} (Thomas) A countable Borel equivalence relation $E$ on a standard Borel space is {\it essentially free} if it is Borel reducible to an equivalence relation arising from the free Borel action of a countable group. 
\end{definition} 

Using the results of \cite{popa}: 

\begin{theorem} (Thomas; \cite{thomas}) Not every countable Borel equivalence relation is 
essentially free. 
\end{theorem} 

Moreover \cite{thomas} also shows there to be no maximum element in the essentially free 
Borel equivalence relations under $\leq_B$. 


However Thomas was prompted to phrase Adams' question in to the new context, asking whether every 
countable Borel, ergodic, measure preserving, equivalence relation is on some conull subset 
Borel reducible to the free Borel action of a countable group. 

\begin{theorem} There is an ergodic, measure preserving, countable 
Borel equivalence relation $E$ on a standard Borel probability 
space $(X, \mu)$ such that 
$E|_C$ is not essentially free on any conull $C\subset X$. 
\end{theorem} 

\section{Acknowledgements} 

I am exceptionally grateful to Simon Thomas, who read through {\it several} earlier 
drafts of this note, pointed out a number off errors, and vastly improved the exposition. 

\section{Proofs} 

\begin{definition} Given $S\subset\Z$ define $G_S$ to be the group with generators 
\[a,\] 
\[b_i, c_i,\]
$i\in \Z$, 
subject to the relations 
\[ab_i a^{-1} = b_{i+1},\] 
\[ac_i a^{-1} = c_{i+1},\] and 
\[b_i c_{i+j} = c_{i+j}b_i\] 
for $j\in S$. 
\end{definition} 

\begin{lemma} \label{1} 
For $i, j\in\Z$, if $b_i c_{i+j} = c_{i+j}b_i$ in $G_S$ then $j\in S$. 
\end{lemma} 

\begin {proof} 
First let $K_S$ be the group generated by just the $b_i$'s and $c_i$'s subject to the 
relation $b_i c_{i+j} = c_{i+j}b_i$ for $j\in S$. Not that for any $i, i+j$ with $j\notin S$ we 
can define 
\[\pi: K_S\rightarrow \F_2 =\langle \sigma, \tau \rangle\] 
\[b_i\mapsto \sigma,\] 
\[c_{i+j}\mapsto \tau,\] 
and all other generators get sent to the identity $e$. It is easily checked that this is a 
homomorphism, and thus $b_i$ and $c_{i+j}$ do not commute in $K_S$. 

Now we can define a homomorphism 
\[\varphi: \Z\rightarrow {\rm Aut} (K_S)\]
\[n\mapsto \varphi_n\] 
where 
\[\varphi_n: h\mapsto a^n h a^{-n}.\] 
This is a homomorphism and we immediately have $G_S$ isomorphic to the semi-direct 
product 
\[\Z \rtimes   K_S, \] 
and hence $b_i c_{i+j} b_i^{-1} c_{i+j}^{-1}$ is not equal to the identity in $G_S$. 
\end{proof} 


\begin{lemma} \label {2} Each $G_S$ is finitely generated. 
\end{lemma} 

\begin{proof} $a, b_0, c_0$ generates. 
\end{proof} 


\begin{lemma} \label{2.5} 
The canonical map $a\mapsto a$, $b_0\mapsto b_0$, $c_0\mapsto c_0$ generates an isomorphism 
from $G_S$ to $G_T$ if and only if $S=T$. 
\end{lemma} 

\begin{lemma} 
\label{2.6} Each $G_S$ has no non-trivial finite normal subgroups. 
\end{lemma} 

\begin{proof} Clearly $a$ is always torsion free, so no finite subgroup can contain this element. However, the other elements in the group all have infinite image under the action of conjugation by 
elements of the cyclic subgroup $\langle a \rangle$. 
\end{proof} 

\begin{proof} From \ref{1}, \ref{2}. 
\end{proof} 

\begin{lemma} \label{3} 
If $H$ is a finitely generated group, then there are only countably many $S$'s for which there is a 
homomorphism 
\[\pi: G_S\rightarrow H\] 
with finite kernel. 
\end{lemma} 


\begin{proof} By \ref{2.6} we need only show that there only countably many $S$'s which admit a monomorphism into $H$. There are only countably many choices for $\pi(a), \pi(b_0), \pi(c_0)$, and 
each of these determine $\pi$ by \ref{2}. If 
\[\pi_S: G_S\rightarrow H,\] 
\[\pi_T: G_T\rightarrow H,\] 
have the same values for 
$\pi_S(a), \pi_S(b_0), \pi_S(c_0)$ and $\pi_T(a), \pi_T(b_0), \pi_T(c_0)$ then \ref{1} yields $S=T$. 
\end{proof} 


\begin{definition} For $S\subset \Z$ and $n\in \Z$ we let 
\[n+S=\{\ell + n: \ell\in S\}.\] 
$S$ is periodic if for some $n$ we have $n+S=S$. 
For $S\subset \Z$ let $H_S$ be the group 
\[{\rm SL}_3(\Z)\times G_S.\] 
Let $X$ be the set of all pairs $(S, f)$ where $f\in 2^{H_S}$ and $S$ is not periodic. 


Given $n\in \Z$, $S, T\subset \Z$ with $n+S=T$, we let 
\[\pi_{S, T}: H_S\cong H_T\] 
induced by the canonical isomorphism 
\[G_S\cong G_T,\] 
\[a\mapsto a,\] 
\[b_0\mapsto b_0,\] 
\[c_0\mapsto c_n.\] 
We will be only considering non-periodic subsets of the integers, and so the choice of the $n$ will likewise be canonical. 

We then let 
\[\sigma_{S, T}: 2^{H_S}\cong 2^{H_T}\] be the canonical isomorphism 
\[\sigma_{S, T} (f)(\gamma) =f(\pi_{S, T}^{-1}(\gamma)).\] 
We then define $E$ on $X$ by 
\[(S, f) E (T, g) \] 
if for some (necessarily unique) $n$ we have $n+S=T$ and 
\[\sigma_{S, T}(f) E_{H_T} g\] 
in the space $2^{H_T}$. 

We let $\mu$ be the canonical measure on $2^\Z$, which we identify with $\p(\Z)$. Note that the 
induced action of $\Z$ on $2^\Z$ given by 
\[(n, S)\mapsto n+S\] 
is $\mu$-ergodic. 

For each $S\subset \Z$ we let $m_S$ be the canonical product measure on $2^{H_S}$. 
Then for $A\subset X$ Borel we let 
\[\nu(A) =\int_{2^\Z}(\int_{2^{H_S}} A_S d m_S)d\mu,\] 
where $A_S=\{f\in 2^{H_S}: (S, f)\in A\}$. In other words, we define a measure on $X$ by 
integrating the $m_S$'s against $\mu$. 
\end{definition} 

\begin{lemma} $E$ is a $\nu$-ergodic, $\nu$-invariant, countable Borel equivalence relation. 
\end{lemma} 



\begin{lemma} \label{conull} 
Given any countable $H$ there is a conull subset of $2^\Z$ on which the resulting 
$H_S$ does not admit a homomorphism to $H$ with finite kernel. 
\end{lemma} 

\begin{proof} By \ref{3}. 
\end{proof} 

\begin{corollary} \label{fubini} For almost every $S$ there is a conull subset of $T\in 2^\Z$ for which the 
resulting $H_T$ does not admit a homomorphism from $H_S$ to $H_T$ with finite kernel. 
\end{corollary} 

\begin{proof} By \ref{conull} and Fubini's theorem. 
\end{proof} 

We will use the following result, which was obtained in \cite{thomas} from 
the Popa superrigidity results of \cite{popa}. 

\begin{theorem} (Thomas; \cite{thomas}) \label{thomas} 
Let $G$ be a countable group and let 
$H= SL_3(\Z)\times G$ act by shift on $2^H$. Let $B\subset 2^H$ be 
a conull set. 
If $K$ is some other group acting freely 
by Borel automorphisms on  a standard Borel space $Y$ with 
\[E_H ^B\leq_B E_K^Y\] 
then there is a homomorphism from $H$ to $K$ with finite kernel. 
\end{theorem} 


\begin{lemma} \label{4} If $B\subset X$ is conull, then $E|_B$ is not essentially free. 
\end{lemma} 



\begin{proof} Suppose 
\[\theta: B\rightarrow Y\] 
reduces $E|_B$ to $E^Y_H$ where $H$ is a countable group annd $Y$ is a free Borel 
$H$-space. Then, by Fubini, we obtain uncountably many $S$'s for which $B_S$, by which I mean 
\[\{f\in 2^{H_S} : (S,f)\in B\},\] 
is conull in $2^{H_S}$. Thus by \ref{thomas} there are 
uncountably many $S$ with a finite to one homomorphism 
\[\phi_S: G_S\rightarrow H,\] 
contradicting \ref{3}. 
\end{proof} 

This answers question 3.23 from \cite{thomas}. 

\begin{lemma} If $X$ is written as the disjoint union smooth union of $E$-invariant sets, 
\[X=\dot{\bigcup_{z\in \R}} X_z,\] 
then some $X_z$ is not essentially free. 
\end{lemma} 

\begin{proof} Instead suppose 
\[f: X\rightarrow \R\] is a Borel $E$-invariant function, with 
\[X_z=f^{-1}[\{z\}.\] 
Since $E$ is $\nu$-ergodic, some 
$f^{-1}[\{z\}]$ is conull, and then the lemma follows from \ref{4}. 
\end{proof} 

Note in particular that our $E$ is Borel reducible to $E_\infty$. Thus we obtain from $E$ not being reducible to a smooth union of disjoint free countable Borel equivalence relations that $E_\infty$ is likewise not reducible to a smooth union of disjoint essentially 
free countable Borel equivalence relations, thereby 
answering question 3.19 of \cite{thomas}. 



There is perhaps something else interesting. 

\begin{lemma} For any conull 
$B\subset X$ there exists a further conull $C\subset B$ with $E|_B$ not Borel reducible to $E|_C$.
\end{lemma} 

\begin{proof} Choose $S$ sufficiently random to ensure $B_S=\{f: (S, f)\in B\}$ is 
conull in $ 2^{H_S}$ and that the conclusion of \ref{fubini} holds 
for $S$. Choose $C_0\subset 2^\Z$ so that $H_S$ does not admit a finite to one homomorphism to 
any $H_T$ for $T\in  C_0$. 

Recall that at each $T\in 2^\Z$ there will be a conull subset of $2^{H_T}$ on which 
$H_T$ acts freely. 
Let $C$ be the 
\[\{(T, f)\in B : T\in C_0, f {\rm \: is \: in \: the \: free \: part \: of \: } 2^{H_T}\}.\] 


Suppose 
\[E|_B\leq_B E|_C\] 
for a contradiction. In particular we can obtain some 
\[\theta: B_S\rightarrow C\] 
witnessing $E_{H_S}^{B_S}\leq_B E|_C$. Use $\theta$ to define 
\[\varphi: B_S\rightarrow 2^\Z\] 
by postcomposing with projection to the first coordinate -- so that 
$\theta(f)$ will always have the form $(\varphi(f), g)$ for some $g$. 
In particular we have $\varphi$ is a homomorphism of equivalence relations, in 
the sense that if $f_1 E_{H_S} f_2$, then $\varphi(f_1)$ is equivalent to 
$\varphi(f_2)$ under the shift action of $\Z$ on $2^\Z$. 

As discussed in $\S$A4 of 
\cite{hjorthkechris}, the orbit equivalence relation induced by 
$H_S$ acting on $B_S$ is $E_0$-ergodic, while the the orbit equivalence relation 
induced by $\Z$ on $2^\Z$ is Borel reducible to $E_0$. Thus we may find a conull 
\[D_S\subset B_S\] and a single 
\[T_0\in 2^\Z\] such that for all $f\in D_S$ 
\[\varphi(f) E_\Z T_0.\] 
Thus after possibly adjusting the Borel reduction $\theta$ we obtain 
\[E_{H_S}^{D_S}\leq_B E_{H_{T_0}}^{C_{T_0}},\] 
with a contradiction to \ref{thomas}. 
\end{proof} 

I did not know of any earlier example of an ergodic countable equivalence relation with this property, 
however this equivalence relation is itself not induced by the free action of a countable group.  It  seems 
then that there is an obvious further question: 

\begin{question} Does there exist a free, Borel,  measure preserving, ergodic action of a countable group 
$H$ on a standard Borel probability space such that for any conull set $B$ there exists a conull $C\subset B$ with $E_H|_B$ not Borel reducible to $E_H|_C$? 
 \end{question} 





\begin{thebibliography}{99} 

\bibitem{adams} S. Adams, 
{\it An equivalence relation that is not freely generated.} 
{\bf Proc. Amer. Math. Soc.} vol. 102 (1988),  565--566

\bibitem{feldmanmoore} J Feldman, C.C. Moore, 
{\it Ergodic equivalence relations, cohomology, and von Neumann algebras. I, II} 
{\bf Trans. Amer. Math. Soc.} 234 (1977),  289--324, 325--359. 

\bibitem{furman} A. Furman,  
{\it Orbit equivalence rigidity,}  {\bf Ann. of Math.} vol. 150 (1999). 

\bibitem{hjorthkechris} G. Hjorth, A.S. Kechris, {\bf Rigidity Theorems for 
Actions of Product Groups and Countable Borel Equivalence Relations,} 
Memoirs of the AMS, vol. 177, 2005. 

\bibitem{jakelo} S. Jackson, A. S. Kechris, A. Louveau, 
{\it Countable Borel equivalence relations, } {\bf J. Math. Log.}, vol. 2 (2002), no. 1, 1--80.

\bibitem{popa} S. Popa, 
{\it Cocycle and Orbit Equivalence Superrigidity for Bernoulli Actions of Kazhdan Groups}, 
preprint. 

\bibitem{thomas} S. Thomas, 
{\it Popa superrigidity and countable Borel equivalence relations}, preprint. 

\bibitem{zimmer} R.J. Zimmer, 
{\bf Ergodic Theory and Semisimple Groups,} 
Birkh
�
�user, Basel, 1984.

\end{thebibliography}

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\leftskip 0.4in 

\noindent {\texttt greg.hjorth@gmail.com} 

\medskip

\noindent http://www.math.ucla.edu/\~{}greg/

\medskip

\noindent Department of Mathematics and Statistics

\noindent University of Melbourne

\noindent Parkville 

\noindent 3010

\noindent Victoria 

\noindent Australia 






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