% LaTeX Article Template - using defaults \documentclass{article}[12pt] \usepackage{amssymb} \usepackage{latexsym} \pagestyle{myheadings} \markright{G. Hjorth} \usepackage{amssymb} \usepackage{latexsym} % Set left margin - The default is 1 inch, so the following % command sets a 1.25-inch left margin. \setlength{\oddsidemargin}{-0.15in} % Set width of the text - What is left will be the right margin. % In this case, right margin is 8.5in - 1.25in - 6in = 1.25in. \setlength{\textwidth}{6.5in} % Set top margin - The default is 1 inch, so the following % command sets a 0.75-inch top margin. \setlength{\topmargin}{-0.1in} % Set height of the header \setlength{\headheight}{0.3in} % Set height of the text \setlength{\textheight}{8.1in} % Set vertical distance between the text and the % bottom of footer \setlength{\footskip}{0.8in} % Set the beginning of a LaTeX document \begin{document} %\large \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{claim}[theorem]{Claim} \newtheorem{subclaim}[theorem]{Subclaim} \newenvironment{proof}[1][Proof]{\begin{trivlist} \item[\hskip \labelsep {\bfseries #1}]}{\hfill$\Box$\end{trivlist}} \newenvironment{definition}[1][Definition]{\begin{trivlist} \item[\hskip \labelsep {\bfseries #1}]}{\end{trivlist}} \newenvironment{example}[1][Example]{\begin{trivlist} \item[\hskip \labelsep {\bfseries #1}]}{\end{trivlist}} \newenvironment{remark}[1][Remark]{\begin{trivlist} \item[\hskip \labelsep {\bfseries #1}]}{\end{trivlist}} \newenvironment{notation}[1][Notation]{\begin{trivlist} \item[\hskip \labelsep {\bfseries #1}]}{\end{trivlist}} \newenvironment{question}[1][Question]{\begin{trivlist} \item[\hskip \labelsep {\bfseries #1}]}{\end{trivlist}} \def\Ubf#1{{\baselineskip=0pt\vtop{\hbox{$#1$}\hbox{$\sim$}}}{}} \def\ubf#1{{\baselineskip=0pt\vtop{\hbox{$#1$}\hbox{$\scriptscriptstyle\sim$}}}{}} \def\R{{\mathbb R}} \def\C{{\mathbb C}} \def\V{{\mathbb V}} \def\N{{\mathbb N}} \def\Z{{\mathbb Z}} \def\K{{\mathbb K}} \def\B{{\mathbb B}} \def\o{{\omega}} \def\oo{{\omega^{\omega}}} \def\lo{{\omega^{<\omega}}} \def\d{{\dot{\bigcup}}} \def\h{{\cal H}} \def\no{\noindent} \def\a{{\mathcal A}} \def\b{{\mathcal B}} \def\l{{\mathcal L}} \def\n{{\mathcal N}} \def\m{{\mathcal M}} \def\p{{\mathcal P}} \def\h{{\mathcal H}} \def\co{{\mathcal O}} \def\t{{\mathcal T}} \def\c{{\mathcal C}} \def\Lck{{L_{\omega_1^{\rm ck}}}} \title{\huge A dichotomy theorem for turbulence} \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} In this note we show: \begin{theorem} \label{main} \label{maintheorem} Let $G$ be a Polish group and $X$ a Polish $G$-space with the induced orbit equivalence relation $E_G$ Borel as a subset of $X\times X$. Then exactly one of the following: \leftskip 0.4in \no (I) There is a countable language $\l$ and a Borel function \[\theta: X\rightarrow {\rm Mod}(\l)\] such that for all $x_1, x_2\in X$ \[x_1 E_G x_2 \Leftrightarrow \theta (x_1) \cong \theta(x_2);\] or \no (II) there is a turbulent Polish $G$-space $Y$ and a continuous $G$-embedding \[\tau: Y\hookrightarrow X.\] \end{theorem} There are various bows and ribbons which can be woven into these statements. We can strengthen (I) by asking that $\theta$ also admit a Borel orbit inverse, that is to say some Borel function \[\rho: B \rightarrow X,\] for some Borel set $B\subset {\rm Mod}(\l)$, such that for all $x\in X$ \[xE_G \rho(\theta(x));\] and then after having passed to this strengthened version of (I) we still obtain the exact same dichotomy theorem, and hence the conclusion that the two competing versions of (I) are equivalent. Similarly (II) can be relaxed to just asking that $\tau$ be a Borel $G$-embedding, or even simply a Borel reduction of the relevant orbit equivalence relations. It is in fact a consequence of \ref{maintheorem} that all the plausible weakenings and strengthenings of (I) and (II) are respectively equivalent to one another. I will not closely examine these possible variations here. The equivalences alluded to above follow from our main theorem and the results of \cite{hjorth}. That monograph had previously shown that (I) and (II) are incompatible, and proved a barbaric forerunner of \ref{maintheorem}, and gone on to conjecture the dichotomy result above. \cite{hjorth} is the proper place to find further discussion of the notation used in the proofs below. Mod$(\l)$ is the space of $\l$-structure on $\N$ equipped with the topology generated by quantifier free formulas. $E_G$ refers to the orbit equivalence relation arising from the indicated action of $G$ on the indicated space. \section{The Gandy-Harrington topology and a derived Polish $G$-space} \label{topology} We recall the notions of {\it recursive} Polish space and {\it recursive} continuous function from \cite{moschovakis}. We assume that all our spaces are recursive Polish spaces, our group is a recursive Polish space, and multiplication viewed as a function \[G\times G\rightarrow G\] is a recursive continuous function, as is the evaluation of the action of $G$ on $X$ viewed as a function \[G\times X\rightarrow X.\] Indeed in the case of $X$ we want to do more than simply assume that $X$ has a recursive basis; we fix from now on a sequence $(U_\ell)_{\ell\in\omega}$ of open subsets of $X$ which forms a basis and such that the assignment \[\ell\mapsto U_\ell\] is suitably effective; for instance, and this is only one method of many to make this precise, we could assume that \[\bigcup_{\ell\in\omega}\{\ell\}\times X\] is a recursive open subset of $\omega\times X$. We also fix a complete compatible metric $d_G(\cdot, \cdot)$ on $G$ and assume it provides a recursive continuous function from $G\times G$ to $\R$. We assume that there is a countable dense subgroup $G_0=\{g_n: n\in\N\}$ of $G$ such that the enumeration $n\mapsto g_n$ is recursive continuous as a function from $\N$ to $G$ and such that $g\cdot U_\ell\in\{U_k: k\in \N\}$ all $g\in G_0$ and $\ell\in\N$. We then fix a basis for $G$ closed under left and right translation by $G_0$ consisting purely of recursive open sets. We assume it includes a neighborhood basis at the identity of symmetric sets. We further assume that $E_G$ is $\Delta^1_1$ as a subset of $X\times X$. Of course in general these assumptions might not be true, and in the arguments below we would need to laboriously replace each instance of ``recursive" with ``recursive in $z$" for some parameter $z$. \def\Xl{X_{\rm low}} \def\X*{X_{*{\rm low}}} \begin{notation} Let $\Xl =\{x\in X: \forall A\in \Sigma^1_1 (x\in A\vee\exists B\in \Sigma^1_1(x\in B\wedge B\cap A =0))\}$. Let $\tau_\infty$ be the topology generated by taking as a basis all $\Sigma^1_1$ sets. \end{notation} In the case that $X=2^\omega$ one actually has $\Xl=\{x\in X: \omega_1^{\rm ck}=\omega_1^{{\rm ck}(x)}\}.$ Note that $\Xl $ is a Borel subset of $X$; on the other hand, while it is lightface $\Sigma^1_1$ it is {\it not} lightface $\Delta^1_1$. Recall that Gandy's low basis theorem states that any non-empty $\Sigma^1_1$ contains an element of $\Xl$. The formulation of the next lemma was influenced by some parallel remarks made by Ramez Sami during the course of a seminar. It has been previously proved in \cite{kechrislouveau} using the notion of strong Choquet space; the proof below is longer but more direct. \begin{lemma} \label{polish} $\tau_\infty$ is a Polish topology on $\Xl $. \end{lemma} \begin{proof} We show $(\Xl , \tau_\infty)$ to be homeomorphic to a $G_\delta$ subset of some well known Polish space. For ease of exposition we assume $X=2^\omega$; it should be clear how to extend the argument to general recursive Polish spaces. Let $S$ be the set of $\Sigma^1_1$ subsets of X. To each $x\in \Xl $ associate \[f_x: S\rightarrow 2\] by the formulation \[f_x(A)=1\] if and only if \[x\in A.\] This map $x\mapsto f_x$ is clearly one-to-one. The preimage of a subbasic open set in $2^S$ of the form \[\{x: f_x(A)=1\}\] is obviously open in $(\Xl , \tau_\infty)$; the preimage of a subbasic open set of the form \[\{x: f_x(A)=0\}\] equals \[\bigcup \{D: D\in \Delta^1_1, D\cap A=0\}\cap \Xl ,\] and is thus again open with respect to $\tau_\infty$. Therefore the map \[x\mapsto f_x\] is a homeomorphism onto its image, and we need only show that $\{f_x: x\in \Xl \}$ is a $G_\delta$ subset of $2^S$ in the product topology to complete the proof. For this purpose let us define at each $n\in\o$ the set $A_n^*$ to be the set of $x\in \Xl $ for which the $n^{\rm th}$ $x$-recursive\footnote{Here for the only time we use $X=2^\omega$; with suitable care the notion of recursive in $x$ can be translated across to arbitrary recursive spaces.} linear order of $\o$ is not a wellorder of rank less than $\omega_1^{\rm ck}$; and then at each $m$ we let $C_{n, m}^*$ be the set of $x\in \Xl $ for which $m$ is in the illfounded part of the $n^{\rm th}$ $x$-recursive linear order of $\o$. These sets are of course all $\Sigma^1_1$. I claim that \[f\in\{f_x: x\in \Xl \}\] if and only if the following four things happen: \begin{enumerate} \item for all $n\in\o$ and basic open $W$ with \[ f(W)=1,\] there is some $U\subset X$ whose diameter is less than $\frac{1}{n}$ with \[\overline{U}\subset W\] and \[f(U)=1;\] \item for all $k\in\o$, $A_0,A_1,...,A_k,C_0, C_1,...,C_k\in\Sigma^1_1$, with \[\forall i\leq k(f(A_i)=1, f(C_i)=0),\] there is some $x\in \Xl$ with \[\forall i\leq k(x\in A_i, x\notin C_i);\] (in particular this rules out grievously absurd choices for $f: S\rightarrow \{0, 1\}$, such as the two constant functions); \item for all $n$ with \[f(A_n^*)=1\] there exists $m$ with \[f(C^*_{n,m})=1;\] \item for each $\Sigma^1_1$ set $A$ we fix a Lusin scheme \[(C_s^A)_{s\in\omega^{<\omega}}\] of $\Pi_0^1$ sets such that \[A=\{x\in X| \exists h\in\omega^\omega\forall n\in \omega (x\in C^A_{h|_n})\};\] we then demand that \[f(A)=1\Rightarrow f(C_0^A)=1,\] and that for all $s\in\omega^\omega$ \[f(C_s^A)=1\Rightarrow \exists k\in\omega(f(C^A_{s^\smallfrown k})=1).\] \end{enumerate} The first, third, and fourth of these conditions correspond to $G_\delta$ sets, while the second is closed. The purpose of the first condition is to ensure that we can indeed find a point $x_f$ in some intersection of basic open $V$ with $f(V)=1$; and in this light the second condition ensures that this $x_f$ meets all the open and closed sets promised by $f$, and from this we can obtain by the fourth condition and the usual kinds of arguments that $x_f$ actually meets every $\Sigma^1_1$ set with $f(A)=1$. The third condition is the key one. It guarantees that no recursive in $x_f$ wellorder has rank exactly $\omega_1^{\rm ck}$. \end{proof} \begin{notation} We let $\X*$ be the set of all $x\in X$ such that there exists some $g\in G$ with \[g\cdot x\in \Xl.\] We endow this with the topology $\tau^*_\infty$ generated by basic open sets of the form \[(A\cap \Xl)^{\Delta V}\] where $A\subset X$ is $\Sigma^1_1$ and $V\subset G$ is open. \end{notation} In fact if $x\in \X*$ one does not just obtain one $g\in G$ with $g\cdot x\in \Xl$ but a comeager set of $g$ which so move $x$ to a real with $\omega_1^{{\rm ck}(g\cdot x)}=\omega_1^{\rm ck}$: We first use that \[\omega_1^{{\rm ck}(x)}=\omega_1^{\rm ck}\] to conclude that there will be a comeager collection of $g$ with $\omega_1^{{\rm ck}(g,x)}=\omega_1^{\rm ck}$, and any such $g$ will do. Thus $\X*$ equals \[(\Xl)^{*G}=(\Xl)^{\Delta G},\] and is consequently Borel. Note moreover that in this space, for any $A\subset \Xl$ which is $\Sigma^1_1$ and for any $V\subset G$ open we in fact have a sequence of $\Delta^1_1$ sets \[(D_\alpha)_{\alpha\in \omega_1^{\rm ck}}\] such that \[A^{*V}=\X*\setminus (\bigcup_{\alpha\in \omega_1^{\rm ck}} (D_\alpha)^{\Delta V}),\] and thus $A^{*V}$ is closed. The next proof crumbles into a sequence of calculations. The first few of these resemble similar points made in the course of the proof of 5.1.5 from \cite{beckerkechris}. \begin{lemma} $(\X*, \tau^*_\infty)$ is a Polish $G$-space. \end{lemma} \begin{proof} The action of $G$ on this space is obviously continuous. \smallskip Claim: $(\X*, \tau^*_\infty)$ is Hausdorff. Proof of claim: $\tau^*_\infty$ includes the subspace topology on $\X*$ induced by the original topology on $X$. Since this original topology is Hausdorff, so is $\tau^*_\infty$. \hfill ($\Box$Claim) \smallskip Claim: $(\X*, \tau^*_\infty)$ is regular. Proof of claim: Let $A^{\Delta V}$ be an open subset of $(\X*, \tau^*_\infty)$ and $x\in A^{\Delta V}$. Then we may find some $W_0\subset V$ open and non-empty with $x\in A^{*W_0}$. Appealing to the continuity of the group operations we may find non-empty open sets \[W_2\subset W_1\subset W_0\] and an open neighborhood $W_3$ of the identity with \[W_1W_3^{-1}\subset W_0\] and \[W_2W_3^{-1}\subset W_1.\] $x$ is then a member of the open set $(A^{*W_1})^{\Delta W_3}$ since for {\it any} $g\in W_3$ we have \[W_1g^{-1}\subset W_0\] and \[\forall^*h\in W_1g^{-1}(h\cdot (g\cdot x)\in A).\] Thus the claim will be proved if we show $(A^{*W_1})^{\Delta W_3}$ is included in the $\tau_\infty^*$-closed set $A^{*W_2}$. But here we have that for any $h\in W_3$ and $z$ with $h\cdot z\in A^{*W_1}$ \[\forall^*g\in W_1(gh\cdot z\in A)\] \[\therefore \forall^*g\in W_1h(g\cdot z\in A).\] But \[W_2W_3^{-1}\subset W_1\] \[\therefore W_2\subset W_1h\] \[\therefore z\in A^{*W_2}.\] \hfill ($\Box$Claim) \smallskip Since $(\X*, \tau^*_\infty)$ is clearly separable, we thus have that it is metrizable by Urysohn. To see that the space is Polish we first define a continuous map \[\pi: G\times \Xl \rightarrow \X*\] by \[(g, x)\mapsto g\cdot x.\] \smallskip Claim: $\pi$ is continuous. Proof of claim: We have $g\cdot x\in A^{\Delta V}$ if and only if $x\in A^{\Delta Vg}$. \hfill ($\Box$Claim) \smallskip Claim: $\pi$ is an open map. Proof of claim: It suffices to show that whenever $A\subset \Xl$ is $\Sigma^1_1$ and $V\subset G$ is open, \[V\cdot A=\bigcup_{W_0, W_1 {\rm \: open,\:} W_0W_1\subset V^{-1}} (\{y: \exists h\in W_0(h\cdot y\in A)\})^{\Delta W_1}.\] There are two parts to demonstrating this. We show these two inclusions separately. \smallskip $\subset$: Let \[g\cdot x=z,\] $g\in V$, $x\in A$. Then considering the continuity of the action we may choose open $V_0$ and $V_1$ containing $g^{-1}$ and $1$ which satisfy: \[V_0(V_1)^{-1}V_1\subset V^{-1}.\] We can then let $W_0=V_0V_1^{-1}$ and $W_1=V_1$. With this arranged we have that {\it for all} $\epsilon \in W_1$ \[(g^{-1}\epsilon^{-1})\epsilon\cdot z=(g^{-1}\epsilon^{-1})\epsilon g\cdot x =x\in A.\] Noting that any such $g^{-1}\epsilon^{-1}$ will be in $W_0$ we have in particular that \[z\in (\{y: \exists h\in W_0(h\cdot y\in A)\})^{* W_1} \subset (\{y: \exists h\in W_0(h\cdot y\in A)\})^{\Delta W_1}.\] \smallskip $\supset$: Consider some \[z\in (\{y: \exists h\in W_0(h\cdot y\in A)\})^{\Delta W_1},\] where $W_0W_1\subset V^{-1}$. Then in particular, we can find $g_1\in W_1$ with \[g_1\cdot z\in \{y: \exists h\in W_0(h\cdot y\in A)\}.\] And then we may find some $g_2\in W_0$ with \[g_2g_1\cdot z\in A.\] But then \[g_2g_1\in V^{-1}\] by assumption on $W_0$ and $W_1$, and hence \[x=g_2g_1\cdot z\] and \[g=(g_2g_1)^{-1}\] witness that $z\in V\cdot A$. \hfill (Claim$\Box$) \smallskip In summary, the above claims have shown \[\pi: G\times \Xl\rightarrow \X*\] to be an open continuous map from a Polish space onto a metric space. Thus by Hausdorff's theorem, $(\X*, \tau^*_\infty)$ is Polish. \end{proof} \section{The split in cases} \label{setup} As above, assume that $G$ is a recursive Polish group and $X$ is a recursive Polish $G$-space. \begin{definition} A sequence \[\vec\b =(\b_\alpha)_{\alpha\in \delta}\] is {\it suitable} if \begin{enumerate} \item each $\b_\alpha$ is a subset of the Borel subsets of $X$; \item $\delta<\omega_1^{\rm ck}$; \item the function \[\alpha\mapsto \b_\alpha\] is $\Delta^1_1$ in the codes for Borel sets; \item \label{a} in particular each $\b_\alpha$ is countable; %\item \label{b} each %\[\tau_{\b_\alpha}=_{\rm df} \{\bigcup C| C\subset \b_\alpha\}\] %generates a finer Polish topology on $X$; \item \label{c} each $\b_\alpha$ includes the basic open sets in the original topology on $X$; %\item each \label{d} $\b_\alpha$ is closed under finite Boolean combinations; \item each \label{e} $\b_\alpha$ is closed under $G_0$-translation. %\item \label{f} if $B\in \b_\alpha$ and $V\subset G$ is basic open, then %\[B^{\Delta V}, B^{* V}\in \b_\alpha.\] \end{enumerate} \end{definition} This definition has glossed over the details of being {\it $\Delta^1_1$ in the codes for Borel sets}; a protracted discussion is unlikely to be helpful. The reader will be able to provide their own coding for Borel sets, and any reasonable understanding of ``in the codes" should suffice; for instance we can ask that there be some recursive wellorder $\prec$ of length $\delta$ and $\Delta^1_1$ function \[F: \omega\times \omega \rightarrow \oo\] such that for any $n\in\omega$ and $\alpha=|\!| n|\!|_\prec$ we have that \[\{F(n, m): m\in\omega\}\] gives a set of codes for the Borel sets appearing in the $\b_\alpha$. %Various closure conditions are placed on the sets $\b_\alpha$. They are %motivated by \cite{beckerkechris}. Among other results %it had been there shown that given a collection of Borel sets $\b$ satisfying %\ref{a}-\ref{f} above, %\[\{B^{\Delta V}: B\in\b, V\subset G {\rm \: basic \: open}\}\] %generates a Polish topology on $X$ under which the $G$-action is still continuous. Given such an admissible sequence, we define a sequence of objects \[\hat{\varphi}_\alpha^{\vec \b}(x, B^{\Delta V_0}, V_1)\] around each point $x\in X$ and at each $\alpha<\delta$. \begin{definition} Given an admissible sequence $\vec\b =(\b_\beta)_{\beta\in \delta}$ as above, $x\in X$, $B\in \b_\alpha$, $V_0, V_1\subset G$ basic open, $V_1$ containing $1_G$, $V_1^{-1}=V_1$, we define $\hat{\varphi}_\beta^{\vec \b}(x, B^{\Delta V_0}, V_1)$ by induction on $\beta$:- \smallskip \no{BASE CASE:} for $x\in B^{\Delta V_0}$ \[\hat{\varphi}_0^{\vec \b}(x, B^{\Delta V_0}, V_1)\] equals the set of $\ell\in \omega$ such that there exists $k\in\o$, $x_0, x_1,..., x_k\in X$ such that \leftskip 0.4in \no $x_0=x$; \no $x'=x_k\in U_\ell$; \no each $x_{i+1}\in V_1\cdot x_i\cap B^{\Delta V_0}$. \leftskip 0in \smallskip \no{INDUCTIVE STEP:} If $\beta$ is an ordinal lying strictly between 0 and $\delta$, then \[\hat{\varphi}_\beta^{\vec \b}(x, B^{\Delta V_0}, V_1)\] equals the set of all \[\langle \hat{\varphi}_\alpha^{\vec \b}(x', C^{\Delta W_0}, W_1), C, W_0, W_1\rangle ,\] where \leftskip 0.4in \no $\alpha< \beta$; \no $C\in \b_\alpha$; \no $W_0, W_1\subset G$ basic open; \no $W_1$ contains $1_G$, $W_1^{-1}=W_1$, \no $x_0=x$; \no $x'=x_k\in C^{\Delta W_0}$; \no each $x_{i+1}\in V_1\cdot x_i\cap B^{\Delta V_0}$. \leftskip 0in \end{definition} \begin{notation} For $\epsilon >0$ we let $W_\epsilon$ be the set \[\{g\in G| d_G(g, 1_G)<\epsilon\}.\] \end{notation} \begin{definition} For $A\subset X$, $A\in \Sigma^1_1$, $V_0, V_1\subset G$ basic open, $V_1^{-1}=V_1, 1_G\in V_1$, and $x\in A^{\Delta V_0}$, we let \[{\cal O}(x, A^{\Delta V_0}, V_1)\] be the set of $x'$ for which there exists $k, x_0, x_1, ..., x_k$ such that \leftskip 0.4in \no $x_0=x$; \no $x'=x_k$; \no each $x_{i+1}\in V_1\cdot x_i\cap A^{\Delta V_0}$. \leftskip 0in \end{definition} \begin{definition} For $\vec\b=(\b_\alpha)_{\alpha<\delta}$ suitable, $\epsilon>0$, $A\in \Sigma^1_1$, $V_0, V_1\subset G$ basic open, $V_1^{-1}=V_1, 1_G\in V_1$, we say that \smallskip \centerline{\it $(A, V_0, V_1)$ is $\epsilon$-stable for $\vec \b$} \smallskip \no if for all $x, x', x'', \alpha<\delta, V_2,V_3, B$ with \leftskip 0.4in \begin{enumerate} \item $x\in A^{\Delta V_0}$, \item $x', x''\in {\cal O}(x, A^{\Delta V_0}, V_1)$, \item $\alpha<\delta$, \item $B\in\b_\alpha$, \item $V_2, V_3\subset G$ basic open, \item $V_3^{-1}=V_3$, \item $1_G\in V_3$, \end{enumerate} \leftskip 0in \no $x'\in B^{\Delta V_2}$ implies \[\exists^* g\in W_\epsilon (g\cdot x''\in B^{\Delta V_2}\wedge \hat{\varphi}^{\vec \b}_\alpha(x', B^{\Delta V_2}, V_3)= \hat{\varphi}^{\vec \b}_\alpha(g\cdot x'', B^{\Delta V_2}, V_3)).\] \end{definition} There are a few remarks about this definition. Firstly that if $(A, V_0, V_1)$ is $\frac{1}{n}$-stable then it is $\epsilon$-stable for any larger $\epsilon$. %; and instead of %considering arbitrary $x', x''\in {\cal O}(x, A^{\Delta V_0}, V_1)$ we can restrict %our attention only to those $x', x''\in G_0\cdot x$ and obtain an equivalent %definition; these points are important only in %that they will later reduce quantifier which seems to range over the reals or some other %large %Polish space by ones ranging over a countable set. The second observation is that we may replace \[\exists^* g\in W_\epsilon (g\cdot x''\in B^{\Delta V_2}\wedge \hat{\varphi}^{\vec \b}_\alpha(x', B^{\Delta V_2}, V_3)= \hat{\varphi}^{\vec \b}_\alpha(g\cdot x'', B^{\Delta V_2}, V_3))\] by \[\exists g\in W_\epsilon (g\cdot x''\in B^{\Delta V_2}\wedge \hat{\varphi}^{\vec \b}_\alpha(x', B^{\Delta V_2}, V_3)= \hat{\varphi}^{\vec \b}_\alpha(g\cdot x'', B^{\Delta V_2}, V_3));\] if one $g\in W_\epsilon$ brings about the desired result then it follows from the structure of the definitions that an open set of them will, and hence certainly a non-meager set will as well. \begin{lemma} \label{reflect} If $\vec\b$ is a suitable sequence, $n\in\o$, $V_0, V_1\subset G$ basic open, $V_1^{-1}=V_1$, $1_G\in V_1$, $A\in\Sigma^1_1$, and if \[(A, V_0, V_1)\] is $\frac{1}{n}$-stable for $\vec \b$, then there is a $\Delta^1_1$ set $D\supset A$ with \[(D, V_0, V_1)\] $\frac{1}{n}$-stable for $\vec \b$. \end{lemma} \begin{proof} For any $\Delta^1_1(x)$ set $C$ we have that the set of $x''$ with \[\exists g^*\in W_{\frac{1}{n}}(g\cdot x'' \in C)\] is uniformly $\Delta^1_1(x)$; in particular, the set of $x''$ with \[\exists^* g\in W_\epsilon (g\cdot x''\in B^{\Delta V_2}\wedge \hat{\varphi}^{\vec \b}_\alpha(x', B^{\Delta V_2}, V_3)= \hat{\varphi}^{\vec \b}_\alpha(g\cdot x'', B^{\Delta V_2}, V_3))\] is uniformly, in the indices for $V_0$, $V_1$, and $B$, $\Delta^1_1(x)$. On the other hand, the set \[{\cal O}(x, A^{\Delta V_0}, V_1)\] is uniformly $\Sigma^1_1(x)$. Thus the collection of $\Sigma^1_1$ sets $A$ with $(A, V_0, V_1)$ $\frac{1}{n}$-stable is uniformly $\Pi^1_1$ in the indices. And thus the lemma follows from $\Pi^1_1$ reflection.\footnote{See \cite{hamash} for the statement and proof of this theorem.} \end{proof} \begin{definition} For $\vec \b=(\b_\alpha)_{\alpha\in\delta}$ suitable and $n\in\o$, let $A_{\vec \b, n}$ be the set of all $x\in X$ for which there is a non-meager collection of $g\in G$ with \[\forall D\in \Delta^1_1, V_0, V_1\subset G{\rm \: basic\: open}\] \[(g\cdot x\in D^{\Delta V_0}\Rightarrow ((D, V_0, V_1) {\rm \: not\:} \frac{1}{n}-{\rm stable\: for \:}\vec \b))).\] \end{definition} For a given $n$ and $\vec \b$, the set $A_{\vec \b, n}$ is $\Sigma^1_1$. \begin{notation} From now on let us endow $\X* \cap A_{\vec \b, n}$ with the topology $\tau^*_{n, \infty}$ generated by all sets of the form \[A^{\Delta V}\] where $A\in\Sigma^1_1$, $A\subset \X* \cap A_{\vec \b, n}$, and $V\subset G$ basic open. Let \[(\hat{U}_\ell)_{\ell\in\o}\] enumerate a basis for $\tau^*_{n, \infty}$. \end{notation} \begin{lemma} $(\X* \cap A_{\vec b, n},\tau^*_{n, \infty})$ is a Polish $G$-space. \end{lemma} \begin{proof} Since it is an open subset of $(\X*, \tau^*_\infty)$. \end{proof} At this point the proof splits into two cases. Case(1) below leads to a turbulent Polish $G$-space $Y$ as in (II) above. Case(2), on the other hand, will finish with a classification of the $G$-orbits by countable models considered up to isomorphism. \section{Case(1)} For some suitable $\vec \b=(\b_\alpha)_{\alpha\in\delta}$ and $n\in\o$, $A_{\vec \b, n}$ is non-empty. \bigskip Let us fix such a $\vec \b$ and $n$. We will press forward, and eventually show that option (II) in \ref{main} holds. Following $\S$6.2 of \cite{hjorth} we may define ``local invariants" for the Polish $G$-space $(\X* \cap A_{\vec\b, n}, \tau^*_{n, \infty})$. \begin{notation} For $x\in \X* \cap A_{\vec\b, n}$ and $U$ $\tau^*_{n,\infty}$-open, $V\subset G$ a symmetric open neighborhood of the identity, we let \[{\cal O}(x, U, V)\] be the set of all $x'$ such that there exists $k, x_0, x_1,..., x_k$ with \[x_0=x,\] \[x_k=x',\] and at each $i0$ the local invariant $\varphi_\alpha(x, U, V)$ encodes the set of all possible $\varphi_\beta(x', U', V')$ we can obtain by reducing $\alpha$ to $\beta$, $U$ to $U'$. $V$ to $V'$, and passing to some other $x'\in {\cal O}(x, U, V)$. Let us fix some particular $x\in \X* \cap A_{\vec \b, n}$; it in fact does not matter which. Still following the relevant $\S$6.2 of \cite{hjorth} we come, for any given $x$ in the space, to some $\gamma(x)<\omega_1$ such that for all $U$ that is $\tau^*_{n,\infty}$-open, for all $V\subset G$ that is a symmetric basic open neighborhood of the identity, and for all $x'. x''\in [x]_G$ \[\varphi_{\omega_1}(x', U, V)=\varphi_{\omega_1}(x'', U, V)\] \[{\rm iff}\] \[\varphi_{\gamma(x)}(x', U, V)=\varphi_{\gamma(x)}(x'', U, V).\] We then obtain a Polish topology on the set \[X^x=_{\rm df}\{y\in \X* \cap A_{\vec \b, n}| \varphi_{\gamma(x)+2}(x, \X* \cap A_{\vec\b, n}, G)= \varphi_{\gamma(x)+2}(y, \X* \cap A_{\vec\b, n}, G)\}\] by taking as a basis all sets of the form \[\{y|\varphi_{\gamma(x)}(y, U, V)=\varphi_{\gamma(x)}(x', U, V)\},\] as $U$ ranges over $\tau^*_{n,\infty}$-open sets, $V$ ranges over symmetric basic open neighborhoods of the identity, and $x'$ ranges over $[x]_G$. From this point on let us understand that $X^x$ always refers to $X^x$ in the indicated Polish topology. \begin{lemma} \label{density} For any $y\in X^x$, \[\overline{[x]_G}=\overline{[y]_G}.\] \end{lemma} Here we understand that the closure operation refers to the indicated topology on the space $X^x$. \begin{proof} An immediate consequence of the definitions. \end{proof} \begin{corollary} \label{meet} For any $y\in X^x$, $A\subset X$ a $\Sigma^1_1$ set, \[[x]_G\cap A\neq 0 {\rm \: iff \:} [y]_G\cap A\neq 0.\] \end{corollary} Of course in this corollary it is important that we restrict ourselves to sets $A$ which are lightface $\Sigma^1_1$ with respect to the original space $X$. \begin{lemma} \label{localdensity} For any $y\in X^x$, \[\gamma(x)=\gamma(y).\] Moreover for any $U$ that is $\tau^*_{n,\infty}$ basic open and $V\subset G$ a symmetric basic open neighborhood of the identity, \[{\cal O}(y, U, V) \] is dense (with respect to the topology on $X^x$) in some basic open set of the form \[\{y|\varphi_{\gamma(x)}(y, U, V)=\varphi_{\gamma(x)}(x', U, V)\}.\] \end{lemma} \begin{proof} These facts can be verified from the definitions in a reasonably routine manner; or see $\S$6.2 \cite{hjorth} for the details. \end{proof} From now until the end of the section, let us just fix some $y\in X^x$. We will try to show that $[y]_G$ is meager in the space $X^x$. If we succeed in this task, for some arbitrarily chosen $y\in X^x$, then we will have obtained that every orbit in the space must be meager. And then by \ref{density} and \ref{localdensity} we have that the definition of turbulence is satisfied. Now turning our attention back towards the fixed, but arbitrary, $y\in X^x$ we have from \ref{meet} that there will be some \[\hat{y}\in [y]_G\] such that for all\footnote{Here, and everywhere else the issue might emerge, $\Sigma^1_1$, $\Delta^1_1$, and $\Pi^1_1$ refer to $\Sigma^1_1(X)$, $\Delta^1_1(X)$, and $\Pi^1_1(X)$ respectively.} $D\in \Delta^1_1$, $V_0, V_1\subset G$ basic open, if \[\hat{y}\in D^{\Delta V_0},\] then $(D, V_0, V_1)$ is not $\frac{1}{n}$-stable for $\vec \b$. Hence by \ref{reflect}, for all $A\in \Sigma^1_1$, $V_0, V_1\subset G$ basic open, if \[\hat{y}\in A^{\Delta V_0},\] then $(A, V_0, V_1)$ is not $\frac{1}{n}$-stable for $\vec \b$. Fix this $\hat{y}$. We need to ``blur" this element of $[y]_G$ making it slightly more generic, without altering its important properties; for this purpose let us fix some $m>n$ such that \[(W_{\frac{1}{m}})^\pm (W_{\frac{1}{m}})^\pm (W_{\frac{1}{m}})^\pm \subset W_{\frac{1}{n}}.\] \begin{claim} There exists $y'\in[y]_G$ such that: \leftskip 0.4in \no (i) for all $A\in \Sigma^1_1$, if $y'\in A$, then \[y'\in A^{*W}\] for some basic open neighborhood $W$ of $1_G$; \no (ii) $\omega_1^{\rm ck}=\omega_1^{{\rm ck}(y')}$; and hence, for all $A\in \Sigma^1_1$, either \[y'\in A\] or there is a $\Delta^1_1$ set $D$ with \[y'\in D\] and \[D\cap A=0;\] \no (iii) for all $A\in \Sigma^1_1$, $V_0, V_1\subset G$ basic open, if \[{y'}\in A^{\Delta V_0},\] then $(A, V_0, V_1)$ is not $\frac{1}{m}$-stable for $\vec \b$. \end{claim} \begin{proof} Note first of all that since $y\in X^x$ we may certainly find $g\in W_{\frac{1}{m}}$ such that \[y'=g\cdot \hat{y}\] satisfies (i) and (ii); any sufficiently generic $g\in W_{\frac{1}{m}}$ will do. For (iii) we need a further subclaim. \smallskip Subclaim: If $(A, V_0, V_1)$ is $\frac{1}{m}$-stable and $g\in W_{\frac{1}{m}}$ with \[y'=g\cdot \hat{y}\in A^{\Delta V_0},\] then there is some $\frac{1}{n}$-stable $(C, V_0', V_1')$ with \[\hat{y}\in C^{\Delta V_0'}.\] Proof of subclaim: Assume not, and then we may assume further that $g\in G_0$, since slightly nudging $g$ into $G_0$ should not affect ${\cal O}(y', A^{\Delta V_0}, V_1)$. Now consider some arbitrary $\hat{z}\in X$ with \[z'=g\cdot \hat{z}\in A^{\Delta V_0},\] and \[\hat{z}^*\in {\cal O}(\hat{z}, A^{\Delta V_0g}, g^{-1}V_1g).\] Let \[z^*=g\cdot \hat{z}^*\in {\cal O}(z', A^{\Delta V_0}, V_1).\] By assumption on $(A, V_0, V_1)$, for any $\alpha<\delta$, $B\in \Sigma^1_1$, and $V_2, V_3$ basic open in $G$, with $B^{\Delta V_2}$ containing $z$, (and hence $B^{\Delta V_2 g^{-1}}$ containing $z^*=g\cdot \hat{z}^*$) we may find \[h\in W_\frac{1}{m}\] with \[\hat{\varphi}^{\vec {\cal B}}_\alpha(hg\cdot \hat{z}, B^{\Delta V_2 g^{-1}}, gV_3 g^{-1})\] (which by definition of $z'$ equals \[\hat{\varphi}^{\vec {\cal B}}_\alpha(h\cdot z', B^{\Delta V_2 g^{-1}}, gV_3 g^{-1}))\] equal to \[\hat{\varphi}^{\vec {\cal B}}_\alpha(z^*, B^{\Delta V_2 g^{-1}}, gV_3 g^{-1})\] \[\therefore \hat{\varphi}^{\vec {\cal B}}_\alpha(g^{-1}hg\cdot \hat{z}, B^{\Delta V_2}, gV_3 g^{-1}) =\hat{\varphi}^{\vec {\cal B}}_\alpha(g^{-1}z^*, B^{\Delta V_2}, gV_3 g^{-1}),\] which in turn by the definition of $z^*$ equals \[\hat{\varphi}^{\vec {\cal B}}_\alpha( \hat{z}^*, B^{\Delta V_2}, gV_3 g^{-1}).\] Thus we have shown $(A, V_0g, g^{-1}V_1g)$ to be $\frac{1}{n}$-stable. \hfill (Claim$\Box$) But this subclaim proves the claim, since $\hat{y}$ was assumed not to be so captured by any $\frac{1}{n}$-stable set. \end{proof} Let us then fix $y'$ as above. \begin{claim} For all $A\in \Sigma^1_1$, $V_0, V_1\subset G$ basic open with $1_G\in V_1$, $V_1^{-1}=V_1$, if \[y'\in A^{\Delta V_0}\] then \[\exists y_1, y_2\in {\cal O}(y', A^{\Delta V_0}, V_1)\] \[\exists \alpha<\delta \exists B\in \b_\alpha\] \[\exists V_2, V_3\subset G {\rm \: basic \: open, } 1_G\in V_3, V_3^{-1}=V_3\] with \[y_1\in B^{\Delta V_2}\] but \[\forall g\in W_{\frac{1}{m}} (g\cdot y_2\in B^{\Delta V_2}\Rightarrow \hat{\varphi}^{\vec {\cal B}}_\alpha(y_1, B^{\Delta V_2}, V_3) \neq \hat{\varphi}^{\vec {\cal B}}_\alpha(g\cdot y_2, B^{\Delta V_2}, V_3)).\] \end{claim} \begin{proof} Otherwise applying (ii) from our assumptions on $y'$ we find $D\in \Delta^1_1$ containing $y'$ such that \[\forall z'\in A^{\Delta V_0}\cap D\] \[\forall z_1, z_2\in {\cal O}(z', A^{\Delta V_0}, V_1)\] \[\forall \alpha<\delta \forall B\in \b_\alpha\] \[\forall V_2, V_3\subset G {\rm \: basic \: open,\: }1_G\in V_3, V_3^{-1}=V_3\] if \[z_1\in B^{\Delta V_2}\] then \[\exists g\in W_{\frac{1}{n}} (g\cdot y_1\in B^{\Delta V_2}\wedge \hat{\varphi}^{\vec {\cal B}}_\alpha(z_1, B^{\Delta V_2}, V_3) = \hat{\varphi}^{\vec {\cal B}}_\alpha(g\cdot z_2, B^{\Delta V_2}, V_3)),\] and hence as prior to \ref{reflect} \[\exists^* g\in W_{\frac{1}{n}} (g\cdot y_1\in B^{\Delta V_2}\wedge \hat{\varphi}^{\vec {\cal B}}_\alpha(z_1, B^{\Delta V_2}, V_3) =\hat{\varphi}^{\vec {\cal B}}_\alpha(g\cdot z_2, B^{\Delta V_2}, V_3)).\] By (i) we obtain \[y'\in D^{*W}\] some basic open neighborhood $W$ of $1_G$. After possibly shrinking $W$ and $V_0$ we can assume \[W\subset V_1\] and that there is some $g_0\in G_0$ with \[g_0W=V_0.\] Here it will be easier if we make a harmless simplifying assumption. I claim that we may assume $V_0=W$ and that $g_0=1_G$. The point is that we can always replace $A$ by the set \[g_0^{-1}A,\] and observe that \[A^{\Delta g_0 W}=(g_0^{-1}A)^{\Delta W}.\] So after making this minor notational reorganization let us assume that \[1_G=g_0,\] \[W=V_0,\] \[1_G\in V_0,\] \[V_0\subset V_1,\] \[y'\in A^{\Delta V_0}\cap D^{*V_0}\subset (A\cap D)^{\Delta V_0}.\] Then for any \[z\in (A\cap D)^{\Delta V_0},\] \[z_1, z_2\in {\cal O}(z, (A\cap D)^{\Delta V_0}, V_1),\] and for any \[\alpha<\delta\] \[B\in \b_\alpha\] \[V_2, V_3\subset G {\rm \: basic \: open}\] with \[1_G\in V_3, V_3^{-1}=V_3\] \[z_1\in B^{\Delta V_2}\] we may find $h_0\in V_0\subset V_1$ with \[z'=_{\rm df}h_0\cdot z\in D\cap (A\cap D)^{\Delta V_0},\] \[\therefore {\cal O}(z', (A\cap D)^{\Delta V_0}, V_1) ={\cal O}(z, (A\cap D)^{\Delta V_0}, V_1),\] \[\therefore z_1, z_2\in {\cal O}(z', (A\cap D)^{\Delta V_0}, V_1) \subset {\cal O}(z', A^{\Delta V_0}, V_1),\] and hence by the assumptions on $D$ and the fact that $z'\in D$ we have \[\exists^*g\in W_{\frac{1}{m}} (g\cdot y_1\in B^{\Delta V_2}\wedge \hat{\varphi}^{\vec {\cal B}}_\alpha(z_1, B^{\Delta V_2}, V_3) = \hat{\varphi}^{\vec {\cal B}}_\alpha(g\cdot z_2, B^{\Delta V_2}, V_3)).\] And thus we have seen that $(A\cap D, V_0, V_1)$ is $\frac{1}{m}$-stable. Which contradicts assumption (iii) on $y'$. \end{proof} But the statement of this lemma is catastrophic with respect to $y$'s hope of having a $G_\delta$ orbit. It implies in particular that for all $A\in \Sigma^1_1$, $V_0, V_1\subset G$ basic open with $1_G\in V_1$, $V_1^{-1}=V_1$, if \[y'\in A^{\Delta V_0}\] then \[\exists y_1, y_2\in A^{\Delta V_0}\cap[y]_G\] with \[{\cal O}(y_1, A^{\Delta V_0}, V_1) ={\cal O}(y_2, A^{\Delta V_0}, V_1),\] \[\therefore \varphi_{\gamma(x)}(y_1, A^{\Delta V_0}, V_1)= \varphi_{\gamma(x)}(y_2, A^{\Delta V_0}, V_1) =\varphi_{\gamma(x)}(y', A^{\Delta V_0}, V_1)\] {\it but} \[\forall g\in G(d_G(g, 1)<\frac{1}{m}\Rightarrow g\cdot y_1\neq y_2).\] But now since the sets of the form \[\{z: \varphi_{\gamma(x)}(z, A^{\Delta V_0}, V_1) =\varphi_{\gamma(x)}(x', A^{\Delta V_0}, V_1)\}\] (as $x'$ ranges over $[x]_G$, $A$ ranges over $\Sigma^1_1$, $V_0, V_1$ range over basic open sets with $V_1$ a symmetric neighborhood of the identity) form a basis of the topology on $X_x$, we have that the induced map \[G\rightarrow [y]_G\] \[g\mapsto g\cdot y\] is not open. Thus by the Effros lemma, $[y]_G$ is not $G_\delta$. Thus we have shown that no orbit in $X_x$ is $G_\delta$. And hence all the orbits are meager. And hence as in $\S$6.2 of \cite{hjorth}, all the orbits in $X_x$ are dense, the local orbits of the form \[{\cal O}(y, A^{\Delta V_0}, V_1)\] are all dense inside the corresponding open set \[\{z: \varphi_{\gamma(x)}(z, A^{\Delta V_0}, V_1) =\varphi_{\gamma(x)}(x', A^{\Delta V_0}, V_1)\}\] for some $x'\in [x]_G$ with $\varphi_{\gamma(x)}(y, A^{\Delta V_0}, V_1) =\varphi_{\gamma(x)}(x', A^{\Delta V_0}, V_1)$, and by the calculation above every orbit is meager. Thus the action is turbulent, and we are done with this case. \section{Case (2)} Not case (1). %And thus in particular we may use transfinite induction and %the number uniformization %theorem\footnote{I.e. that if $A\subset \omega\times\omega$ is $\Pi^1_1$ with %all vertical sections non-empty, then there is $\pi: \omega\rightarrow \omega$ %such that for all $n$ we have $(n, \pi(n))\in A$.} %for $\Pi^1_1$ to find some sequence %\[\vec \b=(\b_\alpha)_{\alpha\in\omega_1^{\rm ck}}\] %such that \begin{claim} There is a sequence \[\vec \b=(\b_\alpha)_{\alpha\in\omega_1^{\rm ck}}\] such that \leftskip 0.4in \noindent (a) $\alpha\mapsto \b_\alpha$ is $\Delta^1_1$ in the codes; \noindent (b) for each $\delta<\omega_1^{\rm ck}$ the sequence $\vec \b|_\delta = (\b_\alpha)_{\alpha\in\delta}$ is suitable; \noindent (c) for each $\delta<\omega_1^{\rm ck}$, $\epsilon>0$, $x\in X$ there is some $D\in \b_\delta$ along with basic open \[V_0, V_1\subset G,\] $V_1$ a symmetric open neighborhood of the identity, $(D, V_0, V_1)$ $\epsilon$-stable for $\vec \b|_\delta$ and \[x\in D^{\Delta V_0}.\] \leftskip 0in \end{claim} \begin{proof} Recalling from Spector-Gandy that a subset of $\Lck$ is $\Pi^1_1$ in the codes if and only if it is $\Sigma_1^{\Lck}$, it suffices to show that there is a total $\Sigma_1^{\Lck}$ function \[\alpha \mapsto B_\alpha,\] with each $B_\alpha$ included in the indices for $\Delta^1_1$ sets, such that if we let $\b_\alpha$ be the corresponding collection of actual $\Delta^1_1$ sets encoded by the $B_\alpha$, then $(\b_\alpha)_{\alpha\in\delta}$ satisfies (c) above and 1, 2, 4, 5, 6 from the definition of suitability all $\delta<\omega_1^{\rm ck}$; note that these requirements are all $\Pi^1_1$ and hence $\Sigma_1^\Lck$. Since the total $\Sigma_1^\Lck$ functions are closed under transfinite $\Sigma_1^\Lck$ recursion, it suffices to show that given any $(B_\alpha)_{\alpha\in \delta}\in \Lck$ which gives rise to $(\b_\alpha)_{\alpha\in \delta}$ with the desired properties we can find some $B_\delta\in \Lck$ such that the slightly extended corresponding $(\b_\alpha)_{\alpha\in \delta+1}$ still has (c) above and 1, 2, 4, 5, 6. We know from the case assumption that given any $x\in X, \epsilon >0$ there will be some $D\in \Delta^1_1$ and basic open $V_0, V_1$ with $x\in D^{\Delta V_0}$ and $(D, V_0, V_1)$ having the $\Pi^1_1(\epsilon)$ property of $\epsilon$-stability. The problem is to find a single set of such indices in $\Lck$ which works for all such $x$ and $\epsilon$. Towards this purpose let $P\subset \omega\times X$ be $\Pi^1_1$ and universal for $\Pi^1_1$ subsets of $X$. We let \[f:\omega\times X\rightarrow {\rm LO}\] be recursive continuous such that $(n, x)\in P$ if and only if $f(n, x)\in $ WO. At each $\beta <\omega_1^{\rm ck} $ and $n\in \omega$ we let $P_{\beta, n}= \{x\in X: f(x) \in {\rm WO}, |\!|f(x)|\!|<\beta\}$\footnote{Here $|\!|w|\!|$ denotes the ordinal isomorphic to $w$ for $w\in$ WO.}. Note that the $\Delta^1_1$ subsets of $X$ are exactly the $\{P_{\alpha, n}: \alpha<\omega_1^{\rm ck}, n\in\N\}$. By the case assumption and the number uniformization theorem\footnote{I.e. that if $A\subset \omega^\omega\times\omega$ is $\Pi^1_1$ with all vertical sections non-empty, then there is $\Delta^1_1$ $\pi: \omega^\omega\rightarrow \omega$ such that for all $z$ we have $(z, \pi(n))\in A$.}, we may find a $\Delta^1_1$ function \[(\epsilon, x)\mapsto (\alpha_{\epsilon, x}, n_{\epsilon, x})\] such that each $(P_{\alpha_{\epsilon, x}, n_{\epsilon, x}}, V_0, V_1)$ is $\epsilon$-stable (some choice of basic open $V_0, V_1$) for $(\b_\alpha)_{\alpha\in\delta}$, and \[x\in P_{\alpha_{\epsilon, x}, n_{\epsilon, x}}^{\Delta V_0}.\] Boundedness gives \[{\rm sup}\{\alpha_{\epsilon, x}: \epsilon > 0, x\in X\}<\omega_1^{\rm ck},\] and so some such recursive ordinal $\gamma> {\rm sup}\{\alpha_{\epsilon, x}: \epsilon > 0, x\in X\}$ will give $\{P_{\beta, n}: n\in \N, \beta\leq \gamma\}$ as required. Under any reasonable system of indexing $\Delta^1_1$ sets we can find $B_\delta\in \Lck$ such that $B_\delta$ exactly gives the indices of the $\Delta^1_1$ sets appearing in $\{P_{\beta, n}: n\in \N, \beta\leq \gamma\}$. \end{proof} \begin{claim} \label{equivalence} For all $x, y\in X$ \[\hat{\varphi}_{\omega_1^{\rm ck}}^{\vec\b}(x, X, G)= \hat{\varphi}_{\omega_1^{\rm ck}}^{\vec\b}(y, X, G)\Rightarrow x E_G y.\] \end{claim} \begin{proof} Suppose not. Then we may find $x, y\in X$ with \[\hat{\varphi}_{\omega_1^{\rm ck}}^{\vec\b}(x, X, G)= \hat{\varphi}_{\omega_1^{\rm ck}}^{\vec\b}(y, X, G)\] but \[x\neg E_G y.\] But then using for the first time that $E_G$ is $\Delta^1_1$ we may apply the low basis theorem\footnote{I.e. any non-empty $\Sigma^1_1$ contains an $x$ with $\omega_1^{{\rm ck}(x)}=\omega_1^{\rm ck}.$} for $\Sigma^1_1$ and find $x, y\in X$ with \[\omega_1^{{\rm ck}(x, y)}=\omega_1^{\rm ck},\] \[\hat{\varphi}_{\omega_1^{\rm ck}}^{\vec\b}(x, X, G)= \hat{\varphi}_{\omega_1^{\rm ck}}^{\vec\b}(y, X, G),\] \[x\neg E_G y.\] Now choose a recursive $\epsilon_0>0$ such that \[\epsilon_0<\frac{1}{2}\] and \[\forall g\in W_{\epsilon_0}(d_X(y, g\cdot y)<\frac{1}{2}).\] \smallskip \noindent{\bf Subclaim:} For all $\beta<\omega_1^{\rm ck}$ there exists \[x_0\in [x]_G\] such that for all $\alpha<\beta$, $B\in \b_\alpha$, $V_2, V_3\subset G$ basic open with $V_3$ a symmetric neighborhood of $1_G$, \[\exists^*h\in W_{\epsilon_0}(\hat{\varphi}_{\alpha}^{\vec\b} (x_0, B^{\Delta V_2}, V_3)= \hat{\varphi}_{\alpha}^{\vec\b}(h\cdot y, B^{\Delta V_2}, V_3)).\] \smallskip \no {\bf Proof of subclaim:} Choose $A\in\b_{\beta}$, $V_0, V_1\subset G$ basic open sets with $V_1$ a symmetric neighborhood of the identity, $(A, V_0, V_1)$ $\epsilon_0$-stable for $\vec\b|_{\beta}$, and \[y\in A^{\Delta V_0}.\] Then choose $x_0E_G x$ with \[\hat{\varphi}_{\beta}^{\vec\b} (x_0, A^{\Delta V_0}, V_1)= \hat{\varphi}_{\beta}^{\vec\b}(y, A^{\Delta V_0}, V_1).\] It follows from the iterative structure of the definition of the $\hat{\varphi}_{\beta}^{\vec\b}(\cdot, \cdot, \cdot)$ invariants that for each $B, V_2, V_3,\alpha$ as in statement of subclaim that \[\exists\hat{y}\in{\cal O}(y, A^{\Delta V_0}, V_1)\] with \[\hat{\varphi}_{\alpha}^{\vec\b} (x_0, B^{\Delta V_2}, V_3)= \hat{\varphi}_{\alpha}^{\vec\b}(\hat{y}, B^{\Delta V_2}, V_3).\] Then by definition of stability we have that \[\exists^*h\in W_{\epsilon_0} (\hat{\varphi}_{\alpha}^{\vec\b}(h\cdot y, B^{\Delta V_2}, V_3)= \hat{\varphi}_{\alpha}^{\vec\b}(\hat{y}, B^{\Delta V_2}, V_3)= \hat{\varphi}_{\alpha^{\rm ck}}^{\vec\b} (x_0, B^{\Delta V_2}, V_3)).\] \hfill (Subclaim$\Box$) Now we may apply {\it overspill}\footnote{I.e. the converse of the usual boundedness theorem for $\Sigma^1_1$ subset of the recursive ordinals.} to find some \[x_0\in [x]_G\] such that for all $\alpha<\omega_1^{\rm ck}$, $B\in \b_\alpha$, $V_2, V_3\subset G$ basic open with $V_3$ a symmetric neighborhood of $1_G$, \[\exists^*h\in W_{\epsilon_0}(\hat{\varphi}_{\alpha}^{\vec\b} (x_0, B^{\Delta V_2}, V_3)= \hat{\varphi}_{\alpha}^{\vec\b}(h\cdot y, B^{\Delta V_2}, V_3)).\] By the low basis theorem we may assume \[\omega_1^{{\rm ck}(x_0, x, y)}=\omega_1^{{\rm ck}(x, y)} =\omega_1^{\rm ck}.\] At this stage we can choose a recursive real $\delta_0$ such that \[\delta_0<\frac{1}{2}\] \[\forall g\in W_{\delta_0}(d_X(x_0, g\cdot x_0)<\frac{1}{2}).\] \smallskip \noindent{\bf Subclaim:} For all $\beta<\omega_1^{\rm ck}$ there exists \[y_0\in W_{\epsilon_0}\cdot y\] such that for all $\alpha<\beta$, $B\in \b_\alpha$, $V_2, V_3\subset G$ basic open with $V_3$ a symmetric neighborhood of $1_G$, \[\exists^*h\in W_{\delta_0}(\hat{\varphi}_{\alpha}^{\vec\b} (y_0, B^{\Delta V_2}, V_3)= \hat{\varphi}_{\alpha}^{\vec\b}(h\cdot x_0, B^{\Delta V_2}, V_3)).\] \smallskip \noindent{\bf Proof of subclaim:} Again we may choose some $(A, V_0, V_1)$ which is $\delta_0$-stable for $\vec\b|_{\beta}$ with $x_0\in A^{\Delta V_0}$. And then we may choose \[y_0\in W_{\epsilon_0}\cdot y\] with \[\hat{\varphi}_{\beta}^{\vec\b} (x_0, A^{\Delta V_0}, V_1)= \hat{\varphi}_{\beta}^{\vec\b}(y, A^{\Delta V_0}, V_1),\] and we may complete the argument as in the proof of the last subclaim. \hfill (Subclaim$\Box$) \smallskip Once more overspill allows us to find $y_0\in W_{\epsilon_0}$ such that for all $\alpha<\omega_1^{\rm ck}$, $B\in \b_\alpha$, $V_2, V_3\subset G$ basic open with $V_3$ a symmetric neighborhood of $1_G$, \[\exists^*h\in W_{\delta_0}(\hat{\varphi}_{\alpha}^{\vec\b} (y_0, B^{\Delta V_2}, V_3)= \hat{\varphi}_{\alpha}^{\vec\b}(h\cdot x_0, B^{\Delta V_2}, V_3)).\] Continuing in a like minded fashion we may produce sequences \[g_0, g_1, \cdots,\] \[x_0, x_1, \cdots,\] \[g'_0, g'_1,\cdots,\] \[y_0, y_1,\cdots,\] \[\delta_0, \delta_1,\cdots,\] \[\epsilon_0, \epsilon_1,\cdots,\] such that: \leftskip 0.4in \noindent (0) for all $j\in\omega$, \[\omega_1^{{\rm ck}(x_j, y_j)}=\omega_1^{{\rm ck}(x_{j+1}, y_j)} =\omega_1^{\rm ck};\] \noindent (i) for all $j\in\omega$, $\alpha<\omega_1^{\rm ck}$, $B\in \b_\alpha$, $V_2, V_3\subset G$ basic open with $V_3$ a symmetric neighborhood of $1_G$, \[\exists^*h\in W_{\delta_j}(\hat{\varphi}_{\alpha}^{\vec\b} (y_j, B^{\Delta V_2}, V_3)= \hat{\varphi}_{\alpha}^{\vec\b}(h\cdot x_j, B^{\Delta V_2}, V_3));\] \noindent (ii) for all $j\in\omega$, $\alpha<\omega_1^{\rm ck}$, $B\in \b_\alpha$, $V_2, V_3\subset G$ basic open with $V_3$ a symmetric neighborhood of $1_G$, \[\exists^*h\in W_{\delta_{j+1}}(\hat{\varphi}_{\alpha}^{\vec\b} (x_{j+1}, B^{\Delta V_2}, V_3)= \hat{\varphi}_{\alpha}^{\vec\b}(h\cdot y_j, B^{\Delta V_2}, V_3));\] \noindent (iii) each $x_j$ equals $g_{j-1}g_{j-2}\cdots g_0\cdot x_0$; \noindent (iv) each $y_j$ equals $g_{j-1}'g_{j-2}'\cdots g_0'\cdot y_0$; \noindent (v) each $g_j\in W_{\delta_j}$; \noindent (vi) each $g_j'\in W_{\epsilon_{j+1}}$; \noindent (vii) at each $j$ we have $\forall g\in W_{\delta_j}$ \[d_G(gg_{j-1}g_{j-2}\cdots g_0, g_{j-1}\cdots g_0)<2^{-j-1};\] \noindent (viii) at each $j$ we have $\forall g\in W_{\epsilon_j}$ \[d_G(gg_{j-1}'g_{j-2}'\cdots g_0', g_{j-1}'\cdots g_0')<2^{-j-1};\] \noindent (ix) at each $j$ \[\forall g\in W_{\delta_j}(d_X(x_j, g\cdot x_j)<2^{-j-1});\] \noindent (x) at each $j$ \[\forall g\in W_{\epsilon_{j+1}}(d_X(y_j, g\cdot y_j)<2^{-j-1}).\] All the relevant sequences are Cauchy, so we can find $g_\infty$, $g_\infty '$ with \[g_j\rightarrow g_\infty,\] \[g_j'\rightarrow g_\infty',\] and hence by the continuity of the action \[x_j\rightarrow g_\infty x_0,\] \[y_j\rightarrow g_\infty ' y_0.\] We will be finished with the proof of the claim if we show that \[g_\infty x_0 = g_\infty ' y_0.\] But note that at any $j$ we can apply (i) to find $h\in W_{\delta_j}$ with \[d_X(h\cdot x_j, y_j)< 2^{-j-1};\] which by (v) yields \[d_X(x_j, y_j)\leq d_X(x_i, h\cdot x_i) + d_X(h\cdot x_i, y_i)< 2^{-j-1}+ 2^{-j-1}=2^{-j}.\] And so \[x_0 E_G y_0\] \[\therefore x E_G y,\] after all. \end{proof} \begin{notation} Let HC denote the collection of hereditarily countable sets. TC$(a)$ denotes the transitive closure of a set $a$. \end{notation} \begin{definition} Let us say that \[\theta: X\rightarrow {\rm HC}\] is $\Delta^1_1$ {\it in the codes} if there is an actual $\Delta^1_1$ function \[\Theta:X\rightarrow 2^{\omega\times\omega}\] such that for all $x\in X$ we have that $\Theta(x)$ is a well founded and extensional relation on $\omega$ whose transitive collapse is isomorphic to \[{\rm TC}(\theta(x))\cup\{\theta(x)\};\] that is to say, $\Theta(x)$ encodes the smallest transitive set containing $\theta(x)$ as an element. \end{definition} \begin{claim} For all $\alpha<\omega_1^{\rm ck}$, $B\in\b_\alpha$, $V_0, V_1\subset G$ basic open with $V_1$ a symmetric neighborhood of the identity, the function \[x\mapsto \hat{\varphi}_\alpha^{\vec\b}(x, B^{\Delta V_0}, V_1)\] is $\Delta^1_1$ in the codes. \end{claim} \begin{proof} This follows routinely by induction on $\alpha$. \end{proof} Thus we may obtain a Borel function \[\theta: X\rightarrow 2^{\omega\times\omega}\] such that for all $x\in X$ the relation $\theta(x)$ on $\omega$ is wellfounded, extensional, and has transitive collapse equal to \[\{\hat{\varphi}_\alpha^{\vec\b}(x, X, G): \alpha<\omega_1^{\rm ck}\} \cup\{{\rm TC}(\hat{\varphi}_\alpha^{\vec\b}(x, X, G)): \alpha<\omega_1^{\rm ck}\}.\] Hence \[xE_G y\] if and only if $\theta(x)$ and $\theta(y)$ are isomorphic as relations on $\omega$. Thus $E_G$ is reducible to \[\cong\!|_{{\rm Mod}(\l)},\] for $\l$ a language generated by a single binary relation. \bigskip Finally some words on how we can extend the conclusion of this case to obtain a reduction that allows an orbit inverse. To start we may apply theorem 5.1.5 of \cite{beckerkechris} to produce a Polish topology $\hat{\tau}$ on $X$ which is compatible with the original Borel structure, includes all sets of the form \[B^{\Delta V}\] where $B\in\b_\alpha$ some $\alpha<\omega_1^{\rm ck}$, and is such that the action of $G$ is still continuous with respect to this space. From here we can consider the Polish $G$-space $(X, \hat{\tau})$. For this space we may directly evoke the generalized Scott invariants \[x\mapsto \varphi_x\] from $\S$6.2 of \cite{hjorth}. Traipsing through the definitions there one sees that \[\varphi_x=\varphi_y\Rightarrow \forall \alpha <\omega_1 (\hat{\varphi}_\alpha^{\vec \b}(x, X, G)= \hat{\varphi}_\alpha^{\vec \b}(y, X, G)),\] and so by \ref{equivalence} \[x\mapsto \varphi_x\] gives a complete invariant. It was shown in $\S$6.2\cite{hjorth} that whenever we have these $\varphi_x$'s serving as complete invariants then we can indeed obtain a Borel reduction to isomorphism on a class of countable structures which admits an orbit inverse. And thus case(2) leads to a strengthened version of (I) as its outcome. \newpage \begin{thebibliography}{99} \bibitem{beckerkechris} H. Becker, A.S. Kechris, {\bf The descriptive set theory of Polish group actions,} London Mathematical Society. \bibitem{hamash} L.A. Harrington, D. Marker, S. Shelah, {\it Borel orders,} {\bf Transactions of the American Mathematical Society,} vol. 310(1988), pp. 293-302. \bibitem{hjorth} G. Hjorth, {\bf Classification and orbit equivalence relations,} American Mathematical Society, Rhode Island, 2000. \bibitem{kechrislouveau} A.S. Kechris, A. Louveau, {\it The classification of hypersmooth Borel equivalence relations,} {\bf Journal of the American Mathematical Society,} vol. 10(1997), pp. 215--242. \bibitem{moschovakis} Y.N. Moschovakis, {\bf Descriptive set theory,} North-Holland, Amsterdam, 1980. \end{thebibliography} \leftskip 0.5in greg@math.ucla.edu \bigskip MSB 6363 UCLA CA 90095-1555 \leftskip 0in \end{document} --=====================_10455774==_--