% LaTeX Report Template - customizing header and footer \documentclass[psfig]{report} \usepackage{oldlfont} \usepackage{amsmath, amssymb} \input{amssymb.def} \input{amssymb.tex} % Set left margin - The default is 1 inch, so the following % command sets a 1.25-inch left margin. %\setlength{\oddsidemargin}{0.25in} \setlength{\oddsidemargin}{0in} % 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}{6in} \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}{-0.25in} % Set height of the header \setlength{\headheight}{0.3in} % Set vertical distance between the header and the text \setlength{\headsep}{0.2in} % Set height of the text \setlength{\textheight}{9in} % Set vertical distance between the text and the % bottom of footer \setlength{\footskip}{0.1in} %\markboth{TFA groups}{TFA groups} % Set the beginning of a LaTeX document \begin{document} \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\V{{\mathbb V}} \def\N{{\mathbb N}} \def\n{{\mathbb N}} \def\Q{{\mathbb Q}} \def\Z{{\mathbb Z}} \def\C{{\mathbb C}} \def\O{{\cal O}} \def\e{\epsilon} \def\h{{\hat{X}}} \def\B{{\cal B}} \def\n{\noindent} %\def\Box{\square} %\def\box{\square} % Redefine "plain" pagestyle \makeatletter % `@' is now a normal "letter' for LaTeX \renewcommand{\ps@plain}{% \renewcommand{\@oddhead}{\textrm{Your Header}\hfil\textrm{\thepage}}% \renewcommand{\@evenhead}{\@oddhead}% \renewcommand{\@oddfoot}{}% empty footer \renewcommand{\@evenfoot}{\@oddfoot}} \makeatother % `@' is restored as a "non-letter" character \Huge \title{\Huge { FINITE RANK TFA GROUPS}} % Enter your title between curly braces \author{{\Huge Greg Hjorth}\\{\LARGE greg@math.ucla.edu}} % Enter your name between curly braces \date{\Huge September 17, 1998, London} % Enter your date or \today between curly braces \maketitle \pagestyle{myheadings} \markright{\large TFA groups} % Set to use the "plain" pagestyle %\pagestyle{plain} \n{\bf $\S$1. A classical result} \bigskip \n {\bf Definition} A {\it TFA group} is an abelian torsion free ($g+g+...g=0\Rightarrow g=0$) group. It has {\it finite} rank $n$ if any $n+1$ many elements are dependent -- that is to say, for any $g_1, ..., g_n, g_{n+1}$ there are $k_1, ...,k_{n+1}\in \Z \setminus \{0\}$ with \[k_1\cdot g_1+...k_{n+1}\cdot g_{n+1}=0.\] \bigskip \bigskip Here \[k\cdot g =_{df}g+g+...^{\:\:\:\:K\:{\rm times}}\:\:+g.\] \bigskip \bigskip \bigskip \bigskip \bigskip The rank $n$ TFA groups are exactly the subgroups of $(\Q^n, +)$. \newpage \noindent {\bf Theorem} (Baer) There is a ``nice" classification of TFA$_1$ groups. \bigskip \bigskip \n{\bf Sketch of Proof.} Let $A\in$ TFA$_1$. For any $g\in A$ not equal to the identity let \[\theta(A, g)=\{(p, n): p^n \:{\rm divides}\: g\}.\] \bigskip \bigskip Claim: If $q$ is a prime, then \leftskip 0.6pt \n (i) $p\neq q$ implies \centerline{$(p, n)\in \theta(A, g)$ iff $(p, n)\in \theta (A, q\cdot g)$;} \n (ii) $p=q$ implies \centerline{$(p, n+1)\in \theta(A, g)$ iff $(p, n)\in \theta (A, q\cdot g)$.} \leftskip 0pt \hfill ($\square$ Claim) \bigskip \bigskip Thus for any two non-zero $g, h\in A$ the symmetric difference $\theta(A, g)\Delta \theta(A, h)$ is finite. Moreover, for any suitably ``convex" \centerline{$S\subset$ Primes$\times \N$} \[|S\Delta \theta(A, g)|<\aleph_0\] we may find $k\in A$ with \[\theta(A, k)=X.\] \newpage Conversely \bigskip Claim: If $A, B\in$ TFA$_1,$ \[g\in A,\: \: h\in B\] \[\theta(A, g)=\theta(B, h)\] then there is an isomorphism between $A$ and $B$ sending $g$ to $h$. \hfill ($\square$ Claim) \bigskip \bigskip Thus \[A\mapsto \{\theta(A, g): g\neq 0\}\] provides a complete invariant for the isomorphism type of $A$, and in fact we can obtain elements of \[{\cal P}({\rm Primes}\times \N)/{\rm Finite}\] (subsets of Primes$\times$ natural numbers considered up to finite difference) as complete invariants for rank one torsion free abelian groups. \hfill $\square$ \newpage \hfill $\empty $ \bigskip \bigskip \bigskip \bigskip \bigskip \n{\bf 1.3 Question} Does a similarly nice classification exist for general finite rank TFA groups? \bigskip \bigskip \bigskip \bigskip \bigskip \bigskip \bigskip We need to make precise the meaning of ``nice". We need definability assumptions on the functions used in a classification theorem -- since unbridled use of the axiom of choice would enable us to ``classify" virtually anything by virtually anything else, subject only to agreement in cardinality. \bigskip \bigskip \bigskip \bigskip \bigskip \newpage %wewewre \bigskip \bigskip \bigskip \bigskip \hfill $\empty $ \bigskip \bigskip \bigskip \bigskip \bigskip Perhaps the most generous class of functions for which we can define and prove theorems without any metamathematical entanglement are the Borel functions. \bigskip \bigskip \bigskip \bigskip \bigskip \n{\bf 1.4 Theorem} In suitable Borel structures, the finite rank TFA groups do not admit a Borel assignment of complete invariants in \[{\cal P}(\N)/{\rm Finite}.\] \bigskip \bigskip Here ${\cal P}(\N)$ (or ${\cal P}(S)$ for any countable set $S$) may be naturally identified with $2^\N$ -- and we obtain a reasonable Borel structure from the product topology. \newpage \n{\bf $\S$2. Borel structure} \bigskip \bigskip \n{\bf 2.1 Definition} A {\it a Polish space} is a separable space allowing a complete metric compatible with its topology. \bigskip \bigskip \bigskip \n{\bf 2.2 Examples} $\R$. $\C$. $2^\N$ in the product topology -- for instance with \[d(f, g)=2^{-\mu n\{f(n)\neq g(n)\}}.\] Any compact metric space. \bigskip \bigskip Given any countable set $S$ we may identify ${\cal P}(S)$ (the set of subsets of $S$) with $2^S$ and endow it with a Polish topology. \newpage \n{\bf 2.3 Fact} Closed (and even $G_{\delta}$) subsets of Polish spaces are again Polish in the subspace topology. \bigskip \bigskip \bigskip \n{\bf 2.4 Definition} Let TFA$_n$ be the set $\{G\subset \Q^n: G$ is a subgroup of $(\Q^n,+)\}$. \bigskip \bigskip \bigskip \bigskip TFA$_n$ is a closed subspace of $\Q^n$, and hence Polish in its own right. Every rank $n$ TFA abelian group can be realized as an element of TFA$_n$. %View $X_0=2^{\N\times \N}$ as a topological space in the %product topology; in this topology it is a {\it Polish space} -- that is to say %a seperable space admitting a complete metric. %\bigskip %\bigskip %Let $X_1$ be the elements $x\in X_0$ such that %\[\{(n, m): x(n, m)=1\}\] %is the graph of a function, $+_x$. $X_1$ is a $G_{\delta}$ %subspace of $X_0$, and therefore again %Polish. %\bigskip %\bigskip %A $G_{\delta}$ subset of a Polish space is always Polish in the subspace %topology. %\bigskip %\bigskip %let $X_2$ be the set of $x\in X_1$ such that $+_x$ defines an associative %and commutative %function on $\N$. %\newpage %Let TFA$_{\infty}$ be the set of $x\in X_2$ such that $+_x$ defines a group %operation on $\N$. %The futher requirements for membership in TFA$_{\infty}$ %are %\bigskip %\leftskip 0.4in %\n (i) $\forall n \exists k (n+_x k= n)$; %\bigskip %\n (ii) $\forall n_1, n_2, k_1, k_2$ %\bigskip %\centerline{$(n_1 +_x k_1=n_1 \wedge %n_2 +_x k_2 = n_2 \Rightarrow k_1 = k_2)$;} %\bigskip %\n (iii) $\forall n\exists k (n +_x (n+_x k)=n)$. %\leftskip 0in %\bigskip %\bigskip %\bigskip %\bigskip %\bigskip %Between them, (i) and (ii) assert existence (and uniqueness) of the identity; %(iii) implies that every element has an inverse. %\bigskip %\bigskip %\bigskip %\bigskip %\bigskip %(ii) is a closed condition; (i) and (iii) correspond to a countable intersection of %open sets -- and thus are $G_{\delta}$. Hence TFA$_{\infty}$ is a Polish %space in the subspace topology. %\newpage %For any $n$, the set of $x\in$ TFA$_{\infty}$ such that $(\N, +_x)$ is a %rank $\leq n$ TFA group is again a relatively $G_{\delta}$ set -- since %it amounts to the requirement that for all $k_1, ...k_n, k_{n+1}\in \N$ there is some %relation of rational dependence. \bigskip \bigskip \bigskip %\bigskip %\bigskip %\bigskip %\n{\bf 2.2 Definition} Let TFA$_n$ be the space of $x\in 2^{\N\times \N}$ such that $+_x$ defines %a rank $n$ TFA group structure on $\N$. \bigskip \bigskip \bigskip \bigskip \bigskip \bigskip In this sense we can obtain a natural Borel structure on the rank $n$ torsion free abelian groups. \newpage \n{\bf $\S$3 Equivalence relations} \bigskip \bigskip \bigskip \n{\bf 3.1 Definition} If $E$ and $F$ are equivalence relations on Polish spaces $X$ and $Y$ then \[E\leq_B F\] indicates that there is a Borel function \[\theta: X\rightarrow Y\] such that for all $x_1, x_2 \in X$ \[x_1 E x_2 \Leftrightarrow \theta(x_1) F \theta(x_2).\] \bigskip \bigskip \bigskip \bigskip A function is {\it Borel} if it pulls open sets back to Borel sets -- that is, sets in the $\sigma$-algebra generated by the open sets. \newpage \n{\bf 3.2 Examples} For $X$ a Polish space, let id$(X)$ be the identity equivalence relation on $X$. \bigskip Let $E_0$ be the equivalence relation on $2^\N$ of eventual agreement -- $x_1 E_0 x_2 $ if there is some $N$ such that for all $n>N$ \[x_1(n)=x_2(n).\] \bigskip Similarly, for any countable set $S$ we can define $E_0(S)$ on \[{\cal P}(S)\sim 2^S\] as the equivalence relation of agreeing on a cofinite set; for any countably infinite set $S$ we have \[E_0\leq_B E_0(S)\leq_B E_0.\] \bigskip Let $\cong_n$ be the equivalence relation on isomorphism on TFA$_n$ (the space of rank $n$ TFA groups with underlying set $\N$). \bigskip \bigskip \bigskip Baer's theorem shows that \[\cong_1\leq_B E_0.\] \newpage $\empty$ \bigskip \bigskip \n{\bf 3.3 Definition} An equivalence relation is {\it countable} if every equivalence class is countable; an equivalence relation $E$ on $X$ is {\it Borel} if it is Borel as a subset of $X\times X$. \bigskip \bigskip \bigskip Each $\cong_n$ (isomorphism on TFA$_n$) is Borel reducible to a countable equivalence relation -- but it is not true that all countable Borel equivalence relations are Borel reducible to $E_0$. \newpage \n{\bf 3.4 Definition} An equivalence relation $E$ is {\it smooth} if $E\leq_B$ id$(\R)$ -- that is to say we may assign reals as complete invariants. A countable equivalence relation is {\it hyperfinite} if we may in a Borel manner assign the structure of a $\Z$ chain to each infinite equivalence class. A countable equivalence relation is {\it treeable} if we may in a Borel fashion assign the structure of a tree (connected graph with no cycles) to each equivalence relation. \bigskip \bigskip \bigskip A countable Borel equivalence relation is smooth if and only if there is a Borel set meeting each orbit in exactly one point. A countable Borel equivalence relation is hyperfinite if and only if it is $\leq_B$ reducible to $E_0$. Smooth $\subset$ hyperfinite $\subset$ treeable $\subset$ ``treeable $\times$ treeable" $\subset$ countable. None of these implications reverse. \newpage $\empty$ \bigskip \bigskip \bigskip \bigskip \n{\bf 3.5 Lemma} If $E$ and $F$ are countable Borel equivalence relations on Polish $X$ and $Y$, with $E\leq_B F$ then \leftskip 0.4in (i) $F$ smooth implies $E$ smooth; (ii) $F$ hyperfinite implies $E$ hyperfinite; (iii) $F$ treeable implies $E$ treeable. \leftskip 0in \bigskip \bigskip \bigskip \bigskip \n{\bf 3.6 Lemma} If $E$ and $F$ are countable Borel equivalence relations on Polish $X$ and $E\subset F$ \leftskip 0.4in (i) $F$ smooth implies $E$ smooth; (ii) $F$ hyperfinite implies $E$ hyperfinite; (iii) $F$ treeable implies $E$ treeable. \leftskip 0in \newpage \n{\bf $\S$4 Results} If $G_1, G_2$ are subgroups of $\Q^n$ and $\pi:G_1\cong G_2$ is an isomorphism, then $\pi$ extends to an automorphism \[\Pi: \Q^n\cong \Q^n.\] Thus we may view $\cong_n$ (isomorphism) on rank $n$ TFA groups as being induced by the automorphism group of $\Q^n$: GL$(n, \Q)$. \bigskip \bigskip \bigskip It is widely believed that: \bigskip \bigskip \bigskip \n{\bf 4.1 Conjecture} If $G$ is an abelian (or even {\it amenable}) countable group acting by Borel automorphisms on a Polish space $X$ with induced orbit equivalence relation $E_G$ then \[E_G\leq E_0.\] (That is, $E_G$ should be hyperfinite.) \bigskip \bigskip Since GL$(1, \Q)=\Q$ is responsible for $\cong_1$, it may be the case that Baer's classification theorem is a special case of this more general conjecture. \newpage Two reasons for expecting the situation could be more complicated in higher dimensions. \bigskip \bigskip \n{\bf 4.2 Fact} (Folklore) (i) For $n\geq 2$, the groups GL$(n, \Q)$, SL$(n, \Z)$, PSL$(n, \Z)=$ SL$(n, \Z)/\{I, -I\}$ all fail to be amenable. (ii) If $G$ is a non-amenable group acting freely by measure preserving transformations on a {\it probability} space then the corresponding orbit equivalence relation $E_G$ is not {\it hyperfinite}. \bigskip \bigskip \n{\bf 4.3 Theorem} (i) For $n\geq 3$ the group SL$(n, \Z)$ is {\it Kazhdan}. (ii) (Adams, Spatzier) If a countable Kazhdan group $G$ acts by measure preserving transformations on a {probability} space then the corresponding orbit equivalence relation $E_G$ is not {\it treeable} whenever there is an infinite subgroup acting freely. \newpage \n{\bf 4.4 Theorem} For $n \geq 3$ the isomorphism relation on rank $n$ TFA groups, $\cong_n$ on TFA$_n$ is not treeable. \bigskip \bigskip \bigskip \n{\bf Outline of proof.} The main issue is to somehow find a measure. Consider the dual group $\Gamma(\Q^n/\Z^n)$ of homomorphism \[\psi: (\Q^n/\Z^n, +) \rightarrow {\mathbb T}=(\R/\Z, +).\] Since $\Q^n/\Z^n$ is countable, the dual group $\Gamma(\Q^n/\Z^n)$ is a compact metric group, and so there is an invariant probability (Haar) measure, $\mu$. \bigskip \bigskip \bigskip Let $\pi_n:\Q^n\twoheadrightarrow \Q^n/\Z^n$ be the canonical projection. For $\psi\in \Gamma$, let \[K(\psi)=\{\vec q\in \Q^n: \psi(\pi_n(\vec q))=0\}.\] \newpage Then push the Haar measure $\mu$ on the dual group along the ``kernel" map $K$ to obtain $\nu=K_*(\mu)$ -- \[\nu(B)=\mu(K^{-1}[B]).\] \bigskip \bigskip \bigskip Here one can show that the action of SL$(n, \Z)$ preserves $\nu$ and there is an infinite subgroup for which the action is free. Thus the orbit equivalence relation induced by SL$(n,\Z)$ is not treeable; since $\cong_n$ (induced by GL$(n, \Q)$) includes $E_{{\rm SL}(n, \Z)}$ we have that $\cong_n$ is non-treeable. \hfill ($\square$) \bigskip \bigskip \bigskip \bigskip A similar argument (with thanks to Alekos Kechris and Simon Thomas) shows that $\cong_2$ is not hyperfinite. However, there is no evidence currently to suggest that $\cong_2$ is non-treeable. % Set the ending of a LaTeX document \end{document}