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\title{\huge Subgroups of abelian Polish groups}         
\author{Greg Hjorth}        % Enter your name between curly braces
\date{\today}          % Enter your date or \today between curly braces
\maketitle




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%\tableofcontents   


\section{Introduction} 
\label{introduction} 

Ilijas Farah and Slawek Solecki showed in 
\cite{farahsolecki} that every Polish group 
contains arbitrarily complicated Borel 
subgroups and go on to ask whether the same can 
be shown for {\it Polishable} subgroups. 
As a partial answer, this notes 
shows that an uncountable abelian Polish 
group contains Polishable subgroups of 
unbounded Borel complexity. 

\bigskip 

{\bf Acknowledgements:} 
These results were largely proved during the course of the 
conference at CRM during the year dedicated to 
Set Theory, and came about after listening to Farah and 
Solecki describe their recent work. I am very grateful for a 
number of conversations with both of them, including one 
lengthy discussion carried out in the taxi on the way out 
to the airport for the flight home. 

I am also grateful to CRM for making 
this meeting possible, 
and I further thank Joan Bagaria and the other members of 
the organizing committee for inviting me to participate. 

I am more than indebted to the anonymous referee for a 
careful and exacting reading of the earlier version of 
this paper. 

\bigskip 


\section{Notation} 


In what follows, $G$ is an uncountable abelian Polish group and $d(\cdot, \cdot)$ a compatible, 
complete, two sided-invariant metric. We write the group operations on $G$ with reference to it 
being abelian -- thus $+$ is the group operation and $n\cdot g$ stands for 
\[g+g+...(n {\rm \: times})..+g.\] 
$0$ is the group identity in $G$. 

We will find it convenient to use the notation of an infinite sum. 
Thus we say that 
\[g=\sum_{n=0}^\infty g_n\] 
if 
\[d(\sum_{n=0}^N g_n, g)\rightarrow 0.\] 
Since $d$ is two sided-invariant metric we obtain that 
\[\sum_{n=0}^\infty g_n =\sum_{n=0}^\infty h_n\] 
if and only if 
\[\sum_{n=0}^\infty (g_n-h_n)=0.\] 

For $g$ a group element we use $o(g)$ to denote the 
{\it order} -- that is to say, the least $\ell $ with 
$\ell \cdot g=0$ if such an $\ell$ exists, and infinity 
otherwise. 

\section{The subgroups} 



We choose a sequence of non-zero elements 
\[(g_i)_i\] 
in $G$ such that: 

\leftskip 0.4in 

\noindent (I) for any $k\in\N$ and $(n_\ell)_{\ell\leq k}$ chosen from $\Z$, with each 
\[|n_\ell|\leq 4\ell k\] 
and 
\[h=n_0\cdot g_0 + n_1\cdot g_1...+n_k\cdot g_k\in G\]
\[h\neq 0\] 
we have that for any choice of $(\hat{n}_\ell)_{\ell > k}$ 
with each $|\hat{n}_\ell|<4\ell k$ that 
\[h\neq \sum_{\ell > k} \hat{n}_\ell\cdot g_\ell;\] 


\noindent (II) for any $\ell$ and $n$ with $|n|\leq \ell^2$ we have 
\[d(0, n\cdot g_\ell)<2^{-\ell};\] 


\noindent (III) either every $g_i$ has infinite order, or the $g_i$'s all have the 
same {\it finite} order, or $o(g_i)\rightarrow \infty$ 
as $i\rightarrow \infty$. 



\leftskip 0in 

\bigskip 
\bigskip 



We obtain (I) and (II) just by taking any sequence which converges to 0 sufficiently fast. (III) then 
follows by refining the sequence as needed. 

\begin{definition} If $h\in G$ then a {\it $k$-representation} of $h$ is an expression of the form 
\[\sum_{\ell\in\N} n_\ell\cdot g_\ell\] 
with 
\[{\rm lim}_{M\rightarrow \infty} \sum_{\ell=0}^M n_\ell\cdot g_\ell = h\] 
and each $|n_\ell|\leq \ell\cdot k.$ 
A {\it representation} is a $k$-representation for some $k$; a group element is 
{\it representable} if it has a representation. 

\end{definition} 

\begin{lemma} 
\label{L1} 
If 
\[\sum_{\ell\in\N} n_\ell\cdot g_\ell,\] 
\[\sum_{\ell\in\N} \hat{n}_\ell\cdot g_\ell,\] are two 
$2k$-representations of $h$ then for all $\ell>k$ 
\[n_\ell\cdot g_\ell = \hat{n}_\ell\cdot g_\ell.\] 
\end{lemma} 

\begin{proof} By the definition of $2k$-representation 
we have $|n_\ell|, |\hat{n}_\ell|\leq 2\ell k$ each $\ell$, and then 
by our assumption (I) on the $(g_i)$ sequence we have that at all $M\geq k$ 
\[\sum_{\ell\leq M} (n_\ell - \hat{n}_\ell)\cdot g_\ell=0.\] 
From this the lemma follows. 
\end{proof} 

\begin{definition} 
We let $H$ be the subgroup of all $h$ which have some representation 
\[\sum_{\ell\in\N} n_\ell\cdot g_\ell\] 
with 
\[\frac{|n_\ell|}{\ell}\rightarrow 0.\] 
\end{definition} 

\begin{definition} 
If 
\[\sum_{\ell\in\N} n_\ell\cdot g_\ell\] 
is a representation of $h$, then we let 
\[\varphi(\sum_{\ell\in\N} n_\ell\cdot g_\ell)={\rm sup}_\ell \frac{|n_\ell|}{\ell}.\] 
If $h\in H$ we let $\psi(h)$ be the infinum of 
\[\varphi(\sum_{\ell\in\N} n_\ell\cdot g_\ell)\] 
as $\sum n_\ell\cdot g_\ell$ ranges over representations of $h$. 
\end{definition} 




\begin{lemma} 
\label{L2} 
In the metric 
\[d_H(h, h')=\psi(h-h')+d(h, h')\] 
we have that $H$ is a Polish group. 
\end{lemma} 

\begin{proof} 
First we need to check that this is indeed a metric, and that follows from the 
observation that given any two representations 
\[\sum n_\ell\cdot g_\ell,\] 
\[\sum \hat{n}_\ell\cdot g_\ell,\] 
for $h-h'$ and $h'-h''$ respectively, we have 
\[\varphi(\sum (n_\ell +\hat{n}_\ell)\cdot g_\ell) \leq 
\varphi(\sum n_\ell\cdot g_\ell)+ \varphi(\sum \hat{n}_\ell\cdot g_\ell).\] 
The metric {\it is} separable since the representations with finite support 
($\forall^\infty \ell (n_\ell =0)$) give rise to a dense subgroup. Continuity of 
the group operations follows since 
\[\psi(h)+\psi(h')\geq \psi(h+h').\] 
This only leaves us with a campaign to show that the metric is complete. 

For this purpose, consider a Cauchy sequence $(h_i)_i$ in $H$, with each 
$h_i$ having  
\[\sum_{\ell\in\N} n_\ell^i\cdot g_\ell\] 
as a representation. Since 
\[|\psi(h_i)-\psi(h_j)|\leq \psi(h_i-h_j)\] 
we get that $(\psi(h_i))_{i\in\N}$ is Cauchy and 
so there is a fixed $k$ such that 
each is a $k$-representation. 
%Thus by \ref{L1} we get that for all $\ell > k$ 
%there $n^\infty_\ell$ such that 
%\[\forall^\infty j (n_\ell^j\cdot g_\ell = n^\infty_\ell\cdot g_\ell);\] 
%by mininizing the choice of the $|n^\infty_\ell|$ and slightly readjusting 
%the original representations we may assume 
Since the space of $k$-representations is precompact, we 
may assume there exist $(n_\ell^\infty)_{\ell\in \N}$ such that 
\[\forall^\infty j(n^j_\ell = n^\infty_\ell).\] 

%For the coordinates less than $k$, we observe that for each $N\geq k$ we have
%for all sufficiently large $j, j'$ and then sufficiently large $M$ 
%\[d(\sum_{\ell\geq k}^M(n^j_\ell\cdot g_\ell -n^{j'}, 
%\sum_{\ell\geq k}^M(\ell\cdot g_\ell)) \]
%is bounded by $2^{-N+1}$, in light of (II) from our list of
%assumptions on the $(g_i)_i$ sequence.
%Thus
%\[d(\sum_{\ell < k}(n^j_\ell\cdot g_\ell), 
%\sum_{\ell <k} (n^{j'}_\ell\cdot g_\ell)) \]
%goes to zero as $j, j'\rightarrow \infty$, and hence at each $\ell<k$ 
%there are $n_\ell^\infty$ such that
%\[\forall^\infty j(n^j_\ell \cdot g_\ell = n^\infty_\ell\cdot g_\ell).\]
%Again we may actually assume
%\[\forall^\infty j(n^j_\ell  = n^\infty_\ell).\]




\medskip 

\noindent{\bf Claim(1):} As 
$j\rightarrow \infty$ and then $M\rightarrow\infty$ 
\[d_H(h_j, \sum_{\ell =0}^M n^\infty_\ell\cdot g_\ell)\rightarrow 0.\] 
 

\medskip 

\noindent{\bf Proof of Claim:} 
Fix $\epsilon > 0$. We need to show that for all sufficiently 
large $j$ we have 
\[\forall^\infty M (d_H(h_j, \sum_{\ell =0}^M n^\infty_\ell\cdot g_\ell) 
<\epsilon).\] 

Choose $j$ large enough that $d_H(h_j, h_{j'})<\epsilon$ all $j'\geq j$ 
and $n^j_\ell = n^\infty_\ell$ all $\ell \leq k$. 
%By (I) we have that at all $j'\geq j$ 
%\[\sum (n^j_\ell - n^{j'}_\ell)\cdot g_\ell\] 
%is the unique 2k-representation of $h_j-h_{j'}$, and hence 
%has $\varphi$ value less than $\epsilon$. 
We may assume without loss of generality that 
$\epsilon < \frac{1}{k}.$ Thus at 
each $j'\geq j$ we have that if 
\[\sum_{\ell\in \N} m_\ell \cdot g_\ell \] 
is a $2k$-representation of $h_j-h_j'$ with 
$\varphi$ value less than $\epsilon$ then 
\[m_\ell =0\] 
at all $\ell\leq k$. Since by \ref{L1} we 
have 
\[\sum(n_\ell^j-n_\ell^{j'})\cdot g_\ell\] 
as the unique $2k$-representation of $h_j-h_{j'}$ with 
coefficient 0 at the $\ell^{\rm th}$ coordinate all $\ell\leq k$, 
and thus 
\[\sum(n_\ell^j-n_\ell^{j'})\cdot g_\ell\]
is the unique $2k$-representation of $h_j-h_{j'}$ with 
$\varphi$ value less than $\epsilon$. 

Therefore if at arbitrarily large $M$ we have 
\[d_H(h_j, \sum_{\ell=0}^M n^\infty_\ell\cdot g_\ell)>\epsilon\] 
then for arbitrarily large $M$ we have 
\[d_H(\sum_{\ell=0}^M n^j_\ell\cdot g_\ell,  
\sum_{\ell=0}^M n^\infty_\ell\cdot g_\ell)> \epsilon,\] 
but then going to some $j'$ with $n^{j'}_\ell =n^\infty_\ell$ 
for all $\ell\leq M$ we have 
\[\varphi(\sum (n^j_\ell -n^{j'}_\ell)\cdot g_\ell)>\epsilon,\] 
with a contradiction. 
\hfill ($\square$Claim) 

\medskip 

\noindent{\bf Claim(2):} As $N\rightarrow \infty$ 
\[\varphi(\sum_{\ell =N}^\infty n^\infty_\ell\cdot g_\ell)
\rightarrow 0.\] 

\medskip 

\noindent{\bf Proof of Claim:} 
Consider $\epsilon>0$. 
Choose $M_0$ such that at all $j\geq M_0$ and $\ell\leq k$ we have $n^j_\ell = n^\infty_\ell$. 
Choose $M_1>M_0$ such that at all $j, j' \geq M_1$ we have $\psi(h_j-h_{j'})<\epsilon.$ 
Choose $N_0$ such that at all $\ell \geq N_0$ 
\[\frac{|n^{M_1}_\ell|}{\ell}\leq \epsilon,\]
\[\therefore \varphi(\sum_{\ell =N_0}^\infty n^{M_1}_\ell\cdot g_\ell)\leq \epsilon.\] 

As in the proof of claim(1) we may assume that $\epsilon<\frac{1}{k}$, 
and again as in that proof we have that for all 
sufficiently large $j'$ 
\[\sum_0^\infty(n^{j'}_\ell -n_\ell^{M_1})\cdot g_\ell\] 
is the only possible 
representation of $h_{j'}-h_{M_1}$ with 
\[\varphi(\sum_0^\infty (n^{j'}_\ell- n^{M_1}_\ell)\cdot g_\ell)
<\epsilon,\] 
and  hence by \ref{L1} for $j'$ sufficiently large we have 
$\sum_0^\infty (n_\ell^{j'} - n_\ell^{M_1})\cdot g_\ell$ is 
{\it the} representation of $h_{j'}-h_j$ with 
$\varphi(\sum_0^\infty (n_\ell^{j'} - n_\ell^{M_1})\cdot g_\ell)
<\epsilon$. Hence at each $N>N_0$ 
\[\varphi(\sum_{\ell = N_0}^{N} n^\infty_\ell\cdot g_\ell) 
\leq \varphi(\sum^{N}_{\ell =N_0}(n^{j'}_\ell - n^{M_1}_\ell)\cdot g_\ell) 
+ \varphi(\sum_{\ell = N_0}^{N} n^{M_1}_\ell\cdot g_\ell)\]
\[\leq \varphi(\sum_0^\infty (n^{j'}_\ell - n^{M_1}_\ell)\cdot g_\ell) 
+\epsilon = \psi(h_{j'}-h_{M_1})+\epsilon <2\epsilon.\] 
Letting $\epsilon\rightarrow 0$ this proves the claim. 
\hfill ($\square$Claim) 


\medskip 


These two claims between them demonstrate that 
\[\sum n^\infty_\ell \cdot g_\ell\] 
represents some $h_\infty\in H$ and moreover 
\[\psi(h_j -h_\infty)\rightarrow 0\] 
as $j\rightarrow \infty$. 




\end{proof} 



\begin{definition} We define the notion of $\alpha$-calibration by induction on 
the countable ordinal $\alpha$. 

A {\it 0-calibration} is an infinite $A\subset\N$. An {\it $\alpha+1$}-calibration 
is an infinite $A\subset \N$ equipped with a partition 
\[A=\dot{\bigcup}B_n,\] 
where each $B_n$ is an $\alpha$-calibration. 
For $\lambda$ a limit, with a specified\footnote{For the purpose of this definition, 
imagine that all our limit ordinals come with a previously decided cofinal sequence.} 
\[\alpha_n\nearrow \lambda,\] we 
say that $A\subset \N$ is $\lambda$-calibration if 
it is partitioned 
\[A=\dot{\bigcup}B_n,\] 
with each $B_n$ an $\alpha_n$-calibration. 

\end{definition} 


\begin{definition} For $A$ a $0$-calibration, a representation 
\[\sum n_\ell\cdot g_\ell\] 
is {\it $A$-$0$-good} if there is an $n_\infty$ such that 
\[\forall^\infty \ell\in A(n_\ell =n_\infty);\] 
that is to say, the $n_\ell$'s, for $\ell\in A$, converge to $n_\infty$. 
We then let $\pi_{A, 0}(\sum n_\ell\cdot g_\ell)$ be this indicated 
$n_\infty$ and 
\[\varphi_{A, 0}(\sum n_\ell\cdot g_\ell) 
=\varphi(\sum n_\ell \cdot g_\ell) + |\pi_{A, 0}(\sum n_\ell\cdot g_\ell)| 
+|\{\ell\in A: n_\ell\neq \pi_{A, 0}(\sum n_\ell\cdot g_\ell)\}|;\] 
in other words, we add into the norm on representations for $H$ information 
about the eventual value $n_\infty$ along with a counter for the number of 
times this value is missed. 

For $A$ an $\alpha+1$-calibration, $A=\dot{\bigcup} B_n$, we say that 
\[\sum n_\ell\cdot g_\ell\] 
is {\it $A$-$\alpha +1$-good} if it is $B_n$-$\alpha$-good at every 
$n$ and there is an $m_\infty$ such that 
\[\forall^\infty n (\pi_{B_n, \alpha}(\sum n_\ell \cdot g_\ell)=m_\infty).\] 
We then let $\pi_{A, \alpha +1}(\sum n_\ell\cdot g_\ell)$ be this eventual 
$m_\infty$ value and let 
\[\varphi_{A, \alpha +1} (\sum n_\ell \cdot g_\ell) 
=\varphi (\sum n_\ell \cdot g_\ell) 
+\sum_n 2^{-n} \frac{\varphi_{B_n, \alpha}(\sum n_\ell \cdot g_\ell)} 
{1+\varphi_{B_n, \alpha}(\sum n_\ell \cdot g_\ell)}\]
\[+ |\pi_{A, \alpha +1}(\sum n_\ell\cdot g_\ell)| 
+|\{n: \pi_{B_n, \alpha}(\sum n_\ell\cdot g_\ell)\neq 
\pi_{A, \alpha +1}(\sum n_\ell\cdot g_\ell)\}|;\] 
that is to say, we merge into a single norm information about the eventual 
value $m_\infty$, the set of exceptional 
$n$'s at which this value is not yet reached, along with a bounded version 
of norms for the various $B_n$ sets. 

For $\lambda$ a limit ordinal, $\alpha_n\nearrow \lambda$, $A$ a $\lambda$-calibration, 
we can define 
{$A$-$\lambda$-goodness} in the exactly parallel manner, replacing each mention of 
$B_n$-$\alpha$-goodness by $B_n$-$\alpha_n$-goodness. 

\end{definition} 

%Note that it follows from \ref{L1} that if one representation of $h\in H$ is 
%$A$-$\alpha$-good, then they all are, and moreover the value of 
%$\pi_{A, \alpha}(\sum n_\ell\cdot g_\ell)$ does not depend on the choice of the 
%representation, and so we write 
%\[\pi_{A, \alpha}(h)\] 
%to indicate the value of 
%\[\pi_{A, \alpha}(\sum n_\ell\cdot g_\ell)\] 
%for any representation. 

We want to now define $\pi_{A, \alpha}(h)$ in a way that 
is independent of the representation. Here there is a split 
in cases, depending on the situation dictated at (III). 

Observe first of all by 
\ref{L1} we have that if $\sum n_\ell\cdot g_\ell$ and 
$\sum n'_{\ell} \cdot g_\ell$ are two 
$A$-$\alpha$-representations of $h\in H$ then the terms 
\[n_\ell\cdot g_\ell, \:\:\:\:\:\:\:\: n'_{\ell}\cdot g_\ell\] 
differ at only finitely many places. 

Therefore in the case of 
every $g_\ell$ having infinite order we get 
\[\pi_{A, \alpha}(\sum n_\ell\cdot g_\ell )
= \pi_{A, \alpha}(\sum n'_\ell\cdot g_\ell)\] 
for any two distinct representations.  
Therefore we just do the natural thing and let 
$\pi_{A, \alpha}(h)$ be the common value realized on 
all the $A$-$\alpha$-good representations. 

The next case is when the $g_\ell$'s {\it all} have the 
{\it same} finite order. Use $M$ to denote this common 
value. We then let $\pi_{A, \alpha}(h)$ be the unique 
$m\in\{0, 1, ..., M-1\}$ for which there is some  
$A$-$\alpha$-good representation 
having 
\[\pi_{A, \alpha}(\sum n_\ell\cdot g_\ell)=m.\] 

In the case that the $g_\ell$'s have strictly increasing but 
finite orders, we let $\pi_{A, \alpha}(h)$ be the unique $m\in 
{\mathbb Z}$ for which there is some $A$-$\alpha$-good representation 
with 
\[\pi_{A, \alpha}(\sum n_\ell\cdot g_\ell)=m.\] 





Note also that the collection of $A$-$\alpha$-good elements 
is closed under the group operations -- a critical fact underpinning the next 
definition. 

\begin{definition} For $A$ an $\alpha$-calibration we let 
\[H_{A, \alpha}\] 
be the subgroup of $H$ consisting 
of all elements which admit an $A$-$\alpha$-good 
representation. 
For $h$ such an element, we let 
\[\psi_{A, \alpha}(h)\] 
be the infinum of 
\[\varphi_{A, \alpha}(\sum_{\ell\in \N} n_\ell\cdot g_\ell)\] 
as $\sum n_\ell\cdot g_\ell$ ranges over all possible representations. 
We then let $d_{A, \alpha}(\cdot, \cdot)$ be defined by 
\[d_{A, \alpha}(h, h')=\psi_{A, \alpha}(h-h') + d_H(h, h').\] 
\end{definition} 


\begin{lemma} 
$d_{A, \alpha}$ defines a complete, separable metric on $H_{A, \alpha}$ for any 
$\alpha$-calibration $A$. 
\end{lemma} 

\begin{proof} 
That this is a metric follows as in the proof for $H$. 
Separability follows by induction on $\alpha$. 
For $\alpha=0$ we observe that the set of 
elements with a representation 
\[\sum n_\ell\cdot g_\ell\] for which there is some $m_\infty$ 
with  
\[\forall^\infty \ell (n_\ell = m_\infty) \]
forms a countable dense subset. For $A=\dot{\bigcup}B_n$ with each 
$B_n$ a $\beta_n$-calibration, we assume that there are countable dense subsets 
$D_n$ of each $H_{B_n,\beta_n}$ and we take as our countable dense subset those 
representations which on finitely many $B_n$'s are an element of a corresponding 
$D_n$, and elsewhere are eventually constant. 

Again, these arguments are quickly dismissed, and 
the main battle is for completeness. 
So we suppose that we have some $d_{A, \alpha}$-Cauchy sequence $(h_j)_{j\in \N}$, 
each $h_j$ having 
\[\sum_{\ell\in \N} n_\ell^j\cdot g_\ell\] 
as a representation. 
Again by the compactness of the space of $k$-representations, as in 
\ref{L2}, we may 
assume there are $(n_\ell^\infty)_{\ell > k}$ such that 
\[\forall^\infty j(n_\ell^j =n_\ell^\infty).\] 


Note that for all sufficiently large $j, j'$ we have 
\[\psi_{A, \alpha}(h_j - h_{j'})<1,\] 
and hence
\[\pi_{A, \alpha}(h_j-h_{j'})=0\] 
by definition of $\psi_{A, \alpha}(\cdot)$. 
 

%\noindent {\bf Proof of Claim:} First consider the case that 
%$\alpha=0$. Let $a$ be the least element of $A$ 
%greater than $k$. Define $M$ so that 
%\[d_{A, \alpha}(h_j, h_{j'})<1\] 
%\[n_{a}^j = n_{a}^{j'};\] 
%by \ref{L2} any representation
%\[\sum n_\ell \cdot g_\ell\] of 
%\[h_j - h_{j'}\]  
%will necessarily have $n_{a} =0$ all $j, j'\geq M$, and hence 
%by the definition of the metric we must have 
%\[\pi_{A, 0}(h_j- h_{j'})=0.\] 


%In the case $\alpha>0$ write 
%\[A=\dot{\bigcup} B_n, \] 
%where each $B_n$ is appropriate $\beta_n$-calibration for some $\beta_n<\alpha$; 
%we may assume inductively that at each $n$ there is $M_n$ such that for all 
%$j, j'> M_n$ 
%\[\pi_{B_n, \beta_n} (h_j-h_{j'})=0.\] 
%Then the definition of the metric $d_{A, \alpha}(h_j, h_{j'})\rightarrow 0$ will 
%require  
%\[\pi_{B_n, \beta_n}(h_j - h_{j'})=0\] 
%for all $n$ and for all $j, j'$ sufficiently large. 
%\hfill ($\square$Claim) 

 

Thus we may find a single $m_\infty$ such that 
\[\forall^\infty j (\pi_{A, \alpha}(h_j)=m_\infty).\] 
Let $h_\infty$ be the limit in the $d_H(\cdot, \cdot)$ metric 
of the $(h_j)_j$ sequence, with representation 
\[\sum_\ell n_\ell^\infty\cdot g_\ell.\] 
%We note that $\varphi(\sum n_\ell^\infty\cdot g_\ell)\leq k$, as in the 
%proof of \ref{L2}. 
Note that each $|n^\infty_\ell|\leq \ell\cdot k$ since 
$\forall^\infty j (n_\ell^j = n^\infty _\ell)$ and 
each $|n^j_\ell|\leq \ell\cdot k$. Thus 
$\sum n^\infty_\ell \cdot g_\ell$ is a 
$k$-representation. 

\medskip 


In the case that $\alpha=0$ we have the following claim: 

\medskip 

\noindent {\bf Claim:} There are only finitely many $a\in A$ with 
\[n_a^\infty\neq m_\infty.\] 


\medskip 

\noindent{\bf Proof of Claim:} We again use the fact that $2k$-representations 
are unique on coordinates bigger than $k$. 
Thus we obtain that if for some $c\geq 2k$ we have $2c$-many $a$'s with 
$n_a^\infty\neq m_\infty$, then for all sufficiently large $j$ we have 
{\it all} $2k$ representations of $h_j$ have at least $c$ many coordinates which 
present a value other than 
\[m_\infty = \pi_{A, 0}(h_j),\] 
and hence we are left with $\psi_{A, 0}(h_j)>c$. 

Thus to recap the argument, we have seen that if at {\it every} $c$ there 
exist more than $c$ many $a's$ with $n_a^\infty\neq m_\infty$ then 
$\psi_{A, 0}(h_j)\rightarrow \infty$ as $j\rightarrow \infty$, 
contradicting the assumption that $(\psi_j)_{j\in\N}$ is 
Cauchy.  
\hfill ($\square$Claim) 

\medskip 

Thus we have $h_\infty\in H_{A,0}$, $\pi_{A, 0}(h_\infty)=m_\infty$; 
the function 
\[j\mapsto \{a\in A: n^j_a\neq m_\infty\}\] 
is eventually constant by the definition of $\psi_{A, 0}(\cdot)$ and 
uniqueness of the $k$-representation on coordinates bigger than $k$. 
Therefore with respect to $d_{A, 0}(\cdot, \cdot)$ we have 
\[h_j\rightarrow h_\infty.\] 

\medskip 

If $\alpha>0$, $A=\dot{\bigcup}B_n$, each $B_n$ a $\beta_n$-calibration, 
we may assume inductively that there are $(p_n)$ such that at every $n$ 
\[\forall^\infty j(\pi_{B_n, \beta_n}(h_j) =p_n),\] 
\[h_\infty\in H_{B_n, \beta_n},\] 
and 
\[h_j\rightarrow h_\infty\] 
in $d_{B_n, \beta_n}(\cdot, \cdot)$. 
By Cauchyness of the $(h_j)$ sequence we must eventually have that for all 
sufficiently large $j, j'$ 
\[\forall n(\pi_{B_n, \beta_n}(h_j)=\pi_{B_n, \beta_n}(h_{j'})),\] 
and hence the $(p_n)$-sequence is eventually constant, with value $m_\infty$. 














\end{proof} 



\begin{lemma} Given any $\Ubf{\Delta}^0_{\alpha }$ set $D\subset \oo$ 
and $\alpha$-calibration $A=\dot{\bigcup}{B_n}$, each $B_n$ a $\beta_n$-calibration, 
there is continuous 
\[f^{A, D}_{\alpha }:\oo \rightarrow G\] 
such that 

\leftskip 0.4in 

\noindent (0) for every $y$ there is a representation 
\[\sum n_\ell\cdot g_\ell\] 
of $f_\alpha^{A, D}(y)$ with each $n_\ell\in \{0, 1\}$. 

\noindent (i) if $y\in D$ then 
$\forall^\infty n (\pi_{B_n, \beta_n}(f_{\alpha}^{A, D}(y))=1)$; 

\noindent (ii) if $y\notin D$ then 
$\forall^\infty n (\pi_{B_n, \beta_n}(f_{\alpha}^{A, D}(y))=0)$; 

\noindent (iii) the representation of each $f_{\alpha}^{A, D}(y)$ has 
 support in $A$ and each $f^{A, D}_\alpha(y)\in H_{A, \alpha}$. 

\leftskip 0in 

\end{lemma} 

\begin{proof} This is proved by induction on $\alpha$, in a manner entirely 
parallel to the similar claims from \cite{hjkelo}. 
First of all one sees that for any clopen $D\subset \oo$ and $A$ a 0-calibration 
we can certainly find continuous $f^{A, D}: \oo\rightarrow G$ with 
\[f^{A, D}(x)=0\] 
if $x\notin D$, but 
\[f^{A, D}(x)={\rm lim}_{N\rightarrow\infty}\sum_{\ell\in A, \ell\leq N} g_\ell,\] 
if $x\in D$. 
This forms the base of our induction. 

Now we do the inductive step, and suppose that $D$ is 
$\Ubf{\Delta}^0_{\alpha}$, $A=\dot{\bigcup}B_n$ is 
an $\alpha$-calibration, each $B_n$ a $\beta_n$-calibration. 
We recall the classical fact (compare \cite{hjkelo}) 
that there must be $\Ubf{\Delta}^0_{\beta_n}$ sets $(E_n)_n$ such that 

\leftskip 0.4in 

\noindent (a) if $y\in D$ then $\forall^\infty n (y\in E_n)$; 

\noindent (b) if $y\notin D$ then $\forall^\infty n (y\notin E_n)$. 

\leftskip 0in 

Appealing to our inductive hypothesis we may find 
\[f_n: \oo \rightarrow H_{B_n, \beta_n},\] 
continuous with respect to the Polish $G$-topology, 
each $f_n(x)$ having support inside $B_n$, 
with $\pi_{B_n, \beta_n}(f_n(x))=1$ if $x\in E_n$, 
$\pi_{B_n, \beta_n}(f_n(x))=0$ if $x\notin E_n$, and 
every $f_n(x)$ having a representation with coordinates taken solely from 
$\{0, 1\}$. 
We can then simply let 
\[f^{A, D}(x)={\rm lim}_{N\rightarrow\infty} f_0(x)+f_1(x)+f_2(x)+...+f_N(x).\] 
Our inductive assumptions on the $f_n$'s suffice to give 
that this is in $H$, and 
then we obtain it will be in $H_{A, \alpha}$ with either (a) or 
(b) above by the choice of the $E_n$ sets. 
It is continuous by condition (II) on the $g_\ell$-sequence 
and the fact that each $f_n(x)$ has support inside $B_n$. 
\end{proof} 



 
\begin{corollary} If $A=\dot{\bigcup}B_n$ is an 
$\alpha+1$-calibration, 
then $H_{A, \alpha+1}$ is not $\Ubf{\Pi}^0_{\alpha}$. 
\end{corollary} 

\begin{proof} Let $C\in \Ubf{\Sigma}^0_{\alpha}$ be of the 
form 
\[C=\bigcup D_n,\] 
where each $D_n\in \Ubf{\Delta}^0_{\alpha}$. Following the last 
lemma we may find continuous 
\[f_n: \oo\rightarrow H_{B_n, \beta_n}\] 
such that 

\leftskip 0.4in 

\noindent (0) for every $y$ there is a representation 
\[\sum_{\ell\in B_n} n_\ell\cdot g_\ell\] 
of $f_n(y)$ with each $n_\ell\in \{0, 1\}$. 

\noindent (i) if $y\in \bigcup_{i\leq n}D_i$ then 
$\pi_{B_{2n}, \beta_{2n}}(f_{2n}(y))=1$; 

\noindent (ii) if $y\notin \bigcup_{i\leq n}D_i$ then 
$\pi_{B_{2n}, \beta_{2n}}(f_{2n}(y))=0$; 

\noindent (iii) the representation of each $f_n(y)$ 
has its support in $B_n$; 

\noindent (iv) for $m$ odd at every $y$ we have 
$\pi_{B_m, \beta_m}(f_m(y))=1$. 

\leftskip 0in 

After this we can simply take 
\[f^C(y)={\rm lim}_{N\rightarrow \infty} f_0(y) + f_1(y) +...f_N(y).\] 
Then in order for the values of the $\pi_{B_n, \beta_n}(f^C(y))$ to settle down 
into a limit we must have that $y\in C$, and thus membership 
$C$ is a necessary condition for $f(y)$ to be in $H_{A, \alpha+1}$. 
But conversely membership in $C$ is also a sufficient condition for $f(y)$ to be 
in this subgroup, and thus we have continuously reduced membership in an 
arbitrary $\Ubf{\Sigma}^0_{\alpha}$ set to membership in $H_{A, \alpha+1}$. 

\end{proof} 

\section{After thoughts} 

As observed by Slawek Solecki, this result for uncountable 
abelian Polish groups implies the same for 
uncountable locally 
compact Polish groups. 

The main point is that we can use approximation by 
Lie groups -- in particular the theorem that every 
connected locally compact group has arbitrarily small 
normal subgroups 
for which the resulting quotient is Lie. (See 
\cite{montgomeryzippin}.) 

One case is that the group is totally disconnected, 
and hence, by local compactness (see \cite{montgomeryzippin}) 
it will have a neighborhood basis of clopen 
compact subgroups. Appealing to \cite{zelmanof} 
we obtain an uncountable abelian compact subgroup, 
and finish by the results of the last section. 

The other case is that there exists a 
closed connected subgroup, which in 
turn will have an uncountable Lie group as a quotient. 
But any Lie group contains one-parametrized 
subgroups, arrived at via the exponential map from the 
tangent bundle at the identity, and hence  a 
continuous image of $\R$ or $\T$, 
and again we fall back into the cases covered by the 
last section. 

The most optimistic conjecture, then, would be that 
every uncountable Polish group contains an uncountable 
abelian subgroup. From this and the results above we 
could obtain arbitrarily complicated Polishable subgroups 
of any uncountable Polish group. 

{\it However} the spirit of fearless scientific 
research demands that we should never take a position 
of optimism when pessimism is unrefuted. Thus: 

\begin{conjecture} 
There is an uncountable Polish group all of whose abelian 
subgroups are discrete. 
\end{conjecture} 

It may even be that some 
minor variation of the 
the surjectively universal two-sided 
invariant group (see \cite{kechris}, 
\cite{beckerkechris}) provides a counterexample. 
Accordingly it seems unlikely that the construction 
above will answer the question raised in 
\cite{farahsolecki}. 





\begin{thebibliography}{99} 

\bibitem{beckerkechris} {\sc H. \ Becker} and 
{\sc A.\ S. \ Kechris}, {\em The descriptive set theory 
of Polish group actions}, 
London Mathematical Society Lecture
Note Series, 232, Cambridge University Press, Cambridge, 1996. 

\bibitem{farahsolecki} 
{\sc I. \ Farah} and 
{\sc S. \ Solecki}, 
Borel subgroups of Polish groups, 
preprint, 
{\texttt{http://www.math.yorku.ca/\~{}ifarah/preprints.html}}. 


\bibitem{hjkelo} {\sc G. \ Hjorth},  
{\sc A.\ S. \ Kechris}, and {\sc A. \ Louveau}, 
The Borel equivalence relations induced actions of the infinite 
symmetric 
group, 
{\it Annals of Pure and Applied Logic}, vol. 92(1998), pp. 63--112. 


\bibitem{kechris} {\sc A. \ S. \ Kechris,} 
{\it Definable equivalence relations and Polish 
group actions,} 
manuscript, Caltech, 1993. 

\bibitem{montgomeryzippin} 
{\sc D. \ Montgomery} and 
{\sc L. \ Zippin}, 
{\it Topological transformation groups,}  
reprint of the 1955 original,  Robert E. Krieger Publishing Co., Huntington, N.Y., 1974. xi+289 pp. 

\bibitem{zelmanof} 
{\sc E. \ I. \ Zelmanov}, 
On periodic compact groups, 
{\it Israel Journal of Mathematics,} 
vol. 77(1992), pp. 83--95.




\end{thebibliography}

\leftskip 0.5in







greg@math.ucla.edu

\bigskip



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UCLA

CA 90095-1555

\leftskip 0in

\end{document}



