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\begin{document}
\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
\newtheorem{theorem}{Theorem}[section]
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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 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
MSB 6363
UCLA
CA 90095-1555
\leftskip 0in
\end{document}