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\date{}
\author{V.~M.~Manuilov}
\title{Almost representations and asymptotic representations of discrete
groups}
\sloppy
\parindent=0em
\begin{document}
\maketitle
\begin{abstract}
We define for discrete finitely presented groups a new property related to
their asymptotic representations. Namely we say that a groups has the
property AGA if every almost representation generates an asymptotic
representation. We give examples of groups with and without this property.
For our example of a group $\G$ without AGA the group
$K^0(B\G)$ cannot be covered by asymptotic representations of $\G$.
\end{abstract}
One of the reasons of attention to almost and asymptotic representations
of discrete groups \cite{Grov,Harp-Kar,c-hig,CGM}
is their relation to $K$-theory
of classifying spaces \cite{c-hig,mish-noor}. It was shown in
\cite{mish-noor} that in the case of finite-dimensional classifying space
$B\G$ in order to construct a vector bundle over it out of an asymptotic
representation of $\G$ it is sufficient to have an $\e$-almost
representation of $\G$ with small enough $\e$. Of course an $\e$-almost
representation contains less information than the whole asymptotic
representation, but it turns out that often the information contained in
an $\e$-almost representation makes it possible to construct the
corresponding asymptotic representation. In the present paper we give the
definition of this property, prove this property for some classes of
groups and finally give an example of a group without this property. We
discuss also this example in relation with its $K$-theory.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Basic definitions}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Let $\G$ be a finitely presented discrete group, and let
$\G=\la F|R\ra=\la g_1,\ldots,g_n|r_1,\ldots,r_k\ra$ be its presentation
with $g_i$ being generators and $r_j=r_j(g_1,\ldots,g_n)$ being relations.
We assume that the set $F=\{g_1,\ldots,g_n\}$ is symmetric, i.e. for every
$g_i$ it contains $g_i^{-1}$ too, and the set $R$ of relations contains
relations of the form $g_ig_i^{-1}$, though we usually will skip these
additional generators and relations.
By $U_\i$ we denote the direct limit of the groups $U_n$
with respect to the natural inclusion $U_n\arr U_{n+1}$ supplied with the
standard operator norm. The unit matrix we denote by $I\in U_\i$.
\begin{dfn}
{\rm
A set of unitaries $\s=\{u_1,\ldots,u_n\}\subset U_\i$ is called an {\it
$\e$-almost representation} of the group $\G$ if after substitution of
$u_i$ istead of $g_i$, $i=1,\ldots,n$, into $r_j$ one has
$$
\norm{r_j(u_1,\ldots,u_n)-I}\leq\e
$$
for all $j=1,\ldots,k$.
}
\end{dfn}
In this case we write $\s(g_i)=u_i$. Remark that this definition depends
on a choice of presentation of the group $\G$, but we will see that this
dependence is not important. Let $\la h_1,\ldots,h_m|s_1,\ldots,s_l\ra$ be
another presentation of $\G$. For an $\e$-almost representation $\s$ with
respect to the first presentation we can define the set of unitaries
$v_1\ldots,v_m\in U_\i$, $v_i=\ov{\s}(h_i)$ putting
$\ov{\s}(h_i)=\s(g_{j_1})\cdot\ldots\cdot\s(g_{j_{n_i}})$,
where $h_i=g_{j_1}\cdot\ldots\cdot g_{j_{n_i}}$.
By the same way starting
from the set $\ov{\s}(h_i)$ we can construct the set $\ov{\ov{\s}}(g_i)$.
\begin{lem}\label{nezavis}
There exist constants $C$ and $D$ (depending on the two presentations)
such that $\ov{\s}$ is a $C\e$-almost representation with respect to the
second presentation of $\G$ and for all $g_i$, $i=1,\ldots,n$, one has
$\norm{\ov{\ov{\s}}(g_i)-\s(g_i)}\leq D\e$.
\end{lem}
{\bf Proof.}
We have to estimate the norms $\norm{s_q(v_1,\ldots,v_m)-I}$,
$q=1,\ldots,l$. To do so notice that every relation $s_q$ can be written
in the form
\begin{equation}\label{kommut}
s_q=a_1^{-1}r_{j_1}^{\ep_1}a_1\cdot\ldots\cdot
a_{m_q}^{-1}r_{j_{m_q}}^{\ep_{m_q}}a_{m_q}
\end{equation}
for some $a_i\in \G$, where $\ep_i=\pm 1$.
Let $M$ be the maximal length of the words
$a_i=a_i(g_1,\ldots,g_n)$ and $a_i^{-1}=a_i^{-1}(g_1,\ldots,g_n)$.
Substitute $u_1,\ldots,u_n$ into these words and
put $b_i'=a_i^{-1}(u_1,\ldots,u_n)\in U_\i$,
$b_i=a_i(u_1,\ldots,u_n)\in U_\i$. Then one has
$$
\norm{b_i'b_i-I}\leq M\e.
$$
It follows from (\ref{kommut}) that
$$
s_q(v_1,\ldots,v_m)=
b_1'r_{j_1}(u_1,\ldots,u_n)b_1\cdot\ldots\cdot
b_{m_q}'r_{j_{m_q}}(u_1,\ldots,u_n)b_{m_q},
$$
but as for every $i$ one has
$$
\norm{b_i'r_{j_i}^{\ep_i}(u_1,\ldots,u_n)b_i-I}\leq
\norm{b_i'b_i-I}+\norm{r_{j_i}^{\ep_i}(u_1,\ldots,u_n)-I}\leq(M+1)\e,
$$
so
$$
\norm{s_q(v_1,\ldots,v_m)-I}\leq m_q(M+1)\e,
$$
which proves the first statement of the Lemma. The second statement is
proved in a similar way. \q
As the number of generators is finite, so the image of every almost
representation lies in finite matrices, $\s\in U_n$ for some $n$. The
minimal such $n$ is called a dimension of $\s$. Usually we will ignore the
remaining infinite unital tail of the matrices $\s(g_i)$ and write
$\s(g_i)\in U_n$ instead of $U_\i$.
The set of all $\e$-almost representations of the group $\G$ we denote by
$R_\e(\G)$.
The deviation of an almost representation can be measured by the
value
$$
\nnn{\s}=\max_j\norm{r_j(u_1,\ldots,u_n)-I}.
$$
Notice that both these definitions also depend on the choice of a
presentation of $\G$.
\medskip
\begin{dfn}[{\rm cf. \cite{c-hig}}]
{\rm
A set of norm-continuous unitary paths
$\s_t=\{u_1(t),\ldots,u_n(t)\}\subset U_\i$, $t\in[0,\i)$, is called an
{\it asymptotic representation} of the group $\G$ if $\nnn{\s_t}$ tends to
zero when $t\to\i$.
}
\end{dfn}
Due to the Lemma \ref{nezavis} this definition does not depend on the
choice of a presentation of $\G$. The set of all asymptotic representations
of the group $\G$ we denote by $R_{asym}(\G)$.
\medskip
Now we are ready to define a new property of finitely generated groups
which we call AGA ({\it Almost} representations {\it Generate Asymptotic}
representations).
\begin{dfn}
{\rm
A group $\G$ possesses the property AGA if for every $\e>0$
one can find a number $\d(\e)$ (with the property $\d(\e)\to 0$
when $\e\to 0$) such that for every
almost representation $\s\in R_\e(\G)$ there exists an asymptotic
representation $\s_t\in R_{asym}(\G)$ such that $\s_0=\s$ and
$\nnn{\s_t}\leq \d(\e)$ for all $t\in [0,\i)$.
}
\end{dfn}
\begin{lem}
The property AGA does not depend on the choice of a presentation of the
group.
\end{lem}
{\bf Proof } immediately follows from the Lemma \ref{nezavis}. \q
\begin{thm}
The following groups have the property AGA:
\begin{enumerate}
\vspace{-\itemsep}
\item
free groups,
\vspace{-\itemsep}
\item
free products of groups having the property AGA,
\vspace{-\itemsep}
\item
finite groups,
\vspace{-\itemsep}
\item
abelian groups,
\vspace{-\itemsep}
\item
fundamental groups of two-dimensional oriented manifolds.
\vspace{-\itemsep}
\end{enumerate}
\end{thm}
{\bf Proof.} The first item is obvious --- free groups have no relations,
so every almost representation is a genuine representation.
The same argument works for the second item too.
The third item was proved in \cite{Grov} --- for finite groups there
exists a genuine representation close to every almost representation.
The fourth and the fifth items we prove in the next sections.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Case of abelian groups}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Let the set
$u_1,\ldots,u_k,v_1,\ldots,v_l\in U_n({\bf C})$ be an $\e$--almost
representation of the abelian group $\G={\bf Z}^k\oplus{\bf
Z}_{n_1}\oplus\ldots\oplus{\bf Z}_{n_l}$ generated by free generators
$a_1,\ldots,a_k$ and by torsion generators $b_1,\ldots,b_l$ with
$b_j^{n_j}=1$. Then one has $\norm{[u_i,u_j]}\leq\e$,
$\norm{[u_i,v_j]}\leq\e$, $\norm{[v_i,v_j]}\leq\e$ and
$\norm{v_j^{n_j}-{\bf 1} }\leq\e$.
\begin{thm}\label{alm->asym}
There exist constants $\e_0>0$ and $C>0$ such that for any
$\e$--almost representation of the group $\G$ with $\e<\e_0$ there exists
a dimension $N$ and
unitary paths $u_i(t),v_j(t)\in U_N({\bf C})$ such that
\begin{enumerate}
\item
$u_i(0)=u_i\oplus {\bf 1} _{N-n}$, $v_j(0)=v_j\oplus {\bf 1} _{N-n}$,
\vspace{-\itemsep}
\item
all the norms $\norm{[u_i(t),u_j(t)]}$,
$\norm{[u_i(t),v_j(t)]}$, $\norm{[v_j(t),v_i(t)]}$,
$\norm{v_j^{n_j}-{\bf 1} }$ do not exceed $C\e^{(1/4)^{k-1}}$,
\vspace{-\itemsep}
\item
$\norm{[u_i(1),u_j(1)]}\leq\e/2$ and $[u_i(1),v_j(1)]=[v_j(1),v_i(1)]=
v_j^{n_j}-{\bf 1} =0$.
\end{enumerate}
\end{thm}
\noindent
{\bf Proof.}
We start with proof of Theorem \ref{alm->asym} in the case of $k=2$,
$l=0$, i.e. for two almost commuting unitaries $u=u_1$ and $v=u_2$
being the images of two free generators of the group ${\bf Z}\oplus{\bf Z}$.
In the special case of the pair of the Voiculescu matrices it was shown in
\cite{mish-noor} that they generate an asymptotic representation of this
group.
\medskip\noindent
Our proof is based on several technical lemmas. We begin with these
lemmas in the first section and afterwards we will give an explicit
construction of the homotopy $u(t)=u_1(t)$ and $v(t)=u_2(t)$. The idea is
that if one of two almost commuting matrices is diagonal then the other
is almost of a block three-diagonal form (cf. \cite{manFA}) and
afterwards we use a variant of the Berg's technique (cf.~\cite{lor-berg}).
%\smallskip\noindent
The general case will easily follow from the case of two unitaries and we
deal with it at the end of the paper.
\medskip\noindent
We do not try to get the best possible estimates and the constants surely
can be chosen in a better way.
%\smallskip\noindent
For convenience we assume that all
$\e$, $\d_1$, $\d_2$ etc. take values in the interval $(0,1)$.
\begin{rmk}
{\rm
Let $A$ be a unital $C^*$--algebra of real rank zero~\cite{bp}.
If we change the complex numbers field ${\bf C}$ by $A$ and consider
almost commuting unitaries in matrix algebras over $A$ then Theorem
\ref{alm->asym} is valid too. The proof is the same since it is possible
to assume these unitaries to be linear combinations of projections.
The only difference is in Lemma \ref{lemma01}, and it is shown in
\cite{manFA} how to reduce elements of general $C^*$--algebras of
real rank zero to elements of the three-diagonal form with respect to an
appropriate choice of their spectral projections.
}
\end{rmk}
\subsection{Technical lemmas}\label{techn.lemmas}
%\setcounter{equation}{0}
We start with two unitaries $u,v\in U_n({\bf C})$ and fix a basis such that
the matrix $u$ is diagonal with diagonal entries $\l_i=e^{2\pi i\ph_i}$,
$i=1,\ldots,n$,
with ordered values: $\ph_i\in [0,1)$, $\ph_{i+1}\geq\ph_i$.
Let $\alpha$ be such that $|e^{2\pi i\alpha}-1|=\sqrt[4]{\e}$,
then the points $e^{2\pi ik\alpha}$, $k=0,1,\ldots$, divide the unit
circle into equal small arcs
$[e^{2\pi ik\alpha},e^{2\pi i(k+1)\alpha})$
(in fact the last arc $[e^{2\pi iK\alpha},1)$ may be shorter
than others but without loss of generality we can assume that $\e$ is such
that this last arc is of the same length too).
Denote those arcs which contain at least one eigenvalue $\l_i$ by
$\D_k$. Then the number $l$ of such arcs is not bigger then
$2\pi/\sqrt[4]{\e}$ and the following properties hold:
\begin{enumerate}
\item
if $\l_i,\l_j\in\D_k$ then $\v \l_i-\l_j\v <\sqrt[4]{\e}$;
\vspace{-\itemsep}
\item
if $\l_i\in\D_{k-1}$ and $\l_j\in\D_{k+1}$ then $\v \l_i-\l_j\v
\geq\sqrt[4]{\e}$.
\end{enumerate}
\noindent
Denote the orthogonal spectral projections of $u$ corresponding to the
arcs $\D_k$ by $p_k$, $p_1\oplus\ldots\oplus p_l={\bf 1} $. Then the matrix
algebra $M_n$ as a module over itself can be decomposed into a direct sum
corresponding to the above projections, $M_n=\oplus_{k=1}^lp_kM_n$ and we
can represent matrices from $M_n$ as smaller matrices of blocks with
regards to this decomposition: $u=\diag(\{u_k\})$ for $u_k=p_kup_k$ and
$v=(v_{km})$ for $v_{km}=p_kvp_m$.
\medskip\noindent
For a matrix $a=(a_{km})$ decomposed with respect to $\oplus_{k=1}^l
p_kM_n$ put
\be\label{tridiag}
d(a)=\left(\begin{array}{cccc}
a_{11}&a_{12}&&a_{1l}\\
a_{21}&a_{22}&\ddots&\\
&\ddots&\ddots&a_{l-1,l}\\
a_{l1}&&a_{l,l-1}&a_{l,l}
\end{array}\right)
\ee
where empty places are zeroes.
We call matrices of the form (\ref{tridiag}) block three-diagonal (we look
at the corner elements $a_{l1}$ and $a_{1l}$ as at continuations of the
diagonals $(a_{12},\ldots,a_{l-1,l})$ and $(a_{21},\ldots,a_{l,l-1})$
respectively).
\medskip\noindent
We begin by proving that the matrix
$v=(v_{km})$ is ``almost'' block three-diagonal.
\begin{lem}\label{lemma01}
If
\begin{equation}\label{ac1}
\norm{[u,v]}\leq\e
\end{equation}
then
$$
\norm{v-d(v)}\leq4\sqrt{\pi}\sqrt[4]{\e}.
$$
\end{lem}
\noindent
{\bf Proof.}
%\medskip\noindent
Divide once more the spectrum of the operator $u$ into smaller (than
$\D_k$) equal arcs $\ov{\D}_s=[d_s,d_{s+1})$ such that
$|d_s-\ov{\l}_s|=|\ov{\l}_s-d_{s+1}|=\e/2$, where $\ov{\l}_s$ are the
middle points of the arcs $\ov{\D}_s$.
Let $\ov{p}_s$ be the spectral projections of $u$ corresponding to
the arcs $\ov{\D}_s$. Then put
$$
\ov{u}=\sum_s \ov{\l}_s\ov{p}_s.
$$
Obviously $\norm{u-\ov{u}}\leq\e/2$, hence in view of (\ref{ac1})
\be\label{ac.2}
\norm{\ov{u}v-v\ov{u}}\leq \norm{\ov{u}v-uv}+
\norm{uv-vu}+\norm{vu-v\ov{u}}\leq 2\e.
\ee
Let $M_n=\oplus_s\ov{p}_sM_n$ be the decomposition of $M_n$
corresponding to the spectral projections $\ov{p}_s$ of $u$. It is a
sub-decomposition of $\oplus_{k=1}^lp_kM_n$ and the matrix $v$ can be
written also as $v=(w_{ij})$,
$w_{ij}=\ov{p}_i v\ov{p}_j$. The matrix entries $v_{km}$
can be viewed as blocks of elements $w_{ij}$.
Denote by $M$ the number of columns of the matrix $(w_{ij})$. Then one has
\be\label{N}
M\leq\frac{4\pi}{\e}.
\ee
\medskip\noindent
%{\bf Proof.}
Consider the matrix $(\ov{u}v-v\ov{u})(\ov{u}v-v\ov{u})^*$.
From the inequality
(\ref{ac.2}) it follows that the norm of this matrix
is less then $4\e^2$,
hence the norm of any element of this matrix is also less than $4\e^2$.
So, as $\ov{\l}_j$ commute with $w_{ij}$, we obtain for every $i$
\be\label{4}
\norm{\sum_{j=1}^M (\ov{\l}_i-\ov{\l}_j)^2 w_{ij}w^*_{ij}}\leq 4\e^2.
\ee
Let $\ov{\l}_i\in\D_{k_i}$, $\ov{\l}_j\in\D_{k_j}$.
As all the summands in (\ref{4}) are positive, so ignoring some
of them we will not increase the norm of the sum, hence
$$
\norm{\sum\nolimits'_j (\ov{\l}_i-\ov{\l}_j)^2 w_{ij}w^*_{ij}}\leq 4\e^2
$$
where the sum $\sum'$ is taken for those $j$ for which
$\v k_j- k_i\v\geq 2$, i.e. we throw away those $w_{ij}$
for which $(v-d(v))_{ij}=0$. As in the sum $\sum'$ we have
$\v\ov{\l}_k-\ov{\l}_j\v\geq\sqrt[4]{\e}$, so
$$
4\e^2>\norm{\sum\nolimits'_j
(\ov{\l}_i-\ov{\l}_j)^2w_{ij}w^*_{ij}}\geq\sqrt{\e}\norm{\sum\nolimits'_j
w_{ij}w^*_{ij}},
$$
hence
\be\label{3}
\norm{\sum\nolimits'_j w_{ij}w^*_{ij}}\leq 4\e^{3/2}.
\ee
To finish the proof we need the following
\begin{lem}\label{NN}
Let $A$ be a $C^*$--algebra and let
$a=(a_{ij})$, $a_{ij}\in A$ be a $M\times M$ matrix such that
for every $i$ one has $\norm{\sum_j a_{ij}a^*_{ij}}\leq\d_1^2$.
Then $\norm{a}\leq\d_1\sqrt{M}$.
\end{lem}
\noindent
{\bf Proof.}
Take $\xi=(\xi_k)$, $k=1,\ldots, M$, $\xi_i\in M_n$.
Then using the generalized
Cauchi-Schwartz inequality~\cite{lance} we get
\begin{eqnarray*}
\norm{a\xi}^2&=&\norm{\sum_{ijk}\xi^*_j a^*_{ij}a_{ik}\xi_k}
\leq\sum_i\norm{\sum_{kj}\xi_j^* a^*_{ij}a_{ik}\xi_k}\\
&\leq&\sum_i\norm{\sum_k a_{ik}a^*_{ik}}\cdot\norm{\sum_k \xi_k^*\xi_k}
\leq M\d_1^2 \norm{\xi}^2,
\end{eqnarray*}
hence we have $\norm{a\xi}\leq\sqrt{M}\d_1\norm{\xi}$. \q
\noindent
Now in view of (\ref{N}) and (\ref{3}) from Lemma \ref{NN}
(for $\d_1=2\e^{3/4}$) we get
$$
\norm{v-d(v)}\leq\sqrt{M}\ 2\e^{3/4}\leq
\sqrt{4\pi}\e^{-1/2}\,2\e^{3/4}= 4\sqrt{\pi}\sqrt[4]{\e}
$$
which ends the proof of Lemma \ref{lemma01}. \q
\noindent
The idea of chopping the space into spectral projections of one
operator and perturbing the other to be three-diagonal has appeared in
\cite{Dav}. Moreover, K. Davidson informed me that using the
Bhatia-Davis-McIntosh method \cite{Bhat} it is possible to get a better
estimate than in Lemma \ref{lemma01}.
\begin{rmk}\label{lacunas}
{\rm
If the spectrum of the matrix $u$ has a gap of length more
than $\sqrt[4]{\e}$ then without loss of generality we can assume that
this gap has zero as its central point. In that case norm of the elements
$v_{1l}$ and $v_{l1}$ is small too and we have
$$
\norm{v-\left(\begin{array}{cccc}
v_{11}&v_{12}&&\\
v_{21}&v_{22}&\ddots&\\
&\ddots&\ddots&v_{l-1,l}\\
&&v_{l,l-1}&v_{l,l}
\end{array}\right)}\leq 4\sqrt{\pi}\sqrt[4]{\e}.
$$
If there exists another gap of length bigger than $\sqrt[4]{\e}$ in
the spectrum of $u$ (say between $\D_r$ and $\D_{r+1}$) then the matrix
$d(v)$ is close to a matrix decomposable into a direct sum of smaller
three-diagonal matrices because the norm of elements $v_{r+1,r}$ and
$v_{r,r+1}$ is small enough.
}
\end{rmk}
%\medskip\noindent
Put
$$
c=c(t)=e^{\pi it/2}\cos{\pi t/2},\qquad s=s(t)=-ie^{\pi it/2}\sin{\pi
t/2},\qquad t\in [0,1].
$$
Then both paths
$\left(\begin{array}{cc}c&s\\s&c\end{array}\right)$ and
$\left(\begin{array}{cc}\ov{c}&\ov{s}\\\ov{s}&\ov{c}\end{array}\right)$
connect matrices $\left(\begin{array}{cc}1&0\\0&1\end{array}\right)$
and $\left(\begin{array}{cc}0&1\\1&0\end{array}\right)$ in the group
$U_2({\bf C})$.
For a matrix
\be\label{3diag}
w=\left(\begin{array}{cccc}
v_{11}&v_{12}&&v_{1l}\\
v_{21}&v_{22}&\ddots&\\
&\ddots&\ddots&v_{l-1,l}\\
v_{l1}&&v_{l,l-1}&v_{l,l}
\end{array}\right)
\ee
and for an integer $m$
define a path $w(t)$ by formula %$w(t)=$
%{\small
\be\label{bigmatr}
w(t){=}\!
\left(\!\!
\begin{array}{ccccccccccccc}
v_{11}&v_{12}&&cv_{1l}&&&&&&&&&sv_{1l}\\
v_{21}&v_{22}&\ddots&&&&&&&&&&\\
&\ddots&\ddots&\!\!v_{l{-}1,l}\!\!&&&&&&&&&\\
\ov{c}v_{l1}&&\!\!v_{l,l{-}1}\!\!&v_{l,l}&\ov{s}v_{l1}&&&&&&&&\\
&&&sv_{1l}&v_{11}&v_{12}&&cv_{1l}&&&&&\\
&&&&v_{21}&v_{22}&\ddots&&&&&&\\
&&&&&\ddots&\ddots&\!\!v_{l{-}1,l}\!\!&&&&&\\
&&&&\ov{c}v_{l1}&&\!\!v_{l,l{-}1}\!\!&v_{l,l}&\ov{s}v_{1l}&&&&\\
&&&&&&&sv_{1l}&\ddots&&&&\\
&&&&&&&&&v_{11}&v_{12}&&cv_{1l}\\
&&&&&&&&&v_{21}&v_{22}&\ddots&\\
&&&&&&&&&&\ddots&\ddots&\!\!v_{l{-}1,l}\!\!\\
\ov{s}v_{l1}&&&&&&&&&\ov{c}v_{1l}&&\!\!v_{l,l{-}1}\!\!&v_{l,l}\\
\end{array}\!\!
\right)
\ee
%}
\noindent
where for $t=0$ the matrix $w$ is repeated $m$ times. Let $N=nm$ be the
dimension of $w(t)$. We show that the path $w(t)$ lies close to the
unitary group $U_N({\bf C})$ and that the corresponding estimate does not
depend on the number $m$.
\begin{lem}\label{bigmatrix}
Suppose that $\norm{v-w}\leq\d_2$ and $\norm{[u,w]}\leq\d_3$. Let
$w(t)=\til{w}(t)h(t)$ be the polar decomposition with unitary $\til{w}(t)$.
Then
$$
\norm{w(t)-\til{w}(t)}\leq 100\d_2 \quad{\it and}\quad \norm{[u\oplus
{\bf 1} _{N-n},w(t)]}\leq \d_3+4\sqrt[4]{\e}.
$$
\end{lem}
\noindent
{\bf Proof.}
Notice that the inequality $\norm{v-w}\leq\d_2$ implies
$\norm{{\bf 1} -v^*w}\leq\d_2$.
Let $w=\til{w}h$ be the polar decomposition. Then we have
\be\label{ner-vo1}
\norm{{\bf 1} -v^*w}=\norm{\til{w}^*v-h}\leq\d_2.
\ee
Passing to adjoints we obtain
\be\label{ner-vo2}
\norm{\til{w}^*v-v^*\til{w}}\leq 2\d_2,
\ee
hence from (\ref{ner-vo1}) we get
$$
\norm{\til{w}^*v-{\bf 1} }\leq 2\d_2,
$$
and in view of (\ref{ner-vo1})
$$
\norm{h-{\bf 1} }\leq 3\d_2.
$$
Therefore we get $\norm{h^2-{\bf 1} }\leq (1+3\d_2)^2-1\leq 15\d_2$, hence
$$
\norm{w^*w-{\bf 1} }\leq 15\d_2.
$$
Notice that the matrix $w^*w-{\bf 1} $ is 9-diagonal (or 5-diagonal, if we
think of the upper right and lower left corners of this matrix as of
continuations of the diagonals close to the main diagonal), therefore the
norm of every diagonal is not more than $15\d_2$.
\medskip\noindent
Now look at the matrix $w^*(t)w(t)-{\bf 1} $. It is a 13-diagonal matrix and
it is easily seen that every diagonal of this matrix is a direct sum of
diagonals of the matrix $w^*w-{\bf 1} $ (possibly with coefficients like
$c(t)$ and $s(t)$), hence we get the estimate
$$
\norm{w^*(t)w(t)-{\bf 1}}\leq 13\cdot 15\d_2=195\d_2,
$$
i.e. $\norm{h^2(t)-{\bf 1} }\leq 195\d_2$.
Then we have $\norm{h(t)-{\bf 1} }\leq 100\d_2$ and finally
$$
\norm{w(t)-\til{w}(t)}=\norm{\til{w}^*(t)w(t)-{\bf 1} }
=\norm{h(t)-{\bf 1} }\leq 100\d_2.
$$
\medskip\noindent
Now we should estimate the commutator $[u\oplus {\bf 1} _{N-n},w(t)]$.
But it is easy to see that
\begin{eqnarray*}
\norm{[u\oplus {\bf 1} _{N-n},w(t)]}&\leq&\norm{[u,w]}\\
&+&\norm{v_{l1}-v_{l1}u_l}+
\norm{u_1v_{l1}-v_{l1}}+\norm{v_{1l}-v_{1l}u_1}+\norm{u_lv_{1l}-v_{1l}}\\
&<&\d_3+2\norm{u_1-{\bf 1} }+2\norm{u_l-{\bf 1} }\leq\d_3+4\sqrt[4]{\e},
\end{eqnarray*}
where $u_1$ and $u_l$ are the first and the last diagonal entries of the
matrix $u$.\q
\begin{lem}\label{2matr.}
Let $w_1,w_2\in U_n({\bf C})$ be almost commuting block 3-diagonal
matrices of the form $($\ref{3diag}$)$, $\norm{[w_1,w_2]}\leq\d_4$.
Define for
both these matrices paths $w_1(t)$ and $w_2(t)$ by $($\ref{bigmatr}$)$.
Then one has
$$
\norm{[w_1(t),w_2(t)]}\leq 13\d_4.
$$
\end{lem}
\noindent
{\bf Proof.}
It can be easily checked that the commutator $[w_1(t),w_2(t)]$ can be
decomposed into sum of the direct sum of smaller commutators $[w_1,w_2]$
and of 12-diagonal matrix with diagonals coinciding with direct sums of
certain diagonals of the matrix $[w_1,w_2]$.\q
\begin{lem}
Let
$$
w'=\left(\begin{array}{cccc}
w_{11}&w_{12}&&w_{1N}\\
w_{21}&w_{22}&\ddots&\\
&\ddots&\ddots&w_{N-1,N}\\
w_{N1}&&w_{N,N-1}&w_{N,N}
\end{array}\right)\in M_N({\bf C})
$$
be a three-diagonal matrix and let
$w'=\til{w}'h'$ be the polar decomposition such that
\be\label{otsen}
\norm{w'-\til{w}'}\leq\d_5.
\ee
Then for any $\d_6>0$ there exist an integer
$r$ $($not depending on $N$$)$ and a decomposition
$$
\til{w}'=w_0+w_1
$$
such that the matrix $w_0$ is $(2r+3)$--diagonal and $\norm{w_1}\leq\d_6$.
\end{lem}
\noindent
{\bf Proof.}
Consider the Taylor formula
\begin{eqnarray*}
\til{w}'&=&w'((w')^*w')^{-1/2}=w'({\bf 1} +((w')^*w'-{\bf 1} ))^{-1/2}\\
&=&w'({\bf 1} +a_1((w')^*w'-{\bf 1} )+a_2((w')^*w'-{\bf 1} )^2+
\ldots+a_r((w')^*w'-{\bf 1} )^r+R_r),
\end{eqnarray*}
where in view of (\ref{otsen}) we have
$$
\norm{R_r}\leq\norm{(w')^*w'-{\bf 1} }^r=\norm{(h')^2-{\bf 1} }^r\leq
(3\d_5)^r.
$$
Take $r$ to satisfy
\be\label{r}
(3\d_5)^r\leq\d_6
\ee
and put
\begin{eqnarray*}
&w_0=w'({\bf 1} +a_1((w')^*w'-{\bf 1} )+a_2((w')^*w'-{\bf 1} )^2+
\ldots+a_r((w')^*w'-{\bf 1} )^r),&\\
&w_1=w'R_r.&
\end{eqnarray*}
Then as $w'$ and $((w')^*w'-{\bf 1} )$ are three-diagonal and
$(2r+1)$--diagonal respectively, so the matrix $w_0$ is
$(2r+3)$--diagonal.\q
\begin{lem}\label{cont}
Let $v'(t)$ be a path in the matrix algebra $M_N({\bf C})$ such that
$$
\dist(v'(t), U_N({\bf C}))\leq\d_7.
$$
Then there exists a unitary path
$v(t)$ such that for every $t$ one has
$$
\norm{v(t)-v'(t)}\leq 3\d_7.
$$
\end{lem}
\noindent
{\bf Proof.}
Let $v'\in M_N({\bf C})$ be a matrix with $\dist(v',U_N({\bf C}))\leq\d_7$,
and let $v'=vh$ be its polar decomposition. By supposition there exists
a unitary $w$ such that $\norm{v'-w}\leq\d_7$. Then $\norm{vh-w}\leq\d_7$,
hence (as in proof of Lemma \ref{bigmatrix}) we get the estimates
\be\label{soprnorm1}
\norm{h-v^{-1}w}\leq\d_7,
\ee
and
\be\label{soprnorm2}
\norm{h-(v^{-1}w)^*}\leq\d_7.
\ee
It follows from (\ref{soprnorm1}) and (\ref{soprnorm2}) that
$$
\norm{v^{-1}w-(v^{-1}w)^*}\leq 2\d_7,
$$
hence
\be\label{soprnorm3}
\norm{v^{-1}w-{\bf 1} }\leq 2\d_7.
\ee
Estimates (\ref{soprnorm1}) and (\ref{soprnorm3}) imply
$\norm{h-{\bf 1} }\leq 3\d_7$,
and finally
$$
\norm{v'-v}=\norm{h-{\bf 1} }\leq 3\d_7.
$$
Now if we have a path $v'(t)$ with the polar decomposition $v'(t)=v(t)h(t)$
then the path $v(t)$ is continuous and $\norm{v(t)-v'(t)}\leq 3\d_7$.\q
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Construction of homotopy}
\subsubsection{Case of two unitaries}\label{paths}
%\setcounter{equation}{0}
In this section we give an explicit construction of the paths $u(t)$ and
$v(t)$ satisfying Theorem \ref{alm->asym}. This construction consists
of four steps.
\medskip\noindent
We start with two unitaries $u,v\in U_n({\bf C})$ with $\norm{[u,v]}\leq\e$
and we assume that $u$ is already diagonal with ordered eigenvalues (cf.
section \ref{techn.lemmas}). Let $N=nm$ where the number $m$ will be
defined later. The homotopy which we are constructing should begin at the
pair of unitaries $u\oplus {\bf 1} _{N-n}$ and $v\oplus {\bf 1} _{N-n}$.
In fact we will
construct paths $u(t)$ and $v'(t)$ with non-unitary $v'(t)$ but this
latter path will be close to the unitary group,
$\dist(v'(t),U_{N})\leq C\sqrt[4]{\e}$ for some constant $C$.
By Lemma \ref{cont} this would imply
existence of exactly unitary path $v(t)$ with the necessary properties.
At every step of the homotopy we should control two estimates --- the
commutator norm $\norm{[u(t),v'(t)]}$ and the distance from the unitary
group $\dist(v'(t),U_N({\bf C}))$.
\subsubsection*{Step 1}
Put $w=d(v)$ defined by (\ref{tridiag}).
While not changing the first matrix $u\oplus {\bf 1} _{N-n}$, we connect the
matrix $v\oplus {\bf 1} _{N-n}$ by the
linear path with the block 3-diagonal matrix
$w\oplus {\bf 1} _{N-n}$. At this step we have
$$
\norm{[u(t),v'(t)]}\leq 4\sqrt[4]{\e}
\quad{\rm and}\quad
\dist(v'(t),U_N({\bf C}))\leq\norm{v-w}\leq 4\sqrt{\pi}\sqrt[4]{\e}.
$$
\subsubsection*{Step 2}
Connect the unit matrix ${\bf 1} _{N-n}$ with the direct sum
$w\oplus\ldots\oplus
w$ of $m-1$ copies of the matrix $w$. The matrix $u\oplus {\bf 1} _{N-n}$ is
still fixed at the second step of homotopy and both the commutator norm
$\norm{[u(t),v'(t)]}$ and $\dist(v'(t),U_N({\bf C}))$ are
not being changed.
\subsubsection*{Step 3}
We still do not change the matrix $u\oplus {\bf 1} _{N-n}$, and define the
path $v'(t)$ on the third step by $v'(t)=w(t)$, where $w(t)$ is given by
(\ref{bigmatr}). Then by Lemma \ref{bigmatrix} (taking there
$\d_1=4\sqrt{\pi}\sqrt[4]{\e}$ and $\d_2=4\sqrt[4]{\e}$) we get the
estimates
$$
\norm{u\oplus {\bf 1} _{N-n},v'(t)}\leq 8\sqrt[4]{\e}
\quad {\rm and}\quad
\dist(v'(t),U_N({\bf C}))\leq\norm{v'(t)-v(t)}\leq 400\sqrt{\pi}\sqrt[4]{\e}
$$
for $v(t)=v'(t)(v'^*(t)v'(t))^{-1/2}$.
\subsubsection*{Step 4}
At the last step we will not change the second matrix $v(1)$.
The matrix $v'(1)$ is 3-diagonal and
$\norm{v'(t)-v(t)}\leq 400\sqrt{\pi}\sqrt[4]{\e}=\d_5$, so by Lemma
\ref{bigmatrix} for $\d_6=\e/8$ we have a decomposition $v(t)=w_0+w_1$
with $\norm{w_1}\leq\e/8$ and with $w_0$ being $(2r+3)$--diagonal for $r$
satisfying the inequality (\ref{r}), i.e.
$$
(1200\sqrt{\pi}\sqrt[4]{\e})^r\leq\e/8.
$$
The choice of such $r$ is possible
if $\e<\e_0$, where $\e_0$ satisfies $1200\sqrt{\pi}\sqrt[4]{\e_0}\leq 1$.
\medskip\noindent
Then $\norm{[u(t),v(1)]}\leq\norm{[u(t),w_0]}+\e/4$ and it will be
sufficient to find a path $u(t)$ starting from $u\oplus {\bf 1} _{N-n}$ such
that
$$
\norm{[u(1),w_0]}\leq\e/4.
$$
For $t\in [0,1]$ define diagonal paths $u_j(t)\in U_n({\bf C})$ by
$$
u^{(1)}(t)=\left(\begin{array}{ccc}
e^{2\pi i((1-t)\ph_1+t\ph_1/m)}&&\\
&\ddots&\\
&&e^{2\pi i((1-t)\ph_n+t\ph_n/m)}
\end{array}\right)
$$
and
$$
u^{(j)}(t)=\left(\begin{array}{ccc}
e^{2\pi it\ph_1j/m}&&\\
&\ddots&\\
&&e^{2\pi it\ph_nj/m}
\end{array}\right)
,\quad j=2,\ldots,m.
$$
Put $u(t)=u^{(1)}(t)\oplus\ldots\oplus u^{(m)}(t)\in U_N({\bf C})$.
Then the diagonal matrix $u(1)$ will consist of all ordered power $m$
roots of eigenvalues $e^{2\pi i\ph_i}$ of the matrix $u$.
\medskip\noindent
Denote by $d_k(w_0)$ the diagonal of the matrix $w_0$ lying $k$ lines above
(or below if $k$ is negative) the main diagonal $d_0(w_0)$.
Then
$$
w_0=\sum_{k=-r-1}^{r+1}d_k(w_0)
$$
and $\norm{d_k(w_0)}\leq 1+\e/8$.
It is easy to see (cf. remark \ref{lacunas} in the case of lacunas in
the spectrum of $u$) that
$$
\norm{[u(t),d_k(w_0)]}\leq (1+\e/8)\cdot 2k\sqrt[4]{\e}
\cdot\left(1-t+\frac{t}{m}\right),
$$
therefore
$$
\norm{[u(t),w_0]}\leq
2\,(1+\e/8)\left(\sum_{k=1}^{r+1}k\right)\sqrt[4]{\e}\
\left(1-t+\frac{t}{m}\right)
\leq 2\,(1+\e/8)\ (r+1)^2\ \sqrt[4]{\e}\ \left(1-t+\frac{t}{m}\right).
$$
So along the whole path we have the necessary estimate
$$
\norm{[u(t),v(t)]}\leq 3(r+1)^2\sqrt[4]{\e}.
$$
To finish the construction of homotopy we have to choose the number $m$ so
that it would satisfy the inequality
$$
2\,(1+\e/8)\ (r+1)^2\ \sqrt[4]{\e}\ \frac{1}{m}<\e/4,
$$
then we will have $\norm{[u(1),w_0]}\leq\e/4$, hence
$\norm{[u(1),v(1)]}\leq\e/2$. The case of two almost commuting unitaries is
proved.
\subsubsection{Case of free abelian group}
Now we can prove the general case of $k$ almost commuting unitaries
$u_1,\ldots,u_k$, $\norm{[u_i,u_j]}\leq\e$. As in the two matrices case
we diagonalize the first matrix $u_1$ and repeat the previous construction
for the other unitaries $u_2,\ldots,u_k$
(using Lemma \ref{2matr.} we can control the commutator norm for these
unitaries). Then we will get the unitary
paths $u_i(t)$ such that $u_i(0)=u_i\oplus {\bf 1} _{N-n}$,
$u_i(1)=u_i^{(1)}$,
with properties
\begin{enumerate}
\item
$\norm{[u_1^{(1)},u_j^{(1)}]}\leq\e'$,\quad $j\neq 1$,
\vspace{-\itemsep}
\item
$\norm{[u_i^{(1)},u_j^{(1)}]}\leq C\sqrt[4]{\e}$,
\vspace{-\itemsep}
\item
$\norm{[u_i(t),u_j(t)]}\leq C\sqrt[4]{\e}$
\end{enumerate}
with small enough $\e'>0$ which we define below and with some constant
$C$. Then at the second step we diagonalize the matrix $u_2^{(1)}$ and
repeat the above construction to the matrices
$u_1^{(1)},u_3^{(1)},\ldots,u_k^{(1)}$. Then we will get the matrices
$u_1^{(2)},\ldots,u_k^{(2)}$ satisfying
\begin{enumerate}
\item
$\norm{[u_2^{(2)},u_j^{(1)}]}\leq\e'$,\quad $j\neq 2$,
\vspace{-\itemsep}
\item
$\norm{[u_1^{(2)},u_j^{(2)}]}\leq C\sqrt[4]{\e'}$,\quad $j\neq 1,2$,
\vspace{-\itemsep}
\item
$\norm{[u_i^{(2)},u_j^{(2)}]}\leq C^2\sqrt[16]{\e}$,\quad $i,j\neq 1,2$,
\vspace{-\itemsep}
\item
$\norm{[u_i(t),u_j(t)]}\leq C^2\sqrt[16]{\e}$.
\end{enumerate}
Repeating this procedure $k-1$ times we finally get the matrices
$u_1^{(k-1)},\ldots,u_k^{(k-1)}$ with properties
\begin{enumerate}
\item
$\norm{[u_1^{(k-1)},u_j^{(k-1)}]}\leq C^{k-2}(\e')^{(1/4)^{k-2}}$,
\quad $j>1$,
\vspace{-\itemsep}
\item
$\norm{[u_2^{(k-1)},u_j^{(k-1)}]}\leq C^{k-3}(\e')^{(1/4)^{k-3}}$,
\quad $j>2$,\quad etc.,
\vspace{-\itemsep}
\item
$\norm{[u_{k-1}^{(k-1)},u_k^{(k-1)}]}\leq\e'$,
\vspace{-\itemsep}
\item
$\norm{[u_i(t),u_j(t)]}\leq C^{k-1}\e^{(1/4)^{k-1}}$.
\end{enumerate}
So, we finally have $\norm{[u_i^{(k-1)},u_j^{(k-1)}]}\leq
C^{k-2}(\e')^{(1/4)^{k-2}}$ for all $i,j$.
Taking $\e'$ to satisfy the inequality
$C^{k-2}(\e')^{(1/4)^{k-2}}\leq\e/2$ we obtain the
statement of the theorem.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{General case}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\noindent
In this section we have to deal with the torsion generators
$v_1,\ldots,v_l$. In the following
we refer to the paper \cite{Grov} though it is possible to give the
explicit construction.
\begin{thm}[\cite{Grov}]
For any $\e$ and for any set of unitaries $v_1,\ldots,v_l\in U_n({\bf C})$
with $\norm{[v_i,v_j]}\leq\e$, $\norm{v_j^{n_j}-{\bf 1} }\leq\e$ there exists
$\d(\e)$ and unitaries $v'_1,\ldots,v'_l\in U_n({\bf C})$ such that
$\d(\e)$ tends to zero when $\e\to 0$ and $[v'_i,v'_j]=0$,
$(v'_j)^{n_j}={\bf 1} $. \q
\end{thm}
\noindent
If $u_1,\ldots,u_k,v_1,\ldots,v_l$ is an $\e$--almost representation
of the abelian group $\G$
then we can connect the unitaries $v_1,\ldots,v_l$ with
$v'_1,\ldots,v'_l$ and afterwards we will have
$\norm{[u_i,v'_j]}\leq\e+2\d=\d'$.
Denote by $p_r$ the spectral projections
for the set of commuting unitaries $v'_j$ and let $N$ be the minimal
common divisor of the degrees $n_j$, $j=1,\ldots,l$. We can assume that
$\e$ is much smaller than the minimal distance between eigenvalues of the
unitaries $v'_1,\ldots,v'_l$ which equals to $|e^{2\pi i/N}-1|$. Let
$u_{i;rs}=p_ru_ip_q$ be the matrix blocks of $u_i$ with respect to the
decomposition $\oplus_r p_r={\bf 1} $. Then it is easy to see from the
estimate $\norm{[u_i,v'_j]}\leq\d'$ that
\be\label{grub}
\norm{u_{i;rs}}\leq\frac{\d'}{|e^{2\pi i/N}-1|},\quad r\neq s.
\ee
Denote now by $\til{u}_i$ the diagonal matrices consisting of the diagonal
entries of $u_i$, $\til{u}_i=\diag\{u_{i;11},u_{i;22},\ldots\}$. Then
by summing the inequalities (\ref{grub})
one gets the following (very rough) estimate:
\begin{lem}
One has $\norm{u_i-\til{u}_i}\leq N^2\frac{\d'}{|e^{2\pi i/N}-1|}$. \q
\end{lem}
\noindent
So the distance from $\til{u}_i$ to the unitary group is small enough, hence
there exist unitaries $u'_{i;rr}\in p_r M_n p_r$ with
$\norm{u_{i;rr}-u'_{i;rr}}\leq N^3\d'=\d''$
(remember that $N$ does not depend on dimension of the almost
represenation). Put $u'_i=\diag\{u'_{i;11},u'_{i;22},\ldots\}$. Then
$\norm{u'_i-u_i}\leq \d''$, so we have $\norm{[u'_i,u'_j]}\leq 2\d''$, and
as the matrices $u'_i$ are block-diagonal, so they exactly commute with
$v'_j$: $[u'_i,v'_j]=0$.
\medskip\noindent
Now we have a $2\d''$--almost representation of the group $\G$ where all
relations hold {\it exactly} except the commutators $[u'_i,u'_j]$.
After connecting $u_i$ with $u'_i$ and $v_j$ with $v'_j$ we can
proceed as in the case of free abelian groups. Let take a basis such that
the matrix $u'_1$ is diagonal with eigenvalues being ordered.
Then as $v'_j$ exactly
commute with $u'_i$, so they are block-diagonal,
$$
v'_j=\diag\{v'_{j;11},v'_{j;22},\ldots\}
$$
and every diagonal entry satisfies $(v'_{j,rr})^{n_j}={\bf 1} $.
After applying our construction of diminishing commutators by
increasing the dimension
we can for any $\e''>0$ get matrices
$u''_i,v''_j$ of bigger dimension with the properties
$$
[u''_i,v''_j]=[v''_i,v''_j]=(v''_j)^{n_j}-{\bf 1} =0,
$$
$$
\norm{[u''_1,u''_i]}\leq\e'',\quad {\rm for}\ \ i>1
$$
and
$$
\norm{[u''_i,u''_j]}\leq C\sqrt[4]{\d''}.
$$
Then proceeding in the same way with $u_2,u_3,$ etc. we get the necessary
paths. \q
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Case of fundamental groups of oriented surfaces}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Let $\G=\la a_1,b_1,\ldots,a_m,b_m|a_1b_1a_1^{-1}b_1^{-1}\cdot\ldots\cdot
a_mb_ma_m^{-1}b_m^{-1}\ra$. Let $\s$ be an $\e$-almost representation of
$\G$, $u_i=\s(a_i)$, $v_i=\s(b_i)$, $u_i,v_i\in U_n$, $i=1,\ldots,m$.
Denote $\g(u,v)=uvu^{-1}v^{-1}$, then we have
$$
\norm{\g(u_1,v_1)\cdot\ldots\cdot \g(u_m,v_m)-I}\leq\e.
$$
Consider the map
\begin{equation}\label{map_gamma}
\g:U_n\times U_n\arr SU_n.
\end{equation}
To prove the property AGA for fundamental groups of oriented
two-dimensional manifolds we have to use the following elementary
statement about the map (\ref{map_gamma}).
\begin{lem}\label{lem:podniat}
Let $(u_0,v_0)\in U_n\times U_n$ and let $c(t)\in SU_n$, $t\in [0,1]$, be
a path such that $\g(u_0,v_0)=c(0)$. Then for any $\d>0$ there
exists a path $(u_t,v_t)\in U_n\times U_n$ such that
$\norm{\g(u_t,v_t)-c(t)}<\d$.
\end{lem}
{\bf Proof.}
Remember that a pair $(u,v)\in U_n\times U_n$ is called {\it irreducible}
if there is no common invariant subspace for $u$ and $v$.
It was shown in \cite{exel} that the set of regular points for the map
$\g$ (\ref{map_gamma}) coincides with the set of irreducible pairs.
Denote the set of reducible pairs $(u,v)\in U_n\times U_n$ by $S$.
For any $k=1,\ldots,n-1$ by $\Sigma_k\subset SU_n$ denote the set of
block-diagonal matrices
$\left(\begin{array}{cc}c_1&0\\0&c_2\end{array}\right)$ with respect to
some invariant subspace $V$, $\dim V=k$, such that $c_1\in SU_k$, $c_2\in
SU_{n-k}$. Put $\Sigma=\cup_k\Sigma_k\subset SU_n$. Then obviously
$\g(S)\subset \Sigma$. Notice that every $\Sigma_k$ is a submanifold
in $SU_n$ with codimension one. So $\Sigma$ divides $SU_n$ into a finite
set of closed path components $M_j$, $\cup_j M_j=SU_n$, and for every point
$c\in M_j$ the set $\g^{-1}(c)$ consists only of regular points. Hence
every path in $M_j$ transversal to its boundary can be lifted up to a path
in $U_n\times U_n$ with a fixed starting point.
Without loss of generality we can assume that the path $c(t)$ is
transversal to every $\Sigma_k$. Let $t_0\in \{c(t)\}\cap\Sigma_k$. It
remains to show that we can lift the path $c(t)$ in some neighborhood of
the point $c_0=c(t_0)$. Let $(u_0,v_0)\in U_n\times U_n$ be such point
that $\g(u_0,v_0)=c_0$. If the point $(u_0,v_0)$ is a regular point then
the statement is obvious. Otherwise we can write
$$
u_0=\left(\begin{array}{cc}u_1&0\\0&u_2\end{array}\right),\qquad
v_0=\left(\begin{array}{cc}v_1&0\\0&v_2\end{array}\right)
$$
with respect to some basis and we can assume that matrices $v_1$ and $v_2$
are diagonal. Let
$$
e^{2\pi i\ph_1},\ldots,e^{2\pi i\ph_k}\quad{\rm and}\quad
e^{2\pi i\ph_{k+1}},\ldots,e^{2\pi i\ph_n}
$$
be the eigenvalues of
$v_1$ and $v_2$ respectively.
Slightly changing $v_0$ we can assume that for all $i=2,\ldots,k$,
$j=k+2,\ldots,n$ the values $\ph_i-\ph_1$, $\ph_j-\ph_{k+1}$ differ from
each other. Multiplying $v_1$ by $e^{-2\pi i\ph_1 t}$ and $v_2$ by
$e^{-2\pi i\ph_{k+1} t}$, $t\in [0,1]$, we connect the matrix $v_0$ with
the matrix
$$
v'_0=\left(\begin{array}{cc}v'_1&0\\0&v'_2\end{array}\right)=
\left(\begin{array}{cc}e^{-2\pi i\ph_1}v_1&0\\
0&e^{-2\pi i\ph_{k+1}}v_2\end{array}\right)
$$
which has two eigenvalues equal to one and all other eigenvalues being
different from each other. Obviously the value
$\g(u_0,v_0)=\g(u_0,v'_0)=c_0$ does not change along this path.
Denote by $e_1,\ldots,e_n$ the basis consisting of the eigenvalues of
$v'_0$ and let $w(t)\in U_n$, $t\in[0,1]$, be a rotation of the vectors
$e_1$ and $e_{k+1}$:
$$
w(t)e_1=\cos t e_1-\sin t e_{k+1},\quad
w(t)e_{k+1}=\sin t e_1+\cos t e_{k+1},\quad
w(t)e_j=e_j \quad {\rm for}\quad j\neq 1,k+1.
$$
Obviously $w(t)$ commutes with $v'_0$. Put $u_t=u_0w(t)$. Then
$$
\g(u_t,v'_0)=u_0w(t)v'_0w^{-1}(t)u_0^{-1}(v'_0)^{-1}=\g(u_0,v'_0)=c_0
$$
and for $\sin t\neq 0$ the pair $(u_t,v'_0)$ is {\it irreducible}
(since $v'_0$ is diagonal with
only two coinciding eigenvalues, so its invariant subspaces are easy to
describe, then it is easy to check that they are not invariant under the
action of $u_t$) with the same value of $\g$. Then it is possible to
extend the path $(u_t,v_t)$ through the point $c_0$. \q
\begin{prop}\label{homot_to_1}
Any $\e$-almost representation of $\G$ is homotopically equivalent in
$R_{2\e}(\G)$ to an $2\e$-almost representation with
$u_2=v_2=\ldots=u_m=v_m=I$.
\end{prop}
{\bf Proof.}
Connect the matrix $\g(u_m,v_m)$ with $I$ by a path $c_m(t)$. Then by the
Lemma \ref{lem:podniat} we can find a path $(u_m(t),v_m(t))\in U_n\times
U_n$ such that
$$
\norm{\g(u_m(t),v_m(t))-c_m(t)}\leq\frac{\e}{2m}.
$$
Notice
that the set $\g^{-1}(I)=\{(u,v):uv=vu\}$ is path-connected, so we can
assume that the end point of the path $(u_m(t),v_m(t))$ is $(I,I)$.
Put
$$
c_{m-1}(t)=\g(u_{m-1},v_{m-1})c^{-1}_m(t).
$$
Again by the Lemma
\ref{lem:podniat} we can find a path $(u_{m-1}(t),v_{m-1}(t))\in
U_n\times U_n$ such that
$$
\norm{\g(u_{m-1}(t),v_{m-1}(t))-c_{m-1}(t)}\leq\frac{\e}{2m}.
$$
Then
$$
\norm{\g(u_1,v_1)\cdot\ldots\cdot\g(u_{m-1}(t),v_{m-1}(t))
\cdot\g(u_m(t),v_m(t))-I}\leq\e+\frac{\e}{m}
$$
and at the end point we have $(u_m(t),v_m(t))=(I,I)$. Proceeding by
induction we finish the proof.\q
It now follows from the proposition \ref{homot_to_1} that the property AGA
for the group $\G$ follows from the same property for the group ${\bf Z}^2$
with generators $u_1,v_1$. \q
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Example of a group without AGA}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Consider the group $\G=\la a,b,c|aca^{-1}c^{-1},b^2,(ab)^2\ra$.
\begin{thm}\label{primergruppy}
The group $\G$ does not possess the property AGA.
\end{thm}
{\bf Proof.}
Put $\o=e^{2\pi i/n}$ and define a
family of almost representations $\s_n$ taking values in $U_n$ by
$$
\s_n(a){=}\!\left(\begin{array}{cccccc}
\o&&&&&\\
&\o^2&&&&\\
&&\cdot&&&\\
&&&\cdot&&\\
&&&&\cdot&\\
&&&&&\o^n
\end{array}\right)\!\!,\
%\quad
\s_n(c){=}\!\left(\begin{array}{cccccc}
0&&&&&1\\
1&0&&&&\\
&1&\cdot&&&\\
&&\cdot&\cdot&&\\
&&&\cdot&\cdot&\\
&&&&1&0
\end{array}\right)\!\!,\
%\quad
\s_n(b){=}\!\left(\begin{array}{cccccc}
&&&&&1\\
&&&&1&\\
&&&\cdot&&\\
&&\cdot&&&\\
&\cdot&&&&\\
1&&&&&
\end{array}\right)\!\!.
$$
Here the matrices $\s_n(a)$ and $\s_n(c)$ are the Voiculescu matrices
\cite{voi} with the winding number \cite{e-l} equal to one, and one has
$$
\s_n(a)\s_n(c)\s_n(a)^{-1}\s_n(c)^{-1}=\o\cdot I,\quad
\s_n(b)^2=I,\quad (\s_n(a)\s_n(b))^2=\o\cdot I,
$$
so
$$
\e_n=\nnn{\s_n}=|\o-1|\to 0 \quad{\rm when}\ \ n\to\i,
$$
hence for every $\e>0$ there exists an $\e$-almost representation $\s$ of
$\G$ such that the winding number of the pair $(\s(a),\s(c))$ equals one.
\smallskip
Suppose the opposite, i.e. that the group $\G$ has AGA. Then there should
be such $\e$ that the number $\d(\e)\leq 1$. Take such $\e$ and an
$\e$-almost representation $\s_0$ with a non-zero winding number of the
pair $(\s_0(a),\s_0(c))$.
\smallskip
By supposition there exists an asymptotic representation
$\s_t\in R_{asym}(\G)$ extending $\s_0$ such that
$\nnn{\s_t}\leq 1$ for all $t\in [0,\i)$ and for any $\e'>0$ there exists
$t_0$ such that $\nnn{\s_{t_0}}\leq \e'$. Fix this $t_0$ and denote
$\s_{t_0}$ by $\s$. Let $n$ and $n+m$ be the dimension of the almost
representation $\s_0$ and $\s$ respectively.
Then one has
$$
\norm{\s(a)\s(c)-\s(c)\s(a)}\leq\e',
\quad
\norm{\s(b)^2-I}\leq\e',
\quad
\norm{\s(a)\s(b)\s(a)\s(b)-I}\leq\e'.
$$
Notice that as along the whole path $\s_t$ one has
$$
\norm{\s_t(b)^2-I}\leq\nnn{\s_t}\leq\d(\e)\leq 1,
$$
so the eigenvalues of $\s_t(b)$ satisfy the estimate $|\l^2-1|\leq 1$,
hence the number of eigenvalues $\l\in\Sp \s_t(b)$ with $\Re\l<0$
does not change along the whole path $\s_t$ in $U_\i$,
therefore the number of eigenvalues $\l\in\Sp\s(b)$ with $|\l+1|\leq \e'$
cannot exceed $n$ (the maximal number of eigenvalues with $\Re\l<0$ of
$\s_0(b)$) and the number of eigenvalues with $|\l-1|\leq \e'$
is not less than $m$.
Then there exists a matrix $\s(b)'\in U_{n+m}$ such that
$\norm{\s(b)-\s(b)'}\leq\e'$ and that the matrix $\s(b)'$ has not more
than $n$ eigenvalues equal to $-1$ and not less than $m$ eigenvalues
equal to $1$. Then we have
\begin{equation}\label{1*}
\norm{\s(a)\s(b)'\s(a)\s(b)'-I}\leq 3\e'.
\end{equation}
Notice that $(\s(b)')^2=I$ and
\begin{equation}\label{1a*}
|\tr(\s(b)')|\geq m-n.
\end{equation}
Let
$$
\s(a)=\left(\begin{array}{ccc}
\o_1&&\\
&\ddots&\\
&&\o_{n+m}
\end{array}\right)
$$
be the matrix of the operator $\s(a)$ in the basis consisting of its
eigenvectors. It was shown in \cite{manFA,BEEK} that if the winding
number of the pair $(\s(a),\s(c))$ is non-zero then for any $\e'>0$ there
exists $\d'(\s')$ such that $\d'(\e')\to 0$ when $\e'\to 0$ and that all
lacunae in $\Sp\s(a)$ do not exceed $\d'(\e')$.
Denote the number of eigenvalues of $\s(a)$ with $|\Im\o_j|>2\e'$ by $N$.
Then we have
\begin{equation}\label{1d*}
N\geq\frac{2\pi-10\e'}{\d'(\e')}.
\end{equation}
As $(\s(b)')^2=I$, so it follows from (\ref{1*}) that
\begin{equation}\label{1c*}
\norm{\s(a)\s(b)'-\s(b)'\s(a)^*}\leq 3\e'.
\end{equation}
Denote by $b_{ij}$ the matrix elements of the matrix $\s(b)'$.
It follows from (\ref{1c*}) that all matrix elements of
$\s(a)\s(b)'-\s(b)'\s(a)^*$ do not exceed $3\e'$, i.e.
\begin{equation}\label{2*}
|b_{ii}(\o_i-\ov{\o}_i)|\leq 3\e', \quad i=1,\ldots,n+m.
\end{equation}
Let us estimate $\tr(\s(b)')$.
We have
$$
|\tr(\s(b)')|=\left|\sum_{i=1}^{n+m}b_{ii}\right|\leq
\sum_{i=1}^{n+m}|b_{ii}|=\sum\nolimits'|b_{ii}+\sum\nolimits''|b_{ii}|,
$$
where $\sum'$ denotes the sum for those numbers $i$ for which one has
$|\Im\o_i|>2\e'$ and $\sum''$ is the sum for the remaining numbers.
As for all $i$ one has $|b_{ii}|\leq 1$, so the last sum do not exceed
the number of summands,
$$
\sum\nolimits''|b_{ii}|\leq n+m-N.
$$
It follows from (\ref{2*}) that for those $i$ which are included into the
first sum we have $|\o_i-\ov{\o}_i|>4\e'$, hence those $b_{ii}$ satisfy
$$
|b_{ii}|<\frac{3}{4},
$$
so
$$
\sum\nolimits'|b_{ii}|<\frac{3}{4}N,
$$
hence we have
$$
|\tr(\s(b)')|<\frac{3}{4}N+n+m-N=n+m-\frac{N}{4}
$$
and it follows from (\ref{1d*}) that
\begin{equation}\label{2a*}
|\tr(\s(b)')|2n}$, then (\ref{1d*}) and
(\ref{2a*}) give a contradiction. \q
\begin{cor}\label{RG=RH}
Let $\s_t\in R_{asym}(\G)$ be an asymptotic representation. Then the
winding number of the pair $(\s_t(a),\s_t(c))$ is zero for big enough $t$.
In particular, it means that $\s_t$ is homotopic to an asymptotic
representation $\rho_t\in R_{asym}(\G)$ with $\rho_t(c)=I$ in the class of
asymptotic representations. \q
\end{cor}
Denote the Grothendieck group of homotopy classes of asymptotic
representations of the group $\G$ by ${\cal R}_{asym}(\G)$. Let $H$ denote
the subgroup $\la a,b|b^2,(ab)^2\ra\cong {\bf Z}_2\ast{\bf Z}_2$. Then
$\G\cong {\bf Z}^2\ast_{\bf Z}H$, so we have
$B\G={\bf T}^2\cup BH$, $S^1={\bf T}^2\cap BH$,
where $B\G$, $BH$, $S^1$ and ${\bf T}^2$ are the classifying spaces of the
groups $\G$, $H$, ${\bf Z}$ and ${\bf Z}^2$ respectively, and the inclusion
$S^1\subset {\bf T}^2$ is the standard inclusion onto the first coordinate.
Then one has an exact sequence
\begin{equation}\label{K6}
\diagram
K^0(B\G)\rto & K^0(BH)\oplus K^0({\bf T}^2)\rto & K^0(S^1)\dto\\
K^1(S^1)\uto & K^1(BH)\oplus K^1({\bf T}^2)\lto & K^1(B\G)\lto
\enddiagram
\end{equation}
and as the maps $K^*({\bf T}^2)\arr K^*(S^1)$ are onto, so the vertical
maps in (\ref{K6}) are zero and the group $K^0(B\G)$ contains an element
$\beta$ which is mapped onto the Bott generator of $K^0({\bf T}^2)$.
Remember that in \cite{mish-noor} a map
\begin{equation}\label{mapasym}
{\cal R}_{asym}(\G)\arr K^0(B\G)
\end{equation}
was constructed, which factorizes \cite{man-mish} through the assembly map
\be\label{sborka}
\alpha:{\cal R}_{\cal Q}(\G\times{\bf Z})\arr K^0(B\G),
\ee
where ${\cal R}_{\cal Q}(\G\times{\bf Z})$ denotes the Grothendieck group of
representations of $\G\times{\bf Z}$ into the Calkin algebra ${\cal Q}$.
It follows from the Corollary \ref{RG=RH} that ${\cal R}_{asym}(\G)={\cal
R}_{asym}(H)$, therefore the element $\beta\in K^0(B\G)$ does not lie in
the image of the map (\ref{mapasym}), hence we obtain
\begin{cor}
The map (\ref{mapasym}) is not a rational epimorphism for the group $\G$.
\q
\end{cor}
We should remark that the element $\b\in K^0(B\G)$ can be obtained as
animage of some representation of the group $\G\times{\bf Z}$
into the Calkin algebra under the mapping $\alpha$ (\ref{sborka}).
\begin{prop}
There exists a representation $\rho\in {\cal R}_{\cal
Q}(\G\times{\bf Z})$ such that $\alpha(\rho)=\b\in K^0(B\G)$.
\end{prop}
{\bf Proof.}
Let us describe how to construct such $\rho$.
Denote by $I_2$ the $2{\times 2}$ unit matrix, and let $J_2$ be the
matrix $\left(\begin{array}{cc}0&1\\1&0\end{array}\right)$. Consider the
sequence $\{\s_n\}$ of almost representations of the group
$\G$ mentioned in the proof of Theorem \ref{primergruppy} and put for
$n\geq 3$
$$
\s'_n(a)=\s_{n-2}(a)\oplus I_2, \quad
\s'_n(c)=\s_{n-2}(c)\oplus I_2, \quad
\s'_n(b)=\s_{n-2}(b)\oplus J_2.
$$
\begin{lem}\label{put}
There exists a homotopy $\{\s_n^t\}_{t\in [0,1]}$ of almost
representations of the group $\G$ connecting $\s_n$ with $\s'_n$ such that
$\nnn{\s_n^t}\leq 3\e_n=3 |e^{2\pi i/n}-1|$.
\end{lem}
{\bf Proof.}
Let $e_1,\ldots,e_n$ be a basis in ${\bf C}^n$ such that the almost
representation $\s_n$ has the above form. Let $u\in M_n({\bf C})$
be the unitary matrix, which transposes the vectors $e_1$
and $e_{n-1}$ and does not move the other vectors of the basis, let
$u_t\in M_n({\bf C})$ be a unitary path connnecting the unit matrix
with the matrix $u$ and such that it does not move these other basis
vectors. Then the path $\s_n^t(b)=u_t^*\s_n(b)u_t$ connects the
matrices $\s_n(b)$ and $\s'_n(b)$. One can directly check that
$\norm{(\s_n^t(b))^2-I}\leq\e_n$ and
$\norm{(\s'_n(a)\s_n^t(b))^2-I}\leq 3\e_n$. It is shown in \cite{mish-noor}
that the Voiculescu pair of matrices $\s_n(a)$ and $\s_n(c)$ can be
connected with the matrices $\s'_n(a)$ and $\s'_n(c)$ by a path such that
$\norm{\s_n^t(a)\s_n^t(c)\s_n^t(a)^{-1}\s_n^t(c)^{-1}-I}\leq \e_n$.
The proof is completed by a direct check of the estimate
$\norm{(\s_n^t(a)\s_n(b))^2-I}\leq \e_n$. \q
We proceed with the construction of the representation $\rho$.
On a Hilbert space $H$ consider the operators given by the matrices
$$
\ov{\s}_n(a)=\s_n(a)\oplus I_2\oplus\ldots\oplus I_2\oplus\ldots,\quad
\ov{\s}_n(c)=\s_n(c)\oplus I_2\oplus\ldots\oplus I_2\oplus\ldots,\quad
$$
$$
\ov{\s}_n(b)=\s_n(b)\oplus J_2\oplus\ldots\oplus J_2\oplus\ldots.
$$
Denote by $F$ the set $\{a,b,c\}$ of generators of the group $\G$.
By Lemma \ref{put} the sequences of operators
$\ov{\s}_n(g)$, $g\in F$, can be included into continuous paths
$\ov{\s}_t(g)$, $t\in [1,\infty)$, such that for
$t\geq n$ all relations of the group $\G$ are satisfied up to
$3\e_n$. Then as in \cite{mish-noor} one can choose an increasing
sequence $\{t_k\}$ of the values for the parameter
$t$ so that the operators $\ov{\s}_{t_k}(g)=\ov{\s}_k(g)$ would satisfy
$$
\lim_{k\to\infty}\norm{\ov{\s}_{k+1}(g)-\ov{\s}_k(g)}=0, \quad g\in F,
$$
$$
\lim_{k\to\infty}\norm{r(\ov{\s}_k(a),\ov{\s}_k(b),\ov{\s}_k(c))-I}=0,
$$
for every relation $r=r(a,b,c)$ of the group $\G$.
Without loss of generality one can assume that the sequence of operators
$\{\ov{\s}_k(g)\}$ contains $\{\ov{\s}_n(g)\}$ as a subsequence.
Put ${\cal H}=\oplus_k H_k$, where $H_k=H$, and for negative $k$
put $\ov{\s}_k(g)=I$, $g\in F$.
Let $T$ be the shift in
${\cal H}$, $T(H_k)=H_{k-1}$. Denote by $d$ a generator for the group
${\bf Z}$. Define the representation $\rho$ on the generators of the
group $\G\times{\bf Z}$ by
$$
\rho(g)=\oplus_k\ov{\s}_k(g),\quad g\in F;\qquad
\rho(d)=T.
$$
It is easy to check that all relations of the group $\G\times{\bf Z}$
are satisfied in the Hilbert space ${\cal H}$ modulo compact operators,
hence $\rho$ is well defined as a representation into the Calkin algebra
(cf. \cite{man-mish}).
As the restriction of this representation onto the subgroup
${\bf Z}^2\times{\bf Z}\subset \G\times{\bf Z}$ is obtained from the
Voiculescu matrices, so this representation defines the Bott generator in
the group $\til{K}^0(B{\bf Z}^2)$, hence the representation
$\rho$ of the group $\G\times{\bf Z}$ into the Calkin algebra defines the
element $\beta\in K^0(B\G)$. \q
\medskip
Remark that in our example the absence of the AGA property is related to
torsion. It would be interesting to know whether torsion-free groups
always have AGA.
\bigskip
{\bf Acknowledgement.}
The present paper was prepared with the partial
support of RBRF (grant N 96-01-00276) and INTAS.
I am grateful to A.~S.~Mishchenko for useful discussions.
%\newpage
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%\end{document}
\vspace{1cm}
{\small
\noindent
V.~M.~Manuilov\\
Dept. of Mech. and Math.,\\
Moscow State University,\\
Moscow, 119899, RUSSIA\\
e-mail: manuilov@mech.math.msu.su
}
\end{document}