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\date{15 April 1996}
\author{V.~M.~Manuilov}
\title{On almost commuting operators}
\sloppy
\begin{document}
\maketitle

 \begin{abstract}
We prove that for some $C^*$-algebras (including the matrix ones) if $h\in
A$ is a selfadjoint, $u\in A$ is a unitary and the norm of their commutant
$[u,h]$ is small enough then one can connect $u$ with the unity by a path
$u(t)$ so that the norm of $[u(t),h]$ would be also small along this path.

 \end{abstract}



 \begin{prop}\label{a.c.}
Let $A$ be a $C^*$-algebra with properties
 \begin{enumerate}
\item
$RR(A)=0$ and $tsr(A)=1$;
\item
for every projection $p\in A$ the unitary group of the $C^*$-algebra $pAp$
is connected.
 \end{enumerate}
Let $h\in A$ be a self-adjoint and let
$u\in A$ be a unitary such that
 \be\label{ac1}
\norm{u^*hu-h}<\d.
 \ee
Then there exists a constant $C$ depending only on $\norm{h}$ and a
path $u(t)$ connecting $u$ with $1$ such that for small enough $\d$
one has for all $t$
 $$
\norm{u^*(t)hu(t)-h}<C\sqrt[4]{\d}.
 $$

\end{prop}
{\bf Proof.}
As $RR(A)=0$ we can assume without loss of generality that the operator
$h$ is a linear combination of mutually orthogonal projections,
$h=\sum_{i=1}^n \l_ip_i\in A$ with real eigenvalues $\l_i$.
We can also assume that these eigenvalues
are ordered, i.e. $\l_1>\ldots >\l_n$. Divide the segment
$[\l_n;\l_1]$ into smaller segments of the length $\sqrt[4]{\d}$ and
denote those segments which contain at least one eigenvalue $\l_i$ by
$\D_k$.  Then the number $m$ of such segments is not bigger then
$(\l_1-\l_n)/\sqrt[4]{\d}+1$ and the following properties hold:
\begin{enumerate}
\item
if $\l_i,\l_j\in\D_k$ then $\v \l_i-\l_j\v <\sqrt[4]{\d}$;
\item
if $\l_i\in\D_{k-1}$ and $\l_j\in\D_{k+1}$ then $\v \l_i-\l_j\v
\geq\sqrt[4]{\d}$.
\end{enumerate}

Let $\mu_k$ be the middle points of the segments $\D_k$. Then if
$\mu_{k+1}-\mu_k>\sqrt[4]{\d}$ then the spectrum of $h$ has a lacuna of
the length not less than $\sqrt[4]{\d}$.

\noindent
Denote the spectral projections of $h$ corresponding to the segments
$\D_k$ by $q_k$, $q_1+\ldots+q_m=1$. Then $A$ as $A$-module can be
decomposed into a direct sum corresponding to the above projections,
$A=\oplus_{k=1}^mq_kA$ and we will represent elements of the algebra $A$
as matrices with regards to this decomposition: $h=\diag(\{h_i\})$ for
$h_i=q_ihq_i$ and $u=(u_{ij})$ for $u_{ij}=q_iuq_j$.

\medskip\noindent
Notice that if $m=1$, i.e. all eigenvalues of $h$ differ from each other
not more than by $\sqrt[4]{\d}$ then we can take any path $u(t)$
connecting $u$ with $1$ because in this case there exists a number $\mu$
such that $\norm{h-\mu} <\sqrt[4]{\d}/2$, hence
 $$
\norm{v^*hv-h}\leq\norm{v^*hv-v^*\mu
v}+\norm{\mu-h}<2\frac{\sqrt[4]{\d}}{2}=\sqrt[4]{\d}
 $$
for any unitary $v$, so further we can assume that $m>1$.

\medskip\noindent
Divide once more the spectrum of the operator $h$ into smaller (than
$\D_k$) segments $\ov{\D}_s$ of length $\d$. Let $\ov{\l}_s$ be the
middle points of the segments $\ov{\D}_s$ and let $\ov{p}_s$ be the
spectral projections of $h$ corresponding to the segments $\ov{\D}_s$.
Then put
 $$
\ov{h}=\sum_s \ov{\l}_s\ov{p}_s.
 $$
Obviously $\norm{h-\ov{h}}<\d/2$, hence in view of (\ref{ac1})
 \be\label{ac.2}
\norm{u^*\ov{h}u-\ov{u}}\leq \norm{u^*\ov{h}u-u^*hu}+
\norm{u^*hu-h}+\norm{h-\ov{h}}< 2\d.
 \ee
Let $A=\oplus_s\ov{p}_sA$ be the decomposition of $A$
corresponding to the spectral projections of $h$. It is a
sub-decomposition of $\oplus_{k=1}^mq_kA$ and the matrix $u$ can be
written as $u=(v_{kl})$,
$v_{kl}=\ov{p}_k u\ov{p}_l$ and the matrix entries $u_{ij}$
can be viewed as blocks of elements $v_{kl}$.
Denote by $N$ the number of columns of the matrix $(v_{kl})$. Then one has
 \be\label{N}
N<\frac{2\norm{h}}{\d}+1
 \ee

Now we turn to construction of the homotopy $u\sim 1$. We begin with a
non-unitary path $u(t)$ which would lie not far from the set of unitaries
$U=U(A)$.

\subsection*{First step of homotopy}
%\medskip\noindent
For a matrix $a=(a_{ij})$ decomposed with respect to $\oplus_{k=1}^m q_kA$
we denote by $d_0(a)$ its main diagonal
$\diag(\{a_{ii}\})$, by $d_k(a)$ denote a diagonal lying $k$ lines above
(or below if $k$ is negative) the main diagonal.
We start by proving that the matrix
$u=(u_{ij})$ is ``almost'' three-diagonal.

\noindent
Let $k_i$ and $k_j$ be the numbers of the segments
$\D_k$ which contain the eigenvalues $\ov{\l}_i$ and $\ov{\l}_j$ of
$\ov{h}$ respectively; $\ov{\l}_i\in\D_{k_i}$, $\ov{\l}_j\in\D_{k_j}$.
Define a three-diagonal matrix $d(a)$
by the following way:
 \begin{enumerate}
\item
if $k_j\geq k_i +2$ or $k_j\leq k_i -2$ then put $(d(a))_{ij}=0$,
\item
if $k_j=k_i\pm 1$ and $\v\mu_{k_j}-\mu_{k_i}\v>\sqrt[4]{\d}$ then put also
$(d(a))_{ij}=0$,
\item
otherwise put $(d(a))_{ij}=a_{ij}$.
 \end{enumerate}

 \begin{lem}\label{lemma01}
It follows from (\ref{ac1}) that for small enough $\d$ one has
 $$
\norm{u-d(u)}<4\norm{h}^{1/2}\sqrt[4]{\d}.
 $$

 \end{lem}
{\bf Proof.}
Consider the matrix $(\ov{h}u-u\ov{h})(\ov{h}u-u\ov{h})^*$.
From the inequality
(\ref{ac.2}) it follows that the norm of this matrix
is less then $4\d^2$,
hence the norm of any element of this matrix is also less than $4\d^2$.
So, as $\ov{\l}_j$ commute with $v_{kj}$, we obtain for every $i$
 \be\label{4}
\norm{\sum_{j=1}^N (\ov{\l}_i-\ov{\l}_j)^2 v_{ij}v^*_{ij}}<4\d^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
 $$
\norm{\sum\nolimits'_j (\ov{\l}_i-\ov{\l}_j)^2 v_{ij}v^*_{ij}}<4\d^2
 $$
where the sum $\sum'$ is taken for those $j$ for which
either $\v k_j- k_i\v\geq 2$ or $\v k_j-k_i\v=1$ and
$\v\mu_{k_j}-\mu_{k_i}\v>\sqrt[4]{\d}$, i.e. we throw away those $v_{ij}$
for which $(u-d(u))_{ij}=0$.  As in the sum $\sum'$ we have
$\v\ov{\l}_k-\ov{\l}_j\v\geq\sqrt[4]{\d}$, so
 $$
4\d^2>\norm{\sum\nolimits'_j
(\ov{\l}_i-\ov{\l}_j)^2v_{ij}v^*_{ij}}\geq\sqrt{\d}\norm{\sum\nolimits'_i
v_{ij}v^*_{ij}},
 $$
hence
 \be\label{3}
\norm{\sum\nolimits'_j v_{ij}v^*_{ij}}<4\d^{3/2}.
 \ee

 \begin{lem}\label{NN}
Let $a=(a_{ij})$, $a_{ij}\in A$ be a $N\times N$ matrix such that
for every $i$ one has $\norm{\sum_j a_{ij}a^*_{ij}}<\e^2$.
Then $\norm{a}<\e\sqrt{N}$.

 \end{lem}
{\bf Proof.}
Take $\xi=(\xi_k)$, $k=1,\ldots, N$, $\xi_i\in A$.
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}
< N\e^2 \norm{\xi}^2,
 \end{eqnarray*}
hence we have $\norm{a\xi}<\sqrt{N}\e\norm{\xi}$. \q

\smallskip\noindent
The sums $\sum'$ correspond to the blocks lying in the
matrix $u-d(u)$.
Now in view of (\ref{N}) and (\ref{3}) from the lemma \ref{NN}
(for $\e=2\d^{3/4}$) we
get for $\d<\norm{h}$
 $$
\norm{u-d(u)}<\sqrt{N}\ 2\d^{3/4}\leq
2\norm{h}^{1/2}\d^{-1/2}\,2\d^{3/4}= 4\norm{h}^{1/2}\sqrt[4]{\d}
 $$
which ends the proof of the lemma \ref{lemma01}.  \q

\medskip\noindent
Define the path $u(t)$ by the formula
 $$
u(t)=(u-d(u))(1-t)+d(u).
 $$
Then $u(0)=u$, $u(1)=d(u)$ and for every $t\in [0,1]$ by the lemma
\ref{lemma01} we have
$\dist(u(t),U)<4\norm{h}^{1/2}\sqrt[4]{\d}$.
Estimate the commutator norm:
 \begin{eqnarray*}
\norm{u(t)\ov{h}-\ov{h}u(t)}&\leq&\norm{u\ov{h}-\ov{h}u}+\norm{(u-d(u))\ov{h}-
\ov{h}(u-d(u))}\\
&<&2\d+2\norm{h}\norm{u-d(u)}<2\d+2\norm{h}4\norm{h}^{1/2}\sqrt[4]{\d}.
 \end{eqnarray*}

\subsection*{Second step of homotopy}

We should connect the matrix $d(u)$ with a diagonal one. To do so we need
the following

 \begin{lem}\label{triang}
{\rm (making matrices almost upper triangular).} \ Let
 $$
a=\left(
\begin{array}{cc}
a_{11}&a_{12}\\
a_{21}&a_{22}
\end{array}
\right)
 $$
be a matrix in $q_{j_1}A\oplus q_{j_2}A$. Then for any $\e>0$ there exists
a unitary path $v(t)$ such that $v(0)=1$ and
 $$
v(1)\cdot a=\left(
\begin{array}{cc}
a'_{11}&a'_{12}\\
a'_{21}&a'_{22}
\end{array}
\right)
 $$
with $\norm{a'_{21}}<\e$.

 \end{lem}
{\bf Proof.} As by assumption $tsr(A)=1$, so for any $\e>0$ we can
find an invertible element $\ov{a}_{11}$ such that $\norm{\ov{a}_{11}-
a_{11}}<\e$. Put $\a=a_{21}(\ov{a}_{11})^{-1}$. Put further
 $$
v(t)=\left(
\begin{array}{cc}
(1+t^2\a^*\a)^{-1/2} & (1+t^2\a^*\a)^{-1/2}\cdot t\a^*\\
-(1+t^2\a\a^*)^{-1/2}\cdot t\a & (1+t^2\a\a^*)^{-1/2}
\end{array}
\right).
 $$
It can be easily seen that $v^*(t)v(t)=v(t)v^*(t)=1$ and $v(1)$ lies in
 the unitary group in the path component of $1$. Estimate the entry
$a'_{21}$ in the product $v(1)a$:
 \begin{eqnarray*}
\norm{a'_{21}}&=&\norm{-(1+\a\a^*)^{-1/2}\a
a_{11}+(1+\a\a^*)^{-1/2}a_{21}}\\
&=&\norm{(1+\a\a^*)^{-1/2}(\a a_{11}-\a \ov{a}_{11})}
\leq \norm{(1+\a\a^*)^{-1/2}\a}\cdot\norm{a_{11}-\ov{a}_{11}}\\
&<& \e\cdot\norm{(1+\a\a^*)^{-1/2}\a}=\e\cdot
\norm{f(\b)}^{1/2}\leq\e
 \end{eqnarray*}
where $\b=\a\a^*$ and $f(\l)=\frac{\l}{1+\l}\leq 1$.\q

\medskip\noindent
Consider the element $u_{21}$
of the matrix $u$. If the segments $\D_1$ and $\D_2$ are
separated then the element $u_{21}$ is already small enough and the
element $(d(u))_{21}$ is already zero.  Put then $v_1=1$. If these
segments are not separated then by the lemma \ref{triang} we can find in
the module $q_1A\oplus q_2A$ a unitary path $v_1(t)$ with $v_1(0)=1$
and
 $$
v_1(1)=\left( \begin{array}{cc}
v^{(1)}_{11}&v^{(1)}_{12}\\
v^{(1)}_{21}&v^{(1)}_{22}
 \end{array}
\right)
 $$
such that
 $$
\left(
\begin{array}{llll}
v^{(1)}_{11}&v^{(1)}_{12}&\vdots & \\
v^{(1)}_{21}&v^{(1)}_{22}&\vdots & \\
\ldots&\ldots&\lefteqn{\ldots}\\
&&\vdots&E
\end{array}
\right)
\left(
\begin{array}{llll}
u_{11}&u_{12}&\phantom{\vdots}&\\
u_{21}&u_{22}&u_{23}\vphantom{\vdots}&\\
&\ldots&\ldots&\ldots\\
&&\ldots&\ldots\phantom{\vdots}
\end{array}
\right)
=
\left(
\begin{array}{llll}
u^{(1)}_{11}&u^{(1)}_{12}&u^{(1)}_{13}\vphantom{\vdots}&\\
u^{(1)}_{21}&u^{(1)}_{22}&u^{(1)}_{23}\vphantom{\vdots}&\\
&\ldots&\ldots&\ldots\\
&&\ldots&\ldots\phantom{\vdots}
\end{array}
\right)
 $$
with $\norm{u^{(1)}_{21}}<\e$.
Here $E$ stands for a unit matrix and empty places stand for zeros.
Notice that then
the third and the following strings of the matrix $u$ do not
change.
So we get a path $u(t)=v_1(t)d(u)$ and as $v_1(t)$ is a unitary, so
 \be\label{est}
\dist(u(t),U)<4\norm{h}^{1/2}\sqrt[4]{\d}.
 \ee
As in the lemma
\ref{triang} the number $\e$ can be taken arbitrarily small, so if we
write zero instead of the element $u^{(1)}_{21}$, the estimate (\ref{est})
would remain valid.

\noindent
Turn now to the element $u_{32}=u^{(1)}_{32}$. If the segments
$\D_2$ and $\D_3$ are separated then this element is already small enough.
Put then $v_2=1$. In the other case we can find a unitary path $v_2(t)$
with $v_2(0)=1$ and
 $$
v_2(1)=\left(
\begin{array}{ll}
v^{(2)}_{22}&v^{(2)}_{23}\\ v^{(2)}_{32}&v^{(2)}_{33}
\end{array}
\right)
 $$
in the module $q_2A\oplus q_3A$ such that
 \begin{eqnarray*}
\left(
\begin{array}{lllll}
1&&&&\\
&v^{(2)}_{22}&v^{(2)}_{23}&\vdots&\\
&v^{(2)}_{32}&v^{(2)}_{33}&\vdots&\\
&\ldots&\ldots&\lefteqn{\ldots}&\phantom{\vdots}\\
&&&\vdots&E
\end{array}
\right)
&\cdot&
\left(
\begin{array}{lllll}
u^{(1)}_{11}&u^{(1)}_{12}&u^{(1)}_{13}&\phantom{\vdots}&\\
&u^{(1)}_{22}&u^{(1)}_{23}&\phantom{\vdots}&\\
&u^{(1)}_{32}&u^{(1)}_{33}&u^{(1)}_{34}&\phantom{\vdots}\\
&&\ldots&\ldots\phantom{\vdots}&\ldots\\
&&&\ldots&\ldots\phantom{\vdots}
\end{array}
\right)
\\
&=&
\left(
\begin{array}{lllll}
u^{(2)}_{11}&u^{(2)}_{12}&u^{(2)}_{13}&&\phantom{\vdots}\\
&u^{(2)}_{22}&u^{(2)}_{23}&u^{(2)}_{24}&\phantom{\vdots}\\
&u^{(2)}_{32}&u^{(2)}_{33}&u^{(2)}_{34}&\phantom{\vdots}\\
&&\ldots&\ldots\phantom{\vdots}&\ldots\\
&&&\ldots&\ldots\phantom{\vdots}
\end{array}
\right)
 \end{eqnarray*}
with $\norm{u^{(2)}_{32}}<\e$. Again we can write zero instead of
$u^{(2)}_{32}$.
Repeating this procedure we obtain a unitary path
 $$
v(t)=v_{m-1}(t)\cdot\ldots\cdot
v_2(t)\cdot v_1(t)
 $$
connecting the unity with the unitary
 $$
v=v(m-1)=v_{m-1}(m-1)\cdot\ldots\cdot v_2(2)\cdot v_1(1).
 $$
Denote $v(t)\cdot d(u)$ by $\ov{u}(t)$ and $v\cdot d(u)$ by $\ov{u}$.
Notice that
 \be\label{dis}
\dist(\ov{u}(t),U)<4\norm{h}^{1/2}\sqrt[4]{\d}
 \ee
for all $t$. Hence we have the estimate
$\norm{\ov{u}_{ij}(t)}<4\norm{h}^{1/2}\sqrt[4]{\d}$ for every $i,j$.
Notice also that for every $t$ the path $\ov{u}(t)$ lies in the
four-diagonal matrices of the form
 $$
\ov{u}(t)=\left(
\begin{array}{llllll}
\ov{u}_{11}&\ov{u}_{12}&\ov{u}_{13}&&&\\
\ov{u}_{21}&\ov{u}_{22}&\ov{u}_{23}&\ov{u}_{24}&&\\
&\ov{u}_{32}&\ov{u}_{33}&\ov{u}_{34}&\ov{u}_{35}&\\
&&\ldots&\ldots&\ldots&\ldots\\
&&&\ldots&\ldots&\ldots\\
&&&&\ldots&\ldots
\end{array}
\right)
 $$
therefore we can easily estimate the commutator
norm along this path. To do so we should deal with the operator $h'=\sum
\mu_k q_k$, $\norm{h-h'}<\frac{1}{2}\sqrt[4]{\d}$. Then
 \begin{eqnarray}\label{com}
\norm{\ov{u}(t)h-h\ov{u}(t)}&<&\norm{[\ov{u}(t),h']}+\sqrt[4]{\d}
\nonumber\\
&\leq&\norm{[d_{-1}(\ov{u}(t)),h']}+\norm{[d_1(\ov{u}(t)),h']}+
\norm{[d_2(\ov{u}(t)),h']}+\sqrt[4]{\d}\nonumber\\
&\leq&\sup_{i,j,k}\norm{\ov{u}_{ij}(t)}(2\v\mu_k-\mu_{k+1}\v+
\v\mu_k-\mu_{k+2}\v)+\sqrt[4]{\d}\nonumber\\
&<&6\sqrt[4]{\d}
 \end{eqnarray}
for small enough $\d$. The resulting matrix $\ov{u}$ is three-diagonal and
upper triangular:
 $$
\ov{u}=
\left(
\begin{array}{lllll}
\ov{u}_{11}&\ov{u}_{12}&\ov{u}_{13}&&\\
&\ov{u}_{22}&\ov{u}_{23}&\ov{u}_{24}&\\
&&\ov{u}_{33}&\ov{u}_{34}&\ov{u}_{35}\\
&&&\ldots&\ldots\\
&&&&\ldots
\end{array}
\right).
 $$

\subsection*{Third step of homotopy}

\noindent
It follows from (\ref{dis}) that the matrix $\ov{u}$ is invertible and
there exists a unitary $w$ such that
 $$
\norm{\ov{u}-w}<4\norm{h}^{1/2}\sqrt[4]{\d}.
 $$
Therefore when $\d$ is small enough we have
 $$
\norm{(\ov{u})^{-1}-w^*}<6\norm{h}^{1/2}\sqrt[4]{\d}.
 $$
Therefore we get the estimate
 $$
\norm{(\ov{u})^{-1}-(\ov{u})^*}<10\norm{h}^{1/2}\sqrt[4]{\d},
 $$
hence
 $$
\norm{d_1(\ov{u})}<10\norm{h}^{1/2}\sqrt[4]{\d} \quad {\rm and} \quad
\norm{d_2(\ov{u})}<10\norm{h}^{1/2}\sqrt[4]{\d}.
 $$
Therefore
 \be\label{dis2}
\dist(d_0(\ov{u}),U)<\dist(\ov{u},U)+\norm{d_1(\ov{u})}+\norm{d_2(\ov{u})}
<24\norm{h}^{1/2}\sqrt[4]{\d}.
 \ee
Connect the matrices $\ov{u}$ and $d_0(\ov{u})$ by a linear path
$\ov{\ov{u}}(t)$. Then along this path one has
 \be\label{dis3}
\dist(\ov{\ov{u}}(t),U)<24\norm{h}^{1/2}\sqrt[4]{\d}
 \ee
and from (\ref{com}) we see that
 $$
\norm{\ov{\ov{u}}(t)h-h\ov{\ov{u}}(t)}<6\norm{h}^{1/2}\sqrt[4]{\d}.
 $$

\subsection*{Fourth step of homotopy}

\noindent
It follows from (\ref{dis2}) that for small enough $\d$ the diagonal
matrix $d_0(\ov{u})$ consists of invertible elements being close to
unitaries and it follows from (\ref{dis3}) that
for every $\ov{u}_{ii}$ one can find a unitary $w_i$ such that
$\norm{\ov{u}_{ii}-w_i}<24\norm{h}^{1/2}\sqrt[4]{\d}$. Take a linear path
connecting the matrices $\ov{u}$ and $\diag\{w_i\}$. Then this path also
lies close to the unitary group $U$. Then we connect the matrix
$\diag\{w_i\}$ with unity. The last path lies in $U$. Let
$\widetilde{u}(t)$ be the path of the fourth step of homotopy. Notice that
it lies within diagonal matrices, hence the commutator norm along this path
is small:
 $$
\norm{\widetilde{u}(t)h-h\widetilde{u}(t)}\leq
\norm{\widetilde{u}(t)h'-h'\widetilde{u}(t)}+\sqrt[4]{\d}=\sqrt[4]{\d}.
 $$
Consider now all four steps of constructed homotopy. We see that along the
whole path $u'(t)$ connecting $u$ with unity we have
 $$
\dist(u'(t),U)<24\norm{h}^{1/2}\sqrt[4]{\d}
 $$
and
 $$
\norm{u'(t)h-hu'(t)}<6\sqrt[4]{\d}.
 $$
Therefore there exists a unitary path $u(t)$ connecting $u$ with unity
such that
 $$
\norm{u(t)-u'(t)}<48\norm{h}^{1/2}\sqrt[4]{\d},
 $$
hence we get the estimate
 \begin{eqnarray*}
\norm{u(t)h-hu(t)}&<&\norm{u'(t)h-hu'(t)}+2\norm{h}\norm{u(t)-u'(t)}\\
&<&6\sqrt[4]{\d}+96\norm{h}^{3/2}\sqrt[4]{\d}=C\sqrt[4]{\d}
 \end{eqnarray*}
which proves the proposition. \q

\bigskip\noindent
Remark that the class of $C^*$-algebras satisfying conditions of the
proposition \ref{a.c.} includes the finite $W^*$-algebras and particularly
the matrix algebras. The constant $C$ is by no means the best one and can
be improved. On the other hand I believe that the power $1/4$ for $\d$ in
our estimate cannot be made bigger. As the commutator norm does not change
if we replace $h$ by $h+\l$, $\l\in{\bf R}$, so the constant $C$ really
depends not on the norm of $h$ but on the length of its spectrum.

\medskip\noindent
Remark also that by the same way one can proof the similar result in the
case when instead of selfadjoint $h$ we deal with a unitary $v$ with a
lacuna in its spectrum. The last restriction is essential as otherwise
there exists an obstruction~\cite{e-l}. The following proposition can be
viewed as a first step in construction of approximately commuting homotopy
between approximately commuting and commuting pairs of unitaries in the
case when the obstruction of~\cite{e-l} is zero.

 \begin{prop}
Let $A$ be a $C^*$-algebra with properties
 \begin{enumerate}
\item
$RR(A)=0$ and $tsr(A)=1$;
\item
for every projection $p\in A$ the unitary group of the $C^*$-algebra $pAp$
is connected.
 \end{enumerate}
Let $u,v\in A$ be unitaries such that
$
\norm{u^*hu-h}<\d.
$
Suppose that the unit circle contains an arc of the length $d$ such that
the intersection of this arc with the spectrum of $v$ is empty.
Then there exists a constant $C$ depending only on $d$ and a
path $u(t)$ connecting $u$ with $1$ such that for
small enough $\d$ and for all $t$ one has
$
\norm{u^*(t)hu(t)-h}<C\sqrt[4]{\d}.
$

 \end{prop}

\medskip\noindent
{\bf Acknowledgement.}  The present paper was prepared with the partial
support of RBRF (grant N 96-01-00182). I am grateful to M.~Frank,
A.~S.~Mishchenko and E.~V.~Troitsky for helpful discussions.



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{\sc R.~Exel, T.~A.~Loring}: Invariants of almost commuting unitaries.
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{\sc E.~C.~Lance}: Hilbert $C^*$-modules -- a toolkit for operator
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\end{thebibliography}

\end{document}

\vspace{1.5cm}
V.~M.~Manuilov\\
Dept. of Mech. and Math.,\\
Moscow State University,\\
Moscow, 119899, RUSSIA\\
e-mail: manuilov@mech.math.msu.su

\end{document}
