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\begin{document}
\title[An invariant for unbounded operators] {An invariant for
unbounded operators}
% Information for first author
\author{Vladimir Manuilov}
% Address of record for the research reported here
\address{Department of Mechanics and Mathematics, Moscow State
University, Leninskie Gory, Moscow, 119992, Russia}
\email{manuilov@mech.math.msu.su}
% \thanks will become a 1st page footnote.
\thanks{}
% Information for second author
\author{Sergei Silvestrov}
\address{Centre for Mathematical Sciences, Department of Mathematics,
Lund Institute of Technology, Lund University, P.O. Box 118,
SE-221 00 Lund, Sweden} \email{sergei.silvestrov@math.lth.se}
\thanks{
The first named author was supported in
part by the RFFI grant No. 02-01-00574 and \mbox{HIII\hspace{-2.3ex}%
\rule{1.9ex}{0.07ex}\,-619.2003.01}, and the second named author
by the Crafoord foundation. Part of the research has been
performed during the Non-commutative Geometry program 2003/2004,
Mittag-Leffler Institute, Stockholm.}
% General info
\subjclass[2000]{Primary 47L60; Secondary 19K14, 46L80}
\date{October 6, 2004}
%and, in revised form, .}
%\dedicatory{This paper is dedicated to our authors.}
\keywords{Bott projections, $K$-group, deformations, invariant}
\begin{abstract}
For a class of unbounded operators, a deformation of a Bott
projection is used to construct an integer-valued invariant
measuring deviation of the non-commutative deformations from the
commutative originals and its interpretation in terms of
$K$-theory of $C^*$-algebras is given. Calculation of this
invariant for specific important classes of unbounded operators is
also presented.
\end{abstract}
\maketitle
We consider a class of unbounded operators on a Hilbert space that
can be viewed as deformations of the complex coordinate on a
complex plane. To distinguish these deformations, we introduce an
integer-valued invariant for such operators using the deformed
Bott projection and calculate it for some examples. We also give a
$K$-theoretical interpretation of this invariant.
Let $x$ be an unbounded (i.e. linear densely defined) operator on
a Hilbert space $H$ with the following properties:
\smallskip
\begin{itemize}
\item[($i$)]
operators $x^*x$ and $xx^*$ are well-defined unbounded
self-adjoint operators on $H$;
\item[($ii$)]
the accumulation points of $\Sp x^*x$ are $0$ and $\infty$ and
outside these two points $\Sp x^*x$ is discrete and of finite
multiplicity;
\item[($iii$)]
operators $x^*x$ and $xx^*$ commute as unbounded operators (i.e.
their spectral measures do).
\end{itemize}
\smallskip
Let $x=uh$ be a polar decomposition with a partial isometry $u$
and an unbounded self-adjoint $h$. In what follows we can work
with isometries or coisometries, but for convenience of notation
we restrict ourselves to the following case:
\smallskip
\begin{itemize}
\item[$(iv)$]
$u$ is unitary.
\end{itemize}
\smallskip
Let $\{\lambda_n^2\}_{n\in\mathbb Z}$ with
$0\leq\ldots\leq\lambda_n\leq\lambda_{n+1}\leq\ldots$ be all the
eigenvalues of $x^*x$. The properties $(ii)$ and $(iii)$ imply
that there exists an orthonormal basis $\{e_n\}_{n\in\mathbb Z}$
in $H$ and a bijection $m:\mathbb Z\to\mathbb Z$ such that this
basis diagonalizes both $x^*x$ and $xx^*$ and one has
\begin{equation}\label{1}
x^*xe_n=\lambda_n^2e_n \quad{\rm and}\quad
xx^*e_n=\lambda_{m(n)}^2e_n.
\end{equation}
Note that if all $\lambda_n$, $n\in\mathbb Z$, are distinct then
the bijection $m$ is unique, otherwise different choices of a
basis above may result in different bijections $m:\mathbb
Z\to\mathbb Z$, but finite multiplicity of all eigenvalues implies
that all bijections satisfying \eqref{1} behave in the same way at
infinity. We will also assume one more property, which does not
depend on choice of $m$ satisfying \eqref{1}:
\smallskip
\begin{itemize}
\item[$(v)$]
$m(-\infty)=-\infty $ and $ m(\infty)=\infty $, i.e.
$\lim_{n\to\pm\infty}m(n)=\pm\infty.$
\end{itemize}
\smallskip
One can look at such an operator $x$ as at a deformed plane: if
$u$ commutes with $h$ and if $\Sp u=\mathbb T$, $\Sp h=[0,\infty)$
then continuous functions of $x$ are in one-to-one correspondence
with continuous functions on a plane. The $K$-group of a plane,
$K^0(\mathbb R^2)$, is a free abelian group generated by the Bott
element, i.e. by the formal difference $[p]-[q]$, $p,q\in
M_2(C_0(\mathbb R^2)^+)$ (here the superscript $+$ denotes the
unitalization), where $p$ is a Bott projection and
$q=\left(\begin{array}{cc}1&0\\0&0\end{array}\right)$. To measure
the deviation of non-commutative deformations of topological
spaces from their `commutative' originals, one deforms the Bott
element (i.e. one changes functions involved in the commuting
projections $p$ and $q$ by their non-commutative analogs ---
expressions in $x^*x$ and $xx^*$, which result in projections $P$
and $Q$) and checks its non-triviality. For the case of compact
spaces like a sphere or a torus, see \cite{Voiculescu,Loring},
where it was shown that the invariant $\tr(P-Q)$ is an integer,
which vanishes in the commutative case and which equals one for
the Voiculescu pair (a non-commutative version of a two-torus).
The same approach works for non-compact spaces as well,
\cite{Nest,Manuilov}, where it was shown that a similar invariant
equals one for a `non-commutative plane' given by a pair of
self-adjoint unbounded operators $(A,B)$ that is a `small'
deformation of the pair $(x,i\frac{d}{dx})$.
For the complex coordinate $z$ on a plane, one can use, for a Bott
projection, the formula
\begin{equation}\label{p}
p=\left(\begin{array}{cc}f(|z|^2)&g(|z|^2)\bar{z}\\
g(|z|^2)z&1-f(|z|^2)\end{array}\right),
\end{equation}
where $f:[0,\infty)\to [0,1)$ is continuous and increasing and $g$
is defined by $g(t)=\sqrt{\frac{f(t)-f^2(t)}{t}}$ (thus $f$ cannot
be arbitrary and should satisfy continuity of the function
$\frac{f(t)-f^2(t)}{t}$ and its vanishing at $0$).
Passing to the non-commutative framework, one should consider, for
an operator $x$ with properties $(i)-(v)$, the operator $$
P=P_f(x)=\left(\begin{array}{cc} p_{11}&p_{12}\\ p_{21}&p_{22}
\end{array}\right)
=\left(\begin{array}{cc}
f(x^*x)&g(x^*x)x^*\\
xg(x^*x)&1-f(xx^*)
\end{array}\right).
$$ As $xh(x^*x)=h(xx^*)x$ for any continuous function $h(t)$ with
$h(0)=\lim_{t\to\infty}h(t)=0$, one easily checks that
$(P_f(x))^2=P_f(x)$, i.e. the operator $P_f(x)$ is a projection.
Let $C(H)$ denote the set of trace class operators on $H$ and let
$\chi_{(\varepsilon,\infty)}$ denote the characteristic function
of the interval $(\varepsilon,\infty)$ for some $\varepsilon>0$.
Put $q_\varepsilon=\chi_{(\varepsilon,\infty)}(x^*x)$. Due to the
property $(ii)$, one can choose the function $f$ such that
$p_{11}-q_\varepsilon\in C(H)$ and $p_{12},p_{21} \in C(H)$, and
due to the properties $(ii)$ and $(v)$, the function $f$ can be
chosen in such a way that $p_{22}-(1-q_\varepsilon)\in C(H)$.
Summing up the properties of an admissible function
$f:[0,\infty)\to [0,1)$, we have:
\smallskip
\begin{itemize}
\item[a)]
$f(0)=0$, $\lim_{t\to\infty}f(t)=1$ and the function
$g(t)=\sqrt{\frac{f(t)-f^2(t)}{t}}$ is continuous with $g(0)=0$;
\item[b)]
the series $\sum_{n>0}(f(\lambda_n^2)-1)$ and
$\sum_{n<0}f(\lambda_n^2)$ are square summable, where
$\lambda_n^2$ are the eigenvalues of $x^*x$.
\end{itemize}
\smallskip
Note that these properties imply $f(x^*x)-q_\varepsilon\in C(H)$
and $p_{12},p_{21}\in C(H)$. Together with $(v)$, they imply
$p_{22}-(1-q_\varepsilon)\in C(H)$).
Define a projection $Q$ by $$
Q=\left(\begin{array}{cc}q_\varepsilon&0\\
0&1-q_\varepsilon\end{array}\right). $$ After an appropriate
choice of $f$ among functions satisfying a) and b) one has
$P_f(x)-Q\in C(H\oplus H)$.
\begin{definition}
Put $\omega(x)=\tr(P_f(x)-Q)$.
\end{definition}
To prove that $\omega(x)$ is well-defined, we require the
following well-known lemma. Its proof can be found, e.g. in
\cite{Davidson}, Lemma VII.8.5.
\begin{lemma}\label{two_proj}
Let $P$ and $Q$ be projections in $B(H)$ with $P-Q\in C(H)$. Then
$\tr(P-Q)$ is an integer.
\end{lemma}
%\samepage{
\begin{theorem}
The number $\omega(x)$ is a well-defined integer and does not
depend on the choice of a bijection $m$ obeying \eqref{1}, of
$\varepsilon$ and of a function $f$ obeying {\rm a)} and {\rm b)}.
\end{theorem}
\begin{proof}
Since $P_f(x)$ and $Q$ are projections with $P_f(x)-Q\in C(H\oplus
H)$, $\tr(P_f(x)-Q)$ is an integer by Lemma~\ref{two_proj}.
Independence from the choice of $m$ and of $\varepsilon$ is
obvious. To check that $\tr(P_f(x)-Q)$ does not depend on our
choice of an admissible function $f$, take two such functions, $f$
and $f'$. Then the function $f_s=sf+(1-s)f'$, $s\in[0,1]$, also
satisfies a) and b), hence $P_{f_s}(x)-Q\in C(H\oplus H)$.
Continuity of $\tr(P_{f_s}(x)-Q)$ with respect to the parameter
$s$ completes the proof.
\end{proof}
%}
We calculate below $\omega(x)$ in some cases. We take
$H=l_2(\mathbb Z)$ and use the standard basis
$\{e_n\}_{n\in\mathbb Z}$ for $H$. Let $L_N\subset H$ be a
subspace generated by $e_n$, $-N\leq n\leq N$. Remark that, since
$P-Q\in C(H\oplus H)$, in order to calculate $\omega(x)$ it
suffices to calculate $\tr((P-Q)|_{L_N\oplus L_N})$ for big enough
$N$.
\begin{example} \label{ex1} Let $x$ be normal of the form
$xe_n=\lambda_ne_n$, where
$0\leq\ldots\leq\lambda_n\leq\lambda_{n+1}\leq\ldots$, $\lim_{n\to
-\infty}\lambda_n=0$, $\lim_{n\to\infty}\lambda_n=\infty$. Then
obviously $\tr(P|_{L_N\oplus L_N})=\tr(Q|_{L_N\oplus L_N})=N$,
hence $\omega(x)=0$.
\end{example}
\begin{example} \label{ex2} Let $x$ be of the form
$xe_n=\lambda_ne_{n+1}$, where
$0\leq\ldots\leq\lambda_n\leq\lambda_{n+1}\leq\ldots$, $\lim_{n\to
-\infty}\lambda_n=0$, $\lim_{n\to\infty}\lambda_n=\infty$. Then
$x^*xe_n=\lambda_n^2e_n$, $xx^*e_n=\lambda_{n-1}e_n$, and
$$
\tr(P|_{L_N\oplus L_N})=\sum_{n=-N}^N f(\lambda_n^2)+
\sum_{n=-N}^N(1-f(\lambda_{n-1}^2))=N-f(\lambda_{-N-1}^2)+f(\lambda_N^2).
$$
On the other hand, $\tr(Q|_{L_N\oplus L_N})=N$. Hence
$$
\lim_{N\to\infty}\tr((P-Q)|_{L_N\oplus L_N})=\lim_{N\to\infty}
f(\lambda_N^2)-f(\lambda_{-N-1}^2)=1,
$$
therefore, $\omega(x)=1$.
\end{example}
\begin{example} \label{ex3} Let $x$ be of the form
$xe_n=\lambda_ne_{n+k}$, where
$0\leq\ldots\leq\lambda_n\leq\lambda_{n+1}\leq\ldots$, $\lim_{n\to
-\infty}\lambda_n=0$, $\lim_{n\to\infty}\lambda_n=\infty$. Then
$x^*xe_n=\lambda_n^2e_n$, $xx^*e_n=\lambda_{n-k}e_n$, and
\begin{eqnarray*}
\tr(P|_{L_N\oplus L_N}) &=& \sum_{n=-N}^N f(\lambda_n^2)+
\sum_{n=-N}^N(1-f(\lambda_{n-1}^2)) \\
&=& N-\sum_{i=1}^k
f(\lambda_{-N-i}^2)+\sum_{i=1}^k f(\lambda_{N-i+1}^2).
\end{eqnarray*}
Once more, $\tr(Q|_{L_N\oplus L_N})=N$. Hence
$$
\lim_{N\to\infty}\tr((P-Q)|_{L_N\oplus L_N})=\lim_{N\to\infty}
\sum_{i=1}^kf(\lambda_{N-i+1}^2)-\sum_{i=1}^k
f(\lambda_{-N-i}^2)=k,
$$
therefore, $\omega(x)=k$.
\end{example}
\begin{example} \label{ex4} Important classes of examples are obtained
from Example \ref{ex2} when $\lambda_{n}^2$ belongs to orbits of
dynamical systems for all $n$, or more precisely when
$\lambda_{n+1}^2 = F(\lambda_{n}^2), n \in \mathbb{Z}$ in Example
\ref{ex2}, where $F:\mathbb{R} \rightarrow \mathbb{R}$ is a Borel
measurable mapping of the real line. Under conditions of Example
\ref{ex2} the operator $x$ satisfies the commutation relation
$x^*x = F(x^*x)$ as
$$ x^*x e_n = \lambda_n^2 e_n = F(\lambda_{n-1}^2)e_n = F(xx^*) e_n
$$
where the spectral mapping theorem was used in the last equality.
For example, if $F(t) = qt$ for $q>1$, then
$\lambda_n=\sqrt{F^{\circ n}(\lambda_0^2)}= \sqrt{q^n \lambda_0},
n\in \mathbb{Z}$ satisfies the conditions of Example \ref{ex2} for
any starting point $\lambda_0>0$. The operator $x$ satisfies the
`quantum plane' commutation relation $x^*x = q xx^*$. By the
result in Example \ref{ex2} we have $\omega(x)=1$.
\end{example}
Now we are going to give an interpretation of $\omega(x)$ in
$K$-theory terms. Let $k$ be the unbounded operator on $H$ given
by $ke_n=\mu_n e_n$, where
$0\leq\ldots\leq\mu_n\leq\mu_{n+1}\leq\ldots$ and $e_n$ is the
eigenbasis for $h$. Put $y=uk$. Then $y$ evidently satisfies the
properties $(i)-(v)$.
\begin{lemma}\label{contin}
One has $\omega(x)=\omega(y)$.
\end{lemma}
\begin{proof}
Put $x(t)=(1-t)x+ty$. Then $\omega(x(t))$ is continuous. But it is
integer-valued, hence constant.
\end{proof}
\begin{lemma}\label{asymp}
There exists a one-parameter family $(\alpha_t)_{t\in(0,\infty)}$
of homeomorphisms of $\mathbb R_+$ such that
\begin{equation}\label{commutn}
\lim_{t\to\infty}\|[u,\varphi(\alpha_t(h))]\|=0.
\end{equation}
for any $\varphi\in C_0([0,\infty))$.
\end{lemma}
\begin{proof}
We give the proof for the case, when all the eigenvalues
$\lambda_n$ of $h$ are distinct. The general case can be reduced
to this one due to Lemma~\ref{contin} by taking $y$ close to $x$
with distinct eigenvalues.
There exists a sequence $(\mu_n)_{n\in\mathbb Z}$ such that
$\ldots<\mu_{-1}<1=\mu_0<\mu_1<\mu_2<\ldots$ and
$|\mu_{m(n)}-\mu_n|\leq
2$ for all $n$. Take a homeomorphism $\psi:\mathbb R_+\to\mathbb
R_+$ such that $\psi(\lambda_n)=\mu_n$ and let
$\beta_t(s)=\frac{s}{t+1}$, $t>0$. Put
$\alpha_t=\psi^{-1}\circ\beta_t\circ\psi$ and
$\widetilde{\varphi}=\psi\circ\varphi\circ\psi^{-1}$. Then
(\ref{commutn}) is equivalent to
$\lim_{\to\infty}\sup_{n\in\mathbb
Z}|\widetilde{\varphi}(\frac{\mu_{m(n)}}{t+1})-\widetilde{\varphi}(\frac{\mu_n}{t+1})|=0$,
which is true because $\widetilde{\varphi}\in C_0([0,\infty))$
(this is obvious for functions with compact support, then one has
to approximate $\varphi$ by such functions).
\end{proof}
\begin{theorem}
$\omega(x)=\ind(uq_\varepsilon)$.
\end{theorem}
\begin{proof}
Without loss of generality we can assume that $f(r)=0$ for
$r\in(0,\varepsilon)$. Then
$\left(\begin{smallmatrix}q_\varepsilon&0\\0&q_\varepsilon\end{smallmatrix}\right)$
commutes with $P_f(x)$ and with $Q$ and
$\left(\begin{smallmatrix}1-q_\varepsilon&0\\0&1-q_\varepsilon\end{smallmatrix}\right)(P_f(x)-Q)=0$,
hence
\begin{equation}\label{11}
\omega(x)=\tr\left(\left(\begin{smallmatrix}q_\varepsilon&0\\0&q_\varepsilon\end{smallmatrix}\right)P_f(x)-
\left(\begin{smallmatrix}q_\varepsilon&0\\0&0\end{smallmatrix}\right)\right).
\end{equation}
Let $y=uk$, where $k=\sum_{n\in\mathbb Z}\mu_ne_n$, where
$\mu_n=0$ for $n<0$ and $\mu_n=\lambda_n$ for $n\geq 0$. By Lemma
\ref{contin}, $\omega(y)=\omega(x)$. Let $H_0=q_0(H)$ and let
$u_0=uq_0$, $k_0=q_0k$, $y_0=u_0k_0$. All these operators and all
further formulas are in the Hilbert space $H_0$. The formula for
$\omega(y_0)$ in $H_0$ is written as
$\omega(y_0)=\tr\left(P_f(y_0)-\left(\begin{smallmatrix}0&0\\0&1\end{smallmatrix}\right)\right)$
and it follows from (\ref{11}) that $\omega(y_0)=\omega(x)$.
Let $C_{0}(\mathbb R^2)$ be the $C^*$-algebra of continuous
functions on $\mathbb R^2$ vanishing at infinity. This algebra is
generated by the functions $e^{2\pi i\theta}$ and $\varphi(r)$,
$\varphi\in C_0([0,\infty))$, where $\theta$ and $r$ are the polar
coordinates on the plane. Note also that the operator
$u\varphi(h)$ is compact for any $\varphi\in C_0([0,\infty))$. In
what follows the main tool will be $E$-theory of Connes and Higson
\cite{CH} based on the notion of asymptotic homomorphisms.
Recall that an asymptotic homomorphism from a $C^*$-algebra $A$ to a
$C^*$-algebra $B$ is a family of maps
$\tau=(\tau_t)_{t\in[0,\infty)}:A\to B$ such that $t\to\tau_t(a)$ is
continuous for each $a\in A$ and
$\tau_t(a_1a_2)-\tau_t(a_1)\tau_t(a_2)$,
$\tau_t(\lambda a_1+a_2)-\lambda\tau_t(a_1)-\tau_t(a_2)$ and
$\tau_t(a^*)-\tau_t(a)^*$ vanish as $t\to\infty$ for all $a,a_1,a_2\in
A$ and all $\lambda\in\mathbb C$.
\begin{corollary}\label{as}
The operator $y_0$ defines, by $e^{2\pi i\theta}\mapsto u_0$,
$\varphi(r)\mapsto \varphi(\alpha_t(k_0))$, the asymptotic
homomorphism $\tau^{y_0}=(\tau_t^{y_0})_{t\in(0,\infty)}$ from
$C_{0}(\mathbb R^2)$ to the $C^*$-algebra $K(H_0)$ of compact
operators.
\end{corollary}
Thus, $x$, and then $y_0$, defines a class $[\tau^{y_0}]$ in the
$E$-theory group $E(C_{0}(\mathbb R^2),K(H))$, which is the same
as $K$-homology group of $C_{0}(\mathbb R^2)$. Let $e$ be a class
in $K$-theory group of $\mathbb R^2$. Then the pairing between
$K$-theory and $K$-homology provides us with an integer
${<}e,[\tau^{y_0}]{>}$.
Recall that the Bott class $\beta$ in $K_0(\mathbb R^2)$ is given
by the formal difference of projections $p$ (\ref{p}) and
$\left(\begin{smallmatrix}0&0\\0&1\end{smallmatrix}\right)$.
\begin{lemma}
One has $\omega(y_0)={<}\beta,[\tau^{y_0}]{>}$, where $\beta\in
K_0(\mathbb R^2)$ is the Bott class.
\end{lemma}
\begin{proof}
Note that $p$ is a two-by-two matrix over the algebra $C_0(\mathbb
R^2)$ with adjoined unit and we can extend $\tau_t^{y_0}$ onto
this algebra by defining $\tau_t^{y_0}(1)=1$. Then it evidently
follows from Corollary \ref{as} that
$\lim_{t\to\infty}\|\tau_t^{y_0}(p)-P_f(y_0)\|=0$. Without loss of
generality we can assume that the function $f$ satisfies $f(r)=1$
for $r\in(N,\infty)$ for sufficiently large $N$. Then the operator
$\tau_t^{y_0}(p)-
\left(\begin{smallmatrix}0&0\\0&1\end{smallmatrix}\right)$ is of
finite rank and one has
${<}\beta,[\tau^{y_0}]{>}=\lim_{t\to\infty}\tr\left(\tau_t^{y_0}(p)-
\left(\begin{smallmatrix}0&0\\0&1\end{smallmatrix}\right)\right)=\omega(y_0)$
(cf. \cite{Loring}).
\end{proof}
To finish the proof of the theorem, note that $u_0$ is Fredholm,
hence it defines a homomorphism $\rho:C(\mathbb T)\to
B(H_0)/K(H_0)$ and (which is the same) an extension of $C(\mathbb T)$
by the $C^*$-algebra of compact operators on $H_0$.
The class of this extension in the $K$-homology
group $K_1(\mathbb T)$ is given by $\ind(u_0)$. On the other hand,
since the Connes--Higson construction \cite{CH} applied to the map
$\rho$ gives the asymptotic homomorphism $\tau^{y_0}$ (this is
because $(1-f)(\alpha_t(k_0))$ is an approximate unit in $K(H_0)$,
asymptotically commuting with $u_0$), we have
$\ind(y_0)={<}\beta,[\tau^{y_0}]{>}$. Finally note that the
Fredholm operators $uq_\varepsilon$ and $u_0$ differ by a finite
rank operator, hence they have the same index.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\end{document}