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\begin{document}

%topmatter
\title{On double commutator relation}
\author{Vasyl Ostrovs'ky\u\i}
\address{Institiute of Mathematics,
  National Acad.\ Sci.\ of Ukraine, Kiev, Ukraine}
\email{vo@imath.kiev.ua}
\date{\today}
\maketitle
% end of topmatter

\noindent
In this paper we consider representations of the $*$-algebra generated
by two self-adjoint elements, $a$, $b$, satisfying the relation 
\begin{equation}\label{double}
[a,[a,b]] =0.
\end{equation}
When considering representations of this algebra by bounded
self-adjoint operators, $A$ and $B$,
notice, first of all, that by virtue of the Kleineke-Shirokov theorem,
for bounded $B$ we have always $[A,B]=0$. 
On the other hand, this algebra has  as partial cases 
such important examples as representations of CCR $[A,B]=iI$, 
and representations of two-dimensional solvable Lie algebra
$[A, B]=iA$, which have ``good'' non-trivial representations by unbounded
operators. Thus, it is natural to  consider representations by 
unbounded operators. But since unbounded operators cannot be defined 
on the whole space, one should specify the operator sense of 
the relation. Probably, the best way here would be to associate 
to relation \eqref{double} a $C^*$-algebra $\mathfrak{A}$ such that
the operators $A$ and $B$ are affiliated with $\mathfrak{A}$ and
generate it (in the sense described in \cite{wor_aff_2}). 
But since it is not clear how  to construct a natural $C^*$-algebra 
associated with \eqref{double},  we  rewrite 
the equation in terms of bounded functions of the operators
$A$, $B$. Note that there is no canonical way of doing it for 
arbitrary algebraic relation; thus, choosing a definition
we select a class of ``good'' representations, which, certainly,
depends on the  accepted definition. 

For example, in \cite{niztur}, the class of representations was
studied for which there exists a dense invariant domain  consisting  
of analytic vectors for $A$ and $B$. It is shown that such class of
representations is ``$*$-wild''.

Another problem is the following. For bounded representations, it is
natural that the operators, $A$, $B$ which represent the generators
are assumed self-adjoint. In the unbounded case, the condition that
the operators $A$ and $B$ are self-adjoint may appear too
restrictive. We assume that the operator $A$, remains self-adjoint,
while  $B$ is are closed (not necessarily self-adjoint) 
extension of a densely defined symmetric operator. We will see that
the considered class contains natural representations for which there
are no invariant domains consisting of analytic vectors. For example,
the relation $[a,b]=ia^3$ which obviously implies \eqref{double} have
a unique representation in the described class, and for this
representation the operator $B$ is symmetric with deficiency indices
(1,0). 



\begin{definition}
We say that a pair, $A$, $B$  is a representation of  relation
\eqref{double}, if $A$ is self-adjoint, and the following relations 
between bounded operators
\begin{equation}\label{bounded}
[E_A(\Delta), U_t^* E_A(\Delta') U_t]=0, 
\quad [E_A(\Delta), U_t E_A(\Delta')U_t^*]=0
\end{equation}
hold for all $t\ge0$, $\Delta$, $\Delta' \in \mathfrak{B}(\mathbb{R})$.
Here and below, $E_A(\cdot)$ is a resolution of the identity 
of the operator $A$, $U_t = e^{itB}$ is a strongly continuous
semigroup of partial isometries.
\end{definition}

\begin{remark}
Since $U_t^* E_A(\Delta)U_t = E_{U_t^*AU_t}(\Delta)$,
we see that \eqref{double} means that the self-adjoint operators
$A$ and $\frac1i\frac d{dt} U^*_tAU_t\big|_{t=0}$ commute in the sense of
commutation of their spectral projections.
On the other hand, if $u\in H$ is a vector on which the operators $AB$ 
and $BA$ are defined, then there exists a derivative
\[
\frac{d}{dt}U_t^* A U_t\Big|_{t=0}f = i(ABf - BAf).
\]
\end{remark}

\begin{remark}
Consider a partial case of relations \eqref{double} --- 
unbounded representations of $*$-algebra generated by a
pair of self-adjoint elements obeying the relation of the
\begin{equation}
\label{poly}
[a,b] = i \,p(a),
\end{equation}
where $p(\cdot)$ is a real polynomial.

Since we are interested in unbounded representations, we need to
discuss the operator sense of the relation  more closely. To do it, we
do some formal calculations.

Let $\phi(\cdot)$ be a polynomial. From \eqref{poly} we have
the following relations 
$$
\phi(a)\, b  = b \,\phi(a) + i \phi'(a)\, p(a),
$$
which can be calculated directly. Then, by induction we get
$$
a\,b^n  = \sum_{k=0}^n i^k C^k_n \,b^k \,\phi_{n-k}(a),
$$
where the polynomials $\phi_i(\cdot)$ are defined as 
 $\phi_0(x) =x$, $\phi_1(x) = p(x)$, $\phi_n(x) = \phi_{n-1}'(x)
\,p(x)$, $n >1$. 


These relations enable one to get some formal relations  for
bounded functions of the generators, which have no sense in the
$*$-algebra but will be assumed for the operators in any
representations. Since the derived relations include only bounded
operators, this eliminates the ambiguity with the sense of the
relation in the unbounded case. 

First, we find, how the formal exponent of $b$ permutes with $a$. According
to the above relations, we have
\begin{align*}
a \, e^{tb} & = a \sum_{n =0 }^\infty \frac{t^n b^n}{n!} =
\sum_{n=0}^\infty \frac {t^n a b^n}{n! } = \sum _{n=0}^\infty
\frac{t^n}{n!} \sum_{k=0}^n i^{n-k} C^k_n b^k \phi_{n-k}(a)
\\
&= \sum_{n=0}^\infty \sum_{k=0}^n \frac{t^k }{k!} b^k
\,\frac{t^{n-k}}{(n-k)!} i^{n-k} \phi_{n-k}(a) 
\\
&= \sum_{k=0}^\infty \frac {t^k b^k}{k!} \sum_{l=0}^\infty \frac{t^l
  i^l}{l!} \phi_l(a) = e^{tb} S_t(a),
\end{align*}
where we write
\[
S_t(x) = \sum_{l=0}^\infty \frac{t^l i^l}{l!} \phi_l(x).
\]

Similarly to \cite{osbook}, we assume that in any representation the
operator $A$ representing $a$ is self-adjoint, and replace it with its
spectral projections. We have the relation for unbounded operators
$A$, $B$ in terms of bounded functions
\begin{equation}
\label{int}
E_A(\Delta) U_t = U_t E_A (S_t^{-1}(\Delta)), \quad E_A(\Delta) U_t^* =
U_t^* E_A (S_{-t}^{-1}(\Delta)),    
\end{equation}
$t \ge 0$,  $\Delta \in
\mathfrak{B}(\mathbb{R})$, which are equivalent to \eqref{bounded} in
the particular case of 
\eqref{poly}.

Operator relations of the form \eqref{int} with a one-parameter group
$U_t$ were studied in \cite{dal} in the framework of abstract
hyberbolic equations. 
\end{remark}

We denote $U^*_t AU_t$ by $A_t$, and put $A_{-t}= U_tAU_t^*$, 
$t\ge0$. Obviously, $A_t$, $t\in
\mathbb{R}$ are self-adjoint operators. We use the following
fact.

\begin{lemma}
Operators $A_t$, $t\in \mathbb{R}$, commute in terms of
their spectral projections.
\end{lemma}

\begin{proof}
First notice that the operators $P_t=U_t^*U_t$, $P_{-t} =U_tU_t^*$ 
commute with $E_A(\Delta)$. Indeed, it is sufficient to put $\Delta'=
\mathbb{R}$ in \eqref{bounded}. Also, since $U_s$ are partial isometries, 
we have $U_s=U_sU_s^*U_s$, $U_s^*=U_s^* U_sU_s^*$. 
Let $\Delta$, $\Delta' \in \mathfrak{B}(\mathbb{R})$. For
any $t>s\ge0$, we have by virtue of \eqref{bounded} and the remarks above
\begin{align*}
&E_{A_t}(\Delta) E_{A_s}(\Delta')= 
\\
&= U_t^* E_A(\Delta)U_t
U_s^* E_A(\Delta') U_s 
= U_s^* (U_{t-s}^*E_A(\Delta)U_{t-s})U_sU_s^*E_A(\Delta')U_s
\\
&= U_s^*(U_{t-s}^* E_A(\Delta)U_{t-s})
E_A(\Delta') U_s 
 = U_s^*E_A(\Delta') (U_{t-s}^* E_A(\Delta)U_{t-s})  U_s
\\
&= U_s^* E_A(\Delta')U_sU_s^* U_{t-s}^*E_A(\Delta)U_t
= U_s^*E_A(\Delta') U_s U_t^* E_A(\Delta) U_t
\\
&=  E_{A_s}(\Delta') E_{A_t}(\Delta).
\end{align*}
The case where $0\le s<t$ is treated similarly. 

Now we show that $E_{A_t}(\Delta)$ commutes with
$E_{A_{-s}}(\Delta')$, $t$, $s\ge0$, for all measurable $\Delta$, $\Delta'$.
First we show that $P_t$ commutes with $A_{-s}$. Indeed, we know that
$P_{t+s} E_A(\Delta) = E_A(\Delta) P_{t+s}$, or
\[
U^*_{t+s}U_{t+s} E_A(\Delta) = E_A(\Delta) U_{t+s}^* U_{t+s}.
\]
Multiplying this equality by $U_s$ from the left and by $U_s^*$ from
the right, we get
\[
P_{-s} P_t U_sE_A(\Delta) U_s^* = U_sE_A(\Delta) U_s^* P_t P_{-s}.
\]
But since $U_t$ is an inverse semigroup, $P_t$ and $P_{-s}$ 
commute (see \cite{inverse}), and $P_{-s}U_s = U_s$,
$U_s^*P_{-s} = U_s^*$, we see that $P_t $ commutes with the spectral
projections of $A_{-s}$.

Similar arguments show that $P_{-s} $ commute with the spectral
projections of $A_p$.

Now for $t$, $s\ge0$ we have
\begin{align*}
&E_{A_t}(\Delta) E_{A_{-s}}(\Delta') =
\\
&= U_t^* E_a(\Delta) U_tU_sE_A(\Delta') U_s^* = U_t^* E_A(\Delta)
U_{t+s} E_A(\Delta ') U_s^*
\\
&= U_t^* E_A(\Delta) P_{-(t+s)}U_{t+s} E_A(\Delta') U_s^* = U_t^*
P_{-(t+s)} E_A(\Delta ) U_{t+s} E_A(\Delta') U_s^*
\\
&=U_t^* U_{t+s} E_{A_{t+s}}(\Delta) E_A(\Delta ' )U_s^* = U_t^*
U_{t+s} E_A(\Delta') E_{A_{t+s}}(\Delta) U_s^*
\\
&= P_t E_{A_{-s}}(\Delta') E_{A_t}(\Delta)P_{-s} =
E_{A_{-s}}(\Delta') E_{A_t}(\Delta),
\end{align*}
since $P_t$ commutes with $A_{-s}$, $P_{-s}$ commutes with $A_t$, and
$P_tA_t = A_t$, $A_{-s}P_{-s} = A_{-s}$,
which completes the proof.
\end{proof}

\begin{proposition}
The space $H$ can be decomposed into a direct sum of invariant
subspaces,
$$
H = H_1 \oplus H_+ \oplus H_- \oplus H_0
$$ 
in such a way that in $H_1$ the
operators $U_s$, $s\ge0$,  are unitary, in $H_+$ they are completely
non-unitary isometries, in $H_-$ they are adjoint to complete
non-unitary isometries, and in $H_0$ we have $U_t \to 0$, $U_t^* \to
0$, $t\to\infty$ strongly (Wold decomposition).
\end{proposition}

\begin{proof}
For the projections $P_t$, $Q_t=P_{-t}$, $t\ge0$, we have $P_t P_s =
P_s$, $Q_tQ_s= Q_s$, $s>t$. Therefore, these families of projections
are monotone and there exist projections $P_\infty = \lim_{t\to\infty}
P_t$, $Q_\infty = \lim_{t\to \infty} Q_t$. These projections commute
with each other, with $P_t$, $t\in \mathbb{R}$, and with the spectral
projections of $A_t$, $t\in \mathbb{R}$, which implies that the
projections $P_1 = P_\infty Q_\infty$, $P_+ = P_\infty - P_1$, $P_- =
Q_\infty - P_1$, $P_0 = I - P_1 - P_+ - P_-$ decompose $H$ into four
invariant with respect to $A_t$, $U_s$, $U_s^*$, $t\in \mathbb{R}$,
$s\ge0$, subspaces $H_1$, $H_+$, $H_-$, $H_0$ which possess the needed
properties.   
\end{proof}
 
Notice that directly from the definition of the operators
$A_t$ it follows that the operators $U_t$ are connected with
the commuting family $A_s$ by the following relations
\[
U_t A_s = U_t U_s^* A U_s = U_{t-s}^* AU_{t-s} U_s = A_{t-s}
U_t
\]


Consider first the unitary case (operators in the subspace $H_1$).
Using the relations above,
we can apply the theorem on commutative models
\cite{bos,berkon,romp} to get the following spectral
decomposition.

\begin{theorem}
Let $A$, $B$ be a pair of self-adjoint operators satisfying
\eqref{bounded}. Then it is unitary equivalent to a pair on
a Hilbert space
\[
H = L_2(\mathbb{R}^\mathbb{R}, d\rho) \otimes \mathfrak{H}
%{\oplus}\int_{\mathbf{R}^\mathbb{R}} H_\lambda)\cdot)\, d\rho(\lambda(\cdot))
\]
by the following formulas
\begin{align*}
(A f)(\lambda(\cdot)) &= \lambda(0) \, f(\lambda(\cdot)),
\\
(U_t f)(\lambda(\cdot)) &= u_t(\lambda(\cdot))\, \tilde
\rho_t(\lambda(\cdot))\, f(\lambda(\cdot +t))
\end{align*}
where $\mathfrak{H}$ is a Hilbert space, $\mathbb{R}^\mathbb{R}$
is a set of all functions on $\mathbb{R}$, $\rho$ is a
probability measure on $\sigma$-algebra of cylinder sets,
quasi-invariant with respect to transformation
$\lambda(\cdot) \mapsto \lambda(\cdot + t)$, $t\in
\mathbb{R}$; $\tilde \rho_t(\lambda(\cdot)) =
\bigl[d\rho(\lambda(\cdot+t))/d\rho(\lambda(\cdot))\bigr]^{1/2}$,
$u_t(\lambda(\cdot))$ is a unitary operator-valued function.
\end{theorem}

In the latter theorem, the spectral measure of the
commuting family $A_s$ can be considered on any
(non-measurable) invariant subset
$\Delta\subset\mathbb{R}^\mathbb{R}$ having full outer
measure (see \cite{ber0} for the corresponding modification
procedure). In particular, to any orbit of $\mathbb{R}$ in
$\mathbb{R}^\mathbb{R}$ there corresponds a unique, up to equivalence,
(quasi-)invariant measure concentrated on the orbit. We consider a
class of such measures and the corresponding irreducible 
representation.

The mapping $\lambda(\cdot) \mapsto \lambda(\cdot +t)$
determines a measurable action of $\mathbb{R}$ on
$\mathbb{R}^\mathbb{R}$. There can be three types of orbits
of this action: free (non-periodic) action, periodic action,
and trivial (fixed-point) action. In each of these cases there
is a unique (up to equivalence) quasi-invariant measure on
an orbit. To obtain a more natural realization of
representations, we pass to a unitary equivalent
realization. Let $\Omega$ be an orbit. Fix $\lambda(\cdot)
\in \Omega$ and consider a mapping $t \mapsto \lambda(\cdot + t)$.
This mapping allows one to equip an orbit with a measure in the
following way: take an image of the Lebesgue measure on the
line for free action, image of the Lebesgue measure on the
interval (one period) for periodic action, and the point
measure for trivial action. Then we have a canonical unitary
isomorphism of $H$ with $L_2(\mathbb{R}, dx)$ for a free
action, with $L_2([0,a], dx)$ ($a$ is the smallest period)
for a periodic case, and $H=\mathbb{C}$ for a  trivial
action. We have $L_2$ instead of a direct sum here due to
the irreducibility. Some more calculations allow one to classify
the corresponding functions $u(\lambda)$ which appear in
Theorem~1.

\begin{theorem}
Any irreducible representation in $H_1$ corresponding to an orbit is
unitary equivalent to one of the following\textup{:}

i\textup{)} representation in $L_2(\mathbb{R}, dx)$ \textup(corresponding to
a free action\textup)\textup{:}
\[
(Af)(x) = \phi(x)\, f(x), \qquad (Bf)(x) = i^{-1}
\frac{d}{dx}\, f(x);
\]
here $\phi(x)$ is a measurable a.e.\ finite non-periodic
function\textup{;} 
the operators are defined on their natural domains\textup{;}

ii\textup{)} representation in $L_2([0,a], dx)$ \textup{(}corresponding to a
periodic action\textup{):}
\[
(Af)(x) = \phi(x)\, f(x), \qquad (Bf)(x) = i^{-1}
\frac{d}{dx}\, f(x);
\]
here $\phi(x)$ is a measurable a.e.\ finite function\textup{;} the
operator $A$ is defined on its natural domain\textup{,} and the domain
of $B$ consists of absolutely continuous functions on
$[0,a]$ satisfying the boundary conditions $f(0) = \alpha
f(a)$\textup{,} $|\alpha|=1$\textup{;} notice that in this family an
additional parameter $\alpha$ appears\textup{;}

iii\textup{)} representation in $\mathbb{C}$ \textup{(}corresponding to a
stationary points\textup{):}
\[
A= x, \qquad B= y,
\]
$x$, $y$ are real numbers.
\end{theorem}

Note that, apart from these representations, there are a lot
of irreducible representations corresponding to ergodic
measures of $\mathbb{R}^\mathbb{R}$ which are not
concentrated on a single orbit. The problem of description
of such representations seems to be very complicated. We
just give an example of such representation.

\begin{example}
Fix a real measurable function $f$ on a two-dimensional
torus $\mathbb{T}^2$. For each $t\in \mathbb{R}$ define a
mapping $S_t\colon \mathbb{T} \to \mathbb{T}$ as follows:
\[
S_t(\alpha, \beta) = (\alpha e^{irt}, \beta e^{it}),
\]
where $r$ is an irrational number.

The mapping
\[
\mathbb{T}^2\ni \lambda \mapsto f_\lambda(x) =f(S_x(\lambda))
\in \mathbb{R} ^\mathbb{R}
\]
allows to construct an invariant measure on
$\mathbb{R}^\mathbb{R}$ as an image of the Lebesgue measure
on $\mathbb{T}^2$. It is easy to show that this measure is
also ergodic.

Consider a Hilbert space $L_2(\mathbb{T}^2, d\lambda)$ and a
pair of operators
\[
(A\phi)(\lambda) = f(\lambda)\, \phi(\lambda), \qquad
(e^{itB}f)(\lambda) = f(S_t(\lambda)).
\]
By an appropriate choice of $f$ one can provide that this
representation be irreducible. On the other hand, it is
obvious that the spectral measure of the family $A_t$ is not
concentrated on a single orbit.
\end{example}

In the case of $H_+$, $H_-$, or $H_0$, only representations similar to
(i) in Theorem~2 arise.

\begin{theorem}
Any irreducible representation of \eqref{bounded} in $H_+$, $H_-$, or
$H_0$ acts in the space $L_2([0,\infty), dx)$, $L_2((-\infty,0],dx)$,
$L_2([0,a],dx)$ with some $a>0$ respectively by the formulas
$$
(Af)(x) = \phi(x)\, f(x), \quad (Bf)(x) = i^{-1} \frac{d}{dx}\, f(x),
$$
where $\phi(x)$ is a measurable a.e.\ finite function, the operator
$A$ is defined on a natural domain, and the domain of $B$ consists of
absolutely continuous functions $f(\cdot)$ with square integrable
derivatives satisfying the following boundary conditions:
$f(0)=0$ for $H_+$, no boundary conditions in $H_-$, $f(0)=0$ and no
boundary conditions at $a$ for $H_0$.
\end{theorem}

\begin{remark}
In the cases listed in the latter theorem, the operator $A$ is
self-adjoint, while $B$ is not. Therefore, there is no invariant
domain consisting of analytic vectors for these operators.
\end{remark}

\begin{proof}
The theorem follows from the following statements.


It is known (see \cite{nafo}) that an irreducible semigroup of
isometries (the case $H_+$) acts on the space $L_2([0,\infty),dx)$ as 
$(U_tf)(x) = \chi_{[t,\infty)}\,f(x-t)$, $t\ge0$, 
and its adjoint semigroup (the case
$H_-$) is unitary equivalent to the semigroup in $L_2((-\infty,0],dx)$
acting as $(U_tf)(x) = f(x-t)$, $t\ge0$. The
action in $H_0$ is given by the following statement \cite{semi}.

\begin{lemma}
Any irreducible semigroup of partial isometries in $H_0$ is unitary
equivalent to the following semigroup in $L_2([0,a],dx)$ with some
$a>0$
$$
(U_tf)(x) = \chi_{[t,a]}(x) \, f(x-t).
$$
\end{lemma}

\begin{lemma}
For any irrducible representation of \eqref{bounded} in $H_+$, $H_-$, or
$H_0$, the semigroup $U_t$ is irreducible, i.e., any bounded operator
commuting with $U_t$, $U_t^*$, $t\ge0$, is a multiple of the identity.
\end{lemma}

\begin{proof}
Consider the case of $H_+$. Then $H= H_+ = L_2([0,\infty),dx) \otimes
H'$, and $U_t =S_t\otimes I$, where $S_t$ is a semigroup of one-sided
shifts in $L_2([0,\infty),dx)$. The operators $Q_t = U_tU_t^*$ act as
multiplication by $\chi_{[t,\infty)}$. Since $A$ commutes with $Q_t$,
$t\ge0$, we have
$$
(Af)(x) = a(x)\, f(x),
$$
where $a(\cdot)$ is a measurable  operator-valued
function, and $a(x)$ is a self-adjoint operator in $H'$. Since $A_t =
U_t^*AU_t$ is a multiplication by $a(x+t)$, and $A_t$ commute for
all $t$, we conclude that $a(x)$, $x\ge0$, form a commuting family in
$H'$. Let $P$ be some joint spectral projection of this commuting
family. Then $I\otimes P$ commutes with $U_t$, $U_t^*$ and $A$;
therefore, by the irreducibility, $H' = \mathbb{C}$, and $H= H_+ =
L_2([0,\infty),dx)$, where $U_t$, $U_t^*$, $t\ge0$, form an
irreducible family.
\end{proof}

It is easy to verify that the corresponding operator $B$ is
$i^{-1}\frac{d}{dx}$ with the domain described in the statement.

To complete the proof, take a standard realization for $U_t$ and
notice that the projections $P_t$, $Q_t$ act as mutiplication by
indicator functions which form a total set in the corresponding space
$L_2([0,\infty),dx)$, $L_2((-\infty,0],dx)$, or $L_2([0,a],dx)$. Since
these projections commute with $A$, we conclude that $A$ is a
multiplication by some real measurable function. 
\end{proof}

%\bibliography{ref}
%\bibliographystyle{amsplain}

\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
\begin{thebibliography}{1}

\bibitem{ber0}
Yu.~M. Berezanski{\u\i}, \emph{Self-adjoint operators in spaces of functions of
  infinitely many variables}, Trans. Math. Monographs, vol.~63, AMS,
  Providence, 1986.

\bibitem{berkon}
Yu.~M. Berezanski\u\i{} and Yu.~G. Kondrat'ev, \emph{Spectral methods in
  infinite-dimensional analysis}, Naukova Dumka, Kiev, 1988, (Russian).

\bibitem{bos}
Yu.~M. Berezanski\u\i{}, V.~L. Ostrovski\u\i{}, and Yu.~S. Samo\u\i{}lenko,
  \emph{Eigenfunction expansion of families of commuting operators and
  representations of commutation relations}, Ukr. Math. Zh. \textbf{40} (1988),
  no.~1, 106--109, (Russian).

\bibitem{dal}
Yu.~L. Daletskii. \emph{Continual integrals related to operator 
evolution equations}, Uspekhi Mat. Nauk \textbf{XVII} (1962), no.~5, 3--115.

\bibitem{nafo}
B.~Sz.-Nagy and C.~Foia\c s, \emph{Harmonic analysis of operators on
  Hilbert space}, North-Holland Publishing Co., Amsterdam---London; 
American Elsevier Publishing Co., Inc., New York; Akad\'emiai Kiad\'o,
Budapest 1970.

\bibitem{niztur}
L.~P.~Nizhnik and L.~B.~Turowska, \emph{Representations of double
  commutator by matrix-differential operators}, Methods
Funct. Anal. Topol. \textbf{3} (1997), no.~3, 75--80.

\bibitem{semi}
V.~Ostrovskyi, \emph{Centered one-parameter semigroups}, in preparation.

\bibitem{romp}
V.~L. Ostrovski\u\i{} and Yu.~S. Samo\u\i{}lenko, \emph{Unbounded operators
  satisfying non-{L}ie commutation relations}, Repts. math. phys. \textbf{28}
  (1989), no.~1, 91--103.

\bibitem{osbook} 
V.~Osrovsky\u{\i} and Yu.~Samo\u{\i}lenko, \emph{Introduction to the
  theory of representations of finitely presented
  $*$-algebras. I.~Representations by bounded operators},
Rev. Math. Math. Phys. \textbf{11} (1999), part~1, 1--261.

\bibitem{inverse}
 A.~L.~T.~Paterson, \emph{Groupoids, inverse semigroups, and their
 operator algebras}. Progress
in Mathematics, \textbf{170}. Birkh\"auser Boston, Inc., Boston, MA, 1999. 

\bibitem{wor_aff_2}
S.~L. Woronowicz, \emph{{$C^*$}-algebras generated by unbounded elements},
  Rewiews in Mathematical Physics \textbf{7} (1995), no.~3, 481--521.

\end{thebibliography}

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