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\begin{center}
{\Large \bf Generalized parabosons and polynomial \\[2mm]
deformations of $osp(1/2)$ }\\[1cm]
{\bf J.\ Van der Jeugt\footnote{Senior Research
Associate of N.F.W.O. (National Fund for Scientific Research of
Belgium).}}\\[5mm] 
Department of Applied Mathematics and Computer Science,
University of Ghent,\\ 
Krijgslaan 281-S9, B-9000 Gent, Belgium.\\[2mm]
E-mail : Joris.VanderJeugt@rug.ac.be.
\end{center}
\vskip 1cm
\noindent {\bf Abstract:}
Inspired by various considerations of generalized boson or paraboson
algebras, we
consider the algebra $R$ generated by three elements $A,B,H$
subject to three relations $[H,A]=A$, $[H,B]=-B$ and
$\{A,B\}=f(H)$. When $f(H)=H$ this coincides with the Lie
superalgebra $osp(1/2)$; when $f$ is a polynomial we speak of
polynomial deformations of $osp(1/2)$. Irreducible
representations of $R$ are described, and in the case
$\deg(f)\leq 2$ we obtain a complete classification, showing
some similarities but also some interesting differences with the
usual $osp(1/2)$ representations. 

\vskip 1cm

\section{Introduction}

This paper has a great amount of overlap with a paper published
in Journal of Mathematical Physics~\cite{VJ}. Nevertheless, I was asked
to write a separate contribution for the
proceedings of the Workshop in Razlog, especially as it might be useful
for the students who have attended the Workshop. In view of this, I
have included a number of exercises in this paper.

In order to introduce parabosons, let us first consider Wigner quantum
systems. For a quantum mechanical system of $n$ particles,
with coordinates $x_k$, momenta $p_k$ and masses $m_k$, the Hamiltonian is
described as
\beq
H=\sum_{k=1}^n {p_k^2\over 2 m_k} + V(x_1,\ldots, x_n)
\eeq
with $V$ a potential. The time evolution of the system is given by the
Heisenberg equations
\beq
\dot p_k = -{i\over \hbar} [p_k,H],\qquad 
\dot x_k = -{i\over \hbar} [x_k,H].\qquad 
\eeq
For a given (analytic) fucntion $F=F(x_1,\ldots,x_n,p_1,\ldots,p_n)$,
the canonical commutation relations
\beq
[x_k,p_l]=i\hbar\delta_{kl},\qquad [x_k,x_l]=[p_k,p_l]=0,
\eeq
imply that
\beq
-{i\over \hbar} [p_k,F]=-{\partial F\over \partial x_k} \ \hbox{ and }
\ -{i\over \hbar} [x_k,F]=-{\partial F\over \partial p_k};
\eeq
thus
\beq
\dot p_k=-{\partial H\over \partial x_k}, \qquad
\dot x_k=-{\partial H\over \partial p_k},
\eeq
i.e.\ the Hamilton equations. Thus the Heisenberg equations and the
canonical commutation relations (CCR's) imply the Hamilton equations.
Similarly one can show that the Hamilton equations and the CCR's imply
the Heisenberg equations. A natural question is whether the Heisenberg
and the Hamilton equations would imply the CCR's. The answer to this
question is no, and the first example is due to Wigner~\cite{Wigner}. 
He showed that
for the one-dimensional harmonic oscillator with $H=p^2/2+x^2/2$ (we
have now put $m=1$ and $\hbar=1$), if one defines as usual
\beq
b^\pm= (x\mp ip)/\sqrt{2},
\label{b}
\eeq
the compatibility of the Hamiltonian and Heisenberg equations only lead
to 
\beq
 [ \{b^+,b^-\},b^\pm] = \pm 2b^\pm.
\label{parabose}
\eeq
This relation is known as the parabose relation. When $x$ and $p$ are
required to satisfy the CCR $[x,p]=i$, then the corresponding operators
(\ref{b}) satisfy the bose relation
\beq
b^-b^+-b^+b^-=1,
\label{bose}
\eeq
and the parabose relation then follows automatically. For the algebra
generated by $b^-$ and $b^+$, subject to (\ref{bose}) and the
``unitarity condition'' $(b^\pm)^\dagger=b^\mp$, there is only one
unitary representation, namely the Fock space with basis vectors
\beq
|n\rangle = {(b^+)^n\over \sqrt{n!}} |0\rangle, \qquad n=0,1,2,\cdots,
\eeq
where $|0\rangle$ is defined by $b|0\rangle=0$. When $x$ and $p$ are
not required to satisfy the CCR, but only (\ref{parabose}), then
(\ref{bose}) is in general not satisfied. The system defined by
(\ref{parabose}) is equivalent to the system 
\beq
H={1\over 2} (b^-b^+ + b^+b^-),\quad [H,b^\pm]=\pm b^\pm.
\eeq
These are the defining relations for the universal enveloping algebra
of the Lie superalgebra $osp(1/2)$. Its irreducible unitary
representations are characterized by a positive real number $\lambda$,
and the spectrum of $H$ is $\lambda +n$ ($n=0,1,2,\cdots$). The lowest
weight state satisfies $b^-|\lambda\rangle=0$ and $H|\lambda\rangle=
\lambda |\lambda\rangle$; the rest of the basis vectors and the action
on them is then easy to calculte. In particular, only for $\lambda=1/2$
do the representatives of $b^\pm$ satisfy the relation $[b^-,b^+]=1$;
in the remaining representations the bose relation is no longer valid,
in other words the CCR is not satisfied, and the system is called a
Wigner quantum system~\cite{PalevWQS}[and references therein].

When considering the bose relation (\ref{bose}), it is common to
introduce a number operator $N$, and then the single relation
(\ref{bose}) is equivalent to the following four relations
\beq
b^+b^-=N,\quad b^-b^+=N+1,\quad [N,b^\pm]=\pm b^\pm.
\eeq
Inspired by this form, and also by so-called $q$-bosons appearing in
Schwinger-type realizations of quantum groups, a number of papers have
introduced the following relations for generalized bosons~:
\beq
a^+a^-=F(N),\quad a^-a^+=F(N+1),\quad [N,a^\pm]=\pm a^\pm,
\eeq
where $F$ is a function satisfying some conditions depending on the
particular application. Similarly for parabosons, the single relation
(\ref{parabose}) is equivalent to the following three relations
\beq
b^+b^-+b^-b^+=2H,\quad [H,b^\pm]=\pm b^\pm,
\eeq
and then generalized parabosons are defined by means of
\beq
a^+a^-+a^-a^+=f(H),\quad [H,a^\pm]=\pm a^\pm.
\label{genpb}
\eeq
Again, $f$ is some function and for simplicity we shall consider the
case that $f$ is of polynomial form. 
The rest of the paper is now devoted to a study of the system
(\ref{genpb}), and in particular to classifying its irreducible
representations. In order to emphasize that from now on we are dealing
with an abstract algebra, the notation is also changed and we
define the algebra $R$ as the complex algebra generated by three
elements $A,B$ and $H$ subject to the relations~: 
\beq
HA-AH=A,\qquad HB-BH=-B,\qquad AB+BA=f(H),
\label{defrel}
\eeq
where $f$ is a fixed polynomial function, $f\in \C[x]$. For many
of the applications that follow, $f$ can also be another
analytic function, but it will be clear that the polynomial case
itself is already a rich structure.

The algebra $R$ is equal to the enveloping algebra of the Lie
superalgebra $osp(1/2)$ when $f(x)=x$, and we shall draw some
similarities and differences with the representation theory of
$osp(1/2)$ for the general case.  In this sense, the
algebra~(\ref{defrel}) can be considered as a polynomial
deformation of $osp(1/2)$. If the last relation
in~(\ref{defrel}) is replaced by $AB-BA=f(H)$, the
representation theory becomes rather different and in that case
one is dealing with polynomial deformations of the enveloping
algebra of $sl(2)$. These algebras have been studied by
Smith~\cite{smith} and were shown to have a surprisingly rich theory of
representations; from the physical point of view they are
non-canonical Heisenberg algebras, and have been investigated by
Brodimas {\em et al}~\cite{brodimas}. For some general literature on
generalized deformed oscillator algebras, we refer to [6-11]; for papers
on polynomial deformations of $sl(2)$ see [12-15]; for relations to
nonlinear quantum models, see [16,17]; the number of papers on
generalized boson or paraboson algebras in the context of quantum
groups is large, and we give only a few references~[18-23].

When studying simple modules (i.e.\ irreducible representations)
of $R$, we shall consider only the
highest weight modules. These are modules $V$ generated by a
vector $v$
satisfying $Av=0$ and $Hv=\la v$. Then $v$ is a highest weight
vector of weight $\la$ ($\la\in\C$). 

\def\deg{\hbox{deg}}
\section{Some general properities of $R$}

First, we shall consider the familiar case when $\deg(f)\leq 1$.
We have already noticed that when $f(x)=x$ the algebra $R$
coincides with the enveloping algebra of the Lie superalgebra
$osp(1/2)$. 
More generally,
let $f(x)=\al x + \be$. There are three cases to be
distinguished.\\ [2mm]
Case (1)~: $\al=0,\; \be\ne 0$.\\
Let $V$ be a highest weight module with highest weight vector
$v_\la$, thus $Av_\la=0$ and $Hv_\la=\la v_\la$. Define
$v_{\la-1}= Bv_\la$; this vector is nonzero because otherwise
the relation $AB+BA=\be$ is not valid when acting on $v_\la$.
Moreover, it follows from the defining relations that
$Hv_{\la-1}= (\la-1)v_{\la-1}$ and $Av_{\la-1}=\be v_\la$. Next,
consider $v_{\la-2}=Bv_{\la-1}$. One finds that $Av_{\la-2}=
ABv_{\la-1}= (\be-BA)v_{\la-1}=0$. Hence if $v_{\la-2}$ were
nonzero, it would be an element of a nonzero submodule of $V$.
But since $V$ is simple, this submodule must be zero and
$v_{\la-2}=0$. Consequently, all simple modules of $R$ are
two-dimensional, and their explicit matrix representations are
given by~:
\beq
A=\left(\begin{array}{cc} 0&\be\\0&0\end{array}\right),\qquad
B=\left(\begin{array}{cc} 0&0\\1&0\end{array}\right),\qquad
H=\left(\begin{array}{cc} \la&0\\0&\la-1\end{array}\right).
\eeq
\vskip 2mm
\noindent
Case (2)~; $\al=\be=0$.\\
{\em Exercise~1.} Similarly as in case (1), show that 
all simple
modules are one-dimensional, and are given by~:
\beq
Av_\la = B v_\la =0,\qquad Hv_\la=\la v_\la.
\eeq
\vskip 2mm
\noindent
Case (3)~: $\al\ne 0$.\\
Let 
\beq
\tilde H=H+\be/\al,\qquad \tilde A=A/\al,\qquad\hbox{and}\qquad\tilde B=B.
\label{tilde}
\eeq
The
relations~(\ref{defrel}) with $f(x)=\al x + \be$ become
\beq
\tilde H\tilde A-\tilde A\tilde H = \tilde A,\qquad
\tilde H\tilde B-\tilde B\tilde H = -\tilde B,\qquad
\tilde A\tilde B+\tilde B\tilde A = \tilde H.
\label{tildeR}
\eeq
These are the defining relations of the enveloping algebra of
$osp(1/2)$. But due to the linear relationship
there is a one-to-one correspondence
between the simple modules of $R$ and the simple modules of the
algebra generated by $\tilde A, \tilde B$ and $\tilde H$,
subject to (\ref{tildeR}).  Since all simple modules of
$osp(1/2)$ are known~\cite{scheunert,hughes}, 
all simple modules of $R$ follow directly.
In particular, for every $j\in\Nat$, there exists (up to
isomorphism) one simple module of $R$ with dimension $2j+1$; and
there exist no simple modules with even dimension.

Next, we consider some general structure results.
The three relations~(\ref{defrel}) allow one to reorder any
polynomial in $A,B$ and $H$ unambiguously. Moreover, using the Diamond
Lemma~\cite{DL} one can verify that all monomials $A^iB^jH^k$ are also
independent. Hence,
\begin{prop}
A basis for $R$ is given by the monomials $A^iB^jH^k$,
$(i,j,k\in\Nat)$.
\label{pbw}
\end{prop}

It is easy to see that $[H,A^iB^jH^k]=(i-j)A^iB^jH^k$. Thus we
have a weight space decomposition for $R$~:
$R=\oplus_{\mu\in\Zah} R_\mu$, where $R_\mu=\{x\in R \;|\;
[H,x]=\mu x\}$. From the defining relations
one can deduce that $R_0=\C[AB,H]$; in other words, the
commutant of $H$ consists of polynomials in $AB$ and in $H$.

\begin{lemm}
Let $f\in\C[x]$.
There exists a unique polynomial $F\in \C[x]$ such that $F(x+1)+F(x)=f(x)$.
\label{lem2}
\end{lemm}
{\em Exercise~2.} Prove the above lemma.

Consider the algebra $R$ with relations (\ref{defrel})
characterized by $f$, and $F$ determined by the previous lemma.

\begin{lemm}
The element $\Om=A^2B^2+AB\left(F(H)-F(H-1)\right)-F(H)^2$ 
is a central element of $R$. 
\label{lemma}
\end{lemm}
{\em Exercise~3.} Verify the above by direct computation of $[A,\Om]$. 

In fact, every other central element of $R$ is a polynomial expression
in $\Om$~:
\begin{prop}
The center of $R$ is equal to $\C [\Om]$.
\end{prop}
There a several ways to prove this. A direct proof, showing that every
central element must be a polynomial in $\Om$, is given
in~\cite{VJ}, and will not be repeated here.

The next result gives the eigenvalue of $\Om$ acting on a
highest weight module.

\begin{prop}
Let $V$ be an $R$ module generated by a highest weight vector
$v_\la$. Then, for all $v\in V$,
\beq
\Om v = -F(\la+1)^2 v,
\label{omeigval}
\eeq
where $F$ is defined by means of Lemma~\ref{lem2}.
\label{omeig}
\end{prop}
{\em Proof.}
The expression of $\Om$ given in Lemma~\ref{lemma} can be
rewritten in the following form~:
\beq
\Om=AB^2A+BA\left(F(H+1)-F(H)\right)-F(H+1)^2.
\eeq
Thus
\beq
\Om v_\la = -F(\la+1)^2 v_\la,
\eeq
and since the module is generated by $v_\la$, $\Om$ has the same
eigenvalue for every $v$ in $V$. \hfill$\Box$

\section{Irreducible representations or simple modules of $R$}

In this section we shall study some general features of simple
modules of $R$, where $f(x)$ is a general polynomial of degree $n$. The {\em
Verma module} $V(\la)$ can be identified with the vector space
with basis
\beq
v_{\la-j} = B^j v_\la,\qquad j\in\Nat,
\eeq
where $v_\la$ is a highest weight vector satisfying $A v_\la =0$
and $H v_\la=\la v_\la$. The Verma module $V(\la)$ contains a
maximal submodule $M$; if $M=\{ 0\}$ then $V(\la)$ is simple,
otherwise the quotient module $L(\la)=V(\la)/M$ is a simple
module of highest weight $\la$. All simple highest weight
modules are obtained this way.

To see whether a vector $v_{\la-j}$ is in a submodule one 
determines $A v_{\la-j}$. \\
{\em Exercise~4.} Show (by induction) that
\beq
AB^jv_\la = \left(f(\la-j+1)-f(\la-j+2)+f(\la-j+3)-\cdots
-(-1)^j f(\la)\right) B^{j-1} v_\la.
\label{avl-j}
\eeq
If $j$ is even, the coefficient in the rhs of (\ref{avl-j}) is
equal to
\beq
F(\la+1-j)-F(\la+1).
\label{eq0}
\eeq
Similarly, using Lemma~\ref{lem2}, one finds that for the case $j$
odd, the rhs of (\ref{avl-j}) is equal to
\beq
F(\la+1-j)+F(\la+1).
\label{eq1}
\eeq
For $\la\in\C$, we define the sets~:
\bea
S_0(\la)&=&\{ j\in 2\Nat^* \;|\; F(\la+1)-F(\la+1-j)=0\} , \label{s0}\\
S_1(\la)&=&\{ j\in (2\Nat+1) \;|\;
F(\la+1)+F(\la+1-j)=0\}, \label{s1} \\
S(\la)&=&S_0(\la)\cup S_1(\la). \label{s}
\eea
We have used the obvious notation $2\Nat^*=\{2,4,6,\cdots\}$ and
$(2\Nat+1)= \{1,3,5,\cdots\}$.
Note that (\ref{eq0}) is of degree $n-1$ in
$\la$, and that (\ref{eq1}) is of degree $n$ in $\la$. As
equations in $j$, (\ref{eq0}) has the trivial solution $j=0$ (to
be excluded from $S_0(\la)$) so that an equation of degree $n-1$
remains, and (\ref{eq1}) is of degree $n$. Of course, for given
$\la$ not all solutions for these equations in $j$ will yield
integer $j$ so that $S_0(\la)$ and/or $S_1(\la)$ are often
empty. 

The primitive vectors of $V(\la)$ are those $v_{\la-j}$ such
that $Av_{\la-j}=0$; they generate the submodules of $V(\la)$.
With the given definitions we have the following lemmas~:

\begin{lemm}
The submodules of $V(\la)$ are $\C[B]B^jV(\la)$ with
$j\in S(\la)$.
\end{lemm}

\begin{lemm}
Let $\la\in\C$. 
\begin{itemize}
\item[(1)] If $S(\la)\ne\emptyset$ then $L(\la)=
V(\la)/B^j V(\la)$, where $j=\min S(\la)$, otherwise
$L(\la)=V(\la)$.
\item[(2)] For even $j$ ($j\in 2\Nat^*$) the number of simple modules
of dimension $j$ is equal to $\#\{\la\in\C \;|\;
F(\la+1)-F(\la+1-j)=0 \hbox{ and } j=\min S(\la)\}$.
\item[(3)] For odd $j$ ($j\in 2\Nat+1$) the number of simple modules
of dimension $j$ is equal to $\#\{\la\in\C \;|\;
F(\la+1)+F(\la+1-j)=0 \hbox{ and } j=\min S(\la)\}$.
\end{itemize}
\end{lemm}
Because of the previous remarks concerning the degrees of the
relevant equations, it follows that for even $j$ there are at
most $n-1$ simple modules of dimension $j$ and that for odd $j$
there are at most $n$ simple modules of dimension $j$. 
Generally there will also be $n-1$ simple modules for even
dimensions and $n$ modules for odd dimensions, apart from
certain exceptions. This will be illustrated in the following
section, where a complete study of $\deg(f)=n=2$ is presented.

\section{The case $\deg(f)=2$}

In the third relation of (\ref{defrel}) $f(H)$ is of the form
$aH^2+bH+c$ with $a\ne 0$. But due to the fact that the representation
theory of $R$ is equivalent to the representation theory of the
algebra generated by (\ref{tilde}) (subject to the corresponding
relations), there are in fact two degrees of freedom that one can delete.
Therefore we shall assume in the rest of this section that the
algebra $R$ is determined by the following relations~:
\beq
HA-AH=A,\qquad HB-BH=-B,\qquad AB+BA=2H^2+2H+1-c,
\label{deg2}
\eeq
where $c\in\C$. The form of $f(H)$ in (\ref{deg2}) is somewhat
arbitrary, but for our purposes the one chosen here seems to be
the most appropriate. We emphasize again that for every given
$f$ of degree $2$ the algebra can be ``rescaled'' by means of a
transformation of type (\ref{tilde}) such that it becomes
(\ref{deg2}). 

With $f(x)=2x^2+2x+1-c$, we have that 
$F(x)=x^2-c/2$. Then the equations (\ref{eq0}) and
(\ref{eq1}) are easy to calculate and one finds respectively
$j(j-2\la-2)=0$ and $(j-2\la-2)^2=2c-j^2$. This implies the
following~: 
\beq
\hbox{If }\la\in\Nat\hbox{ then } S_0(\la)=\{2\la+2\} 
\hbox{ else } S_0(\la)=\emptyset,
\label{s02}
\eeq
\beq
S_1(\la)=\{ j\in(2\Nat+1) \hbox{ such that }
(j-\la-1)^2=c-(\la+1)^2\}. 
\label{s12}
\eeq

The purpose is now twofold~: for given $\la\in\C$, we wish to
give (in terms of $c$) the dimension of $L(\la)$; and for given
$j$, we wish to determine the number of (inequivalent) simple
modules of dimension $j$. The two results follow from a careful
analysis of the equations appearing in (\ref{s02}) and (\ref{s12}). Let
us give one example here, using the notation
\beq
l=\la+1.
\eeq
Suppose $\la\in\C$ is given, and
consider first the case that
$\la\not\in\Nat$. Then $S_0(\la)=\emptyset$. The equation
determining $S_1(\la)$ is given by 
\beq
(j-l)^2=c-l^2. 
\label{eqjl}
\eeq
Can we have
two integer solutions $j_1$ and $j_2$, both odd, for this
equation? From the equation it would then follow that
$j_1+j_2=2l$, which would imply that $l$ or $\la$ is in $\Nat$,
which is not the case. Hence, we can have at most one integer
solution for the equation (\ref{eqjl}). In order to have an
integer and odd solution for (\ref{eqjl}) $l\pm\sqrt{c-l^2}$
must be an element of $2\Nat+1$. This can only happen when $c$
is of the form $2l^2-2lm+m^2$ with $m\in (2\Nat+1)$. If $c$ is
of this form then $L(\la)$ is finite-dimensional with dimension
$m$, otherwise $L(\la)$ is infinite-dimensional.\\
{\em Exercise~5.} Similarly as above, consider the case $\la\in\Nat$. \\
The details of this analysis are somewhat more complicated, and given
in~\cite{VJ}.
The conclusion is~:

\begin{prop}
Given $\la\in\C$ and $l=\la+1$.
\begin{itemize}
\item[(a)] If $\la\not\in\Nat$, let
$\Ga_\la=\{2l^2-2lm+m^2\;|\;m\in(2\Nat+1)\}$. If $c\not\in \Ga_\la$ then
$\dim L(\la)=\infty$; if $c\in \Ga_\la$ then $\dim L(\la)=m$
(where $m$ is the unique odd solution to $c=2l^2-2lm+m^2$).
\item[(b)] If $\la\in\Nat$, let $\De_\la=\{l^2+m^2\;|\; m\in\Nat\}$. If
$c\not\in \De_\la$ then $\dim L(\la)=2l$. If $c\in \De_\la$ and $c$
is even then $\dim L(\la)=2l$; if $c\in \De_\la$ and $c$ is odd
then $\dim L(\la)=2l$ when $c\geq 2l^2$ and $\dim
L(\la)=l-\sqrt{c-l^2}$ when $c< 2l^2$.
\end{itemize}
\label{la}
\end{prop}

The opposite question is even more interesting~: for every given
integer $j$, determine the number of simple modules of dimension
$j$. To state the result, the following sets are introduced~:
\beq
I=\{ p^2+q^2 \;|\; p\in 2\Nat,\quad q\in 2\Nat+1\};
\label{I}
\eeq
\beq
J=\{ k^2/2 \;|\; k\in 2\Nat+1\};
\label{J}
\eeq
and for $c\in I$~:
\beq
K_c=\{ (k,m) \;|\; c=k^2+m^2\hbox{ with } m<k\hbox{ and
}k,m\in\Nat\}. 
\label{Kc}
\eeq

\begin{prop}
Let $c\in\C$ and $I,J,K_c$ be defined by
(\ref{I}), (\ref{J}), (\ref{Kc}) respectively.
\begin{itemize}
\item[(a)] When $c\not\in I$ and $c\not\in J$ then for every even
dimension $j$ there is one simple module $V(\la)$ {\rm [}determined by
$l=\la+1=j/2${\rm ]} and for every odd dimension $j$ there
are two simple modules $V(\la_1)$ and $V(\la_2)$ {\rm [}determined by
the solutions $l_1$ and $l_2$ of $2l^2-2jl+j^2-c=0${\rm ]}.
\item[(b)] When $c\in J$, i.e.\ $c=k^2/2$ with $k\in(2\Nat+1)$, then
for every even dimension $j$ there is one simple module
{\rm [}determined by $l=j/2${\rm ]}, and for every odd dimension there are
two simple modules {\rm [}again determined by the solutions of
(\ref{eqjl}){\rm ]} except when $j=k$~: then there is only one simple
module {\rm [}determined by $l=k/2${\rm ]}.
\item[(c)] When $c\in I$ then for every even dimension $j$ there is one
simple module {\rm [}determined by $l=j/2${\rm ]} except for those $j$ of
the form $j=2k$ with $(k,m)\in K_c$; and for every odd dimension
$j$ there are two simple modules {\rm [}determined by (\ref{eqjl}){\rm ]}
except when $j$ is of the form $j=k+m$ with $(k,m)\in K_c$, in
which case there are no simple modules.
\end{itemize}
\label{j}
\end{prop}

The proof is a somewhat more involved exercise, and can be found
in~\cite{VJ}. 
To illustrate the proposition, consider two examples.\\ 
When $c=9/2$,
$R$ has one simple module in every even dimension, two simple
modules in every odd dimension different from 3, and one simple
module in dimension~3.\\
When $c=1105$, we see that $c\in I$ and in fact
\beq
K_{1105}=\{(33,4), (32,9), (31,12),(24,23)\}.
\label{list}
\eeq
Then $R$ has no modules of dimension 37, 41, 43, 47, 48, 62, 64
and 66. For the remaining even dimensions there is one simple
module, and for the remaining odd dimensions there are two
simple modules.

All this illustrates that the structure of simple modules of $R$
is very rich when $\deg(f)>1$.

\section{Conclusions}

In the previous sections we considered complex representations, and we
gave a complete classification for $\deg(f)=2$.
When faced with a physical problem, one would mainly be interested in
the unitary representations (under the Hermiticity condition
$H^\dagger=H$, $A^\dagger=B$, $B^\dagger=A$). These unitary
representations can be obtained from the set of irreducible
representations given above. A remarkable fact is that the algebras
with $\deg(f)>1$ possess, apart from a class of infinite-dimensional
unitary representations, also an isolated number of finite-dimensional
unitary irreducible representations~\cite{VJ}. This is very different from the
$osp(1/2)$ case, where the only unitary representations are
infinite-dimensional. 

\section*{Acknowledgements}
It is a pleasure to thank the organizers of the 2nd Workshop ``New
Trends in Quantum Field Theory'' for their invitation and for making
the week in Razlog very enjoyable.
I would also like to thank Prof.\ T.D.\ Palev for some stimulating
discussions. 

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\end{document}



