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% Deformed $u(3)$ algebra in an $so_q(3)$ basis
% J. Van der Jeugt
% J. Group Theory in Phys. 2 (1994), 153-160.
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{\LARGE \bf
Deformed $u(3)$ algebra in an $so_q(3)$ basis
}\\[5mm]
J. Van der Jeugt\footnote{Research Associate of the N.F.W.O.
(National Fund for Scientific Research of Belgium)} \\
{\it Vakgroep Toegepaste Wiskunde en Informatica,\\
Universiteit Gent, Krijgslaan 281--S9, B-9000 Gent, Belgium\\
e-mail~: Joris.VanderJeugt@rug.ac.be}\\[5mm]
(Submitted 21 February 1994)\\[5mm]
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\begin{abstract}
Deformations of enveloping algebras are usually defined in terms
of relations between the Chevalley generators. In a physical context,
it may be important to have a deformation with respect to a different
basis. In this paper we define a deformed $u(3)$ algebra in an $so_q(3)$
basis. The deformation is such that the same generators for $u(3)$
in an $so(3)$ basis are used, and only the relations are deformed.
The present approach is compared to previously used methods for the
same problem. PACS~: 02.20, 03.65F, 21.60F.
\end{abstract}
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\section{Introduction}

Since the introduction of quantum groups and quantum enveloping
algebras (quantum algebras, $q$-algebras, deformed algebras)
\cite{kr,dr,ji},
these new mathematical structures are finding
applications in various branches in physics \cite{za}.
There is particularly much interest in realisations of deformed
algebras in terms of $q$-deformed boson creation and annihilation
operators~\cite{bi,mac}, in connection with intermediate statistics.

Quantum algebras associated with simple Lie algebras are usually
defined by means of relations between the Chevalley triples
$\{h_i,e_i,f_i\}$. Considering the algebra generated by
a subset of such triples (and with the corresponding relations),
it is thus rather trivial to construct quantum subalgebras of
the original quantum algebra. One could refer to this as the
construction of subalgebras of quantum algebras in a Chevalley
basis. In this case, the Hopf structure (comultiplication,
counit, and antipode) transfer trivially to the subalgebra.

The construction of subalgebras of quantum algebras in a non-Chevalley
basis is definitely a much harder problem, and so far only some
partial solutions have been presented in the literature.
On the other hand, such embeddings would be the $q$-analogues of
Lie algebra embeddings which are of importance in physics.
For example, chains of Lie algebras $u(n)\supset so(n) \supset
so(3)\supset so(2)$ make their appearance in many algebraic models
in hadronic~\cite{ia}, nuclear~\cite{ell}, atomic~\cite{judd} and
molecular physics~\cite{lev}.
Recently, there have been a number of attempts at obtaining the
$q$-analogue of the principal subalgebra embedding $u(n)\supset so(3)$,
but these were only partially successful, even in the simplest case
$n=3$. The main obstruction seems indeed the fact that one is
dealing with $q$-deformations in a non-Chevalley basis.

A first construction of $u_q(3)\supset so_q(3)$ was given in Ref.~\cite{vdj}.
There, the realisation of $u_q(3)$ in terms of three $q$-bosons
was used, and a subalgebra of type $so_q(3)$ was constructed in
terms of these $q$-bosons. This construction can be written in
a simple form (see Section~3), and one is dealing with a genuine
subalgebra (as an associative algebra)~\cite{sala}. Moreover, the same
construction can be extended to $sp_q(6)$ in order to study the
three-dimensional $q$-deformed harmonic oscillator~\cite{sp6}. However,
there are also a number of drawbacks with this construction~:
the $so_q(3)$ generators are given in terms of the $q$-boson
operators, and thus they can act only on symmetric representations
of $u_q(3)$; the Hopf algebra properties of $u_q(3)$ do not
transfer to $so_q(3)$.

Sciarrino~\cite{sc} approached the problem from a slightly
different viewpoint. He used the $so_q(3)$ construction of Ref.~\cite{vdj},
defined the usual comultiplication, counit and antipode for $so_q(3)$,
and then tried to enlargen this algebra with a $q$-tensor operator~\cite{bt}
of rank~2. It turned out that $so_q(3)$ could be extended by a
number of operators (which, however, do not form a $q$-tensor)~:
this larger algebra is then a $q$-deformation of $u(3)$.
But also here there are some difficulties~: the approach is heuristic
and the operators are still given in terms of the $q$-bosons (thus acting
on symmetric representations only), the $q$-deformed $u(3)$
is different from $u_q(3)$, and it is not clear how to extend the
Hopf algebra properties of $so_q(3)$ to this larger algebra.

A number of authors have tried to use deforming maps~\cite{cz} to tackle this
problem~\cite{pan,sol}. Here, the expressions of $so_q(3)$ generators involve
square roots of $u_q(3)$ generators (hence infinite sums). In this approach,
the drawback are~: one does not obtain a genuine subalgebra, not even as
an associative algebra; one is also restricted to symmetric representations
only; it is not obvious how to define the Hopf algebra structures.

Finally, still another approach was taken by Quesne~\cite{qu}. Here, operators
$B^+_i$ and $B_i$ ($i=-1,0,1$), transforming as vectors under $so_q(3)$,
are constructed, and relations are proposed making use of a $q$-analogue
of coupled commutators. The $q$-couplings of the type $[B^+\times B]^k_m$
($k=0,1,2$) then lead to an interesting $q$-deformation of $u(3)$. This
deformation is again different from $u_q(3)$; however, the operators
$B^+_i$ and $B_i$ can be written in terms of usual $q$-boson operators
which appear in the $u_q(3)$ realisation and thus there is some relation.
On the other hand, also here there are a number of difficulties or
problems~\cite{qu}~: the original $so_q(3)$ algebra
does not seem to appear as a subalgebra
in the deformed $u(3)$, and the existence of a Hopf algebra structure is
not known.

In view of the physical relevance of the Lie algebra chain $u(3)\supset
so(3)$, and the fact that this is one of the simplest examples of a
non-Chevalley embedding, it it worthwhile investigating the $q$-deformation
of this Lie algebra chain by various methods and from different viewpoints.
In the present paper, we present a construction that is closely
related to the classical definition of a $q$-algebra, namely as an
associative algebra generated by a number of generators subject to
a number of relations. The choice of the generators is inspired by
a definition of the $u(3)$ enveloping algebra in the non-deformed
case, in the $so(3)$ basis. The form of the relations is deduced
from the $q$-boson realisation. This leads to a definition of a
deformed $u(3)$ algebra, in which $so_q(3)$ is naturally embedded,
simply in terms of generators and relations. Naturally, one cannot
expect that this approach would solve all problems previously encountered
in the works mentioned here, and we also mention some of the difficulties
of our method in the conclusions.

\section{Generators for $u(3)$ in an $so(3)$ basis}

The Lie algebra of $u(3)$ is usually given in a Chevalley basis,
with generators $E_i$, $F_i$ ($i=1,2$) and $N_i$ ($i=0,1,2$),
subject to the relations~:
\begin{eqnarray}
&& [N_i,N_j]=0, \label{1}\\
&& [N_i,E_j]=(\de_{i,j-1}-\de_{ij})E_j,\qquad
   [N_i,F_j]=-(\de_{i,j-1}-\de_{ij})F_j,\\
&& [E_i,F_j] = \de_{ij} (N_{i-1}-N_i), \label{3}\\
&& E_iE_iE_j-2E_iE_jE_i+E_jE_iE_i=0,\quad(i,j)=(1,2)\hbox{ or }(2,1),
   \label{4}\\
&& F_iF_iF_j-2F_iF_jF_i+F_jF_iF_i=0,\quad(i,j)=(1,2)\hbox{ or }(2,1).
   \label{5}
\end{eqnarray}
A basis of the 9-dimensional Lie algebra $u(3)$ is then $\{N_0,N_1,N_2,
E_1,E_2,[E_1,E_2]$, $F_1$, $F_2$, $[F_1,F_2]\}$, and $H_1=N_0-N_1$, $H_2=N_1-N_2$
are the usual $su(3)$ Cartan elements. The enveloping algebra of $u(3)$
is the associative algebra generated by the $N_i$, $E_i$ and $F_i$,
subject to the relations (\ref{1}--\ref{5}). The quantum enveloping algebra
$u_q(3)$ (in the Chevalley basis) is then generated by the same
generators (in fact, by $E_i$, $F_i$ and $q^{\pm N_i}$),
subject to a $q$-deformation of the relations (\ref{1}--\ref{5}),
i.e.\ in (\ref{3}) $(N_{i-1}-N_i)$ is replaced by $[N_{i-1}-N_i]$ and
in (\ref{4}) and (\ref{5}) $2$ is replaced by $[2]$, where $[x]=
(q^x-q^{-x})/(q-q^{-1})$.

A different basis for $u(3)$ is the set of tensor operators with
respect to the principal $so(3)$ subalgebra of $u(3)$. This
consist of the elements $\{T^k_m,\; k=0,1,2,\; m=-k,-k+1,\ldots,+k\}$.
Herein, $T^0_0$ is a scalar commuting with all other elements,
and the remaining commutation relations can be written in
the following form~:
\begin{eqnarray}%
[T^1_m,T^1_{m'}]&=&-\sqrt{3}(-1)^{m+m'}\3j{1}{1}{1}{m}{m'}{-m-m'}T^1_{m+m'},\\
{ }[T^1_m,T^2_{m'}]&=&\sqrt{15}(-1)^{m+m'}\3j{1}{2}{2}{m}{m'}{-m-m'}T^2_{m+m'},\\
{ }[T^2_m,T^2_{m'}]&=&\sqrt{15}(-1)^{m+m'}\3j{2}{2}{1}{m}{m'}{-m-m'}T^1_{m+m'},
\end{eqnarray}
where the symbol between brackets is a classical 3$j$-coefficient.
The elements $T^1_m$ form the basis of the $so(3)$ subalgebra, and usually
they are rescaled as follows~:
\beq
L_0=\sqrt{2}\,T^1_0,\qquad L_{\pm}=\pm 2 \,T^1_{\pm 1},
\eeq
with commutation relations
\beq
[L_0,L_\pm]=\pm L_\pm, \qquad [L_+,L_-]=2L_0.
\eeq
Also in this basis, it is possible to identify a set of generators
for $u(3)$ including those of $so(3)$. In fact, since $[L_\pm,T^2_{\pm 1}]$
is proportional to $T^2_{\pm 2}$, the Lie algebra of $su(3)$ is generated by
the elements of $so(3)$ together with $T^2_m$ ($m=-1,0,+1$). In order
to write the corresponding relations in an appropriate form, we
define the following rescaling~:
\beq
X_0=\sqrt{2/3} \,T^2_0, \qquad X_\pm=\pm 2 \,T^2_{\pm 1}.
\eeq
Then, the enveloping algebra of $u(3)$ (in an $so(3)$ basis)
is the associative algebra generated by $N$ (a scalar), $L_0,\,L_\pm$
and $X_0,\,X_\pm$ subject to the following relations~:
\begin{eqnarray}
&&[N,L_i]=[N,X_i]=0\quad(i=+,0,-),\qquad [L_0,X_0]=0, \label{11}\\
&& [L_0,L_\pm]=\pm L_\pm,\quad [L_0,X_\pm]=\pm X_\pm,\quad
   [X_0,L_\pm]=\pm X_\pm,\quad [X_0,X_\pm]=\pm L_\pm,\label{12}\\
&& [L_+,L_-]=2L_0, \quad [X_+,X_-]=2L_0,\label{13}\\
&& [L_\pm, X_\mp]= \pm 6 X_0,\label{14} \\
&& X_\pm X_\pm L_\pm - 2 X_\pm L_\pm X_\pm + L_\pm X_\pm X_\pm =0,\label{15}\\
&& L_\pm L_\pm X_\pm - 2 L_\pm X_\pm L_\pm + X_\pm L_\pm L_\pm =0.\label{16}
\end{eqnarray}
The embedding of the $so(3)$ subalgebra is obvious.
To see that (\ref{11}--\ref{16}) are indeed the relations for the
enveloping algebra of $u(3)$, it is sufficient to verify that all
commutation relations for the 9 basis elements of the Lie
algebra $u(3)$ can be determined from (\ref{11}--\ref{16}).
For completeness, we also mention the relation between the
new generators and the Chevalley generators~:
$N=(N_0+N_1+N_2)/3$, $L_0=N_0-N_2$, $X_0=(N_0-2N_1+N_2)/3$,
$L_+=\sqrt{2}(E_1+E_2)$, $L_-=\sqrt{2}(F_1+F_2)$,
$X_+=\sqrt{2}(E_1-E_2)$, $X_-=\sqrt{2}(F_1-F_2)$.

The purpose of the following sections is to determine an
appropriate $q$-deformation of the relations (\ref{11}--\ref{16}),
hence defining a deformed $u(3)$ enveloping algebra with the
same generators as in the non-deformed case, but with $q$-deformed
relations. This runs completely parallel to the situation of
deformation in the Chevalley basis.

\section{The boson realisation of $u_q(3)\supset so_q(3)$}

The quantum enveloping algebra $u_q(3)$  can be realised by
means of three independent $q$-boson operators $(N_i,b_i,b^+_i)$,
where $i=0,1,2$. These satisfy the relations~\cite{bi,mac}~:
\beq
[N_i,b^+_i]=b^+_i,\qquad
[N_i,b_i]=-b_i,\qquad, b_ib_i^+-q^{\pm 1}b_i^+b_i= q^{\mp N_i},
\eeq
and the remaining commutators are zero. The Chevalley generators
of $u_q(3)$ are given by $N_i$ (strictly speaking $q^{\pm N_i}$),
$e_1=b^+_0b_1$, $e_2=b^+_1b_2$, $f_1=b^+_1b_0$ and $f_2=b^+_2b_1$,
satisfying the usual relations (i.e.\ the $q$-deformed
relations (\ref{1}--\ref{5}) ).

In Ref.~\cite{vdj}, an $so_q(3)$ subalgebra was constructed in this realisation.
It can be written in the following form~:
\begin{eqnarray}
l_0 &=& N_{0}-N_{2}, \label{l0}\\
l_{+}&=&q^{N_{2}-{1\over 2}N_1} \sqrt{q^{N_{0}}+q^{-N_{0}}}\; b^+_{0}b_1 +
  b^+_1b_{2} q^{N_{0}-{1\over 2}N_1} \sqrt{q^{N_{2}}+q^{-N_{2}}} ,\label{l+} \\
l_{-}&=&b^+_1b_{0} q^{N_{2}-{1\over 2}N_1} \sqrt{q^{N_{0}}+q^{-N_{0}}}  +
   q^{N_{0}-{1\over 2}N_1} \sqrt{q^{N_{2}}+q^{-N_{2}}}\; b^+_{2}b_1 .
   \label{l-}
\end{eqnarray}
The square root factors appearing here can actually be avoided if
one breaks the symmetry between the expressions for $l_+$ and $l_-$;
then one obtains a genuine subalgebra of $u_q(3)$~\cite{sala}. For the present
purpose, is sufficient to work with (\ref{l0}--\ref{l-}) as they stand.
These operators, defined in terms of the $u_q(3)$ realisation, satisfy
the usual $so_q(3)$ relations, namely
\beq
[l_0,l_\pm]=\pm l_\pm,\qquad [l_+,l_-]=[2l_0].
\eeq
It is possible to simplify the expressions for $l_i$ by making the
following transformation~\cite{qu}~:
\beq
\tilde b^+_i=\left(q^{N_i}+q^{-N_i}\over q+q^{-1}\right)^{1/2} b^+_i,\qquad
\tilde b_i=b_i\left(q^{N_i}+q^{-N_i}\over q+q^{-1}\right)^{1/2}
\hbox{ for } i=0,2.
\eeq
This map transforms a $q$-boson into what could be called a $q^2$-boson,
since
\beq
[N_i,\tilde b^+_i]=\tilde b^+_i,\qquad
[N_i,\tilde b_i]=-\tilde b_i,\qquad,
\tilde b_i\tilde b_i^+-q^{\pm 2}\tilde b_i^+\tilde b_i= q^{\mp 2N_i}.
\eeq
Note that we make this transformation only for $i=0$ and $i=2$.
In terms of the two $q^2$-bosons and the single $q$-boson operators,
the expressions for $l_i$ take the form~:
\begin{eqnarray}
l_0 &=& N_{0}-N_{2}, \\
l_{+}&=&\sqrt{[2]}\left(q^{N_{2}-{1\over 2}N_1} \tilde b^+_{0}b_1 +
  b^+_1\tilde b_{2} q^{N_{0}-{1\over 2}N_1}\right)  ,\label{nl+}\\
l_{-}&=&\sqrt{[2]}\left(b^+_1\tilde b_{0} q^{N_{2}-{1\over 2}N_1}  +
   q^{N_{0}-{1\over 2}N_1} \tilde b^+_{2}b_1 \right).  \label{nl-}
\end{eqnarray}

Next, we intend to construct the $q$-analogues for $X_i$, denoted by
$x_i$, in this boson realisation. Since $x_0$ is a Cartan element,
it is natural to take the same definition as in the non-deformed
case,
\beq
x_0=(N_0-2N_1+N_2)/3.   \label{x0}
\eeq
Similarly, we have $n=(N_0+N_1+N_2)/3$. In order to find a
realisation for $x_+$ and $x_-$, we shall assume that the
relation $[X_0,L_\pm]=\pm X_\pm$ of (\ref{12}) does not
deform. This is a quite natural assumption, since also for
deformations in a Chevalley basis the commutator between
a Cartan element and a non-Cartan element does not deform.
Hence we have $[x_0,l_\pm]=\pm x_\pm$, and thus the realisation
of $x_\pm$ follows from (\ref{nl+}), (\ref{nl-}) and (\ref{x0}).
This means~:
\begin{eqnarray}
x_{+}&=&\sqrt{[2]}\left(q^{N_{2}-{1\over 2}N_1} \tilde b^+_{0}b_1 -
  b^+_1\tilde b_{2} q^{N_{0}-{1\over 2}N_1}\right)  ,\label{x+}\\
x_{-}&=&\sqrt{[2]}\left(b^+_1\tilde b_{0} q^{N_{2}-{1\over 2}N_1}  -
   q^{N_{0}-{1\over 2}N_1} \tilde b^+_{2}b_1 \right).  \label{x-}
\end{eqnarray}

All relevant generators have now been realised in the $q$-boson
representation of $u_q(3)$. In the next section we shall investigate
the relations between these generators. This will give rise to
a $q$-deformation of the relations (\ref{11}--\ref{16}). It should
be emphasized that these relations, although originally deduced
in the $q$-boson representations, are actually independent of
$q$-bosons and one can interpret them as abstract relations defining
a deformation of the enveloping algebra of $u(3)$.

\section{Relations for a deformed $u(3)$}

The $q$-analogues of the relations (\ref{11}--\ref{16})
are systematically deduced, using the realisations of $n$, $l_i$ and
$x_i$ given in the previous section. To simplify the calculations,
one can act with these generators on the states of the Fock space
for the three independent $q$-bosons~\cite{vdj}.

Since $n$, $l_0$ and $x_0$ are in fact the same as $N$, $L_0$ and
$X_0$, it is clear that (\ref{11}) does not change. To construct
$x_\pm$, we have already used $[x_0,l_\pm]=\pm x_\pm$, and similarly
one can verify that the remaining relations in (\ref{12}) also
remain unchanged~:
\beq
[l_0,l_\pm]=\pm l_\pm,\quad [l_0,x_\pm]=\pm x_\pm,\quad
   [x_0,l_\pm]=\pm x_\pm,\quad [x_0,x_\pm]=\pm l_\pm. \label{12q}
\eeq
Next, we already know the commutator of $l_+$ with $l_-$; it follows
that the commutator of $x_+$ with $x_-$ is the same. Hence, for
(\ref{13}) there is a $q$-deformation in the relations~:
\beq
[l_+,l_-]=[2l_0], \quad [x_+,x_-]=[2l_0]. \label{13q}
\eeq
Also the commutator of $l_+$ with $x_-$ is not very difficult
to calculate, although its expression looks rather complex.
We found~:
\beq
[l_\pm, x_\mp]= \pm\left(2q^{3n+3x_0+2}[x_0-n]+q^{n+2x_0+l_0}[n+2x_0-l_0]
+ q^{n+2x_0-l_0}[n+2x_0+l_0]\right).  \label{14q}
\eeq
Note that in the limit $q\rightarrow 1$ we do indeed recover (\ref{13}).
The remaining relations to be constructed are the $q$-analogues
of the Serre-like relations (\ref{15}--\ref{16}), and these seem
to be the most complicated to determine. By explicit actions on
the Fock space, one can verify that no relation exists between
the three operators $x_+x_+l_+$, $x_+l_+x_+$ and $l_+x_+x_+$ separately.
On the other hand, there do exist two relations between the six
operators $x_+x_+l_+$, $x_+l_+x_+$, $l_+x_+x_+$,
$l_+l_+x_+$, $l_+x_+l_+$, and $x_+l_+l_+$ (and similar with the index $-$).
In order to write down these relations in explicit form, it
is convenient to introduce~:
\begin{eqnarray}
v_\pm(q)&=&q x_\pm x_\pm l_\pm -(q+q^{-1})x_\pm l_\pm x_\pm
 +q^{-1} l_\pm x_\pm x_\pm,\\
w_\pm(q)&=&q l_\pm l_\pm x_\pm -(q+q^{-1})l_\pm x_\pm l_\pm
 +q^{-1} x_\pm l_\pm l_\pm.
\end{eqnarray}
Then the relations become~:
\begin{eqnarray}
&&q^{2n-2x_0}\left(v_\pm(q)\mp w_\pm(q)\right)=
[2]\left(v_\pm(q^2)\mp w_\pm(q^2)\right), \label{15q} \\
&&q^{2n-2x_0+2}\left(v_\pm(q^{-1})\pm w_\pm(q^{-1})\right)=
[2]\left(v_\pm(q^{-2})\pm w_\pm(q^{-2})\right). \label{16q}
\end{eqnarray}
Note that in the limit $q\rightarrow 1$ these relations imply
$v_\pm(1)=w_\pm(1)=0$, i.e.\ they coincide with (\ref{15}--\ref{16}).
Using the common notation $[a,b]_{q^\al}=ab-q^\al ba$, it is also possible
to rewrite the relations (\ref{15q}--\ref{16q}) as~:
\begin{eqnarray}
&&q^{2n-2x_0}[x_\pm\pm l_\pm,[x_\pm,l_\pm]_{q^{-2}}]
=(q^2+1) [x_\pm\pm l_\pm,[x_\pm,l_\pm]_{q^{-4}}], \\
&&q^{2n-2x_0+2}[x_\pm\mp l_\pm,[x_\pm,l_\pm]_{q^2}]
=(1+q^{-2}) [x_\pm\mp l_\pm,[x_\pm,l_\pm]_{q^4}].
\end{eqnarray}

We are now in a position to define the deformed $u(3)$ algebra in
an $so_q(3)$ basis. The previous relations have all been deduced
using the explicit forms of the operators in terms of the three
$q$-bosons. At this point we shall ``forget'' about the explicit
expressions of the operators, and simply define a deformed $u(3)$
in terms of a number of generators and a number of relations.
Thus ${\cal G}$, the deformed $u(3)$, is the associative algebra generated
by the elements $n$, $l_0$, $x_0$ (or, more precisely, $q^{\pm n}$,
$q^{\pm l_0}$, $q^{\pm x_0}$), $l_+$, $l_-$, $x_+$ and $x_-$,
subject to the following relations~:
\begin{eqnarray}
&&[n,l_i]=[n,x_i]=0\quad(i=+,0,-),\qquad [l_0,x_0]=0,\label{q1}\\
&&[l_0,l_\pm]=\pm l_\pm,\quad [l_0,x_\pm]=\pm x_\pm,\quad
   [x_0,l_\pm]=\pm x_\pm,\quad [x_0,x_\pm]=\pm l_\pm,\label{q2}\\
&&[l_+,l_-]=[2l_0], \quad [x_+,x_-]=[2l_0],\label{q3}\\
&&[l_\pm, x_\mp]= \pm\left(2q^{3n+3x_0+2}[x_0-n]+
q^{n+2x_0+l_0}[n+2x_0-l_0]+\right. \nn\\
&&\qquad\qquad \left. q^{n+2x_0-l_0}[n+2x_0+l_0]\right),\label{q4}\\
&&q^{2n-2x_0}\left(v_\pm(q)\mp w_\pm(q)\right)=
[2]\left(v_\pm(q^2)\mp w_\pm(q^2)\right),  \label{q5}\\
&&q^{2n-2x_0+2}\left(v_\pm(q^{-1})\pm w_\pm(q^{-1})\right)=
[2]\left(v_\pm(q^{-2})\pm w_\pm(q^{-2})\right). \label{q6}
\end{eqnarray}
These relations are the $q$-analogues of (\ref{11}--\ref{16}).
They define a deformation of the enveloping algebra of $u(3)$ in an
$so(3)$ basis, and the present algebra does contain $so_q(3)$
explicitly. Note that, contrary to the usual case, the relations
are not invariant under the $q\rightarrow q^{-1}$ transformation
(only (\ref{q1}--\ref{q3}) are invariant). Moreover, (\ref{q5})
and (\ref{q6}) mix monomials of the type ``$xxl$'' and ``$llx$''.
These two phenomena are due to the fact that there is not the
same amount of symmetry as in the case of a Chevalley basis, and
seem to be occuring specifically because we
are dealing with a deformation in a non-Chevalley basis.

\section{Comments}

We have presented a deformed $u(3)$ enveloping algebra ${\cal G}$
in an $so_q(3)$
basis. The definition given here is closely related to the definition
of a deformed enveloping algebra in a Chevalley basis. This approach
has a number of advantages, but some of the difficulties that occured
in other methods will also be present here.

Among the advantages, we note that the final definition is only in
terms of generators and relations, and that the relations are rather
simple.  Secondly, the deformed $u(3)$ algebra does contain the algebra
$so_q(3)$ explicitly (namely, it is the subalgebra generated by $l_i$
$(i=+,0,-)$). Note also that the present formulation includes no restriction
to symmetric representations. Furthermore, because of the construction
used in this paper, there is a relation
with the usual $u_q(3)$~: for the class of symmetric representations
the $u_q(3)$ generators and the generators of ${\cal G}$
defined here coincide.

Among the difficulties or open problems, we mention the following.
It would be interesting to see if this algebra can be endowed with
a comultiplication that coincides with the usual one for $so_q(3)$.
Hence the question is~: does there exist a comultiplication $\De$
with $\De(l_0)=l_0\otimes 1 + 1\otimes l_0$,
$\De(l_\pm)=l_\pm\otimes q^{l_0} + q^{-l_0}\otimes l_\pm$,
and $\De(g)\in{\cal G}\otimes{\cal G}$ for the remaining generators
of ${\cal G}$, such that $\De$ is consistent with the relations
(\ref{q1}--\ref{q6})?
This turns out to be a rather difficult problem. Recently,
Sciarrino introduced similar operators and his method
to extend $\De$ would imply that $\De(x_0)=x_0\otimes 1 + 1\otimes x_0$ and
$\De(x_\pm)=x_\pm\otimes q^{l_0} + q^{-l_0}\otimes x_\pm$~\cite{pc}.
This extension is consistent with the relations (\ref{q1}--\ref{q3}),
but not with (\ref{q4}--\ref{q6}). In practice, the comultiplication
is needed to construct new representations out of a set of basic ones.
If all representations needed can be constructed in some other way,
then the neccesity for a comultiplication becomes less important.
In our formulation of the deformed $u(3)$, we are no longer restricted to
symmetric representations, and there may exist a direct way to construct
all finite-dimensional representations without the use of a comultiplication.
This leads us to a second problem~: what are the finite-dimensional
representations of ${\cal G}$ and can they easily be classified?
No doubt this problem is worth investigating,
and our approach has the advantage that at least some of the
representations are already known,
namely the ones corresponding to symmetric representations of $u_q(3)$.


\section{Acknowledgements}
It is a pleasure to thank Dr.\ C.~Quesne (ULB, Brussels), Prof.\ T.D.~Palev
and Dr.\ N.I.~Stoilova (Institute for Nuclear Research and Nuclear Energy,
Sofia), and Prof.\ K.~Sri\-nivasa Rao (IMSc, Madras) for
stimulating discussions.

This work was partially supported by the E.E.C.\ (contract
No.~CI1*-CT92-0101).

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\end{document}